Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

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.

share|improve this answer
    
Why do you call that an accident? –  t.b. Feb 13 '11 at 11:35
1  
@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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.