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This question has been cross posted on MathOverflow with some very interesting answers and discussion.


I'm currently writing a project on the braid groups and their analogues on closed surfaces. It's an easy exercise to show that if $B_n$ is Artin's classical braid group on $n$ strings, then $B_n$ can be embedded in $B_{n+1}$ (the homomorphism is given by 'adding a string' on to the end of any braid in $B_n$ and this can be shown to be a monomorphism). A similar statement can be proved for the pure braid group $P_n$.

Let $P\mathcal{S}_n$ be the pure $n$-string braid group on the sphere $S^2$. Fox's definition of this group is the fundamental group of the configuration space $F_{n}S^2=\prod_n S^2\setminus\{(x_1,\ldots,x_n)|\exists i\neq j, x_i=x_j\}$ with basepoint $\hat{x}=(\hat{x}_1,\ldots,\hat{x}_n)$. The full braid group $\mathcal{S}_n$ is then the fundamental group of the configuration space $B_nS^2=F_nS^2/\sim$ where $x \sim y$ if the coordinates of $y$ are a permutation of the coordinates of $x$. It follows that $\mathcal{S}_n/P\mathcal{S}_n=\Sigma_n$ where $\Sigma_n$ is the symmetric group on $n$ elements.

It's well known that for $n\geq 3$, $P\mathcal{S}_n$ (and so $\mathcal{S}_n$) has torsion elements (given by the solution to the Dirac string problem for instance).

With that framework now built up, my question is, can $\mathcal{S}_n$ be embedded in to $\mathcal{S}_{n+1}$ for $n\geq 3$ (and similarly for their pure counter parts)? The naive 'add a string on the end' map will not work because, for instance, the following trivial braid $\gamma$ here (see image below) becomes non-trivial when a string is added on the end, and so any such map would not respect the equivalence class of isotopic braids.

I would think that the answer is no, but a proof eludes me. I've not been able to find any answer in the extensive literature which leads me to believe that the question is difficult. If a solution does exist, I would prefer a geometric proof as opposed to an algebraic proof, but any proof would be welcomed.

enter image description here

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  • $\begingroup$ What if you add n+1 strand so that there is no interaction with the n-braid? $\endgroup$ Commented Dec 23, 2012 at 14:25
  • $\begingroup$ How would you do that consistently so that the map was an injective homomorphism? That's the essential problem that needs to be solved. My intuition tells me that it can not be done, and so no such map exists. I'd be pleasantly surprised if there was such a non-trivial embedding though. $\endgroup$
    – Dan Rust
    Commented Dec 23, 2012 at 15:33
  • $\begingroup$ Oh, I see, it is not the usual braid group, but on the sphere. $\endgroup$ Commented Dec 23, 2012 at 19:10

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What about a doubling map? Take the $1$st strand and replace it with two parallel copies.

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  • $\begingroup$ That seems like a good idea and looks to sort out the problem with the braid $\gamma$. I'm slightly worried about how you would well-define such a map though, especially as the first strand can take rather complex paths round the sphere. How do you ensure that the new strand doesn't get twisted with the first strand, and in a consistent way. I think my worry is with the word 'parallel'. Could we just take as definition that the braid, when restricted to the first two strands, is trivial, and that the newly added strand is sufficiently close to the first? Thanks for the reply. I'll think on it. $\endgroup$
    – Dan Rust
    Commented Jan 19, 2013 at 1:58
  • $\begingroup$ Yes, that's the definition I'm thinking of. As you say, there are many parallel copies, so you need to pick the unique one that has linking number zero with the original strand. This is the same as your proposal of taking the one that restricts to a trivial link when you delete the other strands. $\endgroup$ Commented Jan 19, 2013 at 2:13
  • $\begingroup$ Now that I think about it, I believe there may be some trouble. Any pure braid with two strings is trivial on the sphere so it's not obvious to me how you would choose a canonical new strand. Especially as any twisting of the new strand really does make a difference when you take the context of the other strings in to account. As an example, how would you choose the second strand if the first spiralled anticlockwise around the inner sphere? $\endgroup$
    – Dan Rust
    Commented Jan 19, 2013 at 2:35
  • $\begingroup$ EDIT: Ignore the example, it's not a very good one. My point about the braid group on two strings being trivial for the sphere is still valid though, and I believe with a little more thought I may be able to find a counter-example. $\endgroup$
    – Dan Rust
    Commented Jan 19, 2013 at 2:49
  • $\begingroup$ Yes, I'm not used to thinking about the spherical braid group. Twisting is definitely a problem! Not sure if there's a way to fix it or not. $\endgroup$ Commented Jan 19, 2013 at 3:01

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