The geometry of discrete groups

Summerschool at Utrecht University

by Martijn Caspers

August, 2016

In this short course we are going to look at groups from a completely newperspective (presumably). Namely, we are going to construct a graph from agroup that is finitely generated. This is the so-called Cayley graph. Thenwe shall look at the geometry of this graph: How fast does it grow? Howmany paths are there between distinct points? What does the growth rate andthe number of paths say about the group itself? This part of mathematics isnormally referred to as geometric group theory and leads to remarkablystrong results and peculiar questions starting from rather elementary objects.One of the highlights is certainly Gromov’s theorem showing that groups ofpolynomial growth are virtually nilpotent and vice versa, just to mention one.Applications of geometric group theory can be found ranging from functionalanalysis, operator algebras, ergodic theory and representation theory (non ofthese shall be covered here).

Figure 1. An example of a Cayley graph.

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Of course we shall only be able to give a mild introduction here. The cornerStone theorem is the Banach Tarski paradox. It states that the solid unitball in in R3 can be decomposed into finitely many pieces such that by justmoving and rotating the pieces one can make 2 balls of the same size (thatis, the volume doubles!). The paradox is not really a paradox, but rather atheorem! It was proved by Stefan Banach and Alfred Tarski in 1924. The keyfact is that the pieces shall have a horrible form: they do not have a well-definedvolume. This is obvious because the volume of an object stays the same if youmove or translate it and our construction increases the volume of our objectby a factor 2. Mathematically this means that the indicator function on thepieces cannot be Riemann integrable (this is the usual way of integrating youlearn in high school or in a first year calculus course). If you follow a course onmeasure theory later on the theorem implies more generally that there exists notranslation and rotation invariant (non-zero) measure on R3 for which each ofthe pieces defined in the proof of the Banach–Tarski theorem are measurable.We mention this only to show why it is important, the proof of the theoremdoes not involve integration theory at all, but relies on group theory.

Outline of these notes. We shall start from the definition of a group andimmediately introduce the Cayley graph and the notion of growth in Section 1.Then we give a proof of the Banach–Tarski paradox in Section 2. In particular weintroduce the notion of a paradoxical group and show how it can be used to proveBanach and Tarski’s theorem. In Section 3 we introduce another importantconcept, namely amenability as a converse to paradoxality. We show that groupsthat grow slowly are amenable. In the lectures we shall most likely skip thecomplete proof of the Banach–Tarski paradox because it takes a little moretime than 1 day. But for the interested reader we did include the proof in thesenotes to make the material complete.

Literature. The reader interested in these and more things is suggested tolook at either [Wag] (for paradoxality), [Pat] or the book in progress [Jus] (foramenability). We assume that the reader is familiar with basic group theory asfor example [Arm].

General notation. We denote |X| for the number of elements in a set X.

1 Growth of groups

For completeness we recall that a group is defined as follows.

Definition 1.1. A group consists of a set G and a multiplication · : G×G→ Gsatisfying the three axioms: (1) the multiplication is associative; (2) there is a

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unit e ∈ G with the property e · a = a = a · e for all a ∈ G; (3) every a ∈ G hasan inverse a−1 such that a−1 · a = e = a · a−1.

We shall omit the dot in the notation of the multiplication. In fact we areonly interested in a particular type of group.

Definition 1.2. A group G is finitely generated if there exists a finite setS ⊆ G that generates G. This means that every element in G can be written asa (finite) product of elements in S or S−1. The set S is called the generatingset. S is called symmetric if S = S−1 := {s−1 | s ∈ S}.

Remark 1.3. Suppose that G is finitely generated. Replacing S by S ∪ S−1

we may always assume that the generating set is symmetric.

Note that finitely generated groups are countable. We list some examples.

Example 1.4. Obvious examples of finitely generated groups are finite groups,Zn or the infinite dihedral group D∞ (recall that D∞ consists of all isometricbijections Z→ Z which is generated by k 7→ −k and k 7→ k+ 1). If G and H arefinitely generated groups, then also the following groups are finitely generated:

1. The Cartesian product G× H;

2. The free product G ∗ H, see Definition 1.11;

3. If HC G (normal subgroup) then the quotient G/H is finitely generated.

Example 1.5. Consider the group:

SL(n,Z) = { group of n× n-matrices with entries in Z and determinant 1} .

In fact it is a small exercise to show that this is a group (use Cramer’s rulefrom linear algebra to show that elements have inverses). For any n ∈ N thegroup SL(n,Z) is finitely generated. In case n ≥ 3 there is a very interestingproof of this fact due to David Kazhdan, see [Kaz], using so-called Kazhdan’sproperty (T). We just mention this is a fact and omit proofs (which are certainlynon-trivial).

Example 1.6. Consider the group S∞ of all bijections f : N→ N for which allbut finitely many points are fixed, i.e. the set Σ(f) := {k ∈ N | f(k) 6= k} isfinite. S∞ is not finitely generated, because if f1, . . . , fn would generated S∞then they can only shuffle the numbers 0 until max∪kΣ(fk).

Definition 1.7. Let G be a finitely generated group with symmetric generatingset S. The Cayley graph of G is the graph with edges the elements of G andvertices the set {(s, t) ∈ G | t−1s ∈ S − {e}}. We will denote this graph byΓ(G) = (V (G), E(G)) where V (G) is the vertex set and E(G) is the edge set.

Note that V (G) is in fact G but as they feel like different objects (verticesverses group elements) we distinguish this in notation. Note that as the set S inDefinition 1.7 is symmetric we have that (s, t) ∈ E(G) if and only if (t, s) ∈ E(G).In this sense the Cayley graphs are non-directed. The most important exampleis the Cayley graph of the free group with 2 generators, see Example 1.8

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Example 1.8. The free group with two generators a and b consists of words ina and b as well as their inverses a−1 and b−1 with the identifications

waa−1v = wa−1av = wv and wbb−1v = wb−1bv = wv,

where v and w are words with letters in {a, b, a−1, b−1}. The multiplication isgiven by concatenation of words. So for example:

(ab3a−5b3) · (b−3a2ba) = ab3a−5b3b−3a2ba = ab3a−5a2ba = ab3a−3ba.

We write F2 for this group from now on. A word in F2 can always be writ-ten in a shortest way in which case we call the word reduced. So the wordsab3a−5b3b−3a2ba and ab3a3ba are the same element of F2 but only the latterone is reduced. The following picture1 illustrates F2 by its Cayley graph. Thevertices are the elements of F2 written as a reduced word and two words (i.e.vertices) v and w are connected iff v ∈ {wa,wa−1, wb, wb−1}. The distance ofa word to the empty word e ∈ F2 is called the word length.

Figure 2. Cayley graph of the free group with 2 generators.

Exercise 1.9. Draw the Cayley graph of Z2 with generating set S = {ek,−ek |1 ≤ k ≤ 2} where e1 = (1, 0), e2 = (0, 1) are the standard basis vectors.

Exercise 1.10. Draw the Cayley graph of the dihedral group Dn with gener-ating set S = {t, r, r−1}. Here t and r are the standard generators having theproperty that trkt = r−k, t2 = e, rn = e.

Definition 1.11 (Free products of groups). Let G and H be groups. The freeproduct G ∗ H is the group consisting of words x1 . . . xk of some length k with

1One needs to read a−1 for A and b−1 for B.

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the property that x2i ∈ G and x2i+1 ∈ H and the rule that xi may only be theidentity if i = 1. Such words are called reduced. If a word x1 . . . xk with lettersxi ∈ G ∪ H is not reduced we may reduce it through the rules: (1) if xi is theidentity then x1 . . . xk = x1 . . . xi−1xi+1 . . . xk; (2) If xi, xi+1 ∈ G or xi, xi+1 ∈ Hthen x1 . . . xk = x1 . . . xi−1yxi+2 . . . xk where y = xixi+1. Note that (2) is morea matter of proper indexation. The product in G ∗ H is given by concatenationof words followed by reduction.

Example 1.12. We have that F2 = Z∗Z, as follows directly from the definitions.

Exercise 1.13. Draw the Cayley graph of Z/4Z ∗ Z/3Z with generating setS = {a, b, a−1, b−1} where a and b are respective generators of Z/4Z and Z/3Z.

Exercise 1.14. Give a group for which Figure 1 is the Cayley graph? Hint:make Exercise 1.13.

We end this section by defining the following important concept.

Definition 1.15. Let G be a group with finite symmetric generating set S.Assume e ∈ S. Let Γ(G) be its Cayley graph and pick some point p ∈ V (G).For k ∈ N let

B(k) = Sk := {g ∈ G | g = s1 . . . sk with si ∈ S}, (1)

i.e. the ball centered at e with radius k.

1. We say that G has polynomial growth if there exists a polynomial Psuch that |B(k)| ≤ P (k);

2. We say that G has exponential growth if |B(k)| > ak for some a > 1;

3. We say that G has intermediate growth if it does not have polynomialgrowth nor it has exponential growth;

4. We say that G has subexponential growth if limk→∞(|B(k)|)1/k = 1.

In these notes we won’t say much about intermediate growth. Polynomialgrowth implies subexponential growth by some elementary calculus.

Proposition 1.16. The growth of a group does not depend on the choice of thegenerating set. More precisely, let G be a group with finite symmetric generatingsets S and S′. Assume e ∈ S and e ∈ S′. Then G has polynomial growth (resp.subexponential growth, exponential growth) with respect to S iff G has polynomialgrowth (resp. subexponential growth, exponential growth) with respect to S′.

Proof. Let B(k) and B′(k) be the balls of (1) with respect to respectively Sand S′. Let l ∈ N be such that S′ ⊆ B(l). Then B′(k) ⊆ B(kl), so that|B′(k)| ≤ |B(kl)|. So if G has polynomial growth for S then |B(k)| ≤ P (k) forsome polynomial P . Then B′(k) ≤ Q(k) := P (kl) and so G has polynomialgrowth for S′. Similarly subexponential or exponential growth for S impliesrespectively subexponential or exponential growth for S′ (easy exercise). Asthe problem is symmetric under changing S and S′ we also get the reverseimplications.

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Obviously finite groups have polynomial growth. We give some more non-trivial examples.

Example 1.17. Let G be a finitely generated abelian group. Let S = {s1, . . . , sn}be a symmetric generating set containing the identity. The set B(k) consistsof elements sl11 . . . s

lnn with

∑i li ≤ k. The number of such elements can be

bounded by P (k) = k|S|. So G has polynomial growth.

Example 1.18. Consider F2. From the Cayley graph we see that if we increaseballs with a distance 1 then for every boundary point we get 3 new points, thatis |B(k+1)|−|B(k)| = 3(|B(k)|−|B(k−1)|) so that we get a recurrence relation|B(k + 1)| = 4|B(k)| − 3|B(k − 1)|. In this case one can write down an explicitformula for |B(k)| namely

|B(k)| = 2 · 3k − 1.

So F2 has exponential growth.

Remark 1.19. The question whether groups that have subexponential growthbut which do not have polynomial growth exist was asked by Milnor in 1968 andremained open for almost 20 years. In 1984 Grigorchuk gave the first exampleof such a group. We do not give his example here.

Exercise 1.20. Determine for each of the groups in Exercises 1.13 and 1.14what kind of growth they have.

Definition 1.21 (Semi-direct product). Suppose we have two groups H and N.Suppose that H acts on N by means of automorphisms, that is there exists agroup homomorphism α : H → Aut(N). The semi-direct product G = N o His the group given as a set by N × H and with multiplication (n, h)(m, k) =(nαh(m), hk).

Exercise 1.22 (The Lamplighter group). Let N be the abelian group of finitelysupported functions Z → Z/2Z. For k ∈ Z, f ∈ H let fk ∈ N be given byfk(l) = f(l − k). So k 7→ (f 7→ fk) makes that Z acts on N by automorphisms.The semi-direct product No Z is called the Lamplighter group. It has naturalgenerators t := 1 ∈ Z and s = δ0 ∈ N (indicator function at 0 ∈ Z). Show thatthe Lamplighter group has exponential growth.

2 Paradoxical groups

In this section we introduce paradoxical groups, show that the free group F2 isparadoxical and how the Banach–Tarski paradox follows from it. Recall fromthe introduction that the latter states that a unit ball in R3 can be chopped upinto finitely many pieces that form together two unit balls with total volumedoubling the original ball. We first state this formally and then give the proof.The only part of this chapter that is really relevant for the next sections is thepart on paradoxical groups, see Section 2.2.

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2.1 Formal statement of the Banach–Tarski paradox

.An affine map on R3 is a map of the form fA,b : x 7→ Ax+a where A ∈ SO(3)

and a ∈ R3. As fA,a ◦ fB,b = fAB,Ab+a and f−1A,a = fA−1,−A−1a we see that the

affine transformations form a group. Geometrically this group consists of allpossible translations and rotations of R3. We call this group Aff(Rn). LetB = {v ∈ R3 | ‖v‖ ≤ 1} be the unit ball in R3 with respect to the Euclidian

distance ‖v‖2 =∑3i=1 v

2i . We let S = {v ∈ R3 | ‖v‖ = 1} be the sphere with

radius 1.

Theorem 2.1. There exists a partition of a subset of B into k + n piecesA1, . . . , An, B1, . . . , Bk and elements f1, . . . , fn, g1, . . . , gk ∈ Aff(R3) in sucha way that the sets

f1(A1), . . . , fn(An),

form a partition of B and (!) the sets

g1(B1), . . . , gk(Bk),

form a partition of B.

Throughout the next sections we prove this theorem, which is known as theBanach-Tarski paradox.

2.2 Paradoxical groups

Definition 2.2. Let G be a group. G is called paradoxical if there exists n+kdisjoint subsets E1, . . . , En, F1, . . . , Fk of G and elements g1, . . . , gn, h1, . . . , hk ∈G such that both g1E1, . . . , gnEn and h1F1, . . . , hkFk form a partition of G.

Example 2.3. Finite groups can never be paradoxical, since then as E1, . . . , En,F1, . . . , Fn are disjoint subsets of G we have:

|G| ≥ |E1|+ . . .+ |En|+ |F1|+ . . .+ |Fk|.

But on the other hand we also have

|G| =|g1E1|+ . . .+ |gnEn|,|G| =|h1F1|+ . . .+ |hkFk|.

As |giEi| = |Ei| and |hiFi| = |Fi|, this leads to the contradiction |G| ≥ 2|G|. Wegive more examples of non-paradoxical groups in Section 3.

Proposition 2.4. F2 is paradoxical.

Proof. Consider the sets:

E1 ={w ∈ F2 | w starts with a},E2 ={w ∈ F2 | w starts with a−1},F1 ={w ∈ F2 | w starts with b},F2 ={w ∈ F2 | w starts with b−1},

(2)

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Then E1, aE2 and F1, bF2 are each partitions of F2.

Exercise 2.5. Show that the free group with 3 generators F3 := Z ∗ Z ∗ Z isparadoxical.

Exercise 2.6. Show that Z/3Z ∗ Z/6Z is paradoxical.

2.3 The Banach-Tarski paradox for the sphere SWe shall now prove that S can be decomposed in finitely many pieces in sucha way that after rotating them they form two copies of S. We first turn thisinto a formal definition. Note that Aff(R3) acts naturally on R3 by evaluationof the mapping.

Definition 2.7. Two subsets A,B ⊆ R3 are called equidecomposable if thereexists a partitions A1, . . . , An of A and f1, . . . , fn ∈ Aff(R3) such that f1(A1),. . ., fn(An) is a partition of B. We shall write A ' B if A and B are equide-composable.

Lemma 2.8. Equidecomposability forms an equivalence relation.

Proof. It is easy to see that ' is symmetric and reflexive. Now we prove trans-itivity of the relation. Suppose that A ' B and B ' C. Let B1, . . . , Bn bea partition of B and f1, . . . , fn ∈ Aff(R3) be such that f(B1), . . . , f(Bn) is a

partition of A. Let B̃1, . . . , B̃k be a partition of B and f̃1, . . . , f̃k ∈ Aff(R3)

be such that f̃(B1), . . . , f̃(Bk) is a partition of C. Consider the partition of B

given by Bi,j = Bi ∩ B̃j . Then the sets

fi(Bi,j), with 1 ≤ i ≤ n, 1 ≤ j ≤ k, (3)

form a partition of A and the sets

f̃j(Bi,j), with 1 ≤ i ≤ n, 1 ≤ j ≤ k,

form a parition of C. In all we have shown that the sets (3) together with the

maps f̃j ◦ f−1i witness that A ' C.

Definition 2.9. A subset C ⊆ R3 is called paradoxical if there are subsetsA,B ⊆ C such that A ∩B = ∅ and A ' C ' B.

Lemma 2.10. Let θ = cos−1( 13 ). The map F2 → SO(3) that assigns

a 7→

cos(θ) − sin(θ) 0sin(θ) cos(θ) 0

0 0 1

, b 7→

1 0 00 cos(θ) − sin(θ)0 sin(θ) cos(θ)

. (4)

determines an injective homomorphism F2 → SO(3).

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Proof. We suggest the reader to skip this proof when reading these notes forthe first time as it is not very interesting and we merely sketch it. See [Wag] fordetails. The non-trivial part of the statement is injectivity of this homomorph-ism. We need to prove that the kernel is {e}. First note that the image of a isin fact given by: 1

3 − 2√

23 0

2√

23

13 0

0 0 1

,

and the matrix of b is obtained by changing the x- and z-axis in the basis.Consider a word w in a and b that ends on a of a−1. By induction on the length

of w one proves that the image of (1, 0, 0) under w is of the form ( a3N ,

b√

23N , c

3N )with a, b, c ∈ Z and N ∈ N (exercise: check this). In fact N may equal theword length of w. In turn one checks by induction to the word length of w thata 6= b mod 3, c 6= b mod 3 and 0 6= b mod 3 (exercise: check this). In particularb 6= 0 and so w can never act trivially on (1, 0, 0). Therefore w cannot be in thekernel of the homomorphism. A symmetry argument allows us to obtain thesame conclusion for words that end on b or b−1.

From now on we suppose that F2 acts on S by means of the homomorphism(4).

Proposition 2.11. There exists a countable subset F ⊆ S such that S − F isparadoxical.

Proof. Let F be the union of the fixed points in S of all elements of F2 − {e}.Note that each element of F2 − {e} acts by a rotation on S and therefore thereare exactly 2 fixed points. Therefore F is countable. Furthermore F2 acts freelyon S −F (why?). The orbits of F2 form a partition of S −F (see [Arm, p. 92]).Let O be the set of all orbits of the action F2 on S − F and for each O ∈ Owe may pick xO ∈ O such that F2xO = O. Let E1, E2, F1, F2 ⊆ F2 be the setsdefined in (2). Set Ai = ∪O∈OEi(xO) and Bi = ∪O∈OFi(xO). As F2 acts freelyon S − F these sets are disjoint (why?). Moreover A1, a(A2) and B1, b(B2) areeach partitions of S − F (why?).

Exercise 2.12. Check each of the statements in the proof of Proposition 2.11after which we wrote (why?).

It remains to repair some cavities, i.e. remove the countable set F fromthe statement of Proposition 2.11. The idea of the following proof is basedon the following principle. How do you show that N ∪ {−1} and N have thesame cardinality? Well, one makes a bijection N ∪ {−1} to N by sending n ton+1. We are going to use the same principle here applied to the set F ⊆ S andcountably many disjoint copies x(F ), x2(F ), x3(F ), . . . for some suitable elementx ∈ SO(3). Here is the formal proof...

Theorem 2.13. The sphere S is paradoxical.

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Proof. It suffices to prove that S ' S−F where F is the countable set from Pro-position 2.11. We claim that there exists an x ∈ SO(3) such that F, xF, x2F, . . .are all disjoint. Suppose for a while that this claim is true, then we may con-clude the proof as follows. We set A1 = ∪∞i=1x

iF and A2 = (S − F ) − A1. Sotogehter they form S − F . We set B1 = ∪∞i=0x

iF and B2 = S − B1. Togetherthey form S. In fact A2 = B2 and A1 = x(B1). This shows that S − C ' S.

It remains to prove our claim. We sketch an abstract argument. Let X1 ⊆ Sbe the union of the fixed points of all elements x ∈ SO(3) for which F∩x(F ) 6= ∅.We claim that X1 is contained in a countable union of unit circles. Indeed,suppose that a, b ∈ F and x(a) = b for x ∈ SO(3). Then the axis of rotation ofx must be equally far from a as from b. This means that the two points thatare fixed by x are equally far from a and b. The space of points on S with equaldistance from a and b form a circle (with radius 1). As there are countably manypairs (a, b) ∈ F×F it follows indeed that X1 is a countable union of circles. Nextone defines Xn as the set of elements x ∈ SO(3) such that F, x(F ), . . . , xn(F ) arenot mutually disjoint. Again Xn turns out to be contained in a countable unionof circles (same argument) and therefore X∞ := ∪∞n=1Xn is so. So X∞ 6= S(why?). Therefore there is x ∈ SO(3) whose fixed points are not in X∞ whichmust then witness the assertion of our claim by construction.

2.4 Conclusion: The Banach-Tarski paradox

We conclude by proving our main theorem. In fact from this point we needalmost no new ingredients.

Proof of Theorem 2.1. By Theorem 2.13 we may find disjoint subsets E1, . . . , En,F1, . . . , Fk of S and f1, . . . , fn, g1, . . . , gk ∈ Aff(Rn) such that f1(E1), . . . , fn(En)is a partition of S and g1(F1), . . . , gk(Fk) is a partition of S. Then the sets

Ai = {x ∈ B\{0} | x

‖x‖∈ Ei}, Bi = {x ∈ B\{0} | x

‖x‖∈ Fi}

again with affine maps gi’s and fi’s witness that B\{0} is paradoxical. As inthe proof of Theorem 2.13 we may show that B\{0} ' B (in this case it is mucheasier to find the affine action x used in the proof) which yields the theorem.

2.5 Final remarks

Three important things need to be said:

1. For the sake of simplicity of our proof we only showed that there aredisjoint subsets A and B of B that are equidecomposable with B. We didnot show that we can also get A∪B = B, but working a bit harder this isalso possible. Another thing we did not optimize is the number of piecesin which A and B are decomposed. It turns out that there is an optimalproof in which A comprises 2 pieces and B comprises 3 pieces.

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2. An intriguing logical axiom states that if {Oi}i∈I is some indexed familyof non-empty sets Oi then there exists a function f : I → ∪i∈IOi withthe property f(i) ∈ Oi. (Imagine a couple of children indexed by I eachhaving a floating balloon in their hands with a non-zero number of candiesOi in it. The axiom states that each child can choose a candy in its ownballoon). This seems sort of obvious, but in fact it cannot be derived fromother axioms and we used it (quite explicitly) in the proof of our paradox.(Exercise: where?). The axiom is called the axiom of choice.

3. There are many generalizations. B may be replaced by any unit ball ofdimension ≥ 3 (and using a proof similar to the one of Section 2.4 it canbe derived from our theorem). Also it is not really important that we dealwith balls but rather with sets in Rn, n ≥ 3 that contain at least a ball ofsome radius. For precise statements we refer to [Wag].

3 Amenable groups

We return to the question which groups are not paradoxical. So far we onlysaw that finite groups cannot be paradoxical (see Example 2.3). The argumentwas a simple counting argument of the number of elements. We are actuallygoing to extend this counting argument and turn it into a density argument.That is given a function f : G → C on a group G is there any way of takingits average/mean value? The following definition shows that for some groupsthere is a sensible notion. We define it only for countable groups to avoidtechnicalities.

Definition 3.1. Let G be a countable group. G is called amenable if thereexists an increasing sequence of finite subsets Fj ⊆ G such that ∪jFj = G andwith the property that,

∀s ∈ G : limj→∞

|sFj ∩ Fj ||Fj |

= 1. (5)

The sets Fj are called Følner sets or a Følner sequence.

Intuitively amenability means that for every s ∈ G for big j the set Fj isalmost invariant under translation with s. Let us give an example.

Example 3.2. The group Z is amenable. Indeed take Fj = [−j, j] (as a discreteinterval!). Then take any k ∈ Z. We have in case j > k,

|(k + Fj) ∩ Fj ||Fj |

=|[k − j, k + j] ∩ [−j, j]|

|[−j, j]|=

2j + 1− k2j + 1

.

And this expression converges to 1 as j →∞.

By a very similar proof we immediately get a whole class of amenable groupsin the following theorem. We write X∆Y = X ∪ Y −X ∩ Y .

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Exercise 3.3. Show that (5) is equivalent to

∀s ∈ G : limj→∞

|sFj∆Fj ||Fj |

= 0.

Exercise 3.4. Let G be a group that is generated by a set S. Show that inorder to conclude (5) it suffices to check (5) for all s ∈ S.

Theorem 3.5. Let G be a finitely generated group with subexponential growth.Then G is amenable.

Proof. Let again S be a symmetric generating set for G containing the identityand consider B(k) = Sk as in (1). As G has subexponential growth we havelimk |B(k + 1)|/|B(k)| = 1. Also limk |B(k − 1)|/|B(k)| = 1. Let s ∈ S. Then,

|sB(k)∆B(k)||B(k)|

≤ |B(k + 1)| − |B(k − 1)||B(k)|

→ 0.

So G is amenable by Exercises 3.3 and 3.4.

Exercise 3.6. Let G be an amenable group with Følner sets Fj . Let s ∈ G.Show that also sFj are Følner sets.

Exercise 3.7. Let G be an amenable group with Følner sets Fj . Let α : G→ Gbe an automorphism. Show that also α(Fj) are Følner sets.

The converse of Theorem 3.5 is not true as the following proposition shows.Recall that the Lamplighter group does not have subexponential growth, seeExercise 1.22.

Proposition 3.8. The Lamplighter group is amenable.

Proof. We in fact prove a stronger statement. Suppose that there is an actionof a group H on a group N by means of automorphisms (we write h ·n for h ∈ Hacting on n ∈ N). Suppose that both H and N are amenable. Then N o H is

amenable. Indeed, let F(H)j and F

(N)j be Følner sets for respectively H and N.

Then we define the set,

F(NoH)i,j := {(n, h) | h ∈ F (H)

i , n ∈ h · F (N)j }.

Note that |F (NoH)i,j | = |F (H)

i ||F(N)j |. For m ∈ N we have,

(m, e)F(NoH)i,j = {(n, h) | h ∈ F (H)

i ,m−1n ∈ h · F (N)j }.

Therefore,

|F (NoH)i,j ∩ (m, e)F

(NoH)i,j |

|F (NoH)i,j |

=1

|F (H)i |

∑h∈F (H)

i

|m(h · F (N)j ) ∩ h · F (N)

j |

|F (N)j |

. (6)

12

Letting j → ∞ and then letting i := i(j) → ∞ shows that (6) goes to 1 (asfollows using Exercise 3.7 and the definition of Følner sets). Also for k ∈ H weget,

(e, k)F(NoH)i,j =

{(k · n, kh) | h ∈ F (H)

i , n ∈ h · F (N)j

}={

(n, h) | k−1h ∈ F (H)i , n ∈ h · F (N)

j

}.

So that|F (NoH)i,j ∩ (e, k)F

(NoH)i,j |

|F (NoH)i,j |

=|F (H)i ∩ k · F (H)

i ||F (H)i |

,

which goes to 1 if we let i, j →∞. The proof then follows from Exercise 3.4.

The following theorem relates amenability to Section 2. In particular it givesexamples of non-amenable groups, such as the free group F2. The usual proofuses Alaoglu’s theorem (=the unit ball in a dual Banach space is compact inthe weak-∗ topology). In fact Alaoglu’s theorem admits a proof that is quitebeautiful (and short) but we have chosen to keep these notes free from functionalanalysis and topology. We recast the proof by using ultrafilters. If one acceptsthat there exists a non-principal ultrafilter then the proof is self-contained. Itis also quite remarkable that they give rise to a notion of limit for which everybounded sequence converges!

Definition 3.9. Let X be a set. An ultrafilter on X is a set ω of subsets of Xwith the following properties. (1) Either A ∈ ω or X−A ∈ ω; (2) If A ⊆ B ⊆ Xand A ∈ ω then B ∈ ω; (3) If A,B ∈ ω then A ∩B ∈ ω; (4) ∅ 6∈ ω.

It follows from the axioms (3) and (4) that we cannot have that both A andX −A are in ω.

Example 3.10. Let x ∈ X and let ω be the set of all sets A ⊆ X thatcontain x. Then it is easy to check that ω is an ultrafilter. Such ultrafiltersare called principal (or trivial). We denote this ultrafilter by ωx. If X is finiteall ultrafilters are trivial. If X is infinite there always exists a non-principalultrafilter (see [Jec]).

Definition 3.11. Let ω be an ultrafilter on N. Let (xi)i be a bounded sequencein C. Then limi,ω xi = x if for every ε > 0 we have {i | |xi − x| < ε} ∈ ω. Wecall x the ultralimit. Note that the limit depends on the choice of ω.

Exercise 3.12. Suppose that limi,ω ai = a and limi,ω bi = b. Prove the follow-ing:

1. limi,ω ai + bi = limi,ω ai + limi,ω bi.

2. If all ai ≥ 0 then limi,ω ai ≥ 0.

3. If ω is principal, say ω = ωx then limi,ω ai = ax.

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So we see already from Exercise 3.12 that the ultralimit does not agree withthe classical limit if the ultrafilter is principal. In general the ultralimit does(always) depend on the choice of the ultrafilter.

Lemma 3.13. Every bounded sequence (xi)i in R has an ultralimit. Supposethat the ultrafilter is non-principal. Then if (xi)i converges in the usual sensethen the limit agrees with the ultralimit.

Proof. Let ω be the ultrafilter. We construct intervals I1 ⊇ I2 ⊇ I3 ⊇ . . . withinterval length converging to 0. The interval Ik is chosen as follows. ConsiderJk,l = [ l

2k ,l+12k ). For every k there is exactly one l for which {i | xi ∈ Jk,l} ∈ ω

(why?). Call this lk. Then set Ik = Jk,lk . We have Ik ⊇ Ik+1 (why?) andclearly the interval length converges to 0. The (usual) limit of the endpointsof the intervals will be called x. Then we claim that limi,ω xi = x. Indeed letε > 0. Consider Bε = {y | |y − x| < ε}. There exists k such that Ik ⊆ Bε(namely for every k with 1

k < ε/2 such interval exists). And therefore {i | xi ∈Bε} ⊇ {i | xi ∈ Ik} ∈ ω and hence (why?) we have {i | xi ∈ Bε} ∈ ω. Thisshows that limi,ω xi = x.

It remains to prove the second statement. Let ε > 0 then as xi convergesto x we see that the set Aε := {i | |x − xi| < ε} has the property that itscomplement is finite, i.e. cofinite. Either Aε or X − Aε is in ω. If the finiteset X − Aε is in ω then it follows (why?) that ω is principal so we must haveAε ∈ ω. This proves the statement.

Exercise 3.14. In the proof of Lemma 3.13: each time we wrote (why?) de-termine which axioms of an ultrafilter are used to draw these conclusions.

Theorem 3.15. Let G be an amenable group. Then G cannot be paradoxical.

Proof. Let V be the vector space given by the linear span of indicator functionsχA with A ⊆ G. Fix a non-principal ultrafilter ω on N. We define a linear mapϕ : V → C by mapping f ∈ V to,

limj,ω

∑s∈Fj

1

|Fj |f(s).

We claim that ϕ is translation invariant. That is, let f ∈ V and for t ∈ G letft(s) = f(ts). We claim that ϕ(f) = ϕ(ft). To prove this, by linearity of ϕ itsuffices to take f = χA for some A ⊆ G. Note that

|A ∩ tFj ||Fj |

≤ |A ∩ Fj ||Fj |

+|A ∩ (tFj∆Fj)|

|Fj |.

The Følner condition gives that|tFj∆Fj ||Fj | → 0 so that

limj,ω

|A ∩ tFj ||Fj |

≤ limj,ω

|A ∩ Fj ||Fj |

. (7)

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The converse inequality of (7) follows from Exercise 3.6. But this just meansthat,

ϕ(ft) = limj,ω

∑s∈Fj

1

|Fj |ft(s) = lim

j,ω

|t−1A ∩ Fj ||Fj |

= limj,ω

|A ∩ tFj ||Fj |

= limj,ω

|A ∩ Fj ||Fj |

= ϕ(f).

Now if G were to be paradoxical then suppose that E1, . . . , En, F1, . . . , Fk aredisjoint subsets witnessing this. That is, there exist g1, . . . , gn, h1, . . . , hk ∈ Gsuch that giEi and hiFi each form a partition of G. Then,

1 ≥ϕ(χ∪1≤i≤nEi∪1≤j≤kFj)

=ϕ(∑

1≤i≤n

χEi+

∑1≤j≤k

χFj)

=∑

1≤i≤n

ϕ(χEi) +

∑1≤j≤k

ϕ(χFj)

=∑

1≤i≤n

ϕ(χgiEi) +

∑1≤j≤k

ϕ(χhjFj)

=ϕ(∑

1≤i≤n

χgiEi) + ϕ(

∑1≤j≤k

χhjFj)

=2ϕ(χG) = 2.

Which is a contradiction, so G cannot be paradoxical.

Remark 3.16. The converse of Theorem 3.15 is also true and is known asTarski’s theorem. The proof would take another chapter to explain and requiressome (light) techniques from functional analysis (namely the Hahn–Banach the-orem).

Remark 3.17 (Thompson group F ). The Thompson group F is defined as thegroup of all continuous increasing piecewise linear bijections f : [0, 1] → [0, 1]with the property that f may only be non-differentiable at the points { k

2l | k, l ∈N} and whose slopes are powers of 2. It is an open question if Thompson’s groupF is amenable or not. Moreover, the last years there have been several wrongproofs of either amenability or non-amenability (in fact there are about equallymany proofs in the positive as there are in the negative, and approximately eachyear one of the sides produces such a proof). Warning: there is hidden lie inthis remark, namely that Thompson’s group F is a topological group (groupwith topology for which multiplication and inversion is continuous) for whichamenability is defined in a different (non-equivalent) way!

Exercise 3.18 (Baumslag–Solitar groups). Let BS(1, n) =< a, b | b−1ab =an >. This is by definition the quotient of the free group F2 with generators a, bmodulo the normal subgroup generated by the element b−1aba−n. Show thatB(1, n) has exponential growth and is amenable.

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References

[Arm] M.A. Armstrong, Groups and symmetry, Springer-Verlag, New York,1988. xii+186 pp.

[Jec] T. Jech, Set Theory, Springer Monographs in Mathematics, 1978.

[Jus] K. Juschenko, Amenability, In preparation. Current version available athttp://www.math.northwestern.edu/~juschenk/book.html.

[Kaz] D. Kazhdan, On the connection of the dual space of a group with thestructure of its closed subgroups, Funct. Anal. Appl. 1, 63–65, 1967.

[Pat] A.L.T. Paterson, Amenability, Mathematical Surveys and Monographs,29. American Mathematical Society, Providence, RI, 1988.

[Wag] S. Wagon, The Banach-Tarski paradox, Cambridge University Press,Cambridge, 1993. xviii+253 pp.

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