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Let $\Omega = \mathbb{C}\setminus\{a, b, c\}$ be the complement of three distinct points in the complex plane. Show that the set of one-to-one holomorphic maps $f : \Omega \to \Omega$ forms a finite group. Is the order of this group independent of the choice $\{a, b, c\}$?

By Casorati-Weierstrass, none of the points can be essential singularities for $f$. However, it is possible that at least one of the points is a pole, for example $f : \mathbb{C}\setminus\{-1, 0, 1\} \to \mathbb{C}\setminus\{-1, 0, 1\}$, $z \mapsto \frac{1}{z}$. For this choice of points, we also have $z \mapsto z$, $z \mapsto -z$, and $z \mapsto -\frac{1}{z}$, so $\mathbb{Z}_2\times\mathbb{Z}_2 \leq \operatorname{Aut}(\mathbb{C}\setminus\{-1, 0, 1\})$. Furthermore, for any $a, b, c$ on a straight line, we can conjugate by a rotation, a translation, and a rescaling to get a corresponding subgroup. If I had to guess, the group will depend on whether or not $a, b, c$ are collinear.

Any hints on how to proceed would be very much appreciated.

1 Answers1

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First normalise the situation. Since the structure of the automorphism group is preserved under biholomorphisms, we can map $\Omega$ to $\Omega' = \mathbb{C}\setminus \{0,1,c'\}$ per $z\mapsto \frac{z-a}{b-a}$.

Now consider the situation from the viewpoint of the Riemann sphere, $\Omega' = \widehat{\mathbb{C}} \setminus \{0,1,c',\infty\}$.

Each of the four points is a removable singularity of $f\in \operatorname{Aut}(\Omega')$ when that is considered as a holomorphic map to the sphere, so $f$ extends to an automorphism of the sphere. Hence

$$\operatorname{Aut}(\Omega') = \left\{f\in \operatorname{Aut}(\widehat{\mathbb{C}}) : f(\{0,1,c',\infty\}) = \{0,1,c',\infty\}\right\},$$

and we see that $\operatorname{Aut}(\Omega')$ can be viewed as a subgroup of $S_4$.

Now investigate what restrictions the location of $c'$ imposes on the permutations of the four points that arise from automorphisms of the sphere.


We always have the four automorphisms

$$T_1 \colon z \mapsto z;\quad T_2\colon z \mapsto \frac{c'}{z};\quad T_3\colon z\mapsto \frac{z-c'}{z-1};\quad T_4\colon z \mapsto c'\frac{z-1}{z-c'}$$

of $\Omega'$. The three non-identity automorphisms among them swap the four points in pairs, and each point in $\{0,1,c',\infty\}$ is mapped to $c'$ by one of these four.

Thus, if $f$ is any automorphism of $\Omega'$, $T_i\circ f$ is an automorphism of $\Omega'$ that leaves $c'$ fixed for exactly one $i\in \{1,2,3,4\}$. So it suffices to see which Möbius transformations permuting $\{0,1,\infty\}$ have $c'$ as a fixed point. Ignoring the identity, we have

  • $z \mapsto \frac{1}{z}$, which swaps $0$ and $\infty$ and has the fixed points $1$ and $-1$, so is an automorphism of $\Omega'$ if and only if $c' = -1$;

  • $z \mapsto 1-z$, which swaps $0$ and $1$ and has the fixed points $\infty$ and $\frac{1}{2}$, so is an automorphism of $\Omega'$ if and only if $c' = \frac{1}{2}$;

  • $z\mapsto \frac{z}{z-1}$, which swaps $1$ and $\infty$ and has the fixed points $0$ and $2$, so is an automorphism of $\Omega'$ if and only if $c' = 2$;

  • $z\mapsto \frac{1}{1-z}$, which cyclically permutes $0 \mapsto 1 \mapsto \infty \mapsto 0$, and its inverse $z \mapsto 1 - \frac{1}{z}$, which have the fixed points $e^{\pm \pi i/3}$, so are automorphisms of $\Omega'$ if and only if $c' = e^{\pi i/3}$ or $c' = e^{-\pi i/3}$.

We see that generically, $\operatorname{Aut}(\Omega')$ is a Klein four-group, but for some special situations, the automorphism group is larger.

If $c' \in \left\{ -1, \frac{1}{2}, 2\right\}$, there is one nontrivial automorphism of $\Omega'$ having $c'$ as a fixed point [an involution], and $\operatorname{Aut}(\Omega')$ is (isomorphic to) the dihedral group ($D_4$ or $D_8$, depending on the used nomenclature) of the rigid motions of a square.

If $c' = e^{\pm \pi i/3}$ (then the three points $0,1,c'$ form an equilateral triangle), there is a nontrivial automorphism of period $3$ of $\Omega'$ having $c'$ as a fixed point, and then $\operatorname{Aut}(\Omega')$ is (isomorphic to) the alternating group $A_4$ of even permutations of $\{0,1,c',\infty\}$.

Daniel Fischer
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  • The automorphisms of the Reimann sphere are Möbius transformations which preserve the cross ratio of any four points, in particular ${0, 1, c', \infty}$. The kernel of the action on ${0, 1, c', \infty}$ by $S_4$ is the Klein four group, so $\operatorname{Aut}(\Omega') \leq \mathbb{Z}_2\times\mathbb{Z}_2$. I imagine the value of $c'$ will effect which subgroup it is. Am I on the right track? – Michael Albanese Aug 16 '14 at 20:20
  • We need not have $\operatorname{Aut}(\Omega') \leq \mathbb{Z}_2\times \mathbb{Z}_2$. If for example $c' = 2$, then $z \mapsto \frac{2}{2-z}$ is a cycle of order $4$, $0 \mapsto 1 \mapsto 2 \mapsto \infty \mapsto 0$. But yes, the value of $c'$ affects the subgroup. – Daniel Fischer Aug 16 '14 at 20:33
  • I understand why $\operatorname{Aut}(\Omega')$ is not necessarily a subgroup of $\mathbb{Z}_2\times\mathbb{Z}_2$. However, I don't know how to determine which elements of $S_4$ correspond to Möbius transformations. According to this, there are six possible values of the cross ratio under permutations, and as Möbius transformations preserve the cross ratio, any permutation which leads to a different value of the cross ratio can't be induced by a Möbius transformation. But even if there is a permutation which doesn't change the . . . – Michael Albanese Aug 16 '14 at 22:58
  • . . . cross ratio, it's not clear to me that there is a Möbius transformation which realises this permutation. – Michael Albanese Aug 16 '14 at 22:58
  • The brute-force way: Take three points in ${0,1,c',\infty}$. Map them to $0,1,\infty$ in that order. See where the transformation maps the fourth point. If it doesn't map it to $c'$, the transformation doesn't belong to $\operatorname{Aut}(\Omega')$, if it does, it belongs. Slightly less brute-forcish, you look at the cross-ratios. $\operatorname{CR}(c',0,1,\infty) = c'$ (or, if you use the other widespread convention for the cross-ratio, $\operatorname{CR}(c',1,0,\infty) = c'$). – Daniel Fischer Aug 16 '14 at 23:23
  • You can express the possible values for the cross-ratios of the permutations in terms of $c'$, and those where the cross-ratio is $c'$ belong to $\operatorname{Aut}(\Omega')$, the others not. – Daniel Fischer Aug 16 '14 at 23:24
  • Using the less brute-forcish way, if you find a permutation which gives the same cross ratio, how do you know that there is indeed a Möbius transformation which realises this permutation? – Michael Albanese Aug 17 '14 at 00:59
  • The cross-ratio is a Möbius transformation. Depending on the convention used $$\operatorname{CR}(z,a,b,c) = \frac{z-a}{z-c}\frac{b-c}{b-a}$$ is the Möbius transformation sending $(a,b,c)$ to $(0,1,\infty)$ (with appropriate interpretation as a limit resulting in the removal of factors containing $\infty$ if one of $a,b,c$ is $\infty$), or $$\operatorname{CR}(z,a,b,c) = \frac{z-b}{z-c}\frac{a-c}{a-b}$$ is the Möbius transformation sending $(a,b,c)$ to $(1,0,\infty)$ (with the same proviso about one of $a,b,c$ being $\infty$). – Daniel Fischer Aug 17 '14 at 09:15