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The Khovanov to Floer spectral sequence — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

The endpoints do not determine intermediate pages

Statement refuted

The false claim is that specifying the isomorphism class of the total homology of a bounded filtered chain complex determines the isomorphism class of its early spectral-sequence pages. More strongly, the false claim is that two filtered complexes with isomorphic total homology must have isomorphic $E_{1}$ pages.

Statement references

Filtered chain complex convention

Homological spectral sequence indexing

Bounded convergence and abutment

Weibel, An Introduction to Homological Algebra, Chapter 5

Sharifi, Homological Algebra, §4

Facts & Assumptions

Given: Chain complexes over $F_{2}$ with an increasing finite filtration preserved by the differential.

[F1]

A filtered chain complex has $d (F_{p} C) \subseteq F_{p} C$ (Filtered chain complex).

[F2]

For the spectral sequence of an increasing filtered chain complex, $E_{0}$ is the associated graded with its filtration-preserving differential, and $E_{r + 1}$ is the homology of $(E_{r}, d_{r})$ (Homological spectral sequence, The next page is the homology of the current page).

[F3]

A finite filtration converges to the associated graded of total homology (Bounded filtered complex spectral sequence abuts to filtered homology).

Proof

1.1F1givenconstruct

Let $A$ be the one-generator complex $F_{2} {z}$ in chain degree $0$ , with $d z = 0$ and filtration degree $0$ . Let $B$ have generators $x$ in degree $1$ and $y, z$ in degree $0$ , differential $d x = y$ , $d y = d z = 0$ , and filtration $F_{0} B = F_{2} {y, z}$ , $F_{1} B = B$ , $F_{- 1} B = 0$ . In both complexes $d$ preserves the filtration, so [F1] applies.

2.1F2step 1.1algebra

In $A$ , $E_{0} = E_{1} = F_{2} {z}$ because the differential is zero. In $B$ , the associated graded has basis $z, y$ in filtration $0$ and $x$ in filtration $1$ ; its $d_{0}$ is zero because $d x = y$ strictly lowers filtration. Therefore $E_{1} (B)$ has dimension $3$ . The $d_{1}$ differential sends the class of $x$ to the class of $y$ and sends $z$ to zero, so $E_{2} (B) = F_{2} {z}$ ; likewise $E_{2} (A) = F_{2} {z}$ .

2.2step 1.1algebra

The total homology of $A$ is $F_{2} {z}$ . In $B$ , the pair $x \mapsto y$ is contractible and the remaining class is $z$ , so $H (B) ≅ F_{2} {z} ≅ H (A)$ . The pair by itself has zero total homology but nonzero $E_{1} = F_{2} {x, y}$ , with its $d_{1}$ killing both classes.

3.1F3step 2.1step 2.2∎

By step 2.1, $\dim E_{1} (A) = 1$ while $\dim E_{1} (B) = 3$ , so their $E_{1}$ pages cannot be isomorphic even though their total homology groups agree by step 2.2. Convergence [F3] identifies only the limiting page with the associated graded of the shared total homology; it does not determine the earlier pages. Thus the stated endpoint-only claim is false.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

The trefoil Khovanov-to-Floer rank calculation

Example

Assume AC and the ACω Heegaard Floer analytic package. Let $T$ be the oriented left-handed trefoil, the closure of the two-strand braid $σ_{1}^{- 3}$ , and choose a marked point away from its three crossings. Its mirror $\overline{T}$ is the closure of $σ_{1}^{3}$ , depicted below. Use the smoothing convention in which the zero-resolution of this mirror diagram is the through pairing and the one-resolution is the cap-cup pairing. The reduced Khovanov cohomology of $\overline{T}$ over $F_{2}$ has total rank $3$ . The double branched cover of $T$ is a lens space $L (3, 1)$ up to orientation, and the hat Floer group of this lens space over $F_{2}$ has total rank $3$ . Thus the $E_{2}$ page and the abutment have equal total rank. Every spectral-sequence differential $d_{r}$ with $r \geq 2$ vanishes; the rank comparison gives no conclusion about the earlier differential $d_{1}$ from $E_{1}$ to $E_{2}$ .

\begin{tikzpicture}[scale=0.55]
\draw (-1,6) .. controls (-3,6) and (-3,0) .. (-1,0);
\draw (1,6) .. controls (3,6) and (3,0) .. (1,0);
\foreach \yt/\yb in {6/4,4/2,2/0} {
  \draw (-1,\yt) -- (1,\yb);
  \draw[white,line width=4pt] (1,\yt) -- (-1,\yb);
  \draw[black,line width=1.1pt] (1,\yt) -- (-1,\yb);
}
\node at (0,7) {$\widehat{\sigma_1^3}$};
\end{tikzpicture}

The drawing depicts the closure of the braid word; the southwest-to-northeast diagonal is drawn over the northwest-to-southeast diagonal, matching the positive-generator convention in the rank calculation. The algebraic calculation uses the smoothing cube and Frobenius maps, not the picture.

Example references

Finite reduced Khovanov calculation

Khovanov-to-branched-cover spectral sequence

Trefoil branched-cover identification

Finite lens-space hat Floer rank

Ozsváth and Szabó, On the Heegaard Floer homology of branched double-covers, §§3 and 6

Lipshitz, Heegaard Floer Homologies, §5.3

Facts & Assumptions

Given: AC, the ACω Heegaard Floer analytic package, and the left-handed trefoil and marked mirror diagram of the Example, with coefficients $F_{2}$ .

[F1]

The reduced Khovanov homology of a trefoil has rank three supplies the exact finite marked reduced cube calculation: the mirror has total reduced Khovanov rank three.

[F2]

The trefoil double branched cover is a lens space of order three supplies the actual annular-cover and meridian-slope identification, up to orientation, with $L (3, 1)$ . Its retained full slope calculation is used without replacing the peripheral basis by an intersection number.

[F3]

The genus-one hat Floer complex of an order-three lens space has rank three supplies rank three for either orientation. The retained ordinary-hat analytic/invariance closure already accepts this whole calculation; no disk-count proof is repeated here.

[L1]

The Khovanov to branched-cover Heegaard Floer spectral sequence is the separate actual spectral-sequence existence input. Its surgery-cube and unlink-map constructions are not consequences of [F1]–[F3].

Proof

1.1F1F2F3L1given

Under the stated assumptions, [L1] supplies the spectral sequence for $T$ . Apply [F1] to its specified marked mirror to obtain $\dim E_{2} = 3$ . Apply [F2] and [F3] to obtain abutment dimension $3$ , regardless of the allowed orientation reversal. The finite image filtration on that abutment has associated graded $E_{\infty}$ , whose total dimension equals the abutment dimension, so $\dim E_{\infty} = 3$ as well.

2.1F1F2F3L1step 1.1algebra∎

For any finite-dimensional page with square-zero differential $d$ , rank-nullity gives $\dim H = \dim E - 2 rank d$ . The rank comparison here is between $E_{2}$ and $E_{\infty}$ , whose total dimensions are both $3$ . A nonzero differential $d_{r}$ for $r \geq 2$ would strictly lower the total dimension, so every such differential vanishes. This argument does not control $d_{1}$ , which is the earlier differential used to form $E_{2}$ . The OS sequence's existence for this diagram still requires [L1]; the dimension argument does not construct it.

The Khovanov to Floer spectral sequence — Examples and Counterexamples

2 · Logical flowchart

3 · Definitions, theorems and proofs

4 · Examples, counterexamples and false statements

The endpoints do not determine intermediate pages

Statement refuted

Statement references

Facts & Assumptions

Proof

The trefoil Khovanov-to-Floer rank calculation

Example

Example references

Facts & Assumptions

Proof

Sources