Product of two elliptic isometries with distincts centers

Question

I'd like to know why is the product of two elliptic isometries of the hyperbolic upper plan (or of the unitary disk) with distincts fixed points is parabolic or hyperbolic?

PS: I only need it for dimension $2$ $ie$ in $PSL(2,\mathbb{R})$

Thank you!

Answer 1

Here is what's happening:

Let $p, q$ be distinct points in the hyperbolic plane $H^2$ and let $\alpha, \beta$ be the angles of rotation of elliptic elements $f, g$ fixing $p$ and $q$ (both $\alpha$ and $\beta$ are computed counterclockwise). Then $h=g\circ f$ is elliptic iff the following holds: There exists a finite hyperbolic triangle with one side $pq$ and the angles $\alpha/2, \beta/2$ at $q, p$ respectively, and lying to the left from the oriented segment $pq$. If we fix the angles $\alpha, \beta$ in advance and take $pq$ sufficiently long then this triangle will not exist, as its sides $s$ (from $p$) and $t$ (from $q$) will be disjoint in $H^2$ (they will intersect either on the boundary or beyond, if you work in the Klein model). In order to prove this statement, consider three isometric reflections $r_1, r_2, r_3$ in the sides $pq, s$ and $t$ of "would be triangle'' defined above. Then $r_2\circ r_1= g, r_1\circ r_3=f$. The composition $h= r_2 \circ r_3$. This transformation is elliptic iff the geodesics $s, t$ intersect in $H^2$.