Proposition 1

Let $(\mathfrak{g},\mathfrak{h})$ be a split semi-simple Lie algebra, R its root system, B a basis of R, and $(n(\alpha , \beta ))_{\alpha ,\beta\in B}$ the corresponding Cartan matrix. For all $\alpha \in B$, let $X_{\alpha}\in \mathfrak{g}^{\alpha}, X_{-\alpha}\in \mathfrak{g}^{-\alpha}$. Then, for $\alpha , \beta \in B$,

$[X_{-\alpha}, X_{\beta}] = 0$ if $\alpha \not=\beta$ (4)

(ad $X_{\alpha}$)$^{1-n(\beta ,\alpha)}X_{\beta}= 0$ if $\alpha \not=\beta$ (5)

(ad $X_{-\alpha}$)$^{1-n(\beta ,\alpha)}X_{-\beta}= 0$ if $\alpha \not=\beta$. (6)

The family $(H_{\alpha})_{\alpha\in B}$ is a basis of $\mathfrak{h}$. If $X_{\alpha}\not= 0$ and $X_{-\alpha}\not= 0$ for all $\alpha \in B$, the Lie algebra $\mathfrak{g}$ is generated by the $X_{\alpha}$ and the $X_{-\alpha}(\alpha \in B)$.

(Recall that, if $\alpha , \beta \in B$ and $\alpha \not=\beta ,n(\beta , \alpha )$ is an integer $\leq 0$, so formulas (5) and (6) make sense.)

Formulas (1), (2) and (3) are clear. If $\alpha \not=\beta ,\beta -\alpha$ is not a root since every element of R is a linear combination of elements of B with integer coefficients all of the same sign (Chap. VI, §1, no. 6, Th. 3). This proves (4). In view of Chap. VI, §1, no. 3, Prop. 9, this also proves that the $\alpha$-chain defined by $\beta$ is

hence $\beta +(1-n(\beta , \alpha ))\alpha \notin R$, which proves (5). The equality (6) is established in a similar way. The family $(H_{\alpha})_{\alpha\in B}$ is a basis of $R^{\vee}$, and hence of $\mathfrak{h}$. If $X_{\alpha}\not= 0$ and $X_{-\alpha}\not= 0$ for all $\alpha \in B$, then $[X_{\alpha}, X_{-\alpha}] =\lambda_{\alpha}H_{\alpha}$ with $\lambda_{\alpha}\not= 0$, so the last assertion follows from §3, no. 3, Prop. 9 (iii).