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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!

WrabbitW
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  • It depends on the fixed points and angles of rotation. The product can also be elliptic. – Moishe Kohan Apr 21 '16 at 00:33
  • Ok so let say one isometry as the center of the disk as fixed point, how could I show that the product isn't elliptic? – WrabbitW Apr 21 '16 at 16:51
  • Knowing one center is not enough, the product could be elliptic. – Moishe Kohan Apr 21 '16 at 17:48
  • Oh ok and if they would have same angle? I´m refering to Case 4 page 136 in the following Paper : Pierre de la Harpe, Free groups in Linear groups, L'enseignement mathématique, t.29 (1983) – WrabbitW Apr 21 '16 at 18:13
  • I do not know that paper but even for equal angles the product can be elliptic (even of the same angle as the other two). – Moishe Kohan Apr 21 '16 at 18:22
  • Really? I´m confused then, because in the paper what they do is: Take 2 elliptic transformation g and h such that g^2 is not the identity and such that they don't share fixed points. Then conjugate g to get a rotation g' with center the origin and consider k:=hg'h^-1. Finally they give some strange argument to say that there is a point in the border of the disc that is sent to same point by k and g'. – WrabbitW Apr 21 '16 at 18:57

1 Answers1

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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 $H^2$.

Moishe Kohan
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  • Ok I get the proof of the statement but I don't really see in my case how they get that we can't find such triangle with k and g'^-1. How could I notice that the centers are "far enough" to get a non elliptic isometry? – WrabbitW Apr 24 '16 at 01:27
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    @WrabbitW Use dual hyperbolic cosine law to show that a triangle with two fixed angles and long enough side between does not exist. – Moishe Kohan Apr 24 '16 at 04:03
  • Hi, I'm so sorry to answer so late... I found and other way to show what I wanted (with more calculus on matrices) I could not see the geometric argument. Thank you a lot for your time! – WrabbitW May 10 '16 at 17:56