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

Question

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.

Answer 1

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.

Answer 2

The answer is No, the graph of an arc-transitive polytope cannot have more symmetries than the polytope. The polytope and its graph have the exact same symmetries!

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In the article "Capturing Polytopal Symmetries by Coloring the Edge-Graph" I prove the following:

(Corollary 5.3.) Let $P\subset \Bbb R^d$ be a polytope with edge-graph $G$. Consider a coloring $G^c$ of the edge-graph where two vertices/edges have the same color if and only if they are in the same orbit w.r.t. the symmetry group $\mathrm{Aut}(P)\subseteq\mathrm O(\Bbb R^d)$. Then $\mathrm{Aut}(G^c)\simeq\mathrm{Aut}(P)$.

In particular, if $P$ is vertex- and edge-transitive, then each vertex/edge of $G^c$ has the same color, that is, the graph is uncolored. Thus $\mathrm{Aut}(G)\simeq\mathrm{Aut}(P)$.

At the core of the proof is a result by Ivan Izmestiev from

I. Izmestiev (2010), "The Colin de Verdière Number and Graphs of Polytopes"

(summarized as Theorem 4.1 in my article), which roughly states that polytope skeleta are spectral embeddings of the edge-graph, when assigning certain vertex/edge weights. These weights can be used as colors, and it then follows from basic spectral graph theory that every combinatorial symmetry of the colored graph extends to a geometric symmetry of the polytope.