I came across something seemingly trivial, but I don't know why this mistake happens.

We have the HOMFLY-polynomial $P(L)\in\mathbb{Z}[l^{\pm1},m^{\pm1}]$ for oriented links $L$, which satisfies:

1) Normalization: $P(unknot)=1$

2) Skein-relation: $lP(L_{+})+l^{-1}P(L{-})=-mP(L_{0})$

for $L_{+,-,0}$ a knot diagram differing in one crossing by over-/under-/no crossing. (Lickorish-Millet-version)

On the other hand, we have the Jones-polynomial $V(L)\in\Bbb{Z}[t^{\pm1/2}]$ with similar properties:

1) Normalization: $V(unknot)=1$

2) Skein-relation: $t^{-1}V(L_{+})-tV(L{-})=(t^{1/2}-t^{-1/2})V(L_{0})$

Now, as the HOMFLY-polynomial is a generalization of the Jones-polynomial, we can (according to my class and Wikipedia) do the following substitution:


When I try to substitute this directly in the HOMFLY Skein-relation, I end up with the wrong sign in front of the $tV(L_{-})$ in the JONES Skein-relation.

Can anybody tell me, whether I have done a stupid Algebra mistake, or have overlooked something that flips the sign during the substitution?

Much obliged


  • $\begingroup$ Btw is anyone can tell me tips on how to make this question more readable, thanks to you too! $\endgroup$
    – Nik Staro
    Jul 11, 2019 at 13:16

1 Answer 1


There are (unfortunately) a number of conventions for the HOMFLY-PT polynomial. The cleanest version is that $P(L)\in\mathbb Z[x^{\pm 1},y^{\pm 1},z^{\pm 1}]$ with $P(\mathrm{unknot}) = 1$ and the skein relation $$ xP(L_+) + yP(L_-) + zP(L_0) = 0.$$ This is a homogeneous polynomial in three variables. By making some choice of projectivization, this can be reduced to a two-variable polynomial. Some common ones are \begin{align*} \alpha P(L_+) - \alpha^{-1} P(L_-) - z P(L_0) &= 0 \\ \ell P(L_+) + \ell^{-1} P(L_-) +m P(L_0) &= 0 \end{align*} (and all three versions show up in the original HOMFLY paper! The $x,y,z$ parameterization is the Main Theorem, the $\ell,\ell^{-1},m$ parameterization is the Lickorish and Millet approach, and the $\alpha,-\alpha^{-1},-z$ parameterization is the Ocneanu approach after a slight substitution.)

The Wikipedia article shows how to get the Jones polynomial from $P(\alpha,z)$, but you have $P(\ell,m)$. Let's arrange the HOMFLY and Jones polynomials against each other to see the equations we need to solve: \begin{align*} \ell P(L_+) + \ell^{-1} P(L_-) &= -m P(L_0) \\ t^{-1}V(L_+) - t V(L_-) &= (t^{1/2}-t^{-1/2}) V(L_0). \end{align*} On the face of it, it seems impossible that simultaneously both $\ell=t^{-1}$ and $\ell^{-1}=-t$ are true! However, we are allowed to scale equations by a nonzero constant $c$: \begin{align*} \ell P(L_+) + \ell^{-1} P(L_-) &= -m P(L_0) \\ ct^{-1}V(L_+) - ct V(L_-) &= c(t^{1/2}-t^{-1/2}) V(L_0). \end{align*} This gives the system of equations \begin{equation*} \begin{cases} \ell=ct^{-1}\\ \ell^{-1}=-ct\\ -m=c(t^{1/2}-t^{-1/2}). \end{cases} \end{equation*} Since I'm more in the business of talking about the knot theory than solving equations by hand, I asked Mathematica for the answer:

In[33]:= Solve[l==c t^-1 && l^-1==-c t && -m==c(t^(1/2)-t^(-1/2)), {c,l,m}]
Out[33]= {{c->I,l->I/t,m->-((I (-1+t))/Sqrt[t])},
          {c->-I,l->-(I/t),m->(I (-1+t))/Sqrt[t]}}

This says we may make either of the two substitutions given by the $\pm$'s: \begin{align*} \ell &= \pm it^{-1} \\ m &= \mp i(t^{1/2}-t^{-1/2}), \end{align*} and the result is the skein relation for the Jones polynomial, but scaled by a factor of $\pm i$.

  • 1
    $\begingroup$ Wow, this makes so much sense. Thank you very much! When you mentioned the Wikipedia articles, I went back to check again, and it turns out, the German page doesn't have the same amount of information as the English one. The part about the different skein-relations is missing! :'D $\endgroup$
    – Nik Staro
    Jul 12, 2019 at 8:53

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