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I would like to understand whether the multivariate Alexander polynomial of a link $L$ does not vanish everywhere for a certain class of links; I don't know the links' diagrams in general but I have some information:

  • $L$ has two components, say $K_1$ and $K_2$;
  • each component is unknotted (considered singularly), and in particular $K_1$ is a simple curve;
  • $K_2$ is contained in the solid torus which is the complement of $K_1$;
  • the linking number of the components vanishes, i.e. $lk(K_1, K_2)=0$. So, $K_2$ is homotopically trivial in the exterior of $K_1$;
  • $L$ is non-split.

An example of such a link is the Whitehead link $W$, whose Alexander polynomial is $\Delta_W(t_1, t_2)=(1-t_1)(1-t_2)$.The Whitehead link

The only general result I was able to find in the literature is that the Alexander polynomial $\Delta_L(t_1, t_2)$ evaluted in $(t_1, 1)$ is equal to zero, i.e. $\Delta_L(t_1, 1)=0$, unfortunately. My guess, just for the records, is that the Alexander polynomials of such links does not vanish everywhere. Any ideas or references are appreciated.

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I believe there are such examples where the Alexander polynomial is zero. Consider the link L11n244, pictured below.

enter image description here

Both the Knot Atlas and Link Info have it listed as having multivariable Alexander polynomial of zero. The picture above satisfies all of your conditions except that neither component is "simple" (which I'm taking to mean a diagram of the unknot with no crossings). But one can just isotope the diagram so that either of the components are simple. I've done that for the blue component below.

enter image description here

It occurs to me that maybe you meant something else with your third condition: $K_2$ is contained in the solid torus which is the complement of $K_1$. The reason I say that is if $K_1$ is an unknot, then its complement is a solid torus. So any two component link $L=K_1\cup K_2$ where $K_1$ is trivial would have $K_2$ contained in the solid torus complement of $K_1$.

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    $\begingroup$ Perfect answer, this link is a counterexample to what I hoped was true! Thank you very much:D $\endgroup$ Commented Jan 29, 2021 at 10:27

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