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.