Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $G$ is a complex semisimple Lie group, $P$ a parabolic subgroup of $G$. What are all of the $P$-invariant subspaces of $\mathfrak{g}/\mathfrak{p}$? In various low dimensional examples, I can calculate them all out, but there should be some easy way to describe each of them in terms of the Dynkin diagram of $G/P$.

share|improve this question
Maybe to make clearer what sort of answer I am looking for, I don't even see how to count how many such subspaces exist, just by looking at the Dynkin diagram. –  Ben McKay Mar 6 '13 at 20:27

3 Answers 3

While I share the doubts of others about finding a magic recipe, I'll add some further remarks about the formulation of the problem, which has already been mostly translated into the language of Lie algebras. Fix a semisimple complex Lie algebra $\mathfrak{g}$ plus a (proper) subset $I$ of the simple roots relative to a Cartan subalgebra $\mathfrak{h}$ and a Borel subalgebra $\mathfrak{b}$ containing it, together with a standard parabolic subalgebra $\mathfrak{p}_I = \mathfrak{l}_I \oplus \mathfrak{u}_I$. Here $\mathfrak{l}_I$ is a reductive Levi subalgebra with simple roots $I$ and $\mathfrak{u}_I$ is spanned by 1-dimensional root spaces belonging to positive roots not in the span of $I$.

Now the Lie algebra problem is to determine all $\mathfrak{p}_I$-stable subspaces of $\mathfrak{u'}_I$, the opposite nilradical. These are sums of negative root subspaces closed under addition of positive roots not in the span of $I$ and also closed under the action of $\mathfrak{l}_I$, hence direct sums of various irreducible modules for the semisimple derived algebra of $\mathfrak{l}_I$.

1) In case $I$ is empty (so $\mathfrak{p}_I = \mathfrak{b}$), such a subspace is a sum of the easiest types of submodules: a single negative root space (say for $\beta<0$) along with all others obtained by adding positive roots to $\beta$. But listing or counting these seems to require brute force enumeration, using data on positive roots for each simple Lie algebra. See the planches at the end of Bourbaki, Lie Groups and Lie Algebras, Chap. IV-VI, or for exceptional types the more convenient explicit lists (ordered by height) at the end of Springer's old paper here. A look at the higher rank exceptional types shows how unpredictable the end results will be.

2) For a larger parabolic $\mathfrak{p}_I$, you also need to keep track of the irreducible modules for $\mathfrak{l}_I$ which occur. This gets complicated, especially if there are numerous simple Lie algebra summands in this reductive Lie algebra. Even for type $G_2$, the Levi subalgebra corresponding to a short simple root requires a little care even though only two irreducible summands of the 5-dimensional opposite nilradical will exist (and their sum is stable under the full parabolic). But in general there are $2^\ell$ non-conjugate types of parabolics if the rank is $\ell$. Lots of work.

share|improve this answer

A $P$-invariant subspace is $T$-invariant, hence it is a sum of weight spaces. So, each such subspace corresponds to a subset of the set of positive roots. The additional condition of $U$-invariance forces these subsets to have the following property: if $\alpha$ is in subset and $\beta - \alpha$ is a sum of positive roots then $\beta$ is in the subset.

EDIt: All this is for the case $P = B$. For arbitrary parabolic the situation is more complicated, see robot's answer.

share|improve this answer
When I said I can calculate the invariant subspaces in examples, that is precisely how I would do it: draw the roots (or get a computer to list them all) and look for all possible sets of roots which are invariant under these sorts of operations. –  Ben McKay Mar 6 '13 at 20:41
@Ben: I am not sure there is a simple answer. For example, among such subsets of roots there are subsets generated jut by one root (take a root $\alpha$ and consider all its sums with sums of positive roots). So, if you could read this from the Dynkin diagram you could read the positive roots as well. But as far as I know there is no recipe for that. By the way for a good list of roots you can look into Bourbaki. –  Sasha Mar 7 '13 at 3:32

This is not an answer. Just a few well known facts.

Each $P$-invariant subset is also invariant with respect to the Levi part $L$ of $P$ and hence it decomposes into irreducibles for $L$.

The representation $\mathfrak{g/p}$ is as a $\mathfrak{p}$-representation isomorphic (via the Killing form) to the nilradical of $\mathfrak{p}$. Now this nilradical is in fact isomorphic to $k$-graded Lie algebra $\bigoplus_{i=1}^k \mathfrak{g}_i$ that is generated (as a Lie algebra) by $\mathfrak{g}_1$. Lie brackets $\mathfrak{g}_i \otimes \mathfrak{g}_j \to \mathfrak{g}_{i+j}$ for $i,j\in \{1,\ldots k\}$ are $L$-equivariant. (See e.g. section 3.1.2 in Parabolic Geometries I Background and General Theory by Čap and Slovák.)

One always have $P$-invariant subspaces given by filtration components $\mathfrak{g}^j = \bigoplus_{i=j}^k \mathfrak{g}_i$.

Consider a $P$-invariant subspace $V$. If $\emptyset \neq (V\cap \mathfrak{g}_i) \neq \mathfrak{g}_i$, then one can perhaps use the generating property of $\mathfrak{g}_1$ and $L$-equivariance of the Lie bracket $\mathfrak{g}_1\otimes (V\cap \mathfrak{g}_i) \to \mathfrak{g}_{i+1}$ to restrict possible $L$-types occurring in $V$.

But I don't know how the $L$-decompositions of $\mathfrak{g}_i$ look like.

share|improve this answer
What is "parabook"? –  Sasha Mar 7 '13 at 3:27
Sorry for the jargon. I've edited my "answer" accordingly. –  Vít Tuček Mar 7 '13 at 13:02
Isn't $\mathfrak{g}/\mathfrak{p}$ isomorphic to the dual of the nilradical of $\mathfrak{p}$? –  MTS Mar 7 '13 at 16:07
@MTS: yes, dual of the nilradical. But classification of the $P$-invariant subspaces of the nilradical gives those of its dual, and vice versa, so good enough. –  Ben McKay Mar 7 '13 at 20:12
@Ben: For applications to tangent (or cotangent) bundles it's important to be clear about where dualization in the Lie algebra is needed. But in the Lie algebra setting the Killing form is helpful, and the duals of the various finite dimensional representations involved are easy enough to handle. –  Jim Humphreys Mar 7 '13 at 21:21

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.