Complex Analysis/Automorphisms of the Unit Disk

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

Theorem

Let $f\colon \mathbb {D} \to \mathbb {D}$ be an automorphism. Then there exists a $z_{0}\in \mathbb {D}$ and a $\lambda \in \mathbb {C}$ with $|\lambda |=1$ such that

$f(z)=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}$

The unit disk $\mathbb {D}$

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

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

Corollary

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

Proof of the Corollary

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

$h(z)=\lambda {\frac {z-z_{1}}{1-{\bar {z}}_{1}z}}$

we have:

$0=h(0)=\lambda (-z_{1})\iff z_{1}=0$

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

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

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

• Existence: Define $g\colon \mathbb {D} \to \mathbb {D}$ by $g(z):=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}$z_0 \in \mathbb D</math>, $|\lambda |=1$ und $f(z)=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}$. Then $f$ is holomorphic, and since

$|f(z)|={\frac {|z-z_{0}|}{|1-{\bar {z}}_{0}z|}}={\frac {|z||1-{\bar {z}}z_{0}|}{|1-{\bar {z}}_{0}z|}}=|z|=1,\quad |z|=1$

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

${\begin{array}{rl}f(z)&=w\\\iff \displaystyle \lambda {\frac {z-z_{0}}{1-z{\bar {z}}_{0}}}&=w\\\iff z-z_{0}&={\bar {\lambda }}w(1-z{\bar {z}}_{0})\\\iff z(1+{\bar {\lambda }}w{\bar {z}}_{0})&={\bar {\lambda }}w+z_{0}\\\iff z&=\displaystyle {\frac {{\bar {\lambda }}w+z_{0}}{1+{\bar {\lambda }}w{\bar {z}}_{0}}}\\\iff z&={\bar {\lambda }}\displaystyle {\frac {w-{(-\lambda z_{0})}}{1-{\overline {(-\lambda z_{0})}}w}}\end{array}}$

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 $|f(z)|\leq |z|$ for all $z\in \mathbb {D}$. Applying the Schwarz Lemma to $f^{-1}$, we similarly obtain $|z|\leq |f(z)|$, so $|f(z)|=|z|$ for all $z\in \mathbb {D}$. 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 $g(z):={\frac {z-z_{1}}{1-{\bar {z}}_{1}z}}$. From the above, $g$ is an automorphism. Then $h:=g\circ f$ is an automorphism of $\mathbb {D}$ with $h(0)=0$, so $h(z)=\lambda z$ for some $|\lambda |=1$. From the calculations above,

$f(z)=g^{-1}(\lambda z)={\frac {\lambda z+z_{1}}{1+{\bar {z}}_{1}\lambda z}}=\lambda {\frac {z+{\bar {\lambda }}z_{1}}{1+{\overline {{\bar {\lambda }}z_{1}}}z}}$

Setting $z_{0}:=-{\bar {\lambda }}z_{1}$, 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