Theorem 2

The following conditions are equivalent:

(i) There exists a Chevalley system $(X_{\alpha})_{\alpha\in R}$ of $(\mathfrak{g},\mathfrak{h})$ such that $\mathscr{G}_{\alpha}=\mathbf{Z}X_{\alpha}$ for all $\alpha \in R$.

(ii) $\mathscr{U}\cap \mathfrak{g}=\mathscr{G}$ and $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$ for all $\alpha \in R$.

(iii) For all $\alpha \in R, x\in \mathscr{G}^{\alpha}, n\in \mathbf{N}$, the endomorphism (ad $x$)$^n/n$! of $\mathfrak{g}$ maps $\mathscr{G}$ to $\mathscr{G}$, and $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$.

(iv) For all $\alpha \in R$ and every basis $x$ of $\mathscr{G}^{\alpha},w_{\alpha}(x)$ maps $\mathscr{G}$ to $\mathscr{G}($that is, maps $\mathscr{G}^{\beta}$ to $\mathscr{G}^{s_{\alpha}(\beta)}$ for all $\beta \in R)$.

(i) $=\Rightarrow$ (ii): let $(X_{\alpha})_{\alpha\in R}$ be a Chevalley system in $(\mathfrak{g},\mathfrak{h})$ such that $\mathscr{G}^{\alpha}=$ $\mathbf{Z}X_{\alpha}$ for all $\alpha \in R$, and let B be a basis of R. For $\alpha \in B$, put $x_{\alpha}=X_{\alpha}, y_{\alpha}=$ $X_{-\alpha}$. Let $\mathscr{U}'$ be the biorder associated by Prop. 3 of no. 6 to $\mathscr{H}$, the $x_{\alpha}$ and the $y_{\alpha}$. It is clear that $\mathscr{U}'\subset \mathscr{U}$. By Cor. 3 and 4 of Prop. $3,x^{(n)}\in \mathscr{U}'$ for all $\alpha \in R,x\in \mathscr{G}^{\alpha}$ and $n\in \mathbf{N}$. Thus $\mathscr{U}=\mathscr{U}'$, which proves (ii).

(ii) $=\Rightarrow$ (iii): this is clear by Lemma 2 of no. 5.

(iii) $=\Rightarrow$ (iv): let $\alpha \in R$ and let $x$ be a basis of $\mathscr{G}^{\alpha}$. Since $[\mathscr{G}^{\alpha},\mathscr{G}^{-\alpha}] =\mathbf{Z}H_{\alpha}$, the unique $y\in \mathfrak{g}^{-\alpha}$ such that $[y, x] =H_{\alpha}$ belongs to $\mathscr{G}^{-\alpha}$. Since exp ad $x$ and exp ad $y$ leave $\mathscr{G}$ stable by (iii), so does $w_{\alpha}(x)$.

(iv) $=\Rightarrow$ (i): let B be a basis of R. Choose a basis $x_{\alpha}$ of $\mathscr{G}^{\alpha}$ for all $\alpha \in B$. Let $y_{\alpha}\in \mathscr{G}^{-\alpha}$ be such that $[y_{\alpha}, x_{\alpha}] =H_{\alpha}$. By §1, no. 5, formulas (5), we have $y_{\alpha}=w_{\alpha}(x_{\alpha}).x_{\alpha}$ so $y_{\alpha}$ is a basis of $\mathscr{G}^{-\alpha}$ by (iv). Let $\mathscr{G}'$ be the order in $\mathfrak{g}$ defined by $\mathscr{H}$, the $x_{\alpha}$ and the $y_{\alpha}$ (no. 6, Cor. 1 of Prop. 3). Then $\mathscr{G}'$ is stable under the (ad $x_{\alpha}$)$^n/n$!, (ad $y_{\alpha}$)$^n/n$! (loc. cit.), and hence under the $w_{\alpha}(x_{\alpha})$.

Now let $\beta \in R$. There exist $\alpha_0, \alpha_1, . . . , \alpha_r\in B$ such that

(Chap. VI, §1, no. 5, Prop. 15). Then $w_{\alpha_r}(x_{\alpha_r}).w_{\alpha_{r-1}}(x_{\alpha_{r-1}}). . . w_{\alpha_1}(x_{\alpha_1})$ maps $\mathscr{G}^{\alpha_0}$ to $\mathscr{G}^{\beta}$ by (iv), and maps $\mathscr{G}'\cap \mathfrak{g}^{\alpha_0}$ to $\mathscr{G}'\cap \mathfrak{g}^{\beta}$ by the preceding. Since $\mathscr{G}'\cap \mathfrak{g}^{\alpha_0}=\mathscr{G}^{\alpha_0}$ (Prop. 3 (ii)), we have $\mathscr{G}'\cap \mathfrak{g}^{\beta}=\mathscr{G}^{\beta}$. Thus

and Cor. 4 of Prop. 3 concludes the proof.