Conjugacy of Cartan involutions

Put $\omega :={\theta }^{\mathrm{\prime }}\theta \in \mathrm{Aut}({\mathfrak{g}}_0)$ and $B_1:=B_{\theta }$. The identity $\theta \omega ={\omega }^{-1}\theta$ holds because both sides equal $\theta {\theta }^{\mathrm{\prime }}\theta$. Since $B$ is invariant under the automorphisms ${\omega }^{-1}$ and $\theta$, $B_1(\omega Z,W)=-B(\omega Z,\theta W)=-B(Z,{\omega }^{-1}\theta W)=-B(Z,\theta \omega W)=B_1(Z,\omega W)$ for all $Z,W$: the automorphism $\omega$ is self-adjoint for the inner product $B_1$.

$\rho :={\omega }^2={\omega }^{\ast }\omega$ is positive definite and self-adjoint for $B_1$: it is self-adjoint because $\omega$ is, and $B_1(\rho Z,Z)=B_1(\omega Z,\omega Z)>0$ for $Z\mathrm{\ne }0$ because $\omega$ is invertible.

Let now $\alpha ,\beta$ be commuting Cartan involutions of ${\mathfrak{g}}_0$ and let $Y$ satisfy $\alpha Y=Y$, $\beta Y=-Y$. Then $0<B_{\alpha }(Y,Y)=-B(Y,\alpha Y)=-B(Y,Y)$ and $0<B_{\beta }(Y,Y)=-B(Y,\beta Y)=+B(Y,Y)$, a contradiction; hence the simultaneous $(+1,-1)$ eigenspace of $(\alpha ,\beta )$ is zero, and exchanging $\alpha$ and $\beta$ shows that the $(-1,+1)$ eigenspace is zero as well. On the two remaining eigenspaces the two involutions agree, so $\alpha =\beta$.

By [L3] choose a $B_1$-orthonormal eigenbasis of $\rho$ with eigenvalues ${\lambda }_j>0$ and define ${\rho }^r$, for real $r$, to act as ${\lambda }_j^r$ on the eigenspace for ${\lambda }_j$. If $X,Y$ are eigenvectors with eigenvalues ${\lambda }_i,{\lambda }_j$, then $\rho \in \mathrm{Aut}({\mathfrak{g}}_0)$ gives $\rho [X,Y]=[\rho X,\rho Y]={\lambda }_i{\lambda }_j[X,Y]$, so $[X,Y]$ lies in the eigenspace for ${\lambda }_i{\lambda }_j$; hence ${\rho }^r[X,Y]=({\lambda }_i{\lambda }_j{)}^r[X,Y]=[{\rho }^{r}X,{\rho }^{r}Y]$ and, by bilinearity, ${\rho }^r\in \mathrm{Aut}({\mathfrak{g}}_0)$. Also ${\rho }^r$ is a function of $\rho$, so it commutes with $\rho$ and with $\omega$.

Let $D$ act as $\log {\lambda }_j$ on the eigenspace for ${\lambda }_j$ of $\rho$, so that $e^D=\rho$ and $D$ is self-adjoint for $B_1$. For eigenvectors $X,Y$ as in step 3.1, $D[X,Y]=(\log {\lambda }_i+\log {\lambda }_j)[X,Y]=[DX,Y]+[X,DY]$, so $D$ is a derivation of ${\mathfrak{g}}_0$ by bilinearity; by [L2] there is a unique $X\in {\mathfrak{g}}_0$ with $D=\mathrm{ad}X$, and then $\rho =e^{\mathrm{ad}X}$.

The powers of $\rho$ satisfy ${\rho }^r\theta =\theta {\rho }^{-r}$ for all real $r$: from $\rho \theta ={\omega }^2\theta =\omega (\theta {\omega }^{-1})=(\omega \theta ){\omega }^{-1}=\theta {\omega }^{-1}{\omega }^{-1}=\theta {\rho }^{-1}$ one gets $\rho \theta Z={\lambda }^{-1}\theta Z$ on an eigenvector of eigenvalue $\lambda$, and iteration gives the identity. Put $\phi :={\rho }^{1\mathrm{/}4}$. Then $\phi \theta {\phi }^{-1}{\theta }^{\mathrm{\prime }}={\rho }^{1\mathrm{/}4}\theta {\rho }^{-1\mathrm{/}4}{\theta }^{\mathrm{\prime }}={\rho }^{1\mathrm{/}2}\theta {\theta }^{\mathrm{\prime }}={\rho }^{1\mathrm{/}2}{\omega }^{-1}={\rho }^{-1\mathrm{/}2}\omega =\omega {\rho }^{-1\mathrm{/}2}={\theta }^{\mathrm{\prime }}\theta {\rho }^{-1\mathrm{/}2}={\theta }^{\mathrm{\prime }}{\rho }^{1\mathrm{/}4}\theta {\rho }^{-1\mathrm{/}4}={\theta }^{\mathrm{\prime }}\phi \theta {\phi }^{-1},$ using ${\rho }^{1\mathrm{/}2}{\omega }^{-1}={\rho }^{-1\mathrm{/}2}\rho {\omega }^{-1}={\rho }^{-1\mathrm{/}2}\omega$, the commutation of $\omega$ with powers of $\rho$, and $\theta {\rho }^{-1\mathrm{/}4}={\rho }^{1\mathrm{/}4}\theta$. Hence $\psi :=\phi \theta {\phi }^{-1}$ is an involution of ${\mathfrak{g}}_0$ commuting with ${\theta }^{\mathrm{\prime }}$, and it is again a Cartan involution: $B_{\psi }(\phi Z,\phi W)=B_1(Z,W)$ is positive definite because $\phi$ is an automorphism and $\theta$ is a Cartan involution.

Applying step 2.2 to the commuting pair $\psi ,{\theta }^{\mathrm{\prime }}$ gives $\psi ={\theta }^{\mathrm{\prime }}$, that is, ${\theta }^{\mathrm{\prime }}=\phi \theta {\phi }^{-1}$ with $\phi ={\rho }^{1\mathrm{/}4}=e^{\mathrm{ad}(X\mathrm{/}4)}$ for the element $X$ of step 4.1. Thus ${\theta }^{\mathrm{\prime }}$ is the conjugate of $\theta$ by the inner automorphism $e^{\mathrm{ad}(X\mathrm{/}4)}$, and the theorem follows.