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I apologize in advance for a somewhat wishy-washy question. I just read the paper "The Braid Group and Other Groups" by F. A. Garside in which he solves the conjugacy problem for the braid group and also obtains a standard form for braids that yields a solution to the word problem and computes the center. The paper is straightforward to follow line by line, however the last section "Other groups" makes me feel like I am completely missing the geometric insight that Garside had while writing the paper - in particular he says that all of the results given above hold for a certain class of more general groups to give the solution to the word/conjugacy problems. He also notes that the diagram that he has drawn for the element $\Delta \in B_4$ is the 2-skeleton of the truncated octahedron and the are similar higher-dimensional polytopes for the other elements $\Delta \in B_n$.

For example, corresponding to the truncated cuboctahedron, Garside tells us that for the group generated by $a_1, a_2, a_3$ subject to the relations $a_1a_2a_1a_2 = a_2 a_1a_2a_1, a_2a_3a_2= a_3a_2a_3, a_1a_3=a_3a_1$, if we consider the element $\Delta = (a_1a_2a_3)^3$, then the results proved in Garside's paper, all suitably modified, still hold. How is this group related to that polytope?

Can someone guide me with the geometric rehydration process for this paper? For concreteness, this might involve relating some of the results in the paper to constructions with a polytope.

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  • $\begingroup$ Footnote: G. S. Makanin solved the conjugacy problem for braid groups in 1968, a year before Garside, though he did not have time to publish a full solution before Garside's paper appeared. You might be interested in Makanin's 1971 paper showing that normalisers in braid groups are f.g. This paper is purely word-combinatorial in nature, though. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 14, 2021 at 10:35

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If you add the relations that the standard generators have order two, then instead of getting the braid group on $n$ strands, you get the symmetric group on $n$ symbols. The generator that corresponded to swapping the $i$th and $(i+1)$st strands maps to the transposition $(i,i+1)$ in the symmetric group. The 1-skeleton of the polytope is the Cayley graph of the symmetric group with respect to this generating set. There is a similar polytope for each finite Coxeter group.

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    $\begingroup$ Since the OP mentions two polytopes, it might be helpful to clarify that the truncated octahedron is the Cayley graph of the symmetric group, while the truncated cuboctahedron is the Cayley graph of the Coxeter quotient of the group described by generators and relations at the end of the post. $\endgroup$
    – Hugh Thomas
    Commented May 14, 2021 at 20:31

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