From: Chris Hillman Subject: Re: Histogramming SO(3) Date: Thu, 18 Feb 1999 16:08:30 -0800 Newsgroups: sci.math.research Keywords: Tiling S^3 with 600 congruent spherical tetrahedra I had assumed Wolfang Huber wanted control over the number of bins, and he asked for cuboidal bins, so I didn't even mention finite subgroups of SU(2). However, Sidney Marshall suggested in private email a simple approach based on the 600-cell which gives 14400 tetrahedral bins in SU(2). His approach has the great advantage that you can explicitly write down the vertices of bins, and moreover the bins are isometric (allowing reflections) tetrahedra distributed in a highly symmetric and well-understood fashion. Moreover, it occurs to me that you can use the group structure to get a more clever test for which rotation is in which bin than testing four simultaneous linear inequalties. At the very least, you can look for the closest 600-cell vertex, translate back to a "standard vertex", and in this way reduce the number of cells you need to actually worry about. Speaking of subgroups, now I am a bit worried that my original scheme might after all imply the existence of a non-existing (abelian!) subgroup :-( Perhaps that idea comes to grief because the flows I mentioned don't commute? Be this as it may, if you don't absolutely insist on cuboidal bins, Marshall's scheme is probably the best approach. I have appended the heart of his email (posted with permission). Chris Hillman ========================================================================= If one wants to partition the 3-sphere of unit quaternions into equal volumes then one wants to look at the symmetry groups on the 3-sphere much like the polyhedral groups on the 2-sphere (i.e., ordinary sphere). These have been extensively studied by H. S. M. Coxeter in "Regular Polytopes" where a tiling of 600 regular spherical tetrahedra are found to tile the 3-sphere. Each tetrahedron can be broken up into 24 congruent pieces for a 14400-piece tiling of the 3-sphere. This results in 7200 bins for O(3) or 14400 bins for SU(2). Another way of putting this is that there is a finite group of rotations of order 14400 on the 3-sphere. It is an "extension" of the icosahedron into 4 dimensions. The vertices of this 600 cell are (+-1, 0, 0, 0) all permutations and signs (8) and (+-1/2, +-1/2, +-1/2, +-1/2) all signs (16) and (+-1, +-t, +-1/t, 0)/2 even permutations (96) where t is (sqrt(5) + 1)/2 = 1.61803... I believe that 14400 is the largest number of pieces a 3-sphere can be disected into that are equivalent under a finite group operation. =================================================================== Hmm... looks to me like 14400 bins for SU(2), 7200 for SO(3) (assuming the tiling of S^3 respects the identification of opposite points), and 14400 again for O(3), but maybe I'm missing something.