6
$\begingroup$

Let $G$ be a reductive complex algebraic group, $P$ a parabolic subgroup, $\mathbb{C}_{\lambda}$ a one-dimensional representation of $P$ and $\mathcal{L}_{\lambda}$ the corresponding line bundle on $G/P$.

Let $T$ be the maximal torus in $P$; then $T$ admits finitely many fixed points on $G/P$, labeled by $W_G/W_P$, where $W_G$ (resp. $W_{P}$) denotes the Weyl group of $G$ (resp. $P$). At each fixed point there is a Borel subgroup containing $T$ which has an open orbit there; write $U_w$ for this open orbit.

Write $\mathfrak{g}$ for the Lie algebra of $G$. My question is, what does the $\mathfrak{g}$-module $\Gamma(U_{w_1}\cap\dots\cap U_{w_k},\mathcal{L})$ look like? Here of course I'm simply taking an arbitrary intersection of the open sets in our cover.

Can we say anything about these modules at all? References would be appreciated.

Improve this question
$\endgroup$
3
  • $\begingroup$ Maybe I'm misunderstanding, but the $B$-orbits don't intersect. Orbits of a fixed group are disjoint almost by definition. $G/P$ doesn't have an open cover by these orbits, it is precisely the disjoint union of the orbits. $\endgroup$
    – Callum
    Commented Jan 3, 2022 at 2:38
  • $\begingroup$ Perhaps you mean the open affine subsets formed by taking the "big" cell for each of a series of Borel subgroups. You can indeed form an open cover this way (and finite as well since $G/P$ is compact). I don't know the general theory of their intersections but the intersection of two complementary ones is a Richardson variety (see mathoverflow.net/questions/193867/…) $\endgroup$
    – Callum
    Commented Jan 4, 2022 at 15:13
  • $\begingroup$ @Callum ah yes sorry that is what I meant--- for each T-fixed point take a Borel subgroup containing T which has an open orbit at that point, and take this as your cover. I will update my question and have a look at your reference. Thanks! $\endgroup$
    – freeRmodule
    Commented Jan 9, 2022 at 20:18

0

Reset to default

You must log in to answer this question.