Published: Apr 1, 2024

Last Modified: Apr 10, 2024

This is a local copy of a post to MathOverflow asking about the Loop Manipulation Group.

Recently, I came across a subgroup of the braid group $B_{2n}$ that I’m calling the “loop manipulation” group $H_n$. The idea is that we treat pairs of adjacent strands in the braid group as “loops” $i = 1, \dots, n$. It is generated by the following elements:

- $T_i$ for $i = 1, \dots, n$,
- $C_i$ for $i = 1, \dots, n-1$
- $P_i$ for $i = 1, \dots, n-1$.

The nomenclature is meant to suggest that $T_i$ is a “twist”, $C_i$ is a “cross”, and $P_i$ is a “pass”. Here is an illustration of the generators.

Generally, $H_n$ is has a lot in common with the braid group $B_n$. The $C_i$ generators are analogous to the the usual generators $\sigma_i$ of the braid group $B_{n}$ so we have:

- The Usual Braid Relation: $C_i C_{i+1} C_i = C_{i+1}C_iC_{i+1}$ for $i = 1, \dots, n-1$.
- The Usual Braid Commutation Relation: If $|i - j| \geq 2$ then $C_iC_j = C_jC_i$

Playing with lots of little diagrams has convinced me of the following relations:

- $T_iT_j = T_jT_i$ for all $i$ and $j$.
- $T_iC_i = C_iT_{i+1}$ for $i = 1, \dots, n-1$.
- $T_iP_i = P_iT_{i+1}$ for $i = 1, \dots, n-1$.

I’ve got two questions about $H_n$:

- Do the relations above determine $H_n$ uniquely? To put it formally: If an abstract group generated by $T_i$, $C_i$, and $P_i$ satisfied these relations, would it be isomorphic to $H_n$?
- Does this group appear in the literature anywhere?

- The question got posted here.
- Relevant Office Camera Photos

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