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Knowing the result of knot surgery is often not enough to determine the knot. Indeed, there are 3-manifolds admitting an infinite number of descriptions as surgery on a (1-component) knot in $S^3$. However, if I know many surgeries, perhaps I can recover the knot? Let me be specific:

Suppose I have a 2-component link $U_1 \cup U_2$ inside the 3-sphere $S^3$ which has linking number $0$ and such that each component $U_1$, $U_2$ is the unknot.

I'm interested in knowing how much surgery tells us in this situation. If I do $1/n$-surgery on $U_2$ I get back $S^3$, but now $U_1$ sits inside $S^3$ as a knot $K(n)$.

Does the sequence $K(1), K(2), K(3), \ldots$ determine the original link $U_1 \cup U_2$ ? Would it even be expected to?.

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2 Answers 2

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Lackenby has a great result (MR1443548) which basically shows that the denominator of the surgery slope, if great enough in absolute value, determines the resulting 3-manifold and knot (subject to a few conditions on the manifold and knot).

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  • $\begingroup$ Isn't one of the conditions for that result that the 3-manifold be closed though? In this case I'm really working with a knot complement - in fact with a solid torus - and doing surgery on a knot inside there, obtaining other knot complements. $\endgroup$
    – Andrew Lobb
    Commented Feb 22, 2011 at 14:18
  • $\begingroup$ I might be wrong - maybe it does do it after all. $\endgroup$
    – Andrew Lobb
    Commented Feb 22, 2011 at 15:46
  • $\begingroup$ It does do it, since the manifolds in Lackenby's theorem may have boundary. You might also be interested in Lackenby's paper (a precursor to the one mentioned in the answer) "Surfaces, Surgery, and Unknotting Operations" which explicitly considers twisting around unknots having linking number zero with some other knot. He doesn't answer your question in that paper, though. $\endgroup$
    – Scott Taylor
    Commented Feb 23, 2011 at 1:14
  • $\begingroup$ There is an assumption in the paper that the link complement be atoroidal. I wonder if in this situation the theorem holds even in the presence of an essential torus (for example if the link is the Bing double of some non-trivial knot)?... $\endgroup$
    – Andrew Lobb
    Commented Feb 23, 2011 at 18:09
  • $\begingroup$ You can work in one half of the complement of the essential torus and conclude something about Dehn surgery on the knot in that complement. By Lackenby's theorem (assuming you chose your essential torus correctly), most of the Dehn surgeries determine the manifold and knot. The question is then whether when you reglue these surgered manifolds to the other side of the essential torus you continue to determine the original manifold and knot. I don't know the answer to that. I would guess that the answer is yes, at least most of the time. $\endgroup$
    – Scott Taylor
    Commented Feb 24, 2011 at 4:11
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If the orginal link $U_1 \cup U_2$ was hyperbolic, the answer is yes. For large enough $n$, $S^3 \setminus K(n)$ will also be hyperbolic, and will approach $S^3 \setminus (U_1 \cup U_2)$ in the Gromov-Hausdorff topology; so the sequence $K(n)$ determines the complement of $U_1 \cup U_2$. The complement doesn't determine the link in general (unlike for knots), but we also have the marking of the component $U_1$ by its meridian, which I believe is enough.

The answer in general is also almost certainly yes, but I haven't thought through all the cases. Note that this operation has a simple geometric description: arrange $U_1 \cup U_2$ so that $U_2$ sits as a flat unknot in a plane. Then to get $K(n)$, remove $U_2$ and twist the bundle of strands that passed through $U_2$ by $n$ full twists.

(This is all much easier than Lackenby's result mentioned above.)

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