*Lecture 11 (2.11.2015)*

We next derive the so-called *recursion formula for the bracket polynomial.*

Proposition 3.3:

- The bracket polynomial satisfies the relation

(This equation has to be read as follows: Let $D$ be a diagram, and pick a crossing of $D$. Call $D’$ be the diagram you obtain by splitting this crossing as type A, and let $D”$ be the diagram that you obtain by splitting this crossing as type $B$. Then the bracket polynomials $\langle D\rangle$,$\langle D’\rangle$,$\langle D”\rangle$ are related by $\langle D\rangle=A\cdot\langle D’\rangle+B\cdot\langle D”\rangle$. But the graphical shorthand equation above is probably much easier to remember.)- Let $D$ be a diagram and let $D_0$ denote a diagram of the unknot with no crossings. Let $D\sqcup D_0$ denote a diagram consisting of the union of $D$ and $D_0$, with no crossings between $D$ and $D_0$. Then $\langle D\sqcup D_0\rangle=d\cdot\langle D\rangle$. Or, as a shorthand

*Proof.* Let $D$ be a diagram, and pick a crossing of $D$. Call $D’$ be the diagram you obtain by splitting this crossing as type A, and let $D”$ be the diagram that you obtain by splitting this crossing as type $B$. Now, if $\sigma$ is a state of $D$, then it gives us a state $\sigma’$ of $D’$ and a state $\sigma”$ of $D”$ by making all the remaining A/B choices listed in $\sigma$. Conversely, if a state $\sigma’$ of $D’$, or a state $\sigma”$ of $D”$ is given, this fixes a state $\sigma$ of $D$ by adding the choice “A” respectively “B” at the distinguished crossing. Thus the states of $D$ are in 1:1 correspondence with the disjoint union of the states of $D’$ and $D”$.

Let $\sigma’$ be a state of $D’$, and let $\sigma$ denote the corresponding state of $D$. Then we have $$a_\sigma=a_{\sigma’}+1,\qquad b_\sigma=b_{\sigma’},\qquad r_\sigma=r_{\sigma’}.$$ Hence the weights of $\sigma$ and $\sigma’$ are related by $$\sigma(D)=A^{a_\sigma}B^{b_\sigma}d^{r_\sigma-1}=A\cdot A^{a_{\sigma’}}B^{b_{\sigma’}}d^{r_{\sigma’}-1}=A\cdot \sigma'(D’).$$ Analogously, if $\sigma”$ is a state of $D”$, and $\sigma$ the corresponding state of $D$, we find $$\sigma(D)=B\cdot \sigma”(D”).$$ Thus $$\langle D\rangle=\sum_\sigma\sigma(D)=\sum_{\sigma’}\sigma(D)+\sum_{\sigma”}\sigma(D)=\sum_{\sigma’}A\sigma'(D’)+\sum_{\sigma”}B\sigma”(D”)=A\langle D’\rangle+B\langle D”\rangle.$$

For the second part, the argument is similar. Here we observe that each state on $D$ fixes a state on $D\sqcup D_0$ and vice versa since $D_0$ has no crossing and therefore no additional A/B choices. If $\sigma$ is a state of $D$, and $\hat{\sigma}$ the corresponding state of $D\sqcup D_0$, then we have $$a_\sigma=a_{\hat{\sigma}},\qquad b_\sigma=b_{\hat{\sigma}},\qquad r_\sigma+1=r_{\hat{\sigma}},$$ because $\hat{\sigma}$ results in one more component, namely $D_0$. Thus $$\hat{\sigma}(D\sqcup D_0)=A^{a_\sigma}B^{b_\sigma}d^{r_\sigma+1}=d\cdot\sigma(D).$$ This implies $$\langle D\sqcup D_0\rangle=\sum_{\hat{\sigma}}\hat{\sigma}(D)=\sum_\sigma d\sigma(D)=d\cdot\langle D\rangle.$$■

It is often times easier and quicker to calculate bracket polynomials with the help of the recursion relation, instead of directly from the definition. The recursion relation relates the bracket polynomial of the diagram under consideration to bracket polynomials of diagrams with one less crossing, which might be known already. Furthermore, it often happens that one ends up with several brackets of identical diagrams, which can then be grouped together, saving computation time.

As an example, we can use the recursion relation to calculate

But the bracket polynomial in the last line we have calculated already, it is $A+dB$. Thus the bracket polynomial of the double twisted unknot is $(A+dB)^2$. See Problem 9b) for an exercise on the recursion relation.

We now want to investigate the behavior of the bracket polynomial under Reidemeister moves. From the definition of the bracket polynomial, it is clear that it is invariant under type 0 moves (planar isotopy) – such moves change only the shape of the arcs, but the structure of the crossings, which is all that matters for the bracket polynomial, remains unchanged.

Let us begin with Reidemeister move II, i.e. we consider

where I have written down the A/B regions at the upper crossing for clarity.

The notation has be understood as above: We mean here the bracket polynomial $\langle D\rangle$ of some diagram $D$, but we only draw a small portion of $D$. In the manipulations to follow, based on Prop. 3.3, all parts of the diagram that are not shown (and which are arbitrary) do not change.

Computing with Proposition 3.3, we find

As usual,we are interested in *invariants*, i.e. we would like to find properties of links/knots that do not depend on the chosen diagram, but are properties of the links/knots themselves. Such properties must be invariant under the Reidemeister moves. From the above calculation, we see that the Reidemeister move

leaves the bracket polynomial invariant if the term $AB$ on the second bracket is one, and the term $A^2+dAB+B^2$ on the first bracket is zero. This is the case when we set $$B=A^{-1}$$ and $$d=-A^2-A^{-2}.$$

For Reidemeister move III, i.e.

it was demonstrated in the lecture that when $B=A^{-1}$ and $D=-A^2-A^{-2}$, then also Reidemeister moves of type III do not change the bracket polynomial. In fact, when these relations hold between $A,B,d$, then move II does not change the bracket polynomial. As the following calculation shows, this also implies invariance of the bracket polynomial under move III:

It remains to consider Reidemeister move I (“inserting a kink”, see lecture 4).

Here we find:

If we put $B=A^{-1}$ and $d=-A^2-A^{-2}$, the prefactor is $Ad+B=-A^3$. Since this is not equal to 1, we do *not* have invariance under Reidemeister moves of type I (similar to writhe, which was only an invariant under *regular *isotopy). We also mention that by a completely analogous calculation, one finds for the opposite kink (the one we used in our definition of move I)

Here we observe that setting $B=A^{-1}$ and $d=-A^2-A^{-2}$ turns the prefactor is into $Bd+A=-A^{-3}$.

We gather all these results in a definition and proposition.

Definition and Proposition 3.4:Given a diagram $D$, we define theKauffman bracketof $D$ as the bracket polynomial of $D$, but with the substitutions $B=A^{-1}$ and $d=-A^2-A^{-2}$ made. Thus the Kauffman bracket only depends on a single variable $A$.As shown above, the Kauffman bracket is an invariant under regular isotopy, that is, it does not change under Reidemeister moves of type 0, II, III. Under move I it behaves as

We will need the Kauffman bracket only in an intermediate stage of our studies, and therefore do not introduce an extra symbol for it. In fact, in many texts, the bracket polynomial with the two substitutions $B=A^{-1}$ and $d=-A^2-A^{-2}$ made is still referred to as the bracket polynomial (see, for example, here). For clarity, I will in this lecture always say “bracket polynomial” when I mean the polynomial in the three independent variables $A,B,d$, and “Kauffman bracket” when I mean the polynomial with the two substitutions made. Note that in contrast to the bracket polynomial, the Kauffman bracket only depends on a single variable, namely, $A$. Because we have replaced $B$ by $A^{-1}$, the Kauffman polynomial is a Laurent polynomial, i.e. it may contains positive and negative powers of $A$ (as the Alexander polynomial). Furthermore, since we substituted $d=-A^2-A^{-2}$, we might have negative coefficients in the Kauffman bracket (which is not possible for the bracket polynomial), and the coefficients do not necessarily sum up to a power of 2, as it was the case for the bracket polynomial.