6.03 Braids: the Wirtinger presentation

Below the video you will find notes and some pre-class questions. Once again, sorry for the gurgling background noises: I can't turn off my office radiator.

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Notes

(0.00) Let $B$ be an $n$ -strand braid inside $D^2\times[0,1]$ . If we take the quotient space $D^2\times S^1=(D^2\times[0,1])/\sim$ , $(x,0)\sim(x,1)$ , then the braid closes up to become a collection of embedded circles $C_B$ in $D^2\times S^1$ (because the component paths $B_k(t)$ start and end in the set of points $z_1,\ldots,z_n$ ). This is called the braid closure $C_B$ of $B$ .

Here is an example: the braid closure of the 2-strand braid $\sigma_1^3$ is the trefoil knot:

(1.59) Let

$X_B=(D^2\times S^1)\setminus C_B$ denote the complement of

$C_B\subset D^2\times S^1$ . Let

$x=[1,0]\in(D^2\times[0,1])/\sim$ (we are thinking of

$D^2\subset\mathbf{C}$ , so

$1\in D^2$ makes sense). We have

$\begin{equation} \label{eq:knotcomp1} \pi_1(X_B,x)=\langle\alpha_1,\ldots,\alpha_n,g\ |\ g\alpha_kg^{-1} =B(\alpha_k)\mbox{ for }k=1,\ldots,n\rangle. \end{equation}$ Here,

$g$ is the loop

$x\times S^1$ and, for

$k\in\{1,\ldots,n\}$ ,

$\alpha_k$ is the element of

$\pi_1(D^2\setminus\{z_1,\ldots,z_n\}$ given by the loop in the figure below and

$B(\alpha_k)$ denotes the Artin action of

$B$ on

$\alpha_k\in\pi_1(D^2\setminus\{z_1,\ldots,z_n\}\cong\mathbf{Z}^{\star n}$ .

(3.01) The space

$X_B$ is the mapping torus of the homeomorphism

$Art_B\colon D^2\setminus\{z_1,\ldots,z_n\}\to D^2\setminus\{z_1,\ldots,z_n\},$ so the lemma follows from the result we proved earlier which gave a presentation for the fundamental group of a mapping torus.

Consider the 2-strand braid

$\sigma_1$ (whose braid closure is an unknot). We have

$\sigma_1(\alpha)=\alpha\beta\alpha^{-1},\quad\sigma_1(\beta)=\alpha,$ so the presentation for

$\pi_1(X_{\sigma_1})$ is:

$\langle\alpha,\beta,g|g\alpha g^{-1}=\alpha\beta\alpha^{-1},g\beta g^{-1}=\alpha\rangle.$

(8.44) If we embed

$D^2\times S^1$ as the standard solid torus in

$\mathbf{R}^3$ then the complement of the braid closure

$C_B\subset \mathbf{R}^3$ has

$\begin{equation} \label{eq:knotcomp2} \pi_1(\mathbf{R}^3\setminus C_B)=\langle\alpha_1,\ldots,\alpha_k\ |\ \alpha_k=B(\alpha_k),\ k=1,\ldots,n\rangle \end{equation}$ where

$B(\alpha_k)$ is the Artin action of

$B$ on the free group

$\langle\alpha_1,\ldots,\alpha_k\rangle$ .

(A homotopy retract of) the complement

$\mathbf{R}^3\setminus C_B$ is obtained from

$X_B$ by attaching a 2-cell along the circle

$x\times S^1$ , which adds the relation

$x=1$ to the presentation from the lemma, yielding the desired presentation.

(11.26) Consider the 2-strand braid

$\sigma_1$ (whose braid closure is an unknot). We have

$\pi_1(X_{\sigma_1})=\langle\alpha,\beta,g|g\alpha g^{-1}=\alpha\beta\alpha^{-1},g\beta g^{-1}=\alpha\rangle$ so the Wirtinger presentation is obtained by setting

$g=1$ :

$\langle\alpha,\beta|\alpha=\alpha\beta\alpha^{-1},\beta=\alpha\rangle.$ We can simplify this to just get

$\langle\alpha\rangle$ , so the fundamental group is

$\mathbf{Z}$ .

I said that this allows us to compute the fundamental group of any knot complement: this is because one can show that any knot is isotopic to a braid closure; a proof of this was first written down by Alexander (1923, ``A lemma on a system of knotted curves'') and it is quite readable.

Pre-class questions

The video claimed that "any braid gives a knot by taking the braid closure". Why was this claim false? What should I have said instead?

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