Groups of conformal isomorphisms of simply connected surfaces

By the uniformization theorem every connected and simply connected surface $$M$$ is conformally equivalent to one of the following three surfaces: open disk $$D$$, complex plane $$\mathbb{C}$$, or $$2$$-sphere $$S^2 = \mathbb{C}\cup \{\infty\}$$.

Let $$H(M)$$ be the group of conformal automorphisms of $$M$$ which is the same as the group of its biholomorphisms.

1. If $$M=S^2$$, then it directly follows from Schwartz lemma that $$H(M)$$ consists of Möbius transformations which leave $$D$$ invariant. This group is the same as the group of Möbius transformations of the upper half-plane (having thus real coefficients). Hence, $$H(D)$$ can be identified with the group of Möbius transformations of the real line. In other words, $$H(D) = \mathrm{PSL}(2,\mathbb{R})$$.

2. Further, it is written in many sources, including Wikipedia, that $$H(S^2)$$ is exactly the group of Möbius transformations, and so it is isomorphic with the group $$\mathrm{PSL}(2,\mathbb{C})$$.

3. It will then follow from 2) that $$H(\mathbb{C})$$ consists of affine maps $$z\mapsto az+b$$, i.e. it is the subgroup of $$\mathrm{PSL}(2,\mathbb{C})$$ fixing $$\infty$$, so $$H(\mathbb{C})=\mathrm{Aff}(\mathbb{C})$$.

My question is where one can find an exact proof of 2) that $$H(S^2) = \mathrm{PSL}(2,\mathbb{C})$$. Thank you in advance. I would very appreciate any information about that.

An elementary proof is the following: first you prove that every holomorphic map $$f:S^2\to S^2$$ is a rational function. Indeed, such map must be a meromorphic function, it has finitely many zeros and poles and is either regular or has a pole at $$\infty$$. So the degree of $$f$$ is defined: the equation $$f(z)=a$$ has $$d$$ solutions, counting multiplicity, for every $$a\in S^2$$. Therefore, if the map is bijective, we must have $$d=1$$ that is $$f\in PSL(2,C)$$.

This proof is essentially elementary algebra, only one fact from Analysis is used: the "Fundamental theorem of algebra".

• How do you prove, using only essentially elementary algebra, that any holomorphic map from sphere to itself is meromorphic? :) Apr 28, 2023 at 16:42
• Very simple. The map $f$ has finitely many zeros and poles in $C$. Let $g$ be the rational function with the same zeros and poles in $C$ (counting multiplicity). Then $h=f/g$ has no zeros or poles in $C$. If $h(\infty)$ is infinite, consider $g/f$ instead. A bounded holomorphic function on the sphere must be constant. (Both the Maximum Principle and Liouville theorem which can be used to make this conclusion have simple algebraic proofs). Apr 29, 2023 at 2:19

See the thesis The Automorphism Groups on the Complex Plane by Aron Persson.

Sketch of the argument:

If we have an injective holomorphic function $$f$$ on $$D\setminus A$$, where $$D\subset\mathbb{C}$$ is a domain and $$A$$ a closed discrete subset, then

1. No point in $$A$$ can be an essential singularity of $$f$$. (via Picard's theorem)
2. If a point of $$A$$ is a pole of $$f$$, then it is of order one. (Otherwise $$1/f$$ is injective with root of order at least two in $$D$$ which is impossible for holomorphic functions.)
3. If every point of $$A$$ is a removable singularity, then the extension of $$f$$ to $$D$$ is holomorphic and injective.

After postcomposing with a suitable Möbius transformation one can assume that f maps infinity to infinity. So f is an entire function on the complex plane. By Casorati-Weierstrass (or Picard), it cannot have an essential singularity at infinity - otherwise it would not be injective. So it is a polynomial, and of degree one, since otherwise it would not be injective.