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When we think of quantum invariants, we usually think of the Jones polynomial or of the coloured HOMFLYPT. But (arguably) the simplest example of a quantum invariant of a knot or link is its Alexander polynomial. From the beginning, the central problem in the study of quantum invariants has been what do they mean topologically? The Alexander polynomial has clear algebraic topological meaning as the order of the Alexander module (first homology of the infinite cyclic cover as a module over the group of deck transformations). Can people conceptually explain (in terms of both physics and mathematics) why the representation theory of certain small quantum groups naturally gives rise to this quantity? Computationally I can understand it, but not conceptually.
A somewhat related question was already asked here.
Update: I posted on this question here and here. See also this question.

edited Feb 3 2010 at 1:16

asked Dec 27 2009 at 2:00

Daniel Moskovich
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This is a very good question! – José Figueroa-O'Farrill Dec 27 2009 at 2:07

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What does it mean for something to be a "quantum" knot invariant? – Kevin Lin Dec 27 2009 at 2:27

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I interpret this as saying that it arises from Chern-Simons theory on some suitable Lie (super)group. Hence the question, the way I interpret it, is whether one can reconcile the original topological definition of the Alexander-Conway invariant with the Chern-Simons point of view and, in particular, explain why one takes the group that one does. – José Figueroa-O'Farrill Dec 27 2009 at 8:06

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A quantum invariant is an operator invariant of a link which comes from a representation of a ribbon Hopf algebra such as a quantum group. You first slice a diagram of your link into basic tangles. Then you assign certain elements in the image of the representation to each basic tangle, and compose in a certain way. The most important of these is the R-matrix, which gets assigned to a crossing, An equivalent definition comes from Chern-Simons theory. The R-matrix for the Alexander polynomial comes classically from the Burau representation. – Daniel Moskovich Dec 27 2009 at 10:22

Ben Webster · Answer 1 · 2009-12-27 14:15:32Z UTC

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This is far from being a full answer, but I think the key is 'how do you see the skein relation from the classical definition?'. Once you know the skein relation, it's easy to show it's a quantum invariant, and the skein relation was discovered in the 1960's, long before anyone knew about quantum groups.

answered Dec 27 2009 at 14:15

Ben Webster♦
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That is certainly one "proof". The skein relation is actually at the end of Alexander's original 1923 paper, although nobody thought of it that way at the time. You can calculate the R-matrix from the Burau representation (or from the Dehn presentation of the knot group as Alexander did), and then reproduce the same matrix from a representation of U_q(U(1|1)) or something. But that gives no conceptual insight into why the Alexander polynomial is a quantum invariant- why is it more than a lucky coincidence? Why that specific quantum group? – Daniel Moskovich Dec 27 2009 at 14:36

In a different direction, I offer the Brandt-Lickorish-Millet-Ho Q-polynomial as an example of an invariant which is defined via a Skein relation but is not a quantum invariant (and is conjectured to contain no quantum invariant which is not a polynomial in the Casson invariant). – Daniel Moskovich Feb 2 2010 at 9:13

Ryan Budney · Answer 2 · 2010-02-03 07:10:17Z UTC

Cimasoni and Turaev have a very natural generalization of the Alexander module that's hovering around your concerns.

I found out about this paper for completely different reasons -- Paolo Salvatore pointed it out when we were trying to come up with an argument that the group completion of the monoid of string links can't be abelian.

Why is the Alexander polynomial a quantum invariant?

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