Is the Alexander polynomial of the 2-cable of a knot zero?

Solution 1:

One definition of the Alexander polynomial of an oriented link $L$ is that you take a Seifert surface $\Sigma$, which is a connected oriented surface whose boundary is $L$ (with the oriented boundary matching the orientations of $L$) and whose interior is disjoint from $L$, then calculating the Seifert form $\alpha:H_1(\Sigma)\times H_1(\Sigma)\to\mathbb{Z}$, where $\alpha([a],[b])=\operatorname{lk}(a,b^+)$ for simple closed curves $a,b\subset \Sigma$ (the curve $b^+$ is from pushing the loop off the surface in the direction of surface normal of $\Sigma$), then letting $A$ be a matrix for $\alpha$ with respect to some basis, and finally computing $\det(tA-A^T)$.

For a 2-cabling $L$ of a knot $K$, we need to decide on how we orient the components. The usual choice is to have both components be co-oriented, but one could potentially choose opposite orientations. These give different answers.

Let's consider the oppositely oriented case first. There is a natural choice of Seifert surface $\Sigma$, which is the annulus that runs along with the two components, parallel to the blackboard for the blackboard framing -- note that the oriented boundary of the annulus is indeed giving the components of $L$ opposite orientations. Thus, $H_1(\Sigma)=\mathbb{Z}$. The core of the annulus $a\subset\Sigma$ (with some orientation) generates $H_1(\Sigma)$. If $L_1$ and $L_2$ are the two components, what we can do is isotope $a$ to $L_1$ and, away from $a$, isotope the pushoff $a^+$ to $L_2$. Thus, $\operatorname{lk}(a,a^+)=\operatorname{lk}(L_1,L_2)$, which is the writhe $w(K)$ of the blackboard framed diagram. Thus, the matrix of the Seifert form can be given as the $1\times 1$ matrix $[w(K)]$ with respect to the $\{a\}$ basis. We calculate that $$ \Delta_L(t) = \det(tA-A^T) = \det(t[w(K)]-[w(k)]^T)=w(k)(t-1). $$ (And remember this is only defined up to multiplication by $\pm t^k$.) So we see that the Alexander polynomial is not telling you very much about the link.

The case where the components are co-oriented is more involved. I point you to Theorem 6.15 of Lickorish's "An Introduction to Knot Theory" (p.60), though as it is stated it is for the case where you make a satellite knot rather than a link, it might need some modification. The main idea is that if you have a Seifert surface for $K$, then you can take two parallel copies $\Sigma_0$ and $\Sigma_1$ (from pushing a copy of the Seifert surface off itself), and then you can add half-twisted strips to the boundaries to get a connected surface $\Sigma$ whose boundary is $L$. One can split up $H_1(\Sigma)$ into $H_1(\Sigma_0)\oplus H_1(\Sigma_1)\oplus H_1(\Sigma')$ where $\Sigma'$ consists of the strips along with a tubular neighborhood of the boundaries of the surfaces. Then it should be possible to work out a Seifert matrix. Sorry to leave it here -- I did calculate some Alexander polynomials of cablings to see that they do not tend to be trivial. The most basic example is $(2,2n)$ torus links (i.e., 2-cablings of the unknot), whose Alexander polynomials are given by $(1-t^{2n})/(1+t)$.

If it turns out to be fine applying the theorem to this case, then the theorem implies that $$\Delta_L(t) = \Delta_T(t)\Delta_K(t^2) = \frac{1-t^{2w(K)}}{1+t}\Delta_K(t^2),$$ where $T$ is the torus link $T(2,2w(K))$ (with both components co-oriented).