3
$\begingroup$

I want to consider (convex) polytopes $P=\mathrm{conv}\{p_1,...,p_n\}\subset\Bbb R^d$ which are both, vertex- and edge-transitive (or maybe stronger: 1-flag-transitive).

Question: Is every such polytope already a uniform polytope?

I know only a few polytopes with such symmetries, all of them are uniform, probably also Wythoffian. Here are some:

The faces of uniform polytopes are uniform again. So far, all I can say about the faces of vertex- and edge-transitive polytopes is, that they are polytopes with all vertices on a sphere and all edges of the same length. While this means that all 2-faces are uniform, it does not immediately follow for the 3-faces (e.g. Pseudorhombicuboctahedron is not uniform but could be a face).

I know that after all, the Wythoffian uniform polytopes are the most well understood. Also, I do not know whethe there is any non-Wythoffian vertex- and edge-transitive polytope. So as a first step, I might ask:

Question: Is every Wythoffian polytopes with such symmetries already uniform?

Update

As mentioned by Dr. Klitzing, the second question seems to be trivial, as a Wythoffian polytope is always uniform as long as its edge lengths are equal everywhere (which surely is the case for edge-transitive polytopes).

Improve this question
$\endgroup$
5
  • $\begingroup$ The cuboctahedron is vertex-transitive and edge-transitive, but not face-transitive. $\endgroup$
    – Richard Stanley
    Jan 30, 2019 at 18:23
  • $\begingroup$ @RichardStanley Face-transitivity is not required for a polytope to be uniform (in the sense here). Please see the linked Wikipedia page for the intended definition. Note that the cuboctahedron is contained in above list as a vertex truncation of the cube. $\endgroup$
    – M. Winter
    Jan 30, 2019 at 18:40
  • $\begingroup$ Oops, I was confusing uniform polytopes with regular polytopes. $\endgroup$
    – Richard Stanley
    Jan 31, 2019 at 4:08
  • $\begingroup$ Any such polytope is CRF. But CRF polytopes generally have little symmetry. $\endgroup$
    – mr_e_man
    May 4, 2022 at 3:02
  • $\begingroup$ ...Not only CRF, but also scaliform. Of the four currently known non-uniform convex scaliform polychora, none are edge-transitive. Of the convex uniform polychora, none are edge-transitive, except those that are also vertex-edge-pair-transitive (there's a reflection symmetry that fixes the edge (setwise, not pointwise) and sends one vertex to the other). I expect that examples can be found for both this question and the linked one, as there's no obvious proof that they don't exist, but of course the problem is to find them.... $\endgroup$
    – mr_e_man
    Jun 13, 2023 at 13:03

2 Answers 2

Reset to default
2
$\begingroup$

Well, your addition on the polytope being forced to be Wythoffian definitely guarantees a positive answer. In fact, every Wythoffian polytope with equal sized edges will be uniform. (Vertex transitivity is already contained when asked to be Wythoffian by means of kaleidoscopical construction.)

But not every uniform polytope is Wythoffian. Examples are eg. the snub figures and Coxeter's grand antiprism. Those cannot be constructed via mere kaleidoscopical constructions.

--- rk

Improve this answer
$\endgroup$
5
  • $\begingroup$ That's interesting. Do you have a source for that claim: "[...] every Wythoffian polytope with equal sized edges will be uniform." ? $\endgroup$
    – M. Winter
    Feb 1, 2019 at 14:51
  • $\begingroup$ Also, is there an easy way to detect whether all edges are of the same length (or equal under automorphisms) from the Wythoff symbol or Coxeter diagram alone? $\endgroup$
    – M. Winter
    Feb 1, 2019 at 14:57
  • $\begingroup$ By Wythoffian polytope I mean a polytope obtained by Wythoff's kaleidoscopical construction, which in turn can be encoded by a decorated Coxeter-Dynkin symbol. That very construction device right ensures all your requieries, whenever the ringed nodes represent "off from the mirror" and the unringed ones an "on the mirror", and when additionally (by usual convention) that "off" is assumed to be the same amount wherever appicable. But infact the decoration is kind a coordination, so you might write the individual offness-length rather than the ringings into the symbol, like x3y4z ... $\endgroup$
    – Dr. Richard Klitzing
    Feb 1, 2019 at 22:39
  • $\begingroup$ Okay thank. But can you still provide me with a source or formal proof of your statement "[...] every Wythoffian polytope with equal sized edges will be uniform."? $\endgroup$
    – M. Winter
    Feb 4, 2019 at 17:58
  • $\begingroup$ Confer here: en.wikipedia.org/wiki/Wythoff_construction. An article on examplified 3D kaleidoscopical constructions with varying, even negative "offness" sizes (aka edge lengths) is available here: bulatov.org/polyhedra/mosaic2000/index.html; a different research on vertex alternations, snubs etc. uses Wythoffian polytopes for starting figures and thus outlines that process too, cf. bendwavy.org/klitzing/pdf/Stott_v8.pdf. $\endgroup$
    – Dr. Richard Klitzing
    Feb 4, 2019 at 19:08
0
$\begingroup$

A vertex- and edge-transitive polyhedron is already a uniform polyhedron.

This is because the vertex-transitivity asures the vertices to be equivalent and therefore the angles of each face (individually) to be the same. The edge-transitivity then asures the edges to be all of the same size (say unity). And so the faces individually to become regular ones.

But be aware, that there are uniform polyhedra, which aren't edge-transitive!

Within higher dimensional spaces, assuming spherical geometry, these axioms only provide scaliformity. That is, you well could use some of the Johnson solids to be cells of your polytope.

The easiest scaliform polychoron (4D polytope) is the lace prism of 2 truncated tetrahedra in inverted orientation. That is the lacing cells here are 8 triangular cupolae and 6 tetrahedra, while the top and bottom base are those truncated tetrahedra.

But again, scaliformity just asks the edges to be of the same size. Not that those have to be equivalent. So your axioms select a subset out of the scaliforms.

--- rk

Improve this answer
$\endgroup$
7
  • $\begingroup$ Thank's for your post, but I think this does not really give an answer. The case is quite trivial for polyhedra (as the faces are regular, this was mentioned in the question already). I think what you call scaliformity (I have a hard time looking that term up except on your personal website) is what I already mentioned in my post about faces with all vertices on a sphere and edges of equal length. I am already aware of the non-triviality of the question. $\endgroup$
    – M. Winter
    Jan 31, 2019 at 14:05
  • $\begingroup$ You also can find a huge section on scaliform polychora on Jonathan Bowers' (aka HedronDude's) website: polytope.net/hedrondude/polychora.htm#scaliform. In fact I for one found that very first above described scaliform polychoron (tut||inv tut) way back in the dawn of the new millenium, then calling it weakly uniform. Later Jonathan provided that positively term. $\endgroup$
    – Dr. Richard Klitzing
    Feb 1, 2019 at 8:24
  • $\begingroup$ In fact the duduction that the polygonal faces have to be regular only works within spherical (or hyperbolical) geometry. Within euclidean spaces you well may have rhombs, shields, and the like stuff. This is why scaliformity adds the requirement for regular polygonal faces in addition. Cf. bendwavy.org/klitzing/explain/scaliform.htm $\endgroup$
    – Dr. Richard Klitzing
    Feb 1, 2019 at 8:28
  • $\begingroup$ I am not sure what you mean by " Within euclidean spaces you well may have rhombs, shields, and the like stuff". My polytope $P$ lives in Euclidean space. All vertices of $P$ are on a sphere because of vertex-transitivity. A face of $P$ lies in a linear subspace. The intersection of a sphere and a linear subspace is a sphere of lower dimension, hence all vertices of the face lie on that lower-dimensional sphere. $\endgroup$
    – M. Winter
    Feb 1, 2019 at 9:06
  • $\begingroup$ Symmetry equivalence of vertices within a point group guarantee a common circumcenter, that is they live on a hypersphere; you are dealing with spherical geometry. But when you consider symmetry equivalence wrt. a space group, you allow for translations too, and you are in euclidean space. And this euclidean space then does not ask for a unique circumradius of each polygonal face. Otoh., within hyperbolical geometry symmetry groups, you might have that additional restriction on the polygonal faces again. $\endgroup$
    – Dr. Richard Klitzing
    Feb 1, 2019 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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