# Jones polynomial invariance

I'm studying the Jones polynomial and I know that it is a knot invariant. I saw that a possible way to define the Jones polynomial is to set the Jones polynomial of the unknot to be 1 and then use the following recursive relation:

$t^{-1}V(L_+)-tV(L_-)+(t^{-1/2}-t^{1/2})V(L_0)=0$

where $L_+$, $L_-$ and $L_0$ stand for 3 diagrams which are everywhere the same but for a neighbourhood where they differ in the following way:

I wanted to prove from this definition the invariance of the Jones polynomial under Reidemeister moves. My attempt was to set the first Reidemeister move as $L_+$, then as $L_-$ I get the same diagram with the exchange of the over-arc and under-arc and as $L_0$ I get a straight line in disjoint union with the unlinked diagram of the unknot.

I don't see how I can proceed further, anyone has any clue?

• That relation isn't likely enough. See Kauffman's paper for more details, but you need to know that $V(K_1 \cup K_2)= V(K_1)V(K_2)$ where $K_1$ and $K_2$ are two disjoint link diagrams (with no shared crossing). Then everything should work out nicely. You are also refering to the unknot when you mean the unique diagram of the unknot with no crossings. You should fix this. – PVAL-inactive Jun 15 '16 at 18:43
• Note that it is not at all clear that the Skein relation can be used to define a knot invariant. If you have a knot invariant satisfying it, it is unique, but existence is another issue. – Hanno Jun 15 '16 at 19:14
• Thank you for your comments, they were helpful. From what I understand I think the best way is using the Kauffman bracket definition for the Jones polynomial and show that the Skein relation is satisfied. – popoolmica Jun 16 '16 at 9:45
• @PVAL The skein relation plus $V(\text{unknot})=1$ is enough. You do not need to refer to crossings for the unknot because this is a skein relation: a relation on isotopy classes of links in $S^3$. – Kyle Miller Nov 22 '17 at 21:30

Suppose $\mathcal{K}$ is the set of all oriented links in $S^3$ up to isotopy, and let $\mathbb{Z}[t^{\pm 1/2}](\mathcal{K})$ denote the set of $\mathbb{Z}[t^{\pm 1/2}]$-linear combinations of isotopy classes of links. Suppose $L_+,L_-,L_0$ are links that outside a given ball are the same, and within the ball the three links each have the respective $2$-tangles you illustrate. Let $M$ be the submodule generated by $\text{unknot}-1$ and by $t^{-1}L_+-tL_--(t^{1/2}-t^{-1/2})L_0$ for all such triples of links. The claim is that the skein module $\mathbb{Z}[t^{\pm 1/2}](\mathcal{K})/M$ is isomorphic to $\mathbb{Z}[t^{\pm 1/2}]$. Then, $V(L)$ is defined to be the image of $L$ in this quotient.
It is not difficult to show that the skein relation is enough to represent each link as a polynomial in this quotient. The first step is to show that a split union of $n$ unknots is represented by $(-t^{1/2}-t^{-1/2})^{n-1}$. The second is to note that every link has finite unlinking number, and then be careful to show that the skein relation allows one to reduce the link to a polynomial. (This is, every link $L$ has at least one $p\in\mathbb{Z}[t^{\pm 1}]$ such that $L-p\in M$.)
The more difficult part is to show that there aren't more relations in $M$. It could very well be that the quotient will be some quotient of $\mathbb{Z}[t^{\pm 1/2}]$. Lickorish gives a proof of the well-definedness of the HOMFLY polynomial (and hence the Jones polynomial) in Introduction to Knot Theory (chapter 15) using the following approach. Let $D_n$ be the set of link diagrams with at most $n$ crossings, and inductively prove that the polynomial associated with links in $D_n$ are unchanged by Reidemeister moves (along with a "Reidemeister IV": passing an unknot under a strand) that involve no more than $n$ crossings, and prove that "ascending diagrams" have the polynomial associated to a disjoint union of unknots. Once this is done, the idea is that any sequence of Reidemeister moves between two equivalent links $L$ and $L'$ can be conducted within $D_n$ for some $n$, hence there are no additional relations.