Intermediate subgroups of braid groups are not bi-orderable

TL;DR

The paper proves that for every $n\ge3$, no intermediate subgroup $H$ with $P_n(N)\subsetneq H\subset B_n(N)$ is bi-orderable, where $N$ is a compact connected surface and $M$ is the disk or a surface with $M\neq S^2,\mathbb{R}P^2$. The approach leverages the known bi-orderability of $P_n(D)$, the non-bi-orderability of $B_n(D)$ for $n\ge3$, and embeddings $B_n(D)\hookrightarrow B_n(M)$ to transfer non-bi-orderability to all intermediate subgroups; a key step is analyzing the cycle type of the permutation $\pi(\beta)$ for braids $\beta$ outside $P_n(D)$ and applying generalized torsion/unique-root arguments. The results yield that $P_n(D)$ is a maximal bi-orderable subgroup of $B_n(D)$ and extend to intermediate subgroups of various surface braid groups, including no bi-orderable intermediate subgroups of $B_\infty(D)$; these findings highlight the intricate interplay between braid group structure, configuration spaces, and orderability properties. The work also connects to torsion phenomena and suggests constraints on how orderability can extend from pure to full braid groups on surfaces.

Abstract

Let $M$ be the disk or a compact, connected surface without boundary different from the sphere $S^2$ and the real projective plane $\mathbb{R}P^2$, and let $N$ be a compact, connected surface (possibly with boundary). It is known that the pure braid groups $P_n(M)$ of $M$ are bi-orderable, and, for $n\geq 3$, that the full braid groups $B_n(M)$ of $M$ are not bi-orderable. The main purpose of this article is to show that for all $n \geq 3$, any subgroup $H$ of $B_n(N)$ that satisfies $P_n(N) \subsetneq H \subset B_n(N)$ is not bi-orderable.

Figures (1)

• Figure 1: The braid $\sigma_i \in B_n(D)$ illustrated geometrically and as an element of $\pi_1(C_n(D))$.

Theorems & Definitions (33)

• Theorem 1
• Theorem 2
• Theorem 3: Artin, A1A2
• Remark 4
• Remark 5
• Lemma 6: Garside, G
• Theorem 7: Artin, A1A2
• Remark 8
• Remark 9
• Example 10