On the complexity of the classification problem for torsion-free abelian groups of finite rank

In 1937, Baer [5] introduced the notion of the type of an element in a torsion-free abelian group and showed that this notion provided a complete invariant for the classification problem for torsion-free abelian groups of rank 1. Since then, despite the efforts of such mathematicians as Kurosh [23] and Malcev [25], no satisfactory system of complete invariants has been found for the torsion-free abelian groups of finite rank n ≥ 2. So it is natural to ask whether the classification problem is genuinely more difficult for the groups of rank n ≥ 2. Of course, if we wish to show that the classification problem for the groups of rank n ≥ 2 is intractible, it is not enough merely to prove that there are 2 such groups up to isomorphism. For there are 2 pairwise nonisomorphic groups of rank 1, and we have already pointed out that Baer has given a satisfactory classification for this class of groups. In this paper, following Friedman-Stanley [11] and Hjorth-Kechris [15], we shall use the more sensitive notions of descriptive set theory to measure the complexity of the classification problem for the groups of rank n ≥ 2. Recall that, up to isomorphism, the torsion-free abelian groups of rank n are exactly the additive subgroups of the n-dimensional vector space Q which contain n linearly independent elements. Thus the collection of torsion-free abelian groups of rank 1 ≤ r ≤ n can be naturally identified with the set S(Q) of all nontrivial additive subgroups of Q. Notice that S(Q) is a Borel subset of the Polish space P(Q) of all subsets of Q, and hence S(Q) can be regarded as a standard Borel space; i.e. a Polish space equipped with its associated σ-algebra of Borel subsets. (Here we are identifying P(Q) with the space 2Qn of all functions h : Q → {0, 1} equipped with the product topology.) Furthermore, the natural action of GLn(Q) on the vector space Q induces a corresponding Borel action on S(Q); and it is easily checked that if A, B ∈ S(Q), then A ∼= B iff there exists an element φ ∈ GLn(Q) such that φ(A) = B. It follows that the isomorphism relation on S(Q) is a countable Borel equivalence relation. (If X is a standard Borel space, then a Borel equivalence relation on X is an equivalence relation E ⊆ X which is a Borel subset of X. The Borel equivalence relation E is said to be countable iff every E-equivalence class is countable.)