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General convex polytopes can not be uniquely reconstructed from their 1-skeleton1, as explained here. Not even the dimension is known from the skeleton, as e.g. the complete graph $K_n,n\ge 5$ is the 1-skeleton of neighborly polytopes that exist in dimensions $\ge 4$.

1By 1-skeleton I mean only the graph (incidence information), with no metric information contained (see comments).

But how well works the reconstruction from the 1-skeleton if we restrict to sufficiently symmetric polytopes. Obviously, vertex-transitivity is not enough as seen from the existence of vertex-transitive neighborly polytopes. But what if we add edge-transitivity, uniformity, arc-transitivity or some requirements on the edge-lenghts? E.g. only simplices are vertex- and edge-transitive polytopes with $K_n$ as their 1-skeleton.

Question: Given a "symmetric" polytope $P$ (replace "symmetric" with the sufficiently strong symmetry requirements of your choice). Can there be a different "symmetric" polytope with the same 1-skeleton as $P$?

I asked a similar question on Math.SE, without much success.

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  • $\begingroup$ hmm, in the cited Math SE question you used "graph" and I took it there may be wrongly to be the 1-skeleton. But now you even call it yourself such. To me a 1-skeleton is metrically rigid, whereas a graph more has just its combinatorical incidence structure. - Thus any 1-skeleton clearly will provide the defining convex polytope back again, simply use the convex hull. But when removing the metrical informations, i.e. speaking of the mere graph, it gets much worse indeed. --- rk $\endgroup$
    – Dr. Richard Klitzing
    Commented Aug 15, 2018 at 21:09
  • $\begingroup$ @Dr.RichardKlitzing These terms are new for me and I am not certain about their exact use in polytope theory. I am sorry for this confusion. Here, and in the linked post on Math.SE, by 1-skeleton I mean the graph with no metric information contained. $\endgroup$
    – M. Winter
    Commented Aug 16, 2018 at 9:15

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