# Can the graph of a symmetric polytope have more symmetries than the polytope itself?

I consider convex polytopes $$P\subseteq\Bbb R^d$$ (convex hull of finitely many points) which are arc-transitive, i.e. where the automorphism group acts transitively on the 1-flags (incident vertex-edge pairs). Especially, this includes that the polytope is vertex- and edge-transitive. The graph of a polytope is the graph isomorphic to its 1-skeleton.

Question: Are there arc-transitive polytopes, where the graph has more symmetries than the polytope?

When weakening the requirement of arc-transitivity, there are examples:

• A rhombus is edge- but not vertex-transitive. However, its graph is vertex-transitive. (Thanks to Henrik for the comment)
• There are vertex-transitive neighborly polytopes other than a simplex, but none of these can be edge-transitive. Their graphs are complete and are therefore edge-transitive.
• If automorphism means an isometry of the surrounding Euclidean space a Rhombus is an example. – HenrikRüping Oct 2 '18 at 14:14
• @HenrikRüping Thank you for your comment. I hoped it would be clear from my explanation that arc-transitivity includes edge- and vertex-transitivity. Unfortunately a rhombus is not vertex-transitive. I will improve my post. – M. Winter Oct 2 '18 at 14:16
• Just to give a trivial illustration, any arc-transitive simplex has edges all of the same length, so it is a regular simplex, so it has the full symmetric group as symmetries. – Gro-Tsen Oct 3 '18 at 8:20

Partial progress: Let $$V$$ be the vertex set of $$P$$, let $$E$$ be the set of directed edges and let $$X$$ be the set of ordered pairs of distinct elements of $$V$$. Let $$G$$ be the group of combinatorial symmetries of the edge graph and let $$\Gamma \subset G$$ be the group of geometric symmetries of the polytope. So it is assumed that $$E$$ is a single orbit for both $$G$$ and $$\Gamma$$ acting on $$X$$. I claim there must be some other $$G$$-orbit on $$X$$ which splits into more than one $$\Gamma$$ orbit. In particular, we must have more than one $$G$$-orbit on $$X$$, which means that neighborly polytopes won't work.

Without loss of generality, we may assume that $$P$$ spans $$\mathbb{R}^d$$ and the centroid of $$P$$ is at $$\vec{0}$$, so action of $$\Gamma$$ extends uniquely to a linear action on $$\mathbb{R}^d$$.

Proof: Suppose to the contrary that $$G$$ and $$\Gamma$$ have the same orbits on $$X$$. Let $$\mathbb{R} V$$ be the permutation representation on $$V$$. It is well known that the dimension of $$\mathrm{Hom}_G(\mathbb{R} V, \mathbb{R} V)$$ is $$|V^2/G| = |X/G| + 1$$, and likewise for $$\mathrm{Hom}_{\Gamma}$$. So the hypothesis on orbits implies that $$\mathrm{Hom}_G(\mathbb{R} V, \mathbb{R} V) = \mathrm{Hom}_{\Gamma}(\mathbb{R} V, \mathbb{R} V)$$. As a corollary, any $$\Gamma$$-subrepresentation $$W$$ of $$\mathbb{R}V$$ is also a $$G$$-subrepresentation, because we can choose a $$\Gamma$$ equivarient projection $$\mathbb{R}V \to W$$, and then this projection will also be $$G$$-equivariant.

The map taking the basis vector $$e_v$$ of $$\mathbb{C} V$$ to the vertex $$v$$ of the polytope $$P$$ gives a $$\Gamma$$-equivariant linear surjection from $$\mathbb{R} V$$ to $$\mathbb{R}^d$$. So $$\mathbb{R}^d$$ can be identified with a $$\Gamma$$ summand of $$\mathbb{R} V$$. But every $$\Gamma$$ summand is also a $$G$$-summand, so the $$\Gamma$$ action extends to a $$G$$ action, contradiction. $$\square$$.

So we want a graph $$(V,E)$$ with arc-transitive symmetry group $$G$$, and a subgroup $$\Gamma$$ of $$G$$ which is still arc-transitive but has more orbits on $$X$$. Such graphs definitely exist. As one example, let $$(V,E)$$ be the Hamming $$n$$-cube, whose symmmetry group is $$S_n \ltimes C_2^n$$ (here $$C_2$$ is the cyclic group of order $$n$$.) If $$H$$ is a transitive but not $$k$$-transitive subgroup of $$S_n$$ for some $$k$$, then $$H \ltimes C_2^n$$ has more orbits on $$X$$, but all edges of $$(V,E)$$ remain a single orbit. But I haven't succeeded yet in embedding an example like this as the edge graph of a polytope.

• How about skewing the Hamming 4-cube in 4 different directions so that every face looks like the same rhombus and we get the same two acute angles and two obtuse angles at every vertex? The graph would keep the same full cube symmetry but the skewed cube would be arc-transitive with fewer symmetries because you now cannot send an acute angle into an obtuse one. – Claude Chaunier Oct 4 '18 at 12:38
• I like this idea, but I can't figure out the details of how to do the skewing to make the polytope stay arc transitive (or even vertex transitive). My first attempt was to take $4$ unit vectors $\vec{v}_1$, \ldots, $\vec{v}_4$ with the same acute angle between each pair and take their Minkowski sum, but that isn't vertex transitive, because some vertices are at the acute angle of $6$ rhombi, some of $3$ and some of $2$. Can you say more? – David E Speyer Oct 4 '18 at 13:53
• @ClaudeChaunier I responded to your comment but forgot to ping you. – David E Speyer Oct 4 '18 at 15:42
• I can show that a Minkowski sum of $n$ linearly independent vectors is vertex transitive (with respect to isometries of $\mathbb{R}^n$) if and only if the $n$ vectors are orthogonal vectors of the same length. So you need to go beyond Minkowski sums to make this work. – David E Speyer Oct 4 '18 at 16:08
• This is what I have in mind. $\begin{matrix}0&0&0&0\\0&0&-1&1\\0&1&1&0\\0&1&0&1\\-1&1&0&0\\-1&1&-1&1\\-1&2&1&0\\-1&2&0&1\\1&0&0&1\\1&0&-1&2\\1&1&1&1\\1&1&0&2\\0&1&0&1\\0&1&-1&2\\0&2&1&1\\0&2&0&2\end{matrix}$. I don't think it's a Minkowski sum. I still need to check the angles. – Claude Chaunier Oct 4 '18 at 18:54