Let $\tilde{B_n}$ be the homotopy braid group; namely, in the deformation of braids, a braid string is allowed to self-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: $\sigma_1,\sigma_2,\cdots, \sigma_{n-1}$


  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 $P_n$ and $g$ is an element of the subgroup of $P_n$ generated by $A_{1,k},A_{2,k},\cdots,A_{k-1,k}$.

My main question is:

What is a presentation of the homotopy pure braid group $\tilde{P_n}$? Is it the usual presentation of the pure braid group $P_n$ plus relation 3 above?

Another question:

The authors of the book said that "it is quite possible that $\tilde{B_n}$ may not be finitely presented". The book was written around 15 years ago; is there any new result proving if $\tilde{B_n}$ is finitely presented?

  • $\begingroup$ I take it that the homotopy pure braid group is the kernel of the natural map to $S_n$? Anyway, the Reidemeister--Schreier process provides an algorithm for computing presentations of finite-index subgroups. $\endgroup$ – HJRW Sep 28 '14 at 6:06
  • $\begingroup$ @HJRW Yes, it is the kernel as you mentioned. Does it make the group to be equal to the usual presentation of the pure braid group $P_n$ plus relation 3 above? $\endgroup$ – Zuriel Sep 28 '14 at 6:50
  • $\begingroup$ Without doing the computation I can't be sure. But the set of relators you add certainly needs to be invariant under the natural $S_n$-action. It's not clear to me that the set of relators of type 3 is invariant. $\endgroup$ – HJRW Sep 28 '14 at 7:10
  • $\begingroup$ Let me try to summarize what the Reidemeister--Schreier process tells you. The group $S_n$ acts by outer automorphisms on $P_n$. For each relator of type 3, you need to add its $S_n$-orbit to the set of relators for $\widetilde{P}_n$. $\endgroup$ – HJRW Sep 28 '14 at 7:12

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