Layla Velazquez

2022-07-01

What is the general form of a rational function which is bijective on the unit disk?

I'm stuck on this problem. If I let $R(z)=\frac{{a}_{0}(z-{a}_{n})\cdots (z-{a}_{1})}{(z-{b}_{m})\cdots (z-{b}_{1})}$, then exactly one of the ${a}_{i}$ must fall in the unit disk and none of the ${b}_{i}$ can be in the unit disk.

Any hints on how to proceed? I see simple functions like $z$ work, but I don't know how to show that higher combinations of them are bijective.

I'm stuck on this problem. If I let $R(z)=\frac{{a}_{0}(z-{a}_{n})\cdots (z-{a}_{1})}{(z-{b}_{m})\cdots (z-{b}_{1})}$, then exactly one of the ${a}_{i}$ must fall in the unit disk and none of the ${b}_{i}$ can be in the unit disk.

Any hints on how to proceed? I see simple functions like $z$ work, but I don't know how to show that higher combinations of them are bijective.

Jake Mcpherson

Beginner2022-07-02Added 23 answers

The only bijective holomorphic functions from the unit disk onto itself are of the form ${e}^{i\theta}{\psi}_{\alpha}$, where $\theta $ is real and

${\psi}_{\alpha}(z)=\frac{z-\alpha}{1-\overline{\alpha}z}$

with $|\alpha |<1$. For a proof, consider any bijective holomorphic function $f$ from the unit disk onto itself. Choose an appropriate value of $\alpha $ in the disk, and then apply the Schwarz lemma to $f\circ {\psi}_{\alpha}$ and its inverse.

We use the fact that a holomorphic bijection has a holomorphic inverse. First, a computation shows that ${\psi}_{\alpha}$ maps the unit disk $\mathbb{D}$ into itself, and since the inverse can be computed explicitly as ${\psi}_{-\alpha}$, it is bijective. Now assume $f:\mathbb{D}\to \mathbb{D}$ which is bijective, and choose the unique point $\alpha \in \mathbb{D}$ satisfying $f(\alpha )=0$. Consider $g=f\circ {\psi}_{-\alpha}$, and note that ${\psi}_{-\alpha}(0)=\alpha $. By the preceding remarks, $g$ is a holomorphic bijection of $\mathbb{D}$ which fixes the origin. By the Schwarz lemma, one has $|{g}^{\prime}(0)|\le 1$ with equality if and only if $g$ is a rotation. The inverse map $h={g}^{-1}$ also maps $\mathbb{D}$ into itself, and satisfies $h(0)=0$, so we have $|{h}^{\prime}(0)|\le 1$ as well. But ${h}^{\prime}(0)=1/{g}^{\prime}(0)$, so we find that $|{g}^{\prime}(0)|=1$ after all. By the case of equality in the Schwarz lemma, $g$ is a rotation, and therefore $f={e}^{i\theta}{\psi}_{\alpha}$ for some real $\theta $.

${\psi}_{\alpha}(z)=\frac{z-\alpha}{1-\overline{\alpha}z}$

with $|\alpha |<1$. For a proof, consider any bijective holomorphic function $f$ from the unit disk onto itself. Choose an appropriate value of $\alpha $ in the disk, and then apply the Schwarz lemma to $f\circ {\psi}_{\alpha}$ and its inverse.

We use the fact that a holomorphic bijection has a holomorphic inverse. First, a computation shows that ${\psi}_{\alpha}$ maps the unit disk $\mathbb{D}$ into itself, and since the inverse can be computed explicitly as ${\psi}_{-\alpha}$, it is bijective. Now assume $f:\mathbb{D}\to \mathbb{D}$ which is bijective, and choose the unique point $\alpha \in \mathbb{D}$ satisfying $f(\alpha )=0$. Consider $g=f\circ {\psi}_{-\alpha}$, and note that ${\psi}_{-\alpha}(0)=\alpha $. By the preceding remarks, $g$ is a holomorphic bijection of $\mathbb{D}$ which fixes the origin. By the Schwarz lemma, one has $|{g}^{\prime}(0)|\le 1$ with equality if and only if $g$ is a rotation. The inverse map $h={g}^{-1}$ also maps $\mathbb{D}$ into itself, and satisfies $h(0)=0$, so we have $|{h}^{\prime}(0)|\le 1$ as well. But ${h}^{\prime}(0)=1/{g}^{\prime}(0)$, so we find that $|{g}^{\prime}(0)|=1$ after all. By the case of equality in the Schwarz lemma, $g$ is a rotation, and therefore $f={e}^{i\theta}{\psi}_{\alpha}$ for some real $\theta $.

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