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Let X be a smooth, projective variety over a field $k \hookrightarrow \mathbb{C}$ and let $g$ be an automorphism of $X$ of finite order. Consider the induced automorphism on the singular cohomology

$g^\ast: H^j(X(\mathbb{C}), \mathbb{Q}) \to H^j(X(\mathbb{C}), \mathbb{Q})$

(or in the De Rham cohomology). Is is true that $g^\ast=id$ when $j \neq \dim X$?

I would really appreciate your help!

Thanks

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No, for example any Riemann surface is a complex projective variety but there are lots of finite-order automorphisms that are not the identity on first homology. Maybe the simplest example is $S^1\times S^1$ and the automorphism exchanging the factors.

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    Hmm, so that can't be realized algebraically? I guess it's like a conjugation, so that makes sense actually.2012-12-19