If $K$ is a finite, $k$-connected, $(k+1)$-dimensional simplicial complex then, by the theorems of Hurewicz and Whitehead, $|K|$ is homotopy equivalent to a point or to a wedge of $(k+1)$-dimensional spheres.

Now suppose that $K$ is also equipped with a fixed-point-free involution $\nu$. Can we say something about the $\mathbb Z_2$-homotopy type of $|K|$?


You can conclude that $|K|$ is a wedge sum of an odd number of $k+1$-dimensional spheres.

Proof: The case when $|K|$ is zero dimensional is obvious, so assume the dimension $k+1$ is greater than zero. By the Lefschetz fixed points theorem, if the action is fixed points free, then the trace of the action of the generator of ${\mathbb Z}/2$ on $H_{k+1}(K;{\mathbb Q})$ must be $(-1)^{k}$. This implies that the rank of $H_{k+1}(K;{\mathbb Q})$ is odd, since any representation of ${\mathbb Z}/2$ is a sum of copies of the trivial representation and the sign representation, so an even-dimensional representation will have an even trace.

On the other hand, it is easy to construct a free action of ${\mathbb Z}/2$ on a wedge sum of an odd number of spheres of any given dimension $d$. Imagine lining up the spheres in a row, letting them touch each other. Embed this union of spheres in ${\mathbb R}^{d+1}$ so that the center of the middle sphere lies at the origin. The antipodal action on ${\mathbb R}^{d+1}$ restricts to a free action on the wedge of spheres.

Added later: I wonder if this action is unique up to equivariant homotopy. My hunch is that it is, but I am not sure. One could begin by proving that the integral homology of $|K|$ is uniquely determined as an integral representation of ${\mathbb Z}/2$.

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    $\begingroup$ If the action is sufficiently nice, then the quotient space has no homology in dimension bigger than $k+1$. Looking at the Serre spectral sequence for the fibration $\lvert K\rvert \to \lvert K \rvert /\mathbb Z /2 \to B\mathbb Z/2$, one can deduce that the $\mathbb Z/2$ representation $H_{k+1}\lvert K\rvert$ must be $\mathbb Z_{(-1)^k}\oplus \Lambda^r$, where $\Lambda$ is the group ring and $r\ge 0$. $\endgroup$ – user83633 Jul 12 '16 at 12:58
  • $\begingroup$ Then it seems possible to me to construct a map $\mathbb{RP}^{k+1}\vee \bigvee_r S^{k+1} \to \lvert K \rvert /\mathbb Z /2$ which is a homotopy equivalence. $\endgroup$ – user83633 Jul 12 '16 at 13:06
  • $\begingroup$ I think this is correct. And once you have the conclusion about homology, I believe you can conclude by obstruction theory that there is a ℤ/2 equivariant equivalence between K and the space that I described. For the relevant obstruction theory see, for example, the paper of Conner and Floyd "Fixed point free involutions and equivariant maps," (2.1) $\endgroup$ – Gregory Arone Jul 12 '16 at 17:08
  • $\begingroup$ This is great, many thanks for your very helpful answers! $\endgroup$ – user1272680 Jul 12 '16 at 17:41

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