*Lecture 10 (27.10.2015)*

# Chapter 3: Polynomial Invariants

In this chapter, we will discover very strong link invariants. Similar to the Alexander polynomials, they will be polynomials. However, the main idea is no longer that of colouring arcs of a diagram (with actual colors, or integers mod $p$), but rather that of *splitting the crossings* of a diagram, and studying the resulting patterns.

By splitting a crossing we mean one of the following two procedures

Here the role of the dashed circle is as with the Reidemeister moves – it is not part of the diagram, but indicates that we change the diagram only locally, with the part outside the circle remaining unchanged. But note that the two operations above are **not **Reidemeister moves! Intuitively, they amount to cutting the over- and understrand and tying them back together in such a way that the crossing is removed.

To distinguish between two different types of splittings, we distinguish the four regions between the arcs that meet at a given crossing. We call “A” the regions that are swept over by the overstrand when this is rotated counter-clockwise, and “B” the other two regions, like in this picture:

Correspondingly, we call a splitting *type $A$* if it joins the two “A” regions, and *type $B$* if it joins the two “B” regions:

When working on a full diagram, we have the choice of a type A or type B split at each of the crossings of the diagram.

Definition 3.1:Astateof a diagram $D$ is a choice of type A or type B split at each crossing of $D$.

Remarks:

- Clearly, a diagram $D$ has exactly $2^{|D|}$ states.
- Applying a state $\sigma$ of $D$ to $D$ means to split all crossings of $D$ according to the choices made in $\sigma$. This results in a different diagram, which has no crossings at all. It will thus be a diagram of some $r$-unlink, where $r$ depends on $\sigma$.
- To each state $\sigma$ we can associate three numbers:
- $a_\sigma$: the number of choices “A” in $\sigma$
- $b_\sigma$: the number of choices “B” in $\sigma$
- $r_\sigma$: the number of components of the diagram obtained from applying $\sigma$ to $D$.

- Since at each crossing, we have to choose either “A” or “B”, we have $a_\sigma+b_\sigma=|D|$ for each state $\sigma$ of $D$.

**Example:** Let us consider the diagram

Here we have 1 crossing, and thus $2^1=2$ states. The first state is type A, and results in

with $a=1$, $b=0$, $r=1$. The second state is type B, and results in

with $a=0$, $b=1$, $r=2$.

It is clear from the pictures that the diagrams resulting from applying different states may be inequivalent. We will see more examples further below.

The next step is inspired from statistical mechanics, where one wants to compute certain properties, like the energy, pressure, etc, of a large system, consisting of many atoms, by an averaging procedure. In this analogy, the large system corresponds to a link diagram, and the many atoms to its crossings.

Definition 3.2:

- Let $D$ be a diagram and $\sigma$ a state of $D$. The
weightof $\sigma$ is defined as $$\sigma(D):=A^{a_\sigma}\cdot B^{b_\sigma}\cdot d^{r_\sigma-1}.$$ Here $A,B,d$ are three commuting variables that are not fixed. (Similar to the variable $t$ of the Alexander polynomial)- The
bracket polynomialof a diagram $D$, written as $\langle D\rangle$, is defined as $$\langle D\rangle:=\sum_\sigma \sigma(D),$$ where the sum runs over all states of $D$.

The bracket polynomial is a polynomial in the three independent variables $A,B,d$. From its definition, it follows that all coefficients are positive, the powers of $A$ and $B$ add in each term to $|D|$, and it contains $2^{|D|}$ terms (some of which may coincide).

**Examples:** 1) In the example considered above, we find the bracket polynomial

to be $$A^1B^0d^0+A^0B^1d^1=A+dB.$$ If the crossing in the original diagram is reversed, we find instead $Ad+B.$

2) Let us next consider the $n$-unlink, presented in a diagram consisting of $n$ circles without any crossings. Then we have $2^0=1$ state, which results in just the same diagram (no crossings, hence no A/B choices at all). Thus $a=b=0$ in this case, and $r=n$. Thus the bracket polynomial of such an $n$-unlink diagram is simply $d^{n-1}$, independent of $A$ and $B$.

In particular, the standard diagram of the unknot (=1-unlink), without any crossings, has constant bracket polynomial $1$.

3) A more interesting example is that of a trefoil diagram, with 3 crossings and consequently, $2^3=8$ states. Here we find

The necessary calculations to arrive at this result can be found in these slides.

As this example shows, calculating the bracket polynomial as a state sum can be tedious for large diagrams due to the large number of terms. Nonetheless, in Exercise 9a), you are asked to do so (for a diagram with crossing number 2).

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 $D$. Let $D’$ 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,

We will give the proof next time.

It is often 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.

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.