2013
DOI: 10.1090/conm/597/11768
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Mapping Class Groups of 3-Manifolds, Then and Now

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Cited by 10 publications
(6 citation statements)
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“…For arbitrary compact 3-manifolds, mapping class groups are only known to be finitely presented, see Conjecture F and the comments at the end of Section 4 of [42]. Notice that there is a subtle difference between the definition of our mapping class group and the mapping class group in [42], where they allow isotopies to move points on the boundary.…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
See 1 more Smart Citation
“…For arbitrary compact 3-manifolds, mapping class groups are only known to be finitely presented, see Conjecture F and the comments at the end of Section 4 of [42]. Notice that there is a subtle difference between the definition of our mapping class group and the mapping class group in [42], where they allow isotopies to move points on the boundary.…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…For arbitrary compact 3-manifolds, mapping class groups are only known to be finitely presented, see Conjecture F and the comments at the end of Section 4 of [42]. Notice that there is a subtle difference between the definition of our mapping class group and the mapping class group in [42], where they allow isotopies to move points on the boundary. But when the boundary is a union of spheres, there is a surjective map from our sphere-permuting mapping class group (Definition 2.1) to their mapping class group, whose kernel is a finite abelian group generated by sphere twists along boundaries.…”
Section: Introduction and Main Resultsmentioning
confidence: 99%
“…Then, the Dehn twist along the meridian of the boundary torus will be trivialized. By a result of McCullough (see for instance [61]), every Dehn twist along the longitude induces a diffeomorphism of the solid torus. Then, the complement of the unknot can carry an electric charge (by a Dehn twist) but no magnetic charge.…”
Section: Electric Charge As a Frame Of The Knot Complementmentioning
confidence: 99%
“…As in the proof of Theorem B, the invertibility of X implies that the V j are distinct smooth manifolds, relative to a fixed identification of ∂V j with N . But since N is a hyperbolic manifold, π 0 (Diff(N )) is finite [17,18] (a thorough discussion of such issues may be found in [23]). By Lemma 5.1, infinitely many of the V j are absolutely distinct.…”
Section: Explicit Examples Of Absolutely Exotic Corks and Anti-corksmentioning
confidence: 99%