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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Lee homology of a knot has two generators

Statement

Assume the Axiom of Countable Choice. Let K be a nonempty oriented knot in the standard oriented three-sphere. Its rational Lee homology is a two-dimensional vector space. For every finite regular diagram of K, the two homology classes obtained from the two orientations by the oriented-resolution parity-of-nesting rule form a basis, and the induced quantum filtration is independent of the diagram.

Facts & Assumptions

Given: A knot diagram and its two possible orientations.

[F1]

For a diagram with c components, the 2c canonical orientation cycles form a basis of Lee homology (Lee filtration, spectral sequence, and orientation classes).

[F2]

Filtered Lee homology is independent of the diagram of the oriented link (Lee filtered homology is an oriented link invariant).

[A1]

ACω is the countable-choice principle used by the finite diagram-move supplier (The Axiom of Countable Choice (ACω)).

Proof

Given: ACω and a finite regular diagram D of K.

1.1F1givenalgebra

A knot has exactly one component, so it has exactly 21=2 orientations. Applying [F1] with c=1 shows that the associated canonical classes are linearly independent and span H(CL(D;Q)). Thus its dimension is two.

2.1A1F2step 1.1∎

If D′ is another diagram of the same oriented knot, [A1] supplies the axiom branch used in [F2], and [F2] gives an isomorphism of Lee homology preserving the induced filtration. Therefore the dimension and filtered vector-space type computed from D agree with those computed from D′. The two canonical classes on each diagram form a basis by step 1.1; no identification of individual basis vectors across diagrams is needed for the claimed rank or filtration invariance.

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