Let $C\_p$ b the cyclic group of order $p$, with $p$ a prime.


Is is possible for $C\_p$ to act (cellularly) on a rationally acyclic finite dimensional CW complex $X$ with no fixed points?


Standard Smith Theory implies that for this to be possible, $H\_\*(X;\mathbb Z/p)$ would have to be infinite dimensional. (So, e.g., with $p=2$, $X$ might be homotopy equivalent to an infinite wedge of $\mathbb RP^2$s.)


I am rather hoping that an example like this exists. But a proof that this can't happen would be fine too!



Can a cyclic group of prime order act on a rationally acyclic finite dimensional complex and have no fixed points?
Yes, I think you can make an example like this (for $p=2$, but it generalizes).


Let $R$ be the group ring $\mathbb Z[C\_2]=\mathbb Z[x]/(x^2-1)$. Make a chain complex of free $R$-modules
$$
M\_0 \leftarrow M\_1 \leftarrow M\_2
$$
as follows.


$M\_0=R$.


$M\_1$ has a basis $(a\_n)$ indexed by $n\ge 0$.


$M\_2$ has a basis $(b\_n)$ also indexed by $n\ge 0$.


$\partial a\_0=x-1$, $\partial a\_n=0$ when $n>0$, $\partial b\_n=(1+x)a\_n+(1-x)a\_{n+1}$.


The homology is such that $H\_0\cong\mathbb Z$ (trivial action), $H\_2=0$, and $H\_1$ has exponent $4$.


Now build a cell complex with free $C\_2$-action, having this as its complex of cellular chains. There is one orbit of $0$-cells, say $e^0$ and $xe^0$. There are $1$-cells $e^1\_n$ and $xe^1\_n$. The cell $e^1\_0$ is attached to $e^0$ and $xe^0$, while for $n>0$ both ends of $e^1\_n$ are attached to $e^0$. There are $2$-cells $e^2\_n$ and $xe^2\_n$, with the attaching map for $e^2\_n$ representing the appropriate $1$-dimensional homology class of this $1$-skeleton.


