By J. A. Hillman
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Extra info for Alexander Ideals of Links
It is clear that for these groups El(G) = O, but more generally E _](G) = 0 for G the group of any b-component n-link (for n ~ 2) as follows from Stallings' theorem and the theorem of the next section. $8 Alexander Ideals and Chen Groups In [|40] Murasugi proved, inter alia, that for G the group of a 2-component link the following conditions are equivalent (I) El(G) = O; (2) the Chen group Q(G;q) = G G"/G G" is isomorphic to Q(F(2);q) q q+l for all q ~ l; (3) N (GqG") " the longitudes of the link are in G(~) = q~1 In the course of his proof, which involved delicate computations in the free differential calculus, he found presentations for the (finitely generated abelian) groups Q(H;q) for H free of finite rank and for H the group of a 2-component link.
There are short exact sequences I + Q(H;q) + H/Hq+IH" § H/HqH" + I by definition of the Chen groups, and so by the five lemma and induction a map f:H § K induces isomorphisms on all Chen groups if and only if all the maps fq are isomorphisms. The arguments below will be in terms of the groups H/HqH" excepting for one appeal to the computation of the Chen groups of a free group by Chen and Murasugi. The qth truncated Alexander module of H is the ~[H/H'~-module A (H) = ~ / ~ 2 ~ + ~ q ; q in ~articular A2(H) = ~/~2 is isomorghic to H/H' (see Chapter IV).
I < ~ and E (H) ~ (1) modulo I. (H) = 0 for i Hence if L is an homology boundary link, so G(L) maps onto H = F(~), then E _I(L) = O. 6 this is also true of ribbon links (taking H = H(R)) and so they provide examples on which to test Smythe's conjecture. Since any epimorphism G(L) + F(~) induces an isomorphism G / G ~ F(~), it must s H(R) if L bounds a ribbon R:~D 2 + S 3. through Thus the criterion of the next theorem may suffice to show that a ribbon link is not an homology boundary link. Theorem !
Alexander Ideals of Links by J. A. Hillman