The Jones polynomial is known to have many different interpretations or definitions, by now. There are connections with QFT, quantum groups, Hilbert schemes, Cherednik algebras, etc.

My question is whether any of these approaches provide a new proof of the fact that the colored Jones polynomials of a knot satisfy a linear recurrence relation with coefficients that are polynomials in $q^n$. This is a theorem of Garoufalidis and Le, in their paper "The colored Jones function is q-holonomic".

My understanding is that some of these approaches mentioned above are a lot more understood for the special case of torus knots, and i would be fine restricting to this special class. Are there any more proofs of the fact that the sequence of colored Jones polynomials is q-holonomic?


1 Answer 1


One statement that would imply that the colored Jones polynomials are q-holonomic involves the Kauffman bracket skein module $S_q(K)$ of the knot complement. This is a module over the skein module of the torus $T^2$, and q-holonomicity follows from the statement "$S_q(K)$ is finitely generated over the subalgebra $\mathbb C [m]$," where $m$ is the meridian of $K$. (This follows from papers of Frohman and Gelca, starting with this one. See also Cor. 1.3 here.) I'm not sure if the statement in quotes is true, although for 2-bridge knots it was proved by Le, and in special cases by Gelca and others.

There is also a conjecture about $S_q(K)$ that can be viewed as a quantization of the statement "$L-1$ divides the $A$-polynomial of $K$," and this conjecture implies q-holonomicity. (This is Thm 5.10 here.)

Unfortunately, as far as I know, people don't know good techniques that can prove statements about skein modules for arbitrary knot complements (although I would love to be proven wrong). So it doesn't seem like these statements will lead to a new proof of q-holonomicity (at least anytime soon).

For torus knots (or, more generally, iterated cables of the unknot), the colored Jones polynomials can be written in terms of a cabling formula involving the double affine Hecke algebra and its polynomial representation. (I think at the moment this is just proved for $sl_2$, or for torus knots and $sl_n$, but it seems likely to be true in general. Some refs are here, here, here). This might imply q-holonomicity because the polynomial representation is holonomic, but I don't know a proof at the moment (I might be able to add one later).

This cabling formula uses the DAHA at $t=q$, but it deforms to arbitrary $t$, and produces polynomials depending on 2 variables $q,t$. I don't know whether these polynomials are $q$-holonomic, though. One might have to change to $(q,t)$-holonomic, which could be defined in terms of the polynomial representation of the DAHA. But this probably wouldn't be necessary if you allow rational functions in the "variable" $q^n$. (I'm not sure if you want to allow this or not.)

  • $\begingroup$ Thanks, this was very helpful! I would be curious to see a proof of holonomicity using the cabling formula involving the polynomial representation of DAHA. I will check the references you gave. $\endgroup$ Sep 10, 2014 at 4:30
  • 2
    $\begingroup$ I'll update this answer if I figure out this proof, it's an interesting question. There's another heuristic reason why the Jones polys are holonomic, although again I'm not sure if it can be turned into a proof. When $q=1$, the skein module of a 3-manifold $M$ is the ring of functions of the scheme $Char(M) := Hom(\pi_1(M), SL_2(\mathbb {C})) / SL_2(\mathbb{C})$ (the character variety). If $M$ is a knot complement, it is known that the image of the restriction map $Char(M) \to Char(\partial M)$ is Lagrangian. One might then expect that its quantization (the skein module) should be holonomic. $\endgroup$ Sep 12, 2014 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.