9

I'm not necessarily looking for a rigorous proof, more an outline of how an undergrad could count the regular 4D polytopes (and perhaps investigate what they look like) as explicably as possible.

Olexandr Konovalov
  • 7,186
  • 2
  • 35
  • 73
Sup
  • 129

2 Answers2

3

Numberphile made a video on exactly this topic a while back. The idea is to consider the dihedral angle between adjacent faces for each of the five Platonic solids – a polychoron will consist of instances of one such solid. For the polychoron to be valid polychoron, at least three cells must meet at an edge and the sum of dihedral angles in 3-dimensional space must be strictly less than 360°.

  • Tetrahedron (dihedral angle 70.5°): three, four or five tetrahedra can share an edge in 3D without overlapping. Bent into 4D space, these configurations give the 5-cell, 16-cell and 600-cell respectively.
  • Cube (90°): three cubes around an edge yields the 8-cell or tesseract. Four around an edge, however, completely fills it up, yielding the ordinary cubic honeycomb and not any polychoron.
  • Octahedron (109.5°): only three octahedra can fit around an edge, yielding the 24-cell.
  • Dodecahedron (116.6°): the situation is the same as for the octahedron, and yields the 120-cell.
  • Icosahedron (138.2°): since this is larger than 120°, three of these cannot fit around an edge in the first place, so no regular polychoron has icosahedral cells.
Parcly Taxel
  • 105,904
  • How can I be sure, for example, that the dodecahedron is regular? It sure looks symmetrical but how do I convince myself it has to be? – Sup Feb 23 '17 at 16:22
  • @Sup Because the same vertex configuration appears at every vertex. The same construction on polygons can be used to show only five Platonic solids, as the Numberphile video shows. – Parcly Taxel Feb 24 '17 at 00:33
1

In any regular $4$-polytope, three or more regular polyhedra meet at each edge. The angles between faces on these polyhedra have to add to less than $2\pi$ (in the same way that angles meeting at a vertex in a polyhedron must add to less than $2\pi$). By calculating the angles between faces in the regular polyhedra, students can prove that the number of tetrahedra meeting at an edge could be $3$, $4$ or $5$, but the number of cubes, octohedra, or dodecahedra meeting at an edge can only be $3$. Icosahedra can't be made to fit around an edge even if there are only $3$ of them.

This shows that there are at most six regular $4$-polytopes, three made of tetrahedra, and one made of each of cubes, octohedra, and dodecohedra. The hardest part is showing that these exist. Clever students might be able to produce a list of $4$-dimensional coordinates and prove that these coordinates make the requisite shape, in the same way that the following coordinates give a dodecahedron: $$(\pm 1,\pm 1,\pm 1)\\ (\pm\phi,\pm1/\phi,0)\\ (0,\pm\phi,\pm1/\phi)\\ (\pm1/\phi,0,\pm\phi)$$

Oscar Cunningham
  • 16,939
  • I'm familiar with the Platonic solid version, where we must have at least 3 faces per vertex. I haven't thought about the 4d version, and I'll admit I'm not sure why we need at least 3 meeting at each edge. Does this continue -- if I wanted $n$ dimensional ones, would at least 3 facets need to meet at each $(n-3)$-face? – pjs36 Feb 21 '17 at 16:20
  • 1
    Yeah. Each $(n-2)$-face lies between two $(n-1)$-faces. So given any $(n-3)$-face, pick one of the $(n-1)$-faces that it belongs to. It meets two $(n-2)$-faces in this $(n-1)$-face, and the two $(n-1)$-faces that meet the original $(n-1)$-face across these $(n-2)$-faces must be distinct. So the $(n-3)$-face lies on at least three $(n-1)$-faces. – Oscar Cunningham Feb 21 '17 at 16:43
  • Got it, thank you, very neat! I wondered if we were essentially picking out a "3D-like" piece of the polytope to think about. I double checked in Ziegler, and intervals in the face lattice of a polytope are (isomorphic to) face lattices of polytopes, which seems to support that idea. – pjs36 Feb 21 '17 at 16:58