Proposition 10

Let $\mathfrak{b},\mathfrak{b}'$ be Borel subalgebras of $\mathfrak{g}$. There exists a splitting Cartan subalgebra of $\mathfrak{g}$ contained in $\mathfrak{b}\cap \mathfrak{b}'$.

Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{b},\mathfrak{n}= [\mathfrak{b},\mathfrak{b}],\mathfrak{n}'= [\mathfrak{b}',\mathfrak{b}'],\mathfrak{p}=\mathfrak{b}\cap \mathfrak{b}'$, and $\mathfrak{s}$ a vector subspace of $\mathfrak{g}$ complementary to $\mathfrak{b}+\mathfrak{b}'$. Denote by $\mathfrak{s}^{\bot},\mathfrak{b}^{\bot},\mathfrak{b}^{'\bot}$ the orthogonal complements of $\mathfrak{s},\mathfrak{b},\mathfrak{b}'$ with respect to the Killing form of $\mathfrak{g}$. Put $l=$ dim$\mathfrak{h}, n=$ dim$\mathfrak{n}, p=$ dim$\mathfrak{p}$. Then dim$\mathfrak{b}=$ dim$\mathfrak{b}'=l+n$,

dim$\mathfrak{s}^{\bot}=$ dim($\mathfrak{b}+\mathfrak{b}'$) $= 2(l+n)-p$,

and so

dim($\mathfrak{s}^{\bot}\cap \mathfrak{p}$)$\geq$ dim$\mathfrak{s}^{\bot}+$ dim$\mathfrak{p}-$ dim$\mathfrak{g}$ (3)

By Prop. 1 of §2, no. $2,\mathfrak{n}\subset \mathfrak{b}^{\bot},\mathfrak{n}'\subset \mathfrak{b}^{'\bot}$. The elements of $\mathfrak{p}\cap \mathfrak{n}$ are nilpotent in $\mathfrak{g}$ (Prop. 9 (ii)), and belong to $\mathfrak{b}'$, and hence to $\mathfrak{n}'$ (Prop. 9 (ii)). Consequently, $\mathfrak{p}\cap \mathfrak{n}\subset \mathfrak{n}\cap \mathfrak{n}'\subset \mathfrak{b}^{\bot}\cap \mathfrak{b}^{'\bot}$, so $\mathfrak{s}^{\bot}\cap \mathfrak{p}\cap \mathfrak{n}= 0$. In view of (3), we see that $\mathfrak{s}^{\bot}\cap \mathfrak{p}$ is a complement of $\mathfrak{n}$ in $\mathfrak{b}$. Let $z$ be an element of $\mathfrak{h}$ regular in $\mathfrak{g}$; there exists $y\in \mathfrak{n}$ such that $y+z\in \mathfrak{s}^{\bot}\cap \mathfrak{p}$; by Prop. 9 (i), ad$_{\mathfrak{g}}(y+z)$ has the same characteristic polynomial as ad$_{\mathfrak{g}}z$, so $x=y+z$ is regular in $\mathfrak{g}$ and a fortiori in $\mathfrak{b}$ and $\mathfrak{b}'$ (Chap. VII, §2, no. 2, Prop. 9). Since $\mathfrak{g},\mathfrak{b},\mathfrak{b}'$ have the same rank, $\mathfrak{b}^0(x) =\mathfrak{g}^0(x) ={\mathfrak{b}'}^0(x)$ is simultaneously a Cartan subalgebra of $\mathfrak{b}$, of $\mathfrak{g}$ and of $\mathfrak{b}'$ (Chap. VII, §3, no. 3, Th. 2). Finally, this Cartan subalgebra of $\mathfrak{g}$ is splitting by the Cor. of Prop. 9.