Clearly the volume of the generalized octant AAAA is 1/16:

*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 4*pi*.

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* + 3*Q* + 2*S* + 2*T* + 2*U* + 2*V* = 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
this table.

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Anthony Phillips

Math Department, SUNY at Stony Brook

September 23, 1996