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Given a non-parabolic transformation which is also an orientation preserving isometry in the hyperbolic upper half plane union the boundary, if I know the two fixed points and they are two different irreducible fractions on the boundary, how can I find the corresponding Möbius transformation?

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I haven't thought about this stuff in a while, but if I recall correctly this only nails down that you have an elliptic Mobius transformation (of which the standard example takes the form $z\mapsto az$ for $a>0$). Check out Alan Beardon's "Geometry of Discrete Groups". –  Aaron Mazel-Gee Feb 13 '11 at 8:01
Small correction - they are hyperbolic, not elliptic, transformations. Beardon's book is indeed a great reference. –  Sam Nead Feb 13 '11 at 22:36
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2 Answers

It is an important and beautiful accident that the orientation preserving isometry group of the upper half plane $\bf{H}$ is $PSL(2,\bf{R})$. The action of the matrix $A = [[a,b],[c,d]]$ is given by $A\cdot z = (az + b)/(cz + d)$. In addition, we allow the point $\infty = 1/0$: this has image $a/c$. Likewise the real number $-d/c$ is sent to infinity.

Now to your question: Suppose that $p, q$ are distinct real numbers, thus two points in $\partial \bf{H}$. Then there is a hyperbolic geodesic connecting them. A calculation using the above reveals that there is a one-parameter family of hyperbolic isometries fixing $p, q$. To get you started here is the first line of the computation supposing that you have explicit, simple, values of $p, q \in \bf{R}$ in mind.

Fix $A$. Then $A \cdot p = p$ implies that $ap + b = cp^2 + dp$. Solve this, do the same for $q$ and remember that $kA = A$ for non-zero $k \in \bf{R}$. (That is what the $P$ stands for in $PSL$.)

As a final note - as the comment suggests it is also possible to find a matrix $B$ sending $p, q$ to $0, \infty$, solve the problem for those very special values, and conjugate back. But perhaps it is important to understand the problem both ways.

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Why do you call that an accident? –  t.b. Feb 13 '11 at 11:35
@Theo - It is not an accident that $PSL(2,R)$ preserves the conformal structure of the upper half-plane. Why conformal transformations (preserving angle) should be hyperbolic transformations (preserving distance) always shocks me a little. For example, it is false in the Euclidean and in the spherical geometries... –  Sam Nead Feb 13 '11 at 22:35
Ah, I see. Thanks for this explanation. –  t.b. Feb 13 '11 at 23:04
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Let $T:z\to z'$ be such a Moebius transform and let $p, q\in{\mathbb R}$, $\ p<q$, be its two fixed points. As $z'=p$ iff $z=p$ and $z'=q$ iff $z=q$, the two variables $z$ and $z'$ have to be related by a formula of the form $${z'-q \over z'-p}=\lambda {z-q \over z- p}$$ for some complex constant $\lambda$. This constant has to fulfill conditions to guarantee that (a) $T$ maps the real axis to itself and (b) $T$ maps, e.g., the point $i$ to a point in the upper half plane.

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