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.)

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    $\begingroup$ 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. $\endgroup$
    – Marty
    Apr 29, 2013 at 5:36
  • $\begingroup$ @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. $\endgroup$ Apr 29, 2013 at 13:07

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – MTS
    Apr 28, 2013 at 23:57
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    $\begingroup$ 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. $\endgroup$ Apr 29, 2013 at 0:21
  • $\begingroup$ @Jim, check your email. $\endgroup$
    – MTS
    Apr 29, 2013 at 1:06
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    $\begingroup$ bearspace.baylor.edu/Markus_Hunziker/www/hermitian.pdf Lemma 2.2 $\endgroup$ Apr 29, 2013 at 8:00

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