8
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

The answer is no.

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.

I suppose the Whitehead link would also work.

Perhaps instead of describing the second link via the Dehn twist it would be more appropriate to describe the homotopy as a "finger move" pushing one crossing over the other.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Carl Apr 7 '16 at 22:21
  • $\begingroup$ I think you can make these examples work as well -- just embed these 3-manifolds in a closed 3-manifold in a way that doesn't allow the knots to become isotopic. The Whitehead example works fine if you double the manifold (the complement of one component of the link). I suspect the Borromean rings one will work similarly. Alternatively do a Dehn filling of slope 0 on the manifolds. $\endgroup$ – Ryan Budney Apr 8 '16 at 2:29
  • 1
    $\begingroup$ I believe that doing a Dehn filling of slope 0, at the very least, does not work. You can slide the knot across the filled meridian disk to get the same effect as a twist would have, and therefore you have an isotopy. I am not sure how doubling works in this case since you need to end up with a closed manifold and it is not clear to me that the complements will still be homeomorphic. $\endgroup$ – Carl Apr 8 '16 at 4:12
  • 1
    $\begingroup$ 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 $X$? $\endgroup$ – Chris Gerig Apr 15 '16 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.