Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 141 / 4

Write $X_{U,P}=S^3\setminus \mathrm{int}(\nu U\bigsqcup \nu K_P)$. Thenwhere the splice gluing exchanges meridian and longitude. Filling the remaining boundary along ${\mu }_P$ restores $\nu K_P$, so the pattern side becomes $X_U=S^3\setminus \mathrm{int}\nu U$, a solid torus. Under the splice gluing its meridian is attached to the meridian of $K_C$, because the meridian and longitude are exchanged twice in the two descriptions. This is the meridional filling of $X_C$, which restores the original three-sphere. Hence

The manifold $X_P$ is obtained from $X_{U,P}$ by filling the $U$-boundary along ${\mu }_U$. The filling core represents ${\iota }_{\ast }^{\partial X_U}[{\lambda }_U]$ in $H_1(X_P)$. The Turaev-torsion Dehn-filling formula therefore givesSince a knot exterior has first Betti number one, its normalized Turaev torsion isSubstitution proves

The Mayer--Vietoris sequence for the splice identifies the companion meridian with the pattern longitude and kills exactly the relation already killed when passing from $X_{U,P}$ to $X_P$. It therefore gives a canonical isomorphismApply multiplicativity of Turaev torsion to the torus union defining $X_{CP}$, and use part (b) to replace the torsion of $X_{U,P}$. The factors $1-[{\lambda }_U]$ cancel against the peripheral factor from the companion exterior. Usingon the three knot exteriors leavesUnder a one-variable identification, ${\iota }_{\ast }^C$ sends the companion variable to $t^w$, where $w=[U]\in H_1(X_P)\cong \mathbb{Z}$ is the winding number of a satellite pattern. Thus this is also the usual Satellite formula for the Alexander polynomial.

The slope-$4/3$ pattern has winding number $4$, so the Satellite formula for the Alexander polynomial givesUsing the stated torus knot formula in each factor,Up to a Laurent unit this may be simplified to $(1-t)(1-t^{24})/((1-t^8)(1-t^3))$.