Not all homogeneous manifolds are on the form $G/H$

Question

Since the (holomorphic) automorphism group of the 2-dimensional ball $B^2\subset \mathbb C^2$ acts transitively on $B^2$. Then $B^2$ is a homogeneous space. Can we write $B^2$ as a quotient of a (complex) lie group and a closed subgroup $G/H$?

Answer 1

There are two ways to interpret your question:

1. Is there a Lie subgroup $G< Aut(B^2)$ such that $G$ admits a complex structure, for which the action $G\times B^2\to B^2$ is holomorphic and transitive.

2. Is there a Lie subgroup $G< Aut(B^2)$ such that $G$ admits a complex structure and for which the action $G\times B^2\to B^2$ is transitive.

Both questions have negative answer. For the first question, more generally, it is easy to prove (using Liouville's theorem) that:

If $X$ is a bounded domain in ${\mathbb C}^n$ and $G\times X\to X$ is a holomorphic action of a connected complex Lie group, then this action is trivial: $gx=x$ for all $g\in G, x\in X$. In particular, a holomorphic action is never transitive on $X$.

For the second question the proof is less than illuminating, it is based on an analysis of connected Lie subgroups of $PU(2,1)$: One first proves that if a connected subgroup admits a complex structure and acts transitively on $B^2$ then this subgroup is solvable. (Use Levi-Malcev theorem.) Then one eliminates solvable subgroups because only one of them has even real dimension $\ge 4$ dimensional and its abelianization is real 1-dimensional, hence, such a subgroup cannot be isomorphic to a complex Lie group. A similar argument applies to $B^n, n\ge 1$.