# Wednesday April 1st

We call $f: U\to V$ biholomorphic if it admits a bijective holomorphic inverse.

Proposition
: If $f: U\to V$ is holomorphic and injective, then $f'(z) \neq 0$, $f$ is biholomorphic onto its image, and thus the inverse of a conformal map is also holomorphic.

Proof
:   We argue by contradiction and suppose $f'(z_0) =0$ for some $z_0 \in U$.
    Then $f(z) - f(z_0) = a(z-z_0)^k + G(z)$ for all $z$ near $z_0$,
    with $a\neq 0, k \geq 2$, and $G$ v anishing to order $k+1$ at $z_0$.
    For $w$ small enough, we write $f(z) - f(z_0) - w = F(z) + G(z)$ where $F(z) = a(z-z_0)^k - w$.

    \

    Since $\abs G < \abs F$ on the boundary of a small disc $D_\eps(z_0)$ and $F$ has at least two zeros in $D_\eps(z_0)$, Rouche's theorem implies that $f(z) - f(z_0) - w$ has at least two zeros in this disc.

    \

    Since $f'(z) \neq 0$ for all $z$ in a small punctured neighborhood of $z_0$, it follows that that roots of $f(z) - f(z_0) - w$ are distinct.
    But then $f$ is not injective, a contradiction.

    \

    Now let $g = f\inv$ be the inverse of $f$ on its range, which we can assume is $V$.
    The second part of the proof follows from a Calculus argument, see slides.


Definition
: Two open sets are *conformally equivalent* iff there exist biholomorphic functions composing to the identity.

Recall the Cayley transform $w = F(z) = {i - z \over i + z}$ with inverse $z = G(w) = i {1-w \over 1 + w}$.

Theorem
: The map $F: \HH \to \DD$ is a conformal map with inverse $G$.

Proof (Sketch)
:   Use the fact that any point in $\HH$ is closer to $i$ than $-i$, so $\abs F < 1$.
    Then show $\Im G(w) > 0$ by expanding $w = u + iv$ and using that fact that $u^2 + v^2 < 1$.

The Cayley transform is a special case of a FLT $f(z) = {az + b \over cz + d}$.
Note that $f(-{d \over c}) = \infty$ is a pole, so $f$ should be defined with values in $\hat \CC \cong \CP^1$.
Since $\lim_{\abs z \to \infty}$ exists, the domain is naturally $\hat \CC$ as well.

Identify $\CP^1$ as the space of lines through $0$ in $\CC^{n+1}$.
Explicitly, take $\vector v, \vector w \in (\CC^{n+1})^\bullet$ and set $\vector v \sim \vector w \iff \vector v = \lambda \vector w$ for some $\lambda \in \CC\units$.
Define open sets using the quotient topology, and this yields a metric defined by
\begin{align*}
d([u], [v]) = \qty{ q - {\abs{ \inner{u}{v}  }} \over \norm{u} \cdot \norm{v} }
.\end{align*}

Let $\infty$ denote the point $(1, 0)^t \in \CP^1$, then every other point corresponds uniquely to an element of the form $(z, 1)^t$.
For any $A\in \GL(2, \CC)$, then $A$ acts on such an element by $L_A(a) = (az+b, cz+d)^t = {az+b \over cz + d} \in \CC$ (by dividing through the second term).

Under function composition, these are Lie groups, and the map $A \mapsto L_A$ is a group morphism.
For any $s\in \CC$, $L_{sA} = L_A$, and $L_A = L_{A_1}$ for some matrix $A_1$ of determinant 1, so $A_1 \in \SL(2, \CC)$.
Finally, $L_{A_1} = L_{A_2} \iff A_2 = \pm A_1$, so the group of FLTs is given by $\PSL(2, \CC) = \SL(2, \CC)/ \theset{\pm I}$.

Now consider taking $a,b,c,d \in\RR$, yielding the subgroup $\SL(2, \RR)$.
Then for any $A\in \GL(2, \RR)$, we have $L_A = L_{A_1}$ iff $A_1 \in \SL(2, \RR)$ and $\det A > 0$.
Moreover, each such $A$ preserves $\HH$.

In particular,
\begin{align*}
\frac{a z+b}{c z+d}=\frac{(a z+b)(z \bar{z}+d)}{(c z+d)(c \bar{z}+d)}=\frac{R}{P}+i y \frac{a d-b c}{P}
.\end{align*}

with
\begin{align*}
R=a c|z|^{2}+b d+(a d+b c) x \in \mathbb{R}, P=|c z+d|^{2}>0, \quad \text { if } y>0
.\end{align*}

Thus $z\in \HH \implies f(z) \in \HH$, and $\aut_\CC(\HH) = \SL(2, \RR)$.
What is $\aut_\CC(\DD)$?

We single out the transform $\phi(z) = {z-i \over z+i}$ represented by
$A_0 \definedas \begin{bmatrix}
1 & -i \
1 & i
\end{bmatrix}$
Conjugating the $\SL(2, \RR)$ action on $\HH$ by $\phi$ yields a map $M_A: \DD \to \DD$.

Letting $A \in \SL(2, \RR)$ and computing $A_0 A A_0\inv$, we find that

\begin{align*}
A_{0} S L(2, \mathbb{R}) A_{0}^{-1}=\left\{\left(\begin{array}{cc}
\alpha & \beta \\
\bar{\beta} & \bar{\alpha}
\end{array}\right) \in G L(2, \mathbb{C}):|\alpha|^{2}-|\beta|^{2}=1\right\}=S U(1,1)
.\end{align*}

It turns out that this is $\aut_\CC(\DD)$.
It can be checked directly that for any matrix $B$ in this group, it in fact send $\bd \DD \to \bd \DD$ by computing the modulus.

Recall that there are four basic types of FLT:

- Translation
- Rotation
- Dilation
- Inversion

Moreover, any FLT can be decomposed as a composition of these 4 types:

- If $c=0$, then $T(z) = {a \over d}z + {b\over d}$ which is a dilation, a rotation, then a translation
- If $c\neq 0$, then long division yields $T(z)=\frac{b c-a d}{c^{2}} \cdot \frac{1}{z+\frac{d}{c}}+\frac{a}{c}$.

Theorem
: If $f$ is holomorphic on $\CC\setminus\theset{z_0}$ and injective, then $f$ is a FLT.


Proof
:   Up to translation, assume $z_0 = 0$.
    Then $f$ has a Laurent series expansion $f(z) = \sum_{n=i\infty}^\infty a_n z^n$.

    \

    **Case 1:**
    If $f$ has an essential singularity at 0, then by Casorati-Weierstrauss, the image of every punctured neighborhood of 0 is dense in $\CC$.

    \

    In particular, then if $B \definedas \theset{z \suchthat \abs{z-1} < {1 \over 2}}$, there exists a $\zeta\not\in B$ with $f(\zeta) \in f(B)$.
    But then there is a $z\in B$ such that $f(z) = f(\zeta)$, contradicting injectivity.

    \

    **Case 2:**
    Suppose $f$ has a pole of order $n$ at 0, then $\frac 1 f$ has a zero of order $n$ at 0.
    Since $1 \over f$ is injective, we must have $n\leq 1$ by a previous proposition showing that the derivative can not be zero.

    \

    Applying the same argument to $f(1/z)$ we conclude that $f(z) = \frac a z + b + cz$ for some constants $a,b,c\in \CC$.
    If both $a\neq 0, c\neq 0$ then $cz^2 + (b-w)z + a = 0$ has two roots (up to multiplicity), contradicting injectivity.

    \

    So either $a=0$ or $c=0$, but not both since $f$ is not constant.
    But in either case, $f$ is an FLT.

Proposition
:   Suppose $f$ is holomorphic at $z_0$.
    Then $f(z) - f(z_0) = \sum_{m=n}^\infty a_m(z-z_0)^m$ with $a_n \neq 0 \iff$ for $\eps$ small enough, there exists $\delta > 0$ such that $f(z) - w$ has distinct roots in $\theset{z\suchthat 0 < \abs{z-z_0} < \eps}$ whenever $0 < \abs{w - f(z_0)} < \delta$.

This states that $f(d_\eps(z_0))$ covers $D_\delta(w_0) \setminus\theset{w_0}$ exactly $n$ times.
In particular, $f$ is injective in a neighborhood of $z_0 \iff f'(z_0) \neq 0$.