∑idiot's Blog

The math fork of sumidiot.blogspot.com

Linear Fractional Transformations

a.k.a. Möbius Transformations, are a type of function. I’ll talk about them as functions from the complex plane to itself. Such functions are given by a formula

$\dfrac{az+b}{cz+d}$

where $a,b,c,d$ are complex values. If $c$ and $d$ are both 0, this isn’t much of a function, so we’ll assume at least one isn’t 0.

I’d like to talk about what these functions do, how to have some hope of picturing them as transformations $\mathbb{C}\to\mathbb{C}$. To do this, let’s consider some easy cases first.

If $c=0$ (and so by assumption $d\neq 0$), then we may write the function $\frac{a}{d}z+\frac{b}{d}$, or simply $a'z+b'$ for some complex values $a',b'$. This is now a linear (some might say affine) transformation of the complex plane. Think about it as the composite $z\mapsto a'z\mapsto a'z+b$, where the first map multiplies by $a'$, and the second adds $b'$. Multiplying by a complex value $a'$ is the same as scaling by the real value $|a'|$ (the “norm” of $a'$, distance from $a'$ to the origin) and then rotating by the “argument” of $a'$. If you think about $a'$ as a point $(r,\theta)$ in polar coordinates, then the argument of $a'$ is $\theta$ (or so), and so multiplication by $a'$ is multiplication by the real value $r$ (which is just a stretching (or shrinking) of the complex plane away from (toward) the origin if $r>1$ (resp. $0\leq r<1$)) and then rotation by the angle $\theta$. The second transformation in the composite, “add $b'$“, just shifts the entire plane (as a “rigid transformation”) in the direction of $b'$.

So the case when $c=0$ is just a linear transformation, which aren’t too difficult to picture. Another important case is $1/z$, so the coefficients are $a=0,b=1,c=1,d=0$. To talk about what this does, let’s first talk about “inversion” with respect to a fixed circle.

Let $C$ be a circle with radius $r$, in the plane, and $z$ any point in the plane. Let $O$ denote the center of the circle and $d$ the distance from $O$ to $z$. The inversion of $z$, with respect to $C$, is the point on the line through $O$ and $z$ (in the direction of $z$ from $O$) whose distance from $O$ is $d'=r^2/d$. This means that points near the center of $C$ are sent far away, and vice versa. Points on $C$ are unchanged. Technically I guess we should say that this function isn’t defined at $O$, but people like to say it is taken to “the point at infinity” and, conversely, that inversion takes $\infty$ to $O$. These things can be made precise.

You might notice that doing inversion twice in a row gets you right back where you started. It also turns out that If $C'$ is another circle in the plane, not passing through $O$, then the inversion of all of its points is another circle. If $C'$ passes through $O$, then the inversion works out to be a line. Since doing inversion twice is the identity, inversion takes lines to circles through $O$. If you’re thinking about the comments about $\infty$ above, this makes sense because every line “goes to $\infty$“, and so the inversion of a line will go through the inversion of $\infty$, which I said should be $O$.

All of this talk about inversion was to describe the function $1/z$. This function is the composite of inversion with respect to the unit circle centered at the origin followed by a reflection across the horizontal axis (real line). Don’t believe me? The equation $d'=r^2/d$ defining the relationship between distances when doing the inversion can be re-written as $dd'=r^2$. If we’re doing inversion with respect to a unit circle, then $dd'=1$. This means that when we multiply $z$ with its inversion with respect to the unit circle, call it $z'$, the result will be a point with norm 1 (i.e., a point on the unit circle). Next up, multiplying $z$ by $z'$ produces a point whose angle from the positive real axis (which I called the argument before, the $\theta$ from polar coordinates) is the sum of the angles for $z$ and $z'$. Since we did the reflection across the horizontal axis, the argument for $z'$ is precisely the negative of the argument for $z$, meaning their sum (the argument of their product) is 0. So $zz'$ is a point on the unit circle making an angle of 0 with the positive real line, i.e., $zz'=1$. That makes $z'=1/z$, as promised.

Let’s get back to the general setup, with the function

$\dfrac{az+b}{cz+d}$

and let’s assume $c\neq 0$ (since we already handled the case $c=0$, it’s just a linear transformation). For some notational convenience, let me let $\alpha=-(ad-bc)/c^2$. Consider the following composite:

$\begin{array}{rcl} z & \xrightarrow{w\mapsto w+\frac{d}{c}} & z+\dfrac{d}{c} \\ {} & \xrightarrow{w\mapsto \frac{1}{w}} & \dfrac{c}{cz+d} \\ {} & \xrightarrow{w\mapsto \alpha w+\frac{a}{c}} & \dfrac{\alpha c}{cz+d}+\dfrac{a}{c} \end{array}$

If you check all of these steps, and then play around simplifying the final expression, then you obtain the original formula above. So we can think of any linear fractional transformation as a composite of some linear functions and an inversion, and we know how to picture all of those steps.

That’s maybe enough for today. It’s certainly enough for me for today. Before I go, I’ll leave you with a video that might be helpful, and is pretty either way.

Like this:

Tags: linear fractional, mablowrimo, mobius transformation

This entry was posted on November 6, 2009 at 8:14 pm and is filed under Play. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Linear Fractional Transformations”

1. LFTs and Ford Circles « ∑idiot's Blog Says:
November 7, 2009 at 8:16 pm | Reply

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