Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Proving that the Jones polynomial is q-holonomic

+ 7 like - 0 dislike
64 views

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?

This post imported from StackExchange MathOverflow at 2014-12-31 12:16 (UTC), posted by SE-user Gjergji Zaimi
asked Sep 9, 2014 in Theoretical Physics by Gjergji Zaimi (35 points) [ no revision ]
retagged Dec 31, 2014

1 Answer

+ 7 like - 0 dislike

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
answered Sep 9, 2014 by Peter Samuelson (70 points) [ no revision ]
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.

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

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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...