Determine the Automorphism group of the unit bidisc.

Question

I'm currently reading through Holomorphic Functions and Integral Representations in Several Complex Variables by R. Michael Range, and I'm stuck on Exercise E.2.4, which states

Let $\Delta^2$ be the unit bidisc in $\mathbb{C}^2$. Show that every $f\in\text{Aut}(\Delta^2)$ is of the form $f=(f_1,f_2)$, where $f_1$ and $f_2$ depend each on only one variable and $f_1,f_2\in\text{Aut}(\Delta)$. (Hint: By using an automorphism of the above simple type, reduce the general case to the case where $f(0)=0$.)

I'm essentially trying to recreate the proof (from Schwarz lemma) of this fact from one variable, but I'm running into trouble. I understand the hint, so I've been looking at $f\in\text{Aut}(\Delta^2)$, such that $f(0)=0$.

What I know so far is that, if $f=(f_1,f_2)$, \begin{align*} \left|\frac{\partial f_1}{\partial z_1}(0,0)\right|&\leq 1, & \left|\frac{\partial f_1}{\partial z_2}(0,0)\right|&\leq 1,\\ \left|\frac{\partial f_2}{\partial z_1}(0,0)\right|&\leq 1, & \left|\frac{\partial f_2}{\partial z_2}(0,0)\right|&\leq 1, \end{align*} by repeated use of the one variable Schwarz lemma. The same thing holds for the inverse $f^{-1}$, and I also know that it must be case that $f'(0)(f^{-1})'(0)=\text{Id}$, but I'm stuck at this point. If I could show that $f_1,f_2$ were dependent on only one of the variables, I'd probably be in business, but alas, I've not been able to do so.

I've looked up the proof in Function Theory of Several Complex Variables by Steven Krantz for the "$n$-disc," and it seems fairly nontrivial, so I suppose there is a more simplistic way for just two variables. Any help is greatly appreciated. Thanks again.

Answer 1

I don't know how Krantz proves it, so maybe what I suggest is exactly the way Krantz uses. The argument works the same for arbitrary $n$, I don't see how special-casing $n = 2$ could lead to a simpler proof.

You are on the right track with using the Schwarz lemma, but you need to use it to get a little stronger result than you have. For a point $p\in \partial_0 \Delta^n$ consider $h \colon \Delta \to \Delta$ given by $h(z) = f_i(z\cdot p)$. The Schwarz lemma says $\lvert h'(0)\rvert \leqslant 1$, so

$$\Biggl\lvert \sum_{k = 1}^n \frac{\partial f_i}{\partial z_k}(0)\cdot p_k\Biggr\rvert \leqslant 1$$

for all $p$ in the distinguished boundary of $\Delta^n$. Thus it follows that

$$\sum_{k = 1}^n \biggl\lvert \frac{\partial f_i}{\partial z_k}(0)\biggr\rvert \leqslant 1$$

for all $i$. The analogous estimate holds for the inverse $g$ of $f$, and since

$$1 = \sum_{k = 1}^n \frac{\partial f_i}{\partial z_k}(0)\cdot \frac{\partial g_k}{\partial z_i}(0)$$

by the chain rule, we have the implications

$$\frac{\partial f_i}{\partial z_k}(0) \neq 0 \implies \biggl\lvert \frac{\partial g_k}{\partial z_i}(0)\}\biggr\rvert = 1,$$

and the same with the roles of $f$ and $g$ interchanged. Hence, for all $i,k$, we have either $\frac{\partial f_i}{\partial z_k}(0) = 0$ or $\bigl\lvert \frac{\partial f_i}{\partial z_k}(0)\bigr\rvert = 1$.

After a permutation and rotations of the components of $f$, we can thus assume $\frac{\partial f_i}{\partial z_k}(0) = \delta_{ik}$, or $J_f(0) = I$.

Then it remains to see that $f$ is the identity. The argument for that uses Montel's theorem and a theorem of Weierstrtaß. By Montel's theorem - the generalisation to dimensions greater than $1$ is straightforward - for a bounded domain $U \subset \mathbb{C}^n$, the family of holomorphic maps $F\colon U \to U$ is normal. By Weierstraß' theorem - again the generalisation to arbitrary (finite) dimensions is straightforward - the families of derivatives (of fixed [multi-]order) of a normal family is again normal.

It follows that if $U\subset \mathbb{C}^n$ is a bounded domain with $0\in U$, and $h\colon U \to U$ is holomorphic with $h(0) = 0$ and $J_h(0) = I$, then $h = \operatorname{id}_U$.

To see that, write

$$h(z) = z + H_k(z) + O(\lVert z\rVert^{k+1})$$

where $H_k$ is a homogeneous polynomial of degree $k > 1$ (with values in $\mathbb{C}^n$). By induction, show that

$$h^m(z) = z + m\cdot H_k(z) + O(\lVert z\rVert^{k+1}),$$

where $h^m$ denotes the $m$-fold iterate of $h$, $h^{m+1} = h^m\circ h$. Use the normality of the family of derivatives of degree $k$ of the functions $h^m$ to deduce $H_k \equiv 0$.

Answer 2

This is the way the Krantz proves it in Function Theory of Several Complex Variables.

Preliminary Proposition

He first proves a Proposition (11.1.2):

Let $\Omega\subseteq\mathbb{C}^n$ be a bounded circular domain. Assume that $0\Omega$. Let $\phi\in\text{Aut}\Omega$ satisfy $\phi(0)=0$. Then $\phi$ is linear.

He proves this by introducing a function $\rho_{\theta}(z)=e^{i\theta}z$, $0\leq \theta <2\pi$, and then defining $$ \psi:=\rho_{-\theta}\circ\phi^{-1}\circ\rho_{\theta}\circ\phi. $$ He proceeds to show that $\psi'=\text{Id}$, and so your result (your result is actually the previous proposition) asserts that $\psi(z)\equiv z$. Since $\phi\circ\rho_{\theta}=\rho_{\theta}\circ\phi$ for all $\theta$, and the only monomials that commute with the $\rho_{\theta}$ are linear polynomials, $\phi$ is linear.

Proof Outline of Main Theorem

To prove the main theorem, he assumes $f(0)=0$, so that the previous proposition immediately applies to say that $f$ is linear. So $f(z)=(b_{ij})(z), 1\leq i,j\leq n$. He lets $$ z^{ik}=\left(\left(1-\frac{1}{k}\right)\overline{\text{sgn }b_{i1}},\ldots ,\left(1-\frac{1}{k}\right)\overline{\text{sgn }b_{in}}\right), $$ where $\text{sgn}=z/|z|$. The $i^{\text{th}}$ component of $f(z^{ij})$ is $$\sum_{j=1}^n\left(1-\frac{1}{k}\right)\big|b_{ij}\big|.$$ This is less than one, so letting $k\to\infty$ shows $\sum_j|b_{ij}|\leq 1$. He then observes the behavior of $w^{jk}=(0,\ldots,0,(1-1/k),0,\ldots ,0)$, where only the $j^{\text{th}}$ entry is nonzero, as $k\to\infty$ (it is shown that $w^{jk}\to\partial \Delta^n$) to conclude each column has precisely one element of modulus 1. He then rules out a few cases to show that $f$ must be a rotation in each component.

Hopefully this is in enough detail. Thanks again.