Lecture 22 (8.12.2015)

In this final lecture we will explain the connection between the braid group and the Jones polynomial. The idea is to use the general scenario described in Lemma 5.7, with the Temperley-Lieb algebra ${\cal A}={\cal T}(\sqrt{N})$ in the concrete realization explained last time, with $d=\sqrt{N}$.

We consider the R-matrices $$R_i=A+A^{-1}U_i,$$ which are familiar from our investigation of the Kauffman bracket via partition functions. As shown in lecture 18, these matrices are invertible and satisfy $R_iR_{i+1}R_i=R_{i+1}R_iR_{i+1}$ if $-A^2-A^{-2}=d=\sqrt{N}.$ Moreover, it follows from the relations of the Temperley-Lieb algebra that $R_iR_j=R_jR_i$ for $|i-j|>1$.

Of course these properties remain true when we multiply all $R_i$ by a fixed number. In the following, it will be convenient to work with $E_i:=U_i/d$, because these matrices satisfy $E_i^2=E_i$ and thus $E_i(1-E_i)=0$. As usual, we set $t=A^{-4}$, which implies $d=-t^{1/2}-t^{-1/2}$, and take our R-matrices to be $$R_i=-\sqrt{t}(1+A^{-2}U_i)=-\sqrt{t}(1+A^{-2}dE_i)=-\sqrt{t}(1-(1+t)E_i)=\sqrt{t}(tE_i-(1-E_i)).$$

We thus have a representation of the braid group in ${\cal T}$, fixed by $$\rho(\sigma_i):=R_i.$$

What is still missing is the functional $\varphi$ from Lemma 5.7. Using the concrete realization of the Temperley-Lieb algebra, we define $\varphi:{\cal T}\to\mathbb C$ by taking the normalised trace ${\rm Tr}(\cdot)/N$ in each tensor factor. To calculate $\varphi$, it is helpful to note that $$\frac{{\rm Tr}(1)}{N}=1$$ (where $1$ denotes the $(N\times N)$ identity matrix), and $$\frac{{\rm Tr}(E_{ij})}{N}=\frac{\delta_{ij}}{N}.$$ For example, for $U_1$ this gives $$\begin{align*}\varphi(U_1)&=\varphi\left(\frac{1}{\sqrt{N}}\sum_{i,j=1}^N E_{ij}\otimes 1\otimes 1\otimes …\right)\\&=\frac{1}{\sqrt{N}}\sum_{i,j=1}^N \frac{{\rm Tr}(E_{ij})}{N}\cdot\frac{{\rm Tr}(1)}{N}\cdot \frac{{\rm Tr}(1)}{N}…\\&=\frac{1}{\sqrt{N}}\sum_{i,j=1}^N \frac{\delta_{ij}}{N}\cdot 1\cdot 1…\\&=\frac{1}{\sqrt{N}}.\end{align*}$$Similarly, for $U_2$ we find $$\begin{align*}\varphi(U_2)&=\varphi\left(\sqrt{N}\sum_{i=1}^N E_{ii}\otimes E_{ii}\otimes 1\otimes …\right)\\&=\sqrt{N}\sum_{i=1}^N \frac{{\rm Tr}(E_{ii})}{N}\cdot\frac{{\rm Tr}(E_{ii})}{N}\cdot\frac{{\rm Tr}(1)}{N}\cdot …\\&=\sqrt{N}\sum_{i=1}^N \frac{1}{N^2}\\&=\frac{1}{\sqrt{N}}.\end{align*}$$

Analogously, $\varphi(T)$ can be computed for any element $T$ of the Temperley-Lieb algebra. Since all but finitely many tensor factors are always identity matrices, which just give factors of one, this is a well-defined functional.

Because $\varphi$ is (up to a factor) the trace, it is also clear that $$\varphi(TS)=\varphi(ST)$$ for all $T,S\in{\cal T}$. Furthermore, one can check ($\star$ Exercise) that $\varphi(1)=1$ and $$\varphi(TU_n)=N^{-1/2}\varphi(T)\,,$$ if $T$ is a linear combination of products involving only $U_1,…,U_{n-1}$.

For such $T$, we then find $$\begin{align*}\varphi(TR_n) &=\varphi(T(-\sqrt{t})(1+A^{-2}U_n))\\ &=-\sqrt{t}\, \left(\varphi(T)+A^{-2}\varphi(TU_n) \right)\\ &=-\sqrt{t}\left(1+\frac{1}{\sqrt{N}A^2} \right)\,\varphi(T)\,.\end{align*}$$ In other words, $\varphi$ satisfies the property required in Lemma 5.7, with $$\alpha=-\sqrt{t}\left(1+\frac{1}{\sqrt{N}A^2} \right)=-\frac{\sqrt{t}}{1+t}\,.$$ Here we have used $\sqrt{N}=d=-A^2-A^{-2}$ and $A^{-4}=t$.

Let us investigate the resulting link invariant at an example — the trefoil $3_1$. We saw before that the trefoil is the closure of $\sigma_1^3$ in $B_2$, the braid group on $n=2$ strands. Since $E_1^2=E_1$ and $E_1(1-E_1)=0$, we have $$R_1^3=(\sqrt{t}(tE_1-(1-E_1)))^3=t^{3/2}(t^3E_1-(1-E_1)).$$

We therefore obtain $$\begin{align*}I(3_1)&=\alpha^{-1}\varphi(R_1^3)\\&=-\frac{1+t}{\sqrt{t}}\varphi(t^{3/2}(t^3E_1-(1-E_1)))\\&=-(1+t)t\varphi(t^3E_1-(1-E_1))\\&=-t^4+t^3+t\end{align*}$$In the last step, we have used $$\varphi(E_1)=d^{-1}\varphi(U_1)=\frac{1}{d^2}=\frac{t}{(1+t)^2}.$$

We thus see that this procedure gives, in the case of the trefoil, the Jones polynomial!

In fact, this is always true — for any link $L$, the invariant calculated along these lines is nothing but the Jones polynomial $V_L(t)$. This is how the Jones polynomial was originally discovered, via von Neumann algebras and braid group representations. The formulation in terms of the bracket polynomial came later.

Proposition 5.9: With the above choices of algebra ${\cal A}={\cal T}(d)$ and functional $\varphi$, the resulting invariant $I$ (see Lemma 5.7) coincides with the Jones polynomial.

Exercise 5.10: Revisit Problem 11 from this new point of view: Choose some of the links for which you already know the Jones polynomial, and compute it again by braid group representation method explained here.