Lecture 19 (30.11.2015)

We proceed with our example for an $R$-matrix and corresponding partition function from the last lecture. Recall our setup: We pick some finite index set $I$ with $|I|=n$ elements, a parameter $A\in\mathbb C$ such that $$A^2+A^{-2}+n=0,$$ and tensors $R,{\bar R}\in M_{I^2}$ which are inspired by the recursion relation of the Kauffman bracket:

kauffman-graph

In formulas, this means $$R^{ab}_{cd}=A\delta^a_c\delta^b_d+A^{-1}\delta^{ab}\delta_{cd}$$ and $${\bar R}^{ab}_{cd}=A\delta^{ab}\delta_{cd}+A^{-1}\delta^a_c\delta^b_d.$$

We checked in the last lecture that due to the special form of these matrices, we do not need arrows (orientation) on their diagrams – they will correspond unambiguously to formulas nonetheless (this is not true for general $R$).

We now have to check that $I,R,\bar R$ satisfy the three conditions of Theorem 4.3. We can do this either graphically, or algebraically. Let us start with the graphical check first.

Considering move II, we have (with explanations below)

check1

In the first step, we considered the upper crossing and realized it to belong to $\bar R$. Now $\bar R$ consists of two terms, one with a horizontal split with factor $A$, and one with a vertical split with factor $A^{-1}$, as indicated. In the next step, we considered the remaining crossing in the two diagrams, which belongs to $R$ instead of $\bar R$, and applied the definition of $R$. This results in four terms, one of which contains a closed loop. But a closed loop is the trace of the identity matrix, which is $n$ because our index set has $n$ elements. This explains the third line. In the last line we used that $n+A^2+A^{-2}=0.$

For move III, we have (with explanations below)

check2

In the first step, we resolved the upper left $R$-crossing similar to what we did for move II. In the second step, we exploited the fact that we already have shown invariance under type II moves, and applied two type II moves to the second diagram. The sum of the two diagrams can then be reassembled to the final diagram, and the graphical proof of invariance under type III moves is complete.

We write down this result as a proposition.

Proposition 4.4: Let $I$ be a finite index set, $A\in\mathbb C$ a parameter such that $$A^2+A^{-2}+|I|=0,$$ and define $R\in M_{I^2}$ by $$R^{ab}_{cd}=A\cdot\delta^a_c \delta^b_d+A^{−1}\cdot\delta^{ab}\delta_{cd}.$$

  • Then $R$ is invertible, with inverse $$(R^{-1})^{ab}_{cd}=A^{-1}\cdot\delta^a_c \delta^b_d+A\cdot\delta^{ab}\delta_{cd},$$ and $I,R,\bar R=R^{-1}$ satisfy the three conditions of Theorem 4.3.
  • The partition function $Z=Z_{I,R,\bar R}$ reproduces the Kauffman bracket $\langle\cdot\rangle$ up to a factor $(-A^2-A^{-2})$,
    $$Z(D)=(-A^2-A^{-2})\cdot\langle D\rangle$$for each oriented diagram $D$.

It is instructive to verify these properties of $R$ also algebraically (that is, in formulas). To this end, we put $n=|I|=-A^2-A^{-2}$ as before, and define a tensor $U\in M_{I^2}$ by $$U^{ab}_{cd}:=\delta^{ab}\delta_{cd}.$$ We have encountered this tensor in Exercise 4.1 before, where we showed that $$U=\sum_{i,j\in I}E_{ij}\otimes E_{ij},$$ with $E_{ij}$ the canonical matrix units of $M_{I^2}$.

Note that $$\begin{align*}U^2&=\left(\sum_{i,j}E_{ij}\otimes E_{ij}\right)\left(\sum_{k,l}E_{kl}\otimes E_{kl}\right)\\&=\sum_{i,j,k,l}E_{ij}E_{kl}\otimes E_{ij}E_{kl}\\&=\sum_{i,j,k,l}\delta_{jk}^2E_{il}\otimes E_{il}\\&=\sum_{i,j,l}E_{il}\otimes E_{il}\\&=n\sum_{i,l}E_{il}\otimes E_{il}\\&=n\cdot U,\end{align*}$$i.e. $U^2=n\cdot U$. We now introduce the shorthand notations $$U_1:=U\otimes 1,\qquad U_2:=1\otimes U,$$where $1$ denotes the identity matrix in $M_I$ (one upper and one lower index).

We claim that with these definitions, $U_1,U_2$ satisfy the relations of the so-called Temperley-Lieb algebra $$\begin{align*}U_1^2&=n\,U_1\,,\qquad U_2^2=n\,U_2\,,\\ U_1U_2U_1&=U_1\,,\qquad U_2U_1U_2=U_2\,.\end{align*}$$ In fact, the first equation follows from $U^2=nU$ because $$U_1^2=(U\otimes 1)(U\otimes 1)=U^2\otimes 1=n\cdot U\otimes 1=n\cdot U_1.$$ The second equation, $U_2^2=n\,U_2$, follows analogously. The third equation was demonstrated in lecture 17, and the last one follows by a completely analogous calculation.

We now come to the algebraic check of the three conditions of Thm. 4.3.

  • $R\bar R=(A+A^{-1}U)(A^{-1}+AU)=1+A^2U+A^{-2}U+U^2=1+(A^2+A^{-2}+n)U=1$, where we have used $U^2=nU$ and $n+A^2+A^{-2}=0$.
  • The second “type II move condition”: Inserting the definitions of $R,\bar R$, and carrying out sums over Kronecker deltas, we find $$\begin{align*}\sum_{i,j}{\bar R}^{bi}_{aj}R^{cj}_{di}&=\sum_{i,j}(A^{-1}\delta^b_a\delta^i_j+A\delta^{bi}\delta_{aj})(A\delta^c_d\delta^j_i+A^{-1}\delta^{cj}\delta_{di})\\&=\sum_{i,j}\left(\delta^b_a\delta^c_d\delta^i_j\delta^j_i+A^2\delta^{bi}\delta_{aj}\delta^c_d\delta^j_i+A^{-2}\delta^b_a\delta^i_j\delta^{cj}\delta_{di}+\delta^{bi}\delta_{aj}\delta^{cj}\delta_{di}\right)\\&=\sum_{i}\delta^b_a\delta^c_d\delta^i_i+A^2\sum_i\delta^{bi}\delta_{ai}\delta^c_d+A^{-2}\sum_i\delta^b_a\delta^{ci}\delta_{di}+\delta^{ca}\delta_{db}\\&=n\delta^b_a\delta^c_d+A^2\delta_{ab}\delta^c_d+A^{-2}\delta^b_a\delta^{cd}+\delta^{ca}\delta_{db}\\&=(n+A^2+A^{-2})\delta^b_a\delta^{cd}+\delta^{ca}\delta_{db}\\&=\delta^{ca}\delta_{db}.\end{align*}$$Here we have used that for a Kronecker delta, the position of the indices is immaterial, i.e. $$\delta^i_j=\delta^j_i=\delta^{ij}=\delta^{ji}=\delta_{ij}=\delta_{ji},$$ in each case, this is 1 for $i=j$ and $0$ for $i\neq j$.
    This shows that the second condition of Thm. 4.3 holds.
  • For the last condition, we use the Temperley-Lieb relations to compute $$\begin{align*}(R\otimes 1)(1\otimes R)(R\otimes 1)&=(A+A^{-1}U_1)(A+A^{-1}U_2)(A+A^{-1}U_1)\\&=A^3+AU_1+AU_2+AU_1+A^{-1}U_1U_2+A^{-1}U_2U_1+A^{-1}U_1^2+A^{-3}U_1U_2U_1\\&=A^3+A^{-1}(U_1U_2+U_2U_1)+AU_1+AU_2+A^{-1}(n+A^2+A^{-2})U_1\\&=A^3+A^{-1}(U_1U_2+U_2U_1)+AU_1+AU_2.\end{align*}$$ The expression in the last line does not change when we exchange $U_1$ with $U_2$. But exchanging $U_1$ and $U_2$ is the same as exchanging $R\otimes 1$ and $1\otimes R$. Thus $$(R\otimes 1)(1\otimes R)(R\otimes 1)=(1\otimes R)(R\otimes 1)(1\otimes R),$$that is, $R$ solves the Yang-Baxter equation.

This completes the algebraic proof of Prop. 4.4.

We have thus found a matrix model for specializations of the Kauffman bracket (specialization because we have to consider the specific values of $A$ for which $A^2+A^{-2}+n=0$, whereas in the original Kauffman bracket, the value of $A$ was unrestricted). This is just one example of $I,R,\bar R$ that leads to an invariant (under regular isotopy) via Theorem 4.3.

Many more matrix models exist. In particular, it is possible to overcome the specialization to $A^2+A^{-2}+n=0$ by a more advanced matrix model. We refer to Kauffman’s book (part I, chapter 9) for a discussion of this point.

From Theorem 4.3, we see that knot theory connects to the study of algebraic structures: To find more invariants, one has to consider certain equations, such as the Yang-Baxter equation, and their solutions. This point of view connects to many fields in mathematics and physics, such as Hopf algebras, Hecke algebras, quantum groups, but also statistical mechanics. Because of lack of time, we can not go into these connections in more depth here.

Finally, there is a deep relation between knot theory, the Yang-Baxter equation, and the so-called braid group. This will be the topic of the next and final chapter.