Every real Cartan subalgebra is conjugate to a theta-stable one

The subspace $\mathfrak{h}={\mathfrak{h}}_0\oplus i{\mathfrak{h}}_0$ is a Cartan subalgebra of $\mathfrak{g}$, and for every conjugate-linear involutive automorphism $\rho$ of $\mathfrak{g}$ the image $\rho (\mathfrak{h})$ is again a Cartan subalgebra of $\mathfrak{g}$. Indeed $\mathfrak{h}$ is the complexification of the nilpotent algebra ${\mathfrak{h}}_0$, so its lower central series is the complexification of that of ${\mathfrak{h}}_0$ and terminates, and $\mathfrak{h}$ is nilpotent; if $X\in \mathfrak{g}$ satisfies $[X,\mathfrak{h}]\subseteq \mathfrak{h}$, writing $X=X_1+i{X}_2$ with $X_j\in {\mathfrak{g}}_0$ and comparing the ${\mathfrak{g}}_0$- and $i{\mathfrak{g}}_0$-components of $[X,{\mathfrak{h}}_0]\subseteq {\mathfrak{h}}_0\oplus i{\mathfrak{h}}_0$ gives $[X_j,{\mathfrak{h}}_0]\subseteq {\mathfrak{h}}_0$, so $X_j\in N_{{\mathfrak{g}}_0}({\mathfrak{h}}_0)={\mathfrak{h}}_0$ and $X\in \mathfrak{h}$; hence $N_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$ and $\mathfrak{h}$ is a Cartan subalgebra by [L7]. For the second claim, $\rho (\mathfrak{h})$ is a complex subspace because $\rho (iH)=-i\rho (H)$, it is closed under brackets because $\rho$ preserves them, it is nilpotent because $\rho$ restricts to an isomorphism of real Lie algebras $\mathfrak{h}\to \rho (\mathfrak{h})$, and $N_{\mathfrak{g}}(\rho \mathfrak{h})=\rho (N_{\mathfrak{g}}\mathfrak{h})=\rho (\mathfrak{h})$: for $X=\rho (X^{\mathrm{\prime }})$ one has $[X,\rho H]=\rho [X^{\mathrm{\prime }},H]$, so $X$ normalizes $\rho (\mathfrak{h})$ exactly when $X^{\mathrm{\prime }}$ normalizes $\mathfrak{h}$.

For a compact real form $\mathfrak{u}$ of $\mathfrak{g}$ with conjugation $\rho$, the map $\rho$ is a Cartan involution of ${\mathfrak{g}}_{\mathbb{R}}$ and $B_{\rho }(Z,W):=-B_{\mathbb{R}}(Z,\rho W)$ is an inner product on ${\mathfrak{g}}_{\mathbb{R}}$: writing $Z=X+iY$, $W=X^{\mathrm{\prime }}+i{Y}^{\mathrm{\prime }}$ with $X,Y,X^{\mathrm{\prime }},Y^{\mathrm{\prime }}\in \mathfrak{u}$ one has $B(Z,\rho W)=B(X,X^{\mathrm{\prime }})+B(Y,Y^{\mathrm{\prime }})+i(B(Y,X^{\mathrm{\prime }})-B(X,Y^{\mathrm{\prime }}))$, and hence $\mathrm{Re}B(Z,\rho W)=B(X,X^{\mathrm{\prime }})+B(Y,Y^{\mathrm{\prime }})$, so $B_{\rho }(Z,Z)=-2(B(X,X)+B(Y,Y))>0$ for $Z\mathrm{\ne }0$ because the Killing form of $\mathfrak{u}$ is negative definite, and $B_{\rho }$ is symmetric and bilinear.

Fix a compact real form ${\mathfrak{u}}_0$ of $\mathfrak{g}$ with conjugation $\tau$, as in [L2], choose a maximal abelian subspace $\mathfrak{a}$ of ${\mathfrak{u}}_0$ and put ${\mathfrak{h}}_c:=\mathfrak{a}\oplus i\mathfrak{a}$. Then ${\mathfrak{h}}_c$ is a Cartan subalgebra of $\mathfrak{g}$ with $\tau ({\mathfrak{h}}_c)={\mathfrak{h}}_c$: it is abelian, its normalizer in $\mathfrak{g}$ is itself because an element $X=Y+iZ$ with $Y,Z\in {\mathfrak{u}}_0$ normalizing ${\mathfrak{h}}_c$ has $[Y,\mathfrak{a}]\subseteq \mathfrak{a}$ and $[Z,\mathfrak{a}]\subseteq \mathfrak{a}$, and for $U\in N_{{\mathfrak{u}}_0}(\mathfrak{a})$ and $A,A^{\mathrm{\prime }}\in \mathfrak{a}$, invariance gives $B([U,A],A^{\mathrm{\prime }})=B(U,[A,A^{\mathrm{\prime }}])=0$. Since $[U,A]\in \mathfrak{a}$ and $B$ is negative definite there, $[U,A]=0$. The centralizer of $\mathfrak{a}$ is $\mathfrak{a}$, because adjoining any centralizing vector gives an abelian subspace and $\mathfrak{a}$ is maximal; so ${\mathfrak{h}}_c$ is nilpotent and self-normalizing, and $\tau$ fixes ${\mathfrak{u}}_0$ and hence $\mathfrak{a}$ and ${\mathfrak{h}}_c$. By step 1.1 both $\mathfrak{h}$ and ${\mathfrak{h}}_c$ are Cartan subalgebras of $\mathfrak{g}$, so by [L3] there is an inner automorphism $\phi$ of $\mathfrak{g}$ with $\phi (\mathfrak{h})={\mathfrak{h}}_c$. Then ${\tau }^{\mathrm{\prime }}:={\phi }^{-1}\tau \phi$ is a conjugate-linear involutive automorphism of $\mathfrak{g}$ whose fixed algebra ${\mathfrak{u}}_0^{\mathrm{\prime }}:={\phi }^{-1}({\mathfrak{u}}_0)$ is again a compact real form of $\mathfrak{g}$, since automorphisms preserve the Killing form, and it satisfies ${\tau }^{\mathrm{\prime }}(\mathfrak{h})={\phi }^{-1}(\tau ({\mathfrak{h}}_c))={\phi }^{-1}({\mathfrak{h}}_c)=\mathfrak{h}$.

Put $\omega :=\sigma {\tau }^{\mathrm{\prime }}$, a complex-linear automorphism of $\mathfrak{g}$ because $\sigma$ and ${\tau }^{\mathrm{\prime }}$ are conjugate-linear involutions; it satisfies $\omega {\tau }^{\mathrm{\prime }}={\tau }^{\mathrm{\prime }}{\omega }^{-1}$ since both sides equal $\sigma$, and it preserves $\mathfrak{h}$, since $\sigma (\mathfrak{h})=\mathfrak{h}$ by [L1] and ${\tau }^{\mathrm{\prime }}(\mathfrak{h})=\mathfrak{h}$ by step 2.1. Put $\rho :={\omega }^2$. Then $\rho$ is an automorphism of ${\mathfrak{g}}_{\mathbb{R}}$ with $\rho (\mathfrak{h})\subseteq \mathfrak{h}$, it is self-adjoint and positive definite for the inner product $B_{{\tau }^{\mathrm{\prime }}}$ of step 1.2, and it satisfies $\rho {\tau }^{\mathrm{\prime }}={\tau }^{\mathrm{\prime }}{\rho }^{-1}$: invariance of $B_{\mathbb{R}}$ under ${\omega }^{-1}$ and ${\tau }^{\mathrm{\prime }}$ gives $B_{{\tau }^{\mathrm{\prime }}}(\omega Z,W)=-B_{\mathbb{R}}(\omega Z,{\tau }^{\mathrm{\prime }}W)=-B_{\mathbb{R}}(Z,{\omega }^{-1}{\tau }^{\mathrm{\prime }}W)=-B_{\mathbb{R}}(Z,{\tau }^{\mathrm{\prime }}\omega W)=B_{{\tau }^{\mathrm{\prime }}}(Z,\omega W)$, so $\omega$ is self-adjoint, $\rho ={\omega }^2={\omega }^{\ast }\omega$ satisfies $B_{{\tau }^{\mathrm{\prime }}}(\rho Z,Z)=B_{{\tau }^{\mathrm{\prime }}}(\omega Z,\omega Z)>0$ for $Z\mathrm{\ne }0$, and $\rho {\tau }^{\mathrm{\prime }}={\omega }^2{\tau }^{\mathrm{\prime }}=\omega (\omega {\tau }^{\mathrm{\prime }})=\omega ({\tau }^{\mathrm{\prime }}{\omega }^{-1})=(\omega {\tau }^{\mathrm{\prime }}){\omega }^{-1}={\tau }^{\mathrm{\prime }}{\omega }^{-1}{\omega }^{-1}={\tau }^{\mathrm{\prime }}{\rho }^{-1}$.

By [L6] applied to the self-adjoint operator $\rho$ and the $\rho$-invariant subspace $\mathfrak{h}$ of step 3.1 (its orthogonal complement is then also $\rho$-invariant), choose a $B_{{\tau }^{\mathrm{\prime }}}$-orthonormal basis of ${\mathfrak{g}}_{\mathbb{R}}$ consisting of eigenvectors of $\rho$ and containing a basis of $\mathfrak{h}$; let ${\rho }^r$, $r\in \mathbb{R}$, act on the eigenspace of $\rho$ for the eigenvalue $\lambda >0$ by multiplication by ${\lambda }^r$. Then ${\rho }^r(\mathfrak{h})\subseteq \mathfrak{h}$ for every real $r$, and ${\rho }^r$ is an automorphism of ${\mathfrak{g}}_{\mathbb{R}}$: if $X,Y$ are eigenvectors with eigenvalues ${\lambda }_i,{\lambda }_j$, then $\rho [X,Y]=[\rho X,\rho Y]={\lambda }_i{\lambda }_j[X,Y]$, so $[X,Y]$ lies in the eigenspace for ${\lambda }_i{\lambda }_j$ (or is $0$) and ${\rho }^r[X,Y]=({\lambda }_i{\lambda }_j{)}^r[X,Y]=[{\rho }^{r}X,{\rho }^{r}Y]$, which extends to all $X,Y$ by bilinearity; moreover ${\rho }^r$ commutes with $\rho$ and with $\omega$ because $\omega \rho =\rho \omega$ and $\omega$ preserves every $\rho$-eigenspace.

Let $D$ be the endomorphism acting as $\log \lambda$ on the eigenspace of $\rho$ for the eigenvalue $\lambda >0$, so that $e^D=\rho$; then $D$ is self-adjoint for $B_{{\tau }^{\mathrm{\prime }}}$, and $D$ is a derivation of ${\mathfrak{g}}_{\mathbb{R}}$, since for eigenvectors $X,Y$ as in step 4.1 one has $D[X,Y]=(\log {\lambda }_i+\log {\lambda }_j)[X,Y]=[DX,Y]+[X,DY]$ and the identity extends by bilinearity. By [L5] there is a unique $X\in {\mathfrak{g}}_{\mathbb{R}}$ with $D={\mathrm{ad}}_X$, so $\rho =e^{{\mathrm{ad}}_X}$ and the automorphism ${\phi }_0:={\rho }^{1\mathrm{/}4}=e^{{\mathrm{ad}}_{X\mathrm{/}4}}$ satisfies ${\phi }_0(\mathfrak{h})\subseteq \mathfrak{h}$ by step 4.1.

Define $\psi :={\phi }_0{\tau }^{\mathrm{\prime }}{\phi }_0^{-1}$. Then ${\psi }^2=\mathrm{i}\mathrm{d}$, and $\psi$ is a Cartan involution of ${\mathfrak{g}}_{\mathbb{R}}$, because $B_{\psi }(Z,W)=B_{{\tau }^{\mathrm{\prime }}}({\phi }_0^{-1}Z,{\phi }_0^{-1}W)$ is positive definite by step 1.2; moreover $\psi (\mathfrak{h})={\phi }_0{\tau }^{\mathrm{\prime }}{\phi }_0^{-1}(\mathfrak{h})=\mathfrak{h}$ because ${\phi }_0^{\pm 1}(\mathfrak{h})\subseteq \mathfrak{h}$ and ${\tau }^{\mathrm{\prime }}(\mathfrak{h})=\mathfrak{h}$. Finally $\psi$ commutes with $\sigma$: using ${\tau }^{\mathrm{\prime }}\sigma ={\omega }^{-1}$ and $\rho {\tau }^{\mathrm{\prime }}={\tau }^{\mathrm{\prime }}{\rho }^{-1}$ from step 3.1, the identities ${\rho }^s\sigma =\sigma {\rho }^{-s}$ and ${\tau }^{\mathrm{\prime }}{\rho }^{-1\mathrm{/}4}={\rho }^{1\mathrm{/}4}{\tau }^{\mathrm{\prime }}$ for all real $s$ (valid on the $\rho$-eigenspaces, on each of which ${\rho }^s\sigma =\sigma (\sigma {\rho }^s\sigma )=\sigma {\rho }^{-s}$ because $\sigma \rho {\sigma }^{-1}={\rho }^{-1}$) give $\psi \sigma ={\phi }_0{\tau }^{\mathrm{\prime }}{\phi }_0^{-1}\sigma ={\rho }^{1\mathrm{/}4}{\tau }^{\mathrm{\prime }}{\rho }^{-1\mathrm{/}4}\sigma ={\rho }^{1\mathrm{/}4}{\rho }^{1\mathrm{/}4}{\tau }^{\mathrm{\prime }}\sigma ={\rho }^{1\mathrm{/}2}{\omega }^{-1}={\rho }^{-1\mathrm{/}2}\omega =\omega {\rho }^{-1\mathrm{/}2}={\rho }^{-1\mathrm{/}4}\omega {\rho }^{-1\mathrm{/}4}={\rho }^{-1\mathrm{/}4}\sigma {\tau }^{\mathrm{\prime }}{\rho }^{-1\mathrm{/}4}=\sigma {\rho }^{1\mathrm{/}4}{\tau }^{\mathrm{\prime }}{\rho }^{-1\mathrm{/}4}=\sigma \psi$.

Since $\psi$ commutes with $\sigma$ and ${\mathfrak{g}}_0$ is the fixed locus of $\sigma$ in ${\mathfrak{g}}_{\mathbb{R}}$ by [L1], $\psi$ preserves ${\mathfrak{g}}_0$: for $Z\in {\mathfrak{g}}_0$ one has $\sigma (\psi Z)=\psi (\sigma Z)=\psi Z$. Hence $\eta :=\psi {\mathrm{\mid }}_{{\mathfrak{g}}_0}$ is an involutive automorphism of ${\mathfrak{g}}_0$, and it is a Cartan involution: for $X\in {\mathfrak{g}}_0\setminus \{0\}$ one has $(B_0{)}_{\eta }(X,X)=-B_0(X,\eta X)=-\frac{1}{2}{B}_{\mathbb{R}}(X,\psi X)=\frac{1}{2}{B}_{\psi }(X,X)>0$, because $B$ restricts to $B_0$ on ${\mathfrak{g}}_0$ and $B_{\mathbb{R}}(X,\psi X)=2\mathrm{Re}B(X,\psi X)=2B(X,\psi X)$ there. Moreover $\psi (\mathfrak{h})=\mathfrak{h}$ with $\psi ({\mathfrak{g}}_0)={\mathfrak{g}}_0$ gives $\eta ({\mathfrak{h}}_0)=\psi (\mathfrak{h}\cap {\mathfrak{g}}_0)=\psi (\mathfrak{h})\cap \psi ({\mathfrak{g}}_0)=\mathfrak{h}\cap {\mathfrak{g}}_0={\mathfrak{h}}_0$.

By [L4] applied to the two Cartan involutions $\eta$ and $\theta$ of ${\mathfrak{g}}_0$ there is $Y\in {\mathfrak{g}}_0$ with $\theta =e^{{\mathrm{ad}}_Y}\eta e^{-{\mathrm{ad}}_Y}$. Put $\chi :=e^{{\mathrm{ad}}_Y}$, an inner automorphism of ${\mathfrak{g}}_0$. Then $\theta (\chi ({\mathfrak{h}}_0))=\chi (\eta ({\mathfrak{h}}_0))=\chi ({\mathfrak{h}}_0)$ by step 7.1, so $\chi ({\mathfrak{h}}_0)$ is a $\theta$-stable subspace of ${\mathfrak{g}}_0$; being the image of a Cartan subalgebra under an automorphism it is again a Cartan subalgebra of ${\mathfrak{g}}_0$. Thus ${\mathfrak{h}}_0$ is conjugate by the inner automorphism $\chi$ to the $\theta$-stable Cartan subalgebra $\chi ({\mathfrak{h}}_0)$, and the theorem follows.