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Khovanov homology is an oriented link invariant

Statement

Assume the Axiom of Countable Choice. For every nonempty oriented link $L$ , the bigraded integral groups $K h^{i, j} (L) : = H^{i} (C^{*, j} (D))$ , where $D$ is any finite regular oriented diagram of $L$ and $C (D)$ is the normalized Khovanov cochain complex, are independent of the choice of $D$ up to bigraded isomorphism.

Statement references

Countable choice axiom

Oriented Reidemeister equivalence

Finite cochain-complex output

Local chain homotopy equivalences

Chain homotopy equivalence convention

Khovanov, A categorification of the Jones polynomial (1999), §§5.1–5.4

Bar-Natan, On Khovanov's categorification of the Jones polynomial (2002), §§3.5.1–3.5.5

Kauffman, An Introduction to Khovanov Homology (2022), §§4–5

Zhang, Notes on Khovanov homology (2025), §§3.6–3.7

Facts & Assumptions

Given: AC $_{ω}$ , two finite regular diagrams of the same oriented link, and their normalized cochain complexes.

[F1]

AC $_{ω}$ is the axiom that every countable family of nonempty sets has a choice function (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F2]

Under AC $_{ω}$ , diagrams of the same oriented link are connected by a finite sequence of planar isotopies and oriented Reidemeister I, II, and III moves (Reidemeister's theorem for oriented diagrams).

[F3]

Every planar isotopy and oriented Reidemeister I, II or III move induces a bigraded chain homotopy equivalence of the normalized complexes (Khovanov complexes are chain homotopy equivalent under Reidemeister moves).

[F4]

A bounded cochain complex has defined cohomology in each bidegree, and chain homotopy equivalent complexes induce isomorphic cohomology (The Khovanov diagram procedure terminates and returns a finite cochain complex, A chain homotopy equivalence).

Proof

1.1F1F2F3givenchoose

By [F1] and [F2], choose a finite sequence $D = D_{0}, D_{1}, \dots, D_{r} = D^{'}$ whose successive diagrams differ by planar isotopy or one oriented Reidemeister move. By [F3], each pair has a bigraded chain homotopy equivalence, with chain maps in both directions and homotopies from their composites to the respective identities. The normalization in that supplier cancels the raw shifts, so these maps have bidegree $(0, 0)$ .

2.1F3F4step 1.1algebra∎

A chain map takes cocycles to cocycles and coboundaries to coboundaries. If two chain maps differ by $d h + h d$ , their values on a cocycle differ by a coboundary, so they induce the same cohomology map. Thus each equivalence in step 1.1 gives mutually inverse maps in every bidegree. Composing the finitely many induced isomorphisms identifies the cohomology of $C (D)$ and $C (D^{'})$ . It follows that the groups $K h^{i, j} (L)$ are independent of the chosen diagram up to bigraded isomorphism. The empty sequence gives the identity, and no assertion of a canonical isomorphism for all move sequences is needed.

Depends on

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Sources

Mikhail Khovanov, A categorification of the Jones polynomial, arXiv:math/9908171v2, §§5.1–5.4 (standard reference, not scraped)
Dror Bar-Natan, On Khovanov's categorification of the Jones polynomial, Algebraic & Geometric Topology 2 (2002), §§3.5.1–3.5.5 (standard reference, not scraped)
Louis H. Kauffman, An Introduction to Khovanov Homology, arXiv:1107.1524v2 (2022), §§4–5 (standard reference, not scraped)
Melissa Zhang, Notes on Khovanov homology, arXiv:2501.03115v1, §§3.6–3.7 (standard reference, not scraped)