2
$\begingroup$

Let $X$ be a smooth projective variety over a finite field $k$. By Grothendieck, the zeta function of $X$ admits the cohomological expression $$ Z(X, t)=\prod_{j=0}^{2\dim X} \det (1-F t \ | \ H^j(X_{\bar{k}}, \mathbb{Q}_\ell))^{(-1)^{j+1}}. $$

When $X$ is an abelian variety, $H^j$ is the $j$-th exterior power of $H^1$, so all the factors should be expressible in terms of the Frobenius action on $H^1$.

Can one write explicit formulas? What are the eigenvalues of $F$ acting on $H^j$ in terms of the eigenvalues on $H^1$?

$\endgroup$

migrated from mathoverflow.net Mar 16 '15 at 9:03

This question came from our site for professional mathematicians.

  • 2
    $\begingroup$ Of course. If Frobenius acts as $F$ on $H^1$, it acts as $\wedge^pF$ on $H^p$. $\endgroup$ – abx Mar 16 '15 at 8:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy