# Biholomorphisms of the polydisk

Let $\mathbb{D}$ denote the unit disk in the complex plane, equipped with the Poincare metric. Let us denote the group of biholomorphisms of $\mathbb{D}$ by $Aut(\mathbb{D})$.

Suppose $F: \mathbb{D}^n \rightarrow \mathbb{D}^n$ is a biholomorphic map. Then it is a standard fact that $F$ must be the composition of $n$ biholomorphisms of the disk $\phi_1,\ldots,\phi_n \in Aut(\mathbb{D})$ and an element of the symmetric group $\sigma\in\mathcal{S}^n$:

$F(z_1,\ldots,z_n) = (\phi_1(z_{\sigma(1)}),\ldots, \phi_n(z_{\sigma(n)}))$

and further that the group of biholomorphisms of $\mathbb{D}^n$, $Aut(\mathbb{D}^n)$ is the semidirect product of $Aut(\mathbb{D})^n$ and $\mathcal{S}^n$.

BUT, while this seems completely plausible to me, I can't quite convince myself that it is true, nor can I find a satisfactory reference. Any suggestions?

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## 1 Answer

$\newcommand{\Aut}{\operatorname{Aut}}$Isn't that covered in any textbook on several complex variables? For instance, Krantz, Function theory of several complex variables, chapter 11 has a few pages on $\Aut(\mathbb{D}^n)$.

An outline of one way to do it (and this is what Krantz does) is to do the following

1. Assume that $\Omega$ is a bounded circular domain with $0 \in \Omega$ and that $\phi \in \Aut(\Omega)$ with $\phi(0) = 0$. Then $\phi$ is linear. (Proof: Look at the Jacobian of $\phi$ and use linear algebra.)

2. Let $\phi \in \Aut(\mathbb{D}^n)$ with $\phi(0) = \alpha$. Compose $\phi$ with a mapping $\psi = \left( \dfrac{\alpha_1-z_1}{1-\bar\alpha_1 z_1}, \ldots, \dfrac{\alpha_n-z_n}{1-\bar\alpha_n z_n} \right)$ so that $f = \psi \circ \phi$ satisfies $f(0) = 0$. By 1., $f$ is linear, and after that it's just some bookkeeping to show that the linear map is basically a permutation.

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Thanks, thats exactly what I was looking for –  Daniel Mckenzie Jan 30 '13 at 9:48