Find the Jones polynomial of the $(2, q)$ torus knot.

It seems like I've made some kind of error in my calculation, because this gives me not quite the correct polynomial for (2, 3), which is the trefoil knot, but I can't figure out what it is. My approach is to think about the writhe and bracket polynomial for the $(2,m)$ torus link ($m \geq 2$), specifically the diagram for it that arises from thinking of it as a closed $(\sigma_1)^m$ braid, and then use $A^{-3w(D)}\langle D \rangle$ with the substitution $A^{-2} = t^{1/2}$.

First of all, using the closed braid diagram I just discussed, there are exactly m crossings, all of which are $\sigma_1$, a right-handed twist of the braid, so the writhe is just exactly m.

Next, I noticed that in formula $\langle$diagram with crossing$\rangle = A \langle$crossing broken one way$\rangle + A^{-1} \langle$crossing broken the other way$\rangle$ where the crossing being considered is the left-most $\sigma_1$ of the $(2, m)$ link, the part left over breaking it that first way is actually just the $(2, m-1)$ link and the part left over breaking it the other way is the unknot with $m-1$ Type I Reidermaster right-handed twists in it. Also, the diagram of the $(2, 2)$ link is just the standard diagram for the Hopf link, which I already know has bracket polynomial $-A^4 - A^{-4}$. Using that I can argue by induction that this bracket polynomial is $-A^{m+2} + \sum_{i=1}^{m-1} (-1)^{m-i} A^{-3 m - 2 + 4 i}$.

Multiplying that by $(-A)^{-3m}$ gives me $(-1)^{-3 m+1} A^{-2 m + 2} + \sum_{i=1}^{m-1} (-1)^{-2 m + i} A^{-3 m - 2 + 4 i - 3 m } $ $= (-1)^{m+1} (A^{-2})^{m - 1} + \sum_{i=1}^{m-1} (-1)^i (A^{-2})^{3 m - 2 i + 1}$, but this can't be right because substituting in $A^{-2} = t^{1/2}$ and $m = 3$ gives me a polynomial which looks like the one for the trefoil knot if I multiplied all of the individual powers by $-1$. Besides that, I've noticed that wikipedia at least has a much more straightforward-looking formula for the Jones polynomial of a $(p,q)$ torus knot that doesn't have some kind of nasty sum in it. Where is my mistake?

  • $\begingroup$ It may help by using the formula for geometric series $\endgroup$
    – wilsonw
    Commented Jan 10, 2019 at 12:02

1 Answer 1


The trefoil as it appears in Rolfsen's knot table is the left-handed trefoil, that is, it is the closure of the two braid $\beta=\sigma_1^{-3}$. Your computation is for the right-handed trefoil. Since the left-handed trefoil is the mirror of the right-handed trefoil, its Jones polynomial can be obtained from the Jones polynomial of the right-handed trefoil by substituting $t^{-1}$ for $t$, just as you've noticed.

The Jones polynomial of the (p,q) torus knot is $$V_{T_{p,q}}(t) = t^{(p-1)(q-1)/2}\frac{1-t^{p+1}-t^{q+1}+t^{p+q}}{1-t^2}.$$ The usual derivation of this formula is via the representation theoretic definition of the Jones polynomial, rather than the Kauffman bracket version. See this note by Jones for an outline of the calculation.

However, one can derive the above formula when p=2 using the Kauffman bracket approach. Specializing to p=2 yields $$V_{T_{2,q}} = t^{(q-1)/2}\frac{1-t^2-t^{q+1}+t^{2+q}}{1-t^2}.$$ Try to prove that your formula for the Jones polynomial is equivalent to the one above.

  • $\begingroup$ Do you know where I can find calculations of the Alexander/HOMFLY polynomials for Torus Knots? Even certain special cases would be great. Thanks. $\endgroup$
    – user637978
    Commented Jun 17, 2020 at 21:04

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