Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

$\newcommand{\Aut}{\operatorname{Aut}}$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.

share|cite|improve this answer
Thanks, thats exactly what I was looking for – Daniel Mckenzie Jan 30 '13 at 9:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.