Nonsingularity of the Casson–Gordon twisted intersection form

The deck group $G=C_d\times \mathbb{Z}$ includes an infinite cyclic factor. Thus its regular cover $Y$ is not a finite-sheeted cover: the free coordinate takes the boundary meridian to one and gives arbitrarily many different sheets. Use the regular cover and the rank-one character projection followed by Laurent-field localization specified in [F2]. Its exactness identifies the module in the Statement with $H_2(W;F)$.

By [F2], $H_i(\mathrm{\partial }W;F)=0$ in every degree. The actual pair sequence therefore identifies $H_2(W;F)$ with $H_2(W,\mathrm{\partial }W;F)$. By [F3] the latter is its conjugate dual, with this map exactly the intersection pairing. It follows that the pairing is nonsingular. The involution in [F2] is precisely complex conjugation with $t\mapsto t^{-1}$, so the pairing is Hermitian in the stated convention.

To spell out the boundary cap, split each finite-dimensional boundary chain group into its cycles and a complement; the differential on that complement is an isomorphism to the preceding boundary space. Its inverses give a contraction of the acyclic boundary complex. The quotient from absolute to relative chains consequently has contractible kernel and preserves the absolute middle homology. This algebraic Poincare cap is enough for the Witt construction; it changes neither its module nor its pairing. The ordinary form, which need not be nonsingular on all ordinary $H_2$, instead uses [F3]'s quotient by its boundary radical. These are distinct operations, and no nonexistent finite $G$-cover or geometric attachment preserving a singular ordinary form has been assumed.