How can one recognize when two Coxeter diagrams represent the same uniform polytope?

If we enumerate uniform polytopes generated by Coxeter diagrams, we find many cases where two apparently distinct Coxeter diagrams yield the same uniform polytope.

A small handful of examples:

• the rectified tetrahedron is the same as the octahedron
• the cantellated 16-cell is the same as the rectified 24-cell
• the cube is the same as a square prism or the product of three segments
• the hypercube is the same as:
• product of two squares
• product of a cube and a segment
• product of a square and two segments
• product of four segments

In none of the above cases is the equivalence obvious to me, from looking at the symbols. How can one recognize it?

One can, of course, use "brute force": that is, construct the two polytopes, and compare their structures.

But is there a simpler / more principled way to recognize equivalence, by just analyzing the graph structure of the two diagrams?

Here's a start.

The uniform polyhedron or planar tiling generated by a 3-node Coxeter diagram can be unambiguously specified by its vertex configuration; that is, the cyclic list of regular polygons surrounding any vertex.

Therefore, to decide whether two 3-node Coxeter diagrams generate the same polyhedron or tiling, it suffices to compare the two vertex configurations.

Translation from 3-node Coxeter diagram to vertex configuration is summarized by the following table, where $$p,q,r \geq 2$$, and any $$2$$'s appearing in the vertex configuration should be removed.

$$\require{HTML}\newcommand{\mypic}[4][]{\style{display: inline-block;background: url(http://i.stack.imgur.com/#4) no-repeat center;#1}{\phantom{\Rule{#2}{#3}{0px}}}} \begin{array}{ccc}\hline\text{Case}&\text{Coxeter diagram}&\text{Vertex configuration}\\\hline \text{1 ringed node}&\mypic{36px}{34px}{hVoci.png}&(p\cdot q)^r\\ \text{2 ringed nodes}&\mypic{36px}{34px}{nDAmp.png}&p\cdot2r\cdot q\cdot2r\\ \text{3 ringed nodes}&\mypic{36px}{34px}{k26a3.png}&2p\cdot2q\cdot2r\\\hline\end{array}\\ \text{}$$

Coxeter diagrams with 4 or more nodes?

It seems that the method described above for 3 node diagrams could, in theory, be used for higher dimensions; but unfortunately the vertex configuration isn't as easy to describe and work with, since it's not just a cyclic list of regular polygons, so it's not clear how to proceed.

Between any two Coxeter groups that are related, what happens is there must be a inactive (unringed) mirror in one group which can be removed which doubles the fundamental domain simplex into the new Coxeter group. Rank 4 groups are harder than Rank 3 since harder to draw fundamental tetrahedra on paper, but the Coxeter graph itself can contain clear symmetry. Lots of examples are here: https://en.wikipedia.org/wiki/Coxeter_notation#Extended_symmetry

Coxeter bracket notation is parallel alternative to the node-branch notation of Coxeter groups. Coxeter and Norman Johnson also have an "extended bracket notation of the form [X[Y]] where [X] is an extending symmetry of [Y]. The simplest extending symmetry, [[X]] is used for reversible Coxeter graphs, like [[n]]=[2n] (doubling dihedral symmetry), and [[3,3]]=[4,3] (tet to oct), while [[3,3,3]] is actually a [2]+ rotational extension, but [[3,3^1,1]] doubles to [4,3,3], and as [3[3^(1,1,1)]] extends to [3,4,3].

And there's a reverse operation, radical subgroups, [3*,4,3] = [3,4,3*] = [3^(1,1,1)] -- removing a [3] mirror, index 6 subgroup goes from F4 goes down to D4. Also a funny [3,3,4,1+]=[3,3^(1,1)] uses 1+ to imply the end mirror being removed, so an index 2 subgroup. You can even do do [4,(3,3) * ], index 24 as order of [3,3], removing 3 mirrors, leaving 4 orthogonal mirrors []x[]x[]x[]=[2,2,2].

The RULE for enumerating these radical subgroups is adjacent odd-order branches must be all removed together, i.e. you need even order branches as delimiters. Like you can't do [5,3*] because 5 is odd. [5] and [3] are both subgroups of [5,3] and Norman Johnson defines a "trionic subgroup", but doesn't help for relating uniform polyhedron because all mirrors would have to be unringed. https://en.wikipedia.org/wiki/Coxeter_notation#Trionic_subgroups

So anyway, these symmetry relations will tell you what uniform polytopes are related. Any mirrors that are removed must be unringed, but otherwise all permutations of rings are allowed and will be related.

• Thanks for this answer! I'm just getting started trying to understand it. One thing right off bat-- you say "Between any two Coxeter groups that are related, [...] which doubles the fundamental domain simplex into the new Coxeter group." This seems to imply that for any any two related Coxeter groups, the size of the fundamental domain simplex of one must be a power of two times that of the other. But, isn't this a counterexample: the cube (•)-4-•---• has char. simplex that is 1/48 of the sphere, whereas when expressed as (•) (•) (•) it has char. simplex that is 1/8 of the sphere. – Don Hatch Sep 28 '20 at 9:44
• Hi Don! Are you the author of a hyperbolic tiler I remember?! Yes, not just power of two. Actually operationally that's a "radical subgroup, [4,3*]=[2,2]. This removes both mirrors around an order 3 point, [3] is dihedral group order 6. While [1+,4,3]=[3,3] is an order 2 subgroup. – Tom R Oct 5 '20 at 21:03
• Here's a table of subgroups of hyperbolic group [6,4]. Because 6 and 4 are both even, there's many subgroups, including chiral/rotational groups. The notations and graphical representation are show with reflection lines colored with the node colors in the Coxeter diagram. Also interesting [6,4*] and [6*,4] both exists as subgroups, but fundamental domain is no longer a triangle. [6,4*] makes a square domain, and [6*,4] has a hexagonal domain. These Coxeter diagrams have red dotted lines for divergent mirrors. en.wikipedia.org/wiki/Truncated_tetrahexagonal_tiling#Symmetry – Tom R Oct 5 '20 at 22:08