I'm working through McMullen's paper "The Alexander polynomial of a 3-manifold and the Thurston norm on cohomology" and have a question concerning the following setup:

Given a link complement $(X, p)$ with $G = \pi_1(X)$, the Alexander polynomial $\Delta$ is the first order of the Alexander module, which is the $\mathbb{Z}[G/G']$-module $H_1(X_{\inf}, \pi^{-1}(p); \mathbb{Z})$ where $\pi: X_{\inf} \to X$ is the maximal free abelian cover of $X$.

Given a homomorphism $\phi: \pi_1(X) \to \mathbb{Z}$ we have another covering space $\pi_{\phi}: X_{\phi} \to X$ characterized by $(\pi_{\phi})_*(\pi_1(X_{\phi})) = \ker \phi$. The first homology of this cover $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ is a $\mathbb{Z}[\mathbb{Z}]$-module under the action of deck transformations, as in the case of the Alexander module.

Now what I don't understand is that McMullen claims that the first elementary ideal of $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ is just the image under "$\phi$" of the first elementary ideal of the Alexander module. (The map $\phi$ is in quotes because I'm referring to the induced map from $\mathbb{Z}[G/G'] \to \mathbb{Z}[G/\ker \phi]$.) As a consequence, if $\Delta_{\phi}$ is the first order of $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$, we have $\phi(\Delta) = \Delta_{\phi}$.

Why is this? McMullen only uses this relation when $\phi$ is primitive, i.e. surjective, but it seems it could be true for any map onto a free abelian group. Perhaps it can be proved using the reduction to algebra by the equation $$ H_1(X_{\phi}, \pi_{\phi}^{-1}(p)) \cong m(G)/m(\ker \phi)m(G) $$ where $m(H)$ is the ideal of $\mathbb{Z}[H]$ generated by $(h - 1: h \in H)$. For example, from here we can fit the Alexander module and $H_1(X_{\phi}, \pi_{\phi}^{-1}(p))$ into an exact sequence of $\mathbb{Z}[G]$-modules, but I still can't come to the conclusion.


1 Answer 1


Going from multivariable Alexander polynomials to onevariable Alexander polynomials is indeed a little awkward using the definition in terms of orders of modules. The situation becomes much clearer if one translates the problem into Reidemeister torsions. The multi/onevariable Alexander polynomials are almost identical with multi/onevariable Reidemeister torsions, and Reidemeister torsion is functorial. The reason for the functoriality of Reidemeister torsion is that it is computed in terms of determinants of square matrices, and these are functorial in the coefficient system. I warmly recommend reading the first chapters of Turaev: Introduction to combinatorial torsions. How to go from (twisted) multivariable Alexander polynomials to onevariable Alexander polynomials is also spelled out in my paper with Taehee Kim: Twisted Alexander norms give lower bounds on the Thurston norm, where we use Reidemeister torsion to prove the result.

  • $\begingroup$ Wow thanks for responding! (I've been looking forward to reading your works for some time :). I think I found another route to this functoriality through the Fox calculus, since a presentation for the Alexander invariant is given by the abelianization map applied to the matrix of Fox derivatives. This also works for the intermediate maps to free abelian groups and the coefficients of this matrix factors through the abelianization. (I didn't put this as an answer because it seems to require the full proof that the Fox matrix provides the needed presentation, which is a bit unwieldy). $\endgroup$ May 7, 2014 at 1:49
  • $\begingroup$ you have to be careful: you are right that Fox derivatives give you a presentation matrix, and you can get the multivariable and the onevariable Alexander polynomial out of it. But the problem is that as a presentation matrix for the Alexander module it is not a square matrix, so you have to take the minors and then the gcd. Taking minors is functorial, but taking gcd's isn't. For example, if the presentation matrix is $(x-1,y-1)$ for two different meridians, then the order is $gcd(x-1,y-1)=1$, but if you abelianize the presentation matrix is $(t-1,t-1)$, so the gcd of the minors is $t-1$. $\endgroup$ May 8, 2014 at 6:43
  • $\begingroup$ the point of considering the Alexander module rel a point is to get to a square presentation matrix, where the issue with the gcd's does not arise. It's a slightly awkward trick, for example it doesn't work for closed 3-manifolds. $\endgroup$ May 8, 2014 at 6:45
  • $\begingroup$ Thanks for that insight. As you point out, the statement in my question that says "if the Alexander ideal is functorial, then the gcd is also" is incorrect. In fact, McMullen only uses $(\Delta_{\phi}) = I_{\phi} = \phi_*(I)$ (these are the ideals generated by the minors). $\endgroup$ May 8, 2014 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.