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Knot Floer genus and fibering detection — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A null-homologous unknot in a homology sphere can have reducible exterior

Statement refuted

Assume the Axiom of Choice. An integral homology 3-sphere is a closed connected oriented 3-manifold Y with H∗(Y;Z)≅H∗(S3;Z); a knot K⊂Y is null-homologous when [K]=0 in H1(Y;Z). The assertion that every smooth null-homologous knot in an integral homology 3-sphere has irreducible complement is false, where irreducible means that every embedded 2-sphere in the complement bounds a 3-ball there.

Diagram

The connected-sum sphere separates the two punctured Poincaré summands; the unknot lies in a smaller ball in the left summand.

diagram failed to render:
\begin{tikzpicture}[scale=0.95]
  \draw[rounded corners,thick] (-2,-1.15) rectangle (1.45,1.15);
  \draw[rounded corners,thick] (2.55,-1.15) rectangle (6,1.15);
  \node at (-0.25,0.55) {$P\setminus\operatorname{int}B^3$};
  \node at (4.25,0.55) {$P\setminus\operatorname{int}B^3$};
  \draw[dashed,thick] (2,-1.15)--(2,1.15);
  \node[above] at (2,1.15) {$S^2$};
  \draw[thick] (-1.2,-0.1) circle (0.55);
  \node at (-1.2,0.15) {$B_0$};
  \draw[thick] (-1.2,-0.1) circle (0.2);
  \node at (-1.2,-0.1) {$U$};
  \node at (-1.2,-0.82) {$N(U)\subset B_0$};
\end{tikzpicture}

Facts & Assumptions

Given: AC, the Poincaré sphere P, the connected sum Y=P#P, a connected-sum sphere S, a standard round unknot U in a 3-ball B0 contained in the first summand and disjoint from S, and a tubular neighborhood N(U)⊂B0.

[F1]

The Poincaré sphere is a nontrivial integral homology sphere: H1(P;Z)=H2(P;Z)=0 and π1(P)≠1.

[F2]

Mayer–Vietoris sequence in singular homology: for an open cover, the sequence H1(U∩V)→H1(U)⊕H1(V)→H1(X)→H0(U∩V)→H0(U)⊕H0(V) is exact.

[F3]

Alexander duality for compact locally contractible subsets of a sphere, Homology of spheres, and The universal coefficient theorem for cohomology over a PID: H1(S2;Z)=0; for an unknot U≅S1 in S3, H~0(S3∖U;Z)=0 and H~1(S3∖U;Z)≅H~1(U;Z)≅Z.

[F4]

Poincaré duality for oriented topological manifolds and The universal coefficient theorem for cohomology over a PID: under AC, a closed connected oriented 3-manifold with H1=0 has H0=H3=Z and H2≅H1=0.

[F5]

Every spherical gluing below is made with a retained product collar S2×(−2ε,2ε), chosen disjoint from U. For closed pieces A− and A+ on the negative and positive sides, replace them by the relative-open sets O−=A−∪(S2×(−2ε,ε)) and O+=A+∪(S2×(−ε,2ε)) in their union. On the added collar of O−, send its parameter t≥0 to (1−s)t; fix A−. The analogous formula on the negative added collar retracts O+ to A+. Their intersection is S2×(−ε,ε) and retracts to S2 by t↦(1−s)t. These homotopies match the identity on the joining sphere and remain valid after deleting U, since the collars miss it. Thus [F2] is applied only to these actual open covers, with the same homology groups as the named closed pieces. The collars are the product collars retained in the chosen smooth ball and connected-sum constructions; no collar for an arbitrary wild sphere is asserted.

[A1]

The Axiom of Choice: AC is the assumption of the local Poincaré-duality suppliers.

Counterexample

1.1givenF1F2F3algebra

The unknot U bounds its spanning disk inside B0, so [U]=0 in H1(Y;Z). For X=P∖int⁡B3, use the collar enlargements [F5] of X and B3 in P. They form an open cover, retract to those pieces, and have intersection retracting to S2; apply [F2] to this open cover. Since H1(S2)=0 by [F3], exactness injects H1(X) into H1(P)=0, so H1(X)=0. The same exact sequence in degree zero, using H1(P)=0 and H0(P)=H0(S2)=H0(B3)=Z, gives 0→Z→H0(X)⊕Z→Z→0. Rank comparison yields H0(X)≅Z, so each punctured summand is connected with H1=0.

1.2F2F3givenalgebra

Put A=B0∖U and let C be the closed ball on the other side of ∂B0 in S3. Identify B0 with the unit ball and U with a round circle in the plane z=0. From every point of A, the vertical segment in the direction of the sign of its z-coordinate reaches ∂B0 without meeting U; when z=0, the starting point is not on U and the upward segment meets the plane only there. Since ∂B0 is path-connected, A is path-connected. Thicken A and C by collars to obtain an open cover of S3∖U whose overlap retracts to S2. The map on H0(S2) is the injection 1↦(1,−1). Since H1(S2)=0 and C is contractible, [F2] identifies H1(A;Z) with H1(S3∖U;Z); [F3] identifies the latter with Z.

2.1A1F2F4givenstep 1.1algebra

Apply [F2] to the open collar enlargements [F5] of the two punctured summands in Y; the pieces and intersection retract to the punctured summands and S2, respectively. By step 1.1 both summands are connected and have zero H1, and the map H0(S2;Z)→H0(P∖int⁡B3;Z)⊕H0(P∖int⁡B3;Z) sends 1 to (1,−1) and is injective. Exactness gives H1(Y;Z)=0. Since Y is closed, connected, and oriented, [F4] gives H2(Y;Z)=0 and H0(Y;Z)=H3(Y;Z)=Z. Hence Y is an integral homology sphere. The duality and UCT uses are under [A1].

2.2F1F2F3step 1.2givenalgebra

The side of S in the second summand is P∖int⁡B3, which is not a 3-ball: capping it with a ball would identify P with S3, contrary to [F1]. The other side is the union of P∖int⁡(B3∪B0) and A, glued along ∂B0. Apply [F5] to obtain an open cover of this side, relative to its remaining boundary, with intersection retracting to that sphere and pieces retracting to the two named sets. Since H1(S2)=0 by [F3], exactness in [F2] makes the map from the direct sum of the two pieces' H1 groups into the union's H1 injective. Step 1.2 therefore makes this side's H1 nonzero. Neither side is a 3-ball, so S cannot bound a 3-ball in Y∖U.

3.1step 2.1step 2.2∎

Therefore Y∖U is reducible and the displayed universal assertion is false. The diagram records the separation and location of U; the proof uses the homology and homotopy arguments above, not the diagram.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

Trefoil and 5_2 knot Floer genus and fibering cases

Example

Assume the Axiom of Choice. For the right-handed trefoil T(2,3), the symmetrized Alexander polynomial is t−1+t−1, the maximum Alexander grading is 1, the integral top knot Floer group is Z, and the knot is fibered. For the standard alternating knot 52, the symmetrized Alexander polynomial is 2t−3+2t−1, the maximum Alexander grading is 1, the integral top knot Floer group is Z2, and the knot is not fibered. The 52 rank calculation uses the alternating-knot formula; its signature determines the Maslov grading shift.

Facts & Assumptions

Given: Full AC, the right-handed trefoil and standard alternating knot 52 with the original integral normalization and classical choice branch declared in the dependencies.

[F1]

External input. Ozsváth–Szabó, Heegaard Floer homology and alternating knots, Geometry & Topology 7 (2003), Theorem 1.3, printed p.230 (PDF p.6 of arXiv:math/0209149v3): if K is alternating and its normalized symmetric Alexander polynomial is a0+∑s>0as(ts+t−s), then HFK^(S3,K,s;Z)≅Z∣as∣, supported in Maslov degree s+σ(K)/2. This is the INTEGRAL theorem, including freeness; no integral conclusion is inferred from the retained mod-two grid comparison. Its signature convention has left trefoil signature +2.

[F2]

External finite table input. Knot Atlas, pages 3_1 and 5_2, section “Polynomial invariants / Alexander polynomial”, records respectively t−1+t−1 and 2t−3+2t−1. These two finite entries and their standard knot diagrams identify the table knots. The same polynomials result from the standard genus-one disk/band Seifert matrices V3=(1101),V5=(1102) (up to simultaneous handedness/sign conventions): det⁡(V3−tV3T)=1−t+t2 and det⁡(V5−tV5T)=2−3t+2t2. Multiplying by t−1 gives the stated symmetric polynomials. The TABLE/diagram identification is external input; the two determinant calculations are finite local checks, not an assertion that an abstract matrix by itself identifies a knot.

[F3]

External input at the exact S3, integral scope. Manolescu, An Introduction to Knot Floer Homology, printed/PDF p.5, Theorem 2.3, gives g(K)=max⁡{s:HFK^(K,s;Z)≠0} for every knot in S3. Theorem 2.5 on that same page gives that a knot in S3 is fibered if and only if its integral top genus group is Z. These exact specializations of the established Ozsváth–Szabó/Ghiggini/Ni/Juhász results are imported. This does not assert completion of the held local general-Y signed comparison or product-criterion proofs. The source explicitly states these as integral groups.

Actual reading limits: the complete Theorem 1.3 statement and its preceding conventions on printed pp.229–230 of the primary alternating paper; the two named Alexander-polynomial table entries (not all table invariants); the full Theorems 2.3 and 2.5 statements and surrounding geometric-applications text on Manolescu p.5. No reading or local reconstruction of the full alternating theorem proof, Ni proof, or entire survey is claimed. These are explicitly authorized external imports under the run-specific source boundary. The original full-choice branch is retained; no integral full-choice theorem is silently cast as a CC theorem.

Verification

1.1F1F2givenalgebra

Both named diagrams are alternating. Apply [F1] with the two polynomials in [F2]. In the trefoil case the coefficients in Alexander degrees 1,0,−1 have absolute values 1,1,1, with zero coefficients elsewhere. Thus the maximum nonzero Alexander grading is 1 and its integral group is Z. In the 52 case the corresponding absolute values are 2,3,2, with zero coefficients elsewhere. Its maximum grading is again 1, with integral top group Z2. The formula for the supporting Maslov degree is exactly that of [F1]; no signature sign is guessed from a drawing.

2.1F1F2F3step 1.1algebra∎

By [F3] the genus is one for both knots. The trefoil's integral top group is Z, so the fiberedness equivalence in [F3] proves that it is fibered. For 52, Z2 is not isomorphic to Z (tensoring with Q already distinguishes their dimensions), so the same equivalence proves that it is not fibered. These arguments prove every Example clause at its full integral coefficient scope. They do not invoke an unearned generic-Y local supplier or weaken a rank over integers to a mod-two rank.

Sources