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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

1 Answers 1

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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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    Sure but a Riemann surface has dimension 1 and I am asking for the action on the cohomology groups in degrees different from 12012-12-19
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    I see, I was thinking of its real dimension. Instead, take a reflection. This acts by $-1$ on $2$-dimensional homology.2012-12-19
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    why is this true? Is your map algebraic?2012-12-19
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    Any orientation-reversing map acts by $-1$ on the top dimensional homology of an orientable manifold. I'm used to thinking in the topological category, but I'm sure this map can be realized algebraically.2012-12-19
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    In coordinates, you could take $(x,y)\mapsto (x,-y)$ as the map on $S^1\times S^1$.2012-12-19
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    An even simpler example maybe is to take a disconnected surface and an automorphism that switches components. This will act nontrivially on $H_0,H_1$ and $H_2$.2012-12-19
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    That's true, but take a connected Riemann surface (that is a complex smooth, algebraic projective curve). Then the action of any automorphism is trivial on $H^0$ and $H^2$2012-12-19
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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