sheaves for which the derived (compact or not) pushforward is zero - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:31:38Z http://mathoverflow.net/feeds/question/88020 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88020/sheaves-for-which-the-derived-compact-or-not-pushforward-is-zero sheaves for which the derived (compact or not) pushforward is zero Dragos Fratila 2012-02-09T18:12:08Z 2012-02-10T09:27:25Z <p>Conventions: sheaf = complex of constructible sheaves (in the l-adic setup with etale tplg or in the complex coefficients setup with analytical tplg).</p> <p>I would like to understand if there is an intuition behind the following property of a sheaf. We consider $X$ a variety (or a scheme, or a gentle alg. stack) not necessarily smooth/proper. What sort of sheaves $\mathcal{F}$ on $X$ satisfy the following property: $Rp_!(\mathcal{F}) = 0$ where $p:X\to {*}$. Or (by Verdier duality) the property:</p> <p>$Rp_*(\mathcal{F}) = 0$.</p> <p>Any (partial) answer in any particular case (say X smooth, or X proper and smooth, etc.) would be nice.</p> <p>------- motivation for the question -----</p> <p>I encountered the notion of "regular sheaf" in the paper of Braverman &amp; Gaitsgory - "Geometric Eisenstein series" which uses the above property. In their context the map $p$ is from $Bun_{T}\to Bun_{T/G_m}$, where $T$ is a torus in a reductive group ($Bun_T$ is over a smooth proj curve); the map $p$ is induced from a cocharacter $G_m\to T$. They call a sheaf regular if $Rp_!\mathcal{F} = 0$ for all the maps $p$ as above that are induced from a coroot. This notion seems to be very important in their paper and I'm trying to understand intuitively this condition and see some simple examples.</p> http://mathoverflow.net/questions/88020/sheaves-for-which-the-derived-compact-or-not-pushforward-is-zero/88022#88022 Answer by Sasha for sheaves for which the derived (compact or not) pushforward is zero Sasha 2012-02-09T18:45:02Z 2012-02-09T18:45:02Z <p>Inspired by Borel-Weil-Bott, The following comes to mind (I don't know enough to put it in context):</p> <p>We can consider a circle $S^1$, and take the local system which corresponds to the sign character of $\mathbb{Z}$. Then its zero'th cohomology is zero, and the first by duality is also zero.</p> <p>Sasha</p> http://mathoverflow.net/questions/88020/sheaves-for-which-the-derived-compact-or-not-pushforward-is-zero/88024#88024 Answer by Donu Arapura for sheaves for which the derived (compact or not) pushforward is zero Donu Arapura 2012-02-09T18:59:12Z 2012-02-10T09:27:25Z <p>This indeed a strong condition! But to get a sense of it, let us look at the simplest case where $X$ is a smooth not necessarily projective complex curve. Also let $\mathcal{F}$ be locally constant, so we can identify it with a representation of $\pi_1(X)$. The vanishing condition will imply that the topological Euler characteristic must be zero. So either $X= G_m$ or $X$ is an elliptic curve. In both cases, nontrivial examples exist: any local system with $$H^0(X,\mathcal{F})= \mathcal{F}^{\pi_1(X)}=0$$ will work. In the first case, this is clear because the Euler characteristic of $\mathcal{F}$ is zero and there is only $H^1$ to contend with. In the second, you use Poincare duality to also kill $H^2$.</p> <p>I think this can be pushed to apply to any local system satisfying the above condition on a semiabelian variety (extension of an abelian variety by a torus)</p>