Consider a vertex-transitive convex polytope with a facet containing more than the half of all vertices. Does it already have to be a simplex or are there other examples?

I am particularly interested in the case where even the affine symmetry group of the polytope acts transitively on its vertices, i.e. we are talking about an orbit polytope.

Thank you in advance!

  • $\begingroup$ Is this in ${\mathbb R}^3$ or in $N$ dimensions? $\endgroup$ Mar 2, 2014 at 12:34
  • $\begingroup$ @ChristianBlatter: The dimension can be arbitrary large. $\endgroup$
    – Dune
    Mar 2, 2014 at 16:54
  • $\begingroup$ For clarification: by 'facet' do you mean $2$-face, or $(n-1)$-face (i.e., 'co-vertex')? $\endgroup$ Mar 4, 2014 at 22:24
  • $\begingroup$ @StevenStadnicki: I mean $(n-1)$-face where $n$ is the dimension of the polytope. $\endgroup$
    – Dune
    Mar 5, 2014 at 8:24
  • $\begingroup$ I have some trouble understanding the second paragraph: isn't a polytope consideres vertex-transitive if and only if its affine symmetry group acts transitively on its vertices? $\endgroup$
    – M. Winter
    Dec 24, 2018 at 13:41

3 Answers 3



Every edge of a triangle contains all but one of the vertices. Every face of a tetrahedron contains all but one of the vertices. Every $(n-1)$-face of an $n$-simplex contains all but one of the vertices.

Every edge of q square contains half of the vertices. Every face of a cube contains half of the vertices. Every $(n-1)$-face of an $n$-cube contains half the vertices.

Is there anything in between the triangle and the square?


Yes! The dual of the cyclic polytope can be an example if parameters are chosen well.

My knowledge of this combinatorial example is due to Carl Lee; the (poor) exposition is due solely to me.

The polytope is 4-dimensional and its combinatorial automorphism group acts vertex transitively. I'm not sure if the standard embedding as the convex hull on points of the moment curve has full combinatorial automorphism group.

Also, I'm not particularly versed in this area, so I describe it in dual form first:

For every pair of positive integers $n$ and $k$ with $k\geq 2n$ define a polytope $P_{n,k}$ as the $2n$-dimensional (abstract) polytope with vertices the integers $\{1,2,\dots,k\}$ mod $k$ and maximal facets ($(2n-1)$-faces) given by $2n$-sets of the form $$\bigcup_{i=1}^n \{ a_i, a_i +1 \}$$ for integers $a_i$ taken mod $k$ such that result really does have $2n$ elements.

It is not hard to count these, there are $\binom{k-n}{n} + \binom{k-n-1}{n-1}$ of them, and exactly $2\binom{k-n-1}{n-1}$ of them contain the vertex $1$. The cyclic group $\mathbb{Z}/k\mathbb{Z}$ acts vertex transitively on the polytope by acting regularly (by addition) on the vertices. This polytope's full symmetry group is usually the dihedral group of order $2k$ acting naturally on the $k$ points, but is sometimes larger.

The polytope in question is the dual polytope, where $n$ and $k$ are chosen so that the inequality works out.

Specifically, $n=2$ (4-dimensional) and $k=6$ gives vertices $\{1,2,3,4,5,6\}$ and maximal facets $\left\{ \{1,2,3,4\}, \{1,2,4,5\}, \{1,2,5,6\}, \\~~~ \{2,3,4,5\}, \{2,3,5,6\}, \{2,3,6,1\}, \\~~~ \{3,4,5,6\}, \{3,4,6,1\}, \{4,5,6,1\} \right\}$

This polytope has 9 facets, and each vertex is contained in 6 maximal facets.

The dual polytope has 9 vertices, and each maximal facet contains 6 vertices. (Yay!)

The combinatorial automorphism group of the cyclic polytope is a wreath product $$S_3 \wr S_2 = \langle (1,3,5), (1,3), (1,2)(3,4)(5,6) \rangle$$ and one can check explicitly that this acts transitively on the maximal facets. Hence in the dual, the combinatorial automorphism group is vertex transitive.

  • $\begingroup$ Let me know if you need the dual polytope described more explicitly, or if you need an embedding into $\mathbb{R}^n$. I also just take it one faith the dual is also vertex transitive; I can check in GAP if you'd like, but I thought I'd ask Carl later to walk me through the dual polytope again. :-) $\endgroup$ Mar 5, 2014 at 16:32
  • $\begingroup$ Thank you for this great answer! This could be a very nice counterexample due to its small dimension (all counterexamples I know are of dimension above 70. But could you please check whether it is vertex-transitive? It is not evident just because of its dual being vertex-transitive. Of course an embedding would also be very nice. :) $\endgroup$
    – Dune
    Mar 5, 2014 at 16:52
  • $\begingroup$ n=3 (dim=6), k=8 also works. I think those are the only 4 and 6 dimensional examples. It appears n=3, k=2n+2 may work in general, same automorphism group structure. I don't know if you need an infinite family (and still don't know about a nice embedding, but surely that can be looked up as the polytope is famous). $\endgroup$ Mar 5, 2014 at 18:38
  • $\begingroup$ The "typical" combinatorial automorphism group, dihedral of order 2k, is realizable as affine symmetries of the "standard" embedding of the cyclic polytope. These aren't isometries (they include shears), and I worry the full combinatorial automorphism group in the k=2n+2 case cannot be geometrically realized in any sense (even projectively). $\endgroup$ Mar 5, 2014 at 18:55
  • $\begingroup$ One reason this example might occur to check is that the dual cyclic polytope is the unique maximizer of the number of vertices given the number of faces, so if we want to stuff vertices into faces, then dual cyclic sounds like a good idea. The face transitivity condition seems pretty unnatural to me so far. $\endgroup$ Mar 5, 2014 at 19:19

Sorry I could not make this a comment because I do not have enough reputation yet, but here is an outline:

For any dimension $n$, start with the simplex of that dimension. The number of vertices in any feature of that simplex will be $n$ with a total number of vertices $n+1$. Now add enough vertices in order to achieve the next simplest vertex-transitive convex polytope. As you continue this process, the total number of vertices increases monotonically and faster than the number of vertices per feature. So for dimension $n=2$ and above, this means you must have a simplex given your conditions. I am unsure about the specifics with $n=0$ or 1.

  • $\begingroup$ Thank you for your answer! I am not sure if I am understanding right. The next simplest vertex-transitive polytopes in my opinion are achieved by placing a vertex on the top of the barycenter of every facet. But in this way every new facet will still be a simplex... On the other hand: do you think you can construct a counterexample in dimension 3 in this way? That would surprise me, I do not think that there is one. $\endgroup$
    – Dune
    Mar 5, 2014 at 8:42
  • $\begingroup$ Hmmm, first let me ask, what precisely are you calling a facet? For example, in 3D, I am accustom to calling vertices, edges, and faces all "facets," but for your problem I thought you were referring to a feature composed of vertices that had dimension of $n-1$ i.e. a face in the 3D case. $\endgroup$
    – Carser
    Mar 5, 2014 at 20:43
  • $\begingroup$ Yes. As mentioned above with "facets" I mean $(n-1)$-dimensional faces. $\endgroup$
    – Dune
    Mar 5, 2014 at 22:53

Let $P\subseteq \Bbb R^d$ be a $d$-dimensional vertex-transitive polytope with $n$ vertices, and a facet containing $m<n$ of these vertices.

An appropriately chosen free join of $P$ with itself will give you a vertex-transitive polytope of dimension $2d+1$, with $2n$ vertices, and a facet containing $n+m$ of these.

The free join construction embedds two polytopes, say $P_1$ and $P_2$, in skew affine subspaces and takes their convex hull. A facet of the free join is spanned by $P_1$ and a facet of $P_2$ or the other way around. If you need more information on this construction, I can elaborate.

Iterating this construction gives you vertex-transitive polytopes with facets containing an arbitrarily large percentage of the vertices. More precisely, after applying the free join $k$ times, you obtain a polytope with $2^kn$ vertices, a facet containing $(2^k-1)n+m$ of these, and therefore the following fraction of the vertices:

$$\frac{(2^k-1)n+m}{2^k n}=(1-2^{-k})+2^{-k}\frac mn \quad\xrightarrow{k\to\infty}\quad 1.$$

Example. The smallest example that I can obtain in this way (for which I am sure that it is not a simplex) is the free join of a square with itself. It will give you a $5$-dimensional (self-dual) polytope with 8 vertices and 8 facets, and each of these contains $6$ vertices.


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