Seifert S-equivalence and the Conway–Arf bridge

By [F1], matrices from any two spanning surfaces are connected by congruences and elementary enlargements. Congruence transports $q(x)=x^{T}Vx(\mathrm{m}\mathrm{o}\mathrm{d}2)$ by an isometry. For an enlargement, write $W$ for the old quadratic space and $u,v$ for the two new basis vectors in the displayed order. The matrices in [F1] give $q(v)=0$, $B(u,v)=1$, and $B(v,W)=0$, where $B$ is the polarization. By its nondegeneracy on $W$ from [F2], there is $w\in W$ with $B(w,x)=B(u,x)$ for every $x\in W$. Then $u^{\mathrm{\prime }}=u+w$ is orthogonal to $W$ and still pairs to $1$ with $v$. Put $u^{\mathrm{\prime }\mathrm{\prime }}=u^{\mathrm{\prime }}+q(u^{\mathrm{\prime }})v$. The refinement identity gives $q(u^{\mathrm{\prime }\mathrm{\prime }})=q(u^{\mathrm{\prime }})+q(u^{\mathrm{\prime }})B(u^{\mathrm{\prime }},v)=0$; also $B(u^{\mathrm{\prime }\mathrm{\prime }},v)=1$ and $u^{\mathrm{\prime }\mathrm{\prime }},v$ are orthogonal to $W$. Their span has quadratic form $q(s{u}^{\mathrm{\prime }\mathrm{\prime }}+tv)=st$, proving that the enlarged space is $W$ orthogonally summed with a hyperbolic plane. Reductions reverse this stable isometry. Composing the finite sequence proves independence of the stable class, and [F2] gives independence of the Arf value. The unstabilized spaces can have different dimensions: the unknot disk has dimension zero, while a tubed disk has a two-dimensional hyperbolic enhancement.

Write ${\mathrm{\nabla }}_K(z)=1+a_2{z}^2+a_4{z}^4+\cdots$. For a knot the matrix size is $2g$, so ${\mathrm{\Theta }}_K(u^{-1})=\det (uV-u^{-1}{V}^T)=\det (u{V}^T-u^{-1}V)=(-1{)}^{2g}{\mathrm{\Theta }}_K(u)={\mathrm{\Theta }}_K(u)$, using transposition and even size. Since replacing $u$ by $u^{-1}$ sends $z=u-u^{-1}$ to $-z$, and substitution into the Laurent polynomial ring is injective (the highest Laurent power of a nonzero polynomial in $z$ cannot cancel), this identity gives ${\mathrm{\nabla }}_K(z)={\mathrm{\nabla }}_K(-z)$ and hence all odd coefficients vanish. Set $u^2=t$. Then $z^2=(u-u^{-1}{)}^2=t+t^{-1}-2\equiv (1+t{)}^2\mathrm{/}t(\mathrm{m}\mathrm{o}\mathrm{d}(2,(1+t{)}^4))$, and $z^4$ is divisible by $(1+t{)}^4$ in that quotient. Hence [F4] gives ${\mathrm{\Delta }}_K^{\mathrm{\#}}(t)\equiv 1+a_2({\mathrm{\nabla }}_K)(1+t{)}^2\mathrm{/}t(\mathrm{m}\mathrm{o}\mathrm{d}(2,(1+t{)}^4))\mathrm{.}$ Comparing with [F5], and using that $(1+t{)}^2\mathrm{/}t$ is nonzero modulo $(1+t{)}^4$, yields $a_2({\mathrm{\nabla }}_K)\equiv \mathrm{Arf}(K)(\mathrm{m}\mathrm{o}\mathrm{d}2)$.