Casson–Gordon signature defects vanish on the slice metabolizer

By [F1], choose an extending character $\tilde{\chi }$ on the branched disk cover, and write its exact order as $D=p^a$. If the original nontrivial character has order $d=p^b$, then $d\mid D$. Restrict $\tilde{\chi }$ to $H_1(X_{D,q})$ and pair it with $\mathrm{\ell }\mathrm{/}q$. Its finite coordinate vanishes on the lifted meridian because that circle bounds the branch-tube disk in the filled cover. Its infinite coordinate takes that circle to one. On the exterior of the boundary knot the finite coordinate is the original $\chi$; on the zero-filling solid torus it is zero. Thus its boundary is precisely the original $\eta$ followed by the inclusion $C_d\times \mathbb{Z}\to C_D\times \mathbb{Z}$, with the root representation unchanged.

The actual oriented boundary of $X_{D,q}$ is $M_{0,q}$. Indeed the product disk tube at the base disk gives a longitude filling of the knot exterior: its framing is zero, since a parallel disk is disjoint and its boundary has linking zero with $K$. Passing to the meridional $q$-cover lifts the same longitude slope, while its normal circle is the closed lift of ${\mu }^q$. These are the zero-surgery cover coordinates in [F2], with the positive boundary orientation retained. Mayer–Vietoris in the rational homology ball $W_{D,q}=X_{D,q}\cup (D^2\times D^2)$, whose overlap is $D^2\times S^1$, gives $H_2(X_{D,q};\mathbb{Q})=0$ and $H_1(X_{D,q};\mathbb{Q})=\mathbb{Q}$ on that overlap circle. The higher groups vanish as well. In particular its ordinary middle nonsingular image form is zero.

The character in step 1.1 is of prime-power order on the actual exterior cover. Hence [F3] gives $H_2(X_{D,q};\mathbb{C}(t))=0$, so its twisted form is also zero. The difference of the two Witt forms of this actual single-boundary smooth filling is therefore zero. By [F4]'s larger-target comparison it computes the original rationalized class even when $D>d$: compare any original multiple bounding datum, viewed over $C_D\times \mathbb{Z}$, with the same number of copies of this filling and use the closed fourth-bordism ambiguity calculation. Thus $\tau ({\mathrm{\Sigma }}_q(K),\chi )=0$ in $W(\mathbb{C}(t))\otimes \mathbb{Q}$. Applying the specified one-sided signature homomorphism gives $\sigma (K,\chi )=0$, as claimed. This proves the full prime-power extension case rather than assuming extension with unchanged finite order.