Algebraic concordance and metabolic forms

If $g=0$, $V$ is the empty matrix and a disk with its empty homology basis realizes it. Assume $g>0$. Put $J=V-V^T$ and $H={\mathbb{Z}}^{2g}$. Choose a primitive vector $e\in H$. Since $J:H\to \mathrm{Hom}(H,\mathbb{Z})$ is an isomorphism, the functional $J(e,-)$ is primitive, so Bezout's identity gives $f\in H$ with $J(e,f)=1$. For every $x\in H$, the vector $x^{\mathrm{\prime }}=x-J(x,f)e+J(x,e)f$ satisfies $J(x^{\mathrm{\prime }},e)=J(x^{\mathrm{\prime }},f)=0$. Thus $H=⟨e,f⟩\oplus ⟨e,f{⟩}^{\perp }$, and the restricted alternating form on the second summand is unimodular. Induction gives a basis with matrix $J_0={⨁}_{k=1}^g\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Let $P\in {\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$ be the corresponding change of basis and put $W=P^{T}VP$; then $W-W^T=J_0$.

To apply [F2] to any disjoint oriented smooth core/push-off pair $J,K$, first use [F4] to obtain a spanning surface $\mathrm{\Sigma }$ for $K$. Apply [F5] to the inclusion of $\mathrm{\Sigma }$ with closed target $J$. Its boundary $K$ is disjoint from $J$, so boundary transversality is vacuous; the isotopy fixes a neighbourhood of $K$ and yields an embedded oriented spanning surface transverse to the unchanged curve $J$. These auxiliary surfaces do not modify the band surface. All constructed curves lie in ${\mathbb{R}}^3\subset S^3$; [F6] supplies a regular diagram of each pair. Thus every hypothesis of [F2] is met, both before and after a local band modification, and the crossing formula computes the topological linking entries in [F3].

Start with a disk and attach $2g$ oriented bands in the standard one-boundary handle pattern. Their core curves form an integral basis with intersection matrix $J_0$. The result is a connected compact smooth oriented surface with one boundary component. Write $W_0$ for its positive-push matrix; by [F1], $W_0-W_0^T=J_0$.

For each band core, inserting a full twist in a small subarc adds two same-sign mixed crossings between that core and its positive push-off, changing its self-linking by $+1$ or $-1$ by step 1.2. The twist box is disjoint from the other core arcs, so the other entries and the surface intersection matrix are unchanged. Perform finitely many such twists to make every diagonal entry of $W_0$ equal the corresponding diagonal entry of $W$.

For each pair $i<j$, choose subarcs of bands $i$ and $j$ in a ball disjoint from every other core and every earlier modification. Insert the oriented local clasp whose two new mixed crossings between $a_i^{+}$ and $a_j$ have the same sign $\epsilon \in \{+1,-1\}$. By step 1.2, their signed sum changes by $2\epsilon$, so $W_{ij}$ changes by $\epsilon$. The clasp preserves the abstract surface intersection form; by [F1], $W_{ij}-W_{ji}$ is unchanged, and therefore $W_{ji}$ also changes by $\epsilon$. Choose the bands' framings through the clasp without adding individual full twists; each band alone is unchanged up to framed isotopy. Thus all diagonal entries stay fixed, and support away from the other cores leaves all other pair entries fixed. The target equation $W-W_0=(W-W_0{)}^T$ off the diagonal holds because both skew parts are $J_0$. A finite number of positive or negative clasps thus sets both entries to their target values for each pair, without changing previously treated pairs. The resulting surface has positive-push matrix exactly $W$.

The basis change $P^{-1}\in {\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$ on this surface changes its matrix from $W$ to $(P^{-1}{)}^{T}W{P}^{-1}=V$. The constructed surface has one boundary component, so its boundary is an oriented knot and the matrix is a Seifert matrix for that knot.

For the positive Hopf bands with unknotted cores, the standard plumbing has the three crossing bands of the positive two-strand three-crossing closure: its two annular strips use successive pairs of the three bands, and their common disk is the plumbing square with its four alternating attaching edges. In oriented braid coordinates $(x,y,t)$ use $(X,Y,Z)=(x,-t,y)$, of determinant $+1$, to draw the closure on the $(X,Y)$ page with viewer $+Z$. The overpass runs down-left and the underpass down-right at each of the three crossings, while the two outside closure arcs are crossing-free. This is precisely the right-hand trefoil diagram of Rolfsen printed146, identified as right-hand on printed201; the comparison fixes the handedness using actual crossing data. The two core classes have intersection number $a{\cdot }_{F}b=1$ by [F2]. Since $F$ has genus one and one boundary component, its first homology has rank two; the intersection matrix in the core classes has determinant $1$. Therefore $a,b$ form an integral basis.

The following calculation also proves the broader matrix assertion: it uses only the stated annulus boundary linkings, positive handle basis, and disjoint disk, so unknotted cores are unnecessary here. By [F1], each diagonal is the self-linking of an annulus core. Equally oriented parallel cores transport to the two boundary components with one orientation reversed, so their linking is minus the boundary linking, namely $-1$. A small positive normal push returns to the first core through curves disjoint from the second. Thus $V_{11}=V_{22}=-1$. The disk specified in [F2] bounds $a^{+}$ in the complement of $b$, hence $\mathrm{lk}(a^{+},b)=0$ and $V_{12}=0$. By [F3], $V_{12}-V_{21}=a{\cdot }_{F}b=1$, so $V_{21}=-1$. These four entries give the displayed matrix.

In $(H,V)\oplus (H,-V)$, the diagonal subgroup $\mathrm{\Delta }H=\{(x,x):x\in H\}$ is a direct summand: it is the kernel of the surjection $(x,y)\mapsto x-y$ and that surjection has section $z\mapsto (z,0)$. Its rank is $\mathrm{rank}H$, half the rank of $H\oplus H$. For $x,y\in H$, $(V\oplus (-V))((x,x),(y,y))=V(x,y)-V(x,y)=0$, so $\mathrm{\Delta }H$ is a metabolizer.

By [F1], the metabolic direct sum in step 1.1 represents zero, while the direct-sum relation identifies its class with $[V]+[-V]$. Hence $[-V]=-[V]$.

For each displayed enlargement, its skew part has block form $E_i-E_i^T=\left(\begin{array}{ccc}J& u& 0\\ -u^T& 0& s_i\\ 0^T& -s_i& 0\end{array}\right),$ where $J=V-V^T$, $u=x-y$, $s_0=-1$, and $s_1=1$. If $V$ is knot-derived, [A1] is the choice hypothesis for its Seifert-pairing convention; the following lattice calculation itself is choice-free. Since $J$ is unimodular, choose the integral vector $c=-J^{-1}u$ and replace the penultimate basis vector by itself plus the old-basis combination $c$. This integral basis change removes the $u$ block and leaves $J\oplus \left(\begin{array}{cc}0& s_i\\ -s_i& 0\end{array}\right)$. Both blocks are unimodular, so each enlargement is an integral Seifert lattice.

In $E_i\oplus (-V)$, take the subgroup generated by $(e_j;e_j)$ for the old basis vectors $e_1,\dots ,e_{2g}$ and by $(b;0)$, where $b$ is the last basis vector of the enlarged lattice. Its rank is $2g+1$, half the total rank. The two displayed enlargement matrices have zero old-to-$b$, $b$-to-old, and $b$-to-$b$ entries. The $V$ and $-V$ blocks cancel on the diagonal generators, while all pairings involving $(b;0)$ vanish. Therefore the form vanishes on this subgroup.

The subgroup in step 1.2 is a direct summand: its generators, together with $(a;0)$ and $(0;e_j)$ for every old basis vector, form an integral basis because subtracting $(0;e_j)$ from $(e_j;e_j)$ gives $(e_j;0)$. It is therefore a metabolizer. By [F2], $[E_i]+[-V]=0$, and [F3] gives $[E_i]=[V]$ in $G_{\mathbb{Z}}$.

The matrix $B=V+V^T$ is symmetric. Modulo $2$, $V+V^T\equiv V-V^T$; by [F1] the determinant of $V-V^T$ is odd, so $\det B$ is odd and in particular nonzero. Thus $B$ is nonsingular over $\mathbb{Q}$.

Since $V$ and $V^T$ are invertible over $\mathbb{Q}$, $T=V^{-1}{V}^T$ is invertible. Direct multiplication gives $T^{T}BT=V(V^T{)}^{-1}(V+V^T)V^{-1}{V}^T=V+V^T=B,$ because the two summands in the middle expression simplify respectively to $V$ and $V^T$. Therefore $T$ is a $B$-isometry.

Finally $I-T=V^{-1}(V-V^T)$, which is invertible by invertibility of $V$ and [F1]. The three defining conditions in [F2] hold, proving the claim.

Both groups in [F1] are countable. A rational structure is specified, after one finite basis choice, by its dimension and two finite rational matrices; the isometry and nonsingularity conditions select a subset of this countable set. An integral Seifert lattice is likewise specified by a finite integer matrix. The corresponding presented groups are quotients of the countable sets of finite signed lists of these matrices: finite lists of rationals or integers have explicit natural-number codes, so a countable union over their lengths is countable. Quotients are countable by the surjection from these codes. This argument does not identify the two types of object. Use the injection [F2] to regard $G_{\mathbb{Z}}$ as a subgroup of $G_{\mathbb{Q}}$ for the rest of the proof.

For each primary group of [F3] other than $t+1$, the involution on $E$ is nontrivial. Indeed if its distinguished generator equalled its inverse, its irreducible polynomial would divide $t^2-1$; $t-1$ is excluded by invertibility of $I-T$ and $t+1$ was separated. If $F=E^{\sigma }$, choose $0\mathrm{\ne }\delta$ with $\sigma \delta =-\delta$. Every $z\in E$ has the unique decomposition $(z+\sigma z)\mathrm{/}2+\delta (z-\sigma z)\mathrm{/}(2\delta )$ with coefficients in $F$, so $E\mathrm{/}F$ is quadratic. Take all hermitian signatures at real places of $F$ where $E$ becomes complex, and take the ordinary rational signature on the $t+1$ component. These form a homomorphism $s:G_{\mathbb{Q}}\to {⨁}_{j\in J}\mathbb{Z}$. The index set $J$ is at most countable: there are countably many rational irreducible polynomials and finitely many real places for each number field. Each individual class has only finitely many primary components, so its total signature has finite support. With [A1] fix an enumeration of this index set, adding zero coordinates if needed, to put the target inside ${⨁}_{n\ge 1}\mathbb{Z}$.

If $s(x)=0$, each of the finitely many nonzero primary components of $x$ has all its signatures zero. By [F4] four times each such component is zero; the primary direct-sum isomorphism [F3] gives $4x=0$. Thus $4\ker s=0$. The same holds for the kernel of $s\iota$ on $G_{\mathbb{Z}}$, since $\iota$ is injective. Applying [F6] separately to these two countable groups gives decompositions of each with at most countable free, order-two and order-four multiplicities. In particular every torsion class has order dividing four; the proof has used signatures to identify all torsion, rather than merely exhibiting some finite-order classes.

In either group, the elements $F_b,P_n,Q_q$ of [F5] meet the three infinitude tests of [F6]. A finite nonzero integer combination of the $F_b$ cannot be torsion: if some nonzero multiple vanished, projection to each distinct primary factor would contradict the infinite order of its $F_b$. The $2Q_q$ are independent order-two elements, by distinct primary factors and exact order four. For the rational quotient by doubles, project to the quadratic primary factor of $P_n$, identified with a hermitian Witt group by [F3], and then take hermitian rank modulo two. This is a homomorphism because metabolic hermitian forms have even rank. The two-dimensional rational structure of $P_n$ has rank one over its quadratic field, so the homomorphism takes value one there; it vanishes on the other $P_m$ by their distinct primary factors, and on every double by its target $\mathbb{Z}\mathrm{/}2$. Hence the $P_n$ are independent modulo doubles in $G_{\mathbb{Q}}$ as well as in $G_{\mathbb{Z}}$, the latter also following directly from [F5]. Thus the $P_n$ are independent in the quotient by doubles in each group. This last condition forces independent order-two summands after splitting; independent order-two elements alone would not suffice, since doubles of order-four elements also have order two. There are infinitely many indicated primes $q$ by the finite Euclidean argument in [F5].

By [F6] and step 4.1, all three multiplicities in both decompositions from step 3.1 are infinite, and their countable upper bounds make them precisely ${\mathrm{\aleph }}_0$. This is the displayed abstract group isomorphism for each of $G_{\mathbb{Q}}$ and $G_{\mathbb{Z}}$. The integral result here follows from its proved injection, the shared exponent bound, the splitting argument and the actual integral families. No assertion that primary summands split integrally, no identification of the rational group with its integral subgroup, and no unspecified integral localization or coupling theorem is needed for this abstract isomorphism. In particular the allowed $t+1$ component of the original rational Definition has been retained throughout.

For a smooth oriented knot $K$, choose a surface and basis from [F1] and define $a(K)=[V_F]\in G_{\mathbb{Z}}$. The class is independent of this choice: [F4] relates any two such matrices by finitely many congruences and elementary enlargements or their inverses. A congruence is an isomorphism of the integral lattice, and each enlargement preserves the class. This also handles every integral basis choice. Applied to a transported surface under an oriented ambient isotopy, the same surface-independence statement and the pairing convention give a function on oriented knot types. This part is proved without a slice-disk premise.

Put representatives of $K,J$ and their chosen surfaces in separated balls, and join the surfaces by the connected-sum band away from the old basis curves. Their old first-homology bases together form a basis of the boundary-connected sum: collapsing the connecting band gives the wedge of the two surfaces, so first homology is their direct sum. Positive push-offs and the curves from opposite balls have linking number zero, since any curve in one ball bounds a rational, indeed integral, singular two-chain there, disjoint from the other ball. Thus the matrix is $V_K\oplus V_J$, and [F2] gives $a(K\mathrm{\#}J)=a(K)+a(J)$. The unknot has its disk surface with empty matrix and hence $a(U)=0$. By step 1.1 these calculations are independent of the separated representatives and chosen surfaces.

Work first in the smooth category. A knot smoothly concordant to $U$ bounds a smooth proper disk by [F6], so [F8] gives $a(K)=0$. The inverse of $[J]$ is represented by $J^{†}$, and therefore $J\mathrm{\#}{J}^{†}$ is concordant to $U$. Step 2.1 gives $a(J^{†})=-a(J)$. If $K$ and $J$ are smoothly concordant, their group difference $K\mathrm{\#}{J}^{†}$ is concordant to $U$, so $0=a(K\mathrm{\#}{J}^{†})=a(K)-a(J)$. Hence $a$ descends to a well-defined map ${\mathrm{\Phi }}_{\mathrm{s}\mathrm{m}}$ on smooth concordance classes. Step 2.1 proves that this map is a group homomorphism. Here [F8] applies the proved locally flat metabolizer argument to the smooth slice disk.

The smooth homomorphism is onto. Each positive generator $[V]$ of $G_{\mathbb{Z}}$ is the image of the realizing knot from [F3]. The image of a group homomorphism is a subgroup, so it also contains the negative of every such generator. Every element of the presented integral group is a finite sum of signed generators, hence is in the image. This proves the entire smooth surjective-homomorphism assertion.

In the locally flat category, [F10] supplies the group law and inverse needed for descent. If $K$ is TOP concordant to $U$, cap the unknot end of its collared annulus with the standard disk in a standard ball. Gluing that ball to $S^3\times I$ adds a radial collar to the ball, and the matching end product charts make the result a locally flat proper slice disk. By [F9] its Seifert form is metabolic, so $a(K)=0$ by [F2]. For TOP-concordant $K,J$, their group difference $K\mathrm{\#}{J}^{†}$ is TOP concordant to $U$. The smooth inverse calculation in step 3.1 already gives $a(J^{†})=-a(J)$, and step 2.1 therefore yields $0=a(K\mathrm{\#}{J}^{†})=a(K)-a(J)$. Hence $a$ descends to ${\mathrm{\Phi }}_{\mathrm{t}\mathrm{o}\mathrm{p}}$. Step 2.1 proves additivity, and the same realization argument as in step 4.1 makes its image contain every signed generator, hence all of $G_{\mathbb{Z}}$. This proves the TOP surjective-homomorphism assertion under the unchanged full-AC hypothesis, using the explicitly external group input and the exact TOP metabolizer interface rather than the former conditional hold.