Classification of Mobius Transformations by Symmetry: 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: