Proposition 3

The image of the composite homomorphism

$T_P\longrightarrow^qT_Q\longrightarrow^f$ Aut($\mathfrak{g},\mathfrak{h}$)

is contained in Aut$_e(\mathfrak{g})$.

Let B be a basis of R. Then $(H_{\alpha})_{\alpha\in B}$ is a basis of $R^{\vee}$, and the dual basis of $(H_{\alpha})_{\alpha\in B}$ in $\mathfrak{h}^*$ is a basis of the group P(R). Hence the group $T_P$ is generated by the homomorphisms $\lambda \rightarrow t^{\lambda(H_{\alpha})}(t\in k^*, \alpha \in B)$. If $\varphi$ is the restriction of such a homomorphism to Q(R), Lemma 1 proves that $f(\varphi )\in$ Aut$_e(\mathfrak{g})$, hence the proposition.

Let $\overline{k}$ be an algebraic closure of $k$. The map which associates to any automorphism $s$ of $\mathfrak{g}$ the automorphism $s\otimes 1$ of $\mathfrak{g}\otimes_k\overline{k}$ is an injective homomorphism from Aut($\mathfrak{g}$) to Aut($\mathfrak{g}\otimes_k\overline{k}$). We denote by Aut$_0(\mathfrak{g})$ the normal subgroup of Aut($\mathfrak{g}$) which is the inverse image of Aut$_e(\mathfrak{g}\otimes_k\overline{k})$ under this homomorphism; this is the set of automorphisms of $\mathfrak{g}$ that become elementary on extending the base field from $k$ to $\overline{k}$. It is clear that Aut$_e(\mathfrak{g})$ is independent of the choice of $\overline{k}$, and that Aut$_e(\mathfrak{g})\subset$ Aut$_0(\mathfrak{g})$. The groups Aut$_0(\mathfrak{g})$ and Aut$_e(\mathfrak{g})$ can be distinct (Chap. VII, §13, no. 1). If $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$, put

Aut$_e(\mathfrak{g},\mathfrak{h}) =$ Aut$_e(\mathfrak{g})\cap$ Aut($\mathfrak{g},\mathfrak{h}$), Aut$_0(\mathfrak{g},\mathfrak{h}) =$ Aut$_0(\mathfrak{g})\cap$ Aut($\mathfrak{g},\mathfrak{h}$).