Question

For a Seifert matrix $V$ of a knot $K$, the Alexander module has presentation matrix $V-tV^T$. The determinant of this matrix is the Alexander polynomial, which is the order of the Alexander module. In particular, the Alexander module is a torsion module, and has a linking form, called the Blanchfield pairing. The S-equivalence class of the Seifert matrix is an invariant of the knot, and uniquely characterizes the Blanchfield pairing. There is a bijective correspondence between S-equivalence classes of Seifert matrices and Blanchfield pairings.
Trotter gave examples of knots with the same Alexander polynomial but non-S-equivalent Seifert matrices. My question is what additional information we need to reconstruct the Blanchfield pairing (Seifert matrix up to S-equivalence) from the Alexander polynomial.

Answer 1

The Blanchfield pairing has many formulations, I like to think of it as a sesquilinear form:

$$ A \otimes A \to \Lambda / \mathbb Z[t^\pm] $$

where $A$ is the Alexander module and $\Lambda$ is the field of fractions of $\mathbb Z[t^\pm]$. This pairing has to be a duality isomorphism, ie: the adjoint

$$ \overline{A} \to Hom_{\mathbb Z[t^\pm]} (A, \Lambda/\mathbb Z[t^\pm]) $$

is an isomorphism of $\mathbb Z[t^\pm]$-modules. $\overline{A}$ is $A$ but given the opposite action of $\mathbb Z[t^\pm]$ (you substitute $t \longmapsto t^{-1}$ before multiplication by a polynomial)

The Blanchfield pairing can be anything of that form. So you take the Alexander module, and soup it up with such an isomorphism between $\overline{A}$ and its ``Ext dual'' $Hom_{\mathbb Z[t^\pm]} (A, \Lambda/\mathbb Z[t^\pm]) $. That is the extra information in the S-equivalence class.

edit: the pairing has a nice geometric interpretation. $A$ is $H_1(\tilde C)$ where $\tilde C \to C$ is the universal abelian cover of the knot complement. Since $A$ is $\mathbb Z[t^\pm]$-torsion, given any $[x] \in A$ let $p$ be such that $px = \partial X$. Then you define the pairing $\langle x, y\rangle = (\sum_i (X \cap t^{i}y)t^i)/p$ provided $X$ and $y$ are transverse representatives when projected to $C$ (in any way that that makes sense). Here $\cap$ is the standard algebraic intersection number of transverse chains.

Answer 2

Clearly the signature, and more generally the Tristram-Levine signatures, would be needed.