MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is asked in the context of (connected) reductive groups.

In the article i'm working on, the author states the following fact (well it's not word to word exact, I simplified it a little) :

If we choose a parabolic subgroup, determined by a simple reflection $s$ in $W$ (the Weyl group, given a maximal torus), then the variety of the Borel subgroups contained in $P$ is one-dimensional, moreover it can be identified with the projective line.

Is there a simple way to prove this ?

What I tried : since the parabolic subgroup $P$ is determined by a simple reflection, I wrote down the Levi decomposition using roots, then I was thinking that the Borel subgroups contained in $P$ are in bijection with the Borel subgroups of the Levi complement.

share|cite|improve this question
By the way, I believe the name is Levi and not Lévi. See – Faisal May 27 '11 at 19:55
Thanks, I edited it. – May 27 '11 at 20:49
up vote 4 down vote accepted

There is some confusion in the way the question is set up. You have to begin with a fixed Borel subgroup $B$ in order to speak about a "simple" reflection in the Weyl group. Then there is a unique "minimal" parabolic subgroup $P \supset B$ corresponding to the specified simple root/reflection. This in turn has a Levi subgroup of rank 1, while $P/B$ is naturally isomorphic to the projective line. Of course, $P$ itself contains many other Borel subgroups besides $B$, but now the problem has shifted. All such Borels contain the radical of $P$. so after factoring that out you are just looking at the flag variety of a rank 1 group which is the desired copy of the projective line.

share|cite|improve this answer
My question might be stupid... are you refering to the semisimple rank ? rather than the (full) rank of the Lévi complement ? – May 27 '11 at 15:16
Which type of rank you consider here makes no difference, since a central torus lies in all Borel subgroups. (In my answer, you could factor out the solvable or the unipotent radical.) Similarly, isogeny type doesn't matter. So in rank 1 you get the same flag variety $\mathbb{P}^1$ for the groups $GL_2, SL_2, PGL_2$, over any algebraically closed field. – Jim Humphreys May 27 '11 at 17:07
thanks a lot again for your answers. – May 27 '11 at 17:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.