The Seifert determinant detects Arf parity near minus one

If $g=0$, the matrix and homology group are empty, ${\mathrm{\Delta }}_K^{\mathrm{\#}}=1$, and $\mathrm{Arf}(K)=0$, so the congruence holds. Assume $g\ge 1$. Choose a symplectic integral basis and, in each pair, arrange $v_{2i-1,2i}\equiv 1$ and $v_{2i,2i-1}\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)$. This is possible because their difference is $1$; if the odd entry is in the reverse position, replace $(a_i,b_i)$ by $(b_i,-a_i)$, which keeps $a_i\cdot b_i=1$. Let $A(t)=V^T-tV$. For indices $r,s$ in a paired position, its entries modulo two are $a_{2i,2i-1}=1$ and $a_{2i-1,2i}=t$; for every other position $a_{rs}=(1+t)v_{rs}$. The diagonal case is included, as $(1-t)v_{rr}=(1+t)v_{rr}$ modulo two.

Let $B$ be the symmetric matrix over ${\mathbb{F}}_2$ with $B_{2i-1,2i}=B_{2i,2i-1}=1$ and all other entries $B_{rs}=v_{rs}$. For a permutation $\varphi$ of $\{1,\dots ,2g\}$, let $\alpha$ count the indices $i$ for which $\varphi (2i)=2i-1$, and let $\beta$ count those for which $\varphi (2i-1)=2i$. Grouping the determinant expansion by these two counts gives $\det A(t)={\sum }_{\alpha ,\beta =0}^g{ϵ}_{\alpha \beta }(1+t{)}^{2g-\alpha -\beta }{t}^{\beta }(\mathrm{m}\mathrm{o}\mathrm{d}2),$ where ${ϵ}_{\alpha \beta }$ is the sum of ${\prod }_r{B}_{r,\varphi (r)}$ over permutations with those counts. Inversion $\varphi \mapsto {\varphi }^{-1}$ preserves the product because $B=B^T$ and swaps $\alpha ,\beta$, so ${ϵ}_{\alpha \beta }={ϵ}_{\beta \alpha }$. There is one permutation with $\alpha =\beta =g$, the product of the $g$ within-pair transpositions, hence ${ϵ}_{gg}=1$. There is no permutation with $(\alpha ,\beta )=(g,g-1)$: if every even row occupies its paired odd column, then every odd row must occupy its paired even column; one exception would leave one even column empty. Symmetry gives ${ϵ}_{g-1,g}=0$. For $(g-1,g-1)$, inversion pairs equal-weight terms and cancels them modulo two unless the permutation is its own inverse. The fixed permutations must have every pair transposed except one pair fixed pointwise; any cross-pair exchange would make at least two pairs exceptional. Therefore ${ϵ}_{g-1,g-1}={\sum }_{i=1}^g{v}_{2i-1,2i-1}{v}_{2i,2i}=\mathrm{Arf}(K)(\mathrm{m}\mathrm{o}\mathrm{d}2)$ by [F2], since the diagonal entries are $q_F(a_i)$ and $q_F(b_i)$. The finite Gauss sum in [F2] is $2^g$ times $(-1)$ to this same parity, so it identifies the determinant coefficient as the Arf sign used in the Conway comparison below.

In the expansion of step 2.1, terms with $2g-\alpha -\beta \ge 4$ vanish modulo $(1+t{)}^4$. For exponent $0$, only $(\alpha ,\beta )=(g,g)$ remains, giving $t^g$. For exponent $1$, the only two terms are the symmetric pair $(g,g-1)$ and $(g-1,g)$, whose coefficient is zero. For exponent $2$, the diagonal term is $\mathrm{Arf}(K)(1+t{)}^2{t}^{g-1}$; the off-diagonal pair $(g,g-2),(g-2,g)$ is divisible by $(1+t{)}^4$, since $t^2+1=(1+t{)}^2$ modulo two. For exponent $3$, symmetry pairs each term with its transpose, and $t^a+t^b$ is divisible by $1+t$, so each paired sum is divisible by $(1+t{)}^4$. Hence $\det (V^T-tV)=t^g+\mathrm{Arf}(K)(1+t{)}^2{t}^{g-1}(\mathrm{m}\mathrm{o}\mathrm{d}(2,(1+t{)}^4))\mathrm{.}$ By [F3], and with the normalization fixed by [F4], multiply by $t^{-g}$ to obtain the asserted congruence for ${\mathrm{\Delta }}_K^{\mathrm{\#}}(t)$.