## Möbius Transformations

Möbius transformations, also called fractional linear transformations, are complex functions of the form $\varphi:z \mapsto \frac{az + b}{cz + d}$ where $a,b,c,d$ are complex numbers and $\det \mmat a b c d \neq 0$.

If we use projective coordinates on $\C$ (i.e. think of $z \in \C$ as the set of all $[u,v] \in \C^2$ with $z = \frac uv$), then the Möbius transformations become matrices $\phi: \vvec z 1 \mapsto \mmat a b c d \vvec z 1 = \vvec {az + b}{cz + d} \sim \frac{az + b}{cz + d}$ You can check that multiplying together these matices gives the composition of the corresponding Möbius transformations. Using this perspective, we see that the nonzero determinant condition tells us that Möbius transformations are invertible.

If we multiply $a,b,c,d$ by the same nonzero complex number, then the Möbius transformation does not change. So Möbius transformations can be identified with the projective general linear group $PGL(2,\C)$. Since we are allowed to rescale the matrix entries, we can use the projective special linear group $PSL(2,\C)$ instead. $PSL(2,\C)$ is simple, so it must precisely capture the Möbius transformations.

Now, the projective coordinates don't just describe the complex plane. They add one point to the plane, $[1, 0]$, which turns the complex plane $\C$ into the Riemann sphere $\hat \C$. So a Möbius transformation maps the complex plane onto the sphere, applies some sort of transformation to the sphere, and maps the sphere back onto the plane. Two natural questions are: "What is the map between the plane and the sphere?" and "What maps do we apply to the sphere?".

### What is the map between the plane and the sphere?

We can break this map from the plane to the sphere into multiple parts. First, we have a map \begin{aligned} f & : \C \to \C^2 \cong \R^4\\ f & : z \mapsto \vvec z 1 \end{aligned} Next, we project from $\R^4$ to $\S^3$ by identifying positive scalar multiples of each other (since the image of $f$ does not include $0$, this is okay). This corresponds to identifying vectors in $\C^2$ which are positive real scalar multiples of each other.

Finally, we project from $\S^3$ to $\S^2$ with the Hopf fibration. This corresponds to identifying vectors in $\C^2$ which differ in phase (i.e. a complex number of unit norm).

Alternatively, we can think of this map as stereographic projection. If we first quotient $\C^2$ by phase, we can identify the image of the complex plane under $f$ with the plane of height-1 in $\C \times \R \cong \R^3$. Then, quotienting out by positive scalar multiplication is precisely stereographic projection onto the unit sphere.

### What maps do we apply to the sphere?

We apply a linear map of determinant 1 to $\C^2$. TODO

## Conformal Maps

A function between Riemannian manifolds $f:(M_1, g_1) \to (M_2, g_2)$ is called conformal if it only distorts the metric by a scalar multiple. That is to say, $f$ is conformal iff there exists a scalar function $h: M_1 \to \R$ such that $f^*g_2 = h \cdot g_1$. Since metrics are positive definite, $h$ must be positive. So it is often convenient to use the conformal scale factor $u: M_1 \to \R$ where $f^* g_2 = e^{2u} g_1$.

Conformal maps have many nice properties. Directly from the definition, we can see that conformal maps preserve angles. I don't currently understand the more complicated nice properties.

## Holomorphic Functions

A function $f: \C \to \C$ is holomorphic if it is complex-differentiable.

The Cauchy-Riemnann equations give a necessary and sufficient condition for a function to be holomorphic. Let $f(x + iy) = u(x,y) + i v(x,y)$ where $u, v : \R^2 \to \R$. Then $f$ is holomorphic if and only if \begin{aligned} \pd u x &= \pd v y\\ \pd u y &= -\pd v x \end{aligned}

We can express the Cauchy-Riemnann equations nicely using the Wirtinger derivative. Let $\pdo{\bar z} := \frac 12 \left(\pdo x + i \pdo y\right)$ Then the Cauchy-Riemann equations are simply the statment $\pd f {\bar z} = 0$

This explains why holomorphic functions are smooth - $\pdo {\bar z}$ is elliptic, and kernels of elliptic operators are smooth. Thus, holomorphic functions must be smooth.

Now, let's look at $\C$ as a 2-dimensional real manifold with the standard metric using the identification $\C \sim \R^2$. The differential of $f = u+iv: \C \to \C$ is given by the jacobian $f_* = \mmat {\pd ux} {\pd vx} {\pd uy} {\pd vy}$ By the Cauchy-Riemann equations, this must have the structure $f_* = \mmat a {-b} b a$ where $a = \pd u x, b = \pd u y$ are real.

Now, we can compute the pullback of the Euclidean metric on $\C$. \begin{aligned} f^*g(v_1, v_2) &= g(f_*v_1, f_*v_2)\\ &= (f_*v_1)^T (f_*v_2)\\ &= v_1^T f_*^T f_* v_2 \end{aligned} So the pullback of the standard metric on $\C$ is given by $f_*^T f_*$. Using our expression for $f_*$, we see that \begin{aligned} f_*^Tf_* &= \mmat a b {-b} a \mmat a {-b} b a\\ &= \mmat{a^2 + b^2} 0 0 {a^2 + b^2}\\ &= (a^2 + b^2)\mathbb{I} \end{aligned} So the pullback of the metric is a scalar multiple of the metric. Thus, the Cauchy-Rimann equations tell us that holomorphic maps are confomal!

Conversely, suppose $f : \C \to \C$ is conformal. Let $f_* = \mmat a b c d$ Again, the pullback of the metric is \begin{aligned} f_*^Tf_* &= \mmat a c b d \mmat a b c d\\ &= \mmat {a^2 + c^2} {ab+cd} {ab+cd} {b^2+d^2} \end{aligned} Since $f$ is conformal, we know that this is a scalar multiple of the identity. So $ab+cd = 0$, and $a^2 + c^2 = b^2 + d^2$.

You can solve this to find two solutions: $a=d, b=-c$ (in which case $f$ is holomorphic), or $a=-d, b=c$ (in which case $f$ is antiholomorphic).

Since $ab+cd=0$, we have that $a = -\frac{cd}{b}$. Thus, $\frac{c^2d^2}{b^2} + c^2 = b^2 + d^2$ Multiplying through by $b^2$, we find that $c^2(b^2+d^2) = b^2(b^2+d^2)$ If $b^2+d^2 = 0$, then $b=d=0$, in which case $a^2+c^2 = 0$, so $f$ is constant. Otherwise, we can divide through by $b^2+d^2$ to find the $b=\pm c$. These two cases give us our two answers.

So the only conformal maps $\C \to \C$ are holomorphic and antiholomorphic functions. In particular, the only orientation-preserving conformal maps are holomorphic functions.

## Automorphisms of the Disk

This machinery of holomorphic functions allows us to nicely characterize the conformal automorphisms of the open unit disk (i.e. invertible conformal maps $D \to D$). As we saw above, it suffices to characterize holomorphic automorphisms of the disk.

(Schwarz) Let $f: D \to D$ be a holomorphic function which fixes the origin. Then $|f(z)| \leq |z|$ for all $z \in D$ and $|f'(0)| \leq 1$. Furthermore, if there exists some nonzero $z$ such that $|f(z)| = |z|$, or if $|f'(0)| = 1$, then $f$ is a rotation.

Since $f$ is holomorphic, we can expand it in a Taylor expansion $f(z) = \sum_{n \geq 0} a_n z^n$. Since $f$ fixes the origin, $a_0 = 0$. So we can define $g(z) = \frac{f(z)}{z}$ by dividing the series expansion by $z$ term by term. This yields the holomorphic function $g(z) := \begin{cases} \frac{f(z)}z & z \neq 0\\ f'(0)&z = 0\end{cases}$ Consider the closed disk $D_r = \{z \;:\; |z| \leq r\}$ for $r < 1$. By the maximum modulus principle, $g$ achieves its maximum on $D_r$ on $\partial D_r$. Let $z_r \in \partial D_r$. Note that \begin{aligned} |g(z_r)| &= \left|\frac{f(z_r)}{z_r}\right|\\ &\leq \frac 1 r \end{aligned}

Taking a limit as $r \to 1$, we see that on the open unit disk, $|g|$ is bounded by 1. And again by the maximum modulus principle, if it achieves its maximum anywhere on the disk, then it is constant.

Any automorphism of the disk which fixes the origin is a rotation.

Let $f$ be an automorphism of the disk. Note that $f^{-1}$ is also an automorphism. So the Schwarz lemma applies to both. Thus,

\begin{aligned} |f(z)| &\leq |z|\\ &= |f^{-1}(f(z))|\\ &\leq |f(z)| \end{aligned} Therefore, $|f(z)| = |z|$ on the disk, so $f$ is a rotation.
Finally, we can classify holomorphic automorphisms of the disk.
Any conformal automorphism of the disk has the form $z \mapsto \lambda \frac{z-a}{\overline a z - 1}$ where $\lambda$ is a complex number of unit norm (i.e. a rotation).
Let $f$ be an automorphism of the disk. If $f$ fixes the origin, then by our corollary $f$ must be a rotation. So suppose $f$ does not fix the origin. Suppose $f(0) = a$. Note that the Möbius transformation $z \mapsto \frac{z-a}{\overline a z - 1}$ maps $a$ to the origin. Furthermore, it maps to disk to itself (TODO: show this). Thus, the composition of this map with $f$ is an automorphism of the disk which fixes the origin. Thus it is a rotation. So $f$ is a rotation composed with the inverse of that Möbius transformation (which is another Möbius transformation of the same form).
We can conclude that all orientation-preserving conformal maps of the unit disk to itself are given by these transformations.

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