= Solution
A genus-one <Seifert matrix> for the <Stevedore knot> is
$$
A=\begin{pmatrix}1&1\\0&-2\end{pmatrix}.
$$
Its Alexander polynomial is
$$
\Delta(t)=\det(tA-A^T)=-2t^2+5t-2,
$$
whose 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,
$$
A+A^T=\begin{pmatrix}2&1\\1&-4\end{pmatrix}
$$
has Smith normal form $\operatorname{diag}(1,9)$. Therefore
$$
H_1(\Sigma_2(6_1))\cong\mathbb Z/9.
$$
If 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 knot>[doubly slice], despite its identically vanishing signature function.
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