Conjugacy of compact real forms

Regarded as a real Lie algebra, ${\mathfrak{g}}_{\mathbb{R}}$ has the same underlying set as $\mathfrak{g}$ and its adjoint operators are the realifications of those of $\mathfrak{g}$: for $Z\in \mathfrak{g}$ the real-linear map ${\mathrm{ad}}_Z$ on ${\mathfrak{g}}_{\mathbb{R}}$ is the realification of the complex-linear ${\mathrm{ad}}_Z$ on $\mathfrak{g}$. Since the real trace of the realification of a complex-linear endomorphism is twice its complex trace, $B_{\mathbb{R}}(Z,W)=2\mathrm{Re}B(Z,W)$ for all $Z,W\in \mathfrak{g}$. If $Z$ lies in the radical of $B_{\mathbb{R}}$, then $\mathrm{Re}B(Z,W)=0$ for every $W$, and substituting $iW$ gives $-\mathrm{Im}B(Z,W)=0$; hence $B(Z,W)=0$ for every $W$ and $Z=0$ by nondegeneracy of $B$. Thus $B_{\mathbb{R}}$ is nondegenerate and ${\mathfrak{g}}_{\mathbb{R}}$ is semisimple by the criterion, so every derivation of ${\mathfrak{g}}_{\mathbb{R}}$ is inner and $Z({\mathfrak{g}}_{\mathbb{R}})=0$.

For each $i$ the map ${\tau }_i$ is real-linear with ${\tau }_i^2=\mathrm{i}\mathrm{d}$ and preserves brackets, and for all $Z,W\in \mathfrak{g}$ one has $B({\tau }_{i}Z,{\tau }_{i}W)=\overline{B(Z,W)}$: writing $Z=X+iY$ and $W=X^{\mathrm{\prime }}+i{Y}^{\mathrm{\prime }}$ with $X,Y,X^{\mathrm{\prime }},Y^{\mathrm{\prime }}\in {\mathfrak{u}}_i$ and using complex bilinearity and symmetry of $B$, both sides equal $B(X,X^{\mathrm{\prime }})-B(Y,Y^{\mathrm{\prime }})-i(B(X,Y^{\mathrm{\prime }})+B(Y,X^{\mathrm{\prime }}))$. Consequently $B_{\mathbb{R}}({\tau }_{i}Z,{\tau }_{i}W)=2\mathrm{Re}\overline{B(Z,W)}=B_{\mathbb{R}}(Z,W)$: the form $B_{\mathbb{R}}$ is ${\tau }_i$-invariant.

Each ${\tau }_i$ is a Cartan involution of ${\mathfrak{g}}_{\mathbb{R}}$. Indeed, for $Z=X+iY\mathrm{\ne }0$ with $X,Y\in {\mathfrak{u}}_i$ one computes $(B_{\mathbb{R}}{)}_{{\tau }_i}(Z,Z)=-B_{\mathbb{R}}(Z,{\tau }_{i}Z)=-2\mathrm{Re}B(Z,{\tau }_{i}Z)=-2(B(X,X)+B(Y,Y))$, using $B(Z,{\tau }_{i}Z)=B(X+iY,X-iY)=B(X,X)+B(Y,Y)$. Both summands are $\le 0$ and each vanishes only at $0$ because the Killing form of ${\mathfrak{u}}_i$ is negative definite, so $(B_{\mathbb{R}}{)}_{{\tau }_i}(Z,Z)>0$ for $Z\mathrm{\ne }0$; the form is symmetric and bilinear, hence positive definite, and ${\tau }_i$ is a Cartan involution.

Put $\omega :={\tau }_2{\tau }_1\in \mathrm{Aut}({\mathfrak{g}}_{\mathbb{R}})$, an invertible automorphism, and put $B_1:=(B_{\mathbb{R}}{)}_{{\tau }_1}$. The identity ${\tau }_1\omega ={\omega }^{-1}{\tau }_1$ holds because both sides equal ${\tau }_1{\tau }_2{\tau }_1$. Invariance of $B_{\mathbb{R}}$ under ${\omega }^{-1}$ and under ${\tau }_1$ therefore gives, for all $Z,W$, $B_1(\omega Z,W)=-B_{\mathbb{R}}(\omega Z,{\tau }_{1}W)=-B_{\mathbb{R}}(Z,{\omega }^{-1}{\tau }_{1}W)=-B_{\mathbb{R}}(Z,{\tau }_1\omega W)=B_1(Z,\omega W),$ so $\omega$ is self-adjoint for the inner product $B_1$. Since $\omega$ is invertible, $\rho :={\omega }^2={\omega }^{\ast }\omega$ satisfies $B_1(\rho Z,Z)=B_1(\omega Z,\omega Z)>0$ for $Z\mathrm{\ne }0$: the automorphism $\rho$ is self-adjoint and positive definite for $B_1$.

Let $\theta ,{\theta }^{\mathrm{\prime }}$ be Cartan involutions of a real semisimple Lie algebra with $\theta {\theta }^{\mathrm{\prime }}={\theta }^{\mathrm{\prime }}\theta$ and let $Y$ satisfy $\theta Y=Y$, ${\theta }^{\mathrm{\prime }}Y=-Y$. Then $0<B_{\theta }(Y,Y)=-B_{\mathbb{R}}(Y,\theta Y)=-B_{\mathbb{R}}(Y,Y)$ and $0<B_{{\theta }^{\mathrm{\prime }}}(Y,Y)=-B_{\mathbb{R}}(Y,{\theta }^{\mathrm{\prime }}Y)=+B_{\mathbb{R}}(Y,Y)$, a contradiction; hence the simultaneous $(+1,-1)$ eigenspace of $(\theta ,{\theta }^{\mathrm{\prime }})$ is zero. Exchanging the roles of $\theta$ and ${\theta }^{\mathrm{\prime }}$ shows that the $(-1,+1)$ eigenspace is zero too, so $\theta ={\theta }^{\mathrm{\prime }}$ on ${\mathfrak{g}}_{\mathbb{R}}$.

By [L3] there is a $B_1$-orthonormal basis of ${\mathfrak{g}}_{\mathbb{R}}$ consisting of eigenvectors of $\rho$, with eigenvalues ${\lambda }_1,\dots ,{\lambda }_n>0$. For real $r$ define ${\rho }^r$ as the endomorphism acting as ${\lambda }_j^r$ on the eigenspace of $\rho$ for ${\lambda }_j$. If $X,Y$ are eigenvectors with eigenvalues ${\lambda }_i,{\lambda }_j$, then from $\rho \in \mathrm{Aut}({\mathfrak{g}}_{\mathbb{R}})$ one gets $\rho [X,Y]=[\rho X,\rho Y]={\lambda }_i{\lambda }_j[X,Y]$, 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 ${\rho }^r$ is an automorphism of ${\mathfrak{g}}_{\mathbb{R}}$. Also ${\rho }^r$ is a function of $\rho$, so it commutes with $\rho$ and with $\omega$.

Define the endomorphism $D$ of ${\mathfrak{g}}_{\mathbb{R}}$ to act as $\log {\lambda }_j$ on the eigenspace for ${\lambda }_j$, so that $\exp (D)=\rho$ and $D$ is self-adjoint for $B_1$; for eigenvectors $X,Y$ as in step 4.1, $D[X,Y]=(\log {\lambda }_i+\log {\lambda }_j)[X,Y]=[DX,Y]+[X,DY]$, and by bilinearity $D$ is a derivation of ${\mathfrak{g}}_{\mathbb{R}}$. By step 1.1 there is a unique $X\in {\mathfrak{g}}_{\mathbb{R}}$ with $D=\mathrm{ad}X$, and then $\rho =\exp (\mathrm{ad}X)$ lies in the subgroup generated by the automorphisms $\exp (\mathrm{ad}W)$, $W\in {\mathfrak{g}}_{\mathbb{R}}=\mathfrak{g}$.

The real powers of $\rho$ satisfy ${\rho }^r{\tau }_1={\tau }_1{\rho }^{-r}$ for all $r$: on an eigenvector $X$ of $\rho$ with eigenvalue $\lambda$ and using ${\tau }_1\rho ={\rho }^{-1}{\tau }_1$ (which follows from $\rho {\tau }_1={\omega }^2{\tau }_1=\omega ({\tau }_1{\omega }^{-1})=(\omega {\tau }_1){\omega }^{-1}={\tau }_1{\omega }^{-1}{\omega }^{-1}={\tau }_1{\rho }^{-1}$, that is, $\rho {\tau }_1={\tau }_1{\rho }^{-1}$) one gets $\rho {\tau }_{1}X={\tau }_1{\rho }^{-1}X={\lambda }^{-1}{\tau }_{1}X$, and iteration gives ${\rho }^r{\tau }_{1}X={\lambda }^{-r}{\tau }_{1}X={\tau }_1{\rho }^{-r}X$. Put $\phi :={\rho }^{1\mathrm{/}4}$, so that $\phi$ is an automorphism of ${\mathfrak{g}}_{\mathbb{R}}$ lying in the subgroup generated by the $\exp (\mathrm{ad}W)$. Then $\phi {\tau }_1{\phi }^{-1}{\tau }_2={\rho }^{1\mathrm{/}4}{\tau }_1{\rho }^{-1\mathrm{/}4}{\tau }_2={\rho }^{1\mathrm{/}2}{\tau }_1{\tau }_2={\rho }^{1\mathrm{/}2}{\omega }^{-1}={\rho }^{-1\mathrm{/}2}\omega =\omega {\rho }^{-1\mathrm{/}2}={\tau }_2{\tau }_1{\rho }^{-1\mathrm{/}2}={\tau }_2{\rho }^{1\mathrm{/}4}{\tau }_1{\rho }^{-1\mathrm{/}4}={\tau }_2\phi {\tau }_1{\phi }^{-1},$ 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 ${\tau }_1{\rho }^{-1\mathrm{/}4}={\rho }^{1\mathrm{/}4}{\tau }_1$.

By step 5.2 the involution $\psi :=\phi {\tau }_1{\phi }^{-1}$ commutes with ${\tau }_2$. It is again a Cartan involution of ${\mathfrak{g}}_{\mathbb{R}}$: $B_{\psi }(\phi Z,\phi W)=B_1(Z,W)$ is positive definite because $\phi$ is an automorphism and ${\tau }_1$ is a Cartan involution by step 2.1. Hence step 3.2 gives $\psi ={\tau }_2$, that is, ${\tau }_2=\phi {\tau }_1{\phi }^{-1}$.

Taking fixed loci and using [L1], ${\mathfrak{u}}_2=\{Z:{\tau }_{2}Z=Z\}=\{Z:\phi {\tau }_1{\phi }^{-1}Z=Z\}=\phi \{Z:{\tau }_{1}Z=Z\}=\phi ({\mathfrak{u}}_1)$. The automorphism $\phi$ is complex-linear because $\rho ={\omega }^2$ is a composition of two conjugate-linear maps and its eigenspaces are therefore complex subspaces, so $\phi$ is complex-linear; it is a finite product of factors $\exp (\mathrm{ad}W)$ with $W\in \mathfrak{g}$, hence an inner automorphism of $\mathfrak{g}$. Restricting $\phi$ to the real form ${\mathfrak{u}}_1$ gives a real Lie-algebra isomorphism onto ${\mathfrak{u}}_2$, and the theorem follows.