2016
DOI: 10.1007/s10474-016-0660-8
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Computing rotation and self-linking numbers in contact surgery diagrams

Abstract: Erratum to: Acta Mathematica Hungarica 150 (2) (2016), 524-540.

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Cited by 13 publications
(25 citation statements)
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“…It is also possible to show that every Legendrian unknot of type U n admits (infinitely many) cosmetic contact surgeries resulting again in (S 3 , ξ st ). By computing explicitly the d 3 -invariants (see [9,10]) of all other contact surgeries one can classify all rational contact Dehn surgeries along a single Legendrian unknot resulting in a S 3 with an arbitrary contact structure [32,Section 5.4]. Moreover, in Section 8 we showed that all topological examples of links not determined by their complements, where one does a composition of (−1)-Rolfsen twist along unknot components, works also for the contact case.…”
Section: Lemma 103 (Criterion For the Legendrian Knot Exterior Problem)mentioning
confidence: 99%
“…It is also possible to show that every Legendrian unknot of type U n admits (infinitely many) cosmetic contact surgeries resulting again in (S 3 , ξ st ). By computing explicitly the d 3 -invariants (see [9,10]) of all other contact surgeries one can classify all rational contact Dehn surgeries along a single Legendrian unknot resulting in a S 3 with an arbitrary contact structure [32,Section 5.4]. Moreover, in Section 8 we showed that all topological examples of links not determined by their complements, where one does a composition of (−1)-Rolfsen twist along unknot components, works also for the contact case.…”
Section: Lemma 103 (Criterion For the Legendrian Knot Exterior Problem)mentioning
confidence: 99%
“…For an even integer n, choose a Legendrian knot L in a Darboux-ball in (M, ξ) with Thurston-Bennequin invariant tb = 1 and rotation number rot = n. Denote the contact manifold obtained by contact (−1)-surgery along L by L(−1). A calculation as for example in [DGS04,DK16] shows that the homology of L(−1) is…”
Section: Properties Of the Contact Surgery Graphmentioning
confidence: 99%
“…It therefore suffices to show that tb(L) = −5 for the Legendrian knot L shown in Figure 4, which can be done with the formula from [28, Lemma 6.6], cf. [21, Lemma 3.1] and [11]. Let M be the linking matrix of the surgery diagram in Figure 4.…”
Section: 2mentioning
confidence: 99%
“…For the rotation number rot(L) of L in the surgered S 3 we also have a formula from [28]; cf. [11]. Write rot 0 for the rotation number of L in the unsurgered copy of S 3 .…”
Section: 2mentioning
confidence: 99%