Let us suppose that we have a ribbon embedding $S^n \rightarrow S^{n+2}$ for $n\geq 3$. Call this knot $K$. By a theorem of Levine (and Trotter for $n=3$ I believe) we know $K$ is unknotted if the complement is homotopy equivalent to $S^1$. The question which I wonder is, does it suffice to show that the fundamental group is $\mathbb{Z}$, given the fact that $K$ is ribbon? For $n=2$ this would be true, as Freedman mentions (that having infinite cyclic fundamental group implies being homotopy equivalent to a circle) but I am not terribly familiar with the high-dimensional case.
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If the fundamental group of the complement is the integers there's still the various Alexander modules for the complement. Ribbon knots aren't immune from having non-trivial Alexander modules in dimensions higher than 1. Take a look at Kearton's papers on "simple n-knots" |
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I found a paper which discusses this issue: Ribbon knots and ribbon disks from Asano, Marumoto, and Yanagawa. They establish that for $n\geq 3$ a ribbon knot with infinite cyclic fundamental group is trivial. http://ir.library.osaka-u.ac.jp/metadb/up/LIBOJMK01/ojm18_01_12.pdf |
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