= Solution
Let $D\subset B^4$ be a <slice disk> for $K$. Its normal bundle is trivial. A small normal push-off $D'$ 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 $D\cup D'$.
Orient $D'$ oppositely to $D$. In the other hemisphere of $S^4$, join their boundary circles by the product annulus supplied by the zero framing. The union
$$
D\cup (K\times[0,1])\cup D'
$$
is a two-sphere. More concretely, it is the rounded boundary of the three-ball $D\times[-\varepsilon,\varepsilon]$, 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.
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