Characterisation of parabolic subalgebras: reference sought

Question

Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{p}$ a subalgebra. As we all know, $\mathfrak{p}$ is parabolic if it contains a Borel (thus maximal solvable) subalgebra. In this case, with $\mathfrak{p}^\perp$ the orthocomplement of $\mathfrak{p}$ with respect to the Killing form of $\mathfrak{g}$, $\mathfrak{p}^\perp$ is the nilradical of $\mathfrak{p}$.

There is a handy converse to this statement which goes as follows: a subalgebra $\mathfrak{p}$ of $\mathfrak{g}$ is parabolic if $\mathfrak{p}^\perp$ is a nilpotent (thus central descending series terminates) subalgebra of $\mathfrak{g}$. Note that there is no a priori demand that $\mathfrak{p}^\perp$ is even contained in $\mathfrak{p}$ (though that is, of course. part of the conclusion).

My question: does anyone know a reference for this (not difficult to prove) fact? (I have, in the past, incorrectly attributed it to Grothendieck.)

Answer 1

I think that this follows from Bourbaki's Éléments de Mathématique. Groupes et algèbres de Lie, Chapitre VIII, §10, Theorem 1 (see below) applied to the adjoint representation. Alas, I cannot provide the google books link because the book that Google Books claims to be this one, is actually Algèbre commutative, Chapitres 5 à 7! (And the "Feedback" link does not allow me to point this out, since in their arrogance, Google does not even allow for the possibility of such an error!)

Théorème 1. --- Soient $V$ un espace vectoriel de dimension finie, $\mathfrak{g}$ une sous-algèbre de Lie réductive dans $\mathfrak{gl}(V)$, $\mathfrak{q}$ une sous-algèbre de Lie de $\mathfrak{g}$ et $\Phi$ la forme bilinéaire $(x,y) \mapsto \mathrm{Tr}(xy)$ sur $\mathfrak{g} \times \mathfrak{g}$. On suppose que l'orthogonal $\mathfrak{n}$ de $\mathfrak{q}$ par rapport à $\Phi$ est une sous-algèbre de Lie de $\mathfrak{g}$ composée d'endomorphismes nilpotents de $V$. Alors, $\mathfrak{q}$ est une sous-algèbre parabolique de $\mathfrak{g}$.

And here's a possible translation:

Theorem 1. --- Let $V$ be a finite-dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra of $\mathfrak{gl}(V)$, $\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x,y) \mapsto \mathrm{Tr}(xy)$ on $\mathfrak{g} \times \mathfrak{g}$. If the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ relative to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of $V$, then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.

Edit

As Fran points out in the comments below, my original translation was incorrect and had $\mathfrak{n}$ nilpotent instead of consisting of nilpotent endomorphisms. Happily, for the case of the adjoint representation, one has Engel's theorem, which says that the the two notions agree.

Answer 2

As alluded to in the comments above, there is a rather old paper by Jacques Dixmier that contains a result in this direction. The reference is

Dixmier, Jacques. Polarisations dans les algèbres de Lie. II. Bull. Soc. Math. France 104 (1976), no. 2, 145--164.

The result is Lemme 1.1, but the proof is attributed to P. Tauvel. The proof proceeds with a judicious application of the invariance of the Killing form and, for the nilpotency of $\mathfrak{p}^{\perp}$, Bourbaki's Groupes et algèbres de Lie, Chapitre I, §5, Lemme 3.

However, there is a caveat: a slightly different notion of co-isotropy is assumed. Dixmier's notion of co-isotropy is to define the orthogonal complement $\mathfrak{p}^f$ with respect to an anti-symmetric bilinear form $B_f$ derived from the Killing form.

Edit: I claimed earlier that Dixmier's notion of co-isotropy implies that $\mathfrak{p}^{\perp}$ is contained in $\mathfrak{p}$. This is not quite correct: you can prove, as is done there, that $\mathfrak{p}^{\perp} = [x,\mathfrak{p}^f]$ is an ideal of $\mathfrak{p}$ without using the assumption of co-isotropy that Dixmier made. The assumption kicks in only for the proof of nilpotency.

Answer 3

If $\mathfrak{p}^\perp$ is a nipotent subalgebra then it must consist of nilpotent elements. Indeed, let $h\in \mathfrak{p}^\perp$ and suppose ${\rm ad}\,h$ is not nilpotent. Let $\mathfrak{g}^0(h)$ be the Fitting null-component of ${\rm ad}\,h$ (it consists of all $x\in\mathfrak{g}$ such that $({\rm ad}\,h)^N(x)=0$ for $N\gg 0$). Since $\mathfrak{p}^\perp$ is nipotent and $[h,\mathfrak{p}]\subseteq \mathfrak{p}^\perp$ we have that $\mathfrak {p}+\mathfrak{p}^\perp\subseteq \mathfrak{g}^0(h)$. If $\mathfrak{g}^0(h)$ is a proper Lie subalgebra of $\mathfrak {g}$, there is a nonzero $a\in \mathfrak{g}^0(h)^\perp$. Then $a\in \mathfrak{p}^\perp\subset \mathfrak{g}^0(h)$ forcing $a\in \mathfrak{g}^0(h)\cap \mathfrak{g}^0(h)^\perp$. However, decomposing $\mathfrak{g}$ into the direct sum of generalised eigenspaces for ${\rm ad}\,h$ it is easy to see that the restriction of the Killing form of $\mathfrak g$ to $\mathfrak{g}^0(h)$ is non-degenerate. So ${\rm ad}\,h$ has to be nilpotent for every $h\in\mathfrak{p}^\perp$.