# Is there a 4-polytope without 3-gonal and 4-gonal faces, other than the 120-cell?

The question is in the title:

Question: Is there any 4-dimensional polytope without 3-gonal and 4-gonal faces (of dimension two), other than the 120-cell?

I consider only convex polytopes (convex hull of finitely many points) that are full-dimensional (not contained in a proper subspace). And I consider a polytope to be distinct from the 120-cell if it has a non-isomorphic face-lattice.

It is known that any 4-polytope must have a 3-gonal, 4-gonal or 5-gonal face of dimension two. The 120-cell has only 5-gonal faces of dimension two.

• "other than" means: not isomorphic as polyhedral complex? (this is a reasonable isomorphism notion; an a priori stronger one would be being isotopic, i.e., have a continuous deformation from one to another) – YCor May 23 at 16:38
• @YCor Yes, thanks. I edited that into the question. – M. Winter May 23 at 16:41
• Did you find who proved that the only possible faces are 3-, 4-, or 5-gonal? (An earlier version requested a citation for that result.) – Brian Hopkins May 23 at 16:54
• @BrianHopkins I still don't have a source, but I realized the following: one can show (via a standard double counting arguments) that a planar graph has a vertex of degree 5 or smaller, or equivalently (considering its dual), a 3-gonal, 4-gonal or 5-gonal face. Since the edge-graph of a (3-dimensional) polyhedron is a planar graph, this proves it in dimension three. This then carries over to higher dimensions by considering the 3-faces of the polytopes. – M. Winter May 23 at 16:57
• I found this highly relevant question with an equivalently relevant answer. – M. Winter May 23 at 18:17

There are other polytopes. To construct one let's do the following. Remember first that in the hyperbolic $$4$$-space there exists a regular compact right-angled 120-cell. Here, right-angled means that any two adjacent faces intersect under angle $$\frac{\pi}{2}$$. Regular means, that all the faces are isomeric, and the polytope has the same group of self-isometries as the Euclidean 120-cell. This polytope is discussed, for example, in
More generally, you can take any compact right-angled hyperbolic polytope in $$\mathbb H^4$$. Since it is hyperbolic and right-angled, it can not have $$2$$-faces that are triangles of quadrilaterals. And there is a infinite number of such polytopes in dimension 4. Each of them gives a Euclidean one as well.
• I added a reference. Yes, starting with such a polytope you get a tiling of $\mathbb H^4$, in the same way as cubes tile $\mathbb R^n$. And what I say is that you need to take the union of two tiles that share a common hyperface to get a new polytope as you want. This can be continued as far as you want. – Dmitri Panov May 23 at 17:13