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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2 Answers
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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"
answered Jun 27, 2012 at 3:55
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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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1$\begingroup$ On the other hand Suciu claims that their result is based on an unproved result erroneously attributed to Lomonaco. maths.ed.ac.uk/~aar/papers/suciu1.pdf $\endgroup$– BlakeCommented Jun 28, 2012 at 12:33
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1$\begingroup$ I guess I did make that claim, back in the day. Surely I must have had a reason, but I forgot by now what that reason was. Nevertheless, that paper of S.J. Lomonaco, cited by Asano, Marumoto, and Yanagawa as "The homotopy groups of knots, II. A solution to Problem 36 of R.H. Fox, to appear", is not listed on MathSciNet, and I could not find a reference to it on GoogleScholar, either. $\endgroup$ Commented Jul 4, 2012 at 23:49