For complex constants $a, b, c,$ and $d$ with $ad - bc \neq 0$, the complex map $z \mapsto \frac{az + b}{cz + d}$ is called a **Möbius transformation**.

These have a surprising amount of structure: writing them as matrices

$\displaystyle \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$

is compatible with composition and inverses, so Möbius transformations are actually a group isomorphic to $PGL_2(\mathbb{C})$ (and $\operatorname{Aut} \hat{\mathbb{C}}$).

A Möbius transformation maps the inside of one circle in the complex plane to the outside of another circle, and so we might expect some self-similarity to arise from iterating them.

Taking a single Möbius transformation $A$ and forming the group $\left< A \right>$ doesn’t get us far, though — the two circles are exchanged, but there just isn’t enough complexity in a free group on one generator to create an interesting picture in the plane.

Instead, we can take *two* Möbius transformations $A$ and $B$ and consider the subgroup $\Gamma = \left< A, B \right>$.

To visualize $\Gamma$, we can plot its set of limit points: moving from a free group on one generator to two drastically increases the number of number of leaves in the Cayley graph.

$\displaystyle A = \frac{1}{2} \left[ \begin{array}{cc} 3 & 3 + 2i \\ 3 + 2i & 3 + 4i \end{array} \right], \displaystyle B \approx \left[ \begin{array}{cc} 0.5 - i & 1.05 - 1.33i \\ -0.18 - 1.18i & 0.5 - i \end{array} \right]$

This may be beautiful, but it’s not quite what we were promised.

Clearly, not any two choices of $A$ and $B$ will do. We’d ideally like to reduce the dimension of the parameter space to only the "good" choices for $A$ and $B$.

And with a treasured family secret, we can do just that...

Long ago, Grandma set out to solve this very problem herself and wrote down a two-parameter recipe for creating only the nicest Möbius transformations.

Scaling all four entries of $A$ or $B$ by a nonzero factor $p \in \mathbb{C}$ scales the determinant by $p^2$ but doesn’t change the map, so we may assume $\det A = \det B = 1$, meaning $A, B \in PSL_2(\mathbb{C})$.

Discrete subgroups of $PSL_2(\mathbb{C})$ are called **Kleinian**, and so $\Gamma = \left< A, B \right>$ will always be a Kleinian group for our purposes.

Next, Grandma only considered groups $\Gamma$ up to conjugation by other elements of $PSL_2(\mathbb{C})$.

Any Möbius transformation acts by rotating the Riemann sphere, so they are completely determined by specifying three corresponding inputs and outputs.

Therefore, considering $\Gamma$ only up to commutation reduces the number of free parameters to just three.

But which three? Grandma chose $\operatorname{tr} A, \operatorname{tr} B,$ and $\operatorname{tr} AB$: they are unaffected by conjugation and independent of $\det A$ and $\det B$ and one another.

For the limit set to be a non-space-filling curve, we want the boundary circles of $A$ and $B$ and their inverses to be tangent, and it turns out this is equivalent to $\operatorname{tr} ABA^{-1}B^{-1} = -2$.

The **Markov identity** for traces states that $\operatorname{tr} A B A^{-1} B^{-1}$ is equal to

$\displaystyle (\operatorname{tr} A)^2 + (\operatorname{tr} B)^2 + (\operatorname{tr} AB)^2 - (\operatorname{tr} A)(\operatorname{tr} B)(\operatorname{tr} AB) - 2,$

so given $\operatorname{tr} A$ and $\operatorname{tr} B$, $\operatorname{tr} AB$ is determined by a quadratic equation.

1. Freely choose $t_A, t_B\in \mathbb{C}$ (specifying $\operatorname{tr} A$ and $\operatorname{tr} B$, respectively).

2. Let $t_{AB}$ be one of the solutions to $x^2 - t_At_Bx + t_A^2 + t_B^2 = 0.$

3. Let $\displaystyle z_0 = \frac{(t_{AB} - 2)t_B}{t_Bt_{AB} - 2t_A + 2it_{AB}}$.

4. Define $A$ and $B$ to be

$\displaystyle A = \left[ \begin{array}{cc} \frac{t_A}{2} & \frac{t_At_{AB} - 2t_B + 4i}{(2t_{AB} + 4)z_0} \\ \frac{(t_At_{AB} - 2t_B + 4i)z_0}{2t_{AB} - 4} & \frac{t_A}{2} \end{array} \right]$

$\displaystyle B = \left[ \begin{array}{cc} \frac{t_B - 2i}{2} & \frac{t_B}{2} \\ \frac{t_B}{2} & \frac{t_B + 2i}{2} \end{array} \right]$.

Matrices baked with Grandma’s Recipe have traces $t_A$ and $t_B$, determinant $1$, satisfy the Markov identity, and some other nice properties: the equal diagonal entries in $A$ ensure $180^\circ$ rotational symmetry in the limit sets, and the factor of $z_0$ is from conjugating by a map that centers them correctly.

If the matrices aren’t working properly, Grandma suggests recalculating everything from scratch twice more and hoping the right answer wins two out of three times.

While discrete subgroups of $PSL_2(\mathbb{C})$ are called **Kleinian**, discrete subgroups of $PSL_2(\mathbb{R})$ are called **Fuchsian**. When the limit set of a Kleinian group is a Jordan curve (as it frequently is with this recipe), the group is called **quasi-Fuchsian**.

These curves are great examples of why the Jordan curve theorem is so hard to prove! In fact, whenever the limit set is a Jordan curve other than a line or circle, it must have Hausdorff dimension strictly between $1$ and $2$.

Grandma’s friend Riley has a much more straightforward recipe that produces related fractals. Here, we simply set

$\displaystyle A = \left[ \begin{array}{cc} 1 & 0 \\ c & 1 \end{array} \right], \qquad B = \left[ \begin{array}{cc} 1 & 2 \\ 0 & 1 \end{array} \right],$

where $c$ is a freely-chosen complex number.

By dropping the Markov identity requirement, $t_{AB}$ becomes a free parameter. Through a staggering amount of computation, we can work out the formulas for $A$ and $B$ in terms of all three traces.

Unfortunately, the results are often akin to letting vinegar be a free parameter when making cake.

And thanks (I think?) to chapter 8 of *Indra’s Pearls* for being an unhinged but half-decent reference.

If you’d like to learn more, chapter 9 is titled "Accidents will happen" (seriously)