LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The singular parameters of the Levine–Tristram form

Statement

Assume full AC. Let $K$ be a smooth oriented knot, let $F$ be any connected compact smooth oriented spanning surface of genus $g$ , let $V$ be its positive-push Seifert matrix, let $Δ_{K} (t)$ be any unit associate of the one-variable Alexander polynomial, and let $ω \in S^{1} ∖ {1}$ . The Hermitian matrix $H_{ω} = (1 - ω) V + (1 - \overset{ˉ}{ω}) V^{T}$ is singular if and only if $Δ_{K} (ω) = 0$ .

Statement references

The positive-push Seifert matrix convention

The determinant presentation of the Alexander polynomial

Facts & Assumptions

Given: Full AC, an oriented knot $K$ , a genus- $g$ spanning surface with Seifert matrix $V$ , and $ω \in S^{1} ∖ {1}$ .

[A1]

Full AC is the ambient choice-function axiom assumed by the Alexander-module and Seifert-pairing suppliers. The Axiom of Choice

[F1]

For this surface, $\det (V - t V^{T})$ is a unit associate of the Alexander polynomial, and its value at $t = 1$ is $\det (V - V^{T}) = \pm 1$ . The Seifert determinant is the knot Alexander polynomial, The square Seifert presentation generates the knot Alexander polynomial, The Alexander polynomial from the zeroth elementary ideal

Proof

1.1A1F1givenalgebra

Since $\overset{ˉ}{ω} = ω^{- 1}$ , put $c = 1 - ω$ . Then $1 - \overset{ˉ}{ω} = - c / ω$ , and therefore $H_{ω} = c (V - ω^{- 1} V^{T})$ . As $c \neq 0$ , taking determinants shows $H_{ω}$ is singular exactly when $\det (V - t V^{T})$ vanishes at $t = ω^{- 1}$ .

2.1F1step 1.1algebra∎

By [F1], the determinant in step 1.1 differs from $Δ_{K} (t)$ by a Laurent unit $\pm t^{m}$ , which never vanishes on $S^{1}$ . For $n = 2 g$ , $M (t)^{T} = V^{T} - t V = - t (V - t^{- 1} V^{T})$ , so taking determinants gives $\det M (t) = t^{n} \det M (t^{- 1})$ ; its zero set is therefore invariant under inversion. Thus the determinant vanishes at $ω^{- 1}$ exactly when $Δ_{K} (ω) = 0$ , proving both implications.

Depends on

Used by

A primitive sixth root makes the trefoil Hermitian form singular Counterexample
Levine–Tristram Hermitian form and signature Definition
One-sided Levine–Tristram signature limits exist Lemma
The classical signature bounds unknotting number Theorem

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Anthony Conway, The Levine–Tristram signature: a survey, 2019–20 MATRIX Annals (2021), pp. 31–56 (arXiv:1903.04477) (standard reference, not scraped)