Lemma 3

Let M be a finite-dimensional vector space, A and B two vector subspaces of $\mathfrak{gl}(M)$ such that $B \subset A$ and T the set of $t \in \mathfrak{gl}(M)$ such that $[t, A] \subset B$. If $z \in T$ is such that $\mathrm{Tr}(zu) = 0$ for all $u \in T$, then $z$ is nilpotent.

It suffices to prove this when K is algebraically closed, which we shall assume henceforth. Let s and n be the semi-simple and nilpotent components of z and let $(e_i)$ be a basis of M such that $s(e_i) = \lambda_i e_i$ ($\lambda_i \in K$). Let $V \subset K$ be the vector space over $\mathbf{Q}$ generated by the $\lambda_i$. We need to prove that $V = \{0\}$. Let f be a $\mathbf{Q}$-linear form on V and let t be the endomorphism of M such that

$te_i = f(\lambda_i)e_i$. If $(E_{ij})$ is the canonical basis of $\mathfrak{gl}(M)$ defined by $E_{ij}e_k = \delta_{jk}e_i$, then

There exists a polynomial P with no constant term and with coefficients in K such that $P(\lambda_i - \lambda_j) = f(\lambda_i) - f(\lambda_j)$ for all i and j (for if $\lambda_i - \lambda_j = \lambda_h - \lambda_k$, then $f(\lambda_i) - f(\lambda_j) = f(\lambda_h) - f(\lambda_k)$ and, if $\lambda_i - \lambda_j = 0$, $f(\lambda_i) - f(\lambda_j) = 0$). Then $\mathrm{ad}\ t = P(\mathrm{ad}\ s)$. On the other hand, $\mathrm{ad}\ s$ is a polynomial with no constant term in $\mathrm{ad}\ z$. Now $(\mathrm{ad}\ z)(A) \subset B$, whence also $(\mathrm{ad}\ t)(A) \subset B$. By the hypothesis $0 = \mathrm{Tr}(zt) = \sum \lambda_i f(\lambda_i)$, whence $0 = f(\mathrm{Tr}(zt)) = \sum f(\lambda_i)^2$. Since the $f(\lambda_i)$ are rational numbers, $f = 0$, which completes the proof.