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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: The loop manipulation generators. (Also, there is a copy on my website here.) Generally, $H_n$ 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:

  1. The usual braid relation: $C_i C_{i+1} C_i = C_{i+1}C_iC_{i+1}$ for $i = 1, \dots, n-1$.
  2. 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:

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

I've got two questions about $H_n$:

  1. 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 $H_n$?
  2. Does this group appear in the literature anywhere?
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  • $\begingroup$ If I've understood things correctly, you need to add the following relations: $P_i C_{i+1} C_i = C_{i+1} C_i P_{i+1}$, and $P_i P_{i+1} C_i = C_{i+1} P_i P_{i+1}$, and $P_i P_{i+1} P_i = P_{i+1} P_i P_{i+1}$. $\endgroup$
    – Sam Nead
    Commented Apr 1 at 17:37
  • $\begingroup$ Thanks. These seem like straightforward generalizations of the usual braid relation. How can I be sure that this list is complete? $\endgroup$
    – pgadey
    Commented Apr 1 at 17:59
  • $\begingroup$ You're thinking of $T_i = \sigma_{2i-1}$ every second braid (twist)? $\endgroup$
    – Jules Lamers
    Commented Apr 1 at 20:10
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    $\begingroup$ I've seen this referred to as the Hilden subgroup of $B_{2n}$. E.g: arXiv:0706.4421 Theorem 5.3 gives a presentation in terms of your generators and 14 relations. $\endgroup$
    – Tuomas Laakkonen
    Commented Apr 2 at 1:27
  • $\begingroup$ If you'd invert one of the $C_i$'s in equation (4) (which I understand you don't want to do) then the subgroup generated by the $C_i$ and $T_j$ is an affine braid group. More precisely, replacing your $C_i$ by something that's usually called $T_i$, which in addition obeys the quadratic relation $(T_i - q)(T_i + q^{-1})=0$, and writing $Y_i$ for your $T_i$, instead by gives (the Bernstein–Zelevinsky presentation of) the affine Hecke algebra of type $GL_n$. (I'm not aware of an analogue of your $P_i$ in that case.) $\endgroup$
    – Jules Lamers
    Commented Apr 7 at 14:19

1 Answer 1

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Your group $H_n$ is (I believe) called the wicket group by Brendle and Hatcher in their paper Configuration spaces of rings and wickets. They provide a presentation in Proposition 3.6. They also provide two references that appear closely related; these are Hilden's 1975 paper and Tawn's 2008 paper.

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