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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!

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  • $\begingroup$ If there were no such example would this not suggest there must be a super fantastic extension of Smith theory? $\endgroup$ Commented Jan 6, 2023 at 19:51
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    $\begingroup$ @PeterKropholler There are now some rather cool new `chromatic' extensions of Smith theory, so far just for actions on finite complexes. But in that spirit, I (with William Balderrama) can show that if a cyclic p group acts on a finite dimensional complex that is acyclic in mod p K-theory (or any higher Morava K-theory), then there will be a fixed point. It would be rather fun to know that this is best possible, and one can't replace mod p K-theory by rational homology. Our collective (lack of) knowledge might suggest that this is the case! $\endgroup$ Commented Jan 6, 2023 at 20:35

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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.

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    $\begingroup$ Very nice. And yes, surely this generalizes to odd primes. $\endgroup$ Commented Jan 7, 2023 at 2:16
  • $\begingroup$ Upon reflection, it would be nice to see a bit more detail about why H_1 is all torsion. $\endgroup$ Commented Jan 7, 2023 at 2:27
  • $\begingroup$ I see. a_1 has order 4 in H_1, and I guess that calculation is typical. I am happy again. $\endgroup$ Commented Jan 7, 2023 at 2:40
  • $\begingroup$ One has lots of choice with those attaching maps for the 2-cells, so one has lots of examples of these Moore spaces of nonabelian groups. $\endgroup$ Commented Jan 7, 2023 at 2:55
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    $\begingroup$ I corrected a small typo. $\endgroup$ Commented Jan 7, 2023 at 18:18
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I needed a slightly more complicated example than this in a paper of mine, so I included a proof there. For any non-trivial finite group $Q$ I give a 3-dimensional rationally acyclic complex with a cocompact action of $Q\times \mathbb{Z}$ such that $Q$ has no fixed points but any reasonable family of subgroups of $Q$ does have fixed points. The article is `On finite subgroups of groups of type VF' Geometry and Topology Vol 9 (2005) 1953--1976 and the relevant statement is Theorem 13. I couldn't get down to 2-dimensional though.

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    $\begingroup$ Thanks, Ian, for this reference. I figured some `old timer' (you are now in that category) might know something about this, since it is on the edge of Smith theory. I wonder if my long-gone colleague Ed Floyd would have known of examples like this. $\endgroup$ Commented Jan 7, 2023 at 22:14
  • $\begingroup$ I'm honoured to be now considered an old timer. I learned a lot of this stuff from Bob Oliver. As you say, Ed Floyd would have been a good source too. $\endgroup$
    – IJL
    Commented Jan 7, 2023 at 23:31

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