 Solution

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.