Are (quasi-)regular polytopes uniquely determined by their edge graph?

Question

I consider polytopes $P\subset\Bbb R^n,n\ge 2$ of arbitrary dimension (intersection of finitely many halfspaces, therefore convex), which are vertex- and edge-transitive (also called quasi-regular).

Question: Can there exist two different such polytopes $P_1\subset \Bbb R^{n_1}$ and $P_2\subset\Bbb R^{n_2}$, maybe even of different dimensions, which have isomorphic edge-graphs (1-skeletons)? If one polytope is just the other one but embedded in a higher dimension, I consider these to be the same.

What if I drop edge-transitivity and instead use some suitable higher-dimensional generalization of uniform polyhedrons, or even weaker, require only that the edges are of the same length.

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Update

I found two statements, relevant for this question:

• Simple polytopes are uniquely determined by their edge-graphs. However, the definition of "simple polytope" fixes the dimension, so there might be higher dimensional realizations too.
• This answer on MO (and the comments) explain that for $K_n,n\ge 5$, there is a polytope of dimension $4\le d\le n-1$ which has $K_n$ as an edge-graph (see neighborly polytopes). So the dimension is not uniquely determined. However, I do not know which of these realizations of $K_n$ is vertex- and/or edge-transitive.

Answer 1

(Quasi-)regular polytopes surely are not uniquely defined by their edge graphs. Just consider the icosahedron x3o5o and the great dodecahedron x5o5/2o. In fact the latter is an edge-faceting of the former (i.e. respecting the same edge graph).

But as soon as you add the (true) convexity constraint, you enforce the edges to be exposed. Thus, again by the very convexity constraint, the only convex figure with that edge skeletton will be the hull polytope thereof.

Neither regularity, uniformity, orbiformity plays any role here nor does transitivity of edges (like quasiregular ones), regularity of faces (like CRF polytopes), or whatever. It is just (true) convexity which ensures that all edges have to be exposed. And that is what is relevant to this argument.

--- rk

Answer 2

During my research, I found out that only vertex-transitivity and convexity is not enough to ensure that the 1-skeleton is uniquely realizable. So the remaining question is whether edge-transitivity, uniformity or a requirement on the edge lengths can make it unique.

The counter-example I found was among the cyclic polytopes, which are a specific way to construct neighborly polytopes in higher dimensions. E.g. the $K_n,n\ge 5$ can be realized as a simplex in $n-1$ dimensions, or as a cyclic polytope in a lower dimension.

The crucial detail I now came across was that cyclic polytopes in even dimensions are combinatorially vertex-transitive and can be realized in the same dimension in a way that realizes all the automorphisms of their face-lattice (see this paper). This means: besides the simplices, there are indeed vertex-transitive (convex) polytopes with $K_n$ as their 1-skeleton. However, I found nothing about whether they can be realized edge-transitive or with any other symmetry.

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Update

I just realized, that the only way to place $n$ points equidistant, we need to arrange them as the vertices of the $(n-1)$-dimensional simplex. Therefore, the only realization of $K_n$ as an edge-transitive polytope (or even as a polytope with equal edge lengths) is the simplex.

Hence, a possible counter-example cannot be neighborly.