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Simon [2] has shown that we can effectively associate to each (tame) knot K in S 3 a finitely presented classifying group CGR(K) so that two knots K and K ' are equivalent if and only if CGR(K) and CGR(K') are isomorphic. CGR(K) is defined in terms of certain cables of K # R, R being some fixed reference knot, and Simon's result depends on several geometrical facts, including Waldhausen's results [3] on sufficiently large 3-manifolds. In this note we produce a simpler classifying group which can be derived from Waldhausen's results in a purely algebraic fashion. THEOREM. Let G = (g ( | rj) be a presentation for the fundamental group of a knot K, and let X denote a longitudinal, and n a corresponding meridional element, properly oriented (see Fig. 1). Then the group GXi ^ with presentation (gita,b\rpa 5 is a classifying group for K. = l, X ~ 1 aX = a 2, b 1 = 1, n ~ l bn = b \ [a, b] = 1

Year: 2011

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