Let $f\colon X' \to X$ be an étale morphism of degree $>1$ between two complex projective manifolds. Suppose $X'$ and $X$ are diffeomorphic to each other and $f$ induces an isomorphism of $\mathbb{Q}$-Hodge structures of $X'$ and $X$. Does $X$ admit a positive degree self-covering, i.e., an étale cover $\phi\colon X\to X$ such that $\operatorname{deg} \phi>1$ (just like the abelian varieties)?



Étale cover of diffeomorphic projective manifolds
Here is a counterexample. Let $E$ be an elliptic curve. It helps if we choose it in such a way that it does not have complex multiplication. Let $L$ be a line bundle on $E$, of degree $0$, corresponding to a divisor class $D$ of infinite order. Let $\pi:X=P(L\oplus 1)\to E$ be the projectivized bundle, a fiber bundle whose fibers are projective lines.


Suppose for contradiction that there is an etale cover $\phi:X\to X$ of degree $n>1$. The map $\phi$ must take each fiber $\pi^{-1}(e)$ of the bundle $\pi$ into some fiber $\pi^{-1}(e')$, since a holomorphic map from a projective line to $E$ must be constant. The resulting map $e\mapsto e'$ must be an etale cover $\psi:E\to E$ of degree $n$, with $\phi$ mapping the fiber $\pi^{-1}(e)$ isomorphically to $\pi^{-1}(\psi(e))$. Necessarily $\psi$ is given by $e\mapsto e\_0+ me$ for some integer $m$, and $n=m^2$.


So $\phi$ gives an isomorphism of bundles between $P(\psi^\ast L\oplus 1)=\psi^\ast P(L\oplus 1)\to E$ and $P(L\oplus 1)\to E$. This implies that the divisor $D$ is equal to $\psi^\ast D= mD$, a contradiction.


On the other hand, we have plenty of etale covers $X'\to X$ with $X'$ diffeomorphic to $X$. In fact, the degree zero line bundle $L$ is trivial topologically, so that $X$ is diffeomorphic to $E\times P^1=S^1\times S^1\times S^2$; and the same is true of $X'$ if $X'$ is the fiber product $E'\times\_E X$ for any etale cover $E'\to E$. And of course each of these maps $X'\to X$ gives an isomorphism in rational homology.


