Complex Analysis/Automorphisms of the Unit Disk

The goal of this article is to characterize all biholomorphic mappings $𝔻\to 𝔻$. We aim to prove:

Theorem

Let $f:𝔻\to 𝔻$ be an automorphism. Then there exists a $z_0\in 𝔻$ and a $\lambda \in ℂ$ with $|\lambda |=1$ such that

$f(z)=\lambda \frac{z-z_0}{1-{\overline{z}}_{0}z}$

The unit disk $𝔻$

Its image under $f(z)=\lambda \frac{z-z_0}{1-{\overline{z}}_{0}z}$ für $z_0=\frac{2}{5}(1+i)$ und $\lambda =2^{-1/2}(1+i)$

Conversely, all such mappings are automorphisms of $𝔻$. Before proving the theorem, we note an important corollary:

Corollary

Let $z_0\in 𝔻$. Then there is exactly one automorphism of $𝔻$ such that $f(0)=z_0$ and $f^{\prime }(0)>0$.

Proof of the Corollary

Uniqueness: If $f$ and $g$ are two such automorphisms, consider $h:=f^{-1}\circ g:𝔻\to 𝔻$. Then $h(0)=0$. By the theorem, there exists $\lambda$ and $z_1$ such that

$h(z)=\lambda \frac{z-z_1}{1-{\overline{z}}_{1}z}$

we have:

$0=h(0)=\lambda (-z_1)⟺z_1=0$

so $h(z)=\lambda z$.Furthermore:

$\lambda =h^{\prime }(0)=(f^{-1}\circ g{)}^{\prime }(0)=\frac{1}{f^{\prime }((f^{-1}\circ g)(0))}\cdot g^{\prime }(0)=\frac{g^{\prime }(0)}{f^{\prime }(0)}>0$

so $\lambda =1$, and hence $h=\mathrm{i}\mathrm{d}$, d. h. $\lambda =1$.

• Existence: Define $g:𝔻\to 𝔻$ by
  $g(z):=\lambda \frac{z-z_0}{1-{\overline{z}}_{0}z}$

where $\lambda$ is chosen so that $g^{\prime }(z_0)>0$ gilt. Then $g$ is an automorphism with $g(z_0)=0$. Let $f:=g^{-1}$. Then $f(0)=z_0$ und $f^{\prime }(0)=1/g^{\prime }(z_0)>0$.

Proof

Step 1: Showing these mappings are automorphisms

Let math>z_0 \in \mathbb D</math>, $|\lambda |=1$ und $f(z)=\lambda \frac{z-z_0}{1-{\overline{z}}_{0}z}$. Then $f$ is holomorphic, and since

$|f(z)|=\frac{|z-z_0|}{|1-{\overline{z}}_{0}z|}=\frac{|z||1-\overline{z}{z}_0|}{|1-{\overline{z}}_{0}z|}=|z|=1,|z|=1$

and $f(z_0)=0$, we have $f(𝔻)\subseteq 𝔻$. To show that $f$ is an automorphism, we prove that $f$ is invertible and its inverse is of the same form. From

$\begin{array}{cc}f(z)& =w\\ ⟺\lambda \frac{z-z_0}{1-z{\overline{z}}_0}& =w\\ ⟺z-z_0& =\overline{\lambda }w(1-z{\overline{z}}_0)\\ ⟺z(1+\overline{\lambda }w{\overline{z}}_0)& =\overline{\lambda }w+z_0\\ ⟺z& =\frac{\overline{\lambda }w+z_0}{1+\overline{\lambda }w{\overline{z}}_0}\\ ⟺z& =\overline{\lambda }\frac{w-(-\lambda z_0)}{1-\overline{(-\lambda z_0)}w}\end{array}$

we see that $f^{-1}$ is of the same form, completing the proof. Step 2: Characterizing all automorphisms

To prove that every automorphism is of the claimed form, consider the special case $f(0)=0$. By the Schwarz's Lemma, we have $|f(z)|\le |z|$ for all $z\in 𝔻$. Applying the Schwarz Lemma to $f^{-1}$, we similarly obtain $|z|\le |f(z)|$, so $|f(z)|=|z|$ for all $z\in 𝔻$. The Schwarz Lemma then implies that $f$ is a rotation, i.e.also, $f(z)=\lambda z$ for some $|\lambda |=1$.

Now let $f(0)=:z_1$. Define $g(z):=\frac{z-z_1}{1-{\overline{z}}_{1}z}$. From the above, $g$ is an automorphism. Then $h:=g\circ f$ is an automorphism of $𝔻$ with $h(0)=0$, so $h(z)=\lambda z$ for some $|\lambda |=1$. From the calculations above,

$f(z)=g^{-1}(\lambda z)=\frac{\lambda z+z_1}{1+{\overline{z}}_1\lambda z}=\lambda \frac{z+\overline{\lambda }{z}_1}{1+\overline{\overline{\lambda }{z}_1}z}$

Setting $z_0:=-\overline{\lambda }{z}_1$, we obtain the claim.

Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Automorphismen der Einheitskreisscheibe - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Automorphismen_der_Einheitskreisscheibe

• Date: 01/08/2024