Conformal automorphism of disk implies fractional linear transformation

Statement

Suppose $D$ is the open unit disc in $\mathbb {C}$. Then, any conformal automorphism (a bijective biholomorphic mapping to itself) of $D$ comes as the restriction to $D$ of a fractional linear transformation.

Proof

Proof outline

Let $G=Aut(D)$ and $H$ be the subgroup of $G$ comprising those automorphisms that arise by restricting fractional linear transformations to $D$. We need to show that $H=G$. We prove this by showing two things:

• $H$ acts transitively on $D$
• $H$ contains the isotropy subgroup of $0$ in $G$

Combining these two facts, we see that $H=G$

Proof of transitivity

We'll show that given any element $w\in D$, there exists an element $\sigma \in H$ such that $\sigma (0)=w$. This'll show transitivity.

Define:

$\sigma :=z\mapsto {\frac {z-w}{{\overline {w}}z-1}}$

This is a fractional linear transformation. First, we see that this transformation takes the unit circle to itself. Indeed, if $|z|=1$, then:

$|z-w|=|{\overline {z}}(z-w)|=|1-{\overline {z}}w|=|1-{\overline {w}}z|$

Hence the ratio has modulus 1.

Second, observe that since it takes 0 to a point inside the disc, it must by definition send the disc to itself (i.e. it cannot send the disc to the exterior of the circle).

Finally, this fractional linear transformation sends $0$ to $w$, as we see by putting $z=0$ in the formula.

Proof of containing the isotropy

Suppose $f:D\to D$ is a biholomorphic mapping such that $f(0)=0$. Then, by the Schwarz Lemma, we have $|f'(0)|le1$. Applying the Schwarz lemma to $f^{-1}$, we see that $|f'(0)|\geq 1$, hence $|f'(0)|=1$, so by the second part of Schwarz lemma, $f$ is a rotation i.e. multiplication by a complex number of unit modulus. Rotations are fractional linear transformations, so we're done.