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

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The Khovanov diagram procedure terminates and returns a finite cochain complex

Statement

For a nonempty finite regular oriented link diagram $D$ with $n$ ordered crossings, the procedure that enumerates all $2^{n}$ smoothing states, assigns one copy of $A = Z [X] / (X^{2})$ per resolution circle, places a generator in degrees $i = ∣ v ∣ - n_{-}$ and $j = ∣ v ∣ + n_{+} - 2 n_{-} + \sum \deg (a)$ , and sums the signed merge or split edge maps terminates after finitely many operations. Its output is a finite direct sum of finite-rank free abelian groups with a differential of bidegree $(1, 0)$ satisfying $d^{2} = 0$ ; hence it is a bounded bigraded cochain complex.

Statement references

State enumeration and edge-map procedure

Square-zero property of the signed differential

Khovanov, A categorification of the Jones polynomial (1999), §§3.3–4.2

Bar-Natan, On Khovanov's categorification of the Jones polynomial (2002), §3

Kauffman, An Introduction to Khovanov Homology (2022), §4

Zhang, Notes on Khovanov homology (2025), §§3.5–3.7

Facts & Assumptions

Given: A nonempty finite regular oriented link diagram $D$ with an ordered crossing set of cardinality $n$ and the finite state procedure in the Theorem.

[F1]

The procedure traverses each function $v : C (D) \to {0, 1}$ once, uses a finite number of edge maps per state, and assigns $A^{\otimes k (v)}$ with the stated gradings (Khovanov chain complex from a diagram).

[F2]

The signed edge sum has square zero for every such finite ordered diagram (The Khovanov cube differential squares to zero).

Proof

1.1F1givenalgebra

If $n = ∣ C (D) ∣$ , there are exactly $2^{n}$ states; each complete smoothing of a finite closed diagram has finitely many circles. Since $A$ is free of rank two with basis $1, X$ , each state module is finite-rank free, and the finite sum of these modules is finite-rank free. The procedure visits each of finitely many states and at most $n$ outgoing edges per state, so it terminates.

2.1F1F2step 1.1algebra∎

Each edge changes $∣ v ∣$ and $i$ by $+ 1$ and changes the algebraic degree by $- 1$ through $m$ or $Δ$ , so its total quantum-degree change is $+ 1 - 1 = 0$ . By [F2] the returned differential squares to zero. The state set is finite, so only finitely many homological degrees occur; this is a bounded bigraded cochain complex.

Depends on

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Sources

Mikhail Khovanov, A categorification of the Jones polynomial, arXiv:math/9908171v2, §§3.3–4.2 (standard reference, not scraped)
Dror Bar-Natan, On Khovanov's categorification of the Jones polynomial, Algebraic & Geometric Topology 2 (2002), §3 (standard reference, not scraped)
Louis H. Kauffman, An Introduction to Khovanov Homology, arXiv:1107.1524v2 (2022), §4 (standard reference, not scraped)
Melissa Zhang, Notes on Khovanov homology, arXiv:2501.03115v1, §§3.5–3.7 (standard reference, not scraped)