Lemma 3

Let g be a Lie algebra, a an ideal of g, M a g-module of finite dimension over

K and $(M_i)_{0 \leq i \leq n}$ a Jordan-Hölder series of the $g$-module $M$. The following conditions are equivalent:

(a) for all $x \in a$, $x_M$ is nilpotent; (b) for all $x \in a$, $x_M$ is in the Jacobson radical of the associative algebra $A$ generated by 1 and the $y_M$ where $y \in g$; (c) for all $x \in a$,

If these conditions are fulfilled, $a$ is orthogonal to $g$ with respect to the bilinear form associated with the $g$-module $M$.

(b) $\Rightarrow$ (a): as $A$ is finite-dimensional over $K$, the Jacobson radical of $A$ is a nilpotent ideal (Algebra, Chapter VIII, § 6, no. 4, Theorem 3) and hence every element of this radical is nilpotent.

(a) $\Rightarrow$ (c): each $Q_i = M_i/M_{i+1}$ ($0 \leq i < n$) is a simple $g$-module. For all $x \in a$, the endomorphism $x_{Q_i}$ (which is derived from $x_M$ by restricting to $M_i$ and passing to the quotient) is nilpotent if condition (a) holds and hence zero by Lemma 2; in other words, $x_M(M_i) \subset M_{i+1}$.

(c) $\Rightarrow$ (b): suppose condition (c) holds; let $x \in a$ and $z \in A$. Then $z(M_i) \subset M_i$ ($0 \leq i < n$) and hence $(zx_M)^n(M) = \{0\}$; thus $Ax_M$ is a left nilideal of $A$ and hence is contained in the Jacobson radical of $A$ (Algebra, Chapter VIII, § 6, no. 3, Corollary 3 to Theorem 1).

Finally, suppose conditions (a), (b) and (c) hold. Let $x \in a$ and $y \in g$. We have just seen that $y_M x_M$ is nilpotent and hence $\operatorname{Tr}(y_M x_M) = 0$, which proves the last assertion of the lemma.