Pushforward of equivariant bundles via the Frobenius morphism - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T13:01:08Z http://mathoverflow.net/feeds/question/53908 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53908/pushforward-of-equivariant-bundles-via-the-frobenius-morphism Pushforward of equivariant bundles via the Frobenius morphism Chuck Hague 2011-01-31T19:38:45Z 2011-01-31T19:38:45Z <p>Let $G$ be a semisimple algebraic group over an algebraically closed field of positive characteristic $p$ and let $B \subseteq G$ be a Borel subgroup. Set $X := G/B$, the flag variety of $G$. Also let $F : X \to X$ denote the (absolute) Frobenius morphism; this morphism is the identity on points, and its comorphism is the $p^{th}$ power map on functions.</p> <p>Let $M$ be a rational $B$-module. Then we have the $G$-equivariant bundle ${\cal L} (M)$ on $X$ with fiber $M$. It is not too hard to verify that there is an isomorphism $F^* \mathcal L(M) \cong \mathcal L(M^{[1]})$, where $M^{[1]}$ denotes the first Frobenius twist of $M$. My question is: Since $F_* \mathcal L(M)$ is a $G$-equivariant bundle on $X$ as well, is there also a nice description of $F_* \mathcal L(M)$? That is, there is some $B$-module $N$ such that $F_* \mathcal L(M) \cong \mathcal L (N)$; what is the relation of $N$ to $M$?</p> <p><strong>Remark</strong>: If we consider the Frobenius morphism not as a map from $X$ to itself but as a map $X \to X^{[1]}$, Haastert has described the $G$-equivariant structure of the pushforward sheaf. However, I am not sure how to use this to describe $F_* \mathcal L(M)$.</p>