A metabolizer forces Levine–Tristram signatures to vanish

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 ${\sigma }_{\omega }(K)$.