By Jonathan A. Hillman

ISBN-10: 0521378125

ISBN-13: 9780521378123

To assault yes difficulties in four-dimensional knot concept the writer attracts on numerous concepts, targeting knots in S^T4, whose basic teams include abelian general subgroups. Their type comprises the main geometrically attractive and most sensible understood examples. in addition, it truly is attainable to use fresh paintings in algebraic easy methods to those difficulties. New paintings in 4-dimensional topology is utilized in later chapters to the matter of classifying 2-knots.

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**Example text**

P: page 146]) and so the localization exists and is flat; if a is a nontrivial element of A then a -1 is in S and annihilates the augmentation module Z. Beyond this his argument follows that of Kaplansky in making use of properties of C· -algebras. (In [Hi 1981] we stated such a lemma for the case when A is central, and tried to derive it algebraically from Kaplansky's Lemma, but there was a gap in our argument). On the evidence of his work on 1-relator groups, Murasugi conjectured that the centre of a finitely presentable group other than Z2 of ~ deficiency 1 is infinite cyclic or trivial, and is trivial if the group has deficiency > 1, and he showed that this is true for the groups of 1-links [Mu 1965].

Finally we shall show that any 2-knot whose group is a PD 4group must be irreducible. Theorem 11 Let G be a PDn -group over 0 witb n > 2. Tben 0 is not a nontrivial free product witb amalgamation over a cyclic subgroup. Proof If G is a nontrivial free product 0 infinite index in O. 22J. dOC ~ 1. OO Mayer-Vietoris ~ max{n -1, 2}. 0 Thus we must have n = 2. Corollary If and B is a PDn -group over 0 and B have homological dimension over 0 tion = K is a 2-knot sucb tbat M(K) is aspberical tben K is not a nontrivial satellite knot.

0 also. ),R [G IT». ) = 0, and suffice duality to show that H· = O. Let S be the multiplicative system R [UIT]-{O} in RlOIT), and let r = R lOIT)S' nonzero homology of C. S then UIT Since H 3(Hom r (C. s ,B» nontrivial is is HS RS = 0 In degree 2. S ) = 0 the so only is any left r-module Poincare by and Kiinneth theorem. 3) HS duality and the is stably free and we may split the boundary maps of the complex C. S to obtain an isomorphism HS(J)C1S(J)C2S = COS(J)C2S(J)C4S' so HS r-module. In particular, X(M) By Rosset's ~ X(M) as has rank a stably free O.