Proposition 4

We have $f(T_Q)\subset$Aut$_0(\mathfrak{g},\mathfrak{h})$ and $\varepsilon^{-1}(W(R)) =$ Aut$_0(\mathfrak{g},\mathfrak{h})$.

a) Let $\varphi \in T_Q$ and let $\overline{k}$ be an algebraic closure of $k$. By Lemma $3,\varphi$ extends to an element of Hom(P(R)$, k^*$). By Prop. 3,

$f(\varphi )\otimes 1\in$ Aut$_e(\mathfrak{g}\otimes_k\overline{k},\mathfrak{h}\otimes_k\overline{k})$.

Hence $f(\varphi )\in$ Aut$_0(\mathfrak{g},\mathfrak{h})$, and Ker $\varepsilon \subset$ Aut$_0(\mathfrak{g},\mathfrak{h})$.

b) The image of Aut$_e(\mathfrak{g},\mathfrak{h})$ under $\varepsilon$ contains W(R) (§2, no. 2, Cor. of Th. 2). In view of a), we see that $\varepsilon^{-1}(W(R))\subset$ Aut$_0(\mathfrak{g},\mathfrak{h})$.

c) It remains to prove that Aut$_0(\mathfrak{g},\mathfrak{h})\subset \varepsilon^{-1}(W(R))$. In view of b), it suffices to prove that $\varepsilon$(Aut$_0(\mathfrak{g},\mathfrak{h})$)$\cap$ Aut(R$,B$), where B denotes a basis of R, reduces to $\{1\}$.

Let $s\in$ Aut$_0(\mathfrak{g},\mathfrak{h})$ be such that $\varepsilon (s)\in$ Aut(R$,B$). The subgroup of A(R) generated by $\varepsilon (s)$ has a finite number of orbits on R. Let U be such an orbit, of cardinal $r$, and $\mathfrak{g}^U=\sum_{\beta\in U}\mathfrak{g}^{\beta}$. Let $\beta_1\in U$, and put $\beta_i=\varepsilon (s)^{i-1}\beta_1$ for

$1\leq i\leq r$, so that $U =\{\beta_1, . . . , \beta_r\}$. Let $X_{\beta_1}$ be a non-zero element of $\mathfrak{g}^{\beta_1}$, and put $X_{\beta_i}=s^{i-1}X_{\beta_1}$ for $1\leq i\leq r$. There exists $c_U\in k^*$ such that $s^rX_{\beta_1}=$ $c_UX_{\beta_1}$, hence $s^rX_{\beta_i}=c_UX_{\beta_i}$ for all $i$, and consequently $s^r|\mathfrak{g}^U=c_U.1$. Let $\varphi \in T_Q$, and $s'=s\circ f(\varphi )$, which by a) is an element of Aut$_0(\mathfrak{g},\mathfrak{h})$. We have ${s'}^r|\mathfrak{g}^U=c'_U.1$, where

Put $B =\{\alpha_1, . . . , \alpha_l\}$ and $\sum_{i=1}^r\beta_i=\sum_{j=1}^lm^U_j\alpha_j$. Since $\varepsilon (s)\in$ Aut(R$,B$), the

$m^U_j$ are integers of the same sign and not all zero. We have

Now $\varphi$ can be chosen so that $c'_U\not= 1$ for every orbit U; indeed, this reduces to choosing elements $\varphi (\alpha_1) =t_1, . . . , \varphi (\alpha_l) =t_l$ of $k^*$ which are not annihilated by a finite number of polynomials in $t_1, . . . , t_l$, not identically zero. For such a choice of $\varphi ,\varepsilon (s') = 1$ by Lemma 2, so