Proposition 6

Let $\mathfrak{a}$ be a semi-simple Lie algebra and $s\in$ Aut($\mathfrak{a}$). The following conditions are equivalent:

(i) $s\in$ Aut$_0(\mathfrak{a})$;

(ii) for any invariant polynomial function $f$ on $\mathfrak{a}$, we have $f\circ s=f$.

By extending scalars if necessary, we can assume that $k$ is algebraically closed. The implication (i) $=\Rightarrow$ (ii) follows from Prop. 3. We assume that condition (ii) is satisfied and prove (i). In view of Prop. 3, and §5, no. 3, Cor. 1 of Prop. 5, we can assume that $s\in$ Aut($\mathfrak{g},\mathfrak{h}$) and that $s$ leaves stable a Weyl chamber C. Let $x\in C\cap \mathfrak{h}_{\mathbf{Q}}$. We have $sx\in C$. If $g$ is a W-invariant polynomial function on $\mathfrak{h}$, we have $g(x) =g(sx)$ (Th. 1 (i)). By Lemma 6, it follows that $sx\in Wx$. Since $sx\in C$, we have $x=sx$ (Chap. V, §3, no. 3, Th. 2). Then $s|\mathfrak{h}=$ Id$_{\mathfrak{h}}$, and $s\in$ Aut$_0(\mathfrak{g},\mathfrak{h}) ($§5, no. 2, Prop. 4).