# Knot invariants from skein relations

Let $k$ be a fixed field. A skein relation defined on $k$ is a local relation on $k$-linear combination of oriented planar link diagrams saying that one can trade an overcrossing for an undercrossing by paying a price in the form of an error term which contains no crossings, and vice versa. More formally, it is a linear relation of the form $$overcrossing = a\, undercrossing + b\, noncrossing$$ with $a,b\in k$, with $a\neq 0$. This last condition is what is needed to get the other way round relation $$undercrossing = a^{-1}\, overcrossing - a^{-1}b\, noncrossing.$$ Moreover, if $b\neq 0$, which is the generic case, we can also write $$noncrossing = b^{-1}overcrossing - ab^{-1}\, undercrossing.$$ The first two relations and Reidemeister relations allow one to reduce any closed link diagram to a $k$-linear combination of disjoint copies of unknot diagrams. Moreover, the third relation shows that two disjoint unknot diagrams are equivalent (under Reidemeister) to a certain scalar multiple of a single unknot, and so inductively that an arbitrary $k$-linear combination of disjoint copies of unknot diagrams is ultimately equivalent to some scalar multiple of a single unknot diagram. Then, adding a single relation of the form $$unknot = c\, \emptyset$$ for some $c\in k$, we see that every closed link diagram is equivalent to a scalar multiple of the empty link diagram. Clearly if $c=0$ then this scalar multiple is zero, so in order to get a nontrivial result, let us assume $c\neq 0$. This shows that there exists at most a unique link invariant $$P: \text{closed links in \mathbb{R}^3}/\text{isotopy} \to k$$ with $P(skein)=0$, $P(unknot)=c$ and $P(\emptyset)=1$. However the skein relation imposes a constraint on $c$ in order for $P$ to possibly exist. Namely, from $$noncrossing = b^{-1}overcrossing - ab^{-1}\, undercrossing$$ we see that the disjoint union of two unknots is equivalent to $b^{-1}(1-a)$ times a single unknot, so we get the equation $$c^2=cb^{-1}(1-a),$$ which, since we are assuming $c\neq 0$, determines $c$ as $$c=\frac{1-a}{b}$$ For instance, for the Jones polynomial, $k=\mathbb{Q}(q)$, $a=q^4$ and $b=q-q^3$ so that $c=q+q^{-1}$ in this case (there are other possible conventions/normalizations for the Jones polynomial, this is the one I use).

The question is the following: is the necessary condition $c=(1-a)b^{-1}$ also sufficient to have a nontrivial closed link invariant $P$ with $P(skein)=0$, $P(unknot)=c$ and $P(\emptyset)=1$?

In other words, can it be shown that the reduction from a closed link diagram to a scalar multiple of an unknot diagram done using a skein relation and Reidmeister moves is always independent of the particular reduction process? (i.e., that two reductions necessarily lead to the same scalar multiple of the the value of the invariant on the unknot diagram?)

• You probably want oriented diagrams; otherwise the Jones polynomial doesn't satisfy the skein relation you've described. Sep 25, 2016 at 0:50
• Yes, absolutely! Thanks for the edit. Sep 25, 2016 at 6:03

Yes, this condition is sufficient.

The $b=0$ case is trivial. For the $b\ne 0$ case, embed $k$ in a larger field $K$ where $-a$ has a square root; then your knot invariant can be derived from the HOMFLY-PT knot invariant $P_H(\ell,m)$ (which lives in $K[\ell,\ell^{-1},m]$), which satisfies the skein relation

$$\ell(\text{over})+\ell^{-1}(\text{under}) + m(\text{none}) = 0$$ and $P_H(\text{unknot})=1$ (with no defined value on the empty link).

Then your invariant $P$ is just $cP_H\left(\frac{1}{\sqrt{-a}},\frac{-b}{\sqrt{-a}}\right)$ with the extra value $1$ for the empty link.

The paper definining HOMFLY-PT is only 8 pages and includes the outlines of four distinct proofs of the well-definedness.

P. Freyd, D. Yetter, J. Hoste, W. B. R. Lickorish, K. Millett, and A. Ocneanu, MR 776477 A new polynomial invariant of knots and links, Bull. Amer. Math. Soc. (N.S.) 12 (1985), no. 2, 239--246.

• Sure! It was immediate to bring back the problem to the classical skein relation for HOMFLY-PT! sometimes the obvious answers are just in front of me and I don't see them. Now I see that the approach I had in mind is the Hoste approach, and that it is actually no so immediate, so I'm reassured. The reason I was lead to my normalization $c=(1-a)b^{-1}$ is that this way the invariant $Q_H= (1-a)b^{-1}P_H\left(\frac{1}{\sqrt{-a}},\frac{-b}{\sqrt{-a}}\right)$ satisfies $Q_H(K_1\sqcup K_2)=Q_H(K_1)Q_H(K_2)$, which I find to be more pleasant for computations. Sep 25, 2016 at 6:32