2015
DOI: 10.1093/imrn/rnv008
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Infinitely Many Knots Admitting the Same Integer Surgery and a Four-Dimensional Extension

Abstract: We prove that for any integer n there exist infinitely many different knots in S 3 such that n-surgery on those knots yields the same 3-manifold. In particular, when |n| = 1 homology spheres arise from these surgeries. This answers Problem 3.6(D) on the Kirby problem list. We construct two families of examples, the first by a method of twisting along an annulus and the second by a generalization of this procedure. The latter family also solves a stronger version of Problem 3.6(D), that for any integer n, there… Show more

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Cited by 12 publications
(21 citation statements)
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“…We outline the proof as follows: Let K 0 and K 1 be the unoriented knots depicted in Figure 1 and give arbitrary orientations on K 0 and K 1 . By using annular twisting techniques developed in [1,2,40,51], we see that 0surgeries on K 0 and K 1 give the same 3-manifold. On the other hand, by Miyazaki's result [36], we can prove that K 0 #K 1 is not ribbon, where K 0 #K 1 denotes the connected sum of K 0 and the mirror image of K 1 .…”
Section: Introductionmentioning
confidence: 92%
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“…We outline the proof as follows: Let K 0 and K 1 be the unoriented knots depicted in Figure 1 and give arbitrary orientations on K 0 and K 1 . By using annular twisting techniques developed in [1,2,40,51], we see that 0surgeries on K 0 and K 1 give the same 3-manifold. On the other hand, by Miyazaki's result [36], we can prove that K 0 #K 1 is not ribbon, where K 0 #K 1 denotes the connected sum of K 0 and the mirror image of K 1 .…”
Section: Introductionmentioning
confidence: 92%
“…In this section, we prove our main theorem. The main tools are Miyazaki's result [36,Theorem 5.5] and annular twisting techniques developed in [1,2,40,51]. For the sake of completeness, we will review these results in Appendices A and B.…”
Section: Proof Of Theorem 16mentioning
confidence: 99%
“…To see this, observe that in the terminology of [AJLO14] the diagram on the left of Figure 2 is a simple annulus presentation for K n . Further, the knot K 0 n,t is obtained from K n by ( * t 2 )-twisting, which is defined by [AJLO14] and is a natural modification of the annulus twisting defined by [Oso06]. Therefore Theorem 3.10 of [AJLO14] implies that X t/2 (K n ) ∼ = X t/2 (K 0 n,t ).…”
Section: Proof Of Theorem 12mentioning
confidence: 99%
“…Further, the knot K 0 n,t is obtained from K n by ( * t 2 )-twisting, which is defined by [AJLO14] and is a natural modification of the annulus twisting defined by [Oso06]. Therefore Theorem 3.10 of [AJLO14] implies that X t/2 (K n ) ∼ = X t/2 (K 0 n,t ). In the notation of the introduction, we take…”
Section: Proof Of Theorem 12mentioning
confidence: 99%
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