Automorphism group of a bounded domain is a Lie group

Remarks

The theorem is due to Henri Cartan. Since $D$ is bounded, Montel's theorem makes families of automorphisms normal, and a limit argument shows that the action is proper. By Cartan's uniqueness theorem, an automorphism $f$ with $f(p) = p$ is determined by its derivative $f'(p)$. So the isotropy group embeds as a compact subgroup of $GL_n(\C)$, and the whole group is locally compact and finite-dimensional.

Example: for the unit disc $\Delta$, $\Aut(\Delta)$ consists of the maps $z \mapsto e^{i\theta}(z - a)/(1 - \bar a z)$ with $\theta \in \R$ and $\abs{a} < 1$. It is a $3$-dimensional Lie group, isomorphic to $PSU(1, 1)$, and the isotropy group of $0$ is the circle of rotations.

Boundedness cannot be dropped. For $n \ge 2$, $\Aut(\C^n)$ contains the maps $(z_1, z_2, \dots) \mapsto (z_1, z_2 + g(z_1), \dots)$ for every entire function $g$, so it is not a finite-dimensional Lie group.

Going deeper

For a student

This is the starting point for the study of homogeneous and symmetric bounded domains. If $\Aut(D)$ acts transitively, then $D \cong \Aut(D)/K$ with $K$ compact, and $D$ becomes a homogeneous space of a Lie group. Combined with the invariant Bergman metric, this leads to Cartan's classification of bounded symmetric domains. For unbounded domains the group can be infinite-dimensional, as for $\C^n$.