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. – Zuriel Dec 5 '14 at 16:24
up vote 2 down vote accepted
+50

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 $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$. – Zuriel Dec 11 '14 at 13:52
  • By the way, your answer looks correct to me; many thanks again! – Zuriel 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!! – Zuriel Dec 11 '14 at 18:14

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.