Fox–Milnor factorization and slice obstructions

If $g=0$, then $L=M=0$, so the empty determinant and $f=1$ give the formula. Suppose $g>0$. Write $L=M\oplus N$ and choose integral bases of the two rank-$g$ summands. In the combined basis, $V=\left(\begin{array}{cc}0& B\\ C& D\end{array}\right)$ with $B,C,D\in M_g(\mathbb{Z})$, because $V$ vanishes on $M\times M$.

A basis change $P\in {\mathrm{GL}}_{2g}(\mathbb{Z})$ replaces $V-t{V}^T$ by $P^T(V-t{V}^T)P$ and leaves its determinant unchanged, since $(\det P{)}^2=1$. Hence it suffices to calculate in the basis of step 1.1.

Put $f(t)=\det (B-t{C}^T)$. In $V-t{V}^T$ the upper-left $g\times g$ block is zero, the upper-right block is $B-t{C}^T$, and the lower-left block is $C-t{B}^T$. Swapping its two row blocks contributes $(-1{)}^{g^2}=(-1{)}^g$ and produces a block-triangular matrix, so $d_V(t)=(-1{)}^g\det (B-t{C}^T)\det (C-t{B}^T)$. Also $t^{g}f(t^{-1})=\det (tB-C^T)=(-1{)}^g\det (C-t{B}^T)$, by factoring $-1$ from each row and transposing. Therefore $d_V(t)=t^{g}f(t)f(t^{-1})$. The assumption on $V-V^T$ gives $d_V(1)=\pm 1$, so the displayed determinant is nonzero.

Since $t^g$ is a unit in $\mathbb{Z}[t^{\pm 1}]$, the exact identity in step 3.1 implies $d_V(t)\doteq f(t)f(t^{-1})$. This proves the factorization, including the rank-zero case of step 1.1.

If the Seifert lattice is metabolic, [F3] supplies a rank-$g$ direct-summand metabolizer for the positive-push matrix [F1]. Its skew part is unimodular by the knot Seifert convention, so [F4] gives $\det (V-t{V}^T)=t^{g}f(t)f(t^{-1})$. By [F2], ${\mathrm{\Delta }}_K(t)$ differs from this determinant by a Laurent unit, proving the algebraic factorization claim. This matrix calculation uses no choice beyond the stated full-AC convention for the Seifert data.

If $K$ bounds the locally flat proper disk of the Statement, [F5] provides the actual integral metabolizer for the positive-push Seifert lattice on the given $F$. Apply step1.1 to obtain ${\mathrm{\Delta }}_K(t)\doteq f(t)f(t^{-1})$. This includes a smooth slice disk as a special case. The conclusion is the stated necessary condition for ${\mathrm{\Delta }}_K$, and does not assert sufficiency or a factorization of $D_K$.

Suppose, to the contrary, that $K$ bounds a locally flat proper disk in $B^4$. The full TOP implication [F1] gives a Laurent polynomial $f(t)$ with ${\mathrm{\Delta }}_K(t)\doteq f(t)f(t^{-1})$. The full-AC hypothesis is the same as that of the supplier.

This contradicts the assumed nonexistence of such an $f$. Therefore no such locally flat slice disk exists. Only the necessary implication was used; nothing asserts that reciprocal-norm factorization is sufficient for sliceness.

If $g=0$, the Hermitian matrix is empty and has signature $0$ by [F4]. Suppose $g>0$. Choose an integral basis of $M$ and extend it by a basis of a complementary summand of $H_1(F;\mathbb{Z})$. In this basis, $V=\left(\begin{array}{cc}0& B\\ C& D\end{array}\right)$ and $H_{\omega }(V)=\left(\begin{array}{cc}0& A\\ A^{\ast }& E\end{array}\right)$, where $A=(1-\omega )B+(1-\bar{\omega })C^T$, $E=E^{\ast }$, and the upper-left block is zero because $V$ vanishes on $M\times M$.

Write $d(t)=\det (V-t{V}^T)$. Transposition and factoring $-t$ from the $2g$ rows give $d(t)=t^{2g}d(t^{-1})$. By [F3], $d(\omega )\mathrm{\ne }0$ since ${\mathrm{\Delta }}_K(\omega )\mathrm{\ne }0$ and the Laurent unit is nonzero at $\omega$; the reciprocal identity gives $d({\omega }^{-1})\mathrm{\ne }0$. Since $H_{\omega }(V)=(1-\omega )(V-{\omega }^{-1}{V}^T)$ and $\omega \mathrm{\ne }1$, the Hermitian matrix is nonsingular.

If $A^{\ast }x=0$, then $H_{\omega }(V)(x,0)=0$. Nonsingularity from step 2.1 forces $x=0$; because $A^{\ast }$ is a square $g\times g$ matrix, it is invertible.

For the Hermitian form with matrix in step 1.1, make the invertible coordinate change $x=x^{\mathrm{\prime }}-\frac{1}{2}(A^{\ast }{)}^{-1}Ey$. This cancels the lower-right term $y^{\ast }Ey$ and leaves $x^{\mathrm{\prime }\ast }Ay+y^{\ast }{A}^{\ast }{x}^{\mathrm{\prime }}$. The further change $z=Ay$ reduces it to the form with matrix $\left(\begin{array}{cc}0& I_g\\ I_g& 0\end{array}\right)$. Its $g$-dimensional subspaces $(u,u)$ and $(u,-u)$ have eigenvalues $+1$ and $-1$, respectively, so its signature is zero. Invertible congruence preserves positive and negative indices, and [F4] identifies this matrix signature with the signature of the chosen matrix; [F5] identifies it with ${\sigma }_{\omega }(K)$.