Proposition 10

Assume that $\mathfrak{g}$ is semi-simple. Let $\mathfrak{h}$ be a splitting Cartan subalgebra of $\mathfrak{g}$, R the root system of $(\mathfrak{g},\mathfrak{h})$, B a basis of R, $R_+$ the set of positive roots relative to B, and $\mathfrak{n}_+=\sum_{\alpha\in R_+}\mathfrak{g}^{\alpha}$. The principal nilpotent elements belonging to $\mathfrak{n}_+$ are the elements of $\mathfrak{n}_+$ having a non-zero component in $\mathfrak{g}^{\alpha}$ for all $\alpha \in B$.

Prop. 8 and Lemma 9 prove that such elements are principal nilpotent. We prove the converse. Evidently we can assume that $\mathfrak{g}$ is simple. Let $h^0$ and $\mathfrak{g}^p$ be as in Prop. 8 and Lemma 9. Let $\omega$ be the highest root, and put $\omega (h^0) = 2q$; we have $q=h-1$, where $h$ is the Coxeter number of R, cf. Chap. VI, §1, no. 11, Prop. 31. Then $\mathfrak{g}^{2q}=\mathfrak{g}^{\omega},\mathfrak{g}^{-2q}=\mathfrak{g}^{-\omega}$, and $\mathfrak{g}^{2k}= 0$ for $|k|> q$. There exists a principal $\mathfrak{s}\mathfrak{l}_2$-triplet $(x^0, h^0, y^0)$. By §1, no. 2, Cor. of Prop. 2, (ad $x^0$)$^{2q}(\mathfrak{g}^{-\omega}) =\mathfrak{g}^{\omega}$, so (ad $x^0$)$^{2q}\not= 0$. Let $x$ be a principal nilpotent element of $\mathfrak{g}$ belonging to $\mathfrak{n}_+$. If $\overline{k}$ is an algebraic closure of $k,x\otimes 1$ and $x^0\otimes 1$ are conjugate under an automorphism of $\mathfrak{g}\otimes_k\overline{k}$ (Prop. 9), so (ad $x$)$^{2q}\not= 0$. There exists $\lambda \in R$ such that (ad $x$)$^{2q}\mathfrak{g}^{\lambda}\not= 0$. Put $x=\sum_{n\geq 1}x_n$, where $x_n\in \mathfrak{g}^{2n}$.

Then

(ad $x$)$^{2q}\mathfrak{g}^{\lambda}\subset$ (ad $x_1$)$^{2q}\mathfrak{g}^{\lambda}+\sum_{k>4q+\lambda(h^0)}\mathfrak{g}^k=$ (ad $x_1$)$^{2q}\mathfrak{g}^{\lambda}$, since $4q+\lambda (h^0)\geq 4q-2q= 2q$. Now (ad $x_1$)$^{2q}\mathfrak{g}^{\lambda}\subset \mathfrak{g}^{4q+\lambda(h^0)}$, where $\lambda =-\omega$. Thus, (ad $x_1$)$^{2q}\mathfrak{g}^{-\omega}=\mathfrak{g}^{\omega}$. We have $\omega =\sum_{\alpha\in B}n_{\alpha}\alpha$ with $n_{\alpha}>0$ for all $\alpha \in$ B (Chap. VI, §1, no. 8, Remark). If there exists $\alpha_0\in B$ such that $x_1\in$

$\sum\mathfrak{g}^{\alpha}$, the relation

$\alpha \in B,\alpha \not=\alpha_0$

implies that $\mathfrak{g}^{\omega}\not\subset$ (ad $x_1$)$^p\mathfrak{g}^{-\omega}$ for all $p$; this is absurd, so the component of $x_1$ in $\mathfrak{g}^{\alpha}$ is non-zero for all $\alpha \in B$.