K = 1/16.
Equations from Lunes
A lune is the volume bounded by three great semi-circles intersecting at two antipodal points. The volume of a lune is easily calculated from its three dihedral angles: it intersects an equatorial great 2-sphere in a geodesic triangle with those three angles; its volume is to 1 (the total volume of the 3-sphere) as the area of that triangle is to 4pi.
Lunes can be used to obtain linear relations among the volumes of the tetrahedra in our collection as follows: whenever two tetrahedra have a congruent face surrounded by complementary dihedrals they can be joined along that face to form a lune; then the sum of their volumes is the volume of that lune. In this way we obtain (writing K for Vol(K), etc.):
O + N = 1/24
Q + P = 1/8
S + R = 1/4
M + L = 7/24
M + N = 1/12
V + S = 1/12
U + V = 1/24
T + U = 1/12.
Equations from Cones
Take 6 points in the group that lie on a great circle, for example 1,a,a^2,-1,-a,-a^2. Cone to that circle from two additional, antipodal group points, for example i and -i. The two cones form a great 2-sphere, which splits the 3-sphere into two equal hemispheres. Coning to these eight points from a point in one of the hemispheres, e.g. d produces twelve tetrahedra whose volumes must add up to 1/2. The elements given as examples here produce the relation
P + 3Q + 2S + 2T + 2U + 2V = 1/2.
Equations from Nesting
If two tetrahedra have a congruent face, and if two of the dihedrals around that face in one tetrahedron are the same as two on the corresponding edges of the other, then the tetrahedra nest one inside the other.
For example, U has the face E with dihedrals gbb, whereas V has the face E with dihedrals gbg. Since g < b, it follows that V fits inside U. The left-over space must be another tetrahedron on the list, in fact it is Q. Applied to the volumes, this gives the equation:
U = V + Q.
Similarly, O has the face A with dihedrals hhh; N has the face A with dihedrals hhi. Since h < i, we know O is contained inside N; the difference is V:
N = O + V.
This analysis yields 11 independent linear equations
(besides K = 1/16)
which can be solved for the 11 unknown volumes. The results are in
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