Proposition 2

The sequence of homomorphisms

1 $\longrightarrow T_Q\longrightarrow^f$ Aut($\mathfrak{g},\mathfrak{h}$)$\longrightarrow^{\varepsilon}$ A(R) $\longrightarrow$ 1

is exact.

a) Let $\varphi \in$ Ker $f$. Then $\varphi (\alpha ) = 1$ for all $\alpha \in R$. Since R generates the group $Q(R),\varphi$ is the identity element of $T_Q$.

b) Let $\varphi \in T_Q$. The restriction of $f(\varphi )$ to $\mathfrak{h}=\mathfrak{g}^0$ is the identity, so

Im $f\subset$ Ker$\varepsilon$.

c) Let $s\in$ Ker$\varepsilon$. Then $s|\mathfrak{h}=$ Id$_{\mathfrak{h}}$. For all $\alpha \in R$, we have $s(\mathfrak{g}^{\alpha}) =\mathfrak{g}^{\alpha}$, and there exists a $t_{\alpha}\in k^*$ such that $sx=t_{\alpha}x$ for all $x\in \mathfrak{g}^{\alpha}$. Writing down the condition that $s\in$ Aut($\mathfrak{g}$), we obtain the relations

$t_{\alpha}t_{-\alpha}= 1$ for all $\alpha \in R$

$t_{\alpha}t_{\beta}=t_{\alpha+\beta}$ when $\alpha , \beta , \alpha +\beta \in R$.

Under these conditions, there exists $\varphi \in T_Q$ such that $\varphi (\alpha ) =t_{\alpha}$ for all $\alpha \in R$ (Chap. VI, §1, no. 6, Cor. 2 of Prop. 19). Then $s=f(\varphi )$. Hence, Ker $\varepsilon \subset$ Im $f$.

d) The image of Aut($\mathfrak{g},\mathfrak{h}$) under $\varepsilon$ contains W(R) by §2, no. 2, Cor. of Th. 2, and contains Aut(R$,B$) by Prop. 1. Hence this image is equal to A(R).