# Does the cyclic group $\Bbb Z/4 \Bbb Z$ acts freely on $S^{2k} \times \Bbb CP^n$?

I was wondering whether the cyclic group $$\mathbb Z/4\Bbb Z$$ acts freely on $$S^{2k} \times \Bbb CP^n$$ where $$n>1$$? It seems to me that it does not act freely. In case it acts freely then the induced action on cohomology must be non-trivial as the Euler characteristic is non-zero. I was trying to prove using Lefschetz fixed point theorem. But I could not able to derive any contradiction.

Thank you so much for your help.

• Was this asked before and deleted? I'm sure I saw it recently, but can't seem to find it now. – user44191 Jan 6 at 8:43
• yes, I deleted it before as I thought I solved it. But in the prove there was mistake. – student Jan 6 at 8:47

Given a self-homeomorphism $$\sigma: S^{2k} \times \Bbb{CP}^n$$, you want to know what the induced map is on cohomology, written via the Kunneth decomposition as $$\Bbb Z[c, S]/(c^{n+1}, S^2)$$, where $$|c| =2$$ and $$|S| = 2k$$.
Then for any $$k$$, the classes $$c$$ and $$-c$$ are the only primitive degree 2 classes $$x$$ which do not square to zero and for which $$x^{n+1} = 0$$, as $$(ac + bS)^{n+1} = a^n b c^n S$$ (which is only relevant if $$k = 1$$), so we have $$\sigma^*(c) = \pm c$$. For any $$k$$, because $$\pm S$$ are the unique primitive classes in $$H^{2k}$$ which have non-trivial products with $$c^n$$ (which is preserved by $$\sigma$$ up to a sign) but trivial cup-square, we may write $$\sigma^*(S) = \pm S$$.
Therefore $$\sigma^2$$ fixes both $$c$$ and $$S$$, and hence acts by the identity on cohomology. The Lefschetz number is then $$\chi(S^{2k} \times \Bbb{CP}^n) = 2n+2 \neq 0$$. So for any self-homeomorphism $$\sigma$$, the map $$\sigma^2$$ has fixed points, and thus there is no free $$\Bbb Z/2j$$ action on these spaces for any $$j > 1$$.
• actually, no free $\Bbb Z/n$ action for $n > 2$. – Mike Miller Jan 6 at 15:40
• Thanks a lot. Yes, there is no free action of $\Bbb Z/n$ for $n>2$. – student Jan 6 at 18:01