## Lecture 13 (9.11.2015)

Starting from the bracket polynomial, we introduced the Kauffman bracket$\langle\cdot\rangle$ and the normalized bracket polynomial $X$. Now we do one more step, a simple change of variables, to arrive at the Jones polynomial. It is named after its inventor Vaughan Jones (1984), and therefore called $V$.

Definition 3.8: The Jones polynomial $V_L$ of an oriented link $L$ is defined as $$V_L(t):=X_L(t^{-1/4}).$$

Thus the Jones polynomial is basically the same as the normalized bracket polynomial, up to the change of variables $$A=t^{-1/4}\;\;\Longleftrightarrow\;\; t=A^{-4}.$$ This change of variables has its roots in the original invention of the Jones polynomial, which was via a very different route, involving operator algebras and the braid group.

As examples, we calculate the Jones polynomials of Hopf links and trefoils. One way of doing this is to go via the Kauffman bracket, then to the normalized bracket polynomial, and then via the change of variables $t=A^{-4}$ to the Jones polynomial. (But we will see a better way of doing it below.)

We begin with the Hopf link $H$
and first compute the Kauffman bracket, using its recursion relation. Since the Kauffman bracket does not depend on the orientation, we drop the orientation in this step.

Here we have used the recursion relation for the Kauffman bracket two times, and then the fact that the standard unknot diagram has Kauffman bracket 1, and the the standard diagram of the 2-unlink has Kauffman bracket $(-A^2-A^{-2})$ (cf. the previous lectures).

To go from this polynomial to the normalized bracket polynomial, we need the writhe of our oriented link, which we easily see to be $-2$ (two negative crossings). Hence the normalized bracket polynomial differs from the Kauffman bracket by a factor of $(-A^3)^2=A^6$, and we get $$X_H(A)=A^6\cdot(-A^4-A^{-4})=-A^{10}-A^2.$$ To obtain the Jones polynomial of $H$, we have to make the substitution $A^{-4}=t$, which results in $$V_H(t)=-(t^{-1/4})^{10}-(t^{-1/4})^2=-t^{-5/2}-t^{-1/2}.$$

Alternatively, we could equip the Hopf link with a different orientation, like this one
We call this Hopf link $\tilde{H}$. Since $H$ and $\tilde{H}$ only differ in their orientation, they have the same Kauffman bracket $\langle H\rangle=\langle\tilde{H}\rangle=-A^4-A^{-4}$, as just computed. To find $X_{\tilde{H}}(A)$ and $V_{\tilde{H}}(t)$, we check that the writhe of $\tilde{H}$ is $w(\tilde{H})=+2$, and thus $$X_{\tilde{H}}(A)=(-A^3)^{-2}(-A^4-A^{-4})=-A^{-2}-A^{-10},$$ and $$V_{\tilde{H}}(t)=-t^{5/2}-t^{1/2}.$$

As another example, we consider an oriented trefoil, called $T$,
Note that there exist two different trefoils (this one and its mirror image), so it is important to be clear about which one you mean. We recall from the previous lecture that the normalized bracket polynomial of $T$ is $$X_T(A)=A^{-4}+A^{-12}-A^{-16}.$$ Changing variables, we find its Jones polynomial to be $$V_T(t)=X_T(t^{-1/4})=t+t^3-t^4.$$

These calculations give examples how to compute Jones polynomials. They are, however, a bit tedious because one first has to compute the Kauffman bracket, then the writhe, then the normalized bracket polynomial, and only then the Jones polynomial. It would be better if one could compute $V$ directly. This is indeed possible, as we will learn (among other things) from the following theorem.

Theorem 3.9: The Jones polynomial $V$ has the following properties.

1. $V$ is an invariant of oriented links, and a Laurent polynomial in $t^{1/2}$.
2. The Jones polynomial of the unknot is 1.
3. $V$ satisfies the skein relation
4. The Jones polynomial of the disconnected union of an arbitrary link $L$ and the unknot $U$ is $$V(L\sqcup U)=(-t^{1/2}-t^{-1/2})\cdot V(L).$$
5. The Jones polynomial of the mirror image $L^*$ of a link $L$ is $$V_{L^*}(t)=V_L(t^{-1}).$$

Moreover, the Jones polynomial is uniquely characterized by properties 1)-3).

Partial proof and remarks: 1) The first part of the first statement is clear: We already know that the normalized bracket polynomial is an invariant of oriented links. As the Jones polynomial differs from that one simply by changing variables, it also is an invariant of oriented links. The second part of 1), claiming that $V_L$ is always a Laurent polynomial in $t^{1/2}$, means that it contains only powers of the form $t^{s}$, where $s\in\{0,\pm\frac{1}{2},\pm 1,\pm\frac{3}{2},\pm 2,…\}$. From the substitution $A=t^{-1/4}$, one could expect also quarter powers like $\pm\frac{1}{4}$ to appear, but that is not the case. We do not give a proof of this, see, for example, the lecture notes by Roberts, p. 55.

2) This is clear because we already know that the normalized bracket polynomial of the unknot is 1.

3) Before we give a proof of this relation, let us point out some differences to the recursion relation of the Kauffman bracket,
The first difference is that the Kauffman bracket does not care about orientations, whereas for the Jones polynomial, this is important. Both, the recursion relation of the Kauffman bracket, and the skein relation of the Jones polynomial, consider three different configurations at a single crossing of a diagram. In the case of the Kauffman bracket, these configurations are the crossing and its two different splits. In the case of the Jones polynomial, these configurations are the crossing (with orientation), the flipped crossing (i.e. over- and understrand are exchanged, but the orientation is kept), and one split. The split is distinguished by the orientation, it is that split of the crossing in which the arrows match over the splitting. In the picture below, we have the oriented crossing on the left, which in the Jones skein relation is replaced by the first split crossing (picture in the middle), whereas the second split crossing (picture on the right, where the arrows do not match) does not appear.

After these remarks, we turn to the proof of the skein relation. We begin by writing down the Kauffman recursion relation for splitting an over- and an undercrossing. At this stage, no orientation is required:

Notice how the splittings of these crossings differ because the A and B regions are exchanged between the first and second line. We next multiply the first equation by $A$, the second equation by $A^{-1}$, and substract them. This yields
We now switch to the normalized bracket polynomial, for which we need to include an orientation. Since the Kauffman bracket does not depend on the orientation, we can also write
Let us denote by $w$ the writhe of the diagram on the right hand side, where the crossing is split, by $w’$ the writhe of the first diagram on the left hand side, and by $w”$ the writhe of the second diagram of the left hand side. Since the first crossing is positive, and the second one is negative, we have $$w’=w+1,\qquad w”=w-1.$$ We now multiply our equation by $(-A^3)^{-w}=(-A^3)^{-(w’-1)}=(-A^3)^{-(w”+1)}$. This gives
Finally, we multiply the equation by $-1$ and substitute $A^{-4}=t$. This gives the claimed skein relation for the Jones polynomial.

4) This relation follows from the analogous equation for the Kauffman polynomial and the observation that the writhe of $L\sqcup U$ coincides with the writhe of $L$ because the crossings of these two links coincide.

5) This is clear because $$V_{L^*}(t)=X_{L^*}(t^{-1/4})=X_L((t^{-1/4})^{-1})=X_L((t^{-1})^{-1/4})=V_L(t^{-1}).$$

Finally, the claim that the Jones polynomial is uniquely characterized by 1)-3), is the hardest part of this theorem. We do not give a proof here. ■

We now give examples how to work with the skein relation to compute Jones polynomials. See also problem 11 for further examples. It is important to remember that in the Jones’ skein relation,

• the positive crossing comes with a factor $t^{-1}$
• the negative crossing comes with a factor $-t$
• the split crossing comes with a factor $(t^{1/2}-t^{-1/2})$, and is split is made according to the orientation.

We begin with the Hopf link $H$ we considered earlier in this lecture,

and would like to verify its Jones polynomial $V_H(t)=-t^{-5/2}-t^{-1/2}$ that we computed earlier. To this end, we insert this link into the skein relation, picking one crossing. The upper crossing of the shown diagram of $H$ is negative. Hence applying the skein relation to the upper crossing results in

Now it is important to remember that the Jones polynomial is a link invariant. This means that when evaluating the Jones polynomial of an (oriented) diagram, we are allowed to change the diagram by any (oriented) Reidemeister move (of type 0,I,II,III), without changing the Jones polynomial. Looking at the equation above, we see on the right hand side that the shown diagram is a diagram of the unknot: Applying a type I move to the middle part show the equivalence

Hence

Similarly, the second diagram on the left hand side of the skein relation can be simplified by a type II move. We therefore have, making use of part 4) of Theorem 3.9,

If we insert these expressions into our skein relation for the Hopf link and then solve the equation for $V_H$, we find $$V_H(t)=-t^{-1}\cdot((t^{1/2}-t^{-1/2})-t^{-1}\cdot(-t^{1/2}-t^{-1/2}))=-t^{-5/2}-t^{-1/2},$$ which agrees with the result we had obtained earlier with the other method.

As one more example, we again consider the trefoil $T$ with the diagram
Also in this case, we want to verify the Jones polynomial $V_T(t)=t+t^3-t^4$, as calculated earlier with the other method. We may apply the skein relation at any of the three crossings of this diagram. Let us choose the crossing in the upper left, which we see to be positive. Hence the skein relation reads here
Carefully observe which crossing is positive /negative, receiving the factor $t^{-1}$ and $-t$, respectively. Also pay attention to keeping the right orientations for new components that might appear when splitting crossings (term on the right).

When we look at this equation, we observe that the second diagram on the left hand side is a diagram of the unknot (pull down the upper part of the diagram with a type II move, then unfold the kink with a type I move), and consequently has Jones polynomial 1. The diagram on the right hand side, on the other hand, represents an oriented Hopf link. Observe the equivalence
which becomes apparent when you flip over the loop on the right. We have computed the Jones polynomial of this link $\tilde{H}$ already, it was $V_{\tilde{H}}(t)=-t^{5/2}-t^{1/2}$. Inserting into the skein relation for the trefoil and solving for $V_T$, this gives $$V_T(t)=t^2+t\cdot(t^{1/2}-t^{-1/2})\cdot(-t^{5/2}-t^{1/2})=t+t^3-t^4,$$which agrees with our earlier result.

Calculating with the skein relation of the Jones polynomial needs some practice (problem 11). You can also check your results by using software such as the Wolfram knot explorer (see lecture 1).