Proposition 4

Let $\alpha$ and $\beta$ be linearly independent roots. Let $p$ (resp. $q$) be the largest integer $j$ such that $\beta +j\alpha$ (resp. $\beta -j\alpha$ ) is a root.

(i) The vector subspace $\sum_{-q\leq j\leq p}\mathfrak{g}^{\beta+j\alpha}$ of $\mathfrak{g}$ is a simple $\mathfrak{s}_{\alpha}$-module of dimension $p+q+ 1$.

(ii) If $\alpha +\beta$ is a root, then $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] =\mathfrak{g}^{\alpha+\beta}$.

Let $X_{\alpha}$ (resp. $x$) be a non-zero element of $\mathfrak{g}^{\alpha}$ (resp. $\mathfrak{g}^{\beta+p\alpha}$). Then

Thus, $x$ is primitive of weight $p+q$ for the representation of $\mathfrak{s}\mathfrak{l}(2, k)$ on $\mathfrak{g}$ associated to $X_{\alpha}$; but the $\mathfrak{s}\mathfrak{l}(2, k$)-module $\sum_{-q\leq j\leq p}\mathfrak{g}^{\beta+j\alpha}$ is of dimension $p+q+ 1$; hence it is simple (§1, no. 2, Prop. 2). If $\alpha +\beta \in R$, then $p\geq 1$, so the elements of $\mathfrak{g}^{\beta}$ are not primitive, and hence $[X_{\alpha},\mathfrak{g}^{\beta}]\not= 0$. Since $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}]\subset$ $\mathfrak{g}^{\alpha+\beta}$, we see finally that $[\mathfrak{g}^{\alpha},\mathfrak{g}^{\beta}] =\mathfrak{g}^{\alpha+\beta}$.