*Lecture 21 (7.12.2015)*

In the last lecture, we discussed how to pass from braids to links by closing them. We defined a closure map from the braid group $B_n$ on $n$ strands to links, denoted $b\mapsto L(b)$ or $b\mapsto L_n(b)$ if we want to emphasize the number of strands $n$.

In this context, there are two important theorems.

Theorem 5.3 (Alexander’s Theorem):Any link is the closure of some braid. In more detail: Given a link $L$, there exists a number of strands $n\in\mathbb N$ and a braid $b\in B_n$ such that $L=L_n(b)$.

We will not give a proof of Alexander’s Theorem here. The algorithm to find a closed braid representation of a given link $L$ works as follows:

- pick a point inside a diagram of $L$, and give $L$ an orientation.
- follow the strands of $L$ and “throw them over the fixed point” whenever they change the direction (mathematically positive/negative) in which they circulate around the fixed point.
- See these slides for an example, realizing the figure of eight knot as a closed braid.

Example 5.4:We realize a Hopf link $H$ as a closed braid.

These equivalences show $$H=L_2(\sigma_1^2).$$

Exercise 5.5:Pick some knots from a knot catalogue and realize them as closed braids.

It is useful to represent links as closed braids because in this way, tools from group theory become available. However, the representation of a link as a closed braid is *not* unique. To illustrate this, we consider two operations on braids (called *Markov moves*) that do change braids, but do not change the link closures of these braids.

To discuss the first Markov move, we first realize that the closure map has the property $$L(bb’)=L(b’b)\,,\qquad b,b’\in B_n.$$ This has been explained in the lecture, and can be seen by rotating the diagram representing the closed braid. As a consequence, whenever we have two braids $b,g\in B_n$, then $$L(gbg^{-1})=L(g(bg^{-1}))=L((bg^{-1})g)=L(be)=L(b),$$ that is, $b$ and $gbg^{-1}$ close to the same link. However, $b$ and $gbg^{-1}$ can look very different as braids (draw some examples to illustrate this). The operation $b\mapsto gbg^{-1}$ is called *Markov move of type I.*

Whereas the first Markov move takes place on braids of a fixed number $n$ of strands, one increases the number of strands by one for the second Markov move. In fact, let $b\in B_n$ be a braid of $n$ strands, which we can close to a braid as in this picture:

We may also consider $b$ as a braid of $n+1$ strands, by applying the braiding to the first $n$ strands, and keeping the last strand fixed. We can then compose it with the elementary braid $\sigma_n$, flipping the strands $n$ and $n+1$. Closing the resulting braid $b\sigma_n$ to a link looks as in this picture:

Since the kink in the inner loop can be removed by a type I Reidemeister move, we see that $b\sigma_n$ has the same link closure as $b$, i.e. $$L_n(b)=L_{n+1}(b\sigma_n).$$ The operation $b\mapsto b\sigma_n$ (with $b\in B_n$) is called *Markov move of type II.*

As discussed, both Markov moves do not change the link closure of braids. Markov’s Theorem says that also the converse is true.

Theorem 5.6 (Markov’s Theorem):Two braids $b,b’$ close to the same link, i.e. $L(b)=L(b’)$, if and only if $b$ can be transformed into $b’$ by a finite sequence of Markov moves of types I and II.

As the last topic of these lectures, we want to explain how these facts can be used to produce further link invariants. Similar to our procedure in Chapter 4, we will first consider a general scenario (just as we considered the general concept of a partition function) and then a specific example (just as we considered the example of the Kauffman bracket being recovered from the partition function).

The main idea is to consider suitable *representations of the braid groups.* By this we mean the following: One picks invertible matrices $R_i$ (one matrix per elementary braid), such that $$R_iR_j=R_jR_i\,,\qquad |i-j|>1,$$ and $$R_iR_{i+1}R_i=R_{i+1}R_iR_{i+1}.$$ One can then form a “concrete realization” of braids in terms of matrices by assigning matrices to braids: For the elementary braids, one assigns $$\sigma_i\mapsto R_i\,,\qquad \sigma_i^{-1}\mapsto R_i^{-1}\,,\qquad e\mapsto 1,$$ where $1$ denotes the identity matrix. Products of elementary matrices are then mapped to matrices according to $$b:=\sigma_{i_1}^{\pm 1}\sigma_{i_2}^{\pm 1}\cdots\sigma_{i_p}^{\pm 1}\mapsto R_{i_1}^{\pm 1}R_{i_2}^{\pm 1}\cdots R_{i_p}^{\pm 1}=:\rho(b).$$ This gives us a map $\rho$ from braids to matrices, which has the properties $$\rho(bb’)=\rho(b)\rho(b’)\,,\qquad\rho(b^{-1})=\rho(b)^{-1}.$$ Such a map is called a *representation*, it plays a fundamental role in group theory.

In this context, the general situation for obtaining link invariants from braids is the following one.

Lemma 5.7:Suppose the $R_i$ lie in some algebra $\cal A$ (that means, with $A,B\in\cal A$, also $AB$, $A+B$, and $c\cdot A$, where $c$ is an arbitrary complex number, all lie in $\cal A$), and there exists a linear map $$\varphi:{\cal A}\to\mathbb C$$ such that $$\varphi(AB)=\varphi(BA)\qquad\text{for all } A,B\in\cal A$$ and $$\varphi(\rho(b)R_n)=\alpha\cdot\varphi(\rho(b))\qquad\text{for all} b\in B_n,$$ with some fixed parameter $\alpha$. Then $$M(b):=\alpha^{-(n-1)}\cdot\varphi(\rho(b))\,,\qquad b\in B_n\,,$$ is invariant under both Markov moves. Consequently, the assignment $$I(L):=M(b_L),$$ where $L$ is a link and $b_L$ some braid which closes to $L$, is well-defined and an invariant of links under ambient isotopy.

*Proof:* We check invariance under both Markov moves. For $b,g\in B_n$, we find $$M(gbg^{-1})=\alpha^{-(n-1)}\varphi(\rho(gbg^{-1}))=\alpha^{-(n-1)}\varphi(\rho(g)\rho(b)\rho(g)^{-1})=\alpha^{-(n-1)}\varphi(\rho(b))=M(b),$$ where we have used the properties of $\varphi$ and $\rho$. For $b\in B_n$, we have $b\sigma_n\in B_{n+1}$, and thus find $$M(b\sigma_n)=\alpha^{-n}\varphi(\rho(b)R_n)=\alpha^{-n}\alpha\varphi(\rho(b))=M(b).$$ This shows that $M$ is indeed invariant under both Markov moves. Now let $L_1,L_2$ be two equivalent topological links, arising from braids as $L_1=L(b_1)$, $L_2=L(b_2)$. By Markov’s theorem, this implies that there is a finite sequence of Markov moves taking $b_1$ to $b_2$. Since $M$ is invariant under Markov moves, this implies $M(b_1)=M(b_2)$, and thus $I(L_1)=I(L_2)$, i.e. $I$ is a link invariant. ■

We next have to show that the assumptions made in this lemma can actually be realized, and lead to interesting invariants. To do so, we take the algebra $\cal A$ to be the Temperley-Lieb algebra ${\cal T}(d)=\cal A$ which we have encountered before. In an abstract formulation, this is an algebra which consists of all sums and products of elements $U_i$, $i\in\mathbb N$, satisfying the relations $$\begin{align*}U_iU_j&=U_jU_i\,,\qquad|i-j|>1\\U_iU_{i\pm 1}U_i&=U_i\\U_i^2&=d\cdot U_i,\end{align*}$$ where $d$ is some complex parameter.

We have seen explicit realizations of such relations before (at least for a small number of $U_i$’s). For example, as in Problem 16, we can set $$\begin{align*} U_1&=\frac{1}{\sqrt{N}}\sum_{i,j=1}^N E_{ij}\otimes 1\otimes 1\otimes …\\U_2&=\sqrt{N}\sum_{i=1}^NE_{ii}\otimes E_{ii}\otimes1\otimes 1\otimes …\\U_3&=\frac{1}{\sqrt{N}}\sum_{i,j=1}^N 1\otimes E_{ij}\otimes 1\otimes 1\otimes …\\U_4&=\sqrt{N}\sum_{i=1}^N1\otimes E_{ii}\otimes E_{ii}\otimes1\otimes 1\otimes …\\ U_5&=…\end{align*}$$Here $1,E_{ij}$ denote the identity matrix and canonical matrix units of the $N\times N$ matrices. (We use $N$ instead of $n$ in order not to confuse the matrix size with the number of strands.)

The $\otimes …$ in these formulas means that to the right of this symbol, we have infinitely many more factors of $\otimes 1$. To discuss some subleties of this construction would require to talk about von Neumann algebras, which would be very beautiful, but also beyond the scope of this course. For actual calculations with expressions of the above form, however, note that for a finite product of $U_i$’s, we are effectively working with a finite number of tensor factors because far enough to the right we only have tensor factors of identity matrices which do not matter for multiplications.

A slightly enhanced version of Problem 16 is the following.

Exercise 5.8:Verify that the matrices $U_i$ satisfy the relations of the Temperley-Lieb algebra, with $$d=\sqrt{N}.$$