Proof that the Szilassi polyhedron is not abstractly regular

Question

The Szilassi polyhedron has seven hexagonal faces and 14 trihedral vertices. This is enough to make it a "regular toroid" in according to Szilassi.^[1] However abstract polytopes are required to have a flag-transitive automorphism group, which is a more restrictive definition of regular.

So the question arises:

Is the Szilassi polyhedron abstractly regular?

Since the Szilassi polyhedron has 7 hexaongal faces it should have 84 flags, and thus, if it were regular, it would have an automorphism group of size 84. When I check the Atlas of Small Abstract Regular Polytopes there is no Abstract Regular polytope of Schläfli type {6,3} with a automorphism group of size 84.

The pages on the AoSARP are generated by a program exhaustively up to 1024 flags, so either there's a bug in that program or the Szilassi polyhedron is not abstractly regular, and I'm inclined to believe the latter over the former. (Or I have made a mistake in using the AoSARP, that is always a possibility)

But I'd like to prove it. I could construct the adjacency matrix and brute force check every pair of flags for symmetry, but this is time consuming and not very enlightening at the end.

I know a couple of easy checks that can show a polyhedron is not abstractly regular:

1. Check every face has the same number of sides (Yes n=6)
2. Check every vertex has the same number of faces (Yes n=3)
3. Check every k-zigzag (here k=2) has the same number of edges (Yes n=14)
4. Check the above for the dual. (Yes)

None of these are sufficient to be abstractly regular, but they are all necessary, and the Szilassi polyhedron seems (the last one is a bit tough to compute since the dual has such high degree vertices), to pass all of them.

Is there a nice proof that the Szilassi polyhedron is not abstractly regular?

Answer 1

Here's one construction of the (abstract) Szilassi polyhedron. Take a regular hexagon, and surround it with six more hexagons, to make a larger hexagonal patch with $18$ loose edges, or $6$ zigzags of $3$ edges each. Attach each zigzag to the one on the opposite side (by a translation, not a $180^\circ$ rotation), as in the image:

For comparison, see Wolfram MathWorld's net. (It seems to be missing a label $n''$ on the lower right patch.)

For any flag $x$, there's exactly one other flag $x^v$ that has the same face and edge as $x$ but a different vertex. Likewise, there's one other flag $x^e$ that has only a different edge, and $x^f$ that has only a different face.

In the Szilassi polyhedron, there's a flag $x$ that makes a closed loop after a certain sequence of $16$ moves, but the adjacent flag $y=x^v$ doesn't make a closed loop after the same sequence of moves:

$$x^{vfe\,vfe\,vfe\,ve\,vfe\,ve}=x$$ $$y^{vfe\,vfe\,vfe\,ve\,vfe\,ve}\neq y$$ $$\,$$

$$\,$$

Thus $x$ and $y$ are not equivalent, and the polyhedron is not regular.

Answer 2

Note that the faces of the Szilassi polytope are all hexagons. It therefore is easy to alternate the vertices on each single face. This however induces a valid vertex alternation rule, applicable to the full polyhedron. (E.g., within the labeling of this picture of your mentioned wiki page on the Szilassi polyhedron, those vertex subsets would be {A, C, E, G, I, K, M} and {B, D, F, H, J, L, N}.) But then, the existance of such a subdivision is exactly what makes the according abstract polytope to a chiral one.

Whereas the mentioned atlas clearly states that it lists the regular abstract polytopes.

The incidence matrix (by equivalence types - here: all elements of every type) of the Szilassi polyhedron clearly reads: $$\begin{array}{l|c|c|c}V & 14 & 3 & 3\\\hline E&2 & 21 & 2\\\hline F&6 & 6 & 7\end{array}$$

The atlas at least provides e.g. the polytope {6,3}*588, then having 98 vertices, 147 edges, 49 faces instead - i.e. a regular polytope where the Szilassi polytope seems to be a chiral modwrap (or quotient: division by 7) thereof.

But with respect to the afore mentioned chirality the incidence matrix could well be given instead likewise by: $$\begin{array}{l|cc|c|c}V&7&*&3&3\\ &*&7&3&3\\\hline E&1&1&21&2\\\hline F&3&3&6&7\end{array}$$

Aside note:
A similar effect also occurs for the cube. Here likewise the vertices can be alternated, then providing a pair of dual, vertex inscribed tetrahedra. However, the compounds of dual simplices (in any dimension) are known to be flag transitive, i.e. regular themselves. (Sure, a compound is not a polytope, but the cube, as the hull of that stella octangula, again is.) - A likewise "compound effect" in the Szilassi polyhedron case seems not to work out, at least is this exactly your observation.

--- rk