A Projection of the Six-Dimensional Cube

Layers of the 6-Cube Projection

Introduction. These four polyhedron models comprise a model of a projection of the six-dimensional cube into three-dimensional space. There are several ways to define the n-dimensional cube in general. The most direct is probably to say that it is the convex hull of the 2^n points in n-dimensional space whose coordinates are either +1 or -1. The familiar square in 2-dimensions, with vertices {(-1,-1),(-1,1),(1,-1),(1,1)}, is thus the 2-cube. The ordinary cube in three dimensions is the 3-cube. It follows from this definition that the the number of k-cells in an n-dimensional cube is the coefficient of x^k in the polynomial


Thus, the 6-cube has 64 0-cells (vertices), 192 1-cells (edges), 240 2-cells (squares), and so on.

One may specify this projection very efficiently if one is familiar with the rotation symmetries of the regular icosahedron. Recall that the icosahedron has six axes of pentagonal symmetry. If B is an orthonormal basis for a real 6-dimensional inner-product space, one merely maps each each of these basis elements to a unit vector parallel to each of the six pentagonal rotation axes of the regular icosahedron, and extend linearly. Naturally, the projection must possess icosahedral symmetry.

The four models should be regarded as different layers of the image of the projection. The largest model is that of a rhombic triacontahedron, a convex polyhedron comprised of 30 identical rhombii. The second largest model is obtained by removing this outer layer of rhombii, thus revealing a set of 60 rhombii underneath. Removing these 60 rhombii reveals another polyhedron again comprised of 60 rhombii. Finally, removing this layer reveals a small great rhombic triacontahedron. The great rhombic triacontahedron is also comprised of 30 identical rhombii, although, possessing many self-intersections, it is not a convex polyhedron. For definiteness, let (0) denote the outer convex layer and (A), (B), and (C) the subsequent deeper layers.

The four polyhedra {(0),(A),(B),(C)} are stellations of the rhombic triacontahedron, meaning that each is built from the same set of 30 planes of a rhombic triacontahedron. Obviously (0) is the rhombic triacontahedron itself. It is also worth noting that the polyhedron (0) is the dual to the icosidodecahedron and the polyhedron (C) is dual to the great icosidodecahedron.

Some Topology. The polyhedra {(0),(A),(B),(C)} may be studied in the context of topological surfaces and branched coverings. Each of the four polyhedra represents an immersion of a compact closed oriented 2-dimensional manifold into 3-dimensional space. Suppose a polyhedron (X) and a large sphere S^2 are both centered at the origin in 3-space. Then any ray emanating from the origin intersects the polyhedron in one or more places on (X) and at exactly one point on the sphere. This therefore yields a map


This map is well-defined because the surface of (X) is closed and the map is continuous because (X) is the image of an immersion in 3-space. Given a generic point p on the sphere, the preimage of p under f consists of n points, where n is the covering index of f. Under these circumstances, one has the equation

where b is the total branching number of f and k is the Euler characteristic of (X). With these considerations, one may quickly produce the following table:
| polyhedron |     v    |   e |  f |  k | n |  b |
|     (0)    | 12+20    |  60 | 30 |  2 | 1 |  0 |
|     (A)    | 12+2(20) | 120 | 60 | -8 | 2 | 12 |
|     (B)    | 12+2(20) | 120 | 60 | -8 | 6 | 20 |
|     (C)    | 12+20    |  60 | 30 |  2 | 7 | 12 |
(Recall that if v is the number of edges, e the number of edges, and f the number of faces, then the Euler characteristic is k=v-e+f.)

Some (Speculative) Algebra. Evidently there is a duality between the four different polyhedra in this model of the 6-cube. Notice that the surfaces represented by (0) and (C) are homeomorphic to each other, as are the surfaces represented by (A) and (B). Moreover, the sum of the covering indices for (0) and (C) is 8, and the same is true of the covering indices for (A) and (B). More concisely, one may write


However, more is true. Using an appropriate group addition law, the four polyhedra {(0),(A),(B),(C)} is isomorphic to the Klein four-group, the direct product of two 2-element groups. Notice that, as subsets of 3-space, (C) may be obtained as the union of scaled-down versions of (A) and (B). Also, (A) may be obtained by subtracting a scaled-up version of (B) from a scaled-up version of (C). In general, the group operation is defined to be "symmetric difference mod dilation". One can thus check that
(A)+(B)=(C), (A)+(C)=(B), and (B)+(C)=(A).

Notice that the polyhedron (0) does not interact with any of the polyhedra {(A),(B),(C)} under this group operation, so it serves as the identity element.

Zometool. All of the rhombii in this projection of the 6-cube are congruent. Moreover, one may describe them as "Golden rhombii" because the ratio of their diagonals is the Golden Ratio b=(1+sqrt(5))/2. Finally, all of these polyhedra are "zonohedra" because all of their faces are parallelograms.

Since these polyhedra are zonohedra, one may wonder if these are constructible using Zome. Indeed this is the case since the projected edges lie in the red zones of Zome; that is, it is conceivable that one may construct them with red struts. Building a 2-skeleton of the entire 6-cube model is impossible with commonly available parts, because there are so many false intersections that one would require too many different lengths of red struts. Nevertheless, one may build models of each the polyhedra {(0),(A),(B),(C)} individually. For example, the Zome models of (0) and (C) are pictured below.

The Rhombic Triacontahedron, (0)

The Great Rhombic Triacontahedron, (C)

A Research Question. What can one say about the analogous surfaces embedded in the 10-cube? One can use a construction similar to that of the four polyhedra {(0),(A),(B),(C)} in order to obtain a projection of the 10-cube into 3-dimensional space. This construction uses the 10 trihedral axes instead of the 6 pentagonal axes of the regular icosahedron to correspond to the 10 basis elements of 10-dimensional space. The outer layer of the corresponding projection of the 10-cube is shown below. Notice that the structure is much more complicated, as there are two orbits of rhombii, two orbits of edges, and three orbits of vertices under the action of the icosahedral group.

Outer Layer of the 10-Cube Projection

It is likely that the answer to this research question may be a sort of dual to the problem of stellating the regular icosahedron. There are only four compact stellations of the regular dodecahedron, and these appear to have some correspondence to the four polyhedra {(0),(A),(B),(C)}. After all, the regular dodecahedron possesses 6 pairs of opposite faces, each having the symmetry of a regular pentagon. To each pair of opposite faces of the regular dodecahedron, there is exactly one red zone perpendicular to these faces. This goes likewise for the icosahedron. The regular icosahedron has 10 pairs of opposite faces, each with trihedral symmetry, and to each of these pairs corresponds exactly one perpendicular yellow zone.

The Medial Rhombic Triacontahedron


H. S. M. Coxeter. Regular Polytopes. 3rd ed. Dover Publications, Inc., New York, 1973.

Index to Polyhedra