Proposition 14

Let $\mathfrak{g}$ be a semi-simple Lie algebra, A a group and $r$ a homomorphism from A to Aut($\mathfrak{g}$). Let $\mathfrak{m}$ be the subalgebra of $\mathfrak{g}$ consisting of the elements invariant under $r(A)$. Assume that the linear representation $r$ is semi-simple. Then $\mathfrak{m}$ satisfies conditions (1) and (2) of Lemma 2 of no. 3; it is reductive in $\mathfrak{g}$.

The proof is analogous to that of the preceding proposition:

Let $\mathfrak{g}^+$ be the vector subspace of $\mathfrak{g}$ generated by the $r(a)x-x,a\in A, x\in \mathfrak{g}$. The vector space $\mathfrak{g}'=\mathfrak{m}+\mathfrak{g}^+$ is stable under $r(A)$. Let $\mathfrak{n}$ be a complement of $\mathfrak{g}'$ in $\mathfrak{g}$ stable under $r(A)$. If $x\in \mathfrak{n}, a\in A,r(a)x-x\in \mathfrak{n}\cap \mathfrak{g}^+= 0$, so $x\in \mathfrak{m}$ and then $x= 0$ since $\mathfrak{m}\cap \mathfrak{n}= 0$. Thus, $\mathfrak{g}=\mathfrak{g}'=\mathfrak{m}+\mathfrak{g}^+$. Let B be the Killing form of $\mathfrak{g}$ and let $y\in \mathfrak{m}, a\in A, x\in \mathfrak{g}$. Then

Thus $\mathfrak{m}$ and $\mathfrak{g}^+$ are orthogonal relative to B. It follows that the restriction of B to $\mathfrak{m}$ is non-degenerate; hence condition (1) of Lemma 2. Condition (2) is immediate by transport of structure.