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So this is from Ratcliff's book on Hyperbolic Manifolds, and I don't understand why $\beta_i$ are parabolic translations. Note that a parabolic translation is a map $\varphi\in \text{M}(U^n)$ conjugate to a translation $x+a$, where $a\in E^{n-1}$.

EDIT: Also $M_0(\ast)$ are the set of orientation preserving Mobius transformations. And $M(\ast)$ are the set of Mobius transformations.

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  • $\begingroup$ A hyperbolic translation is not a map "conjugate to a translation $x+a$ where $a \in E^{n-1}$." $\endgroup$ – Lee Mosher Feb 6 '17 at 14:16
  • $\begingroup$ My bad it should say parabolic translation. $\endgroup$ – Enigma Feb 6 '17 at 16:14
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Consider the spheres $\Sigma$, $\Sigma_1$. They are tangent at some point $p \in E^{n-1}$.

Choose $\psi \in M(U^n)$ such that $\psi(p) = \infty$. It follows that $\psi(\Sigma)$, $\psi(\Sigma_1)$ are vertical planes and they are parallel.

The map $\psi \sigma\psi^{-1}$ is reflection in $\psi(\Sigma)$ and the map $\psi\sigma_1\psi^{-1}$ is reflection in $\psi(\Sigma_1)$, and their product $\psi \sigma\psi^{-1} \psi\sigma_1\psi^{-1} = \psi\sigma\sigma_1\psi^{-1}$ is a parabolic transformation of the form $x \mapsto x + a$ where $\frac{1}{2} a$ is a vector perpendicular to those two planes having one endpoint on one plane and opposite endpoint on the other.

Thus, $\sigma\sigma_1$ is conjugate to $x \mapsto x+a$, and is therefore parabolic.

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