Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k \subset \mathbb{C}$ be a number field and $X$ a smooth algebraic variety over $k$. Asssume $X$ is equipped with an automorphism $h$ of order $n$. Then $h$ induces linear maps in singular cohomology $H^\ast(X(\mathbb{C}), \mathbb{Q})$ and de Rham cohomology $H_{dR}^\ast(X/k)$ whose eigenvalues are roots of unity of order $n$. We will have a decomposition according to these eigenvalues. However, when working with singular cohomology, one needs to extend first the scalar to the cyclotomic field $\mathbb{Q}(\mu_n)$. Then:

$H^j(X(\mathbb{C}), \mathbb{Q}(\mu_n))=\bigoplus_{a=0}^{n-1} H^j_{\zeta^a}(X)$

where $\zeta$ is a primitive root of unity and $H_{\zeta^a}^j$ is the subspace where $h^\ast$ acts by multiplication by $\zeta^a$.

My question is: when working with de Rham cohomology, is it also necessary to extend the scalars or do ones has already such a decomposition over $k$?

Thanks

share|improve this question
1  
It is also necessary to extend the scalars. Consider for example the product of a smooth alg. var. $X$ over $k$ with itself $n$-times. Let $h$ act by cyclic permutation of the factors. Then the action of $h$ on $H^1_{\rm dR}$ on the product is up to a sign also by permutation of the $H^1_{\rm dR}$ cohomology groups of $X$. This action cannot be diagonalised on $k$ in general unless you extend the scalars. –  Damian Rössler Feb 15 '13 at 14:39
    
Thanks for your answer Damian! –  autof Feb 15 '13 at 15:35
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.