# Equivalence of two relations in Braid groups

Let $B_n$ be the braid group; that is, a group generated by $\sigma_1,\cdots,\sigma_{n-1}$ with relations

1. $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$ for $i=1,\cdots,n-2$;
2. $\sigma_i\sigma_j=\sigma_j\sigma_i$ if $i,j\in\{1,\cdots,n-1\}$ and $|i-j|\geq 2$.

For $1\leq i<j\leq n$, let the usual generator $A_{i,j}$ of pure braid groups be defined as $$A_{i,j}=(\sigma_{j-1}\sigma_{j-2}\cdots\sigma_{i+1})\sigma_i^2(\sigma_{i+1}^{-1}\cdots\sigma_{j-2}^{-1}\sigma_{j-1}^{-1}).$$

I need to prove that the following two sets have the same normal closure:

1. The set of all $[A_{j,k},h^{-1}A_{j,k}h]$, where $1\leq j<k\leq n$ and $h$ is an element of the subgroup generated by $A_{j,j+1},A_{j,j+2},\cdots,A_{j,n}$.

2. The set of all $[A_{j,k},g^{-1}A_{j,k}g]$, where $1\leq j<k\leq n$ and $g$ is an element of the subgroup generated by $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$.

My attempt: I believe once we proved that $1\Rightarrow2$, the other direction should be similar. Then I am planning to prove that assuming 1 is true, for each $i=1,\cdots,k-1$, $[A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1$. But the difficulty is :

• I do not see any obvious way to prove this. Should I use the ususal presentation of pure braid group? If yes, how?
• Even if I can prove $[A_{j,k},A_{i,k}^{-1}A_{j,k}A_{i,k}]=1$, does this imply 2? Anyway, generally in a group $G$ where $a,b,g\in G$, $g$ commutes with $g^a$ and $g^b$ does not imply that $g$ commutes with $g^{ab}$, where $g^a=a^{-1}ga$.
• In your definition of $A_{i,j}$ there is a typo: $\sigma_{i+1}^2$ should be $\sigma_{i+1}^{-1}$. I tried to correct it, but "Edits must be at least 6 characters long".
– j.p.
Dec 5 '14 at 14:47
• Thank you so much @j.p. for your sharp eyes! I have corrected the typo. Dec 5 '14 at 16:24

I claim the "strand reversing" automorphism of the braid group defined on generators as $\sigma_i \mapsto \sigma_{n-i}^{-1}$ sends $A_{i,j} \mapsto A_{n-j+1,n-i+1}^{-1}$.

This shows the normal closures of the two sets of commutators you mention agree, as this automorphism sends the subgroup $h$ is quantified over to the subgroup $g$ is quantified over (for the corresponding indices).

The basic idea is that $A_{i,j}$ is "pulling the $j$-th strand around the $i$-th." This is equivalent to "pulling the $i$-th strand around the $j$-th". Let's show this algebraically.

Let $i < j$. We'd like to define $A_{j,i}$ similarly, show these are equal, and then use an automorphism of the braid group to relate your statements.

$A_{j,i} = (\sigma_i^{-1}\cdots \sigma_{j-2}^{-1})\sigma_{j-1}^2(\sigma_{j-2}\cdots \sigma_i)$

$A_{i,j} = (\sigma_{j-1}\sigma_{j-2}\cdots \sigma_{i+1}) \sigma_i^2 (\sigma_{i+1}^{-1}\cdots \sigma_{j-2}^{-1}\sigma_{j-1}^{-1})$

Helpful will be the intermediates: Pull the $i$-th strand up to the $j-2$ spot, then move the $j$-th strand down one to the $j-1$ spot, then use $\sigma_{j-2}^2$, and move them back to where they started:

$H_{i,j,1} = (\sigma_i^{-1}\cdots \sigma_{j-3}^{-1}\sigma_{j-1})\sigma_{j-2}^2(\sigma_{j-1}^{-1}\sigma_{j-3}\cdots \sigma_i)$

This is the same as our $A_{j,i}$ as follows:

$H_{i,j,1} = (\sigma_i^{-1}\cdots \sigma_{j-3}^{-1}\sigma_{j-2}^{-1}\sigma_{j-2}\sigma_{j-1})\sigma_{j-2}^2(\sigma_{j-1}^{-1}\sigma_{j-2}^{-1}\sigma_{j-2}\sigma_{j-3}\cdots \sigma_i)$

$H_{i,j,1} = (\sigma_i^{-1}\cdots \sigma_{j-3}^{-1}\sigma_{j-2}^{-1}\sigma_{j-1}\sigma_{j-2})\sigma_{j-1}\sigma_{j-2}(\sigma_{j-1}^{-1}\sigma_{j-2}^{-1}\sigma_{j-2}\sigma_{j-3}\cdots \sigma_i)$

$H_{i,j,1} = (\sigma_i^{-1}\cdots \sigma_{j-3}^{-1}\sigma_{j-2}^{-1})\sigma_{j-1}\sigma_{j-1}\sigma_{j-2}\sigma_{j-1}(\sigma_{j-1}^{-1}\sigma_{j-2}^{-1}\sigma_{j-2}\sigma_{j-3}\cdots \sigma_i)$

$H_{i,j,1} = (\sigma_i^{-1}\cdots \sigma_{j-3}^{-1}\sigma_{j-2}^{-1})\sigma_{j-1}^2(\sigma_{j-2}\sigma_{j-3}\cdots \sigma_i) = A_{j,i}$

Now define $H_{i,j,2}$ to be pulling the $i$-th strand up to the $j-3$ spot, the $j$-th down to the $j-2$ spot, and using $\sigma_{j-3}^2$, and putting them back. Then $H_{i,j,2} = H_{i,j,1}$. In fact, by induction you can prove $A_{j,i} = H_{i,j,1} = H_{i,j,2} = \cdots = A_{i,j}$

Now consider a "strand reversing" isomorphism given by $\sigma_i \mapsto \sigma_{n-i}^{-1}$.

Under this isomorphism, $A_{i,j} \mapsto (\sigma_{n-j+1}^{-1}\sigma_{n-j+2}^{-1}\cdots \sigma_{n-i-1}^{-1}) \sigma_{n-i}^{-2} (\sigma_{n-i-1}\cdots \sigma_{n-j+2}\sigma_{n-j+1}) = A_{n-i+1,n-j+1}^{-1} = A_{n-j+1,n-i+1}^{-1}$

• Thank you for your answer. I am sorry about the confusion; yes, I mean all $i$, $j$ and $h$. Basically I mean the normal closure of all elements of the form $[A_{j,k},h^{-1}A_{j,k}h]$, where $1\leq j<k\leq n$ and $h$ is an element of the subgroup generated by $A_{j,j+1},A_{j,j+2},\cdots,A_{j,n}$ is equal to the normal closure of all elements of the form $[A_{j,k},g^{-1}A_{j,k}g]$, where $1\leq j<k\leq n$ and $g$ is an element of the subgroup generated by $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$. Dec 11 '14 at 13:52
• By the way, your answer looks correct to me; many thanks again! Dec 11 '14 at 14:09
• @Zuriel Great. I've submitted an edit to your question to reflect what you say in your comment. Let me know if that's what you meant.
– aes
Dec 11 '14 at 16:53
• Exactly!　Thank you so much for everything!! Dec 11 '14 at 18:14