MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Perhaps I am missing something here. But my quick reaction is that surely they are the same. Assuming that $D(\mathcal{F})$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 $\mathbb{F}_q$ to 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 the the action of 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$1/(q^nt)$. 1 Perhaps I am missing something here. But my quick reaction is that surely they are the same. Assuming that $D(\mathcal{F})$ 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 $\mathbb{F}_q$ to 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 the the action of $1-t\sigma$ (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.