# Monday April 6th

## Automorphisms of the Disc

Lemma (Schwarz)
:   Let $f: \DD \to \DD$ be holomorphic with $f(0) = 0$, then

    1. $\abs{f(z)} \leq \abs{z}$ for all $z\in \DD$.
    2. If for some $z_0 \neq 0$ we have $\abs{f(z_0)} = \abs{z_0}$, then $f$ is a rotation (i.e. $f(z) = e^{i\theta}z$)
    3. $\abs{f'(0)} \leq 1$ and is equality holds, $f$ is a rotation.

Proof
:   \hfill

    1. Expand $f(z) = a_1 z + a_2 z^2 + \cdots$ using $f(0) = 0 \implies a_0 = 0$.
      Then $f(z)/z$ is holomorphic in $\DD$ and bounded in modulus by $1/r$ for $r = \abs{z} < 1$.
      By the maximum modulus principle, this is true for any $\abs{z} < r$, so letting $r\to 1$ yields (1).

    2. Since $\abs{f(z)} < 1$, the modulus of $f(z)/z$ attains its maximum in $\DD^\circ$ and thus is constant.
      So $f(z) = cz$ and at $z_0$ we have $\abs{f(z) / (z-0)} = 1$, so $\abs{c} = 1$ and $f(z)$ is a rotation.

    3. If $g(z) = f(z)/z$ then $\abs{g(z)} < 1$ for all $z\in \DD$ and $g(0) = \lim_{z\to 0} {f(z) f(0) \over z} = f'(0)$.
      If $\abs{f'(0)} = 1$ then $\abs{g(0)} =1$, ???.

We proved that $\aut \DD = \SU(1, 1)$, which preserves a certain Hermitian form and is a non-compact group.
We also showed that $\aut \HH = \SL(2, \RR)$.

Note that the rotations $r_\theta: z \mapsto e^{i\theta} z \in \Aut \DD$.
There are also functions of the form $\psi_\alpha: z \mapsto {\alpha - z \over 1 - \bar \alpha z}$ where $\alpha \in \DD$.
It is holomorphic in $\DD$ with a simple pole at $z= 1/\bar \alpha$.

Writing $z= e^{i\theta}$ yields

\begin{align*}
\psi_{\alpha}
\left(e^{i \theta}\right)=\frac{\alpha-e^{i \theta}}{1-\bar{\alpha} e^{i \theta}}=-e^{-i \theta} \frac{\alpha-e^{i \theta}}{\bar{\alpha}-e^{-i \theta}} \quad \Longrightarrow\left|\psi_{\alpha}\left(e^{i \theta}\right)\right|=1
.\end{align*}

By the MMP, $\abs{\psi_\alpha(z)} < 1$ for all $z\in \DD$.
The inverse can be solved for, and it turns out that $\psi_\alpha\inv = \psi_\alpha$ where $0 \tofrom \alpha$.

Theorem
: \hfill
  \begin{align*}
  \operatorname{Aut}(\mathbb{D})=\left\{f(z)=e^{i \theta} \frac{\alpha-z}{1-\bar{\alpha} z}: \theta \in \mathbb{R},|\alpha|<1\right\}
  .\end{align*}

Note that this group is not compact, since it's homeomorphic to $\RR \cross \DD$.

Proof
:   Given $f\in \Aut(\DD)$, there must be some $\alpha$ such that $f(\alpha) = 0$.
    Consider $g = f\circ \psi_\alpha$, then $g(0) = 0$ and applying the Schwarz lemma yields $\abs{g(z)} \leq \abs z$ for all $z \in \DD$.
    Since $g\inv(0) = 0$, applying this lemma again to $g\inv$ shows $\abs{g\inv(w)} \leq \abs{w}$ for all $w\in \DD$.
    Letting $w = g(z)$ we obtain $\abs{g(z)} = \abs{z}$ for all $z\in \DD$, so by Schwarz $g(z) = e^{i\theta}z$ is a rotation.
    Thus $f(z) = e^{i\theta} \psi_\alpha(z)$.

Recall that $F(z) = {i-z \over i+z}: \HH \to \DD$ has inverse $i {1 - w \over 1 + w}$, so we can define a map $\Gamma: \aut(\DD) \to \aut(\HH)$ where $\phi \mapsto F\inv \phi F$.
This can be shown directly using algebra