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Is it true that for prime $p\neq 2 $, $k > 1$ and $n_1,n_2,\dots,n_k\geq 1$, the cyclic group $\mathbb{Z}_p$ has no continuous free action on $ \mathbb{C}P^{n_1} \times \mathbb{C}P^{n_2} \times \cdots \times \mathbb{C}P^{n_k}$?

How to prove it?

Thank you so much in advance.

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  • $\begingroup$ Do you assume $p$ prime (it matters if $p=4$) $\endgroup$ – YCor Nov 19 '18 at 14:26
  • $\begingroup$ @YCor Yes $p$ is a prime. $\endgroup$ – Shivani Sengupta Nov 19 '18 at 14:28
  • $\begingroup$ Related (case $k=1$) mathoverflow.net/questions/315449 $\endgroup$ – YCor Nov 19 '18 at 16:06
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    $\begingroup$ Doesn’t this follow quickly from the Lefschetz fixed point formula? $\endgroup$ – Will Sawin Nov 19 '18 at 16:26
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    $\begingroup$ @ Nick L: These automorphisms have order 2, and this is excluded in the question. LFPT tells you immediately that an an automorphism of $\mathbb{CP}^n$ of odd prime order has fixed points. However this is not so clear (to me) in the product case. $\endgroup$ – abx Nov 19 '18 at 18:43
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Consider the action of the automorphism on $H^2(\prod_i \mathbb C P^{n_i} , \mathbb Z)$ by linear automorphisms and $H^*(\prod_i \mathbb C P^{n_i} , \mathbb Z)$.

Inside $H^2(\prod_i \mathbb C P^{n_i} , \mathbb Z)$ consider the set of nonzero integral multiples of hyperplane classes of the different factors. This set is consists exactly of the elements of $H^2$ that cannot be written as a sum of two elements of smaller nilpotence order in the ring $H^*$. (This follows from the fact that the nilpotence index of an element $x$ is the sum of $n_i$ over all $i$ such that $x$ contains a nonzero multiple of the hyperplane class of $\mathbb CP^{n_i}$, which can be checked by binomial coefficients).

Because this set has a characterization, it is preserved by any automorphism. Hence any automorphism of the product of projective spaces acts by permutation of the hyperplane classes and scalar multiplication. Thus the action on $H^2$ is by the group of signed permutation matrices. Conjugacy classes in their group are characterized by their cycle type, and, for each cycle, the product of the nonzero entries, which is $\pm 1$.

Hence elements of order $p$ consist of a union of $p$-cycles and fixed points. The products of the nonzero entries should have order p and thus should be $+1$. Moreover, the $p$-cycles consist must permute tuples of $i$ where $n_i$ is constant.

The Lefschetz number of such an automorphisms (i.e. its trace on cohomology) is the product over the cycles and fixed points of the trace on the corresponding tensor factor of the cohomology. For a fixed $\mathbb CP^{n_i}$, this is simply $n_i+1$. For a $p$-cycle acting on $(\mathbb CP^{n_i})^p$, we can form a basis of the cohomology consisting of products of powers of the hyperplane classes of the individual $\mathbb CP^{n_i}$s, and the only monomials fixed by this action are those with equal powers of each class, of which there are $n_i+1$.

Hence the product is positive, so the Lefschetz number is nonzero, so the number of fixed points is nonzero.

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  • $\begingroup$ Could you please tell me the meaning of "Inside $H^2(\prod_i \mathbb C P^{n_i}, \mathbb Z)$ consider the set of nonzero integral multiples of hyperplane classes of the different factors". I don't have knowledge about it. If I take the generator of the cohomology ring $\mathbb C P^{n_i}$ is $b_i$, then in terms of generator what is the set? $\endgroup$ – Shivani Sengupta Nov 21 '18 at 5:06
  • $\begingroup$ @ShivaniSengupta It is the set of integer multiples of the $b_i$s. "Hyperplane class" is just the algebraic geometry word for the generators of this ring. $\endgroup$ – Will Sawin Nov 21 '18 at 5:36

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