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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!
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.
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.
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.
