Geometric way to view the truncated braid groups? 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 $(n,k)\in\{(3,6),(4,4),(6,3)\}$).
 A: Here is a suggestion towards a geometric interpretation of the group $B_3/(\sigma_1^k)$ with $3\le k\le 5$.
First note that $B_3$ is generated by $\sigma_1$ and $x=\sigma_1\sigma_2$. The braid relation is equivalent to $(\sigma_1x)^2=x^3$, and $\Delta=x^3$ is the Garside element. It generates the center of $B_3$, but I will not use this fact.
Let $K$ be the set of oriented edges of the platonic solid. Each such edge $e$
determines two vertices (source $s(e)$ and target $t(e)$) and two faces ($l(e)$
on the left of $e$ and $r(e)$ on the right of $e$). For any oriented edge e, let
$\sigma_1\cdot e$ and $x\cdot e$ be the oriented edges with
$$ s(\sigma_1\cdot e)=s(e),\quad r(\sigma_1\cdot e)=l(e) $$
and
$$ s(x\cdot e)=t(e),\quad l(x\cdot e)=l(e). $$
It follows that
$$ x^3\cdot e=e,\quad \sigma_1\cdot (x\cdot (\sigma_1\cdot (x\cdot e)))=e,
$$
and $k$-fold action of $\sigma_1$ on $e$ gives $e$ for each oriented edge $e$.
It is not very difficult to show that this defines a free and transitive action of
$B_3/(\sigma_1^k,\Delta)$ on $K$.
Since $\Delta $ is central in $B_3$, we may extend this action to an action of
$B_3/(\sigma_1^k)$ on $K\times C_N$ with $N$ the order of $\Delta $ in
$B_3/(\sigma_1^k)$ and $C_N$ the cyclic group of order $N$.
Note that $x^3=\Delta $. In order to implement this action, let us fix a set $L$
of oriented edges such that each face is adjacent to exactly one edge in $L$ in
counterclockwise orientation. (There may be other edges in $L$ adjacent to the
face in clockwise orientation.) Moreover, we add to each oriented edge an integer $l'(e)$. Now we define
$$ \sigma_1\cdot (e,m)=(\sigma_1\cdot e,m),\quad
x\cdot (e,m)=(x\cdot e,m+l'(e)+\varepsilon_e),$$
where $\varepsilon_e\in \{0,1\}$, and
$$ \varepsilon_e=1 \quad \Leftrightarrow e\in L. $$
It turns out that this defines an action of $B_3/(\sigma_1^k)$ on $K\times C_N$
if and only if for each face $F$,
$$ \sum_{e\in K,l(e)=F}l'(e)=0 $$
and for each oriented edge $e$,
$$ l'(e)+l'(-e)+\varepsilon_e+\varepsilon_{-e}=1. $$
Here $-e$ is the edge $e$ with opposite orientation.
Such a labeling exists if and only if $N|f/2$ (with $f$ the number of faces).
It follows that $N=f/2$ and
$$ |B_3/(\sigma_1^k)|=\frac f2|K|=6\left(\frac f2\right)^2 $$
since $|K|=3f$.
