Let $L_1$ and $L_2$ be two nonintersecting picewise-linear or smooth knots in $\mathbb R^3$. Suppose they are ambient isotopic. Does there exist an embedded surface $f: S^1\times[0,1]\to \mathbb R^3$ such that $L_1=f(\cdot,0)$ and $L_2=f(\cdot,1)$. This is similar to concordance but stronger — this "concordance" must be realised in $\mathbb R^3$ instead of $\mathbb R^3 \times [0,1]$.
If necessary, assume that $L_1,L_2$ are not linked or separated far from each other.
In other words is there a Seifert surface that is homeomorphic to a cylinder for a link consisting of two isotopic knots.
Best Answer
Take a knot, and another copy of the same knot far away. It is then a nice instructive exercise to prove that if they are not unknots they do not cobound annuli.