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.
 

The Classification!

Sorry about the delay in posting--this post is going to be long, meticulous, and introducing new topics, so it took me a long time to write out.

In my last post, I talked about how we know every Mobius Transformation T has at least 1 and potentially 2 fixed points. If T has only 1 fixed point, then we know it's conjugate (meaning equivalent) to a translation on the sphere. If T has 2, then it can either be loxodromic or elliptic, based on what kinds of fixed point they are, and how the points spiral. These transformations are conjugate to a transformation T'(z)=kz. A special case of the loxodromic transformation is hyperbolic, when k is real.

This all becomes a bit easier to conceptualize and think about in hyperbolic geometry, which I'll address later in this post. But first, I'll go through the types of transformations in terms we've talked about so far. Pretty much the logic I'm doing here is given a certain Mobius Transformation T, we conjugate it by another transformation S and then look at its action and see T'(z)=STS^-1(z). S isn't any arbitrary transformation though; it is determined by the fixed points of T. The underlying motivation behind creating S is wanting T' to send ∞ to ∞. This means we need to map the fixed point of T to infinity by S. If there is no other fixed point, we can think of the translation around infinity. If there is another fixed point, again, for ease, we map the other fixed point of T to 0 by S. This makes T' a linear mapping because the only places a line T(z)=az is fixed is at 0 and ∞.

For reference later on. If instead there is only one fixed point, P, then S(z) = Az/(z-P).

How do we know if there's one or two fixed points? Well remember Mobius Transformations are of the form T(z)=(az+b)/(cz+d), and its fixed points are of the form z= (az+b)/(cz+d). If we rearrange and expand, we get cz² + (d-a)z -b =0. Of course you all remember your quadratic formula of z = (-b +/- Sqrt(b² -4ac))/ 2a, right? Well when we plug in what we just rearranged for our fixed points, you'll see it's (a-d +/- Sqrt((d-a)² +4cb))/ 2c. What determines how many fixed points there are is the discriminant, (d-a)² +4cb. If the discriminant is 0, then we have only one fixed point, (a-d)/2c. But if it's nonzero, then we'll get 2 fixed points, because of the plus or minus sign.

Let's go a step further. If we expand (d-a)² +4cb, you get d² +a²-2ad+4cd. We want to factor this, so we do the following: d² +a²-2ad -2ad +2ad +4cd and factoring, now, we get (a+d)² -4(ad-bc). Two things to remember for this to be geometrically significant is 1) the trace of a matrix is defined as a+d; and 2) for all Mobius Transformations, ad-bc=1. (If you have any questions on either of those, comment.) Thus, we know the amount of fixed points a Mobius Transformation has is determined by its trace squared minus four: (TrT)² -4.

Off the bat, if the discriminant is 0, or there's only one fixed point, P, you know that your function is parabolic. We create our S as some Az/(z-P), as then S(P) will be mapped to ∞. Then we can conjugate T by S^-1 and now we have our T'(z)= z+b. This may seem a bit fast, but it'll get clearer when we do more work for other transformations. Parabolic ones are probably the simplest--their fixed point is both. An example is T(z) = (9z-2)/(8z+1). Feel free to verify that. Notice how this transformation when normalized has a trace of 2. This is crucial, because it is a necessary condition for a parabolic transformation to have a trace of +/-2. This is because the only way the discriminant will equal 0

So that's what happens if we have one fixed point, but what about for 2? As you can see in the reference picture above, the difference between a loxodromic and an elliptic Mobius Transformation is what the modulus of k is equal to: if |k|=1, then T is elliptic, if not, T is loxodromic. Or, we can more easily see the difference if we look again to the discriminant. If the discriminant is positive, then you'll get a loxodromic transformation, and if it's negative, you'll get an elliptic one. We know what the discriminant will be based on the trace: (TrT)² -4, so we can easily classify loxodromic transformations as those with -∞< TrT<-2 and 2<TrT<∞ and elliptic transformations as -2<TrT<2.

To clarify this and to check our processes, let's look to some examples. First, let T(z) = (2z+5)/(-3z-1). So, -5/2 maps to 0, -1/3 maps to ∞, -6/5 maps to 1, and -1/2 + Sqrt(69)i = P and -1/2 - Sqrt(69)i = Q are our fixed points. Thus, we construct an S(z) = A(z-Q)/(z-P). Now P maps to ∞, Q maps to 0, and 0 maps to AQ/P. So now we have a conjugation of T by S, T'(z)= STS^-1(z), such that T'(z)=kz. Whatever |k| is equal to will tell us if T is elliptic or loxodromic. That can be done by finding |T'(1)|=|k(1)|=|STS^-1(1)|, because |T'(1)|=|k|, but that is really arduous and tedious work, so I'm just going to quickly demonstrate to you this T I've selected is elliptic by showing one |T'(z)|=|z|. Obviously this isn't mathematically rigorous, but given that I'm not going to choose any particular number to ensure it comes out that way, this is useful to a more casual reader for a heuristic understanding, but I still do promise (and please verify it yourself if you justly don't take me at my word) that |T'(1)|=1). I've already said in this paragraph that S(0) is AQ/P, and T maps -5/2 to 0. We can easily figure out S^-1maps A((-5/2)-Q)/((-5/2)-P) to -5/2. So, T'(A((-5/2)-Q)/((-5/2)-P))= AQ/P. Because A is a constant, we can remove it when considering the modulus. Having calculated the modulus of both, I have found that |((-5/2)-Q)/((-5/2)-P)| = |Q/P| = 1. Thus T is elliptic.

Now we should find out if T being elliptic coheres to the set of conditions we've also derived, so let's now check if the discriminant of T is negative, which would indicate T being elliptic. And indeed it is, as TrT = 1/13, and -2<(1/13)-4 <2!

You can verify that T(z)=(4z+5)/(3z+2) is indeed loxodromic, using either of the ways above.

To sum thus far up, if TrT =2 or -2, T is parabolic, if -∞<TrT<-2 or 2<TrT<∞, T is loxodromic, and if -2<TrT<2, then T is elliptic. That is completely exhaustive of all the possibilities of T.

Next, let's turn to the geometric interpretations of the different kinds of Mobius Transformations.
A loxodromic Mobius transformation has two fixed points--one attracting and one repelling. Naturally, points here spiral out from the source (repelling point) and into the sink (attracting point).

These pictures are, again, from Indra's. Notice how everything is flowing out of the fixed point on the left, and flowing into the fixed point on the right.

And, a special kind of loxodromic Mobius Transformation, as I said at the top of this post, is a hyperbolic Mobius Transformation, when |k| and TrT are real. For instance, having conjugated by S just as we did before, T'(z) could end up equaling something like (i+9)z, or T'(z) could end up equaling something like 9z. In the latter case, T would be hyperbolic. Hyperbolic Mobius Transformations don't have points spiral into or out of either fixed point, but instead move in circles through them:

Finally, elliptic transformations don't have any sources or sinks, meaning neither of their fixed points are attracting or repelling, that both of them are neutral. Here, the points simply move in circles around the fixed points:

Actually, I'm going to break the post here for the sake of consistency of post-theme, but this will immediately continue in the next post.

How Attracting

Sorry about the hiatus on posting, I was in San Diego for a bit of debate prep for the upcoming Tournament of Champions. But fear not, the project lived on, even if it wasn't chronicled on blogger. Today I'm going to post about fixed points and their importance in the classification of Mobius Transformations.

A fixed point is pretty simple. For a transformation T(z), a fixed point is any point z such that T(z)=z. See? It's fixed, as if we imagine the transformation moves or changes the domain, that point z stays put. Fixed points are important, as they allow us to understand the transformations around a dynamic focused on the point.

Let's first consider a linear function T(z)=az. If we want to find the fixed points, we just have to find anywhere z=az. The only two places where this happens are 0=0z and ∞ = ∞z. Geometrically, this looks like a line from 0 to ∞, pretty intuitive, ie, it's a linear function, but on the Riemannian Sphere, it looks like a spiral:

From Indra's. Also known as a loxodrome from the Greek "running obliquely." Here, 0 acts as a "source" and ∞ as a "sink." I'll clarify some of the terminology more later.

This gets even more interesting. 0 and ∞ seem like the most special numbers, but on the Riemannian Sphere, because 1/0 and 1/∞ are well defined, they are just like any other numbers. This means we can conjugate by other transformations to have the same spiral on the sphere, just with different sinks and sources! Let's say we wanted to make our sinks and sources -1 and 1 (which are arbitrary, we could make them whatever), then we just need another Mobius Transformation R such that R(z) maps 0 to -1 and ∞ to 1. Fortunately, we know how to do this from the last post, and we create R(z) = (z-1)/(z+1). Check: R(0) = -1/1 = -1 and R(∞) = ∞/∞ = 1. Now because we have to conjugate T to find RTR^-1(z), so we need to find R^-1. That's not difficult if we imagine Mobius Transformations as matrices, and then we know that R^-1 is (z+1)/(-z+1). To check this, we should verify that -1 is sent to 0 and 1 is sent to ∞. That is the case, so we are good.

T^*= RTR^-1(z) has the properties we wanted. It has 2 fixed points at -1 and 1. T^*=
RTR^-1(-1)= RT(0) = R(0) = -1, by construction. Interestingly, though T = az was linear, after conjugating by two non-linear Mobius Transformations, T^* is distinctly nonlinear. If you were to calculate RTR^-1 out, you would get ( z(1+a)+a-1 )/( z(a-1)+a+1).

If T looked like:

Then T* would look like:

Recall that Mobius Transformations are of the form T(z)= (az+b)/(cz+d). Thus, a fixed point under any arbitrary Mobius Transformation is any z such that z= (az+b)/(cz+d). If we do some rearrangement, we see that cz² + z(d-a) - b = 0. This is a quadratic, and we can solve for z to find the fixed points: ( (a-d) ± sqrt( (a-d)² + 4bc) ) / 2c. This means that every Mobius Transformation has either one or two fixed points, depending on the discriminant. I'll talk more about this in the next post, which will come soon.

Off the bat, we can see some interesting things. If c=0 (and d isn't, else that'd just describe the above case), for some Mobius Transformation T(z)= (az+b)/d = (a/d)z + (b/d), then T is affine (affine means a linear function translated from 0. Essentially, linear functions look like T(z) = az, and affine functions look like T(z) = az+b). If T is affine, then it only has one fixed point at infinity-- T(∞)= a(∞)+b =∞, and not at 0 (T(0) = b). This makes sense because then our quadratic formula from above would be ( (a-d) ± sqrt( (a-d)²) ) / 0, which equals ∞. 
 
Now I'd like to take a step back and talk about what it means to be a source or a sink. A source is a fixed point that is repelling, and a sink is a fixed point that is attracting. Alternatively, a fixed point can be neither, in which case it is a neutral fixed point. 

The definition of an attracting fixed point x₀ is for some L in the open interval (0,1), there exists an open interval U containing x₀, such that |f(x)-x₀| ≤ L|x-x₀|, for all x in U. Conversely, a repelling point is a fixed point x₀ such that there's a L>1 and an open interval U containing x₀ such that |f(x)-x₀| ≥ L|x-x0|, for all x in U.

Pretty much, what that means is that if you have a sink or an attracting point, then if you take some space, then undergoing iterations of the function, it will be bijectively mapped to a smaller and smaller space arbitrarily close to the point. Of course by "smaller" I don't mean "less points", but instead a smaller looking area. The reason this still works, though, is because there are uncountably many real numbers (there are as many numbers in the interval (0,1) as there are between (-∞,∞) ). The opposite is the case for repelling points.

This will come in handy to think about when next post I'll classify all types of Mobius Transformations into three categories based on how many and which kinds of fixed points they have.