The skew part of a Seifert matrix is the surface intersection form

Construct the smooth handle basis and its intersection matrix. Cap the single boundary component by a disk and apply [F1] to obtain the oriented polygon normal form ${\prod }_{h=1}^g[a_h,b_h]$; remove the capping disk to recover the one-boundary surface. In this model the $h$th handle has two standard core curves, one meridional and one longitudinal, which meet once. Choose their orientations so the ordered tangent pair is positive at that crossing; curves from distinct handles are disjoint. The finite chart and disk-band construction in [F1] realizes this model on the given smooth surface: draw the curves as finite piecewise-smooth arcs in its bands and smooth their finitely many corners in disjoint coordinate disks, leaving their crossings fixed. In the model the surface retracts to the graph of these $2g$ core generators, so their homology classes form an integral basis; the same is true on $F$. Their intersection matrix is therefore the block sum $J_0={⨁}_{h=1}^g\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with zero entries between distinct handles.

Compare positive and negative pushes in the constructed handle basis. Use the curves $b_i,b_j$ constructed in step 1.1 and let $C_i=b_i\times [-ϵ,ϵ]$ be the product strip, oriented by $(n,u)$ so that $\mathrm{\partial }{C}_i=b_i^{+}-b_i^{-}$. Write $V^0$ for the Seifert matrix in this basis. By [F5] choose a compact oriented spanning surface ${\mathrm{\Sigma }}_j$ for $b_j$, and choose regular diagrams for the resulting link pairs. For $i=j$ the desired skew-entry identity is $0=0$, and the diagonal surface-intersection entry is zero by the handle model in step 1.1; hence take $i\mathrm{\ne }j$ in the intersection construction. The boundary curves $b_i^{+},b_i^{-},b_j$ are pairwise disjoint. First let $X=\mathrm{Int}{\mathrm{\Sigma }}_j$ with inclusion $f:X\to S^3\subset {\mathbb{R}}^4$ and closed target $\mathrm{\partial }{C}_i$. Choose an open collar $G$ of $b_j$ disjoint from this target and protect a smaller collar. Its complement $K=X\setminus G$ is compact. By [F7] choose a smooth $\chi :X\to [0,1]$, equal to one near $K$, compactly supported in a relatively compact open neighbourhood of $K$ avoiding the smaller collar. Then $\chi =0$ only where $f$ is already transverse. For $\mathrm{\parallel }v\mathrm{\parallel }<1\mathrm{/}2$ define $\mathcal{F}(p,v)=(f(p)+\chi (p)v)\mathrm{/}\mathrm{\parallel }f(p)+\chi (p)v\mathrm{\parallel }$. Where $\chi >0$ its parameter derivative is $\chi \mathrm{/}\mathrm{\parallel }f+\chi v\mathrm{\parallel }$ times the tangent projection onto $T{S}^3$, hence surjective; where $\chi =0$, $d\chi =0$ and old transversality applies. Thus the family is transverse. By [F7] every small parameter ball has a transverse slice. Compact support and direct differentiation give uniform $C^1$ convergence to the inclusion. A sufficiently small slice therefore extends by the unchanged boundary to an embedded spanning surface: finite compact coordinate subcharts keep a differential lower bound near the diagonal, and the remaining compact pairs keep positive image separation. Denote this surface again by ${\mathrm{\Sigma }}_j$; it is transverse to $\mathrm{\partial }{C}_i$. Let $P=C_i\cap b_j$. These points are finite, and the tangent of $b_j$ is not in $T{C}_i$ there because the handle curves cross transversely in $F$. Consequently $C_i$ is already transverse to ${\mathrm{\Sigma }}_j$ near $P$. It is also transverse near $\mathrm{\partial }{C}_i\cap {\mathrm{\Sigma }}_j$, since ${\mathrm{\Sigma }}_j$ was made transverse to $\mathrm{\partial }{C}_i$. Put $U=S^3\setminus b_j$ and $X=\mathrm{Int}{C}_i\setminus P$. The target $Z={\mathrm{\Sigma }}_j\setminus b_j$ is a closed embedded boundaryless submanifold of $U$: ${\mathrm{\Sigma }}_j$ is compact and all its boundary was removed. The inclusion $X\to U$ is a map between boundaryless manifolds. Protect a closed inner collar of $\mathrm{\partial }{C}_i$ and, at every point of $P$, a sufficiently small punctured closed disk neighbourhood. These are closed subsets of $X$, and the inclusion is already transverse to $Z$ on their neighbourhoods. Choose these neighbourhoods small enough to retain all the original endpoint data. For the inclusion $f:X\to U\subset S^3\subset {\mathbb{R}}^4$, choose an open protected region $G$ containing all boundary ends and punctured disks at $P$, on which $f$ is transverse to $Z$; protect smaller such regions. Then $K=X\setminus G$ is compact. Choose a relatively compact open $W\subset X$ containing $K$ and missing the smaller protected regions, and use [F7] to choose $\chi \in [0,1]$ equal to one near $K$, with compact support in $W$. Where $\chi =0$ old transversality holds. Use the explicit family $\mathcal{F}(p,v)=(f(p)+\chi (p)v)\mathrm{/}\mathrm{\parallel }f(p)+\chi (p)v\mathrm{\parallel }$. As above, where $\chi >0$ the parameter derivative surjects onto $T{S}^3$ and where $\chi =0$ the old derivative is unchanged, so the total family is transverse to $Z$. The compact set $f(\mathrm{supp}\chi )$ has positive distance from $b_j$; choosing $v$ small enough that the uniform displacement is less than half that distance keeps the entire family in $U$. Parametric transversality and the dense-complement statement [F7] now give an arbitrarily small transverse parameter. Direct differentiation on the compact support gives uniform $C^1$ convergence. Finite compact coordinate subcharts and off-diagonal separation, as in the preceding phase, ensure embeddedness. The family is identically the original inclusion on smaller punctured endpoint disks and boundary collars, so extends by the identity at $P$ and across $\mathrm{\partial }{C}_i$, fixes all endpoint intersection germs, and creates no new intersection with $b_j$. Thus the original compact strip and spanning surface have been made transverse, preserving exactly the required endpoint sets. To apply the boundary-preimage theorem within its hypotheses, remove pairwise disjoint small open disks $D_p^{\circ }\subset \mathrm{Int}{C}_i$ about the points $p\in P$, small enough that their punctured parts lie in the protected neighborhoods, and put $C_i^0=C_i\setminus {\bigcup }_{p\in P}{D}_p^{\circ }$. The inclusion $C_i^0\to U$ is defined on a compact surface with boundary; it is transverse to $Z$, and its restriction to each boundary component is transverse to $Z$ by the protected endpoint germs on $\mathrm{\partial }{C}_i$ and the local half-arc at each $p$. The boundary-preimage theorem gives a compact oriented one-manifold $C_i^0\cap Z$. Adjoining the protected half-arcs from the inner circles to their centers gives a compact oriented one-manifold $C_i\cap {\mathrm{\Sigma }}_j$ with boundary exactly $(\mathrm{\partial }{C}_i\cap {\mathrm{\Sigma }}_j)\cup (C_i\cap b_j)$. No added double half or disjointness assertion about one is used. Orient its tangent line by requiring $(n_{C_i},n_{{\mathrm{\Sigma }}_j},\mathrm{\ell })$ to be positive in $S^3$, where the first two vectors are the positive normals of the oriented surfaces. The local determinant check at an endpoint on $\mathrm{\partial }{C}_i$ gives boundary sign minus the ordered boundary-curve/surface sign; at an endpoint on $\mathrm{\partial }{\mathrm{\Sigma }}_j$ it gives boundary sign plus the ordered surface/boundary-curve sign. Explicitly, the first local model is $C_i=\{z=0,x\ge 0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ and boundary tangent $-{\mathrm{\partial }}_y$, and ${\mathrm{\Sigma }}_j=\{y=0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_z)$; the intersection tangent is inward $+{\mathrm{\partial }}_x$. The second model is ${\mathrm{\Sigma }}_j=\{z=0,x\ge 0\}$ with basis $(-{\mathrm{\partial }}_x,{\mathrm{\partial }}_y)$ and boundary tangent $+{\mathrm{\partial }}_y$, and $C_i=\{y=0\}$ with basis $({\mathrm{\partial }}_x,{\mathrm{\partial }}_z)$; the intersection tangent is again inward $+{\mathrm{\partial }}_x$. Reversing either surface orientation reverses both signs in its endpoint identity. Thus the zero signed boundary sum is $0=-(\mathrm{\partial }{C}_i\cdot {\mathrm{\Sigma }}_j)+(C_i\cdot b_j),$ and therefore $b_i^{+}\cdot {\mathrm{\Sigma }}_j-b_i^{-}\cdot {\mathrm{\Sigma }}_j=C_i\cdot b_j\mathrm{.}$ If $u,v$ are the positive tangents to $b_i,b_j$ at one of their crossings and $n$ is the positive normal to $F$, then the strip tangent basis is $(n,u)$ and adjoining $v$ gives $(n,u,v)$. By [F3] this has the ambient sign of $(u,v)$ in $F$, and [F4] fixes the ordered intersection convention; hence $C_i\cdot b_j=b_i{\cdot }_F{b}_j$. Applying [F5] to the transverse link pairs yields $V_{ij}^0-\mathrm{lk}(b_i^{-},b_j)=b_i{\cdot }_F{b}_j\mathrm{.}$

The pair $(b_i^{-},b_j)$ can be translated through the normal collar by $+ϵ$ while keeping the two curves disjoint; it becomes $(b_i,b_j^{+})$. Hence $\mathrm{lk}(b_i^{-},b_j)=\mathrm{lk}(b_i,b_j^{+})$ by [F2]. Symmetry of linking in [F2] changes the latter to $\mathrm{lk}(b_j^{+},b_i)=V_{ji}^0$. Therefore step 2.1 gives $V_{ij}^0-V_{ji}^0=b_i{\cdot }_F{b}_j$.

Identify the intersection matrix in every basis. Step 3.1 proves $V^0-(V^0{)}^T=J_0$. For any other integral basis $a_j={\sum }_i{b}_i{P}_{ij}$, the change matrix $P$ lies in ${\mathrm{G}\mathrm{L}}_{2g}(\mathbb{Z})$. Bilinearity of the Seifert pairing [F2] and of surface intersection gives $V=P^T{V}^{0}P$ and $J_F=P^T{J}_{0}P$. Hence $V-V^T=J_F$, $J_F^T=-J_F$, and $\det J_F=(\det P{)}^2\det J_0=1$. If $g=0$, the matrices are empty and the determinant is the empty product $1$.