## Thursday, September 2, 2010

### a 3d mandelbrot

This is an idea for a mandelbrot set taken to the next dimension.
The boundary of the mandelbrot set can be thought of as a set of bifurcation points, each bifurcation point represents a stable orbit, the stable orbit takes up a small space so variations of it allow an expanded attraction basin, these are seen as the disks that sit off the main cardoid. The stable orbits can be equally considered to be polygons.
From the first disk clockwise towards the real axis the largest disks come off at the main bifurcation points that represent orbits that are: a line, a triangle, a quadrilateral, a pentagon, hexagon etc. They aren't regular polygons. Other smaller disks appear at stars and more complex polygons.

A 3d equivalent of the boundary would be a set of points that represent the stable polyhedrons that exist. Thus you may expect the bulbs to appear at a plane, tetrahedron, cube, octahedron etc.
In order for an iteration to trace out a plane it needs to operate with a full orientation and the iteration needs to branch, in fact produce two children each iteration to represent the 2 dimensions of the surface of a polyhedron. Inside would be if all branches stay bounded.
An equivalent of the mandelbrot would also have universal properties, one necessary condition seems to be the doubling of the angle.

What would such a set look like? well you would expect offshoot bulbs at distorted versions of all the regular polyhedrons and stellated polyhedrons. It seems like the number of bulbs will be far fewer as stable polyhedrons are rarer than polygons. It also seems like the large bulbs will be finite in size, so the surface will be quite different to the mandelbrot set.

Here is an attempt at the formula, which at least gives an indication of how such a mandelbrot might be constructed:

Julia set:
The julia branching iteration requires a rotation for each branch, which is 2 quaternions, so the map is 8D.
branch1: q1 = q1*q1 + c; q2 = q2*q1 + c;
branch2: q1 = q1*q2 + c; q2 = q2*q2 + c;

The idea is that each iteration the orientation (representing an polyhedron face) is rotated in two different directions, deformed versions of all platonic solids should be possible from this if c=0. If not then this formula is wrong.

Mandelbrot set:
the 4d set of points c for which the corresponding julia set is connected

A problem with this is that the set is intractable, the computation doubles each iteration.

Another construction may be mathematically simpler and in this way may be more likely to be the correct extension...
Instead of quaternion pairs you operate on octonions O, so the julia set needs to produce two versions of O^2 (for branch1 and branch2), the first uses just the octonion multiply. The second is the octonion multiply but with the first 4 and last 4 components swapped. This recognises the fact that the octonion is two quaternions that have equal precedence.