Download 2-knots and their groups by Jonathan A. Hillman PDF

By Jonathan A. Hillman

To assault definite difficulties in four-dimensional knot concept the writer attracts on various concepts, concentrating on knots in S^T4, whose primary teams comprise abelian common subgroups. Their category includes the main geometrically beautiful and most sensible understood examples. furthermore, it truly is attainable to use fresh paintings in algebraic the way 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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61. Sphericity of the eJ:terior The oustanding property of the exterior of a classical knot is that it is asphericaJ. In contrast, the exterior of a higher dimensional knot is aspherical only when it has the homotopy type of a circle, in which case the knot must be trivial [OV 19731. The proof that we shall give is due to Eckmann, who also showed that the exterior of a higher dimensional link with more than one component is never aspherical [Ec 19761. (The exterior of a I-link is aspherical if and only if the link is unsplittable).

With coefficients Then O. ) = 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.

Then def P = l-X(C(P» PI (C(P>;Q)- P2 (C(P);Q) " PI (O;Q)- P2(O;Q), and the = necessi ty of the condi- tions is clear. G = 2, the image of Z[G]r lemma. G = 2, the in Z[G]g ZIG] - Z is projective, by Schanuel's into Z[G]r splits, and L Hattori-Stallings rank O. of L is projective. is concentrated on the conjugacy class <1> of the identity [Ec 1986], and so the Kaplansky rank of L is Q~Z[G]L the dimension = 0 and so L of = Q~Z[G)L. If def P = Pl(G;Q)-P2(G;Q) then 0, by a theorem of Kaplansky. (See Section 2 of [Dy 1987] for more details on the properties and interrelations of the various notions of rank).

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