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Jul 28, 2020 at 8:54 comment added Henri @vrz the rational cohomology is enough
Jul 28, 2020 at 7:43 comment added user145520 @Henri do you need equality in the rational singular cohomology or in the Picard group for the argument to work?
Jul 28, 2020 at 7:31 comment added Henri Alternative proof once one has $c_1(V)=0$: the same argument shows $c_2(V)\cdot H^{n-2}=0$ hence by Yau's uniformization theorem, $V$ is a torus quotient.
Jul 27, 2020 at 22:55 history became hot network question
Jul 27, 2020 at 21:40 comment added Henri Here's a attempt to prove Piotr's statement. As $V$ has no rational curve (otherwise $f$ would be isomorphic over it, but $H$ has necessarily degree $0$ over it). So $K_V$ is nef, but $K_V\cdot H^{n-1}=0$ since $f^*K_V=K_V$. Hence $c_1(V)=0$. By Beauville-Bogomolov, $V$ admits an étale cover $A\times W\to V$, with $A$ is abelian and $\pi_1(W)=0$. Let $g$ be the étale cover obtained on $A\times W$ by fiber product. One checks that the fundamental group acts diagonally on $\mathbb C^{\dim A}\times W$ hence $g=g_A\times g_W$ is product. Since ${\rm Pic}(A\times W)$ is a product, one has $W=0$.
Jul 27, 2020 at 16:34 comment added Piotr Achinger On second thought, the assertion perhaps should be that $V$ admits a finite etale (Galois) cover $A\to V$ with $A$ an abelian variety. And of course $k>1$.
Jul 27, 2020 at 16:26 comment added Piotr Achinger Complementing Will's excellent answer - I'm unable to find a reference right now, but I imagine that the following is true and known: if $V$ is a smooth projective algebraic variety admitting an etale endomorphism $f\colon V\to V$ which is polarized in the sense that $f^* H = kH$ for some ample class $H$ and some integer $k$, then $V$ is an abelian variety.
Jul 27, 2020 at 15:33 vote accept CommunityBot
Jul 27, 2020 at 15:27 answer added Will Sawin timeline score: 5
Jul 27, 2020 at 14:53 history asked user145520 CC BY-SA 4.0