impact of Poincaré duality on functional equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:55:30Z http://mathoverflow.net/feeds/question/43507 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43507/impact-of-poincare-duality-on-functional-equation impact of Poincaré duality on functional equation norondion 2010-10-25T12:04:58Z 2010-11-15T23:23:16Z <p>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 $L(X,D(\mathcal{F}),T)$?</p> http://mathoverflow.net/questions/43507/impact-of-poincare-duality-on-functional-equation/46160#46160 Answer by profilesdroxford54 for impact of Poincaré duality on functional equation profilesdroxford54 2010-11-15T22:54:34Z 2010-11-15T23:23:16Z <p>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 <code>$H^i(\overline{X},\mathcal{F})^*\simeq H^{2n-i}(\overline{X},D(\mathcal{F}))$</code> (Here base change from <code>$\mathbb{F}_q$</code> t<code>$D(\mathcal{F})$</code>o its algebraic closure commutes with duality).</p> <p>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.</p> <p>One (possibly esoteric) way of thinking about the (cohomological) L-function, is as action induced by <code>$1-t\sigma$</code> (where $\sigma$ is Frobenius) on the determinant line (tensored with <code>$\mathbb{Q}_l(t)$</code>, strictly speaking) of the perfect complex $R\pi_!(\mathcal{F})$ of <code>$\mathbb{Q}_l$</code> 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.</p> <p>PS Perhaps what you really want is the proof of the functional equation along these lines. The key point is to not only replace <code>$\mathcal{F}$</code> by <code>$D(\mathcal{F})$</code> but also to replace $t$ by $1/(q^nt)$.</p>