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Grid homology and combinatorial knot Floer theory — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: Not suppliedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Worked grid-complex cases for the unknot

Example

Assume countable choice. Let $U$ be the oriented unknot and let $G_{2}, G_{3}$ be the specified $2$ -by- $2$ and $3$ -by- $3$ grids. Their tilde homology ranks are $2$ and $4$ . For each of these grids, the grid-to-knot-Floer identification gives $\tilde{G H} (G_{n}) ≅ \hat{H F K} (U) \otimes V^{\otimes (n - 1)}$ , where $n = 2$ or $3$ , $V$ has one generator in bidegree $(0, 0)$ and one in bidegree $(- 1, - 1)$ , and $\hat{H F K} (U)$ has rank one by the standard genus-zero diagram and pointed-diagram invariance. Thus the respective ranks are exactly the dimensions supplied by the stable factors $V$ and $V^{\otimes 2}$ .

Example references

The 2-by-2 finite rectangle calculation

The 3-by-3 finite rectangle calculation

The genus-zero one-pointed unknot calculation

The grid-to-HFK identification and stable factor

Ozsvath, Stipsicz, and Szabo, Grid Homology for Knots and Links, Exercise 4.4.3 and Proposition 4.6.15

Manolescu, Ozsvath, and Sarkar, “A combinatorial description of knot Floer homology,” §§3–4

CounterexampleConstruction: Literature-sourcedVerification: Not suppliedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Raw tilde grid homology depends on the unknot grid

Statement refuted

Assume countable choice. The raw tilde grid homology is not a knot invariant by itself. The two specified grid diagrams both represent the oriented unknot, but their tilde homology ranks are $2$ and $4$ . For these grids the grid-to-knot-Floer comparison gives $\tilde{G H} (G_{n}) ≅ \hat{H F K} (U) \otimes V^{\otimes (n - 1)}$ , where $n = 2, 3$ , ${rank}_{F_{2}} V = 2$ , and ${rank}_{F_{2}} \hat{H F K} (U) = 1$ by the standard genus-zero diagram and pointed-diagram invariance. Removing the factor $V^{\otimes (n - 1)}$ would therefore falsely identify different raw ranks with the same rank-one knot invariant.