by wendy » Thu Oct 09, 2008 9:14 am
You can't use the 5d cube (pentaract) to do this. You need at least the 6d cube.
Put simply, the N-cube is a regular rhombotope, (in the loose sense, ie made of parallelograms, parallelohedra &c). Any normal projection must itself be some kind of parallelotope (including polygons of even side). The outcome here is that one can make such figures out of the sum of every combination of N vectors, to give an N-cube. Every edge of the n-cube is parallel to exactly one of these vectors.
The rhombic icosahedron has six different directions of edges, and therefore requires six different vectors: ie a 6-cube.
Note first that if a given vertex appears on the surface of a projection, so must the one directly opposite. This means that every surtope of a cube-projection must have equal opposite parallel sides. Such figures are what we might call rhombous. [Coxeter's name for these are zonotopes: he devotes an entire chapter of 'regular polytopes' to this].
The assorted edges of the 4d pentagonal figures have sixty different orientations, so we would expect at least 60 dimensions for the basis of a cube-projection. One could take the omnitruncate, for example, x3x3x5x, which does exist as a projection of the 60-cube, in much the same way that x3x5x is a projection of the 6-cube. However, the 60-cube has 1,152 921 504,606 846 976 vertices.
The rhombic tricontahedron style projection does not occur in 4d. Only one figure (m3o3o5m) has suitable faces in the topological sense, but the parallelohedra faces are not rhombous: the opposite edges are not equal. All of the others involve polygons with an odd number of sides: ie not rhombous