Take the 2-minute tour ×

Given a variety $X/\mathbf{F}_q$ and a sheaf $\mathcal{F}$ on it, what is the relation of $L(X,\mathcal{F},T)$ and $L(X,D(\mathcal{F}),T)$?

share|improve this question
Is $X$ smooth and projective? Is $\mathcal{F}$ constructible with prime-to-q torsion? –  S. Carnahan Oct 25 '10 at 15:31
Yes, you can assume this. –  Timo Keller Nov 8 '10 at 20:12

1 Answer 1

Perhaps I am missing something here. But my quick reaction is that surely they are the same. Assuming that $D$ is the full Grothendieck-Verdier duality functor, then $H^i(\overline{X},\mathcal{F})^*\simeq H^{2n-i}(\overline{X},D(\mathcal{F}))$ (Here base change from $\mathbb{F}_q$ t$D(\mathcal{F})$o its algebraic closure commutes with duality).

Then since the characteristic polynomial of an endomorphism of a vector space is the same as that of its transpose, and n is even, nothing changes.

One (possibly esoteric) way of thinking about the (cohomological) L-function, is as action induced by $1-t\sigma$ (where $\sigma$ is Frobenius) on the determinant line (tensored with $\mathbb{Q}_l(t)$, strictly speaking) of the perfect complex $R\pi_!(\mathcal{F})$ of $\mathbb{Q}_l$ vector spaces. Then the determinant of the dual is the dual of the determinant -- but these are both scalars acting on dual lines and so are the same.

PS Perhaps what you really want is the proof of the functional equation along these lines. The key point is to not only replace $\mathcal{F}$ by $D(\mathcal{F})$ but also to replace $t$ by $1/(q^nt)$.

share|improve this answer

Your Answer


