Proposition 5

Assume that $\mathfrak{g}$ is splittable semi-simple. Let $\mathfrak{h}$ be a splitting Cartan subalgebra of $\mathfrak{g}$, R the set of roots of $(\mathfrak{g},\mathfrak{h})$, and B a basis of R. Let $h$ be a simple element of $\mathfrak{g}$ belonging to $\mathfrak{h}$. Then $h$ is conjugate under Aut$_e(\mathfrak{g},\mathfrak{h})$ to an element $h'$ of $\mathfrak{h}$ such that $\alpha (h')\in \{0,1,2\}$ for all $\alpha \in B$.

The eigenvalues of ad$_{\mathfrak{g}}h$ belong to $\mathbf{Z}($§1, no. 2, Cor. of Prop. 2). Hence $h\in \mathfrak{h}_{\mathbf{Q}}$. There exists an element $w$ of the Weyl group of $(\mathfrak{g},\mathfrak{h})$ such that $\alpha (wh)\geq 0$ for all $\alpha \in B$ (Chap. VI, §1, no. 5, Th. 2 (i)). In view of §2, no. 2, Cor. of Th. 2, we are reduced to the case in which $\alpha (h)\in \mathbf{N}$ for all $\alpha \in B$. Let $R_+$ be the set of positive roots relative to B, and $R_-=-R_+$. There exists an $\mathfrak{s}\mathfrak{l}_2$-triplet in $\mathfrak{g}$ of the form $(x, h, y)$. Let T be the set of roots $\beta$ such that $\beta (h) =-2$. Then $T\subset R_-$ and $y\in \sum_{\beta\in T}\mathfrak{g}^{\beta}$. Assume that there

exists $\alpha \in B$ such that $\alpha (h)>2$. For all $\beta \in T$, we have $(\alpha +\beta )(h)>0$, so $\alpha +\beta \notin R_-$ and $\alpha +\beta \not= 0$; on the other hand, since $\beta \in R_-$ and $\alpha \in B$, we have $\alpha +\beta \notin R_+$; hence $\alpha +\beta \notin R\cup \{0\}$, so $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] = 0$. Thus, $[y,\mathfrak{g}^{\alpha}] = 0$. But ad$_{\mathfrak{g}}y|\mathfrak{g}^{\alpha}$ is injective since $\alpha (h)>0 ($§1, no. 2, Cor. of Prop. 2). This contradiction proves that $\alpha (h)\leq 2$ for all $\alpha \in B$.