For a Seifert matrix $V$ of a knot $K$, the Alexander module has presentation matrix $VtV^T$. 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 Sequivalence class of the Seifert matrix is an invariant of the knot, and uniquely characterizes the Blanchfield pairing. There is a bijective correspondence between Sequivalence classes of Seifert matrices and Blanchfield pairings.
Trotter gave examples of knots with the same Alexander polynomial but nonSequivalent Seifert matrices. My question is what additional information we need to reconstruct the Blanchfield pairing (Seifert matrix up to Sequivalence) from the Alexander polynomial.
2 Answers
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 Sequivalence 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.

$\begingroup$ Is there a good reference for the statement in your last paragraph? $\endgroup$ Jun 12, 2011 at 20:04

1$\begingroup$ There's a bunch. Hillman's "Algebraic invariants of links" is a good one. Usually the transversality is expressed by representing one of $X$ and $Y$ in simplicial and the dual polyhedral coordinates (like in Poincare's proof of Poincare duality). But the basic idea of representing Poincare duality torsion pairings like that goes back at least as far as Seifert and Threlfall's textbook. And it falls out of the mathematics of Poincare duality and the Universal Coefficient Theorem very naturally. $\endgroup$ Jun 12, 2011 at 20:46

$\begingroup$ Thanks! I have Hillman's book already; I'll read the section on the Blanchfield pairing more carefully. $\endgroup$ Jun 13, 2011 at 2:37
Clearly the signature, and more generally the TristramLevine signatures, would be needed.

1$\begingroup$ I think it was Trotter that showed the TL signatures (equivalently the Milnor signatures) determine the blanchfield pairing rationally. But integrally I doubt the answer is that clean. $\endgroup$ Dec 30, 2009 at 22:53

$\begingroup$ Can you give a reference for that? Integrally, I expect some sort of Minkowski units would be needed, but I don't know whether even that is enough. $\endgroup$ Dec 31, 2009 at 0:46

1$\begingroup$ Trotters's "On Sequivalence of Seifert Matrices" (1973) shows Sequivalence is the same as isometry of the Blanchfield pairing. In the same paper he composes the Blanchfield pairing with what's now called the "Trotter trace" to show that rationally Blanchfield pairing classification reduces to the classification of a skewsymmetric rationalvalued bilinear form on the Alexander module. From here the reference is Milnor's "On Isometries of inner product spaces". Scanning through the paper I think you might need real coefficients for the classification. $\endgroup$ Jan 1, 2010 at 22:05


1$\begingroup$ @Ian Agol It was 10 years ago and I don't remember what I meant. 2021 Daniel wishes 2011 Daniel had made good on his word and updated the answer! $\endgroup$ Feb 11, 2021 at 21:09