# Ioana's Superrigidity Theorem and Orbit Equivalence Relations

Authors: Samuel Coskey
Journal: ISRN Algebra
 (2013)

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## Abstract

We give a survey of Adrian Ioana's cocycle superrigidity theorem for profinite actions of Property (T) groups and its applications to ergodic theory and set theory in this expository paper. In addition to a statement and proof of Ioana's theorem, this paper features the following: (i) an introduction to rigidity, including a crash course in Borel cocycles and a summary
of some of the best-known superrigidity theorems; (ii) some easy applications of superrigidity, both to ergodic theory (orbit equivalence)
and set theory (Borel reducibility); and (iii) a streamlined proof of Simon Thomas's theorem that the classification of torsion-free abelian groups of finite rank is intractable.


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## Body

## 1. Introduction

In the past fifteen years superrigidity theory has had a boom in the number and variety of new applications. Moreover, this has been coupled with a significant advancement in techniques and results. In this paper, we survey one such new result, namely, Ioana’s theorem on profinite actions of Property (T) groups and some of its applications in ergodic theory and in set theory. In the concluding section, we highlight an application to the classification problem for torsion-free abelian groups of finite rank. The narrative is strictly expository, with most of the material being adapted from the work of Adrian Ioana, mine, and Simon Thomas.Although Ioana’s theorem is relatively recent, it will be of interest to readers who are new to rigidity because the proof is natural and there are many immediate applications. Therefore, we were keen to keep the nonexpert in mind. We do assume that the reader is familiar with the notion of ergodicity of a measure-preserving action and with unitary representations of countable groups. We will not go into great detail on Property (T), since for our purposes it is enough to know thatSLn(ℤ) satisfies Property (T) when n>2. Rather, we will introduce it just when it is needed, and hopefully its key appearance in the proof of Ioana’s theorem will provide some insight into its meaning.The concept of superrigidity was introduced by Mostow and Margulis in the context of studying the structure of lattices in Lie groups. Here,Γ is said to be a lattice in the (real) Lie group G if it is discrete and G/Γ admits an invariant probability measure. Very roughly speaking, Margulis showed that if Γ is a lattice in a simple (higher-rank) real Lie group G, then any homomorphism from Γ into an algebraic group H lifts to an algebraic map from G to H. This implies Mostow’s theorem, which states that any isomorphic lattices Γ,Λ in a simple (higher-rank) Lie group G must be conjugate inside G.We will leave this first form of rigidity on the back burner and primarily consider instead a second form, initially considered by Zimmer, which is concerned with groupactions. (The connection between the two forms of rigidity is that both can be cast in terms of measurable cocycles, which will be introduced in the next section. For the connection between cocycles and lifting homomorphisms, see [1, Example 4.2.12].) The basic notions are as follows. Two probability measure-preserving actions Γ↷X and Λ↷Y are said to be orbit equivalent if there exists a measure-preserving almost bijection f:X→Y such that Γx=Γx′ if Λf(x)=Λf(x′). They are said to be isomorphic if additionally there exists an isomorphism ϕ:Γ→Λ such that f(γx)=ϕ(γ)f(x). Essentially, Zimmer showed that any (irreducible) ergodic action Γ↷X of a lattice in a (higher rank) simple Lie group is superrigid in the sense that it cannot be orbit equivalent to another action of an algebraic group Λ↷Y without being isomorphic to it. (For elementary reasons it is necessary to assume that Λ acts freely on Y.) See [1, Theorem 5.2.1] for a weak statement of this result and [2, Section 1] for further discussion.It is natural to ask whether there exists an analog of Zimmer’s theorem in the context of general measure-preserving actions, that is, with the algebraic hypothesis onΛ removed. Many rigidity results have been established along these lines (for instance, see [2–4]). One of the landmark results in this direction was obtained recently by Popa [5], who found a large class of measure-preserving actions Γ↷X which are superrigid in the general sense that Γ↷X cannot be orbit equivalent to another (free) action without being isomorphic to it. In particular his theorem states that if Γ is a Property (T) group, then the free part of its left-shift action on X=2Γ (the so-called Bernoulli action) is an example of a superrigid action. Following on Popa’s work, Ioana’s theorem gives a second class of examples of superrigid actions, namely, the profinite actions of Property (T) groups.This paper is organized as follows. The second section gives some background on Borel cocycles, a key tool in rigidity theory. A slightly weakened version of Ioana’s theorem is stated in the third section. The proof itself is split between Section4, which contains a general purpose lemma, and Section 5, which contains the heart of the argument. Although these are largely unchanged from Ioana’s own account, I have inserted many additional remarks to smooth the experience for the newcomer.In Section6, we give a couple of the easier applications of the main theorem. First, we show how to obtain many orbit inequivalent profinite actions of SLn(ℤ). We also explore applications to logic and set theory by considering Borel reducibility. In particular, we point out some of the extra challenges one faces when working in the purely set-theoretic (i.e., Borel) context, as opposed to the more familiar measure context.Finally, in the last section, we use Ioana’s theorem to give a self-contained and slightly streamlined proof of Thomas’s theorem that the complexity of the isomorphism problem for torsion-free abelian groups of finite rank increases strictly with the rank.


## 2. Rigidity via Cocycles

We begin by introducing a slightly more expansive notion of orbit equivalence rigidity. IfΓ↷X and Λ↷Y are arbitrary Borel actions of countable groups, then a function f:X→Y is said to be a homomorphism of orbits if Γx=Γx′ implies Λf(x)=Λf(x′). It is said to be a homomorphism of actions if additionally there exists a homomorphism ϕ:Γ→Λ such that f(γx)=ϕ(γ)f(x). (Note that these terms are not exactly standard.) Informally, we will say that Γ↷X is superrigid if whenever Λ↷Y is a free action and f:X→Y is a homomorphism of orbits, then f in fact arises from a homomorphism of actions (i.e., f is equivalent to a homomorphism of actions in a sense defined below).Following Margulis and Zimmer, we will require the language of Borel cocycles to describe and prove superrigidity results. Acocycle is an object which is associated with a given homomorphism of orbits f:X→Y as follows. Observe that for every (γ,x)∈Γ×X, there exists a λ∈Λ such that f(γx)=λf(x). Moreover, Λ acts freely on Y if and only if this λ is always uniquely determined by the data f, γ, and x. In other words, in this case, f determines a function α:Γ×X→Λ which satisfies
(1)f(γx)=α(γ,x)f(x).
This map is called the cocycle corresponding tof, and it is easy to see that it is Borel whenever f is. Moreover, the cocycle α satisfies the composition law α(γ′γ,x)=α(γ′,γx)α(γ,x); this is called the cocycle condition. See Figure 1 for a visual depiction of the cocycle condition.Figure 1
The cocycle condition:α(γ′γx)=α(γ′,γx)α(γ,x).Whenf is actually action-preserving, that is, f(γx)=ϕ(γ)f(x) for some homomorphism ϕ:Γ→Λ, then we have α(γ,x)=ϕ(γ), so that α is independent of the second coordinate. Conversely, if α is independent of the second coordinate, then one can define ϕ(γ)=α(γ,·) and the composition law implies that ϕ is a homomorphism. In this situation, the cocycle is said to be trivial.In practice, when establishing rigidity, one typically shows that an arbitrary cocycle (arising from a homomorphism of orbits) is equivalent to a trivial cocycle (which therefore arises from a homomorphism of actions). Here, we say that homomorphisms of orbitsf,f′:X→Y are called equivalent if there exists a Borel function b:X→Λ such that f′(x)=b(x)f(x), a.e. (i.e., they lift the same function on the quotient spaces X/Γ→Y/Λ). In this case, the corresponding cocycles α,α′ are said to be cohomologous. It is easy to check that f,f′ are equivalent via b if and only if the corresponding cocycles α,α′ satisfy the relation α′(γ,x)=b(γx)α(γ,x)b(x)-1   a.e.; this is called the cohomology relation. The easiest way to see that this is the case is to glance at Figure 2.Figure 2
The cohomology relation for cocycles:α′(γ,x)=b(γx)α(γ,x)b(x)-1.We close this section by remarking that not all cocycles arise from orbit-preserving maps. An abstractcocycle is any Borel function satisfying the cocycle condition a.e., and two cocycles are said to be cohomologous if there exists a Borel function b satisfying the cohomology relation a.e. The most powerful superrigidity results often have the conclusion that “every cocycle is cohomologous to a trivial cocycle.” However, for most applications there is no need for the extra strength gained by using the abstract cocycle formulation.


## 3. Ioana’s Theorem

Cocycle superrigidity results were first established by Margulis and Zimmer for cocyclesΓ↷X→Λ where Γ is a lattice in a higher rank Lie group acting ergodically on X. These results carried the additional hypothesis that Λ is contained in an algebraic group. The first example of the most general form of cocycle superrigidity, with the target Λ arbitrary, was Popa’s result concerning Bernoulli actions. In this section, we will discuss Ioana’s theorem, which establishes similar conclusions for profinite actions.Here,Γ↷X is said to be profinite if as a Γ-set, X is the inverse limit of a family of finite Γ-sets Xn. In particular, there exist equivariant projections πn:X→Xn and each element x∈X can be identified with the thread (πn(x)). We are interested in the ergodic case; here, each Xn is equipped with the uniform probability measure and Γ↷Xn is transitive.Theorem 1 (Ioana).
LetΓ↷(X,μ) be an ergodic, measure-preserving, profinite action, with invariant factor maps πn:X→Xn. Assume that Γ has Property (T). Then for any cocycle α:Γ↷X→Λ, there exists n and a∈Xn such that the restriction of α to the action Γa↷πn-1(a) is cohomologous to a trivial cocycle.In other words, the conclusion is thatΓ↷X is “virtually superrigid” in the sense that any orbit preserving map, after it is restricted to a finite index component of the left-hand side, comes from an action preserving map. Ioana’s theorem is interesting when contrasted with Popa’s theorem; while Bernoulli actions are strongly mixing, profinite actions are highly nonmixing. Indeed, for each n, Γ just permutes the blocks πn-1(a), for a∈Xn, and it follows that ⋃a∈Xn‍Xa×Xa is a Γ-invariant subset of X×X.We remark that although our variant of Ioana’s theorem is sufficient for most applications, it is weaker than the state of the art in several ways. First, Ioana requires only thatΓ have the relative Property (T) over some infinite normal subgroups N such that Γ/N is finitely generated. Second, Ioana also shows that α is equivalent to a cocycle defined on all of X. Last, Furman has generalized the statement by replacing profinite actions with the more general class of compact actions. (Γ↷X is said to be compact if when regarded as a subset of (X,μ) it is precompact in a suitable topology.)


## 4. Cocycle Untwisting

We begin with the following preliminary result, which roughly speaking says that ifα:Γ↷X→Λ is a cocycle and if for each γ it has a “very likely” value, then α is cohomologous to the map which always takes on this likely value. In particular, in this case, α is cohomologous to a trivial cocycle.Theorem 2.
LetΓ↷(X,μ) be ergodic and measure-preserving, and let α:Γ↷X→Λ be a cocycle. Suppose that for all γ∈Γ there exists λγ∈Λ such that
(2)μ{x∣α(γ,x)=λγ}≥C>78.
Then the map ϕ(γ)=λγ is a homomorphism and α is cohomologous to it.It is easy to see thatϕ must be a homomorphism: indeed, the hypothesis guarantees that there is a nonnull set of x for which ϕ(γ′γ)=α(γ′γ,x)=α(γ′,γx)α(γ,x)=ϕ(γ′)ϕ(γ). Hence it remains only to establish the following result.Lemma 3.
LetΓ↷(X,μ) be ergodic and measure-preserving, and let α,β:Γ↷X→Λ be cocycles. Suppose that for all γ∈Γ(3)μ{x∣α(γ,x)=β(γ,x)}≥C>78.
Then α is cohomologous to β.We understand this result to say that ifα and β are close in an L∞ sense, then they are cohomologous. It follows upon similar results of Popa and Furman, which draws the same conclusion in the case that α and β are close in an appropriate L1 sense (for instance, see [6, Theorem 4.2]). Ioana’s proof, given below, may be safely skipped until reading the next section.Proof of Lemma3.
LetΓ↷X×Λ be the action given by
(4)γ(x,λ)=(γx,α(γ,x)λβ(γ,x)-1)
(this is an action thanks to the cocycle condition), and consider the corresponding left-regular representation. The reason for using this representation is that α is close to β if and only if a particular vector is close to being invariant. Namely, let
(5)ξ=χX×e
(read: the characteristic function of X×e) and notice that
(6)〈γξ,ξ〉=μ{x∣α(γ,x)=β(γ,x)}.
Using this together with the law of cosines, the hypothesis now translates to say that ∥γξ-ξ∥≤C<1/2 for all γ∈Γ. It is not difficult to see that this implies that there is an invariant vector η such that ∥η-ξ∥<1/2. (Indeed, letting S denote the convex hull of Γ·ξ¯, it is easily seen that there exists a unique vector η∈S of minimal norm; this η is necessarily invariant.)
The idea for the conclusion of the proof is as follows. If we hadη=χgraph(b) for some function b:X→Λ, then we would be done. Indeed, in this case, the invariance of η would mean that b(x)=λ if and only if b(γx)=α(γ,x)λβ(γ,x)-1, so that b(γx)=α(γ,x)b(γx)β(γ,x)-1. In other words, b would witness the fact that α is cohomologous to β. The fact that ∥η-ξ∥<1/2 implies that this is close to being the case.
We actually defineb(x)= the λ such that |η(x,λ)|>1/2, if it exists and is unique. The above computation shows that when b(x),b(γx) are both defined, the cohomology relation holds. Moreover, the set where b is defined is invariant, so by the ergodicity of Γ↷X, it suffices to show that this set is nonnull. In fact, since η and ξ are close, b must take value e on a nonnull set:
(7)μ{x:|ξ(x,e)-η(x,e)|≥12}≤4∫{x:|ξ(x,e)-η(x,e)|≥1/2}‍|ξ-η|2≤4∥ξ-η∥2<1.
This shows that {x:|η(x,e)|>1/2} is non-null, as desired. A similar computation is used to show that, with probability 1, e is the unique such element of λ.


## 5. Ioana’s Proof

What We Want. We wish to find some n and a∈Xn such that for all γ∈Γa,
(8)(μa×μa){x,x′∣α(γ,x)=α(γ,x′)}≥C>78,
where μa denotes the normalized restriction of μ to πn-1(a). This would imply, by a straightforward computation, that for each γ∈Γa there exists a λ∈Λ such that
(9)μa{x∣α(γ,x)=λ}≥C>78
and this would complete the proof thanks to Theorem 2.What We Have. Unfortunately, it is only immediately possible to obtain that the quantities in (8) tend to 1 on average, at a rate depending on γ. That is, for each γ∈Γ, we have
(10)limn→∞1|Xn|∑a∈Xn‍(μa×μa){x,x′∣α(γ,x)=α(γ,x′)}=1.
To see this, first note that it is equivalent to
(11)limn→∞∑λ∈Λ‍(1|Xn|∑a∈Xn‍μa{x∣α(γ,x)=λ}2)=1.
Now, we generally have that for any subset S⊂X,
(12)limn→∞1|Xn|∑a∈Xn‍μa(S)2=μ(S).
This is because the family {χπn-1(a)∣a∈Xn,n∈ω} is dense in L2, and while the right-hand side is the norm-squared of χS, the left-hand side is the norm-squared of χS projected onto the span of {χπm-1(a)∣a∈Xm,m≤n}. Finally, just apply (12) to each set S={x∣α(γ,x)=λ} and use the dominated convergence theorem to pass the limit through the sum over all λ∈Λ.Proof.
The gap between what he has (the asymptotic information) and what we want (the uniform information) is bridged by Property (T). Once again the first step is to consider an appropriate representation; this time one which compares the values ofα(γ,x) as x varies. That is, let Γ↷X×X×Λ by
(13)γ(x,x′,λ)=(γx,γx′,α(γ,x)λα(γ,x′)-1),
and consider the left-regular unitary representation corresponding to this action. The idea, very roughly, is that the degree to which α(γ,x) is independent of x will be measured by how close a particular vector is to being Γ-invariant.
More precisely, for eachn define an orthonormal family of vectors ξa for a∈Xn by
(14)ξa=|Xn|·χπn-1(a)×πn-1(a)×e,
and consider their normalized average
(15)ξn=1|Xn|∑a∈Xn‍ξa.
Then a simple calculation shows that
(16)〈γξa,ξa〉=(μa×μa){x,x′∣α(γ,x)=α(γ,x′)},〈γξn,ξn〉=1|Xn|∑a∈Xn‍(μa×μa){x,x′∣α(γ,x)=α(γ,x′)}.
So now, “what we have” and “what we want” can be translated as follows: we have the ξn form a family of almost invariant vectors and we want a single n and a∈Xn such that ξa is nearly invariant, uniformly for all γ∈Γa.
The remainder of the argument is straightforward. Since theξn forms a family of almost invariant vectors, Property (T) implies that there exist n and an invariant vector η such that ∥η-ξn∥≤δ. Let η′ be the restriction of η to the set ∪a∈Xn(πn-1(a)×πn-1(a)×Λ). Since this set is invariant, we have the fact that that η′ is invariant as well. Since ξn is supported on this set, we retain the property that ∥η′-ξn∥≤δ.
Now, we simply expressη′ as a normalized average of orthogonal Γa-invariant vectors. More specifically, write
(17)η′=1|Xn|∑a∈Xa‍ηa,
where ηa is the appropriately rescaled restriction of η′ to the set πn-1(a)×πn-1(a)×Λ. Then by the law of averages, we must have somea∈Xn such that ∥ηa-ξa∥≤δ. Moreover, ηa is Γa-invariant, so that for all γ∈Γa we have 〈γηa,ηa〉=1. It follows that by an appropriate choice of δ, we can make 〈γξa,ξa〉≥C>7/8 for all γ∈Γa.


## 6. Easy Applications

In this section, we use Ioana’s theorem for one of its intended purposes: to find many highly inequivalent actions. The results mentioned here are just meant to give the flavor of applications of superrigidity; they by no means demonstrate the full power of the theorem. In the next section we will discuss the slightly more interesting and difficult application to torsion-free abelian groups. For further applications, see for instance [7–10].In searching for inequivalent actions, one might of course consider a variety of inequivalence notions. Here, we focus on just two of them: orbit inequivalence and Borel incomparability. Recall from the introduction thatΓ↷X and Λ↷Y are said to be orbit equivalent if there exists a measure-preserving and orbit-preserving almost bijection from X to Y. Notice that this notion depends only on the orbit equivalence relation arising from the two actions and not on the actions themselves. When this is the case, we will often conflate the two, saying alternately that certain actions are orbit equivalent or that certain equivalence relations are “orbit equivalent.”Borel bireducibility is a purely set-theoretic notion with its origins in logic. The connection is that ifE is an equivalence relation on a standard Borel space X, then we can think of E representing a classification problem. For instance, if X happens to be a set of codes for a family of structures, then studying the classification of those structures amounts to studying the isomorphism equivalence relation E on X. We refer the reader to [11] for a complete introduction to the subject.IfE and F are equivalent relations on X and Y, then E is said to be Borel reducible to F if there exists a Borel function f:X→Y satisfying xEx′ if and only if f(x)Ff(x′). We think of this saying that the classification problem for elements of X up to E is no more complex than the classification problem for elements of Y up to F. Thus, if E and F are Borel bireducible (i.e., there is a reduction both ways), then they represent classification problems of the same complexity.It is elementary to see that neither orbit equivalence or Borel bireducibility implies the other. For instance, given anyΓ-space X one can form a disjoint union X⊔X′, where X′ is a Γ-space of very high complexity which is declared to be of measure 0. Conversely, if X is an ergodic and hyperfinite Γ-space, then it is known that it is bireducible with X⊔X, but the two cannot be orbit equivalent. It is even possible, without much more difficulty, to find two ergodic actions which are bireducible but not orbit equivalent.We are now ready to begin with the following direct consequence of Ioana’s theorem. It was first established by Simon Thomas in connection working on classification problem for torsion-free abelian groups of finite rank. His proof used Zimmer’s superrigidity theorem and some additional cocycle manipulation techniques; with Ioana’s theorem in hand, the proof will be much simpler.Corollary 4.
Ifn≥3 is fixed and p,q are primes such that p≠q, then the actions of SLn(ℤ) on SLn(ℤp) and SLn(ℤq) are orbit inequivalent and Borel incomparable.Here,ℤp denotes the ring of p-adic integers. It is easy to see that SLn(ℤ)↷SLn(ℤp) is a profinite action, being the inverse limit of the actions SLn(ℤ)↷SLn(ℤ/piℤ) together with their natural system of projections.Proof.
Letp≠q and suppose that f is either an orbit equivalence or a Borel reduction from SLn(ℤ)↷SLn(ℤp) to SLn(ℤ)↷SLn(ℤq). We now apply Ioana’s theorem together with the understanding of cocycles gained in the previous section. The conclusion is that we can suppose without loss of generality that there exists a finite index subgroup Γ0≤SLn(ℤ), a Γ¯0-coset X⊂SLn(ℤp), and a homomorphism ϕ:Γ0→SLn(ℤ) which makes f into an action-preserving map from Γ0↷X into SLn(ℤ)↷SLn(ℤq).
Now, in the measure-preserving case, it is not difficult to conclude thatf is a “virtual isomorphism” between the two actions. We claim that this can be achieved even in the case that f is just a Borel reduction. First, we can assume that ϕ is an embedding. Indeed, by Margulis’s theorem on normal subgroups [1, Theorem 8.1.2], either im(ϕ) or ker(ϕ) is finite. If ker(ϕ) is finite, then we can replace Γ0 by a finite index subgroup (and X by a coset of the new Γ¯0) to suppose that ϕ is injective. On the other hand, if im(ϕ) is finite, then we can replace Γ0 by a finite index subgroup to suppose that ϕ is trivial. But this would mean that f is Γ0-invariant, and so by ergodicity of Γ0↷X, f would send a conull set to a single point, contradicting that f is countable-to-one.
Second,ϕ(Γ0) must be a finite index subgroup of SLn(ℤ). Indeed, by Margulis’s superrigidity theorem, ϕ can be lifted to an isomorphism of SLn(ℝ), and it follows that ϕ(Γ0) is a lattice of SLn(ℝ). But then it is easy to see that any lattice which is contained in SLn(ℤ) must be commensurable with SLn(ℤ).
Third, by the ergodicity ofΓ0↷X, we can assume that im(f) is contained in a single ϕ(Γ0)¯ coset Y0. And now because ϕ(Γ0) preserves a unique measure on Y0 (the Haar measure) and because ϕ(Γ0) preserves f*(Haar), we actually conclude that f is measure-preserving. In summary, we have shown that (ϕ,f) is a measure and action-preserving isomorphism between Γ0↷X0 and ϕ(Γ0)↷Y0, which establishes the claim.
Finally, a short computation confirms the intuitive, algebraic fact that the existence of such a map is ruled out by the mismatch in primes between the left-and right-hand sides. We give just a quick sketch; for a few more details see [12, Section 6]. Now, it is well-known that there are constants Ap such that for any Δ≤SLn(ℤ) of finite index, the index of Δ¯ in SLn(ℤp) divides Appr for some r. It follows that if Δ≤Γ0, then X breaks up into some number N of ergodic Δ-sets with N|Appr. Since (ϕ,f) is a measure and action-preserving isomorphism, we also have that Y breaks up into N ergodic ϕ(Δ) sets, and hence N|Aqqs also. But it is not difficult to choose Δ small enough to ensure that N is large enough for this to be a contradiction.This argument can be easily generalized to give uncountably many incomparable actions ofSLn(ℤ). Given an infinite set S of primes with increasing enumeration S={pi}, we can construct a profinite SLn(ℤ)-set
(18)KS=lim←SLn(ℤ/p1⋯piℤ).
It is not much more difficult to show (as Ioana does) that when |SΔS′|=∞, the actions SLn(ℤ)↷KS and SLn(ℤ)↷KS′ are orbit inequivalent. In fact, this shows that there are “E0 many” orbit inequivalent profinite actions of SLn(ℤ). Of course, it is known from different arguments (exposited in [13, Theorem 17.1]) that the relation of orbit equivalence on the ergodic actions of SLn(ℤ) is very complex (for instance not Borel). But the methods used here give us more detailed information: we have an explicit family of inequivalent actions, the actions are special (they are classical and profinite), and what is more they are Borel incomparable.So far, we have considered only free actions ofSLn(ℤ). But if one just wants to use Ioana’s theorem to find orbit inequivalent actions, it is enough to consider actions which are just free almost everywhere. Here, a measure-preserving action Γ↷X is said to be free almost everywhere if the set {x∣γ≠1→γx≠x} is conull (i.e., the set where Γ acts freely is conull).Unfortunately, in the purely Borel context it is not sufficient to work with actions which are free almost everywhere, since in this case we are not allowed to just delete a null set on the right-hand side. The next result shows how to get around this difficulty. Once again, it was originally obtained by Simon Thomas using Zimmer’s superrigidity theorem.Corollary 5.
Ifn≥3 is fixed and p,q are primes with p≠q, then the actions of SLn(ℤ) on ℙ(ℚpn) and ℙ(ℚqn) are orbit inequivalent and Borel incomparable.Here,ℙ(ℚpn) denotes projective space of lines through ℚpn. Since ℙ(ℚpn) is a transitive SLn(ℤp)-space, this result is quite similar to the last one. We note also that while SLn(ℤ) does not act freely on ℙ(ℚpn), it does act freely on a conull subset [12, Lemma 6.2].Proof.
First suppose thatf:ℙ(ℚpn)→ℙ(ℚqn) is a measure-preserving and orbit-preserving map. Then we can simply restrict the domain of f to assume that it takes values in the part of ℙ(ℚqn)  where SLn(ℤ) acts freely. Afterwards, we can obtain a contradiction using essentially the same combinatorial argument as in the proof of Corollary 4.
The proof in the case of Borel reducibility requires an extra step. Namely, we cannot be sure thatf sends a conull set into the part of ℙ(ℚqn) where SLn(ℤ) acts freely. However, if it does not, then by the ergodicity of SLn(ℤ)↷ℙ(ℚpn) we can assume that f sends a conull set into the part of ℙ(ℚqn) where SLn(ℤ) acts nonfreely. Our aim will be to show that this assumption leads to a contradiction.
First, let us assume that there exists a conull subsetX⊂ℙ(ℚpn) such that, for all x∈X, there exists γ≠1 such that γf(x)=f(x). Then for all x∈X, f(x) lies inside a nontrivial eigenspace of some element of SLn(ℤ). Hence, if we let Vx denote the minimal subspace of ℚqn which is defined over ℚ¯ such that f(x)⊂Vx, then Vx is necessarily nontrivial.
Note that sinceℚ¯ is countable, there are only countably many possibilities for Vx. Hence, there exists a non-null subset X′ of X and a fixed subspace V of ℚqn such that for all x∈X′, we have Vx=V. By the ergodicity of SLn(ℤ)↷X, the set X′′=SLn(ℤ)·X′ is conull, and it follows that we can adjust f to assume that for all x∈X′′ we have Vx=V. (More precisely, replace f(x) by f′(x)=f(γx), where γ is the first element of SLn(ℤ) such that γx∈X′′.)
Now, letH≤GL(V) denote the group of projective linear transformations induced on V by SLn(ℤ){V}. It is an easy exercise, using the minimality of V, to check that H acts freely on ℙ(V) and that f is a homomorphism of orbits from SLn(ℤ)↷X′′ into H↷ℙ(V). Admitting this, we can finally apply Ioana’s theorem to suppose that there exists a finite index subgroup Γ0≤Γ and a nontrivial homomorphism ϕ:Γ0→H. As in the proof of Corollary 4, we can suppose that ϕ is an embedding. We thus get a contradiction from the next result, below.Theorem 6.
IfΓ0≤SLn(ℤ) is a subgroup of finite index and G is an algebraic ℚ¯-group with dim(G)<n2-1, then Γ0 does not embed G(ℚ¯).The idea of the proof is to apply Margulis’s superrigidity theorem. That is, one wishes to conclude that such an embedding lifts to some kind of rational mapSLn(ℝ)→G, a clear dimension contradiction. However, a little extra work is needed to handle the case of a ℚ¯-group on the right-hand side (see [10, Theorem 4.4]).


## 7. Torsion-Free Abelian Groups of Finite Rank

The torsion-free abelian groups of rank 1 were classified by Baer in 1937. The next year, Kurosh and Malcev expanded on his methods to give classifications for the torsion-free abelian groups of ranks 2 and higher. Their solution, however, was considered inadequate because the invariants they provided were no easier to distinguish than the groups themselves.In 1998, Hjorth proved, using methods from the study of Borel equivalence relations, that the classification problem for rank 2 torsion-free abelian groups isstrictly harder than that for rank 1 (see [14]). However, his work did not answer the question of whether the classification problem for rank 2 groups is as complex as for all finite ranks or whether there is more complexity that is to be found by looking at ranks 3 and higher.LetR(n) denote the space of torsion-free abelian groups of rank exactly n, that is, the set of full-rank subgroups of ℚn. Let ≅n denote the isomorphism relation on R(n). In this section we will give a concise and essentially self-contained proof of Thomas’s theorem:Theorem 7 (Thomas [15]).
Forn≥2, one has that ≅n lies properly below ≅n+1 in the Borel reducibility order.Thomas’s original argument used Zimmer’s superrigidity theorem. In this presentation, we have essentially copied his argument verbatim, with a few simplifications stemming from the use of Ioana’s theorem instead of Zimmer’s theorem.The first connection between this result and the results of the last section is that forA,B∈R(n), we have A≅B if and only if there exists g∈GLn(ℚ) such that B=g(A). Hence, the isomorphism relation ≅n is given by a natural action of the linear group GLn(ℚ). Unfortunately, even restricting to just the action of SLn(ℤ), the space R(n) is nothing like a profinite space.The Kurosh-Malcev Invariants.Although I have said that the Kurosh-Malcev invariants do not adequately classify the torsion-free abelian groups of finite rank, we will get around our difficulties by working with the Kurosh-Malcev invariants rather than with the original space R(n). The following is the key result concerning the invariants; see [16, Chapter 93] for a full account.Theorem 8 (Kurosh, Malcev).
The mapA↦Ap=ℤp⊗A is a GLn(ℚ)-preserving bijection between the (full rank) p-local subgroups of   ℚn and the (full rank) ℤp-submodules of ℚpn. The inverse map is given by Ap↦A=Ap∩ℚn.Here, a subgroup ofℚn is said to be p-local if it is infinitely q-divisible for each prime q≠p. Kurosh and Malcev proved that a subgroup A≤ℚn is determined by the sequence (Ap); this sequence is said to be the Kurosh-Malcev invariant corresponding to A. It follows of course that A is determined up to isomorphism by the orbit of (Ap) under the coordinatewise action of GLn(ℚ). (It is now easy to see why these invariants serve as a poor classification: such orbits can be quite complex.) All that we will need from this classification is the following corollary.Proposition 9.
There exists a Borel reduction fromGLn(ℚ)↷ℙ(ℚpn) to ≅n.SinceGLn(ℚ)↷ℙ(ℚpn) is closely related to a profinite action, Proposition 9 will eventually enable us to apply Ioana’s theorem in the proof of Theorem 7.Sketch of Proof. Given a linear subspace V≤ℚpn, let V⊥ denote its orthogonal complement. Then there exists a vector v such that V⊥⊕ℤpv is a full-rank submodule of ℚpn. By Theorem 8, this module corresponds to an element f(V)∈R(n). This is how the Kurosh-Malcev construction is used.To verify that it works, one uses the fact that the Kurosh-Malcev construction isGLn(ℚ)-preserving, together with the technical fact: if dimW=dimW′=n-1 and W⊕ℤpw,W′⊕ℤpw′ are full-rank modules, then W′=gW for some g∈GLn(ℚ) actually implies that W′⊕ℤpw′=g(W⊕ℤpw) for some g∈GLn(ℚ).The Problem of Freeness.Suppose now that n≥2 and that there exists a Borel reduction from ≅n+1 to ≅n. By Proposition 9, there exists a profinite, ergodic SLn+1(ℤ)-space X (namely X=ℙ(ℚpn+1)) and a countable-to-one homomorphism of orbits f from SLn+1(ℤ)↷X to ≅n. We can almost apply Ioana’s theorem, except that unfortunately ≅n is not induced by a free action of any group. The following simple observation gives us an approach for getting around this difficulty.Proposition 10.
Letf be a homomorphism of orbits from Γ↷X into Λ↷Y. Suppose that there exists a fixed K≤Λ such that for all x∈X, stabΛ(f(x))=K. Then NΛ(K)/K acts freely on f(X), and f is a homomorphism of orbits from Γ↷X into NΛ(K)/K↷f(X).Proof.
By definition, we have thatNΛ(K)/K acts on f(X) by λK·y=λy. The action is free because λy=y implies that λ∈K. To see that f is still a homomorphism of orbits, just note that if f(x′)=λf(x) then since stabf(x)=stabf(x′)=K, it follows that λ normalizes K.One can now formulate a strategy for proving Thomas’s theorem along the following lines:Claim 1.
By passing to a conull subset ofX, we can assume without loss of generality that for all x we have stabGLn(ℚ)(f(x))= some fixed K.Claim 2.
There cannot exist a nontrivial homomorphism from (a finite index subgroup of)SLn+1(ℤ) into NGLn(ℚ)(K)/K.This would yield a contradiction, since by Proposition9 and Claim 1, Ioana’s theorem would provide the nontrivial homomorphism ruled out in Claim 2. Unfortunately, this approach does not turn out to be a good one. The reason is that Claim 1 seems to be as difficult to be proved as Theorem 7 itself. Moreover, Claim 2 is not known to be true in this generality. (In fact, Claim 1 has recently been established by Thomas in [10], but his proof actually requires all of the arguments below and more).Use Quasi-Isomorphism Instead.To reduce the number of possibilities for stab(f(x))=Aut(f(x)), we change categories from isomorphism to quasi-isomorphism. We say that groups A,B≤ℚn are quasi-isomorphic, written as A~nB, if and only if B is commensurable with an isomorphic copy of A. Of course, ~n is a courser relation than ≅n, but it is easy to check that it is still a countable Borel equivalence relation (indeed, the commensurability relation is a countable relation in this case, see [15, Lemma 3.2]). Hence, the map f from above is again a countable-to-one Borel homomorphism from SLn+1(ℤ)↷X to ~n.Now, rather than attempting to fix the automorphism group off(x), we will fix the quasiendomorphism ring QEnd(A) of f(x). Here, if A≤ℚn then g∈GLn(ℚ) is said to be a quasiendomorphism of A if ϕ(A) is commensurable with a subgroup of A. (Equivalently, nϕ(A)⊂A for some n∈ℕ.) Then unlike End(A), it is clear that QEnd(A) is a ℚ-subalgebra of Mn×n(ℚ). It follows that there are just countably many possibilities for QEnd(f(x)), since an algebra is determined by any  ℚ-vector space basis for it. Hence, there exists K such that QEnd(f(x))=K for a nonnull set of x. Arguing as in the proof of Corollary 5, we may replace X by a conull subset and adjust f to assume that for all x∈X, we have QEnd(f(x))=K.Thus, we have successfully obtained our analog of Claim1 for quasi-isomorphism. Indeed, copying the arguments in the proof of Proposition 10, we see that f is a homomorphism
(19)f:SLn+1(ℤ)↷X⟶NGLn(ℚ)(K)K×↷f(X)
and that NGLn(ℚ)(K)/K× acts freely on f(X). We may therefore apply Ioana’s theorem to suppose that there exists a finite index subgroup Γ0≤PSLn+1(ℤ), a positive measure X0⊂X, and a homomorphism ϕ:Γ0→NGLn(ℚ)(K)/K× such that for x∈X0 and γ∈Γ, we have
(20)f(γx)=ϕ(γ)f(x).
Note that ϕ must be nontrivial; since if ϕ(Γ0)=1, then this says that f is Γ0-invariant. But then, by ergodicity of Γ0↷X0, f would send a conull set to one point, contradicting that f is countable-to-one.A Dimension Contradiction.The set theory is now over; we have only to establish the algebraic fact that the analog of Claim 2 holds: there does not exist a nontrivial homomorphism from Γ0 into NGLn(ℚ)(K)/K×. Again by Margulis’s theorem on normal subgroups, we can suppose that ϕ is an embedding. Then using Margulis’s superrigidity theorem, it suffices to show that NGLn(ℚ)(K)/K× is contained in an algebraic group of dimension strictly smaller than dim(PSLn+1)=(n+1)2-1.To see this, first note that since the subalgebraK of Mn×n(ℚ) is definable from a vector space basis, we have that K=K(ℚ), where K is an algebraic ℚ-group inside Mn×n. Basic facts from algebraic group theory imply that NGLn(ℚ)(K)=N(ℚ) and K×=K′(ℚ), where again N,K′ are algebraic ℚ-groups inside Mn×n. Finally, NGLn(ℚ)(K)/K× is exactly N(ℚ)/K′(ℚ), which is contained in the algebraic ℚ-group N/K′. Since the dimension of an algebraic group decreases when passing to subgroups and quotients, we have
(21)dim(NK′)≤dim(Mn×n)=n2<(n+1)2-1,
as desired. This completes the proof.



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