By Jonathan Hillman

This publication is meant as a reference on hyperlinks and at the invariants derived through algebraic topology from overlaying areas of hyperlink exteriors. It emphasizes positive factors of the multicomponent case now not in most cases thought of by means of knot theorists, equivalent to longitudes, the homological complexity of many-variable Laurent polynomial jewelry, loose coverings of homology boundary hyperlinks, the truth that hyperlinks usually are not often boundary hyperlinks, the reduce primary sequence as a resource of invariants, nilpotent final touch and algebraic closure of the hyperlink crew, and disc hyperlinks. Invariants of the kinds thought of right here play a vital function in lots of functions of knot idea to different parts of topology.

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HOMOLOGY AND DUALITY IN COVERS If M is a finitely generated A^-module, let TM be its A^-torsion submodule and let zM be the submodule of TM generated by elements whose annihilator ideal is not contained in a proper principal ideal. (This is the maximal pseudonull submodule of M, cf. Chapter 3). Let tM = TM/zM. Let L be a /x-component 1-link and let p : X' —> X be the maximal abelian cover of the exterior X — X(L). The meridians of L determine an isomorphism of the covering group TT/TT' = Aut(X'/X) with Z^, and so we may identify Z[7r/7r'] with A^.

Ext^Gm{Hp{X-,R[G/H\lR[G/H\) =» H^(X;R[G/H}), whose rth differential dr has bidegree (1 — r, r). 2. [G/H\Hq(X;R[G/H]),R[G/J]) th 29 =• Hp+q(X; R[G / J]), r whose r differential d has bidegree (—r,r — 1). There are similar spectral sequences for the equivariant homology and cohomology of pairs of cell complexes. (See [McC] for more details on these spectral sequences). 2. Covers of link exteriors As the abelianization of a link group is free abelian, link exteriors have canonical free abelian covers.

Then ak(M/pM) = 0 if k < q and aq+i(M/pM) = R/p, so akM < p if and only if k < q. Moreover M has a presentation with q generators, by Nakayama's Lemma. Since M/pM has dimension q, all the entries of the presentation matrix are in p, and hence Ek(M) < p if and only if k < q, that is, if and only if ak+\M < p. In other words, y/Ek(M) - ^Jak+\M. • In fact Ek(M) < ak+1M < (Ek(M) : Ek+1(M)), for all k > 0 [BE77]. (This paper also gives sufficient conditions for these ideals to be equal). Is Er(M) = EQ(TM)Er(M/TM) always?

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