6
$\begingroup$

Math Overflow seems to have a dearth of low dimensional topology, but this seems like an interesting question. Let $L$ be a $2$-component link in $S^3$. Suppose that there is a framed arc joining the two components such that surgery along that arc yields an unknot. (In other words, fatten the arc to a long thin rectangle where two of its edges are subarcs of the link components. Delete these arcs and replace them with the other two edges of the rectangle.) I wonder how restrictive this condition is. For example, can one prove that there are links which don't have this property? I don't really have a good feel for what is possible.

One way to construct an example is to start with an unknotted surface with one boundary component. Now attach a band to the boundary of this surface that travels through all of the surfaces holes. This is now a 2-component link. Snipping the band can be realized by surgery along a transverse arc, and yields the boundary of the original surface, which is an unknot.

$\endgroup$

3 Answers 3

6
$\begingroup$

One source of restrictions is the Montesinos trick: if you take the branched double cover of $L$, then a small neighborhood of the framed arc lifts to a solid torus because it intersects $L$ in two small arcs. Replacing those small arcs with the ones coming from the surgery produces another solid torus in the new branched cover, so the two branched covers are related by a Dehn surgery. If the surgery produces the unknot, then this means that the branched double cover $\Sigma(L)$ has to be Dehn surgery on a knot in $S^3$.

For example, the above implies that $H_1(\Sigma(L))$ must be cyclic. If $K$ is a knot with determinant $d > 1$, then $H_1(\Sigma(K))$ has order $d$. The connected sum $L = K \# T(2,2d)$ is a 2-component link with branched double cover $\Sigma(K) \# L(2d,1)$, where $T(2,2d)$ is the $(2,2d)$-torus link and $L(2d,1)$ a lens space. Its first homology is $H_1(\Sigma(K)) \oplus \mathbb{Z}/(2d)$, and since this is not cyclic, no surgery on $L$ will produce the unknot.

This also reproves the result of Scharlemann which Ian Agol mentioned, because if L is a split link $L_1 \sqcup L_2$ then its branched cover is $\Sigma(L_1) \# \Sigma(L_2) \# S^1\times S^2$. Then by Gabai's proof of the Poenaru conjecture (which came after Scharlemann's paper), we must have $\Sigma(L_1)=\Sigma(L_2)=S^3$, and so $L_1$ and $L_2$ are both unknots.

$\endgroup$
2
  • 1
    $\begingroup$ I think this observation underlies the other papers I linked to. $\endgroup$
    – Ian Agol
    Jul 30, 2015 at 18:30
  • $\begingroup$ It's certainly used in the paper by Bao. I didn't see how the other papers use it to prove their results about band sums, although Eudave-Muñoz does invoke it afterward to apply his result to the cabling conjecture. $\endgroup$ Jul 30, 2015 at 19:29
6
$\begingroup$

The main theorem of this paper of Scharlemann implies that a split two component link which is not the unlink cannot have a band sum giving the unknot. See also Eudave-Muñoz and Ishihara-Motegi and Bao. This move is called "band surgery" or "H(2) move" (after Hoste-Nakanishi-Taniyama).

$\endgroup$
3
  • $\begingroup$ The funny thing is that I read Scharlemann's paper in grad school, and you were in the audience when I presented it in the topology seminar! Are you saying that a band sum of a nontrivial split link which is the unknot would give a counterexample to the 4D Schönflies conjecture? $\endgroup$
    – Jim Conant
    Jul 30, 2015 at 14:11
  • $\begingroup$ Hey Jim - the paper has nothing to do with Schoenfies - it's about 2-spheres embedded in $\mathbb{R}^4$ with 4 critical points. Saying such a sphere is standard is equivalent to saying that any band sum of the 2-component unlink which gives the unknot is standard. Scharlemann proves this actually for split links. A simpler proof was found later by Thompson. dx.doi.org/10.1016/0040-9383(87)90060-7 $\endgroup$
    – Ian Agol
    Jul 30, 2015 at 17:41
  • $\begingroup$ Oh oops. That's what happens when I comment when I first get up in the morning. 2-spheres, 3-spheres same difference. $\endgroup$
    – Jim Conant
    Jul 30, 2015 at 18:12
3
$\begingroup$

A recentish paper of mine (which generalizes portions of Scharlemann's and Eudave-Munoz's work) also addresses this question.

MR3192616 Taylor, Scott A. Comparing 2-handle additions to a genus 2 boundary component. Trans. Amer. Math. Soc. 366 (2014), no. 7, 3747--3769.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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