Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is perhaps a vague question, but hopefully there exists literature on the subject. The question is motivated by an answer I gave to this question. I also asked a related question on MO, although hopefully this question should be easier.

There exists a rather remarkable relationship between the 5 platonic solids, and the factor groups of the $n$-string braid groups $B_n$ by adjoining the relation $\sigma_i^k=1$ (here the $\sigma_i$, $1\leq i\leq n-1$ are the usual generators of $B_n$). We'll call these groups $B_n(k)$ the truncated braid groups of type $(n,k)$ where $B_n(k)=B_n/\langle \sigma_i^k\rangle$.

Theorem *: For $n\geq 3$, the group $B_n(k)$ is finite if and only if $k=2$ or $(n,k)$ is the Schläfli type of one of the 5 platonic solids. For these cases, $$|B_n(k)|=\left(\frac{f(n,k)}{2}\right)^{n-1}n!$$ where $f(n,k)$ is the number of faces of the platonic solid of type $(n,k).$

The 5 platonic solids correspond to the pairs $(n,k)\in\{(3,3),(3,4),(4,3),(3,5),(5,3)\}$. This is equivalent to the pair $(n,k)$ being a solution to the inequality $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}.$$

*It appears that this theorem was proved by Coxeter in

H. S. M. Coxeter, Factor groups of the braid group, Proceedings of the Fourth Can. Math. Cong., Banff 1957, University of Toronto Press (1959), 95–122.

although it is proving difficult for me to find a copy of this online or in my institution's library. From what I can gather, the proof is rather algebraic/combinatorial, although without access to a copy I can't say for sure. My question is:

Question Can one view the finite truncated braid groups in a geometric way as some action (in the vague, not necessarily strict group-action, sense) on the corresponding platonic solid or related objects?

Some of the approaches I have taken have included:

  • considering the isometry group on the corresponding solid (group orders don't match up),
  • considering paths on the surface which 'remember' the side that a face was entered from (relations don't match up),
  • considering labellings of the edges of the faces of the solid so that no face has a pair of edges with the same label, with group elements being permutations of edges which preserve this property (not sure if order matches up - seems difficult to calculate but may be the best approach so far),
  • considering 'rolling' the solid along a surface until it reaches its starting point again (no obvious way of making this a group via some kind of homotopy)

It seems that there should be some nice geometric interpretation of these groups, especially as every face has $n$ edges and every vertex is shared by $k$ faces.

It would be especially nice if the usual generators can be realised in a nice geometric way. Ultimately it would be nice to extend such a geometric interpretation of these finite truncated braid groups to the infinite cases as well (which correspond to regular tilings of the hyperbolic plane in most cases, and the complex plane in the case $(n,k)\in\{(3,6),(4,4),(6,3)\}$).

share|cite|improve this question
This is a really interesting problem. I know it has been a long time since you posted this, but I found this. Perhaps it has what you were looking for. – N. Owad Feb 12 '14 at 18:07
For n=3 there's definitely a link to the Platonic solids; I can probably give an answer tying it together on a representation-theoretic level. For n>3 you might be able to find a link but it's probably going to require you do something like cut out a particular slice; consider k=2, where the resulting group is the good old symmetric group, and the dihedral symmetry that comes from the corresponding "solid" is much smaller than the full symmetric group. – W. Schlieper Sep 29 '15 at 17:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.