## Mobius Transformations and Holomorphic Maps

## 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

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

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 \[\pd f {\bar z} = 0\]

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).

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.

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.

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.