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.)
This post imported from StackExchange MathOverflow at 2014-12-31 12:16 (UTC), posted by SE-user Peter Samuelson