Bracket relations and Killing signs in a Cartan decomposition

Given: A real semisimple Lie algebra ${\mathfrak{g}}_0$ with Cartan involution $\theta$, Cartan decomposition ${\mathfrak{g}}_0={\mathfrak{k}}_0\oplus {\mathfrak{p}}_0$, and Killing form $B$.

$\theta$ is an involutive Lie-algebra automorphism and $B_{\theta }(X,Y)=-B(X,\theta Y)$ is positive definite (Cartan involution of a real semisimple Lie algebra, Cartan decomposition of a real semisimple Lie algebra).

The Killing form is symmetric and invariant: $B([X,Y],Z)=B(X,[Y,Z])$, and $B$ is nondegenerate on ${\mathfrak{g}}_0$ (Killing form, Cartan's semisimplicity criterion).

Proof technique: direct.

1.1 Let $X,Y$ be eigenvectors of $\theta$ with eigenvalues $\epsilon ,\delta \in \{1,-1\}$. Since $\theta$ is an automorphism, $\theta [X,Y]=[\theta X,\theta Y]=\epsilon \delta [X,Y]$, so $[X,Y]$ lies in the $\epsilon \delta$-eigenspace. This gives the three bracket relations at once. [L1, algebra]

2.1 For $X\in {\mathfrak{k}}_0$ and $Y\in {\mathfrak{p}}_0$, invariance and the eigenvalue relation give $B(X,Y)=B(\theta X,\theta Y)=B(X,-Y)=-B(X,Y)$, hence $2B(X,Y)=0$ and $B(X,Y)=0$: the two summands are orthogonal. [L1, L2, step 1.1, algebra]

3.1 For $X\in {\mathfrak{k}}_0$ we have $B(X,X)=-B(X,\theta X)=-B_{\theta }(X,X)\le 0$, with equality only for $X=0$; hence $B$ is negative definite on ${\mathfrak{k}}_0$. For $Y\in {\mathfrak{p}}_0$ we have $B(Y,Y)=B(Y,-\theta Y)=-B_{\theta }(Y,-Y)=B_{\theta }(Y,Y)\ge 0$ with equality only for $Y=0$; hence $B$ is positive definite on ${\mathfrak{p}}_0$. [L1, step 1.1, step 2.1, algebra]

4.1 By step 1.1, ${\mathfrak{k}}_0$ is closed under brackets and is therefore a Lie subalgebra; the restriction of $\theta$ to ${\mathfrak{k}}_0$ is the identity and to ${\mathfrak{p}}_0$ is minus the identity by definition of the eigenspaces. Together with steps 2.1 and 3.1 this proves all the assertions. [L1, step 1.1, step 2.1, step 3.1, algebra] ∎