TITLE: High-dimensional ribbon knots
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: 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