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.
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.
