TITLE: Characterisation of parabolic subalgebras: reference sought
QUESTION [8 upvotes]: 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.)

REPLY [2 votes]: 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 $h\in\mathfrak{p}^\perp$.