§4.4.2 The Group of Meromorphic Automorphisms on $\hat{ℂ}$

4.4.2 The Group of Meromorphic Automorphisms on $\hat{ℂ}$

It is generally common to consider a meromorphic function as a function in the form of $𝑓 : 𝑈 \to \hat{ℂ}$ . Let $Aut (\hat{ℂ})$ denote the group of meromorphic automorphisms on $\hat{ℂ}$ .

To make more profound conclusions on the structure of $Aut (\hat{ℂ})$ , we will introduce certain concepts from group theory.

Definition 4.4.2.1 (Coset).

Let $𝐺$ be a group, and let $𝐻 \leq 𝐺$ be a subgroup (operation denoted by juxtaposition). Then the left coset of $𝐻$ in $𝐺$ with respect to $𝑔 \in 𝐺$ is defined as

𝑔 𝐻 = {𝑔 ℎ : ℎ \in 𝐻} .

The right coset is defined as

𝐻 𝑔 = {ℎ 𝑔 : ℎ \in 𝐻} .

The subgroup

𝐻

is normal iff the left and right cosets are equal. The notation

𝐻 ⊴ 𝐺

is used to represent a normal subgroup. Cosets, like groups and sets, are unordered.

Theorem 4.4.2.2.

Let $𝐺$ be a group and $𝑁 \leq 𝐺$ a subgroup. The set of left cosets

𝐺 / 𝑁 = {𝑔 𝑁 : 𝑔 \in 𝐺}

admits a group structure with operation

(𝑔 𝑁) (ℎ 𝑁) = (𝑔 ℎ) 𝑁

if and only if

𝑁

is a normal subgroup of

𝐺

(i.e.

𝑁 ⊴ 𝐺

Proof.

We prove the two implications separately.

If $𝑁 ⊴ 𝐺$ then $𝐺 / 𝑁$ is a group.
Assume $𝑁$ is normal, $𝑁 ⊴ 𝐺$ , so $𝑔 𝑁 𝑔^{- 1} = 𝑁$ for every $𝑔 \in 𝐺$ (equivalently $𝑔^{- 1} 𝑁 𝑔 = 𝑁$ ). Define a product on $𝐺 / 𝑁$ by
$(𝑔 𝑁) (ℎ 𝑁) = (𝑔 ℎ) 𝑁 .$
We now verify that this product is well-defined: if $𝑔 𝑁 = 𝑔^{'} 𝑁$ and $ℎ 𝑁 = ℎ^{'} 𝑁$ then we need $(𝑔 ℎ) 𝑁 = (𝑔^{'} ℎ^{'}) 𝑁$ . Since $𝑔 𝑁 = 𝑔^{'} 𝑁$ , there exists $𝑛_{1} \in 𝑁$ with $𝑔^{'} = 𝑔 𝑛_{1}$ , and since $ℎ 𝑁 = ℎ^{'} 𝑁$ there exists $𝑛_{2} \in 𝑁$ with $ℎ^{'} = ℎ 𝑛_{2}$ . Then
$𝑔^{'} ℎ^{'} = (𝑔 𝑛_{1}) (ℎ 𝑛_{2}) = 𝑔 (𝑛_{1} ℎ) 𝑛_{2} = 𝑔 ℎ (ℎ^{- 1} 𝑛_{1} ℎ) 𝑛_{2} .$
Because $𝑁$ is normal we have $ℎ^{- 1} 𝑛_{1} ℎ \in 𝑁$ , and $𝑛_{2} \in 𝑁$ , so $(ℎ^{- 1} 𝑛_{1} ℎ) 𝑛_{2} \in 𝑁$ . Hence $𝑔^{'} ℎ^{'} \in (𝑔 ℎ) 𝑁$ , meaning that $(𝑔^{'} ℎ^{'}) 𝑁 = (𝑔 ℎ) 𝑁$ . Thus the product is well-defined.
Associativity follows from associativity in $𝐺$ :
$((𝑔 𝑁) (ℎ 𝑁)) (𝑘 𝑁) = (𝑔 ℎ 𝑘) 𝑁 = (𝑔 𝑁) ((ℎ 𝑁) (𝑘 𝑁)) .$
The identity is $𝑒 𝑁 = 𝑁$ , since $(𝑒 𝑁) (𝑔 𝑁) = (𝑒 𝑔) 𝑁 = 𝑔 𝑁$ and similarly on the other side. The inverse of $𝑔 𝑁$ is $𝑔^{- 1} 𝑁$ , because $(𝑔 𝑁) (𝑔^{- 1} 𝑁) = (𝑔 𝑔^{- 1}) 𝑁 = 𝑁$ . Thus $𝐺 / 𝑁$ is a group.
If $𝐺 / 𝑁$ can be given a group structure via the coset multiplication, then $𝑁 ⊴ 𝐺$ .
Fix $𝑔 \in 𝐺$ and $𝑛 \in 𝑁$ arbitrarily. By assumption, we have
$(𝑔 𝑁) (𝑛 𝑁) (𝑔^{- 1} 𝑁) = (𝑔 𝑛 𝑔^{- 1}) 𝑁 .$
Because $𝑛 𝑁 = 𝑒 𝑁$ , we also have
$(𝑔 𝑁) (𝑛 𝑁) (𝑔^{- 1} 𝑁) = (𝑔 𝑁) (𝑒 𝑁) (𝑔^{- 1} 𝑁) = (𝑔 𝑔^{- 1}) 𝑁 = 𝑁,$
implying that $(𝑔 𝑛 𝑔^{- 1}) 𝑁 = 𝑁$ , and hence $𝑔 𝑛 𝑔^{- 1} \in 𝑁$ for any $𝑔 \in 𝐺$ , $𝑛 \in 𝑁$ . Hence, $𝑔 𝑁 𝑔^{- 1} \subseteq 𝑁$ . Now replacing $𝑔$ with $𝑔^{- 1}$ and rearranging yields $𝑛 \in 𝑔 𝑁 𝑔^{- 1}$ , or that $𝑁 \subseteq 𝑔 𝑁 𝑔^{- 1}$ . Therefore, $𝑁$ is normal.

⁠ $□$

Under the normality of $𝑁$ , the group $𝐺 / 𝑁$ is known as the quotient group of $𝐺$ by $𝑁$ .

Remark.

Every subgroup of an abelian group is normal.

Definition 4.4.2.3 (Group Homomorphism).

Let $(𝐺, \cdot)$ and $(𝐻, *)$ be groups. A function $𝜑 : 𝐺 \to 𝐻$ is said to be a group homomorphism if

𝜑 (𝑔_{1} \cdot 𝑔_{2}) = 𝜑 (𝑔_{1}) * 𝜑 (𝑔_{2}) \forall 𝑔_{1}, 𝑔_{2} \in 𝐺 .

Definition 4.4.2.4 (Group Isomorphism).

A group homomorphism

𝜑 : 𝐺 \to 𝐻

is called an isomorphism if it is bijective.

If there exists an isomorphism between two groups $𝐺$ and $𝐻$ , they are said to be isomorphic, denoted by $𝐺 ≅ 𝐻$ . The utility of groups allows us to classify them according to their structure: if two groups are isomorphic, they are essentially the same from a group-theoretic perspective. This viewpoint lets us replace complicated groups with simpler, isomorphic ones, and study their properties without loss of generality.

Let us now examine $Aut (\hat{ℂ})$ . Let $𝑓 (𝑧) \in Aut (\hat{ℂ})$ such that $𝑓 (\infty) = \infty$ . It follows that $𝑓$ maps $ℂ$ to $ℂ$ bijectively and $𝑓 \in Aut (ℂ) < Aut (\hat{ℂ})$ . Therefore, $𝑓 (𝑧)$ has the form $𝑎 𝑧 + 𝑏$ , where $𝑎 \in ℂ^{*} = ℂ ∖ {0}$ and $𝑏 \in ℂ$ are constants.

Let $𝑓 (𝑧) \in Aut (\hat{ℂ})$ such that $𝑓 (\infty) \neq \infty$ . Then,

𝑔 (𝑧) = \frac{1}{𝑓 (𝑧) - 𝑓 (\infty)}

is in $Aut (\hat{ℂ})$ and $𝑔 (\infty) = \infty$ . By the property above, $𝑔 (𝑧) = 𝑐 𝑧 + 𝑑$ for some complex $𝑑$ and nonzero $𝑐$ . Hence,

𝑓 (𝑧) = \frac{𝑓 (\infty) (𝑐 𝑧 + 𝑑) + 1}{𝑐 𝑧 + 𝑑} .

Let $𝑎 = 𝑐 𝑓 (\infty)$ and $𝑏 = 𝑑 𝑓 (\infty) + 1$ . Then

𝑓 (𝑧) = \frac{𝑎 𝑧 + 𝑏}{𝑐 𝑧 + 𝑑} .

In this specific construction, $𝑎 𝑑 - 𝑏 𝑐 = - 𝑐 \neq 0$ . Let the matrix $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ correspond to this transformation, where for any nonzero scalar $𝑘$ , $𝑘 (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ corresponds to $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ . Therefore, we can arbitrarily pick $𝑎 𝑑 - 𝑏 𝑐$ to be $1$ .

Therefore, there exists a one-to-one correspondence between $Aut (\hat{ℂ})$ and the group under matrix multiplication of

{(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) : det ((\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})) = 1} / {\pm 𝐈} .

The quotient group is taken because the matrix $(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ corresponds to the same transformation as $(\begin{matrix} - 𝑎 & - 𝑏 \\ - 𝑐 & - 𝑑 \end{matrix})$ . This group, denoted by

PSL (2, ℂ) = SL (2, ℂ) / {\pm 𝐈} ≅ Aut (\hat{ℂ}),

is known as the projective special linear group of order $2$ , and is isomorphic to the group of Möbius transformations, consisting of all complex linear fractional transformations.

Therefore, any meromorphic automorphism on $\hat{ℂ}$ is a composition of rotations, dilations, translations, and inversions. We will now state this formally:

Theorem 4.4.2.5 (The Meromorphic Automorphism Group on $\hat{ℂ}$ ).

$\forall 𝑓 \in Aut (\hat{ℂ})$ , $𝑓$ is a Möbius transformation. In other words, $\exists 𝑎, 𝑏, 𝑐, 𝑑 \in ℂ$ satisfying $𝑎 𝑑 - 𝑏 𝑐 \neq 0$ such that

𝑓 (𝑧) = \frac{𝑎 𝑧 + 𝑏}{𝑐 𝑧 + 𝑑} .

Moreover, every such Möbius transformation is in

Aut (\hat{ℂ})

The group of holomorphic automorphisms on $𝔻$ , or $Aut (𝔻)$ , is also a subgroup of $Aut (\hat{ℂ})$ .

Proposition 4.4.2.6.

Suppose we have two Möbius transformations represented by the matrices

(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})

and

(\begin{matrix} 𝑒 & 𝑓 \\ 𝑔 & ℎ \end{matrix})

. Then their composition is a Möbius transformation represented by

(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) (\begin{matrix} 𝑒 & 𝑓 \\ 𝑔 & ℎ \end{matrix})

Proof.

From simple substitution, we have

\frac{𝑎 \frac{𝑒 𝑧 + 𝑓}{𝑔 𝑧 + ℎ} + 𝑏}{𝑐 \frac{𝑒 𝑧 + 𝑓}{𝑔 𝑧 + ℎ} + 𝑑} = \frac{𝑎 𝑒 𝑧 + 𝑎 𝑓 + 𝑏 𝑔 𝑧 + 𝑏 ℎ}{𝑐 𝑒 𝑧 + 𝑐 𝑓 + 𝑑 𝑔 𝑧 + 𝑑 ℎ} = \frac{(𝑎 𝑒 + 𝑏 𝑔) 𝑧 + (𝑎 𝑓 + 𝑏 ℎ)}{(𝑐 𝑒 + 𝑑 𝑔) 𝑧 + (𝑐 𝑓 + 𝑑 ℎ)},

which corresponds to the product

(\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}) (\begin{matrix} 𝑒 & 𝑓 \\ 𝑔 & ℎ \end{matrix})

. ⁠

□

We have now introduced three of the most important regions in complex analysis: $𝔻$ , $ℂ$ , and $\hat{ℂ}$ . Their importance will be later explained by the Uniformization Theorem (@ thm:uniformization).

§4.4.2 The Group of Meromorphic Automorphisms on ℂ̂

4.4.2 The Group of Meromorphic Automorphisms on ℂ̂

§4.4.2 The Group of Meromorphic Automorphisms on $\hat{ℂ}$

4.4.2 The Group of Meromorphic Automorphisms on $\hat{ℂ}$