Maximal abelian subspaces of p are conjugate by K

The family $\{{\mathrm{ad}}_H:H\in \mathfrak{a}\}$ consists of pairwise commuting diagonalizable endomorphisms of ${\mathfrak{g}}_0$ by [L3], hence is simultaneously diagonalizable: ${\mathfrak{g}}_0={⨁}_{\lambda }{\mathfrak{g}}_0^{\lambda }$, where $\lambda$ runs over the real-linear functionals on $\mathfrak{a}$ and ${\mathfrak{g}}_0^{\lambda }=\{X:[H,X]=\lambda (H)X\text{ for all }H\in \mathfrak{a}\}$. Only finitely many functionals with ${\mathfrak{g}}_0^{\lambda }\mathrm{\ne }0$ occur, because each such $\lambda$ is determined by its values on a basis of $\mathfrak{a}$, and each of those values is an eigenvalue of some ${\mathrm{ad}}_{H_i}$.

There exists $H\in \mathfrak{a}$ with $\lambda (H)\mathrm{\ne }0$ for every $\lambda \mathrm{\ne }0$ occurring in step 1.1: the kernels of the finitely many nonzero $\lambda$ are proper subspaces of the real vector space $\mathfrak{a}$, and a finite union of proper subspaces cannot exhaust $\mathfrak{a}$. For such an $H$ one has $Z_{{\mathfrak{g}}_0}(H)={\mathfrak{g}}_0^0$, because the eigenvalues of ${\mathrm{ad}}_H$ on ${\mathfrak{g}}_0^{\lambda }$ are the nonzero numbers $\lambda (H)$.

For such a regular $H$ one has $Z_{{\mathfrak{p}}_0}(H)=\mathfrak{a}$: the inclusion $\mathfrak{a}\subseteq Z_{{\mathfrak{p}}_0}(H)$ is clear, and if $X\in Z_{{\mathfrak{p}}_0}(H)$ then $[H,X]=0$, so $X\in {\mathfrak{g}}_0^0$; the subspace $\mathfrak{a}+\mathbb{R}X$ of ${\mathfrak{p}}_0$ is abelian and contains the maximal $\mathfrak{a}$, hence equals $\mathfrak{a}$ and $X\in \mathfrak{a}$. The same argument with ${\mathfrak{a}}^{\mathrm{\prime }}$ in place of $\mathfrak{a}$ produces $H^{\mathrm{\prime }}\in {\mathfrak{a}}^{\mathrm{\prime }}$ with $Z_{{\mathfrak{p}}_0}(H^{\mathrm{\prime }})={\mathfrak{a}}^{\mathrm{\prime }}$.

By compactness of $K$ and continuity of $\mathrm{Ad}$ the function $f(k):=B({\mathrm{Ad}}_k{H}^{\mathrm{\prime }},H)$ attains a minimum at some $k_0\in K$.

For every $Z\in {\mathfrak{k}}_0$ the smooth function $r\mapsto B({\mathrm{Ad}}_{\exp (rZ)}{\mathrm{Ad}}_{k_0}{H}^{\mathrm{\prime }},H)$ is minimized at $r=0$, so its derivative vanishes there: with $Y:={\mathrm{Ad}}_{k_0}{H}^{\mathrm{\prime }}\in {\mathfrak{p}}_0$ one has $0=B([Z,Y],H)=-B(Y,[Z,H])$ for all $Z\in {\mathfrak{k}}_0$, since $B$ is invariant. As $[Y,H]\in [{\mathfrak{p}}_0,{\mathfrak{p}}_0]\subseteq {\mathfrak{k}}_0$ and $B$ is nondegenerate on ${\mathfrak{k}}_0$, it follows that $[Y,H]=0$.

Hence $Y\in Z_{{\mathfrak{p}}_0}(H)=\mathfrak{a}$ by step 3.1, so $\mathfrak{a}\subseteq Z_{{\mathfrak{p}}_0}(Y)$. Since $Z_{{\mathfrak{p}}_0}(Y)={\mathrm{Ad}}_{k_0}{Z}_{{\mathfrak{p}}_0}(H^{\mathrm{\prime }})={\mathrm{Ad}}_{k_0}{\mathfrak{a}}^{\mathrm{\prime }}$ by step 3.1 and equivariance of the centralizer, and the right-hand side is abelian, the maximal abelian subspace $\mathfrak{a}$ equals ${\mathrm{Ad}}_{k_0}{\mathfrak{a}}^{\mathrm{\prime }}$.

Finally, if $X\in {\mathfrak{p}}_0$, then $\mathbb{R}X$ lies in a maximal abelian subspace ${\mathfrak{a}}^{\mathrm{\prime }\mathrm{\prime }}$ of ${\mathfrak{p}}_0$ (existence by finite-dimensionality and the ascending chain condition on subspaces), and step 6.1 gives ${\mathfrak{a}}^{\mathrm{\prime }\mathrm{\prime }}={\mathrm{Ad}}_k\mathfrak{a}$ for some $k\in K$; hence $X\in {\bigcup }_{k\in K}{\mathrm{Ad}}_k\mathfrak{a}$ and ${\mathfrak{p}}_0={\bigcup }_{k\in K}{\mathrm{Ad}}_k\mathfrak{a}$. In particular $\dim \mathfrak{a}=\dim {\mathfrak{a}}^{\mathrm{\prime }}$ for any two maximal abelian subspaces, so the real rank is well defined.