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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.

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3 Answers 3

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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".

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  • $\begingroup$ How do you prove, using only essentially elementary algebra, that any holomorphic map from sphere to itself is meromorphic? :) $\endgroup$ Apr 28, 2023 at 16:42
  • $\begingroup$ 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). $\endgroup$ Apr 29, 2023 at 2:19
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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.
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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.

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