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Algebraic concordance and metabolic forms — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Worked cases for metabolic Seifert forms

Example

Assume full AC. The disk Seifert surface of the oriented unknot has the rank-zero Seifert lattice and is metabolic. In the positive-push convention, the standard positive Hopf-band plumbing for the right-handed trefoil has matrix V=(−10−1−1). Since V is invertible, its rational isometric structure has symmetric form B=V+VT=(−2−1−1−2), which is negative definite; the integral Seifert lattice is not metabolic.

Facts & Assumptions

Given: Full AC, the disk Seifert surface of the oriented unknot, and the standard positive Hopf-plumbing Seifert surface of the right-handed trefoil.

[A1]

Full AC is the choice-function axiom used by the knot Seifert-pairing interface. The Axiom of Choice

[F1]

The rank-zero lattice is allowed; its zero subgroup is a primitive rank-zero metabolizer. Metabolic Seifert forms and algebraic knot concordance

[F2]

The positive Hopf-plumbing basis has matrix V=(−10−1−1). A positive Hopf plumbing gives the trefoil Seifert matrix

[F3]

Since V is invertible, (V+VT,V−1VT) satisfies the rational isometric-structure definition. Integral Witt classes and rational Seifert isometric structures, An invertible Seifert matrix gives a rational isometric structure

[F4]

A metabolic lattice must have a rank-half direct summand on which the Seifert pairing vanishes. Metabolic Seifert forms and algebraic knot concordance

Proof

1.1A1F1givenalgebra

The disk has H1=0, so its Seifert matrix is the empty matrix. The subgroup M=0 has rank zero, is a direct summand, and the pairing vanishes on it. By [F1], this is metabolic and represents the identity class.

2.1F2F3F4algebra∎

By [F2] and [F3], the trefoil's rational structure has symmetric form B=(−2−1−1−2). For a vector (m,n)∈Q2, B((m,n),(m,n))=−2(m2+mn+n2)=−2(m+n/2)2−3n2/2, which is negative whenever (m,n)≠(0,0). Thus the rational structure has no nonzero isotropic line. For an integral vector the Seifert self-pairing is −(m2+mn+n2)<0, so no rank-one subgroup has identically zero Seifert pairing, and [F4] rules out an integral metabolizer.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Unimodular skew part does not force metabolicity

Statement refuted

Assume full AC. The false claim is: every positive-rank integral Seifert lattice with unimodular alternating part is metabolic.

Facts & Assumptions

Given: Full AC and the positive-push Seifert matrix of the right-handed trefoil.

[A1]

Full AC is the choice-function axiom used by the knot Seifert-pairing interface. The Axiom of Choice

[F1]

The positive Hopf plumbing has matrix V=(−10−1−1). A positive Hopf plumbing gives the trefoil Seifert matrix

[F2]

A metabolic rank-two lattice must have a rank-one subgroup on which V vanishes, and hence every vector in that subgroup has zero self-pairing. Metabolic Seifert forms and algebraic knot concordance

[F3]

Since V is invertible, (V+VT,V−1VT) satisfies the rational isometric-structure definition. Integral Witt classes and rational Seifert isometric structures, An invertible Seifert matrix gives a rational isometric structure

[F4]

A rational isotropic line for an integral bilinear form saturates to a rank-one integral direct summand with zero restriction. A rational isotropic subspace saturates to an integral direct summand

Counterexample

The right-handed trefoil's positive-push Seifert lattice has V=(−10−1−1) and V−VT=(01−10). Its symmetric form B=V+VT=(−2−1−1−2) is negative definite, so the lattice is not metabolic, as the calculation below proves.

1.1A1F1givenalgebra

From [F1], V−VT=(01−10) has determinant 1, so this is a positive-rank integral Seifert lattice satisfying the false claim's hypothesis.

2.1F2F3F4algebra∎

By [F3], the associated rational form is B=(−2−1−1−2). For every nonzero (m,n)∈Q2, B((m,n),(m,n))=−2(m2+mn+n2)=−2(m+n/2)2−3n2/2<0, so there is no rational isotropic line to which [F4] applies. In particular, for every nonzero integral vector the Seifert self-pairing −(m2+mn+n2) is negative; [F2] rules out a rank-one metabolizer. This lattice refutes the claim.

Sources