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.~~

`$\mathfrak{g}$`

. A Cartan subalgebra is also nilpotent, for example, but consists of "semisimple" elements. An arbitrary nilpotent subalgebra could involve both types. Unless you assume $\mathfrak{n}$` consists of nilpotent elements, the discussion gets more subtle (and the orthocomplement need not even be a subalgebra of`$\mathfrak{g}$`

) – Jim Humphreys Jun 22 '10 at 12:35notalready implied by the Bourbaki theorem is that the orthocomplement of a nilpotent subalgebra containing nonzero semisimple elements is never a Lie subalgebra (since if it were, it would have to be parabolic and thus its orthocomplement in turn would be a nil algebra)? This is somewhat roundabout to state though probably true. I haven't seen it in print, however. – Jim Humphreys Jun 22 '10 at 22:16