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A primitive sixth root makes the trefoil Hermitian form singular

Statement refuted

Assume full AC. The false statement is: “For every $ω \in S^{1} ∖ {1}$ , 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.

Statement references

The form and root-average convention

The positive trefoil matrix and Alexander polynomial

Alexander roots are precisely singular parameters

Existence of the one-sided limits at the root

Facts & Assumptions

Given: Full AC, the positive trefoil matrix $V_{T} = (\begin{array}{cc} - 1 & 0 \\ - 1 & - 1 \end{array})$ , and a primitive sixth root $ω = e^{i π / 3}$ .

[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]

$V_{T}$ is the positive-push trefoil matrix and its Alexander polynomial is $1 - t + t^{2}$ . Trefoil and figure-eight Alexander polynomials from Seifert matrices

[F2]

A root of the Alexander polynomial is exactly a singular parameter for $H_{ω}$ , 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 $ω + \overset{ˉ}{ω} = 1$ , $∣ c ∣^{2} = (1 - ω) (1 - \overset{ˉ}{ω}) = 2 - (ω + \overset{ˉ}{ω}) = 1$ . Direct substitution gives $H_{ω} (V_{T}) = (\begin{array}{cc} - 1 & - \overset{ˉ}{c} \\ - c & - 1 \end{array}), \det H_{ω} = 1 - ∣ c ∣^{2} = 0$ . The matrix is nonzero (its diagonal entries are $- 1$ ), so it is singular. Also $1 - ω + ω^{2} = 0$ , 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 $H_{ω}$ may be used at the root itself.

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Sources

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