Equivalent knots have same Alexander polynomial

Question

We have defined the Alexander polynomial for an oriented link $L$ via Seifert matrices, i.e. if $A$ is a Seifert matrix for $L$, then $\det (tA-A^T)$ is the Alexander polynomial for $L$. This is a generator for the first elementary ideal of the $\mathbb{Z}[t^{-1},t]$-module $H_1(X_{\infty})$. So the Alexander polynomial is unique up to multiplication with units $\pm t^{\pm n}$. We also know that the matrix $tA-A^T$ is a presentation matrix for the $\mathbb{Z}[t^{-1},t]$-module $H_1(X_{\infty})$ and any other presentation matrix would give the same determinant.

The Alexander polynomial for the unknot is $t$ and one can show that the pretzel knot $P(-3,5,7)$ has also Alexander polynomial $t$ but, i.e. via the Jones polynomial, one can show that this is a non-trivial knot.

That is that knots with same Alexander polynomial need not to be equivalent. But knots being equivalent have the same Alexander polynomial, I have read.

Can somebody explain why? Have equivalent knots the same Seifert matrix? Are the homology groups $H_1(X_{\infty})$ equal for equivalent knots?

Answer 1

The Seifert matrix for a link is determined up to S-equivalence. What this means is that if $F_1$ and $F_2$ are surfaces whose boundaries are equivalent links, then their corresponding Seifert matrices are S-equivalent. Another property of S-equivalent matrices is that they give the same Alexander polynomial. I suggest reading about S-equivalence in which ever text you are using.

Answer 2

Ignoring Seifert surfaces: the Alexander polynomial of a group $G$ whose abelianization is $\mathbb{Z}$ is the generator of the elementary ideal of the $\mathbb{Z}[t,t^{-1}]$-module which is the abelianization of $[G,G]$, where $t$ is one of the two generators of the abelianization of $G$. (That is, the Alexander polynomial of a space $X$ such that $H_1(X)=\mathbb{Z}$ is the generator of the elementary ideal of the $\mathbb{Z}[t,t^{-1}]$-module $H_1(X_\infty)$, where $X_\infty$ is the infinite cyclic cover of $X$, where $t$ is one of the two generators of $H_1(X)$.)

Reidemeister moves do not change the orientation of the meridian of a knot $K$, and they also do not change $\pi_1(X)$, so neither $H_1(X)$ nor $H_1(X_\infty)$ change, hence the Alexander polynomial does not change.