Real Cartan subalgebras need not be conjugate

The operators of the adjoint representation in the basis $(h,e,f)$ are computed from [L1] to be ${\mathrm{ad}}_h=\mathrm{diag}(0,2,-2)$, ${\mathrm{ad}}_e=\left(\begin{array}{ccc}0& 0& 1\\ -2& 0& 0\\ 0& 0& 0\end{array}\right)$ and ${\mathrm{ad}}_f=\left(\begin{array}{ccc}0& -1& 0\\ 0& 0& 0\\ 2& 0& 0\end{array}\right)$; hence $B(h,h)=\mathrm{tr}({\mathrm{ad}}_h^2)=8$, $B(e,f)=\mathrm{tr}({\mathrm{ad}}_e{\mathrm{ad}}_f)=4$ with ${\mathrm{ad}}_e{\mathrm{ad}}_f=\mathrm{diag}(2,2,0)$, and $B(h,e)=B(h,f)=B(e,e)=B(f,f)=0$. The matrix of $B$ in the basis $(h,e,f)$ is therefore $\left(\begin{array}{ccc}8& 0& 0\\ 0& 0& 4\\ 0& 4& 0\end{array}\right)$, of determinant $-128\mathrm{\ne }0$, so $B$ is nondegenerate and ${\mathfrak{g}}_0$ is semisimple by [L4].

For every automorphism $\alpha$ of ${\mathfrak{g}}_0$ one has ${\mathrm{ad}}_{\alpha X}=\alpha {\mathrm{ad}}_X{\alpha }^{-1}$ for all $X$, because $\alpha$ preserves brackets; consequently $B(\alpha X,\alpha Y)=\mathrm{tr}(\alpha {\mathrm{ad}}_X{\alpha }^{-1}\alpha {\mathrm{ad}}_Y{\alpha }^{-1})=\mathrm{tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)=B(X,Y)$ by [L3]. Thus $\alpha$ preserves the Killing form, and the sign of $B(X,X)$ for $X\mathrm{\ne }0$ is an invariant of the automorphism orbit of $X$.

In the basis $(h,e+f,e-f)$ of ${\mathfrak{g}}_0$ one computes ${\mathrm{ad}}_k(h)=[e-f,h]=-2(e+f)$, ${\mathrm{ad}}_k(e+f)=[e-f,e+f]=2h$ and ${\mathrm{ad}}_k(e-f)=0$, so ${\mathrm{ad}}_k^2(h)=-4h$, ${\mathrm{ad}}_k^2(e+f)=-4(e+f)$ and ${\mathrm{ad}}_k^2(e-f)=0$; therefore $B(k,k)=\mathrm{tr}({\mathrm{ad}}_k^2)=-4-4+0=-8$. In particular $B(h,h)=8>0$ and $B(k,k)=-8<0$.

The involution $\theta (X)=-X^{\mathsf{T}}$ is a Cartan involution of ${\mathfrak{g}}_0$: it is an involutive automorphism because the transpose is an anti-automorphism of the associative algebra $M_2(\mathbb{R})$ while the bracket is $[X,Y]=XY-YX$, and the form $B_{\theta }(X,Y)=-B(X,\theta Y)$ is positive definite. Indeed $\theta (h)=-h$, $\theta (e+f)=-(e+f)$ and $\theta (e-f)=e-f$; in the basis $(h,e+f,e-f)$ this gives $B_{\theta }(h,h)=B(h,h)=8$, $B_{\theta }(e+f,e+f)=B(e+f,e+f)=2B(e,f)=8$ and $B_{\theta }(e-f,e-f)=-B(e-f,e-f)=8$, using $B(e,e)=B(f,f)=0$ from step 1.1 and $B(e-f,e-f)=-8$ from step 2.1, while $B_{\theta }(e+f,e-f)=-B(e+f,e-f)=0$ and $B_{\theta }$ vanishes on the pairs involving $h$ because $B(h,e+f)=B(h,k)=0$; the three basis vectors are pairwise $B_{\theta }$-orthogonal, so $B_{\theta }$ has matrix $8\cdot \mathrm{i}\mathrm{d}$ in this basis and is positive definite.

Each of the lines $\mathbb{R}h$ and $\mathbb{R}k$ is a Cartan subalgebra of ${\mathfrak{g}}_0$. Both are abelian, hence nilpotent. For the normalizer of $\mathbb{R}h$, write $X=ah+be+cf$; then $[X,h]=-2be+2cf\in \mathbb{R}h$ forces $b=c=0$, so $N_{{\mathfrak{g}}_0}(\mathbb{R}h)=\mathbb{R}h$. For the normalizer of $\mathbb{R}k$, use the basis $(h,e+f,k)$: writing $X=ah+b(e+f)+ck$ one has $[X,k]=a[h,k]+b[e+f,k]=2a(e+f)-2bh$ by step 2.1 and $[k,k]=0$, and $2a(e+f)-2bh\in \mathbb{R}k$ forces $a=b=0$, because $\{h,e+f,k\}$ is a basis and $\mathbb{R}k\cap \mathrm{span}\{h,e+f\}=0$; hence $N_{{\mathfrak{g}}_0}(\mathbb{R}k)=\mathbb{R}k$. By [L5] both lines are Cartan subalgebras.

Suppose towards a contradiction that $\alpha$ is an automorphism of ${\mathfrak{g}}_0$ with $\alpha (\mathbb{R}h)=\mathbb{R}k$. Then $\alpha (h)=ck$ for some $c\mathrm{\ne }0$, since $\alpha (h)\mathrm{\ne }0$ spans $\alpha (\mathbb{R}h)$ and $\alpha (h)\mathrm{\notin }\{0\}$. Applying step 1.2 with $X=Y=h$ gives $B(h,h)=B(\alpha h,\alpha h)=B(ck,ck)=c^{2}B(k,k)$, that is $8=-8c^2$, which is impossible for real $c\mathrm{\ne }0$.

The line $\mathbb{R}h$ lies in ${\mathfrak{p}}_0$ and the line $\mathbb{R}k$ lies in ${\mathfrak{k}}_0$: $\theta (h)=-h$ and $\theta (k)=-k^{\mathsf{T}}=k$. Hence, for the $\theta$-stable Cartan structure, $\mathbb{R}h={\mathfrak{a}}_0$ is a split part with ${\mathfrak{t}}_0=0$, and $\mathbb{R}k={\mathfrak{t}}_0$ is a compact part with ${\mathfrak{a}}_0=0$; in particular each of the two lines is a $\theta$-stable one-dimensional subspace.

Consequently no automorphism of ${\mathfrak{g}}_0$ carries $\mathbb{R}h$ onto $\mathbb{R}k$, so the two Cartan subalgebras are not conjugate by any real inner automorphism either, since every inner automorphism is an automorphism; the compact and split Cartan subalgebras $\mathbb{R}k$ and $\mathbb{R}h$ of ${\mathfrak{s}\mathfrak{l}}_2(\mathbb{R})$ are therefore not conjugate.