# Is there any edge- but not vertex-transitive polytope in $d\ge 4$ dimensions?

I consider convex polytopes $$P\subset\Bbb R^d$$. The polytope is called vertex- resp. edge-transitive, if any vertex resp. edge can be mapped to any other by a symmetry of the polytope.

I am looking for polytopes which are edge- but not vertex-transitive. There are infinitely many of these for $$d=2$$, and exactly two for $$d=3$$ (rhombic dodecahedron and rhombic tricontrahedron, see below).

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I do not know a single example for $$d\ge 4$$.

I believe it is easy to see that the edge-graph of such a polytope must be bipartite, and thus, zonotopes might be a good place to start looking. But my constructions fail for $$d\ge 4$$.

• To echo your bipartite point: Wikipedia says "Every edge-transitive graph that is not vertex-transitive must be bipartite and either semi-symmetric or biregular." – Joseph O'Rourke Aug 14 at 13:51
• @JosephO'Rourke That was my thought. But one has to be careful: the symmetry group of the polytope might be smaller than the one of its edge-graph. "Bipartite" should still hold, but I do not immediately know about "semi-symmetric or biregular". Update: I have read the definition of these terms, and they should probably still hold. – M. Winter Aug 14 at 13:54
• Both your examples in d=3 are the convex hull of the union of a regular polyhedron and its dual (appropriately scaled). Have you tried this construction in d=4? – Yoav Kallus Aug 14 at 14:02
• @YoavKallus That's a great idea. Thank you. What else comes to my mind now is to let a vertex grow out of each facet of a (regular) polytope and at some point the original edges will vanish in the inside of a facet. This might give examples. This corresponds to your idea of looking for the "appropriate scaling". And no, have not tried this, but will do so now. – M. Winter Aug 14 at 14:08
• @YoavKallus At least the construction using the 4-cube and the 4-crosspolytope does not work (no matter the scaling, their are at least two orbits of edges). – M. Winter 2 days ago