Take the tour ×  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.  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 $M$?
Remark: If we consider the Frobenius morphism not as a map from
$X$ to itself but as a map $X \to X^{[