# Are the canonical actions on Schubert Cells Linearizable?

G. Schwarz constructed a (counter)example for an action of a simple algebraic group on an affine space that is not linearizable (i.e., it is not a representations).

Natural examples of affine spaces that are not readily vector spaces are Schubert cells. So it was tempting to look for reductive group actions on them and see if they can lead to more counter-examples.

For a parabolic subgroup $P$ of a linear algebraic group $G$, (say $G$ semi-simple) we can take a Schubert cell $C\subset G/P$. By definition $C$ is the orbit for a (maximal) solvable subgroup
which is far from a reductive group. However one can look at the largest parabolic subgroup of $G$ acting on $C$, and restrict the action to a Levi part $L$ and ask if the action of $L$ on $C$ is linearizable. (Easy to see examples where $L$ is more than a maximal torus of $G$).

In a 5-minute meeting with M. Brion I asked this question and he said `Yes, it would follow from the slice theorem'. Can any one elaborate on his brief answer?

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I should be able to work this out, but not off the top of my head. The slice theorem follows almost immediately from stuff in Chapter 28 of Humphreys's book on Linear Algebraic Groups. That's where I'd start to look. – Alexander Woo Nov 1 '12 at 20:36
@Woo. Thanks. I'd like to make one more point. As $C$ is not a closed subvariety of $G/P$ the largest subgroup of $G$ stabilizing it need not be an algebriac subgroup of $G$. However if we don't work in $G/B$ we can find non-Borel parabolics stablizing $C$. – P Vanchinathan Nov 29 '12 at 23:07