The braid group on 3 strands has the presentation $\langle x,y \;|\; xyx=yxy\rangle$. A group $G$ is called right-orderable if there is a total order $<$ on the set $G$ such that if $a<b$ then $ac<bc$ for all $c\in G$. It is known that braid groups are right-orderable.

Is there a non-right-orderable torsion-free quotient group of the braid group on 3 strands?

  • $\begingroup$ What do you mean by a factor? $\endgroup$ – HJRW Mar 15 '15 at 21:06
  • $\begingroup$ @HJRW: factor means quotient. $\endgroup$ – Alireza Abdollahi Mar 15 '15 at 21:19

Yes, there are many such examples. The braid group is isomorphic to the fundamental group of the trefoil knot complement. The trefoil knot $T$ admits many Dehn fillings, parameterized by $r\in \mathbb{Q} \cup \{\infty\}$. If $|r|\geq 1$, then the Dehn filling $S^3_r(T)$ is an L-space. Moreover, it is usually Seifert-fibered and has torsion-free fundamental group. By Theorem 4 of Boyer-Gordon-Watson, these Dehn fillings do not have orderable fundamental group (the notion of L-space is actually not relevant to the answer to this question, it just gave the quickest way to cite the literature).

  • $\begingroup$ What is $\hat{\mathbb{Q}}$? $\endgroup$ – Qiaochu Yuan Mar 16 '15 at 4:55
  • $\begingroup$ $\mathbb{QP}^1$ - I changed the notation, since this is probably only used by 3-manifold topologists. $\endgroup$ – Ian Agol Mar 16 '15 at 5:00
  • $\begingroup$ @IanAgol: Is it known a group presentation (not necessarily finite) for $\pi_1(S_r^3(T))$ for some $r$ with $|r|\geq 1$? $\endgroup$ – Alireza Abdollahi May 26 '15 at 20:03
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    $\begingroup$ Yes, it is easy to obtain such representations. One adds a relation of the form m^a*l^b, where r=a/b and m is the longitude, l the meridian. $\endgroup$ – Ian Agol May 26 '15 at 20:32
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    $\begingroup$ Yes, those two relations (I didn't compute m and l, but this should be easy). $\endgroup$ – Ian Agol May 27 '15 at 6:06

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