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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 $g_{ds}(K)=0$.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.