May 18, 2017
Consider the following instructions for constructing Sierpinski's triangle. Start with an equilateral triangle, (we'll call it \(E_0\)). Make 3 copies. Contract each copy by a factor of 1/2, and then translate each copy into one of the corners of \(E_0\). Call the resulting union of these 3 copies \(E_1\). Follow the same instructions using \(E_1\) to get \(E_2\), and continue out to infinity.
The rule that takes \(E_0\) to \(E_1\).
We can rephrase the instructions that take \(E_0\) to its \(i\)th copy as a mapping \(\phi_i\), and the three functions \(\{\phi_1,\phi_2,\phi_3\}\) taken together form an IFS (iterated function system), and the Sierpinski triangle is its attractor.
What if we were to define another iterated function system \(\{\psi_1,\psi_2,\psi_3\}\) such that \(\phi_i(E_0) = \psi_i(E_0)\) for every \(1\leq i \leq 3\). Would it necessarily have the same attractor?
The first case we will look at is probably the most obvious. What if instead of merely contracting each copy of \(E_0\), we also performed a dihedral group operation, (i.e a rotation or reflection) on it. Since there are 6 elements in the dihedral group of an equilateral triangle, we have \(6^3\) possible IFS's.
Each \(\phi_i\) can be modified to include a dihedral group operation.
Unfortunately, they all have the same attractor.
Each of the IFS's has the same attractor. (How boring.)
At this point, it's not exactly clear what other mappings there are to choose. Ideally, we might like to see what happens when we rotate each copy by an arbitrary angle, but since triangles only have 3-fold rotational symmetry, that would violate our rule that \(\phi_i(E_0) = \psi_i(E_0)\). Of course, we could just define some new operation that looks like a rotation but under which an equilateral triangle has full rotational symmetry.
Consider the following operation:
Consider a curve \(C\) of the form \(r = \rho(\theta)\). In our case, the curve will be the boundary of an equilateral triangle. Let \(R(\beta)\) denote a rotation by an angle \(\beta\) with respect to \(C\). (Here we do not mean a rotation in the normal sense.) Then, in polar coordinates,
$$ R(\beta) (r,\theta) = (r \rho(\theta + \beta) / \rho(\theta), \theta + \beta) $$Essentially, the angular component changes as normal, but instead of tracing out a circle, a point undergoing a continuous rotaion traces out a path similar to \(C\). In fact, if we took \(C\) to be a circle, we would see that the rotation operator reduces to that for a normal rotation.
If we take \(C\) to be the boundary of \(E\), then we can apply any arbitrary "rotation" with respect to \(C\) to any of our copies and still maintain our rule that \(\phi_i(E) = \psi_i(E)\).
Each \(\phi_i\) can be modified to include a "rotation".
So what kind of attractors do we get now? Well, it's not entirely sure if the resulting IFS's technically have attractors, since the \(\phi_i\) aren't necessarily contraction mappings anymore. (I'll have to figure this out some time.) That's not to say we can't just plot the \(E_n\)'s for various IFS's.
Here's an animation I made of \(E_6\) for various IFS's. Each \(\phi_i\) is the composition of a clockwise "rotation" by an angle \(\beta\) with a contraction by a factor of 1/2 and an appropriate translation. The rotation angle \(\beta\) increases by 1 degree each frame, pausing at 60 degrees, ending at 120 degrees.