Low Dimensional Topology

September 26, 2010

What is the right context for the Alexander polynomial?

Filed under: Knot theory,Quantum topology — dmoskovich @ 1:45 pm

I’m now a postdoc at the University of Toronto, and I have a new baby daughter (named Chana or Ann), hence I haven’t posted for a while.
A common theme in topology, and in mathematics in general I suppose, is that it is important to work out the natural setting of a problem. This would be the setting in which the problem is stated most naturally, and in which it’s solution makes most sense conceptually.

For instance, in knot theory, does your problem make more sense for knots or for links? (Example: skein theoretic invariants) Is there a reason that you are restricting to knotted $S^1$‘s in $S^3$, or would the problem make more sense for knotted $S^n$‘s in $S^{n+2}$ (Example: Seifert surfaces), or perhaps for more general knotted objects in more general manifolds (Example: fundamental group of the complement)? Do you need the manifold at all? Maybe your problem is one which involves group presentations, and is most naturally stated as a combinatorial group theory problem regarding groups with presentations of deficiency 1? Maybe your problem is homological, and makes sense for any homology manifold? (Example: homological invariants) Maybe it’s a problem which makes most conceptual sense in the context of symmetric chain complexes and L-theory? (Example: linking forms)
Sometimes the search for the right setting for a problem might also involve restricting the class of objects you might want to consider. And sometimes it might involve changing it completely.
In this post, I’d like to discuss the natural setting for the Alexander polynomial.

Setting 1: Modules over unique factorization domains

A knot complement has first homology $\mathbb{Z}$, so the homology of its infinite cyclic cover carries the natural structure of a finitely presented $\mathbb{Z}[t^{\pm 1}]$-module. This module, which we call $A$, has a presentation matrix
$F^{n+1}\overset{M}{\to}F^{n}\to A\to 0,$
which is called the Alexander matrix. The $0$th elementary ideal $E$ is the ideal generated by the $n\times n$ minors of $M$. The ring $\mathbb{Z}[t^{\pm 1}]$ is a unique factorization domain. The maximal common divisor of the elements of $E$ (which are polynomials) is an element in $\mathbb{Z}[t^{\pm 1}]$, i.e. a polynomial. This is the Alexander polynomial.
In this approach, the Alexander polynomial is an invariant of a finitely presented module over a unique factorization domain ($A$ could have been any such module). This was Alexander’s original approach. It was not conducive to proving theorems about the form of Alexander polynomials, and therefore I don’t think that this is the natural setting in which Alexander polynomials should be considered.

Setting 2: Modules over principal ideal domains

The next approach, historically, was to localize $\mathbb{Z}[t^{\pm 1}]$ and to consider $A$ as a module over its ring of fractions Λ. The ring Λ is a principal ideal domain. The module $A$ is torsion when considered as a module over Λ (Lemma 4 in Milnor’s A duality theorem for Reidemeister torsion), and the generator of its annihilator is (roughly) the Alexander polynomial.
Localizing (the process of considering $A$ as a module over Λ instead of over $\mathbb{Z}[t^{\pm 1}]$) was a process which took me a while to conceptually understand… but now I think I can even understand it well enough to explain it! Let $S$ be the set of polynomials which we are inverting to get from $\mathbb{Z}[t^{\pm 1}]$ to Λ, which are those polynomials which have constant term 1. The conceptual idea behind localizing is that the Alexander polynomial should really be thought of as an invariant of the non-$S$-torsion part of $A$. In other words, the Alexander polynomial is really an invariant of finitely presented modules over principal ideal domains, not an invariant of modules over unique factorization domains at all… or is it?

Setting 3: (Non)planar algebras

The third approach which I’d like to discuss is the quantum topology approach. I don’t know the conceptual reason that the Alexander polynomial is a quantum invariant (I blogged about it, though here, here, and here), but I have a better grip of the question now. A quantum invariant is not really an invariant of a knot or a link, as far as I can make out- conceptually it’s an invariant of a planar algebra, or of a non-planar generalization of such. You can project a knot in $\mathbb{R}^3$ to a plane to obtain a knot diagram. Knot diagrams are made up of over- and under- crossings (the generators for our planar algebra), which are joined together in a plane. Two knot diagrams represent the same knot if and only if they are related by a finite sequence of Reidemeister moves. The planar algebra generated by crossings modulo Reidemeister moves is what the Alexander polynomial (of a knot), in its quantum aspect, is an invariant of.
Via Melvin-Morton-Rozansky, the Alexander polynomial occurs as a certain part (the “wheels part”) of the universal quantum invariant, a very complicated invariant called the Kontsevich invariant. So to understand the Alexander polynomial, one has to drag it out of the mud- the Kontsevich invariant mud. But now, because we’re thinking of the Alexander polynomial as an invariant of planar algebras and suchlike, maybe we could find a nicer planar algebra for which the Alexander polynomial itself was the universal finite-type invariant? This would allow us to directly attempt to understand the conceptual reason why the Alexander polynomial is a quantum invariant.
Dror Bar-Natan has done just this (click on any of his newer talks), showing that the Alexander polynomial is the universal finite type invariant of $W$-knots. A $W$-knot topologically corresponds to a knotted torus in $\mathbb{R}^4$. But from the point of view of quantum topology, we think of it as a diagram in a (non-planar) algebra generated by over- and under- crossings, whose endpoints are connected by strands. These strands are now no longer required to live in the plane- they are just associations between endpoints of crossings, and are free to intersect. Two such diagrams are considered equivalent if they are related by a certain extended family of Reidemeister moves.

The whole setup is designed to make the quantum picture as simple as possible. And indeed, the universal quantum invariant for such objects turns out to be the Alexander polynomial. Knots map into $W$-knots by inclusion (remember that a knot is now a certain sort of diagram).
So, from a quantum perspective, are $W$–knots in fact the natural setting for thinking about the Alexander polynomial?

Something completely different?

There is great scope for irresponsible fantasies. It isn’t hard to come up with a silly dream for a unified perspective which makes it clear that the Alexander polynomial is quantum. Perhaps the day will come when one of these silly dreams will come true.

1. Congratulations!

Comment by Richard Kent — September 27, 2010 @ 7:00 pm

2. Isn’t is confusing to give your child both a Hebrew (Chana) and Anglican (Ann) name?

On another note, I don’t mean to carp here, but I wanted to hijack this opportunity to bring up another topic that for some reason has been ignored on this blog. Peter Shalen has been posting a bunch of preprints on the ArXiv recently on the Margulis number for hyperbolic 3-manifolds. The latest preprint of his has a generic Margulis number at 0.29. Are any experts here able to comment on this, specifically is this useful? Do we care what the actual number is, or is it simply enough to know that there is a Margulis number?

Comment by Mayer A. Landau — September 29, 2010 @ 10:36 pm

• I would argue that they are variants of the same Hebrew name (http://en.wikipedia.org/wiki/Anna_(given_name)). English lacks a voiceless velar frictive such as the Hebrew “heth”, hence Chana becomes Hanna or Anna, and Ann or Anne in England. My grandma z”l was Ann on her passport, and she’s Chana bat Chaim on her gravestone…
So my daughter will be Chana ben Daniel David on her Ketuba (a few years down the road, b”h), and is Ann Moskovich on her passport.
Peter Shalen’s preprints look very interesting! I’d also be interested in one of my cobloggers giving us an idea of what was done!

Comment by dmoskovich — October 2, 2010 @ 7:30 pm

3. Small point, but your question asks for “the right context”. This pre-supposes there is a right context. What if there isn’t? I mean, we are free to interpret the Alexander polynomial any way we want. Perhaps there’s no compelling reason to depart from this relativist position.

At present my preference when thinking about the Alexander polynomial is to think of it as an order ideal of the Alexander Module. The Alexander Module is a “genuinely interesting thing” to me, while the Alexander polynomial is just some odd projected-out data that is partially informative — by its very nature sometimes it tells me a lot, sometimes it tells me nothing interesting. I think this is very compatible with this relativist position — since the Alexander polynomial is a coarse projection of a geometrically inspired idea, it is bound to appear as a mysterious object as it simultaneously informs on and hides geometric information.

Of course, some of my work is very much an attempt to contradict this position, but as academics it’s in our job description to keep seemingly contradictory hopes alive.

Comment by Ryan Budney — October 6, 2010 @ 1:10 pm

• One can argue a relativistic position, but taking absolutist positions allows me to make bolder statements!
The Alexander polynomial, in any aspect of it which you wish to study, is a fascinating subject.

Comment by dmoskovich — October 21, 2010 @ 12:51 pm

Blog at WordPress.com.