Knot Floer genus and fibering detection

By [F7], the order-$120$ subgroup of unit quaternions acts freely by left multiplication, so $S^3\to P$ is a covering with deck group $I^{\ast }$. Each left multiplication is isotopic to the identity because $S^3$ is path-connected; hence the action preserves orientation and the quotient is oriented. Since $S^3$ is simply connected by [F1], this cover is universal by [F2]. Then [F3] gives ${\pi }_1(P)\cong I^{\ast }$, which is nontrivial because its order is $120$.

In the abelianization, write the images of $s,t$ additively. The presentation in [F7] gives the relation $(st{)}^2=s^3=t^5$, hence $2s+2t=3s=5t$. The first equality yields $s=2t$; substituting into $3s=5t$ gives $6t=5t$, hence $t=0$ and $s=0$. Thus $(I^{\ast }{)}^{\mathrm{a}\mathrm{b}}=0$, and [F4] gives $H_1(P;\mathbb{Z})=0$.

The quotient is connected, compact, and oriented, so $H_0(P;\mathbb{Z})=H^0(P;\mathbb{Z})=\mathbb{Z}$. By [F6], $H^1(P;\mathbb{Z})$ is an extension of $\mathrm{Hom}(0,\mathbb{Z})=0$ by ${\mathrm{Ext}}^1(\mathbb{Z},\mathbb{Z})=0$, so $H^1(P;\mathbb{Z})=0$. The AC assumption is the one in [A1], needed for [F6].

By [F5] and step 2.1, $H_2(P;\mathbb{Z})=0$ and $H_3(P;\mathbb{Z})\cong H^0(P;\mathbb{Z})=\mathbb{Z}$. For $i>3$, [F5] identifies $H_i(P;\mathbb{Z})$ with negative-degree cohomology, which is zero. Together with steps 1.2 and 2.1, this is exactly the integral homology of $S^3$. Step 1.1 shows that $P$ is not simply connected, so it is not $S^3$. The use of [F5] requires [A1].

Construct the actual marked closure/adapted surface diagram from the relative Morse handles, carrying the page longitude and meridional boundary frame. Positive boundary fingers, the uniform relative windings, and permissible niceification give the admitted diagram. The Chern obstruction formula evaluates on minus F as minus 2g plus twice the number of interior polygon coordinates. Consequently its degree-g generators are exactly its outer generators. Each index-one disk is an exhausted embedded bigon or rectangle and lifts uniquely in the cut/double surface. The literal Fredholm conjugacies and the pullback of the specified periodic-domain orientation character align every coefficient after generator signs. Thus this generator correspondence identifies the full disk-count differentials over the integers. Signed relative moves transport the identification to the defining complexes, with their markings and coherent orientation classes.

For a horizontal decomposition, the cut components are balanced. In its adapted diagram there is no reverse cross-family intersection from side two to side one; the equal row and column counts force every perfect-matching generator to avoid the polygon. Thus the whole complex is the outer complex. The cut surface is a disjoint union, so each index-one disk has exactly one nonconstant factor. Ordering determinant factors gives the Koszul differential ∂(x1⊗x2)=∂x1⊗x2+(−1)^|x1|x1⊗∂x2. This is the horizontal chain map and the complete integral chain formula. Taking homology over any field gives Ni's horizontal formula. Taking homology over the integers instead gives the Künneth exact sequence including Tor; no unqualified integral homology tensor formula is asserted.

For an essential separating product annulus with connected R+ and R− and two balanced cut pieces, form the two rounded horizontal surfaces from opposite boundary faces and the annulus. Their actual cobordism regions give the required relative class and their Euler sums give the horizontal equality. This reduces the operation to attachment of a product region. A rooted cut-arc forest removes a product region meeting the sutures by literal product-disk identifications. For an interior essential product region, the explicit once-punctured-torus block forces its two fixed coordinates and isolates the only possible extra genus-one two-corner domain, whose Maslov index is minus one; it cannot contribute. Both comparison diagrams therefore have precisely the same supported disks and conjugate oriented operators in the unchanged complement. The resulting integral chain identification, followed by the remaining product-disk cuts, gives the full separating-annulus graded tensor operation. It gives Ni's field formula and the integral Künneth/Tor formulation as above. All constructions keep the stated basepoint zero regions, actual periodic lattices and essential-annulus hypotheses. They apply when the specified characteristic product region supplies those actual surfaces; they do not infer its geometry from generic disk existence.

Choose a Seifert surface F realizing the least genus g=g(K), and let E be the compact exterior. The exterior is irreducible: a sphere in its interior bounds two balls in S³ by F1; the connected knot neighborhood lies entirely on one side, so the other ball is disjoint from it and lies in E. A small oriented meridian loop meets F once, so F is nonseparating: a separating two-sided surface would have algebraic intersection zero with every closed loop. Cut E along the connected F to obtain connected M with the product annular suture and the two F faces. Their Euler characteristics agree and the boundary meets that suture, so M is balanced. A sphere in M is disjoint from F in E. Its E-ball is also disjoint from F: since F is connected, disjoint from the sphere, and has its boundary on ∂E outside that interior ball, it lies entirely on the other side. Thus that ball remains a ball in M. This proves irreducibility of the actual cut without a general-Y irreducibility assertion.

The two faces are incompressible. A compression of F along a nonseparating essential circle produces a connected Seifert surface of genus g−1. A separating compression produces a boundary-bearing Seifert component and a closed component. The latter has positive genus, since otherwise the original compression circle bounded a disk on F; discarding it again gives a Seifert surface of genus strictly less than g. Both contradict minimality. A face compression in M would be such a compression in E. For g=0 the disk has no essential compression circle.

To prove the actual sutured norm-minimality, take any compact oriented properly embedded competitor T representing the positive face class in H₂(M,γ;Z). Its total oriented boundary in H₁(γ;Z) is one copy of the longitude. Put its boundary circles in the interior of the annular suture. Cap an innermost contractible boundary circle by its disk in the annulus, pushed into the boundary collar; this operation increases Euler characteristic and cannot increase the Thurston complexity. Continue until every boundary circle is essential. Whenever both orientations remain, choose an adjacent opposite pair. The annulus between them contains no other boundary circle. Glue it to T in a thin boundary collar and round the corners. This gives an embedded oriented surface and removes the pair. Euler characteristic is unchanged; if two components merge, the inequality max(0,−(χ₁+χ₂))≤max(0,−χ₁)+max(0,−χ₂) shows that total complexity cannot increase. For two circles on one component its Euler characteristic, hence complexity, is unchanged. The finite sequence ends with exactly one essential boundary circle, since its total winding is one. Delete all closed components, which also cannot increase complexity. The remaining connected surface embeds in E and extends across the knot collar to an actual Seifert surface for K of the same genus. Its genus is at least g by minimality. For g≥1 its complexity is therefore at least 2g−1=x(F); for g=0 it is at least zero=x(F). It follows that x(T)≥x(F). The same argument applies to the negative face. Thus M is the actual taut balanced sutured manifold, by irreducibility and norm-minimality. The boundary cancellations were used only to derive a lower bound and need not preserve the competitor's relative class after it is mapped to E.

Work under the full-Choice analytic framework F6 on the actual fixed finite hierarchy and marked adapted diagram. Its admitted sutured operator problems retain the boundary/divisor barriers, periodic-domain and area conditions and the specified coherent orientation class. With the regular data and compactification/gluing construction supplied at that scope, apply F3 to the marked Seifert surface F. Its meridional frame, rather than an unspecified closure framing, gives an admissible adapted representative. By F4 every generator has Alexander degree g−n_P(x), with n_P(x) a nonnegative integer. Hence its whole integral chain complex has no generator, and thus no homology, in a grading greater than g.

The entire degree-g summand is the outer summand, and F3 identifies its signed differential with the actual cut complex of M. Since M is taut, F2 gives that this cut homology contains a split copy of Z for the orientation class induced by the given knot complex. Consequently the integral knot Floer group at degree g is nonzero. This also covers g=0: the taut cut argument and the same signed chain map apply to the minimum disk, without a genus-one connected-sum or field-coefficient substitution.

Degree g is nonzero and every larger degree is zero, so the maximum of the stated nonvanishing degree set is exactly g(K). The proof uses only the literal S³ ball-side and minimum-Seifert-genus arguments above; it makes no claim of arbitrary general-Y Thurston norm detection.

Let $E$ be the compact exterior and $(M,\gamma )$ its cut along $F$. Its two faces are copies of $F$, and its vertical boundary is the annulus $\mathrm{\partial }F\times I$. It is balanced: both faces have Euler characteristic $1-2g$, no component is closed, and every boundary component meets the vertical annulus. A meridian intersects $F$ algebraically once, so $F$ is nonseparating and the cut is connected.

The signed cut comparison gives $\mathrm{SFH}(M,\gamma ;\mathbb{Z})\cong \widehat{\mathrm{H}\mathrm{F}\mathrm{K}}(Y,K,[F],g;\mathbb{Z})\cong \mathbb{Z}$. The grading uses the supplied class $[F]$ and genus of this actual surface; no genus-detection theorem for a different ambient manifold is substituted.

The irreducibility assumption applies to $E$: pushing away the deleted-knot end identifies every compact sphere question with that in $Y\setminus K$. A sphere in the interior of $M$ is therefore a sphere disjoint from $F$ in $E$ and bounds a ball in $E$. Since $F$ is connected and disjoint from this sphere, if it entered the ball it would lie wholly in its interior, contradicting $\mathrm{\partial }F\subset \mathrm{\partial }E$. The ball is consequently disjoint from $F$; shrink the cut collar if necessary to obtain the same ball in $M$. Thus $M$ is irreducible.

If $(M,\gamma )$ were not taut, its integral sutured Floer homology would vanish by the external vanishing input, contradicting step 1.2. Therefore it is taut. This is the contrapositive of F2, not a reversal of taut hierarchy nonvanishing.

If the taut balanced $(M,\gamma )$ were not a product, F3 would give ${\mathbb{Z}}^2\hookrightarrow \mathrm{SFH}(M,\gamma ;\mathbb{Z})$. Such an injection into $\mathbb{Z}$ is impossible: images $a,b$ of its two basis vectors satisfy the nontrivial integer relation $ba-ab=0$ when either is nonzero, and an image zero itself contradicts injectivity. The two disjoint outer summands in the source argument therefore contradict step 1.2. Hence the actual $F$-cut exterior is a product sutured manifold. This includes $g=0$ and uses the corrected characteristic-pair qualification F4.

Use the actual binding bridge and drilling construction to represent the exterior with two meridional sutures, carrying the actual page longitude and the class of F. The adapted surface cut is literally the result of cutting this mapping torus along its fiber, hence is F times an interval with its product sutures. The fiber has no closed component and its boundary meets both signed boundary regions, so it is a good decomposing surface. First apply the adapted relative winding of F6, away from the protected boundary marking, to obtain an admissible good surface diagram of the same marked exterior and fiber. Now apply the finite permissible niceification, which preserves this established admissibility. Signed relative diagram moves compare the resulting complex to the chosen defining complex, with its specified coherent orientation class transported. No restriction on Y or change of coefficient ring is made.

In this marked adapted diagram the local Euler obstruction is minus 2g plus twice the number of interior polygon coordinates when evaluated on minus F. Therefore the Alexander degree is g minus that number: degree g is precisely the outer generator set, and no higher degree occurs. The product cut has zero periodic-domain group, either directly from its empty product diagram and the relative handle description, or from its vanishing relative integral homology. The signed cut comparison consequently identifies the whole degree-g differential with the cut differential, after harmless changes of generator signs.

The empty diagram of the actual product has one empty-tuple generator and no nonconstant disk, so its integral chain complex is Z with zero differential. The signed relative move equivalences identify its homology with that of the actual cut diagram. Combining with the full top-summand chain identification gives the asserted integral group isomorphism. The exterior orientation character restricts trivially to the zero-periodic cut, so this conclusion holds for the specified defining class without asserting equivalence of arbitrary orientation systems on the uncut exterior.

Apply F1 to the actual class $[F]$, the actual genus $g$ and the given integral group. The cut exterior is a product $F\times [0,1]$ as a sutured manifold. In particular the two horizontal faces correspond to the two copies of the original $F$, rather than to a surface selected only by a genus or homology calculation.

To restore the exterior, identify those two faces using their actual gluing map. In product coordinates this is $(x,1)\sim (h(x),0)$ for a self-homeomorphism $h:F\to F$. Thus the exterior is the mapping torus of $h$. Its projection to $[0,1]\mathrm{/}(0\sim 1)=S^1$ is a surface bundle, and the seam is precisely the original embedded $F$.

The vertical boundary is the mapping torus of $\mathrm{\partial }F$. Its boundary collar extends this bundle over the deleted-knot end, giving the fibration of $Y\setminus K$ with fiber the interior of $F$, or equivalently with compact fiber $F$ in the compact exterior. This proves the stated fiberedness with the prescribed surface, for arbitrary $Y$ and also for genus zero.