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Levine–Tristram signatures and the Arf invariant — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

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

A primitive sixth root makes the trefoil Hermitian form singular

Statement refuted

Assume full AC. The false statement is: “For every ω∈S1∖{1}, the Levine–Tristram Hermitian matrix of every knot is nonsingular.” The positive trefoil matrix at a primitive sixth root is singular. At that root the nondegenerate signature calculation does not apply; the defined knot invariant is the one-sided-average value.

Facts & Assumptions

Given: Full AC, the positive trefoil matrix VT=(−10−1−1), and a primitive sixth root ω=eiπ/3.

[A1]

Full AC is the ambient choice-function axiom. The signature and root-average conventions are specified separately by the form definition; the trefoil matrix is supplied by [F1]. The Axiom of Choice, Levine–Tristram Hermitian form and signature

[F1]

VT is the positive-push trefoil matrix and its Alexander polynomial is 1−t+t2. Trefoil and figure-eight Alexander polynomials from Seifert matrices

[F2]

A root of the Alexander polynomial is exactly a singular parameter for Hω, and the invariant at such a root is the average of one-sided signatures. The singular parameters of the Levine–Tristram form, Levine–Tristram Hermitian form and signature

[F3]

Each one-sided limit at this root exists as an integer, so the arithmetic mean prescribed by the definition is defined. One-sided Levine–Tristram signature limits exist

Proof

1.1A1F1givenalgebra

Let c=1−ω. Since ω+ωˉ=1, ∣c∣2=(1−ω)(1−ωˉ)=2−(ω+ωˉ)=1. Direct substitution gives Hω(VT)=(−1−cˉ−c−1),det⁡Hω=1−∣c∣2=0. The matrix is nonzero (its diagonal entries are −1), so it is singular. Also 1−ω+ω2=0, confirming this is an Alexander root.

2.1F2F3step 1.1algebra∎

The zero determinant refutes the universal nonsingularity statement. By [F2], this root is precisely where the definition uses the arithmetic mean of the two one-sided signature limits; [F3] guarantees that this average is defined. No signature formula requiring an invertible Hω may be used at the root itself.

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

Trefoil and figure-eight Levine–Tristram and Arf calculations

Example

Assume full AC. Use the positive-push Seifert matrices VT=(−10−1−1),VF=(−10−11) for the positive trefoil and the figure-eight knot, respectively. At ω=−1, their Levine–Tristram signatures are −2 and 0. In the displayed symplectic bases both quadratic forms have q(a)=q(b)=1, so both Arf invariants are 1. Their normalized Conway polynomials are 1+z2 and 1−z2, respectively.

Facts & Assumptions

Given: Full AC and the two displayed oriented knot Seifert matrices in the positive-push convention.

[A1]

Full AC is the ambient hypothesis for the Seifert pairing. The Axiom of Choice

[F1]

The trefoil matrix VT is the positive-push matrix for the positive trefoil. Trefoil and figure-eight Alexander polynomials from Seifert matrices

[F2]

The figure-eight matrix VF is a positive-push Seifert matrix; its skew part is the standard symplectic block. Seifert pairing and matrix of an oriented knot, The skew part of a Seifert matrix is the surface intersection form

[F3]

The Arf invariant is q(a)q(b) for a genus-one symplectic basis and equals a2 modulo two. Knot Arf invariant from the Seifert form, The Arf invariant is the second Conway coefficient modulo two

[F4]

For a Seifert matrix V, the Seifert determinant ΘK(u)=det⁡(u−1V−uVT) gives the normalized Conway polynomial ΘK(u)=∇K(u−u−1). Seifert determinant constructs the normalized Conway polynomial

[F5]

The Levine–Tristram signature of the knot is independent of the chosen surface and basis, including the one-sided-average value at an Alexander root. Seifert stabilization preserves the Levine–Tristram signature

Proof

1.1A1F1F2F5givenalgebra

Since 1−(−1)=2, H−1(V)=2(V+VT). For the trefoil, VT+VTT=(−2−1−1−2) has eigenvalues −1,−3, so the signature is −2. For the figure-eight, VF+VFT=(−2−1−12) has determinant −5<0, hence one positive and one negative eigenvalue, so the signature is 0.

2.1F1F2F3F4step 1.1algebra∎

The standard intersection block is (01−10) for both matrices. The quadratic values on the basis vectors are their diagonal entries modulo two, namely (1,1) for each matrix. Therefore Arf⁡=1⋅1=1 for each knot. Their Alexander determinants are det⁡(VT−tVTT)=1−t+t2 and det⁡(VF−tVFT)=−1+3t−t2. Substituting t1/2−t−1/2=z with the symmetric normalization gives ∇T=1+z2 and ∇F=1−z2; both z2 coefficients are odd, agreeing with the Arf values.

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