Mobius Transformations in Hyperbolic Geometry: The Upper Half-Plane and the Poincare Disc Model

So much for the trend of lighthearted titles. Maybe this one is just an ironic deconstruction on what it means to be a lighthearted title.

As long as we are considering geometric interpretations of each type of Mobius Transformation, I feel like using hyperbolic geometry gives a more illuminating correspondence of the geometric implications of T. So, let me introduce some new terms, but before I do, I'll broadly explain what hyperbolic geometry is. Hyperbolic geometry is Non-Euclidean geometry, where Euclidean is the normal, everyday space you're used to working with. An example of a Non-Euclidean geometry is the Earth--if you wanted to get from New York to LA , the fastest way to get there might look something like:

But obviously if we're in R³, the fastest way would just be a straight line, which would mean you'd have to go inside the Earth's crust. What's more is that hyperbolic geometry has a lot richer properties and depth of study than Euclidean geometry. Remember from the first blog post about Symmetry, that there are limited ways you can tile a Euclidean plane. But in hyperbolic geometry, there are infinite ways.

Now, let's define some terms. H is the upper half-plane. This is the set of all complex numbers with a positive imaginary part. dH is the boundary of the upper half-plane, also known as the "circle at infinity." dH, then, is just any number that has imaginary part value 0 or is infinity. If you were to stereographically project dH back down to C², you'd get the real number line, plus infinity. So you can see that topologicallydH is a circle.

Now it's useful to see that if we're in H, our Mobius Transformations can be classified by fixed points still, but more elegantly. If T has 2 fixed points, and both are in dH. If T has one fixed point in dH, and none in H, then T is hyperbolic (note: there are no non-hyperbolic loxodromic Mobius Transformations when in H). If T has no fixed points in dH and one in H, then T is elliptic. 

Define D to be the Poincare Disc Model. This will be useful to imagine our elliptic Mobius Transformations. Similar to H, D, is the disc of all complex numbers z such that |z|<1. The boundary of D,dD, or the circle at infinity, is the set of complex numbers with |z|=1. Using h(z) = (z-i)/(iz-1), we can bijectively map H to D and dH to dD, allowing us to transfer the logic we've established thus far and apply it to D and dD. I won't show how they are bijective or how we calculate distances, as that will be unnecessary to a causal reader. But what we can do, knowing h exists, is conjugate a Mobius Transformation T in H by h to get a Mobius Transformation in D, of the form hTh^-1.

This means all Mobius Transformations in D, for some complex numbers e and f, and their conjugates e-bar and f-bar, are of the form: T(z) = (ez+f)/(fbar z + ebar). The way I figured that out is by using an arbitrary Mobius Transformation in H, (az+b)/(cz+d) and then composing and simplifying hTh^-1. Protip--if you want to check that for yourself, remember that composing two Mobius Transformations is the same as multiplying the matrices of their coefficients.

Within D, the classification of a Mobius Transformation T willlook similar to the way it did in H, but a bit different, so be careful not to just mentally replace "H" in your head with "D". If T has 2 fixed points, and both are in dD, then T is hyperbolic. If T has one fixed point in dD, and none in D, then T is parabolic. If T has no fixed points in dD and one in D, then T is elliptic. 

If we take an e value to be e^iθ/2 and an f value to be 0, our Mobius Transformation in D will be T(z)= e^iθ, and acts by rotating the Poincare disc around 0 by an angle of θ. We know that all rotations have one fixed point--the point about which they're rotating--so now we can just think of elliptic transformations as rotations! But let's not stop there. If we know that rotations are elliptic transformations in D, then what do they look like in H? This time, instead of conjugating hTh^-1, we want to go from D to H, so we're instead conjugating h^-1Th. The resulting Mobius Transformation now looks like T(z) = (cos(θ/2)z + sin(θ/2))/(-sin(θ/2)z + cos(θ/2)). And because we remember that all rotations have a fixed point where they rotate around, we need to find the fixed point. Rearranging, we find that -sin(θ/2)z² + cos(θ/2)z = cos(θ/2)z + sin(θ/2), and cancelling the cos(θ/2)z and then dividing by -sin(θ/2), we get that z² =-1, which means the fixed point is at i! i is in H, not dH, which means T is elliptic in H.

There is a lot more cool stuff to do with Mobius Transformations in hyperbolic geometry, and I really regret that I hadn't worked with them in hyperbolic geometry sooner and that I have to debate at the Tournament of Champions this week, so I can't devote much more time to going deeper into concepts like Fuchsian groups. Oh well, I'll see what I can do. At least I finally classified these Mobius Transformations, and at least I finally feel like I've done some real math.