Disk-band Seifert surfaces give actual cyclic zero-surgery bounding manifolds

Work in an inward collar $S^3\times [0,ϵ)$ of the standard ball. Push $F$ to the graph of a smooth positive interior depth function, zero on its boundary and with positive inward derivative there. Such a function is obtained from the supplied finite band and disk charts and positive finite bump sums, retaining the prescribed boundary collar. Round its finitely many seams in those charts. Denote this neat pushed surface by $F^{\mathrm{\prime }}$, and remove the interior of a small product normal disk neighborhood. The result $V_F$ is a compact smooth manifold after the displayed corner roundings. Its boundary is the knot exterior $E_K$ glued to $F\times S^1$ along their common torus: a normal disk to $F^{\mathrm{\prime }}$ has boundary the positive meridian, and the surface boundary direction is the preferred zero-linking longitude. This already fixes the peripheral marking, before any new handle is attached.

Here is a useful cut model, including why its chamber is a ball. The trace of the inward push gives a three-dimensional sheet from $F^{\mathrm{\prime }}$ back to its original position in $S^3$. Cut $V_F$ open along the portion of this sheet outside the removed normal neighborhood. The two banks are copies of a thickening of $F$, and the resulting chamber $Q$ is a ball with these marked banks. This last assertion has a collar construction. In each normal two-plane the removed disk and the cut trace are a rounded slit reaching the outside boundary. Cutting the slit opens it into an indentation attached to that boundary. Over a small three-dimensional neighborhood of $F$ in $S^3$, straighten that indentation to a depth graph $s=h(x)$; the graph has support in that neighborhood. The straightening in a normal section is the increasing reparametrization between two rounded intervals, and agrees with the identity at the outer section endpoints. The product coordinates along $F$ paste these reparametrizations; at its boundary use the supplied half-disk collar, so the slit endpoint is the same rounded half-section on both sides. All widths can be chosen common over the finitely many compact disk/band pieces. For a collar of depth $L>\max h$, the formula $(x,s)\mapsto (x,h(x)+(L-h(x))s\mathrm{/}L)$ is a diffeomorphism from the original collar onto the graph indentation, agrees with the unchanged interior along a smaller interpolated collar, and has strictly positive normal derivative. It is identity outside the graph support. Undoing this indentation identifies $Q$ with the standard ball. The banks retain precisely the positive and negative normal push copies of $F$ in its boundary. This is a cut-and-collar parameterization, not a claim that an arbitrary smooth four-ball is standard.

We now change the boundary to actual zero surgery. Let $H$ be a smooth genus-$g$ handlebody whose oriented boundary is the union of a copy of $F$ and a disk along $\mathrm{\partial }F$. Such a marked handlebody is supplied by the finite standard disk with $2g$ bands on the boundary of the standard genus-$g$ three-dimensional handlebody. The given disk-band surface has a finite smooth triangle presentation by subdividing its finitely many rectangles and disks in their coordinates. Both surfaces have Euler characteristic $1-2g$ and one boundary, so [F2] matches them with the exact given boundary map and collar. This transfers the handlebody parameter to the given $F$. Glue $H\times S^1$ to $V_F$ along $F\times S^1$, using the normal circle as the second factor and opposite boundary orientations. Round the product corners in those same collars, and call the result $W_F$. The remaining boundary is exactly $E_K\cup (D^2\times S^1)$, where the disk meridian is the actual preferred longitude and the circle core is the actual positive meridian. Its orientation restricts to the original oriented exterior. Therefore the boundary identification is the oriented [F1] zero surgery, rather than a homology-equivalent substitute. Equivalently, this gluing attaches the $g$ round handles which compress $F$ to the displayed disk. It is not the simply connected two-handle trace of zero surgery.

Form the infinite cyclic cut space by taking chambers $Q_k$, $k\in \mathbb{Z}$, and identifying the positive bank of $Q_k$ with the negative bank of $Q_{k-1}$. Fix the indexing so the positive meridian crosses to the next chamber; its deck motion is $T$. Finite consecutive unions of chambers are simply connected by [F3]: each chamber is a ball and each joining thickened $F$ is connected, so the tree of trivial vertex groups has trivial pushout. A compact loop or homotopy meets only finitely many chambers, since the chamber pattern is locally finite. Hence the full cut space is simply connected and is the universal cover, with deck group generated by $T$. Thus ${\pi }_1(V_F)=\mathbb{Z}$, generated by the actual positive meridian. The cover's Mayer–Vietoris sequence, or its degreewise short exact small-chain sequence, has the form of the mapping cone of the difference of the two bank inclusions, from $C_{\ast }(F)\otimes \mathrm{\Lambda }$ to $C_{\ast }(Q)\otimes \mathrm{\Lambda }$. The same assertion holds directly with a specified character in place of $\mathrm{\Lambda }$: coefficients are trivial within each bank and chamber, and the second bank map has the character factor. Since $Q$ is a ball and $F$ has one boundary component, it gives $H_2(V_F;\mathrm{\Lambda })\cong H_1(F;\mathbb{Z})\otimes \mathrm{\Lambda },H_3(V_F;\mathrm{\Lambda })=0.$ On degree-zero homology the difference map is $1-t$ up to its invertible sign/deck factor. For every $z\mathrm{\ne }1$ this factor is invertible, while the degree-one chamber homology is zero. Directly using the character small-chain sequence therefore gives $H_2(V_F;{\mathbb{C}}_z)\cong H_1(F;\mathbb{C})$, without a possibly nonflat specialization argument.

We calculate the ordinary part separately over $\mathbb{R}$. Mayer–Vietoris for the original decomposition of the ball into $V_F$ and $F\times D^2$ identifies the rim-torus map $H_1(F)\otimes H_1(S^1)=H_2(F\times S^1)⟶H_2(V_F)$ as an isomorphism: its adjacent ball homology groups vanish and $H_2(F\times D^2)=0$. It also gives $H_1(V_F)=\mathbb{R}$, generated by the meridian, since the other degree-one summand is $H_1(F)$ and the overlap map to it is the identity. One has $H_3(V_F)=0$ since the adjacent degree-three product group and ball groups are zero. For $W_F=V_F\cup (H\times S^1)$ the degree-two overlap map is $(\mathrm{i}\mathrm{d},j_{\ast })$ in this rim identification, where $j_{\ast }:H_1(F)\to H_1(H)$ is onto of ranks $2g\to g$. Its cokernel has dimension $g$. The degree-one overlap map is $(x,n)⟼(n,j_{\ast }x,n)\in H_1(V_F)\oplus H_1(H)\oplus H_1(S^1),$ with an immaterial minus sign on one gluing summand. Its kernel has dimension $g$. Exactness therefore gives $\dim H_2(W_F)=2g$. The image of $H_2(H\times S^1)$ is injective: the graph image $(x,j_{\ast }x)$ meets $0\oplus H_2(H\times S^1)$ only in zero. It has dimension $g$. Choose the $g$ disjoint framed longitude curves of the explicit three-dimensional handlebody. Their products with $S^1$ are disjoint embedded tori with disjoint normal pushes. They represent this image, so it is isotropic.

We calculate the form with its signs. Represent the finitely many basis cycles by oriented smooth one-cycles on $F$, with transverse intersections, and choose independent normal levels for any coinciding pieces. For $a_i$ take a cap of its positive push in $Q_1$ and a cap of its negative push with reversed orientation in $Q_0$. Their union across the banks is an oriented smooth two-cycle ${\mathrm{\Sigma }}_i$ representing the connecting class in step3.1. Caps can be obtained by coning a smooth cycle in the supplied ball parameter and smoothing the cone as a map relative to its boundary; immersion is unnecessary. For intersection counts take two independent copies. Cover their compact interior overlap by finitely many coordinate charts, add coordinate-vector perturbations with supported bumps, and apply [F5] to the product evaluation map and the diagonal of the product ball. Its derivative in those independent target-translation parameters spans the normal to the diagonal. Thus a sufficiently small parameter makes the two maps transverse, while their already separated boundary collars remain fixed. The intersection points are finite, and self-intersections of either separate map do not affect the resulting intersection of cycles. The ordinary intersection number of two such caps in a ball is the linking number of their boundary cycles: choose a spanning two-chain for one boundary in the boundary three-sphere and push its interior inward, where its intersections with the other cap are exactly the signed crossings defining linking. Changing either cap adds a closed two-cycle in the ball and hence changes the count by zero. At each transverse intersection its chart orientation sign is exactly the local orientation evaluation in cap duality, so this geometric count is the homological intersection form of [F4]. Thus the same rule applies to these cycle representatives and to an independent copy for a self-pairing.

Every loop of $F$ is trivial in ${\pi }_1(V_F)$: its pushed bank copy lies in the ball $Q$. The map ${\pi }_1(F)\to {\pi }_1(H)$ is onto, as is seen in the standard handlebody by its $g$ longitude curves on the punctured boundary. Its kernel kills only these boundary relations. Applying [F3] to $V_F\cup (H\times S^1)$ therefore kills all of the free handlebody group and identifies its circle factor with the positive meridian of $V_F$. No relation kills any power of that meridian: the maps to $\mathbb{Z}$ which are zero on ${\pi }_1(H)$ and one on the circle and meridian agree on the overlap. Hence ${\pi }_1(W_F)=\mathbb{Z}$ with exactly the specified generator. Reduction modulo $d$ gives the claimed extensions on the actual boundary.

For completeness the four cap pairings give the following finite table, in our first-positive-push convention: $\begin{array}{cc}k& {\mathrm{\Sigma }}_i\cdot T^k{\mathrm{\Sigma }}_j\\ 0& V_{ij}+V_{ji}\\ 1& -V_{ij}\\ -1& -V_{ji}\\ \mathrm{\mid }k\mathrm{\mid }>1& 0.\end{array}$ Indeed the two same-chamber upper-upper and lower-lower boundary pairings are respectively $\mathrm{lk}(a_i,a_j^{+})=V_{ji}$ and $\mathrm{lk}(a_j,a_i^{+})=V_{ij}$, by moving the two chosen normal levels simultaneously without crossing either curve. The lower cap has opposite boundary orientation; at a neighboring chamber exactly one of the two caps is lower, giving the minus sign. The positive neighboring translate pairs the positive push of $a_i$ with $a_j$, and the negative neighboring translate pairs the opposite order. Nonneighboring chamber interiors are disjoint. This also proves the diagonal entries by using an independent parallel cycle before taking the same class. Hence the table sums to $(1-t)V+(1-t^{-1})V^T$. It is Hermitian for $t^{\ast }=t^{-1}$. With this displayed pairing, $\lambda (Tx,y)=t\lambda (x,y)$ and $\lambda (x,Ty)=t^{-1}\lambda (x,y)$ by shifting the summation index, explicitly fixing the coefficient convention.

For coefficients with $1-t$ invertible, the complexes of both $F\times S^1$ and $H\times S^1$ are contractible. Explicitly the cellular circle complex is $0\to R\stackrel{t-1}{\to }R\to 0$ (or its inverse-deck version, differing by a unit); its contraction sends degree zero to degree one by $(t-1{)}^{-1}$. In the product total complex the contraction is $(-1{)}^p$ times this map on a degree-$p$ base chain; direct substitution in the total differential gives $ds+sd=1$. The finite product cells on $F,H$ give the same result for local singular homology by subdivision and their ordinary cellular comparison. The same calculation works directly for each $z\mathrm{\ne }1$, with $(z-1{)}^{-1}$, including every singular Alexander root. Mayer–Vietoris in [F3] now makes $H_2(V_F)\to H_2(W_F)$ an isomorphism for these coefficients. The cycles of the cap calculation can be pushed into the interior of $V_F$, so their intersections computed in $W_F$ have exactly the same orientation and signs. This proves the localized and every nontrivial-character form assertions. Nothing here divides by $\det H(z)$.

The ordinary first homology of $M_K$ is one-dimensional, generated by the meridian. One can compute this with the same supplied exterior/tube: Mayer–Vietoris for $S^3=E_K\cup (S^1\times D^2)$ says that the torus first homology maps isomorphically to the direct sum of the two first homologies, because $H_2(S^3)=H_1(S^3)=0$. The tube kills the meridian, so the exterior homology is generated by it; its kernel slope is the preferred longitude since $F$ bounds it in the exterior. The zero filling kills precisely that longitude and retains the meridian. By step4.2 the map $H_1(M_K)\to H_1(W_F)$ is consequently an isomorphism. The pair exact sequence and [F4] then show that $H_2(W_F)\to H_2(W_F,M_K)$ is onto between vector spaces of equal dimension, hence an isomorphism. Its intersection form is therefore nonsingular. Its $g$-dimensional isotropic subspace from step3.2 has half its dimension. In a basis beginning with that subspace its matrix is $\left(\begin{array}{cc}0& B\\ B^{\ast }& D\end{array}\right)$ with $B$ invertible; the linear-first basis change $G=\left(\begin{array}{cc}I& 0\\ -\frac{1}{2}D{B}^{-1}& I\end{array}\right)$ gives $GH{G}^{\ast }=\left(\begin{array}{cc}0& B\\ B^{\ast }& 0\end{array}\right)$, removing $D$. The resulting off-diagonal form has equal positive and negative inertia (negating one summand negates the form). Thus its signature is zero. The same argument covers $g=0$ with empty matrices. This establishes the ordinary part without substituting $t=1$ into the localized basis, which would be invalid. All constructions used the supplied smooth disk-band geometry; no spin, Arf-zero, finite-cover nonsingularity or TOP smoothing premise was added.