Criterion for homogeneity

Question

Let $\Omega$ be a bounded domain in $\mathbb{C}^{n}$ and let $G=Aut(\Omega)$ be the full group of self-biholomorphisms of $\Omega$. Assume that there is $z\in \Omega$, such that the orbit of $z$ is somewhere dense, i. e. $\mathrm{int}~ \overline{G\cdot z}\ne\varnothing$. Does it follow that $\Omega$ is homogeneous, i.e. $G$ acts transitively on $\Omega$?

Answer 1

Edit: (21 May 2017) I have modified my answer to cover the case that the OP meant to ask, i.e., the assumption is that the closure of an orbit has nonempty interior.

Now that you have added the assumption of boundedness (and, I assume, connectedness, since, without it, the answer would clearly be 'no'), we can answer the question: Yes, a connected, bounded domain $\Omega\subset \mathbb{C}^n$ is homogeneous as long as there is any point $z\in\Omega$ such that $\mathrm{int}\bigl(\,\overline{G{\cdot}z}\,\bigr)$ is not empty.

The argument is as follows: By a result of Kobayashi $\Omega$ is hyperbolic, and hence $G=\mathrm{Aut}(\Omega)$ is a Lie group that, in particular, preserves the Bergman metric on $\Omega$ (cf., Theorem 9.1 of S. Kobayashi, Intrinsic distances, measures, and geometric function theory Bulletin of the AMS 82 (1976), 357–416).

Suppose that $z\in \Omega$ is such that the closure of its orbit contains an $\epsilon$-ball $B_\epsilon(p)\subset \Omega$ for some positive $\epsilon$ (using, say, the Bergman metric to define distances). Then the closure of this orbit contains $B_{\epsilon/2}(g{\cdot}z)$ whenever $g{\cdot}z \in B_{\epsilon/2}(p)$, and hence the closure of the orbit of $z$ contains $B_{\epsilon/2}(g{\cdot}z)$ for all $g\in G$. Consequently, the closure of the orbit of $z$ is both open and closed and hence, by connectedness, is all of $\Omega$.

Now, let $p\in\Omega$ be arbitrary and let $g_k\in G$ be a sequence such that $g_k{\cdot}z$ is a Cauchy sequence that converges to $p$. Since the maps $\phi_k:\Omega\to\Omega$ defined by $\phi_k(w) = g_k{\cdot}w$ are isometries of the Bergman metric, it follows that we can, by reducing to a subsequence, assume that, not only does $\phi_k(z)$ converge to $p$, but the sequence $\phi_k'(z):T_z\Omega\to T_{\phi_k(z)}\Omega$ converges to an isometry $f:T_z\Omega\to T_p\Omega$. It then follows by a standard argument using the path-connectedness of $\Omega$, that the sequence of Kähler isometries $\phi_k:\Omega\to\Omega$ converges on compact sets to a Kähler isometry $\phi_\infty:\Omega\to\Omega$.

It is easy to show that $\phi_\infty$ is a bijection, and, hence, it must belong to $G$. Thus, $p$ is in the $G$-orbit of $z$. Since $p$ was arbitrary, it follows that $G{\cdot}z = \Omega$, as was to be shown.