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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.

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  • $\begingroup$ "If necessary, assume that $L_1$, $L_2$ are not linked or separated far from each other." What does this sentence mean? $\endgroup$
    – Sam Nead
    Commented Mar 17, 2022 at 17:50
  • $\begingroup$ @SamNead $L_1$ sits in a ball $B_1$, $L_2$ sits in a ball $B_2$ and $B_1$ and $B_2$ are disjoint. $\endgroup$
    – Dmitrii Korshunov
    Commented Mar 17, 2022 at 17:54
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    $\begingroup$ This rarely ever occurs. You realize all such pairs by taking a Seifert surface for a knot, then taking a parallel copy of the knot in the Seifert surface as your 2nd knot. Perhaps this should be called the "parallel cable" of the knot. I think the only time this generates an unlinked pair is when you start off with the unknot. $\endgroup$
    – Ryan Budney
    Commented Mar 17, 2022 at 18:14
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    $\begingroup$ Take a knot, and another copy of the same knot far away. If they are not unknots, they do not cobound annuli (exercise). $\endgroup$
    – Bruno Martelli
    Commented Mar 17, 2022 at 20:18
  • $\begingroup$ @BrunoMartelli--Given the clarification by the OP, your comment seems to answer the question. Why don't you write it as an answer. $\endgroup$
    – Danny Ruberman
    Commented Mar 18, 2022 at 12:15

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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.

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    $\begingroup$ thank you! (take a plane that separates the first knot from the second one and intersect it with the annulus. move it, if necessary, to turn the intersection into a smooth submanifold. one of its connected components must be a circle that is non-trivial in the annulus. this component is by construction not tangled. shrink the annulus to a nontangled equator and untie -- this would give the triviality of the initial knot) $\endgroup$
    – Dmitrii Korshunov
    Commented Mar 26, 2022 at 0:31
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As far as I understand your question, the answer is "no". Consider the Whitehead link.

https://en.wikipedia.org/wiki/Whitehead_link

The two components of the Whitehead link are unknots (thus they are isotopic) but they do not cobound a cylinder.

Thank you for clarifying that you wish to assume that the given link is "split": the components are separated by a two-sphere. In this case, the link is a "split unlink": a distant union of unknots. As Bruno points out, this is an exercise.

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  • $\begingroup$ the two components are linked though $\endgroup$
    – alesia
    Commented Mar 17, 2022 at 17:55
  • $\begingroup$ @alesia The components of the Whitehead link are homologically unlinked... You are correct that Dmitry K assumed that the components are "are not linked or separated". I found that sentence to be very confusing (as I pointed out). $\endgroup$
    – Sam Nead
    Commented Mar 18, 2022 at 10:23

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