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Let $X \subset \mathbb{P}^N$ be a smooth complete intersection, say over the complex numbers, and let $g$ be a finite order automorphism of $X$.

I would like to prove that $g^\ast$ acts trivially on $H^j(X(\mathbb{C}), \mathbb{Q})$ for all $i \neq \dim X$.

Can someone help me?

I guess what is needed is to relate $H^j(X(\mathbb{C}), \mathbb{Q})$ to $H^i(\mathbb{P}^N(\mathbb{C}), \mathbb{Q})$ which is very simple.

I know that automorphisms of the projective space act trivially on cohomology, but to use that it would be necessary to know that $g$ is induced by an automorphism of $\mathbb{P}^N$. Is that true?

Or maybe the reason is simply that $H^i(X(\mathbb{C}, \mathbb{Q})$ is 0 or 1-dimensional for $i \neq \dim X$?

Thanks for your help!

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Your second last line is indeed correct. Look up the "Lefschetz hyperplane theorem" for the relevant general statement. – ulrich Oct 29 '13 at 7:26
Thanks ulrich! I actually know that $H^j(X)=H^j(\mathbb{P}^N)$ for $j \neq \dim X$. But how can use this information to get the statement about action on cohomology? – loureedmath Oct 29 '13 at 7:54
If the space is one dimensional, then any automorphism acts by a root of unity, so $\pm 1$. It cannot be $-1$ on $H^2$ since automorphisms preserve ample classes. By taking powers you get the same result for all degrees. – ulrich Oct 29 '13 at 9:30
It's not very clear to me why this question is being downvoted so. – Todd Trimble Oct 29 '13 at 15:36
@Todd: Not clear to me either -- maybe the username? – Jason Starr Oct 30 '13 at 13:03

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