Proposition 2

$a)$ The real vector subspace $\mathfrak{a}_0=\mathfrak{h}_0+\sum_{\alpha\in R}\mathbf{R}X_{\alpha}$ of $\mathfrak{a}$ is a

real form of $\mathfrak{a}$, of which $\mathfrak{h}_0$ is a Cartan subalgebra. The pair $(\mathfrak{a}_0,\mathfrak{h}_0)$ is a split semi-simple real Lie algebra, of which $(X_{\alpha})$ is a Chevalley system.

b) Let $\sigma$ be the conjugation of $\mathfrak{a}$ relative to $\mathfrak{a}_0$. Then $\sigma \circ \theta =\theta \circ \sigma$. The set of fixed points of $\sigma \circ \theta$ is a compact real form $\mathfrak{a}_u$ of $\mathfrak{a}$, of which $i\mathfrak{h}_0$ is a Cartan subalgebra.

Part a) follows immediately from the preceding. We prove b). Since $\sigma \circ \theta$ and $\theta \circ \sigma$ are two semi-linear maps from $\mathfrak{a}$ to $\mathfrak{a}$ that coincide on $\mathfrak{a}_0$, they coincide. Now $\sigma \circ \theta$ satisfies conditions (2) of no. 1, hence is the conjugation of $\mathfrak{a}$ relative to the real form $\mathfrak{a}_u$ consisting of the $x\in \mathfrak{a}$ such that $\sigma \circ \theta (x) =x$ (Prop. 1). For all $\alpha \in R$ put

Then the $\mathbf{R}$-vector space $\mathfrak{a}_u$ is generated by $i\mathfrak{h}_0$, the $u_{\alpha}$ and the $v_{\alpha}$. More precisely, if we choose a chamber C of R, then

It is clear that $i\mathfrak{h}_0$ is a Cartan subalgebra of $\mathfrak{a}_u$, and it remains to prove that the restriction of B to $\mathfrak{a}_u$ is negative. Now $i\mathfrak{h}_0$ and the different subspaces of the form $\mathbf{R}u_{\alpha}\oplus \mathbf{R}v_{\alpha}$ are orthogonal with respect to B, by (5); the restriction of B to $i\mathfrak{h}_0$ is negative and

hence the conclusion.