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.

  • 1
    $\begingroup$ "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) $\endgroup$
    – YCor
    May 23, 2020 at 16:38
  • $\begingroup$ @YCor Yes, thanks. I edited that into the question. $\endgroup$
    – M. Winter
    May 23, 2020 at 16:41
  • $\begingroup$ 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.) $\endgroup$ May 23, 2020 at 16:54
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    $\begingroup$ @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. $\endgroup$
    – M. Winter
    May 23, 2020 at 16:57
  • 1
    $\begingroup$ I found this highly relevant question with an equivalently relevant answer. $\endgroup$
    – M. Winter
    May 23, 2020 at 18:17

2 Answers 2


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


at the bottom of page 65. The references to more classical articles are given there.

Now, if we double it in one face then we get a new convex polytope, and it is not hard to see, that it doesn't have 2-faces that are triangles and quadrilaterals. But any convex hyperbolic polytope is also combinatorially equivalent to a Euclidean one.

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.

  • $\begingroup$ Thank you for your answer. I am not familiar with some terminology: 1. what is this "regular compact right-angled 120-cell" in hyperbolic 4-space. Is it like a tiling of 4-space? I wasn't able to find out what a "convex hyperbolic polytope" is via google. 2. What does it mean to "double it on one face"? $\endgroup$
    – M. Winter
    May 23, 2020 at 17:06
  • $\begingroup$ 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. $\endgroup$ May 23, 2020 at 17:13

Dmitri's answer is definitely correct. I just want to add my geometric intuition, and a generalization, which, in hindsight, is quite obvious.

All in all, we can have the following:

If $P\subset\Bbb R^d$ is a polytope with $n$ facets, each of which is combinatorially (or projectively) equivalent to $Q\subset\smash{\Bbb R^{d-1}}\!$, then for each $k\ge 1$ there also exists a polytope $P_k\subset\Bbb R^d$ with $k(n-2)+2$ facets, all of which are combinatorially (or projectively) equivalent to $Q$.

With this, it should be clear that there are many 4-polytopes with only 5-gonal 2-faces.

The main idea is visualized below.


  1. Fix a face $\sigma\subset P$.
  2. Let $P'$ be the polytope obtained from $P$ by applying a certain projective transformation that a) fixes $\sigma$, and b) moves all vertices of $P$ "beyond" $\sigma$ (see the image). This construction is related to the idea behind the Schlegel diagram, in particular, this transformations always exists.
  3. Glue $P'$ and $P$ on their common face isomorphic to $\sigma$ (if we have chosen the correct transformation in 2., then this is a convex polytope).

Repeat this to obtain as many $Q$-facets as you like.

Still, it might be interesting to determine the atomic $Q$-facetted polytopes, i.e. those, which are not "stacked" in the sense above.


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