The slice-ribbon conjecture asserts that all slice knots are ribbon.
This assumes the context:
1) A `knot' is a smooth embedding $S^1 \to S^3$. We're thinking of the 3-sphere as the boundary of the 4-ball $S^3 = \partial D^4$.
2) A knot being slice means that it's the boundary of a 2-disc smoothly embedded in $D^4$.
3) A slice disc being ribbon is a more fussy definition — a slice disc is in ribbon position if the distance function $d(p) = |p|^2$ is Morse on the slice disc and having no local maxima. A slice knot is a ribbon knot if one of its slice discs has a ribbon position.
My question is this. All the above definitions have natural generalizations to links in $S^3$. You can talk about a link being slice if it's the boundary of disjointly embedded discs in $D^4$. Similarly, the above ribbon definition makes sense for slice links. Are there simple examples of $n$-component links with $n \geq 2$ that are slice but not ribbon? Presumably this question has been investigated in the literature, but I haven't come across it. Standard references like Kawauchi don't mention this problem (as far as I can tell).
Best Answer
Ryan, I think this is an open problem. The best related result I know is a theorem of Casson and Gordon [A loop theorem for duality spaces and fibred ribbon knots. Invent. Math. 74 (1983)] saying that for a fibred knot that bounds a homotopically ribbon disk in the 4-ball, the slice complement is also fibred.
More precisely, they are assuming that the knot K bounds a disk R in the 4-ball such that the inclusion
$S^3 \smallsetminus K \hookrightarrow D^4 \smallsetminus R$
induces an epimorphism on fundamental groups. If one glues R to a fibre of the fibration $S^3 \smallsetminus K \to S^1$ to obtain a closed surface F, then the statement is that the monodromy extends from F to a solid handlebody which is a fibre of a fibration $D^4 \smallsetminus R \to S^1$ extending the given one on the boundary.