Reflections generating isometry group I was reading an article and it states that every isometry of the upper half plane model of the hyperbolic plane is a composition of reflections in hyperbolic lines, but does not seem to explain why this is true. Could anyone offer any insight? Thanks.
 A: An isometry $\phi:M\to N$ between connected Riemannian manifolds $M$ and $N$ is completely determined by its value at a single point $p$ and its differential at $d\phi_p$.
Take any isometry $\phi$ of $\mathbb{H}^2$.  Connect $i$ and $\phi(i)$ by a (unique) shortest geodesic and let $C$ be a perpendicular bisector of the connecting geodesic.  Then the reflection $r_C$ across $C$ maps $i$ to $\phi(i)$.  Now take an orthonormal basis $e_j$ at $i$.  It is mapped to $d\phi e_j\in T_{\phi(i)}\mathbb{H}^2$.  Linear algebra tells us that in $T_{\phi(i)}\mathbb{H}^2$ we can map $d\phi e_j$ to $dr_C e_j$ by a reflection across a line (if $\phi$ is orientation-preserving) or by a rotation (if $\phi$ is orientation-reversing).  A two-dimensional rotation can be written as a composition of two reflections.
By the exponential map, the lines across which we're reflecting map to geodesics in $\mathbb{H}^2$ and the reflections extend to reflections of $\mathbb{H}^2$ across those geodesics.
We have therefore written $\phi$ as a composition of reflections.
In some cases, e.g. in Thurston, Three-Dimensional Geometry and Topology, the isometry group of $\mathbb{H}^2$ is defined as the group generated by reflections across circles (e.g., in the Poincare model).  Chapter two of that book has a discussion of hyperbolic geometry and exercises comparing the various perspectives on isometries, e.g., as $$Sl(2;\mathbb{R}) = SO(1,1) = \mbox{Möbius transformations with real coefficients} = \langle\mbox{refl. across circles}\rangle.$$You might also check Chapter B (I think) of Benedetti and Petronio, Lectures on Hyperbolic Geometry.
A: Reference: See section 4.3 "The three reflections theorem" in Geometry of Surfaces by John Stillwell.
