LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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

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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.

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A metabolic Seifert matrix has Fox–Milnor factorization

Statement

Let $g$ be a nonnegative integer, let $L$ be a free abelian group of rank $2 g$ , and let $V : L \times L \to Z$ be a bilinear form whose matrix in an integral basis satisfies $\det (V - V^{T}) = \pm 1$ . Let $M \subseteq L$ be a direct summand of rank $g$ such that $V (x, y) = 0$ for every $x, y \in M$ . Write $d_{V} (t) = \det (V - t V^{T})$ , using the empty determinant $1$ when $g = 0$ . Then there is a polynomial $f (t) \in Z [t]$ such that

$d_{V} (t) = t^{g} f (t) f (t^{- 1}) .$

In the Laurent polynomial ring, define $p ≐ q$ to mean $p = \pm t^{k} q$ for some integer $k$ . Thus $d_{V} (t) ≐ f (t) f (t^{- 1})$ .

Facts & Assumptions

Given: A finite-rank integral bilinear form $(L, V)$ with the stated unimodular skew part, and a rank- $g$ direct summand $M$ on which $V$ vanishes.

[A1]

The stated hypotheses are the entire input: $L ≅ Z^{2 g}$ , $\det (V - V^{T}) = \pm 1$ , $M$ is a direct summand of rank $g$ , and $V ∣_{M \times M} = 0$ .

Proof

1.1A1givenchooseconstruct

If $g = 0$ , then $L = M = 0$ , so the empty determinant and $f = 1$ give the formula. Suppose $g > 0$ . Write $L = M \oplus N$ and choose integral bases of the two rank- $g$ summands. In the combined basis, $V = (\begin{array}{cc} 0 & B \\ C & D \end{array})$ with $B, C, D \in M_{g} (Z)$ , because $V$ vanishes on $M \times M$ .

2.1A1step 1.1algebra

A basis change $P \in {GL}_{2 g} (Z)$ replaces $V - t V^{T}$ by $P^{T} (V - t V^{T}) P$ and leaves its determinant unchanged, since $(\det P)^{2} = 1$ . Hence it suffices to calculate in the basis of step 1.1.

3.1A1step 2.1algebra

Put $f (t) = \det (B - t C^{T})$ . In $V - t V^{T}$ the upper-left $g \times g$ block is zero, the upper-right block is $B - t C^{T}$ , and the lower-left block is $C - t B^{T}$ . Swapping its two row blocks contributes $(- 1)^{g^{2}} = (- 1)^{g}$ and produces a block-triangular matrix, so $d_{V} (t) = (- 1)^{g} \det (B - t C^{T}) \det (C - t B^{T})$ . Also $t^{g} f (t^{- 1}) = \det (t B - C^{T}) = (- 1)^{g} \det (C - t B^{T})$ , by factoring $- 1$ from each row and transposing. Therefore $d_{V} (t) = t^{g} f (t) f (t^{- 1})$ . The assumption on $V - V^{T}$ gives $d_{V} (1) = \pm 1$ , so the displayed determinant is nonzero.

4.1step 1.1step 3.1algebra∎

Since $t^{g}$ is a unit in $Z [t^{\pm 1}]$ , the exact identity in step 3.1 implies $d_{V} (t) ≐ f (t) f (t^{- 1})$ . This proves the factorization, including the rank-zero case of step 1.1.

Used by

Locally flat sliceness implies Fox–Milnor factorization Theorem

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Charles Livingston, A Survey of Classical Knot Concordance (standard reference, not scraped)
Anthony Conway, The Levine–Tristram Signature: A Survey (standard reference, not scraped)