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Reading through a book of Farb & Magalit (p. 271/in pdf 288 shortly before Prop 10.3) I stumbled upon (after some cleaning) the following equation for hyperbolic metric:

$dist(z, Az)=cosh^{-1}(trace(A)/2)$

for some non-elliptic $A \in PSL_2(\mathbb R)$ and $A(\cdot)$ the typical möbius transformation. I have some problems proving that equation (mod ignoring some constants). My first steps first regarding the case $z=i$ (this should suffice for non-elliptic transformations?) and using this equation were quite fruitless. I would be happy for some help/hint/link were this equation comes from.

Edit: I found the mistake (if someone happens do struggle at the same point and visits this page): The $dist(z, Az)$ should actually be the translation length of the möbius transformation, so the minimum of all these possible $z \in \mathbb H$ or in the context where this question arose the length of the geodesic arc connecting the two corresponding sides of the fundamental area. This is exactly what this equation says for hyperbolic $A$ (s. wiki). Not sure, why $A$ couldn't be parabolic in this context, but will figure this out too (this is outside the question).

ctst
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  • Your formula cannot possibly be correct, because the right hand side is independent of $z$ but the left hand side is not. – Lee Mosher Oct 10 '16 at 16:55
  • @LeeMosher By analogy with $dist(x, x + v)$ is independent of $x$ in flat Euclidean space with the usual metric, hyperbolic $dist(z, Az)$ could be independent of $z$ if $A$ moves every $z$ the same hyperbolic distance? I don't know if this is possible in curved space, however. – Claude Oct 10 '16 at 19:08
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    The only isometry $A$ of the hyperbolic plane which has the property that $dist(z,Az)$ is independent of $z$ is the identity, for which $d(z,Az)=0$. – Lee Mosher Oct 10 '16 at 20:15
  • @LeeMosher Thanks for the information. Seems I have to do this a different way or overlooked something else. There is no chance this equation is true for some special $z$ (independent of $A$)? – ctst Oct 11 '16 at 09:15
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    There are no “special $z$”: all points of the hyperbolic plane are equivalent, since there is always an isometry to map one to the other. – MvG Oct 11 '16 at 11:09
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    @MvG Oh, yes, you are right. Hence this must be messed up quite some time earlier. I think we can consider this question as closed (unless someone who struggled at the same point and didn't make the same mistakes as me stumbles in here, but then the question should be heavily changed). Thanks for the help people! – ctst Oct 11 '16 at 12:13

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