$$B_{n}(S^2)=\langle \sigma_1,\sigma_2,...\sigma_{n-1}\mid \sigma_{i}\sigma_{j}=\sigma_{j}\sigma_{i} \text{ if } |i-j|>1;\qquad$$ $$\qquad \sigma_{i}\sigma_{j}\sigma_{i}=\sigma_{j}\sigma_{i}\sigma_{j} \text{ if } |i-j|=1,\quad \sigma_1\sigma_2...\sigma_{n-1}^2...\sigma_2\sigma_1=1\rangle$$ For $n\ge 4$, $B_{n}(S^2)$ is an infinite group. According to J.V Buskirk this group contain an element of finite order, consequently it is not orderable, i'm interested to the subgroups of $B_{n}(S^2)$ that are orderable, one way to solve this problem is to find subgroups that are locally indicable, so commutator subgroup of $B_{n}(S^2)$ is a candidate. Is there any way for finding orderable subgroups of $B_{n}(S^2)$?

  • $\begingroup$ orderable means bi-orderable or left-orderable for you? $\endgroup$
    – YCor
    Jul 17, 2017 at 8:01
  • 1
    $\begingroup$ It seems to me that torsion is basically the only problem. Let's consider the pure braid groups $PB_n(S^2)$, which of course have finite index in $B_n(S^2)$. The Birman Exact Sequence asserts that the kernel of the "strand forgetting map" $PB_n(S^2)\to PB_{n-1}(S^2)$ is $\pi_1$ of the $(n-1)$-punctured sphere, hence free and non-trivial for $n>2$. Furthermore, if I recall correctly, $PB_3(S^2)$ is trivial. Hence, by induction on $n$, for $n\geq 4$, by induction on $n$, $PB_n(S^2)$ should be locally indicable. $\endgroup$
    – HJRW
    Jul 17, 2017 at 9:06
  • $\begingroup$ YCor, thank you , orderable means left orderable. $\endgroup$ Jul 18, 2017 at 13:20

1 Answer 1


A google search for "braid groups orderable" gives a wealth of information. In particular, Juan Gonzalez-Meneses (arXiv preprint here) proved that pure braid groups on closed orientable surfaces are biorderable.

(Recall that the pure braid group $PB_n(S^2)$ is the kernel of the natural homomorphism $B_n(S^2)\to S_n$ induced by the action on the punctures. In particular, it has finite index in the braid group.)


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