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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 on math.SE.

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 $\sigma_i$, $1\leq i\leq n-1$ is a generator 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 $$\frac{1}{n}+\frac{1}{k}>\frac{1}{2}.$$

This naturally leads to the question though of what happens when $(n,k)$ satisfies $$\frac{1}{n}+\frac{1}{k}=\frac{1}{2}$$ or $$\frac{1}{n}+\frac{1}{k}<\frac{1}{2}.$$ In the case of equality, $(n,k)\in\{(3,6),(4,4),(6,3)\}$ corresponds to a regular tiling of the complex plane by triangles, squares and hexagons respectively.

In the case of $\frac{1}{n}+\frac{1}{k}<\frac{1}{2}$, $(n,k)$ corresponds to a regular tiling of the hyperbolic plane by regular hyperbolic $n$-gons with $k$ tiles meeting at each vertex.

My question is

Does this classification of truncated braid groups by tiling type correspond to any properties of the groups other than finiteness in the case of tilings of the sphere (platonic solids)?

Certainly one could imagine that there should be some distinction between those $(n,k)$ which tile the complex plane, and those which tile the hyperbolic plane. One might expect that those groups which correspond to tilings of the complex plane aren't as infinite as those corresponding to tilings of the hyperbolic plane (for some definition of 'as infinite').

One might also expect there to be some relationship between the truncated braid groups and the discrete isometric group actions on the respective space which fix the corresponding tiling.

For $(n,k)$ corresponding to a hyperbolic tiling, does the type of the orbifold given by quotienting the hyperbolic plane by the Fuchsian group which fixes the tiling tell us anything about the corresponding truncated braid group?

I can't help but feel this material must have been covered somewhere in the literature, but it's hard to know what to search for.

*I can't recall who this theorem is attributed to off the top of my head (possibly Coxeter) but I believe it's referenced in

K. Murasugi & B. Kurpita, A Study of Braids, Kluwer Academic Publishers, 1999

of which I do not have a copy at hand.

EDIT : 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.

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    $\begingroup$ The groups you are considering are commensurable to the orbifold fundamental groups of the orbifold with the singular locus given by the hyperplanes $z_i=z_j$ in $C^n$, where you declare that the structure group of every hyperplane is $C_k$ (cyclic group of order $k$). Such groups (where you add more relators though) appear prominently in the work of Deligne and Mostow, and others on complex-hyperbolic lattices. For the more recent work, see Dmitry Panov's paper front.math.ucdavis.edu/1010.1448 and references therein. Maybe Dmitry will write an answer... $\endgroup$
    – Misha
    May 28, 2013 at 19:17

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