Proposition 7

Let $\mathfrak{g}$ be a splittable semi-simple Lie algebra, $\mathfrak{b}$ a Borel subalgebra of $\mathfrak{g}$, and $\mathfrak{p}_1$ and $\mathfrak{p}_2$ distinct parabolic subalgebras of $\mathfrak{g}$ containing $\mathfrak{b}$. Then $\mathfrak{p}_1$ and $\mathfrak{p}_2$ are not conjugate under Aut$_0(\mathfrak{g})$.

We can assume that $k$ is algebraically closed. Let $s\in$ Aut$_0(\mathfrak{g})$ be such that $s(\mathfrak{p}_1) =\mathfrak{p}_2$. Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{b}\cap s(\mathfrak{b}) ($§3, no. 3, Prop. 10). Since $\mathfrak{h}$ and $s(\mathfrak{h})$ are Cartan subalgebras of $s(\mathfrak{b})$, there exists $u\in [\mathfrak{b},\mathfrak{b}]$ such that $e^{adu}(\mathfrak{h}) =s(\mathfrak{h})$ (Chap. VII, §3, no. 4, Th. 3). Replacing $s$ by $e^{-adu}s$, we are reduced to the case in which $s(\mathfrak{h}) =\mathfrak{h}$, and $s$ then induces on $\mathfrak{h}$ an element $\sigma$ of the Weyl group W of $(\mathfrak{g},\mathfrak{h})$ (Prop. 4). Let C be the Weyl chamber corresponding to $\mathfrak{b}$. Then $\mathfrak{p}_1$ and $\mathfrak{p}_2$ correspond to facets $F_1$ and $F_2$ of $\mathfrak{h}_{\mathbf{R}}$ contained in the closure of C. We have $\sigma (F_1) = F_2$. Since $\sigma \in W$, this implies that $F_1= F_2$ (Chap. V, §3, no. 3, Th. 2) so $\mathfrak{p}_1=\mathfrak{p}_2$.