Let $X$ be a projective integral scheme over an algebraically closed field $k$. Does $\mathrm{Aut}^0_{X/k}(k)$ act trivially on $NS(X)$?
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5$\begingroup$ Yes. $NS(X)$ is a finitely generated abelian group. The homomorphism $\operatorname{Aut}(X)\rightarrow \operatorname{Aut}(NS(X)) $ is algebraic, it maps a connected subgroup to the identity. $\endgroup$– abxCommented Feb 15, 2018 at 9:30
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2$\begingroup$ The question's already answered, but it's perhaps worth mentioning the more interesting direction: the kernel of $\operatorname{Aut}(X) \to \operatorname{Aut}(NS(X))$ is an algebraic group, in the sense of having only finitely many components. So something that acts trivially is almost in $\operatorname{Aut}^0$ (e.g. some positive iterate is in $\operatorname{Aut}^0(X)$). $\endgroup$– user47305Commented Feb 15, 2018 at 16:15
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Yes, as abx already said. The morphism $\mathrm{Aut}(X) \to \mathrm{Aut}(\mathrm{NS}(X))$ is algebraic. Thus, it maps the connected component of the algebraic group $\mathrm{Aut}(X)$ to the trivial connected component of $\mathrm{Aut}(\mathrm{NS}(X))$. QED