6.01 Braids: Introduction

Below the video you will some pre-class questions and notes to accompany the video.




Fix a collection of \(n\) points \(z_1,\ldots,z_n\) in \(\mathbf{C}\). An \(n\)-strand braid \(F\) is a collection of \(n\) continuous maps \(F_1,\ldots,F_n\colon[0,1]\to\mathbf{C}\) such that:
We can draw a picture of a braid as a collection of pairwise disjoint paths \(\gamma_1,\ldots,\gamma_n\) in \(\mathbf{C}\times[0,1]\): \[\gamma_k(t)=(F_k(t),t).\] In fact, since \([0,1]\) is compact and the image of a compact set by a continuous map is compact, the images of the paths \(F_k\) are contained in some compact set in the plane, and we can always homotope everything (by a family of rescalings depending on \(t\)) to assume that our braids are contained in \(D\times[0,1]\), where \(D\) is the unit disc.

The picture above shows a 3-strand braid in red which permutes the three black points via \((13)\) (if they are numbered \(1,2,3\) left-to-right).

Equivalence of braids

(3.23) We say that two \(n\)-strand braids \(F\) and \(G\) are equivalent if there is a collection of homotopies \(H_k(s,t)\), \(k=1,\ldots,n\), such that \(\{H_k(s,t)\}_{k=1}^n\) is a braid for each fixed value of \(s\) and such that \[H_k(0,t)=F_k(t),\ H_k(1,t)=G_k(t),\]

Group law

(5.20) If \(F\) and \(G\) are two \(n\)-strand braids with associated permutations \(\sigma\) and \(\tau\) respectively then their product \(G\cdot F\) is the braid \[ (G\cdot F)_i(t) =\begin{cases} F_i(2t)&\mbox{ if }t\in[0,1/2]\\ G_{s(i)}(2t-1)&\mbox{ if }t\in[1/2,1]. \end{cases} \]

Pictorially, we multiply braids by stacking them:

The set of equivalence classes of \(n\)-strand braids form a group \(B_n\) under this stacking product.
This is an exercise.

Much of the proof of this theorem should look a little bit like the proof that the fundamental group is a group. This is not a coincidence: the \(n\)-strand braid group is the fundamental group of a particular space, the unordered configuration space of \(n\) points in the disc.

Configuration space

(8.20) Let \(OC_n\) be the subset of \(\mathbf{C}^n\) defined by \[OC_n:=\{(x_1,\ldots,x_n)\in \mathbf{C}^n\ :\ x_i\neq x_j\mbox{ for }i\neq h\}.\] We call a point \((x_1,\ldots,x_n)\in OC_n\) an ordered configuration of points in the disc and \(OC_n\) is called the ordered configuration space of \(n\) points in the plane. There is an action of the permutation group \(S_n\) on \(OC_n\); a permutation \(s\) acts as \[(x_1,\ldots,x_n)s=(x_{s(1)},\ldots,x_{s(n)}).\] The quotient \(UC_n:=OC_n/S_n\) is called the space is called the unordered configuration space of \(n\) points in the plane.

The fundamental group \(\pi_1(UC_n,[z_1,\ldots,z_n])\) is isomorphic to the \(n\)-strand braid group.
This should be clear from the definition of a braid: a braid is a collection of paths \(F_1(t),\ldots, F_n(t)\) with \(F_i(t)\neq F_j(t)\) if \(i\neq j\) and such that \(F_i(0)=z_i\), \(F_i(1)=z_{s(i)}\). Such a collection of paths defines a loop: \[ [F_1(t),\ldots,F_n(t)]\] in the unordered configuration space based at \([z_1,\ldots,z_n]\) and conversely. A homotopy of braids gives a homotopy of loops in the unordered configuration space (again, just by definition). Stacking braids corresponds to concatenating loops.

Presentation of the braid group

(11.38) We will assume that the points \(z_i\) are equally spaced along a line. For each \(i=1,2,\ldots,n-1\) there is an elementary braid:



The braid group \(B_n\) is generated by the elementary braids subject to the following relations: \begin{gather*} \sigma_i\sigma_j=\sigma_j\sigma_i\mbox{ if }|i-j|\geq 1\\ \sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}. \end{gather*}
Proof not included! It is an exercise to check that the braid relations hold. Later, I will give \(\epsilon\) more explanation for how one would go about checking that these relations suffice.

Pre-class questions

  1. Why do braids form a group under stacking?


Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.