As it is known from the classification of finite subgroups of $SO(3)$ (for example, here on page 3), there are such groups $G$ and their orders $e_i$ of stabilizer subgroups of point of set $S$ which contains points that are fixed by some nontrivial element of $G$:

  • Cyclic: n, n
  • Dihedral: 2, 2, n/2
  • Tetrahedral: 2, 3, 3
  • Octahedral: 2, 3, 4
  • Icosahedral: 2, 3, 5

It is easy to check that there are all possible cases of orders $e_i$ using Riemann-Hurwitz formula for ramified covering $X \rightarrow X/G$ by this subgroup ($n=|G|$ and $g=g(X/G)$):

$$2 = n (2-2g) - \sum \frac{n}{e_i}(e_i-1)$$

There is $\frac{n}{e_i}$ is size of orbit and $e_i$ is ramification index equal to size of stabilizer subgroup.

I am interesting whether it is possible to prove that there is only one group of defined orders of stabilizers - my plan is like this:

  1. Let X, Y be two Riemann spheres with finite subgroups G, H with equal orbit sizes
  2. Then both $X/G$ and $Y/H$ are of genus 0 so Riemann spheres
  3. If there are three orbits, there are three branch points on $X/G$ and $Y/H$ so one can identify this points using 3-transitivity of $PSL(2,C)$
  4. I suppose that one can also identify two ramified coverings of Riemann sphere by other Riemann sphere with equal ramification, but I have no idea how
  5. Then $G$ is the group of deck transformations of this covering

The classical Galois theory of covering spaces probably applies to do what you want in part 4.

Pop out the singular points on the base, and their inverse images up stairs, and now you have a classical covering space. The group of deck transformation of the covering $p:E\rightarrow B$ where $B,E$ are connected, locally path connected and Hausdorff, is isomorphic to $N(p_{\#}\pi_1(E,e))/p_{\#}\pi_1(E,e)$ where the normalizer $N(p_{\#}\pi_1(E,e))$ means the normalizer in $\pi_1(B,b)$, where of course we chose a basepoint $b\in B$ and a basepoint $e\in p^{-1}(b)$.

The fundamental group of the base in this case is an $n$-punctured sphere. It has presentation, $$\pi_1(S^2-\{n \ \mathrm{points}\})=\langle x_i|\prod_{i=1}^nx_i=e\rangle.$$
The generators correspond to a loop coming from the basepoint, and going around a deleted point. If you choose them carefully the last loop is the inverse of the product of the first $n-1$, leading to the presentation above.

Naming the ramification index of each point you popped out allows you to name a normal subgroup of $\pi_1(S^2-\{n \ \mathrm{points}\})$. Specifically the normal subgroup generated by $x_i^{e_i}$.

Since this subgroup is normal, it's normalizer is all of $$\pi_1(S^2-\{n \ \mathrm{points}\})=\langle x_i|\prod_{i=1}^nx_i=e\rangle.$$

So the group of deck transformations is going to be,

$$G(p,E)=\langle x_i|\prod_{i=1}^nx_i=e, x_i^{e_i}=e\rangle.$$

In the cases where you are interested except for the cyclic group, you have classical triangle groups, and the presentation above from the branching data is exactly the classical presetation so they are isomorphic.

For instance in the case of ramification indices $2,3,5$ the presentation is $$ \langle x_1,x_2,x_3| x_1x_2x_3=e, x_1^2=e,x_2^3=e,x_3^5=e\rangle,$$ the standard presentation of the icosahedral group.

The cyclic group also gets the classical presentation by this procedure.

Get careful with the theory of covering spaces, and be careful with your branch points and you will have a proof.

  • $\begingroup$ This is pretty classical stuff. There is a lecture notes in math by Vogt, Coldeway, and Zieschang that deals with a lot of these issues. $\endgroup$ – Charlie Frohman Jan 5 '14 at 18:04

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