Bordism independence of the Casson–Gordon Witt class

Put $F=\mathbb{C}(t)$. For any smooth four-manifold $V$ over $B$ let $D(V)$ be its twisted nonsingular image Witt form minus the scalar extension to $F$ of its ordinary complex nonsingular image form. The representation $(a,n)\mapsto {\zeta }_d^a{t}^n$ is unitary for the stipulated involution. By [F3], the two terms vanish on a closed five-bordism boundary with their respective extending systems. Hence on closed four-manifolds their difference is a homomorphism from ${\mathrm{\Omega }}_4^{SO}(B)$ to $W(F)$. For a constant map the two coefficient systems are the same constant $F$ system, so $D(Z,\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t})=0$.

Applying that homomorphism to [F2] gives $dD(Z,f)=0$ for every closed $Z$ over $B$. In particular every closed ambiguity has torsion Witt difference and becomes zero after tensoring with $\mathbb{Q}$. This conclusion concerns the entire Witt class, not just its signature. No arbitrary-group analytic index theorem is required.

If the bounding multiplicities are $r_0,r_1$, take $r_1$ disjoint copies of $W_0$ and $r_0$ copies of $W_1$. Their whole marked boundaries are the same $r_0{r}_1$ copies. Glue them, reversing the second orientation, with the same boundary character maps. Whole-boundary Witt gluing [F3] gives $D(Z)=r_{1}D(W_0)-r_{0}D(W_1)\mathrm{.}$ This uses the ordinary nonsingular quotient even when the ordinary absolute form has a radical. The twisted boundary is acyclic by [F1], so that term uses precisely the absolute module of the Statement. By step 2.1 the left side is zero rationally. Dividing by $r_0{r}_1$ proves $D(W_0)\mathrm{/}{r}_0=D(W_1)\mathrm{/}{r}_1$. Disjoint union, repeated multiples and all marked gluing choices are therefore compatible.

We also record the target enlargement needed for a disk extension. Embed $C_d$ in $C_D$ for $d\mid D$ by $a\mapsto (D\mathrm{/}d)a$, and keep the infinite coordinate unchanged. With positive roots of unity, ${\zeta }_D^{(D\mathrm{/}d)a}={\zeta }_d^a$, so the two rank-one $F$ coefficient systems and their forms are literally the same. Regard any original bounding datum as a datum over the larger target. Even if its boundary map is not onto that target, the proof of steps 1.1–3.1 still applies: [F2] and [F3] allow arbitrary continuous maps to $B$, and acyclicity concerns the actual coefficient system. One may use the induced principal cover, which can be disconnected; its local coefficient module is unchanged. Thus a smooth filling over the larger finite cyclic target computes the same rational class.

Finally the specified one-sided signature is a well-defined homomorphism on this rational Witt group. The entries of a Hermitian rational matrix, its nonzero determinant, and a rational change-of-basis matrix have only finitely many zeros and poles on the unit circle unless the relevant function is identically zero. For a nonsingular representative choose a short interval $0<\theta <ϵ$ avoiding those finite exceptional sets. Its evaluations are nonsingular Hermitian matrices, and their eigenvalue signs cannot change without a zero determinant, so the signature is constant there. Congruent representatives have congruent evaluations on a common such interval. A hyperbolic form evaluates to a hyperbolic form and has zero signature. Direct sums add signatures, and rationalization extends this homomorphism by rational linearity. This verifies exactly the one-sided limit specified in the invariant Definition; it does not substitute another averaging convention. Together with step 3.1 this completes the claimed rationalized Witt construction.