A Factorization of the Conway Polynomial and Covering Linkage Invariants

J.P. Levine showed that the Conway polynomial of a link is a product of two factors: one is the Conway polynomial of a knot which is obtained from the link by banding together the components; and the other is determined by the μ̄-invariants of a string link with the link as its closure. We give another description of the latter factor: the determinant of a matrix whose entries are linking pairings in the infinite cyclic covering space of the knot complement, which take values in the quotient field of Z[t, t]. In addition, we give a relation between the Taylor expansion of a linking pairing around t = 1 and derivation on links which is invented by T.D. Cochran. In fact, the coefficients of the powers of t − 1 will be the linking numbers of certain derived links in S. Therefore, the first non-vanishing coefficient of the Conway polynomial is determined by the linking numbers in S. This generalizes a result of J. Hoste. 1. Statement of Results Let L be an oriented (m + 1)-component link (m ≥ 1) in the 3-sphere S3. Throughout the paper knots and links are assumed to be oriented. It is interesting but difficult in general to figure out the structure of a link invariant. The Conway polynomial ∇L(z) of L , the most popular polynomial link invariant, was not an exception, either. However, J.P. Levine [6] made a breakthrough. Namely he showed that there is a following relationship between ∇L(z) and ∇K(z), where K is a knot obtained from L by banding together the components and Γ(z) is a power series in z which depends on the choice of bands: ∇L(z) = ∇K(z) Γ(z). Viewing the choice of bands as the choice of a string link represenation of L, Levine descrived Γ(z) by μ̄-invariants of the string link as follows. Theorem 1.1. ([6] Theorem 1) Let S be a string link, with closure LS and knot closure KS. Then we have that ∇LS (z) = ∇KS(z)ΓS(z), where ΓS(z) is a power series given by the formula: ΓS(z) = (u+ 1) e/2 det(λij(u)) with z = u/ √ u+ 1, e = { 0 if m is even 1 if m is odd and :