Does every nilpotent element in a real Lie algebra lie in some minimal parabolic Lie algebra?

Definitions: Let $$\mathfrak{g}$$ be a finite-dimensional, real, semisimple Lie algebra with some Cartan involution $$\theta$$. Let $$\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$$ be the usual Cartan decomposition arising from $$\theta$$.

Then, following Knapp's book on Lie algebras, we say a subalgebra $$\mathfrak{q} \subset \mathfrak{g}$$ is minimal parabolic if it is, via the adjoint group, conjugate to a Lie algebra of the form $$\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n},$$ where $$\mathfrak{a}$$ denotes a choice of maximal abelian subalgebra of $$\mathfrak{p}$$, the space $$\mathfrak{m}:= Z_\mathfrak{k}(\mathfrak{a})$$ denotes the centralizer of $$\mathfrak{a}$$ in $$\mathfrak{k}$$, and $$\mathfrak{n}$$ denotes a choice of positive restricted root space with respect to the restricted root space decomposition of $$(\mathfrak{g},\mathfrak{a})$$, i.e. every element in $$\mathfrak{n}$$ lies, for some $$\alpha \in \mathfrak{a}^*$$ in some

$$\mathfrak{g}_\alpha = \{X \in \mathfrak{g} : \text{ad}_H(x) = \alpha(H) \cdot x \quad \forall H \in \mathfrak{a} \},$$

and the possible $$\alpha$$ are restricted to a certain choice of positivity, as with the usual root space decomposition.

First (smaller) question: I believe this should be equivalent to the common definition arising from "a subalgebra is parabolic if its complexification is parabolic in $$\mathfrak{g}_\mathbb{C}$$", where a parabolic algebra is simply one containing a Borel subalgebra (a maximal solvable one). Knapp does not go into detail on this. Is there an easy way to see this?

Second question: With respect to the given definition, does every nilpotent element $$n \in \mathfrak{g}$$ lie in some minimal parabolic subalgebra?

The second question is clearly true for compact real forms, where no nontrivial nilpotents exist, and there are some arguments from Bourbaki which are highly specific to split real froms. But apart from that, I'm not really sure...

• Good questions. I believe what you believe in your first question, admittedly without having a rigorous or short argument. I would also think that the second question is answered "yes" by Bourbaki's Lie Groups and Algebras, ch. VIII §10 cor. 2 to theorem 2, if at the end for the non-split case one replaces "Borel" by "minimal parabolic". Of course that relies on a positive answer to the first question. Maybe you want to have a closer look at that corollary and its surroundings? Unfortunately I'm not sure if I have time right now to think that through. Jun 16 at 21:49
• I believe the answer to the second question is yes, but I am not so sure about the first. In Knapp there is a result stating that all minimal parabolic are conjugate to eachother by the adjoint action of $K$ (the subgroup of the adjoint group with Lie algebra $\mathfrak{k}$ if you really want to start from the Lie algebra, although Knapp of course starts from the group.) So one thing to check is if the same is true for Borell subalgebras or that there are in some sense 'too many' of those (e.g. multiple $K$-orbits) Jun 29 at 12:56
• For the second question. If I recall correctly the argument is something similar to this: start with your nilpotent $X$, define $Y = \theta(X)$ and $H = [X, Y]$ and $X, Y, H$ will form an sl2-triple. In particular we have that $H$ is semi-simple and contained in the -1 eigenspace of $\theta$, so it is contained in some maximal space $\mathfrak{a}$ of pairwise commuting elements with these two properties. Having $\mathfrak{a}$ we can construct $\mathfrak{m}$ and $\mathfrak{n}$ as in your post and by construction we get for free that $X \in \mathfrak{n}$. Jun 29 at 12:59
• Warning: I wrote this from memory without checking if any part of it is wrong or 'not as easy as it looks' Jun 29 at 13:01
• It's an interesting idea! But is it really clear that this forms an sl2-triple, i.e. how does one show $[H,X] = 2X$? It's not immediately clear to me, I guess this has to use nilpotence of $X$ in some nontrivial way. Jun 29 at 14:06

Took some work and some scouring through the literature, but we got there in the end. Both questions are answered positively — though my brain still needs some time to digest the answer for question one, and make sure that it's really really true, so, eh, approach with care.

First question: Whenever I say parabolic here, I mean not the Knapp definition, but the one in terms of complexification, see the OP.

From Lemma in Section 3.2 from Wolf, Koranyi, we can extract the following (heavily paraphrased, but hopefully equivalent):

Let $$\mathfrak{g}$$ be a real semisimple Lie algebra and $$\mathfrak{q}$$ a parabolic subalgebra. Then there is some Cartan decomposition $$\mathfrak{g} = \mathfrak{t} \oplus \mathfrak{p}$$ some ​maximally noncompact ("maximally split") Cartan subalgebra $$\mathfrak{h} = \mathfrak{t} \oplus \mathfrak{a}$$ of $$\mathfrak{g}$$, where $$\mathfrak{t} \subset \mathfrak{k} \text{ (the "totally nonsplit" part)}, \quad \mathfrak{a} \subset \mathfrak{p} \text{ (the "totally split" part) },$$ a subspace $$\mathfrak{a}' \subset \mathfrak{a}$$ and a choice of positive roots $$P$$ in the restricted root space decomposition of $$(\mathfrak{g}, \mathfrak{a}')$$ so that $$\mathfrak{q} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in P} \mathfrak{g}_\alpha, \quad \mathfrak{g}_\alpha = \{x \in \mathfrak{g} : [a,x] = \alpha(a) \cdot x \quad \forall a \in \mathfrak{a}'\}.$$

(Careful: In the source, the root space decomposition is carried out in the complexification $$\mathfrak{g}_\mathbb{C}$$, but we can also carry it out in the real setting, since the ad-action of elements in $$\mathfrak{a}' \subset \mathfrak{a}$$ is real diagonalizable. This is always necessary for the restricted root space decomposition.)

Very verbose, but in the end, in the above notation, every parabolic subalgebra contains $$Z_\mathfrak{k}(\mathfrak{a}) \oplus \mathfrak{a}$$ in the $$\mathfrak{g}_0$$-component and some choice of positive restricted roots of $$\mathfrak{a}$$ in the $$\bigoplus_{\alpha \in P} \mathfrak{g}_\alpha$$-component. Hence every parabolic subalgebra contains some subalgebra of the form $$\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$$. And indeed, the subalgebras $$\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$$ are parabolic, since their complexification contains a Borel algebra associated to the complexification of the Cartan subalgebra $$\mathfrak{a} \oplus \mathfrak{t}$$.

Hence, Knapp's minimal parabolic subalgebras are exactly the minimal parabolic subalgebras in the usual sense.

Second question:

In Bourbaki, Chapter VIII, Exercise 3a for §5, we learn:

If $$\mathfrak{q}, \mathfrak{p}$$ are parabolic subalgebras of a semisimple (real or complex) Lie algebra $$\mathfrak{g}$$, and $$\mathfrak{q} \subset \mathfrak{p}$$, then the radical of $$\mathfrak{p}$$ is contained in the radical of $$\mathfrak{q}$$.

And in Bourbaki, Chapter VIII, §10, Corollary 2, we learn:

Every subalgebra $$\mathfrak{n}$$ of a (real or complex) Lie algebra $$\mathfrak{g}$$, consisting only of nilpotent elements of $$\mathfrak{g}$$, is contained in the nilradical of a parabolic subalgebra $$\mathfrak{q}$$.

As a corollary of the two: Given a subalgebra $$\mathfrak{n} \subset \mathfrak{g}$$, contained in the radical of some parabolic subalgebra $$\mathfrak{q}$$. But then there is some minimal parabolic $$\mathfrak{q}_0 \subset \mathfrak{q}$$ with $$\mathfrak{n} \subset \text{rad}(\mathfrak{q}) \subset \text{rad}(\mathfrak{q}_0)$$.

Wolf, J. A.; Koranyi, A., Generalized Cayley transformation of bounded symmetric domains, Am. J. Math. 87, 899-939 (1965). ZBL0137.27403.

Bourbaki, Nicolas, Elements of mathematics. Lie groups and Lie algebras. Chapters 7 and 8, Berlin: Springer (ISBN 3-540-33939-6). 271 p. (2006). ZBL1181.17001.