# Monday March 30th

## Conformal Equivalence

We'll be following Stein, around Chapter 8 currently but skip section 4.
We'll discuss Kobe's proof of the Riemann Mapping Theorem.

Definition
: A bijective holomorphic function $f: U \to V$ is called a *conformal map* or *biholomorphism*.
  Given such a map, we say that $U$ and $V$ are *conformally equivalent* or *biholomorphic*.

Note that this gives an equivalence relation on subsets of $\CC$.

Proposition (1.1)
: If $f: U \to V$ is holomorphic and injective, then for every $z\in U$, $f'(z) \neq 0$.
  In particular, the inverse of $f$ defined on its image is holomorphic, and thus the inverse of a conformal map is holomorphic.

This is not an iff, take $f(z) = z^2$, then $2z\neq 0$ on $\CC\setminus 0$ but $\sqrt{z}$ has two values for every $z$, failing to be injective.

Proof
:   Toward a contradiction, suppose $f'(z_0) = 0$ for some $z_0 \in U$.
    Then since $f$ is holomorphic, we can expand about $z_0$ to obtain
    $$
    f(z) - f(z_0) = a_k(z-z_0)^j + G(z), \quad\quad a_k \neq 0,~k\geq 2
    $$
    for all $z$ near $z_0$, where $G(z)$ vanishes to order $k+1$ at $z_0$.

    For sufficiently small $w$, we write
    $$
    f(z) - f(z_0) - w = F(z) + G(z) \quad\text{where}\quad F(z) = a_k (z-z_0)^k - w
    $$
    Because $\abs{G(z)} < \abs{F(z)}$ on a small disc centered at $z_0$ and $G(z)$ vanishes to order $k+1$ at $z_0$ (where $k> 1$),  we can conclude that $F(z)$ h as at least two distinct zeros inside this disc.
    Applying Rouche's theorem, we conclude that $f(z) - f(z_0) - w$ has at least two zeros as well.
    But this contradicts the injectivity of $f$.

    Now let $g = f\inv$ on the range of $f$, which we can assume is $V$.
    For $w$ close to $w_0$, write $w = f(z)$ and $w_0 = f(z_0)$, yielding

    ????? See notes

Example (Cayley Transform, Important)
: The unit disc and upper half plane are mapped to each other via $F(z) = {i - z \over i + z}$ with inverse $G(w) = i {1 - w \over 1 + w}$.

Note that this is a fractional linear transformation.


Theorem (1.2)
: The map $F: \HH \to \DD$ is a conformal map with inverse $D$.

Proof
:   Both maps are holomorphic on their domains.
    Any point in $\HH$ is closer to $i$ than $-i$, and we want to show that $\abs{F(z)} < 1$ for every $z\in \HH$.
    We also need to show $\Re G(z) > 0$.
    \begin{align*}
    \im G(w)
    &= \Re\qty{ 1 + u -iv \over 1 + u + iv } \\
    &= \cdots \\
    &= {1 - v^2 - u^2 \over \qty{1+u}^2 + v^2} \\
    &> 0 \quad \text{ if } u^2 + v^2 < 1
    .\end{align*}

    This also shows that $u^2  +v^2 = 1 \implies G(w) = 0$.

Definition (Fractional Linear Transform)
:   A function of the form $f(z) = {az + b \over cz + d}$ where $ad-bc\neq 0$ is called a**fractional linear transformation**.
    The determinant condition here is to insure injectivity; otherwise this just yields the constant map.
    Note that $d = {bc \over a}$

Example
: $\GL(2, \CC), \SL(2, \CC)$, etc

Fact
:   Fractional linear transformations are determined by their values on $0, 1, \infty$.
    Thus they form a group $\SL(2, \CC) / \theset{\pm 1} = \PSL(2, \CC)$.


We can define $\PP\CC^1$ by taking $(u, v) \in \CC^2\setminus 0$ and setting $u\sim v$ iff $u = \lambda v$ for $\lambda \neq 0$.
In this case, FLTs are linear maps $\PP\CC^1 \selfmap$ where we send $f(z)$ to $[a, b; c, d] \cdot [z, 1]$.
Note that $f(-d \over c) = \infty = [1, 0]$.

The four basic types of FLT are

- Translation: $T(z) = z+b$
- Rotation: $T(z) = e^{i\theta} z$
- Dilation $T(z) = az$ for $a> 0$.
- Inversion $T(z) = {1 \over z}$.

In the special case that $c=0$, we have $T(z) = {a \over d}z + {b\over d}$ which is a rotation followed by a translation.
If $c=0$, ???

Proposition
: LFTs form a group under function composition.

Proof
: Just need to check that $w\inv = {aw - b \over -cw + a}$.

> Qual alert.

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

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

    If $f$ has an essential singularity at zero, then by Casorati-Weierstrass implies that the image of over punctured disc is dense in $\CC$.
    In particular, $B \definedas \theset{z \suchthat \abs{z-1} < {1 \over 2}}$ and there exists a $\zeta \not\in B$ such that $f(\zeta)$ in $f(B)$.
    But then there exists a $z\in B$ such that $f(z) = f(\zeta)$, contradicting injectivity of $f$.

    If $f$ has a pole, it must be order at most 1, otherwise the reciprocal will have a zero of order 2 and fail injectivity, and $f$ is injective iff $1/f$ is injective.
    Then the Laurent series has at most 3 terms, $az\inv + b + cz$.
    But $a=0$, otherwise solving by the quadratic formula yields two roots.