I need to understand the following proposition:
The real form $G_0$ has only finitely many orbits on the flag manifold $Z=G/P$ where G is a semisimple complex lie group and $G_0$ is its real form and $P$ is a parabolic subgroup.
- Does $G_0$ always exist for semisimple complex Lie groups $G$?
- Why $G_0$ has at least one open orbit in $Z$?
- Is there any example which explains the proposition and my questions?
A complex semisimple Lie group has many real forms. For example $SL(n,\mathbb R)$ and $SU(p,q)$ with $p+q=n$ are real forms of $SL(n,\mathbb C)$.
If you accapt the fact that $G_0$ acts with finitely many orbits on $G/P$, then the fact that there are open orbits follows easily: Since any orbit is a homogeneous space under the action of $G$, it either has to be open or it has to have empty interior. But the manifold $G/P$ cannot be written as a union of finitely many subsets with empty interior.
An instructive example is given by $G=SL(n,\mathbb C)$, $G_0=SU(p,q)$ and $G/P$ the Grassmannian of $k$-dimensional subspaces in $\mathbb C^n$. The group $SU(p,q)$ is defined by a Hermitian form $h$ of signature $(p,q)$ on $\mathbb C^n$. Given a $k$-dimensional complex subspace $V\subset\mathbb C^n$ you can look at the restriction of $h$ to $V$. This is a Hermitian form of signature $(p',q')$ with $p'\leq p$, $q'\leq q$ and $p'+q'\leq k$. It is obvious that for two subspaces in the same $G_0$-orbit the signatures of the restriction of $h$ agree. With a bit of linear algebra, one shows that the converse is true, too, so the $G_0$-orbits are parametrized by the finite set $\{(p',q'):p'\leq p,q'\leq q,p'+q'\leq k\}$. The open $G_0$-orbits are exactly those on which the restriction of $h$ is non-degenerate, i.e. for which $p'+q'=k$.