6.01 Braids: Introduction

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

Next video: Braids: Artin action.
Index of all lectures.

Notes

Definitions

(0.00)

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:

$F_i(t)\neq F_j(t)$ if $i\neq j$
$F_i(0)=z_i$ , $F_i(1)=z_{s(i)}$ for some permutation $s\colon\{1,\ldots,n\}\to\{1,\ldots,n\}$ .

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:

$\vdots$

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

Why do braids form a group under stacking?

Navigation

Next video: Braids: Artin action.
Index of all lectures.