Levine–Tristram signatures and the Arf invariant

Choose a basepoint $x_0\in X\setminus F$, which exists on a local side of any interior point of $F$. Represent any class of ${\pi }_1(X,x_0)$ by a continuous based loop, reparametrize it to be constant near the basepoint, and apply relative Whitney approximation followed by relative transversality from [F2]. This gives a smooth based representative transverse to $\mathrm{Int}F$, still constant near $x_0$. Define $I_F$ on that class to be its signed intersection count. If two such loops are homotopic in $X$, reparametrize the homotopy to be constant on endpoint collars. It is then smooth on a neighbourhood of the closed collars, so [F2] smooths it relative to a smaller collar and makes it transverse to $\mathrm{Int}F$. The inverse image of $\mathrm{Int}F$ is a compact oriented one-manifold in the homotopy annulus; it is compact because the annulus image is disjoint from $L=\mathrm{\partial }F$. Orient it using the orientations of the annulus and the normal of $F$. Its oriented boundary signs at the two ends are the intersection signs of the endpoint loops with opposite signs on the two boundary circles. The total signed boundary count is zero, so the endpoint intersection counts agree. Concatenation adds counts, hence $I_F:{\pi }_1(X,x_0)\to \mathbb{Z}$ is a homomorphism.

Near a point of $K_i$, choose oriented local coordinates $(x,y,z)$ in which $K_i$ is the $x$-axis and $F$ is the half-plane $y\ge 0$, $z=0$. The boundary orientation and the stated positive-meridian convention make $(x,y,z)$ ambient-positive. The small positive meridian crosses this half-plane exactly once, at $(0,\rho ,0)$ for some $\rho >0$, in the $+z$ direction. Thus ${\bar{I}}_F([{\mu }_i])=1$ for every $i$.

By [F3], this homomorphism factors through a homomorphism ${\bar{I}}_F:H_1(X;\mathbb{Z})\to \mathbb{Z}$. The signed intersection convention in the Statement agrees with this induced map on the class represented by $J$.

By [F1], the classes $[{\mu }_i]$ form a basis of $H_1(X;\mathbb{Z})$. The homomorphisms ${\bar{I}}_F$ and $\phi$ agree on every basis element by step 1.2 and [F1], so they are equal. Evaluating this equality on $[J]$ gives $J\cdot F=\phi ([J])$.

Here is the boundary-relative arrangement required in [F2]. Frame the normal disks of each link component $K$ by the inward tangent of $F_0$ and its oriented normal. The inward ray of $F_1$ has winding zero relative to this framing. Indeed a parallel copy $K_i$ lying just inside either $F_i$ satisfies $\mathrm{lk}(K,K_i)=-{\sum }_{J\mathrm{\ne }K}\mathrm{lk}(K,J)$: the surface obtained by removing the collar at $K$ is disjoint from $K$ and has boundary $K_i$ together with the other link components, so linking with that boundary is zero. The difference of these two self-linkings is the ray winding: in $S^1\times (D^2\setminus 0)$ the difference of the two sections is their angular degree times the positive meridian, which has linking one. Thus the angular difference has a smooth real lift $\theta (k)$. In the tubular coordinates the maps $(k,z)\mapsto (k,\exp (it\chi (\mathrm{\mid }z\mathrm{\mid })(\delta -\theta (k)))z)$, where $\chi =1$ near zero and is zero near the tube boundary, give an ambient isotopy fixing $L$; their inverses use the negative angle. Apply it to $F_1$, with a fixed $0<\delta <\pi$. Its inward ray is now everywhere at angle $\delta$ from that of $F_0$. Compactness and the smooth collar expansions make the two surfaces disjoint on a common sufficiently small punctured collar of $L$. On the remaining compact part, choose a generic sufficiently $C^1$-small perturbation of $F_1$ transverse to $F_0$, fixed on these collars, by [F2]. The straight small perturbation path consists of embeddings: locally this follows by projecting to the tangent planes and the inverse function theorem, and away from a fixed diagonal neighborhood compactness gives a positive separation between distinct source points. Extend this surface isotopy ambiently by [F2]; its velocity is zero on $L$, so the ambient flow fixes $L$. The remaining intersection is a compact transverse one-manifold, hence a finite union of circles. Choose a closed tubular neighborhood $N(L)$ sufficiently small that the trimmed surfaces $G_i=F_i\cap X$, where $X=S^3\setminus \mathrm{Int}N(L)$, have disjoint boundary curves on $\mathrm{\partial }X$.

By [F6], the inclusion of $X$ into $S^3\setminus L$ is a deformation-retraction equivalence, so the total-linking character $\phi$ of [F1] restricts to ${\pi }_1(X)$. By [F8], $X$ is path-connected, locally path-connected and semilocally simply connected, so the covering classification applies. For either connected two-sided surface $G_i$, its positive and negative collar regions are each connected, and every component of $X\setminus G_i$ has frontier in $G_i$; therefore there is at most one component on each side and at most two in total. A positive meridian ${\mu }_j\subset \mathrm{Int}X$ has signed intersection ${\mu }_j\cdot G_i=1$ by [F7], so $G_i$ does not separate $X$, since every closed loop has algebraic intersection zero with a separating surface. Thus each cut exterior is connected. Cutting $X$ along $G_0$ and stacking the resulting cobordism $Y$ over the integers, gluing the positive face of one copy to the negative face of the next, gives the connected cover classified by $\ker \phi$: a lifted loop changes its index by its algebraic intersection with $G_0$, which equals its total-linking value by [F7]. The same construction using $G_1$ therefore gives the same cover $p:\tilde{X}\to X$ by [F1] and [F8]. Write $P_n=t^n{P}_0$ for the $G_0$ lifts and $Q_r=t^r{Q}_0$ for the $G_1$ lifts. The closed slabs $Y_n$ between $P_n$ and $P_{n+1}$ have disjoint interiors, and $p$ is injective on $Y_n\setminus P_{n+1}$. The second stacking description shows that $Q_0$ separates $\tilde{X}$ into a negative and a positive component $C_{-},C_{+}$, with $Q_r\subset C_{-}$ precisely for $r<0$. Indeed the negative slabs in the $G_1$ stacking form $C_{-}$ and the positive slabs form $C_{+}$. Thus the slabs and both half-stacks are connected. Compactness of $Q_0$ and local finiteness of the slabs imply that $Q_0$ meets only finitely many $P_n$.

Suppose there is an interior intersection and choose the largest $n$ with $Q_0\cap P_n\mathrm{\ne }\mathrm{\varnothing }$. The positive end is the same in the two stackings, since a positive meridian increases both indices. Thus every higher face $P_j$, $j>n$, lies on the positive side $C_{+}$ of $Q_0$. Choose a component of $C_{-}\cap \mathrm{Int}{Y}_n$ incident to a curve of $Q_0\cap P_n$ and let $\tilde{M}$ be its closure. It misses $P_{n+1}$, so lies in the half-open slab on which $p$ is injective. No other $Q_r$ cuts it. For $r>0$, $Q_r\subset C_{+}$. For $r<0$, maximality implies $Q_r\cap P_n=\mathrm{\varnothing }$, since such an intersection would translate to $Q_0\cap P_{n-r}$. But translating a point of $Q_0\cap P_n$ shows that $Q_r$ meets $P_{n+r}$, strictly below $P_n$. The connected $Q_r$, disjoint from the separating cut face $P_n$, therefore stays entirely below that face and misses $Y_n$. Consequently $M=p(\tilde{M})$ is the closure of a component of $X\setminus (G_0\cup G_1)$, and restoring the link collars gives a compact three-manifold with corners abutting each surface from only one side. At least one interior intersection curve belongs to its boundary.

Triangulate the finitely many seam circles and collar edges, extend these triangulations over the compact surface faces by the two-dimensional case of [F3], and then triangulate this $M$ relative to its faces ${\mathrm{\partial }}_{i}M=M\cap F_i$ by [F3]. In the second barycentric subdivision let $A$ be the simplicial neighborhood of ${\mathrm{\partial }}_{0}M$ together with the primal one-skeleton, and put $B=\overline{M\setminus A}$. In each tetrahedron the part left outside that neighborhood is the cone from its barycenter towards the barycenters of its triangular faces not on ${\mathrm{\partial }}_{0}M$; gluing these pieces exhibits $B$ as a neighborhood of the dual graph, rooted on its boundary disks $B\cap F_1$. Replace the face ${\mathrm{\partial }}_{0}M$ of $F_0$ by $T=\overline{\mathrm{\partial }A\setminus {\mathrm{\partial }}_{0}M}$. For $F_1$ remove only $B\cap F_1$, not its entire $M$-face, and insert $\overline{\mathrm{\partial }B\setminus (B\cap F_1)}$. These are allowed finite moves: push $F_0$ through the initial face collar and then through a neighborhood of the primal graph; push $F_1$ through a neighborhood of the dual graph rooted on $B\cap F_1$. For each graph take a spanning forest rooted on its attaching face. Crossing a forest edge together with its new vertex is a ball isotopy. Every remaining edge attaches a product solid cylinder to two disks of the current surface, exactly [F4]. The graphs have finitely many edges and each root lies on the current surface; all feet approach from the selected side in $M$. Thus orientation and connectedness of each spanning surface are preserved. All moves are relative to $L$: in its product collar taper every displacement to zero at the core $L$, using the normal disk radial coordinate, and round the new collar arcs to join $L$ with the shifted frontier. The graphs and their surgery feet are away from $L$. This changes only a collar and is an isotopy fixing its core.

The two modified surfaces coincide on $T$, including the part $A\cap F_1$ of the old second face, and, before rounding, retain their old portions outside $M$. Perform the following rounding in arbitrarily small bicollars straddling $\mathrm{\partial }M$, so the old intersection circles are not pointwise protected. Push the two copies of $T$ to opposite sides in its product neighborhood, with the collar taper just described. At an old intersection circle the normal cross-section consists of two coordinate axes bounding the selected quadrant $M$; the replacement rounds the quadrant corner, and the two small opposite pushes are disjoint there. Thus every old intersection circle incident to $M$ disappears. All other old circles are untouched, and compact separation away from these corners and the common frontier lets the pushes be chosen small enough to create no new intersection. The number of interior circles strictly decreases. Repeat steps 2.1–4.1 while it is positive. The process terminates in finitely many moves with disjoint interiors, always fixing $L$.

In this last case their union, with the orientation on the second surface reversed and their common link collars rounded, is a connected embedded closed oriented surface $\mathrm{\Sigma }$ in $S^3$. Its second integral cohomology is $\mathbb{Z}$, so [F5] says its complement has exactly two components. The local two-sided surface charts show that the closure of either component has the whole $\mathrm{\Sigma }$ as boundary: each boundary subset is open and closed in the connected surface, and each side occurs in every chart. Choose one such closure $M$. Its faces are the whole of the two original surfaces, with the small common link collar restored. Apply the construction of $A$ and $B$ once more. Both modified surfaces are now the common replacement surface $T=\overline{\mathrm{\partial }A\setminus {\mathrm{\partial }}_{0}M}$ (up to the opposite small parallel pushes), so they are ambient isotopic through that product neighborhood, fixed on $L$. This final step does not assert that the original disjoint surfaces were parallel. Concatenate the finite moves from $F_0$ to its last modified surface, this last isotopy, and the reversed finite moves from $F_1$ to its last modified surface. All are ambient isotopies fixing $L$, embedded arc surgeries, or their reverses, and all spanning surfaces are connected and oriented. This is the required finite sequence.

Fix a finite cellulation of $F$ from [F1]. An integral cellular $1$-cycle represents each class of $H_1(F;\mathbb{Z})$. Replace each edge with coefficient $n$ by $\mathrm{\mid }n\mathrm{\mid }$ parallel oriented copies, reversing orientation when $n<0$. At each vertex the cycle condition says that incoming and outgoing copies occur in equal number. Around the boundary of a small vertex disk there is therefore an adjacent incoming/outgoing pair; join that pair by a small arc and delete its endpoints, repeating finitely until all endpoints are paired. The resulting arcs can be chosen disjoint in the disk and oriented consistently with the copies. Each smoothing changes the chain only by the boundary of a $2$-chain in that disk. The result is a finite disjoint union of oriented embedded circles representing the original homology class.

For representatives $X,X^{\mathrm{\prime }}$ of the same class in $H_1(F;\mathbb{Z})$, there is an integral singular $2$-chain $A$ in $F$ with $\mathrm{\partial }A=X-X^{\mathrm{\prime }}$. Push the chain through a small positive normal level of $F$; its boundary is $X^{+}-X^{\mathrm{\prime }+}$ and it is disjoint from any representative $Y\subset F$. Thus $X^{+}$ and $X^{\mathrm{\prime }+}$ represent the same class in $H_1(S^3\setminus Y;\mathbb{Z})$, so their meridian-coordinate linking sums with $Y$ agree. Varying the small positive level traces an annulus in the positive bicollar, also disjoint from $Y$, and gives the same class.

If $Y,Y^{\mathrm{\prime }}$ represent the same class, choose an integral $2$-chain $B$ in $F$ with $\mathrm{\partial }B=Y-Y^{\mathrm{\prime }}$. For each circle component of $X^{+}$ and each circle component of $Y-Y^{\mathrm{\prime }}$, [F4] supplies a spanning surface and [F5] permits a homotopy of the other circle in its complement to make it transverse to that surface while remaining an embedded circle. By [F3] choose a regular diagram of the pair. By [F2], linking is the half-sum of mixed crossing signs; exchanging the two components leaves the same mixed crossings with the same signs, so linking is symmetric. Bilinearity then gives $\mathrm{lk}(X^{+},Y-Y^{\mathrm{\prime }})=\mathrm{lk}(Y-Y^{\mathrm{\prime }},X^{+})$. The cycle $Y-Y^{\mathrm{\prime }}$ bounds $B$ in $S^3\setminus X^{+}$, so its total meridian coefficient there is zero. Hence the two linking sums agree in the second variable as well. Additivity of meridian coefficients gives bilinearity in both variables.

The value in the Statement is therefore independent of the chosen cycle representatives and push-off level and is bilinear on homology. In any ordered integral basis its matrix entries are $V_{ij}=\mathrm{lk}(a_i^{+},a_j)$, as required.

The two surgery disks and a narrow strip in $F$ joining them have a disk neighbourhood $D$. Replacing the disks by the lateral annulus of the cylinder changes this neighbourhood into a once-punctured torus $P$ with the same boundary $\mathrm{\partial }D$. Write $F_0=\overline{F\setminus D}$. Any old homology class can be represented in $F_0$: replace each of its finitely many arcs through $D$ by an arc along $\mathrm{\partial }D$, which changes the cycle by a boundary in the disk. Take open collar neighbourhoods in $F$ and $F^{\mathrm{\prime }}$ so that $F_0$ and respectively $D$ or $P$ form open covers whose intersection retracts onto $\mathrm{\partial }D$. In the Mayer–Vietoris sequence [F3], the map from $H_0(\mathrm{\partial }D;\mathbb{Z})=\mathbb{Z}$ to the direct sum of the two $H_0$ groups is $n\mapsto (n,-n)$ and is injective. For $F=F_0\cup D$, the $H_1$ map from the intersection circle has image generated by $([\mathrm{\partial }D],0)$, since $H_1(D;\mathbb{Z})=0$. For $F^{\mathrm{\prime }}=F_0\cup P$, the same image is generated by $([\mathrm{\partial }D],0)$ because the oriented boundary of $P$ is zero in $H_1(P;\mathbb{Z})$, while $H_1(P;\mathbb{Z})\cong {\mathbb{Z}}^2$ on its longitudinal and meridional curves. Thus $H_1(F^{\mathrm{\prime }})\cong H_1(F)\oplus {\mathbb{Z}}^2$, with the old classes as a direct summand. Choose the added basis classes $b,c$ from those two curves; together with the old basis they are an integral basis.

The surgery retains the old oriented surface outside the two attaching disks, so its positive normal there is unchanged and the old matrix block remains $V$. Use product coordinates $(t,w)$ on the cylinder, with $w\in D^2$, and take $c=\{t_0\}\times \mathrm{\partial }{D}^2$. Let $b$ follow one longitudinal side of the tube and return through $F_0$; it crosses each cross-section disk once. The disks $\{t_0\}\times D^2$ and their sufficiently close parallel disks miss every old basis curve and its push-off, so [F4] gives zero for the two Seifert pairings between $c$ and every old basis curve. Along the lateral annulus, the positive normal points either out of or into the solid cylinder. If it points out, the radius-one cross-section disk misses $b^{+}$, giving $V(b,c)=0$, while the disk of slightly larger radius bounds $c^{+}$ and meets $b$ once; it gives $\mathrm{lk}(b,c^{+})=+1$ after orienting $c$, hence $V(c,b)=+1$ by the symmetry in [F4]. If it points in, the radius-one disk meets $b^{+}$ once, giving $V(b,c)=+1$ after orienting $c$, while the disk of slightly smaller radius bounds $c^{+}$ and misses $b$, giving $\mathrm{lk}(b,c^{+})=0$ and therefore $V(c,b)=0$. These intersection statements follow from the explicit single transverse point in the product coordinates and [F4]. In either case the concentric parallel meridians have $V(c,c)=0$. The remaining entries involving $b$ are integers $x$, $y$, and $z=\mathrm{lk}(b^{+},b)$. Thus in the ordered basis $(a_1,\dots ,a_m,b,c)$ the matrix has exactly one of the two displayed forms.

At an isotopy step in the finite surface sequence of [L1], [F2] gives equality of the matrices in the transported basis. If the target basis is changed by $P\in {\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$, bilinearity gives $V^{\mathrm{\prime }}=P^{T}VP$, which is one of the defining S-equivalence moves by [F4].

Consider one arc surgery in that sequence. Choose the old basis curves away from the two surgery disks. The added product cylinder contributes a longitudinal class $b$ and a meridian class $c$; together with the old basis these form an integral basis of the new surface because the surgery adds one orientable handle and hence two first-homology generators. In a cross-section $\{t_0\}\times D^2$ of the cylinder, $c=\{t_0\}\times \mathrm{\partial }{D}^2$ bounds the disk $D=\{t_0\}\times D^2$ and the longitudinal class $b$ meets $D$ once. The disk and its parallel copies miss the old curves and their positive push-offs, so both pairings between $c$ and every old basis curve vanish. If the positive normal of the lateral annulus points outward from the cylinder, $b^{+}$ misses $D$ and the slightly larger disk bounded by $c^{+}$ meets $b$ once; hence [F6] gives $V(b,c)=0$ and $V(c,b)=\pm 1$. If that normal points inward, the smaller disk bounded by $c^{+}$ misses $b$, while $b^{+}$ crosses $D$ once, giving $V(c,b)=0$ and $V(b,c)=\pm 1$. Orient $c$ so the single nonzero entry is $+1$. The same disks show $V(c,c)=0$. The tube core has integer self-linking $z=V(b,b)$, and its pairings with the old curves give integer column and row vectors $x$ and $y^T$. By [F7], the difference between the two new cross entries equals the oriented intersection of the same ordered basis curves, so this is exactly one of $E_0(V;x,y,z)$ or $E_1(V;x,y,z)$, up to an integral basis change.

Apply [L1] to obtain a finite sequence of isotopies and arc surgeries or their reverses from $F$ to $F^{\mathrm{\prime }}$. Steps 1.1 and 1.2 turn every isotopy or forward surgery into a defining S-equivalence move, and reverse surgery gives its inverse. Their finite composition, including initial and final basis changes, proves the statement.

If two integral coordinate vectors $x,x^{\mathrm{\prime }}$ represent the same class modulo two, then $x^{\mathrm{\prime }}=x+2y$ for an integral vector $y$, and $(x+2y{)}^{T}V(x+2y)-x^{T}Vx=2x^{T}Vy+2y^{T}Vx+4y^{T}Vy$ is even. Thus $x^{T}Vx(\mathrm{m}\mathrm{o}\mathrm{d}2)$ is well-defined. Expanding the quadratic expression gives $q_F(x+y)-q_F(x)-q_F(y)=x^T(V+V^T)y(\mathrm{m}\mathrm{o}\mathrm{d}2)=x^T(V-V^T)y=x{\cdot }_{F}y,$ where plus and minus agree in ${\mathbb{F}}_2$. By [F2], this alternating form is nondegenerate.

In a symplectic basis, every vector is uniquely $x={\sum }_i({\alpha }_i{a}_i+{\beta }_i{b}_i)$ with ${\alpha }_i,{\beta }_i\in {\mathbb{F}}_2$. The refinement identity gives $q_F(x)={\sum }_i({\alpha }_i{q}_F(a_i)+{\beta }_i{q}_F(b_i)+{\alpha }_i{\beta }_i)\mathrm{.}$ Therefore the finite Gauss sum factors as ${\sum }_{x\in H_1(F;{\mathbb{F}}_2)}(-1{)}^{q_F(x)}={\prod }_i{\sum }_{a,b\in {\mathbb{F}}_2}(-1{)}^{a{q}_F(a_i)+b{q}_F(b_i)+ab}=2^g(-1{)}^{{\sum }_i{q}_F(a_i)q_F(b_i)}\mathrm{.}$ The left side depends only on $q_F$, not on a basis, so the exponent defining the Arf value is basis-independent.

Under either enlargement in [F3], the new basis vectors $u,v$ have $u\cdot v=1$, $q(u)=z(\mathrm{m}\mathrm{o}\mathrm{d}2)$, and $q(v)=0$; moreover $v$ is orthogonal to all old basis vectors because its encircling disk misses them. If $u$ pairs nontrivially with old basis vectors, add a linear combination of the old symplectic basis to $u$ to obtain $u^{\mathrm{\prime }}$ orthogonal to all of them; the nondegenerate old intersection matrix gives the required coefficients, and $u^{\mathrm{\prime }}\cdot v=1$ remains unchanged. The old basis together with $u^{\mathrm{\prime }},v$ is symplectic; its new Arf contribution is $q(u^{\mathrm{\prime }})q(v)=0$, while the old $q$-values are unchanged. More precisely, put $u^{\mathrm{\prime }\mathrm{\prime }}=u^{\mathrm{\prime }}+q(u^{\mathrm{\prime }})v$. The refinement identity and $q(v)=0$, $u^{\mathrm{\prime }}\cdot v=1$ give $q(u^{\mathrm{\prime }\mathrm{\prime }})=q(u^{\mathrm{\prime }})+q(u^{\mathrm{\prime }})=0$, while $u^{\mathrm{\prime }\mathrm{\prime }}\cdot v=1$ and both $u^{\mathrm{\prime }\mathrm{\prime }},v$ remain orthogonal to the old subspace. Thus the enlarged quadratic space is isometric to the old one orthogonally summed with the hyperbolic plane $q_H(s,t)=st$: on $s{u}^{\mathrm{\prime }\mathrm{\prime }}+tv$ the value is $st$. An integral congruence reduces to an invertible mod-two coordinate change and hence an isometry of quadratic spaces. Consequently, the stable isometry class is invariant under congruences, enlargements and their reverses. Here stable isometry means isometry after adjoining finitely many copies of $q_H$ on either side. Thus the Gauss-sum Arf value is unchanged by enlargement or reduction. By the knot-matrix S-equivalence lemma, all spanning surfaces give the same value.

Since $\bar{\omega }={\omega }^{-1}$, put $c=1-\omega$. Then $1-\bar{\omega }=-c\mathrm{/}\omega$, and therefore $H_{\omega }=c(V-{\omega }^{-1}{V}^T)$. As $c\mathrm{\ne }0$, taking determinants shows $H_{\omega }$ is singular exactly when $\det (V-t{V}^T)$ vanishes at $t={\omega }^{-1}$.

By [F1], the determinant in step 1.1 differs from ${\mathrm{\Delta }}_K(t)$ by a Laurent unit $\pm t^m$, which never vanishes on $S^1$. For $n=2g$, $M(t{)}^T=V^T-tV=-t(V-t^{-1}{V}^T)$, so taking determinants gives $\det M(t)=t^n\det M(t^{-1})$; its zero set is therefore invariant under inversion. Thus the determinant vanishes at ${\omega }^{-1}$ exactly when ${\mathrm{\Delta }}_K(\omega )=0$, proving both implications.