I think it is true that there is no free-action of $\mathbb{Z}_p(p\neq 2$) on product of $\mathbb{CP}^n(n$ odd) and $\mathbb{S}^{2m}$. But I don't know how to prove it. Any solution will be helpful.

Thanks in advance.

  • $\begingroup$ When $n=1$, may be this linked paper contains usefull information link.springer.com/article/10.1007%2FBF02571395 $\endgroup$ Jul 9, 2018 at 19:54
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    $\begingroup$ A form of the Hilbert-Smith conjecture states that there is no continuous, effective action of the p-adics on a connected manifold for any prime p. As far as I know, this is only settled in dimensions < 4. $\endgroup$
    – K.K.
    Jul 30, 2018 at 15:16
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    $\begingroup$ The most recent edit seems to have removed a lot of information and made the title much less informative. If you have a new question about p-adic actions, I recommend reverting the edit and asking that in a new post. $\endgroup$
    – j.c.
    Jul 30, 2018 at 19:26

1 Answer 1


I think the following will work for $p > 3$. I am not sure if it can be made to work for $p =3$, but maybe it can. I will the appeal to the representation theory of $\mathbb{Z}/p\mathbb{Z}$, but probably this is overkill.

Given a continuous action of $\mathbb{Z}/p\mathbb{Z}$ on your space $X = \mathbb{CP}^n\times S^{2m}$, the rational cohomology groups are all representations of $\mathbb{Z}/p\mathbb{Z}$. There are two irreducible representations of $\mathbb{Z}/p\mathbb{Z}$ over $\mathbb{Q}$, the trivial representation and a non-trivial representation of dimension $p-1$. Each cohomology group can be expressed as a direct sum of copies of these representations.

On the other hand, the Künneth theorem shows that the only non-zero cohomology groups of $X$ have dimension 1 or 2, and that these occur in even degree. Since $p-1 > 2$, the cohomology groups must be direct sums of copies of the trivial representation. In other words, each element $g$ of $\mathbb{Z}/p\mathbb{Z}$ acts trivially on the cohomology of $X$. It follows easily now that the Lefschetz number of any such element is positive, and therefore that $g$ has a fixed point.

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    $\begingroup$ I think if you take into account the cup product structure, you can directly see that the action on cohomology is trivial. $\endgroup$ Jul 9, 2018 at 19:03
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    $\begingroup$ @AchimKrause Indeed :) I think my argument actually shows that if $X$ is a sufficiently nice compact space all of whose Betti numbers are $< p-1$ and with non-zero Euler characteristic, then $\mathbb{Z}/p\mathbb{Z}$ cannot act freely. $\endgroup$
    – K.K.
    Jul 9, 2018 at 22:14

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