Does $A_{j,k}$ commute with all its conjugates in homotopy braid groups?

Let $\tilde{B_n}$ be the homotopy braid group; namely, in the deformation of braids, a braid string is allowed to intersect itself. Similarly let $\tilde{P_n}$ be the homotopy pure braid group.

I am reading A study of Braids by K. Murasugi and B. I. Kurpita. In Chapter 7, Theorem 3.3, a presentation of $\tilde{B_n}$ is given as follows:

Generators: $\sigma_1,\sigma_2,\cdots, \sigma_{n-1}$

Relations:

1. $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$;

2. $\sigma_i\sigma_j=\sigma_j\sigma_i$ for $|i-j|\geq 2$;

3. $A_{j,k}$ commutes with $gA_{j,k}g^{-1}$, where $A_{j,k}$ are the usual generators of the pure braid group $P_n$ defined by $$A_{j,k}=(\sigma_{k-1}\sigma_{k-2}\cdots\sigma_{j+1})\sigma_j^2(\sigma_{j+1}^{-1}\cdots\sigma_{k-2}^{-1}\sigma_{k-1}^{-1})$$ and $g$ is an element of the subgroup of $P_n$ generated by $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$.

Exercise 3.4 of the book says that the Relations 3 above can be replaced by

$A_{j,k}$ commutes with $gA_{j,k}g^{-1}$ where $g$ is an element of the subgroup of $P_n$ generated by $A_{j,j+1},A_{j,j+2},\cdots,A_{j,n}$.

I am wondering what if we replace the Relation 3 by

$A_{j,k}$ commutes with $gA_{j,k}g^{-1}$ where $g$ is an element of $P_n$.

Is it still the same definition as the original one? If no, that means that there exists a pure braid $g$ such that $A_{j,k}$ doe not commute with $gA_{j,k}g^{-1}$. Anyone know such an example?

Edit: My last paragraph was not clear. It should be that "there exists a pure homotopy braid $g$ such that $A_{j,k}$ (regarded as a homotopy braid) doe not commute with $gA_{j,k}g^{-1}$ in $\tilde{P_n}$".

• There is a nice characterization of when two elements of the braid group commute in terms of the Thurston curve reduction system for the braids. It's a good way to approach these types of concerns. – Ryan Budney Nov 28 '14 at 18:29
• I don't understand the given description of the homotopy braid group. Why don't you just get the symmetric group this way? – Qiaochu Yuan Nov 29 '14 at 9:06
• @QiaochuYuan, here each strand is allowed to self-intersect, but different strands are not allowed to intersect. See Link groups (maths.ed.ac.uk/~aar/papers/milnorlink.pdf) by J. Milnor. – Zuriel Nov 29 '14 at 12:12

1 Answer

Consider $A_{2,3}=\sigma_2^2$. We have that $\sigma_2^2$ does not commute with $\sigma_1^2.\sigma_2^2.(\sigma_1^2)^{-1}$ in the homotopy pure braid group, i.e, $[\sigma_1^2,\sigma_2^2]$ is non trivial.

Habegger and Lin showed that the homotopy pure braid group is classified by Milnor invariants without repeated indices (more precisely, they classified string links up to homotopy, and showed that any string link is homotopic to a pure braid). We have that Milnor invariant $\mu_{123}$ detects the commutator $[\sigma_1^2,\sigma_2^2]$. Notice that the closure of $[\sigma_1^2,\sigma_2^2]$ is isotopic to Borromean rings, which is typically detected by Milnor link invariants $\overline{\mu}_{123}$; see Milnor's paper.

• Thank you for the answer! By the way, why did you edit and say that this does not seem to answer the question? I think it is exactly what I was asking. – Zuriel Nov 29 '14 at 12:14
• By the way, if I replace Relation 3 by "$A_{j,k}$ commutes with $gA_{j,k}g^{-1}$ where $g$ is an element of $P_n$", what is the factor group then? – Zuriel Nov 29 '14 at 12:16
• The pure braid group is generated by the $A_{j,k}$'s, isn't it ? So it seems to me that you get the abelianized homotopy pure braid group. – jbm Nov 30 '14 at 7:55
• It is true that the pure braid group is generated by the $A_{j,k}$'s but taking $[A_{j,k},g^{-1}A_{j,k}g]=1$ does not necessarily abelianize the group. It does abelianize if we take $g^{-1}A_{j,k}g=1$, but now we are only requiring each generator commutes with all its conjugates. – Zuriel Nov 30 '14 at 12:00