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

Let $G$ be a simple group (say $SL_n$) and let $B$ be a Borel subgroup (say upper triangular matrices). Then all irreducible representations of $G$ are induced from one-dimensional representations of $B$, i.e characters of $B$. However, only some of the weights will induce to non-zero representations. Relative to the standard maximal torus of diagonal matrices, these weights appear to be the anti-dominant weights. However, $B$ contains other tori as well, and no one choice is canonical.

My question is: since induction from $B$ didn't require a choice of torus, how are some characters of $B$ already deemed to be anti-dominant? Where is the symmetry broken?

share|cite|improve this question
up vote 13 down vote accepted

One of the sneaky tricks that Lie theorists play on students is that they tell them about Cartan subalgebras, and then at some point, they pull the rug out and say "just joking; really you should think about the abstract Cartan." The abstract Cartan is a Borel mod its radical. You might say "which Borel?" but it doesn't matter; all these quotients are canonically isomorphic (any two Borels are conjugate, and any two ways of making them conjugate differ by an element of the Borel).

Note, a Cartan subgroup of your Lie algebra isn't canonically isomorphic to the abstract Cartan; you have to choose a Borel containing that Cartan subgroup first. That's where the symmetry is broken.

The abstract Cartan has a canonical notion of positive weight: you look at the action of $B/N$ on the quotient of the Lie algebra $n/[n,n]$, and the weights that appear there are your simple roots (all the positive roots can be obtained by taking the associated graded for the filtration $n\supset [n,n]\supset [n,[n,n]]$). Again, any way of making Borels conjugate carries these weights to the corresponding ones for the other Borel. So once you've picked a Borel, you can use the isomorphism of the abstract Cartan to your chosen one to get a root system on your chosen one.

EDIT: As Allen notes below: I learned most of this from the book of Chriss and Ginzburg.

You get fundamental coroots by taking minimal parabolics over your Borel (those obtained by adding one negative simple root). There's a canonical map from the abstract Cartan of $P/[P,P]$ to that of $G$, and there's unique coweight of $P/[P,P]$ (which is $SL_2$ or $PSL_2$) so that the action on the unique simple root space has eigenvalue 2. The image of that is the simple coroot.

share|cite|improve this answer
Ah, I see. How would you then get the simple coroots (to eventually get your fundamental weights)? – Dinakar Muthiah Nov 11 '09 at 5:25
So what are the roots? (You've given the simple roots) – Rob Harron Nov 11 '09 at 5:41
I added answers to both of these above. – Ben Webster Nov 11 '09 at 15:33
Cool, thanks. Are there good references for this? – Rob Harron Nov 11 '09 at 16:45
None I know of. It's easy to prove; you just note that if you pick a Cartan subgroup, the action on $n$ is isomorphic (as a T-rep) to the associated graded, and the action of your chosen Cartan on the associated graded matches with the action of the abstract Cartan under the isomorphism induced by picking the Borel. – Ben Webster Nov 11 '09 at 16:54

B is what determines which weights are dominant. Given a choice of maximal torus T, you get a root system. A choice of Borel, B, containing T provides a choice of positive roots. This defines the dominant weights.

share|cite|improve this answer
But without choosing a maximal T, how would you say which roots are dominant? – Dinakar Muthiah Nov 11 '09 at 4:22
B has a maximal abelian quotient T, given a short exact sequence 0 -> N -> B -> T -> 0. This sequence is semi-direct, and the positive weights are the weights of T acts on N. There are no choices here. – David Speyer Nov 11 '09 at 4:25
That said, I would be interested in knowing what happens when you induce a non-dominant weight form B to G. – David Speyer Nov 11 '09 at 4:25
But the splitting is not unique. How do you get B/N to act on N? If you induce from a non-anti-dominant weight you'll get 0. – Dinakar Muthiah Nov 11 '09 at 4:43
B/N acts on the Lie algebra n/[n,n] by the adjoint action. The weights of that action are the positive roots. – Ben Webster Nov 11 '09 at 5:12

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.