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
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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.
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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.