most recent 30 from http://mathoverflow.net 2013-05-22T03:46:53Z http://mathoverflow.net/feeds/question/56239 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56239/do-the-results-of-1-n-surgery-determine-the-link Andrew Lobb 2011-02-22T02:36:21Z 2011-04-02T20:04:11Z

<p>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:</p> <p>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.</p> <p>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 $K(n)$.</p> <p>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?.</p>

http://mathoverflow.net/questions/56239/do-the-results-of-1-n-surgery-determine-the-link/56241#56241 Scott Taylor 2011-02-22T02:58:07Z 2011-02-22T02:58:07Z

<p>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).</p>

http://mathoverflow.net/questions/56239/do-the-results-of-1-n-surgery-determine-the-link/60383#60383 Dylan Thurston 2011-04-02T20:04:11Z 2011-04-02T20:04:11Z

<p>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.</p> <p>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.</p> <p>(This is all much easier than Lackenby's result mentioned above.)</p>