## LFTs and Ford Circles

Given 4 complex numbers, $a,b,c,d$, we may consider the linear fractional transformation (LFT)

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

Well, 4 numbers are enough to make a matrix, $\left(\begin{smallmatrix} a & b\\c & d\end{smallmatrix}\right)$. Is there any better reason to relate the linear fractional transformation with this matrix?

Suppose you have two matrices, $\left(\begin{smallmatrix}a&b \\ c&d\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}a'&b'\\c'&d'\end{smallmatrix}\right)$. Then the product is as follows:

$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}=\begin{pmatrix}aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd'\end{pmatrix}$.

If you take the composite of the two linear fractional transformations, i.e.,

$\dfrac{a\cdot \frac{a'z+b'}{c'z+d'}+b}{c\cdot \frac{a'z+b'}{c'z+d'}+d}$

and then play around simplifying that expression for a few minutes, you obtain the LFT

$\dfrac{(aa'+bc')z+(ab'+bd')}{(ca'+dc')z+(cb'+dd')}$,

which is precisely the LFT corresponding to the product matrix above. So, if nothing else, writing LFTs as matrices this way won’t lead us astray when thinking about composites.

This idea is not without its confusion, for me anyway. Generally when you think about a 2×2 matrix of complex values, you are thinking about that matrix as a linear map $\mathbb{C}^2\to\mathbb{C}^2$, which is not what we are doing above. Instead, I guess we are saying that the “monoid” (group without inverses) of 2×2 matrices, $M_2(\mathbb{C})$, acts (in the technical sense) on $\mathbb{C}$ as linear fractional transformations. My guess is that there are even better ways to say what is going on.

I think it is also important to keep in mind that two different matrices may correspond to the same LFT. For example, $\left(\begin{smallmatrix}1&2\\ 0&2\end{smallmatrix}\right)$ represents the same LFT as $\left(\begin{smallmatrix} 1/2 & 1\\ 0 & 1\end{smallmatrix}\right)$. More generally, if $\lambda$ is any complex value (nonzero), then $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ represents the same LFT as $\left(\begin{smallmatrix}\lambda a&\lambda b\\ \lambda c & \lambda d\end{smallmatrix}\right)$. I guess one can think of $M_2(\mathbb{C})$ as a $\mathbb{C}$-vector space (isomorphic to $\mathbb{C}^4$), and then think of its projective space (the quotient where two “vectors” (matrices here) are the same when they differ by a scalar (complex) multiple), which I’ll denote $P(M_2(\mathbb{C}))$. Then I think I’m saying that the action of $M_2(\mathbb{C})$ on $\mathbb{C}$ actually is an action of the quotient, $P(M_2(\mathbb{C}))$. I’m not sure if this is a useful viewpoint (or, indeed, correct).

Yesterday, when I was talking about how to picture what an LFT does to $\mathbb{C}$, I wrote down a factorization of the LFT as a composite. Our new notation gives us another way to write that factorization (recall $\alpha=-(ad-bc)/c^2$, and that we had assumed $c\neq 0$):

$\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\alpha & a/c\\ 0&1\end{pmatrix}\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\begin{pmatrix}1&d/c\\ 0&1\end{pmatrix}$.

As is frequently useful, we will assume that $ad-bc\neq 0$ (indeed, this factorization seems to require it – I think I’m missing something somewhere, anybody see it?). Notice that $ad-bc$ is the determinant of the matrix representing our LFT. We may then multiply all entries in our matrix (without changing the LFT, as discussed above) by $1/(ad-bc)$, and obtain a matrix with determinant 1. Let’s do that, making $\alpha=-1/c^2$.

Yesterday, when I was working on the factorization above, I only had something like $\epsilon$ idea where I was going. I think today I’ve got about twice that, so I want to re-write the factorization. Let me write it as

$\begin{pmatrix}1 & a/c\\ 0 & 1\end{pmatrix}\begin{pmatrix}1&0 \\ 0& c^2\end{pmatrix}\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & d/c\\ 0 & 1\end{pmatrix}$.

So what’s the connection with Ford circles? Recall that for a reduced fraction $h/k$, the associated Ford circle is the circle centered at $(h/k,1/(2k^2))$ with radius $1/(2k^2)$. Following Rademacher (and, presumably, others), let us say that the “fraction” $1/0$ also gets a Ford “circle”, the line $y=1$ in the plane. This isn’t such a nasty thing to do, as it has the tangency properties I talked about when talking about Ford circles. Anyway, let us think about applying our transformation $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$, as the composite given above, and see what happens to this line $y=1$. We’ll assume that $a,b,c$, and $d$ are all integers.

The first step, $\left(\begin{smallmatrix}1&d/c\\ 0 &1\end{smallmatrix}\right)$ is the linear translation $z+d/c$. Since $d/c$ is real (since $c,d$ are integers), this translation is a horizontal shift, which leaves $y=1$ unchanged.

Next up, $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$, which is $-1/z$. Thinking of a point on the line $y=1$, you can quickly determine that its polar coordinates are $(\csc \theta,\theta)$. The transformation $-1/z$ is the composite of: (1) inversion with respect to the unit circle (the point becomes $(\sin \theta,\theta)$), (2) reflection across the horizontal axis (giving $(\sin \theta,-\theta)$), and finally (3) multiplication by -1 (giving $(-\sin \theta,-\theta)$, since these are polar coordinates). This final point is $(\sin \theta,\pi-\theta)=(\sin \pi-\theta,\pi-\theta)$. As $\theta$ varies through $(0,\pi)$, $\pi-\theta$ also varies through this interval, and so we get the graph of the polar curve $r=\sin \theta$. If you know your polar curves, you know what this looks like…

So, the first two transformations take the line $y=1$ to the circle with center (0,1/2) and radius 1/2. The next in our composite is multiplication by $1/c^2$, which is just a scaling (since $c\in \mathbb{R}$). This scaling takes our circle to the circle with center $(0,1/(2c^2))$ and radius $1/(2c^2)$. Finally, the last transformation is another horizontal translation, leaving our circle centered at $(a/c,1/(2c^2))$. We recognize this as the Ford circle for the fraction $a/c$ (as long as that fraction is reduced).

Wasn’t that fun? If you want to think about it some more, you might convince yourself that any point above the line $y=1$ will get moved to a point inside the Farey circle resulting from this process.

Anyway, enough out of me. Hopefully tomorrow I’ll have slightly more of an idea what I’m talking about. Don’t count on it though.

### 3 Responses to “LFTs and Ford Circles”

1. The Only LFTs You’ll Ever Need « ∑idiot's Blog Says:

[…] ∑idiot's Blog The math fork of sumidiot.blogspot.com « LFTs and Ford Circles […]

2. Qiaochu Yuan Says:

There’s no mystery in the relationship between fractional linear transformations and matrices. The action of $M_2(\mathbb{C})$ on $\mathbb{C}^2$ respects scalar multiplication and fixes the identity, so it gives a corresponding action on $\mathbb{P}^1_{\mathbb{C}}$ defined as the quotient of $\mathbb{C}^2 - \{ (0, 0) \}$ by scalar multiplication, and this gives the desired homomorphism.

• sumidiot Says:

I figured it was some projective thing, but hadn’t thought about it enough. Your comments inspired me to, and I think I get it now…

Every point $(z_1,z_2)$ in $\mathbb{P}^1_{\mathbb{C}}$, where $z_2\neq 0$, has a representative of the form $(z,1)$, obtained by setting $z=z_1/z_2$. The matrix $\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)$ acts on this point, taking it to $(az+b,cz+d)$, and then this then the same point as $((az+b)/(cz+d),1)$ in the projective space (assuming $cz+d\neq 0$). And then you can say whatever needs to be said about the case when that second coordinate is 0 corresponding to the point at $\infty$

Thanks for bringing this up and thereby encouraging me to look at it again.