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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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1 Answer 1

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 1 –  autom Dec 19 '12 at 8:02
    
I see, I was thinking of its real dimension. Instead, take a reflection. This acts by $-1$ on $2$-dimensional homology. –  Grumpy Parsnip Dec 19 '12 at 12:55
    
why is this true? Is your map algebraic? –  autom Dec 19 '12 at 13:11
    
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. –  Grumpy Parsnip Dec 19 '12 at 13:53
    
In coordinates, you could take $(x,y)\mapsto (x,-y)$ as the map on $S^1\times S^1$. –  Grumpy Parsnip Dec 19 '12 at 13:54

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