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

This question has some overlap with previous ones but doesn't seem to have a well-documented answer. I recall some literature (mostly involving Lie groups and hermitian symmetric pairs, etc.) which concerns maximal parabolic subalgebras of a simple Lie algebra $\mathfrak{g}$ over the field $\mathbb{C}$ or related parabolic subgroups of Lie groups. Here $\mathfrak{p}$ can be taken as standard relative to some fixed Borel subalgebra and Cartan subalgebra, with a decomposition as direct sum of a Levi subalgebra (involving all but one simple root) along with the nilradical $\mathfrak{n}$.

In this situation, is there a necessary and sufficient criterion in the literature for $\mathfrak{n}$ to be abelian, which can then be checked easily case-by-case for the simple types?

For example, neither of the two types of maximal (=minimal) parabolic subalgebra in the Lie algebra $G_2$ turns out to have an abelian nilradical. It's also true that $G_2$ has no minuscule highest weights for its irreducible finite dimensional representations.

On the other hand, one of the equivalent conditions for a dominant integral weight to be (co)minuscule implies that the nilradical of the parabolic subalgebra stabilizing a highest weight vector in the corresponding representation must be abelian. (See my previous question here for some references on minuscule weights.) But I don't recall now exactly how precise a criterion exists in the literature for $\mathfrak{n}$ to be abelian when $\mathfrak{p}$ is maximal.

[I was just thinking about this in connection with a newer question here which is not precisely formulated but apparently involves a similar setting.]

ADDED: The answer (plus email) and references given are quite helpful though somewhat intertwined with Lie groups and differential geometry or algebraic groups and algebraic geometry. I was looking for a self-contained approach via roots and weights within the traditional Lie algebra setting. Anyway, a uniform elementary statement seems to emerge as follows. With $\mathfrak{g}$ simple, take $\mathfrak{p}$ to be a maximal parabolic corresponding to the set of simple roots excluding $\alpha$. Then $\mathfrak{n}$ is abelian iff $\alpha$ has coefficient 1 in the highest root, or iff $\mathfrak{p}$ is the stabilizer of a highest weight vector in the irreducible representation whose highest weight is "cominuscule" relative to $\alpha$ (involving interchange of types B, C). (These criteria are then easy to check case-by-case.)

share|cite|improve this question
The reference of (my) choice is Richardson, Rohrle, Steinberg, "Parabolic subgroups with abelian unipotent radical," in Inventiones v.110, no. 3 (1992), p. 649-671. – Marty Apr 29 '13 at 5:36
@Marty: Yes, that comes close to being a self-contained algebraic treatment (Lemma 2.2 especially) even though it's written in the language of algebraic groups and focuses on orbits, etc. I'd forgotten how their arguments work. – Jim Humphreys Apr 29 '13 at 13:07
up vote 4 down vote accepted

Denote by $\mathfrak{l}$ the Levi factor of the parabolic, so that $\mathfrak{p} = \mathfrak{l} \oplus \mathfrak{n}$, and note that this is a splitting as $\mathfrak{l}$-modules. Also denote by $\mathfrak{n}_-$ the nilradical of the opposite parabolic subalgebra; this is the dual of $\mathfrak{n}$ via the Killing form of $\mathfrak{g}$.

Here are the equivalent conditions that I know:

  1. $\mathfrak{n}$ is abelian.
  2. $\mathfrak{n}_-$ is abelian.
  3. $\mathfrak{n}$ is an irreducible representation of $\mathfrak{l}$.
  4. $\mathfrak{n}_-$ is an irreducible representation of $\mathfrak{l}$.
  5. $\mathfrak{p}$ is maximal and the simple root of $\mathfrak{g}$ that is removed from $\mathfrak{l}$ has coefficient 1 in the highest root of $\mathfrak{g}$.
  6. $[\mathfrak{n},\mathfrak{n}] \subseteq \mathfrak{l}$.
  7. $(\mathfrak{g},\mathfrak{l})$ is a symmetric pair, i.e. there is an involutive Lie algebra automorphism of $\mathfrak{g}$ whose fixed-point subalgebra is precisely $\mathfrak{l}$.

Clearly condition 5 is the easiest way to check this, assuming you have handy a table of highest roots of the simple Lie algebras. One can be found in, e.g. Table 2 in the exercises of Chapter 12 of Introduction to Lie Algebras and Representation Theory, by... well, you.

I have not seen this entire collection of equivalent criteria written up in one place, although many of them are discussed in Lemma 7.3.1 of Multiplicity-free Theorems of the Restrictions of Unitary Highest Weight Modules with respect to Reductive Symmetric Pairs, by Toshi Kobayashi.

share|cite|improve this answer
I should say that some people refer to this property by saying that $\mathfrak{p}$ is of cominuscule type. It is related to the property of minusculity, but is not quite the same. See the beginning of Chapter 9 of Billey and Lakshmibai, Singular Loci of Schubert Varieties for a little more information on that connection. – MTS Apr 28 '13 at 23:57
Yes, I was being careless in not writing cominiscule (so I've edited a bit). The criterion 5 you list does seem optimal for checking and agrees with cases I recollect from various sources. I'm still wondering if such equivalent conditions have been written down explicitly with proofs, but at least the bits and pieces make sense to me. I will have to get acquainted with the Kobayashi paper. – Jim Humphreys Apr 29 '13 at 0:21
@Jim, check your email. – MTS Apr 29 '13 at 1:06

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.