countable abelian group actions and hyperfinite equivalence relations

1 23

Inventiones mathematicae ISSN 0020-9910Volume 201Number 1 Invent. math. (2015) 201:309-383DOI 10.1007/s00222-015-0603-y

Countable abelian group actions andhyperfinite equivalence relations

Su Gao & Steve Jackson

1 23

Your article is protected by copyright and

all rights are held exclusively by Springer-

Verlag Berlin Heidelberg. This e-offprint is

for personal use only and shall not be self-

archived in electronic repositories. If you wish

to self-archive your article, please use the

accepted manuscript version for posting on

your own website. You may further deposit

the accepted manuscript version in any

repository, provided it is only made publicly

available 12 months after official publication

or later and provided acknowledgement is

given to the original source of publication

and a link is inserted to the published article

on Springer's website. The link must be

accompanied by the following text: "The final

publication is available at link.springer.com”.

Invent. math. (2015) 201:309–383DOI 10.1007/s00222-015-0603-y

Countable abelian group actions and hyperfiniteequivalence relations

Su Gao1 · Steve Jackson1

Received: 16 August 2008 / Accepted: 28 August 2013 / Published online: 28 May 2015© Springer-Verlag Berlin Heidelberg 2015

Abstract An equivalence relation E on a standard Borel space is hyperfiniteif E is the increasing union of countably many Borel equivalence relationsEn where all En-equivalence classs are finite. In this article we establish thefollowing theorem: if a countable abelian group acts on a standard Borel spacein a Borel manner then the orbit equivalence relation is hyperfinite. The proofuses constructions and analysis of Borel marker sets and regions in the space2Z

<ω. This technique is also applied to a problem of finding Borel chromatic

numbers for invariant Borel subspaces of 2Zn.

Contents

1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3112 Clopen marker sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3143 Regular marker regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3194 An application of regular marker regions . . . . . . . . . . . . . . . . . . . . . . . 3315 Orthogonal marker regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

S. Gao acknowledges United States NSF grants DMS-0501039, DMS-0901853 andDMS-1201290 for the support of his research. S. Jackson acknowledges United States NSFgrants DMS-0901853 and DMS-1201290 for the support of his research.

B Steve [email protected]

Su [email protected]

1 Department of Mathematics, University of North Texas,1155 Union Circle #311430, Denton, TX 76203, USA

123

Author's personal copy

http://crossmark.crossref.org/dialog/?doi=10.1007/s00222-015-0603-y&domain=pdf

310 S. Gao, S. Jackson

6 Hyperfiniteness of F(Z<ω) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3417 The non-free part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348

7.1 Preliminaries about the space XH . . . . . . . . . . . . . . . . . . . . . . . . 3497.2 Marker regions in XH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3547.3 Hyperfiniteness of EH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

8 Actions of countable abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . 3689 Continuous embeddings into E0 . . . . . . . . . . . . . . . . . . . . . . . . . . . 37010 Open problems and further remarks . . . . . . . . . . . . . . . . . . . . . . . . . . 381References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

This paper is a contribution to the study of countable Borel equivalencerelations on Polish spaces. A topological space X is Polish if it is separableand completely metrizable. An equivalence relation E on X is Borel if Eis a Borel subset of X × X ; it is countable if each E-equivalence class iscountable. By a well known theorem of Feldman andMoore [5] any countableBorel equivalence relation E on a Polish space X is the orbit equivalencerelation of a Borel action of a countable group G.

Among all countable Borel equivalence relations the class of hyperfiniteequivalence relations is best understood. An equivalence relation E is hyper-finite if E = ⋃

n En where En ⊆ En+1 and each En is a Borel equivalencerelation with all equivalence classes finite. A canonical example of a hyperfi-nite equivalence relation is the relation E0 defined on the Cantor space 2ω asfollows:

xE0y ⇐⇒ ∃m ∀n ≥ m (x(n) = y(n)).

It is also well known (c.f. [4]) that any hyperfinite equivalence relation isthe orbit equivalence relation of a Borel action of Z. In fact, any Borel Z-action gives rise to a hyperfinite equivalence relation. Other characterizationsof hyperfinite equivalence relations and complete abstract classifications ofhyperfinite equivalence relations can also be found in [4].

However, the following has been an outstanding open problem since 1980s:For which countable groups G do we have that any Borel G-action gives riseto only hyperfinite equivalence relations? Based on results in the measuretheory context [3,10] Weiss [12] asked if this is true for any amenable group.He also proved it for Z

n. A more general result by Jackson et al. [8] showsthis for any finitely generated group with polynomial growth (by a famoustheorem of Gromov this is exactly the class of finitely generated groups thatare nilpotent-by-finite).

The main theorem of this paper is to prove that a Borel action of any count-able abelian group gives rise to a hyperfinite equivalence relation. To achievethis it suffices to show the hyperfiniteness of a specific orbit equivalence rela-tion, that is, the shift action of Z

<ω on the space of its subsets 2Z<ω

. Much ofthe effort is to reproveWeiss’ result that the shift action ofZ

n on 2Zngives rise

to a hyperfinite equivalence relation. In the new proof a theory of Borel marker

123


Countable abelian group actions and hyperfinite… 311

sets and regular Borel marker regions is developed. This allows us to provestronger results such as that there is a continuous (even computable) reductionfrom the shift action of Z

n into E0. We also present another application of thistheory to a problem on Borel chromatic numbers.

1 Preliminaries

In this first section we recall some general definitions and fix the notation tobe used throughout the paper. More ad hoc definitions and notation will beintroduced in later sections.

If E is an equivalence relation on the Polish space X, and F an equivalencerelation on the Polish space Y, then we say E is Borel reducible to F, writtenE ≤B F, if there is a Borel function f : X → Y such that

∀x, y ∈ X (xEy ↔ f (x)F f (y)).

The function f is called a Borel reduction from E to F, and we also say that fBorel reduces E to F. Any Borel reduction from E to F induces an injectionfrom the quotient space X/E into Y/F. We say E is Borel embeddable intoF, written E �B F, if there is a one-to-one Borel function f : X → Ywhich reduces E to F. In this case we call f a Borel embedding of E intoF. The notion of Borel reduction is fundamental in the theory of definableequivalence relations. Sometimes the reduction function f can be continuousor even computable (or recursive). In these cases we will speak of continuous(resp. computable) reductions and continuous (resp. computable) embeddings.

A particular case of interest is when the equivalence relation is generatedby a Polish group action. Let G be a Polish group, X a Polish space anda : G × X → X a continuous action of G on X. In this context X is called aPolish G-space. We use the abbreviation g · x to denote a(g, x) if there is nodanger of confusion.

The action is free if for any g ∈ G with g = 1G and any x ∈ X, g · x = x .For any Polish G-space X the free part of the action is the subset of X definedby

FG(X) = {x ∈ X : ∀g ∈ G (g = 1G → g · x = x)}.In general FG(X) is an invariant �1

1 subset of X. In case G is countable (andthe action is continuous) FG(X) is an invariant Gδ subset of X, and thereforeFG(X) naturally becomes a Polish G-space.If X is a Polish G-space, the orbit equivalence relation, denoted by EX

G , isdefined as follows: for x, y ∈ X,

xEXG y ⇐⇒ ∃g ∈ G (g · x = y).

123



In case the action is free, EXG is a Borel equivalence relation (as a subset of

X × X ). Alternatively, if G is countable (and the action is continuous), theorbit equivalence relation EX

G is Fσ . If Y is a subset of X, then we denote theequivalence relation EX

G � (Y × Y ) simply by EXG � Y.

If X is a Polish G-space and x ∈ X, we denote the orbit or the equivalenceclass of x by

[x]G = {g · x : g ∈ G}.In general, if E is an equivalence relation on X and x ∈ X, we let [x]E denotethe E-equivalence class of x :

[x]E = {y ∈ X : yEx}.In case there is no danger of confusion we will omit all the subscripts.

Throughout the paper we will consider only countable groups G, which arePolish groups with the discrete topology. In fact, we will mostly consider aspecific Polish G-space. Let 2G denote the power set of G, or equivalentlythe product space

∏g∈G{0, 1} with the product topology, where 2 = {0, 1} is

equipped with the discrete topology. 2G is a Polish space. If D ⊆ G is a finitesubset then a function s : D → 2 determines a basic (cl)open set

Ns = {x ∈ 2G : ∀g ∈ D (x(g) = s(g))}.For x ∈ Ns we also write s ⊆ x . We also denote the set

{s : D → 2 : D ⊆ G finite}by 2<G . For s, t ∈ 2<G, we say that s and t are incompatible if there isg ∈ dom(s) ∩ dom(t) with s(g) = t (g); otherwise s and t are said to becompatible. We write s ⊆ t and say that t extends s if s and t are compatibleand dom(s) ⊆ dom(t).

Consider the natural shift action of G on 2G : for A ⊆ G and g ∈ G,

g · A = {gh : h ∈ A}.This action is continuous. For conveniencewewill denote the orbit equivalencerelation E2G

G simply by E(G). We also denote by F(G) the free part of thisaction as well as the equivalence relation E(G) � F(G).

We will also work with the natural shift action of G on 2<G : if g ∈ G ands : D → 2 where D ⊆ G is finite, then g · s : g · D → 2 and for h ∈ g · D,

(g · s)(h) = s(g−1h).

123



Most of the time we work with groups G = Zn for some positive integer n.

They will be expressed as additive groups as usual. When n is fixed we denoteby e1, e2, . . . , en the group elements

e1 = (1, 0, . . . , 0),

e2 = (0, 1, . . . , 0),

· · · · · ·en = (0, 0, . . . , 1).

We will refer to them as the generators of Zn. Any g ∈ Z

n can be uniquelyexpressed as (g1, g2, . . . , gn) or g1e1+g2e2+· · ·+gnen for g1, g2, . . . , gn ∈Z. We define a norm on Z

n by

‖g‖ = ‖(g1, g2, . . . , gn)‖ = max{|g1|, |g2|, . . . , |gn|}.

The norm induces a distance function on Zn: for g, h ∈ Z

n,

ρ(g, h) = ‖g − h‖.

When G = Zn acts on X = F(G) = F(Zn) the above distance function

induces further a distance function on X as follows:

ρX (x, y) ={‖g‖, if g · x = y,

∞, if (x, y) ∈ E(G).

Note that this is well defined since the action is free. When there is no dangerof confusion we will drop the subscript X in denoting ρX . When M ⊆ X andx ∈ X, we also define, as usual,

ρ(x, M) = inf{ρ(x, y) : y ∈ M}.

In addition, if S1, S2 ⊆ X, we also let

ρ(S1, S2) = inf{ρ(x, y) : x ∈ S1, y ∈ S2}.

This distance function is not to be confused with the genuine metric d on2Z

ndefined by

d(x, y) ={0, if x = y,2−min{‖g‖: x(g) =y(g)}, otherwise.

123



Note that d is in fact an ultrametric, i.e.,

d(x, y) ≤ max{d(x, z), d(y, z)}

for any x, y, z ∈ 2Zn.

The set of natural numbers (nonnegative integers) will be denoted as ω. Wealso use the interval notation: if a, b ∈ Z and a < b, then [a, b] denotes theset {c ∈ Z : a ≤ c ≤ b}.

Throughout the paper we also fix a linear order<n onZn so that ‖g‖ < ‖h‖

implies g <n h. For instance, elements of Z can be enumerated according to<1 as

0,−1, 1,−2, 2, . . . ,−k, k, . . . .

The equivalence relation E0 on 2ω defined previously is the prototypicalhyperfinite equivalence relation. All of the proofs of hyperfiniteness in thispaper will proceed by showing that the equivalence relation E reduces to (orembeds into) E0. In this regard we note the simple fact that it does not matterif we use E0 = E0(2ω) on 2ω or the variation E0(ω

ω) on ωω, defined as theeventual equality relation on sequences of natural numbers. We state this inthe following proposition.

Proposition 1.1 There is a continuous (in fact computable or recursive)embedding from E0(ω

ω) to E0(2ω).

Proof Let π : ω → ω × ω be a computable bijection between ω and ω × ω.

Let π(n) = (π0(n), π1(n)). For every k ∈ ω, let ((k)0, (k)1, . . .) be the base-2expansion of k, that is, k = ∑

i (k)i ·2i (so for large enough i, (k)i = 0). Definef : ωω → 2ω by: f (x)(n) = (x(π0(n)))π1(n). f is easily an embedding, andis computable. ��

2 Clopen marker sets

Throughout this section we fix a positive integer n.

Lemma 2.1 (Basic clopen marker lemma) Let d be a positive integer. Thenthere is a relatively clopen set S ⊆ F(Zn) such that

(i) if x, y ∈ S are distinct, then ρ(x, y) > d;(ii) for any x ∈ F(Zn), ρ(x, S) ≤ d.

Proof Let s0, s1, . . . enumerate all elements s of 2<Znsuch that g ·Ns∩Ns = ∅

for all g ∈ Zn with 0 < ‖g‖ ≤ d. Define a sequence of sets Si by induction

on i. For i = 0 let S0 = Ns0 . For i > 0 let

123



Si = Si−1 ∪⎛

⎝Nsi −⋃

‖g‖≤d

g · Si−1

⎞

⎠ .

Finally let S = ⋃i Si ∩ F(Zn).

By induction on i we show that if x, y ∈ Si are distinct, then ρ(x, y) > d. Ifi = 0 and x, y ∈ S0 then x, y ∈ Ns0 . Suppose ρ(x, y) ≤ d. In particular thereis g ∈ Z

n with g · x = y and ‖g‖ ≤ d. This implies that g · Ns0 ∩ Ns0 = ∅,

contradicting our choice of s0. Thus we have ρ(x, y) > d. Now supposei > 0 and x, y ∈ Si . A similar argument as above shows that x and y cannotbe both in Nsi . By induction we may also assume that x and y are not bothin Si−1. Thus without loss of generality we may assume that x ∈ Si−1 − Nsiand y ∈ Nsi − Si−1. Assume ρ(x, y) ≤ d. In particular there is g ∈ Z

n withg · x = y and ‖g‖ ≤ d. It follows that y ∈ ⋃

‖g‖≤d g · Si−1, contradictingy ∈ Si . Thus again we have ρ(x, y) > d.

To verify clause (ii) note that for any x ∈ F(Zn) there is si with x ∈ Nsi .

In fact, if no such si exists then there is g ∈ Zn with 0 < ‖g‖ ≤ d such

that g · Ns ∩ Ns = ∅ for all s ⊆ x . The continuity of the action then impliesthat g · x = x, contradicting the assumption that the action is free. Now letx ∈ F(Zn) and let i be the smallest such that x ∈ Nsi . If x ∈ Si then x ∈ Sand ρ(x, S) = 0, therefore we are done. Otherwise, assume x ∈ Si . By thedefinition of Si it follows that x ∈ ⋃

‖g‖≤d g · Si−1. This is to say that thereare g ∈ Z

n with ‖g‖ ≤ d and y ∈ Si−1 such that g · y = x, and thusρ(x, S) ≤ ρ(x, y) ≤ d.

Finally we check that S is clopen in F(Zn). By induction on i it is easyto see that each Si is clopen. Thus it follows immediately from the definitionthat S is open in F(Zn). The argument in the preceding paragraph gives thatif x ∈ F(Zn)− S then x ∈ ⋃

0<‖g‖≤d g · S. Conversely if x ∈ ⋃0<‖g‖≤d g · S

then by (i) we have that s ∈ S. Thus x ∈ F(Zn) − S iff x ∈ ⋃0<‖g‖≤d g · S.

Since S is open and the action is continuous, this shows that F(Zn) − S isopen. ��Remark 2.1 (1) In the above proof the set S′ = ⋃

i Si is in fact an open setin 2Z

nsatisfying both clauses (i) and (ii). A closer scrutiny shows that

S′ is �01 and S is �0

1 in F(Zn). In [2] it was already shown that thereis a continuous reduction from E(Z) to E0 on 2ω. Enhanced by thesecomputability considerations one can show that there is a computablereduction from E(Z) to E0. Later in Sect. 9 we generalize this to E(Zn)

for all finite n ≥ 1.(2) Clause (ii) implies that S is in fact a complete section for the action ofZn on

F(Zn), that is, S meets every equivalence class (i.e., for any x ∈ F(Zn),

S ∩ [x] = ∅). Elements of S are customarily called markers or markerpoints. The set S itself is called a marker set.

123



(3) In the remainder of this article we will refer to the properties (i) and (ii)as basic marker properties. The integer d will be referred to as a markerdistance and the set S the basic marker set for the marker distance d.

Marker arguments started from Slaman–Steel’s proof of hyperfiniteness ofE(Z) in [11]. In this proof an infinite descending and vanishing sequence ofBorel marker sets was constructed and then used to show that E(Z) is anincreasing union of finite Borel equivalence relations. In our context if F(Z)

admits a sequence

S0 ⊇ S1 ⊇ · · · ⊇ Sk ⊇ · · ·

of descending clopen marker sets so that⋂

k Sk = ∅, then the Slaman–Steelargument would give another continuous reduction from F(Z) to E0 (differentfrom the one constructed in [2]). Thus it is a natural question whether such asequence of clopen marker sets exists.

We first note that basic clopen marker sets can always be thinned down byspecifying a sufficiently larger marker distance.

Lemma 2.2 Let d be a positive integer and S0 be a basic clopen marker setin F(Zn) for the marker distance d. Let D > 2d be a positive integer. Thenthere is a clopen set S1 ⊆ S0 such that

(i) if x, y ∈ S1 are distinct, then ρ(x, y) > D − 2d;(ii) for any x ∈ F(Zn), ρ(x, S1) ≤ D + d.

Proof Let S′ be a basic clopen marker set for the marker distance D. Let <

be a linear order of the (finite) set B = {g ∈ Zn : ‖g‖ ≤ d}. Then for x ∈ S0

let x ∈ S1 iff there is g ∈ B such that g · x ∈ S′ and that for all h ∈ B withh < g, h−1g · x ∈ S0. It is clear from the definition and the continuity of theaction that S1 is clopen. It is useful to note that for each z ∈ S′ there is exactlyone x ∈ S1 with ρ(x, z) ≤ d. In fact, by the basic marker properties of S0, foreach z ∈ S′ there is at least one x ∈ S0 with ρ(x, z) ≤ d, and thus the set

{g ∈ B : g−1 · z ∈ S0}

is nonempty. Let g be the <-least element of this set and let x = g−1 · z.Then for all h ∈ B with h < g, h−1g · x = h−1 · z ∈ S0. Therefore x is theunique element of S1 with ρ(x, z) ≤ d. Conversely, for any x ∈ S1 there isalso exactly one z ∈ S′ with ρ(x, z) ≤ d, by a similar proof.

Now the conditions (i) and (ii) are easy to check. Let x, y ∈ S1 be distinctand let u, v ∈ S′ be such that ρ(x, u), ρ(y, v) ≤ d. Then by the basic markerproperties of S′, ρ(u, v) > D. Thus ρ(x, y) ≥ ρ(u, v)−ρ(x, u)−ρ(y, v) >

D−2d.On the other hand, let x ∈ F(Zn) and z ∈ S′ be such that ρ(x, z) ≤ D.

123



Let y ∈ S1 be such that ρ(y, z) ≤ d. Then ρ(x, S1) ≤ ρ(x, y) ≤ ρ(x, z) +ρ(y, z) ≤ D + d. ��

Note that we cannot guarantee that S1 has the exact basic marker propertiesfor any distance. However, when D � d, the terms D−2d and D+d are bothfairly close to D. For practical purposes these approximate marker propertiesare good enough for our arguments. For notational convenience we will usethe following terminology. Let d be a positive integer and 0 < ε < 1 be areal number. A marker set S ⊆ F(Zn) is said to satisfy the (d, ε)-markerproperties if

(i) if x, y ∈ S are distinct, then ρ(x, y) > (1 − ε)d;(ii) for any x ∈ F(Zn), ρ(x, S) ≤ (1 + ε)d.

This will allow us to iterate the argument in the above proof and generate aninfinite descending sequence of clopen marker sets with arbitrarily prescribedapproximate marker properties.

Lemma 2.3 Let d be a positive integer, 0 < ε < 1 a real number and S0a clopen marker set in F(Zn) satisfying the (d, ε)-marker properties. Let0 < δ < 1 be a real number. Then there is a positive integer D and a clopenset S1 ⊆ S0 satisfying the (D, δ)-marker properties.

Proof It is easy to see if we choose D so that δD > 2d(1 + ε) and S′ to bea basic clopen marker set for the marker distance D, then the proof of thepreceding lemma gives the desired S1 ⊆ S0 with appropriate approximatemarker properties. ��

In this lemma it is obvious that S1 will be a proper subset of S0 if D issufficiently large. Better yet, we can guarantee that S1 is sparse in S0 (on eachequivalence class under an appropriately defined density notion). By iteratingthe lemmawe can thus create an infinite descending sequence of clopenmarkersets each of which is sparse in the previous set. If S is the intersection of allthese clopen marker sets then for each x ∈ F(Zn) either S∩[x] = ∅ or S∩[x]is a singleton. We have thus proved the following fact.

Proposition 2.4 There exists an infinite descending sequence of clopen com-plete sections of F(Zn)

S0 ⊇ S1 ⊇ · · · ⊇ Sk ⊇ · · ·such that for each x ∈ F(Zn), |⋂k Sk ∩ [x]| ≤ 1.

Note that⋂

k Sk ∩ [x] = ∅ must hold for some x ∈ F(Zn), since theequivalence relation F(Zn) is non-smooth. The following theorem shows that⋂

k Sk ∩ [x] = ∅ cannot hold for every x ∈ F(Zn), in other words,⋂

k Sk ∩

123



[x] = ∅ must also hold for some x ∈ F(Zn). This gives a negative answerto the question following Remark 2.1, and hence a continuous reduction fromF(Z) to E0 has to come from an argument different than Slaman–Steel’s.

Theorem 2.5 (c.f. [6]) Let S0 ⊇ S1 ⊇ · · · ⊇ Sk ⊇ · · · be an infinite descend-ing sequence of closed complete sections of F(Zn). Then

⋂k Sk = ∅.

The theorem is a special case of the main theorem proved in [6] as theauthors’ joint work with B. Seward, which we state below.

Theorem 2.6 (Gao et al. [6]) Let G be a countable group and let S0 ⊇ S1 ⊇· · · ⊇ Sk ⊇ · · · be an infinite descending sequence of closed complete sectionsof F(G). Then

⋂k Sk = ∅.

A more direct proof of Theorem 2.5 can be found in the recent [7],Section 3.3.

The following two propositions rule out the existence of marker sets whichare “too regular”. The proof essentially uses the fact that the Z

n-action onF(Zn) is strongly mixing and therefore the action of any subgroup is ergodic.

Proposition 2.7 There is no Borel marker set M ⊆ F(Z) such that for anyx ∈ F(Z) there is some d > 1 such that for all y, z ∈ [x] ∩ M, ρ(y, z) is amultiple of d.

Proof Let μ be the product measure on 2Z. Then μ is a Z-invariant, non-atomic, ergodic, Borel probability measure. Its restriction to F(Z), which westill denote by μ, continues to be such a measure on F(Z). Now assume Mis a Borel marker set with the stated property. Then for each x ∈ F(Z) letdx > 1 be the largest integer such that ρ(y, z) is a multiple of dx for ally, z ∈ M ∩ [x]. For each integer d > 1 let Xd = {x ∈ F(Z) : dx = d}.Also let X1 = {x ∈ F(Z) : |[x] ∩ M | = 1}. Then {Xd : d ≥ 1} form acountable partition of M into Z-invariant Borel subsets. Since M ∩ X1 is aBorel transversal on X1, μ(X1) = 0. By ergodicity there is some d > 1 suchthat μ(Xd) = 1.

Let Md = (dZ) · (M ∩ Xd). Then Md is a Borel complete section ofF(Z) � Xd so that for any x ∈ Md and y ∈ [x], y ∈ Md iff ρ(x, y) is amultiple of d. It follows that Xd = Md ∪ 1 · Md ∪ · · · ∪ (d − 1) · Md .

Now consider the action of dZ on Xd . Then μ � Xd is dZ-invariant, non-atomic, and ergodic. Moreover, the sets Md , . . . , (d−1) ·Md are dZ-invariantBorel subsets with the same μ-measure. This is a contradiction since on theone hand each of them should have measure 1/d, and on the other hand eachshould have measure either 0 or 1. ��

The following generalization can be established by a similar proof.

123



Proposition 2.8 There is no Borel marker set M ⊆ F(Zn) such that for anyx ∈ M there is a proper subgroup G of Z

n such that [x] ∩ M ⊆ G · x .In contrast to this we note that it is possible to construct marker sets with

a moderately regular pattern. This is made precise by the following result.The proof of the following lemma also provides a preview of a technique,called big-marker-little-marker argument, which will be important in laterconstructions.

Lemma 2.9 For any integer d > 0 there is a clopen marker set Md ⊆ F(Z)

such that

(i) for all z ∈ F(Z) there is x ∈ Md with ρ(x, z) ≤ d, and(ii) for all x, y ∈ Md , if ρ(x, y) < 2d + 1 then ρ(x, y) ∈ {d + 1, d + 2}.Proof Let D = (d + 1)2 and MD be the clopen marker set obtained fromapplying the Basic Clopen Marker Lemma. For each i ∈ [D + 1, 2D + 1] let

MD,i = {x ∈ MD : i · x ∈ MD}.Then {MD,i | i ∈ [D + 1, 2D + 1]} form a partition of MD into clopen sets.Now for each i ∈ [D + 1, 2D + 1] let

i = qi (d + 1) + ri , where qi , ri ∈ ω and 0 ≤ ri < d + 1.

Here qi ≥ d since i > D = (d + 1)2. Thus one can write

i = (qi − ri )(d + 1) + ri (d + 2) = ai (d + 1) + bi (d + 2)

for some integer ai , bi ≥ 0. Let

Gi = {k(d + 1) | k ≤ ai } ∪ {ai (d + 1) + k′(d + 2) | k′ < bi }.Define

Md,i = MD,i ∪ Gi · MD,i ,

and Md = ⋃i Md,i . Then it is easy to check that Md is as required. ��

A generalization of this is the main theorem of the next section.

3 Regular marker regions

Recall that in general a marker set is a Borel complete section of F(Zn).

In the preceding section we studied the existence of marker sets for F(Zn)

123



· · · · · ·

Fig. 1 Marker intervals with marker points highlighted

with various additional properties. The main new property considered is theclopenness. Marker sets constructed in earlier research have been successfullyapplied to show hyperfiniteness of equivalence relations. Slaman and Steel[11] constructed Borel marker sets for arbitrary countable Borel equivalencerelations and used the marker sets for F(Z) to show that E(Z) is hyperfinite.Later several proofs of hyperfiniteness of E(Zn) for n > 1 were found, but allof them are based on constructions of Borel marker sets for F(Zn). In all ofthese proofs, an important technique is to decompose the equivalence classesinto finite marker regions. We recall two examples below, but first we makea notational point to be used throughout. When we represent an equivalenceclass [x] ∈ F(Zn) pictorially, we arrange the elements of [x] in a Z

n gridaccording to the group action. For example, for n = 2, the element (1, 0) · x isthe point immediately to the right of x and (0, 1) · x is the point immediatelyabove x . We cannot, in a Borel manner, pick a point x0 ∈ [x] to serve asan “origin” for this representation, but we can nevertheless represent [x] (orsubsets of [x], etc.) in this manner.

Example 3.1 Let M ⊆ F(Z) be a Borel marker set. For each x ∈ F(Z) define

lM(x) �{

(−n) · x, if n ∈ ω is the least such that (−n) · x ∈ M,

undefined, if no such n exists.

Then define a subequivalence relation RM of F(Z) by

x RM y ⇐⇒ [x] = [y] and lM(x) � lM(y).

(Recall that if f and g are partial functions, then f (x) � g(x) if both f (x)and g(x) are defined and f (x) = g(x), or both f (x) and g(x) are undefined.)It can be easily seen that for each x ∈ F(Z), the RM -equivalence class of xis of the form I · x for an interval I in Z (which is possibly infinite). For thisreason we say that RM gives a partition of all F(Z)-equivalence classes intomarker intervals (see Fig. 1).

Conversely, suppose R is a Borel subequivalence relation of F(Z) so thatfor every x ∈ F(Z), [x]R = [x] and [x]R = I · x for some interval I in Z.

Then by taking all the left end points of the intervals I we recover a markerset MR . It is also easy to see that the above procedures are inverses of eachother, i.e., RMR = R and MRM = M.

Example 3.2 Let M ⊆ F(Zn) be a Borel marker set. For each x ∈ F(Zn)

define cM(x) = g ·x,where g is the<n-least element ofZn such that g ·x ∈ M.

123



· · · · · ·

Fig. 2 An alternative construction of marker intervals

markers and perpendicular bisectors markers and Dirichlet regions

Fig. 3 Two-dimensional Dirichlet regions

Note that since M is a complete section of F(Zn), cM is always defined. Thenwe can again define a subequivalence relation RM of F(Zn) by

x RM y ⇐⇒ cM(x) = cM(y).

When n = 1, each of the RM -equivalence classes is still a marker interval;however they are different from the ones constructed in the preceding example(see Fig. 2). When n > 1, the RM -equivalence classes are called Dirichletregions.

The terminology comes from a geometric construction in Rn, which we

describe informally below. For this discussion, we use the standard Euclideanmetric onR

n and Zn (although we could use the sup norm we fixed earlier; the

description is a little cleaner if we use the standard metric). Fix x ∈ F(Zn).

Write M ∩ [x] = J · x for J ⊆ Zn. Regard J as a subset of R

n. For eachpair of distinct elements (p, q) of J find the perpendicular bisector (in theusual metric of R

n) of the line segment pq,which is a hyperplane in Rn. Then

the Dirichlet regions in Rn (see Fig. 3) are polygonal regions with segments

of these perpendicular bisectors as boundaries. They are characterized by theproperty that each region J0 contains a unique element p0 of J in its interiorso that J0 consists of points whose distance to p0 is smaller than its distanceto any other points in J .

In particular let J0 be the region containing the origin. Then (J0∩Zn)·x is the

Dirichlet region of x given by RM .We remark that the construction is invariant.The precise meaning of this invariance is that, if we instead fix x ′ = g · x withg = 0, then the Dirichlet regions constructed in R

n for x ′ are the result of

123



applying a shift by g−1 to the Dirichlet regions constructed in Rn for x . Thus

by following the construction for x we will be able to recover the Dirichletregion for x ′. We also remark that the discussion in this paragraph only givesthe main idea of the construction. The details are incomplete, because we didnot address what to do if a lattice point in R

n falls exactly on the boundary of aregion. In this case the lattice point will be regarded as being contained in onlyone of the Dirichlet regions, and the conflict (among all the Dirichlet regionshaving this element on the boundary) is resolved according to the linear order<n . We omit the precise details for now. We note that the arguments in thispaper will not use Dirichlet regions, but regions which are “rectangles”. Wediscuss the meaning of this below.

These examples illustrate how the geometries of Rn come into play in the

study of hyperfiniteness of countable equivalence relations. In order to explorethis connection more easily we fix some terminology to be used for the restof the paper. In general bymarker regionswe mean the R-equivalence classesfor some Borel subequivalence relation R of F(Zn). Now fix an x0 ∈ F(Zn).

From the obvious correspondence [x0] = Zn · x0 the F(Zn)-equivalence class

[x0] can be given a geometric structure identical to that of Zn. Moreover, for

any x ∈ [x0], there is a unique J ⊆ Zn so that [x]R = J · x0, and thus

the marker region of x, [x]R, also inherits a geometric structure identical tothat of J in Z

n. If x1 ∈ [x0], say x1 = g · x0, then from [x0] = Zn · x1

and [x]R = J ′ · x1 we get another copy of Zn and a subset J ′ representing

the corresponding geometric structures of [x0] and the marker region of x .However, since J = gJ ′ the said geometric structures are identical up to ashift.

Very often the marker regions we consider come from a marker set M ofF(Zn). For this reason the Borel subequivalence relation is often denoted RM .

As in the example of Dirichlet regions it is often the case that RM comes froma geometric construction on the background geometric structure of R

n. Andwe will employ the usual terminology of solid geometry when we describethese geometric constructions in R

n. For example, the Dirichlet regions arereferred to as polygonal regions. Also, we say a marker region is a rectangle(or a square, or a cube, etc.) if it is such inZ

n, or more generally it comes fromsuch in R

n in the sense of our description of this paragraph and the precedingone.

Ourmain objective of this section is to constructmarker regionswith regulargeometric shapes. Our main theorem belowwill be a construction of rectangu-lar marker regions. Moreover, we also restrict the choices for edge lengths ofthe rectangles as much as we can. By doing all this we obtain marker regionswith a regular geometry. This turns out to be crucial in further constructionsin hyperfiniteness proofs.

123



Theorem 3.1 Let d > 0 be an integer. Then there is a subequivalence relationRnd of F(Zn) such that Rn

d is relatively clopen and the Rnd -marker regions are

n-dimensional rectangles with edge lengths either d or d + 1.

Remark 3.3 When we say that Rnd is (relatively) clopen, we actually mean

that {(x, g) ∈ F(Zn) × Zn : g · x Rn

d x} is a clopen subset of F(Zn) × Zn.

Equivalently, for every g ∈ Zn, the set {x ∈ F(Zn) : g · x Rn

d x} is a clopensubset of F(Zn). We will use this terminology throughout.

The proof of the theorem uses the big-marker-little-marker method. Wewill present the proof in several steps. As a first step we show that, in order toobtain the almost cubical marker regions of edge lengths approximately d, itis enough to create marker regions which are n-dimensional rectangles eachof which has edge lengths much bigger than d. The basic idea of the proof ofthe following lemma is similar to that of Lemma 2.9.

Lemma 3.2 Let d > 0 and D > d2 be integers. Let RD be a subequivalencerelation of F(Zn) so that the RD-marker regions are n-dimensional rectangleswith edge lengths greater than D. Then there is a subequivalence relationRd ⊆ RD so that every RD-marker region is partitioned into Rd-markerregions, which are n-dimensional rectangles with edge lengths either d ord + 1. Moreover, if RD is clopen and there is � > D so that each RD-markerregion has edge lengths ≤ �, then Rd can also be clopen.

Proof As we have seen in the proof of Lemma 2.9, every integer l > d2 canbe written as ad + b(d + 1) with a, b ≥ 0. It follows that an n-dimensionalrectangle, for instance [0, l1] × · · · × [0, ln] with l1, . . . , ln > d2, can betiled by n-dimensional rectangles with edge lengths either d or d + 1. Thistiling corresponds to a finite equivalence relation Rl1,...,ln on the finite set[0, l1] × · · · × [0, ln].

Now for each x ∈ F(Zn) let

�l(x) = (l1(x), . . . , ln(x))

where for each 1 ≤ i ≤ n,

li (x) = |{a ∈ Z : (aei · x)RDx}|.Since the RD marker regions are rectangles, the function �l(x) measures thelengths of the sides the region containing x . It is clear that for any x RDy,�l(x) = �l(y). Similarly we can also measure the relative position of eachelement in its marker region. For each x ∈ F(Zn) define

�p(x) = (p1(x), . . . , pn(x))

123



where for each 1 ≤ i ≤ n,

pi (x) = − inf{a ∈ Z : (aei · x)RDx}.Finally we define Rd by

x Rd y ⇐⇒ x RDy and �p(x)R�l(x) �p(y).It follows immediately that the Rd -marker regions are n-dimensional rectan-gles of edge lengths d or d + 1.

Now suppose RD is clopen and � > D is such that each RD-marker regionhas edge lengths ≤ �. Then the values of �l(x) and �p(x) are all bounded by�. This implies that Rd as we defined above is clopen. ��

The essence of the above proof is a finitary algorithm being applied uni-formly and invariantly to all equivalence classes. This illustrates a generalmethod we will use again and again. Very often we will focus on the descrip-tion of the finitary algorithm and not explicitly state the formal definitions.

The rectangularmarker regionswe need in the condition of the above lemmawill come from polyhedra with faces perpendicular to the coordinate axes.Let us define precisely the meaning of these terms as we will be using themthroughout. By a rectangular polyhedronwe mean a finite union of rectanglesin Z

n. This is equivalent to saying it is a bounded set which is a finite booleancombination of rectangles. By a face F of a rectangular polyhedron P wemean a set F ⊆ P such that for some 1 ≤ i ≤ n we have that F is a maximalsubset of P satisfying the following:

(i) for any x, y ∈ F and g · x = y, the i th coordinate of g is zero, and(ii) either ei · F ∩ P = ∅ or −ei · F ∩ P = ∅.

We call such an F a face of R perpendicular to ei , or simply an i -face. Asimple example is shown in Fig. 4a. Note that a polyhedral region R may havemany faces perpendicular to ei . Also, the i-faces need not be “connected”subsets of P in the sense that for x, y ∈ F there is a sequence

g1, . . . , gm ∈ {±e1, . . . ,±ei−1,±ei+1, . . . ,±en}such that y = (g1 + · · · + gm) · x and (g1 + · · · + gl) · x ∈ F for all1 ≤ l ≤ m. For P a rectangle, this definition of face is the usual one, but forP a rectangular polyhedron this definition is perhaps not completely standard(the more customary notion would add the requirement that F be connectedin the sense mentioned above). The current definition, however, is the mostuseful for our purposes. Fig. 4b illustrates the current definition of face.

Two faces are parallel if they are perpendicular to the same ei , that is, if theyare both i-faces for some 1 ≤ i ≤ n.Wealso define the notion of perpendicular

123



Fig. 4 Examples of faces.a A polyhedral face Fperpendicular to ei . b Oneface or two faces? Under ourdefinition, F1 and F2 are partof the same face

ei

F

(A)

F1

F2

(B)

distance between two parallel faces as follows. If F1 and F2 are both i-faces,then their perpendicular distance is the absolute value of the unique integerai whenever there are a j ∈ Z for all 1 ≤ j ≤ n with

(a1e1 + · · · + aiei + · · · + anen) · F1 ∩ F2 = ∅.

Note that the perpendicular distance between two parallel faces F1 and F2 isno more than ρ(F1, F2), but it is possible for ρ(F1, F2) to be large while theirperpendicular distance remains small, or even zero.

Lemma 3.3 Let D > 0 be an integer. Let R0 be a subequivalence relation ofF(Zn) so that the R0-marker regions are n-dimensional polyhedra with facesperpendicular to the coordinate axes. Suppose that for each R0-marker regionevery pair of parallel faces have a perpendicular distance greater than D.Thenthere is a subequivalence relation R1 ⊆ R0 so that every R0-marker region ispartitioned into R1-marker regions, which are n-dimensional rectangles withedge lengths greater than D. Moreover, if R0 is clopen and there is � > Dso that each R0-marker region is contained in an n-dimensional cube of edgelengths �, then R1 can also be clopen.

Proof It is enough to describe a finitary algorithm to decompose each n-dimensional polyhedron with the described property into n-dimensionalrectangles with edge lengths greater than D. Let P be a finite polyhedralregion in Z

n with each of its faces perpendicular to some coordinate axis.Note that the faces of P are finite subsets of P definable from P.

Let F1, . . . , Fk be all the faces of P. For each 1 ≤ j ≤ k, the face Fjpartitions P into at most two parts as follows. Suppose ei is perpendicular toFj and without loss of generality assume−ei ·Fj ∩ P = ∅.By our assumptionon the perpendicular distances of parallel faces, for every 0 ≤ c ≤ D, cei ·Fj ⊆P. We define F+

j ⊆ P to be the set of all x ∈ P such that for any y ∈ Fj ,

letting g be the unique element of Zn with g · y = x, the i th coordinate of g

is non-negative. Let also F−j = P − F+

j . F−j could be empty.

123



Fig. 5 A partition of apolyhedron into rectangularregions

Finally define a subequivalence relation RP on P by

x RP y ⇐⇒ ∀1 ≤ j ≤ k(x ∈ F+

j ↔ y ∈ F+j

).

Then the equivalence classes of RP are rectangles whose faces are parts oflinear expansions of the faces of P. These rectangles have edge lengths greaterthan D exactly because the parallel faces of P have perpendicular distancesgreater than D. This construction is illustrated in Fig. 5. ��

To finish the proof of the theorem we only need to create polyhedral markerregionswith the abovementioned property. To achieve thiswe start with a basicclopen marker set M for a marker distance � much bigger than D. Considerthe equivalence relation RM defined by

x RM y ⇐⇒ ∀z ∈ M (ρ(x, z) ≤ � ↔ ρ(y, z) ≤ �).

The basic marker properties ofM imply that RM is in fact clopen. Note that theRM -marker regions are already polyhedra with faces perpendicular to coordi-nate axes; this follows from the choice of the distance function ρ on F(Zn).

Therefore we only need to modify these marker regions so that the perpendic-ular distance between parallel faces is larger than D. There are two potentialdifficulties. First, we need to see that there is no geometric obstacle to imple-ment this idea, that is, we should be able to do this on any given F(Zn)

equivalence class. Second and more seriously, we need to be able to performthe modifications in an invariant Borel manner. To address these issues weemploy again the big-marker-little-marker method.

123



Proof of Theorem 3.1 Let �2 � �1 � D > d2. Let M1 be a clopen markerset given by Lemma 2.1 for the marker distance�1 and M2 ⊆ M1 be given byLemma 2.2 for the marker distance�2. Let g1, . . . , gk enumerate all elementsg ∈ Z

n with ‖g‖ ≤ �2 + �1. Then F(Zn) = ⋃1≤i≤k gi · M2. We define a

sequence A0, A1, . . . , Ak of subsets of M1 as follows:

A0 = M2,

Ai = M1 ∩ (gi · M2) −⋃

j<i

A j , for 1 < i ≤ k.

Then A0, A1, . . . , Ak is a partition of M1 into disjoint clopen sets with theproperty that, for any 0 ≤ i ≤ k and x = y ∈ Ai , ρ(x, y) > �2 − 2�1.

We now define RM1 as promised

x RM1 y ⇐⇒ ∀z ∈ M1 (ρ(x, z) ≤ �1 ↔ ρ(y, z) ≤ �1).

In order to describe the modifications we will perform on the RM1-markerregionswewill use the following auxiliary notation. Let J1 = {g ∈ Z

n : ‖g‖ ≤�1}. For each x ∈ M1 let Rx = J1 · x . Then Rx is the cubic region with centerx and edge length 2�1.When there is no danger of confusion, we will slightlyabuse the terminology of marker regions and refer to each Rx also as a markerregion. With this notation RM1 can be equivalently expressed as

x RM1 y ⇐⇒ ∀z ∈ M1 (x ∈ Rz ↔ y ∈ Rz).

In this sense we also say that RM1 is generated by the collection {Rx | x ∈ M1}of (possibly overlapping) marker regions. To create marker regions with thedesired property we will adjust the collection of cubes {Rx | x ∈ M1} to acollection of rectangles {R′

x | x ∈ M1} so that(i) Rx ⊆ R′

x for each x ∈ M1,

(ii) the corresponding faces of Rx and R′x have a perpendicular distance no

more than 110�1, and

(iii) for each face F1 of R′x and any parallel face F2 of R

′y withρ(x, y) ≤ 3�1,

the perpendicular distance between F1 and F2 is at least D.

The subequivalence relation generated by the collection {R′x | x ∈ M1} clearly

satisfies the condition of the preceding lemma.We define the collection {R′

x | x ∈ M1} by giving an inductive definition of{R′

x | x ∈ Ai } for 0 ≤ i ≤ k. For x ∈ A0, let Rx = R′x . Consider i > 0 and

assume that for all j < i and x ∈ A j , R′x has been defined to satisfy (i)-(iii).

Now let x ∈ Ai . Let R′y1, . . . , R

′ym enumerate all surrounding rectangles with

y1, . . . , ym ∈ ⋃j<i A j and ρ(x, yl) ≤ 3�1 for 1 ≤ l ≤ m. Note that there

123



is a fixed upper bound N for the number m which depends only on n. For

example, a volume argument shows that N ≤(

(8�1)n

�n1

)= 8n.

It follows that, for each face F of Rx , there are at most 2N many faces ofR′y1, . . . , R

′ym which are parallel to F. Moreover these faces are the only ones

that might have a perpendicular distance ≤D from the adjusted face F ′ of R′x

corresponding to F. Therefore, as long as 110�1 > 2N (2D), the face F can be

shifted away from the center x to a suitable F ′ so that (ii) and (iii) are satisfiedfor F ′. To define R′

x we need only adjust each of its 2n faces in turn. It is clearthat (i) holds with this construction. Note that if x = y ∈ Ai , then

ρ(x, y) > �2 − 2�1 � �1,

thus the constructions of R′x and R′

y do not affect each other. It follows that theresulting regions {R′

x | x ∈ M1} satisfy properties (i)–(iii). By the precedinglemmas the proof of the theorem is complete. ��

Note that Theorem 3.1 is a generalization of Lemma 2.9. In view of thecontrast between Lemma 2.9 and Proposition 2.8, there are also limitations onhow regular the higher dimensional marker regions can be. For instance, in thecase n = 2 Theorem 3.1 guarantees the existence of Borel tilings of F(Z2)

by four kinds of almost square tiles, with dimensions d × d, d × (d + 1),(d + 1) × d and (d + 1) × (d + 1), respectively. Proposition 2.8 impliesimmediately that there is no Borel tiling with only one kind of these tiles used.A more tedious argument shows that there is no Borel tiling which uses onlytiles of dimension d × d and d × (d + 1). Similar arguments also rule out thecombinations {d × d, (d + 1) × d}, {d × (d + 1), (d + 1) × (d + 1)} and{(d + 1) × d, (d + 1) × (d + 1)}. The other cases are not clear. In particularit is not clear whether there are Borel tilings which use just two or three kindsof the above tiles. We conjecture that the number 4 is optimal.

We make some comments on the role marker regions are going to play inour proof of hyperfiniteness. Since marker regions correspond to finite Borelequivalence relations, a natural attempt to prove hyperfiniteness is to create asequence of such finite subequivalence relations so that they form an increasingsequence

R0 ⊆ R1 ⊆ · · · ⊆ Rk ⊆ · · ·In terms of the geometry of marker regions when we require Rk ⊆ Rk+1it is equivalent to requiring that the Rk+1-marker regions be unions of Rk-marker regions. In this case we say that the marker regions (of different levels)cohere with each other. If increasing levels of coherent marker regions can beproduced, then the union of these equivalence relations is hyperfinite. Thusthe question is whether the union is the entire equivalence relation F(Zn).

123



In Weiss’ proof of the hyperfiniteness of F(Zn), increasing levels of coher-ent regions were used. These were obtained bymodifying the Dirichlet regionsso as to get coherence. Using the clopen marker of Lemma 2.1, this construc-tion produces an increasing sequence of (relatively) clopen marker regions forF(Zn). The union En of the corresponding finite subequivalence relations isa hyperfinite subequivalence relation of F(Zn). However, En is not equal toF(Zn), but rather has finite index in it (but this suffices to show the hyperfinite-ness of F(Zn)). Moreover, the index of En in F(Zn) grows exponentially withn. Thus, an application of this method to study F(Z<ω) would only produce ahyperfinite subequivalence relation of F(Z<ω) with infinite index. This doesnot seem to be progress since it is easy to prove abstractly that any aperiodiccountable Borel equivalence relation contains a hyperfinite subequivalencerelation.

One can ask, then, whether there is any chance for this method to succeed.Specifically, can we create increasing finite equivalence relations

R0 ⊆ R1 ⊆ · · · ⊆ Rk ⊆ · · ·

with each Ri relatively clopen in F(Zn) such that⋃

k Rk = F(Zn)? Thefollowing theorem gives a negative answer in a somewhat stronger sense. LetR be a subequivalence relation (not necessarily finite) of F(Zn).We say that Ris nondegenerate if for all x ∈ F(Zn) there are y, z ∈ [x]with (y, z) ∈ R, thatis to say, there are at least two R-marker regions in each F(Zn)-equivalenceclass.

Theorem 3.4 There is no increasing sequence of nondegenerate, relativelyopen subequivalence relations of F(Zn)

R0 ⊆ R1 ⊆ · · · ⊆ Rk ⊆ · · ·

such that⋃

k Rk = F(Zn).

Proof Assume there were such a sequence. For each k ∈ ω let Uk be an opensubset of F(Zn)2 so that Rk = Uk ∩ F(Zn). Define the set Ck of boundarypoints of marker regions for Rk by

x ∈ Ck ⇐⇒ x ∈ F(Zn) and ∃g ∈ Zn (‖g‖ = 1 and (g · x, x) ∈ Rk).

Since Rk is nondegenerate, Ck is a complete section of F(Zn). Also, since(g · x, x) ∈ F(Zn) we have that

x ∈ Ck ⇐⇒ x ∈ F(Zn) and ∃g ∈ Zn(‖g‖ = 1 and (g · x, x) ∈ Uk).

123



Note that there are only finitely many g ∈ Zn with ‖g‖ = 1, hence it follows

that Ck is a relatively closed subset of F(Zn). Now it is clear since the Rk areincreasing that

C0 ⊇ C1 ⊇ · · · ⊇ Ck ⊇ · · ·Thus by Theorem 2.5

⋂k Ck = ∅. Let x ∈ ⋂

k Ck . There is g ∈ Zn with

‖g‖ = 1 so that for infinitely many k, (g · x, x) ∈ Rk . This contradicts⋃k Rk = F(Zn). ��This shows that the marker regions have to be used in a more sophis-

ticated way than to serve directly as witnesses for hyperfiniteness. Thatis, we must go in a different direction from using clopen markers toproduce coherent marker regions. Indeed, one of the main ideas of thecurrent proof will be to use the clopen marker sets to produce markerregions which are as “anti-coherent” as possible. This will be made pre-cise in Sect. 5 where we will construct orthogonal marker regions, whichwill be one of our main technical tools. Our strategy will be to con-struct the anti-coherent marker regions Rk and then produce an embed-ding π : F → E0 (here F could be F(Zn) or F(Z<ω)) where π(x)will code the sequence of marker regions that x belongs to. For this towork, we will need that the boundaries of the Rk regions leave finitesets (that is, for each finite set A ⊆ [x], for x ∈ F, we have thatfor large enough k that A misses the boundary of all Rk regions). Itwill be the anti-coherence of the marker regions which guarantees thisproperty.

We mention one more negative result concerning attempts to witnesshyperfiniteness by marker regions. The next theorem says that, even usingnon-coherent marker regions, hyperfiniteness cannot be witnessed by markerregions which are too geometrically regular (for example, nearly cubical rec-tangular regions as constructed in Theorem 3.1).

Theorem 3.5 There does not exist a sequence of Borel subequivalence rela-tions Rk of F(Z2) such that F2 = ⋃

k Rk where on each F(Z2) class each Rkinduces a partition into “bounded geometry” polygons, and such that for allx, y ∈ F(Z2), x F2y iff ∃n ∀k ≥ n x Rk y. By “bounded geometry polygons”we mean: there are numbers ak < bk with limk→∞ ak = ∞ such that eachedge of a polygon in Rk has length between ak and bk .

The proof of Theorem 3.5 is a measure or probability argument which weomit. The theorem shows that themarker regions Rk must become increasingly“fractal-like” as k gets larger. Indeed, themarker regionswe eventually producewill have this fractal-like property.

123



4 An application of regular marker regions

In this section we present an application of regular marker regions to a Borelcoloring problem. General Borel coloring problems were studied by Kechriset al. [9] and Ben Miller. The specific question we deal with below was com-municated to us by Miller. Case n = 2 of Theorem 4.2 was joint work withMiller, again by a different proof.

We use the following notation. LetC be a set and X be an invariant subset of2Z

n. A function c : X → C is a coloring of X if for all x ∈ X and 1 ≤ i ≤ n,

c(ei · x) = c(x) unless ei · x = x . In this case we also say that c is a |C |-coloring of X. If X is an invariant Borel subset of 2Z

n, we define the Borel

chromatic number for X, denoted by χB(X), to be the smallest integer k > 1such that there exists a Borel k-coloring of X.

Kechris et al. [9] have shown that χB(2Z) = 3 and that for all n > 1 andany invariant Borel X ⊆ 2Z

non which the action of Z

n is non-smooth, 3 ≤χB(X) ≤ 2n+ 1. We show below that for all n > 1 we have χB(F(Zn)) ≤ 4.Concerning the non-free part, the authors and independently R. Muchnik (wethank A. S. Kechris for communicating this to us) have shown the following.

Theorem 4.1 χB(2Zn) = 2n + 1 for each n > 1.

Wewill present the proof of Theorem 4.1 elsewhere, and give here the proofof the 4-colorability of the free part F(Zn) in the following strong form.

Theorem 4.2 For each n > 1 there is a continuous 4-coloring of F(Zn).

Proof Fix an integer d which is a multiple of 10 so that d > 10 · 2n+1. LetR = Rn

d be the clopen subequivalence relation of F(Zn) given byTheorem3.1.We need to work with the geometry of R-marker regions.

To begin with, we define a new distance function on F(Zn). First note thatthe following defines a norm on Z

n: for g = (g1, . . . , gn),

‖g‖ν = |g1| + · · · + |gn|.This norm induces a distance function ν on F(Zn), which we call the NewYork distance: for x, y ∈ F(Zn),

ν(x, y) ={‖g‖ν, if g · x = y,

∞, if (x, y) ∈ F(Zn).

Two elements x and y are of New York distance 1 exactly when there is some1 ≤ i ≤ n such that ei · x = y or ei · y = x (in which case −ei · x = y). A 4-coloring of F(Zn) is a function c : F(Zn) → {0, 1, 2, 3} such that c(x) = c(y)whenever ν(x, y) = 1.

123



Consider the sets

B1 = {x ∈ F(Zn) : ∃1 ≤ i ≤ n (ei · x, x) ∈ R or (−ei · x, x) ∈ R},B2 = {x ∈ F(Zn) : ∃1 ≤ i ≤ n (2ei · x, x) ∈ R or (−2ei · x, x) ∈ R}.

The set B1 consists of the boundary points of all the R-marker regions, and B2consists of all points of B1 as well as points of New York distance 1 to thosein B1. Both B1 and B2 are clopen subsets of F(Zn).

Define

C0 = {x ∈ F(Zn) : ∀1 ≤ i ≤ n, (ei · x, x) ∈ R}.

Then C0 is a transversal for R and it consists of exactly one extremal point ofeach R-marker region. Furthermore, let

C2 = {x ∈ B2 : ν(x, x0) is even, where x0 ∈ C0 and x0Rx},

and

C1 = C2 ∩ B1.

Note that C0 ⊆ C1 ⊆ C2.

Finally, define A1 = B1 − C1. Note that all sets defined so far are clopen.Next define a binary relation E by

xEy ⇐⇒ ν(x, y) = 1 and (x, y ∈ A1 or x, y ∈ C1).

Let ∼ be the transitive closure of E . Then ∼ is an equivalence relation. Thefollowing fact will be crucial for our further construction.

Claim 1 Each ∼-equivalence class has no more than 2n2many elements.

Proof of Claim 1 First note that if xEy then (x, y) ∈ R. To see this, assumex Ry and let x0 be the unique element inC0 with x0Rx ; since ν(x, y) = 1, thenν(x, x0) differs from ν(y, x0) by 1, and thus x and y cannot both be in A1 orC1, contradicting xEy. Thus if xEy, the R-marker regions containing x and yare different. Moreover, if xEy and letting Rx and Ry be the R-marker regionscontaining x and y respectively, then there is a face F1 of Rx containing x anda face F2 of Ry containing y so that F1 and F2 are parallel. In fact, if σ · x = y,then both F1 and F2 are perpendicular to the vector σ . The four possible casesfor a fixed vector σ in dimension n = 2 are demonstrated in Fig. 6.

123



x y

F1 F2

Rx Ry

σx

y

F1 F2

Rx Ry

σx

y

F1 F2

Rx Ry

σx y

F1 F2

Rx Ry

σ

Fig. 6 E-neighboring points and perpendicular faces

Next we show that if x ∼ y and g · x = y with g = (g1, . . . , gn) ∈ Zn,

then for all 1 ≤ i ≤ n, |gi | < 2n. Toward a contradiction suppose there is achain

x = x0 E x1 E · · · E xk = y

with |g j� | ≥ 2n for some j ≤ k, 1 ≤ � ≤ n, where g j = (g j

1 , . . . , gjn ) ∈ Z

n

is such that g j · x = x j . By taking k minimal, we may assume that |g j� | ≤ 2n

for all j, �. Fix i so that |gki | ≥ 2n (so actually |gki | = 2n). It follows that thereare distinct

j0, j1, . . . , j2n−1 < k

such that for each l < 2n, g jli = l and g jl+1

i = l + 1. It follows that for eachl < 2n there are faces Fl and F ′

l such that Fl contains x jl , F′l contains x jl+1,

and both Fl and F ′l are perpendicular to ei . Also, every point in Fl is of the

form h ·x with hi = l, and every point in F ′l is of the form h ·x with hi = l+1.

This implies that there are at least 2n +1 faces from various R-marker regionseach of which is perpendicular to ei , and all of these 2n + 1 faces have a pointthat is within ρX distance 2n < d

10 of the point x . This, however, contradictsthe following fact. ��Fact 1 Let R be a rectangular partition of Z

n with each rectangle having edgelengths in {d, d + 1}. Then for any x ∈ Z

n, the ball

Bρ

(

x,d

10

)

={

y ∈ Zn : ρ(x, y) ≤ d

10

}

of radius d10 about x can intersect at most 2n many parallel faces of rectangles

in R.

Proof Without loss of generality we may assume x is the origin. Suppose thatB = Bρ(�0, d

10 ) intersects more that 2n many faces of rectangles, and all ofthese faces are parallel to a given faceF .Clearly each rectangle in the partitionR can have at most one face parallel to F which intersects S. Thus, we have

123



more than 2n rectangles R1, . . . , R2n+1, each of which has a face parallel toFwhich intersects B. Let c1, . . . , c2n+1 be the centers of these rectangles. Sincethere are only 2n “quadrants” about the origin inZ

n, there must be two of theserectangles, say R1 and R2, such that their centers c1 and c2 lie in the samequadrant about �0. Without loss of generality, say c1 and c2 both have non-negative values in their coordinate representations. Say c1 = (a1, a2, . . . , an),c2 = (a′

1, a′2, . . . , a

′n). Then, 0 ≤ ai ≤ d

10 + d2 = 3

5d for all i. It followsthat ( 3

10d, 310d, . . . , 3

10d) ∈ R1. Similarly, ( 310d, 3

10d, . . . , 310d) ∈ R2, which

contradicts the fact that R1 and R2 are disjoint. ��Now it follows immediately from the above fact that for any x ∈ B1, the

∼-equivalence class of x is confined to a cubical regionwith all edges of length2n. Thus there are at most (2n)n = 2n

2many points in each ∼-equivalence

class. This completes the proof of the claim. ��For any finite S ⊆ Z

n, let πS be the lexicographically least element of S.

Note that for any finite S ⊆ Zn and g ∈ Z

n, πgS = gπS. It follows that forany finite subset of an F(Zn)-class, in general of the form S · x for a finiteS ⊆ Z

n, the lexicographically least element πS · x, is well defined (that is, itdoes not depend on the choice of x). Given each subset S of [0, d]n in Z

n wefix a 2-coloring cS of S into {0, 1} so that c(πS) = 0.

We are finally ready to define a continuous 4-coloring c of F(Zn). Letx ∈ F(Zn). If x ∈ C2−C1, then let c(x) = 0. Since distinct points inC2−C1have ν distance > 1, the function c defined so far is a coloring of C2 − C1.

Next for x ∈ C1, let x0 be the lexicographically least element of [x]∼ and let[x]∼ = S · x0. By the proof of the above claim, S ⊆ [0, 2n − 1]n ⊆ [0, d]n.We define c : [x]∼ = S · x0 → {0, 1} so that

c(g · x0) = cS(g).

At this point note that if x1 ∈ C1 and x2 ∈ C2 − C1 then ν(x1, x2) > 1. Thusthe function c defined so far is a coloring on C2.

Continuing the definition, for x ∈ A1, we also consider [x]∼ and let x0be its lexicographically least element and S ⊆ Z

n so that [x]∼ = S · x0. Wedefine c : [x]∼ = S · x0 → {2, 3} so that

c(g · x0) = cS(g) + 2.

It is obvious that the function c defined so far is a coloring of C2 ∪ A1.

Now we claim that if x ∈ C2 ∪ A1 and ν(x, y) = 1, then y ∈ A1. To seethis, let y ∈ A1 and let ν(x, y) = 1. Let y0 ∈ C0 be the unique point withy0Ry. Then ν(y, y0) is odd. If (x, y) ∈ R, then x ∈ B1 = C1∪ A1 ⊆ C2∪ A1.

If x Ry, then x ∈ B2 and, since ν(x, y) = 1, we have that ν(x, y0) is even.This implies that x ∈ C2. We have thus shown the claim.

123



Finally for x ∈ C2 ∪ A1 we let Dx = [x]R − (C2 ∪ A1). Let x0 be thelexicographically least element of Dx and let Dx = S · x0. Define c : Dx =S · x0 → {2, 3} so that

c(g · x0) = cS(g) + 2.

By the above claim, if x1 ∈ Dx , y1 ∈ Dx and ν(x1, y1) = 1, then it must bethat y1 ∈ C2; thus c(y1) ∈ {0, 1} and therefore c(y1) = c(x1). This shows thatc is a coloring of all of F(Zn).

To see that c is a continuous coloring just note that for each colorκ ∈ {0, 1, 2, 3}, c−1(κ) is clopen in F(Zn). This completes the proof ofTheorem 4.2. ��

An interesting question which remains open is whether F(Zn) admits acontinuous (or just Borel) 3-coloring. Investigation of this problem suggeststhat the answer is closely related towhetherwemay improveour regularmarkerregions so that they exhibit further regularity properties. In particular, we areinterested in the following intuitive question: can we create regular markerregions so that not only each of them is a rectangle but also they almost lineup with each other? The precise question for n = 2 is as follows.

Question 4.1 Let d > 0 be an integer. Is there a Borel subequivalence relationRd of F(Z2) such that

(a) each Rd -marker region is a rectangle with edge lengths either d or d + 1,and

(b) if R1 and R2 are two adjacent Rd-marker regions and F1 and F2 are paralleledges of R1 and R2 respectively, then the perpendicular distance betweenF1 and F2 is either 0, 1, d or d + 1?

5 Orthogonal marker regions

In Sect. 3we constructed regularmarker regions for F(Zn).Recall that for eachd > 0we constructed clopen subequivalence relation Rn

d of F(Zn) such that oneach equivalence class of F(Zn), the Rn

d classes are n-dimensional rectangleswith each edge length either d or d + 1. We refer to Rn

d as a rectangularpartition of F(Zn). For such a partition Rn

d , we show how to construct anorthogonal partition Rn

d . This will also be a clopen subequivalence relationof F(Zn) with rectangular regions, and moreover will be orthogonal to Rn

d .

By this we mean that every face of a rectangle in Rnd will be fairly far away

from any parallel face of a nearby rectangle in Rnd . The precise statement is in

Lemma 5.1.We adopt a running convention throughout Sects. 5 and 6. In the statements

of all results involving the quantity d,we always assume d is sufficiently large

123



Fig. 7 Before adjustment,the balls Bρ(x, s

2 ), x ∈ M,

could have parallel faces thatare close

s

so that all expressions involving d have value at least 1. For example, in thestatement of Lemma 5.1 it should be assumed that d is greater than 9000n16n

2.

This avoids certain trivialities associated with small values of d.Now we state our basic lemma on orthogonal marker regions. We will not

directly use this lemma in the proof of our main result (in Sect. 6), but rathera technical strengthening of it which we give immediately afterward. Never-theless, this lemma shows in a simpler setting the ideas involved. We note thatthe values of the various constants appearing in the lemma are not critical.

Lemma 5.1 Let Rnd be a clopen rectangular partition of F(Zn),with each rec-

tangle having edge lengths in {d, d+1}. Then there is a clopen subequivalencerelation Rn

d of F(Zn) satisfying:

(1) Each Rnd class is a rectangular region with edge lengths between 9d and

12d.

(2) Every face of a region R1 ∈ Rnd is at least 1

9000n16n2d from any parallel

face of a region R2 ∈ Rnd .

Proof Let s = Cd, where C = 110 · 4 · 8 · 16n. We first let M ⊆ F(Zn) be aclopenmarker set for distance s

2 .Wemayassume as in the proof ofTheorem3.1that M = A0 ∪ A1 ∪ · · · ∪ Ak, and for any x, y ∈ Ai , ρ(x, y) � s. We letR be the collection of cubes with edge length s which are centered about thepoints of M. As in the proof of Theorem 3.1, we proceed in k steps to adjustthese cubes, at step i adjusting the faces of the cubes with centers in Ai . Wedo this adjustment in two steps. At the end of both steps, we will have movedeach face outward by no more than s

4 .

Figure 7 shows an initial arrangement of the rectangles before adjustment. Inthe figure each dot represents a point x ∈ M, each solid square represents a ballBρ(x, s

2 ) = {y ∈ F(Z2) : ρ(x, y) ≤ s2 },which is a cube centered about x ∈ M

123



with edge length s, and each dashed square represents a ball Bρ(x, s4). Since

M has marker distance s2 , the balls Bρ(x, s

4) are disjoint for distinct x ∈ M.

In the first step of the adjustment, we move each face F of a cube R withcenter x ∈ Ai outward by no more than s

8 to avoid each parallel face F ′ ofany previously adjustedR cubes with 0 < ρ(F,F ′) ≤ 2s. This adjustment isillustrated in Fig. 8. Note that there are possibly multiple such faces F ′.

SupposeF ′ is a parallel face of a previously adjustedR cube R′ with centermarker point x ′ ∈ M.We note that R′ must lie entirely inside the ball of radiuss2 + 2s + 3

2s = 4s centered at x, that is, R′ ⊆ Bρ(x, 4s). Note also that thevolume of Bρ(x, 4s) is (8s)n.

Consider all previously adjusted regions with a parallel face F ′ such thatρ(F,F ′) ≤ 2s. Each of them contains a smaller ball Bρ(x ′, s

4) with volume( s2 )

n. These smaller balls are disjoint as we noted before, and they remainunchanged throughout the adjustment (see Fig. 9). It follows that there areat most 16n such smaller balls within the ball Bρ(x, 4s). Thus there are atmost 16n previously adjusted cubes that we have to consider. Each of such

x

12s 2s

32s

R

F Fs8

x R

F

Fig. 8 The first step of the adjustment

Fig. 9 The balls Bρ(x, s4 ),

x ∈ M, remain unchangedand disjoint throughout theadjustment

s/4

s/4

s/2 s/4

123



x

< 32s

R∗

F∗

s4·8·16n + d

74s + 10d 7

4s + 10d

Fig. 10 In the second step of the adjustment, the face F∗ will be adjusted further by no morethan s/8 to avoid parallel faces from Rn

d cubes in a large region

previously adjusted cubes has at most two faces parallel to F, but only one ofthem can have a perpendicular distance within s

8 to the face F . Therefore thetotal number of faces F ′ we have to avoid is at most 16n. So, by moving theface F of R no more than s

8 , we can ensure that its perpendicular distance toall parallel faces of previously adjusted cubes is at least s

2·8·16n .In the second step of the adjustment, we move each of these faces outward

by no more than s4·8·16n < s

8 . This will keep their perpendicular distances fromthe previously adjusted faces considered in the first step at least s

4·8·16n = 110d.

The objective now is to avoid all of the parallel faces of rectangles in Rnd within

74s + 10d of the adjusted face.This adjustment is shown in Fig. 10. In the figure, R∗ represents a R cube

after the first step of adjustment and after the second step of adjustment forsome of its faces. Its edge lengths are now between s and 3

2s. F∗ represents aface of R∗ under consideration.

Suppose N is an Rnd region with a face F such that ρ(F,F∗) ≤ 7

4s + 10d(not shown in Fig. 10). In Fig. 10 a dashed rectangle represents a region thatN must intersect if F satisfies the above inequality and F is to be withinparallel distance d of the face obtained by moving F∗ outward by no morethat s

4·8·16n (note that if the parallel distance from F to the adjustedF∗ is≥ d,

then the conclusion of the lemma automatically holds). It follows that N mustlie entirely in another rectangular region whose side lengths are bounded asfollows: in the direction of the movement, the side length is at most

s

4 · 8 · 16n + 2(d + 1) <s

4 · 8 · 16n + 3d;

in the other directions the side length is at most

3

2s + 2

(7

4s + 10d + (d + 1)

)

< 5s + 23d.

123



Thus the volume of this rectangular region is at most

( s

4 · 8 · 16n + 3d)

(5s + 23d)n−1.

Note that each Rnd region has volume at least dn. Therefore the number of Rn

dregions that can be present in this rectangular region is at most

( s4·8·16n + 3d

)(5s + 23d)n−1

dn.

Since each Rnd region contains at most two faces parallel to F∗, the total

number of parallel faces from the Rnd regions inside this rectangular region is

at most

2 ·( s4·8·16n + 3d

)(5s + 23d)n−1

dn. (E1)

This number is bounded above by

2 ·s

2·8·16n · (6s)n−1

dn= 1

48

(3

8

)n

Cn ≤ 9000n16n2.

So, by moving F∗ outward by no more than s4·8·16n we may avoid all of these

faces by at at least

s

4 · 8 · 16n · 9000−n · 16−n2 ≥ 1

9000n16n2d.

We have finished the description of two steps of adjustment to cubes in R.

Recall that the adjustment is applied to all R cubes in k stages. At the end ofthis adjustment we obtain a collection R of (possibly overlapping) rectangularregions. This is a situation we have encountered in the proof of Theorem 3.1.As in that proof we will apply Lemma 3.3 (depicted by Fig. 5) to obtain aclopen subequivalence relation R of F(Zn).

To bemore specific, the R regions are obtained by extending the faces of theR regions throughout any R regions they intersect. Consider a particular Rregion. Suppose F1 and F2 are parallel faces of R regions which both intersectthe current region. Then F1 and F2 must lie within 3

2s of each other. Say F1 andF2 came from regions thatwere adjusted at stages i1 < i2, respectively. Then atstage i2 the face F2 corrresponding to F2 before the adjustment was at distanceat most 32s+ s

4 < 2s from F1. Thus, at stage i2 we made the adjustment so thatthe perpendicular distance of F2 and F1 is at least s

4·8·16n . So, when we applyLemma 3.3 and extend F1 and F2 throughout the current region, the extendedfaces will still have a perpendicular distance at least s

4·8·16n = 110d from each

123



other. It follows that each rectangular region obtained by applying Lemma 3.3will have edge lengths at least 110d. Similarly, the extension of F2 came froma face F∗

2 that was present at the end of the first step of adjustment. So, theextension of F2 can lie at most s

4 + 32s = 7

4s from F∗2 . So, a parallel face of

an Rnd region that can have a perpendicular distance ≤ d + 1 to the extension

of F2 must be within 74s + 2d of F∗

2 . In the second step of the adjustment, weguaranteed that these extended faces will be at least 1

9000n16n2d from a parallel

face of Rnd .

To summarize,wehave defined a clopen subequivalence relation R of F(Zn)

such that on each F(Zn) class, R induces a partition into rectangular regionseach having edge lengths between 110d and 3s

2 . Furthermore, each face of

one of these rectangles are at least 9000−n16−n2d from any parallel face of arectangle in Rn

d .

Finally, we subdivide each of these rectangular regions into smaller rectan-gular regions as follows. Divide each edge into intervals of size between 91

2dand 111

2d (use the fact that any positive integer ≥ 110 can be written as apositive integral linear combination of 10 and 11). Consider one of the corre-sponding faces (that is, perpendicular to the edge and passing through one ofthe endpoints of these intervals). We move it by no more than d

2 to avoid theparallel faces of rectangles in Rn

d within 10d. There are at most

2 ·(3s2 + 20d

)n−1 (d2 + 20d

)

dn≤ 2n · 21 · Cn−1

faces to avoid, so we may avoid them by at least

(d2

)

2n · 21 · Cn−1 ≥ 1

9000n16n2d.

The resulting rectangular regions have edge lengths between 9d and 12d.

Each face of one of these regions is at least 19000n16n2

d from a parallel face of

a rectangle in Rnd which is within 10d of the face. Thus, we have satisfied the

requirements of the lemma. ��The next lemma is the technical strengthening of Lemma5.1 thatwe actually

need. The main difference is that we now avoid finitely many rectangular sub-equivalence relations instead of just one. The reader should think of the boundb which appears in the statement as a fairly small integer (its precise role willbecome clearer in the next section).

Lemma 5.2 Let R1, . . . , Rk be a sequence of clopen subequivalence relationsof F(Zn) satisfying the following:

123



(1) On each F(Zn) class, each Ri induces a partition into polyhedral regionsR which are unions of rectangles r such that each edge length of one ofthe r rectangles is between d and 12d.

(2) In any ball B of radius 100,000 · 16nd in some F(Zn) equivalence class,there are at most b integers i such that one of the Ri regions has a face Fwhich intersects B.

Then there is a clopen subequivalence relation Rnd ⊆ F(Zn) satisfying:

(1) Each Rnd class is a rectangular region with edge lengths between 9d and

12d.

(2) Every face of an Rnd region R is at least 1

9000n16n2bd from any parallel face

of an Ri region R′ (for any i).

Proof The proof is similar to that of Lemma 5.1. The only difference is thatin step two of the adjustments, we must avoid all the parallel faces of not justthe one family Rn

d , but of all the families R1, . . . , Rk .However our hypothesisgives that for each rectangle of size

(5s + 23d) × · · · × (5s + 23d) ×( s

4 · 8 · 16n + 3d)

,

which lies in a sphere of radius < 100,000 ·16nd, there are at most b many isuch that some face of a region in Ri intersects this rectangle. So, there are atmost

(2b) ·[

(5s + 23d)n−1 · ( s4·8·16n + 3d

)

dn

]

≤ 9000n16n2b

many faces to avoid. Thus we may avoid them by at least

s

4 · 8 · 16n · 9000−n · 16−n2 · 1b

≥ 1

9000n16n2bd.

��We remark that in hypothesis (1) of Lemma 5.2 we use d and 12d rather

than d and d+1 so that formally Lemma 5.2 is a strengthening of Lemma 5.1.

6 Hyperfiniteness of F(Z<ω)

In this section we show that the free part F(Z<ω) of 2Z<ω

with the usualaction of Z

<ω is hyperfinite. For notational clarity we let F denote the setF(Z<ω) and Fω denote the orbit equivalence relation on F from the action ofZ

<ω. Each subgroup Zn also acts on F and induces a subequivalence relation

123



Fig. 11 The inductiveconstruction of Ri

j

R11 R2

1 R31 R4

1

R22 R3

2 R42

R33 R4

3

R44

...

...

· · · · · ·

which we denote by Fn. So, F1 ⊆ F2 ⊆ . . . form an increasing sequence, andFω = ⋃

n Fn. Of course, all the Zn act freely on F as well.

To show Fω is hyperfinite, we will construct finite subequivalence relationsRi and use their structure to define a reduction from Fω to E0.Actually, the Riwe build will be relatively clopen subequivalence relations. Recall this meansthat {(x, g) ∈ F × Z

<ω : g · x Ri x} is relatively clopen in F × Z<ω (which is

equivalent to saying that for each g ∈ Z<ω, {x ∈ F : g · x Ri x} is relatively

clopen in F). Then we will be able to obtain a continuous embedding of Fω

into E0(ωω). The rough strategy for building the Ri is as follows. We start

with a sufficiently fast growing sequence of marker distances d1 � d2 � . . . .

For each i, let Rii ⊆ Fi be a clopen subequivalence relation of Fi given by

Theorem 3.1 for marker distance di . In i − 1 steps we will then successivelymodify the rectangles induced by Ri

i to regions giving Rii−1, . . . , R

i1. Each Ri

jwill be a clopen subequivalence relation of Fω, such that on each Fj class eachregion induced by Ri

j will be a finite union of rectangles with edge lengths onthe order of d j . Furthermore, the rectangles contributing to the boundary of

the region will be “orthogonal” to the regions induced by R jj , R

j+1j , . . . , Ri−1

j(that is, wewill maintain a certain distance between nearby parallel faces). Theadjusted regions defining Ri

1 will witness the hyperfiniteness of Fω. Figure 11illustrates the order of construction of the subequivalence relations Ri

j .

As we successively adjust the regions in the definitions going from Rii to

Ri1, the boundaries of the regions will become increasingly “fractal like.” That

is, successive adjustments will be done on increasingly small scales.We note two points about our final regions corresponding to Ri

1. First, theywill not cohere. That is, we will not have that Ri

1 ⊆ Ri+11 . Indeed, since

all the Ri1 will be clopen, coherence is impossible to achieve by Theorem 3.4.

Second, according to Theorem 3.5 it is in some sense necessary that the bound-aries of the Ri

1 regions become increasingly fractal. So, for example, even forthe finite-dimensional relation Fn on F(Zn), the original rectangular sube-quivalence relations Rn

d of Theorem 3.1 can never by themselves witness thehyperfiniteness of Fn.

123



We now turn to the construction of Rij . The next lemma will provide the

argument for passing from Rij+1 to Ri

j .

Lemma 6.1 Let j < i. Let d j+1, d j be positive integers with d j+1 � d j

(the exact condition necessary is specified below). Suppose Rij+1 is a clopen

subequivalence relation of Fi ⊆ Fω. Assume that the restriction of Rij+1 to

each Fj+1 class induces a partition into polyhedral regions each of which isa union of rectangles with edge lengths between d j+1 and 12d j+1. Suppose

clopen equivalence relations R jj , R

j+1j , . . . , Ri−1

j and R j+1j+1, . . . , R

i−1j+1 have

been defined and satisfy:

(1) R jj ⊆ Fj ; Rk

j ⊆ Fk and Rkj+1 ⊆ Fk for all j < k ≤ i − 1.

(2) R jj induces a partition of each Fj class into rectangular regions with edge

lengths in {d j , d j + 1}.(3) For j < k ≤ i −1, the restriction of Rk

j to each Fj class gives a partitioninto polyhedral regions R each of which is a union of rectangles with edgelengths between 9d j and 12d j .

(4) On each Fj class, for each region R induced by the restriction of Rkj ,

there is a region R′ induced by the restriction of Rkj+1 such that each face

of R is within 12d j of a face of R′.(5) In any ball B of radius 100,000 ·16 j d j contained in an Fj class, there are

at most j + 1 many k with j < k ≤ i − 1 such that some region inducedby the restriction of Rk

j has a face intersecting B.

(6) For any j < k1 < k2 ≤ i − 1, and regions R1, R2 contained in an Fj

class induced by the restrictions of Rk1j , Rk2

j respectively, if F1, F2 are

parallel faces of R1, R2, then ρ(F1,F2) > 19000 j16 j2 ( j+1)

d j .

(7) For any j < k1 < k2 ≤ i, and regions R1, R2 contained in an Fj+1

class induced by the restrictions of Rk1j+1, R

k2j+1 respectively, ifF1,F2 are

parallel faces of R1, R2, then ρ(F1,F2) > 19000 j+116( j+1)2 ( j+2)

d j+1.

Then there is a clopen subequivalence relation Rij ⊆ Fi satisfying the

following:

(1) On each Fj class, Rij induces a partition into polyhedral regions R each

of which is a union of rectangles with edge lengths between 9d j and 12d j .

(2) On each Fj class, for each region R induced by Rij , there is a region R′

induced by Rij+1 restricted to the Fj class such that each face of R is

within 12d j of a face of R′.(3) Condition (5) continues to hold, where now j < k ≤ i.(4) Condition (6) continues to hold, where now j < k ≤ i.

123



Proof Let Rij+1 and the R j

j , R j+1j , . . . , Ri−1

j be given satisfying the above

hypotheses. By assumption, the restriction of Rij+1 to each Fj+1 class induces

a partition into polyhedral regions each of which is a union of rectangles withedge lengths between d j+1 and 12d j+1.When we further restrict Ri

j+1 to eachFj class, note that we still obtain a partition into polyhedral regions each ofwhich is a union of rectangles with edge lengths between d j+1 and 12d j+1,

because a cross section of a rectangular region is still a rectangular regionwhose edge lengths are a subset of the original edge lengths.

We apply Lemma 5.2 as follows. Let the d of Lemma 5.2 be the current d j .

Let the R1, . . . , Rk of that lemma be the current R j+1j , . . . , Ri−1

j . Hypothesis(1) of Lemma 5.2 follows from the current hypothesis (3). Hypothesis (2) ofLemma 5.2 follows from the current hypothesis (5), with j + 1 as the b of thatlemma. Let R be the clopen subequivalence relation of Fj given by Lemma 5.2(i.e., the Rn

d of that lemma). Then each R class is a rectangular region withedge lengths between 9d j and 12d j , and every face of an R region is at least

19000 j16 j2 ( j+1)

d j from any parallel face of any Rkj region with j < k < i.

For each R class A let c(A) be the center point of this rectangular region(uniquely defined in some reasonable manner if A has even edge lengths). Wedefine Ri

j by, for all x, y ∈ F,

x Rij y ⇐⇒ c([x]R)Ri

j+1c([y]R).

Thus, each Rij class is a union of R classes (which are rectangular regions), and

is obtained from an Rij+1 class by adding or subtracting certain R rectangular

regions, which must be near the boundary of the Rij+1 region.

Figure 12 illustrates the definition of Rij . In Fig. 12a the small regions

are R rectangles with edge lengths between d j and 12d j . Their centers arerepresented by the black dots. The solid line represents a face of an Ri

j+1region. Since such faces have edge lengths at least d j+1, with d j+1 � d j ,

they typically look like straight lines at the scale of this picture. Figure 12bshows the boundary of Ri

j regions with the above definition. The boundary

becomes “fractalized” because of the addition and subtraction of R regions tothe original Ri

j+1 region.Conclusion (1) of the current lemma follows from the fact that each of the

R region is a rectangle with edge lengths between 9d j and 12d j .

Conclusion (2) is immediate from the definition of the Rij and the fact that

each R class is a rectangle with edge lengths ≤ 12d j .

That (6) continues to hold follows immediately from conclusion (2) ofLemma 5.2.

123



(A) (B)

Fig. 12 The definition of Rij . a The boundary of R

ij+1 regions. b The boundary of Ri

j regions

Finally we come to the key point; that condition (5) continues to hold.Toward a contradiction assume not, and let B be a ball in an Fj class of radius100,000 ·16 j d j . Suppose j < k1 < · · · < k j+2 ≤ i are such that for each

l = 1, . . . , j + 2, there is an Rklj region with a face intersecting B. By our

hypothesis (4), each of these faces is within 12d j of a face of a region induced

by Rklj+1 restricted to the current Fj class. Let A be the collection of these

Rklj+1 regions restricted to the Fj class.Now consider the Fj+1 class containing the current Fj class. Each A ∈ A

is a cross section of an Rklj+1 region A′. Moreover, every face of an A ∈ A is

a cross section of a face of A′. Thus we obtain j + 2 faces of various Rklj+1

regions each of which has a face with a cross section intersecting B. Sincewe are in ( j + 1)-dimensional space now, at least two of these j + 2 facesmust be parallel. Thus, we have k1, k2 with j < k1 = k2 ≤ i, a face F1 of aregion induced by the restriction of Rk1

j+1 on this Fj+1 class, and a face F2 of

a region induced by the restriction of Rk2j+1 on the same Fj+1 class, so that F1

and F2 are parallel and ρ(F1,F2) ≤ 200, 000 · 16 j d j + 24d j . However, byour hypothesis (7) we also have

ρ(F1,F2) >1

9000 j+116( j+1)2( j + 2)d j+1 > 200, 000 · 16 j d j + 24d j

123



since d j+1 � d j . This contradiction shows that (5) continues to hold andcompletes the proof of the lemma. ��

We are now in a position to prove themain result of this section, that the shiftequivalence relation Fω on F = F(Z<ω) is hyperfinite. In fact, we prove thisin a strong form, namely the reduction to E0 we are seeking can be continuousand injective. Recall from Proposition 1.1 that the two forms of E0, E0(ω

ω)

and E0(2ω), are continuously embeddable into each other. Thus we may usewhichever one is most convenient.

Theorem 6.2 There is a continuous embedding from Fω into E0.

Proof Fix a sufficiently fast growing sequence of natural numbers

1 � d1 � d2 � · · ·

For each i, let Rii be the clopen subequivalence relation of Fi ⊆ Fω from The-

orem 3.1 inducing a rectangular partition of the Fi classes with edge lengths in{di , di +1}. Inductively on i we define the clopen subequivalence relations Ri

i ,

Rii−1, . . . , R

i1. The relation Ri

i has already been defined, and we assume thatRij+1 and all the Rk

l for 1 ≤ l ≤ k < i have been defined. In particular, all of

the relations R jj , R

j+1j , . . . , Ri−1

j have been defined. Moreover, if i > j + 1,

all the R j+1j+1, . . . , R

i−1j+1 have also been defined. We assume inductively that

Rij+1 and the subequivalence relations R

jj , . . . , R

i−1j and R j+1

j+1, . . . , Ri−1j+1 sat-

isfy the hypotheses of Lemma 6.1. We then get Rij from Lemma 6.1. From

Lemma 6.1, we are able to define Rij so that the resulting sequences of sube-

quivalence relations continue to satisfy these hypotheses.So, we may assume the clopen subequivalence relations Ri

j are defined forall i ≥ 1 and all 1 ≤ j ≤ i, and they satisfy the hypotheses of Lemma 6.1. Weshow that if x, y ∈ F(Z<ω) and x Fωy, then for all large enough i we havex Ri

1y.Fix k0 such that x Fk0 y. Toward a contradiction assume that the set

I = {i ≥ 1 : ¬(x Ri1y)}

is infinite. Let x0 = x, x1, . . . , xN = y be a sequence such thatρ(xl, xl+1) = 1and ρ(x, xl) ≤ ρ(x, y). For each i ∈ I there is 0 ≤ l < N such that xl is aboundary point for the Ri

1 region containing x, that is, xl Ri1x and¬(xl Ri

1xl+1).

Thus there is a fixed l such that the set

J = {i ∈ I : xl is a boundary point of the Ri1 region containing x}

123



is infinite. Let z denote this fixed xl . Fix k > k0, and consider k + 1 manyelements of J which are greater than k, say i1 < . . . < ik+1. Consider the Fkclass containing z. Applying conclusion (2) of Lemma 6.1 iteratively, we getthat for each of these im there is a point um on a boundary face Fm of a regioninduced by the restriction of Rim

k so that

ρ(z, um) ≤ 12(d1 + d2 + · · · + dk−1).

This implies that any two of these faces are within 24(d1 + d2 + · · · + dk−1)

of each other. Each of the Fm is a (k − 1)-dimensional hyperplane in k-dimensional space. Since there are k + 1 many of them, at least two of themmust be parallel. However, conclusion (4) of Lemma 6.1 implies that any twosuch parallel faces must be at least

1

9000k16k2(k + 1)dk > 24(d1 + d2 + · · · + dk−1)

apart, using here dk � dk−1. This contradiction shows that x Ri1y for all large

enough i.Finally, we define our embedding π from Fω to E0(ω

ω). For x ∈ F we letπ(x) = (π1(x), π2(x), . . .), where πi (x) ∈ ω is an integer code for the Ri

1class containing x defined as follows. Let ci = ci (x) be the “center point”of the Ri

1 class containing x . By this we mean the center of the rectangularRii class “closest” to the Ri

1 class of x ; for instance, it can be defined as theunique Ri

i class within Hausdorff distance 12d1+· · ·+12di−1 to the Ri1 class

containing x . Fix bijections

κ : P =⋃

{2S : S ⊆ Z<ω is finite} → ω

and

τ : ω × ω → ω.

Let

Ni = Ni (x) = κ(ci � {g ∈ Z

i ∧ ||g|| ≤ 20di })

.

Here we regard Zi ⊆ Z

<ω in the natural way. Let also

hi = hi (x) = the unique element hi ∈ Zi such that hi · ci−1 = ci ,

123



where c0 = c0(x) = x . Let πi (x) = τ(Ni (x), hi (x)). This finishes the defini-tion of πi as well as of π. We have π is continuous since all of the Ri

1 regionsare clopen.

If x, y ∈ F and x Fωy, then for large enough i we have x Ri1y, and so

ci (x) = ci (y). It follows that π(x)E0π(y). On the other hand, from the E0class of π(x) we may recover the Fω class of x using the hi and the fact thatNi (x) codes

ci (x) � {g ∈ Zi ∧ ||g|| ≤ 20di },

while the distance between ci−1(x) and ci (x) is at most 13di . Likewise, π isone-to-one since from π(x) we can in a similar manner reconstruct x . Note inparticular that in the reconstruction of x we need to use the fact that c0(x) = xand that π1(x) encodes h1(x). We have thus proved that π is an embeddingfrom Fω into E0. ��

7 The non-free part

So far in this paper we have considered the free part F of X = 2Z<ω

with theshift action of Z

<ω. As in Sect. 6 we continue to denote the resulting orbitequivalence relation Fω.We showed that Fω is hyperfinite. In fact, we showedthis in a strong way, namely there is a continuous embedding from Fω into E0.

In this section we show that the orbit equivalence relation on the entire spaceX is hyperfinite.Throughout this section let Eω denote the equivalence relation on X gener-

ated by the shift action of Z<ω. For each n, viewing Z

n as a subgroup of Z<ω

in the natural way, we obtain also an action of Zn on X. Let En be the sube-

quivalence relation of Eω generated by this action of Zn. Then Eω = ⋃

n En.

We will actually show that there is a continuous embedding from En into E0.

However, we do not know if there is a continuous embedding from Eω intoE0.

Our study of the non-free part is set up as follows. For x ∈ X, let

Gx = {g ∈ Z<ω : g · x = x}

be the stabilizer of x . The map x �→ Gx is a Borel map from X to P(Z<ω)

(which is essentially ωω). For a subgroup H of Z<ω, let

XH = {x ∈ X : Gx = H}.Then XH is an invariant Borel subset of X. Let EH denote the equivalencerelation Eω restricted to XH . We will define our reduction from Eω to E0 bydefining it on each XH separately. More precisely, we will define a Borel map

123



(x, H) �→ π(x, H) ∈ ωω

with domain the Borel subset of X × P(Z<ω) consisting of those (x, H)

for which Gx = H. For each H, this map restricted to XH (more preciselyXH ×{H}) will be a reduction of (XH , EH ) to E0. This will suffice as we canthen easily modify π to π ′ which codes H into the output as well as π(x, H).

Then π ′ will be a reduction of (X, Eω) to E0. Since x determines H, we willusually write π(x) instead of π(x, H). In the argument below we will in factfix the subgroup H and just define π on XH . The reader can check that themap we define is actually Borel as a function of both arguments x and H. Thefact that the map x �→ Gx is only Borel (and not continuous), however, iswhat prevents us from getting a continuous reduction.

7.1 Preliminaries about the space XH

We let

e1 = (1, 0, 0, . . .), e2 = (0, 1, 0, . . .), . . . , en, . . .

be the standard set of generators for the groupZ<ω.Every element g ∈ Z

<ω canbewritten uniquely in the form g = a1e1+· · ·+akek = (a1, . . . , ak, 0, 0, . . .),where ak = 0. We say that k = supp(g) is the support of g (and if g =(0, 0, . . .) we say that supp(g) = ∅). We view Z

n as the subgroup of Z<ω

generated by e1, . . . , en. We let �0 denote the identity element (0, 0, . . .) ofZ

<ω. For g = a1e1 + · · · + akek ∈ Z<ω and n ≥ 1, let πn(g) = an for n ≤ k

and πn(g) = 0 for n > k.

Definition 7.1 Let H be a subgroup of Z<ω. We say that B = {h1, h2, . . .} is

a normal basis for H if the following are satisfied:

(1) H = 〈h1, h2, . . .〉.(2) supp(h1) < supp(h2) < · · ·(3) For every i, if k = supp(hi ), then πk(hi ) > 0.(4) For every n, H ∩ Z

n = 〈hi : supp(hi ) ≤ n〉.We have the following standard algebraic fact. For the sake of completeness

we include a proof.

Lemma 7.2 Every subgroup H of Z<ω has a normal basis.

Proof Let h1, h2, . . . enumerate the non-�0 elements of the sequence h′1,

h′2, . . . , where h

′n is defined as follows. If H ∩ Z

n = H ∩ Zn−1, let h′

n = �0.Otherwise, let In = {πn(h) : h ∈ H ∩ Z

n}. In is an ideal in Z, and so

123



is principal. Let h′n ∈ H be such that In = 〈πn(h′

n)〉 and πn(h′n) > 0.

It is straightforward to check that H ∩ Zn = 〈h′

1, . . . , h′n〉, and the result

follows. ��It is clear from the above proof that one may choose a normal basis in a

Borel manner from H.

Henceforth we fix an arbitrary subgroup H of Z<ω and a normal basis

B = {h1, h2, . . .} for H.

Definition 7.3 We say n is an essential dimension if H ∩ Zn = H ∩ Z

n−1.

Otherwise we say n is an inessential dimension.

So, n is an inessential dimension precisely when we have n = supp(hi ) forsome i (necessarily i ≤ n). Moreover, if n is an inessential dimension thenthere is a unique i ≤ n such that n = supp(hi ).

Definition 7.4 For each n, we define

s(n) ={∞, if n is an essential dimension,

πn(hi ), otherwise if n = supp(hi ) for some i ≥ 1.

We call s(n) the size of the nth dimension.

Thus, the values of s(n) are finite for the inessential dimensions. The followingfact records the significance of the size function.

Lemma 7.5 For every g ∈ Z<ω, there is a unique g′ ∈ Z

<ω such that g−g′ ∈H and such that for all inessential dimensions n, 0 ≤ πn(g′) < s(n).

Proof We first show the existence of such a g′. Let n0 be the largest inessentialn ≤ supp(g) such that πn(g) < 0 or πn(g) ≥ s(n). Fix i ≥ 1 so thatsupp(hi ) = n0. From the definition of s(n) there is an integer a with 0 ≤πn0(g−ahi ) < s(n0).Also,πn(g−ahi ) = πn(g) for n > n0.By induction onn0, there is an h ∈ H such that g′ = (g−ahi )−h satisfies 0 ≤ πn(g′) < s(n)

for all inessential n. So, g = g′ + (ahi + h) with ahi + h ∈ H, and we aredone.

To see the uniqueness, suppose g′′ also satisfied the requirements, and g′′ =g′. Note that g′′ − g′ ∈ H. Let n1 denote the largest n such that πn(g′′) −πn(g′) = 0. Since g′′ − g′ ∈ H, we must have that n1 is inessential. However,note also 0 < |πn1(g

′′)−πn1(g′)| < s(n1).Without loss of generality we may

assume 0 < πn1(g′′)−πn1(g

′) < s(n1). This, however, violates the definitionof s(n1) since g′′ − g′ ∈ H and s(n1) is the minimal value of |πn1(h)| forh ∈ H with supp(h) ≤ n1. ��

In view of Lemma 7.5, we can make the following definition.

123



Definition 7.6 We say that g ∈ Z<ω is in standard form if for all inessential

dimensions n we have 0 ≤ πn(g) < s(n). Denote the set of all g ∈ Z<ω in

standard form by �. For each g ∈ Z<ω let σ(g) denote the unique g′ ∈ �

such that g − g′ ∈ H.

In general, if g and h are in standard form, then −g and g + h are notnecessarily in standard form. We need an estimate for the norm of σ(−g) −(−g) and that ofσ(g+h)−(g+h).Recall that for g = a1e1+· · ·+akek ∈ Z

<ω,

the norm of g is defined as ‖g‖ = max{|ai | : i ≤ k}.Wewill use the followingquantity throughout the rest of the section.

Definition 7.7 For each inessential dimension k, define ck = ‖hi‖ wheresupp(hi ) = k. For each n ≥ 1 let Dn be the set of all inessential dimensionsk ≤ n. Let s = max{s(k) : k ∈ Dn}. Define

β(n) =∏

k∈Dn

(ck + 2s) .

Lemma 7.8 Let g, h ∈ � ∩ Zn. Then

(a) ‖σ(−g) + g‖ ≤ β(n), and(b) ‖σ(g + h) − g − h‖ ≤ β(n).

Proof (a) Suppose g = a1e1 + · · · + anen ∈ �. Let k1 < k2 < . . . < kmenumerate the inessential dimensions ≤ n. We have supp(h j ) = k j for all1 ≤ j ≤ m. Since g is in standard form, we have 0 ≤ ak j = πk j (g) ≤s(k j ) = πk j (h j ) for all 1 ≤ j ≤ m. Let s = max{s(k j ) : 1 ≤ j ≤ m}.

Now −g = (−a1)e1 + · · · + (−an)en. Since σ(−g) + g ∈ H ∩ Zn, there

is a unique sequence α1, . . . , αm of integers such that

σ(−g) + g = α1h1 + α2h2 + · · · + αmhm .

This gives

σ(−g) = α1h1 + α2h2 + · · · + αmhm + (−a1)e1 + · · · + (−an)en.

Consider the projection πkm . Since σ(−g) is in standard form, we have0 ≤ πkm (σ (−g)) < s(km). This gives 0 ≤ αms(km)−akm < s(km).However,0 ≤ akm < s(km). So 0 ≤ αm ≤ 1.

Next we consider the projection πkm−1 . Note that

|αmπkm−1(hm) − akm−1 | ≤ ckm + s.

The expression appearing in between the absolute value signs is the coefficientof σ(−g) at the coordinate ekm−1 before αm−1hm−1 is taken into account. Itfollows that |αm−1| ≤ ckm + s.

123



In a similar fashion, we next consider the projection πkm−2 . Note that

|αm−1πkm−2(hm−1) + αmπkm−2(hm) − akm−2 |≤ (ckm + s)ckm−1 + ckm + s

≤ (ckm + s)(ckm−1 + s).

The expression appearing in between the absolute value signs above is thecoefficient of σ(−g) at the coordinate ekm−2 before αm−2hm−2 is taken intoaccount. From this we get that |αm−2| ≤ (ckm + s)(ckm−1 + s). Continuing inthis fashion, we get that

|α j | ≤∏

j+1≤l≤m

(ckl + s)

for all 1 ≤ j ≤ m and

‖α1h1 + · · · + αmhm‖ ≤ |α1|ck1 + · · · + |αm |ckm ≤∏

1≤ j≤m

(ck j + s) < β(n).

(b) Suppose g = a1e2 + · · · + anen ∈ � and h = b1e1 + · · · + bnen ∈ �.

Again let k1 < k2 < · · · < km enumerate the inessential dimensions≤ n. Let

s = max{s(k j ) : 1 ≤ j ≤ m}.

Let α1, . . . , αm be such that

σ(g + h) = α1h1 + · · · + αmhm + (a1 + b1)e1 + · · · + (an + bn)en.

Consider successively πkm , πkm−1, . . . , πk1 as in the proof of (a). First weobtain

0 ≤ αms(km) + akm + bkm < s(km),

which implies |αm | ≤ 1. Then we have

|αmπkm−1(hm) + akm−1 + bkm−1 | ≤ ckm + 2s,

which implies |αm−1| ≤ ckm + 2s. In general, we have

|α j | ≤∏

j+1≤l≤m

(ckl + 2s)

123



for all 1 ≤ j ≤ m, and so

‖α1h1+· · ·+αmhm‖≤|α1|ck1 + · · · + |αm |ckm ≤∏

1≤ j≤m

(ck j + 2s) = β(n).

��Note that Lemma 7.5 implies that � is a set of coset representatives for H

in Z<ω. For x ∈ XH , we have [x]E = {g · x : g ∈ �}. If g, h ∈ � and g = h,

then g ·x = h ·x . So, for a fixed x we can identify [x]E with the set of g ∈ Z<ω

in standard form. Note that this identification is not invariant, but depends onx .

Belowwe define a notion of distance on XH , although it will not correspondto a true metric.

Definition 7.9 For x, y ∈ XH with xEn y, let ρnx (y) = ‖g‖, where g ∈ Z

n isthe unique element in standard form such that g · x = y.

Thus, ρnx (y) is the “distance” from x to y as seen in the local (i.e., centered

about x) rectangular coordinate systemabout x .This does not give a truemetricsince ρn

x (y) = ρny (x) in general. However, in some sense these functions are

close to a metric. We make this precise in the following lemma.

Lemma 7.10 For any x, y ∈ XH with x En y we have |ρnx (y)−ρn

y (x)| ≤ β(n).

Proof Let g ∈ � ∩ Zn be such that g · x = y. Then ‖g‖ = ρn

x (y). We alsohave σ(−g) · y = (−g) · y = x and ‖σ(−g)‖ = ρn

y (x). Thus

|ρnx (y) − ρn

y (x)| = |‖g‖ − ‖σ(−g)‖| ≤ ‖σ(−g) + g‖ ≤ β(n)

by Lemma 7.8(a). ��With this notion of distance, we can now extend our basic marker result to

the space XH .

Lemma 7.11 Fix n and amarker distance d � β(n).Then there is a relativelyclopen set Mn

d ⊆ XH satisfying the following:

(i) For every x, y ∈ Mnd with x En y we have ρn

x (y) ≥ d − β(n).

(ii) For every y ∈ XH , there is an x ∈ Mnd such that ρn

x (y) ≤ d.

Proof The proof is similar to that of Lemma 2.1. Let s0, s1, . . . enumerate theelements s of 2<Z

nsatisfying:

(1) The basic neighborhood Ns ∩ XH of XH is nonempty.(2) For every �0 = g ∈ Z

<ω in standard form with ‖g‖ ≤ d + β(n), g · Ns ∩Ns = ∅.

123



As in the proof of Lemma 2.1 we define by induction on i a sequence of clopensets Si ⊆ XH . Let S0 = Ns0 ∩ XH . Note that if x, y ∈ Nsi ∩ XH and x = y,then ρn

x (y), ρny (x) > d +β(n). In particular, (i) holds for the set S0. For i > 0

we define

Si = Si−1 ∪(Nsi −

⋃{g · Si−1 : ‖g‖ ≤ d and g is in standard form}

).

Let Mnd = ⋃

i Si . If x = y are both in Si−1, then ρnx (y), ρn

y (x) > d by induc-tion. If x = y are both in Nsi , then ρn

x (y), ρny (x) ≥ d+β(n). Finally, suppose

x ∈ Si−1 and y ∈ Nsi −⋃{g ·Si−1 : ‖g‖ ≤ d and g is in standard form}.Then

ρnx (y) > d. From Lemma 7.10 we also have ρn

y (x) > d − β(n). It followsthat (i) holds for Mn

d .

To see (ii), let y ∈ XH . Let i be the least such that y ∈ Nsi . This existssince for every �0 = g ∈ Z

<ω in standard form, g · y = y, and so for someneighborhood N of y, g · N ∩ N = ∅. If (ii) failed for y, then

y /∈⋃

{g · Si−1 : ‖g‖ ≤ d and g is in standard form}.

It follows from the definition of Si that y ∈ Si ⊆ Mnd , which contradicts the

failure of (ii). ��

7.2 Marker regions in XH

As in the case of the free part, we consider marker regions in XH that arerectangular and almost rectangular. These notions are defined below. In thissectionwewill use the vector notion �a to denote a sequence {ai } for the essentialdimensions i ≤ n for some fixed n. We use the following conventions aboutvectors:

(1) �a < �b if for ai < bi for all essential dimensions i ≤ n; similarly for≤, >,≥, etc. between vectors;

(2) �a + �b denotes the sequence {ai + bi }; similarly for −.

(3) |�a| denotes the sequence {|ai |}.(4) If k is a number then k�a denotes the sequence {kai }.(5) If k is a number then �a > k means ai > k for all essential dimensions

i ≤ n; similarly for ≤, >,≥, etc.(6) If k is a number then �a + k denotes the sequence {ai + k}; similarly

for −.

Definition 7.12 Given n ≥ 1 and integers ai < bi for the essential dimensionsi ≤ n, let

123



�(�a, �b) ={g ∈ � ∩ Zn : for all essential dimensions i≤n, ai ≤ πi (g) ≤ bi }

= {g ∈ Zn : for all essential dimensions i ≤ n, ai ≤ πi (g) ≤ bi∧ for all inessential dimensions i ≤ n, 0 ≤ πi (g) ≤ s(i)}

A rectangle in an En equivalence class of XH is a set of the form �(�a, �b) · xfor some x and �a < �b. We denote it by R(x; �a, �b).

In view of the reference point x,we also say that R(x; �a, �b) is a rectangle inthe x-coordinate system. We refer to �b− �a as the edge lengths of the rectangleR(x; �a, �b). These are explicitly defined only for the essential dimensions; foran inessential dimension i ≤ n, the implicit edge length is s(i).

Definition 7.13 Wesay that R(x; �a, �b) is centeredabout the point x if �a = −�b.In this case we say x is the center of the rectangle R(x; �a, �b).

In general, if yEnx but y = x, then a rectangle R in the x-coordinatesystem may no longer be a rectangle in the y-coordinate system. However, byLemma 7.10 R will be close to a rectangle in the y-coordinate system withβ(n) as a bound for the error. In other words, R will still be an almost rectangle(the precise definition and statement follow). In particular, if the edge lengthsof R are large compared with β(n), then in the y-coordinate system, R willstill be almost a rectangle of the same edge lengths. This will be a key pointbehind the arguments in the rest of this section.

We now make the notion of an almost rectangle precise.

Definition 7.14 Let � be a positive integer. By an �-almost rectangle in an Enequivalence class of XH we mean a set R such that there is x ∈ XH and �a < �bwith

R(x; �a + �, �b − �) ⊆ R ⊆ R(x; �a − �, �b + �).

In this casewe say that R is an �-almost rectangle in the x-coordinate systemwith edge lengths �b − �a (again specified only for the essential dimensions).Note that the edge lengths of an almost rectangle are not well defined.

Lemma 7.15 Let R be a rectangle in an En equivalence class [x]En of XH .

Let yEnx . Then R is a β(n)-almost rectangle in the y-coordinate system.

Proof Without loss of generality assume R = R(x; �a, �b) = �(�a, �b) · x . Leth ∈ � ∩ Z

n be such that h · y = x . Then

R = (�(�a, �b) + h) · y = {σ(g + h) : g ∈ �(�a, �b)} · y.For each essential dimension i ≤ n, let a′

i = ai + πi (h) and b′i = bi + πi (h).

Then we claim that

R(y; �a′ + β(n), �b′ − β(n)) ⊆ R ⊆ R(y; �a′ − β(n), �b′ + β(n)). (E2)

123



To show this it suffices to verify that

�(�a′ + β(n), �b′ − β(n)) ⊆ {σ(g + h) : g ∈ �(�a, �b)}⊆ �(�a′ − β(n), �b′ + β(n)).

Let g ∈ �(�a, �b). By Lemma 7.8(b) we have ‖σ(g + h) − g − h‖ ≤ β(n). Itfollows that for any essential dimension i ≤ n,

πi (g) + πi (h) − β(n) ≤ πi (σ (g + h)) ≤ πi (g) + πi (h) + β(n).

From this the second inclusion follows.To prove the first inclusion, let g ∈ �(�a′ + β(n), �b′ − β(n)). Then for all

essential dimensions i ≤ n,

ai + πi (h) + β(n) ≤ πi (g) ≤ bi + πi (h) − β(n),

and thus

ai + β(n) ≤ πi (g − h) ≤ bi − β(n).

Again by Lemma 7.8(b) we have ‖σ(g− h)− g+ h‖ ≤ β(n). It follows that

ai ≤ πi (g − h) − β(n) ≤ πi (σ (g − h)) ≤ πi (g − h) + β(n) ≤ bi .

This shows that σ(g − h) ∈ �(�a, �b). The inclusion follows upon noting thatg = σ(σ(g − h) + h) since g ∈ �. ��

Wenext consider center points for almost rectangles. Intuitively, such centerpoints are not unique but they should be close to each other. Our next definitionand lemma make this precise.

Definition 7.16 Let � be a positive integer and R be an �-almost rectangle insome En equivalence class. By an �-almost center for R we mean a point xsuch that there is �b > 0 with

R(x;−�b + �, �b − �) ⊆ R ⊆ R(x;−�b − �, �b + �).

In this case we say that R is an �-almost rectangle centered about x .

In the degenerate case that bi < � for some i, any point x with

R ⊆ R(x;−�b − �, �b + �)

is an �-almost center for R. In particular, an �-almost center need not becontained in R. Of course, when the edge lengths of the �-almost rectangle

123



R are sufficiently large, then the situation is nondegerate, and every �-almostcenter for R is necessarily contained in R.

Lemma 7.17 Let R be an �-almost rectangle in an En equivalence class. Letx be an �-almost center for R.

(a) If yEnx, then y is an (� + ρny (x) + β(n))-almost center for R.

(b) If y is also an �-almost center, then ρny (x) ≤ 2� + β(n).

Proof Assume R(x;−�b+�, �b−�) ⊆ R ⊆ R(x;−�b−�, �b+�).Let h ∈ �∩Zn

be such that h · y = x .Then ρny (x) = ‖h‖. Similar to the proof of Lemma 7.15,

if we define ci = πi (h) for all essential dimensions i ≤ n, then by the Eq. (E2)in that proof, we have

R(y;−�b + �c + � + β(n), �b + �c − � − β(n)) ⊆ R(x;−�b + �, �b − �) ⊆ R

and

R ⊆ R(x;−�b − �, �b + �) ⊆ R(y;−�b + �c − � − β(n), �b + �c + � + β(n)).

It follows that

R(y;−�b + |�c| + � + β(n), �b − |�c| − � − β(n))

⊆ R ⊆ R(y;−�b − |�c| − � − β(n), �b + |�c| + � + β(n)).

Since |ci | ≤ ρny (x), we get that y is an (� + ρn

y (x) + β(n))-almost center forR. This proves (a).For (b), assume y is an �-almost center for R. Thus for some �d > 0 we

have

R(y;−�d + �, �d − �) ⊆ R ⊆ R(y;−�d − �, �d + �).

It follows that

R(y;−�d + �, �d − �) ⊆ R(y;−�b + �c − � − β(n), �b + �c + � + β(n))

and

R(y;−�b + �c + � + β(n), �b + �c − � − β(n)) ⊆ R(y;−�d − �, �d + �).

These in turn imply that for all essential dimensions i ≤ n,

− di + � ≥ −bi + πi (h) − � − β(n),

di − � ≤ bi + πi (h) + � + β(n),

123



− bi + πi (h) + � + β(n) ≥ −di − �,

bi + πi (h) − � − β(n) ≤ di + �.

These imply

bi + |πi (h)| − 2� − β(n) ≤ di ≤ bi − |πi (h)| + 2� + β(n),

and so |πi (h)| ≤ 2� + β(n). ��It is not clearwhether every �-almost rectanglemust have an �-almost center.

But we have the following fact.

Lemma 7.18 Let R be an �-almost rectangle in an En equivalence class. ThenR has an (� + β(n) + 1)-almost center.

Proof Assume R = R(x; �a, �b). Let ci be the integer part of 12 (ai + bi ) for all

essential dimensions i ≤ n. Let h ∈ � be such that πi (h) = ci for all essentialdimensions i ≤ n and πi (h) = 0 for all inessential dimensions i ≤ n. Lety = h · x . Then (−h) · y = x and −h ∈ �. By the Eq. (E2) in the proof ofLemma 7.15 we have

R(y; �a − �c + � + β(n), �b − �c − � − β(n))

⊆ R ⊆ R(y; �a − �c − � − β(n), �b − �c + � + β(n)).

Since �c − �a ≤ �b − �c ≤ �c − �a + 1, we get that y is an (� + β(n) + 1)-almostcenter for R. ��

In practice, the almost rectangles we construct will come from actual rectan-gles and will have almost centers with possibly a slightly different parameter�. By Lemma 7.17(b), the set of �-almost centers for an �-almost rectangle Ris finite. This will allow us to define a finitary algorithm (which we will notspecify) to identify a “lexicographically least” �-almost center of R, whichwe denote by cn(R). As we noted above, if R has sufficiently large edgelengths then cn(R) ∈ R. As usual it follows that the function cn is contin-uous in the sense that for each finite set {g1, . . . , gm} ⊆ � ∩ Z

n, the mapx �→ cn({g1 · x, . . . , gm · x}) from XH to XH is continuous.

For further discussions on the construction of regular marker regions in XH ,

we are going to use the following notion of distance between two sets.

Definition 7.19 Given two sets A and B in an En equivalence class of XH ,

we define the Hausdorff distance between A and B by

ρn(A, B) = max

{

supx∈A

infy∈B ρn

x (y), supy∈B

infx∈A

ρny (x)

}

.

123



The definition is motivated by the observation that, if R is an �-almostrectangle in the x-coordinate system, then there is a rectangle R′ in the x-coordinate system such that ρn(R, R′) ≤ �.

We now make more remarks on the almost rectangles we will construct. Inpractice, the almost rectangles we use will be regions whose boundaries arefaces of actual rectangles. In fact, the almost rectangles themselves will not beactual rectangles because the different faces will come from rectangles havingdifferent centers. If x = y are two such centers, a “flat face” of a rectangle inthe x-coordinate system will not be flat in the y-coordinate system.

More specifically, for each essential dimension i ≤ n of a rectangleR(x; �a, �b), there are two i-faces of R, that is, the points of R of the form{g · x : πi (g) = ai } and {g · x : πi (g) = bi }. Note that these faces are definedwith x as a necessary parameter. From the proof of Lemma 7.15 we have thatif R is a rectangle in the x-coordinate system and F is an i-face of R (withrespect to x), and yEnx, thenF is within Hausdorff distance β(n) of an i-faceof a rectangle R′ in the y-coordinate system.

Before closing this subsection we define a notion of simultaneous almostrectangles for all dimensions and prove a strengthening of Lemma 7.15 forthis notion. The following definition and lemma will be important for ourhyperfiniteness proof. In stating the definitionwe deviate slightly from the con-ventions we adopted earlier for the notion of vectors. Henceforth �β denotes thesequence {β(m)} for allm ≤ n.We also refer to various other vectors of differ-ent lengths.Wewill define them explicitly when there is a danger of confusion.

Definition 7.20 A region R in an En equivalence class [x]En is said to be a�β-almost rectangle in En if for every m ≤ n, and every yEnx, R ∩ [y]Em is aβ(m)-almost rectangle in Em .

Lemma 7.21 Let R be a rectangle in an En-class [x]En . Let yEnx . Then Ris a �β-almost rectangle in En in the y-coordinate system.

Proof Let R = �(�a, �b) · x for �a < �b. Let m ≤ n. Let a′i = ai and b′

i = bi fori ≤ m. Let h ∈ � ∩ Z

n be such that h · y = x . Let

hl = π1(h)e1 + · · · + πm(h)emand

hr = h − hl = πm+1(h)em+1 + · · · + πn(h)en.

Then both hl and hr are still in standard form. Now for g ∈ �(�a, �b) we have

g · x ∈ R ∩ [y]Em ⇐⇒ g ∈ �(�a, �b) ∧ g · x ∈ [y]Em

⇐⇒ g ∈ �(�a, �b) ∧ ∃g′ ∈ Zm (g · x = g′ · y)

123



⇐⇒ g ∈ �(�a, �b) ∧ ∃g′ ∈ Zm (g + h = g′)

⇐⇒ g ∈ �(�a, �b) ∧ ∃g′ ∈ Zm (g + hr = g′).

Note that if g′ ∈ Zm and g′ = g + hr , then g′ ∈ �(�a′, �b′). Thus R ∩ [y]Em is

of the form:

{(g′ − hr ) · x : g′ ∈ �(�a′, �b′)}and also of the form

{(g′ + hl) · y : g′ ∈ �(�a′, �b′)}.Since both g′ and hl are in standard form, the proof of Lemma 7.15 shows thatthis is a β(m)-almost rectangle in the y-coordinate system. ��

7.3 Hyperfiniteness of EH

Our proof of the hyperfiniteness of EH will follow closely the outline of proofsin the free part.

We next generalize our rectangular marker region lemma to the non-freecase.

Lemma 7.22 Fix n, and let d � β(n). Then there is a relatively clopenpartition R of XH such that each R ∈ R is a �β-almost rectangle in En withedge lengths d.

Proof The proof is similar to that of Theorem 3.1. Fix �2 � �1 � D > d2.Let M1 = Mn

D1⊆ XH be the clopen marker set from Lemma 7.11 for distance

�1. Let N ⊆ XH be a relatively clopen marker set for distance �2. For eachx ∈ N , let gx ∈ � ∩ Z

n be lexicographically least such that gx · x ∈ M1. ByLemma 7.10, ‖gx‖ ≤ �1 +β(n). Let M2 = {gx · x : x ∈ N } ⊆ M1. Then M2is also relatively clopen.

If x = y are both in M2, we claim that ρnx (y) ≥ �2 − (2�1 + 6β(n)). To

see this, let g1 ∈ � and x ′ ∈ N with g1 · x ′ = x and ‖g1‖ ≤ �1 + β(n),

let g2 ∈ � with g2 · x = y and ‖g2‖ = ρnx (y), and finally let g3 ∈ � and

y′ ∈ N with g3 · y′ = y and ‖g3‖ ≤ �1 + β(n). Then σ(−g3) · y = y′and ‖σ(−g3)‖ ≤ �1 + 2β(n) by Lemma 7.8(a). Let g = g1 + g2 + σ(−g3).Then g · x ′ = y′ and ‖g‖ ≤ ρn

x (y) + 2�1 + 3β(n). Since g is a sum of threeelements in standard form, Lemma 7.8(b) gives ‖σ(g)‖ ≤ ‖g‖ + 2β(n). So,ρnx ′(y′) = ‖σ(g)‖ ≤ ρn

x (y) + 2�1 + 5β(n). Since ρnx ′(y′) ≥ �2 − β(n), this

shows ρnx (y) ≥ �2 − (2�1 + 6β(n)) � �1.

A similar argument shows that for every y ∈ XH , there is an x ∈ M2 suchthat ρn

x (y) ≤ �2 + �1 + 3β(n). Let g1, . . . , gk enumerate the elements g ∈

123



�∩Zn with ‖g‖ ≤ �2+�1+3β(n).We define the sequence A0, A1, . . . , Ak

of relatively clopen subsets of XH as follows. Let A0 = M2. For i > 0 letAi = M1 ∩ (gi · M2) − ⋃

ι<i Aι. So, M1 is the disjoint union of the Ai , andfor every i and every x = y in Ai we have ρn

x (y) ≥ �2 − (2�1 + 6β(n)).

For each x ∈ M1, let Rx ⊆ [x]En be the rectangle centered about x withedge lengths 4�1. Thus, XH = ⋃

x∈M1Rx . In k + 1 steps we adjust the

rectangles Rx for x in A0, . . . , Ak respectively to rectangular regions R′x . We

will move each face of a rectangle Rx at each step by no more than 110�1.

At each step, as we define R′x we also describe a partition of R′

x into smalleralmost rectangular regions.

To begin, for x ∈ A0 let R′x = Rx . Suppose the adjusted rectangles R′

x havebeen defined for x ∈ A j , j < i. Consider x ∈ Ai . We move each face of Rxto avoid parallel faces of nearby R′

y for y ∈ ⋃ι<i Aι. To be precise, consider

an e-face F of Rx , where e ≤ n is an essential dimension. A face of R′y is

parallel to F if it is an e-face of R′y . Without loss of generality, assume that

F = {g · x : πe(g) = 2�1 ∧ |π j (g)| ≤ 2�1 for all essential dimensions

j = e ∧ 0 ≤ π j (g) < s( j) for all inessential dimensions j}.

(The case where πe(g) = −2�1 is similar.) We define an integer a with|a| ≤ 1

10�1, and shift the face to

F ′ = {g · x : πe(g) = 2�1 + a ∧ |π j (g)| ≤ 2�1 for all essential dimensions

j = e ∧ 0 ≤ π j (g) < s( j) for all inessential dimensions j}.

Shifting these faces will define the new rectangle R′x .Wewish to find a so that

the shifted face F ′ will satisfy:

for any parallel faceG of a rectangle R′y for y ∈

⋃

ι<i

Aι, (*)

and any u ∈ F ′, v ∈ G, ρnu (v) ≥ 2D + β(n).

There is clearly a bound N = N (n) depending only on n, for the number ofy ∈ ⋃

ι<i Aι with ρnx (y) ≤ 6�1. In fact a volume argument in the x-coordinate

system shows this number to be bounded by

(13�1)m

(�1/2 − β(n))m≤ 52m ≤ 52n,

where m is the number of essential dimensions ≤ n. In fact, let Dn be the setof all inessential dimensions j ≤ n. For any distinct y, z ∈ M1,

123



ρny (z) ≥ �1 − β(n) > �1 − 2β(n).

Thus the rectangles centered about y with edge lengths 12 (�1 − 2β(n)) are

pairwise disjoint for distinct y ∈ M1 if �1 > 4β(n), which follows from theassumption that�1 � β(n), and there are atmost (12�1−β(n))m ·∏ j∈Dn

s( j)many elements in any one of these rectangles. However, if ρn

x (y) ≤ 6�1,

then the rectangle centered about y with edge lengths 12�1 − β(n) is entirely

contained in the rectangle centered about x with edge lengths

2 ·(

6�1 + (1

2�1 − β(n)) + β(n)

)

= 13�1,

and the latter rectangle has (13�1)m ·∏ j∈Dn

s( j)many elements. The estimatenow follows from again �1 ≥ 4β(n).

Note that if R′y is a previously adjusted rectangle having a face G parallel

to F and such that there are u ∈ F, v ∈ G with ρnu (v) ≤ �1, then

ρnx (y) ≤ (5

1

10)�1 + 3β(n) < 6�1.

Thus, there are at most 2N faces G to avoid. In the x-coordinate system, eachof these faces consists of points whose eth coordinates vary by at most β(n),

by Lemma 7.8(b). The interval of possible values for a has size 210�1, and

there are at most 2N many intervals of size 2β(n) to avoid. We may find ana in the interval of possible values which avoids all of the smaller intervalsby at least ε as long as 2�1

10 > 2N (2β(n) + 2ε). Given that �1 is sufficientlylarge compared to β(n), we may assume that 2�1

10 > 2N (4D + 4β(n)) (withε = 2D + β(n)), and it follows that (∗) holds. This completes the definitionof the R′

x for x ∈ Ai .

Finally, we partition each new adjusted rectangle R′x . For each essential

dimension e ≤ n, let Ge0, . . . ,Ge

Leenumerate the e-faces of the R′

y, y ∈⋃

ι<i Aι, which intersect R′x . Each of these faces Ge

l is a rectangular facein the y-coordinate system for some y with yEnx . That is, Ge

l ⊆ Hel =

{g · y : πe(g) = cl} for some cl . The faces {Hel } partition R′

x into smallerregions, each of which is a �β-almost rectangle in En with edge lengths atleast 2D. Note that a finite intersection of �β-almost rectangles is a �β-almostrectangle. From this observation it follows that all of the regions produced are�β-almost rectangles. The set S = R′

x − ⋃ι<i

⋃y∈Aι

R′y is a union of a set of

such �β-almost rectangular regions.At the end of step k, we have partitioned each En class into �β-almost

rectangular regions, each having edge lengths of at least 2D− 2β(n). That is,for any u, v on parallel faces of the almost rectangular subregion, ρn

x (y) ≥

123



2D − 2β(n). Finally, since 2D − 2β(n) > D > d2, it is straightforwardto divide each such �β-almost rectangular region T into �β-almost rectangularsubregions with edge lengths almost d, that is, for u, v on parallel faces wehave d−β(n)−1 ≤ ρn

x (y) ≤ d+β(n)+1 Namely, work in the x-coordinatesystem. For each essential dimension i ≤ n, let [ai , bi ] be an interval inZ approximating within β(n) the values of πi (g) for the set of g such thatg · x ∈ T .Divide each such interval into subintervals each of length d or d+1.This defines the partition of T into �β-almost rectangles of edge length d. ��

We next give the analog of Lemma 5.1 in the non-free case.

Lemma 7.23 Fix n and d � β(n). Suppose Rnd is a clopen partition of

(XH , En) into �β-almost rectangular regions of edge length d. Then thereis a clopen subequivalence relation Rn

d on (XH , En) satisfying:

(1) Each Rnd class is a �β-almost rectangle with edge lengths between 9d and

12d.

(2) If F1 is a face of a Rnd region R1, and F2 a parallel face of a Rn

d regionR2, then for any u ∈ F1, v ∈ F2, ρn

u (v) > 19000n16n2

d.

Proof As in Lemma 5.1, let s = Cd where C = 110 · 4 · 8 · 16n. As inLemma 7.22, there is a clopen set M = M0 ∪ M1 ∪ · · · ∪ Mk such that the setR of rectangles of edge lengths s centered at the points of M covers XH , andfor each l, and any x = y in Ml, ρn

x (y) � s. As in the proof of Lemma 5.1,we proceed in k steps. At step l we adjust the rectangles with centers in Ml .

As in that proof we do each adjustment in two steps. The set we are adjustingis a rectangle Rx in the x-coordinate system about the center point x . Foreach essential dimension i ≤ n, we adjust each of the two i-faces of Rx . Theprecise notion of “face” and “adjustment” is as in the proof of Lemma 7.22.As in Lemma 5.1, in the first step we adjust by no more than s

8 , and in thesecond step by no more than s

4·8·16n < s8 . The various faces to be avoided are

now no longer true faces as seen in the x-coordinate system, but within β(n)

of a face of a rectangle in the x-coordinate system. The key point is that aslong as d is sufficiently large compared to β(n), then the inequalities in theproof of Lemma 5.1 continue to hold.

Consider, for example the second step of the adjustment. Each of the sur-rounding almost rectangles to be avoided can be approximated within β(n) bya rectangle with edge lengths d (for the essential dimensions). Say there arem ≤ n essential dimensions i ≤ n. Then the estimate (E1) from Lemma 5.1is modified by replacing occurrences of d in the numerator by d + β(n), andin the denominator by d − β(n). So, we have:

123



2 ·( s4·8·16n + 3(d + β(n))

) · (5s + 23(d + β(n)))m−1

(d − β(n))m

as the upper bound, which is still less than 9000n ·16n2 if d is sufficiently largecompared to β(n).As in the proof of Lemma 7.22, we can avoid parallel facesfrom these almost rectangles by ε as long as

s

4 · 8 · 16n ≥ 9000n · 16n2 · (2β(n) + 2ε).

With ε = 19000n ·16n2 d, this requires

110d ≥ 2 · 9000n · 16n2 · β(n) + 2d,

which follows from our assumption that d � β(n). The rest of the proof issimilar to the argument for Lemma 5.1. ��

In an exactly similar fashion we also have the following analog ofLemma 5.2.

Lemma 7.24 Fix n and d � β(n). Let R1, . . . , Rk be a sequence of clopensubequivalence relations of (XH , En) satisfying the following:

(1) On each En class, each Ri induces a partition into regions R which areunions of �β-almost rectangles r such that each edge length of one of ther rectangles is between d and 12d.

(2) In any ball B = {y ∈ [x]E j : ρjx (y) ≤ 10, 000 ·16 j d j } of radius 100,000 ·

16nd in some En equivalence class, there are at most b integers i suchthat one of the Ri regions has a face F which intersects B.

Then there is a clopen subequivalence relation Rnd ⊆ En satisfying:

(1) Each Rnd class is a �β-almost rectangular region with edge lengths between

9d and 12d.

(2) If F is a face of an Rnd region R, and F ′ is a parallel face of an Ri region

R′ (for any i), then for any u ∈ F, v ∈ F ′, ρnu (v) > 1

9000n16n2bd.

To prove that EH is hyperfinite, we inductively define finite subequivalencerelations Ri in a similar fashion as for the free part. We need the analog ofLemma 6.1 which we now state. In this lemma all equivalence relations aresubequivalence relations of (XH , Ei ).

Lemma 7.25 Let j < i. Let d j+1, d j be positive integers with d j+1 � d j ,

d j � β( j), and d j+1 � β( j + 1). Suppose Rij+1 is a clopen subequivalence

relation of Ei . Assume that the restriction of Rij+1 to each E j+1 class induces

123



a partition into regions each of which is a union of �β-almost rectangles withedge lengths between d j+1 and 12d j+1. Suppose the clopen subequivalence

relations R jj , R

j+1j , . . . , Ri−1

j and R j+1j+1, . . . , R

i−1j+1 of Eω have been defined

and satisfy:

(1) R jj ⊆ E j ; Rk

j ⊆ Ek and Rkj+1 ⊆ Ek for all j < k ≤ i − 1.

(2) R jj induces a partition of each E j class into �β-almost rectangular regions

with edge lengths d j .

(3) For j < k ≤ i −1, the restriction of Rkj to each E j class gives a partition

into regions R each of which is a union of �β-almost rectangles with edgelengths between 9d j and 12d j .

(4) On each E j class, for each region R induced by the restriction of Rkj , there

is a region R′ induced by the restriction of Rkj+1 such that the Hausdorff

distance between R and R′ is at most 12d j .

(5) In any ball B = {y ∈ [x]E j : ρjx (y) ≤ 10, 000 ·16 j d j } of radius 100,000 ·

16 j d j contained in an E j class, and for each essential dimension e ≤ jthere are at most j + 1 many k with j < k ≤ i − 1 such that some regioninduced by the restriction of Rk

j has an e-face intersecting B.

(6) For any j < k1 < k2 ≤ i − 1, and regions R1, R2 in an E j class induced

by restrictions of Rk1j , Rk2

j respectively, ifF1,F2 are parallel faces of R1,

R2, then for any u ∈ F1, v ∈ F2, ρju (v) > 1

9000 j16 j2 ( j+1)d j .

(7) For any j < k1 < k2 ≤ i and regions R1, R2 in an E j+1 class

induced by the restrictions of Rk1j+1, Rk2

j+1 respectively, if F1, F2 are

parallel faces of R1, R2, then for any u ∈ F1, v ∈ F2, ρj+1u (v) >

19000 j+116( j+1)2 ( j+2)

d j+1.

Then there is a clopen subequivalence relation Rij ⊆ Ei satisfying the

following:

(1) On each E j class, Rij induces a partition into regions R each of which is

a union of �β-almost rectangles with edge lengths between 9d j and 12d j .

(2) On each E j class, for each region R induced by the restriction of Rij , there

is a region R′ induced by the restriction of Rij+1 such that the Hausdorff

distance between R and R′ is at most 12d j .

(3) Condition (5) continues to hold, where now j < k ≤ i.(4) Condition (6) continues to hold, where now j < k ≤ i.

Proof The proof is almost identical to that of Lemma 6.1. We first applyLemma 7.24 to produce the partition into �β-almost rectangular regions R ofedge lengths d j . The R1, . . . , Rk of that lemma are the restrictions to E j of

123



the current R jj , . . . R

i−1j . The hypotheses of Lemma 7.24 are guaranteed by

assumption (5) above together with d j � β( j). Our assumption is that therestriction of Ri

j+1 to each E j+1 class induces a partition into regions each of

which is a union of �β-almost rectangles of edge lengths at least d j+1 � d j .

When we further restrict each Rij+1 region to each E j class, we still have

regions which are finite unions of �β-almost rectangles of edge lengths betweend j+1 and 12d j+1. This follows from the definition of a �β-almost rectangle. Inparticular, these restricted regions are unions of β( j)-almost rectangles (thisis the main point; when restricted to E j the error is bounded by β( j), ratherthan β( j + 1)).

We next use Rij+1 and R to define Ri

j .As in Lemma 6.1, letA denote the col-

lection of j-dimensional �β-almost rectangles of edge length d j induced by R.

In particular, each of these regions is a β( j)-almost rectangle. By Lemma 7.18each β( j)-almost rectangle R ∈ A has a (2β( j) + 1)-almost center c j (R).

Since d j � β( j) we still have c j (R) ∈ R. We define x Rij y iff x ∈ R1 ∈ A,

y ∈ R2 ∈ A, and c j (R1)Rij+1c j (R2). Clearly each Ri

j class is a union of�β-almost rectangles of edge lengths d j .

For any R ∈ A and x ∈ R, we have

ρjc j (x)

(x) ≤ 6d j + 2β( j) + 1 < 6d j + 3β( j)

and, by Lemma 7.10, ρjx (c j (x)) ≤ 6d j + 4β( j). Thus any point in an Ri

j

region is within 6d j + 4β( j) of an Rij+1 region. Moreover, we claim that

every point on the boundary of an Rij region is within 18d j + 16β( j) of a

point on the boundary of an Rij+1 region. To see this, let x be a boundary point

of an Rij region, let R ∈ A with x ∈ R, and u = c j (R). Since x is a boundary

point, there is ‖g‖ = 1 such that y = g · x is not contained in the Rij region

containing x . In particular y ∈ R. Let R′ ∈ Awith y ∈ R′ and let v = c j (R′).Note that either g or −g is in standard form. Then by Lemma 7.8(b), we have

ρju (v) ≤ ρ

ju (x) + ρ

jx (y) + ρ

jy (v) + 2β( j)

≤ (6d j + 3β( j)) + (1 + β( j)) + (6d j + 4β( j)) + 2β( j)

≤ 12d j + 11β( j).

Since ¬(uRij+1v), there is a boundary point w in the Ri

j+1 region containing

u with ρju (w) ≤ 12d j + 11β( j). Therefore

ρjx (w) ≤ ρ

jx (u) + ρ

ju (w) + β( j)

123



≤ (6d j + 4β( j)) + (12d j + 11β( j)) + β( j)

= 18d j + 16β( j),

as we claimed.In view of these estimates, the inequalities used in the proof of Lemma 6.1

continue to hold if d j+1 � d j � β( j). The rest of the proof is thus completedas in Lemma 6.1. ��

We now state our main result for the non-free part.

Theorem 7.26 Let H be a subgroup of Z<ω and XH = {x ∈ X : Gx = H},

where Gx = {g ∈ Z<ω : g · x = x}. Let EH be the equivalence relation on

XH from the shift action of Z<ω. Then there is a continuous embedding from

(XH , EH ) into E0.

Proof The proof is similar to that of Theorem 6.2. We start with a sufficientlyfast growing sequence 1 � d1 � d2 � · · · , where di � β(i) for all i. Foreach i, let Ri

i be the clopen subequivalence relation of Ei whose classes are �β-almost rectangles given by Lemma 7.22. We define inductively the sequenceRii , Ri

i−1, . . . , Ri1. Assume Ri

j+1 has been defined as have all the Rkj for

j ≤ k < i. Assume also that each Rij+1 region restricted to each E j+1 class

is a union of �β-almost rectangles of edge lengths between d j+1 and 12d j+1,

and that the hypotheses of Lemma 7.25 are satisfied. Lemma 7.25 then givesRij and maintains the hypotheses. Thus, we define all the Ri

j .

Suppose xEy, and fix k0 so that xEk0 y. Let η = ρk0x (y). Suppose

I = {i ≥ k0 : ¬x Ri1y}

is infinite. For each i ∈ I, the Ri1 region containing x must have a boundary

point in the ball in Ek0 of radius η about x (in the x-coordinate system), sowe may assume that some fixed point z lies on all their boundaries. From theestimates we made in the proof of Lemma 7.25, the point z is within

(18d1+16β(1))+(18d2+16β(2))+· · ·+(18dk0−1+16β(k0 − 1))+2β(k0)

of a point on a boundary face of an approximating rectangle in Rik0

for eachi ∈ I. Since β(i) � di and dk0−1 � dk0, we still contradict conclusion (4) ofLemma 7.25 as before.

Thus a continuous embedding from EH into E0(ωω) can be defined by

encoding the Ri1 regions by integers, in a similar fashion as we did in the proof

of Theorem 6.2. This completes the proof of Theorem 7.26. ��

123



8 Actions of countable abelian groups

In this section we summarize the results obtained in the previous two sectionsand deduce some corollaries. First we have the following basic theorem, whichis a straightforward consequence of Theorem 7.26.

Theorem 8.1 Let E be the orbit equivalence relation induced by the shiftaction of Z

<ω on 2Z<ω

. Then E is hyperfinite.

This has the following immediate corollary.

Corollary 8.2 Let G be a countable abelian group acting in a Borel manneron a standard Borel space and let E be the induced orbit equivalence relation.Then E is hyperfinite.

Proof Since every countable abelian group is a homomorphic image of Z<ω,

any action of G is also an action of Z<ω. By a result of Becker–Kechris ([1,

Theorem 3.5.3]), the shift action of Z<ω on 2Z

<ωis universal among all Borel

actions of Z<ω. For the reader unfamiliar with this result, a complete proof is

contained in the proof of Theorem 8.5 below. ��This (partially) answers a question ofWeiss (c.f. [4]) and also gives a positive

answer to the following question of Kechris. Consider the equivalence relation∼P on the set R+ of positive real numbers defined by

x ∼P y ⇐⇒ x/y ∈ Q.

This is knownas the commensurability equivalence relation or thePythagoreanequivalence relation. Kechris asked if it is hyperfinite. Since the equivalencerelation is induced by the product action of the multiplicative group of thepositive rational numbers, we have the affirmative answer.

Corollary 8.3 The commensurability equivalence relation ∼P is hyperfinite.

In contrast, recall that the Vitali equivalence relation∼V on R is defined by

x ∼V y ⇐⇒ x − y ∈ Q.

It is much easier and straightforward to see that ∼V is hyperfinite. A token ofits simplicity is that the usual reduction used to show that∼V≤B E0 is a Baireclass 1 function and in fact continuous on the set R − Q of irrationals.

In the rest of this section we give some further corollaries of our maintheorem. First note that the proof of themain theorem also shows the followingslight variation.

123



Theorem 8.4 Let 2 ≤ k ≤ ω. Let E be the orbit equivalence relation inducedby the shift action of Z

<ω on X = kZ<ω

. Then E is hyperfinite.Moreover, letting H be a subgroup of Z

<ω and XH = {x ∈ X : Gx = H},then there is a continuous embedding of E � XH into E0.

Next we consider continuous free actions of countable abelian groups. Inthe special case where the underlying space is 0-dimensional we always havecontinuous embeddings into E0.

Theorem 8.5 Let G be a countable abelian group acting continuously andfreely on a 0-dimensional Polish space X. Let E be the induced orbit equiva-lence relation. Then there is a continuous embedding of E into E0.

Proof This follows from the techniques used in [4], specifically those used inthe proofs of Propositions 1.2, 1.4 and 1.6 of [4], which we recall below. First,fixing a countable clopen base {Ui } for the topology of X, we can define areduction f0 : X → (2ω)G by

f0(x)(g)(i) = 1 ⇐⇒ g−1 · x ∈ Ui .

f0 is indeed an injective continuous G-map, i.e., g · f0(x) = f0(g · x). Itfollows then that the image of f0 is an invariant Borel subset of the free partof (2ω)G . Next, via any bijection between ω and Z−{0} we identify the latterspace with (2Z−{0})G . Then f0 is essentially a continuous reduction from X to(2Z−{0})G . In the third step define a further reduction f1 : (2Z−{0})G → 3G×Z

by

f1(p)(g, n) ={p(g)(n), if n = 0,2, if n = 0.

Then f1 is still injective and continuous, and sends the free part of (2Z−{0})Ginto the free part of 3G×Z (by the proof of [4, Proposition 1.6]). In the fourthstep we fix an onto homomorphism π : Z

<ω → G×Z and note that the trivialreduction f2 : 3G×Z → 3Z

<ωdefined by

f2(p)(g) = p(π(g))

is injective and continuous and sends the free part of 3G×Z into XH , whereH = ker(π).NowbyTheorem8.4 the last equivalence relation is continuouslyembeddable into E0. ��

The 0-dimensionality condition on X is of course necessary. In case X isconnected every continuous function from X into 2ω is constant. Regardingthe commensurability equivalence relation we have the following corollary.

123



Corollary 8.6 There is a Baire class 1 reduction from ∼P into E0, which iscontinuous on the set of positive irrational numbers.

Proof Repeat the aboveproof verbatimassuming {Ui } to be the usual countableopen base forR+, i.e., eachUi is an open intervalwith rational endpoints. Thenf0 is Baire class 1 and is continuous on the set of positive irrational numbers.The proof then gives aBaire class 1 reduction of∼P to E0,which is continuouson the irrational part. ��

In general, without assuming that the underlying space is 0-dimensional thefollowing theorem is all we know.

Theorem 8.7 Let G be a countable abelian group acting continuously on aPolish space X. Let E be the induced orbit equivalence relation. Let E∗

ω be theorbit equivalence relation induced by the shift action of Z

<ω on RZ

<ω. Then

there is a continuous embedding of E into E∗ω.

Proof This is again achieved by composing several reductions. First we definea reduction f ∗

0 : X → XG by

f ∗0 (x)(g) = g−1 · x .

Then f ∗0 is injective and continuous, and it is a reduction of the orbit equiv-

alence relation on X to that on XG . Next we note that any Polish space ishomeomorphic to a Gδ subset of [0, 1]ω. Thus by identifying X with itshomeomorphic copy in [0, 1]ω we may regard f ∗

0 to be a reduction fromX to ([0, 1]ω)G .Now the reduction functions f1 and f2 in the preceding proofcan be similarly redefined to get reductions f ∗

1 : ([0, 1]ω)G → RG×Z and

f ∗2 : R

G×Z → RZ

<ω. The composition f ∗

2 ◦ f ∗1 ◦ f ∗

0 is a continuous embed-ding of E into E∗

ω. ��

9 Continuous embeddings into E0

In this section we show that there is a continuous, in fact computable (orrecursive), embedding from the shift equivalence relation on 2Z

ninto E0. For

simplicity we use the following notation in this section. Let n ≥ 1 be fixed. LetX = 2Z

nand let π be the shift map on X. Let E denote the shift equivalence

relation on X, and F the shift equivalence relation on F(X), where F(X)

denotes the free part of X. Later in this section we will make use of the proofsgiven in Sect. 7. However, there will be no confusion since we do not workwith 2Z

<ωin this section, and the results we use from Sect. 7 are all about the

shift action of Zn. We state the main theorem of this section below.

Theorem 9.1 For each n ≥ 1, there is a computable one-to-one f : 2Zn → 2ω

such that for all x, y ∈ 2Znwe have xEy iff f (x)E0 f (y).

123



The proof of Theorem 9.1 will use the ideas of Theorem 7.26 along withthose of [2] where it was shown that there is a continuous embedding from 2Z

into E0. The rest of this section is devoted to the proof of Theorem 9.1. We fixthe dimension n for the remainder of the section.

Suppose that H = 〈h1, . . . , h p〉 is a subgroup of Zn, with h1, . . . , h p a

normal basis for H (see Definition 7.1). Let X H = {x ∈ X : ∀h ∈ H h · x =x}. In particular, X = X H0, where H0 = {�0} is the identity subgroup ofZn. Recall that XH denotes the set of x ∈ X such that Gx = H, where

Gx = {g ∈ Zn : g · x = x}. Thus, XH ⊆ X H . Note that XH is the “free part”

of X H , that is, the part of X H whose stabilizer group is exactly equal to H.

We write FH for E � XH and EH for E � X H . Clearly X H is closed in X andXH is a Gδ in X H . It is also obvious that if dim(H) = n then XH is finite.We note the following fact which will be used substantially in the rest of thesection.

Lemma 9.2 If dim(H) < n then XH is dense in XH .

Proof Let x ∈ X H and F ⊆ Zn be finite. We need to find y ∈ XH such that

y � F = x � F. Let K be the quotient group Zn/H and q be the natural

homomorphism from Zn to K . Since dim(H) < n, K is infinite. Since x ∈

X H , we have that for all g ∈ Zn and h ∈ H, x(g) = (h−1 · x)(g) = x(g+ h).

Thus if we let x∗ : K → {0, 1} be given by x∗(q(g)) = x(g) for all g ∈ Zn,

then x∗ is well defined.We claim that there is a y∗ : K → {0, 1} such that y∗ � q(F) = x∗ � q(F)

and for all non-identity q ∈ K , q · y∗ = y∗. In other words, y∗ is in thefree part of 2K and y∗ extends x∗ � q(F). This follows immediately from thedensity of the free part of 2K because K is infinite. We give a quick argumentto keep our exposition self-contained. Let q1, . . . , qk, . . . enumerate all non-identity elements of K . Let F∗

0 = q(F). Inductively we define finite setsF∗k+1 ⊇ F∗

k and y∗ � F∗k+1 so that qk ∈ F∗

k+1 and so that there is q ∈ F∗k+1

with qk + q ∈ F∗k+1 and y∗(q) = y∗(q + qk). Suppose F∗

k and y � F∗k have

been defined. Since F∗k is finite and K is infinite, there must be some q ∈ F∗

k .

We just make sure q, q + qk ∈ F∗k+1 and y∗(q) = y∗(q + qk). This finishes

the construction of y∗ and proves the claim.Now define y : Z

n → {0, 1} by letting y(g) = y∗(q(g)) for all g ∈ Zn.

Then it is easy to verify that y � F = x � F. Clearly h · y = y for all h ∈ H.

To see that y ∈ XH , assume toward a contradiction that g ∈ Zn − H is such

that g−1 · y = y. Then q(g) is not the identity in K , but y∗(q+q(g)) = y∗(q)

for all q ∈ K . This means q(g)−1 · y∗ = y∗, a contradiction. ��Relatively open subsets of XH can be extended to X H systematically as

follows. Fix a standard countable base {Vi } for the topology on X. Then{Vi ∩ X H } is a base for the topology of X H and {Vi ∩ XH } is a base for

123



the topology of XH . We regard both of them as the standard bases. Now forany sequence (im)m∈ω, if A = ⋃

m∈ω(Vim ∩ XH ) is an open subset of XH ,

we let A = ⋃m∈ω(Vim ∩ X H ). Then A is an open subset of X H such that

A ∩ XH = A. Note that A is not canonically determined from the set A.

Rather, it depends on the sequence (im). For notational simplicity we will justwrite A, assuming implicitly a neighborhood sequence (im) has been specified.This notation will also be used for subsets of XH × XH , etc., unless specifiedotherwise.

We also need a refinement of the above notion. Recall that a set A ⊆ XH isrelatively �0

1 if there is a computable total function f : ω → ω such that

A =⋃

m∈ω

(V f (m) ∩ XH ).

In this case we also let A = ⋃m∈ω(V f (m) ∩ X H ). Again, it is understood that

this depends on the enumerating function f, and we have implicitly chosenone. We call A the �0

1 extension of A to X H .

We will fix computable bijections between ω and various relevant setssuch as Z, Z

n, ω<ω, etc. For each of the relevant sets N , the computablebijective function from N onto ω will be denoted by #(·) and referred to asthe coding function. For instance, if (k0, . . . , kl) ∈ ω<ω is a sequence, then#(k0, . . . , kl) ∈ ω is the code for the sequence. The inverse of a coding func-tion is called a decoding function. For instance, the decoding function fromω to Z

n gives a computable enumeration of all group elements in Zn, and we

use the notation k �→ gk for it. Via the coding and decoding functions we willbe able to speak of computable functions between any two relevant sets.

The coding function # : Zn → ωwill also induce a one-one correspondence

between X = 2Znand 2ω. This allows us to refer to a lexicographic order for

elements of X.

We note the following fact about the computability of a normal basis.

Lemma 9.3 There is a computable total function t : ω<ω → ω<ω such thatfor any sequence (k0, . . . , kl) ∈ ω<ω, if (m0, . . . ,mq) = t (k0, . . . , kl), thengm0, . . . , gmq is a normal basis for 〈gk0, . . . , gkl 〉.Proof As in Sect. 7 for each 1 ≤ i ≤ n we use Z

i to denote the subsetZi × {0}n−i of Z

n. Given gk0, . . . , gkl let H = 〈gk0, . . . , gkl 〉. By a Gaussianelimination algorithm, one can find gk′

1, . . . , gk′

nsuch that for any 1 ≤ i ≤ n,

〈gk′1, . . . , gk′

i〉 = H ∩ Z

i . Then the proof of Lemma 7.2 gives an algorithmthat can be applied to the non-zero elements of gk′

1, . . . , gk′

nto obtain a normal

basis gm0, . . . , gmq . ��Let B be the set of all normal bases for subgroups of Z

n. It is easily seenfrom Definition 7.1 that B is a computable subset of (Zn)<ω.

123



For each subgroup H with normal basis B = (h1, . . . , h p) ∈ B we shalldefine a computable embedding πH from EH into E0. In fact, πH will be atotal computable function from X to 2ω such that πH � X H is an embeddingfrom EH into E0. To do this, we define a total computable function fromB × ω × ωω to ω. We denote the value of the function at (B, k, x) by π B

k (x).If H = 〈B〉, we will also write πH

k (x), with the understanding that a normalbasis is fixed. We think of πH

k (x) as the stage k output of πH (x). That is, wewill have that πH (x) = (πH

0 (x), πH1 (x), . . .) is the sequence of the stage k

outputs. We will have that πH0, with H0 = {�0}, is then the desired embeddingfor Theorem 9.1.

We shall define πHk (x) by a reverse induction on H. That is, we will define

πHk (x) assuming πH ′

k (x) is defined for all subgroups H ′ with H ′� H. This is

clearly a legitimate definition as there is no infinite strictly increasing sequenceof subgroups of Z

n.

Consider first the case dim(H) = n, so H has a normal basis B =(h1, . . . , hn). In this case Z

n/H is finite, and a finite set of representatives forall cosets of H inZ

n is computable from B (for instance, the set of all elementsinZ

n in standard formwith respect to H in the sense ofSect. 7 is such a set). Let-ting F0 be such a finite set of coset representatives, any x ∈ X H is determinedby x � F0. It follows that each EH class is finite, and in fact contains no morethan k0 = 2|F0| many elements. Let F1 = F0 + [−k0, k0]n. Then, given anyx, y ∈ X H , it is decidable whether y ∈ [x]EH using information only aboutx � F1 and y � F1.Let k1 be large enough so that F1 ⊆ {g0, g1, . . . , gk1} and letF2 = {g0, g1, . . . , gk1}. Then, given any x, y ∈ X H , the lexicographic orderbetween x and y is decidable using information only about x � F2 and y � F2.Combining these procedures, we may compute from any x ∈ X H the lexico-graphically least element of [x]EH with information only about x � F2. Suchan algorithm can be applied to any element of X, and we obtain σ B : X → X,

a total computable function uniformly in B, with the property that for allx ∈ X H , σ B(x) is the lexicographically least member of [x]E = [x]EH .

σ B � X H is a reduction from EH to the equality relation on X. There is easilya computable function τ from X to 2ω which embeds (X,=) into (2ω, E0).

Finally, let γ B : X → ωω be a computable function defined uniformly in Bsuch that for x ∈ X H , γ B(x)(0) codes the least γ ∈ G such that γ ·x = σ B(x),and γ B(x)(k) = 0 for k > 0. Let πH

k (x) = #(τ (σ B(x))(k), γ B(x)(k)). It iseasy to check that πH � X H is an embedding from EH into E0(ω

ω).

Next we handle the inductive step. Assume henceforth that dim(H) = p <

n and that B = (h1, . . . , h p) is a normal basis for H.Assume we have definedπ B′k (x) for x ∈ X, k ∈ ω, and B ′ ∈ B with 〈B ′〉 � H. We proceed to define

π Bk (x).

123



Tobegin, fix a sufficiently fast growing sequencedB1 � dB

2 � · · · ofmarkerdistances. We may assume that the map (B, i) �→ dB

i is computable. We willuse the terminology from Sect. 7. In particular, let bB denote the value of β(n)

from Definition 7.7 with respect to H. Here, of course, n is fixed, so bB is justa single number. A �bB-almost rectangle will be as in Definition 7.20, usingnow the constant function β(k) = bB for all 1 ≤ k ≤ n. We may assumethat dB

1 is sufficiently large compared to bB . We now repeat the proofs ofLemmas 7.22, 7.23, and 7.24 in this context, that is, for n fixed. To fix notation,we summarize the main steps of these proofs. All of the constructions beloware done uniformly in the normal basis B = (h1, . . . , h p) for H. To ease thenotation, however, we sometimes suppress mentioning this dependence on B.

For each i, let Rii = Ri,B

i be the subequivalence relation of FH obtainedfrom Lemma 7.22 for edge length di = dB

i . Each Rii class is a b-almost

rectangle with edge lengths di . Let Ci = CBi ⊆ XH denote the set of centers

of these b-almost rectangular regions. Recall that a center of such a b-almostrectangular region is the lexicographically least element of the set of all ofits (2b + 1)-almost centers. The proofs of Lemmas 7.11, 7.22 and 7.18 showthat Ci is relatively clopen in XH . In fact, the Ci are uniformly �0

1 in Band i. That is, there are �0

1 relations A1, A2 ⊆ B × X × ω such that for allB ∈ B and x ∈ XH we have x ∈ CB

i ↔ A1(B, x, i) ↔ ¬A2(B, x, i). Tosee this, note that the dB

i and the sequences sn = sBn (defined for each i) asin the proof of Lemma 7.11 are all given by computable functions. The restof the proof of Theorem 3.1 (and Lemma 7.22) involves effective operationson these sets. So, there is a computable f such that for all B and i we havethat CB

i = ⋃k(V f (B,i,k) ∩ XH ). We let C B

i = ⋃k(V f (B,i,k) ∩ X H ) be the �0

1extension of CB

i to X H .

The proof of Theorem 3.1 (and so also Lemma 7.22) then shows that therelations Ri,B

i are, uniformly in B and i, �01 on XH . It follows that there are

�01 relations A3, A4 ⊆ B×X×Z

n×ω such that for B ∈ B, x ∈ XH , g ∈ Zn,

and i ∈ ω we have

Ri,Bi (x, g · x) ↔ A3(B, x, g, i) ↔ ¬A4(B, x, g, i).

Note that for all B, the sets AB3 and AB

4 (the sections of A3, A4 at B) areactually disjoint on X H × Z

n × ω. This is because, if (x, g, i) ∈ AB3 ∩ AB

4with x ∈ X H , then by the density of XH in X H shown by Lemma 9.2, therewould be an x ′ ∈ XH with (x ′, g, i) ∈ AB

3 ∩ AB4 . Let Ri,B

i be the�01 extension

of AB3 to X H .

As in the proof of Lemma 7.25 and Theorem 7.26 (which in turn are modi-fications of the proof of Theorem 6.2), we proceed in i −1 steps to modify thesubequivalence relations Ri

i to relations Rii−1, R

ii−2, . . . , R

i1, all of which are

123



subequivalence relations of FH . Thus, the Rij are subequivalence relations of

FH and satisfy the following:

(1) On each FH class, Rij induces a partition into regions R each of which is

a union of b-almost rectangles with edge lengths between 9d j and 12d j .

(2) For each region R induced by Rij , there is a region R′ induced by Ri

j+1such that the Hausdorff distance between R and R′ is at most 12d j .

(3) For any j < k1 = k2 ≤ i − 1, and R1, R2 regions induced by Rk1j ,

Rk2j respectively, if F1, F2 are parallel boundary faces of R1, R2, then

ρ(F1,F2) > 19000 j16 j2 ( j+1)

d j .

The proof of Lemma7.25 is effective, and thus the Ri1 = Ri,B

1 are, uniformlyin B and i, �0

1 on XH . So, there are �01 relations A5, A6 ⊆ B × X × Z

n × ω

such that for all B ∈ B, x ∈ XH , g ∈ Zn, and i ∈ ω,

x Ri,B1 g · x ↔ A5(B, x, g, i) ↔ A6(B, x, g, i).

In Lemma 9.5 below we show that each Ri,B1 class contains exactly one point

of CBi . The proof of Theorem 7.26 shows that for each B the Ri,B

1 give acontinuous embedding of FH into E0, and by our remarks this embedding isactually computable on XH . Our embedding πH , however, will be defined onthe entire X and, more importantly, must be a reduction on the “non-free” partof X H as well.

Let A5 and A6 denote the respective �01 extensions of A5 and A6 to B ×

X H × Zn × ω. Let Ri,B

1 be the extension of Ri,B1 to X H × X H defined as

follows.

x Ri,B1 y ↔ ∃z ∈ X H ∃g1, g2 ∈ Z

n (z ∈ C Bi ∧ ‖g1‖, ‖g2‖ < 20dB

i

∧ g1 · z = x ∧ g2 · z = y ∧ A5(B, z, g1, i) ∧ A5(B, z, g2, i)).

It is straightforward to check that Ri,B1 ∩ (XH × XH ) = Ri,B

1 and that Ri,B1 is

�01 in the sense that {(x, g) : x Ri,B

1 g · x} is �01 on X H × Z

n. Also, Ri,B1 is a

symmetric, transitive relation on X H (but not necessarily reflexive). We givea proof of the transitivity below.

Lemma 9.4 Ri,B1 is a transitive relation on XH × X H .

Proof Suppose x Ri,B1 y and y Ri,B

1 w. Let z1, g1, g2 witness x Ri,B1 y, and z2,

g3, g4 witness y Ri,B1 w. We first claim that z1 = z2. Otherwise by the density

of XH in X H , there would be z′1, z′2, y′ ∈ XH with z′1 = z′2, z′1, z′2 ∈ CBi

123



and Ri,B1 (z′1, y′), Ri,B

1 (z′2, y′). Since Ri,B1 is an equivalence relation, we have

z′1Ri,B1 z′2, a contradiction as z′1 and z′2 are distinct points of CB

i .

With z1 = z2, we now have z1, g1, g4 witness that x Ri,B1 w. ��

It follows from the above lemma that, if x ∈ X H , then on each EH class[x]EH , each Ri,B

1 gives a partition of a subset of [x]EH . We next note that for

every B ∈ B and i ∈ ω, if x ∈ X H is in the domain of Ri,B1 then there is

a unique point of C Bi in the Ri,B

1 equivalence class of x . This follows froma density argument similar to the one in the above proof, provided that thecorresponding uniqueness statement in XH holds. The uniqueness in XH inturn follows from our construction of the Ri

1 regions in Lemma 7.25. Sincethis argument was not presented in Sect. 7 (it was not needed there), we givethe proof below.

Lemma 9.5 For every B ∈ B and i ∈ ω, there is a unique point in CBi in each

Ri,B1 equivalence class. Also, if x ∈ X H is in the domain of Ri,B

1 then there is

a unique point of C Bi in the Ri,B

1 equivalence class of x .

Proof For each B, i and almost rectangular region R in Ri,Bi , since dB

i � bB,

it follows easily that there is a unique point c(R) of CBi in R. For B ∈ B,

x ∈ XH , and i ∈ ω there is a unique Ri,Bi class R ⊆ [x]E such that the Ri,B

1class R′ of x satisfies

ρB(R, R′) ≤ 12(dB1 + · · · + dB

i−1)

(here ρB denotes the Hausdorff distance as defined using the basis B, seeDefinition 7.19). The fact that R exists follows by repeated application of (2).To see uniqueness, suppose R and S were two such regions for R′. Then

ρB(R, S) ≤ 24(dB1 + · · · + dB

i−1) < dBi − bB .

This is a contradiction as ρB(c(R), s) ≥ dBi − bB for any s ∈ S. For x ∈ XH ,

let cBi (x) = c(R) where R is the unique Ri,Bi class in [x]E such that

ρ(R, [x]Ri,B1

) ≤ 12(d1 + · · · + di−1).

Then cBi (x) ∈ CBi and cBi (x) is in the unique R as above. We must have

cBi (x) ∈ R′ = [x]Ri,B1

as otherwise there is a district Ri,Bi class S = R such

that

ρB(cbi (x), S) ≤ 12(dB1 + · · · + dB

i−1) < dBi − bB,

123



and this is not the case. The same argument shows that there cannot be twodistinct Ri,B

i classes R and S such that c(R) and c(S) both lie in R′.Suppose now that x ∈ X H is in the domain of Ri,B

1 . From the definition of

Ri,B1 it is immediate that that there is a z ∈ C B

i with z Ri,B1 x (take g1 = 0 in the

definition of Ri,B1 ). Suppose z1, z2 ∈ C B

i were two such points. The density ofXH in X H would then give points z′1, z′2, x ′ in XH with z′1 = z′2, z1, z2 ∈ CB

i ,

and z′1Ri,B1 x ′, z′2R

i,B1 x ′. This contradicts the previous paragraph. ��

For each x ∈ X and m ≥ 1, let Nm(x) denote the basic open set in X H :

Nm(x) = {y ∈ X H : ∀g ∈ Zn (‖g‖ ≤ m → y(g) = x(g))}.

If x ∈ X H , then Nm(x) = ∅.Wemay identify Nm(x)with x � {g : ‖g‖ ≤ m}.The following definition will be important for our construction. For �0

1sets C B

i , A5, and A6 defined above, we define �01 approximations for them

in the usual manner. That is, if f is a computable function such that C Bi =

⋃k(V f (B,i,k) ∩ X H ) we set Cm,B

i = ⋃k≤m(V f (B,i,k) ∩ X H ). We likewise

define Am5 and Am

6 .

Definition 9.6 For B ∈ B, i,m ∈ ω, and x ∈ X H , we say x is i -determinedby depth m (with respect to B) if there is a g ∈ Z

n with ‖g‖ < 20dBi such that

z.= g−1 · x satisfies the following:

(1) Nm(z) ⊆ Cm,Bi

(2) {B} × Nm(z) × {g} × {i} ⊆ Am5 .

(3) For every h ∈ Zn with‖h‖ < 20dB

i wehave {B}×Nm(z)×{h}×{i} ⊆ Am5

or {B} × Nm(z) × {h} × {i} ⊆ Am6 .

Roughly speaking, this says that, by the mth stage of the computation, it isalready determined that the neighborhoodof depthm about zmust be containedin C B

i , x = g · z must be Ri,B1 equivalent to z, and also for all ‖h‖ < 20dB

i

whether or not h · z is Ri,B1 equivalent to z has been decided. In other words,

this neighborhood which is determined by the mth stage of the computationcompletely determines the points which are Ri,B

1 equivalent to z.We denote D(B, x, i,m) ↔ x is i-determined by depth m relative to B.

This is a �01 relation on B × X H × ω × ω.

Note that if x ∈ XH , then for each i there is a large enough m so that xis i-determined by depth m (relative to B which generates H ). On the otherhand, if x ∈ X H − XH , then for any given i there may or may not be an msuch that x is i-determined by depth m. If x is i-determined by depth m, weclearly have that x is in the domain of Ri,B

1 . Finally, note that for any B ∈ B,

123



x, y ∈ X H and i,m ∈ ω, if x is i-determined by depth m and x Ri,B1 y, then

y is also i-determined by depth m. To see this, let g and z = g−1 · x witnessthat x is i-determined by depth m, and let z′, g1, g2 witness that x Ri,B

1 y (so

z′ = g−11 · x). Since z, z′ ∈ C B

i and z Ri,B1 x, z′ Ri,B

1 x, it follows that z = z′.We have therefore z Ri,B

1 y and also from (3) of Definition 9.6 it follows that{B} × Nm(z) × {g2} × {i} ⊆ Am

5 or {B} × Nm(z) × {g2} × {i} ⊆ Am6 . Since

z Ri,B1 y we must have the first alternative. This verifies (2) of Definition 9.6

for y being i-determined by stage m, and (1), (3) for y follow from z = z′.We now proceed to the definition of πH

k (x). The output πHk (x) will be

effectively computable from x, k, and the�01 relation D(B, x, i,m). This will

give that πH is computable.For B ∈ B, x ∈ X H , and i,m ≥ 1, if x is i-determined by depth m, we let

xmi be the unique point in C Bi with x Ri,B

1 xmi . Note that this point, if it exists,does not depend onm. That is, if xmi is defined then so is x p

i for all p ≥ m andx pi = xmi . If x is not i-determined by depth m, then we leave xmi undefined.

Definition 9.7 For x ∈ X H and k ∈ ω, let ik(x) be the largest i ≤ k suchthat x is i-determined by depth k, if such an i exists. If no such i exists setik(x) = 0 (these notions are defined relative to B, which we suppress in thenotation).

We say x is active at stage k if ik(x) > ik−1(x), and otherwise say x ispassive at stage k.

Fix now B ∈ B, x ∈ X H and assume πH1 (x), . . . , πH

k−1(x) have beendefined. To define πH

k (x) we consider two cases according to whether x isactive or passive at stage k. The idea is that when x is active at stage k we“guess” that x is the “free” part XH and define πH

k (x) accordingly. When xis passive at stage k, we “guess” that x is in the “non-free” part X H − XHand define πH

k (x) accordingly. The key points are that if x ∈ X H − XH thenwe will eventually guess correctly, while if x ∈ XH then we guess correctlyinfinitely often.

Case I: x is active at stage k.In this case ik(x) > ik−1(x). By definition, x is ik(x)-determined by depth

k. Let zk(x) be the unique point of Cik(x) such that x Rik(x)zk . Similarly definezk−1(x). If zk−1(x) is defined (that is, ik−1(x) > 0), let ak(x) ∈ Z

n be theleast a in standard form with

a · zk−1(x) � {g : ‖g‖ ≤ 20dBik } = zk(x) � {g : ‖g‖ ≤ 20dB

ik }.If zk−1(x) is not defined, set ak(x) to be the least a in standard form such that

a · x � {g : ‖g‖ ≤ 20dBik } = zk(x) � {g : ‖g‖ ≤ 20dB

ik }.

123



An easy density argument shows that ‖ak(x)‖ ≤ dBk−1 + dB

k + 2bB < 2dBk .

Let

Nk(x) = N20dBik(zk(x)) = zk(x) � {g : ‖g‖ ≤ 20dB

ik }.

We let πHk (x) = #(ak(x), Nk(x)), that is, πH

k (x) is an integer coding(ak(x), Nk(x)).

Case II x is passive at stage k.Let zk(x) be as in the previous case, with the provision that if ik(x) = 0

then set zk(x) = x . Let uk(x) be the least element u of Zn − H such that

(u · zk(x))(g) = zk(x)(g)

for all g ∈ Zn with ‖g‖, ‖u + g‖ < 20k. It is easy to see that there is always

such a u using the fact that Zn − H is infinite. Let Hk(x) = 〈H, uk(x)〉 with

normal basis Bk(x) = t (B, uk(x)), where t is as defined in Lemma 9.3. LetπHk (x) code Bk(x) and π

Hk(x)k (zk(x)), where π

Hk(x)k is the stage k output for

the map πHk(x), which is defined by induction since Hk(x) � H. Finally, ifk − 1 was an active stage or if the value of uk−1(x) is different from uk(x)then we code also π

Hk(x)0 (zk(x)), . . . , π

Hk(x)k−1 (zk(x)) into πH

k (x).This completes the definition ofπH

k (x) and so of themapπH .The definitionabove, and our comments along the way, show that the map (B, k, x) �→πHk (x) is computable; the only other point to observe is that the recursion

occuring in Case II (defining πHk (x) in terms of π

Hk(x)k (x)) is legitimate as

there is no infinite strictly increasing chain of subgroups of Zn. It remains to

show that πH is a reduction of EH to E0(ωω), and that πH is one-to-one on

X H .

Lemma 9.8 Suppose x, y are in XH and xEH y. If x ∈ XH (and so also y ∈XH) then for infinitely many k we have that x and y are active at stage k, andfor large enough k, x is active at stage k iff y is active at stage k. Furthermore,for large enough active k we have zk(x) = zk(y). If x ∈ X H − XH (and soalso y ∈ X H − XH) then for all large enough k we have that x and y arepassive at stage k.

Proof Suppose first that x ∈ XH . Then for all i there is an m such that x isi-determined by depthm. For k ≥ m we have ik(x) ≥ i, and so the ik(x) → ∞as k → ∞. Thus, there are infinitely many k such that x is active at stage k.Since the Ri,B

1 witness hyperfiniteness on XH , there is an i ′ such that for all

i ≥ i ′, x Ri,B1 y. For such i and anym it follows that x is i-determined by depth

m iff y is i-determined by depthm. From this and the fact that the ik(x) → ∞it follows that for all large enough k, ik(x) = ik(y). Hence, for large enough

123



k we have that x is active at stage k iff y is active at stage k. For these large k(i.e., with ik(x) ≥ i ′) we also have that zk(x) = zk(y) since x R

ik ,B1 y.

Suppose now that x, y ∈ X H − XH . Thus there is an u ∈ Zn − H such that

u · x = x (and also u · y = y). Since u /∈ H, there is an u′ ∈ u + H with u′ instandard form (with respect to B), and u′ = �0. From the marker constructionof Lemma 7.11 it follows that for any i with dB

i > ‖u′‖ that C Bi ∩ [x]E = ∅.

In particular, for any such i, x and y will never be i-determined by depth m(for any m). Say this happens for all i ≥ i ′. Let k0 be large enough so that forany i < i ′, if x or y is i-determined by depth m, for some m, then x or y isi-determined by depth k0. Thus, for all k ≥ k0 the values of ik(x) and ik(y)are stabilized, that is, x and y are passive at stage k. ��Lemma 9.9 πH is a reduction from EH to E0(ω

ω).

Proof Suppose first that x, y ∈ XH and xEy. From Lemma 9.8 there areinfinitelymany k such that x and y are both active at stage k, and zk(x) = zk(y).In Case I, the output consists of ak together with Nk . Note that Nk dependsonly on zk . Thus, Nk(x) = Nk(y). Since ak only depends on zk−1 and zk, wehave ak(x) = ak(y). Thus, πH

k (x) = πHk (y) for large enough active k. In case

II, the values of Hk = 〈H, uk〉 and πHkk (zk) being output also depend only on

zk, and so will be the same for x and y.Next suppose that x, y ∈ X H − XH and xEH y. From Lemma 9.8, for

large enough k both x and y will be passive at stage k. So we are in Case IIin the definition of πH

k for both x and y. It follows that for large enough k thevalues of uk and thus Hk will be the same for both x and y, and both will beeventually constant. By induction π

Hk(x)k (x) = π

Hk(y)k (y) for large enough k.

So, πHk (x) = πH

k (y) for large enough k.For the converse, suppose next that x, y ∈ X H and πH (x)E0π

H (y). Firstnote that either x, y are both in XH or both in X H − XH . This followsfrom Lemma 9.8 and the fact that we can distinguish the outputs in the activeversus the passive cases. So, assume first that x, y ∈ XH . By consideringthe output values πH

k (x) for the infinitely many k where x is active, we seethat the E0 class of πH (x) determines a tail of a sequence (ak(x), Nk(x))as in Case I. Since ‖ak(x)‖ < 2dik , and the definition of Nk(x) involvesall g with ‖g‖ < 20dik , it follows that from the E0 class of πH (x) wecan reconstruct the EH class of x . Suppose next that x, y ∈ X H − XH .

From Lemma 9.8 both z(x) = limk zk(x) and z(y) = limk zk(y) exist. Ofcourse, z(x)EH x and z(y)EH y. The outputs πH

k (x) are eventually of the form

(B ′(x), πH ′(x)k (z(x))), where H ′(x) = 〈H, u(x)〉, u(x) is the least element

u of Zn − H such that u · x = x, and B ′(x) is the normal basis for H ′(x).

Likewise for πHk (y). Since πH (x)E0π

H (y), we have B ′(x) = B ′(y) andtherefore u(x) = u(y). Let B ′ = B(x) = B(y) and H ′ = H ′(x) = H ′(y) be

123



the common values. It follows that πH ′(z(x))E0π

H ′(z(y)). If dim(H ′) < n,

then by induction it follows that z(x)EH ′z(y). Since EH ′ ⊆ EH , we havexEH y. If dim(H ′) = n, then the E0 class of πH ′

(x) determines the E0 classof τ ◦σ B′

(x), where τ, hB′are as defined in the p = n case. From the proper-

ties of τ, this determines σ B′(x), which is the lexicographically least element

of the finitely many elements of [x]E . So, πH (y) determines also this sameelement, and it follows that [x]E = [y]E . ��Lemma 9.10 πH is one-to-one from XH to ωω.

Proof Suppose that x, y ∈ X H and πH (x) = πH (y). From Lemma 9.9 weknow that xEH y. Suppose first that x, y ∈ XH . Consider the sequence k0 <

k1 < · · · of k which are active for x (and hence also y, since this determinedfrom the output πH (x) = πH (y)). For each kq , the output πH

kq(x) encodes

the unique ak(x) in standard form (relative to H ) such that ak(x) · zkq−1(x) =zkq (x). Thus, the sequence of zkq (x) is determined from the output πH (x).From this it follows as in Lemma 9.9 that x is also determined from πH (x),and so x = y.

Suppose next that x, y ∈ X H −XH .Let z(x), z(y) again denote the limitingvalues of the zk(x) and zk(y) as in Lemma 9.9. The value of zk(x) only changesfor active k, and the outputπH

k (x) records the change, that is, anak(x) such thatak(x) · zk−1(x) = zk(x). Also, the output πH (x) determines an a′ ∈ Z

n suchthat a′ · x = zk0(x), where k0 is least so that ik0 > 0. So, πH (x) determinesan element g such that g · x = z(x). Since πH (x) = πH (y), we also haveg · y = z(y). Let u ∈ Z

n be the least element of Zn − H such that u · x = x .

Since xEH y we have u · y = y as well. Let H ′ = 〈H, u〉. Let k be the leaststage such that for all k ≥ k, x is k is passive at stage k and ak = u.ThenπH

k (x)

encodes not onlyπH ′k

(z(x)) but alsoπH ′0 (z(x)), . . . , πH ′

k−1(z(x)) aswell. Thus,

the output πH (x) determines completely πH ′(z(x)). If dim(H ′) < n, then

by induction z(x) is determined from πH ′(z(x)), and thus x = g−1 · z(x) is

determined as well. If dim(H ′) = n, then we use the earlier observation thatπH ′

is one-to-one in this base case of our induction. ��We have thus completed the proof of Theorem 9.1.

10 Open problems and further remarks

In this final section we summarize the problems left open by this article forfuture research. First of all the main result of this paper is a small step towardthe solutions of the following hierarchy of fascinating open questions in thefield.

123



Problem 10.1 (The Increasing Union Problem) Let E = ⋃n En where each

En is a hyperfinite equivalence relation and En ⊆ En+1. Is E hyperfinite?

Problem 10.2 (Weiss) Is every amenable equivalence relation hyperfinite?In particular, is every orbit equivalence relation induced by an action of acountable amenable group hyperfinite?

The following is a less ambitious problemalong the line.However, a positivesolution would already surpass all known results to date.

Problem 10.3 Is every orbit equivalence relation of a countable nilpotentgroup action hyperfinite?

We do not know if the method used in this article will continue to play a rolein these problems. Nevertheless, some of our results seem to be applicable in amore general context. For instance the basic marker lemmas are all adaptableto the case of general countable groups.

Another problem on which we made some initial progress in this article butis still open is the computation of the Borel chromatic numbers.

Problem 10.4 What is the Borel chromatic number of

XH = {x ∈ 2Zn : Gx = H},

where Gx = {g ∈ Zn : g · x = x}, for H a subgroup of Z

n, n ≥ 2?

This question has not been completely answered even for the free part(corresponding to the trivial G). Our Theorem 4.2 showed that 4 colors sufficebut it is not clear if it is optimal. Recycling an old piece of terminology wepose the following question.

Problem 10.5 (The Borel Four-Color Problem) Is there a Borel 3-coloring ofF(Zn) for n ≥ 2?

We also conjecture that the answer is uniform for all n ≥ 2. Closely relatedto this problem is the intuitive question of how regular we can make the Borelmarker regions. We already formulated a precise question as Question 4.1.Here we pose a general vague question.

Problem 10.6 Do there existBorelmarker regions for F(Zn)which are almostcubes and are almost lined up?

Finally we reiterate that in contrast to our results in the preceding sectionwe do not know if Eω (the shift equivalence relation on 2Z

<ω) continuously

embeds into E0. In view of the proof of Theorem 8.5 this is equivalent to thefollowing problem.

Problem 10.7 Let G be a countable abelian group acting continuously on a0-dimensional Polish space X. Is EX

H continuously reducible to E0?

123



References

1. Becker, H., Kechris, A.S.: The descriptive set theory of Polish group actions. In: LondonMathematical Society Lecture Note Series, vol. 232. Cambridge University Press, Cam-bridge (1996)

2. Boykin, C., Jackson, S.: Some applications of regular markers. In: Logic Colloquium ’03.Lecture Notes in Logic, vol. 24. Association for Symbolic Logic, La Jolla (2006)

3. Connes, A., Feldman, J., Weiss, B.: An amenable equivalence relation is generated by asingle transformation. Ergodic Theory Dynamical Syst. 1, 430–450 (1981)

4. Dougherty, R., Jackson, S., Kechris, A.S.: The structure of hyperfinite Borel equivalencerelations. Trans. Am. Math. Soc. 341(1), 193–225 (1994)

5. Feldman, J., Moore, C.C.: Ergodic equivalence relations, cohomology and von Neumannalgebras, I. Trans. Am. Math. Soc. 234, 289–324 (1977)

6. Gao, S., Jackson, S., Seward, B.: A coloring property for countable groups. Math. Proc.Camb. Philos. Soc. 147(3), 579–592 (2009)

7. Gao, S., Jackson, S., Seward, B.: Group colorings and Bernoulli subflows (2011). (Preprint)8. Jackson, S., Kechris, A.S., Louveau, A.: Countable Borel equivalence relations. J. Math.

Log. 2(1), 1–80 (2002)9. Kechris, A.S., Solecki, S., Todorcevic, S.: Borel chromatic numbers. Adv. Math. 141(1),

1–44 (1999)10. Ornstein, D., Weiss, B.: Ergodic theory of amenable group actions I. The Rohlin lemma.

Bull. Am. Math. Soc. 2, 161–164 (1980)11. Slaman, T., Steel, J.: Definable functions on degrees. In: Cabel Seminar. Lecture Notes in

Mathematics, vol. 1333, pp. 81–85, 37–55. Springer, Berlin (1988)12. Weiss, B.: Measurable dynamics. In: Conference in Modern Analysis and Probability (R.

Beals et al. eds.). Contemporary Mathematics, vol. 26, pp. 395–421. American Mathemat-ical Society, RI (1984)

123


Documents

countable abelian group actions and hyperfinite equivalence relations