Complex Analysis/Automorphisms of the Unit Disk

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

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

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

we have:

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

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

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

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

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

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

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

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

we see that ${\displaystyle f^{-1}}$ 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 ${\displaystyle f(0)=0}$. By the Schwarz's Lemma, we have ${\displaystyle |f(z)|\leq |z|}$ for all ${\displaystyle z\in \mathbb {D} }$. Applying the Schwarz Lemma to ${\displaystyle f^{-1}}$, we similarly obtain ${\displaystyle |z|\leq |f(z)|}$, so ${\displaystyle |f(z)|=|z|}$ for all ${\displaystyle z\in \mathbb {D} }$. The Schwarz Lemma then implies that ${\displaystyle f}$ is a rotation, i.e.also, ${\displaystyle f(z)=\lambda z}$ for some ${\displaystyle |\lambda |=1}$.

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

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

Setting ${\displaystyle z_{0}:=-{\bar {\lambda }}z_{1}}$, we obtain the claim.

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