Given a 3-manifold $M$, one can define the Kauffman bracket skein module $K_t(M)$ as the $C$-vector space with basis "links (including the empty link) in $M$ up to ambient isotopy," modulo the skein relations, which can be found in the second paragraph of section two of http://arxiv.org/abs/math/0402102. (Side question - how can I draw these relations in Latex?)

If $S$ is a surface, then $K_t(S\times [0,1])$ has an algebra structure given by stacking one link on top of another. If $S$ is a boundary component of $M$, then $K_t(M)$ is a (left) $K_t(S\times [0,1])$ module, where the left module structure is given by gluing $S\times \{1\}$ to the copy of $S$ in the boundary of $M$. In this situation, we can define a left module map $K_t(A\times [0,1]) \to K_t(M)$ which is uniquely defined by "(empty link in $S\times [0,1]$) maps to (empty link in $M$)." The "peripheral ideal" is the kernel of this module map, and is a left ideal of $K_t(S\times [0,1])$.

The motivation for these definitions comes from knot theory - if $K$ is a knot in $S_3$, then the complement of a small tubular neighborhood of $K$ is a manifold with a torus boundary, and the algebra $K_t(T^2\times [0,1])$ and module $K_t(S^3 \setminus K)$ give information about the knot $K$.

Now I can ask my question: Is there a manifold $M$ with a torus boundary such that the peripheral ideal is trivial?

I've just recently started learning about knot theory, and I'm having a hard time trying to figure this out. One thing that I do know is that $M$ cannot be of the form $S^3 \setminus K$, because of propositions 7 and 8 in http://arxiv.org/abs/math/9812048. I also suspect that $M$ will actually have two boundary components which are a torus, but I don't really have a good reason for this.

Also, I suspect this might be a hard question, so any hints about one might approach it would be helpful.

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    $\begingroup$ Knot complements always have non-abelian $SU(2)$ representations, as proven by Kronheimer-Mrowka. msp.warwick.ac.uk/gt/2004/08/p007.xhtml ams.org/mathscinet-getitem?mr=2106239 This implies that the $SL_2(C)$ character variety is non-trivial, and in fact the peripheral ideal is non-trivial. ams.org/journals/proc/2005-133-09/S0002-9939-05-07814-7/… msp.warwick.ac.uk/agt/2004/04/p050.xhtml I think this implies that the peripheral ideal is non-trivial, by specialization to $t=-1$. $\endgroup$ – Ian Agol Jun 25 '11 at 2:46
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    $\begingroup$ In general, though, it's not known that for a manifold with torus boundary, there is a non-abelian $SL_2(C)$ representation. This holds for geometric manifolds, and their connect sums. However, it hasn't been shown in general for manifolds which have a non-trivial JSJ decomposition. In many cases, though, you can prove that there is a non-abelian $SL_2(C)$ rep. even in the presence of incompressible tori. $\endgroup$ – Ian Agol Jun 25 '11 at 2:50

There is a gap in the proof that the peripheral ideal is nontrivial in that paper.

Thang Le and Stavros came up with a more algebraic way of definining a closely related ideal that they could prove was nontrivial.

I think its a great problem. A good starting point might be to prove it for Torus knots. There is a recent paper of Julien Marche that computes the Kauffman bracket skein module of all torus knots, but stop short of understanding the module structure over the skein module of the torus. You might start there.

I am willing to conjecture that the peripheral ideal is always nontrivial for any link. In fact, Thang has recently proved a weak form of this.

We defined the peripheral ideal to be the extension to the noncommutative torus of the kernel of the inclusion map of skein algebra of the torus into the skein module of the complement of the knot. Via an identification of the skein algebra of the torus with the symmetric part of the noncommutative torus the ideal corresponds to the ideal of the image of the $SL_2\mathbb{C}$-characters of the knot group in the characters of $\mathbb{Z}\times \mathbb{Z}$. We found a way of seeing the colored Jones polynomial of the knot as lying in the dual to the $SL_2\mathbb{C}$-characters of the knot group, and we found that the colored Jones polynomial is in the annihilator of the peripheral ideal.

Thang and Stavros stepped back from the picture, and found a formal connection between the Jones polynomial and the noncommutative torus, and then just defined their ideal to be the annihilator of the Jones polynomial. Using formal properties of the $R$-matrix they were able to give an axiomatic proof that their ideal was nontrivial.

The conjecture is about the relation between the formal definition of quantum invariants and their concrete realization. The Kauffman bracket skein module of a knot complement is a deformation quantization of the unreduced scheme of the $SL_2\mathbb{C}$-characters of its fundamental group. The conjecture that the peripheral ideal is nontrivial is motivated by this idea, and the fact that the $SL_2\mathbb{C}$-character variety of a nontrivial knot, is nontrivial, meaning the $A$-ideal is nontrivial. This should mean that the peripheral ideal is nontrivial.

The orthogonality between the peripheral ideal and the colored Jones polynomial should lead to data about the $SL_2\mathbb{C}$-character variety of the knot being expressed in the aggregate behavior of the colored Jones polynomial of the knot.

  • $\begingroup$ By peripheral idea of a link, do you mean this: fix a torus component of the boundary of $S^3 \setminus L$, which gives $K_t(S^3 \setminus L$ a module structure, and take the kernel of the map determined by "empty link in the torus goes to empty link in $S^3 \setminus L$? Which paper of Le and Stavros did you mean? They have several together. Also, the paper by Marche definitely looks interesting. $\endgroup$ – Peter Samuelson Jul 10 '10 at 3:18
  • $\begingroup$ The natural thing to do from the viewpoint of representations is to look at the skein module of the manifold with boundary a disjoint union of tori as a module over the tensor product of the skein algebras of the tori. Poincare duality will guarantee that if the classical object is nonempty then the ideal will be nontrivial. There is an earlier paper of Razvan and I where we identify the skein algebra of the torus with the symmetric part of the noncommutative torus. Stavros and Thang work in a variation of the noncommutative torus that is a PID. I con't have access to the citation now. $\endgroup$ – Charlie Frohman Jul 11 '10 at 12:21
  • $\begingroup$ I found the paper by Stavros and Thang, and it looks interesting. The two ideals definitely look very related. Thanks for the suggestions, and I'll comment again if we make any progress. $\endgroup$ – Peter Samuelson Jul 15 '10 at 23:16

This is not an answer, more like one comment and one suggestion for an approach to this problem.

The comment is that this looks like a 4-dimensional TQFT. You have an algebra associated to a surface, a module associated to a 3-manifold, and a vector associated to a 4-manifold. The reason it is not usually presented this way is that the dependence on the 4-manifold is not interesting; it only depends on the signature (i.e. the cobordism class).

The suggestion for an approach is to look at $q=1$, the classical limit. Here the algebra associated to a surface is the coordinate ring of the character variety. The key observation is that a skein determines a function on the space of flat connections by taking traces of holonomy and the skein relation corresponds to $tr(A^{-1}B)+tr(AB)=tr(B)$. This was written up by Doug Bullock.

My suggestion then is for you to look at your question in this context. I don't know if this will help but it seems more likely to be a question that an expert can answer.

  • $\begingroup$ This is a good suggestion and a nice paper - thanks. Also, I noticed you were mentioned in the introduction to the paper :-) I didn't quite follow the second paragraph, in this setting, what's the vector associated to a 4-manifold? $\endgroup$ – Peter Samuelson Jul 15 '10 at 23:08
  • $\begingroup$ Thanks. I explained the idea to Doug Bullock at the Banach Centre, Warsaw and he wrote it up. I should probably fess up and admit I am not clear about the 4-manifold part of this story. $\endgroup$ – Bruce Westbury Jul 16 '10 at 2:33

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