Proposition 13

Let $\mathfrak{g}$ be a semi-simple Lie algebra, $\mathfrak{a}$ a subalgebra of $\mathfrak{g}$ reductive in $\mathfrak{g}$, and $\mathfrak{m}$ the commutant of $\mathfrak{a}$ in $\mathfrak{g}$. The subalgebra $\mathfrak{m}$ satisfies conditions (1) and (2) of Lemma 2 of no. 3; it is reductive in $\mathfrak{g}$.

By Prop. 6 of Chap. I, §3, no. 5, applied to the $\mathfrak{a}$-module $\mathfrak{g}$, we have $\mathfrak{g}=\mathfrak{m}\oplus [\mathfrak{a},\mathfrak{g}]$. Let B be the Killing form of $\mathfrak{g}$, and let $x\in \mathfrak{a}, y\in \mathfrak{m}, z\in \mathfrak{g}$. Then,

$B([z, x], y) = B(z,[x, y]) = 0$ since $[x, y] = 0$,

which shows that $\mathfrak{m}$ is orthogonal to $[\mathfrak{a},\mathfrak{g}]$ relative to B. Since B is non-degenerate, and since $\mathfrak{g}=\mathfrak{m}\oplus [\mathfrak{a},\mathfrak{g}]$, this implies that the restriction of B to $\mathfrak{m}$ is non-degenerate; condition (1) of Lemma 2 is thus satisfied.

Now let $x\in \mathfrak{m}$ and let $s$ and $n$ be its semi-simple and nilpotent components. The semi-simple component of ad $x$ is ad $s$, cf. Chap. I, §6, no. 3. Since ad $x$ is zero on $\mathfrak{a}$, so is ad $s$, by Prop. 4 (i). Thus $s\in \mathfrak{m}$, so $n=x-s\in \mathfrak{m}$, and condition (2) of Lemma 2 is satisfied.