# “Totally transitive” polytopes which are not regular

Is it possible to have an abstract polytope which is vertex-transitive, edge-transitive, face-transitive, etc. (individually transitive on faces of each particular dimension) and yet not flag-transitive?

• Archimedean solids, maybe? – Oscar Lanzi Aug 24 '17 at 0:00

The rhombus tiling of the plane! (Not the "rhombille", but simply a "tilted" square tiling.)

I don't know if this counts, because it is an infinite polytope, and it is not purely abstract, but here it is.

A single rhombus has two different angles, so it is not vertex-transitive, but the tiling is, because each vertex has two of each angle. Specifically, one vertex in one rhombus maps to the opposite vertex by reflection across a diagonal, and it maps to the adjacent vertices by 180° rotation around an edge.

An edge in one rhombus maps to the other three by reflections across the diagonals. Combining this with translations to map to different rhombi, the tiling is edge-transitive.

One rhombus maps to any other by lattice translations, or rotations, reflections, or glide-reflections, so it is face-transitive.

Yet the rhombus tiling is not flag-transitive. Consider two flags, both of which contain the same rhombus and edge, but opposite vertices of the edge. Any symmetry which preserves the rhombus cannot be a translation or glide-reflection, so we are left with reflections and rotations centered on the rhombus. But they all (except the identity) map the edge to a different edge. So no symmetry of the tiling can map this flag to that flag.

EDIT :

This can be made finite by considering it as a tiling of a flat torus instead of a plane. As before, its geometric structure is not regular, but its abstract structure is regular, and both are "totally transitive".

I just now realized that my previous answer is still valid, even though it was about geometry. This answer will focus on combinatorics.

I said before that a rhombic tiling of a torus must be abstractly regular. That is only true if the number of rhombi in the horizontal direction is the same as in the vertical direction. If they are different, I think it is irregular. Here is a picture of a tiling with $3 \times 2$ rhombi, and more rhombi in between. It has a total of $12$ vertices, $24$ edges, and $12$ faces. (Note that the Euler characteristic is $0$, as it should be for a torus.)

I think my Hasse diagram is upside down, and I have no idea how to calculate its automorphisms.

Any geometric symmetry is also a combinatorial symmetry (though the converse is false), so my previous answer shows that this toroidal dodecahedron is abstractly vertex-, edge-, and face-transitive.

Again, if we consider two flags with the same face and edge, but different vertices, we can see that they are distinct. Starting from face $AUBX$ and travelling diagonally across vertices and other faces ("diagonal" meaning "not directly connected by an edge", not referring to angles, lengths, etc.), moving toward vertex $U$ will return us to the starting point after crossing $2$ vertices, but moving toward $B$ will cross $3$ vertices. So the flag containing {$AUBX$, $UB$, $U$} is different from the flag containing {$AUBX$, $UB$, $B$}.

To be extra sure that this is correct, someone should find all of the dodecahedron's automorphisms, and see which of the dodecahedron's elements and flags get sent to each other.

It seems I am late to the party, but the canonical example are chiral abstract polytopes. They are, by definition, full-transitive but not regular, as they have 2 orbits of flag. There is a paper by Pellicer, available in the arxiv, where a construction for chiral polytopes, for all ranks greater than 2, is provided. The catch is that these are hard to picture (so forget about a Hasse diagram or a simple "intuitive" description), as they are very, very large. I hope this helps