# Are knots determined by their complements within a homotopy class?

Suppose $M$ is a closed 3-manifold and $K_1,K_2$ are two homotopic knots in $M$. That is, they are two embeddings $f_1,f_2\colon S^1 \to M$ such that there exists a homotopy $h\colon S^1 \times [0,1] \to M$ with $h(x,0) = f_1(x)$ and $h(x,1) = f_2(x)$. Suppose also that there is an orientation preserving homeomorphism $M - K_1 \to M - K_2$. Must $K_1$ and $K_2$ be isotopic in $M$?

A positive result seems like it would have fairly large implications, not to mention being necessarily more difficult than Gordon and Luecke's result (and is therefore unreasonable to expect here) so I am mainly wondering if there are any known or simple counterexamples.

It stems from the fact that links are not determined by their complements. If you take the Borromean rings (for example), and think of one component as being a knot in the exterior of the other two components, i.e. a knot in the connect sum of two copies of $S^1 \times D^2$, then this knot is homotopic to another knot, which when put together with the other two components of the Borromean rings, is not isotopic to the Borromean rings, but the complement is diffeomorphic, preserving orientation.

You get these homotopic but not isotopic knots by doing a Dehn twist along a spanning disc for one of the other two components -- or you could do it on both simultaneously.

• Thanks for the observation. The Whitehead link (among others) does seem to work for the case where $M$ has boundary. However, I'm mainly interested in when $M$ is closed, preventing that kind of twisting operation.
• I am very confused on how doubling works here, because 1) we change the complements, and 2) I don't see what stops the knots from moving through the double to become isotopic. Am I correct in the setup that we first take $X:=S^3-\text{nbhd}(J)$ where $J$ is one component of the Whitehead link $J\cup K$ and then consider $K$ (along with some homotopic knot $K'$) inside $M=D(X)$? If so, then granted $J\cup K$ is not isotopic to $J\cup K′$ in $S^3$, I don't see how to use this to prove that $K$ is not isotopic to $K′$ in $M$. That is, why $K\cong K'$ in $M\;\Rightarrow\;K\cong K'$ in $X$? Apr 15 '16 at 23:32