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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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An invertible Seifert matrix gives a rational isometric structure

Statement

Assume full AC for the knot Seifert-pairing convention. Let V∈M2g(Z) be an invertible Seifert matrix and put B=V+VT and T=V−1VT on Q2g. Then B is a nonsingular symmetric bilinear form, T is an invertible B-isometry, and I−T is invertible; hence (Q2g,B,T) is a rational Seifert isometric structure.

Facts & Assumptions

Given: Full AC and an invertible integral Seifert matrix V.

[A1]

Full AC is the stated hypothesis of the Seifert-pairing interface. The Axiom of Choice

[F1]

For a knot Seifert matrix, V−VT is the unimodular surface intersection matrix. The skew part of a Seifert matrix is the surface intersection form

[F2]

A rational Seifert isometric structure has a nonsingular symmetric form B, an invertible B-isometry T, and invertible I−T. Integral Witt classes and rational Seifert isometric structures

Proof

1.1A1F1givenalgebra

The matrix B=V+VT is symmetric. Modulo 2, V+VT≡V−VT; by [F1] the determinant of V−VT is odd, so det⁡B is odd and in particular nonzero. Thus B is nonsingular over Q.

2.1F2step 1.1algebra

Since V and VT are invertible over Q, T=V−1VT is invertible. Direct multiplication gives TTBT=V(VT)−1(V+VT)V−1VT=V+VT=B, because the two summands in the middle expression simplify respectively to V and VT. Therefore T is a B-isometry.

3.1F1F2step 2.1algebra∎

Finally I−T=V−1(V−VT), which is invertible by invertibility of V and [F1]. The three defining conditions in [F2] hold, proving the claim.

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