Joint mathematics meetings, San Diego, January 2008 (slides)

Abstract: Up to isomorphism, the torsion-free abelian groups of rank $n$ are members of the following standard Borel space: $R(n)=$ the subgroups of $\mathbb Q^n$ which contain a basis for $\mathbb Q^n$. Hjorth and Thomas have shown that the isomorphism equivalence relations on the $R(n)$ increase in complexity with $n$.

Here is the relevant complexity notion, due to H. Friedman. For $E,F$ Borel equivalence relations on standard Borel spaces $X,Y$, write $E\leq_BF$ iff there exists a Borel function $f\colon X\to Y$ such that $x\mathrel{E}x’\iff f(x)\mathrel{F}f(x’)$.

We compare the isomorphism equivalence relation on $R(n)$ with that of quasi-isomorphism. The definition is: $A,B\in R(n)$ are quasi-isomorphic iff $A$ is commensurable with an isomorphic copy $B’\in R(n)$ of $B$. The advantage is that unique decomposition holds with respect to quasi-isomorphism, but fails with respect to isomorphism. This leads one to ask whether quasi-isomorphism is truly less complex than isomorphism.

Our result is a “no” answer, namely that the isomorphism and quasi-isomorphism relations on $R(n)$ are incomparable with respect to $\leq_B$. The proof relies on a recent superrigidity theorem, due to A. Ioana, regarding profinite actions of Kazhdan groups.