1989
DOI: 10.1016/0166-8641(89)90080-1
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Manifold covers of 3-orbifolds with geometric pieces

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Cited by 17 publications
(17 citation statements)
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“…By [11], & has a geometric decomposition and so also a finite covering by a manifold. That every automorphism of u x & is induced by a diffeomorphism of & follows as Theorem 1 in [15] or from [7]. That diffeomorphisms inducing the same outer automorphism are isotopic is a consequence of the main results in [16,17].…”
Section: J? N (K)\k 0^s \Kmentioning
confidence: 81%
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“…By [11], & has a geometric decomposition and so also a finite covering by a manifold. That every automorphism of u x & is induced by a diffeomorphism of & follows as Theorem 1 in [15] or from [7]. That diffeomorphisms inducing the same outer automorphism are isotopic is a consequence of the main results in [16,17].…”
Section: J? N (K)\k 0^s \Kmentioning
confidence: 81%
“…Now by Thurston's orbifold geometrization theorem (see [11]; see also [9]) the orbifold ££ n (K), having non-empty singular set, either has itself a geometric structure or has a decomposition into geometric pieces (along euclidean 2-suborbifolds). Then by results of McCullough and Miller [7], it has a finite covering by a manifold. By [15], Theorem 1 or [7], closed irreducible geometrically decomposable 3-orbifolds with infinite orbifold fundamental groups (and finite coverings by manifolds) are determined, up to orbifold homeomorphism, by their fundamental groups.…”
Section: Theorem 2 the Symmetry Group Sym(s Z K): = N 0 T>iff(s 3 mentioning
confidence: 93%
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“…It is a consequence of the Geometrization Theorem (cf. [5,32]) that if a 3-orbifold does not contain any bad 2-suborbifolds, then it must be very good, i.e., it is the quotient of a 3-manifold by a finite group action. For simplicity, we shall call such a 3-orbifold good.…”
Section: Recollections and Preliminary Lemmasmentioning
confidence: 99%
“…Then there is an orientable 3-manifold Y ′ equipped with a finite group action of G, such that Y 0 = Y ′ /G (cf. [5,32]). Since π orb 1 (Y 0 ) = π orb 1 (Y ), z(π orb 1 (Y 0 )) is also infinite, and consequently, z(π 1 (Y ′ )), which contains π 1 (Y ′ ) ∩ z(π orb 1 (Y 0 )), is infinite.…”
Section: Injectivity Of S 1 -Actions When π 1 Has Infinite Centermentioning
confidence: 99%