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Seifert matrices and the Alexander module — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The unknot has a zero-order Seifert matrix

Statement refuted

Every Seifert matrix of an oriented knot has positive order, meaning that its number of rows and columns is at least one.

Statement references

A Seifert matrix has order equal to the rank of the first homology of the chosen surface

Facts & Assumptions

Given: Full AC, the oriented round unknot, its oriented spanning disk, and the positive-push Seifert-matrix definition.

[A1]

Full AC supplies the countable-choice interface used by the surface definition and normal-form supplier. The Axiom of Choice, AC implies DC implies countable choice

[L1]

The standard countable-choice principle obtained from full AC is equivalent to the surface normal-form supplier's countable-choice formulation for foliated chart pairs. The Axiom of Countable Choice ( ${A C}_{ω}$ ), The countable-choice principle used in the foliation pair, Countable choice is equivalent to nonempty countable products

[F1]

A Seifert surface is connected, compact, smooth, oriented, and has boundary the oriented knot; for a one-boundary-component surface its genus satisfies $χ (F) = 1 - 2 g$ . Oriented Seifert surfaces in the three-sphere

[F2]

A genus-zero compact surface with one boundary component has $H_{1} (F; Z) ≅ Z^{0} = 0$ . Finite surface normal forms, Jordan disks, and torsion control

[F3]

The Seifert matrix is indexed by an ordered integral basis of $H_{1} (F; Z)$ and therefore has order equal to the number of basis elements. Seifert pairing and matrix of an oriented knot

Counterexample

technique · direct

1.1A1L1F1givenconstruct

Take the round unknot $K \subset S^{3}$ with its standard oriented flat disk $F$ in a coordinate ball. The disk is connected, compact, and smooth, and its Euler characteristic is $1$ , so [F1] gives genus $g = 0$ .

2.1F2F3step 1.1givenalgebra∎

By [F2], $H_{1} (F; Z) = 0$ and its integral basis is empty. The matrix in [F3] has one row and one column for each basis element, hence is $0 \times 0$ . Its order is zero, contradicting the refuted positive-order claim.

Sources

The genus-zero surface has free first homology of rank zero

ExampleConstruction: Literature-sourcedVerification: AI-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Trefoil Seifert matrix and basis changes

Example

Assume the Axiom of Choice and take the genus-one Seifert surface for the trefoil described as $K_{1}$ in Cattabriga–Francaviglia, Example 6.2.11. In their basis $f_{1}, f_{2}$ , their negative-first Seifert form has matrix $A = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ , and they identify $K_{1}$ as the trefoil. Their convention is $α (x, y) = lk (x^{-}, y)$ ; the positive-first convention here is $θ (x, y) = lk (x^{+}, y)$ . In the swapped basis $a_{1} = f_{2}$ , $a_{2} = f_{1}$ , the positive-first Seifert matrix is

V = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array}), V - V^{T} = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array}), \det (V - t V^{T}) = 1 - t + t^{2} .

The determinant-one basis change $P_{0} = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ gives $P_{0}^{T} V P_{0} = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ . It changes the off-diagonal entries but not the diagonal entries. The determinant-one basis change $P_{1} = (\begin{array}{cc} 1 & 1 \\ 2 & 3 \end{array})$ gives $P_{1}^{T} V P_{1} = (\begin{array}{cc} 3 & 5 \\ 4 & 7 \end{array})$ , changing every entry. For either $P$ , the transformed Alexander presentation matrix is $P^{T} (V - t V^{T}) P$ , so it presents an isomorphic $Λ$ -module.

Example references

Cattabriga–Francaviglia, Example 6.2.11 and Theorem 6.2.9

The positive-push convention used for $V$

The matrix presentation of the one-variable Alexander module

Facts & Assumptions

Given: Full AC, the cited genus-one oriented trefoil surface, its negative-first matrix in the source basis, and the positive-first convention of this run.

[A1]

Full AC supplies the choice hypotheses of the Seifert-surface and normal-form interfaces. The Axiom of Choice, AC implies DC implies countable choice

[F1]

The source's genus-one example gives $A = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ for $α (x, y) = lk (x^{-}, y)$ and identifies its $K_{1}$ as a trefoil. Cattabriga–Francaviglia, Example 6.2.11, printed p. 45

[F2]

The positive-first matrix is defined by $V_{i j} = lk (a_{i}^{+}, a_{j})$ , linking is symmetric, and moving both curves through a disjoint collar preserves linking. Seifert pairing and matrix of an oriented knot, Well-definedness of the Seifert pairing

[F3]

For an integral basis change $P$ , the new Seifert matrix is $P^{T} V P$ and the Alexander presentation matrix is $V - t V^{T}$ . Seifert matrices of ambient-isotopic pairs are S-equivalent, A Seifert matrix presents the one-variable Alexander module

Proof

1.1A1F1F2givenalgebra

Under [A1], apply the Seifert-pairing convention to the source's supplied oriented surface and basis. In the original source basis, $A_{i j} = lk (f_{i}^{-}, f_{j})$ . By symmetry and a simultaneous positive normal translation, this equals $lk (f_{j}^{+}, f_{i})$ , which is the $(j, i)$ entry of the positive-first matrix in the same basis. Thus that matrix is $A^{T}$ . Let $S = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ be the basis swap. Then $S^{T} A^{T} S = A$ , so the positive-first matrix in the swapped basis is the displayed $V$ .

2.1F2F3step 1.1algebra

Direct multiplication gives $V - V^{T} = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ . Also $V - t V^{T} = (\begin{array}{cc} 1 - t & t \\ - 1 & 1 - t \end{array})$ , whose determinant is $(1 - t)^{2} + t = 1 - t + t^{2}$ . The skew-part identity agrees with the oriented genus-one handle intersection block in the stated positive-normal convention.

2.2F3step 1.1algebra

For $P_{0}$ , $V P_{0} = (\begin{array}{cc} 1 & 1 \\ - 1 & 0 \end{array})$ and $P_{0}^{T} (V P_{0}) = (\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array})$ . For $P_{1}$ , $V P_{1} = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array})$ and $P_{1}^{T} (V P_{1}) = (\begin{array}{cc} 3 & 5 \\ 4 & 7 \end{array})$ . Both determinants are $1$ , so both are integral basis changes; the second matrix differs from $V$ in all four entries.

3.1F3step 2.1step 2.2algebra∎

For either unimodular $P$ , $(P^{T} V P) - t (P^{T} V^{T} P) = P^{T} (V - t V^{T}) P$ . Right multiplication by $P$ does not change the image of the map on $Λ^{2}$ , and left multiplication by the automorphism $P^{T}$ identifies the two cokernels. The module presentation is therefore independent of these basis changes, as asserted.

Remarks

The following schematic is the closure of a two-strand braid with three crossings. Gaps mark under-strands and the uninterrupted local arc at each crossing is the over-strand. It identifies the knot picture only; the matrix and determinant calculations above use the cited source matrix and the explicit convention translation in step 1.1.

\begin{tikzpicture}[scale=0.85,>=stealth]
  \draw[thick] (-1,3)--(-1.35,2.8).. controls (-1.55,2) and (-1.55,1) ..(-1.35,0.2)--(-1,0);
  \draw[thick] (1,3)--(1.35,2.8).. controls (1.55,2) and (1.55,1) ..(1.35,0.2)--(1,0);
  \draw[thick] (-1,3)--(0,2.5)--(1,2)--(0,1.5)--(-1,1)--(0,0.5)--(1,0);
  \draw[thick] (1,3)--(0,2.5)--(-1,2)--(0,1.5)--(1,1)--(0,0.5)--(-1,0);
  \draw[white,line width=4pt] (0.28,2.64)--(-0.28,2.36);
  \draw[thick] (-0.28,2.64)--(0.28,2.36);
  \draw[white,line width=4pt] (0.28,1.64)--(-0.28,1.36);
  \draw[thick] (-0.28,1.64)--(0.28,1.36);
  \draw[white,line width=4pt] (0.28,0.64)--(-0.28,0.36);
  \draw[thick] (-0.28,0.64)--(0.28,0.36);
  \node at (0,3.35) {$\text{two braid strands}$};
  \node at (0,-0.35) {$\text{three crossings; closure arcs at the sides}$};
\end{tikzpicture}