What are examples for convex polytope $P\subset \Bbb R^d,d\ge 3$ for which holds

  • $P$ is 2-face transitive (that is, all 2-faces are equivalent under the symmetries of $P$), and
  • all 2-faces of $P$ are 4-gons (not necessarily squares, or rectangles).

I know the $d$-cubes, rhombic dodecahedron and rhombic triacontahedron. Are there any others?

I expect other such a polytope, if at all, then only in $d\ge 4$. Maybe a slight modification of a neighborly cubical polytope as constructed by Ziegler here, but as they are, they have two 2-face orbits.

  • $\begingroup$ Rhombic dodecahedron? $\endgroup$ – Ilya Bogdanov Sep 16 '19 at 20:27
  • $\begingroup$ @IlyaBogdanov You are absolutely right. And also the rhombic triacontahedron. I have overlooked these. So I guess my question now is whether there are any more? $\endgroup$ – M. Winter Sep 16 '19 at 20:31
  • $\begingroup$ What about Cartesian products of $d$-cubes, rhombic dodecahedrons and rhombic triacontahedrons? $\endgroup$ – LeechLattice Sep 16 '19 at 23:30
  • $\begingroup$ @Bullet51 I think the cartesian product of a $d_1$-cube and a $d_2$-cube is a $(d_1+d_2)$-cube. You are right about the latter two examples. I already mentioned these in my comment, but I should edit them into my post. $\endgroup$ – M. Winter Sep 17 '19 at 5:13

In fact, there are many to be found on Wikipedia under isogonal figures, even in three dimensions.

Examples in dimension four are obtained as dual polytope of runcinated 4-simplex or runcinated 24-cell. This works, because these both runcinations are

  1. edge-transitive, and
  2. each edge is contained in exactly four facets.

Since in the dual edges become 2-faces and facets become vertices, the resulting dual polytopes will be 2-face-transitive, and each 2-face will have four vertices, that is, they will be 4-gons.

If I am not mistaken, the cells of these two examples are trigonal trapezohedron (aka. elongated cubes), and these are 3-dimensional examples also found in the Wikipedia list mentioned above.

| cite | improve this answer | |

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.