Theorem 1

Let V be a finite dimensional vector space, $\mathfrak{g}$ a reductive Lie subalgebra in $\mathfrak{g}\mathfrak{l}(V),\mathfrak{q}$ a Lie subalgebra of $\mathfrak{g}$ and $\Phi$ the bilinear form $(x, y) \rightarrow$ Tr($xy$) on $\mathfrak{g}\times \mathfrak{g}$. Assume that the orthogonal complement $\mathfrak{n}$ of $\mathfrak{q}$ with respect to $\Phi$ is a Lie subalgebra of $\mathfrak{g}$ consisting of nilpotent endomorphisms of V. Then $\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}$.

$a)\mathfrak{q}$ is the normalizer of $\mathfrak{n}$ in $\mathfrak{g}:$ let $\mathfrak{p}$ be this normalizer. Let $x\in \mathfrak{q}$ and $y\in \mathfrak{n}$; for all $z\in \mathfrak{q}$, we have $[z, x]\in \mathfrak{q}$, so

in other words, $[x, y]\in \mathfrak{n}$. Hence $\mathfrak{q}\subset \mathfrak{p}$. Since $\mathfrak{n}$ is an ideal of $\mathfrak{p}$ consisting of nilpotent endomorphisms of V, P is orthogonal to $\mathfrak{n}$ with respect to $\Phi$ (Chap. I, no. 3, Prop. $4d$)$)$. Since $\Phi$ is non-degenerate$^4,\mathfrak{p}\subset \mathfrak{q}$, hence our assertion.

b) There exists a reductive Lie subalgebra $\mathfrak{m}$ in $\mathfrak{g}\mathfrak{l}(V)$ such that $\mathfrak{q}$ is the semi-direct product of $\mathfrak{m}$ and $\mathfrak{n}:$ let $\mathfrak{n}_V(\mathfrak{q})$ be the largest ideal of $\mathfrak{q}$ consisting of nilpotent endomorphisms of V. Then $\mathfrak{n}_V(\mathfrak{q})$ contains $\mathfrak{n}$, and it is orthogonal to $\mathfrak{q}($loc. cit.); hence $\mathfrak{n}=\mathfrak{n}_V(\mathfrak{q})$. Moreover, $\mathfrak{g}$ is reductive in $\mathfrak{g}\mathfrak{l}(V)$ by hypothesis, hence decomposable (Chap. VII, §5, no. 1, Prop. 2); since $\mathfrak{q}$ is the intersection of $\mathfrak{g}$ with the normalizer of $\mathfrak{n}$ in $\mathfrak{g}\mathfrak{l}$(V), it is a decomposable Lie algebra (loc. cit., Cor. 1 of Prop. 3). Thus, our assertion follows from Prop. 7 of Chap. VII, §5, no. 3.

Choose a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{m}$; denote by $\mathfrak{g}_1$ the commutant of $\mathfrak{h}$ in $\mathfrak{g}$, and put $\mathfrak{q}_1=\mathfrak{q}\cap \mathfrak{g}_1,\mathfrak{n}_1=\mathfrak{n}\cap \mathfrak{g}_1$.

c) The Lie algebras $\mathfrak{g}_1,\mathfrak{q}_1$ and $\mathfrak{n}_1$ satisfy the same hypotheses as $\mathfrak{g},\mathfrak{q}$ and $\mathfrak{n}:$ since $\mathfrak{m}$ is reductive in $\mathfrak{g}\mathfrak{l}$(V), $\mathfrak{h}$ is commutative and is composed of semi-simple endomorphisms of V (Chap. VII, §2, no. 4, Cor. 3 of Th. 2). Thus $\mathfrak{g}_1=\mathfrak{g}^0(\mathfrak{h})$ is reductive in $\mathfrak{g}$ (Chap. VII, §1, no. 3, Prop. 11), hence also in $\mathfrak{g}\mathfrak{l}(V)$ (Chap. I, §6, no. 6, Cor. 2 of Prop. 7). It is clear that $\mathfrak{n}_1$ is composed

$^4$ Let $\mathfrak{z}$ be the orthogonal complement of $\mathfrak{g}$ with respect to $\Phi$; this is an ideal of $\mathfrak{g}$ contained in $\mathfrak{n}$, so every element of $\mathfrak{z}$ is nilpotent. The identity representation of $\mathfrak{g}$ is semi-simple (Chap. I, §6, Cor. 1 of Prop. 7). Hence $\mathfrak{z}= 0$ (Chap. I, §4, no. 3, Lemma 2). of nilpotent endomorphisms of V. Since $\mathfrak{h}$ is a subalgebra of $\mathfrak{q}$, reductive in $\mathfrak{g}\mathfrak{l}$(V), the adjoint representation of $\mathfrak{h}$ on $\mathfrak{q}$ is semi-simple; by construction, $\mathfrak{q}_1$ is the set of invariants of ad$_{\mathfrak{q}}(\mathfrak{h})$, so $\mathfrak{q}=\mathfrak{q}_1+ [\mathfrak{h},\mathfrak{q}]$ (Chap. I, §3, no. 5, Prop. 6). Since

an element of $\mathfrak{g}_1$ is orthogonal to $\mathfrak{q}_1$ if and only if it is orthogonal to $\mathfrak{q}$; consequently, $\mathfrak{n}_1=\mathfrak{g}_1\cap \mathfrak{n}$ is the orthogonal complement of $\mathfrak{q}_1$ in $\mathfrak{g}_1$.

d) The Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{m}$ is a Cartan subalgebra of $\mathfrak{g}:$ We have $\mathfrak{q}=\mathfrak{m}\oplus \mathfrak{n}$ and $\mathfrak{h}=\mathfrak{m}\cap \mathfrak{g}_1$, so it is immediate that $\mathfrak{q}_1=\mathfrak{h}\oplus \mathfrak{n}_1$. Moreover, $[\mathfrak{h},\mathfrak{n}_1] = 0,\mathfrak{h}$ is commutative and $\mathfrak{n}_1$ is nilpotent, so the Lie algebra $\mathfrak{q}_1$ is nilpotent. By a) and $c),\mathfrak{q}_1$ is the normalizer of $\mathfrak{n}_1$ in $\mathfrak{g}_1$; a fortiori$,\mathfrak{q}_1$ is equal to its normalizer in $\mathfrak{g}_1$, hence is a Cartan subalgebra of $\mathfrak{g}_1$. Since $\mathfrak{g}_1$ is reductive in $\mathfrak{g}\mathfrak{l}$(V), it follows from Cor. 3 of Th. 2 of Chap. VII, §2, no. 4, that $\mathfrak{q}_1$ is composed of semi-simple endomorphisms of V; thus, since $\mathfrak{n}_1$ is composed of nilpotent endomorphisms of V, we have $\mathfrak{n}_1= 0$. Consequently, $\mathfrak{h}=\mathfrak{q}_1$ is a Cartan subalgebra of $\mathfrak{g}_1$, and since $\mathfrak{g}_1$ normalizes $\mathfrak{h}$, we have $\mathfrak{h}=\mathfrak{g}_1$. Thus, we have proved that every element of $\mathfrak{h}$ is a semi-simple element of $\mathfrak{g}$, and that the commutant of $\mathfrak{h}$ in $\mathfrak{g}$ is equal to $\mathfrak{h}$; it follows that $\mathfrak{h}=\mathfrak{g}^0(\mathfrak{h})$, so $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$.

$e)\mathfrak{q}$ is a parabolic subalgebra of $\mathfrak{g}:$ by the preceding, $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g},\mathfrak{n}$ consists of nilpotent elements of $\mathfrak{g}$, and $[\mathfrak{h},\mathfrak{n}]\subset \mathfrak{n}$. Let $\overline{k}$ be an algebraic closure of $k$; by definition, $\mathfrak{q}$ is parabolic in $\mathfrak{g}$ if and only if $\overline{k}\otimes_k\mathfrak{q}$ is a parabolic subalgebra of $\overline{k}\otimes_k\mathfrak{g}$. The properties stated above being preserved by extension of scalars, for the proof we can restrict ourselves to the case in which $\mathfrak{h}$ is splitting. Let R be the root system of $(\mathfrak{g},\mathfrak{h})$; by Prop. 2 (v) of §3, no. 1, there exists a subset P of R such that $P\cap (-P) =\emptyset$ and $\mathfrak{n}=\sum_{\alpha\in P}\mathfrak{g}^{\alpha}$.

Let $P'$ be the set of roots $\alpha$ such that $-\alpha \notin P$; we have $P'\cup (-P') = R$, and the orthogonal complement $\mathfrak{q}$ of $\mathfrak{n}$ in $\mathfrak{g}$ is equal to $\mathfrak{h}+\sum_{\alpha\in P'}\mathfrak{g}^{\alpha}$. We have

proved that $\mathfrak{q}$ is parabolic. Q.E.D.