CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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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.

Statement references

The positive-push matrix convention

The definition of metabolicity

The rational isometric-structure convention

The trefoil matrix from positive Hopf plumbing

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 $V = (\begin{array}{cc} - 1 & 0 \\ - 1 & - 1 \end{array})$ . 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, $(V + V^{T}, V^{- 1} V^{T})$ 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 $V = (\begin{array}{cc} - 1 & 0 \\ - 1 & - 1 \end{array})$ and $V - V^{T} = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ . Its symmetric form $B = V + V^{T} = (\begin{array}{cc} - 2 & - 1 \\ - 1 & - 2 \end{array})$ is negative definite, so the lattice is not metabolic, as the calculation below proves.

1.1A1F1givenalgebra

From [F1], $V - V^{T} = (\begin{array}{cc} 0 & 1 \\ - 1 & 0 \end{array})$ 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 $B = (\begin{array}{cc} - 2 & - 1 \\ - 1 & - 2 \end{array})$ . For every nonzero $(m, n) \in Q^{2}$ , $B ((m, n), (m, n)) = - 2 (m^{2} + m n + n^{2}) = - 2 (m + n / 2)^{2} - 3 n^{2} / 2 < 0$ , so there is no rational isotropic line to which [F4] applies. In particular, for every nonzero integral vector the Seifert self-pairing $- (m^{2} + m n + n^{2})$ is negative; [F2] rules out a rank-one metabolizer. This lattice refutes the claim.

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Sources

Dale Rolfsen, Knots and Links (standard reference, not scraped)
Charles Livingston, A Survey of Classical Knot Concordance (standard reference, not scraped)