Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 112 / 3

A slice knot is a knot $K\subset S^3=\partial B^4$ that bounds a smooth properly embedded slice disk in $B^4$. A doubly slice knot is a transverse equatorial cross-section of an unknotted smooth two-sphere in $S^4$.

The slice genus $g_4(K)$ is the minimum genus of a smooth compact connected oriented surface properly embedded in $B^4$ with boundary $K$. The double slice genus $g_{ds}(K)$ is the minimum genus of an unknotted closed connected oriented surface in $S^4$ whose transverse intersection with an equatorial $S^3$ is $K$. Thus slice and doubly slice mean respectively $g_4(K)=0$ and $g_{ds}(K)=0$.

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Remove a small ball meeting $K$ in an unknotted arc. The remaining knotted arc lies in a three-ball. Rotate that ball once around its boundary two-sphere in $S^4$ and rotate the arc through $m$ additional full twists during the revolution. The trace, capped along the fixed endpoints, is the twist-spun knot ${\mathrm{Tw}}_m(K)$.

The Zeeman theorem on twist-spun knots says that the complement fibers over $S^1$ with fiber the punctured $m$-fold cyclic branched cover of $S^3$ over $K$. For $m=1$ or $-1$ this cover is $S^3$, so ${\mathrm{Tw}}_{\pm 1}(K)$ is unknotted.

A projection-independent specification of the requested banded-unlink diagram for ${\mathrm{Tw}}_0(T_{2,3})$ is as follows. Draw a reflection-symmetric diagram of $T_{2,3}\#(-T_{2,3})$ with the connected-sum neck on the symmetry axis. Cut at the neck, perform the oriented smoothing in each reflected crossing pair to obtain the lower unlink, and retain the two dual rectangular bands in each pair, one above and one below the projection plane. Simultaneous surgery on all these paired bands gives the reflected upper unlink. Capping the two unlinks produces exactly the spinning movie: the lower half rotates the cut trefoil through one semicircle and the upper half supplies its mirror semicircle. The paired placement of the bands records zero twisting; adding one full relative twist to every band pair gives the corresponding $m$-twist-spun diagram.

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A genus-one Seifert matrix for the Stevedore knot isIts Alexander polynomial iswhose roots are $2$ and $1/2$. There are no unit roots, so Question 1(a) proves that every Levine-Tristram signature of the Stevedore knot vanishes.

On the other hand,has Smith normal form $\mathrm{diag}(1,9)$. ThereforeIf a knot is doubly slice, the linking form on the first homology of its two-fold branched cover of a knot is hyperbolic: it has two complementary metabolizers, arising from the two sides of the unknotted sphere. A cyclic group of order nine has a unique subgroup of order three, so its linking form of a branched cover cannot have two complementary metabolizers. The Stevedore knot is consequently not doubly slice, despite its identically vanishing signature function.

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In either diagram of the Stevedore knot, the evident ribbon band can be cut to leave a two-component unlink. Cap those components by disjoint disks in $B^4$ and restore the band. This constructs a ribbon disk, hence proves that the Stevedore knot is slice.

The two displayed Stevedore diagrams give two distinct ribbon-band presentations. Use one below the equator and the reverse of the other above it. In banded-unlink diagram language, draw their common unlink and include the two ribbon bands, one from each presentation. Reading the movie from bottom to top gives the indicated birth level, the two saddle bands, and the death level. The resulting surface is knotted: the two-band movie is the standard presentation obtained by gluing the two Stevedore ribbon disks, and a van Kampen calculation retains a noncyclic quotient coming from the trefoil group, whereas the complement of the unknotted model has cyclic fundamental group.

There is a terminology issue in the question. If “2-knot” means an embedded $S^2$, as it normally does, one minimum, two index-one saddles, and one maximum have Euler characteristicso the resulting orientable surface knot is a torus, not a two-sphere. The construction above answers the question under the broader usage in which “2-knot” means a connected knotted surface. Under the standard narrow definition, the requested critical-point data are impossible.

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Let $D\subset B^4$ be a slice disk for $K$. Its normal bundle is trivial. A small normal push-off $D^{\prime }$ is disjoint from $D$, and its boundary is the zero-framed Seifert longitude ${\nu }_K$. Hence the two-component link $K_{2,0}=K\cup {\nu }_K$ bounds the pair of disjoint disks $D\cup D^{\prime }$.

Orient $D^{\prime }$ oppositely to $D$. In the other hemisphere of $S^4$, join their boundary circles by the product annulus supplied by the zero framing. The unionis a two-sphere. More concretely, it is the rounded boundary of the three-ball $D\times [-\epsilon ,\epsilon ]$, so it is unknotted. Its equatorial intersection is $K\cup {\nu }_K$, and both link components lie on the same sphere. Thus $K_{2,0}$ is doubly slice as a colored link with the trivial coloring.

Solved by gpt-5.6-sol high.