Currently I am reading a paper "Infinite group actions on spheres" by Gaven Martin. I am a first year graduate students and I got lots of questions, so one of them is about the following example: (sorry in advance, if that question would be trivial)

In Example 4.2. of that paper, we let $G$ be a nonelementary Fuchsian group acting on $\mathbb{S}^2$. Then $G$ leaves the disk $D^2$ invariant, and we identify the disk to a point $x_0$ and extend the action of $G$ over this point by agreeing that every element of $G$ fixes $x_0$. This produces a group homeomorphism $H$ of $\mathbb{S}^2 / D \cong \mathbb{S}^2$ acting properly discontinuously in the complement of the point $x_0$. Thus $H$ is a convergence group and is not conjugate to any Mobius group since the stabilizer of a point in a Mobius group is virtually abelian and the stabilizer of $x_0$ is $H \cong G$.

$\textbf{I have a question regarding the last sentence:}$

Why is the stabilizer of a point in a Mobius group virtually abelian in this case?

Is that true in general for $n$-sphere?

Thank you in advance!

  • 1
    $\begingroup$ Yes, this is true in all dimensions. See e.g. Ahlfors, "Möbius transformations in several dimensions" which can be easily found online. $\endgroup$ – Igor Belegradek Aug 4 '16 at 1:50
  • $\begingroup$ @IgorBelegradek thank you very much! $\endgroup$ – Jane Aug 4 '16 at 1:52

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