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).

$\quad$$\quad$ $\quad$$\quad$

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 $d\ge 4$.

  • $\begingroup$ 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." $\endgroup$ – Joseph O'Rourke Aug 14 '19 at 13:51
  • $\begingroup$ @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. $\endgroup$ – M. Winter Aug 14 '19 at 13:54
  • 3
    $\begingroup$ 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? $\endgroup$ – Yoav Kallus Aug 14 '19 at 14:02
  • $\begingroup$ @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. $\endgroup$ – M. Winter Aug 14 '19 at 14:08
  • $\begingroup$ @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). $\endgroup$ – M. Winter Aug 15 '19 at 7:24

The answer is No, there are no other such polytopes, as I was able to show in this recent preprint.

Theorem. In dimension $d\ge 4$, an edge-transitive polytope is vertex-transitive.

The idea is as follows: first, show that every edge-transitive polytope $P$ that is not vertex-transitive has the following three properties:

  1. all edges of $P$ are of the same length,
  2. $P$ has an edge in-sphere, and
  3. the edge-graph of $P$ is bipartite.

Call a polytope with these three properties bipartite. One then tries to classify these polytopes instead. This is easier, because every face of a bipartite polytope is again bipartite (not true for edge- or vertex-transitive polytopes).

The second step is to deal with all inscribed bipartite polytopes. It is not hard to see that these are zonotopes. By a result from another preprint of mine (see also this question), inscribed zonotopes with all edges of the same length are vertex-transitive. We can therefore exclude all the inscribed bipartite polytopes.

In the third step one classifies all the 3-dimensional non-inscribed bipartite polyhedra. This is quite tedious. Here is one example of a polyhedron which satisfies 2. and 3., but fails to have all edges of the same length. The deviation is so miniscule, that it cannot be spotted visually.

The result is then that there are only two such polyhedra: exactly those that I already mentioned in the question.

The final step is then to show that no 4-dimensional non-inscribed bipartite polytope can be built if we can use only these two polyhedra as facets. This uses a straight-forward argument on dihedral angles (see also Nick's answer).

| cite | improve this answer | |

If you consider a tiling of 3-space to be a 4-dimensional polytope, then the Rhombic dodecahedral honeycomb would work.

Other possibilities are limited by the potential 3-faces. Because every edge has one endpoint in each of two vertex orbits, the 2-faces must all have evenly many sides.

If the edge-transitivity descends to the 3-faces, then the 3-faces must be among the 9 isotoxal polyhedra: the five Platonic solids, the cuboctahedron, the icosidodecahedron, the rhombic dodecahedron, or the rhombic triacontahedron. The only ones of these with only even-length faces are the 3-cube, the rhombic dodecahedron, and the rhombic triacontahedron. With dihedral angles of 90°, 120°, and 144° respectively, these can only build up the 4-cube, the cubic tiling of 3-space, and the above-mentioned rhombic dodecahedral honeycomb. (A subgroup of the 4-cube's symmetry group acts in an edge-transitive but not vertex-transitive manner; you can color alternate vertices in two colors).

On the other hand, perhaps the 3-faces are not isotoxal: this occurs if, for some pair of edges $e$ and $e'$ of a 3-face $G$, every symmetry mapping $e$ to $e'$ also maps $G$ to a different 3-face incident to $e'$. In this case we can still say that the 3-cells are equilateral polyhedra with all even-length faces. There's also fairly strong requirements on the vertex figures, which must be vertex-transitive.

| cite | improve this answer | |

You can 1-subdivide any regular polytope to obtain edge-transitive but not vertex-transitive bipartite graph. More generally, you can take an incidence graph between k-faces and r-faces in a regular polytope. By incidence I mean putting an edge if one contains the other.

If you don't want a geometric property, e.g., convexity, then you can easily do so by taking any favourite finite group of yours. Namely, take a group G, take two subgroups H1 and H2 that have different indices, and consider the incidence graph with cosets, i.e., gH1 and gH2 are adjacent. The group G acts transitively on each side and on edges, but one cannot map a vertex to the other side.

The face-incidence example is indeed a special case of this algebraic construction, by taking two parabolic subgroups that correspond to faces.

| cite | improve this answer | |
  • 2
    $\begingroup$ Sorry, but I do not understand how any of these gives a convex edge-transitive polytope. E.g. the first part: I cannot just subdivide the edges of a regular polytope, I have to change the geometry for that too, and by that, probably create other edges. $\endgroup$ – M. Winter Aug 20 '19 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.