# How to prove that a subalgebra $\mathfrak p$ in a semisimple lie algebra $\mathfrak g$ is parabolic if $\mathfrak p^\perp=\mathfrak{rn}(\mathfrak p)$?

Let $\mathfrak g$ be a semissimple Lie algebra over an algebraically closed field of characteristic zero. Suppose $\mathfrak p$ is a subalgebra of $\mathfrak g$ such that $\mathfrak p^\perp=\mathfrak{rn}(\mathfrak p)$, where $\mathfrak p^\perp$ denote the subspace of the elements of $\mathfrak g$ which are perpendicular to $\mathfrak p$ in relation to the Killing form of $\mathfrak g$ and $\mathfrak{rn}(\mathfrak p)$ is the nilpotent radical of $\mathfrak p$. I would like to have a proof or a tip that, in this case, $\mathfrak p$ must be a parabolic subalgebra.

This question arose from an exercise from a book about Lie algebras (San Martin, "Álgebras de Lie"). I've been thinking about it for some time and have already tried the references in http://en.wikipedia.org/wiki/Parabolic_Lie_algebra .

Thank you.

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Presumably you want the characteristic of the base field to be zero? –  Stephen Mar 2 '12 at 19:37
You're right. Thank you. I've just edited. –  Júlio César Mar 2 '12 at 22:59

(1) Given a Borel subalgebra $\mathfrak{b}$ of a semisimple Lie algebra $\mathfrak{g}$, the nilradical $\mathfrak{n}^+$ of $\mathfrak{b}$ is the orthogonal complement of $\mathfrak{b}$ in $\mathfrak{g}$ with respect to the Killing form, and is moreover equal to the set of elements of $\mathfrak{b}$ that act ad-nilpotently on $\mathfrak{g}$.
(2) Given an element $x$ of a Lie algebra with an associative bilinear form (such as the Killing form), the $\mathrm{ad}(x)$ generalized eigenspaces for eigenvalues $\lambda$ and $\mu$ are orthogonal unless $\mu=-\lambda$.
Let $\mathfrak{b}$ be a Borel subalgebra of $\mathfrak{g}$ containing the nilradical $\mathfrak{n}=\mathfrak{rn(p)}$ of $\mathfrak{p}$. Our aim is to prove that $\mathfrak{b} \subseteq \mathfrak{p}$, since a parabolic subalgebra is by definition a subalgebra that contains a Borel subalgebra.
Choose $x \in \mathfrak{n}$. Since $\mathfrak{n}$ is a nilpotent ideal of $\mathfrak{p}$, the linear map $\mathrm{ad}(x)$ is nilpotent on $\mathfrak{p}$, and writing $\mathfrak{g}=E_0 \oplus E$ where $E_0$ is the $0$ generalized eigenspace and $E$ is the sum of the non-zero generalized eigenspaces of $\mathrm{ad}(x)$, we must have $E_0$ and $E$ orthogonal with respect to the Killing form (fact 2), so that $E$ is orthogonal to $\mathfrak{p}$. The latter contains its own orthogonal complement so $E$ is zero and it follows that $\mathrm{ad}(x)$ is nilpotent on $\mathfrak{g}$. Combining this with fact 1, $\mathfrak{n} \subseteq \mathfrak{n}^+=\mathfrak{b}^\perp$, implying that $b \subseteq \mathfrak{n}^\perp=(\mathfrak{p}^\perp)^\perp=\mathfrak{p}$.