Chapter 3

Paradoxical decompositions ofgroups and their actions

In this Chapter we discuss classical paradoxical decompositions: Hausdorff,Banach-Tarski and strong Banach-Tarski paradoxes. We refer to the book ofWagen, [141], for more on paradoxes.

The starting point in the developments of paradoxical decompositionsof group actions was the famous paradoxical decomposition of the spheredue to Hausdorff, [71]. He showed that the unit sphere can be decomposedinto finitely many pieces in a way that rotating these pieces we can obtaintwo unit spheres. The main idea on why this paradox exists is that thegroup of rotations of the sphere contains the free group on two generators F2.Hausdorff was the first who specified exact rotations that generate the freegroup. It is an easy exercise to prove that F2 has paradoxical decompositionwhile acting on itself. Moreover, if a group action is free and the left actionis paradoxical, then the action itself is paradoxical. Since the action of F2 onthe complement of the set of fixed points of the action of F2 on the sphere isfree, we have that this set (and thus the whole sphere) can be paradoxicallydecomposed.

By projecting points of the unit ball onto the sphere this immediatelyimplies the same paradox for the unit ball without its center. Banach andTarski, [9], showed that there exists a decomposition of a ball into finitelymany pieces such that rotating and shifting each piece we can obtain twoballs of the sizes of the original ball. The pieces are not solid in the usualsense, but rather a sets of points and some of the pieces are non-Lebesguemeasurable. Their work is based on Hausdorff paradox and on an earlier work

17

of Giuseppe Vital, [140], who constructed the first example of non-Lebesguemeasurable set. Vital’s and Hausdorff’s results are dependent on Zermelo’saxiom of choice, which is also crucial for Banach-Tarski paradox. We willstate precisely the parts of the proof that involve the axiom of choice.

A stronger version of Banach-Tarski says that any bounded set with non-empty interior can be decomposed and rearranged into any other boundedset with non-empty interior. This paradox is also called pea-to-Sun paradox:one can obtain the Sun by cutting and moving around the pieces of a tinypea.

The paradoxical decompositions was an essential step in the developmentof amenability. In the later chapter we will see that the existence of para-doxical decomposition is equivalent to non-amenability of the underlininggroup.

3.1 Paradoxical actions

Definition 3.1.1. Let G be a group acting on a set X. A subset E ⊂ X isparadoxical if there exist a pairwise disjoint subsets A1, . . . , An, B1, . . . , Bm

in E and there exist g1, . . . , gn, h1, . . . , hm in G such that

E =n⋃

i=1

giAi =m⋃j=1

hjBj.

An action of a group G on a set X is paradoxical if X is paradoxical.The group itself is called paradoxical if the action of the group on itself byleft multiplication is paradoxical. Later on we will see that the group isparadoxical if and only if it is non-amenable.

The very first example and the only know explicit construction of para-doxical decomposition of a group is the free group on two generators F2.Let a, b ∈ F2 be the free generators of F2. Denote by ω(x) the set of allreduced words in F2 that start with x. Thus the group can be decomposedinto pairwise disjoint sets as follows

F2 = {e} ∪ ω(a) ∪ ω(a−1) ∪ ω(b) ∪ ω(b−1).

Since F2\ω(x) = xω(x−1) for all x in {a, a−1, b, b−1} we have a paradoxicaldecomposition:

F2 = ω(a) ∪ aω(a−1) = ω(b) ∪ bω(b−1).

18

In fact with few additional assumptions one can push a paradoxical de-composition of a group to a set on which it acts. This is exactly the placewhere the Axiom of Choice is needed.

Theorem 3.1.2. If a group a G is paradoxical then any free action of G ona set X is paradoxical.

Proof. Let G =n⋃

i=1

giAi =m⋃j=1

hjBj be a paradoxical decomposition of G. By

Axiom of Choice we can select a subset M of X which contains exactly oneelement from each orbit of G. We have that

⋃g∈G

gM is disjoint partition of

M . Indeed, if gx = hy for some g, h ∈ G and x, y ∈ X, then x = y by thechoice of M . Since the action is free, we have g = h . Define now

Ai =⋃g∈Ai

gM and Bj =⋃g∈Bj

gM.

Obviously these sets remain disjoint, on the other hand we have

X =n⋃

i=1

giAi =m⋃j=1

hjBj.

This implies, in particular, that all free actions of F2 admit paradoxicaldecomposition. Note that if the action of G on X is transitive then it is notnecessary to use the axiom of choice.

We also have the following general fact.

Theorem 3.1.3. Suppose that G acts on sets X and Y and that f : Y → Xis a G-equivariant map. Then if the action of G on X is paradoxical, thisimplies that the action of G on Y is paradoxical. In particular, if G has aparadoxical action, then G is a paradoxical group.

Proof. It suffices to notice that if A1, . . . , An, B1, . . . , Bm ⊂ X give a para-doxical decomposition of X then their preimages under f give a paradoxicaldecomposition of Y .

Now, if the action of G on X is paradoxical, take any x ∈ X and considerthe map g 7→ gx. This map is G-equivariant. Therefore by above G isparadoxical.

19

3.2 Hausdorff paradox

Since all free actions of the free group on two generators give rise to para-doxical actions, the idea of paradoxical decomposition in Euclidean spaceis build up on the existence of free subgroups in the group of rotations intree-dimensional Euclidean space SO(3).

It is well known fact that SO(3) contains a lot of copies of the free groupon two generators. There many non-constructive proofs of this. For example,one can invoke Tit’s alternative to show this. In fact, a stronger statementis true. If we consider SO(3)×SO(3) with product topology, then the set ofall pairs that generate F2 is dense in SO(3)× SO(3).

The first explicit construction of the free subgroup in SO(3) goes backto Hausdorff, [71]. Here we give a simplified construction of Swierczkowski,[136].

Theorem 3.2.1. There are two rotations in SO(3) which generate the freegroup on two generators.

Proof. We define this rotation explicitly, by matrices

T±1 =

13

∓2√2

30

±2√2

313

00 0 1

; R±1 =

1 0 0

0 13

∓2√2

3

0 ±2√2

313

.

These are rotations by angle arccos(13) around z-axis and x-axis. Let ω

be a non-trivial reduced word in the free group on two generators. We willshow that q(ω) is a non-trivial rotation, where q is a homomorphism of F2

that sends generators to T and R. To simplify notations we will denote q(ω)again by ω. Conjugating ω by T we may assume that ω ends by T±1 on theright. To prove the theorem it would suffice to show that

ω

100

=1

3k

a

b√

2c

,

where a, b, c are integers, b is not divisible by 3 and k is the length of ω.

20

In order to show this we will proceed by induction on the length of ω. If|ω| = 1, then ω = T±1 and

ω

100

=1

3

1

±2√

20

.

Now let ω be equal to T±1ω′ or R±1ω′, where ω′ satisfies

ω′

100

=1

3k−1

a′

b′√

2c′

.

Applying matrices we see that either a = a′ ∓ 4b′, b = b′ ± 2a′, c = 3c′

or a = 3a′, b = b′ ∓ 2c′, c = c′ ± 4b′ and thus a, b and c are integers. It isleft to show that b is not divisible by 3. Now ω can be written as T±1R±1v,R±1T±1v, T±1T±1v or R±1R±1v, for some possibly empty word v. Thus wehave 4 cases to consider:

1. ω = T±1R±1v. In this case b = b′ ∓ 2a′ and a′ is divisible by 3. By as-sumptions, b′ is not divisible by 3 and thus b is not divisible by 3 as well.

2. ω = R±1T±1v. In this case b = b′ ∓ 2c′ and c′ is divisible by 3. But b′

is not divisible by 3, thus b is not divisible by 3 as well.

3. ω = T±1T±1v. By assumptions v(1, 0, 0) = 13k−2 (a′′,

√2b′′, c′′). It fol-

lows that b = b′±2a′ = b±2(a′′∓4b′′) = b′+ b′′±2a′′−9b′′ = 2b′−9b′′.Therefore, b is not divisible by 3.

4. ω = R±1R±1v. This case is similar to the previous one.

Thus, in all possible cases b is not divisible by 3, which implies the statement.

Now the proof of Hausdorff paradox is almost straightforward.

Theorem 3.2.2 (Hausdorff paradox). There exists a countable subset in asphere S2 such that its complement in S2 is SO(3)-paradoxical.

21

Proof. We can not apply Theorem 3.1.2 right away, since each non-trivialrotation fixes two points on the sphere. For a fixed free subgroup of SO(3),let M be the set of all points in S2, which are fixed by some elements of thisgroup. Obviously, M is countable and S2\M is invariant under the action ofour free group. Thus by Theorem 3.1.2 the statement follows.

In the next section we will show that the sphere S2 itself has a SO(3)-paradoxical decomposition.

3.3 Banach-Tarski paradox

The classical Banach-Tarski paradox amounts to a decomposition of a unitball into finitely many pieces, rearranging this pieces into unit balls. Sincethe group of rotations preserves the origin, it would not be sufficient toobtain Banach-Tarski paradox. For this purpose we add translations intothe picture.

Definition 3.3.1. Let G be a group acting on a set X. Two subsetsA,B ⊂ X are equidecomposable if there exist a pairwise disjoint subsetsA1, . . . , An ⊂ A and a pairwise disjoint subsets B1, . . . , Bn ⊂ B and g1, . . . , gnin G such that

A =n⋃

i=1

Ai, B =n⋃

j=1

Bj

and gi(Ai) = Bi for all 1 ≤ i ≤ n.

It is straightforward to check that equidecomposibility is an equivalencerelation. We denote it by A ∼G B or by A ∼ B when the ambient group isclear.

Proposition 3.3.2. Let G be a group that acts on a set X. If F ⊂ X isparadoxical and F is equidecomposable to E, then E is paradoxical.

Proof. Let F =n⋃

i=1

giAi =m⋃j=1

hjBj be a paradoxical decomposition of E.

Thus, by transitivity, E is equidecomposable to bothn⋃

i=1

Ai andm⋃j=1

Bj. This

implies that E is paradoxical.

22

It turns out that for countable sets of the sphere the situation is evenmore paradoxical then we might expect.

Proposition 3.3.3. Let D be a countable subset of S2. Then S2 and S2\Dare SO(3)-equidecomposable.

Proof. Since D is countable we can find a line L which does not intersect D.Let Λ be a collection of all α ∈ [0, 2π] for which there exist a natural numbern and a point x in D such that both x and ρ(x) are in D, where ρ is therotation around L by an angle nα. Obviously, Λ is countable and we can findθ ∈ [0, 2π)\Λ. Let ρ be a rotation by θ around L, then ρn(D) ∩ ρk(D) = ∅for all natural numbers k 6= n. Thus for D′ =

⋃n≥0

ρn(D) we have

S2 = D′ ∪ (S2\D′) ∼ ρ(D′) ∪ (S2\D′) = S2\D,

which proves the claim.

The direct corollary of this proposition and Hausdorff paradox is thefollowing.

Corollary 3.3.4. S2 is SO(3)-paradoxical.

Now we are ready to state and prove Banach-Tarski paradox, [9]. Denoteby E(3) the group of all Euclidean transformations of R3, that is the groupgenerated by all rotations and translations.

Theorem 3.3.5 (Banach-Tarski Paradox). Every ball in R3 is E(3)-paradoxical.

Proof. Let B1 be a unit ball around the origin. By Corollary 3.3.4, we canfind A1, . . . , An, B1, . . . , Bn ⊂ S2 and g1, . . . , gn, h1, . . . , hm ∈ SO(3) whichsatisfy Definition 3.1.1. Define

Ai = {tx : t ∈ (0, 1], x ∈ Ai} and Bj = {tx : t ∈ (0, 1], x ∈ Bj}.

Then we have A1, . . . , An, B1, . . . , Bn ⊂ B1\{0} are pairwise disjoint and

B1 =⋃

1≤j≤ngiAi =

⋃1≤j≤m

hjAj. Thus B1\{0} is paradoxical. We will show

that it is equidecomposable to B1.

23

Let x = (0, 0, 12). Let L be a line which does not contain {0} with x ∈ L.

Fix any rotation ρ of infinite order around this line. Let D = {ρn(0) : n ≥ 1},then 0 /∈ D and

B1 = D ∪ (B1\D) ∼ ρ(D) ∪ (B1\D) = B1\{0}.

Thus by Proposition 3.3.2 we have the statement.

24

3.4 Pea to Sun paradox

In this section we present the strongest version of Banach-Tarski paradox,which says that every two bounded sets with non-empty interior are equide-composable with respect to the group of Euclidean transformations of R3.This paradox is called Pea to Sun Paradox, since even a small pea can bedecomposed and rearranged to form the Sun.

Let a group G be acting on a set X and let A,B be two subsets of X. Wewill write A �G B is there exists a subset in B, which is G-equidecomposablewith A.

It is trivial to check that if two sets A and B are G-equidecomposable,then there exists a bijection φ : A → B such that for every C ⊂ A we haveC is G-equidecomposable with φ(C).

Lemma 3.4.1. Let G be a group that acts on a set X and A,B be twosubsets of X. If A �G B and B �G A, then A ∼G B.

Proof. Let A′ ⊆ A and B′ ⊆ B be such that A′ ∼G B and B′ ∼G A. Letφ : A → B′ be a bijection with property that for every C ⊂ A we have Cis G-equidecomposable with φ(C). Similarly, let ψ : B → A′ be a bijectionwith property that for every C ⊂ B we have C is G-equidecomposable withφ(C).

Define C0 = A\A′, and Cn+1 = ψ(φ(Cn)). Consider the set C =∞⋃n=0

Cn.

We have ψ−1(A\C) = B\φ(C). This implies that the sets A\C and B\φ(C)are G-equidecomposable. Similarly, C ∼G φ(C). Hence, we have

A = (A\C) ∪ C ∼G (B\φ(C)) ∪ φ(C) = B.

Using this lemma, we provide the following reformulation of Banach-Tarski paradox.

Theorem 3.4.2. Let B,B1, . . . , Bn be the balls in R3 of the same radius,such that B1, . . . , Bn are pairwise disjoint. Then ∪iBi ∼E(3) B.

Proof. Using the Lemma it suffices to show that ∪iBi �E(3) B. This, in turn,will follow from the case n = 2 and induction. So it suffices to show that

25

B1 ∪B2 �E(3) B. So let us take pairwise disjoint A1, . . . , As, C1, . . . , Ck ⊂ Band g1, . . . , gs, h1, . . . , hk ∈ E(3) such that B = ∪igiAi = ∪jCj. Then

B1 ∼ B = ∪igiAi � ∪iAi,

B2 ∼ B = ∪jgjCj � ∪jCj

Now we are ready to use Banach-Tarski paradox to prove its strong ver-sion.

Theorem 3.4.3 (Pea to Sun Paradox). Let A and B be bounded subsetswith non-empty interior in R3. Then A and B are E(3)-equidecomposable.

Proof. By Lemma 3.4.1 it is sufficient to prove that A � B. Since A isbounded we can find r > 0 such that A ⊂ Br(0). Let x be an interiorpoint of B. Then there is ε > 0, such that Bε(x) ⊂ B. Let g1, . . . , gn betranslations of R3 such that

Br(0) ⊂ g1Bε(x) ∪ . . . gnBε(x).

Choose translations h1, . . . , hn of R3, such that

hiBε(x) ∩ hjBε(x) = ∅, for alli 6= j.

Consider the set Y =⋃hjBε(x). By the reformulation of Banach-Tarski

Paradox, Theorem 3.4.2, we have Y � Bε(x). Hence,

A ⊂ Br(0) ⊂ g1Bε(x) ∪ . . . ∪ gnBε(x) � Y � Bε(x) ⊂ B,

which implies the statement.

26

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