amenable groups - tu dresdengournay/amenability-note.pdf · amenability cirm luminy - 2014/1/14...

Amenability CIRM Luminy - 2014/1/14 (version of 2016/7/18)

Amenable groupswritten1 by Antoine Gournay

Throughout this note, the convention is right multiplication and right convolution to becoherent with the notation in other lectures.

1 How to build amenable groups

𝐴△𝐵 = (𝐴 ∖𝐵) ∪ (𝐵 ∖ 𝐴). Groups are assumed countable and discrete.

Definition 1.1. (Følner, 1955) A group Γ is amenable if and only if there exists a Følner

sequence, i.e. a sequence of finite sets {𝐹𝑛}𝑛≥1 such that for all 𝛾 ∈ Γ, lim𝑛→∞

|𝐹𝑛𝛾△𝐹𝑛|

|𝐹𝑛|= 0 F

Example 1.2. Here are a few basic examples:

· A finite group is amenable, take 𝐹𝑛 = Γ for all 𝑛. Then 𝐹𝑛𝛾 = 𝐹𝑛 and|𝐹𝑛𝛾△𝐹𝑛|

|𝐹𝑛|= 0.

· A direct union of finite groups is amenable Γ = ⊕𝑛∈N𝐺𝑖 (where 𝐺𝑖 are finite groups).An element 𝛾 ∈ Γ is a sequence 𝛾𝑖 ∈ 𝐺𝑖, so that for some 𝑛0 (depending on 𝛾), ∀𝑛 ≥ 𝑛0,𝛾𝑛 = 𝑒𝐺𝑖

. Take 𝐹𝑛 = ⊕𝑛𝑖=0𝐺𝑖, then for any 𝛾 ∈ Γ, 𝐹𝑛𝛾 = 𝐹𝑛 for 𝑛 large enough. So


|𝐹𝑛|= 0 for 𝑛 large enough.

· Z is amenable. Take 𝐹𝑛 = [1, 𝑛]. Without loss of generality, one can compute only𝛾 > 0. Then


|𝐹𝑛|= 1

𝑛|[1, 𝛾] ∪ [𝑛+ 1, 𝑛+ 𝛾]| = 2|𝛾|/𝑛→ 0. ♣

When the graph is finitely generated by a set 𝑆, 𝜕𝑆𝐹𝑛 = {edges (in the Cayley graph of Γfor 𝑆) between 𝐹𝑛 and 𝐹 c

𝑛 := Γ ∖𝐹𝑛. A first useful lemma that reduces some computation is:

Lemma 1.3

Assume Γ is finitely generated. A sequence {𝐹𝑛} is Følner if and only if lim𝑛→∞

|𝜕𝐹𝑛|

|𝐹𝑛|= 0.

The proof is left as an exercise. An isoperimetric profile for the group is an increasing functionF : R>0 → R>0 such that ∃𝐶 > 0 so that for any finite set 𝐹 , F (|𝐹 |) ≤ |𝜕𝐹 |. Amenablegroups are groups with sublinear F .

In L. Saloff-Coste’s lectures it was shown that growth of balls which is ≥ 𝐶 ′𝑛𝑑 impliesF (𝑥) = 𝑥(𝑑−2)/𝑑 works. However putting 𝑑 = ∞ is a risky business: there are groups whichare amenable but have growth faster than any polynomial (e.g. 𝐵𝑆(1, 2) from C.Pittet’slecture). It’s an easy exercise to show F (𝑥) = 𝑥(𝑑−1)/𝑑 is optimal for Z𝑑 (see exercises).

Actually one can easily show that groups with “slow” volume growth are amenable. Recalla group is of subexponential growth if 𝛽𝑆(𝑛) := |𝐵(𝑛)| = 𝑒𝑓(𝑛) for some (increasing) function

𝑓 : N→ R>0 satisfying lim𝑛→∞

𝑓(𝑛)

𝑛= 0.

1without much preparation! These notes are in an incomplete state. Please email comments / corrections /improvements / references / insults / etc... to [email protected]


Proposition 1.4 (Hulanicki, 1971?2)

Let Γ be a [finitely generated] group of subexponential growth, then Γ is amenable.

Proof: The number of edges in 𝜕𝐵𝑛 is at most the number of vertices in the sphere ofradius 𝑛+ 1 times the maximal degree of a vertex, |𝑆|.

|𝜕𝐵𝑛|

|𝐵𝑛|≤ |𝑆|

|𝐵𝑛+1| − |𝐵𝑛|

|𝐵𝑛|= |𝑆|(𝑒𝑓(𝑛+1)−𝑓(𝑛) − 1).

The claim is that there is some sequence of integers {𝑛𝑘}𝑘≥0 so that 𝑓(𝑛𝑘 + 1)− 𝑓(𝑛𝑘) tendsto 0. If this is the case, then 𝐹𝑘 = 𝐵𝑛𝑘 would be a Følner sequence.

So assume there are no such sequences, or, in other words that ∃𝜖 > 0 such that ∀𝑛 >0, 𝑓(𝑛+1)−𝑓(𝑛) > 𝜖 (since 𝑓(𝑛) is increasing, absolute values are not required). This impliesthat

𝑓(𝑘) = 𝑓(0) +𝑘∑𝑖=1

(𝑓(𝑖)− 𝑓(𝑖− 1)

)> 𝑘𝜖.

This a contradiction with lim𝑛→∞

𝑓(𝑛)

𝑛= 0. Hence the desired subsequence exists and Γ is

amenable. �

Note that it is highly non-trivial to see whether the sequence of all balls works (and not justa subsequence). This turns out to be true (but it’s difficult) in nilpotent groups. It is openin a group of intermediate growth (intermediate = superpolynomial and subexponential).

The converse of proposition 1.4 is not true: 𝐵𝑆(1, 2) is amenable but has exponentialgrowth (see C. Pittet’s lecture).

Proposition 1.4 gives many groups without too much effort. The next theorem is usefulto build groups out of known amenable groups.

Theorem 1.5 (“The closure properties”3)

Let Γ, 𝑁 and {Γ𝑖}𝑖≥0 be amenable groups.

(a) If 𝐻 is a subgroup4 of Γ then 𝐻 is amenable “Subgroup”

(b) If 𝐻 is an extension 𝑁 by Γ ( i.e. 1→ 𝑁 → 𝐻 → Γ→ 1 is an exact sequence) then𝐻 is amenable “Extension”

(c) If 𝑁 C Γ then 𝐻 = Γ/𝑁 is amenable “Quotient”

(d) If 𝐻 is a direct limit of the Γ𝑖 then 𝐻 is amenable “Direct limit”

It’s perfectly possible to prove these properties from the current definition of amenability. Itturns out to be much easier to use the most convenient of the many equivalent definition ofamenability to do this.

But before moving to these considerations, it’s nice to wander a bit.

2Conflicting references: Pier [see p.114 and p.117] attributes it to Hulanicki 1971. In the book by Greenleaf(1969), the proof of theorem 3.3.6 clearly contains a very similar argument. Some people told me they weretold that it is in a paper in Russian... but I have no clear references, possibly Adel’son-Vel’skiı & Šreıder (1957in Uspehi Mat. Nauk.)...

3(a)-(c) are in von Neumann’s article [1929]. (d) is also in there for direct unions.4When dealing with groups which are not with the discrete topology, one should require that the subgroups

be closed. Otherwise, it is possible to have a free group as a subgroup of some compact groups.


Definition 1.6. A group is called elementarily amenable (short notation: EA) if it is ob-tained by (many) applications of the closure properties (a)-(d) starting from the followingclass of groups: finite groups and Z. F

Day asked whether “amenable” = EA (“Day’s conjecture”). Here are two important factsabout EA groups:

Theorem 1.7 (Chou, 1980 )

If Γ is elementarily amenable, then Γ has either polynomial or exponential growth.

Since there are groups of intermediate growth [Grigorchuk, recall J. Brieussel’s talk], amenable= EA. In fact, it’s even known that throwing in all groups of subexponential growth in the“construction via closure properties” is not sufficient to get all amenable groups [Bartholdi-Virag].

Theorem 1.8 (Chou, 1980 )

If Γ is elementarily amenable, then Γ can be obtained by using only (b) and (d).

There are finitely generated simple amenable groups [Juschenko-Monod, 2013]. To see thata [fin.gen.] simple amenable groups is not EA, one needs to know that the only necessaryproperties are (b) and (d). If Γ is simple, clearly the last operation was not an extension.But since Γ is finitely generated, the direct limits is unimportant: the generators live in afinite part of the limit. Thus one cannot get a simple group.

Example 1.9. Solvable groups are given by successive extensions of Abelian group, hencethey are EA (and, in particular, amenable). ♣

Mixing extensions and direct unions can be useful to create groups with “strange” prop-erties . See “More examples” in §4.A (e.g. EA groups which are neither linear nor residuallyfinite) or Reiter and Liouville §5.A (e.g. amenable groups with non-constant bounded har-monic functions) below.

Projective limits [a.k.a. inverse limits] of amenable groups are not amenable. I lack aconvincing example at the moment5.

2 Equivalent definitions and proof of “closure properties”

Much like hyperbolic groups, the great thing about amenable groups is that there are lots ofways to see them which turn out to be equivalent.

Although many of these equivalent definitions do not require the axiom of choice, it willbe used without warning6.

Some preliminary notations and terminology... A map (of a vector space into another

vector space) is affine if it preserves mid-points (see §5.B), i.e. 𝑓(𝑣+𝑤2

) = 12

(𝑓(𝑥) + 𝑓(𝑦)

).

Denote by 1𝐴 the characteristic function of 𝐴 and 1 := 1Γ. A function satisfies 𝑓 ≥ 0 if∀𝑥, 𝑓(𝑥) ≥ 0. The operator 𝑅𝜇 is defined by right convolution by 𝜇, i.e. 𝑅𝜇𝑓 = 𝑓 * 𝜇. Themeasure 𝜇(𝑘) is the 𝑘th iterated convolutions of 𝜇. The function 𝛿𝛾 is the Dirac mass at 𝛾.

5A naive idea would be to look at the projective limits of SL2(Z𝑛). Indeed, since F2 embeds in SL2(Z) itwill embed in there. However, the natural topology on such a projective limit makes it into a compact group(and not a discrete one)... and compact groups are always amenable.

6Indeed, it is not possible to find an element in (ℓ∞Γ)* ∖ ℓ1Γ without the axiom of choice. Of course, suchan element exists only if Γ is infinite.


Theorem 2.1

Assume Γ is countable (and discrete). The following are equivalent:

(i) There exists a Følner sequence, i.e. {𝐹𝑛} finite sets such that lim𝑛→∞

|𝐹𝑛𝛾△𝐹𝑛||𝐹𝑛|

= 0.

[Følner 1955]

(ii) If Γ acts by affine transformations on 𝐾 a (non-empty) compact and convex set ina LCTVS7, then there is a fixed point. [Day 1961]

(iii) There is an invariant mean, i.e. a linear map 𝑚 : ℓ∞Γ → R such that 𝑚(1) = 1,𝑚(𝑓) ≥ 0 whenever 𝑓 ≥ 0 and 𝑚(𝑓 * 𝛿𝛾) = 𝑚(𝑓). [von Neumann 1929]

(iv) For any 𝜇 finitely supported probability measure on Γ, ‖𝑅𝜇‖ℓ2→ℓ2 = 1, i.e. theprobability of return decays subexponentially: 𝜇(2𝑛)(𝑒)1/2𝑛 → 1. [Kesten 1959]

(v) There is a sequence 𝑓𝑛 ∈ ℓ2Γ of finitely supported elements so that ‖𝑓𝑛‖ℓ2 = 1 and,∀𝛾 ∈ Γ, ‖𝑓𝑛 * 𝛿𝛾 − 𝑓𝑛‖ℓ2 → 0. [Dixmier 1950?]

(vi) There is a Reiter sequence, i.e. a sequence 𝜉𝑛 of finitely supported probabilitymeasures such that, ∀𝛾 ∈ Γ, ‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 → 0. [Reiter 1964∼6]

Definition (i) is saying that these groups have sublinear isoperimetry. Strangely (at first), thisis equivalent to a very powerful fixed point property: (ii) (the case Γ = Z is the Kakutani[-Markov] theorem)8. (iii) is the original definition and, as shown in I. Chatterji’s lecture, isan obstruction to paradoxical decompositions ([Hausdorff-]Banach-Tarski paradox). Maps 𝑚as in (ii) are called invariant means. (iv) bridges (surprisingly) these properties with thoseof the random walk.

(v), in the original language, is to say that the right-regular representation (the “natural”action of Γ on ℓ2Γ) has almost fixed vectors9. (vi) is a variant of (v) with 2 replaced by 1;it remains actually true for any 𝑝. The importance of 𝑝 = 1 is obvious for those who likeBanach algebras (make ℓ1Γ an algebra by taking convolution for the product)10.

Funnily, the fact that the Følner condition (i) implies amenability (the existence of anaverage) is already done in Ahlfors11. Følner does not refer to Ahlfors, but uses a differentmethod (and his result is stronger; see section “More equivalences; combinatorial” in §4.Bbelow) which goes through the “Dixmier condition” (to which he also does not refer; seesection “More equivalent conditions; paradoxical”). The Dixmier condition is already fairlypresent in von Neumann’s original paper, which leaves a possibility that Følner was simplyunaware of these works.

One could think that 𝜇(𝑛) or its “lazy version” or 1𝑁

∑𝑁𝑖=0 𝜇

(𝑖) would be good candidatesfor 𝜉𝑛 in (vi). It turns out this is related to bounded harmonic functions, see “Reiter and

7LCTVS = Locally convex topological vector space8The first statement of this I could locate is by Day (1961, corrected in 1964). The statement in its full

generality is in Rickert 1967. Day refers to Kakutani 1938 and Markov 1936.9The official formulation of this is to say that the trivial representation is weakly contained in the left-regular

representation. In fact, all irreducible representations are weakly contained in the regular representation (theHulanicki-Reiter theorem).

10(vi) =⇒ (v) was known to Dieudonné (1960), and (v) =⇒ (vi) is due to Reiter (1964). The full variationsfor all 𝑝 could be due to Stegeman (1965). The equivalence with amenability seems to be only noticed inHulanicki (1966). In any, case I could not find this statement in the 1950 paper of Dixmier...

11see “Zur Theorie der Uberlagerungsflächen”, Acta Math. 65:157–194, 1935.


Liouville” in §5.A below.The proof of these equivalence can be tedious. In fact, it is much more difficult to do this

for the general case of locally compact groups. So let’s start slow by looking at how easy theclosure properties are if one picks the right definition of amenability.

Proof of (iv) =⇒ (a): If 𝐻 is a subgroup of Γ then a measure 𝜇 on 𝐻 is also a measureon Γ (irreducibility is not required!). The random walk will be identical, so there is nothingelse to say. �

Proof of (ii) =⇒ (b): Recall we have a quotient map 𝜋 : 𝐻 → Γ and 𝑁 = ker𝜋.We wish to show 𝐻 is amenable. Take any affine action on a compact convex 𝐾 = ? onwhich 𝐻 acts. Then there is an action by 𝑁 . Since 𝑁 is amenable, the points fixed by 𝑁 ,𝐾𝑁 = {𝑘 ∈ 𝐾 | ∀𝑛 ∈ 𝑁, 𝑘 · 𝑛 = 𝑘}, is a non-empty subset. Because the action is affine, it isstill convex. It’s also clearly closed in 𝐾, hence compact.

There is no way in general to define an action of Γ on 𝐾, but it’s possible on 𝐾𝑁 . Define,for 𝑘 ∈ 𝐾𝑁 , 𝑘 · 𝛾 := 𝑘 · ℎ where ℎ is some element in 𝜋−1(𝛾). This is well-defined sinceℎ = 𝑛ℎ𝑖 for a fixed ℎ𝑖 representative of the coset and some 𝑛: the choice of 𝑛 is irrelevantsince 𝐾𝑁 is fixed by 𝑁 .

Thus 𝐾𝑁 has a non-empty fixed point set by 𝐻, (𝐾𝑁)Γ. This is easily seen equal to𝐾𝐻 . �

Proof of (iii) =⇒ (c): Let 𝜋 : Γ → 𝐻 (and 𝑁 = ker𝜋). Let 𝑚Γ be the invariant meanon Γ.12 Given a 𝑓 ∈ ℓ∞𝐻 we wish to compute an invariant mean. Define 𝜋*𝑓 ∈ ℓ∞Γ by𝜋*𝑓(𝛾) = 𝑓

(𝜋(𝛾)

). Put 𝑚𝐻(𝑓) := 𝑚Γ(𝜋*𝑓). To see this defines an invariant mean on 𝐻,

observe that 𝜋*𝑓 ∈ (ℓ∞Γ)𝑁 . As in the previous proof, 𝜋*(𝑓 *𝛿ℎ) = 𝜋*𝑓 *𝛿𝜋−1(ℎ) is well-defined,since it does not depend on the actual element chosen in 𝜋−1(ℎ). Hence, 𝑚𝐻 is an invariantmean on 𝐻. �

Proof of (i) =⇒ (d): For simplicity, restrict to direct unions, 𝐻 = ⊕𝑖Γ𝑖. For directlimits, you can either modify this argument to direct limits or try a different approach in theexercises.

Let 𝐹 (𝑖)𝑛 be Følner sequences for Γ𝑖. Try the following sequence for 𝐻: 𝐹𝑛 = ⊕𝑛

𝑖=0𝐹(𝑖)𝑛 .

Given 𝛾 ∈ 𝐻, let 𝑘 be the index starting from which it is trivial, i.e. 𝛾𝑖 = 𝑒Γ𝑖 for 𝑖 > 𝑘.Leaving brutally many terms out of the inclusion-exclusion principle,


|𝐹𝑛|≤

𝑘∑𝑖=1

|𝐹 (𝑖)𝑛 𝛾𝑖△𝐹 (𝑖)

𝑛 |

|𝐹(𝑖)𝑛 |

→ 0 �

2.1 Some details on (iv)

For the convenience of the reader (which might have not benefited from L. Saloff-Coste’slectures), it might be good to recall why the two conditions in (iv) are equal. A very simple(but important) remark to begin with is that 𝜇(𝑛)(𝑒)1/𝑛 need not converge. Indeed, if Γ admitsa Cayley graph which is bipartite and 𝜇 is supported on the generators for this Cayley graph,then 𝜇(𝑛)(𝑒) is alternatively 0 and > 0. The basic example is the infinite line (the classicalCayley graph for Z). In this example, one may compute hands on that 𝜇(2𝑘)(𝑒)1/2𝑘 tends to1 while the limit on odd numbers is clearly 0.

12The proof does not require that 𝑁 be amenable, although this is automatic thanks to (a).


Note that, for any 𝑓, 𝑔 ∈ ℓ2Γ,

⟨𝛿𝑥 * 𝑓 | 𝑔⟩ =∑𝛾∈Γ

𝑓(𝑥−1𝛾)𝑔(𝛾)𝜂:=𝑥−1𝛾=

∑𝜂∈Γ

𝑓(𝜂)𝑔(𝑥𝜂) = ⟨𝑓 | 𝛿𝑥−1 * 𝑔⟩

Similarly, ⟨𝑓 * 𝛿𝑥 | 𝑔⟩ = ⟨𝑓 | 𝑔 * 𝛿𝑥−1⟩. This implies that for a symmetric 𝜇,

(SA) ⟨𝜇 * 𝑓 | 𝑔⟩ = ⟨𝑓 | 𝜇 * 𝑔⟩ and ⟨𝑓 * 𝜇 | 𝑔⟩ = ⟨𝑓 | 𝑔 * 𝜇⟩.

The second equality is exactly saying that 𝑅𝜇 is self-adjoint. Let us explore some easyconsequences. Recall 𝑅𝜇 is the operator defined by 𝑅𝜇𝑓 = 𝑓 * 𝜇.

Lemma 2.2

For any 𝜇 ∈ ℓ1Γ symmetric, one has

· 𝜇(2𝑛)(𝑒) = ‖𝜇(𝑛)‖2ℓ2 = ‖𝑅𝑛𝜇𝛿𝑥‖

2ℓ2;

· sup𝑥∈Γ

𝜇(2𝑛)(𝑥) = 𝜇(2𝑛)(𝑒);

· and 𝑛 ↦→ 𝜇(2𝑛)(𝑒) is non-increasing.

Proof: For the first one, just note that

𝜇(2𝑛)(𝑒) = ⟨𝜇(2𝑛) | 𝛿𝑒⟩ = ⟨𝜇(𝑛) | 𝜇(𝑛)⟩ = ‖𝜇(𝑛)‖2ℓ2.

Then remark that

‖𝑅𝑛𝜇𝛿𝑥‖

2ℓ2 = ⟨𝛿𝑥 * 𝜇

(𝑛) | 𝛿𝑥 * 𝜇(𝑛)⟩ = ⟨𝛿𝑥−1 * 𝛿𝑥 * 𝜇

(𝑛) | 𝜇(𝑛)⟩ = ⟨𝜇(𝑛) | 𝜇(𝑛)⟩.

The second comes from a slight variation:

𝜇(2𝑛)(𝑥) = ⟨𝜇(2𝑛) | 𝛿𝑥⟩ = ⟨𝜇(𝑛) | 𝛿𝑥 * 𝜇(𝑛)⟩

𝐶−𝑆≤ ‖𝜇(𝑛)‖ℓ2‖𝛿𝑥 * 𝜇

(𝑛)‖ℓ2 = ‖𝜇(𝑛)‖2ℓ2 = 𝜇(2𝑛)(𝑒).

The last one can be obtained as a corollary:

𝜇(2𝑛+2)(𝑒) =∑𝑥

𝜇(2𝑛)(𝑥)𝜇(2)(𝑥−1) ≤ 𝜇(2𝑛)(𝑒)∑𝑥

𝜇(2)(𝑥−1) = 𝜇(2𝑛)(𝑒). �

Lemma 2.3

For any 𝜇 ∈ ℓ1Γ symmetric, the operator norm of 𝑅𝜇 : ℓ2 → ℓ2 is ≤ 1.

Proof: Indeed, by invoking Young’s inequality13 : ‖𝑅𝜇𝑓‖ℓ2 = ‖𝑓 *𝜇‖ℓ2 ≤ ‖𝑓‖ℓ2‖𝜇‖ℓ1 . If youdon’t like Young’s inequality, a direct computation (where C-S stands for the Cauchy-Schwarzinequality) also works. First, note that

|𝑅𝜇𝑓(𝑥)|2 = |

∑𝑠

𝑓(𝑥𝑠)𝜇(𝑠)|2𝐶−𝑆≤

(∑𝑠

|𝑓(𝑥𝑠)|2𝜇(𝑠))(∑

𝑠

𝜇(𝑠))=∑𝑠

|𝑓(𝑥𝑠)|2𝜇(𝑠).

Then by rearranging:

‖𝑅𝜇𝑓‖2ℓ2 ≤

∑𝑥

∑𝑠

|𝑓(𝑥𝑠)|2𝜇(𝑠) =∑𝑦

(|𝑓(𝑦)|2

∑𝑡

𝜇(𝑡−1))=∑𝑦

|𝑓(𝑦)|2 = ‖𝑓‖2ℓ2. �

13In fact, this shows that 𝑅𝜇 has norm ≤ 1 as an operator from ℓ𝑝 to ℓ𝑝. To avoid this inequality for 𝑝 = 2,use Hölder instead of Cauchy-Schwarz.


Another easy consequence of (SA) is that, for any 𝑓 ∈ ℓ2Γ, the map 𝑛 ↦→‖𝑅𝑛+1

𝜇 𝑓‖ℓ2‖𝑅𝑛

𝜇𝑓‖ℓ2is

increasing:

‖𝑅𝑛𝜇𝑓‖

2ℓ2 = ⟨𝑓 * 𝜇(𝑛) | 𝑓 * 𝜇(𝑛)⟩ = ⟨𝑓 * 𝜇(𝑛−1) | 𝑓 * 𝜇(𝑛+1)⟩

𝐶−𝑆≤ ‖𝑅𝑛+1

𝜇 𝑓‖ℓ2‖𝑅𝑛−1𝜇 𝑓‖ℓ2

Hence the limit 𝜌𝑓 := lim𝑛→∞

‖𝑅𝑛+1𝜇 𝑓‖ℓ2

‖𝑅𝑛𝜇𝑓‖ℓ2

exists (these ratios form an increasing sequence bounded

by the operator norm of 𝑅𝜇, which is = 1).A bound on all the 𝜌𝑓 would be a bound on the operator norm: as the ratio is increasing,

(R) ‖𝑅𝜇𝑓‖ℓ2 ≤ ‖𝑓‖ℓ2‖𝑅2

𝜇𝑓‖ℓ2

‖𝑅1𝜇𝑓‖ℓ2

≤ ‖𝑓‖ℓ2‖𝑅3

𝜇𝑓‖ℓ2

‖𝑅2𝜇𝑓‖ℓ2

≤ . . . ≤ ‖𝑓‖ℓ2‖𝑅𝑛+1

𝜇 𝑓‖ℓ2

‖𝑅𝑛𝜇𝑓‖ℓ2

≤ ‖𝑓‖ℓ2𝜌𝑓 .

Remember that if a ratio test converge, then the root test also converges to the same value(but it is possible that the root test converges and not the ratio test). Hence,

𝜌𝑓 = lim𝑛→∞ ‖𝑅𝑛

𝜇𝑓‖1/𝑛ℓ2 .

Note that the ratio is easy to compute for any Dirac mass:

‖𝑅𝑛+1𝜇 𝛿𝑥‖ℓ2

‖𝑅𝑛𝜇𝛿𝑥‖ℓ2

=‖𝜇(𝑛+1)‖ℓ2

‖𝜇(𝑛)‖ℓ2=

⎯⎸⎸⎷𝜇(2𝑛+2)(𝑒)

𝜇(2𝑛)(𝑒).

Here is a corollary:

Lemma 2.4

The function 𝑛 ↦→𝜇(2𝑛+2)(𝑒)

𝜇(2𝑛)(𝑒)is increasing; consequently, so is 𝑛 ↦→ 𝜇(2𝑛)(𝑒)1/2𝑛.

Turning14 back to our problem, 𝜌𝛿𝑥 is the same as 𝜇(2𝑛)(𝑒)1/2𝑛.For 𝑓 =

∑𝑎𝛾𝛿𝛾 of finite support, note that

‖𝑅𝑛𝜇𝑓‖

1/𝑛ℓ2

𝑇.𝐼.≤(∑

|𝑎𝛾| · ‖𝑅𝑛𝜇𝛿𝛾‖ℓ2

)1/𝑛

= ‖𝑓‖1/𝑛ℓ1 𝜇(2𝑛)(𝑒)1/2𝑛,

where “T.I.” stands for the triangle inequality. So that 𝜌𝑓 ≤ lim𝑛→∞ 𝜇(2𝑛)(𝑒)1/2𝑛. From this, one

actually sees

Lemma 2.5

‖𝑅𝑛𝜇‖ℓ1→ℓ2 = 𝜇(2𝑛)(𝑒)1/2.

Proof: The inequality ≤ was just proved (and extends to ℓ1 by density of finitely supportedfunctions) and the inequality ≥ is obtained by looking at a Dirac mass (or in fact, any𝑓 =

∑𝑎𝛾𝛿𝛾 with 𝑎𝛾 ≥ 0). �

Lemma 2.6

‖𝑅𝜇‖ℓ2→ℓ2 = lim𝑛→∞ 𝜇(2𝑛)(𝑒)1/2𝑛.

14If you want to see directly in this case that the root test converges, note that 𝜇(𝑛+𝑚)(𝑒) ≥ 𝜇(𝑛)(𝑒)𝜇(𝑚)(𝑒).This means that 𝐹 (𝑛) = − ln𝜇(2𝑛)(𝑒) is a sub-additive function. Invoke Fekete’s lemma to conclude.


Proof: Let 𝜌 = lim𝑛→∞ 𝜇(2𝑛)(𝑒)1/2𝑛.

Until now we have shown that, for any function of finite support, one has 𝜌𝑓 ≤ 𝜌. Using(R), one also gets that for such functions ‖𝑅𝜇𝑓‖ℓ2 ≤ 𝜌‖𝑓‖ℓ2. The result then follows by densityof finitely supported functions. Let us nevertheless put it in full details as approximatingelements of the spectrum by finitely supported functions will come up again later.

Suppose there is a 𝑓 ∈ ℓ2Γ with ‖𝑅𝜇𝑓‖ℓ2 = 𝜌‖𝑓‖ℓ2 for some 𝜌 > 0 and some 𝜖 > 0.WLOG, ‖𝑓‖ℓ2 = 1. Then, for any 𝜖 ∈]0, 1[, one may15 write 𝑓 = ℎ + 𝑔 where 𝑔 and ℎ havedisjoint support, ℎ is finitely supported, ‖𝑔‖ℓ2 ≤ 𝜖 and 1− 𝜖 ≤ ‖ℎ‖ℓ2 ≤ 1.

On one hand,

‖𝑅𝜇𝑓‖ℓ2 = ‖𝑅𝜇(ℎ+ 𝑔)‖ℓ2𝑇.𝐼.≤ ‖𝑅𝜇ℎ‖ℓ2 + ‖𝑅𝜇𝑔‖ℓ2

‖𝑅𝜇‖=1

≤ ‖𝑅𝜇ℎ‖ℓ2 + ‖𝑔‖ℓ2 ≤ ‖𝑅𝜇ℎ‖ℓ2 + 𝜖

while, on the other,

‖𝑅𝜇𝑓‖ℓ2 = 𝜌‖𝑓‖ℓ2 = 𝜌√‖ℎ‖2ℓ2 + ‖𝑔‖2ℓ2 ≥ 𝜌‖ℎ‖ℓ2

Putting this together one has

‖𝑅𝜇ℎ‖ℓ2 ≥ 𝜌‖ℎ‖ℓ2 − 𝜖 ≥ (𝜌−𝜖

1− 𝜖)‖ℎ‖ℓ2.

Since one has ‖𝑅𝜇ℎ‖ℓ2 ≤ 𝜌‖ℎ‖ℓ2, for any finitely supported ℎ, one gets that

𝜌−𝜖

1− 𝜖≤ 𝜌.

Since 𝜖 is arbitrarily small , this shows ‖𝑅𝜇𝑓‖ℓ2 ≤ 𝜌‖𝑓‖ℓ2.

For the reverse inequality, simply recall that‖𝑅𝜇𝜇(𝑛)‖ℓ2‖𝜇(𝑛)‖ℓ2

tends to this limit. �

Remark 2.7. The above applies more generally to bounded self-adjoint operators 𝑅 on

ℓ2N. Namely, the operator norm of 𝑅 is equal to sup𝑥∈N

lim𝑛→∞ ‖𝑅𝑛𝛿𝑥‖

1/𝑛ℓ2 . ♠

3 Proof of equivalences

Most of these proofs are significantly more technical in the general locally compact case. Hereis the pattern that will be followed here:

(iv) ⇐ (i) ⇒ (ii)⇓ ⇑ ⇓(v) ⇒ (vi) ⇐ (iii)

Proof of (i) =⇒ (ii): This proof is very intuitive. We are given a Følner sequence 𝐹𝑛 forΓ and an action Γ 𝐾. The aim is to find a fixed point in 𝐾.

Pick any 𝑣 ∈ 𝐾 and define 𝑣𝑛 = 1|𝐹𝑛|

∑𝑥∈𝐹𝑛 𝑣 · 𝑥. This sequence is infinite so it has

accumulation points in 𝐾 (by compactness). It remains to check these accumulation pointsare fixed by Γ. Notice that |𝐹𝑛𝛾 ∖ 𝐹𝑛| = |𝐹𝑛 ∖ 𝐹𝑛𝛾| =

12|𝐹𝑛𝛾△𝐹𝑛|. Compute

𝑣𝑛 · 𝛾 − 𝑣𝑛 =1

|𝐹𝑛|

( ∑𝑥∈𝐹𝑛𝛾∖𝐹𝑛

𝑣 · 𝑥−∑

𝑥∈𝐹𝑛∖𝐹𝑛𝛾

𝑣 · 𝑥

)

=|𝐹𝑛𝛾△𝐹𝑛|

2|𝐹𝑛|

(1

|𝐹𝑛𝛾 ∖ 𝐹𝑛|

∑𝑥∈𝐹𝑛𝛾∖𝐹𝑛

𝑣 · 𝑥−1

|𝐹𝑛 ∖ 𝐹𝑛𝛾|

∑𝑥∈𝐹𝑛∖𝐹𝑛𝛾

𝑣 · 𝑥

)

15𝑔 is just 𝑓 restricted to the complement of some large enough finite set.


This seems nasty but notice that the two sums in parenthesis are actually convex combina-tions. Since 𝐾 is convex,

𝑣𝑛 · 𝛾 − 𝑣𝑛 ∈|𝐹𝑛𝛾△𝐹𝑛|

2|𝐹𝑛|(𝐾 −𝐾).

Since 𝐾 is compact, it’s quite intuitive this tends to 0. To see it, here are the required factsabout LCTVS: 1- the topology is given by a family of semi-norms; 2- a sequence tends to 0if and only if it tends in to 0 in any semi-norm compatible with the topology. Since 𝐾 iscompact, it is bounded in any semi-norm. Thus 𝑣𝑛 · 𝛾 − 𝑣𝑛 tends to 0 in any semi-norm (it’sa scalar tending to 0 times a bounded quantity).

This insures invariance of accumulation points and shows the existence of a fixed pointin 𝐾. �

Before moving on, some background on means is required.

Definition 3.1. A mean is a linear map 𝜇 : ℓ∞Γ → R such that 𝜇(1) = 1 and 𝜇(𝑓) ≥ 0whenever 𝑓 ≥ 0. F

Denote by 𝑀 the set of means. Let Pr𝑐(Γ) be the set of finitely supported probabilitymeasures on Γ. Then Pr𝑐(Γ) ⊂𝑀 . Here is a small lemma which will be useful

Lemma 3.2

Means are elements of (ℓ∞Γ)* of norm 1. Hence 𝑀 is a convex set which is bounded innorm (hence compact in the weak* topology). Furthermore, Pr𝑐(Γ) is weak

* dense in 𝑀 .

Proof of the lemma: To show a mean belongs to (ℓ∞Γ)* it suffices to show it is bounded.But, ∀𝑓 ∈ ℓ∞Γ,

−‖𝑓‖ℓ∞1 ≤ 𝑓 ≤ ‖𝑓‖ℓ∞1.

Using positivity and linearity, 𝑚(‖𝑓‖ℓ∞1+𝑓) = ‖𝑓‖ℓ∞+𝑚(𝑓) ≥ 0. Similarly, 𝑚(𝑓)−‖𝑓‖ℓ∞ ≥0. Thus |𝑚(𝑓)| ≤ ‖𝑓‖ℓ∞ . This implies ‖𝑚‖ℓ∞→R ≤ 1. Since 𝑚(1) = 1, we actually haveequality.

For convexity, it’s quite easy to check that the conditions are stable under convex combi-nations.

Assume there is a 𝑚 ∈ 𝑀 ∖Pr𝑐(Γ)w*. By Hahn-Banach, this means 𝑚 can be separated

from Pr𝑐(Γ) by a careful evaluation at some 𝑓 : ∃𝑓 ∈ ℓ∞Γ such that 𝑚(𝑓) = 0 but 𝜇(𝑓) = 0for all 𝜇 ∈Pr𝑐(Γ). But if𝑚(𝑓) = 0 then there must be a 𝑥 ∈ Γ for which 𝑓(𝑥) = 0 (otherwise‖𝑓‖ℓ∞ = 0, and by continuity 𝑚(𝑓) = 0). Pick 𝛿𝑥 ∈Pr𝑐(Γ), then 𝛿𝑥(𝑓) = 0. This shows thatPr𝑐(Γ) is weak

* dense in 𝑀 . �

Proof of (ii) =⇒ (iii): By the lemma 𝑀 is a convex set which is compact in the weak*

topology. It sits in (ℓ∞Γ)* a LCTVS. There is a natural action of Γ on 𝑀 : (𝑚 · 𝛾)(𝑓) =𝑚(𝑓 * 𝛿𝛾−1).

16 This action is the action dual to the action on ℓ∞Γ.To conclude, observe that a fixed point by this action is exactly an invariant mean. Since

the action is affine17, the conclusion follows. �

16This is just a “change of variable” in the integration. If 𝜇 ∈Pr𝑐(Γ), 𝜇(𝑓) =∫𝑓(𝑥)d𝜇(𝑥). Then act as usual

on Pr𝑐(Γ) ⊂ ℓ1Γ by 𝜇 · 𝛾 = 𝜇 * 𝛿𝛾 . A change of variable gives (𝜇 * 𝛿𝛾)(𝑓) = 𝜇(𝑓 * 𝛿𝛾−1).17One does not need to write the action down for this. Indeed the Mazur-Ulam theorem insures an isometric

action is affine.


Proof of (iii) =⇒ (vi): We are given an invariant mean 𝑚. By the lemma, Pr𝑐(Γ) is

weak* dense in 𝑀 . This means that there is a sequence 𝜇𝑖 ∈ Pr𝑐(Γ) with 𝜇𝑖𝑤*−→ 𝑚. It

remains to show they are Reiter. For any 𝑓 ∈ ℓ∞Γ and any 𝛾,

𝜇𝑖 * 𝛿𝛾−1(𝑓)− 𝜇𝑖(𝑓) = 𝜇𝑖(𝑓 * 𝛿𝛾)− 𝜇𝑖(𝑓)→ 𝑚(𝑓 * 𝛿𝛾)−𝑚(𝑓) = 0.

So at least 𝜇𝑖 * 𝛿𝛾−1 − 𝜇𝑖𝑤*−→ 0.

To improve this to a norm convergence, fix some finite set 𝐹 ⊂ Γ and consider 𝑋 =⊕𝑔∈𝐹 ℓ

1Γ with the norm ‖(𝜉𝑔)‖𝑋 = max𝑔∈𝐹

‖𝜉𝑔‖ℓ1. The set

𝐾 = {⊕𝑔∈𝐹 (𝜉 * 𝛿𝑔 − 𝜉) | 𝜉 ∈Pr𝑐(Γ)}

is (again) a convex set. By the previous argument, it contains 0 in its weak* closure. In termsof separation by evaluation (Hahn-Banach), we actually get 0 in the weak convergence (sincewe are looking at elements in ℓ1 and 0 ∈ ℓ1). A wonderful corollary of Hahn-Banach18 comesto the rescue: the closure of a convex sets in the weak topology is the same as the closure inthe norm topology19. This yields a sequence 𝜉𝑖,𝐹 so that max

𝑔∈𝐹‖𝜉𝑖,𝐹 * 𝛿𝑔 − 𝜉𝑖,𝐹‖ℓ1 → 0. Taking

a well-chosen subsequence 𝜉𝑖𝑘,𝐹𝑘 will give the desired Reiter sequence. �

Proof of (vi) =⇒ (i): The idea is to find Følner sets in the level sets of 𝜉𝑛. Define for𝜇 ∈Pr𝑐(Γ), 𝐹 (𝜇, 𝑡) = {𝑥 ∈ Γ | 𝜇(𝑥) > 𝑡}. Then for a fixed 𝑥

|𝜉(𝑥)− 𝜇(𝑥)| =∫ 1

0|1𝐹 (𝜉,𝑡)(𝑥)− 1𝐹 (𝜇,𝑡)(𝑥)|d𝑡.

Summing over 𝑥 one gets,

‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 =∫ 1

0|𝐹 (𝜉𝑛 * 𝛿𝛾, 𝑡)△𝐹 (𝜉𝑛, 𝑡)|d𝑡.

Assume, there is no way to find a Følner sequence in 𝐹 (𝜉𝑛, 𝑡) (as 𝑡 and 𝑛 vary). This impliesthat there is some 𝜖 > 0 such that |𝐹 (𝜉𝑛 * 𝛿𝛾, 𝑡)△𝐹 (𝜉𝑛, 𝑡)| ≥ 𝜖|𝐹 (𝜉𝑛, 𝑡)|. But then

‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 ≥ 𝜖∫ 1

0|𝐹 (𝜉𝑛, 𝑡)|d𝑡 = 𝜖‖𝜉𝑛‖ℓ1 = 𝜖.

This contradicts the fact that 𝜉𝑛 is a Reiter sequence. �

In fact, looking closer at the previous argument shows that “most” level sets are goodin the following sense. Let 𝑣1 > 𝑣2 > . . . > 𝑣𝑘 > 𝑣𝑘+1 = 0 the values taken by 𝜉𝑛 and𝐹𝑖 = {𝑥 ∈ Γ | 𝜉𝑛(𝑥) ≥ 𝑣𝑖}. Assume ‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 ≤ 𝜖, then

𝜖 ≥ ‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 =𝑛∑𝑖=1

𝛽𝑖|𝐹𝑟𝛾△𝐹𝑟|

|𝐹𝑟|

where 𝛽𝑖 = (𝑣𝑖 − 𝑣𝑖+1)|𝐹𝑖| (so that∑𝛽𝑖 = ‖𝜉𝑛‖ℓ1). Hence 𝛽 can be seen as a measure on the

indices. Then 𝛽({𝑖 | |𝐹𝑖𝛾△𝐹𝑖| ≥ 𝐶𝜖|𝐹𝑖|}) ≤ 1/𝐶. One can also define 𝜇 to be the measureon [0, 𝑣1] which has uniform density |𝐹𝑖| on each interval [𝑣𝑖+1, 𝑣𝑖].

𝜇({𝑡 ∈ [0, 𝑣1] | |𝐹 (𝜉𝑛 * 𝛿𝛾, 𝑡)△𝐹 (𝜉𝑛, 𝑡)| ≥ 𝐶𝜖|𝐹 (𝜉𝑛, 𝑡)|}) ≤ 1/𝐶

18This consequence is sometimes referred to Mazur’s theorem or the Mazur trick.19Actually, in ℓ1 weak and norm convergence coincide, see Lemma 5.1. But this argument passes more easily

to non-discrete groups.


Proof of (i) =⇒ (iv): Take 𝑓𝑛 = 1𝐹𝑛/√|𝐹𝑛| (the ℓ

2-normalised characteristic function

of 𝐹𝑛). Suppose the measure 𝜇 is supported on 𝑈 ⊂ Γ (finite). Then a direct computationgives

‖𝑅𝜇𝑓𝑛‖2 ≥|𝜕−𝑈𝐹𝑛|

|𝐹𝑛|→ 1,

where 𝜕−𝑈𝐹𝑛 = {𝛾 ∈ 𝐹𝑛 | 𝛾𝑈 ⊂ 𝐹𝑛}. (To see why this should converge to 1, prove lemma1.3!) By definition, of the operator norm, one has ‖𝑅𝜇‖ = 1. �

Note that one can also prove (i) ⇒ (v) and (i) ⇒ (vi) by the same argument. In fact, onecan get elements of ℓ𝑝Γ which are almost invariant in ℓ𝑝-norm by taking the ℓ𝑝-normalisedcharacteristic functions of the 𝐹𝑛.

Proof of (iv) =⇒ (v): Recall that 𝑅𝜇 is contracting, i.e. its operator norm is ≤ 1 (see§2.1). Next, 𝑅𝜇 has no eigenvector of eigenvalue 1, i.e. no 𝑓 ∈ ℓ2Γ such that 𝑓 * 𝜇 = 𝑓 .Indeed, such a function would be harmonic and, hence, satisfy a maximum principle. Butsince 𝑓 ∈ ℓ2Γ it must decrease to 0 at infinity. This implies 𝑓 must be 0 everywhere.

If you are not familiar with harmonic functions, just think of it as follows. Pick someelement 𝑥. The equation 𝑓 = 𝑓 * 𝜇 tells you you can express the value at 𝑥 in terms of aconvex combination of values taken on the set 𝑆 = Supp𝜇. Now repeat this for each elementof 𝑆. This might make 𝑓(𝑥) pop-up again but you can cancel it using the first step. Repeatingthis process over and over again, you will have an expression of 𝑓(𝑥) in term of a convexcombination of values of 𝑓 which are as far away as you like. However, these need to tend to0 because 𝑓 ∈ ℓ2Γ. Hence, at any point the value is 0

Repeating the same argument with 12(Id + 𝜇) instead of 𝜇, one sees that 1 is in the

spectrum20 (since ‖𝑅𝜇‖ℓ2→ℓ2 = 1) but there are no such eigenvectors. This means there arealmost eigenvectors: for any 𝜖 > 0 there is a 𝑓 of norm 1 which decomposition 21 is made ofeigenvectors of eigenvalue ≥ 1 − 𝜖, i.e. ⟨𝑅𝜇𝑓𝑛 | 𝑓𝑛⟩ ≥ 1 − 𝜖. One may also assume this 𝑓 isfinitely supported by cutting off very far away.

Thus ∑𝑥

(1− ⟨𝑓 * 𝛿𝑥 | 𝑓⟩)𝜇(𝑥) = 1− ⟨𝑅𝜇𝑓 | 𝑓⟩ ≤ 𝜖

But for any 𝑥, ‖𝑓 * 𝛿𝑥‖ℓ2 = ‖𝑓‖ℓ2 = 1 so (1− ⟨𝑓 * 𝛿𝑥 | 𝑓⟩) ≥ 0. This implies that, for any 𝑥 inthe support of 𝜇, 1− ⟨𝑓 * 𝛿𝑥 | 𝑓⟩ ≤ 𝜖/inf

𝑠∈𝑆𝜇(𝑠).

Recall that ‖𝑎− 𝑏‖2ℓ2 = ⟨𝑎− 𝑏 | 𝑎− 𝑏⟩ = 2− 2⟨𝑎 | 𝑏⟩. Hence, for all 𝑥 in the support of 𝜇,12‖𝑓 * 𝛿𝑥 − 𝑓‖2ℓ2 = 1− ⟨𝑓 * 𝛿𝑥 | 𝑓⟩ ≤ 𝜖/inf

𝑠∈𝑆𝜇(𝑠)

To conclude, let 𝐹𝑘 be an increasing sequence of finite sets so that ∪𝐹𝑘 = 𝐺. Define 𝑓𝑘 bylooking at 𝜇𝑘 with uniform support on 𝐹𝑘 and then take 𝜖 = 1/|𝐹𝑘|𝑘. This gives a sequence𝑓𝑘 which satisfies ‖𝑓 * 𝛿𝑥 − 𝑓‖2ℓ2 ≤ 2/𝑘 for any 𝑥 ∈ 𝐹𝑘 and concludes the proof.

�

Proof of (v) =⇒ (vi): Take 𝜉𝑛 = 𝑓2𝑛. Then

‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 =∑𝑥∈Γ

|𝑓𝑛(𝑥𝛾)2 − 𝑓𝑛(𝑥)

2|

=∑𝑥∈Γ

|𝑓𝑛(𝑥𝛾)− 𝑓𝑛(𝑥)| · |𝑓𝑛(𝑥𝛾) + 𝑓𝑛(𝑥)|

≤∑𝑥∈Γ

|𝑓𝑛(𝑥𝛾)− 𝑓𝑛(𝑥)| · (|𝑓𝑛(𝑥𝛾)|+ |𝑓𝑛(𝑥)|).

20With more efforts, one can show contractions on ℓ2 always have vectors which are “almost” of eigenvalue 1.21This can, of course, be said more rigorously with spectral decomposition of self-adjoint operators.


Split in two sums and use Cauchy-Schwarz to get

‖𝜉𝑛 * 𝛿𝛾 − 𝜉𝑛‖ℓ1 ≤ ‖𝑓𝑛 * 𝛿𝛾 − 𝑓𝑛‖ℓ2(‖𝑓𝑛‖+ ‖𝑓𝑛 * 𝛿𝛾‖ℓ2) = 2‖𝑓𝑛 * 𝛿𝛾 − 𝑓𝑛‖ℓ2.

This tends to 0 as 𝑛→∞. �

4 More...

This section gathers up many things which would have been nice to say if I would have hadthe time and energy to do it.

4.A ... examples

Here is a classical example:

Example 4.1 (Lamplighter group). Z and Z2 are amenable. So is 𝐿 = ⊕𝑔∈ZZ2. Elements of𝐿 can be seen as finitely supported functions from Z to Z2. Consequently there is a “natural”action by Z on these functions by shifting. This turns out to be an automorphism of 𝐿. LetΓ = 𝐿o Z be the semi-direct product obtained from this action:

({𝑎𝑔}𝑔∈Z, 𝑘) · ({𝑏𝑔}𝑔∈Z, ℓ) = ({𝑎𝑔 + 𝑏𝑔+𝑘}𝑔∈Z, 𝑘 + ℓ)

It’s easier to describe this in words. The Z coordinate describes where the lamplighter is.The first coordinate describes which lamp are on or off. If one multiplies (on the right) by1 ∈ Z, the lamplighter moves. If one multiplies by 𝛿0 (the function taking value 0 ∈ Z2

everywhere but at 0 ∈ Z where it takes value 1 ∈ Z2), then the lamp at the current positionof the lamplighter is turned on or off.

Through this description, one sees that the group is actually finitely generated, eventhough it is made of an infinitely generated one (and a finitely generated one). It’s not toohard to check this group has exponential growth as well. A Følner sequence can be explicitlygiven by taking elements (𝜉, 𝑘) so that 𝜉 (seen as a function from Z → Z2) is supported on[1, 𝑛] and 𝑘 belongs to [1, 𝑛]. ♣

It’s an exercise to show these groups are of exponential growth.If one starts from finitely generated Abelian groups and uses only extensions, the class

obtained is called polycyclic groups. These groups are much more nicely behaved than a“generic” solvable group.

There are many other groups of this type, if one changes Z2 by some group 𝐻 and Z by 𝐺,the notation for the resulting group is 𝐻 ≀𝐺 (this notation is not unanimous, it’s sometimes𝐺 ≀𝐻). The following nice example was mentioned to me by Maxime Gheysens and AdrienLe Boudec. It’s an EA group which is not virtually solvable and not linear.

Example 4.2. Take 𝐴5 (the alternating group on 5 elements). Let Γ = 𝐴5 ≀Z. Γ is EA butit is not linear (i.e. is not isomorphic to a subgroup of some matrix group). To see this theTits alternative is required: a linear group either contains a free group or is virtually solvable(i.e. has a solvable subgroup of finite index).

Assume Γ is linear. Since it is amenable it cannot contain a free group. Thus it should bevirtually solvable. It takes some playing around to see that subgroups of Γ are either finite,isomorphic to Z or isomorphic to 𝐵 ≀Z for some subgroup 𝐵 of 𝐴5. In all cases, there are not


of finite index (unless they are isomorphic to Γ). So Γ should be solvable. But a computationof its commutator subgroup leaves ⊕Z𝐴5. This group is stable under further commutators,so Γ is not solvable.

The group is however residually finite. ♣

Another interesting example is the following (it is neither linear, residually finite norsolvable).

Example 4.3. Take the group of all finitely supported permutation of the integers, 𝑆∞. Itis a direct limit of 𝑆2𝑛+1 (all permutations supported on [−𝑛, 𝑛]), hence amenable (but notfinitely generated. To make it into a finitely generated group, add the shift: the infinite cyclicpermutation which sends 𝑘 to 𝑘 + 1. It fairly easy to see the resulting group is an extensionof 𝑆∞ by Z: 1→ 𝑆∞ → Γ→ Z→ 1. [In fact, it’s a semi-direct product.] So Γ is EA.

Γ is finitely generated: take one transposition (12), and, upon conjugation by the shift,one may get any other transposition. [It’s easy to believe that Γ is not finitely presented.]It’s also fairly easy to check that Γ is neither virtually-solvable nor residually finite. Thesame argument as in the previous example shows it is not linear. ♣

4.B ... equivalences; combinatorial

Here are a few less-known equivalent conditions to amenability. The third is (surprisingly)due to Følner (in the same 1955) paper.

Theorem 4.4

The following are equivalent:

(i) There exists a sequence {𝐹𝑛} of finite sets such that lim𝑛→∞


= 0.[Følner 1955]

(i-2) There exists 𝑐 ∈ [0, 2[ and a sequence {𝐹𝑛} such that ∀𝛾 ∈ Γ, lim sup𝑛→∞


≤ 𝑐. -

- - - [Nagnibeda-Juschenko 2013]

(i-3) There exists 𝑐 ∈ [0, 2[ such that ∀𝜇 ∈Pr𝑐(Γ), ∃𝐹 ⊂ 𝐺 finite with∑𝛾𝜇(𝛾) |𝐹𝛾△𝐹 |

|𝐹 |≤ 𝑐.

- - - - [Følner 1955]

(i-4) There exists 𝑐 ∈ [0, 2[ such that 𝑆 ⊂ 𝐺 finite symmetric generating, ∃𝐹 ⊂ 𝐺 finite

with∑𝛾∈𝑆

|𝐹𝛾△𝐹 ||𝑆||𝐹 |

≤ 𝑐. [Thom 2013]

(i-3) has been somehow largely forgotten (which explains the strange chronology). It actuallysuffices to consider 𝜇 with rational values. It’s easy to do (i) =⇒ (i-2) =⇒ (i-3) =⇒ (i-4).However, it seems at first difficult to believe that (i-4) or even (i-2) imply (i). To show (i-3)=⇒ (iii), we will follow Følner’s original argument22. He uses a preliminary result:

22In a modern perspective, the underlying idea in this proof is to show that inf𝜈‖𝜈 − 𝜇𝑖 * 𝜈‖ℓ1 = 0 for any

finite number of 𝜇𝑖 ∈Pr𝑐(Γ). From there, one can use the arguments in (iii) =⇒ (vi).


Lemma 4.5 (Dixmier 195023)

Γ is amenable if and only if sup𝑥

𝐻(𝑥) ≥ 0 for any 𝐻 of the form

(D) 𝐻(𝑥) =𝑛∑𝑖=1

ℎ𝑖(𝑥)− ℎ𝑖(𝑥𝛾𝑖)

where ℎ𝑖 ∈ ℓ∞Γ (for 𝑖 = 1, . . . , 𝑛) and 𝛾𝑖 ∈ Γ

Proof: If there exists a function 𝐻 which is written as in (D) and sup𝐻(𝑥) ≤ 𝑐 for some𝑐 < 0, then there can be no invariant mean: 𝑚(𝐻) = 0 but by positivity 𝑚(𝐻) ≤ 𝑐 < 0.

On the other hand, if 𝐻 satisfies (D) implies sup𝐻(𝑥) ≥ 0, one can define “explicitly” an

invariant mean as follows. Consider ��(𝑓) = inf𝐻

sup𝑥

(𝑓(𝑥) + 𝐻(𝑥)

). Then ��(𝜆𝑓) = 𝜆��(𝑓)

(for 𝜆 ≥ 0) and ��(𝑓 + 𝑔) ≤ ��(𝑓) + ��(𝑔). Furthermore, �� is invariant (��(𝑓) is clearly 0 if𝑓 = 𝑔 − 𝑔 * 𝛿𝛾). Thus, by Hahn-Banach’s extension, there exists a linear functional (whichwill still be invariant) such that 𝑚(𝑓) ≤ ��(𝑓). �

The function �� is called upper mean value. The idea stems from a simpler upper meandefined (on Abelian groups) via ��(𝑓) = inf

𝜇∈Pr𝑐(Γ)sup𝑥

𝑓 * 𝜇(𝑥). It is worth mentioning that

Banach proved the Hahn-Banach extension theorem to show that �� can be turned into alinear map 𝑚 (i.e. he proved Abelian groups are amenable). There are recent work (byCannon, Floyd and Parry) in which functions of the form 𝐻 are reinterpreted as coolingfunctions.

Proof of (i-3) =⇒ (iii): Introduce

( D𝜖) ∃𝐻 such that 𝐻 =𝑛∑𝑖=1

ℎ𝑖 − ℎ𝑖 * 𝛿𝛾𝑖, 2𝑛max𝑖‖ℎ𝑖‖ℓ∞ ≤ 1 and sup

𝑥𝐻(𝑥) ≤ −𝜖.

and

(I𝑘) ∀𝜇, ∃𝐹 such that∑𝛾∈Γ

𝜇(𝛾)|𝐹𝛾 ∩ 𝐹 |

|𝐹 |≥ 𝑘.

The proof by goes contradiction: ( D𝜖) holds (for some 𝜖 > 0) even though (I𝑘) is true. So,take 𝐻0 as in (D𝜖). Let 𝐹 associated to the measure 𝜇 = 1

𝑛

∑𝑛𝑖=1 𝛿𝛾𝑖. Define 𝐻1 as follows:

𝐻1 =1|𝐹 |

𝑛∑𝑖=1

ℎ𝑖 * 1𝐹 − ℎ𝑖 * 1𝐹 * 𝛿𝛾𝑖.

This sum is apparently a sum on 𝑛|𝐹 | terms of functions with ‖ℎ‖ℓ∞ ≤ 1/|𝐹 |, and sup𝑥𝐻1(𝑥) ≤

−𝜖. However, by the choice of 𝐹 , many terms in the sum cancel out: there are at most(1 − 𝑘)𝑛|𝐹 | in the sum. This means, ‖𝐻1‖ℓ∞ ≤ (1 − 𝑘). This implies −(1 − 𝑘) ≤ −𝜖, i.e.𝜖 ≤ 1− 𝑘.

But one can iterate this process: define 𝐻2 from 𝐻1 as 𝐻1 was defined from 𝐻0. Thisimplies 𝜖 ≤ (1− 𝑘)2. Constructing a sequence 𝐻𝑚 in a similar fashion implies 𝜖 ≤ (1− 𝑘)𝑚.This contradicts (D𝜖) (for any 𝜖 > 0). By Lemma 4.5, there is an invariant mean. �

23Følner seems to have been unaware of the result of Dixmier and proved this result in a previous paper in1954. Both were very probably inspired by von Neumann 1929 (Hilfsatz 2, p.90 in §1 ¶1). Greenleaf calls itthe “Dixmier condition”.


For the record, here is a proof of (i-2) =⇒ (v). It requires some familiarity with ultra-powers.

Proof of (i-2) =⇒ (v): Let 𝜉𝑛 be the ℓ2-normalised characteristic function of the sets 𝐹𝑛,

i.e. 𝜉𝑛 = 1𝐹𝑛/√|𝐹𝑛|. So by assumption:

⟨𝜆𝛾𝜉𝑖 − 𝜉𝑖 | 𝜆𝛾𝜉𝑖 − 𝜉𝑖⟩ = ‖𝜆𝛾𝜉𝑖 − 𝜉𝑖‖2 =

|𝛾𝐹𝑖△𝐹𝑖|

|𝐹𝑖|≤ 𝑐

Using this equality and ⟨𝜉𝑖 | 𝜉𝑖⟩ = ‖𝜉𝑖‖2 = 1, one gets

⟨𝜆𝛾𝜉𝑖 − 𝜉𝑖 | 𝜆𝛾𝜉𝑖 − 𝜉𝑖⟩ = 2− 2⟨𝜆𝛾𝜉𝑖 | 𝜉𝑖⟩.

For 𝑖 big enough, this implies that

0 <2− 𝑐

2≤ ⟨𝜉𝑖 | 𝜆𝛾𝜉𝑖⟩

Consider (𝜉𝑖)𝑖∈N as an element in some ultrapower ℓ2(Γ)𝜔 of ℓ2(Γ). Then the above inequalityimplies that any translate of this element is separated by a hyperplane away from 0 ∈ ℓ2(Γ)𝜔.Thus, if 𝑋 is the closure of the convex hull of the Γ-orbit of (𝜉𝑖)𝑖∈N, then 𝑋 is also separatedaway from 0 by this hyperplane. Since the space is uniformly convex, there is a closest pointprojection on closed convex sets. Thus the nearest point to 0 in 𝑋 is a fixed point underthe action of Γ. This means that the left regular representation in ℓ2(Γ) admits almost fixedpoints and implies the equivalent condition (v). �

4.C ... equivalences; paradoxical

Definition 4.6. A group Γ is said to have a paradoxical decomposition if there exists

· 𝑠, 𝑡 ∈ N,

· {𝐴𝑖}𝑠𝑖=1 and {𝐵𝑖}

𝑡𝑗=1 collections of mutually disjoint subsets of Γ

· {𝛾𝑖}𝑠𝑖=1 and {𝜂𝑗}

𝑡𝑗=1 elements of Γ

so that

Γ =(⊔𝑠𝑖=1 𝐴𝑖

)⊔(⊔𝑡𝑗=1 𝐵𝑗

)= ⊔𝑠

𝑖=1𝐴𝑖𝛾𝑖 = ⊔𝑡𝑗=1𝐵𝑗𝜂𝑗

where ⊔ stresses that the unions are disjoint. The Tarski number of a group, say 𝜏 (Γ) is theminimum of 𝑠 + 𝑡 over all paradoxical decompositions of Γ (and ∞ if Γ has no paradoxicaldecompositions). F

Note the convention whether 𝛾𝑖 and 𝜂𝑗 are on the left is not important (one could put italso on the right by sending all elements to their inverse). Since the Cayley graphs are heredefined by multiplication on the right, the convention on the left would be more geometric.Indeed, multiplying on the left is a graph automorphism. Hence one cuts the Cayley graphin finitely many pieces (the 𝐴𝑖s and 𝐵𝑗s). Upon moving (by automorphisms) the 𝐴𝑖s onerecovers the full Cayley graph. Same goes for the 𝐵𝑗s

It is fairly easy to check that 𝑠, 𝑡 ≥ 2 (see exercises) so 𝜏 (Γ) ≥ 4. A well-known result isthat 𝜏 (Γ) = 4 if and only if Γ admits ℱ2 (the free group on 2 generators) as a subgroup (seeexercises)24.

The Banach-Tarski paradox (which actually goes back to Hausdorff) stems from the factthat ℱ2 has a paradoxical decomposition and that it is a subgroup of SO3(R).

24This is an unpublished work of Jonsson in the 40’s and a published work of Dekker in the 50’s.


Theorem 4.7

Γ is amenable if and only if(vii) Γ does not have a paradoxical decomposition, i.e. 𝜏 (Γ) =∞ (Tarski, 1938)

To be more precise (iii) =⇒ (vii) is due to von Neumann (1929). The other direction is dueto Tarski25.

Proof of (iii) =⇒ (vii): Assume Γ has a paradoxical decomposition and let 𝑚 be theinvariant mean, then

1𝑛𝑜𝑟𝑚.= 𝑚(1Γ) = 𝑚(

∑𝑖

1𝐴𝑖+∑𝑗

1𝐵𝑗)𝑙𝑖𝑛.=∑𝑖

𝑚(1𝐴𝑖) +

∑𝑗

𝑚(1𝐵𝑗)

where “norm.” stands for normalisation and “lin.” for linearity (it is crucial that the sumsare finite!). On the other hand,

1𝑛𝑜𝑟𝑚.= 𝑚(1Γ) = 𝑚(

∑𝑖

1𝐴𝑖𝛾𝑖)𝑙𝑖𝑛.=∑𝑖

𝑚(1𝐴𝑖𝛾𝑖)𝑖𝑛𝑣.=∑𝑖

𝑚(1𝐴𝑖)

where “inv.” stands for the invariance. The same goes with the 𝐵𝑖s. Putting these 3 equationstogether gives 1 = 2 which is a contradiction. �

For the other direction, the following theorem will be required:

Theorem 4.8 (Hall’s marriage Lemma)...

Proof of (vii) =⇒ (i): The proof will actually be ¬(i) =⇒ ¬(vii), i.e. the absence ofFølner sequence implies paradoxical decomposition. Indeed, ¬(i) means (using Lemma 1.3)

∃𝐶 > 0 so that ∀𝐹 ⊂ Γ finite, |𝐹 | ≤ 𝐶|𝐹+1 ∖ 𝐹 |

where 𝐹+𝑟 = {𝛾 ∈ Γ | 𝑑(𝛾, 𝐹 ) ≤ 𝑟}. So |𝐹+1| ≥ (1+𝐶)|𝐹 |. But by a relative simple iterationargument,

|𝐹+𝑟| ≥ (1 + 𝐶)|𝐹+(𝑟−1)| ≥ . . . ≥ (1 + 𝐶)𝑟|𝐹 |.

Just pick 𝑟 so that (1 + 𝐶)𝑟 ≥ 2. Consider the bipartite graph with vertices on one side𝐹 and the other 𝐹+𝑟. Put an edge between vertices if the two elements are related by anelement in 𝐵𝑟 = {𝛾 ∈ Γ | 𝑑(𝛾, 𝑒Γ) ≤ 𝑟}.

Using Hall’s marriage Lemma, one finds two way of sending 𝐹 to 𝐹+𝑟. These partial mapsuse at most finitely many elements (because 𝐵𝑟 is finite). By looking at all those maps foran increasing family of sets 𝐹𝑘,

26 one gets a family of such maps.Now this sequence could be quite chaotic. Note for each 𝐵𝑖 there is however a subsequence

which is eventually constant when restricted to 𝐵𝑖. Restrict further this subsequence forlarger ball and use a diagonal argument to conclude. �

It might be good to underline that one does not need the axiom of choice27 to see that(vii) =⇒ (i).

25Since there was no other criterion for amenability at the time of Tarski, his proof was certainly non-trivial.I could not get my hands on the original paper...

26To use Hall’s marriage Lemma, we really need that all subsets 𝐹 of 𝐹𝑘 have the property that there areenough elements in 𝐹+𝑟.

27Of course, the axiom of countable choice is used, but that one was never controversial.


4.D ... equivalences; fixed point

In construction....

5 Varia

5.A Reiter and Liouville

As mentioned before, one could think that some 𝜇(𝑛) could be good choices for the Reitersequence. Recall that a function 𝑓 is 𝜇-harmonic (for some 𝜇 ∈ Pr𝑐(Γ)) if 𝑓 * 𝜇 = 𝑓 . Γ issaid to be 𝜇-Liouville if there are no 𝜇-harmonic functions in ℓ∞Γ except constant functions.

Before moving on to the proof, a small lemma will be required:

Lemma 5.1

In ℓ1N, weak and strong convergence coincide.

Proof : It is classical that strong convergence implies weak convergence. Assume thesequence 𝑥𝑛 converges weakly but not strongly. WLOG, one may assume it converges to 0(if it converges to 𝑥 look at the sequence 𝑥𝑛 − 𝑥0 instead). WLOG, one may also assumethat ‖𝑥𝑛‖ℓ1 ≥ 1 (because it does not converge strongly, this is ≥ 𝛿 for some subsequence; justrestrict to this subsequence and multiply by a scalar to get ≥ 1).

Since all function with values ± on a finite set belong to ℓ∞N, weak convergences impliesthat the 𝑥𝑛 tend to 0 on finite sets. On the other hand, for each 𝑥𝑛 there is a finite set 𝐴 sothat ‖𝑥𝑛‖ℓ1(𝐴c

𝑛) < 𝜖 (for some fixed and small 𝜖 → 0. Hence one may find sequences 𝑛𝑖 and𝑁𝑖 so that

∀𝑖‖𝑥𝑛𝑖‖ℓ1([𝑁𝑖−1,𝑁𝑖]c) < 𝜖

Pick 𝑦 ∈ ℓ∞N, given by

for 𝑛 ∈ [𝑁𝑖−1, 𝑁𝑖], 𝑦(𝑛) = sgn 𝑥𝑛𝑖(𝑛)

with sgn𝑟 = 𝑟|𝑟|

if = 0 and 0 if 𝑟 = 0. Then, notice that, ∀𝑖,

⟨𝑦 | 𝑥𝑛𝑖⟩ ≥ ‖𝑥𝑛𝑖‖ℓ1N − 2𝜖 ≥ 1− 2𝜖.

Hence, the sequence cannot converge weakly to 0. �

Recall that Pr𝑐(Γ) is the space of finitely supported probability measures. Pr(Γ) is thenorm-closure of Pr𝑐(Γ) in ℓ1Γ, i.e. probability measures in ℓ1Γ.

Proposition 5.2

Let Γ be finitely generated and let 𝜇 ∈ Pr(Γ) be symmetric and irreducible. Γ is𝜇-Liouville if and only if the sequence 𝜉𝑁 = 1

𝑁

∑𝑁−1𝑖=0 𝜇(𝑖) is a Reiter sequence.

Proof: (⇐) Assume 𝑓 is 𝜇-harmonic and 𝜉𝑛 is Reiter. Since 𝜉𝑛 are convex combinations of𝜇(𝑛), 𝑓 * 𝜉𝑛 = 𝑓 . So for any 𝑥, 𝑠 ∈ Γ

|𝑓(𝑥𝑠)− 𝑓(𝑥)| = |𝑓 * 𝛿𝑠(𝑥)− 𝑓(𝑥)| = |𝑓 * 𝜉𝑛 * 𝛿𝑠(𝑥)− 𝑓 * 𝜉𝑛(𝑥)|= |𝑓 * (𝜉𝑛 * 𝛿𝑠 − 𝜉𝑛)(𝑥)| ≤ ‖𝑓‖ℓ∞‖𝜉𝑛 * 𝛿𝑠 − 𝜉𝑛‖ℓ1


Since 𝑛 is arbitrary, 𝑓(𝑥𝑠) = 𝑓(𝑥). But 𝑥 and 𝑠 are also arbitrary so 𝑓 is constant.(⇒) Assume 𝜉𝑁 = 1

𝑁

∑𝑁−1𝑖=0 𝜇(𝑖) (or any of its subsequence) is not Reiter. Pick some 𝑠 ∈ Γ.

Then28 ∃𝜖 > 0 so that, ∀𝑁 , ‖𝛿𝑠 * 𝜉𝑁 − 𝜉𝑁‖ℓ1 > 𝜖. Since this is norm-separated from 0, thereis an element in the dual of ℓ1Γ (i.e. ℓ∞Γ) which also does29. So, there exists 𝑔 ∈ ℓ∞Γ suchthat, ∀𝑛,

(**)

∫

Γ𝑔d(𝛿𝑠 * 𝜉𝑁)−

∫Γ𝑔d𝜉𝑁

> 𝜖.

Put 𝑓(𝑥) = lim𝑛

∫Γ 𝑔(𝑦)d(𝛿𝑥 * 𝜉𝑛)(𝑦) where the limit is given by some diagonal subsequence30

or a weak* accumulation point31. Then 𝑓 is bounded (obvious) and harmonic:

|(𝑓 − 𝑓 * 𝜇)(𝑥)| = | lim𝑁

∫Γ 𝑔(𝑦)d(𝛿𝑥 * 𝜉𝑁 − 𝛿𝑥 * 𝜉𝑁 * 𝜇)(𝑦)|

= | lim𝑁

∫Γ 𝑔(𝑦)d(

1𝑁[𝛿𝑥 − 𝛿𝑥 * 𝜇

(𝑁)])(𝑦)|≤ ‖𝑔‖ℓ∞ lim𝑁 ‖

1𝑁(𝛿𝑥 − 𝛿𝑥 * 𝜇

(𝑁))‖ℓ1≤ ‖𝑔‖ℓ∞ lim𝑁

2𝑁

≤ 0.

Note that it is crucial that the limit tends to 0 whatever the subsequence chosen. Lastly,|𝑓(𝑠−1)− 𝑓(𝑒)| > 𝜖 thanks to equation (**), so we have that 𝑓 is not a constant. �

The first proof that ℎ(𝜇) = 0 (the entropy) gives a Reiter sequence is due to Avez 1974.He used it to show that ℎ(𝜇) = 0 is equivalent to 𝜇-Liouville. This as been revisited byKaimanovich & Vershik: ℎ(𝜇) = 0 implies that 𝜉𝑁 = 1

2𝜇(𝑁) + 1

2𝜇(𝑁+1) is Reiter.

Note that, even on Z and for a symmetric finitely supported 𝜇, 𝜇(𝑛) is in general not aReiter sequence. Indeed, if the Cayley graph generated by 𝑆 = Support(𝜇) is bipartite, it iseasy to show that 𝜇(𝑛) is supported on elements 𝛾 such that |𝛾|𝑆 ≡ 𝑛 mod 2. In particular,for any 𝑠 ∈ 𝑆, the support of 𝛿𝑠 * 𝜇

(𝑛) and 𝜇(𝑛) are disjoint.Proposition 5.2 also implies that non-amenable groups are not Liouville (for symmetric

irreducible probability measures). Here is another argument (taken from U. Bader’s lectures)which does not require symmetry:

Proposition 5.3

If Γ is not amenable, then it is not 𝜇-Liouville for any irreducible 𝜇 ∈Pr(Γ).

Proof : Recall 𝑀 is the convex set of means inside the LCTVS (ℓ∞Γ)*, and, as Γ is notamenable, none of them is invariant. 𝜇 acts on function in ℓ∞Γ by convolution and hencedefines a dual action on its dual. Because 𝜇 is a probability measure this action (is givenby taking a limit of convex combinations, and hence) preserves 𝑀 . Now, by the Markov-Kakutani theorem32 this action will have a fixed point.

Pick 𝑚 to be such a fixed point. Since it is not invariant, ∃𝛾 ∈ Γ so that 𝛾 · 𝑚 = 𝑚(again the action on means is dual to the right-regular action on ℓ∞Γ). ∃𝑔 ∈ ℓ∞Γ so that

28We use the absence of invariance on the left instead of the right. But since 𝜇 is symmetric, this is equivalent.29This is where Lemma 5.1 comes in: the sequence is also weakly separated from 0.30Since 𝑔 is bounded by 𝑐 = ‖𝑔‖ℓ∞ , this is a sequence of functions in [−𝑐, 𝑐]Γ. Enumerate the vertices and take

a first subsequence 𝑛[1]𝑘 which converges at the first vertex. Refine this subsequence at a second vertex to get asubsequence 𝑛[2]𝑘 where the function converges on both vertices, etc... One obtains a sequence of subsequences𝑛[𝑖]𝑘 . The diagonal subsequence 𝑛𝑘 = 𝑛

[𝑘]𝑘 will converge everywhere.

31In this case, see the comments after Proposition 5.3.32The semi-group N acts on 𝑀 by applying 𝑛 times 𝜇. The proof goes exactly as in (i) =⇒ (ii).


(𝛾 · 𝑚)(𝑔) = 𝑚(𝑔). Define ℎ(𝛾) = (𝛾 · 𝑚)(𝑔). Then ℎ is non-constant (because 𝑚 is notinvariant) but is harmonic (because 𝑚 is fixed by 𝜇). �

The previous proof also indicates how to modify the proof of (⇒) in Proposition 5.2 tomore general groups. One needs to replace the particular convex combination 𝜉𝑁 by 𝑋, theconvex set generated by the sequence 𝜇(𝑛). Separation in the weak topology follows fromHahn-Banach (instead of Lemma 5.1). 𝑌 is then defined as an weak* accumulation point of𝑓𝑛(𝑥) :=

∫Γ 𝑔(𝑦)d(𝛿𝑥 *𝜇

(𝑛))(𝑦). The Markov-Kakutani theorem is applied to find a 𝜇-invariantelement in 𝑌 .

Note that neither the Proposition 5.2 nor Proposition 5.3 require that 𝜇 is finitely sup-ported. A construction of Kaimanovich & Vershik 1983 shows that, in an amenable group,one can always construct a (in general, not finitely supported) probability measure 𝜇 so thatthe group is 𝜇-Liouville (this was a conjecture of Furstenberg). Hence

Theorem 5.4 (Kaimanovich & Vershik 1983 )

A group Γ is amenable if and only if there is a symmetric and irreducible probabilitymeasure 𝜇 so that Γ is 𝜇-Liouville.

Proof : to do... �

However, there are amenable groups which are not 𝜇-Liouville (for a symmetric finitelysupported measure 𝜇). The simplest example is the lamplighter group on Z3 : Γ = Z2 ≀ Z

3.This was shown by A.Erschler

Example 5.5. Let Γ = Z2 ≀ Z3 be the lamplighter group on Z3 (there is a lamp at each

vertex in Z3). To define an explicit bounded harmonic function, first notice that since thespace where the “lamplighter” moves is Z3. Up to forgetting the lamps, one has a (lazy)random walk on Z3. It is transient, so with probability 1 the lamplighter will visit finitelymany times a finite subset 𝐹 of Z3. Fix some lamp states on 𝐹 , i.e. fix an element 𝑎 of⊕𝑥∈𝐹Z2. Define, for 𝑥 ∈ Γ,

ℎ(𝑥) = P(for 𝑛 large enough, a SRW starting at 𝑥 has lamp state equal to 𝑎 on 𝐹 ).

Transience is crucial for this probability to be defined. It’s obviously bounded. By construc-tion (and elementary probability), it is 𝜇-harmonic. Remains to check it is non-constant. Tosee this take 𝑥 and 𝑦 ∈ Γ so that the lamplighter position 𝑧 ∈ Z3 is the same in 𝑥 and 𝑦 butthe lamp states on 𝐹 are equal to 𝑎 in 𝑥 and not equal to 𝑎 in 𝑦. Let 𝑧 go to infinity, thenℎ(𝑥)→ 1 while ℎ(𝑦)→ 0. Hence ℎ is a non-constant bounded harmonic function. ♣

One can also extend these harmonic functions by using any 𝑓 ∈ ℓ∞𝐿 (where 𝐿 = ⊕𝑔∈Z𝑑Z2)and then ℎ being the expected value of 𝑓 after the walker has gone to ∞. These functionsare dubbed “limit configurations”. A. Erschler33 showed these are actually the only boundednon-constant harmonic functions on Z2 ≀ Z

𝑑 if 𝑑 > 4. From this description, can you find the𝑔 in proposition 5.2 (and 𝑠, depending on 𝑎) which separates the convex hull of the 𝜇(𝑖) fromtheir translates (it might be hard to prove it actually does!).

33This has recently been proved for any 𝑑 > 2 by Y. Peres.


5.B The Mazur-Ulam theorem

This proof follows [a presentation of N. Monod of] a proof due to B. Nica [12].

Definition 5.6. Let (𝑋, 𝑑) be a metric space. A mid-point is a map m : 𝑋 × 𝑋 → 𝑋satisfying

(A) 𝑑(𝑥,m𝑥,𝑦

)= 𝑑

(𝑦,m𝑥,𝑦

)= 1

2𝑑(𝑥, 𝑦

).

(B) m𝑥,𝑦 = m𝑦,𝑥.

A symmetry is an isometry 𝜎 ∈ Isom(𝑋) such that ∃𝑝 ∈ 𝑋 with ∀𝑧 ∈ 𝑋,m𝑧,𝜎𝑧 = 𝑝. A convexsymmetric space is a space with mid-point (𝑋, 𝑑,m) such that

(C) 𝑑(m𝑥,𝑦, 𝑧) ≤12

(𝑑(𝑥, 𝑧) + 𝑑(𝑦, 𝑧)

).

(D) ∀𝑥, 𝑦 ∈ 𝑋, there is a symmetry 𝜎 such that 𝜎𝑥 = 𝑦, 𝜎𝑦 = 𝑥. F

Note that the symmetry in (D) corresponding to 𝑥 and 𝑦 has necessarily 𝑝 = m𝑥,𝑦. Insome Banach space, there are many mid-point maps different from 𝑥+𝑦

2(although this is the

simplest one).

Theorem 5.7

If 𝑋 is a LCTVS which is also a convex symmetric space, then every isometry preservesm.

Proof: Let 𝑔 ∈ Isom(𝑋). Fix 𝑥, 𝑦 ∈ 𝑋 and let Δ𝑔 := Δ𝑔(𝑥, 𝑦) := 𝑑(m𝑔𝑥,𝑔𝑦, 𝑔m𝑥,𝑦). Then

(*) Δ𝑔

(C)

≤ 12

(𝑑(𝑔m𝑥,𝑦, 𝑔𝑥) + 𝑑(𝑔m𝑥,𝑦, 𝑔𝑦)

)= 1

2

(𝑑(m𝑥,𝑦, 𝑥) + 𝑑(m𝑥,𝑦, 𝑦)

) (A)= 1

2𝑑(𝑥, 𝑦).

Using (D), take 𝜎 a symmetry corresponding to 𝑔𝑥 and 𝑔𝑦 (so that 𝑝 = m𝑔𝑥,𝑔𝑦) and letℎ = 𝑔−1𝜎𝑔 ∈ Isom(𝑋), then ℎ𝑥 = 𝑦 and ℎ𝑦 = 𝑥. As a consequence,

Δℎ = 𝑑(mℎ𝑥,ℎ𝑦, 𝑔−1𝜎𝑔m𝑥,𝑦) = 𝑑(m𝑦,𝑥, 𝑔

−1𝜎𝑔m𝑥,𝑦)(B)= 𝑑(m𝑥,𝑦, 𝑔

−1𝜎𝑔m𝑥,𝑦) = 𝑑(𝑧, 𝜎𝑧

)where 𝑧 = 𝑔m𝑥,𝑦

= 2𝑑(𝑧, 𝑝) = 2𝑑(𝑔m𝑥,𝑦,m𝑔𝑥,𝑔𝑦) since 𝑝 = m𝑔𝑥,𝑔𝑦

= 2Δ𝑔.

Now, the upper bound (*) is independent of the isometry chosen. Thus, one gets a contra-diction (by repeating this “doubling”) unless Δ𝑔 = 0. �

A map is affine is it preserves convex combinations. Actually, if the vector space is over R, 𝑓is affine if and only if 𝑓(𝑣+𝑤

2) = 1

2

(𝑓(𝑥) + 𝑓(𝑦)

). Since m𝑥,𝑦 =

12(𝑥+ 𝑦) is always a mid-point

map in a Banach space, one gets:

Corollary 5.8 (Mazur-Ulam, 1932 )

Every isometry of a Banach space is affine.


5.C More fixed points properties

6 Exercises

Everywhere: Γ is a finitely generated (discrete) group. For hints look at the end.

Exercise 1 Show that a group Γ is finite if and only if every affine action on a vectorspace admits a fixed point. [It might make life easier to use the Mazur-Ulam theorem: anisometric action on a Banach space is affine.]

Exercise 2 Show that a sequence {𝐹𝑛} of finite sets is Følner (i.e. ∀𝑈 ⊂ Γ finite,𝐹𝑛𝑈△𝐹𝑛|

|𝐹𝑛|→ 0) if and only

|𝜕𝐹𝑛|

|𝐹𝑛|→ 0.

Exercise 3 The aim is to compute the isoperimetric profile of Z𝑑 with a non-optimalconstant34: |𝐹 |(𝑑−1)/𝑑 ≤ 1

2|𝜕𝐹 |. To do so, let 𝜋𝑖 : Z

𝑑 → Z𝑑−1 be the projection orthogonal to

the 𝑖th coordinate. Let 𝑝 : Z𝑑 → Z be the projection on the first coordinate. Take 𝐹 ⊂ Z𝑑

finite and let 𝐹𝑖 := 𝜋𝑖(𝐹 ). Let 𝑎𝑥 = |𝑝−1(𝑥)| and (for 𝑗 ≥ 2) 𝑎𝑥𝑗 = |𝜋𝑗(𝑝−1(𝑥))|.

1. Assuming the Loomis-Whitney inequality, i.e.

(LW) |𝐹 |𝑑−1 ≤𝑑∏

𝑖=1

|𝐹𝑖|,

deduce that |𝐹 |(𝑑−1)/𝑑 ≤ 12|𝜕𝐹 |.

2. When 𝑑 = 2, show directly that |𝐹 | ≤ |𝐹1| · |𝐹2|.

3. To show (LW) for 𝑑 > 2, induction will be done. First, show that |𝐹 |*=∑

𝑥∈Z 𝑎𝑥,

𝑎𝑥†

≤ |𝐹1| and∑

𝑥∈Z 𝑎𝑥𝑗‡= |𝐹𝑗|.

4. Use induction and † to get 𝑎𝑑−1𝑥

S

≤ |𝐹1|∏𝑑𝑗=2 𝑎𝑥𝑗.

5. Use *, S, Hölder’s inequality and ‡ to conclude that (LW) holds.

6. Show that the profile is sharp (up to the constant).

Exercise 4 The infinite permutation group is

𝑆∞ = {𝜎 : N→ N | 𝜎 is bijective and there is a finite set 𝐹 such that 𝜎|N∖𝐹 = Id}.

Show that it is amenable and is not finitely generated. Show that 𝐴∞, the even permutationsin 𝑆∞, is simple.

Exercise 5 Prove the “closure properties” as follows:

1. “Subgroup” (a) using the “Følner sequences” (i) definition.

34For many application, having the optimal constant is not so important. This argument is taken fromLoomis-Whitney 1940’s.


2. “Extensions” (b) using the “invariant means” (iii) definition.

3. “Quotients” (c) using the “probability of return” (iv) definition [Kesten’s criterion]

4. “Direct limits” (d) using the “fixed point property” (ii) definition.

5. Show that a semi-direct product of amenable groups is amenable (a subcase of (b))using the “Følner sequences” definition (i).

Exercise 6 Assume Γ is infinite. Let 𝑐𝑜Γ be the norm closure of finitely supportedfunctions in ℓ∞Γ. Show that if 𝑚 is an invariant mean, then 𝑚(𝑓) = 0 for all 𝑓 ∈ 𝑐0Γ.Conclude that the kernel of any35 invariant mean contains (the norm closure of)

𝑐0Γ + {𝑓 * 𝛿𝛾 − 𝑓 | 𝑓 ∈ ℓ∞Γ and 𝛾 ∈ Γ}.

Can you show this is indeed the intersection of the kernel of all means?

Exercise 7 The goal is to show there always exists Følner sequences which are right ANDleft-invariant.

1. Show that if there is a right-Reiter sequence, then there is a left-Reiter sequence. Deduceany amenable group has a sequence which is both left- and right-Reiter.

2. Conclude that there is always a left- and right-Følner sequence.

Exercise 8 Show directly that the “Følner sequence” definition (i) implies the “invariantmean” definition (iii). [You will need to know what an ultralimit is...]

Exercise 9 Show that lim inf𝑛→∞|𝐵𝑛+1||𝐵𝑛|

= 0 implies the group is amenable.

Exercise 10 Here are steps to show: 𝜏 (Γ) = 4 ⇐⇒ Γ has a subgroup isomorphic to ℱ2.

1. Prove that, WLOG, 𝛾1 = 𝜂1 = 𝑒Γ.

2. Deduce that 𝑠 ≥ 2, 𝑡 ≥ 2 and 𝜏 (Γ) ≥ 4.

3. Use the Ping-Pong-Lemma to show that

𝜏 (Γ) = 4 =⇒ Γ has a subgroup isomorphic to ℱ2.

4. Show that 𝜏 (ℱ2) = 4.

5. If 𝐻 is a subgroup of Γ, show that 𝜏 (𝐻) ≥ 𝜏 (Γ). [This is also a way of doing (vii) =⇒(a).]

Exercise 11 Assume 𝐻 is a subgroup of Γ and 𝑄 = Γ/𝑁 for some normal subgroup 𝑁 .Show that 𝜏 (𝐻) ≥ 𝜏 (𝐺) and 𝜏 (𝑄) ≥ 𝜏 (𝐺). [This is also a way of doing (vii) =⇒ (c).]

Exercise 12 Show that any group with 𝜏 (Γ) < ∞ has a finitely generated subgroup 𝐻with 𝜏 (𝐻) = 𝜏 (Γ).

Hint[s] for 1: “finite”⇒ “fixed”: The average of the orbit of a vector is fixed. Other direction: look atthe right-regular action of Γ on 𝑃 = {𝑓 ∈ ℓ1Γ |

∑𝑥 𝑓 = 1}. Be careful: 𝑃 is not a vector space. Take

35Even on Z, there are uncountably many invariant means...


𝑄 = {𝑓 ∈ ℓ1Γ |∑

𝑥 𝑓 = 0}, take an affine map 𝜑 it into 𝑃 (i.e. add 𝛿𝑒 to each element). Act on 𝑣 ∈ 𝑄 asfollows: 𝑣 · 𝛾 = 𝛼−1

(𝛼(𝑣) · 𝛾

). Either compute this map to see it’s affine, or use Mazur-Ulam (both 𝛼 and the

action are isometric).

Hint[s] for 2: Let 𝜕𝑟𝐹 be the 𝑟-boundary of 𝐹 : look at all edges with some extremity [or all vertices] atdistance ≤ 𝑟 from 𝐹 . Give a rough bound: |𝜕𝑟𝐹 | ≤ (|𝑆|+ 1)𝑟|𝜕𝐹 |.

Hint[s] for 3: (1) for any 𝑖, 2|𝐹𝑖| ≤ |𝜕𝐹 |. (5) the form of Hölder’s inequality required here is∑

𝑥 |∏𝑑

𝑗=2 𝑎𝑥𝑗 | ≤∏𝑑𝑗=2

(∑𝑥 |𝑎𝑥𝑗 |

𝑑−1)1/(𝑑−1)

.

Hint[s] for 5: (1) For each coset, one gets a sequence (the intersection of the Følner sequence with thecoset). What happens if none of these satisfy the Følner condition. (2) Average over cosets to be able to projectthe function to the quotient, then average there. (3) Show that the probability of return is bigger in the quotientgroup than in the quotiented group. (4) Look at the limit of the fixed point sets. (5) If Γ = 𝐺o𝐻, 𝐹𝑛 is Følnerfor 𝐺 and 𝐸𝑛 is Følner for 𝐻, try either ({𝑒𝐺}, 𝐹𝑛)(𝐸𝑛, {𝑒𝐻}) or (𝐸𝑛, {𝑒𝐻})({𝑒𝐺}, 𝐹𝑛) (to get either a left- ora right-Følner sequence).

Hint[s] for 6: 𝑚 is continuous and the kernel is closed. To find the kernel of a mean it is easier to thinkin terms of Følner sequences. Let 𝑓𝑛 = |𝐹𝑛|

−1𝑓 * 1𝐹𝑛 (this is a convex combination of translates of 𝑓). 𝑓𝑛 hasthe same invariant mean as 𝑓 and 𝑚(𝑓) = 0 if and only if lim

𝑛→∞‖𝑓𝑛‖ℓ∞ → 0.

Hint[s] for 7: If 𝜉𝑛 is right-Reiter, 𝜉−1𝑛 is left-Reiter and 𝜉−1𝑛 * 𝜉𝑛 is both.

Hint[s] for 8: Define 𝑓𝑛(𝑥) =1

|𝐹𝑛|

∑𝛾∈𝐹𝑛

𝑓(𝑥𝛾) = 𝑓 * 1𝐹𝑛(𝑥). Show this function becomes essentially

constant and define 𝑓 = lim 𝑓𝑛 (where this is an ultralimit). Put 𝑚(𝑓) = 𝑓(𝑥) for any 𝑥.

Hint[s] for 9: |𝐵𝑛+1△𝐵𝑛||𝐵𝑛|

= |𝐵𝑛+1||𝐵𝑛|

− 1.

Hint[s] for 10: 3: Ping-pong lemma: Let 𝐻 be a group, generated by two elements 𝑎 and 𝑏 of infiniteorder. Suppose there is a 𝐻-action on a set 𝑋 such that there are non-empty subsets 𝐴𝐵 ⊂ 𝑋 with 𝐵 notincluded in 𝐴 and such that for all 𝑛 ∈ Z ∖ {0} one has 𝑎𝑛 ·𝐵 ⊂ 𝐴 and 𝑏𝑛 ·𝐴 ⊂ 𝐵. Then 𝐻 is freely generatedby {𝑎, 𝑏}.

4: Say the generators of the free group are 𝑎 and 𝑏. A first try would be 𝐴1 = all words beginning in 𝑎,𝐴2 = all words beginning in 𝑎−1, 𝐵1 = all words beginning in 𝑏 and 𝐵2 = all words beginning in 𝑏−1. But theidentity is not included. Add the identity to one set (and change the place of infinitely many other elements)in order to fix this.

5: Each [right- or left- ?] cosets of the subgroup has a paradoxical decomposition with the same translatingelements. Put them together.

Hint[s] for 11: Very simliar to 10.5. If 𝜏 (𝑄) = ∞ there is nothing to prove. Say 𝑄 has a paradoxicaldecompositions with 𝑞𝑖 and 𝑟𝑗 as translating elements and 𝐴𝑖, 𝐵𝑖 as sets. For the paradoxical decompositionof Γ, pull the sets back, i.e. use 𝜋−1(𝐴𝑖) where 𝜋 : Γ � 𝑄 and pick the translating elements in 𝜋−1(𝑞𝑖) and𝜋−1(𝑟𝑗) (it does not matter which one you pick).

Hint[s] for 12: Look at the subgroup generated by the elements used to make the paradoxical decompo-sition. One inequality comes from exercise 10.

7 References

There are many book which treat these equivalences, the two standard references are Pier“Amenable locally compact groups” and Paterson “Amenability”. Greenleaf “Invariant meanson topological groups” is actually quite nice to read (but not so easy to find). The appendixin the book by Bekka, de la Harpe & Valette “Kazhdan property (T)” (appendix G) is alsoa great source. Some lecture notes by Anne Thomas can be another good source. All these


do the “general case” of locally compact groups (which might make a first reading harder).The above proofs were obtained by successive simplifications of the proofs in these varioussources.

Further references

[1] A. Avez, Théorème de Choquet-Deny pour les groupes à croissance non exponentielle,C. R. Acad. Sci. Paris Sér. A 279:25–28, 1974.

[2] I. Benjamini and O. Schramm, Percolation beyond Z𝑑, many questions and a few answers,Electron. Comm. Probab. 1(8):71–82, 1996.

[3] J. W. Cannon, W. J. Floyd and W. R. Parry, Amenability, Folner sets, and coolingfunctions, Ar𝜒i𝜈:1201.0132, 30 Dec 2011.

[4] T. Ceccherini-Silberstein, R. Grigorchuk, de la Harpe, Amenability and paradoxicaldecompositions for pseudogroups and discrete metric spaces, Tr. Mat. Inst. Steklova

224:68–111, 1999; translation in Proc. Steklov Inst. Math. 224(1):57–97, 1999.

[5] J. Dixmier, Les moyennes invariantes dans les semi-groups et leurs applications, ActaSci. Math. Szeged 12:213–227, 1950.

[6] A. Erschler, Poisson-Furstenberg boundary of random walks on wreath products andfree metabelian groups, Comment. Math. Helv., 86:113–143, 2011.

[7] E. Følner, A generalisation of a theorem of Bogoliouboff with an appendix on Banachmean value in non-Abelian groups, Math. Scand. 2:5–18, 1954.

[8] E. Følner, On groups with full Banach mean value, Math. Scand. 3:243–254, 1955.

[9] F. P. Greenleaf, Amenable actions of locally compact groups. J. Functional Analysis4:295–315, 1969.

[10] K. Juschenko and T. Smirnova-Nagnibeda, Small spectral radius and percolation con-stants on non-amenable Cayley graphs, Ar𝜒i𝜈:1206.2183, 22 Oct 2012.

[11] H. Kesten, Full Banach mean values on countable groups, Math. Scand., 7:146–156,1959.

[12] B. Nica, The Mazur-Ulam Theorem, Expo. Math. 30(4):397–398, 2012. See alsoAr𝜒i𝜈:1306.2380 (June 2013).

[13] A. Tarski, Algebraische Fassung des Massproblems, Fund. Math. 31:47–66, 1938. Seealso Collected papers (Birkhäuser), vol. 2, pp. 453–472, 1986.

[14] A. Thom, A remark about the spectral radius, Ar𝜒i𝜈:1306.1767, 7 Jun 2013.

[15] J. von Neumann, Zur allgemein Theorie des Masses, Fund. Math. 13:73–116, 1929.

[16] S. Wagon, The Banach-Tarski paradox, Cambridge university press, 1986.

amenable groups - tu dresdengournay/amenability-note.pdf · amenability cirm luminy - 2014/1/14...

Documents