§3.5 The Group of Holomorphic Automorphisms on the Unit Disk

3.5 The Group of Holomorphic Automorphisms on the Unit Disk

The following important result can be directly obtained from the Maximum Modulus Principle.

Lemma 3.5.1 (Schwarz).

If $𝑓 : 𝔻 \to 𝔻$ is holomorphic and $𝑓 (0) = 0$ , then

| 𝑓 (𝑧) | \leq | 𝑧 |, | 𝑓^{'} (0) | \leq 1 .

Any one of the inequalities becomes equalities iff $𝑓 (𝑧)$ is in the form of $𝑧 e^{i 𝜏}$ , where $𝜏 \in ℝ$ . In other words, $𝑓$ is a pure rotation.

Proof.

Define the auxiliary function

𝑔 (𝑧) = {\begin{cases} \frac{𝑓 (𝑧)}{𝑧} & if 𝑧 \neq 0, \\ 𝑓^{'} (0) & if 𝑧 = 0 . \end{cases}

Because ${lim}_{𝑧 \to 0} \frac{𝑓 (𝑧)}{𝑧} = 𝑓^{'} (0)$ , $𝑔 (𝑧)$ is holomorphic on $𝔻$ . Since $𝑓$ is an automorphism on the open disk, $\forall | 𝑧 | < 1$ , $| 𝑓 (𝑧) | < 1$ . By the Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ), $\forall 0 < 𝜀 < 1$ , $\forall 𝑧 \in 𝐷 (0, 𝜀)$ ,

| 𝑔 (𝑧) | \leq max_{𝑧_{𝜀} \in 𝜕 𝐷 (0, 𝜀)} \frac{| 𝑓 (𝑧_{𝜀}) |}{𝜀} < \frac{1}{𝜀} .

As $𝜀 \to 1^{-}$ , we obtain that $\forall 𝑧 \in 𝔻$ , $| 𝑔 (𝑧) | \leq 1$ , or that $| 𝑓 (𝑧) | \leq | 𝑧 |$ . Let $𝑧 = 0$ . Then we get $| 𝑔 (0) | = | 𝑓^{'} (0) | \leq 1$ .

For the sake of the equality condition, assume $| 𝑓 (𝑧) | = | 𝑧 |$ . Then $| 𝑔 (𝑧) | \equiv 1$ on the unit open disk. By Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ , $𝑔 (𝑧) = e^{i 𝜏}$ where $𝜏 \in ℝ$ and $𝑓 (𝑧) = 𝑧 e^{i 𝜏}$ on $𝔻$ .

Next, assume only that $| 𝑓^{'} (0) | = 1$ . It follows that $| 𝑔 (0) | = 1$ . Since $| 𝑔 (𝑧) | \leq 1$ for all $𝑧 \in 𝔻$ , it follows from Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ that $𝑔$ is constant with magnitude $1$ , or in the form of $e^{i 𝜏}$ , where $𝜏 \in ℝ$ is a constant. Consequently, $𝑓 (𝑧) = 𝑧 e^{i 𝜏}$ . ⁠ $□$

To discuss the main topic of this section, we will first introduce the concept of a group.

Definition 3.5.2 (Group).

A group is a nonempty set $𝐺$ and a binary operation (we will denote this as $*$ ) satisfying the four group axioms:

Closure: $\forall 𝑎, 𝑏 \in 𝐺$ , $𝑎 * 𝑏 \in 𝐺$ .
Associativity: $\forall 𝑎, 𝑏, 𝑐 \in 𝐺$ , $(𝑎 * 𝑏) * 𝑐 = 𝑎 * (𝑏 * 𝑐)$ .
Identity Element: $\exists 𝑒 \in 𝐺$ such that $\forall 𝑎 \in 𝐺$ , $𝑎 * 𝑒 = 𝑒 * 𝑎 = 𝑎$ . Note that $𝑒$ is unique; if $𝑒, 𝑓 \in 𝐺$ were both identity elements, then $𝑒 * 𝑓 = 𝑓 * 𝑒 = 𝑒 = 𝑓$ , and are equal.
Inverse Element: $\forall 𝑎 \in 𝐺$ , $\exists 𝑎^{- 1} \in 𝐺$ such that $𝑎 * 𝑎^{- 1} = 𝑒 = 𝑎^{- 1} * 𝑎$ , where $𝑒$ is the identity element. Note that $𝑎^{- 1}$ is unique. Assume $𝑏, 𝑐$ were both inverses of $𝑎$ . Then, $𝑏 = 𝑏 * 𝑒 = 𝑏 * (𝑎 * 𝑐) = (𝑏 * 𝑎) * 𝑐 = 𝑐$ , and are equal.

A subgroup $𝐻$ of $𝐺$ is a subset of $𝐺$ that is also a group under the same operation as $𝐺$ . This relationship is denoted by $𝐻 \leq 𝐺$ or $𝐻 < 𝐺$ for proper subgroups.

Group operations are not necessarily commutative. In the case that they are, (specifically if $𝑎, 𝑏 \in 𝐺 ⟹ 𝑎 * 𝑏 = 𝑏 * 𝑎$ ), then $𝐺$ is an abelian group.

If $𝑈 \subseteq ℂ$ is connected and $𝑓 : 𝑈 \to 𝑈$ is holomorphic on $𝑈$ and bijective, $𝑓$ is a holomorphic automorphism on $𝑈$ . The group of holomorphic automorphisms on $𝑈$ is denoted by $Aut (𝑈)$ , which is the set of all holomorphic automorphisms such as $𝑓$ , with the operation of composition $(𝑓 \circ 𝑔)$ .

First we will show that $\forall 𝑎 \in 𝔻$ ,

𝜑_{𝑎} (𝑧) = \frac{𝑧 - 𝑎}{1 - \overline{𝑎} 𝑧} \in Aut (𝔻) . (3.5 . 1)

Firstly, the function is holomorphic on $𝔻$ as $| 𝑧 | \leq 1$ , $| \overline{𝑎} | < 1$ , the denominator never vanishes. Additionally, $𝜑_{𝑎} (𝑎) = 0$ .

First, we will observe the image of $𝜕 𝔻$ . Let $| 𝑧 | = 1$ . Then,

| 𝜑_{𝑎} (𝑧) | = | \frac{1}{𝑧} | | \frac{𝑧 - 𝑎}{\frac{1}{𝑧} - \overline{𝑎}} | = | \frac{𝑧 - 𝑎}{\overline{𝑧} - \overline{𝑎}} | = 1 .

Therefore, the image of $𝜕 𝔻$ lies on $𝜕 𝔻$ , and since $𝑓$ is holomorphic and non-constant, by the Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ), for any $| 𝑧 | < 1$ , $| 𝜑_{𝑎} (𝑧) | < 1$ . Therefore, $𝑓$ maps $𝔻$ to $𝔻$ . We next aim to show that $𝑓 : 𝔻 \to 𝔻$ is bijective.

Let us first confirm injectivity. For all $𝑧_{1}, 𝑧_{2} \in 𝔻$ , we will observe when

\frac{𝑧_{1} - 𝑎}{1 - \overline{𝑎} 𝑧_{1}} = \frac{𝑧_{2} - 𝑎}{1 - \overline{𝑎} 𝑧_{2}}

is satisfied. It follows that

\begin{matrix} (𝑧_{1} - 𝑎) (1 - \overline{𝑎} 𝑧_{2}) = (𝑧_{2} - 𝑎) (1 - \overline{𝑎} 𝑧_{1}), \\ 𝑧_{1} - 𝑎 - \overline{𝑎} 𝑧_{1} 𝑧_{2} + {| 𝑎 |}^{2} 𝑧_{2} = 𝑧_{2} - 𝑎 - \overline{𝑎} 𝑧_{1} 𝑧_{2} + {| 𝑎 |}^{2} 𝑧_{1} . \end{matrix}

Then,

{| 𝑎 |}^{2} (𝑧_{2} - 𝑧_{1}) = 𝑧_{2} - 𝑧_{1} ⟺ ({| 𝑎 |}^{2} - 1) (𝑧_{2} - 𝑧_{1}) = 0 .

Since $| 𝑎 | < 1$ , then ${| 𝑎 |}^{2} - 1 \neq 0$ , and we get $𝑧_{2} - 𝑧_{1} = 0$ . This proves the univalence of $𝜑_{𝑎} (𝑧)$ .

Next, we will solve for the inverse of $𝜑_{𝑎}$ . Let $𝑧 = 𝜑_{𝑎} (𝑤) = \frac{𝑤 - 𝑎}{1 - \overline{𝑎} 𝑤}$ . Then,

𝑧 - \overline{𝑎} 𝑧 𝑤 = 𝑤 - 𝑎 ⟺ 𝑤 = \frac{𝑧 + 𝑎}{1 + \overline{𝑎} 𝑧} . (3.5 . 2)

It follows that $𝜑_{- 𝑎} = {(𝜑_{𝑎})}^{- 1}$ . Thus $𝜑_{𝑎}$ is surjective and a bijective automorphism. It follows that (3.5.1) $(3.5.1) (3.5.1)$ is true. Functions in the form of $𝜑_{𝑎}$ (where $𝑎 \in 𝔻$ ) are known as Möbius transformations, and the group of all such transformations is known as the Möbius transformation group on the unit disk, which is a subgroup of $Aut (𝔻)$ . Functions in the form of $𝜌_{𝜏} (𝑧) = 𝑧 e^{i 𝜏}$ , where $𝜏 \in ℝ$ is constant, form a group known as the rotation group, which is also a subgroup of $Aut (𝔻)$ .

Theorem 3.5.3 (The Holomorphic Automorphism Group on $𝔻$ ).

$\forall 𝑓 \in Aut (𝔻)$ , $𝑓$ is a composition between a Möbius transformation and a rotation. In other words, $\exists | 𝑎 | < 1$ and $\exists 𝜏 \in ℝ$ such that

𝑓 (𝑧) = 𝜑_{𝑎} \circ 𝜌_{𝜏} (𝑧) .

Moreover, all such functions are in

Aut (𝔻)

Proof.

Define the auxiliary function

𝜓 (𝑧) = 𝜑_{𝑓 (0)} \circ 𝑓 (𝑧) .

It follows that $𝜓 \in Aut (𝔻)$ . Furthermore,

𝜓 (0) = 𝜑_{𝑓 (0)} \circ 𝑓 (0) = 0 .

By the Schwarz Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ), $| 𝜓^{'} (0) | \leq 1$ . Since $𝜓^{- 1} \in Aut (𝔻)$ with $𝜓^{- 1} (0) = 0$ , $| {(𝜓^{- 1})}^{'} (0) | \leq 1$ . Then,

| {(𝜓^{- 1})}^{'} (0) | = | \frac{1}{𝜓^{'} (𝜓^{- 1} (0))} | = | \frac{1}{𝜓^{'} (0)} | \leq 1 .

Then, $| 𝜓^{'} (0) | = 1$ , and by the equality statement of Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ,

𝜓 (𝑧) = 𝑧 e^{i 𝜏} = 𝜌_{𝜏} (𝑧)

for some constant $𝜏 \in ℝ$ , and

𝑓 (𝑧) = 𝜑_{𝑓 (0)}^{- 1} \circ 𝜌_{𝜏} (𝑧) .

By (3.5.2) $(3.5.2) (3.5.2)$ , it follows that

𝑓 (𝑧) = 𝜑_{- 𝑓 (0)} \circ 𝜌_{𝜏} (𝑧) .

⁠ $□$

As a direct consequence of Theorem 3.5.3 $Theorem 3.5.3 Theorem 3.5.3$ , we have the following result:

Lemma 3.5.4 (Schwarz–Pick).

Let $𝑓 : 𝔻 \to 𝔻$ be holomorphic. For all $𝑧_{1}, 𝑧_{2} \in 𝔻$ , let $𝑤_{1} = 𝑓 (𝑧_{1})$ and $𝑤_{2} = 𝑓 (𝑧_{2})$ . Then,

| \frac{𝑤_{1} - 𝑤_{2}}{1 - 𝑤_{1} \overline{𝑤_{2}}} | \leq | \frac{𝑧_{1} - 𝑧_{2}}{1 - 𝑧_{1} \overline{𝑧_{2}}} | . (3.5 . 3)

and

\frac{| d 𝑤 |}{1 - {| 𝑤 |}^{2}} \leq \frac{| d 𝑧 |}{1 - {| 𝑧 |}^{2}} . (3.5 . 4)

The equalities hold iff $𝑓 \in Aut (𝔻)$ .

Proof.

Let

𝜑_{- 𝑧_{1}} (𝑧) = \frac{𝑧 + 𝑧_{1}}{1 + \overline{𝑧_{1}} 𝑧} \in Aut (𝔻), 𝜑_{𝑤_{1}} (𝑧) = \frac{𝑧 - 𝑤_{1}}{1 - \overline{𝑤_{1}} 𝑧} \in Aut (𝔻) .

It follows that

𝜑_{𝑤_{1}} \circ 𝑓 \circ 𝜑_{- 𝑧_{1}} (0) = 𝜑_{𝑤_{1}} (𝑤_{1}) = 0 .

Then by the Schwarz Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ), for $𝑧 \in 𝔻$ ,

| 𝜑_{𝑤_{1}} \circ 𝑓 \circ 𝜑_{- 𝑧_{1}} (𝑧) | \leq | 𝑧 | .

Let $𝑧_{2} = 𝜑_{- 𝑧_{1}} (𝑧)$ . Then,

| 𝜑_{𝑤_{1}} \circ 𝑓 (𝑧_{2}) | \leq | 𝜑_{𝑧_{1}} (𝑧_{2}) | ⟺ | 𝜑_{𝑤_{1}} (𝑤_{2}) | \leq | 𝜑_{𝑧_{1}} (𝑧_{2}) |,

confirming (3.5.3) $(3.5.3) (3.5.3)$ . By the second statement of the Schwarz Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ), $| {(𝜑_{𝑤_{1}} \circ 𝑓 \circ 𝜑_{- 𝑧_{1}})}^{'} (0) | \leq 1$ .

By the chain rule,

| 𝜑_{{(𝑤_{1})}^{'}} (𝑤_{1}) 𝑓^{'} (𝑧_{1}) 𝜑_{{(- 𝑧_{1})}^{'}} (0) | \leq 1 .

Let us now calculate the derivatives of $𝜑_{𝑤_{1}}$ and $𝜑_{- 𝑧_{1}}$ . By the quotient rule,

𝜑_{𝑤_{1}}^{'} (𝑧) = \frac{1 - \overline{𝑤_{1}} 𝑤_{1}}{{(1 - \overline{𝑤_{1}} 𝑧)}^{2}}, 𝜑_{𝑤_{1}}^{'} (𝑤_{1}) = \frac{1}{1 - \overline{𝑤_{1}} 𝑤_{1}},

and

𝜑_{- 𝑧_{1}}^{'} (𝑧) = \frac{1 - \overline{𝑧_{1}} 𝑧_{1}}{{(1 + \overline{𝑧_{1}} 𝑧)}^{2}}, 𝜑_{- 𝑧_{1}}^{'} (0) = 1 - \overline{𝑧_{1}} 𝑧_{1} .

Since both derivatives are positive,

| 𝑓^{'} (𝑧_{1}) | \leq \frac{1 - \overline{𝑤_{1}} 𝑤_{1}}{1 - \overline{𝑧_{1}} 𝑧_{1}} .

Since $𝑧_{1} \in 𝔻$ is arbitrary, it follows that

| \frac{d 𝑤}{d 𝑧} | \leq \frac{1 - \overline{𝑤} 𝑤}{1 - \overline{𝑧} 𝑧} ⟺ \frac{| d 𝑤 |}{1 - \overline{𝑤} 𝑤} \leq \frac{| d 𝑧 |}{1 - \overline{𝑧} 𝑧} . (3.5 . 5)

By the Schwarz Lemma (Lemma 3.5.1 $Lemma 3.5.1 Lemma 3.5.1$ ), under the equality condition that

| 𝜑_{{(𝑤_{1})}^{'}} (𝑤_{1}) 𝑓^{'} (𝑧_{1}) 𝜑_{{(- 𝑧_{1})}^{'}} (0) | = 1,

we have that

𝜑_{𝑤_{1}} \circ 𝑓 \circ 𝜑_{- 𝑧_{1}} = e^{i 𝜏},

where $𝜏 \in ℝ$ is constant. It follows that

𝑓 = 𝜑_{- 𝑤_{1}} \circ e^{i 𝜏} \circ 𝜑_{𝑧_{1}} \in Aut (𝔻) . □

Remark.

In Section 8 $Section 8 Section 8$ , we will introduce the hyperbolic metric on $𝔻$ , defined as

d 𝑠^{2} = \frac{4 {| d 𝑧 |}^{2}}{{(1 - {| 𝑧 |}^{2})}^{2}} .

From (3.5.5)

(3.5.5) (3.5.5)

, we get that the hyperbolic metric does in.notcrease under a holomorphic mapping of

𝔻

to itself. This metric is invariant (the equality condition) under all functions in

Aut (𝔻)

. This gives a geometric explanation for Lemma 3.5.1

Lemma 3.5.1 Lemma 3.5.1