1

I'm not as mathy as I wish I were, so wrt the advice stackexchange just presented me with: to "explain what I don't know," I'm afraid that's a rather long list, stretching deep into the "unknown unknowns" territory where be dragons.

But I do know that rhombic dodecahedra are a favorite polyhedron of mine, packing the way spheres pack, and while lately I got the crazy idea to write a basic little card game that takes place on the surface of a 3-sphere (that's the one in four dimensions, right? or does it depend who I'm talking to?), I was trying to figure out how to divide it into bins. Sure I could just intersect the surface of the 3-sphere with a matrix of tesseracts like a simpleton, but … could I have more fun than that?

It would be so much better for my ego, and the snobbery I'm planning, to map the 3-sphere onto a polytope with a more elegant net. The 16-cell is fine, the 24-cell is pretty hot and tempting, but I can't quite envision a 4-dimensional polytope with rhombidodecahedral (is that a word?) cells. Does it exist?

Travis Well
  • 141
  • 4

1 Answers1

1

Sadly, such a 4-polytope cannot exist.

You might know that in a polyhedron, at every vertex the sum of interior angles of incident faces adds up to less than $360^\circ$. This is the reason why there are no polyhedra in which all faces are, say, regular hexagons. This is because a regular hexagon has an interior angle of $120^\circ$, and so no three of these can fit around a vertex.

An equivalent statement holds in higher dimensions. In a 4-polytope, around each edge the dihedral angles of incident cells must add up to less than $360^\circ$. But the dihedral angle of the rhombic dodecahedron is also $120^\circ$, and so no three can fit around an edge. Therefore no 4-polytope can have only cells that are rhombic dodecahedra. At least when you want to have them in their most symmetric version.

I do not know whether there can be a 4-polytope in which all cells are "combinatorially equivalent" to the rhombic dodecahedron.

M. Winter
  • 29,928