# 6.01 Braids: Introduction

## Video

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

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