# A Newly Discovered Compound Polytope

Coxeter, in Regular Polytopes, Section 14.3, undertakes a systematic search for regular compound polytopes in four dimensions. In his definition of "regular compound polytope", the vertices of the compound must be those of an enclosing regular polytope Π, and the compound must share the symmetries of Π. His method involves considering the set of vertices at a distance a from an initial vertex 00. These vertices lie in a section of the polytope Π determined by a hyperplane perpendicular to the diameter through 00. If these vertices include those of a tetrahedron of side a, then this tetrahedron is one cell of a {3,3,3} (which Coxeter usually denotes α4) inscribed in Π. This polytope, and its images under all the symmetries of Π, are a compound polytope inscribed in Π.

Coxeter examined the sections of {5,3,3}; in one of them, which he denoted 180, there are 28 vertices. He found a set of four forming a tetrahedron of the appropriate size, and accordingly listed one compound of 120 α4's.

Recently, Dik Winter re-examined Coxeter's work, and found that all 28 vertices belonged to tetrahedra of the right size. The vertex group, that is, the symmetries of Π which leave 00 fixed, has two orbits on this set of seven tetrahedra. One orbit is a singleton, containing the tetrahedron that Coxeter found; the second orbit comprises the other six tetrahedra. The latter tetrahedra determine six α4's inscribed in Π and sharing the vertex 00. Together with their images under the symmetries of Π they make up a compound of 720 α4's whose vertices, in groups of 6, are the vertices of Π. The notation for this compound is 6{5,3,3}[720 α4].

One may ask whether this new compound can be divided into two enantiomorphous forms, as the familiar compound of ten tetrahedra can be divided into two compounds of five. To answer this question we must look more closely at the seven tetrahedra in section 180, and the action on them of the vertex group.

If 00 = (0, 0, 0, 1) then the vertex group acts on three-space by coordinate permutations with an even number of changes of sign. This is the group of the tetrahedron which Coxeter selected, whose vertices are

 0 = (√5, √5, √5), 1 = (√5, -√5, -√5), 2 = (-√5, √5, -√5), 3 = (-√5, -√5, √5).

I will call this one the "primary tetrahedron." Its symmetry group, of course, acts as the full group of permutations of (0, 1, 2, 3). An element of the group can be denoted by its action on (0123); thus, (1230) is an element of order 4. The matrix representation of an element can easily be deduced from the observation that 0+1, 0+2, and 0+3 are proportional to the basis vectors, while 0+1+2+3 = 0.

To find the secondary tetrahedra, we may start with one of the remaining vertices in the section 180, such as (-3, √5, 1). The other vertices which are at the requisite distance of 2√10 from this, and from one another, are (1, -√5, 3), (3, √5, -1), and (-1, -√5,-3). The stabilizer of this tetrahedron is the cyclic group of order four generated by (1230).

The stabilizer of a secondary tetrtahedron contains two direct, and two opposite, transformations. Thus in the action of the direct subgroup on the six secondary tetrahedra, the stabilizer of one is of order 2, and the six tetrahedra form a single orbit; they cannot be separated into two enantiomorphous subsets.

Here is an applet with which to visualize the seven tetrahedra of this note. The primary tetrahedron has larger dots at the vertices, which are labeled with numbers as in the discussion above. You can use the checkboxes to add or remove the primary and the six secondary tetrahedra from the display. You can select an element of the symmetry group in a pop-up menu; to the right of this is a button to "Apply" the symmetry. This will act on the selected set of secondary tetrahedra. The check box "Show Axes" reveals or conceals the coordinate axes. The whole display may be rotated by selecting and dragging any part of the display; the "Reset" button returns the display to its unrotated position.

Last modified on \$Date: 2010-12-09 14:45:44 -0500 (Thu, 09 Dec 2010) \$

Christopher J. Henrich