Versteh’s wer’s kann – schönes Pic jedenfalls.
Can a nontrivial 3D version of the Mandelbrot set be defined such that the z = 0 section would yield the original 2D set? Extending complex numbers with a third component looks like a natural approach. Unfortunately, Frobenius‘ and Hurwitz’s theorems exclude the possibility of an extension that would inherit their nice properties (division and norm multiplicativity). Nonetheless, the fact that there is no perfectly correct solution in my opinion only makes the problem more tempting.
Let’s consider 3D hypercomplex („triplex“) numbers x + i y + j z. Since we want to extend the complex numbers, i² = -1. Assuming commutativity (a b = b a), we only need to fill two multiplication table entries to define the multiplication rule: j² and i j. If we limit the possible values to ±1, ±i, ±j and 0, there are 49 possible combinations. Only 5 of them produce distinct symmetric and bounded M-sets. One of them (i j = 0, j² = -1) is a solid of revolution. The image depicts 4 remaining nontrivial M-sets.
(via nerdcore)