Hindawi Publishing CorporationISRN AlgebraVolume 2013 Article ID 387540 8 pageshttpdxdoiorg1011552013387540
Research ArticleIoanarsquos Superrigidity Theorem and Orbit Equivalence Relations
Samuel Coskey
Department of Mathematics Boise State University 1910 University DR Boise ID 83725 USA
Correspondence should be addressed to Samuel Coskey scoskeynylogicorg
Received 9 October 2013 Accepted 10 November 2013
Academic Editors M Przybylska and A Rapinchuk
Copyright copy 2013 Samuel Coskey This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
We give a survey of Adrian Ioanarsquos cocycle superrigidity theorem for profinite actions of Property (T) groups and its applications toergodic theory and set theory in this expository paper In addition to a statement and proof of Ioanarsquos theorem this paper featuresthe following (i) an introduction to rigidity including a crash course in Borel cocycles and a summary of some of the best-knownsuperrigidity theorems (ii) some easy applications of superrigidity both to ergodic theory (orbit equivalence) and set theory (Borelreducibility) and (iii) a streamlined proof of SimonThomasrsquos theorem that the classification of torsion-free abelian groups of finiterank is intractable
1 Introduction
In the past fifteen years superrigidity theory has had a boomin the number and variety of new applications Moreoverthis has been coupled with a significant advancement intechniques and results In this paper we survey one suchnew result namely Ioanarsquos theorem on profinite actions ofProperty (T) groups and some of its applications in ergodictheory and in set theory In the concluding section wehighlight an application to the classification problem fortorsion-free abelian groups of finite rank The narrative isstrictly expository with most of the material being adaptedfrom the work of Adrian Ioana mine and SimonThomas
Although Ioanarsquos theorem is relatively recent it will beof interest to readers who are new to rigidity because theproof is natural and there are many immediate applicationsTherefore we were keen to keep the nonexpert in mindWe do assume that the reader is familiar with the notion ofergodicity of a measure-preserving action and with unitaryrepresentations of countable groupsWewill not go into greatdetail on Property (T) since for our purposes it is enough toknow that SL
119899(Z) satisfies Property (T) when 119899 gt 2 Rather
we will introduce it just when it is needed and hopefully itskey appearance in the proof of Ioanarsquos theorem will providesome insight into its meaning
The concept of superrigidity was introduced by Mostowand Margulis in the context of studying the structure of
lattices in Lie groups Here Γ is said to be a lattice inthe (real) Lie group 119866 if it is discrete and 119866Γ admitsan invariant probability measure Very roughly speakingMargulis showed that if Γ is a lattice in a simple (higher-rank)real Lie group 119866 then any homomorphism from Γ into analgebraic group119867 lifts to an algebraic map from 119866 to119867 Thisimplies Mostowrsquos theorem which states that any isomorphiclattices Γ Λ in a simple (higher-rank) Lie group 119866 must beconjugate inside 119866
We will leave this first form of rigidity on the backburner and primarily consider instead a second form initiallyconsidered by Zimmer which is concerned with groupactions (The connection between the two forms of rigidity isthat both can be cast in terms of measurable cocycles whichwill be introduced in the next section For the connectionbetween cocycles and lifting homomorphisms see [1 Exam-ple 4212]) The basic notions are as follows Two probabilitymeasure-preserving actions Γ 119883 andΛ 119884 are said to beorbit equivalent if there exists a measure-preserving almostbijection 119891 119883 rarr 119884 such that Γ119909 = Γ119909
1015840 if Λ119891(119909) = Λ119891(1199091015840)They are said to be isomorphic if additionally there exists anisomorphism 120601 Γ rarr Λ such that 119891(120574119909) = 120601(120574)119891(119909)Essentially Zimmer showed that any (irreducible) ergodicaction Γ 119883 of a lattice in a (higher rank) simple Lie groupis superrigid in the sense that it cannot be orbit equivalent toanother action of an algebraic group Λ 119884 without beingisomorphic to it (For elementary reasons it is necessary to
2 ISRN Algebra
assume that Λ acts freely on 119884) See [1 Theorem 521] fora weak statement of this result and [2 Section 1] for furtherdiscussion
It is natural to ask whether there exists an analog of Zim-merrsquos theorem in the context of general measure-preservingactions that is with the algebraic hypothesis on Λ removedMany rigidity results have been established along these lines(for instance see [2ndash4]) One of the landmark results in thisdirection was obtained recently by Popa [5] who found alarge class of measure-preserving actions Γ 119883 which aresuperrigid in the general sense that Γ 119883 cannot be orbitequivalent to another (free) action without being isomorphicto it In particular his theorem states that if Γ is a Property(T) group then the free part of its left-shift action on 119883 = 2
Γ
(the so-called Bernoulli action) is an example of a superrigidaction Following on Poparsquos work Ioanarsquos theorem gives asecond class of examples of superrigid actions namely theprofinite actions of Property (T) groups
This paper is organized as follows The second sectiongives some background on Borel cocycles a key tool inrigidity theory A slightly weakened version of Ioanarsquos the-orem is stated in the third section The proof itself is splitbetween Section 4 which contains a general purpose lemmaand Section 5 which contains the heart of the argumentAlthough these are largely unchanged from Ioanarsquos ownaccount I have inserted many additional remarks to smooththe experience for the newcomer
In Section 6 we give a couple of the easier applicationsof the main theorem First we show how to obtain manyorbit inequivalent profinite actions of SL
119899(Z)We also explore
applications to logic and set theory by considering Borelreducibility In particular we point out some of the extrachallenges one faces whenworking in the purely set-theoretic(ie Borel) context as opposed to the more familiar measurecontext
Finally in the last section we use Ioanarsquos theorem to givea self-contained and slightly streamlined proof of Thomasrsquostheorem that the complexity of the isomorphism problem fortorsion-free abelian groups of finite rank increases strictlywith the rank
2 Rigidity via Cocycles
We begin by introducing a slightly more expansive notionof orbit equivalence rigidity If Γ 119883 and Λ 119884 arearbitrary Borel actions of countable groups then a function119891 119883 rarr 119884 is said to be a homomorphism of orbits if Γ119909 = Γ119909
1015840
implies Λ119891(119909) = Λ119891(1199091015840) It is said to be a homomorphismof actions if additionally there exists a homomorphism 120601
Γ rarr Λ such that 119891(120574119909) = 120601(120574)119891(119909) (Note that theseterms are not exactly standard) Informally we will say thatΓ 119883 is superrigid if whenever Λ 119884 is a free action and119891 119883 rarr 119884 is a homomorphism of orbits then 119891 in factarises from a homomorphism of actions (ie 119891 is equivalentto a homomorphism of actions in a sense defined below)
Following Margulis and Zimmer we will require thelanguage of Borel cocycles to describe and prove superrigidityresults A cocycle is an object which is associated with a given
x 120574x
120574998400120574x
f(x)120572(120574 x)
f(120574x)
120572(120574 998400120574 x)
f(120574998400120574x)
120572(120574
998400120574x)
Γ X Λ Y
Figure 1 The cocycle condition 120572(1205741015840120574119909) = 120572(1205741015840 120574119909)120572(120574 119909)
homomorphism of orbits 119891 119883 rarr 119884 as follows Observethat for every (120574 119909) isin Γ times 119883 there exists a 120582 isin Λ suchthat 119891(120574119909) = 120582119891(119909) Moreover Λ acts freely on 119884 if andonly if this 120582 is always uniquely determined by the data 119891 120574and 119909 In other words in this case 119891 determines a function120572 Γ times 119883 rarr Λ which satisfies
Thismap is called the cocycle corresponding to119891 and it is easyto see that it is Borel whenever 119891 is Moreover the cocycle120572 satisfies the composition law 120572(120574
1015840120574 119909) = 120572(1205741015840 120574119909)120572(120574 119909)this is called the cocycle condition See Figure 1 for a visualdepiction of the cocycle condition
When 119891 is actually action-preserving that is 119891(120574119909) =
120601(120574)119891(119909) for some homomorphism 120601 Γ rarr Λ thenwe have 120572(120574 119909) = 120601(120574) so that 120572 is independent of thesecond coordinate Conversely if 120572 is independent of thesecond coordinate then one can define 120601(120574) = 120572(120574 sdot) and thecomposition law implies that 120601 is a homomorphism In thissituation the cocycle is said to be trivial
In practice when establishing rigidity one typicallyshows that an arbitrary cocycle (arising from a homomor-phism of orbits) is equivalent to a trivial cocycle (whichtherefore arises from a homomorphism of actions) Here wesay that homomorphisms of orbits 119891 1198911015840 119883 rarr 119884 are calledequivalent if there exists a Borel function 119887 119883 rarr Λ suchthat 1198911015840(119909) = 119887(119909)119891(119909) ae (ie they lift the same functionon the quotient spaces 119883Γ rarr 119884Λ) In this case thecorresponding cocycles 120572 1205721015840 are said to be cohomologous It iseasy to check that 119891 1198911015840 are equivalent via 119887 if and only if thecorresponding cocycles 120572 1205721015840 satisfy the relation 1205721015840(120574 119909) =
119887(120574119909)120572(120574 119909)119887(119909)minus1 ae this is called the cohomology relation
The easiest way to see that this is the case is to glance atFigure 2
We close this section by remarking that not all cocyclesarise from orbit-preserving maps An abstract cocycle is anyBorel function satisfying the cocycle condition ae and twococycles are said to be cohomologous if there exists a Borelfunction 119887 satisfying the cohomology relation ae The mostpowerful superrigidity results often have the conclusion thatldquoevery cocycle is cohomologous to a trivial cocyclerdquo Howeverfor most applications there is no need for the extra strengthgained by using the abstract cocycle formulation
ISRN Algebra 3
x 120574x
f(x)
b(x) b(120574x)
120572(120574 x)f(120574x)
f998400(x) f
998400(120574x)
120572998400(120574 x)
Γ X Λ Y
Figure 2 The cohomology relation for cocycles 1205721015840(120574 119909) =
Cocycle superrigidity results were first established by Mar-gulis and Zimmer for cocycles Γ 119883 rarr Λ where Γ
is a lattice in a higher rank Lie group acting ergodicallyon 119883 These results carried the additional hypothesis thatΛ is contained in an algebraic group The first example ofthe most general form of cocycle superrigidity with thetarget Λ arbitrary was Poparsquos result concerning Bernoulliactions In this section wewill discuss Ioanarsquos theoremwhichestablishes similar conclusions for profinite actions
Here Γ 119883 is said to be profinite if as a Γ-set X is theinverse limit of a family of finite Γ-sets119883
119909 isin 119883 can be identified with the thread (120587119899(119909)) We are
interested in the ergodic case here each119883119899is equipped with
the uniform probability measure and Γ 119883119899is transitive
Theorem 1 (Ioana) Let Γ (119883 120583) be an ergodic measure-preserving profinite action with invariant factor maps 120587
119899
119883 rarr 119883119899 Assume that Γ has Property (T)Then for any cocycle
120572 Γ 119883 rarr Λ there exists 119899 and 119886 isin 119883119899such that the
restriction of 120572 to the action Γ119886 120587minus1119899
(119886) is cohomologous to atrivial cocycle
In other words the conclusion is that Γ 119883 is ldquovirtuallysuperrigidrdquo in the sense that any orbit preserving map afterit is restricted to a finite index component of the left-handside comes from an action preserving map Ioanarsquos theoremis interesting when contrasted with Poparsquos theorem whileBernoulli actions are strongly mixing profinite actions arehighly nonmixing Indeed for each 119899 Γ just permutes theblocks 120587
minus1
119899(119886) for 119886 isin 119883
119899 and it follows that ⋃
119886isin119883119899
119883119886times 119883119886
is a Γ-invariant subset of 119883 times 119883We remark that although our variant of Ioanarsquos theorem
is sufficient for most applications it is weaker than the stateof the art in several ways First Ioana requires only thatΓ have the relative Property (T) over some infinite normalsubgroups 119873 such that Γ119873 is finitely generated SecondIoana also shows that 120572 is equivalent to a cocycle definedon all of 119883 Last Furman has generalized the statement byreplacing profinite actions with the more general class ofcompact actions (Γ 119883 is said to be compact if whenregarded as a subset of (119883 120583) it is precompact in a suitabletopology)
4 Cocycle Untwisting
We begin with the following preliminary result whichroughly speaking says that if 120572 Γ 119883 rarr Λ is acocycle and if for each 120574 it has a ldquovery likelyrdquo value then 120572
is cohomologous to the map which always takes on this likelyvalue In particular in this case 120572 is cohomologous to a trivialcocycle
Theorem 2 Let Γ (119883 120583) be ergodic and measure-preserv-ing and let 120572 Γ 119883 rarr Λ be a cocycle Suppose that for all120574 isin Γ there exists 120582
Then the map 120601(120574) = 120582120574is a homomorphism and 120572 is
cohomologous to it
It is easy to see that 120601must be a homomorphism indeedthe hypothesis guarantees that there is a nonnull set of 119909 forwhich 120601(120574
1015840120574) = 120572(1205741015840120574 119909) = 120572(1205741015840 120574119909)120572(120574 119909) = 120601(1205741015840)120601(120574)Hence it remains only to establish the following result
Lemma 3 Let Γ (119883 120583) be ergodic and measure-preservingand let 120572 120573 Γ 119883 rarr Λ be cocycles Suppose that for all120574 isin Γ
We understand this result to say that if 120572 and 120573 are closein an 119871infin sense then they are cohomologous It follows uponsimilar results of Popa and Furman which draws the sameconclusion in the case that 120572 and 120573 are close in an appropriate1198711 sense (for instance see [6 Theorem 42]) Ioanarsquos proofgiven below may be safely skipped until reading the nextsection
Proof of Lemma 3 Let Γ 119883 times Λ be the action given by
(this is an action thanks to the cocycle condition) andconsider the corresponding left-regular representation Thereason for using this representation is that120572 is close to120573 if andonly if a particular vector is close to being invariant Namelylet
120585 = 120594119883times119890
(5)
(read the characteristic function of 119883 times 119890) and notice that
Using this together with the law of cosines the hypothesisnow translates to say that 120574120585minus120585 le 119862 lt 12 for all 120574 isin Γ It isnot difficult to see that this implies that there is an invariantvector 120578 such that 120578 minus 120585 lt 12 (Indeed letting 119878 denote
4 ISRN Algebra
the convex hull of Γ sdot 120585 it is easily seen that there exists aunique vector 120578 isin 119878 of minimal norm this 120578 is necessarilyinvariant)
The idea for the conclusion of the proof is as follows Ifwe had 120578 = 120594graph(119887) for some function 119887 119883 rarr Λ then wewould be done Indeed in this case the invariance of 120578wouldmean that 119887(119909) = 120582 if and only if 119887(120574119909) = 120572(120574 119909)120582120573(120574 119909)
minus1 In other words 119887 wouldwitness the fact that 120572 is cohomologous to 120573 The fact that120578 minus 120585 lt 12 implies that this is close to being the case
We actually define 119887(119909) = the 120582 such that |120578(119909 120582)| gt 12if it exists and is unique The above computation shows thatwhen 119887(119909) 119887(120574119909) are both defined the cohomology relationholds Moreover the set where 119887 is defined is invariant so bythe ergodicity of Γ 119883 it suffices to show that this set isnonnull In fact since 120578 and 120585 are close 119887 must take value 119890
on a nonnull set
120583 119909 1003816100381610038161003816120585 (119909 119890) minus 120578 (119909 119890)
1003816100381610038161003816 ge1
2
le 4int119909|120585(119909119890)minus120578(119909119890)|ge12
1003816100381610038161003816120585 minus 12057810038161003816100381610038162
le 41003817100381710038171003817120585 minus 120578
10038171003817100381710038172
lt 1
(7)
This shows that 119909 |120578(119909 119890)| gt 12 is non-null as desired Asimilar computation is used to show that with probability 1119890 is the unique such element of 120582
5 Ioanarsquos Proof
WhatWeWant We wish to find some 119899 and 119886 isin 119883119899such that
where 120583119886denotes the normalized restriction of 120583 to 120587minus1
119899(119886)
This would imply by a straightforward computation that foreach 120574 isin Γ
119886there exists a 120582 isin Λ such that
120583119886119909 | 120572 (120574 119909) = 120582 ge 119862 gt
7
8(9)
and this would complete the proof thanks to Theorem 2
WhatWeHave Unfortunately it is only immediately possibleto obtain that the quantities in (8) tend to 1 on average at arate depending on 120574 That is for each 120574 isin Γ we have
| 119886 isin 119883119899 119899 isin 120596 is dense in
1198712 and while the right-hand side is the norm-squared of 120594119878
the left-hand side is the norm-squared of 120594119878projected onto
the span of 120594120587minus1
119898(119886)
| 119886 isin 119883119898 119898 le 119899 Finally just apply (12)
to each set 119878 = 119909 | 120572(120574 119909) = 120582 and use the dominatedconvergence theorem to pass the limit through the sum overall 120582 isin Λ
Proof Thegap betweenwhat he has (the asymptotic informa-tion) and what we want (the uniform information) is bridgedby Property (T) Once again the first step is to consider anappropriate representation this time onewhich compares thevalues of 120572(120574 119909) as 119909 varies That is let Γ 119883 times 119883 times Λ by
and consider the left-regular unitary representation corre-sponding to this action The idea very roughly is that thedegree to which 120572(120574 119909) is independent of 119909 will be measuredby how close a particular vector is to being Γ-invariant
More precisely for each 119899 define an orthonormal familyof vectors 120585
So now ldquowhat we haverdquo and ldquowhat we wantrdquo can be translatedas follows we have the 120585
119899form a family of almost invariant
vectors and we want a single 119899 and 119886 isin 119883119899such that 120585
119886is
nearly invariant uniformly for all 120574 isin Γ119886
The remainder of the argument is straightforward Sincethe 120585119899forms a family of almost invariant vectors Property
(T) implies that there exist 119899 and an invariant vector 120578 suchthat 120578 minus 120585
119899 le 120575 Let 1205781015840 be the restriction of 120578 to the set
cup119886isin119883119899
(120587minus1119899
(119886)times120587minus1119899
(119886)timesΛ) Since this set is invariant we havethe fact that that 1205781015840 is invariant as well Since 120585
119899is supported
on this set we retain the property that 1205781015840 minus 120585119899 le 120575
Now we simply express 1205781015840 as a normalized average oforthogonal Γ
119886-invariant vectors More specifically write
1205781015840
=1
radic1003816100381610038161003816119883119899
1003816100381610038161003816
sum119886isin119883119886
120578119886 (17)
ISRN Algebra 5
where 120578119886is the appropriately rescaled restriction of 1205781015840 to the
set 120587minus1119899
(119886) times 120587minus1119899
(119886) timesΛ Then by the law of averages we musthave some 119886 isin 119883
119899such that 120578
119886minus 120585119886 le 120575 Moreover 120578
119886is Γ119886-
invariant so that for all 120574 isin Γ119886wehave ⟨120574120578
119886 120578119886⟩ = 1 It follows
that by an appropriate choice of 120575 we can make ⟨120574120585119886 120585119886⟩ ge
119862 gt 78 for all 120574 isin Γ119886
6 Easy Applications
In this section we use Ioanarsquos theorem for one of its intendedpurposes to find many highly inequivalent actions Theresults mentioned here are just meant to give the flavor ofapplications of superrigidity they by no means demonstratethe full power of the theorem In the next section we willdiscuss the slightly more interesting and difficult applicationto torsion-free abelian groups For further applications seefor instance [7ndash10]
In searching for inequivalent actions onemight of courseconsider a variety of inequivalence notions Here we focuson just two of them orbit inequivalence and Borel incom-parability Recall from the introduction that Γ 119883 andΛ 119884 are said to be orbit equivalent if there exists ameasure-preserving andorbit-preserving almost bijection from119883 to119884Notice that this notion depends only on the orbit equivalencerelation arising from the two actions and not on the actionsthemselves When this is the case we will often conflate thetwo saying alternately that certain actions are orbit equivalentor that certain equivalence relations are ldquoorbit equivalentrdquo
Borel bireducibility is a purely set-theoretic notion withits origins in logic The connection is that if 119864 is an equiv-alence relation on a standard Borel space 119883 then we canthink of119864 representing a classification problem For instanceif 119883 happens to be a set of codes for a family of structuresthen studying the classification of those structures amountsto studying the isomorphism equivalence relation 119864 on 119883We refer the reader to [11] for a complete introduction to thesubject
If 119864 and 119865 are equivalent relations on 119883 and 119884 then119864 is said to be Borel reducible to 119865 if there exists a Borelfunction 119891 119883 rarr 119884 satisfying 119909119864119909
1015840 if and only if119891(119909)119865119891(1199091015840) We think of this saying that the classificationproblem for elements of 119883 up to 119864 is no more complex thanthe classification problem for elements of 119884 up to 119865 Thus if119864 and 119865 are Borel bireducible (ie there is a reduction bothways) then they represent classification problems of the samecomplexity
It is elementary to see that neither orbit equivalence orBorel bireducibility implies the other For instance given anyΓ-space 119883 one can form a disjoint union 119883 ⊔ 119883
1015840 where 1198831015840
is a Γ-space of very high complexity which is declared to beof measure 0 Conversely if 119883 is an ergodic and hyperfiniteΓ-space then it is known that it is bireducible with119883⊔119883 butthe two cannot be orbit equivalent It is even possible withoutmuch more difficulty to find two ergodic actions which arebireducible but not orbit equivalent
We are now ready to begin with the following direct con-sequence of Ioanarsquos theorem It was first established by SimonThomas in connection working on classification problem for
torsion-free abelian groups of finite rank His proof usedZimmerrsquos superrigidity theorem and some additional cocyclemanipulation techniques with Ioanarsquos theorem in hand theproof will be much simpler
Corollary 4 If 119899 ge 3 is fixed and 119901 119902 are primes such that119901 = 119902 then the actions of SL
119899(Z) on SL
119899(Z119901) and SL
119899(Z119902) are
orbit inequivalent and Borel incomparable
Here Z119901denotes the ring of 119901-adic integers It is easy to
see that SL119899(Z) SL
119899(Z119901) is a profinite action being the
inverse limit of the actions SL119899(Z) SL
119899(Z119901119894Z) together
with their natural system of projections
Proof Let 119901 = 119902 and suppose that 119891 is either an orbit equiv-alence or a Borel reduction from SL
119899(Z) SL
119899(Z119901) to
SL119899(Z) SL
119899(Z119902) We now apply Ioanarsquos theorem together
with the understanding of cocycles gained in the previoussection The conclusion is that we can suppose without lossof generality that there exists a finite index subgroup Γ
0le
SL119899(Z) a Γ
0-coset 119883 sub SL
119899(Z119901) and a homomorphism
120601 Γ0
rarr SL119899(Z) which makes 119891 into an action-preserving
map from Γ0 119883 into SL
119899(Z) SL
119899(Z119902)
Now in the measure-preserving case it is not difficult toconclude that 119891 is a ldquovirtual isomorphismrdquo between the twoactions We claim that this can be achieved even in the casethat 119891 is just a Borel reduction First we can assume that 120601is an embedding Indeed by Margulisrsquos theorem on normalsubgroups [1 Theorem 812] either im(120601) or ker(120601) is finiteIf ker(120601) is finite then we can replace Γ
0by a finite index
subgroup (and 119883 by a coset of the new Γ0) to suppose that
120601 is injective On the other hand if im(120601) is finite then wecan replace Γ
0by a finite index subgroup to suppose that 120601 is
trivial But this would mean that 119891 is Γ0-invariant and so by
ergodicity of Γ0
119883 119891 would send a conull set to a singlepoint contradicting that 119891 is countable-to-one
Second 120601(Γ0) must be a finite index subgroup of SL
119899(Z)
Indeed byMargulisrsquos superrigidity theorem 120601 can be lifted toan isomorphism of SL
119899(R) and it follows that 120601(Γ
0) is a lattice
of SL119899(R) But then it is easy to see that any lattice which is
contained in SL119899(Z) must be commensurable with SL
119899(Z)
Third by the ergodicity of Γ0
119883 we can assumethat im(119891) is contained in a single 120601(Γ
0) coset 119884
0 And
now because 120601(Γ0) preserves a unique measure on 119884
0(the
Haar measure) and because 120601(Γ0) preserves 119891
lowast(Haar) we
actually conclude that 119891 is measure-preserving In summarywe have shown that (120601 119891) is a measure and action-preservingisomorphism between Γ
0 119883
0and 120601(Γ
0) 119884
0 which
establishes the claimFinally a short computation confirms the intuitive alge-
braic fact that the existence of such a map is ruled out by themismatch in primes between the left-and right-hand sidesWe give just a quick sketch for a few more details see [12Section 6] Now it is well-known that there are constants 119860
119901
such that for any Δ le SL119899(Z) of finite index the index of Δ in
SL119899(Z119901) divides 119860
119901119901119903 for some 119903 It follows that if Δ le Γ
0
then 119883 breaks up into some number 119873 of ergodic Δ-setswith119873|119860
119901119901119903 Since (120601 119891) is ameasure and action-preserving
6 ISRN Algebra
isomorphism we also have that 119884 breaks up into 119873 ergodic120601(Δ) sets and hence 119873|119860
119902119902119904 also But it is not difficult to
choose Δ small enough to ensure that 119873 is large enough forthis to be a contradiction
This argument can be easily generalized to give uncount-ably many incomparable actions of SL
119899(Z) Given an infinite
set 119878 of primes with increasing enumeration 119878 = 119901119894 we can
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
assume that Λ acts freely on 119884) See [1 Theorem 521] fora weak statement of this result and [2 Section 1] for furtherdiscussion
It is natural to ask whether there exists an analog of Zim-merrsquos theorem in the context of general measure-preservingactions that is with the algebraic hypothesis on Λ removedMany rigidity results have been established along these lines(for instance see [2ndash4]) One of the landmark results in thisdirection was obtained recently by Popa [5] who found alarge class of measure-preserving actions Γ 119883 which aresuperrigid in the general sense that Γ 119883 cannot be orbitequivalent to another (free) action without being isomorphicto it In particular his theorem states that if Γ is a Property(T) group then the free part of its left-shift action on 119883 = 2
Γ
(the so-called Bernoulli action) is an example of a superrigidaction Following on Poparsquos work Ioanarsquos theorem gives asecond class of examples of superrigid actions namely theprofinite actions of Property (T) groups
This paper is organized as follows The second sectiongives some background on Borel cocycles a key tool inrigidity theory A slightly weakened version of Ioanarsquos the-orem is stated in the third section The proof itself is splitbetween Section 4 which contains a general purpose lemmaand Section 5 which contains the heart of the argumentAlthough these are largely unchanged from Ioanarsquos ownaccount I have inserted many additional remarks to smooththe experience for the newcomer
In Section 6 we give a couple of the easier applicationsof the main theorem First we show how to obtain manyorbit inequivalent profinite actions of SL
119899(Z)We also explore
applications to logic and set theory by considering Borelreducibility In particular we point out some of the extrachallenges one faces whenworking in the purely set-theoretic(ie Borel) context as opposed to the more familiar measurecontext
Finally in the last section we use Ioanarsquos theorem to givea self-contained and slightly streamlined proof of Thomasrsquostheorem that the complexity of the isomorphism problem fortorsion-free abelian groups of finite rank increases strictlywith the rank
2 Rigidity via Cocycles
We begin by introducing a slightly more expansive notionof orbit equivalence rigidity If Γ 119883 and Λ 119884 arearbitrary Borel actions of countable groups then a function119891 119883 rarr 119884 is said to be a homomorphism of orbits if Γ119909 = Γ119909
1015840
implies Λ119891(119909) = Λ119891(1199091015840) It is said to be a homomorphismof actions if additionally there exists a homomorphism 120601
Γ rarr Λ such that 119891(120574119909) = 120601(120574)119891(119909) (Note that theseterms are not exactly standard) Informally we will say thatΓ 119883 is superrigid if whenever Λ 119884 is a free action and119891 119883 rarr 119884 is a homomorphism of orbits then 119891 in factarises from a homomorphism of actions (ie 119891 is equivalentto a homomorphism of actions in a sense defined below)
Following Margulis and Zimmer we will require thelanguage of Borel cocycles to describe and prove superrigidityresults A cocycle is an object which is associated with a given
x 120574x
120574998400120574x
f(x)120572(120574 x)
f(120574x)
120572(120574 998400120574 x)
f(120574998400120574x)
120572(120574
998400120574x)
Γ X Λ Y
Figure 1 The cocycle condition 120572(1205741015840120574119909) = 120572(1205741015840 120574119909)120572(120574 119909)
homomorphism of orbits 119891 119883 rarr 119884 as follows Observethat for every (120574 119909) isin Γ times 119883 there exists a 120582 isin Λ suchthat 119891(120574119909) = 120582119891(119909) Moreover Λ acts freely on 119884 if andonly if this 120582 is always uniquely determined by the data 119891 120574and 119909 In other words in this case 119891 determines a function120572 Γ times 119883 rarr Λ which satisfies
Thismap is called the cocycle corresponding to119891 and it is easyto see that it is Borel whenever 119891 is Moreover the cocycle120572 satisfies the composition law 120572(120574
1015840120574 119909) = 120572(1205741015840 120574119909)120572(120574 119909)this is called the cocycle condition See Figure 1 for a visualdepiction of the cocycle condition
When 119891 is actually action-preserving that is 119891(120574119909) =
120601(120574)119891(119909) for some homomorphism 120601 Γ rarr Λ thenwe have 120572(120574 119909) = 120601(120574) so that 120572 is independent of thesecond coordinate Conversely if 120572 is independent of thesecond coordinate then one can define 120601(120574) = 120572(120574 sdot) and thecomposition law implies that 120601 is a homomorphism In thissituation the cocycle is said to be trivial
In practice when establishing rigidity one typicallyshows that an arbitrary cocycle (arising from a homomor-phism of orbits) is equivalent to a trivial cocycle (whichtherefore arises from a homomorphism of actions) Here wesay that homomorphisms of orbits 119891 1198911015840 119883 rarr 119884 are calledequivalent if there exists a Borel function 119887 119883 rarr Λ suchthat 1198911015840(119909) = 119887(119909)119891(119909) ae (ie they lift the same functionon the quotient spaces 119883Γ rarr 119884Λ) In this case thecorresponding cocycles 120572 1205721015840 are said to be cohomologous It iseasy to check that 119891 1198911015840 are equivalent via 119887 if and only if thecorresponding cocycles 120572 1205721015840 satisfy the relation 1205721015840(120574 119909) =
119887(120574119909)120572(120574 119909)119887(119909)minus1 ae this is called the cohomology relation
The easiest way to see that this is the case is to glance atFigure 2
We close this section by remarking that not all cocyclesarise from orbit-preserving maps An abstract cocycle is anyBorel function satisfying the cocycle condition ae and twococycles are said to be cohomologous if there exists a Borelfunction 119887 satisfying the cohomology relation ae The mostpowerful superrigidity results often have the conclusion thatldquoevery cocycle is cohomologous to a trivial cocyclerdquo Howeverfor most applications there is no need for the extra strengthgained by using the abstract cocycle formulation
ISRN Algebra 3
x 120574x
f(x)
b(x) b(120574x)
120572(120574 x)f(120574x)
f998400(x) f
998400(120574x)
120572998400(120574 x)
Γ X Λ Y
Figure 2 The cohomology relation for cocycles 1205721015840(120574 119909) =
Cocycle superrigidity results were first established by Mar-gulis and Zimmer for cocycles Γ 119883 rarr Λ where Γ
is a lattice in a higher rank Lie group acting ergodicallyon 119883 These results carried the additional hypothesis thatΛ is contained in an algebraic group The first example ofthe most general form of cocycle superrigidity with thetarget Λ arbitrary was Poparsquos result concerning Bernoulliactions In this section wewill discuss Ioanarsquos theoremwhichestablishes similar conclusions for profinite actions
Here Γ 119883 is said to be profinite if as a Γ-set X is theinverse limit of a family of finite Γ-sets119883
119909 isin 119883 can be identified with the thread (120587119899(119909)) We are
interested in the ergodic case here each119883119899is equipped with
the uniform probability measure and Γ 119883119899is transitive
Theorem 1 (Ioana) Let Γ (119883 120583) be an ergodic measure-preserving profinite action with invariant factor maps 120587
119899
119883 rarr 119883119899 Assume that Γ has Property (T)Then for any cocycle
120572 Γ 119883 rarr Λ there exists 119899 and 119886 isin 119883119899such that the
restriction of 120572 to the action Γ119886 120587minus1119899
(119886) is cohomologous to atrivial cocycle
In other words the conclusion is that Γ 119883 is ldquovirtuallysuperrigidrdquo in the sense that any orbit preserving map afterit is restricted to a finite index component of the left-handside comes from an action preserving map Ioanarsquos theoremis interesting when contrasted with Poparsquos theorem whileBernoulli actions are strongly mixing profinite actions arehighly nonmixing Indeed for each 119899 Γ just permutes theblocks 120587
minus1
119899(119886) for 119886 isin 119883
119899 and it follows that ⋃
119886isin119883119899
119883119886times 119883119886
is a Γ-invariant subset of 119883 times 119883We remark that although our variant of Ioanarsquos theorem
is sufficient for most applications it is weaker than the stateof the art in several ways First Ioana requires only thatΓ have the relative Property (T) over some infinite normalsubgroups 119873 such that Γ119873 is finitely generated SecondIoana also shows that 120572 is equivalent to a cocycle definedon all of 119883 Last Furman has generalized the statement byreplacing profinite actions with the more general class ofcompact actions (Γ 119883 is said to be compact if whenregarded as a subset of (119883 120583) it is precompact in a suitabletopology)
4 Cocycle Untwisting
We begin with the following preliminary result whichroughly speaking says that if 120572 Γ 119883 rarr Λ is acocycle and if for each 120574 it has a ldquovery likelyrdquo value then 120572
is cohomologous to the map which always takes on this likelyvalue In particular in this case 120572 is cohomologous to a trivialcocycle
Theorem 2 Let Γ (119883 120583) be ergodic and measure-preserv-ing and let 120572 Γ 119883 rarr Λ be a cocycle Suppose that for all120574 isin Γ there exists 120582
Then the map 120601(120574) = 120582120574is a homomorphism and 120572 is
cohomologous to it
It is easy to see that 120601must be a homomorphism indeedthe hypothesis guarantees that there is a nonnull set of 119909 forwhich 120601(120574
1015840120574) = 120572(1205741015840120574 119909) = 120572(1205741015840 120574119909)120572(120574 119909) = 120601(1205741015840)120601(120574)Hence it remains only to establish the following result
Lemma 3 Let Γ (119883 120583) be ergodic and measure-preservingand let 120572 120573 Γ 119883 rarr Λ be cocycles Suppose that for all120574 isin Γ
We understand this result to say that if 120572 and 120573 are closein an 119871infin sense then they are cohomologous It follows uponsimilar results of Popa and Furman which draws the sameconclusion in the case that 120572 and 120573 are close in an appropriate1198711 sense (for instance see [6 Theorem 42]) Ioanarsquos proofgiven below may be safely skipped until reading the nextsection
Proof of Lemma 3 Let Γ 119883 times Λ be the action given by
(this is an action thanks to the cocycle condition) andconsider the corresponding left-regular representation Thereason for using this representation is that120572 is close to120573 if andonly if a particular vector is close to being invariant Namelylet
120585 = 120594119883times119890
(5)
(read the characteristic function of 119883 times 119890) and notice that
Using this together with the law of cosines the hypothesisnow translates to say that 120574120585minus120585 le 119862 lt 12 for all 120574 isin Γ It isnot difficult to see that this implies that there is an invariantvector 120578 such that 120578 minus 120585 lt 12 (Indeed letting 119878 denote
4 ISRN Algebra
the convex hull of Γ sdot 120585 it is easily seen that there exists aunique vector 120578 isin 119878 of minimal norm this 120578 is necessarilyinvariant)
The idea for the conclusion of the proof is as follows Ifwe had 120578 = 120594graph(119887) for some function 119887 119883 rarr Λ then wewould be done Indeed in this case the invariance of 120578wouldmean that 119887(119909) = 120582 if and only if 119887(120574119909) = 120572(120574 119909)120582120573(120574 119909)
minus1 In other words 119887 wouldwitness the fact that 120572 is cohomologous to 120573 The fact that120578 minus 120585 lt 12 implies that this is close to being the case
We actually define 119887(119909) = the 120582 such that |120578(119909 120582)| gt 12if it exists and is unique The above computation shows thatwhen 119887(119909) 119887(120574119909) are both defined the cohomology relationholds Moreover the set where 119887 is defined is invariant so bythe ergodicity of Γ 119883 it suffices to show that this set isnonnull In fact since 120578 and 120585 are close 119887 must take value 119890
on a nonnull set
120583 119909 1003816100381610038161003816120585 (119909 119890) minus 120578 (119909 119890)
1003816100381610038161003816 ge1
2
le 4int119909|120585(119909119890)minus120578(119909119890)|ge12
1003816100381610038161003816120585 minus 12057810038161003816100381610038162
le 41003817100381710038171003817120585 minus 120578
10038171003817100381710038172
lt 1
(7)
This shows that 119909 |120578(119909 119890)| gt 12 is non-null as desired Asimilar computation is used to show that with probability 1119890 is the unique such element of 120582
5 Ioanarsquos Proof
WhatWeWant We wish to find some 119899 and 119886 isin 119883119899such that
where 120583119886denotes the normalized restriction of 120583 to 120587minus1
119899(119886)
This would imply by a straightforward computation that foreach 120574 isin Γ
119886there exists a 120582 isin Λ such that
120583119886119909 | 120572 (120574 119909) = 120582 ge 119862 gt
7
8(9)
and this would complete the proof thanks to Theorem 2
WhatWeHave Unfortunately it is only immediately possibleto obtain that the quantities in (8) tend to 1 on average at arate depending on 120574 That is for each 120574 isin Γ we have
| 119886 isin 119883119899 119899 isin 120596 is dense in
1198712 and while the right-hand side is the norm-squared of 120594119878
the left-hand side is the norm-squared of 120594119878projected onto
the span of 120594120587minus1
119898(119886)
| 119886 isin 119883119898 119898 le 119899 Finally just apply (12)
to each set 119878 = 119909 | 120572(120574 119909) = 120582 and use the dominatedconvergence theorem to pass the limit through the sum overall 120582 isin Λ
Proof Thegap betweenwhat he has (the asymptotic informa-tion) and what we want (the uniform information) is bridgedby Property (T) Once again the first step is to consider anappropriate representation this time onewhich compares thevalues of 120572(120574 119909) as 119909 varies That is let Γ 119883 times 119883 times Λ by
and consider the left-regular unitary representation corre-sponding to this action The idea very roughly is that thedegree to which 120572(120574 119909) is independent of 119909 will be measuredby how close a particular vector is to being Γ-invariant
More precisely for each 119899 define an orthonormal familyof vectors 120585
So now ldquowhat we haverdquo and ldquowhat we wantrdquo can be translatedas follows we have the 120585
119899form a family of almost invariant
vectors and we want a single 119899 and 119886 isin 119883119899such that 120585
119886is
nearly invariant uniformly for all 120574 isin Γ119886
The remainder of the argument is straightforward Sincethe 120585119899forms a family of almost invariant vectors Property
(T) implies that there exist 119899 and an invariant vector 120578 suchthat 120578 minus 120585
119899 le 120575 Let 1205781015840 be the restriction of 120578 to the set
cup119886isin119883119899
(120587minus1119899
(119886)times120587minus1119899
(119886)timesΛ) Since this set is invariant we havethe fact that that 1205781015840 is invariant as well Since 120585
119899is supported
on this set we retain the property that 1205781015840 minus 120585119899 le 120575
Now we simply express 1205781015840 as a normalized average oforthogonal Γ
119886-invariant vectors More specifically write
1205781015840
=1
radic1003816100381610038161003816119883119899
1003816100381610038161003816
sum119886isin119883119886
120578119886 (17)
ISRN Algebra 5
where 120578119886is the appropriately rescaled restriction of 1205781015840 to the
set 120587minus1119899
(119886) times 120587minus1119899
(119886) timesΛ Then by the law of averages we musthave some 119886 isin 119883
119899such that 120578
119886minus 120585119886 le 120575 Moreover 120578
119886is Γ119886-
invariant so that for all 120574 isin Γ119886wehave ⟨120574120578
119886 120578119886⟩ = 1 It follows
that by an appropriate choice of 120575 we can make ⟨120574120585119886 120585119886⟩ ge
119862 gt 78 for all 120574 isin Γ119886
6 Easy Applications
In this section we use Ioanarsquos theorem for one of its intendedpurposes to find many highly inequivalent actions Theresults mentioned here are just meant to give the flavor ofapplications of superrigidity they by no means demonstratethe full power of the theorem In the next section we willdiscuss the slightly more interesting and difficult applicationto torsion-free abelian groups For further applications seefor instance [7ndash10]
In searching for inequivalent actions onemight of courseconsider a variety of inequivalence notions Here we focuson just two of them orbit inequivalence and Borel incom-parability Recall from the introduction that Γ 119883 andΛ 119884 are said to be orbit equivalent if there exists ameasure-preserving andorbit-preserving almost bijection from119883 to119884Notice that this notion depends only on the orbit equivalencerelation arising from the two actions and not on the actionsthemselves When this is the case we will often conflate thetwo saying alternately that certain actions are orbit equivalentor that certain equivalence relations are ldquoorbit equivalentrdquo
Borel bireducibility is a purely set-theoretic notion withits origins in logic The connection is that if 119864 is an equiv-alence relation on a standard Borel space 119883 then we canthink of119864 representing a classification problem For instanceif 119883 happens to be a set of codes for a family of structuresthen studying the classification of those structures amountsto studying the isomorphism equivalence relation 119864 on 119883We refer the reader to [11] for a complete introduction to thesubject
If 119864 and 119865 are equivalent relations on 119883 and 119884 then119864 is said to be Borel reducible to 119865 if there exists a Borelfunction 119891 119883 rarr 119884 satisfying 119909119864119909
1015840 if and only if119891(119909)119865119891(1199091015840) We think of this saying that the classificationproblem for elements of 119883 up to 119864 is no more complex thanthe classification problem for elements of 119884 up to 119865 Thus if119864 and 119865 are Borel bireducible (ie there is a reduction bothways) then they represent classification problems of the samecomplexity
It is elementary to see that neither orbit equivalence orBorel bireducibility implies the other For instance given anyΓ-space 119883 one can form a disjoint union 119883 ⊔ 119883
1015840 where 1198831015840
is a Γ-space of very high complexity which is declared to beof measure 0 Conversely if 119883 is an ergodic and hyperfiniteΓ-space then it is known that it is bireducible with119883⊔119883 butthe two cannot be orbit equivalent It is even possible withoutmuch more difficulty to find two ergodic actions which arebireducible but not orbit equivalent
We are now ready to begin with the following direct con-sequence of Ioanarsquos theorem It was first established by SimonThomas in connection working on classification problem for
torsion-free abelian groups of finite rank His proof usedZimmerrsquos superrigidity theorem and some additional cocyclemanipulation techniques with Ioanarsquos theorem in hand theproof will be much simpler
Corollary 4 If 119899 ge 3 is fixed and 119901 119902 are primes such that119901 = 119902 then the actions of SL
119899(Z) on SL
119899(Z119901) and SL
119899(Z119902) are
orbit inequivalent and Borel incomparable
Here Z119901denotes the ring of 119901-adic integers It is easy to
see that SL119899(Z) SL
119899(Z119901) is a profinite action being the
inverse limit of the actions SL119899(Z) SL
119899(Z119901119894Z) together
with their natural system of projections
Proof Let 119901 = 119902 and suppose that 119891 is either an orbit equiv-alence or a Borel reduction from SL
119899(Z) SL
119899(Z119901) to
SL119899(Z) SL
119899(Z119902) We now apply Ioanarsquos theorem together
with the understanding of cocycles gained in the previoussection The conclusion is that we can suppose without lossof generality that there exists a finite index subgroup Γ
0le
SL119899(Z) a Γ
0-coset 119883 sub SL
119899(Z119901) and a homomorphism
120601 Γ0
rarr SL119899(Z) which makes 119891 into an action-preserving
map from Γ0 119883 into SL
119899(Z) SL
119899(Z119902)
Now in the measure-preserving case it is not difficult toconclude that 119891 is a ldquovirtual isomorphismrdquo between the twoactions We claim that this can be achieved even in the casethat 119891 is just a Borel reduction First we can assume that 120601is an embedding Indeed by Margulisrsquos theorem on normalsubgroups [1 Theorem 812] either im(120601) or ker(120601) is finiteIf ker(120601) is finite then we can replace Γ
0by a finite index
subgroup (and 119883 by a coset of the new Γ0) to suppose that
120601 is injective On the other hand if im(120601) is finite then wecan replace Γ
0by a finite index subgroup to suppose that 120601 is
trivial But this would mean that 119891 is Γ0-invariant and so by
ergodicity of Γ0
119883 119891 would send a conull set to a singlepoint contradicting that 119891 is countable-to-one
Second 120601(Γ0) must be a finite index subgroup of SL
119899(Z)
Indeed byMargulisrsquos superrigidity theorem 120601 can be lifted toan isomorphism of SL
119899(R) and it follows that 120601(Γ
0) is a lattice
of SL119899(R) But then it is easy to see that any lattice which is
contained in SL119899(Z) must be commensurable with SL
119899(Z)
Third by the ergodicity of Γ0
119883 we can assumethat im(119891) is contained in a single 120601(Γ
0) coset 119884
0 And
now because 120601(Γ0) preserves a unique measure on 119884
0(the
Haar measure) and because 120601(Γ0) preserves 119891
lowast(Haar) we
actually conclude that 119891 is measure-preserving In summarywe have shown that (120601 119891) is a measure and action-preservingisomorphism between Γ
0 119883
0and 120601(Γ
0) 119884
0 which
establishes the claimFinally a short computation confirms the intuitive alge-
braic fact that the existence of such a map is ruled out by themismatch in primes between the left-and right-hand sidesWe give just a quick sketch for a few more details see [12Section 6] Now it is well-known that there are constants 119860
119901
such that for any Δ le SL119899(Z) of finite index the index of Δ in
SL119899(Z119901) divides 119860
119901119901119903 for some 119903 It follows that if Δ le Γ
0
then 119883 breaks up into some number 119873 of ergodic Δ-setswith119873|119860
119901119901119903 Since (120601 119891) is ameasure and action-preserving
6 ISRN Algebra
isomorphism we also have that 119884 breaks up into 119873 ergodic120601(Δ) sets and hence 119873|119860
119902119902119904 also But it is not difficult to
choose Δ small enough to ensure that 119873 is large enough forthis to be a contradiction
This argument can be easily generalized to give uncount-ably many incomparable actions of SL
119899(Z) Given an infinite
set 119878 of primes with increasing enumeration 119878 = 119901119894 we can
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
Cocycle superrigidity results were first established by Mar-gulis and Zimmer for cocycles Γ 119883 rarr Λ where Γ
is a lattice in a higher rank Lie group acting ergodicallyon 119883 These results carried the additional hypothesis thatΛ is contained in an algebraic group The first example ofthe most general form of cocycle superrigidity with thetarget Λ arbitrary was Poparsquos result concerning Bernoulliactions In this section wewill discuss Ioanarsquos theoremwhichestablishes similar conclusions for profinite actions
Here Γ 119883 is said to be profinite if as a Γ-set X is theinverse limit of a family of finite Γ-sets119883
119909 isin 119883 can be identified with the thread (120587119899(119909)) We are
interested in the ergodic case here each119883119899is equipped with
the uniform probability measure and Γ 119883119899is transitive
Theorem 1 (Ioana) Let Γ (119883 120583) be an ergodic measure-preserving profinite action with invariant factor maps 120587
119899
119883 rarr 119883119899 Assume that Γ has Property (T)Then for any cocycle
120572 Γ 119883 rarr Λ there exists 119899 and 119886 isin 119883119899such that the
restriction of 120572 to the action Γ119886 120587minus1119899
(119886) is cohomologous to atrivial cocycle
In other words the conclusion is that Γ 119883 is ldquovirtuallysuperrigidrdquo in the sense that any orbit preserving map afterit is restricted to a finite index component of the left-handside comes from an action preserving map Ioanarsquos theoremis interesting when contrasted with Poparsquos theorem whileBernoulli actions are strongly mixing profinite actions arehighly nonmixing Indeed for each 119899 Γ just permutes theblocks 120587
minus1
119899(119886) for 119886 isin 119883
119899 and it follows that ⋃
119886isin119883119899
119883119886times 119883119886
is a Γ-invariant subset of 119883 times 119883We remark that although our variant of Ioanarsquos theorem
is sufficient for most applications it is weaker than the stateof the art in several ways First Ioana requires only thatΓ have the relative Property (T) over some infinite normalsubgroups 119873 such that Γ119873 is finitely generated SecondIoana also shows that 120572 is equivalent to a cocycle definedon all of 119883 Last Furman has generalized the statement byreplacing profinite actions with the more general class ofcompact actions (Γ 119883 is said to be compact if whenregarded as a subset of (119883 120583) it is precompact in a suitabletopology)
4 Cocycle Untwisting
We begin with the following preliminary result whichroughly speaking says that if 120572 Γ 119883 rarr Λ is acocycle and if for each 120574 it has a ldquovery likelyrdquo value then 120572
is cohomologous to the map which always takes on this likelyvalue In particular in this case 120572 is cohomologous to a trivialcocycle
Theorem 2 Let Γ (119883 120583) be ergodic and measure-preserv-ing and let 120572 Γ 119883 rarr Λ be a cocycle Suppose that for all120574 isin Γ there exists 120582
Then the map 120601(120574) = 120582120574is a homomorphism and 120572 is
cohomologous to it
It is easy to see that 120601must be a homomorphism indeedthe hypothesis guarantees that there is a nonnull set of 119909 forwhich 120601(120574
1015840120574) = 120572(1205741015840120574 119909) = 120572(1205741015840 120574119909)120572(120574 119909) = 120601(1205741015840)120601(120574)Hence it remains only to establish the following result
Lemma 3 Let Γ (119883 120583) be ergodic and measure-preservingand let 120572 120573 Γ 119883 rarr Λ be cocycles Suppose that for all120574 isin Γ
We understand this result to say that if 120572 and 120573 are closein an 119871infin sense then they are cohomologous It follows uponsimilar results of Popa and Furman which draws the sameconclusion in the case that 120572 and 120573 are close in an appropriate1198711 sense (for instance see [6 Theorem 42]) Ioanarsquos proofgiven below may be safely skipped until reading the nextsection
Proof of Lemma 3 Let Γ 119883 times Λ be the action given by
(this is an action thanks to the cocycle condition) andconsider the corresponding left-regular representation Thereason for using this representation is that120572 is close to120573 if andonly if a particular vector is close to being invariant Namelylet
120585 = 120594119883times119890
(5)
(read the characteristic function of 119883 times 119890) and notice that
Using this together with the law of cosines the hypothesisnow translates to say that 120574120585minus120585 le 119862 lt 12 for all 120574 isin Γ It isnot difficult to see that this implies that there is an invariantvector 120578 such that 120578 minus 120585 lt 12 (Indeed letting 119878 denote
4 ISRN Algebra
the convex hull of Γ sdot 120585 it is easily seen that there exists aunique vector 120578 isin 119878 of minimal norm this 120578 is necessarilyinvariant)
The idea for the conclusion of the proof is as follows Ifwe had 120578 = 120594graph(119887) for some function 119887 119883 rarr Λ then wewould be done Indeed in this case the invariance of 120578wouldmean that 119887(119909) = 120582 if and only if 119887(120574119909) = 120572(120574 119909)120582120573(120574 119909)
minus1 In other words 119887 wouldwitness the fact that 120572 is cohomologous to 120573 The fact that120578 minus 120585 lt 12 implies that this is close to being the case
We actually define 119887(119909) = the 120582 such that |120578(119909 120582)| gt 12if it exists and is unique The above computation shows thatwhen 119887(119909) 119887(120574119909) are both defined the cohomology relationholds Moreover the set where 119887 is defined is invariant so bythe ergodicity of Γ 119883 it suffices to show that this set isnonnull In fact since 120578 and 120585 are close 119887 must take value 119890
on a nonnull set
120583 119909 1003816100381610038161003816120585 (119909 119890) minus 120578 (119909 119890)
1003816100381610038161003816 ge1
2
le 4int119909|120585(119909119890)minus120578(119909119890)|ge12
1003816100381610038161003816120585 minus 12057810038161003816100381610038162
le 41003817100381710038171003817120585 minus 120578
10038171003817100381710038172
lt 1
(7)
This shows that 119909 |120578(119909 119890)| gt 12 is non-null as desired Asimilar computation is used to show that with probability 1119890 is the unique such element of 120582
5 Ioanarsquos Proof
WhatWeWant We wish to find some 119899 and 119886 isin 119883119899such that
where 120583119886denotes the normalized restriction of 120583 to 120587minus1
119899(119886)
This would imply by a straightforward computation that foreach 120574 isin Γ
119886there exists a 120582 isin Λ such that
120583119886119909 | 120572 (120574 119909) = 120582 ge 119862 gt
7
8(9)
and this would complete the proof thanks to Theorem 2
WhatWeHave Unfortunately it is only immediately possibleto obtain that the quantities in (8) tend to 1 on average at arate depending on 120574 That is for each 120574 isin Γ we have
| 119886 isin 119883119899 119899 isin 120596 is dense in
1198712 and while the right-hand side is the norm-squared of 120594119878
the left-hand side is the norm-squared of 120594119878projected onto
the span of 120594120587minus1
119898(119886)
| 119886 isin 119883119898 119898 le 119899 Finally just apply (12)
to each set 119878 = 119909 | 120572(120574 119909) = 120582 and use the dominatedconvergence theorem to pass the limit through the sum overall 120582 isin Λ
Proof Thegap betweenwhat he has (the asymptotic informa-tion) and what we want (the uniform information) is bridgedby Property (T) Once again the first step is to consider anappropriate representation this time onewhich compares thevalues of 120572(120574 119909) as 119909 varies That is let Γ 119883 times 119883 times Λ by
and consider the left-regular unitary representation corre-sponding to this action The idea very roughly is that thedegree to which 120572(120574 119909) is independent of 119909 will be measuredby how close a particular vector is to being Γ-invariant
More precisely for each 119899 define an orthonormal familyof vectors 120585
So now ldquowhat we haverdquo and ldquowhat we wantrdquo can be translatedas follows we have the 120585
119899form a family of almost invariant
vectors and we want a single 119899 and 119886 isin 119883119899such that 120585
119886is
nearly invariant uniformly for all 120574 isin Γ119886
The remainder of the argument is straightforward Sincethe 120585119899forms a family of almost invariant vectors Property
(T) implies that there exist 119899 and an invariant vector 120578 suchthat 120578 minus 120585
119899 le 120575 Let 1205781015840 be the restriction of 120578 to the set
cup119886isin119883119899
(120587minus1119899
(119886)times120587minus1119899
(119886)timesΛ) Since this set is invariant we havethe fact that that 1205781015840 is invariant as well Since 120585
119899is supported
on this set we retain the property that 1205781015840 minus 120585119899 le 120575
Now we simply express 1205781015840 as a normalized average oforthogonal Γ
119886-invariant vectors More specifically write
1205781015840
=1
radic1003816100381610038161003816119883119899
1003816100381610038161003816
sum119886isin119883119886
120578119886 (17)
ISRN Algebra 5
where 120578119886is the appropriately rescaled restriction of 1205781015840 to the
set 120587minus1119899
(119886) times 120587minus1119899
(119886) timesΛ Then by the law of averages we musthave some 119886 isin 119883
119899such that 120578
119886minus 120585119886 le 120575 Moreover 120578
119886is Γ119886-
invariant so that for all 120574 isin Γ119886wehave ⟨120574120578
119886 120578119886⟩ = 1 It follows
that by an appropriate choice of 120575 we can make ⟨120574120585119886 120585119886⟩ ge
119862 gt 78 for all 120574 isin Γ119886
6 Easy Applications
In this section we use Ioanarsquos theorem for one of its intendedpurposes to find many highly inequivalent actions Theresults mentioned here are just meant to give the flavor ofapplications of superrigidity they by no means demonstratethe full power of the theorem In the next section we willdiscuss the slightly more interesting and difficult applicationto torsion-free abelian groups For further applications seefor instance [7ndash10]
In searching for inequivalent actions onemight of courseconsider a variety of inequivalence notions Here we focuson just two of them orbit inequivalence and Borel incom-parability Recall from the introduction that Γ 119883 andΛ 119884 are said to be orbit equivalent if there exists ameasure-preserving andorbit-preserving almost bijection from119883 to119884Notice that this notion depends only on the orbit equivalencerelation arising from the two actions and not on the actionsthemselves When this is the case we will often conflate thetwo saying alternately that certain actions are orbit equivalentor that certain equivalence relations are ldquoorbit equivalentrdquo
Borel bireducibility is a purely set-theoretic notion withits origins in logic The connection is that if 119864 is an equiv-alence relation on a standard Borel space 119883 then we canthink of119864 representing a classification problem For instanceif 119883 happens to be a set of codes for a family of structuresthen studying the classification of those structures amountsto studying the isomorphism equivalence relation 119864 on 119883We refer the reader to [11] for a complete introduction to thesubject
If 119864 and 119865 are equivalent relations on 119883 and 119884 then119864 is said to be Borel reducible to 119865 if there exists a Borelfunction 119891 119883 rarr 119884 satisfying 119909119864119909
1015840 if and only if119891(119909)119865119891(1199091015840) We think of this saying that the classificationproblem for elements of 119883 up to 119864 is no more complex thanthe classification problem for elements of 119884 up to 119865 Thus if119864 and 119865 are Borel bireducible (ie there is a reduction bothways) then they represent classification problems of the samecomplexity
It is elementary to see that neither orbit equivalence orBorel bireducibility implies the other For instance given anyΓ-space 119883 one can form a disjoint union 119883 ⊔ 119883
1015840 where 1198831015840
is a Γ-space of very high complexity which is declared to beof measure 0 Conversely if 119883 is an ergodic and hyperfiniteΓ-space then it is known that it is bireducible with119883⊔119883 butthe two cannot be orbit equivalent It is even possible withoutmuch more difficulty to find two ergodic actions which arebireducible but not orbit equivalent
We are now ready to begin with the following direct con-sequence of Ioanarsquos theorem It was first established by SimonThomas in connection working on classification problem for
torsion-free abelian groups of finite rank His proof usedZimmerrsquos superrigidity theorem and some additional cocyclemanipulation techniques with Ioanarsquos theorem in hand theproof will be much simpler
Corollary 4 If 119899 ge 3 is fixed and 119901 119902 are primes such that119901 = 119902 then the actions of SL
119899(Z) on SL
119899(Z119901) and SL
119899(Z119902) are
orbit inequivalent and Borel incomparable
Here Z119901denotes the ring of 119901-adic integers It is easy to
see that SL119899(Z) SL
119899(Z119901) is a profinite action being the
inverse limit of the actions SL119899(Z) SL
119899(Z119901119894Z) together
with their natural system of projections
Proof Let 119901 = 119902 and suppose that 119891 is either an orbit equiv-alence or a Borel reduction from SL
119899(Z) SL
119899(Z119901) to
SL119899(Z) SL
119899(Z119902) We now apply Ioanarsquos theorem together
with the understanding of cocycles gained in the previoussection The conclusion is that we can suppose without lossof generality that there exists a finite index subgroup Γ
0le
SL119899(Z) a Γ
0-coset 119883 sub SL
119899(Z119901) and a homomorphism
120601 Γ0
rarr SL119899(Z) which makes 119891 into an action-preserving
map from Γ0 119883 into SL
119899(Z) SL
119899(Z119902)
Now in the measure-preserving case it is not difficult toconclude that 119891 is a ldquovirtual isomorphismrdquo between the twoactions We claim that this can be achieved even in the casethat 119891 is just a Borel reduction First we can assume that 120601is an embedding Indeed by Margulisrsquos theorem on normalsubgroups [1 Theorem 812] either im(120601) or ker(120601) is finiteIf ker(120601) is finite then we can replace Γ
0by a finite index
subgroup (and 119883 by a coset of the new Γ0) to suppose that
120601 is injective On the other hand if im(120601) is finite then wecan replace Γ
0by a finite index subgroup to suppose that 120601 is
trivial But this would mean that 119891 is Γ0-invariant and so by
ergodicity of Γ0
119883 119891 would send a conull set to a singlepoint contradicting that 119891 is countable-to-one
Second 120601(Γ0) must be a finite index subgroup of SL
119899(Z)
Indeed byMargulisrsquos superrigidity theorem 120601 can be lifted toan isomorphism of SL
119899(R) and it follows that 120601(Γ
0) is a lattice
of SL119899(R) But then it is easy to see that any lattice which is
contained in SL119899(Z) must be commensurable with SL
119899(Z)
Third by the ergodicity of Γ0
119883 we can assumethat im(119891) is contained in a single 120601(Γ
0) coset 119884
0 And
now because 120601(Γ0) preserves a unique measure on 119884
0(the
Haar measure) and because 120601(Γ0) preserves 119891
lowast(Haar) we
actually conclude that 119891 is measure-preserving In summarywe have shown that (120601 119891) is a measure and action-preservingisomorphism between Γ
0 119883
0and 120601(Γ
0) 119884
0 which
establishes the claimFinally a short computation confirms the intuitive alge-
braic fact that the existence of such a map is ruled out by themismatch in primes between the left-and right-hand sidesWe give just a quick sketch for a few more details see [12Section 6] Now it is well-known that there are constants 119860
119901
such that for any Δ le SL119899(Z) of finite index the index of Δ in
SL119899(Z119901) divides 119860
119901119901119903 for some 119903 It follows that if Δ le Γ
0
then 119883 breaks up into some number 119873 of ergodic Δ-setswith119873|119860
119901119901119903 Since (120601 119891) is ameasure and action-preserving
6 ISRN Algebra
isomorphism we also have that 119884 breaks up into 119873 ergodic120601(Δ) sets and hence 119873|119860
119902119902119904 also But it is not difficult to
choose Δ small enough to ensure that 119873 is large enough forthis to be a contradiction
This argument can be easily generalized to give uncount-ably many incomparable actions of SL
119899(Z) Given an infinite
set 119878 of primes with increasing enumeration 119878 = 119901119894 we can
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
the convex hull of Γ sdot 120585 it is easily seen that there exists aunique vector 120578 isin 119878 of minimal norm this 120578 is necessarilyinvariant)
The idea for the conclusion of the proof is as follows Ifwe had 120578 = 120594graph(119887) for some function 119887 119883 rarr Λ then wewould be done Indeed in this case the invariance of 120578wouldmean that 119887(119909) = 120582 if and only if 119887(120574119909) = 120572(120574 119909)120582120573(120574 119909)
minus1 In other words 119887 wouldwitness the fact that 120572 is cohomologous to 120573 The fact that120578 minus 120585 lt 12 implies that this is close to being the case
We actually define 119887(119909) = the 120582 such that |120578(119909 120582)| gt 12if it exists and is unique The above computation shows thatwhen 119887(119909) 119887(120574119909) are both defined the cohomology relationholds Moreover the set where 119887 is defined is invariant so bythe ergodicity of Γ 119883 it suffices to show that this set isnonnull In fact since 120578 and 120585 are close 119887 must take value 119890
on a nonnull set
120583 119909 1003816100381610038161003816120585 (119909 119890) minus 120578 (119909 119890)
1003816100381610038161003816 ge1
2
le 4int119909|120585(119909119890)minus120578(119909119890)|ge12
1003816100381610038161003816120585 minus 12057810038161003816100381610038162
le 41003817100381710038171003817120585 minus 120578
10038171003817100381710038172
lt 1
(7)
This shows that 119909 |120578(119909 119890)| gt 12 is non-null as desired Asimilar computation is used to show that with probability 1119890 is the unique such element of 120582
5 Ioanarsquos Proof
WhatWeWant We wish to find some 119899 and 119886 isin 119883119899such that
where 120583119886denotes the normalized restriction of 120583 to 120587minus1
119899(119886)
This would imply by a straightforward computation that foreach 120574 isin Γ
119886there exists a 120582 isin Λ such that
120583119886119909 | 120572 (120574 119909) = 120582 ge 119862 gt
7
8(9)
and this would complete the proof thanks to Theorem 2
WhatWeHave Unfortunately it is only immediately possibleto obtain that the quantities in (8) tend to 1 on average at arate depending on 120574 That is for each 120574 isin Γ we have
| 119886 isin 119883119899 119899 isin 120596 is dense in
1198712 and while the right-hand side is the norm-squared of 120594119878
the left-hand side is the norm-squared of 120594119878projected onto
the span of 120594120587minus1
119898(119886)
| 119886 isin 119883119898 119898 le 119899 Finally just apply (12)
to each set 119878 = 119909 | 120572(120574 119909) = 120582 and use the dominatedconvergence theorem to pass the limit through the sum overall 120582 isin Λ
Proof Thegap betweenwhat he has (the asymptotic informa-tion) and what we want (the uniform information) is bridgedby Property (T) Once again the first step is to consider anappropriate representation this time onewhich compares thevalues of 120572(120574 119909) as 119909 varies That is let Γ 119883 times 119883 times Λ by
and consider the left-regular unitary representation corre-sponding to this action The idea very roughly is that thedegree to which 120572(120574 119909) is independent of 119909 will be measuredby how close a particular vector is to being Γ-invariant
More precisely for each 119899 define an orthonormal familyof vectors 120585
So now ldquowhat we haverdquo and ldquowhat we wantrdquo can be translatedas follows we have the 120585
119899form a family of almost invariant
vectors and we want a single 119899 and 119886 isin 119883119899such that 120585
119886is
nearly invariant uniformly for all 120574 isin Γ119886
The remainder of the argument is straightforward Sincethe 120585119899forms a family of almost invariant vectors Property
(T) implies that there exist 119899 and an invariant vector 120578 suchthat 120578 minus 120585
119899 le 120575 Let 1205781015840 be the restriction of 120578 to the set
cup119886isin119883119899
(120587minus1119899
(119886)times120587minus1119899
(119886)timesΛ) Since this set is invariant we havethe fact that that 1205781015840 is invariant as well Since 120585
119899is supported
on this set we retain the property that 1205781015840 minus 120585119899 le 120575
Now we simply express 1205781015840 as a normalized average oforthogonal Γ
119886-invariant vectors More specifically write
1205781015840
=1
radic1003816100381610038161003816119883119899
1003816100381610038161003816
sum119886isin119883119886
120578119886 (17)
ISRN Algebra 5
where 120578119886is the appropriately rescaled restriction of 1205781015840 to the
set 120587minus1119899
(119886) times 120587minus1119899
(119886) timesΛ Then by the law of averages we musthave some 119886 isin 119883
119899such that 120578
119886minus 120585119886 le 120575 Moreover 120578
119886is Γ119886-
invariant so that for all 120574 isin Γ119886wehave ⟨120574120578
119886 120578119886⟩ = 1 It follows
that by an appropriate choice of 120575 we can make ⟨120574120585119886 120585119886⟩ ge
119862 gt 78 for all 120574 isin Γ119886
6 Easy Applications
In this section we use Ioanarsquos theorem for one of its intendedpurposes to find many highly inequivalent actions Theresults mentioned here are just meant to give the flavor ofapplications of superrigidity they by no means demonstratethe full power of the theorem In the next section we willdiscuss the slightly more interesting and difficult applicationto torsion-free abelian groups For further applications seefor instance [7ndash10]
In searching for inequivalent actions onemight of courseconsider a variety of inequivalence notions Here we focuson just two of them orbit inequivalence and Borel incom-parability Recall from the introduction that Γ 119883 andΛ 119884 are said to be orbit equivalent if there exists ameasure-preserving andorbit-preserving almost bijection from119883 to119884Notice that this notion depends only on the orbit equivalencerelation arising from the two actions and not on the actionsthemselves When this is the case we will often conflate thetwo saying alternately that certain actions are orbit equivalentor that certain equivalence relations are ldquoorbit equivalentrdquo
Borel bireducibility is a purely set-theoretic notion withits origins in logic The connection is that if 119864 is an equiv-alence relation on a standard Borel space 119883 then we canthink of119864 representing a classification problem For instanceif 119883 happens to be a set of codes for a family of structuresthen studying the classification of those structures amountsto studying the isomorphism equivalence relation 119864 on 119883We refer the reader to [11] for a complete introduction to thesubject
If 119864 and 119865 are equivalent relations on 119883 and 119884 then119864 is said to be Borel reducible to 119865 if there exists a Borelfunction 119891 119883 rarr 119884 satisfying 119909119864119909
1015840 if and only if119891(119909)119865119891(1199091015840) We think of this saying that the classificationproblem for elements of 119883 up to 119864 is no more complex thanthe classification problem for elements of 119884 up to 119865 Thus if119864 and 119865 are Borel bireducible (ie there is a reduction bothways) then they represent classification problems of the samecomplexity
It is elementary to see that neither orbit equivalence orBorel bireducibility implies the other For instance given anyΓ-space 119883 one can form a disjoint union 119883 ⊔ 119883
1015840 where 1198831015840
is a Γ-space of very high complexity which is declared to beof measure 0 Conversely if 119883 is an ergodic and hyperfiniteΓ-space then it is known that it is bireducible with119883⊔119883 butthe two cannot be orbit equivalent It is even possible withoutmuch more difficulty to find two ergodic actions which arebireducible but not orbit equivalent
We are now ready to begin with the following direct con-sequence of Ioanarsquos theorem It was first established by SimonThomas in connection working on classification problem for
torsion-free abelian groups of finite rank His proof usedZimmerrsquos superrigidity theorem and some additional cocyclemanipulation techniques with Ioanarsquos theorem in hand theproof will be much simpler
Corollary 4 If 119899 ge 3 is fixed and 119901 119902 are primes such that119901 = 119902 then the actions of SL
119899(Z) on SL
119899(Z119901) and SL
119899(Z119902) are
orbit inequivalent and Borel incomparable
Here Z119901denotes the ring of 119901-adic integers It is easy to
see that SL119899(Z) SL
119899(Z119901) is a profinite action being the
inverse limit of the actions SL119899(Z) SL
119899(Z119901119894Z) together
with their natural system of projections
Proof Let 119901 = 119902 and suppose that 119891 is either an orbit equiv-alence or a Borel reduction from SL
119899(Z) SL
119899(Z119901) to
SL119899(Z) SL
119899(Z119902) We now apply Ioanarsquos theorem together
with the understanding of cocycles gained in the previoussection The conclusion is that we can suppose without lossof generality that there exists a finite index subgroup Γ
0le
SL119899(Z) a Γ
0-coset 119883 sub SL
119899(Z119901) and a homomorphism
120601 Γ0
rarr SL119899(Z) which makes 119891 into an action-preserving
map from Γ0 119883 into SL
119899(Z) SL
119899(Z119902)
Now in the measure-preserving case it is not difficult toconclude that 119891 is a ldquovirtual isomorphismrdquo between the twoactions We claim that this can be achieved even in the casethat 119891 is just a Borel reduction First we can assume that 120601is an embedding Indeed by Margulisrsquos theorem on normalsubgroups [1 Theorem 812] either im(120601) or ker(120601) is finiteIf ker(120601) is finite then we can replace Γ
0by a finite index
subgroup (and 119883 by a coset of the new Γ0) to suppose that
120601 is injective On the other hand if im(120601) is finite then wecan replace Γ
0by a finite index subgroup to suppose that 120601 is
trivial But this would mean that 119891 is Γ0-invariant and so by
ergodicity of Γ0
119883 119891 would send a conull set to a singlepoint contradicting that 119891 is countable-to-one
Second 120601(Γ0) must be a finite index subgroup of SL
119899(Z)
Indeed byMargulisrsquos superrigidity theorem 120601 can be lifted toan isomorphism of SL
119899(R) and it follows that 120601(Γ
0) is a lattice
of SL119899(R) But then it is easy to see that any lattice which is
contained in SL119899(Z) must be commensurable with SL
119899(Z)
Third by the ergodicity of Γ0
119883 we can assumethat im(119891) is contained in a single 120601(Γ
0) coset 119884
0 And
now because 120601(Γ0) preserves a unique measure on 119884
0(the
Haar measure) and because 120601(Γ0) preserves 119891
lowast(Haar) we
actually conclude that 119891 is measure-preserving In summarywe have shown that (120601 119891) is a measure and action-preservingisomorphism between Γ
0 119883
0and 120601(Γ
0) 119884
0 which
establishes the claimFinally a short computation confirms the intuitive alge-
braic fact that the existence of such a map is ruled out by themismatch in primes between the left-and right-hand sidesWe give just a quick sketch for a few more details see [12Section 6] Now it is well-known that there are constants 119860
119901
such that for any Δ le SL119899(Z) of finite index the index of Δ in
SL119899(Z119901) divides 119860
119901119901119903 for some 119903 It follows that if Δ le Γ
0
then 119883 breaks up into some number 119873 of ergodic Δ-setswith119873|119860
119901119901119903 Since (120601 119891) is ameasure and action-preserving
6 ISRN Algebra
isomorphism we also have that 119884 breaks up into 119873 ergodic120601(Δ) sets and hence 119873|119860
119902119902119904 also But it is not difficult to
choose Δ small enough to ensure that 119873 is large enough forthis to be a contradiction
This argument can be easily generalized to give uncount-ably many incomparable actions of SL
119899(Z) Given an infinite
set 119878 of primes with increasing enumeration 119878 = 119901119894 we can
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
where 120578119886is the appropriately rescaled restriction of 1205781015840 to the
set 120587minus1119899
(119886) times 120587minus1119899
(119886) timesΛ Then by the law of averages we musthave some 119886 isin 119883
119899such that 120578
119886minus 120585119886 le 120575 Moreover 120578
119886is Γ119886-
invariant so that for all 120574 isin Γ119886wehave ⟨120574120578
119886 120578119886⟩ = 1 It follows
that by an appropriate choice of 120575 we can make ⟨120574120585119886 120585119886⟩ ge
119862 gt 78 for all 120574 isin Γ119886
6 Easy Applications
In this section we use Ioanarsquos theorem for one of its intendedpurposes to find many highly inequivalent actions Theresults mentioned here are just meant to give the flavor ofapplications of superrigidity they by no means demonstratethe full power of the theorem In the next section we willdiscuss the slightly more interesting and difficult applicationto torsion-free abelian groups For further applications seefor instance [7ndash10]
In searching for inequivalent actions onemight of courseconsider a variety of inequivalence notions Here we focuson just two of them orbit inequivalence and Borel incom-parability Recall from the introduction that Γ 119883 andΛ 119884 are said to be orbit equivalent if there exists ameasure-preserving andorbit-preserving almost bijection from119883 to119884Notice that this notion depends only on the orbit equivalencerelation arising from the two actions and not on the actionsthemselves When this is the case we will often conflate thetwo saying alternately that certain actions are orbit equivalentor that certain equivalence relations are ldquoorbit equivalentrdquo
Borel bireducibility is a purely set-theoretic notion withits origins in logic The connection is that if 119864 is an equiv-alence relation on a standard Borel space 119883 then we canthink of119864 representing a classification problem For instanceif 119883 happens to be a set of codes for a family of structuresthen studying the classification of those structures amountsto studying the isomorphism equivalence relation 119864 on 119883We refer the reader to [11] for a complete introduction to thesubject
If 119864 and 119865 are equivalent relations on 119883 and 119884 then119864 is said to be Borel reducible to 119865 if there exists a Borelfunction 119891 119883 rarr 119884 satisfying 119909119864119909
1015840 if and only if119891(119909)119865119891(1199091015840) We think of this saying that the classificationproblem for elements of 119883 up to 119864 is no more complex thanthe classification problem for elements of 119884 up to 119865 Thus if119864 and 119865 are Borel bireducible (ie there is a reduction bothways) then they represent classification problems of the samecomplexity
It is elementary to see that neither orbit equivalence orBorel bireducibility implies the other For instance given anyΓ-space 119883 one can form a disjoint union 119883 ⊔ 119883
1015840 where 1198831015840
is a Γ-space of very high complexity which is declared to beof measure 0 Conversely if 119883 is an ergodic and hyperfiniteΓ-space then it is known that it is bireducible with119883⊔119883 butthe two cannot be orbit equivalent It is even possible withoutmuch more difficulty to find two ergodic actions which arebireducible but not orbit equivalent
We are now ready to begin with the following direct con-sequence of Ioanarsquos theorem It was first established by SimonThomas in connection working on classification problem for
torsion-free abelian groups of finite rank His proof usedZimmerrsquos superrigidity theorem and some additional cocyclemanipulation techniques with Ioanarsquos theorem in hand theproof will be much simpler
Corollary 4 If 119899 ge 3 is fixed and 119901 119902 are primes such that119901 = 119902 then the actions of SL
119899(Z) on SL
119899(Z119901) and SL
119899(Z119902) are
orbit inequivalent and Borel incomparable
Here Z119901denotes the ring of 119901-adic integers It is easy to
see that SL119899(Z) SL
119899(Z119901) is a profinite action being the
inverse limit of the actions SL119899(Z) SL
119899(Z119901119894Z) together
with their natural system of projections
Proof Let 119901 = 119902 and suppose that 119891 is either an orbit equiv-alence or a Borel reduction from SL
119899(Z) SL
119899(Z119901) to
SL119899(Z) SL
119899(Z119902) We now apply Ioanarsquos theorem together
with the understanding of cocycles gained in the previoussection The conclusion is that we can suppose without lossof generality that there exists a finite index subgroup Γ
0le
SL119899(Z) a Γ
0-coset 119883 sub SL
119899(Z119901) and a homomorphism
120601 Γ0
rarr SL119899(Z) which makes 119891 into an action-preserving
map from Γ0 119883 into SL
119899(Z) SL
119899(Z119902)
Now in the measure-preserving case it is not difficult toconclude that 119891 is a ldquovirtual isomorphismrdquo between the twoactions We claim that this can be achieved even in the casethat 119891 is just a Borel reduction First we can assume that 120601is an embedding Indeed by Margulisrsquos theorem on normalsubgroups [1 Theorem 812] either im(120601) or ker(120601) is finiteIf ker(120601) is finite then we can replace Γ
0by a finite index
subgroup (and 119883 by a coset of the new Γ0) to suppose that
120601 is injective On the other hand if im(120601) is finite then wecan replace Γ
0by a finite index subgroup to suppose that 120601 is
trivial But this would mean that 119891 is Γ0-invariant and so by
ergodicity of Γ0
119883 119891 would send a conull set to a singlepoint contradicting that 119891 is countable-to-one
Second 120601(Γ0) must be a finite index subgroup of SL
119899(Z)
Indeed byMargulisrsquos superrigidity theorem 120601 can be lifted toan isomorphism of SL
119899(R) and it follows that 120601(Γ
0) is a lattice
of SL119899(R) But then it is easy to see that any lattice which is
contained in SL119899(Z) must be commensurable with SL
119899(Z)
Third by the ergodicity of Γ0
119883 we can assumethat im(119891) is contained in a single 120601(Γ
0) coset 119884
0 And
now because 120601(Γ0) preserves a unique measure on 119884
0(the
Haar measure) and because 120601(Γ0) preserves 119891
lowast(Haar) we
actually conclude that 119891 is measure-preserving In summarywe have shown that (120601 119891) is a measure and action-preservingisomorphism between Γ
0 119883
0and 120601(Γ
0) 119884
0 which
establishes the claimFinally a short computation confirms the intuitive alge-
braic fact that the existence of such a map is ruled out by themismatch in primes between the left-and right-hand sidesWe give just a quick sketch for a few more details see [12Section 6] Now it is well-known that there are constants 119860
119901
such that for any Δ le SL119899(Z) of finite index the index of Δ in
SL119899(Z119901) divides 119860
119901119901119903 for some 119903 It follows that if Δ le Γ
0
then 119883 breaks up into some number 119873 of ergodic Δ-setswith119873|119860
119901119901119903 Since (120601 119891) is ameasure and action-preserving
6 ISRN Algebra
isomorphism we also have that 119884 breaks up into 119873 ergodic120601(Δ) sets and hence 119873|119860
119902119902119904 also But it is not difficult to
choose Δ small enough to ensure that 119873 is large enough forthis to be a contradiction
This argument can be easily generalized to give uncount-ably many incomparable actions of SL
119899(Z) Given an infinite
set 119878 of primes with increasing enumeration 119878 = 119901119894 we can
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
It is not much more difficult to show (as Ioana does) thatwhen |119878Δ1198781015840| = infin the actions SL
119899(Z) 119870
119878and
SL119899(Z) 119870
1198781015840 are orbit inequivalent In fact this shows
that there are ldquo1198640manyrdquo orbit inequivalent profinite actions
of SL119899(Z) Of course it is known from different arguments
(exposited in [13 Theorem 171]) that the relation of orbitequivalence on the ergodic actions of SL
119899(Z) is very complex
(for instance not Borel) But the methods used here giveus more detailed information we have an explicit family ofinequivalent actions the actions are special (they are classicaland profinite) and what is more they are Borel incomparable
So far we have considered only free actions of SL119899(Z)
But if one just wants to use Ioanarsquos theorem to find orbitinequivalent actions it is enough to consider actions whichare just free almost everywhere Here a measure-preservingaction Γ 119883 is said to be free almost everywhere if the set119909 | 120574 = 1 rarr 120574119909 = 119909 is conull (ie the set where Γ acts freelyis conull)
Unfortunately in the purely Borel context it is not suffi-cient to work with actions which are free almost everywheresince in this case we are not allowed to just delete a null set onthe right-hand side The next result shows how to get aroundthis difficulty Once again it was originally obtained by SimonThomas using Zimmerrsquos superrigidity theorem
Corollary 5 If 119899 ge 3 is fixed and 119901 119902 are primes with 119901 = 119902then the actions of 119878119871
119899(Z) on P(Q119899
119901) and P(Q
119899
119902) are orbit
inequivalent and Borel incomparable
HereP(Q119899119901)denotes projective space of lines throughQ119899
119901
Since P(Q119899119901) is a transitive SL
119899(Z119901)-space this result is quite
similar to the last one We note also that while SL119899(Z) does
not act freely on P(Q119899119901) it does act freely on a conull subset
[12 Lemma 62]
Proof First suppose that 119891 P(Q119899
119901) rarr P(Q119899
119902) is a measure-
preserving and orbit-preserving map Then we can simplyrestrict the domain of 119891 to assume that it takes values inthe part of P(Q119899
119902) where SL
119899(Z) acts freely Afterwards
we can obtain a contradiction using essentially the samecombinatorial argument as in the proof of Corollary 4
The proof in the case of Borel reducibility requires anextra step Namely we cannot be sure that 119891 sends a conullset into the part of P(Q119899
119902) where SL
119899(Z) acts freely However
if it does not then by the ergodicity of SL119899(Z) P(Q119899
119901) we
can assume that 119891 sends a conull set into the part of P(Q119899119902)
where SL119899(Z) acts nonfreely Our aim will be to show that
this assumption leads to a contradictionFirst let us assume that there exists a conull subset 119883 sub
P(Q119899119901) such that for all 119909 isin 119883 there exists 120574 = 1 such that
120574119891(119909) = 119891(119909) Then for all 119909 isin 119883 119891(119909) lies inside a nontrivialeigenspace of some element of SL
119899(Z) Hence if we let 119881
119909
denote the minimal subspace of Q119899119902which is defined over Q
such that 119891(119909) sub 119881119909 then 119881
119909is necessarily nontrivial
Note that since Q is countable there are only countablymany possibilities for 119881
119909 Hence there exists a non-null
subset 1198831015840 of 119883 and a fixed subspace 119881 of Q119899119902such that for
all 119909 isin 1198831015840 we have119881
119909= 119881 By the ergodicity of SL
119899(Z) 119883
the set 11988310158401015840 = SL119899(Z) sdot 119883
1015840 is conull and it follows that we canadjust 119891 to assume that for all 119909 isin 11988310158401015840 we have119881
119909= 119881 (More
precisely replace 119891(119909) by 1198911015840(119909) = 119891(120574119909) where 120574 is the firstelement of SL
119899(Z) such that 120574119909 isin 11988310158401015840)
Now let 119867 le GL(119881) denote the group of projectivelinear transformations induced on 119881 by SL
119899(Z)119881
It is aneasy exercise using the minimality of 119881 to check that119867 actsfreely on P(119881) and that 119891 is a homomorphism of orbits fromSL119899(Z) 11988310158401015840 into 119867 P(119881) Admitting this we can
finally apply Ioanarsquos theorem to suppose that there exists afinite index subgroup Γ
0le Γ and a nontrivial homomorphism
120601 Γ0
rarr 119867 As in the proof of Corollary 4 we can supposethat 120601 is an embedding We thus get a contradiction from thenext result below
Theorem 6 If Γ0
le SL119899(Z) is a subgroup of finite index and
G is an algebraic Q-group with dim(G) lt 1198992 minus 1 then Γ0does
not embed G(Q)
The idea of the proof is to apply Margulisrsquos superrigiditytheorem That is one wishes to conclude that such anembedding lifts to some kind of rational map SL
119899(R) rarr G
a clear dimension contradiction However a little extra workis needed to handle the case of a Q-group on the right-handside (see [10 Theorem 44])
7 Torsion-Free Abelian Groups of Finite Rank
The torsion-free abelian groups of rank 1 were classified byBaer in 1937 The next year Kurosh and Malcev expanded onhis methods to give classifications for the torsion-free abeliangroups of ranks 2 and higher Their solution however wasconsidered inadequate because the invariants they providedwere no easier to distinguish than the groups themselves
In 1998 Hjorth proved using methods from the studyof Borel equivalence relations that the classification problemfor rank 2 torsion-free abelian groups is strictly harder thanthat for rank 1 (see [14]) However his work did not answerthe question of whether the classification problem for rank 2
groups is as complex as for all finite ranks or whether there ismore complexity that is to be found by looking at ranks 3 andhigher
Let119877(119899)denote the space of torsion-free abelian groups ofrank exactly 119899 that is the set of full-rank subgroups ofQ119899 Letcong119899denote the isomorphism relation on 119877(119899) In this section
ISRN Algebra 7
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
we will give a concise and essentially self-contained proof ofThomasrsquos theorem
Theorem 7 (Thomas [15]) For 119899 ge 2 one has that cong119899lies
properly below cong119899+1
in the Borel reducibility order
Thomasrsquos original argument used Zimmerrsquos superrigiditytheorem In this presentation we have essentially copiedhis argument verbatim with a few simplifications stemmingfrom the use of Ioanarsquos theorem instead of Zimmerrsquos theorem
The first connection between this result and the results ofthe last section is that for 119860 119861 isin 119877(119899) we have 119860 cong 119861 if andonly if there exists 119892 isin GL
119899(Q) such that 119861 = 119892(119860) Hence
the isomorphism relationcong119899is given by a natural action of the
linear group GL119899(Q) Unfortunately even restricting to just
the action of SL119899(Z) the space 119877(119899) is nothing like a profinite
space
The Kurosh-Malcev Invariants Although I have said thatthe Kurosh-Malcev invariants do not adequately classify thetorsion-free abelian groups of finite rank we will get aroundour difficulties byworkingwith theKurosh-Malcev invariantsrather than with the original space 119877(119899) The following is thekey result concerning the invariants see [16 Chapter 93] fora full account
Theorem 8 (Kurosh Malcev) Themap119860 997891rarr 119860119901= Z119901otimes119860 is
a GL119899(Q)-preserving bijection between the (full rank) 119901-local
subgroups of Q119899 and the (full rank)Z119901-submodules ofQ119899
119901The
inverse map is given by 119860119901997891rarr 119860 = 119860
119901cap Q119899
Here a subgroup of Q119899 is said to be 119901-local if it isinfinitely 119902-divisible for each prime 119902 = 119901 Kurosh andMalcevproved that a subgroup 119860 le Q119899 is determined by thesequence (119860
119901) this sequence is said to be the Kurosh-Malcev
invariant corresponding to 119860 It follows of course that 119860 isdetermined up to isomorphism by the orbit of (119860
119901) under
the coordinatewise action of GL119899(Q) (It is now easy to
see why these invariants serve as a poor classification suchorbits can be quite complex) All that we will need from thisclassification is the following corollary
Proposition 9 There exists a Borel reduction fromGL119899(Q)
P(Q119899
119901) to cong119899
Since GL119899(Q) P(Q
119899
119901) is closely related to a profinite
action Proposition 9will eventually enable us to apply Ioanarsquostheorem in the proof of Theorem 7
Sketch of Proof Given a linear subspace 119881 le Q119899119901 let 119881perp
denote its orthogonal complementThen there exists a vectorV such that 119881perp oplus Z
119901V is a full-rank submodule of Q119899
119901 By
Theorem 8 this module corresponds to an element 119891(119881) isin
119877(119899) This is how the Kurosh-Malcev construction is usedTo verify that it works one uses the fact that the Kurosh-
Malcev construction is GL119899(Q)-preserving together with the
technical fact if dim119882 = dim1198821015840 = 119899minus1 and119882oplusZ1199011199081198821015840oplus
Z1199011199081015840 are full-rank modules then 1198821015840 = 119892119882 for some
119892 isin GL119899(Q) actually implies that 1198821015840 oplus Z
1199011199081015840 = 119892(119882 oplus Z
119901119908)
for some 119892 isin GL119899(Q)
The Problem of Freeness Suppose now that 119899 ge 2
and that there exists a Borel reduction from cong119899+1
to cong119899 By Proposition 9 there exists a profinite
ergodic SL119899+1
(Z)-space 119883 (namely 119883 = P(Q119899+1
119901))
and a countable-to-one homomorphism oforbits 119891 from SL
119899+1(Z) 119883 to cong
119899 We can almost apply
Ioanarsquos theorem except that unfortunately cong119899
is notinduced by a free action of any group The following simpleobservation gives us an approach for getting around thisdifficulty
Proposition 10 Let119891 be a homomorphism of orbits from Γ
119883 into Λ 119884 Suppose that there exists a fixed 119870 le Λ suchthat for all 119909 isin 119883 stab
Λ(119891(119909)) = 119870Then119873
Λ(119870)119870 acts freely
on 119891(119883) and 119891 is a homomorphism of orbits from Γ 119883 into119873Λ(119870)119870 119891(119883)
Proof By definition we have that 119873Λ(119870)119870 acts on 119891(119883) by
120582119870 sdot 119910 = 120582119910 The action is free because 120582119910 = 119910 implies that120582 isin 119870 To see that 119891 is still a homomorphism of orbits justnote that if 119891(119909
1015840) = 120582119891(119909) then since stab119891(119909) =stab119891(1199091015840) =
119870 it follows that 120582 normalizes 119870
One can now formulate a strategy for proving Thomasrsquostheorem along the following lines
Claim 1 By passing to a conull subset of 119883 we can assumewithout loss of generality that for all 119909 we havestabGL
119899(Q)(119891(119909)) = some fixed 119870
Claim 2 There cannot exist a nontrivial homomorphismfrom (a finite index subgroup of) SL
119899+1(Z) into 119873GL
119899(Q)(119870)
119870
This would yield a contradiction since by Proposition 9and Claim 1 Ioanarsquos theorem would provide the nontrivialhomomorphism ruled out in Claim 2 Unfortunately thisapproach does not turn out to be a good one The reason isthat Claim 1 seems to be as difficult to be proved asTheorem 7itself Moreover Claim 2 is not known to be true in thisgenerality (In fact Claim 1 has recently been established byThomas in [10] but his proof actually requires all of thearguments below and more)Use Quasi-Isomorphism Instead To reduce the number ofpossibilities for stab(119891(119909)) = Aut(119891(119909)) we change categoriesfrom isomorphism to quasi-isomorphismWe say that groups119860 119861 le Q119899 are quasi-isomorphic written as 119860sim
119899119861 if and
only if 119861 is commensurable with an isomorphic copy of 119860Of course sim
119899is a courser relation than cong
119899 but it is easy to
check that it is still a countable Borel equivalence relation(indeed the commensurability relation is a countable relationin this case see [15 Lemma 32]) Hence the map 119891 fromabove is again a countable-to-one Borel homomorphismfrom SL
119899+1(Z) 119883 to sim
119899
Now rather than attempting to fix the automorphismgroup of 119891(119909) we will fix the quasiendomorphism ring
8 ISRN Algebra
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
QEnd(119860) of 119891(119909) Here if 119860 le Q119899 then 119892 isin GL119899(Q) is said
to be a quasiendomorphism of 119860 if 120601(119860) is commensurablewith a subgroup of 119860 (Equivalently 119899120601(119860) sub 119860 for some119899 isin N) Then unlike End(119860) it is clear that QEnd(119860)
is a Q-subalgebra of 119872119899times119899
(Q) It follows that there arejust countably many possibilities for QEnd(119891(119909)) since analgebra is determined by any Q-vector space basis for itHence there exists 119870 such that QEnd(119891(119909)) = 119870 for anonnull set of 119909 Arguing as in the proof of Corollary 5 wemay replace119883 by a conull subset and adjust 119891 to assume thatfor all 119909 isin 119883 we have QEnd(119891(119909)) = 119870
Thus we have successfully obtained our analog of Claim 1for quasi-isomorphism Indeed copying the arguments in theproof of Proposition 10 we see that 119891 is a homomorphism
119891 SL119899+1
(Z) 119883 997888rarr119873GL
119899(Q) (119870)
119870times 119891(119883)(19)
and that 119873GL119899(Q) (119870)119870times acts freely on 119891(119883) We may
therefore apply Ioanarsquos theorem to suppose that there existsa finite index subgroup Γ
0le PSL
119899+1(Z) a positive measure
1198830
sub 119883 and a homomorphism 120601 Γ0
rarr 119873GL119899(Q)(119870)119870times
such that for 119909 isin 1198830and 120574 isin Γ we have
Note that 120601 must be nontrivial since if 120601(Γ0) = 1 then this
says that 119891 is Γ0-invariant But then by ergodicity of Γ
0 1198830
119891 would send a conull set to one point contradicting that 119891is countable-to-oneA Dimension Contradiction The set theory is now overwe have only to establish the algebraic fact that the analogof Claim 2 holds there does not exist a nontrivial homo-morphism from Γ
0into 119873GL
119899(Q)(119870)119870times Again by Margulisrsquos
theorem on normal subgroups we can suppose that 120601 is anembedding Then using Margulisrsquos superrigidity theorem itsuffices to show that 119873GL
119899(Q)(119870)119870times is contained in an alge-
braic group of dimension strictly smaller than dim(PSL119899+1
) =
(119899 + 1)2
minus 1To see this first note that since the subalgebra 119870 of
119872119899times119899
(Q) is definable from a vector space basis we havethat 119870 = K(Q) where K is an algebraic Q-group inside119872119899times119899
Basic facts from algebraic group theory imply that119873GL
119899(Q)(119870) = N(Q) and 119870times = K1015840(Q) where again NK1015840
are algebraic Q-groups inside 119872119899times119899
Finally 119873GL119899(Q)(119870)119870times
is exactly N(Q)K1015840(Q) which is contained in the algebraicQ-group NK1015840 Since the dimension of an algebraic groupdecreases when passing to subgroups and quotients we have
dim(NK1015840
) le dim (119872119899times119899
) = 1198992
lt (119899 + 1)2
minus 1 (21)
as desired This completes the proof
References
[1] R J Zimmer Ergodic Theory and Semisimple Groups vol 81ofMonographs in Mathematics Birkhauser Basel Switzerland1984
[2] A Furman ldquoOrbit equivalence rigidityrdquoAnnals of MathematicsSecond Series vol 150 no 3 pp 1083ndash1108 1999
[3] N Monod and Y Shalom ldquoOrbit equivalence rigidity andbounded cohomologyrdquo Annals of Mathematics vol 164 no 3pp 825ndash878 2006
[4] Y Kida ldquoOrbit equivalence rigidity for ergodic actions of themapping class grouprdquoGeometriae Dedicata vol 131 pp 99ndash1092008
[5] S Popa ldquoCocycle and orbit equivalence superrigidity for mal-leable actions of120596-rigid groupsrdquo InventionesMathematicae vol170 no 2 pp 243ndash295 2007
[6] A Furman ldquoOn Poparsquos cocycle superrigidity theoremrdquo Interna-tional Mathematics Research Notices IMRN no 19 2007
[7] A Ioana ldquoCocycle superrigidity for profinite actions of prop-erty (T) groupsrdquo Duke Mathematical Journal vol 157 no 2 pp337ndash367 2011
[8] S Coskey ldquoThe classification of torsion-free abelian groups offinite rank up to isomorphism and up to quasi-isomorphismrdquoTransactions of the AmericanMathematical Society vol 364 no1 pp 175ndash194 2012
[9] S Coskey ldquoBorel reductions of profinite actions of 119878119871119899(Z)rdquo
Annals of Pure and Applied Logic vol 161 no 10 pp 1270ndash12792010
[10] S Thomas ldquoThe classification problem for 120575-local torsion-freeabelian groups of finite rankrdquo Advances in Mathematics vol226 no 4 pp 3699ndash3723 2011
[11] S Gao Invariant Descriptive Set Theory vol 293 of Pure andApplied Mathematics CRC Press Boca Raton Fla USA 2009
[12] S Thomas ldquoSuperrigidity and countable Borel equivalencerelationsrdquo Annals of Pure and Applied Logic vol 120 no 1ndash3pp 237ndash262 2003
[13] A S KechrisGlobal Aspects of Ergodic GroupActions vol 160 ofMathematical Surveys and Monographs American Mathemati-cal Society Providence RI USA 2010
[14] G Hjorth ldquoAround nonclassifiability for countable torsion freeabelian groupsrdquo in Abelian Groups and Modules pp 269ndash292Birkhauser Basel Switzerland 1999
[15] S Thomas ldquoThe classification problem for torsion-free abeliangroups of finite rankrdquo Journal of the American MathematicalSociety vol 16 no 1 pp 233ndash258 2003
[16] L Fuchs Infinite Abelian Groups Academic Press New YorkNY USA 1973 Pure and Applied Mathematics Vol 36-II
