arXiv:1408.5805v2 [] 4 Oct 2016

GROUPS, ORDERS, AND DYNAMICS
Bertrand Deroin
CNRS and École Normale Supérieure

Andrés Navas
Universidad de Santiago de Chile

Cristóbal Rivas
Universidad de Santiago de Chile

Contents
INTRODUCTION

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1 SOME BASIC AND NOT SO BASIC FACTS
1.1 General Definitions . . . . . . . . . . . . . . . . . .
1.1.1 Positive and negative cones . . . . . . . . .
1.1.2 A characterization involving finite subsets .
1.1.3 Left-orderable groups and actions on ordered
1.1.4 Semiconjugacy in Homeo+ (R) . . . . . . . .
1.2 Some Relevant Examples . . . . . . . . . . . . . . .
1.2.1 Abelian and nilpotent groups . . . . . . . .
1.2.2 Subgroups of the affine group . . . . . . . .
1.2.3 Free and residually free groups . . . . . . . .
1.2.4 Thompson’s group F . . . . . . . . . . . . .
1.2.5 Some relatives of F . . . . . . . . . . . . . .
1.2.6 Braid groups . . . . . . . . . . . . . . . . .
1.3 Other Forms of Orderability . . . . . . . . . . . . .
1.3.1 Lattice-orderable groups . . . . . . . . . . .
1.3.2 Locally-invariant orders and diffuse groups .
1.4 General Properties . . . . . . . . . . . . . . . . . .
1.4.1 Left-orderable groups are torsion-free . . . .
1.4.2 Unique roots and generalized torsion . . . .
1.4.3 The Unique Product Property (U.P.P.) . . .
1.4.4 More Combinatorial Properties . . . . . . .
1.4.5 Isoperimetry and Left-Orderable Groups . .

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2 A PLETHORA OF ORDERS
2.1 Producing New Left-Orders . . . . . . . . . . . . . . . . . . . . .
2.1.1 Convex extensions . . . . . . . . . . . . . . . . . . . . . .

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ii

CONTENTS
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3 ORDERABLE GROUPS AS DYNAMICAL OBJECTS
3.1 Hölder’s Theorem . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The Conrad Property . . . . . . . . . . . . . . . . . . . . . .
3.2.1 The classical approach revisited . . . . . . . . . . . .
3.2.2 An approach via crossings . . . . . . . . . . . . . . .
3.2.3 An extension to group actions on ordered spaces . . .
3.2.4 The Conradian soul of a left-order . . . . . . . . . . .
3.2.5 Approximation of left-orders and the Conradian soul
3.2.6 Groups with finitely many Conradian orders . . . . .
3.3 An Application: Ordering Solvable Groups . . . . . . . . . .
3.3.1 The space of left-orders of finite-rank solvable groups
3.3.2 The general case . . . . . . . . . . . . . . . . . . . .
3.4 Verbal Properties of Left-Orders . . . . . . . . . . . . . . . .
3.5 A Non Left-Orderable Group, and More . . . . . . . . . . .
3.5.1 No left-order on finite-index subgroups of SL(n, Z) . .
3.5.2 A canonical decomposition of the space of left-orders

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4 PROBABILITY AND LEFT-ORDERABLE GROUPS
4.1 Amenable Left-Orderable Groups . . . . . . . . . . . . . . . .
4.2 Non-Amenable, Left-Orderable Groups with no Free Subgroup
4.2.1 (Non-)amenable actions . . . . . . . . . . . . . . . . .
4.2.2 A finitely-presented version . . . . . . . . . . . . . . .
4.3 Almost-Periodicity . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Almost-periodic representations . . . . . . . . . . . . .
4.3.2 A bi-Lipschitz conjugacy theorem . . . . . . . . . . . .
4.3.3 Actions almost having fixed points . . . . . . . . . . .
4.3.4 Indicability of amenable left-orderable groups revisited
4.4 Random Walks on Left-Orderable Groups . . . . . . . . . . .
4.4.1 Harmonic actions and Derriennic’s property . . . . . .
4.4.2 Properties of stationary measures . . . . . . . . . . . .
4.4.3 Recurrence . . . . . . . . . . . . . . . . . . . . . . . .

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2.2

2.1.2 Free products . . . . . . . . . . . . . . . . . . .
2.1.3 Left-orders from bi-orders . . . . . . . . . . . .
The Space of Left-Orders . . . . . . . . . . . . . . . . .
2.2.1 Finitely many or uncountably many left-orders .
2.2.2 The space of left-orders of the free group . . . .
2.2.3 Finitely-generated positive cones . . . . . . . .

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CONTENTS

iii

4.4.4

Existence of stationary measures . . . . . . . . . . . . . . 196

INTRODUCTION
The theory of bi-orderable groups is a venerable subject in Algebra which has
been extensively developed during the whole last century, starting from seminal
works of Dedekind, Hölder, and Hilbert. Though less developed, the theory of
left-orderable groups has reemerged in the recent years, mainly by the discovery of
many new examples of (at least partially) left-orderable groups of geometric origin
(braid groups [57]; some groups of contact diffeomorphisms [70, 83]; right-angled
Artin groups [68], hence virtually the fundamental groups of closed hyperbolic 3manifolds by the combination of several recent remarkable results [2, 11, 89, 109]).
Moreover, the question of knowing whether certain groups are left-orderable or
not (Garside groups [57]; lattices in higher-rank simple Lie groups [122, 192];
groups with Kazhdan’s property (T) [9, 156], with the subsequent question of
whether bi-orderable groups are a-(T)-menable [41]), the discovery of relevant
examples of groups in the framework of orderable groups (non-amenable groups
[128, 144]), as well the use of orderability in the solution of long-standing problems (H. Neumann’s conjecture [142]) have attracted the interest to the subject of
people coming from different branches of mathematics. Furthermore, the introduction of new techniques –mainly dynamical, analytic and probabilistic– have
revitalized the theory.
This is an almost self-contained monograph (containing some new results) on
left-orderable groups which mostly rely on dynamical and probabilistic aspects,
but also on geometric, combinatorial, analytic, and topological ones.
In Chapter 1, we review the basic definitions and treat several relevant examples, as for example solvable groups, Thompson’s groups, and free groups. Also,
we discuss some of the general properties of groups admitting orders with different invariance properties, as well as certain combinatorial issues that are closely
related. We close the chapter with a result by Gromov concerning the (linear)
isoperimetric profile of left-orderable groups.
In Chapter 2, we show that, besides the fact that many groups admit leftorders, in general they actually admit a lot. To better study this phenomenon,
we introduce the notion of the space of left-orders associated to a left-orderable
group, and we discuss some of its properties. As a concrete example, we treat
the case of the free group from several points of view. Moreover, we present
examples of left-orderable groups having uncountably many left-orders but whose
associated spaces of left-orders contain isolated points, and we give a description
of the groups admitting only finitely many left-orders.

2

CONTENTS

In Chapter 3, we put in a dynamical framework some classical results of the
theory, and we present some new developments via this approach. We begin
with the classical Hölder Theorem characterizing group left-orders satisfying an
Archimedean type property. We then pass to the theory of Conradian left-orders.
We first review the classical approach of Conrad, and then we give an alternative dynamical approach leading to applications in the study of the topology of
the space of left-orders. In particular, we give a complete characterization of
the groups admitting finitely many Conradian left-orders, as well as a description of the space of left-orders of countable solvable groups. We close the chapter with a general decomposition of the space of left-orders of finitely-generated
left-orderable groups into three canonical subsets according to their dynamical
properties. We also discuss the algebraic relevance of this classification, in particular in regard with the open problem of left-orderability of lattices in higher-rank
simple Lie groups.
Chapter 4 is devoted to several recent results relying on techniques with a
probabilistic flavour. We begin with Morris-Witte’s theorem asserting that leftorderable, amenable groups are locally indicable. We next consider actions by
almost-periodic homeomorphisms, and we provide a construction of a space involving all of them which somewhat replaces the space of left-orders; using this, we
give an alternative proof of Morris-Witte’s theorem above. We close this chapter
by considering random walks on finitely-generated left-orderable groups, showing
recurrence type properties and existence of harmonic functions of dynamical origin. More importantly, we explain how probabilistic arguments provide canonical
coordinates for actions on the line.
There are several classical references on the topic of (left-)orderable groups,
as for example [17, 84, 121]. Quite naturally, in many sections, there is a large
intersection of this monograph with these books. However, our presentation is
new in many aspects. In particular, the order we have chosen for the topics
is not the historical one, in opposition to most (all ?) known references of the
theory. Though this may cause some minor problems of lecture (our exposition
is not always “linearly ordered”), we think that, definitively, this presentation is
more appropriate for our main purpose, namely, to put the classical results of the
theory in a modern framework which allows integrating them with the new results
as well as with some of the recent developments in the conjunction between group
theory and dynamics. Besides, there are many other texts that may be considered
for complementary reading. In particular, let us mention those concerning orders
on braid groups [57] and low-dimensional topology [48], as these two subjects are
not deeply treated in this monograph. Another topic not treated here is that of

CONTENTS

3

circular orders on groups. This is well developed in [5, 33, 36, 79, 135], yet many
open questions remain in the subject.
This text is complemented with many exercises which sometimes correspond
to little results in the literature. More importantly, several open problems are
spread out along the text. (A complementary list of open questions, mostly
concerning classical achievements of the theory, may be found in [15]; see also
[57, Chapter XVI]).
This text started growing from Notes that the second-named author wrote for
mini-courses at the Third Latinoamerican Congress of Mathematicians (2009),
the Uruguayan Colloquium of Mathematics (2009), and the School Young Geometric Group Theory II (Haifa, 2013). All authors would like to express their
gratitude to L. Bartholdi, D. Calegari, M. Calvez, A. Clay, Y. de Cornulier, P. Dehornoy, A. Erschler, É. Ghys, A. Glass, R. Grigorchuk, F. Haglund, T. Hartnick,
S. Hurtado, T. Ito, D. Kielak, V. Kleptsyn, T. Koberda, K. Mann, I. Marin,
G. Metcalfe, D. Morris-Witte, L. Paris, F. Paulin, D. Rolfsen, F. le Roux, Z.
Šunić, R. Tessera, L. Vendramin and B. Wiest, as well as all the parcipants of
the meeting “Orderable Groups” held at Cajón del Maipo (September 2014), for
valuable discussions, comments, and suggestions, and E. Trucco for informatics
technical support.
All the three authors acknowledge the funding from the CONICYT PIA 1103
Project DySyRF (Center of Dynamical Systems and Related Fields). B. Deroin
was also partially supported by ANR-08-JCJC-0130-01, ANR-09-BLAN-0116,
and ANR-13-BS01-0002. A.Navas would like to acknowledge the support/hospitality
of the ERC starting grant 257110 RaWG / Institut Henri Poincaré, as well as
the support of the the CONICYT PIA 1415 Project (Geometry at the Frontier),
both during the final stage of this work. C. Rivas was also partially supported
by a CONICYT’s grant Inserción 79130017.

4

CONTENTS

To Anaı̈s, Raphaël,
Corita, Chı́o, Nachito
Adela, and Benito.

CONTENTS

5

Notation
Because of our dynamical approach, group elements will be often denoted by
the letters f, g, h (sometimes, by u, v, w as well). Nevertheless, a, b, c are used
when dealing with specific groups (free groups, fundamental groups of surfaces,
braid groups, etc). Following the classical notation, when dealing with braid
groups, we also use the letter σ to denote elements. Groups are generally denoted
by Γ, though G, H are sometimes used, as well as C (for convex subgroups), R (for
nilpotent radicals) and T (for Tararin groups). In many cases, we implicitly assume that the groups we are dealing with are nontrivial; for left-orderable groups,
this is equivalent to being infinite. Similarly, when we consider actions on the line,
we will implicitly assume that these are actions by orientation-preserving homeomorphisms. In general, real-valued function will be denoted by φ, ψ, whereas
group representations by Φ, Ψ. This notation is coherent in that certain functions
to be constructed will appear to be representations, that is, homomorphisms into
the additive group of reals.
Below, we list some notation used throughout this text:
hg1 , g2 , . . .i+ : the semigroup generated by g1 , g2 , . . .
Γ1 ⋆ Γ2 : the (non-Abelian) free product of Γ1 and Γ2 .
LO(Γ): the space of left-orders of Γ.
BO(Γ): the space of bi-orders of Γ.
CO(Γ): the space of Conradian orders of Γ.
C (Γ): the Conradian soul of a left-ordered group (Γ, ).
P+ : the positive cone of a left-order .
N := {1, 2, . . .}.
N0 := {0, 1, 2, . . .}.
(R, +): the group of reals under addition.
R∗ : the group of positive reals under multiplication.
Aff(R), Aff + (R): the group of affine homeomorphisms of the real line and the
subgroup of orientation-preserving ones, respectively.
]
PSL(2, R), PSL(2,
R): the group of orientation-preserving projective homeomorphisms of the circle and the group of the lifts to the real line, respectively.
Homeo+ (R): the group of orientation-preserving homeomorphisms of the line.
F: Thompson’s group of piecewise-affine, dyadic, orientation-preserving homeomorphisms of the interval.

6

CONTENTS

Fn : the free group on n generators (we will implicitly assume that n ≥ 2).
PAff + (R): the group of orientation-preserving homeomorphisms of the real line
that are piecewise affine.
PP+ (R): the group of orientation-preserving homeomorphisms of the real line
that are piecewise in PSL(2, R).
Bn : the braid group in n strands.
PBn : the pure braid group in n strands.
BS(1, ℓ): the Baumslag-Solitar group ha, b : bab−1 = aℓ i.
Gm,n : the torus-knot group ha, b : am = bn i.

Chapter 1
SOME BASIC AND NOT SO
BASIC FACTS
1.1

General Definitions

An order relation  on a group Γ is left-invariant (resp. right-invariant)
if for all g, h in Γ such that g  h, one has f g  f h (resp. gf  hf ) for all f ∈ Γ.
The relation is bi-invariant if it is simultaneously invariant by the left and by
the right. To simplify, we will use the term left-order (resp. right-order) for
referring to a left-invariant total order on a group. We will say that a group Γ is
left-orderable (resp. right-orderable, bi-orderable) if it admits a total order
which is invariant by the left (resp. by the right, simultaneously by the left and
right).
Example 1.1.1. Clearly, every subgroup of a left-orderable group is left-orderable.
More interestingly, an arbitrary product Γ of left-orderable groups Γλ is left-orderable.
(This also holds for bi-orderable groups with the very same proof.) Indeed, fixing an
total well-order on the set of indices Λ and a left-order λ on each Γλ , let  be the
associated lexicographic order. This means that (gλ ) ≺ (hλ ) if the smallest λ ∈ Λ such
that gλ 6= hλ satisfies gλ ≺λ hλ . It is easy to check that  is total and left-invariant.

1.1.1

Positive and negative cones

If  is an order on Γ, then f ∈ Γ is said to be positive (resp. negative) if
f ≻ id (resp. f ≺ id). Notice that if  is total, then every nontrivial element is
either positive or negative, and f ≻ id if and only if f −1 ≺ id. Moreover, if  is
left-invariant and P + = P+ (resp. P − = P− ) denotes the set of positive (resp.
7

8

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

negative) elements in Γ (usually called the positive (resp. negative) cone),
then P + and P − are semigroups, and Γ is the disjoint union of P + , P − and {id}.
Conversely, to every decomposition of Γ as a disjoint union of semigroups
P + , P − and {id} such that P − = (P + )−1 := {f : f −1 ∈ P + }, it corresponds a
left-order  defined by f ≺ g whenever f −1 g ∈ P + . Notice that Γ is bi-orderable
exactly when these semigroups may be taken invariant by conjugacy (that is,
when they are normal subsemigroups).
Remark 1.1.2. The characterization in terms of positive and negative cones shows
immediately the following: If  is a left-order on a group Γ, then the reverse order
¯ id if and only if g ≺ id is also left-invariant and total.
 defined by g ≻
Remark 1.1.3. Given a left-order  on a group Γ, we may define an order ∗ by
letting f ∗ g whenever f −1 ≻ g−1 . Then the order ∗ turns out to be right-nvariant.
One can certainly go the other way around, producing left-orders from right-orders. As
a consequence, a group is left-orderable if and only if it is right-orderable. Since our
view is mostly dynamical, we prefer to work with left-orders, yet most of the literature
on the subject is written for right-orders.
Remark 1.1.4. It is worth mentioning that f ≺ g for a left-order  does not imply
that f −1 ≻ g−1 . Actually, it is easy to check that this holds for all f, g as above if and
only if  is a bi-order.

1.1.2

A characterization involving finite subsets

A group Γ is left-orderable if and only if for every finite family G of non-identity
elements, there exists a choice of (compatible) exponents η : G → {−1, +1} such
that id does not belong to the semigroup generated by the elements g η(g) , g ∈ G.
Indeed, the necessity of the condition is clear: it suffices to fix a left-order 
on Γ and choose each exponent η(g) so that g η(g) becomes a positive element.
Conversely, assume that for each finite family G of elements in Γ different from
the identity there is a choice of compatible exponents η : G → {−1, +1}, and let
X (G, η) denote the (nonempty and closed) subset of {−1, +1}Γ\{id} formed by the
functions sign satisfying
sign(h) = +1 and sign(h−1 ) = −1

for every h ∈ g η(g) , g ∈ G

+

.

(Here and in what follows, given a family of group elements F , we let hF i+ be
the semigroup spanned by them.) We then let X (G) be the union of all the sets
of the form X (G, η) for some choice of compatible exponents η on G. Notice that,
if {Xi := X (Gi ), 1 ≤ i ≤ n} is a finite family of subsets of this form, then the

1.1. GENERAL DEFINITIONS

9

intersection X1 ∩ . . . ∩ Xn contains the (nonempty) set X (G1 ∪ . . . ∪ Gn ), and
it is therefore nonempty. Since {−, +}Γ\{id} is compact, a direct application of
the Finite Intersection Property shows that the intersection X of all sets of the
form X (G) is (closed and) nonempty. Finally, each point in X corresponds in an
abvious way to a left-order on Γ.
Analogously, one may show that a group is bi-orderable if and only if for
every finite family G of non-identity elements, there exists a choice of exponents
η : G → {−1, +1} such that id does not belong to the smallest semigroup which
simultaneously satisfies the next two properties:
– It contains all the elements g η(g) ;
– For all f, g in the semigroup, both f gf −1 and f −1 gf also belong to it.
We leave the proof to the reader. As a corollary, we obtain that left-orderability
and bi-orderability are local properties, that is, if they are satisfied by every
finitely-generated subgroup of a given group, then they are satisfied by the whole
group. Similarly, these are residual properties: if for every nontrivial element
there is a surjective group homomorphism into a group with that property mapping the prescribed element into a nontrivial one, then the group inherits the
property. (Notice that this follows more easily from Example 1.1.1.)

1.1.3

Left-orderable groups and actions on ordered spaces

If Γ is a left-orderable group, then Γ acts faithfully on a totally ordered space by
order-preserving transformations. Indeed, fixing a left-order  on Γ, we may consider the action of Γ by left-translations on the ordered space (Γ, ). Conversely,
if Γ acts on a totally ordered space (Ω, ≤) by order-preserving transformations,
then we may fix an arbitrary well-order 6wo on Ω and define a left-order  on Γ by
letting f ≻ id if and only if f (wf ) > wf , where wf = min6wo {w : f (w) 6= w}.
More generally, if we also have a function sign : Ω → {−, +}, we may associate to it the left-order  for which f ≻ id whenever sign(wf ) = + and
f (wf ) > wf , or sign(wf ) = − and f (wf ) < wf . These left-orders will be referred
to as dynamical-lexicographic ones.
Left-orders obtained from preorders. Recall that a preorder  on a group
Γ is a reflexive and transitive relation for which both f  g and g  f may hold
for different f, g. The existence of a total, left-invariant preorder is equivalent to
the existence of a semigroup P (containing the identity) such that P ∪ P −1 = Γ.
Indeed, having such a P , one may declare f  g if and only if f −1 g ∈ P .
Conversely, a preorder  as above yields the semigroup P := {g : g  id}.

10

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Using the dynamical characterization of left-orders we next show that, if Γ admits
sufficiently many total preorders so that different elements can be “distinguished”,
then it is left-orderable.
Proposition 1.1.5. Let Γ be a group and {Pλ , λ ∈ Λ} a family of subsemigroups
such that:
(i) Pλ ∪ Pλ−1 = Γ, for all λ T
∈ Λ;
(ii) The intersection P := λ∈Λ Pλ satisfies P ∩ P −1 = {id}.
Then Γ is left-orderable.

Proof. For each λ ∈ Λ, let Γλ = Pλ ∩ Pλ−1 . Fix a total order on the set of indices
Λ, and let Ω be the space of all cosets gΓλ, where g ∈ Γ and λ ∈ Λ. Define an
order ≤ on Ω by letting gΓλ ≤ hΓλ′ if either λ is smaller than λ′ , or λ = λ′ and
g −1 h ∈ Pλ (this does not depend on the chosen representatives g, h). By property
(i), this order is total. The group Γ acts on Ω by f (gΓλ) = f gΓλ. This action
preserves ≤. Moreover, ifTf acts trivially, then f lies in Γλ for all λ. Hence, by
property (ii) above, f ∈ λ∈Λ Pλ ∩ Pλ−1 = P ∩ P −1 = {id}. This shows that
the action is faithful, hence Γ is left-orderable.


Exercise 1.1.6. Let P := {g : g  id} be the semigroup of non-positive elements of a
left-invariant total preorder  on a group Γ. For each h ∈ Γ, let Ph := {h−1 gh : g ∈ P }.
(i) Show that T
each Ph induces a total preorder on Γ.
(ii) Let H := h∈Γ (Ph ∩ Ph−1 ). Show that H is a normal subgroup of Γ.
(iii) Show that Γ/H is a left-orderable quotient of Γ, which is nontrivial provided there
are at least two non-equivalent for .
Hint. Although everything can be directly checked, a dynamical view proceeds as follows: the quotient space Γ/ ∼ obtained by idendification of -equivalent points (i.e.
elements f, g such that f  g  f ) is totally ordered, and Γ acts on it by left-translations
preserving this order; the kernel of this action corresponds to the subgroup H.

The analogue of the preceding proposition for partial left-orders does not hold.
Indeed, in §1.4.1, we will see many examples of torsion-free groups that are not
left-orderable, yet the following holds:
Exercise 1.1.7. Show that a group is torsion-free if and only if it admits a family
{λ : λ ∈ Λ} of partial, left-invariant orders such that, for each f 6= g, there exists
λ ∈ Λ satisfying g ≺λ f .
Hint. If Γ acts (faithfully) on a set X and Y ⊂ X has trivial stabilizer, then one
may define a partial, left-invariant left-order  on Γ by letting h ≻ id if and only if
h(Y ) ⊂ Y . If Γ is torsion-free and f 6= g, then for X := Γ and Y := {h, h2 , . . .}, where
h := g−1 f , this procedure yields a partial left-order for which f ≻ g.

1.1. GENERAL DEFINITIONS

11

On group actions on the real line. For countable left-orderable groups, one
may take the real line as the ordered space on which the group acts. (The first
reference we found on this is [98]; see also [79].)
Proposition 1.1.8. Every left-orderable countable group acts faithfully on the
real line by orientation-preserving homeomorphisms.
Proof. Let Γ be a countable group admitting a left-invariant total order .
Choose a numbering (gi )i≥0 for the elements of Γ, put t(g0 ) = 0, and define
t(gk ) by induction in the following way: assuming that t(g0 ), . . . , t(gi ) have been
already defined, if gi+1 is bigger (resp. smaller) than g0 , . . . , gi then let t(gi+1 ) be
max{t(g0 ), . . . , t(gi )} + 1 (resp. min{t(g0 ), . . . , t(gi )} − 1), and if gm ≺ gi+1 ≺ gn
for some m, n in {0, . . . , i} and gj is not between gm and gn for any 0 ≤ j ≤ i
then put t(gi+1 ) := (t(gm ) + t(gn ))/2.
Notice that Γ acts naturally on t(Γ) by g(t(gi )) = t(ggi ). We leave to the
reader to check that this action extends continuously to the closure of the set
t(Γ). Finally, one can extend the action to the whole line by extending the maps
g affinely to each interval of the complement of the closure of t(Γ).

Remark 1.1.9. There is a subtle issue involving the choice of midpoints in the construction above to ensure continuity. Actually, many other choices do work, but not
arbitrary ones. The important property is that for each increasing sequence of elements
g1 ≺ g2 ≺ . . ., all of them smaller than a certain g so that any h ≺ g is eventually smaller
than some gn , we must have that t(gn ) converges to t(g).

It is worth analyzing the preceding proof carefully. If  is an left-order on a
countable group Γ and (gi )i≥0 is a numbering of the elements of Γ, then we will
call the (associated) dynamical realization the action of Γ on R constructed in
this proof. It is easy to see that this realization has no global fixed point unless Γ
is trivial. Moreover, if f is an element of Γ whose dynamical realization has two
fixed points a < b (which may be equal to ±∞) and has no fixed point in ]a, b[, then
there must be some point of the form t(g) inside ]a, b[. Finally, it is not difficult
to show that the dynamical realizations associated to different numberings of the
elements of Γ are all topologically conjugate.1 Therefore, we can speak of any
dynamical property of the dynamical realization without referring to a particular
numbering.
1

A group representation (action) Φ1 : Γ → Homeo+ (R) is topologically conjugate to Φ2
if there exists an orientation-preserving homeomorphism ϕ of the real line into itself such that
ϕ ◦ Φ1 (g) = Φ2 (g) ◦ ϕ, for all g ∈ Γ. Notice that conjugacy classes yield an equivalence relation;
see §1.1.4 for more on this.

12

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Remark 1.1.10. Throughout the text, if not stated otherwise, we will always assume
that t(id) = 0 in the construction of the dynamical realization.
Exercise 1.1.11. Show that for any dynamical realization of a left-order, a Gδ -dense
subset of points in the real line have a free orbit.
Remark 1.1.12. Notice that, to define a dynamical-lexicographic left-order on the
group Homeo+ (R) of orientation-preserving homeomorphisms of the real line, it is not
necessary to well-order all the points in R: it is enough to consider a well-order on
a dense set; in particular, a dense sequence suffices. Clearly, Homeo+ (R) admits uncountably many left-orders of this type. However, there are left-orders that do not arise
this way; see Example 2.2.2.
Remark 1.1.13. The group G+ (R, 0) of germs at the origin of orientation-preserving
homeomorphisms of the real line is left-orderable. Perhaps the easiest way to show
this is by using the characterization in terms of finite subsets above. Let ĝ1 , . . . , ĝk be
nontrivial elements in G+ (R, 0), and let g1 , . . . , gk be representatives of them. Take a
sequence (xn,1 ) of points converging to the origin in the line so that for each n at least
one of the gi ’s moves xn,1 . Passing to a subsequence if necessary, we may assume that,
for each i ∈ {1, . . . , k}, either gi (xn,1 ) > xn,1 for all n, or gi (xn,1 ) < xn,1 for all n, or
gi (xn,1 ) = xn,1 for all n. In the first case we let ηi := +1, and in the second case we
let ηi := −1. In the third case, ηi is still undefined. However, this may happen only for
k − 1 of the gi ’s above. For these elements, we may repeat the procedure by considering
another sequence (xn,2 ) converging to the origin... In at most k steps, all the ηi ’s will
be thus defined. We claim that this choice is compatible. Indeed, given an element
ηi
ηi
ηi
ηi
ĝ = ĝi1 1 · · · ĝiℓ ℓ , the choice above implies that gi1 1 · · · giℓ ℓ (xn,1 ) ≥ xn,1 for all n, where
the inequality is strict if some of the gij ’s moves some of (equivalently, all) the points
xn,1 . If this is the case, then this implies that ĝ cannot be the identity. If not, then we
may repeat the argument with the sequence (xn,2 ) instead of (xn,1 )... Proceeding this
way, we conclude that ĝ is nontrivial.
A nice consequence of the claim above is that every countable group of germs at the
origin of homeomorphisms of the real line admits a realization (but not necessarily an
“extension”!) as a group of homeomorphisms of the interval. Notice that, conversely,
Homeo+ ([0, 1]) embeds into G+ (R, 0). (This embedding is not obtained by looking
at the germs of elements of Homeo+ ([0, 1]) near the origin –the homomorphism thusobtained is not injective–, but by taking infinite copies of Homeo+ ([0, 1]) on intervals
accumulating the origin). Despite of this, the groups G+ (R, 0) and Homeo+ ([0, 1]) are
non-isomorphic; see [134]. Actually, there is no nontrivial homomorphism from G+ (R, 0)
into Homeo+ ([0, 1]). (See Example 2.2.24 for another –much simpler– example of an
uncountable left-orderable group that has no nontrivial action on the real line.)

We do not know whether there is an analogue of the preceding example in
higher dimensions.

1.1. GENERAL DEFINITIONS

13

Question 1.1.14. Does there exist a finitely-generated group of germs at the
origin of homeomorphisms of the plane having no realization as a group of homeomorphisms of the plane ?
Notice that the results and techniques of [33] show that such a group cannot
arise as a group of germs of C 1 diffeomorphisms. We should point out, however,
that imposing regularity conditions for a group action may lead to very serious
algebraic restrictions; see [150] for a general panorama on this topic (see also
[16, 32, 152], as well as [35] for examples of a related nature).
Let us close this discussion with a nice open question raised by Calegari [31].
Question 1.1.15. Is the group of orientation-preserving homeomorphisms of the
disc that are the identity on the boundary left-orderable ?

1.1.4

Semiconjugacy in Homeo+(R)

Two (non necessarily injective) group representations Φ1 , Φ2 : Γ → Homeo+ (R)
will be said to be semiconjugate if there is a non-decreasing map ϕ : R → R
which is proper (i.e. the preimage of every compact set is bounded or, equivalently -since ϕ is monotone-, ϕ(R) is unbounded in both directions), and such
that for all g ∈ Γ,
ϕ ◦ Φ1 (g) = Φ2 (g) ◦ ϕ.
(1.1)
In most of the literature, in the definition above, one also requires the continuity of the map ϕ. However, this extra condition causes more problems than those
it solves. For instance, if we insist on continuity, then actions without global fixed
points admitting a discrete minimal invariant set may be non semiconjugate to
a Z-action by translations. However, with our definition of semiconjugacy, these
actions are always semiconjugate. More importantly, (if we drop the continuity
assumption, then) we have
Proposition 1.1.16. Semiconjugacy is an equivalence relation.
Proof. Reflexivity is obvious and transitivity is easy to check. Below we prove
symmetry.
Suppose (1.1) holds. Since ϕ is proper, we may define ψ(x) := sup ϕ−1 (x) ∈ R.
Since ϕ is monotone, we have
ψ(x) = sup ϕ−1 ((−∞, x]) = sup{y | ϕ(y) ≤ x}.

14

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

From the last equality, the fact that ψ is non-decreasing is obvious. Further,
properness of ψ easily follows from the properness of ϕ. Finally, as
Φ1 (g)(ψ(x)) =
=
=
=
=

sup{Φ1 (g)(y) | ϕ(y) ≤ ψ(x)}
sup{z | ϕ(Φ1 (g)−1(z)) ≤ ψ(x)}
sup{z | Φ2 (g)−1(ϕ(z)) ≤ x}
sup{z | ϕ(z) ≤ Φ2 (g)(x)}
ψ(Φ2 (g)(x))

holds for all g ∈ Γ and all x ∈ R, we see that ψ satisfies the semiconjugacy
relation.

Exercise 1.1.17. Let Γ be a countable group of orientation-preserving homeomorphisms of the real line. Using this action, produce a dynamical-lexicographic order 
on Γ. Show that the original action is semiconjugate to the dynamical realization of .
Give examples where this semiconjugacy is not a conjugacy.
Exercise 1.1.18. Let (Γ, ) be a countable left-ordered group and Γ0 a subgroup.
Show that the restriction to Γ0 of the dynamical realization of  is semiconjugate to
the dynamical realization of the restriction of  to Γ0 .
Remark 1.1.19. Since our definition of semiconjugacy still involves non injective representations, it applies to actions of different groups, provided these actions factor
throughout the action of the same group.

1.2

Some Relevant Examples

At first glance, it appears as surprising that many (classes of) torsion-free
groups turn out to be left-orderable. Here we give a brief discussion on some of
them.

1.2.1

Abelian and nilpotent groups

The simplest bi-orderable groups are the torsion-free, Abelian ones. Clearly,
there are only two bi-orders on Z. The case of Z2 is more interesting. According
to [175, 181, 188], there are two different types of bi-orders on Z2 . Bi-orders of
irrational type are completely determined by an irrational number λ: for such
an order λ an element (m, n) is positive if and only if λm + n is a positive
real number. Bi-orders of rational type are characterized by two data, namely a

15

1.2. SOME RELEVANT EXAMPLES

pair (x, y) ∈ Q2 up to multiplication by a positive real number, and the choice of
one of the two possible bi-orders on the subgroup {(m, n) : mx + ny = 0} ∼ Z.
Thus, an element (m, n) ∈ Z2 is positive if and only if either mx + ny is a
positive real number, or mx + ny = 0 and (m, n) is positive with respect to the
chosen bi-order on the kernel line (isomorphic to Z). The set of left-orders on Z2
naturally identifies to the Cantor set (see §2.2 for more on this).
The description of all bi-orders on Zn for larger n continues inductively. (A
good exercise is to show all of this using the results of §3.2.3.) For a general
torsion-free, Abelian group, recall that the rank is the minimal dimension of
a vector space over Q in which the group embeds. The reader should have no
problem to show in particular that a torsion-free, Abelian group of rank ≥ 2
admits uncountably many left-orders.

id •
• • •

..
........
.......
........
.
.
.
.
.
.
.
...
........
........
........
.
.
.
.
.
.
.
..
........
........
........
.
.
.
.
.
.
.
........

•
•

•

•

•

•

•

•

......
.......
........
.
.
.
.
.
.
...
........
........
.......
.
.
.
.
.
.
.
..
........
.......
........
.
.
.
.
.
.
.
......
........

•

•

•

•

•

•

•

•

•

•

•

•

•

•
•
•

•

•
•
•

•

•
•
•
•
•
•
•

Figure 1: The positive cone of a left-order on Z2 .
Torsion-free, nilpotent groups are also bi-orderable. Indeed, let Γi denote
the ith -term of the lower central series of a group Γ (that is, Γ1 := Γ and
Γi+1 := [Γ, Γi ]), and let Hi (Γ) be the isolator of Γi defined by

Hi (Γ) := g ∈ Γ : g n ∈ Γi for some n ∈ N .

If Γ is nilpotent (i.e. if Γk+1 = {id} for a certain k), then each Hi (Γ) is a
normal subgroup of Γ, and Hi (Γ)/Hi+1 (Γ) is a torsion-free, central subgroup of
Γ/Hi+1 (Γ) (see [113] for the details). Notice that, if Γ is also torsion-free, then
Hk+1 (Γ) = {id}.

16

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Let Pi be the positive cone of any left-order on (the torsion-free Abelian group)
Hi (Γ)/Hi+1 (Γ), and let Gi be the set of elements in Hi (Γ) which project to an
element in Pi when taking the quotient by Hi+1 (Γ). Using the fact that each
Hi (Γ)/Hi+1 (Γ) is central in Γ/Hi+1 (Γ), one may easily check that the semigroup
P := Gk−1 ∪ Gk−2 ∪ . . . ∪ G1 is the positive cone of a bi-order on Γ.
Example 1.2.1. The Heisenberg group
H = f, g, h : [f, g] = h−1 , [f, h] = id, [g, h] = id
is a non-Abelian nilpotent group of nilpotence degree 2. It may also
group of lower triangular matrices with integer entries so that each
equals 1 via the identifications





1 0 0
1 0 0
1 0 0





f=
1 1 0
, g=
0 1 0
, h=
0 1 0
0 0 0
0 1 1
1 0 1

be seen as the
diagonal entry


.

Notice that the linear action of H on Z3 fixes the hyperplane {1} × Z2 and preserves
the lexicographic order on it. The left-orders on H induced from this restricted action
(c.f. §1.1.3) are (total but) not bi-invariant. This example can be seen as a kind of
evidence of the following nice result due to Darnel, Glass, and Rhemtulla [54]: If all
the left-orders of a left-orderable group are bi-invariant, then the group is Abelian.
Both the space of left-orders and bi-orders of countable, torsion-free, nilpotent
groups which are not rank-1 Abelian are homeomorphic to the Cantor set; see Theorem
3.2.20 for the former and [143] for the latter. Moreover, there is the next remarkable
theorem of Malcev [132] (resp. Rhemtulla [17, Chapter 7]): Every bi-invariant (resp.
left-invariant) partial order on a torsion-free, nilpotent group can be extended to a
bi-invariant (resp. left-invariant) total order.
Exercise 1.2.2. Let H̄ be the subgroup of the Heisenberg group formed by the matrices
of the form


1 0 0
 2x 1 0  , x, y, z in Z.
z 2y 1

(i) Show that the commutator subgroup

1 0
 0 1
4z 0

is formed by the matrices of the form

0
0  , z ∈ Z.
1

(ii) Conclude that H̄/H̄ ′ is isomorphic to Z2 × Z/4Z, hence has torsion.

1.2. SOME RELEVANT EXAMPLES

17

Exercise
1.2.3. Show that a group Γ is residually torsion-free nilpotent if and only
T
if
(Since bi-orderability is a residual property, such a group is
i Hi (Γ) = {id}.
necessarily bi-orderable.)
Remark. Quite surprisingly, torsion-free, residually nilpotent groups do not necessarily
satisfy this property. Actually, such a group may fail to be bi-orderable; see [10].

1.2.2

Subgroups of the affine group

Let Aff + (R) denote the group of orientation-preserving affine homeomorphisms of the real line (the affine group, for short). For each ε 6= 0, a partial
order ε may be defined by declaring that f is positive if and only if f (1/ε) > 1/ε.
This means that


o
n
u v
+
: u + vε > 1 .
Pε = f =
0 1

These orders were introduced (in a more algebraic way) by Smirnov in [182].
For a finitely-generated subgroup Γ of Aff + (R), the corresponding action on
the line has free orbits. Thus, one may choose ε so that ε is a total order. In the
general case, one may “complete” the partial order into a total one (c.f. §1.1.3).
As a consequence, non-Abelian subgroups of Aff + (R) admit uncountably many
left-orders.
As a concrete and relevant example, for each integer ℓ ≥ 2, the BaumslagSolitar group BS(1, ℓ) = hg, h : hgh−1 = g ℓ i embeds into the affine group by
identifying g and h to x 7→ x + 1 and x 7→ ℓx, respectively. (See Exercise 1.2.4
below.) Notice that, for an irrational ε 6= 0, the associate order ε is total. If one
chooses a rational ε, then it may happen that ε is only a partial order. However,
in this case, the stabilizer of the point 1/ε is isomorphic to Z, and thus ε can be
completed to a total left-order of BS(1, ℓ) in exactly two different ways. Notice
that the reverse orders ε may be retrieved by the same procedure but starting
with the embedding g : x 7→ x − 1 and h : x 7→ ℓx, and changing ε by −ε.
There is still another way to order BS(1, ℓ). Namely, BS(1, ℓ) can be thought
of as the semidirect product Z[ 1ℓ ] ⋊ Z coming from the exact sequence
h1i
−→ BS(1, ℓ) −→ Z −→ 0.
0 −→ Z
ℓ

Using this, one may define the bi-orders , ′ by letting ( ℓmn , k) ≻ id (resp.
( ℓmn , k) ≻′ id) if and only if either k > 0, or k = 0 and ℓmn > 0 (resp. k > 0, or
′
k = 0 and ℓmn < 0). Together with the reverse orders  and  , this completes
the list of all bi-orders on BS(1, ℓ) (see Example 3.2.55).

18

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

A similar description applies to all non-Abelian subgroups of Aff + (R). In
the terminology of §2.2, the associated spaces of left-orders are homeomorphic to
the Cantor set. The study of more general solvable left-orderable groups is more
involved, yet it crucially relies on the case of affine groups. We will come back to
this point in §3.3.1 and §3.3.2.
Exercise 1.2.4. Prove that the map from BS(1, ℓ) = hg, h : hgh−1 = g ℓ i into the
affine group that makes correspond g and h to x 7→ x + 1 and x 7→ ℓx, respectively, is
an embedding.
Hint. Prove that the conjugates of g commute, and then write every element of BS(1, ℓ)
in normal form as a power of h followed by products of conjugates of g.
Exercise 1.2.5. Show that every embedding of BS(1, ℓ) into Aff + (R) is obtained by
letting g, h correspond, respectively, to any nontrivial translation and an homothethy
of ratio ℓ.

1.2.3

Free and residually free groups

The free group F2 (hence every non-Abelian free group) is bi-orderable. Although this result is originally due to Shimbireva [180], it is sometimes attributed
to Vinogradov, and in general to Magnus, as it arises as a consequence of his famous “expansions”. Below we sketch Magnus’ construction, which covers Shimbivera approach; a variation of Vinogradov’s original proof (which actually applies
to free products of arbitrary bi-orderable groups) will be treated in §2.1.2.
Consider the (non-Abelian) ring A = ZhX, Y i formed by the formal power
series with integer coefficients in two independent variables X, Y . Denoting by
o(k) the subset of A formed by the elements all of whose terms have degree at
least k, one easily checks that

F := 1 + o(1) = 1 + S : S ∈ o(1)

is a subgroup (under multiplication) of A. Moreover, if f, g are (free) generators
of F2 , the map Φ sending f (resp. g) to the element 1 + X (resp. 1 + Y ) in A
extends in a unique way into an injective homomorphism Φ : F2 → F. Now fix
a lexicographic type order relation on F that is bi-invariant under multiplication
by elements in F. (Notice that this order is not invariant under multiplication
by certain elements in A.) Using this order and the homomorphism φ, the free
group F2 may be easily endowed with a bi-invariant left-order.
The above technique –called the Magnus expansion– actually shows that
F2 is residually torsion-free nilpotent. Indeed, it is easy to check that Φ(Γnil
i ) is

19

1.2. SOME RELEVANT EXAMPLES

contained in 1 + o(i + 1), for every i ≥ 0. We refer to [131] for more details on
all of this.
Surface groups. Surface groups are residually free, hence bi-orderable. Actually, as we show below, these groups are fully residually free, which is a stronger
property (see Remark 1.2.6 below). Recall that, if P is some group property, then
a group Γ is said to be fully residually P if for every finite subset G ⊂ Γ \ {id},
there exists a surjective group homomorphism from Γ into a group ΓG satisfying
P such that the image of every g ∈ G is nontrivial. Equivalently, for every finite
subset G ⊂ Γ, there is an homomorphism Φ into a group satisfying P whose
restriction to G is injective.
Remark 1.2.6. Obviously, the direct product F2 × F2 is residually free. However,
it is not fully residually free, because given any f, g, h in F2 , no homomorphism from
F2 × F2 to a free group maps the elements id, (f, id), (g, id), ([f, g], id), and (id, h), into
five different ones. Indeed, as (id, h) commutes with (f, id) and (g, id), a separating
homomorphism should send these three elements into some cyclic subgroup. However,
if this is the case, then ([f, g], id) is mapped to the identity.

Below we deal with the case of surfaces of even genus (the case of odd genus
easily follows from this). The following lemma, due to Baumslag, will be crucial
for us. The geometric proof we give appears in [6].
Lemma 1.2.7. Let g1 , . . . , gk be elements in a free group Fn , and let f be another
element which does not commute with any of them. Then there exists N ∈ N such
that, for every |ni | ≥ N, m ∈ N, and ji ∈ {1, . . . , k},
gj1 f n1 gj2 f n2 . . . gjm f nm 6= id.
Proof. Let a− , a+ be the endpoints (in the boundary at infinity ∂Fn ) of the axis
axis(f ) determined by f in the Cayley graph of Fn . Since gi does not commute
with f , one must have {gi (a− ), gi (a+ )} ∩ {a− , a+ } = ∅, for every i ∈ {1, . . . , k}.
Let U − , U + be neighborhoods in ∂Fn of a− and a+ , respectively, satisfying
gi (U − ∪ U + ) ∩ (U − ∪ U + ) = ∅

for each i.

There exists N ∈ N such that, for all r ≥ N,
f r (∂Fn \ U − ) ⊂ U + ,

f −r (∂Fn \ U + ) ⊂ U − .

A ping-pong type argument (see [93]) then shows the lemma.



20

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Let Γ = Γ2n be the π1 of an orientable surface S2n of genus 2n (n ≥ 1). Let
us consider the standard presentation
Γ = gi , gi′ , hi , h′i , 1 ≤ i ≤ n : [g1 , g1′ ] · · · [gn , gn′ ] · [h′n , hn ] · · · [h′1 , h1 ] = id .
Following [21], let σ be the automorphism of Γ that leaves the gi ’s and gi′ ’s
fixed while sending every hi to f hi f −1 and every h′i to f h′i f −1 , where f :=
[g1 , g1′ ] · · · [gn , gn′ ]. (Geometrically, this corresponds to the Dehn twist along the
closed curve obtained from a simple curve that joins the first and 2nth vertices
of the hyperbolic 4n-gon that yields S2n .) Finally, let ϕ be the surjective homomorphism from Γ to the free group F2n with free generators a1 , . . . , an , a′1 , . . . , a′n
defined by ϕ(gi ) = ϕ(hi ) = ai and ϕ(gi′ ) = ϕ(h′i ) = a′i . We claim that the sequence of homomorphisms ϕ ◦ σ k is eventually faithful, in the sense that given
f1 , . . . , fm in Γ, there exists N ∈ N so that for all k ≥ N, the image under ϕ ◦ σ k
of any of the fi ’s is nontrivial (thus showing that Γ is fully residually free).
To show the claim above, given g ∈ Γ \ {id}, let us write it in the form
g = w1 (gi , gi′ ) · w2 (hi , h′i ) · · · w2p−1 (gi , gi′ ) · w2p (hi , h′i ),
where each wi is a reduced word in 2n letters (the first and/or the last wi may
be trivial). Up to modifying the w2j−1 ’s, we may assume that each w2j (where
1 ≤ j ≤ p) is such that w2j (hi , h′i ) is not a power of f . Notice that the centralizer
of f in Γ is the cyclic group generated by f . By regrouping several wj ’s into a
longer word if necessary, unless g itself is a power of f , we may also assume that
w2j−1 (gi , gi′ ) is not a power of f . Let f¯ be the image of f under ϕ. We have
ϕ ◦ σ k (g) = w1 f¯k w2 f¯k · · · w2p−1 f¯k w2p c̄k ,
where wj = wj (ai , a′i ). Since f¯ does not commute with any of the wj ’s, Lemma
1.2.7 implies that ϕ ◦ σ k (g) is nontrivial.

1.2.4

Thompson’s group F

Thompson’s group F is perhaps the simplest example of a bi-orderable group
that is not residually nilpotent. Recall that this is the group of the orientationpreserving, piecewise-affine homeomorphisms f of the interval [0, 1] such that:
– The derivative of f on each linearity interval is an integer power of 2;
– f induces a bijection of the set of dyadic rational numbers in [0, 1].
This group is not residually nilpotent because its commutator subgroup F’ is
simple (see [38]). To see that it is bi-orderable, for each nontrivial f ∈ F we

1.2. SOME RELEVANT EXAMPLES

21

+
−
+
denote by x−
f (resp. xf ) the leftmost point x (resp. the rightmost point x )
for which Df+ (x− ) 6= 1 (resp. Df− (x+ ) 6= 1), where Df+ and Df− stand for the
corresponding lateral derivatives. One can immediately visualize four different
−
+
bi-orders on (each subgroup of) F, namely the bi-order +
x− (resp. x− , x+ ,
−
−
−
x+ ) for which f is positive if and only if Df+ (xf ) > 1 (resp. Df+ (xf ) < 1,
+
Df− (x+
f ) < 1, Df− (xf ) > 1). Although F admits many more bi-orders than
these (see theorem 1.2.9 below), the case of F′ is quite different. The result below
is essentially due to Dlab [65] (see also [157]).
−
+
−
Theorem 1.2.8. The only bi-orders on F′ are +
x− , x− , x+ and x+ .

Remark that there are four other “exotic” bi-orders on F, namely:
−
– The bi-order +,−
0,x− for which f is positive if and only if either xf = 0 and
−
Df+ (0) > 1, or x−
f 6= 0 and Df+ (xf ) < 1;

−
– The bi-order −,+
0,x− for which f is positive if and only if either xf = 0 and
Df+ (0) < 1, or x−
6 0 and Df+ (x−
f =
f ) > 1;

+
– The bi-order +,−
1,x+ for which f is positive if and only if either xf = 1 and
Df+ (1) < 1, or x+
6 1 and Df− (x+
f =
f ) > 1;

+
– The bi-order −,+
1,x+ for which f is positive if and only if either xf = 1 and
Df+ (1) > 1, or x+
6 1 and Df− (x+
f =
f ) < 1.

−,+
+,−
Notice that, when restricted to F′ , the bi-order +,−
0,x− (resp. 0,x− , 1,x+ , and
−
+
−
+
−,+
1,x+ ) coincides with x− (resp. x− , x+ , and x+ ). Let us denote the set of
the previous eight bi-orders on F by BO Isol (F).

There is another natural procedure to create bi-orders on F. For this, recall
the well-known (and easy to check) fact that F′ coincides with the subgroup of F
formed by the elements f satisfying Df+ (0) = Df− (1) = 1. Now let Z2 be any
bi-order on Z2 , and let F′ be any bi-order on F′ . It readily follows from Dlab’s
theorem that F′ is invariant under conjugacy by elements in F. Hence, one may
define a bi-order  on F by declaring
that
/ F′ and
 f ≻ id if′ and only if either f ∈

log2 (Df+ (0)), log2 (Df− (1)) ≻Z2 0, 0 , or f ∈ F and f ≻F′ id (see §2.1.1 for
more details on this type of construction).
All possible ways of left-ordering finite-rank, Abelian groups were described
in §1.2.1. Since there are only four possibilities for F′ , the preceding procedure
−
+
−
gives us four sets (which we will coherently denote by Λ+
x− , Λx− , Λx+ , and Λx+ )
naturally homeomorphic to the Cantor set (in the sense of §2.2) inside the set of
bi-orders of F. The main result of [157] establishes that these bi-orders, together

22

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

with the eight special bi-orders previously introduced, are all possible bi-orders
on F. The proof is an elementary application of Conrad’s theory (c.f. §3.2.1.)
Theorem 1.2.9. The set of all bi-orders of F consists of the disjoint union of
−
+
−
BO Isol (F) and the sets Λ+
x− , Λx− , Λx+ , and Λx+ .
Thompson’s group F is remarkable on many aspects. Among its most relevant
properties, we can mention that it is finitely presented (this is well explained in
[38]; see also Exercise 1.2.10 below for an sketch of proof) and contains no free
subgroup (this was first proved in [24]; see the next section for a proof of a slightly
generalized version of this fact). At the time of this writing, the challenging
question of the amenability of this group remains open.
Exercise 1.2.10. The goal of this exercise is to provide the main steps of proof of the
next two presentations of Thompson’s group F:
F1 = a, b : [a−1 b, aba−1 ] = [a−1 b, a2 ba−2 ] = id ,
F2 = c0 , c1 , c2 , . . . : ck cn c−1
k = cn+1 for all k < n .
Here, a and b are in correspondence to c0 and c1 , respectively, and correspond to the
elements f0 , f1 in F whose graphs are drawn below.
.........
..........
..........
.
.
.
.
.
.
.
.
.
.....
..........
..........
....
.
.
.
f0.........
...
....
.
...
...
.
..
..
....
.
..
..
...
.
..

..
..........
..........
..........
.
.
.
....
....
...
...
.
..
..
.
.
.
....
....
....
.
.
.
f1......
....
....
.
.
.
.
.
.
.
.....
....
....

Figure 2: The graphs of f0 and f1 .
(i) Let Φ : F1 → F2 be the map sending a to c0 and b to c1 . Show that Φ extends to a
group homomorphism.
Hint. What is to be checked is that the following relations are satisfied in F2 :
−1
−1
2
−2
[c−1
0 c1 , c0 c1 c0 ] = [c0 c1 , c0 c1 c0 ] = id.

To do this, just notice that from
−1
c0 c2 c−1
0 = c3 = c1 c2 c1

−1
(resp. c0 c3 c−1
0 = c4 = c1 c3 c1 ),

we get that c−1
0 c1 conmutes with c2 (resp. c3 ).

23

1.2. SOME RELEVANT EXAMPLES

(ii) Let Φ̂ : F2 → F1 be the map sending c0 to a and c1 to b. Show that Φ̂ extends to a
group homomorphism (hence, by (i), to a group isomorphism, with Φ̂ = Φ−1 ).
Hint. Set a0 := a and an := an−1 ba−(n−1) for n ≥ 1. The task is to show that
ak an a−1
k = an+1

for all k < n.

(1.2)

One can show this by simultaneously proving that
[ba−1 , aj ] = id for all j ≥ 3.

(1.3)

First notice that the second condition above holds for j = 3 and j = 4, since
[a−1 b, aba−1 ] = id =⇒ [ba−1 , a2 ba−2 ] = id =⇒ [ba−1 , a3 ] = id
and
[a−1 b, a2 ba−2 ] = id =⇒ [ba−1 , a3 ba−3 ] = id =⇒ [ba−1 , a4 ] = id.
Assume that (1.2) holds for k ≤ n ≤ k + i − 3 and that (1.3) holds for 3 ≤ j ≤ i. Then
ba−1 commutes with both a3 and ai , hence with ai+1 = a3 ai a−1
3 , and therefore (1.3)
holds for j = i + 1. Moreover,
an+1 ak = an ba−n ak−1 ba−(k−1) = ak−1 an−k+1 ba−(n−k+1) ba−1 a−k+2
= ak−1 an−k+2 (ba−1 )a−(k+2) = ak−1 (ba−1 )an−k+2 a−(k+2)
= (ak−1 ba−(k−1) )(ak−2 an−k+2 a−(k−2) ) = ak an ,
and therefore (1.2) holds for n = k + i− 2. The proof can be completed via an induction
argument.
(iii) Every element f ∈ F can be identified in an obvious way with a map that sends in
an ordered way the terminal points (called leaves) of a dyadic rooted tree into those of
another dyadic rooted tree having the same number of leaves, and conversely. In this
view, elements f0 and f1 correspond to the following diagrams:
.......
..... ........
....
.....
.
.
.
.
....
... ...
... .....
..
.

f0
−→

.......
..... ........
....
.....
.
.
.
..
....
... ...
... ....
..
..
..
.. ..
... .....
.. .

.......
..... ........
....
.....
.
.
.
..
....
... ...
... ....
..
..

f1
−→

.......
..... ........
....
.....
.
.
.
..
....
... ...
... ....
...
..
.
.....
... .....
.. .

Figure 3: The diagrams of the elements f0 and f1 .
−(n−1)

Given n ≥ 2, let fn := f0n−1 f1 f0
fn is

. Check that the tree diagram associated to

24

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS
.....
.... .....
.... ........ ........
.
.
.
..... ...
....
.....
....
.
n.
.... .........
.
.
.
.
..... ........
.
.
.
.....
.
....
......
.
.... ......... ......
....
.
.
.
.
.
.
.....
...
.
.... .........
....
.....
....
...
........
. .
... ....
..
..

fn

.........................................................

.....
.... .....
.... ........ ........
.
.
.
..... ...
....
.....
....
.
n.
.... .........
.
.
.
.
..... ........
.
.
.
.....
.
....
......
.
.... ......... ......
....
.
.
.
.
.
.
.....
...
.
.. ...
.... ........
.....
....
.....
...
.. ...
... .....
..
..

−(n−1)

Figure 4: The diagram of the element fn := f0n−1 f1 f0

.

(the right hand side tree above will be denoted by Tn ). Notice that, although the
diagram representing an element f ∈ F is not unique, there is a unique reduced one,
in the sense that
V any other representative diagram can be obtained from this just by
adding carets ( ) at the leaves of the source tree and the image leaves of the target
tree. (We keep right-to-left notation for group multiplication, which is opposite to most
of the literature on the subject, including [38].)
(iv) Show that f0 , f1 , f2 , f3 , . . . (hence f0 , f1 ) generate F. To do this, show that every
element f ∈ F may be written in the form
f0−r0 f1−r1 · · · fn−rn fnsn · · · f1s1 f0s0 ,

where ri ≥ 0, si ≥ 0. Besides, for a nontrivial element, such a writing is unique when
respecting the next two properties: exactly one of rn , sn is zero, and if rk , sk are both
positive for a certain k < n, then at least one of rk+1 , sk+1 is positive.
Hint. Let f ∈ F be an element represented by a tree diagram in which the target tree
is Tn . For each leaf vi of the source tree (which are numbered starting from 0), let si
be the length of the largest path along (unit) left branches that starts at vi and does
sn+2 sn+1
not touch the right side of the tree. Show that f = fn+2
fn+1 · · · f1s1 f0s0 . (Notice that
sn+1 = sn+2 = 0.) See the figure below for an example.
......
.... ........
.....
....
....
.....
.
.
....
....
.. ..
... .....
... .....
.. ..
..
..
..
.
... .
v0 v1 ...... ....... v5
. ..
.....
.. ...
... .... v4
. .

f22 f0

............................................................................

v2 v3

↓ ↓ ↓ ↓↓ ↓

..........
.... ........
....
.....
......
...
.... ....
.
.
. . .........
.
.
.
.....
..
....
........
.... ........
.
.
.
.
.....
.
.
.
.
....
.......
.... ........
.
.
.
.
.....
.
.
.
.....
....
.. ..
.. ...
... ....
..
..

1 0 2 00 0
Figure 5: The map f22 f0 as predicted by the claim above.
(v) Show that F is isomorphic to both F1 and F2 .

25

1.2. SOME RELEVANT EXAMPLES

Hint. Show that the map a 7→ f0 and b 7→ f1 extends to a surjective group homomorphism Φ1 : F1 → F, hence to a surjective group homomorphism Φ2 : F2 → F. To show
that the latter is injective, use the relations
−1
cn c−1
k = ck cn+1 ,

−1
ck c−1
n = cn+1 ck ,

ck cn = cn+1 ck ,

for n > k,

to transform an arbitrary expression in the ci ’s into one where only negative exponents
appear on the left, only positive exponents appear on the right, and the subindices are
ordered, say
s1 s0
0 −r1
n sn
c−r
· · · c−r
n cn · · · c1 c0 .
0 c1
Besides, if both rk , sk are positive and both rk+1 , sk+1 are zero for a certain k, then
using the relation c−1
k cn+1 ck = cn for n > k, one may decrease the subindex of each
k
and
cskk . Proceeding this way as much as possible, we get either the
entry between c−r
k
empty word or an expression as in (iv) above, which was shown to correspond (under
Φ1 ) to a nontrivial element of F.
Exercise 1.2.11. Let us consider the (binary) Cantor set {0, 1}N endowed with the
product topology.
(i) Let F̂ be the group of homeomorphisms of {0, 1}N generated by the maps

ξ
if ξ = 0η,





0η
if ξ = 00η,








if ξ = 100η,

 10η
â(ξ) :=
10η
if ξ = 01η,
and
b̂(ξ) :=




if ξ = 101η,



 110η




11η
if ξ = 1η,


111η
if ξ = 11η.

Show that F̂ is isomorphic to F.
Hint. Notice that â and b̂, respectively, may be represented by the same tree diagrams
of the elements a ∼ f0 and b ∼ f1 of F1 ∼ F.
(ii) Let φ2 : {0, 1}N → [0, 1] be defined by
φ2 (ξ) =

X ij
,
2j
j≥1

where ξ = (i1 , i2 , . . .), ij ∈ {0, 1}.

Check that φ2 is one-to-one except at points that correspond to dyadic rational numbers. Besides, show that φ2 semiconjugates the action of F̂ on {0, 1}N to that of F on
[0, 1], in the sense that
f (φ2 (ξ)) = φ2 (fˆ(ξ))
holds for each f ∈ F and all ξ ∈ {0, 1}N , where fˆ∈ F̂ denotes the element corresponding
to f .

26

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Exercise 1.2.12. Given a finite binary sequence s, we let âs be the map that consists
of the action of â localized at the subtree starting at the terminal vertex of the path s.
In precise terms,

if ξ = sη,
 sâ(η)
âs (ξ) :=

ξ
otherwise.

Notice that â1 = b̂.
(i) Prove that all elements âs ∈ F are conjugate to one of â, â0 , â1 , â10 , and that none
2
of these elements is conjugate to another one in this list. (Notice that â0 = â−1 â−1
1 â .)
(ii) For each pair of finite binary sequences s, t, we let ât (s) be the image of s under
ât in case this is defined, which happens either when s and t are incompatible (which
means that none of them extends the other one, in which case ât (s) = s) or when s
starts with t00, t01 or t1. Show that F, viewed as a group generated by the elements
âs , admits the presentation
âs : ât âs â−1
t = aât (s) for all s, t such that ât (s) is defined .
Hint. First check that all these relations are satisfied in F. Moreover, by identifying
F ∼ F2 , notice that the presentation above contains that of F2 , as c0 identifies with â
and ck to â1k for k ≥ 1 (where 1k stands for a 1 repeated k times), and â1k (1n ) = 1n+1
for all k < n.

1.2.5

Some relatives of F

The group of piecewise real-analytic, orientation-preserving diffeomorphisms
of the interval is bi-orderable. (Notice that this group contains F.) Indeed, we
may let f to be positive if and only if the point x−
f := inf{x : f (x) 6= x} is such
that f (y) > y for every y > x sufficiently close to x−
f . Restricted to F, this
+
−
+
bi-order coincides with x− . Extensions of x− , x+ and −
x+ can be defined
in an analogous way. Similarly, groups of piecewise real-analytic, orientationpreserving diffeomorphisms of the real line that behave nicely close to infinity are
also bi-orderable.
The group of piecewise analytic diffeomorphisms contains a remarkable subgroup, namely that of piecewise projective diffeomorphisms. Recall that the
projective line P1 (R) is the set of lines of R2 passing throughout the origin. It
identifies with R ∪ {∞}, where a real y ∈ R corresponds to the line R(1, y), and
the point ∞ to the line R(0, 1). The group PGL(2, R) acts on P1 (R): via the
identification P1 (R) ≃ R ∪ {∞}, this action is given by


ay + b
a b
y=
.
c d
cy + d

1.2. SOME RELEVANT EXAMPLES

27

Notice that this action factors throughout PSL(2, R).
Let PP+ (R) be the subgroup of Homeo+ (R) consisting of the homeomorphisms
that coincide with a projective map on each piece of a subdivision of R into
finitely many intervals. This group contains the group PAff + (R) of piecewise
affine homeomorphisms, which itself contains F. The next classical theorem was
established by Brin and Squier [24] for the group PAff + (R). The extension to
PP+ (R) appears in the work of Monod [144].
Theorem 1.2.13. The group PP+ (R) does not contain any non abelian free
subgroup.
Proof. For each f ∈ PP+ (R), denote by suppo (f ) the open support of f , that
is, the set of points x such that f (x) 6= x. This is a finite union of disjoint
open intervals. Given g, h in PP+ (R), the union suppo(g) ∪ suppo (h) is also a
finite number of disjoint open intervals I1 , . . . , In . We claim that the following
property holds: If [a, b] is a compact interval contained in one of these intervals
Ik , with 1 ≤ k ≤ n, then there exists a word w in g and h for which w([a, b]) is
disjoint from [a, b] (observe that w([a, b]) is still contained in Ik ). Otherwise, the
supremum of the orbit of a under hg, hi would be ≤ b, which is absurd due to the
definition of open supports.
Assume for a contradiction that there exist two elements g, h in PP+ (R) such
that any reduced word in g and h is nontrivial. Observe that the map f0 :=
[[g, h], [g 2, h]] is the identity close to the endpoints of each Ik . Indeed, if we
change the projective coordinates on a neighborhood of an endpoint of Ik so that
this is moved to infinity, then the maps g and h become affine on each half of
this neighborhood. Thus, [g, h] and [g 2, h] become translations, hence commute.
However, the reduced expression of f0 in g and h is
f0 = ghg −1h−1 g 2 hg −1h−1 g −1hg 2 h−1 g −2,
hence f0 is nontrivial.
Let f be a nontrivial element in hg, hi which is the identity on neighborhoods of
the endpoints of each Ij , and such that the number of components Ij intersecting
the support of f is minimal among all elements verifying these properties. Choose
one of these components Ik , and let [a, b] ⊂ Ik be a compact subinterval of Ik
such that f is the identity on Ik \ [a, b]. By the claim above, there exists a word
w in g and h such that w([a, b]) is disjoint from [a, b]. Inside Ik , the support of
wf w −1 is hence disjoint from the support of f , and thus the restrictions of f
and wf w −1 to Ik generate a subgroup isomorphic to Z2 . As a consequence, the

28

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

number of components Ij in restriction to which [f, wf w −1] is not the identity
is strictly smaller that the corresponding number for f . Since [f, wf w −1] is the
identity on a neighborhood of the endpoints of each Ij , we conclude by minimality
that [f, wf w −1] is the identity everywhere, and therefore f and wf w −1 generate
a group isomorphic to Z2 . Nevertheless, such a subgroup cannot arise inside a
non-Abelian free group on two generators.

We we will see in §4.2 that the group PP+ (R) contains many interesting
finitely-generated subgroups. Among them, the most remarkable are those that
are both non-amenable and finitely presented. The existence of these groups
has been recently proved in [128]. Actually, these are the first examples of nonamenable, finitely-presented, torsion-free groups containing no free subgroup in
two generators. (Examples of finitely-presented, non-amenable groups without
free subgroups but containing a lot of torsion were already known; see [161].)

1.2.6

Braid groups

Perhaps the most relevant examples of left-orderable groups are the braid
groups Bn . Recall that Bn has a presentation of the form
Bn = σ1 , . . . , σn−1: σi σi+1 σi = σi+1 σi σi+1 for 1 ≤ i ≤ n−2, σi σj = σj σi for |i−j| ≥ 2 .

Following Dehornoy [56], for i ∈ {1, . . . , n − 1}, an element of Bn is said to be ipositive if it may be written as a word of the form w1 σi w2 σi · · · wk σi wk+1, where
±1
±1
the wi ’s are (perhaps trivial) words on σi+1
, . . . , σn−1
(and σi appears at least
once). An element in Bn is said to be D-positive if it is i-positive for some
i ∈ {1, . . . , n − 1}. A remarkable result of Dehornoy establishes that the set of
D-positive elements form the positive cone of a left-order order D on Bn . In
other words:
– For every nontrivial σ ∈ Bn , either σ or σ −1 is i-positive for some i. (Actually, Dehornoy provides an algorithm, called handle reduction, for recognizing
positive elements and putting them in the form above.)
– If σ ∈ Bn is nontrivial, then σ and σ −1 cannot be simultaneously D-positive.
Notice that Bn is not bi-orderable, as it contains nontrivial elements which are
conjugate to their inverses: (σ1 σ2 σ1 )−1 (σ1 σ2−1 )(σ1 σ2 σ1 ) = (σ1 σ2−1 )−1 . Despite of
this, D satisfies an important weak property of bi-invariance called subword
property: All conjugates of the generators σi are D -positive.
None of the statements above is easy to prove; see for example [57]. In §2.2.3,
we will give a short proof for the case of B3 .

1.3. OTHER FORMS OF ORDERABILITY

29

Pure braid groups. According to Falk and Randell [71], pure braid groups PBn
are residually torsion-free nilpotent, hence bi-orderable. An alternative approach
to the bi-orderability of PBn using the Magnus expansion was proposed by Rolfsen
and Zhu in [115] (see also [138]). Let us point out, however, that these bi-orders
are quite different from the Dehornoy left-order. Indeed, we will see in §3.2.1 that
no bi-order on PBn can be extended to a left-order of Bn .
For nice bi-orderable groups which are a mixture of pure braid groups and
Thompson’s groups, see [29].

1.3
1.3.1

Other Forms of Orderability
Lattice-orderable groups

A lattice-ordered group (or ℓ-ordered group) is a partially ordered group
(Γ, ) such that  is left and right invariant, and for each pair of group elements
f , g, there is a minimal (resp. maximal) element f ∨g (resp. f ∧g) simultaneously
larger (resp. smaller) than f and g. (Notice that f ∧ g = (f −1 ∨ g −1 )−1 .) For
instance, the group A(Ω, ≤) of all order automorphisms of a totally ordered space
(Ω, ≤) is ℓ-orderable, as one may define f  g whenever f, g in A(Ω, ≤) satisfy
f (w) ≥ g(w) for all ω ∈ Ω. In this case, we have f ∨ g (w) = max≤ {f (w), g(w)}
and f ∧ g (w) = min≤ {f (w), g(w)}.
Example 1.3.1. For the group Homeo+ (R), this and its reverse order (which is also
an ℓ-order) are the only possible ℓ-orders; see [97].

The converse to the preceding remark is an important theorem due to Holland
[98]. (The proof below is taken from [17, Chapter VII]; see also [84, Chapter 7],
[85, Appendix I], and [96].)
Theorem 1.3.2. Every ℓ-ordered group (Γ, ) acts by automorphisms of a totally
ordered space (Ω, ≤) in such a way that f  g implies f (w) ≤ g(w) for all w ∈ Ω,
and f ∨ g (w) = max≤ {f (w), g(w)} and f ∧ g (w) = min≤ {f (w), g(w)}. In
particular, every ℓ-orderable group is left-orderable.
For the proof, let (Γ, ) be an ℓ-subgroup of A(Ω, ≤) for a totally ordered
space (Ω, ≤), and let P be the set of non-negative elements for the associate order.
For each
T w ∈ Ω, denote by Pw the semigroup {f ∈ Γ : f (w) ≥ w}. Then:
(i)
w∈Ω
S Pw−1= P ,
(ii) Pw Pw = Γ, for all w ∈ Ω.
This turns natural the following version of Proposition 1.1.5:

30

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Lemma 1.3.3. Let (Γ, ) be an ℓ-ordered group with set of non-negative elements
P . Assume that Γ contains a family of subsemigroups Pλ , λ ∈ Λ, satisfying (i)
and (ii) above. Then the conclusion of Theorem 1.3.2 is satisfied.
Proof. Proceed as in the proof of Proposition 1.1.5. Since  is bi-invariant, for
each f ∈ P , g ∈ Γ, and λ ∈ Λ, we have g −1f g ∈ P ⊂ Pλ . By definition, this
implies that f (gΓλ) ≥ gΓλ . Finally, if f ∈
/ P , then by (ii) we have f ∈ Pλ−1 \ Pλ
for some λ, which yields f Γλ < Γλ . The claims concerning f ∨ g and f ∧ g are
left to the reader.

Proof of Theorem 1.3.2. Denoting by P the set of non-negative elements of
, for each h ∈ Γ \ P choose a maximal ℓ-subsemigroup Ph of Γ containing P but
not h. We obviously have
\
Ph = P,
h∈Γ\P

so that condition (i) above is satisfied. The proof of condition (ii) is by contradiction. Assume throughout that for certain h ∈ Γ \ P and g ∈ Γ, we have g ∈
/ Ph
−1
and g ∈
/ Ph .
Claim (i). Neither id ∧ g nor id ∧ g −1 belong to Ph .

Indeed, notice that g = [g(id ∧ g)−1](id ∧ g) = (id ∨ g)(id ∧ g). Since id ∨ g
belongs to P ⊂ Ph , if id ∧ g were contained in Ph , then this would imply that
g also belongs to Ph , contrary to our hypothesis. A similar argument applies to
id ∧ g −1.

Claim (ii). There exist n1 , n2 in N and h1 , h2 in Ph such that [(id ∧ g)h1 ]n1  h
and [(id ∧ g −1 )h2 ]n2  h.

By the maximality of Ph , the element h belongs to the smallest ℓ-subsemigroup
hPh , id∧giℓ (resp. hPh , id∧g −1iℓ ) containing Ph and id∧g (resp. Ph and id∧g −1).
Thus, the claim follows from the following fact: For each f ≺ id, the semigroup
hPh , f iℓ is the set S of elements which are larger than or equal to (f f¯)n for some
f¯ ∈ Ph and some n ∈ N. To show this, first notice that this set is an ℓ-semigroup.
Indeed, if (f f¯1 )n1  g1 and (f f¯2 )n2  g2 , with f¯1 , f¯2 in Ph and n1 , n2 in N, then
both g1 and g2 are larger than or equal to (f f¯)n , where f¯ := id ∧ f¯1 ∧ f¯2 ∈ Ph
and n := max{n1 , n2 }. Hence, (f f¯)2n  g1 g2 and (f f¯)n  g1 ∧ g2 , and therefore
g1 g2 and g1 ∧ g2 (as well as g1 ∨ g2 ) belong to S. Since f ∈ S and Ph ⊂ S,
this shows that hPh , f iℓ ⊂ S. Finally, we also have S ⊂ hPh , f iℓ . Indeed, if
ḡ ≻ (f f¯)n for some f¯ ∈ Ph and n ∈ N, then since (f f¯)n ∈ hPh , f iℓ , we have
ḡ := ḡ(f f¯)−n (f f¯)n ∈ P ·hPh , f iℓ = hPh , f iℓ . This shows the claim.

1.3. OTHER FORMS OF ORDERABILITY

31

Claim (iii). Let n := max{n1 , n2 } and f := id ∧ h1 ∧ h2 , where h1 , h2 and n1 , n2
2n−1
are as in (ii). Then the element fˆ := [(id ∧ g)f ] ∨ [(id ∧ g −1)f ]
is smaller
than or equal to h.
Indeed, since f , id ∧ g, and id ∧ g −1 lie in P −1 , we have [(id ∧ g)f ]n  h and
[(id ∧ g −1 )f ]n  h. Now, as id ∧ g and id ∧ g −1 commute (their product equals
id ∧ g ∧ g −1), we easily check that fˆ may be rewritten as


[(id∧g)f ]2n−1 ∨ [(id∧g)f ]2n−2 [(id∧g −1 )f ] ∨ [(id∧g)f ]2n−3 [(id∧g −1 )f ]2 ∨. . .∨[(id∧g −1 )f ]2n−1.

Each term of this ∨-product contains either [(id ∧ g)f ]n or [(id ∧ g −1)f ]n together
with non-positive factors. The claim follows.

To conclude the proof, notice that from (id∧g)∨(id∧g −1 ) = id it follows that
fˆ = f 2n−1 . Since f ∈ Ph , the same holds for fˆ. Nevertheless, as fˆ−1 h ∈ P ⊂ Ph ,
this implies that h = fˆ(fˆ−1 h) ∈ Ph , which is a contradiction.

Left-orderable groups v/s ℓ-orderable groups. Let us point out that ℓorderability is a stronger property than left-orderability. For instance, ℓ-orderability
is a non-local property [84, Theorem 2.D]. A more transparent difference concerns
roots of elements, as all ℓ-orderable groups satisfy the property of conjugate
roots (C.R.P.): Any two elements f, g satisfying f n = g n for some n ∈ N are
conjugate. Indeed, if for such f, g we let
h := f n−1 ∨ f n−2 g ∨ f n−3g 2 ∨ . . . ∨ g n−1,
then we have
f h = f n ∨ f n−1 g ∨ f n−2 g 2 ∨ . . . ∨ f g n−1 = f n−1 g ∨ f n−2 g 2 ∨ . . . ∨ f g n−1 ∨ g n = hg.
This property fails to be true for left-orderable groups, as shown by the next
Exercise 1.3.4. The π1 of the Klein bottle may be presented in the form ha, b : bab = ai.
(This is nothing but the infinite dihedral group.) This group is easily seen to be leftorderable (see §2.2.3 for a discussion on this). Prove that this group is not ℓ-orderable
by showing that the elements x = ba and y = a satisfy x2 = y 2 but are not conjugate.
Remark. Despite this example, notice that every left-orderable group Γ embeds into a
lattice-orderable group, namely the group of all order permutations of Γ endowed with
a left-order.
Remark 1.3.5. Left-orderable groups satisfying the C.R.P. are not necessarily ℓorderable. Concrete examples are braid groups: in §1.2.6, we will see that these groups
are left-orderable, the C.R.P. for them is shown in [86], and the fact that Bn is not
ℓ-orderable (for n ≥ 3) is proved in [141].

32

1.3.2

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Locally-invariant orders and diffuse groups

Following [42] and the references therein, a partial order relation  on a group
Γ is said to be locally invariant if for every f, g in Γ, with g 6= id, either f g ≻ f
or f g −1 ≻ f . Obviously, every left-order is a locally-invariant order. Examples of
non left-orderable groups admitting a locally-invariant order have been recently
given: see §1.4 (see also Theorem 4.1.10 for the case of amenable groups).
Exercise 1.3.6. Show that a group Γ admits a locally-invariant order if and only if
there exist a partially ordered space (Ω, ≤) and a map ϕ : Γ → Ω such that for every
f, g in Γ, with g 6= id, either ϕ(f g) > ϕ(f ) or ϕ(f g −1 ) > ϕ(f ).
Example 1.3.7. Based on [18, 58, 90], it is shown in [42] that many groups with
hyperbolic properties admit locally-invariant orders. More precisely, let (X, d) be a
geodesic δ-hyperbolic metric space [81] and Γ a group acting on X by isometries so that
d(x, g(x)) > 6δ holds for all x ∈ X. Then fixing x0 ∈ X, the function g 7→ d(x0 , g(x0 ))
satisfies the property of the preceding exercise. In particular, Γ admits a locallyinvariant order.
This construction applies to many groups. In particular, if Γ is a residually finite
Gromov-hyperbolic group (as for instance the π1 of a compact hyperbolic manifold),
then Γ contains a finite-index subgroup admitting a locally-invariant order. Similarly, a
group acting isometrically and without inversions on a real-tree has a locally-invariant
order.

At first glance, the notion of locally-invariant order may look strange. Perhaps
a more clear view is provided by an equivalent formulation in terms of cones. More
precisely, given a group Γ, denote by P (Γ) the family of subsets (cones) P ⊂ Γ
such that id ∈
/ P and, for all g 6= id, at least one of the elements g, g −1 lies in P .
A field of cones is a map f → Pf from Γ into P (Γ). This field will be said to be
equivariant if the following condition holds (see Figure 6):
if g ∈ Pf

and h ∈ Pf g ,

then

gh ∈ Pf .

33

1.3. OTHER FORMS OF ORDERABILITY

Figure 6: The cone condition (above) and its negation (below) for
locally-invariant orders.
It turns out that locally-invariant orders and equivariant fields of cones are
equivalent notions. Indeed, assume that  is a locally-invariant order on a group
Γ. For each f ∈ Γ, define Pf by letting
g ∈ Pf

if and only if

f g ≻ f.

By definition, each Pf belongs to P (Γ). We claim that the field f → Pf is
equivariant. Indeed, the conditions g ∈ Pf and h ∈ Pf g mean, respectively, that
f g ≻ f and f gh ≻ f g. Hence, by transitivity of , we have f gh ≻ f , that is,
gh ∈ Pf , as desired.
Conversely, let f 7→ Pf be an equivariant field of cones. Define a relation  on
Γ by letting f ≻ g whenever g −1f ∈ Pg . We claim that this is a locally-invariant
order. To see that  is antisymmetric, assume f ≻ g and g ≻ f . Then g −1 f ∈ Pg
and f −1 g ∈ Pf . By equivariance, this implies that id = (g −1f )(f −1 g) ∈ Pg , which
is a contradiction. To see that  is transitive, assume f ≻ g and g ≻ h. Then
g −1 f ∈ Pg and h−1 g ∈ Ph . By equivariance, h−1 f = (h−1 g)(g −1f ) ∈ Ph , which
means that f ≻ h. Finally, given f ∈ Γ and g 6= id, we have either f −1 gf ∈ Pf ,
or f −1 g −1 f ∈ Pf . In the former case, gf ≻ f , and in the latter, g −1 f ≻ f .
Exercise 1.3.8. Associated to each ℓ ∈ Z there is a locally-invariant order ℓ on
Z defined by m ≻ℓ n if and only if either n > m ≥ ℓ or n < m ≤ ℓ − 1. (See
Figure 7.) Show that every locally-invariant order on Z either is the canonical one, its
reverse, or contains one of the orders ℓ . (Notice that we way enlarge ℓ by defining
non-contradictory inequalities between integers m, n such that m > ℓ > n.)

34

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS
•
ℓ−4

...................

•
ℓ−3

...........
...........
•
•
•
........
........
ℓ+1
ℓ+2
ℓ
Figure 7: A locally-invariant order on Z.

...................

•
ℓ−2

...................

•
ℓ−1

...........
........

•
ℓ+3

There is a closely related notion to locally-invariant orders introduced by
Bowditch in [18]. Namely, given a subset A of a group Γ, an extremal point
of A is a point f ∈ A such that, if f g ∈ A and f g −1 ∈ A for some g ∈ Γ, then
g = id. A group Γ is said to be weakly diffuse if every nonempty finite subset
has an extremal point.
Proposition 1.3.9. A group admits a locally-invariant order if and only if it is
weakly diffuse.
Proof. Let Γ be a group admitting a locally-invariant order . Given a nonempty,
finite subset A of Γ, let f be a maximal element (with respect to ) of A. We
claim that f is an extremal point of A. Indeed, let g ∈ Γ such that f g ∈ A and
f g −1 ∈ A. If g were nontrivial then we would have either f g ≻ f or f g −1 ≻ f .
However, this contradicts the maximality of f ∈ A.
For the proof of the converse implication, see Exercise 2.2.5.

Exercise 1.3.10. According to [126], every weakly diffuse group is diffuse, that is,
every finite subset of cardinality larger than one has at least two extremal points. Show
this by contradiction.
Hint. Assume that A is a finite subset having only the identity as an extremal point.
Then the same holds for A−1 . If A has more than one point, show that the set B :=
A ∪ A−1 has no extremal point.

1.4
1.4.1

General Properties
Left-orderable groups are torsion-free

Indeed, if f ≻ id (resp. f ≺ id) for some left-order , then for all n ∈ N we
have
f n ≻ . . . f 2 ≻ f ≻ id
(resp. f n ≺ . . . ≺ f 2 ≺ f ≺ f ≺ id).

As we have seen in §1.2.1, the converse is true for Abelian and more generally
for nilpotent groups, but does not hold for Abelian-by-finite groups: a classical
relevant example is the crystallographic group
Γ = ha, b : a2 ba2 = b, b2 ab2 = ai.

(1.4)

35

1.4. GENERAL PROPERTIES

Here we give some properties of this group (for further details, see [18, 167] as
well as [163, Chapter 13]). If we let c := (ab)−1 , then the subgroup ha2 , b2 , c2 i is
torsion-free, rank-3 Abelian, and normal. The corresponding quotient is isomorphic to the 4-Klein group. (An order-4, non-Abelian group). The crystallographic
action on R3 is given by
a(x, y, x) = (x + 1, 1 − y, −z),
b(x, y, z) = (−x, y + 1, 1 − z),

c(x, y, z) = (1 − x, −y, z + 1).
To see that Γ is torsion-free, first notice that, since every element in the 4-Klein
group has order 2, a nontrivial, finite-order element of Γ must have order 2. Now
let w ∈ Γ be nontrivial, say w = a2i b2j c2k a (the cases where the last factor is
either b or c are similar). Then
w 2 = a2i b2j c2k aa2i b2j c2k a = a2i b2j c2k a2i b−2j c−2k a2 = a4i+2 6= id.
Finally, notice that for any choice of exponents ε, δ in {−1, +1}, the defining
relations of Γ yield
(aε bδ )2 (bδ aε )2 = aε b−δ b2δ aε b2δ aε bδ aε = aε b−δ a2ε bδ a2ε a−ε = aε b−δ bδ a−ε = id.
Obviously, this implies that no compatible choice of signs for a, b exists, hence Γ
is not left-orderable. (For more conceptual proofs of a different nature, see either
§1.4.3 or Example 3.2.11.)
Exercise 1.4.1. Consider the set G of triplets of the form (u, v, w), where each u, v, w
is either an integer or of the form m̂, with m ∈ Z.
(i) Show that the rule
(u1 , v1 , w1 )(u2 , v2 , w2 ) := (u1 ⊕ u2 , v1 ⊕ v2 , w1 ⊕ w2 ),
where for m, n in Z,
m ⊕ n := m + n,

\
m ⊕ n̂ := m
+ n,

\
m̂ ⊕ n := m
− n,

m̂ ⊕ n̂ := m − n,

endows G with a group structure.
(ii) Show that the group Γ above identifies to the subgroup of G generated by a := (10̂0̂)
and b := (0̂11̂).
Exercise 1.4.2. Show that every group having a locally-invariant order is torsion-free.

36

1.4.2

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Unique roots and generalized torsion

Bi-orderable groups have a stronger property than absence of torsion, namely
they have no generalized torsion: If f 6= id, then no nontrivial product of
conjugates of f is the identity. (In particular, no nontrivial element is conjugate
to its inverse.) These groups also have the unique root property: If f n = g n for
some integer n, then f = g. Once again, none of these properties characterizes
bi-orderability (see [17, Example 4.3.1] and [8, 14], respectively). It seems to
be unknown whether the former implies left-orderability. The latter does not: a
concrete example (taken from [17, Chapter VII]) is
Γn = g, h : f gf g 2 · · · f g n = f −1 gf −1g 2 · · · f −1 g n = id ,

where n is “large”.

Example 1.4.3. As we saw in Example 1.3.4, the Klein-bottle group is left-orderable
but does not satisfy the C.R.P., hence it is not bi-orderable. Another way to contradict bi-orderability consists in noticing that it has generalized torsion: (a−1 ba)b = id.
Moreover, the unique root property fails: (ba)2 = a2 , though ba 6= a.
Exercise 1.4.4. Prove that in any bi-orderable group, the following holds: If f commutes with a nontrivial power of g, then it commutes with g. Show that this is no
longer true for left-orderable groups.
Exercise 1.4.5. Show that for bi-orderable groups, the normalizer of any finite subset
coincides with its centralizer. Again, show that this is no longer true for left-orderable
groups.

1.4.3

The Unique Product Property (U.P.P.)

A group Γ is said to have the U.P.P. if given any two finite subsets {gi }, {hj },
there exists f ∈ Γ that may be written in a unique way as a product gi hj .
Every left-orderable group has the U.P.P. Indeed, given two finite subsets
A := {g1 , . . . , gn } and B := {h1 , . . . , hm }, let f := gi hj be the element of AB
that is maximal with respect to a fixed left-order on Γ. If f were equal to gi′ hj ′
for some i′ , j ′ , then hj  hj ′ , as otherwise f = gi hj ≺ gi hj ′ would contradict the
maximality of f . Similarly, hj ′  hj . Thus hj = hj ′ , which yields gi = gi′ .
Notice that the minimum element in AB has also a unique expression as
above. This is coherent with a result from [184] contained in the following
Exercise 1.4.6. Show that U.P.P. implies a “double” U.P.P., in the sense that given
any two finite subsets A, B such that |A|+|B| > 2, there exist at least two elements
in AB which may be written in a unique way as a product ab, with a ∈ A and b ∈ B.
(Compare Exercise 1.3.10.)

1.4. GENERAL PROPERTIES

37

Hint. Assume that a group Γ has the U.P.P. but only ab ∈ AB has a unique representation in AB, and let C := a−1 A, D := Bb−1 , E := D −1 C, and F := DC −1 . Using the
fact that, in CD, only id = id · id has a unique representation, show that, in EF , no
element has unique representation.
Exercise 1.4.7. Show that a group satisfies the U.P.P. if and only if for any finite
subset A there exist at least one (actually, two) elements in A2 which may be written
in a unique way as a product ab, with a, b in A.

Let us remark that groups with the U.P.P. are torsion-free. Indeed, if f n =
id for some f 6= id, then the U.P.P. fails for the finite subsets A = B =
{id, f, . . . , f n−1}. The converse to this remark is false. Indeed, as Promislow
showed in [167], the crystallographic group of §1.4.1 does not satisfy the U.P.P.
(see Exercise 1.4.11 below for the details; see also [170] for a different example
using small cancellation techniques, and [116, 183] for more recent developments.)
The following question was raised by Linnell; the recent (negative) answer from
[116] will be discussed below.
Question 1.4.8. Are U.P.P.-groups left-orderable ?
Locally-invariant orders, diffuse groups, and the U.P.P. The U.P.P. is satisfied by all weakly diffuse groups (hence, by groups admitting a locally-invariant
order; see Proposition 1.3.9). Indeed, given nontrivial finite subsets A, B of a
weakly diffuse group Γ, let f ∈ AB be an extremal point of AB. We claim that
f may be written in a unique way as gh, with g ∈ A and h ∈ B. Indeed, if
f = g1 h1 = g2 h2 , with g1 , g2 in A and h1 , h2 in B, then letting h := h−1
1 h2 we
have f h = g1 h2 ∈ AB and f h−1 = g2 h1 ∈ AB. Since f is an extremal point of
AB, this implies that h = id, which yields h1 = h2 and g1 = g2 .
Below we elaborate on an example that shows that the answer to Linnell’s
question above is negative for a certain “large” group. (For amenable groups the
situation is unclear, due to Theorem 4.1.10.) First notice that, by Example 1.3.7,
isometry groups of hyperbolic metric spaces with “large displacement” admit
locally-invariant orders, and hence satisfy the U.P.P. (Actually, a combination
of remarkable recent results establishes that the π1 of every closed, hyperbolic 3manifold contains a finite-index group that is bi-orderable; see [2, 11, 68, 89, 109].)
This motivates the following
Question 1.4.9. Does there exist a sequence of compact, hyperbolic 3-manifolds
whose injectivity radius converges to infinity and whose π1 are non left-orderable ?
(Examples of non left-orderable 3-manifold groups appear in [34, 53].)

38

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

This question seems to have an affirmative but difficult solution. Indeed,
it is not very hard to prove that, if Γ is the π1 of a compact, hyperbolic 3manifold with nontrivial first Betti number, then Γ is left-orderable (it is actually
Conrad-orderable, in the terminology of §3.2; see [20]). A sequence of compact,
hyperbolic 3-manifolds with trivial first Betti number and whose injectivity radius
converges to infinite appears in [37]. However, it seems hard to adapt the methods
therein to show that the π1 of infinitely many of these manifolds are non leftorderable. Actually, an obvious difficulty comes from the fact that they are
virtually orderable, as was mentioned above.
Despite the above, it was cleverly noticed by Dunfield and included in the
work of Kionke and Raimbauld (see [116]) that there is an hyperbolic 3-manifold
whose π1 is known to be non left-orderable and for which a lower estimate of
its injectivity radius allows applying the results described in Example 1.3.7, thus
concluding that it admits a locally-invariant order, hence it satisfies the U.P.P.
This yields a negative answer to Question 1.4.8 above. As Kionke and Raimbauld
point out, the next question remains open:

Question 1.4.10. Does there exist a U.P.P.-group that is not weakly diffuse ?

Exercise 1.4.11. Consider the subset

A = B = (ba)2 , (ab)2 , a2 b, aba−1 , b, ab−1 a, b−1 , aba, ab−2 , b2 a−1 , a(ba)2 , bab, a, a−1
of the crystallographic group Γ = ha, b : a2 ba2 = b, b2 ab2 = ai introduced in §1.4.1.
(i) Show that, via the identification of Exercise 1.4.1, this set becomes
(002),(002),(2̂11̂),(2̂11̂),(0̂11̂),(0̂11̂),(0̂11̂),(0̂11̂),(12̂0̂),(12̂0̂),(10̂2̂),(10̂2̂),(10̂0̂),(10̂0̂),
d
where m (resp. m̂) is written instead of −m (resp. −m).

(ii) In the (partial) multiplication table below, check that each value corresponding to
the product of a pair of elements appears at least twice.

1.4. GENERAL PROPERTIES

39

(002) (002) (2̂11̂) (2̂11̂) (0̂11̂) (0̂11̂) (0̂11̂) (0̂11̂) (12̂0̂) (12̂0̂) (10̂2̂) (10̂2̂) (10̂0̂) (10̂0̂)
(002) (004) (000) (2̂13̂) (2̂11̂) (0̂13̂) (0̂11̂) (0̂13̂) (0̂11̂) (12̂2̂) (12̂2̂) (10̂0̂) (10̂4̂) (10̂2̂) (10̂2̂)
(002) (000) (004) (2̂11̂) (2̂13̂) (0̂11̂) (0̂13̂) (0̂11̂) (0̂13̂) (12̂2̂) (12̂2̂) (10̂4̂) (10̂0̂) (10̂2̂) (10̂2̂)
(2̂11̂) (2̂11̂) (2̂13̂) (020) (002) (220) (222) (200) (202) (1̂3̂1) (3̂3̂1) (1̂1̂3) (3̂1̂1) (1̂1̂1) (3̂1̂1)
(2̂11̂) (2̂13̂) (2̂11̂) (002) (020) (202) (200) (222) (220) (1̂1̂1) (3̂1̂1) (1̂1̂1) (3̂1̂3) (1̂1̂1) (3̂1̂1)
(0̂11̂) (0̂11̂) (0̂13̂) (220) (202) (020) (022) (000) (002) (1̂3̂1) (1̂3̂1) (1̂1̂3) (1̂1̂1) (1̂1̂1) (1̂1̂1)
(0̂11̂) (0̂13̂) (0̂11̂) (222) (200) (022) (020) (002) (000) (1̂3̂1) (1̂3̂1) (1̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂1)
(0̂11̂) (0̂11̂) (0̂13̂) (200) (222) (000) (002) (020) (022) (1̂1̂1) (1̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂1) (1̂1̂1)
(0̂11̂) (0̂13̂) (0̂11̂) (202) (220) (002) (000) (022) (020) (1̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂1)
(12̂0̂) (12̂2̂) (12̂2̂) (3̂1̂1) (3̂3̂1) (1̂1̂1) (1̂1̂1) (1̂3̂1) (1̂3̂1) (200) (000) (222) (022) (220) (020)
(12̂0̂) (12̂2̂) (12̂2̂) (1̂1̂1) (1̂3̂1) (1̂1̂1) (1̂1̂1) (1̂3̂1) (1̂3̂1) (000) (200) (022) (222) (020) (220)
(10̂2̂) (10̂4̂) (10̂0̂) (3̂1̂3) (3̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂3) (1̂1̂1) (222) (022) (200) (004) (202) (002)
(10̂2̂) (10̂0̂) (10̂4̂) (1̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂3) (1̂1̂1) (1̂1̂3) (022) (222) (004) (200) (002) (202)
(10̂0̂) (10̂2̂) (10̂2̂) (3̂1̂1) (3̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂1) (220) (020) (202) (002) (200) (000)
(10̂0̂) (10̂2̂) (10̂2̂) (1̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂1) (1̂1̂1) (020) (220) (002) (202) (000) (200)

Remark 1.4.12. By the preceding exercise, the crystallographic group Γ, though being
torsion-free, does not admit a locally-invariant order. It is worth mentioning that this
is actually the case of “most” finitely-presented groups in a very precise random model
for groups; see [162].
Remark 1.4.13. As it was shown by Morris-Witte, finite-index subgroups of SL(3, Z)
are non left-orderable (see Theorem 3.5.1). For large index, these groups are torsionfree, and it seems to be unknown whether they satisfy the U.P.P. By Exercise 1.3.6,
the following question makes sense: Does there exist a norm on SL(3, Z) such that for
finite but “large” index subgroups Γ one has either kf gk > kf k or kf g −1 k > kf k for
every f, g in Γ, with g 6= id ?

On Kaplansky’s conjecture. A famous question due to Kaplansky (commonly
referred to as the Kaplansky conjecture) asks whether the group algebra of a
torsion-free group over a ring A has no zero-divisors provided A has no zerodivisors. (Even the case where A = Z is open.) The restriction on the torsion is
natural. Indeed,
f n = id

=⇒

(f − 1)(f n−1 + f n−2 + · · · + f + 1) = f n − 1 = 0.

(1.5)

It easily follows from the definitions that every group satisfying the U.P.P. also
satisfies the conclusion of the Kaplansky conjecture. For the crystallographic

40

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

group considered above, Kaplansky’s conjecture is known to be true by different
methods (see for instance [27, 72, 74]; see also [163, 130]).
Example 1.4.14. Consider the free Burnside group
B(m, n) := a1 , . . . , am : W n = id for every word W .
It is known that for m ≥ 2 and n odd and large-enough, B(m, n) is infinite (actually, it is
non-amenable; see [1]). Of course, every element in this group has finite order. However,
it is still interesting to look for zero-divisors in its group algebra that are “nontrivial”
(i.e. do not arise from an identity of the form (1.5)). For instance, according to [105],
this is the case for
A := (1 + c + . . . + cn−1 )(1 − aba−1 ),

B := (1 − a)(1 + b + . . . + bn−1 ),

where a := a1 , b := a2 , and c := aba−1 b−1 . (Checking that AB = 0 is an easy exercise.)

1.4.4

More Combinatorial Properties

Recently, orderable groups have been considered as a natural framework to
extend certain basic results of Additive Combinatorics (see [75, 147, 186] as general references). One of the most elementary ones is the inequality for product
sets
|AB| ≥ |A| + |B| − 1,
(1.6)
which holds for any finite subsets A, B of the integers (this is an easy exercise).
In this regard, it is worth mentioning that this readily extends to finite subsets of
left-orderable groups. Indeed, modulo multiplying B on the right by the largest
possible element of type h−1 , where h ∈ B, we may assume that id is the smallest
element of B. Then, if we order the elements in A (resp. B) in the form g1 ≺
. . . ≺ gn (resp. id = h1 ≺ . . . ≺ hm ), we have
g1 ≺ g2 ≺ . . . ≺ gn ≺ gn h2 ≺ . . . ≺ gn hm .
Less trivially, (1.6) still holds for finite subsets of torsion-free groups, as it was
proved by Kemperman in [112].
Theorem 1.4.15. For all finite subsets A, B of a torsion-free group, we have
|AB| ≥ |A| + |B| − 1.

1.4. GENERAL PROPERTIES

41

Proof. To begin with, notice that the claim of the theorem trivially holds if
either |A| or |B| equals 1. Moreover, changing A by g −1 A and B by Bh−1 for
g ∈ A and h ∈ B, we reduce the general case to that where id ∈ A ∩ B. Assume
for a contradiction that A, B are finite subsets that do not satisfy (1.6) and for
which the value of m := |AB| is minimal, that of n := |A| + |B| is maximal while
attaining m, and that of |A| is maximal while attaining both m, n (the extremal
properties being realized among subsets containing id).
As id ∈ A ∩ B, we also have
|AB| ≥ |A| + |B| − |A ∩ B|.
Hence, |A∩B| ≥ 2. Let H be the subsemigroup generated by A∩B. We consider
two different cases.
Case I. We have Af ⊂ A for all f ∈ A ∩ B.
Then, as id ∈ A, this implies H ⊂ A. Therefore, H is a finite subsemigroup
of a group, hence a (finite) subgroup. As |H| ≥ 2, this produces torsion elements.

Case II. There exists f ∈ A ∩ B such that Af is not contained in A.
Fixing such an f , let A′ := {g ∈ A : gf ∈
/ A} and B ′ := {h ∈ B : f h ∈
/ B}.
There are two subcases to consider.
If |A′ | ≥ |B ′ |, then let A∗ := A ∪ A′ f and B ∗ := B \ B ′ . (Notice that B \ B ′ 6= ∅
since id ∈
/ B ′ .) One easily checks that A∗ B ∗ ⊂ AB, hence |A∗ B ∗ | ≤ |AB|.
Moreover, |A∗ | = |A| + |A′ f | = |A| + |A′| and |B ∗ | = |B| − |B ′ |, thus |A∗ | + |B ∗ | ≥
|A| + |B|. Finally, |A∗ | > |A|, as A′ is nonempty. Therefore, by the choice of A, B,
we must have
|A∗ B ∗ | ≥ |A∗ | + |B ∗ | − 1,

hence

|AB| ≥ |A∗ B ∗ | ≥ |A∗ | + |B ∗ | − 1 ≥ |A| + |B| − 1,

which is a contradiction.
If |A′ | < |B ′ |, then let A∗ := A \ A′ (which is nonempty as id ∈
/ A′ ) and B ∗ :=
∗ ∗
∗ ∗
B ∪ f B. Again, A B ⊂ AB, hence |A B | ≤ |AB|. Moreover, |A∗ | = |A| − |A′ |
and |B ∗ | = |B| + |B ′ | yield |A∗ | + |B ∗ | > |A| + |B|. By the choice of A, B, this
implies
|A∗ B ∗ | ≥ |A∗ | + |B ∗ | − 1,
hence

|AB| ≥ |A∗ B ∗ | ≥ |A∗ | + |B ∗ | − 1 > |A| + |B| − 1,

which is again a contradiction.



42

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Example 1.4.16. By pursuing on the technique of proof above, Brailovsky and Freiman
proved in [19] that equality arises if and only if A and B are geometric progressions
on different sides. More precisely, there exist group elements f, g, h and non-negative
integers n, m such that
A = {g, gf, . . . , gf n−1 },

B = {h, f h, . . . , f m−1 h}.

Showing such a claim for left-orderable groups is an straighforward exercise.

Below we present another proof of Theorem 1.4.15 following the ideas of Hamidoune [91] that is somewhat closer to the techniques of the next section. We refer
to [92] for more details and furter developments, including an alternative proof
of the Brailovsky-Freiman theorem above.
Another proof of Theorem 1.4.15. Given a finite subset B of a group Γ, for
each finite subset A ⊂ Γ we let ∂ B A := AB \ A (compare (1.9)). Given a positive
integer k, we say that a subset C is (B, k)-critical if |C| ≥ k and

|∂ B C| = min |∂ B A| : |A| ≥ k .

We say that C is a (B, k)-atom if it is a (B, k)-critical set of smallest cardinality.

Claim (i). If C is a (B, k)-atom and C ′ is (B, k)-critical, then either C ⊂ C ′ or
|C ∩ C ′ | ≤ k − 1.
Indeed, assume C is not contained in C ′ and |C ∩C ′| ≥ k. Then, by definition,
|∂ B C| < |∂ B (C ∩ C ′ )|.
Let C∗ (resp. C∗′ ) be the complement of C ∪ ∂ B C (resp. C ′ ∪ ∂ B C ′ ). On the one
hand, we have
|∂ B C ∩ C ′ |+|∂ B C ∩ ∂ B C ′ |+|∂ B C ∩ C∗′ | = |∂ B C|

< |∂ B (C ∩ C ′ )|

≤ |C ∩ ∂ B C ′ |+|∂ B C ∩ C ′ |+|∂ B C ∩ ∂ B C ′ |,

hence |∂ B C ∩ C∗′ | < |C ∩ ∂ B C ′ |. On the other hand, we have
|∂ B C ′ ∩ C|+|∂ B C ′ ∩ ∂ B C|+|∂ B C ′ ∩ C∗ | ≤ |∂ B C ′ |

≤ |∂ B (C ′ ∩ C)|

≤ |C∗′ ∩ ∂ B C|+|∂ B C ′ ∩ C∗ |+|∂ B C ′ ∩ ∂ B C|,

hence |∂ B C ′ ∩ C| ≤ |C∗′ ∩ ∂ B C|. These two conclusions are certainly in contradiction.

43

1.4. GENERAL PROPERTIES

Claim (ii). If C is a (B, k)-atom and g 6= id, then |C ∩ gC| ≤ k − 1.
Indeed, the set gC is a (B, k)-atom as well. Moreover, we cannot have gC ⊂ C,
otherwise g would be a torsion element. (If gC ⊂ C, then gC = C, so that g acts
as a permutation of C and therefore g n h = h for all h ∈ C.)
Claim (iii). For all finite sets A, B, we have |AB| ≥ |A| + |B| − 1.
Indeed, we may assume that B contains id. Let C be a (B, 1)-atom. Again,
we may assume id ∈ C. If C contains another element g, then |C ∩ gC| ≥ 1, a
contradiction to (ii). Hence, C = {id}. Therefore, for every (nonempty) finite
subset A,
|AB| − |A| = |AB \ A| ≥ |CB \ C| = |B \ {id}| = |B| − 1,
which shows the claim.



Remark 1.4.17. It is conjectured that k-atoms have cardinality equal to k for torsionfree groups. This holds for instance for groups satisfying the U.P.P. (this is an easy
exercise; see [92, Lemma 4] in case of problems).

A direct consequence of the preceding theorem is the inequality |A2 | ≥ 2|A|−1
for all finite subsets A of torsion-free groups. The next result from [76] improves
this inequality for non-Abelian bi-orderable groups.
Theorem 1.4.18. Let A be a finite subset of a bi-orderable group. If |A2 | ≤
3|A| − 3, then the subgroup generated by the elements in A is Abelian.
Proof. The proof is by induction on |A|. If |A| = 2, say A = {f1 , f2 }, then
|A2 | = 3|A| − 3 = 2 implies A2 = {f12 , f1 f2 = f2 f1 , f22 }, thus the group generated
by f1 , f2 is Abelian. Assume that the theorem holds for subsets of cardinality
≤ k, and let A := {f1 , . . . , fk+1}, where fi ≺ fj holds whenever i < j for a fixed
bi-order  on the underlying group. We let i be the maximal index for which the
subgroup generated by B := {f1 , . . . , fi } is Abelian, and we assume that i ≤ k.
Then fi+1 does not belong to the subgroup generated by B. Moreover, there is
f ∈ B not commuting with fi+1 ; we let fj be the maximal such element. We also
let C := {fi+1 , . . . , fk+1 }. Assume throught that |A2 | ≤ 3|A| − 3.
Claim (i). We have |C 2 | ≤ 3|C| − 3.
Indeed, using  (see also Exercise 1.4.5), one readily checks that
B 2 ∩(fi+1 B∪Bfi+1 ) = ∅,

fi+1 B 6= Bfi+1 ,

C 2 ∩(B 2 ∪fi+1 B∪Bfi+1 ) = ∅. (1.7)

44

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Therefore,
|C 2 | ≤ |A2 | − |B 2 | − |fi+1 B ∪ Bfi+1 |
≤ (3|A| − 3) − (2|B| − 1) − (|B| + 1) = 3(|A| − |B|) − 3 = 3|C| − 3,
as claimed.
By the inductive hypothesis, the group generated by C is Abelian. As fj and
fi+1 ∈ C do not commute, we must have
C 2 ∩ (fj C ∪ Cfj ) = ∅.

(1.8)

Claim (ii). We have Bfi+1 ∩ fj C = {fj fi+1 }. In particular, |Bfi+1 ∪ fj C| = k.
Indeed, assume fm fi+1 = fj fn , with fm ∈ B and fn ∈ C. If fm ≺ fj , then
fi+1 ≻ fn , which is impossible since fi+1 is the smallest element in C. If fm ≻ fj ,
then fm commutes with fi+1 , and so does fj = fm fi+1 fn−1 , which is absurd.
Claim (iii). We have B 2 ∩ (fj C ∪ Cfj ) = ∅.
Indeed, assume fm fn = fj fℓ holds for fm , fn in B and fℓ ∈ C. Since fn ≺ fℓ ,
we must have fm ≻ fj . Moreover, as B generates an Abelian group, fm fn = fn fm ,
hence also fn ≻ fj . Therefore, both fm , fn commute with fi+1 , and so does
fj = fm fn fℓ−1 , which is absurd. This shows that B 2 ∩fj C = ∅. That B 2 ∩Cfj = ∅
is proved similarly.
Claim (iv). We have A2 = B 2 ∪ C 2 ∪ Bfi+1 ∪ fj C.
It follows from the above that
|B 2 ∪ C 2 ∪ Bfi+1 ∪ fj C| = |B 2 | + |C 2 | + |Bfi+1 ∪ fj C|
≥ (2i − 1) + (2(k − i + 1) − 1) + k = 3(k + 1) − 3.
By the hypothesis |A2 | ≤ 3|A| − 3, this implies the claim.
Notice that fi+1 fj ∈
/ B 2 and fi+1 fj ∈
/ C 2 , by (1.7). A contradiction is then
provided by the next two claims below.
/ Bfi+1 .
Claim (v). We have fi+1 fj ∈
Indeed, assume fi+1 fj = fm fi+1 for fm ∈ B. If fm ≻ fj , then it commutes
with fi+1 . Thus, fi+1 fj = fi+1 fm , hence fj = fm , which is absurd. Suppose
fm ≺ fj . By Exercise 1.4.5, there exists fn ∈ B such that fi+1 fn ∈
/ Bfi+1 . Thus,
necessarily, fn 6= fj . We cannot have fn ≻ fj , otherwise fi+1 fn = fn fi+1 ∈ Bfi+1 ,
a contradiction. Therefore, fn ≺ fj .

1.4. GENERAL PROPERTIES

45

By (1.7), fi+1 fn ∈
/ B 2 ∪ C 2 . Hence, by Claim (iv), we have fi+1 fn ∈ fj C,
so that there is fℓ ∈ C such that fi+1 fn = fj fℓ . As B generates an Abelian
subgroup,
fj fℓ fj = fi+1 fn fj = fi+1 fj fn .
As fn ≺ fj , this implies fj fℓ ≺ fi+1 fj = fm fi+1 . However, this is impossible,
since fj ≻ fm and fℓ  fi+1 .

Claim (vi). We have fi+1 fj ∈
/ fj C.
Assume fi+1 fj = fj fm holds for a certain fm ∈ C. By Exercise 1.4.5, there
exists fn ∈ C such that fn fj ∈
/ fj C. As fi+1 fj ∈ fj C, it holds fi+1 ≺ fn .
Moreover, by (1.8) and Claim (iii), fn fj ∈
/ B 2 ∪ C 2 . Thus, by Claim (iv), we have
fn fj ∈ Bfi+1 . Let fℓ ∈ B be such that fn fj = fℓ fi+1 .
Notice that fℓ 6= fj , otherwise fn fj would belong to fj C. If fℓ ≻ fj , then it
commutes with fi+1 , and so does fj = fn−1 fi+1 fℓ , which is absurd. If fℓ ≺ fj ,
then, as fi+1 and fn commute,
fi+1 fℓ fi+1 = fi+1 fn fj = fn fi+1 fj .
As fi+1 ≺ fn , this implies fℓ fi+1 ≻ fi+1 fj = fj fm . However, this is impossible,
since fℓ ≺ fj and fi+1  fm .

Example 1.4.19. Following [76], let A = Ak be the subset of the Baumslag-Solitar
group BS(1, 2) := ha, b : aba−1 = b2 i given by A := {a, ab, ab2 , . . . , abk−1 }. Check that
|A2 | = 3|A| − 2, yet BS(1, 2) is bi-orderable and non-Abelian.
Exercise 1.4.20. Let Γ be the Klein-bottle group ha, b : aba−1 = b−1 i (c.f. Example
1.3.4). Check that the set A = Ak := {a, ab, ab−1 , . . . abk−1 } satisfies |A2 | = 2|A| − 1,
yet Γ is left-orderable and non-Abelian.
Exercise 1.4.21. Using Brailovsky-Freiman’s theorem (c.f. Example 1.4.16), prove
that if A is a subset of a torsion-free group satisfying |A2 | = 2|A| − 1, then A generates
either an Abelian subgroup or a group isomorphic to the Klein-bottle group.

1.4.5

Isoperimetry and Left-Orderable Groups

The aim of this section is to develop some ideas recently introduced by Gromov
in [88]. Let us begin with the notion of isoperimetric profile, due to Vershik.
Let Γ be a finitely-generated group acting on a set X, and let G be a finite
generating system containing id. For Y ⊂ X, its boundary (with respect to G)
is defined as
∂G Y := GY \ Y,
(1.9)

46

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

where GY := {g(y) : g ∈ G, y ∈ Y }. The maximal function I : N → N satisfying

|∂G Y | ≥ I |Y |

for all finite Y ⊂ X is called the combinatorial isoperimetric profile of the
Γ-action, and will be denoted by I(X;Γ,G) .
An important case arises when X = Γ is endowed with the action by lefttranslations. In this case, the isoperimetric profile is denoted I(Γ,G) . We list
below some important properties.
Subaditivity. If Γ is infinite, then for all r1 , r2 we have
I(Γ,G) (r1 + r2 ) ≤ I(Γ,G) (r1 ) + I(Γ,G) (r2 ).
Indeed, choose Yi ⊂ Γ such that |Yi | = ri and ∂G Yi = I(Γ,G) (ri ), with i ∈
{1, 2}. Since Γ is infinite, after “moving” Y2 keeping Y1 fixed,2 we may assume
that Y1 and Y2 are disjoint and ∂G (Y1 ⊔ Y2 ) = ∂G Y1 ⊔ ∂G Y2 . This yields
I(Γ,G) (r1 + r2 ) ≤ ∂G (Y1 ⊔ Y2 ) = ∂G (Y1 ) + ∂G (Y2 ) = I(Γ,G) (r1 ) + I(Γ,G) (r2 ).
I is non-decreasing under extensions. If Γ ⊂ Γ1 are infinite groups and
G ⊂ G1 , then for all r,
I(Γ1 ,G1 ) (r) ≥ I(Γ,G) (r).

Indeed,
any finite subset Y ⊂ Γ1 may be decomposed as a disjoint union
Fk
Y = i=1 Yi , where the points in each Yi are in the same class modulo Γ. Since
G ⊂ Γ,
k
G
∂G1 Y ⊃ ∂G Y =
∂G Yi .
i=1

Thus,

∂G1 Y ≥

k
X
i=1

∂G Yi ≥

k
X
i=1



I(Γ,G) |Yi | ≥ I(Γ,G)

k
X
i=1



|Yi | = I(Γ,G) |Y | .

I is non-increasing under homomorphisms. If Φ : Γ → Γ is a surjective
group homomorphism and G = Φ(G), then, for all r,
I(Γ,G) (r) ≥ I(Γ,G) (r).
2

Notice that h(∂G Y ) = ∂G (hY ), for all h ∈ Γ and all Y ⊂ Γ.

47

1.4. GENERAL PROPERTIES


:=
f ∈ Γ : |Φ−1 (f )∩Y | ≥ m .
Indeed, given
a
finite
subset
Y
⊂
Γ,
we
let
Y
m
P
Clearly, |Y | = m≥1 |Y m |. If we are able to show that
X
∂G Y m ,
|∂G Y | ≥
(1.10)
m≥1

then this would yield
X

X


|Y m | = I(Γ,G) |Y | ,
|∂G Y | ≥
I(Γ,G) |Y m | ≥ I(Γ,G)
m≥1

m≥1

thus showing our claim. Now, to show (1.10), every f in ∂G Y m may be written
as Φ(g)h for some g ∈ G and h ∈ Y m . By definition, Φ−1 (h) ∩ Y ≥ m, and
Φ−1 (f ) ∩ Y < m. Thus, there must be some gf in G and h ∈ Φ−1 (h) such
that f = Φ(gf h), with h ∈ Y and gf h ∈
/ Y . The correspondence f 7→ gf h from
S
into
∂
Y
is
injective,
because
Φ(gf h) = f . This shows (1.10).
Y
∂
G
m G m
Suppose that Γ acts on a linear space V. Given a subspace D ⊂ V and a
finite generating set G ⊂ Γ containing id, we define its boundary as the quotient
space

∂G D := G · Y D,

where G · D is the subspace generated by {g(v) : g ∈ G, v ∈ D}. Using now the
notation | · | for the dimension of a vector space, we define the linear isoperimetric profile of the Γ-action on V as the maximal function I satisfying, for all
finite dimensional subspaces D ⊂ V,

|∂G D| ≥ I |D| .

∗
We denote this function by I(V;Γ,G)
. In the special case where V is the group algebra R(Γ) (viewed as the vector space of finitely-supported, real-valued functions
∗
on Γ), we simply use the notation I(Γ,G)
.
As is the case of I(Γ,G) , the function I ∗ (Γ, G) is subadditive, as well as nonincreasing under group extensions. It is unclear whether it is non-increasing
under group homomorphisms. However, we will see that if the target group is
left-orderable, then this property holds (c.f. Proposition 1.4.25).
There is a simple relation between I and I ∗ for all finitely-generated groups.

Proposition 1.4.22. For every finitely-generated group Γ and all r ≥ 0,
∗
I(Γ,G) (r) ≥ I(Γ,G)
(r).

48

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Proof. To each finite subset Y ⊂ Γ we may associate the subspace DY := Y R
formed by all the functions whose support is contained in Y . We clearly have
|Y | = |DY | and |∂G Y | = |∂G DY |, which easily yields the claim.

The opposite inequality is not valid for all groups, as the following example
shows.
Example 1.4.23. If Γ is a group containing a nontrivial finite subgroup Γ0 and Γ0 ⊂ G,
then for any finite dimension subspace D ⊂ R(Γ) of finitely-supported

 functions which
∗
are constant along the cosets of Γ0 , we have I(Γ,G) |D| < I(Γ,G) |D| . If Γ is infinite,
∗
this yields I(Γ,G)
(r) < I(Γ,G) (r), for all r ≥ 0.

Despite the preceding example, the equivalence between I and I ∗ holds for leftorderable groups. (It is an open question whether this remains true for torsion-free
groups; for groups with torsion, see Example 1.4.26.)
Theorem 1.4.24. If Γ is a finitely-generated left-orderable group, then for every
∗
.
finite generating system G containing id, one has I(Γ,G) = I(Γ,G)
To show this theorem, we will use an argument which is “dual” to that of
Proposition 1.4.22.
Isoperimetric Domination (ID). Let Γ be a group acting on a set X and
on a vector space V. Suppose there exists an equivariant map D 7→ YD from the
Grassmanian GrV of finite dimensional subspaces of V into the family of subsets
of X such that:
(i) |D| = |YD |, for all D ∈ GrV ;
S
S
(ii) |span( i Di )| ≥ | i YDi |, for every finite family {Di } ⊂ GrV .
We claim that, in this case, for every finite generating set G containing id and all
r ≥ 0,
∗
I(X;Γ,G) (r) ≥ I(V;Γ,G)
(r).
(1.11)
Indeed, taking any D so that |D| = r, we have |YD | = r and
|∂G D| = G · D/D = span

[

g∈G

≥
which easily yields (1.11).

gD
[

g∈G



− |D| ≥

YgD − |D| ≥

[

g∈G

g(YD ) − |YD | = |∂G YD |,

49

1.4. GENERAL PROPERTIES

ID for left-ordered groups. In view of the above discussion, in order to prove
Theorem (1.4.24) it suffices to exhibit an ID from GrR(Γ) to 2Γ . The construction
proceeds as follows. Fix a left-order  on Γ. To each finitely-supported, realvalued function ϕ on Γ, we may associate the minimum g ∈ Γ in its support
(where the minimum is taken with respect to ). Denote this point by gϕ . Now,
if D ⊂ V is a finitely-dimensional subspace, then the number of points gϕ which
may appear for some ϕ ∈ D is finite. In fact, a simple “passing to a triangular
basis” argument using the left-order shows that the cardinality of this subset
YD ⊂ Γ equals |D|, so property (i) above is satisfied. Property (ii) is also easily
verified, thus concluding the proof.
Proposition 1.4.25. Let Φ : Γ → Γ be a surjective group homomorphism. If Γ
is left-orderable, then denoting G = Φ(G) we have, for all r ≥ 0,
∗
∗
I(Γ,G)
(r) ≥ I(Γ,G)
(r).

Proof. Fix a left-order  on Γ, and for each g ∈ Γ denote
A≺g := {ϕ ∈ A(Γ) : gϕ ≺ g},

Ag := {ϕ ∈ A(Γ) : gϕ  g}.

Given a finitely dimensional subspace D ⊂ A(Γ), define U = UD : Γ → N0 by



U(g) := dim AΦ(g) ∩ D A≺Φ(g) ∩ D .
Let SU be the subgraph of U, that is,

SU := (g, n) ∈ Γ × N : U(g) ≥ n .

Since Γ naturally acts on Γ × N and the action is free on each level, we have
X
X


|∂G SU | =
∂G SU ∩(Γ × {m}) ≥
I(Γ,G) SU ∩(Γ × {m})
m≥1

m≥1

≥ I(Γ,G)

X

m≥1



|SU ∩(Γ × {m})| = I(Γ,G) |SU | .

Moreover, one easily convinces that |SU | = |D|. Putting all of this together, we
obtain



∗
|∂G D| = |∂G SU | ≥ I(Γ,G) |SU | = I(Γ,G) |D| = I(Γ,G)
|D| ,

where the last equality comes from Theorem 1.4.24.



50

CHAPTER 1. SOME BASIC AND NOT SO BASIC FACTS

Example 1.4.26. Following [7], we consider the lamplighter group Γ := Z ≀ Z/2Z =
Z ⋉ ⊕i∈Z Z/2Z, where the action of Z consists in shifting coordinates. We view elements
of Γ as pairs (t, f ), where t ∈ Z and f is a finitely-supported function from Z into Z/2Z.
As a generating set we consider G := {id, (0, δ0 ), (±1, 0)}, where δ0 stands for the Dirac
delta at 0. The subspaces
+
*
X
(t, f ) : t ∈ {1, . . . , n}
Dn :=
supp(f )⊂{1,...,n}

satisfy |Dn | = n and |∂G Dn | = 2. However, every finite subset Y ⊂ Γ for which
|∂G Y |/|Y | ≤ 2/n must have at least 2λn points for a certain constant λ > 0. Indeed, this
follows from that the ball of radius 2n+2 in Γ has more than 2n points as an application
of the Saloff-Coste’s isoperimetric inequality [190]: If Y satisfies |∂G Y |/|Y | ≤ 1/n, then
its cardinal is greater than or equal to a half of the cardinal of a ball of radius n/2.
(See [88] for an elementary proof of this inequality.)
Remark 1.4.27. In the example above, the group Γ not only contains torsion elements
but is also amenable. In this direction, let us point out that a nice theorem due to
Bartholdi [7] establishes that for non-amenable groups, the linear isoperimetric profile
cannot behave sublinearly along any subsequence (see [88] for an alternative proof using
orderings !).

Chapter 2
A PLETHORA OF ORDERS
2.1
2.1.1

Producing New Left-Orders
Convex extensions

A subset S of a left-ordered group (Γ, ) is said to be convex (with respect
to ) if, for all f ≺ g in S, every element h ∈ Γ satisfying f ≺ h ≺ g belongs to
S. If S is a subgroup, this is equivalent to that g ∈ S for all g ∈ Γ such that
id ≺ g ≺ f for some f ∈ S.
The family of -convex subgroups is linearly ordered (by inclusion). More
precisely, if Γ0 and Γ1 are convex (with respect to ), then either Γ0 ⊂ Γ1
or Γ1 ⊂ Γ0 . Moreover, the union and the intersection of any family of convex
subgroups is a convex subgroup.
Example 2.1.1. For each g ∈ Γ it is usual to denote Γg (resp. Γg ) the largest (resp.
smallest) convex subgroup which does not (resp. does) contain g. The inclusion Γg ⊂ Γg
is referred to as the convex jump associated to g. In general, Γg fails to be normal
in Γg . Normality holds for bi-orders, and in such a case the quotient Γg /Γg is Abelian
(see §3.2.3 for the study of left-orders for which this holds for every g).
Example 2.1.2. It is not difficult to produce examples of group left-orders without
maximal proper convex subgroups: consider for instance a lexicographic left-order on
ZN . Nevertheless, if the underlying group Γ is finitely generated, such a maximal
subgroup always exists. Indeed, given a system of generators id ≺ g1 ≺ . . . ≺ gk , let Γ0
be the maximal convex subgroup that does not contain gk . Then Γ0 ( Γ. Moreover, if
Γ1 is a convex subgroup containing Γ0 , then, by definition, gk ∈ Γ1 . By convexity, all
the gi ’s belong to Γ1 , hence Γ1 = Γ. Thus, Γ0 is the maximal proper convex subgroup.

51

52

CHAPTER 2. A PLETHORA OF ORDERS

In the dynamical terms of §1.1.3, convex subgroups are characterized by the
following
Proposition 2.1.3. Let (Γ, ) be a countable left-ordered group, and let Γ∗ be
a convex subgroup. Then, in the dynamical realization of , there is a bounded,
Γ∗ -invariant interval I with the property that g(I) ∩ I = ∅ for every g ∈ Γ \ Γ∗ .
Conversely, let Γ be a group acting by orientation-preserving homeomorphisms
of the real line without global fixed points. Suppose that there is an interval I
with the property that, for all g ∈ Γ, the intersection g(I) ∩ I either is empty
or coincides with I. Then, in any dynamical-lexicographic order induced from
a sequence (xn ) starting with x1 ∈ I, the stabilizer StabΓ (I) is a proper convex
subgroup.
Proof. Suppose (Γ, ) is a countable left-ordered group having Γ∗ as a proper
convex subgroup. Consider its dynamical realization, and let a := inf{h(0) | h ∈
Γ∗ } and b := sup{h(0) | h ∈ Γ∗ }. Then I := (a, b) is a bounded interval fixed by
Γ∗ . Moreover, if g ∈ Γ is such that g(I) ∩ I 6= ∅, then there is h ∈ Γ∗ such that
gh(0) ∈ I. By convexity, this implies that gh ∈ Γ∗ , which yields g ∈ Γ∗ .
Conversely, suppose that for a Γ-action on the line, there is a bounded interval
I satisfying that g(I)∩I either is empty or coincides with I, for each g ∈ Γ. Let 
be a order induced from a sequence (xn ), with x1 ∈ I. If g ∈ Γ satisfies id ≺ g ≺ h
for some h ∈ StabΓ (I), then by definition we have x1 ≤ g(x1 ) ≤ h(x1 ), hence
g(I) = I. Therefore, StabΓ (I) is -convex.

The convex extension procedure. Let Γ∗ be a -convex subgroup of Γ, and
let ∗ be any left-order on Γ∗ . The extension of ∗ by  is the order relation
′ on Γ whose positive cone is (P+ \ Γ∗ ) ∪ P+∗ .
One easily checks that ′ is also a left-invariant total order relation, and that
Γ∗ remains convex in Γ with respect to ′ . Moreover, the family of ′ -convex
subgroups of Γ is formed by the ∗ -convex subgroups of Γ∗ and the -convex
subgroups of Γ that contain Γ∗ .
Example 2.1.4. Let (Γ, ) be a left-ordered group, and Γ∗ a -convex subgroup. The
extension of (the restriction to Γ∗ of)  by  will be referred as the left-order obtained
by flipping the convex subgroup Γ∗ . An important case of this seemingly innocuous
construction arises for braid groups; see the end of §2.2.3.
Remark 2.1.5. As we have already pointed out, convex subgroups are not necessarily
normal. In the case it is, the left-order passes to the quotient. Conversely, if Γ contains
a normal subgroup Γ∗ such that both Γ∗ and Γ/Γ∗ are left-orderable, then Γ admits a

2.1. PRODUCING NEW LEFT-ORDERS

53

left-order for which Γ∗ is convex. Indeed, letting ∗ and 0 be left-orders on Γ∗ and
Γ/Γ∗ , respectively, we may define  on Γ by letting f ≺ g if either f Γ∗ ≺0 gΓ∗ , or
f Γ∗ = gΓ∗ and f −1 g is ∗ -positive.
Thus, the extension of a left-orderable group by a left-orderable one is left-orderable.
L
Using Example 1.1.1, this implies that the wreath product Γ1 ≀ Γ2 := Γ2 Γ1 ⋊ Γ2
of two left-orderable groups is left-orderable.

In dynamical terms, convex subgroups are relevant because of the following
Remark 2.1.6. Let (Γ, ) be a left-ordered group, and let Γ∗ be a -convex subgroup.
The space of left cosets Ω = Γ/Γ∗ carries a natural total order ≤, namely f Γ∗ < gΓ∗ if
f h1 ≺ gh2 for some h1 , h2 in Γ∗ (this definition is independent of the choice of h1 , h2
in Γ∗ ). Moreover, the action of Γ by left-translations on Ω preserves this order. An
important case (to be treated in §3.5) arises when Γ∗ is the maximal proper convex
subgroup (whenever it exists); see Example 2.1.2.

The preceding construction allows showing the following useful
Proposition 2.1.7. Let Γ be a left-orderable group, and let {Γλ : λ ∈ Λ} be a
family of subgroups each of which is convex with respect
T to a left-order λ . Then
there exists a left-order on Γ for which the subgroup λ Γλ is convex.
For the proof, we need a lemma that is interesting by itself.

Lemma 2.1.8. Let Γ be a group acting faithfully on a totally ordered space (Ω, ≤)
by order-preserving transformations. Then for every Ω ⊂ Ω, there is a left-order
on Γ for which the stabilizer of Ω is a convex subgroup.
Proof. Proceed as in §1.1.3 using a well-order ≤wo on Ω for which Ω is an initial
segment.

Proof of Proposition 2.1.7. As we saw in Example 2.1.6, each space of cosets
Γ/Γλ inherits a total order ≤λ thatQis preserved by the left action of Γ. Fix a
well-order 6wo on Λ, and let Ω := λ∈Λ Γλ × Γ be endowed with the associate
dynamical-lexicographic total order ≤. This means that ([gλ ], g) ≤ ([hλ ], h) if
either the smallest (according to 6wo ) index λ such that [gλ ] 6= [hλ ] is such that
[gλ ] >λ [hλ ], or the classes of gλ and hλ (with respect to Γλ ) are equal for every
λ and g ≺ h. The left action of Γ on Ω is faithful
T and preserves this order. Since
the stabilizer of ([id]λ )λ∈Λ × Γ coincides with λ Γλ , the proposition follows from
the preceding lemma.


54

2.1.2

CHAPTER 2. A PLETHORA OF ORDERS

Free products

As we have seen in §1.2.3, free groups are bi-orderable. Actually, a much more
general statement involving free products holds.
Theorem 2.1.9. The free product of an arbitrary family of bi-orderable groups
is bi-orderable. Moreover, given bi-orders on each of the free factors, there is a
bi-order on the free product that extends these bi-orders.
Let us point out that a similar statement holds for left-orderability. However,
the proof is much simpler. Indeed, let Γ = ⋆Γλ be a free product of left-orderable
groups. Then the direct sum ⊕λ Γλ is left-orderable. Moreover, the kernel of the
natural homomorphism from Γ to ⊕λ Γλ is known to be a free group. Since free
groups are left-orderable, Γ itself is left-orderable. (An alternative –dynamical–
argument is contained in the proof of Theorem 2.2.30.)
The statement concerning bi-orderability is more subtle. For instance, the
argument above does not apply, as the bi-orders in the free kernel are not necessarily invariant under conjugacy by elements of Γ. Although we are mostly
concerned with left-orders here, we next reproduce the proof of Theorem 2.1.9
given in [17] in order to give some insight on Question 2.2.40 further on. The
reason is that this proof yields, for the case of free groups, bi-orders that are
different from those obtained using the lower central series (c.f. Example 2.1.11).
An elegant alternative proof, which is somewhat close to Vinogradov’s original
approach, is given in [13]; see also [107].
Exercise 2.1.10. Show that the free product of groups with the U.P.P. has the U.P.P.
(See [184] in case of problems.) Show an analogous statement for groups admitting a
locally-invariant order.

Bi-ordering free products via an inductive argument. Given a free product
of groups Γ = ⋆Γλ , every element may be written in a unique way as a product
f = f1 · · · fℓ , where each fi belongs to some Γλ and no consecutive fi′ s belong to
the same Γλ . We will call the integer ℓ = ℓ(f ) the length of the element f in the
free product ⋆Γλ .
Assume now that each Γλ is bi-orderable, and let λ be a bi-order on it. The
strategy will consist in defining a normal positive cone P on Γ by looking at the
lengths of the elements. Nevertheless, along this proof, we will need to consider
more general free products than Γ.
Elements of length 1. Such an element g belongs to one of the factors Γλ . We
then let f belong to P if and only if f ≻λ id.

2.1. PRODUCING NEW LEFT-ORDERS

55

Now, to deal with elements of length larger than 1, we begin by fixing a
left-order ≤ on Γ, and for each λ, we denote Gλ = ⋆ω>λ Γω .
Elements of larger length. Let f := f1 · · · fℓ be an element of length ℓ larger than
1, where fi ∈ Γλi for each i. Let λ(f ) := min≤ {λi }. Then f = g0 h1 g1 h2 · · · hk gk ,
where the h′j s correspond to the factors of f not lying in Γλ(f ) , and each gj belongs
to Γλ(f ) (g0 and/or gk+1 may be trivial). Notice that
f¯ := (h1 · · · hk )−1 f =

k
Y
(hi+1 · · · hk )−1 gi (hi+1 · · · hk ).
i=0

There are two cases:
(i) If h1 . . . hk 6= id, then since the length of h = hf := h1 · · · hk is smaller than
ℓ, we may assume inductively that we know whether h belongs to P or not. In
the case it belongs, we let f belong to P ; otherwise, we let f −1 belong to P .
Q
(ii) If h = id, then we think of f = ki=0 (hi+1 · · · hk )−1 gi (hi+1 · · · hk ) as an
element of length smaller than ℓ in the free product ⋆h∈Gλ hΓλ h−1 . By using the
above rules inside this free product, we may determine whether f should belong
to P or not.
One easily convinces that the above procedure yields a subset P of Γ that is
disjoint from P −1 and such that Γ \ {id} = P ∪ P −1 . We now proceed to check
that P is a normal semigroup.
To see that f1 f2 belongs to P for each f1 , f2 in P , we proceed by induction on
the length of f1 f2 in an appropriate free product. More precisely, if ℓ(f1 f2 ) = 1,
then both f1 , f2 belong to the same Γλ (and are λ -positive), hence f1 f2 ∈ P .
Otherwise, there are two cases:
– If λ(f1 ) < λ(f2 ) (resp. λ(f1 ) > λ(f2 )), then writing f1 f2 = hf1 f2 (f2−1 f¯1 f2 )
(resp. f1 f2 = f1 hf2 f¯2 ) we see that hf1 f2 = hf1 f2 (resp. hf1 f2 = f1 hf2 ). Since the
length of hf1 f2 (resp. f1 hf2 ) is (nonzero and) strictly smaller than that of f1 f2 ,
we may argue by induction to conclude that f1 f2 ∈ P .
¯
¯
– If λ(f1 ) = λ(f2 ) = λ, then writing f1 f2 = hf1 hf2 (h−1
f2 f1 hf2 )f2 we see that
hf1 f2 = hf1 hf2 . If either hf1 or hf2 is nontrivial, then this yield hf1 hf2 ∈ P , hence
f1 f2 ∈ P . If not, then the length of f1 f2 = f¯1 f¯2 viewed in the free product
⋆h∈Gλ hΓλ h−1 is smaller than its length in ⋆Γλ , which still allows concluding by
induction.
Finally, to show that P is normal, it is enough to show that gP g −1 = P for
all g of length 1. Fix f ∈ P . There are three cases:

56

CHAPTER 2. A PLETHORA OF ORDERS

– If λ(g) < λ(f ), then hgf g−1 = f . Since f ∈ P , by the definition of P we must
have gf g −1 ∈ P .
– If λ(g) = λ(f ), then hgf g−1 = hf . If hf 6= id, then it belongs to P , hence gf g −1
belongs to P as well. If hf = id, then the length of gf g −1 viewed in ⋆h∈Gλ hΓλ h−1
is smaller than its length in ⋆Γλ . Thus, an inductive argument applies in this
case.
– If λ(g) > λ(f ), then hgf g−1 = ghf g −1. If hf 6= id, then, as its length is smaller
than that of f , an inductive argument applies. If hf = id, then f may be written
−1
as f = (h1 g1 h−1
1 ) · · · (hk gk hk ), with g1 . . . , gk in Γλ(f ) and h1 , . . . , hk in Gλ(f ) .
Therefore,
k
Y
−1
(ghi )gi (ghi )−1 .
gf g =
i=1

Now recall that ≤ was chosen to be left-invariant, hence hi < hj holds if and
only if ghi < ghj . The construction of P then shows that f ∈ P if and only if
gf g −1 ∈ P .

Example 2.1.11. By performing the construction above appropriately, one may obtain
a bi–order on a free group F2 = hf, gi for which both f, g are positive but f [f, g] is
negative. Notice that this cannot happen for the Magnus left-order, as well as for any
bi–order on F2 obtained via the lower central series.

2.1.3

Left-orders from bi-orders

As we already pointed out, a left-orderable group all of whose left-orders
are bi-invariant is necessarily Abelian [54]. This suggests the existence of natural
procedures to create left-orders starting with bi-orders on groups. Here we briefly
discuss two of them.
Left-orders from the sequence of convex subgroups. Let {Γi } be the family
of convex subgroups for a bi-order  on a group Γ. Since  is bi-invariant, for
every g ∈ Γ, each subset of the form gΓi g −1 is also convex. Given any well-order
≤wo on the set of indices i, we may define a left-order ′ on Γ as follows: Given
g ∈ Γ, we look for the minimal (with respect to ≤wo ) i such that gΓi g −1 6= Γi ,
and we let j(i) so that gΓi g −1 = Γj(i) . If j(i) >wo i (resp. j(i) <wo i), then we let
g ≻′ id (resp. g ≺′ id); if g fixes each Γi , then we let g ≻′ id if and only if g ≻ id.
One easily checks that ′ is well-defined and left-invariant. Notice that, if
every convex subgroup is normal, then ′ coincides with the original bi-order .
on Thompson’s group F (c.f.
Example 2.1.12. Let us consider the bi-order −
x+
§1.2.4). Let (xi ) be a numbering of all dyadic, rational numbers of ]0, 1[. Each xi

2.1. PRODUCING NEW LEFT-ORDERS

57

gives raise to a convex subgroup Γi formed by the elements g such that g(x) = x for
all x ∈ [xi , 1]. Although there are more convex subgroups than these, this family is
invariant under the conjugacy action. By performing the construction above, we get
the left-order  on F for which f ≻ id if and only if f (xi ) > xi holds for the smallest
integer i such that f (xi ) 6= xi . (Compare §1.1.3.)

Combing elements with trivial conjugacy action on a certain left-order.
Proposition 2.1.14 below appears in [127], yet it was already implicit in [54].
Lemma 2.1.13. Suppose  is a left-order on a group Γ admitting a normal,
convex subgroup Γ∗ , and let g ∈ Γ \ Γ∗ . If conjugation by g preserves the leftorder on Γ∗ (that is, g(P+ ∩ Γ∗ )g −1 = P+ ∩ Γ∗ ), then there exists a left-order on
the subgroup hg, Γ∗i that has g as minimal positive element and coincides with 
on Γ∗ .
Proof. Since Γ∗ is normal in Γ, every element in hg, Γ∗i may be written in a
unique way in the form g n h, with n ∈ Z and h ∈ Γ∗ . Define ∗ on hg, Γ∗i by
letting g n h ∗ id if and only if either h ∈ P+ or h = id and n > 0. Invariance
of  under conjugacy by g shows that this is a well-defined left-order on hg, Γ∗i.
That ∗ coincides with  on Γ∗ follows from the definition. Finally, the fact that
g is the minimal positive element of ∗ follows from the definition.

Combined with the convex extension technique, this lemma allows producing
many interesting left-orders. Invoking Example 2.1.1, this is summarized in
Proposition 2.1.14. Let (Γ, ) be a bi-ordered group, and let Γg ⊂ Γg be the
convex jump associated to an element g ∈ Γ. Assume that the quotient Γg /hg, Γg i
is torsion-free. Then there exists a left-order ′ on Γ having g as minimal positive
element.
Proof. First notice that both Γg and Γg are invariant under conjugacy by g. As
 is bi-invariant, conjugacy by g preserves the positive cone of Γg . Thus, we are
under the hypothesis of the preceding lemma, which allows to produce a left-order
on hg, Γg i having g as minimal positive element. This left-order may be extended
to a left-order ∗ on Γg , as Γg /hg, Γg i is assumed to be torsion-free (recall that
Γg /Γg is Abelian; see Example 2.1.1). Finally, we let ′ be the extension of ∗
by .

Example 2.1.15. Given an element g in the free group Γ := Fn , let k = k(g) ∈ N be
such that g ∈ Γk \ Γk+1 , where Γi denotes the ith -term of the lower central series. If

58

CHAPTER 2. A PLETHORA OF ORDERS

gΓk has no nontrivial root in Γk−1 /Γk , then we are under the hypothesis of Proposition
2.1.14 for any bi-order on Fn obtained from the series Γi . Thus, g appears as the
minimal positive element for a left-order on Fn .

2.2

The Space of Left-Orders

Following Ghys [78] and Sikora [181], given a left-orderable group Γ, we denote
by LO(Γ) the set of all left-orders on Γ. This space of left-orders carries a
natural (Hausdorff and totally disconnected) topology whose sub-basis is the
family of sets of the form Uf,g = {: f ≺ g}. Because of left-invariance, another
sub-basis is the family of sets Vf = {: id ≺ f }.
To better understand the topology on LO(Γ), one may proceed as in §1.1.2 by
identifying left-orders on Γ to certain points in {−1, +1}Γ\{id} . Nevertheless, to
cover also the case of partial left-orders, it is better to model LO(Γ) as a subset
of {−1, +1}Γ×Γ\∆ , namely the one formed by the functions ϕ satisfying:
– (Reflexivity) ϕ(g, h) = +1 if and only if ϕ(h, g) = −1;
– (Transitivity) if ϕ(f, g) = ϕ(g, h) = +1, then ϕ(f, h) = +1;
– (Left-invariance) ϕ(f g, f h) = ϕ(g, h) for all f and g 6= h in Γ.
Indeed, every left-order  on Γ leads to such a function ϕ , namely ϕ (g, h) = +1
if and only if g ≻ h. Conversely, every ϕ with the above properties induces a
left-order ϕ on Γ, namely g ≻ϕ h if and only if ϕ(g, h) = +1. Now, if we
endow {−1, +1}Γ×Γ\∆ with the product topology and the subset above with the
subspace one, then the induced topology on LO(Γ) coincides with the one previously defined by prescribing the sub-basis elements. As a consequence, since
{−1, +1}Γ×Γ\∆ is a compact space and the subspace above is closed, the topological space LO(Γ) is compact.
As a matter of example, in regard to the convex extension procedure (c.f.
§2.1.1), the reader should have no problem in showing the next
Proposition 2.2.1. Let  be a left-order on Γ and Γ∗ a -convex subgroup.
Then the map from LO(Γ∗ ) into LO(Γ) that sends ∗ into its convex extension
by  is a continuous injection. If, in addition, Γ∗ is normal, then there is a
continuous injection from LO(Γ∗ ) × LO(Γ/Γ∗ ) into LO(Γ) having  in its image.
Example 2.2.2. The subspace of dynamical-lexicographic left-orders on Homeo+ (R)
(c.f. §1.1.3) is not closed inside LO(Homeo+ (R)) (it is unknown whether it is dense).
To show this, let (yk ) be a dense sequence of real numbers, and let (xn ) be a monotone

2.2. THE SPACE OF LEFT-ORDERS

59

sequence converging to a point x ∈ R. For each n define a sequence (yn,k )k by yn,1 = xn
and yn,k = yk−1 for k > 1. This gives raise to a sequence of left-orders n (the sign
of each point yn,k is chosen to be +). Passing to a subsequence if necessary, we may
assume that n converges to a left-order  on Homeo+ (R). We claim that  is not
of type. Indeed, let ′ be an arbitrary left-order, and let x′ be the first point different
from x for the well-order leading to ′ (thus, x′ may be the first or the second term
of this well-order). Let f ∈ Homeo+ (R) be such that f (x) = x and f (y) > y for all
y 6= x. Let g, h be elements in Homeo+ (R) that coincide with f in a neighborhood of x
and g(x′ ) > x′ > h(x′ ). By definition, the signs of g, h with respect to ′ are different.
However, since g(yn,1 ) = g(xn ) > xn = yn,1 and h(yn,1 ) > yn,1 for all n large enough,
both g, h are n -positive. Passing to limits, both g, h become -positive, thus showing
that  cannot coincide with ′ .

If Γ is a countable left-orderable group, then the natural topology of LO(Γ)
is metrizable. Indeed, if G0 ⊂ G1 ⊂ . . . is a complete exhaustion of Γ by finite
sets, then we can define the distance between two different left-orders ≤ and 
by letting d(≤, ) = 2−n , where n is the maximum non-negative integer such
that ≤ and  coincide on Gn . An equivalent metric d′ is obtained by letting
′
d′ (≤, ) = 2−n , where n′ is the maximum non-negative integer such that the
positive cones of ≤ and  coincide on Gn′ , that is, P≤+ ∩ Gn′ = P+ ∩ Gn′ . One
easily checks that these metrics are ultrametric. Moreover, the fact that LO(Γ)
is compact becomes more transparent in this case, as it follows from a Cantor
diagonal type argument.
When Γ is finitely generated, one may choose Gn as being the ball of radius
n with respect to some finite, symmetric system of generators G of Γ (centered
at id), usually denoted by Bn = Bn (id). Here, symmetric means that g −1 ∈ G
for all g ∈ G, and the ball Bn is the set of elements having word-length at most
n, where the word-length of g ∈ Γ is the minimum m for which g which can be
written in the form g = gi1 gi2 · · · gim , with gij ∈ G. One easily checks that the
metrics on LO(Γ) resulting from two different finite systems of generators are not
only topologically but also Hölder equivalent.
Exercise 2.2.3. Given a bi-orderable group Γ, denote by BO(Γ) the space of biorders of Γ. Show that BO(Γ) is closed inside LO(Γ), hence compact.
Exercise 2.2.4. Given a group Γ admitting a locally-invariant order (c.f. §1.3.2),
denote by LIO(Γ) the set of all locally-invariant orders on Γ. Consider the topology
on LIO(Γ) having as a sub-basis the family of sets Uf,g = {: f ≺ g}. Show that,
endowed with this topology, LIO(Γ) is compact. Conclude that a group Γ admits a

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CHAPTER 2. A PLETHORA OF ORDERS

locally-invariant order if and only if each of its finitely-generated subgroups admits such
an order. (Compare [42, Theorem 2.4].)
Hint. As a model of LIO(Γ) consider the subset of {−1, 0, +1}Γ×Γ\∆ formed by the
functions ϕ such that ϕ(g, h) = +1 if and only if ϕ(h, g) = −1, and such that for every
g 6= id and h ∈ Γ one has either ϕ(hg, h) = +1 or ϕ(hg−1 , h) = −1. (Two elements g, h
that are incomparable for a locally invariant order will then satisfy ϕ (g, h) = 0...)
Exercise 2.2.5. Complete the proof of Proposition 1.3.9 showing that weakly diffuse
groups admit a locally-invariant order. (See [126] in case of problems.)
Hint. By a compactness type argument, it is enough to show the following: For each
finite subset A of Γ, there exists a partial order  such that for all f ∈ A and each
nontrivial element g ∈ Γ such that both f g and f g−1 lie in A, either f g ≻ f or
f g−1 ≻ f . To construct such a , proceed by induction, the case where A is a single
element being evident. Now, given an arbitrary A, by the weakly diffuse property there
is h ∈ A such that for each nontrivial element g ∈ Γ, either hg ∈
/ A or hg −1 ∈
/ A. By
the induction hypothesis, A \ {h} admits an order as requested. Extend  to all A by
declaring h being larger than all other elements.

The group Γ (continuously) acts on LO(Γ) by conjugacy (equivalently, by
right multiplication): given an order  with positive cone P + and an element
f ∈ Γ, the image of  under f is the order f whose positive cone is f P + f −1 . In
other words, one has g f h if and only if f −1 gf  f −1 hf , which is equivalent
to gf  hf . Also notice that the map sending  to  from Example 1.1.2 is a
continuous involution of LO(Γ).
Example 2.2.6. If a group left-order is obtained via an action on an totally ordered
space Ω, then the conjugacy action corresponds to changing the order of the comparison
points. More precisely, in the notation of §1.1.3, if  comes from a well-order ≤wo on
Ω, then f is obtained from the same action using the well-order f∗ (≤wo ) given by
ω1 f∗ (≤wo ) ω2 whenever f (ω1 ) ≤wo f (ω2 ). In particular, for countable subgroups of
Homeo+ (R), if  is induced from a dense sequence (xn ) in R, then f is induced from
the sequence (f (xn )).
Remark 2.2.7. If Γ is a left-orderable group, then the whole group of automorphisms
of Γ (and not only the group of internal automorphisms) acts on LO(Γ). This may be
useful to study bi-orderable groups. Indeed, since the fixed points for the right action
of Γ on LO(Γ) correspond to the bi-invariant left-orders, the group Out(Γ) of outer
automorphisms of Γ acts on the corresponding space of bi-orders BO(Γ). The reader
is referred to [117] for some applications of this idea.

In general, the study of the dynamics of the action of Γ on LO(Γ) should
reveal useful information. Let us formulate a very simple questions on this.

2.2. THE SPACE OF LEFT-ORDERS

61

Question 2.2.8. For which countable left-orderable groups having an infinite
space of left-orders is the action of Γ on LO(Γ) uniformly equicontinuous or
distal ? The same question makes sense for minimality1, or for having a dense
orbit (the latter is the case of free groups, as we will see in §2.2.2; for the former,
we do not know any example).

2.2.1

Finitely many or uncountably many left-orders

We now state the first nontrivial general theorem concerning the space of leftorders of left-orderable groups. This result was first obtained by Linnell [123] by
elaborating on previous ideas of Smirnov, Tararin, and Zenkov. Let us point that
no analogue for spaces of bi-orders holds [30]; see however §3.2.6.
Theorem 2.2.9. If the space of left-orders of a left-orderable group is infinite,
then it is uncountable.
The starting point to show this result is the following. Let Γ be a left-orderable
group and M a minimal subset of LO(Γ), that is, a nonempty, closed subset
that is invariant under the conjugacy action of Γ and does not properly contain
any nonempty, closed, invariant set. Since the set M ′ of accumulation points of
M is both closed and invariant, one has either M ′ = M or M ′ = ∅. In other
words, either M has no isolated point, or it is finite. In the former case, a wellknown result in General Topology asserts that M must be uncountable (see [95,
Theorem 2-80]). In the latter case, the stabilizer of any point  of M is a finiteindex subgroup of Γ restricted to which  is bi-invariant. Theorem 2.2.9 then
follows from the following
Proposition 2.2.10. Let (Γ, ) be a left-ordered group containing a finite-index
subgroup Γ0 restricted to which  is bi-invariant. If  has a neighborhood in
LO(Γ) containing only countably many left-orders, then LO(Γ) is finite.
The proof of this proposition uses results and techniques from the theory of
Conradian orders. Hence, we postpone the (end of the) proof of Theorem 2.2.9
to §3.2.6.
Notice that the argument above “distinguishes” conjugate left-orders, though
one would like to consider them as being “equal” in many senses (for instance,
they share all dynamical features). This naturally leads to many questions, as
for instance:
1

Recall that an action is said to be minimal if every orbit is dense.

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CHAPTER 2. A PLETHORA OF ORDERS

Question 2.2.11. Can the space of orbits LO(Γ)/Γ be a non-standard Borelian
space for a left-orderable group Γ ?
Besides, K. Mann and the third author asked the next
Question 2.2.12. Does there exist a left-orderable group Γ with a left-order
which neither is isolated nor belongs to a Cantor subset of LO(Γ) ?
In the rest of this section, we give a beautiful characterization (due to Tararin
[187]) of groups having finitely many left-orders. (We will refer to them as
Tararin groups.) Recall that a rational series for a group Γ is a finite
sequence of subgroups
{id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ

that is subnormal (that is, each Γi is normal in Γi−1 , but not necessarily in Γ),
and such that each quotient Γi−1 /Γi is torsion-free rank-1 Abelian. Such a series
is said to be normal if each Γi is normal in Γ. Notice that every group admitting
a rational series is left-orderable.
Theorem 2.2.13. Every left-orderable group having only finitely many left-orders
admits a unique rational series
{id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ.

This series is normal and no quotient Γi−2 /Γi is bi-orderable. Conversely, if
a group Γ admits a normal rational series such that no quotient Γi−2 /Γi is biorderable, then (Γ is left-orderable and) its space of left-orders LO(Γ) is finite.
In such a situation, for every left-order on Γ, the convex subgroups are exactly
Γ0 , Γ1 , . . . , Γk , the number of left-orders on Γ is 2k , and each left-order is uniquely
determined by the sequence of signs of any family of elements gi ∈ Γi−1 \ Γi .
Example 2.2.14. The Klein-bottle group K2 = ha, b : a−1 ba = b−1 i admits exactly
four left-orders, having as positive cones ha, bi+ , ha, b−1 i+ , ha−1 , bi+ , and ha−1 , b−1 i+ ,
respectively (see §2.2.3 for details). The associate rational series is {id} ⊳ hbi ⊳ K2 .
More generally, let us consider the group
−1
Kk = ha1 , . . . , ak : a−1
i+1 ai ai+1 = ai , ai aj = aj ai for |i − j| ≥ 2i.

One can easily check (either using Theorem 2.2.13 above or by a direct computation)
that Kk admits 2k left-orders, each of which being determined by the signs of the ai ’s.
The corresponding rational series is
{id} ⊳ ha1 i ⊳ ha1 , a2 i ⊳ . . . ⊳ ha1 , a2 , . . . , ak i.

2.2. THE SPACE OF LEFT-ORDERS

63

Example 2.2.15. A dynamical counterpart of having finitely many left-orders for a
group is that, up to semiconjugacy, only a few actions on the real line may arise. For
the case of the group K2 above, this translates into the two items below. (See Example
2.2.24 for an application.)
Claim (i). Suppose K2 = ha, b : a−1 ba = b−1 i acts on the real line and there is x ∈ R
such that x ≤ a(x). Then b has a fixed point in I = [x, a(x)].

Otherwise, changing b by its inverse if necessary, we may assume that b(z) > z for all
z ∈ I. In particular, a(x) < ba(x), hence x < a−1 ba(x) = b−1 (x). Therefore, b(x) < x,
a contradiction.
As a consequence, every open interval I fixed by a on which a acts freely is also
fixed by b. Moreover, b has infinitely many fixed points in I.

Claim (ii). For every open interval J fixed by b and containing no fixed point of b inside,
we have a(J) ∩ J = ∅.

Indeed, as hbi is normal in K, we have that a(J) ∩ J is either J or empty. But the
first possibility cannot occur, since in that case b would have fixed points in J, due to
Claim (i).

The proof of Theorem 2.2.13 will be divided into several parts, some of which
involve notions and results contained in the beginning of the next chapter.
Lemma 2.2.16. If a left-orderable group admits only finitely many left-orders,
then all of them are Conradian.
Proof. Let Γ be a left-orderable group whose space of left-orders is finite. For
a finite-index subgroup Γ∗ of Γ, the conjugacy action on LO(Γ) is trivial. This
means that every left-order of Γ is bi-invariant (hence Conradian) when restricted
to Γ∗ . The lemma then follows from Proposition 3.2.9.

We may now proceed to show the first claim contained in Theorem 2.2.13.
Proposition 2.2.17. Let Γ be a left-orderable group admitting only finitely many
left-orders. Then, for every left-order  on Γ, the chain of -convex subgroups
is a finite rational series.
Proof. To show finiteness, let us fix n ∈ N such that the number of left-orders on
Γ is strictly smaller than 2n . Following Zenkov [194], we claim that the family of
-convex subgroups has cardinality ≤ n. Otherwise, if
{id} = Γ0 ( Γ1 ( . . . ( Γn = Γ

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CHAPTER 2. A PLETHORA OF ORDERS

is a chain of distinct -convex subgroups, then for each ι = (i1 , . . . , in ) ∈ {−1, +1}n
we may define the left-order ι as being equal to n , where 1 , 2 , . . . , n are
the left-orders on Γ1 , . . . , Γn , respectively, which are inductively defined by:
– If i1 = 1 (resp. i1 = −1), then 1 is the restriction of  (resp. ) to Γ1 .
– For n ≥ k ≥ 2, if ik = 1 (resp. ik = −1), then k is the extension of k−1 by
the restriction of  (resp. ) to Γk .
Clearly, the left-orders ι are different for different choices of ι, which shows the
claim.
Now let
{id} = Γk ( Γk−1 ( . . . ( Γ0 = Γ
be the chain of all -convex subgroups of Γ. In the terminology of Example 2.1.1,
the inclusion Γi ( Γi−1 is the convex jump associated to any element in Γi−1 \ Γi .
By Theorem 3.2.27, Γi is normal in Γi−1 , and the induced left-order on Γi−1 /Γi
is Archimedean. By Hölder’s theorem (c.f. §3.1), this quotient is torsion-free
Abelian. Finally, its rank must be 1, as otherwise it would admit uncountably
many left-orders (c.f. §1.2.1), which would allow to produce –by convex extension
(c.f. Proposition 2.2.1)– uncountably many left-orders on Γ.

Proposition 2.2.18. A left-orderable group admitting finitely many left-orders
has a unique (hence normal) rational series.
Proof. If {id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ is a rational series for a group Γ, then
for every h ∈ Γ, the conjugate series
{id} = hΓk h−1 ⊳ hΓk−1 h−1 ⊳ . . . ⊳ hΓ0 h−1 = Γ
is also rational. Therefore, the uniqueness of such a series implies its normality.
To show the uniqueness, let us consider two rational series
{id} = Gk ⊳Gk−1 ⊳. . .⊳G1 ⊳G0 = Γ,

′

′

{id} = H k ⊳H k −1 ⊳. . .⊳H 1 ⊳H 0 = Γ,

where Γ is supposed to admit only finitely many left-orders. Both G1 and H 1 are
normal in Γ, and the quotients Γ/H 1 and Γ/G1 are torsion-free Abelian. This
easily implies that G1 ∩ H 1 is also normal in Γ and the quotient Γ/(G1 ∩ H 1 ) is
torsion-free Abelian. Since G1 ∩ H 1 is convex with respect to some left-order on
Γ (c.f. Proposition 2.1.7), the rank of Γ/(G1 ∩ H 1 ) must be 1; otherwise, this
quotient would admit uncountably many left-orders, thus yielding –by convex
extension– uncountably many left-orders on Γ. We conclude that, for every g ∈ G1

2.2. THE SPACE OF LEFT-ORDERS

65

(resp. h ∈ H 1 ), one has g n ∈ G1 ∩ H 1 (resp. f n ∈ G1 ∩ H 1 ) for some n ∈ N.
However, both G1 and H 1 are stable under roots, hence g ∈ H 1 (resp. h ∈ G1 ).
This easily implies that G1 = H 1 .
Arguing similarly but with G1 = H 1 instead of Γ, we obtain G2 = H 2 .
Proceeding in this way finitely many times, we conclude that the rational series
above coincide.

The structure of the quotients Γi−2 /Γi is given by the (proof of the) next
Proposition 2.2.19. Let Γ be a left-orderable group having finitely many leftorders. If
{id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ
is the unique rational series of Γ, then no quotient Γi−2 /Γi is bi-orderable.

Proof. The group Γi−1 /Γi is normal in Γi−2 /Γi . Hence, Γi−1 /Γ acts by conjugacy on the torsion-free, rank-1, Abelian group Γi−1 /Γi . Now it is easy to see
that every automorphism of a torsion-free, rank-1, Abelian group is induced by
the multiplication by a real number. As a consequence, the non-Abelian group
Γi−2 /Γi embeds into Aff(R). The non bi-orderability of Γi−2 /Γi is thus equivalent
to that the image of this embedding is not contained in Aff + (R). (This is also
equivalent to that some element is conjugate to a negative power of itself.) But
if this were not the case, then, according to §1.2.2, the quotient Γi−2 /Γi (hence
Γ) would admit uncountably many left-orders.

We next proceed to show the converse statements.
Proposition 2.2.20. Let Γ be a group admitting a normal rational series
{id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ
such that no quotient Γi−2 /Γi is bi-orderable. For each i ∈ {1, . . . , k}, let us choose
gi ∈ Γi−1 \ Γi . Then every left-order on Γ is completely determined by the signs
of the gi ’s. Moreover, for any such choice of signs, there exists a left-order on Γ
realizing it.
Proof. The realization of signs ι ∈ {−1, +1}k proceeds as the proof of the first
claim of Proposition 2.2.17, and we leave the details to the reader. As before, we
will denote by ι the left-order that realizes the corresponding signs.
Now let  be a left-order on Γ, and let ι = (i1 , . . . , in ) be the associate
sequence of signs of the gi ’s. To prove that the positive cones of  and ι

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CHAPTER 2. A PLETHORA OF ORDERS

coincide, it suffices to show that P+ι ⊂ P+ (c.f. Exercise 2.2.42). As we saw in
the proof of the preceding proposition, after changing gi by a root if necessary,
−1
we may assume that gi−1
gi gi−1 = giri for a negative rational number ri . Since
gkik ∈ Γk−1 belongs to both P+ι and P+ , and since Γk−1 is rank-1 Abelian, we
have
P+ι ∩ Γk−1 ⊂ P+ .
ti

k−1
Now every element g ∈ Γk−2 \ Γk−1 may be written as g = gksik gk−1
for some
rational numbers s and t 6= 0; such an element is ι -positive if and only if
t > 0. If besides s ≥ 0, then g is also -positive. Otherwise, s < 0, and g may be
ri tik−1 sik
gk , and since ri t > 0, this shows that g is still -positive.
rewritten as g = gk−1
Therefore, we have

P+ι ∩ Γk−2 \ Γk−1 ⊂ P+ ,

hence

P+ι ∩ Γk−2 ⊂ P+ .

Proceeding in this way finitely many times, one concludes that P+ι ⊂ P+ .



Exercise 2.2.21. Show that every Tararin group admits a unique nontrivial torsionfree Abelian quotient, namely the quotient with respect to the maximal proper convex
subgroup.
Exercise 2.2.22. Let Γ be a left-orderable group for which the whole family of subgroups that are convex for some left-order on Γ is finite. Show that Γ admits only
finitely many left-orders.
Remark. This result is also due to Tararin; see [121, §5.2] in case of problems.

We next provide a quite clarifying result on the dynamics of the action of a
Tararin group on its space of left-orders.
Proposition 2.2.23. The action of a Tararin group Γ on its space of left-orders
LO(Γ) has two orbits. Moreover, for any two left-orders  and ′ on Γ, there
is g ∈ Γ such that g and ′ coincide on the maximal convex subgroup of any of
its left-orders (namely Γ1 , in the notation of Theorem 2.2.13). Moreover, if we
let h be any element in Γ \ Γ1 acting on Γ1 /Γ2 as the multiplication by a negative
number (c.f. Proposition 2.2.19 and its proof ), then g can be taken either in Γ1
or in hΓ1 .
Proof. Choose elements gi ∈ Γi−1 \ Γi , where i ∈ {2, . . . , k}. In case the signs of
g2 under  and ′ are the same, let h2 := id; otherwise, let h2 be the element h
above. Then the sign of g2 for h2 is the same as that for ′ .

2.2. THE SPACE OF LEFT-ORDERS

67

In case the signs of g3 for h2 and ′ coincide, let h3 := id. Otherwise, let
h3 be an element in Γ1 \ Γ2 acting on Γ2 /Γ3 as the multiplication by a negative
number. Then the signs of both g2 , g3 for h3 h2 and ′ are the same.
Continuing this way, we obtain an element g := hk · · · h2 such that the signs
of all gi ’s for g and ′ coincide, where i ∈ {2, . . . , k}. This certainly implies
that g and ′ are the same when restricted to Γ1 .

Example 2.2.24. Let us consider the group
−1
Γ := as , s ∈ R : a−1
s at as = at whenever t < s .

We claim that Γ is left-orderable but has no nontrivial action on the line. Proving that
Γ is left-orderable is easy. Indeed, every g ∈ Γ may be written in normal form as
g = ans11 · · · anskk , with s1 > s2 > . . . > sk , ni 6= 0.
We may then declare such a g ∈ Γ to be positive if n1 > 0, thus getting a left-order on Γ
(details are left to the reader). Next, assume for a contradiction that Γ acts nontrivially
on the real line. Then there is t ∈ R such that at acts nontrivially. Let It be an open
interval fixed by at containing no fixed point of at . By Example 2.2.15, for each s > t,
we have that as has no fixed point in the closure of It , and that as (It ) ∩ It = ∅. Let Is
be the minimal open interval fixed by as that contains It . Example 2.2.15 again implies
that for each pair of real numbers s1 > s2 larger than t, we have as1 (Is2 ) ∩ Is2 = ∅.
We thus obtain that {as (It )}s>t is an uncountable collection of disjoint open intervals,
which is absurd.

2.2.2

The space of left-orders of the free group

The space of left-orders of the free group is known to be homeomorphic to the
Cantor set. This is a result of McCleary essentially contained in [136], though
an alternative (dynamical) proof appears in [153]. The general strategy of [153]
proceeds as follows:
– Associated to a given left-order on Fn , let us consider the corresponding dynamical realization.
– If we perturb the generators of this realization (as homeomorphisms of the
line), we still have an action of the free group, which is “in general” faithful, thus
yielding a new left-order on Fn .
– If the perturbation above is “small”, then the new left-order is close to the
original one.

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CHAPTER 2. A PLETHORA OF ORDERS

– Finally, the perturbation can be made so that the new left-order is different
from the original one, as otherwise the original action would be “structurally
stable” (i.e. actions which are “close” to it are semiconjugate), which is easily
seen to be impossible.
Remark 2.2.25. The fact that isolated left-orders induce structurally stable actions
holds in full generality. The converse is, however, false, as it is shown for example by the
Baumslag-Solitar group (c.f. §3.3.1). Nevertheless, under certain natural assumptions,
the converse is still true. See [135] for more on all of this.

As we will see below, a similar but more careful argument shows that the space
of left-orders of the free product of two finitely-generated left-orderable groups is
a Cantor set.
More interestingly, using results from [119, 120, 136], Clay has shown the
existence of a left-order on Fn whose orbit under the conjugacy action is dense
[44]. Using this, he deduces that LO(Fn ) is homeomorphic to the Cantor set by
means of the argument contained in the following
Exercise 2.2.26. Let Γ be a countable group having a left-order whose orbit under
the conjugacy action is dense. Show that LO(Γ) is a Cantor set.
Hint. If there is an isolated left-order , then its reverse left-order  is also isolated.
If there is a left-order of dense orbit, this forces the existence of g ∈ Γ so that g = .
However, this is impossible, since the signs of g for both  and  coincide.
Remark. For the Baumslag-Solitar group BS(1, ℓ), no left-order has dense orbit. However, from the description given in §1.2.3, it readily follows that, for each irrational
number ε 6= 0, the orbit of ε under the action of the whole group of automorphisms
is dense.

Actually, the fact that an orbit is dense is not rare but “generic” in the space
of left-orders of the free group, as follows from the next proposition that relies on
a quite classical argument.
Proposition 2.2.27. Let Γ be a countable left-orderable group. If Γ admits a
dense orbit under the conjugacy action, then this is the case for the orbits of a
Gδ -dense set of points in LO(Γ).
Proof. Consider an arbitrary finite family of elements f1 , . . . , fk in Γ for which
the basic open set Vf1 ∩ . . . ∩ Vfk is nonempty. (Recall that Vf := {: id ≺ f }.)
Let LO(Γ; f1 , . . . , fk ) be the subset of LO(Γ) formed by the left-orders  for
which there exists g ∈ Γ such that g belongs to Vf1 ∩ . . . ∩ Vfk . Then the set

2.2. THE SPACE OF LEFT-ORDERS

69

LO(Γ; f1 , . . . , fk ) is a union of open basic sets, hence open. Moreover, since we
are assuming the existence of a dense orbit, LO(Γ; f1 , . . . , fk ) is also dense.
Now let LO∗ (Γ) be the (countable) intersection of all the sets LO(Γ; f1 , . . . , fk )
obtained above. On the one hand, Baire’s category theorem implies that LO∗ (Γ)
is a Gδ -dense subset of LO(Γ). On the other hand, the definition easily yields
that every left-order in LO∗ (Γ) has a dense orbit.

In [171], a dynamical proof of the existence of a left-order with dense orbit
in LO(Fn ) is given. This proof is based on the following construction (which is
closely related to the ideas of [136]):
– Choose a countable dense set k of LO(Fn ), and for each k consider the dynamical realization Φk of k .
– Fix a sequence of positive integers n(k) converging to infinity very fast, and a
family of disjoint intervals [r(k), s(k)] that is unbounded in both directions.
– For each k, take a conjugate copy on [r(k), s(k)] of the restriction of Φk to
[−n(k), n(k)].
– Take extensions of the generators of Fn so that they become homeomorphisms
of the line.
Roughly, the resulting action contains all the possible “finite information” of
all left-orders of Fn . By performing the construction carefully, one can ensure that
there is a single orbit containing the “center” of every [r(k), s(k)]. It therefore
yields to a new left-order  on Fn , and by suitable conjugacies inside Fn , this
left-order “captures” all the information above. In concrete terms, the orbit of 
under the conjugacy action is dense.
Example 2.2.28. For Thompson’s group F, no description of all left-orders is available.
(For bi-orders, see §1.2.4). Actually, it is unknown whether its space of left-orders is a
Cantor set. This question is actually open for all non-solvable groups of piecewise-affine
homeomorphisms of the interval.
Example 2.2.29. It can be checked in many ways that the π1 of orientable surfaces
are left-orderable. For instance, these groups are torsion-free and 1-relator, and all
these groups are locally indicable, a property discussed in §3.2.1 which is stronger
than left-orderability (c.f. Remark 3.2.6). Following the lines of the proof above, it
has been recently showed in [3] that the spaces of left-orders of compact hyperbolic
surface groups are homeomorphic to the Cantor set; actually, these groups also admit
left-orders whose orbits under the conjugacy action are dense.

The case of free products. Following a short argument from [171], we next
show the following

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CHAPTER 2. A PLETHORA OF ORDERS

Theorem 2.2.30. The space of left-orders of the free product of any two leftorderable groups has no isolated point.
Proof. We assume that the factors Γ1 , Γ2 of our free product Γ := Γ1 ⋆ Γ2 are
finitely generated; for the general case, see Exercise 2.2.31. Given a left-order  on
Γ, we consider the associated dynamical realization (see Proposition 1.1.8 and the
comments after it). Fix a finite system of generators of Γ, and for each n ∈ N let
fn , gn be, respectively, the smallest and largest element (for ) in the ball of radius
n in Γ. Let ϕn an orientation-preserving homeomorphism of the real line which is
the identity on [t(fn ), t(gn )]. Consider the following action of Γ on the real line:
for g ∈ Γ1 , the action is the conjugate of its -dynamical realization under ϕ; for
g ∈ Γ2 , the action is its -dynamical realization. (Notice that this yields an action
since Γ is a free product; however, this action may fail to be faithful). We claim
that if (xn ) is a dense sequence of points starting at x1 := t(id), then the positive
cone of the induced dynamical lexicographic left-order ϕn coincides with that
of  on the ball of radius n. Indeed, by construction, an element h ∈ Γ belongs
to Pϕn if and only if ϕ−1
n hϕn (t(id)) > t(id). Now, since ϕn |[t(fn ),t(gn )] ≡ Id, this
−1
is equivalent to ϕn h(t(id)) > t(id), that is ϕ−1
n (t(h)) > t(id). Now, if h belongs
to the ball of radius n, then t(h) lies in [t(fn ), t(gn )], hence ϕ−1
n (t(h)) = t(h), and
this is bigger than t(id) if and only if h is -positive.
Now fix n ∈ N and let us perform the preceding construction with a map ϕn
such that ϕn (s2 ) = s1 for s1 , s2 satisfying t(gn ) < s1 < t(h1,n ) < t(h2,n ) < s2 ,
where hi,n is in Γi . Since t(h1,n ) < t(h2,n ), we have h1,n ≺ h2,n . Now, from
−1
−1
ϕ−1
n (t(h1,n )) > ϕn (s1 ) = s2 > t(h2,n ) we obtain ϕn h1,n ϕn (t(id)) > h2,n (t(id)),
which by construction is equivalent to h1,n ≻ϕn h2,n .
Although the left-order ϕn may be partial (this arises when the new action
of Γ is unfaithful), it can be extended (using the convex extension procedure) to
a left-order n . By construction, the positive cones of  and n coincide on the
ball of radius n, though n and  are different. This concludes the proof.

Exercise 2.2.31. Provide the details of the proof of the preceding theorem for non
finitely-generated factors.
Hint. Use a compactness type argument.
Remark 2.2.32. The preceding theorem does not hold for direct products. Indeed,
using dynamical methods, it is shown in [135] that the space of left-orders of F2 × Z
has isolated points. However, we do not know of any example of left-orderable groups
Γ1 and Γ2 having the Cantor set as their spaces of left-orders but for which the space
of left-orders of the product Γ1 × Γ2 is not a Cantor set.

2.2. THE SPACE OF LEFT-ORDERS

71

A geometric/combinatorial proof. The fact that the space of left-orders of
a free group is a Cantor set can be established by means of different arguments.
Roughly, this is done in two steps:
Step I. If a left-order is an isolated point in the space of left-orders of the free
group, then its positive cone must be finitely generated as a semigroup.
Step II. There is no finitely-generated positive cone in the free group.
Concerning Step I, it is not hard to see that a finitely-generated positive cone
yields an isolated point in the space of left-orders (see Proposition 2.2.41 below),
yet the converse is not necessarily true (c.f. Example 2.2.43). The issue here
is that the converse can be directly established for free groups. This is due to
Smith and Clay, and we reproduce their (quite involved) proof from [49] below.
(It would be desirable to get more geometric/transparent arguments that apply
to other groups.)
Theorem 2.2.33. If a left-order on Fn is isolated in the space of left-orders, then
its positive cone must be finitely generated as a semigroup.
Proof. Let BN := BN (id) denote the ball of radius N with respect to the
canonical system of generators. Say that a subset S ⊂ Fn is total at length N if
it is antisymmetric (i.e. g ∈ S =⇒ g −1 ∈
/ S) and for all g ∈ BN \ {id}, either
−1
g ∈ S or g ∈ S. (Notice that id ∈
/ S.) The crucial point of the proof is the
following
Claim (i). If S ⊂ BN is total at length N −1 and satisfies S = hSi+ ∩BN , then for
every element g of length N not lying in S ∪ S −1 , the semigroup hS, gi+ remains
antisymmetric.
Let us assume this for a while, and let  be an isolated left-order on Fn . Let
f1 , . . . , fk be finitely many -positive elements such that  is the only left-order
on Fn for which all these elements are positive. If P is not finitely generated as
a semigroup, then there must exist an increasing sequence of integers Nm such
that each set S := P ∩ BNm is total at length Nm though there is g = gm of
length Nm + 1 that is not contained in S ∪ S −1 . By Claim (i), the semigroup
hS, gi+ is antisymmetric. Since it is total of length Nm , using Claim (i) in an
inductive way we may extend it to an antisymmetric semigroup which together
with its inverse covers Fn \ {id}, thus inducing a left-order on Fn . Obviously, the
same procedure can be carried out starting with g −1 instead of g. Now, if Nm
is large enough so that f1 , . . . , fk are all contained in BNm , then the procedure
above would give at least two different left-orders with all these elements positive
(one with g positive, the other with g negative). This is a contradiction.

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CHAPTER 2. A PLETHORA OF ORDERS

Let us now proceed to the proof of Claim (i). To do this, say that a finite
subset S ⊂ Fn is stable if for all f, g in S, the product f g lies in S whenever
kf gk ≤ max{kf k, kgk}. (Here and in what follows, k · k stands for the wordmetric on Fn .)
Claim (ii). If S ⊂ Fn is stable and g ∈ hSi+ is written in the form g = h1 · · · hk ,
with each hi ∈ S and k minimal, then for each 1 ≤ i ≤ k − 1,
khi+1 · · · hk k < khi hi+1 · · · hk k.
Write hi = fi f¯i , hi+1 = f¯i−1 fi+1 , with no cancellation in fi fi+1 = hi hi+1 . We
claim that kf¯i k < khi k/2 and kf¯i k < khi+1 k/2. Indeed, if kf¯i k ≥ khi k/2 then


khi hi+1 k = kfi k + kfi+1 k = khi k − kf¯i k + khi+1 k − kf¯i k

khi k  
khi k 
≤ khi k −
+ khi+1 k −
= khi+1 k,
2
2

which forces hi hi+1 ∈ S, thus contradicting the minimality of k. The proof of the
inequality kf¯i k < khi+1 k/2 proceeds similarly.
−1
We may hence write hi = f¯i−1
gi f¯i , where gi is not the empty word and
−1
hi hi+1 = f¯i−1 gi gi+1 f¯i+1 , without cancelation for all i. It follows that hi+1 · · · hk =
f¯i−1 gi+1 · · · gk f¯k , without cancelation. Since kf¯i k < khi k − kf¯i k = kf¯i−1 k + kgi k,
finally we have
khi+1 · · · hk k = kf¯i k + kgi+1 k + . . . + kgk k + kf¯k k
< kf¯i−1 k + kgi k + kgi+1 k + . . . + kgk k + kf¯k k = khi · · · hk k.
Claim (iii). For every subset S ⊂ BN , the equality S = hSi+ ∩ BN holds if and
only if for each f, g in S such that kf gk ≤ N, the element f g lies in S.
The forward implication is obvious. For the converse, given g ∈ hSi+ ∩ BN ,
write it in the form g = h1 · · · hk , with each hi in S and k ≤ N. Since the
hypothesis implies that S is stable, we may apply Claim (ii), thus yielding
khk k ≤ khk−1hk k ≤ . . . kh2 · · · hk k ≤ kh1 · · · hk k = kgk ≤ N.
Again, since S is stable, this implies hk−1 hk ∈ S; hence by induction hk−2 hk−1 hk ∈
S,..., h2 · · · hk−1 hk ∈ S, and finally g = h1 · · · hk−1 hk ∈ S, as claimed.

Claim (iv). If f, g are reduced words in Fn , with kf gk = N, kf k ≤ N, kgk = N,
then kf k must be even and exactly half of f must cancel in the product f g.

2.2. THE SPACE OF LEFT-ORDERS

73

Moreover, after cancelation, at least the right half of f g must be the same as the
right half of g.
Indeed, write f = h1 h̄ and g = h̄−1 h2 , so that f g = h1 h2 , without cancelation.
Then
kh1 k + kh̄k ≤ N, kh2 k + kh̄k = N, kh1 k + kh2 k = N.
The last two eaqualities yield kh1 k = kh̄k. Therefore, kf k = 2kh1 k is even, and
kh1 k = kf k/2, so that exactly half of f disappears in the product f g. Moreover,
from the first two relations we obtain kh1 k ≤ kh2 k, hence kh̄k ≤ kh2 k, which
shows that at least the right half of g survives in the product f g.
We may finally finish the proof of Claim (i). Let us begin by letting S1 :=
S ∪ {g} and, for i > 0,
Si+1 = Si

[

{f g : f, g in Si and kf gk ≤ N, f g ∈
/ Si }.

Obviously, there must exist an index j such that Sj = Sj+1 . By Claim (iii), for
such a j we have Sj = hS, gi+ ∩ BN , and Sj is stable.
Assume for a contradiction that hS, gi is not antisymmetric. Since Claim
(ii) easily implies that the semigroup generated by an stable set excluding id is
antisymmetric, we must have id ∈ Sj , hence there is a smallest index k such that
both h, h−1 belong to Sk for a certain element h.
Suppose that h ∈ S and h−1 ∈ Sk , and write h−1 = h1 h2 , with h1 , h2 in Sk−1 .
Either h1 ∈
/ S or h2 ∈
/ S (otherwise, h−1 would be in S). Let us consider the first
case (the other is analogous). Then h−1
1 = h2 h belongs to Sk , as h2 ∈ Sk−1 and
−1
h ∈ S ⊂ Sk−1 . However, h1 ∈
/ S, otherwise h1 , h−1
1 would be both in Sk−1 , thus
contradicting the minimality of k. Summarizing, we have that h1 , h−1
1 are both
−1
in Sk , though h1 ∈
/ S and h1 ∈
/ S.
The preceding argument allows reducing the general case to that where h ∈
/S
and h−1 ∈
/ S. Since S is total at length N − 1, by the minimality of k, every
element in Sk−1 \ S must have length N.

Claim (v). Every element in Sk−1 \ S, as well as h and h−1 , may be written in the
form h1 gh2 , where h1 , h2 lie in S ∪ {id} and have both even length, and where
exactly the left (resp. right) half of h2 (resp. h1 ) cancels in the product h1 gh2
above. (Notice that this implies kh1 gh2k = kh1 gk = kgh2 k = N.)
The proof is made by induction on i ≤ k for elements f ∈ Si \S with kf k = N.
In the case i = 1, such an element f corresponds to g, which is written in the
desired form. For the induction step, we must consider three different cases:

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CHAPTER 2. A PLETHORA OF ORDERS

– Assume f is a product f = h1 gh2 h̄1 g h̄2 , with both h1 gh2 and h̄1 g h̄2 in Si−1 \S of
length N. By Claim (iv), N must be even, and exactly the right half of h1 gh2 must
cancel with the left half of h̄1 g h̄2 in the product. By the induction hypothesis,
the former is nothing but the right half of g followed by h1 , and the later is h̄1
followed by the left half of g. Thus h1 gh2 h̄1 g h̄2 = h1 g h̄2 after cancelation, so that
f has the desired form.
– Suppose f = f h1 gh2 , where f ∈ S, h1 gh2 ∈ Si \ S, kf k ≤ N, kh1 gh2 k = N.
By Claim (iv), exactly the right half of f cancels in f h1 gh2 . If kf k ≤ kh1 k, this
implies that this cancelation happens in the product f h1 , so that kf h1 k = kf k ≤
N, thus yielding f h1 ∈ S because S is stable. If kh1 k ≤ kf k, then the entire
left half of h1 cancels in f h1 , so that kf h1 k ≤ kf k ≤ N, yielding again f h1 ∈ S.
That f h1 has even length and half of it cancels in the product f h1 gh2 now follows
from Claim (iv).
– Finally, the case where f = h1 gh2 f , with f ∈ S, h1 gh2 ∈ Si \ S, kf k ≤ N,
kh1 gh2 k = N, can be treated in a similar way to that of the preceding one.
To conclude the proof of Theorem 2.2.33, let us finally write h = h1 gh2 and
h = h̄1 g h̄2 as in Claim (v). There are two cases to consider:
– If N is even, write g = g1 g2 , where kg1 k = kg2 k = kgk/2. Then
−1

id = hh−1 = h1 g1 g2 h2 h̄1 g1 g2 h̄2 .
In this product, the right half of h must cancel against the left half of h−1 ,
that is, g2 h2 h̄1 g1 = id. Therefore, id = h1 g1 g2 h̄2 = h1 g h̄2 . But this implies
g −1 = h̄2 h1 ∈ hSi+ ∩ BN = S, which is a contradiction.
– If N is odd, write g = g1 f g2 , where f is the generator of Fn that appears in
the central position when writing g in reduced form. Proceeding as before, we
get id = hh−1 = h1 g1 f 2 g2 h̄2 with no further cancelation. This is absurd.


Remark 2.2.34. There are uncountably many left-orders on Fn for which the canonical generators fi are positive. Indeed, this is a direct consequence of the preceding
theorem, though it can be proved directly in a much more elementary way. What is
less trivial is that there are left-orders on Fn that extend the lexicographic order on
hf1 , . . . , fn i+ . This is proved in [185] via a concrete realization of the free group as
a group of homeomorphisms of the real line. The order thus obtained is perhaps the
simplest left-order that can be defined on Fn , and can be described as follows: Letting

2.2. THE SPACE OF LEFT-ORDERS

75

ϕ : Fn → R be defined by
ϕ(f ) =

{subwords of f of the form fj fi−1 , j > i}

− {subwords of f of the form fj−1 fi , j > i}

1
if f ends with f1 , f2 , . . . fn ,
1
+
−1 if f ends with f1−1 , f2−1 , . . . , fn−1 ,
2
0
if f is trivial,

where f is a reduced word on fi±1 , one has f ≻ id if and only if ϕ(f ) > 0.
A slight modification of the method above actually provides a Cantor set of leftorders, each of which extends the lexicographic order.

Step II. above, due to Kielak [114], is of a different nature. In a much more
general way, it concerns groups of fractions of finitely-generated semigroups (inside groups with infinitely many ends). Recall that given a semigroup P inside
a group Γ, we say that Γ is the group of fractions of P if every element in Γ
can be written in the form gh−1, where both g, h lie in P ∪ {1}. An illustrative
example, which covers that of nilpotent groups, is given by the next
Proposition 2.2.35. Let Γ be a group generated by a finite set of elements
f1 , . . . , fk , and let P be the semigroup generated by them together with id. If
Γ has no free sub-semigroup, then each of its elements may be written in the form
f g −1 for certain f, g in P .
Proof. We first claim that, given any f, g in P , there exist f and g such that
g −1 f = f¯ḡ −1 , that is f ḡ = g f¯. Otherwise, we would have f P ∩ gP = ∅, and this
implies that the sub-semigroup generated by f and g is free. Indeed, if h1 and h2
are different words in positive powers of f, g, to see that h1 6= h2 we may assume
that h1 begins with f and h2 with g. Then the condition f P ∩ gP = ∅ implies
that h1 6= h2 , since h1 ∈ f P and h2 ∈ gP .
Now let h := f1 g1−1f2 g2−1 · · · fk gk−1 be an arbitrary element in Γ, where all
−1
−1
fi , gi belong to P . By the discussion above, we may replace gk−1
fk by f¯k ḡk−1
,
thus obtaining
−1 −1
h = f1 g1−1 f2 g2−1 · · · fk−1 f¯k ḡk−1
gk .
−1
−1
Now, we may replace gk−2 fk−1 f¯k by an expression of the form f¯k−1 ḡk−2
, thus
obtaining
−1 −1 −1
h = f1 g1−1f2 g2−1 · · · fk−2 f¯k−1 ḡk−2
ḡk−1 gk .

Repeating this argument no more than k − 1 times, we finally get an expression
of f of the form f g −1, where f, g belong to P .


76

CHAPTER 2. A PLETHORA OF ORDERS

Exercise 2.2.36. Given Γ := Z ≀ Z = Z ⋉ ⊕Z Z, let a be a generator of the (left) factor
Z, and let b be a generator of the 0th factor Z in the right. Show that a, b generate Γ,
though the semigroup P generated by them and id satisfies aP ∩ bP = ∅.

Proposition 2.2.35 is false in a very strong way for free groups. This is the
content of the next result from [114], which actually applies to any group having
infinitely many ends.
Theorem 2.2.37. If P is a finitely-generated, proper subsemigroup of Fn , then
Fn is not the group of fractions of P .
Proof. Let P be a finitely-generated semigroup for which Fn is the group of fractions. Let us consider the finitely many generators of P as a system of generators
of Fn , and let us look at the corresponding Cayley graph. Since free groups have
infinitely many ends, there exists a radius N such that the complement of BN (id)
has at least three connected components. For simplicity, let us denote just by B
the ball centered at the identity and radius N. We will show that P = Fn .
Claim (i). One has B(P −1 ∪ {id}) = Fn .
The claim is equivalent to that P ∪ {id} intersects every ball BN (f ) = f B.
Let us thus suppose that (P ∪ {id}) ∩ f B = ∅ for a certain f ∈ Fn . Let E0 be a
connected component of the complement of f B not containing the identity. Let
h be an arbitrary element in the complement of B, and let E be the connected
component of Fn \ B containing h. Using the dynamical properties of the action
of Fn on its space of ends (roughly, transitivity and local contraction), one easily
convinces oneself that there exists g ∈ Fn such that gh ∈ E0 , gB ⊂ E0 , and gE
does not contain B.
Write gh in the form h1 h−1
2 , with both h1 , h2 in P ∪ {id}. Since f B does not
intersect P ∪ {id}, the element h1 must lie in the connected component of Fn \ f B
containing id. Starting from the point h1 , the path obtained by concatenation
−1
with h−1
2 must cross f B as well as gB. In particular, there is an element in P
−1
(namely, a terminal subword of h−1
= gh.
2 ) joining some point in gB to h1 h2
−1
Thus, there is an element of P joining an element of B to h, which shows that
h belongs to BP −1 .
The preceding conclusion was established for all elements h ∈ Fn \ B. This
obviously implies that Fn = B(P −1 ∪ {id}), as desired.
Claim (ii). One has P = Fn .

Let A be a subset of minimal cardinality such that A(P −1 ∪ {id}) = Fn . We
claim that A must be a singleton. Indeed, if A contains two elements f 6= g,

2.2. THE SPACE OF LEFT-ORDERS

77

then we may write f −1 g = h1 h−1
2 for certain h1 , h2 in P ∪ {id}. Hence f, g both
−1
belong to f h1 P . Therefore, letting A′ := A ∪ {f h1 } \ {f, g}, we still have
A′ (P −1 ∪ {id}) = Fn . However, this contradicts the minimality of the cardinality
of A.
We thus conclude that for a certain element h, we have h(P −1 ∪ {id}) = Fn .
Certainly, this implies that P −1 ∪ {id} = Fn . In particular, letting f be any
nontrivial element, both f, f −1 must belong to P −1 , hence their product f f −1 =
id is also in P −1 . We thus obtain P −1 = Fn , and taking inverses this yields
P = Fn .

Notice that the proof above still leaves open the next
Question 2.2.38. Do there exist k > 2 and a finitely-generated, proper subsemigroup P of Fn such that every element of Fn can be written in the form
(−1)k+1
f1 f2−1 f3 · · · fk
, with all f1 , . . . , fk belonging to P ∪ {id} ?
A direct corollary of Theorem 2.2.37 is that Fn does not admit an order with a
finitely-generated positive cone. Together with Theorem 2.2.33, this yields again
that LO(Fn ) has no isolated point, hence it is a Cantor set.
So far we have seen two different ways to show that LO(F2 ) is a Cantor set.
A natural problem that arises is whether these can be pursued or combined to
obtain finer information of this space for different metrics of geometric origin.
For instance, what is the Hausdorff dimension of this space ? In order to solve
this question one would need explicit estimates on the speed of approximation of
a given left-order.
Another direction of research concerns algorithmic properties for orders. Indeed, by using recursive functions (in the sense of computability theory), it is
not hard to construct left-orders on Fn such that the problem of comparison of
elements is undecidable, i.e. there is no algorithm that, on every input consisting of a pair of elements g, h in Fn , decides whether f ≺ g or not. This is of
course not the case for higher-rank Abelian groups, yet in both cases the spaces
of left-orders are just Cantor sets.
Left-orders v/s bi-orders. Despite the good understanding of LO(F2 ), the
next question remains open.
Question 2.2.39. Does there exist a partial left-order on F2 whose positive cone
is closed under taking roots and that cannot be extended into a total left-order ?
As was pointed out by George Metcalfe, the positive cone generated by a2 , b2 ,
a−1 b−1 in F2 = ha, bi cannot be extended into that of a total left-order; however,

78

CHAPTER 2. A PLETHORA OF ORDERS

it is not closed under taking roots of elements. Let us also point out that the
analogous question for bi-orders has a negative answer; see for instance [121,
Corollary 3.6.7] and [164]. All of this is to be compared with the case of nilpotent
groups, as described at the end of Example 1.2.1. In another direction, there is
a major question that remains open for bi-orders on Fn , first raised by McCleary
in [193, page 127].
Question 2.2.40. Are the spaces of bi-orders of (finitely-generated, non-Abelian)
free groups homeomorphic to the Cantor set ?
Notice that, using the lower central series, one may provide a Cantor set of
“standard” bi-orders on free groups. Nevertheless, according to Example 2.1.11,
there are more bi-orders on free groups than these. Actually, it is shown in [117]
that, although the action of Out(Fn ) on BO(Fn ) is faithful, its restriction to the
set of standard left-orders is not (the kernel coincides with the Torelli subgroup,
that is, the one whose elements act trivially on homology).

2.2.3

Finitely-generated positive cones

We first recall a short argument due to Linnell showing that, if a left-order
 on a group Γ is non-isolated in LO(Γ), then its positive cone is not finitely
generated as a semigroup.
Proposition 2.2.41. If the positive cone of a left-order  on a group Γ is finitely
generated as a semigroup, then  is isolated in LO(Γ).
Proof. If g1 , . . . , gk generate P+ , then the only left-order on Γ that coincides
with  on any set containing these generators and the identity element is  itself
(see the exercise below).

Exercise 2.2.42. Show that if two left-orders  and ′ on the same group satisfy
P+ ⊂ P+′ , then they coincide.

The converse to the preceding proposition is not true. For instance, the dyadic
rationals admit only two left-orders, though none of them has a finitely-generated
positive cone. One may easily modify this example in order to obtain a finitelygenerated one (see the example below). However, the search for conditions ensuring that isolated left-orders must have a finitely-generated positive cone is
interesting by itself.

79

2.2. THE SPACE OF LEFT-ORDERS

Example 2.2.43. The group of presentation Γ := ha, b : bab−1 = a−2 i is covered
by Theorem 2.2.13: it has exactly four left-orders (compare Example 2.2.14), which
depend only on the signs of the generators a, b. However, the positive cone of none of
these orders is finitely generated since they all contain a copy of Z[ 21 ] inside hhaii (the
smallest normal subgroup containing a).

Besides the obvious case of Z, perhaps the simplest example of a finitely
generated positive cone for a group left-order occurs for the Klein-bottle group
K2 = ha, b : bab = ai: one may take ha, bi+ as such a cone (see Figure 8).
Actually, K2 admits exactly four left-orders, each of which having a finitelygenerated positive cone (the other cones are ha, b−1 i+ , ha−1 , bi+ , and ha−1 , b−1 i+ ).
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• ......a...

b .....

• ......a...

b .....

• ......a...

b .....

Figure 8: The positive cone P + = ha, bi+ on K2 = ha, b : a−1 ba = b−1 i.
Rather surprisingly, finitely-generated positive cones also occur on braid groups,
according to a beautiful result due to Dubrovina and Dubrovin [67]
Theorem 2.2.44. For each n ≥ 3, the braid group Bn admits the decomposition
Bn = a1 , . . . , an−1

+

−1
⊔ a−1
1 , . . . , an−1

+

⊔ {id},

80

CHAPTER 2. A PLETHORA OF ORDERS

where a1 := σ1 · · · σn−1 , a2 := (σ2 · · · σn−1 )−1 , a3 := σ3 · · · σn−1 , a4 := (σ4 · · · σn−1 )−1 ,
(−1)n−1
. . ., and an−1 := σn−1 .
Notice that this theorem also holds for n = 2, yet it is trivial in this case as
B2 is isomorphic to Z.

Figure 9: The DD-positive cone on B3 .

As a particular case, for B3 , the semigroup PDD = hσ1 σ2 , σ2−1 i+ is the positive
cone of a left-order DD . This can be visualized in Figure 9, where we depict
the Cayley graph of B3 (essentially, a product of a quasi-isometric copy of Z2
by a dyadic rooted tree). Notice that for the generators a = a1 := σ1 σ2 and
b = a2 := σ2−1 , the presentation of B3 becomes B3 = a, b : ba2 b = a . Thus, in
the picture above, an arrow pointing from left to right should be added to every
diagonal edge of the graph. These arrows represent multiplications by a, while
all arrows explicitly appearing represent multiplications by b. Starting at the
identity (in red), every positive (green) element can be reached by a path that
follows the direction of the arrows. Conversely, every negative (blue) element can
be reached by a path starting at the identity following a direction opposite to
that of the arrows. Finally, no (nontrivial) element can be reached both ways.

81

2.2. THE SPACE OF LEFT-ORDERS

Quitte remarkably, this particular example was already known (and seems to
be folklore, at least for a certain community) for a long time; see [66].
Retrieving the DD-order from the D-order. The proof of Theorem 2.2.44
strongly uses Dehornoy’s theorem, as described in §1.2.6. To begin with, notice
that it readily follows from the definition that for each j ∈ {1, . . . , n − 1}, the
subgroup hσj , . . . , σn−1 i ∼ Bn−j+1 of Bn is D -convex.
In particular, for the case of B3 , the cyclic subgroup hσ2 i is D -convex.
One can hence define the order 3 on B3 as being the extension by D of the
restriction to hσ2 i of the reverse order D . (This corresponds to the flipping of
D on hσ2 i, as introduced in Example 2.1.4.) We claim that the positive cone
of 3 is generated by the elements a1 := σ1 σ2 and a2 := σ2−1 , thus showing the
theorem in this particular case. Indeed, by definition, these elements are positive
with respect to 3 . Thus, it suffices to show that for every c 6= id in B3 , either
c or c−1 belongs to ha1 , a2 i+ . Now, if c or c−1 is 2-positive, then there exists an
+
+
integer m 6= 0 such that c = σ2m = a−m
2 , and therefore c ∈ ha2 i ⊂ ha1 , a2 i
if m < 0, and c−1 ∈ ha2 i+ ⊂ ha1 , a2 i+ if m > 0. If c is 1-positive, then for a
certain choice of integers m′′1 , . . . , m′′k′′ +1 , one has
m′′

m′′

m′′
k′′

c = σ2 1 σ1 σ2 2 σ1 · · · σ2

m′′
k′′ +1

σ1 σ2

.

Using the identity σ1 = a1 a2 , this allows us to rewrite c in the form
m′

m′

m′

m′ ′

c = a2 1 a1 a2 2 a1 . . . a2 k′ a1 a2 k +1
for some integers m′1 , . . . , m′k′ +1 . Now, using several times the (easy to check)
identity a2 a21 a2 = a1 , one may easily express c as a product
m

mk
m2
k+1
1
c = am
2 a1 a2 a1 . . . a2 a1 a2

in which all the exponents mi are non-negative. This shows that c belongs to
ha1 , a2 i+ . Finally, if c−1 is 1-positive, then c−1 belongs to ha1 , a2 i+ .
The extension of the preceding argument to the general case proceeds inductively as follows. Let us see Bn−1 = hσ̃1 , . . . , σ̃n−2 i as a subgroup of Bn =
hσ1 , . . . , σn−1 i via the homomorphism σ̃i 7→ σi+1 . Then n−1 induces an order
on hσ2 , . . . , σn−1 i ⊂ Bn , which we still denote by n−1 . We then let n be the
extension of n−1 by the D-order D . Then, using the inductive hypothesis as
well as the remarkable identities (that we leave to the reader)
(−1)n−1

(a2 a−1
3 · · · an−1

(−1)n−1

)a1n−1 (a2 a−1
3 · · · an−1

) = a1 ,

(−1)n−1 2

(a2 a−1
3 · · · an−1

) = a2n−1 ,

82

CHAPTER 2. A PLETHORA OF ORDERS

one may check as above that the positive cone of the order n coincides with the
semigroup ha1 , . . . , an−1 i+ , thus showing the theorem.
Exercise 2.2.45. Prove that the only convex subgroups of Bn for both the D-order
and the DD-order are C 0 := Bn , C 1 := ha2 , . . . , an−1 i = hσ2 , . . . , σn−1 i, . . . , C n−1 :=
han−1 i = hσn−1 i and C n := {id}.

Let us emphasize that assuming Theorem 2.2.44, we can follow the arguments
above backwards and retrieve the Dehornoy’s order on Bn . (Details to the reader.)
A more conceptual approach to this phenomenon was proposed by Ito in [104],
and it is developed in the next
Exercise 2.2.46. Let g1 , . . . , gk be finitely many generators of a group Γ. For each
k+1
i ∈ {1, . . . , k}, let hi := (gi gi+1 · · · gk )(−1) , and denote by Pi the semigroup generated
by gi , . . . , gk . Assume that the following condition (called Property (F) in [104])
holds: For each i ∈ {1, . . . , k − 1}, both gi Pi+1 gi−1 and gi−1 Pi+1 gi are contained in the
semigroup Pi− made of the inverses of the elements in Pi .
(i) Prove that h1 , . . . , hk generate the positive cone of a left-order on Γ if and only if
the gi ’s define a Dehornoy-like order, which means that every nontrivial element
may be written as a product of elements gi , . . . , gk so that gi appears with only positive
exponents, and no g ∈ Γ is such that both g and g −1 may be written in such a way.
(ii) Referring to Theorem 2.2.44, check that gi := σi and hi := ai satisfy property (F).

Torus-knot groups. We next give an elementary proof of that the torus-knot
groups Gm,n := hc, d : cm = dn i do admit left-orders with finitely-generated positive cones. This is closely related to what was previously shown for braid groups,
since for (m, n) = (3, 2) we retrieve the braid group B3 for the generating set
c ∼ σ1 σ2 and d ∼ σ1 σ2 σ1 . In this case, the positive cone given by Theorem2.2.44
is generated by a := c ∼ σ1 σ2 and b := c−2 d ∼ σ2−1 , respect to which the presentation becomes G3,2 = ha, b : ba2 b = ai. Also, notice that for (m, n) = (2, 2)
we retrieve the Klein-bottle group. In this case, the generating system of the
positive cone is made of a := c and b := c−1 d, for which the presentation becomes
K4 = ha, b : bab = ai.
After some computations, one easily convinces oneself that the natural extension of this corresponds to the presentation
Gm,n = a, b : (bam−1 )n−1 b = a ,
where a := c and b := c−(m−1) d. The following result appears in [151] for n = 2
and in [104] for the general case.

2.2. THE SPACE OF LEFT-ORDERS

83

Theorem 2.2.47. For each m > 1 and n > 1, the group Gm,n can be decomposed
as
Gm,n = ha, bi+ ⊔ ha−1 , b−1 i+ ⊔ {id}.
Quite naturally, the proof of this theorem involves two issues:
Step I. Every nontrivial element lies in ha, bi+ ∪ ha−1 , b−1 i+ .

Step II. No nontrivial element lies in both ha, bi+ and ha−1 , b−1 i+ .

In what follows, we only consider the case (m, n) 6= (2, 2), because (m, n) =
(2, 2) corresponds to the Klein’s bottle group K4 as previously described. (Some
of the arguments below do not apply in this case.) For Step I, we begin with
Claim (i). The element ∆ := am belongs to the center of Gm,n .
Indeed, from (bam−1 )n−1 b = a it follows that (bam−1 )n = (am−1 b)n = am . Thus,
b∆ = bam = b(am−1 b)n = (bam−1 )n b = am b = ∆b.
Moreover, a∆ = am+1 = ∆a.
A word in (positive powers of) a, b (resp. a−1 , b−1 ) will be said to be positive
(resp. negative). It is non-positive (resp. non-negative) if it is either trivial or
negative (resp. either trivial or positive).
Claim (ii). Every element w ∈ Gm,n may be written in the form ū∆ℓ for some
non-negative word ū and ℓ ∈ Z.
Indeed, in any word representing w, we may rewrite the negative powers of a
and b using the relations
a−1 = am−1 ∆−1 ,

b−1 = a−1 (bam−1 )n−1 = am−1 ∆−1 (bam−1 )n−1 ,

and then use the fact that ∆ belongs to the center of Gm,n .
Notice that since am = ∆, every u ∈ ha, bi+ may be written in the form
u = bs0 ar1 bs1 · · · bsk−1 ark ∆ℓ ,
where si > 0 for i ∈ {1, . . . , k −1}, s0 ≥ 0, ri ∈ {1, . . . , m−1} for i ∈ {1, . . . , k −1},
rk ≥ 0, and ℓ ≥ 0. Therefore, by Claim (ii), every w ∈ Gm,n may be written as
w = bs0 ar1 bs1 · · · bsk−1 ark ∆ℓ = bs0 ar1 bs1 · · · bsk−1 ark +mℓ ,

(2.1)

84

CHAPTER 2. A PLETHORA OF ORDERS

where the properties of ri , si above are satisfied, and ℓ ∈ Z. Such an expression
will be said to be a normal form for w if k is minimal. (Notice that, a priori,
these normal forms may be non-unique for a given w.)
There are two cases to consider. If rk + mℓ ≥ 0, then w is obviously nonnegative. Therefore, Step I is concluded by the next
Claim (iii): If rk + mℓ < 0, then w is negative.
The proof is by induction on the length k of the normal form. To begin with,
notice that (bam−1 )n−1 b = a yields ba−1 = (a−m+1 b−1 )n−1 . Thus, for k = 1, that
is, for w = bs0 ar1 +mℓ , we have

s

s
w = bs0 a−1 ar1 +mℓ+1 = a−1 aba−1 0 ar1 +mℓ+1 = a−1 a(a−m+1 b−1 )n−1 0 ar1 +mℓ+1 ,
and the last expression is easily seen to be negative just by noticing that
a(a−m+1 b−1 )n−1 = a−m+2 b−1 (a−m+1 b−1 )n−2 .
Assume the claim holds up to k −1. Proceeding as before, expression (2.1)
becomes
w = bs0 ar1 · · · bsk−1 −1 [ba−1 ]ark +mℓ+1 = bs0 ar1 · · · bsk−1 −1 [(a−m+1 b−1 )n−1 ]amk +mℓ+1 .
If sk−1 > 1, then writing
w = bs0 ar1 · · · bsk−1 −2 [ba−1 ]a−m+2 b−1 (a−m+1 b−1 )n−2 ark +mℓ+1 ,
we see we can repeat the process changing ba−1 by (a−m+1 b−1 )n−1 . Otherwise,
w = bs0 ar1 · · · ark−1 [(a−m+1 b−1 )n−1 ]ark +mℓ+1
= bs0 ar1 · · · bsk−2 ark−1 −m+1 b−1 (a−m+1 b−1 )n−2 ark +mℓ+1
= bs0 ar1 · · · bsk−2 −1 [ba−1 ]ark−1 −m+2 b−1 (a−m+1 b−1 )n−2 ark +mℓ+1 ,
so that in case rk−1 < m − 1 (and sk−2 > 0), we may repeat the procedure.
Continuing in this way, one easily convinces that, unless
sk−1 = . . . = sk−(n−1) = 1,

sk−n > 0,

rk−1 = . . . = rk−(n−1) = m − 1,

the expression for w above may reduced into one of the form
w = bs0 · · · bsi a−1 ω

(2.2)

2.2. THE SPACE OF LEFT-ORDERS

85

for certain si > 0, i < k, and a non-positive word ω. As i < k, the induction hypothesis applies to bs0 · · · bsi a−1 , which is hence negative, and so does w. Assume
otherwise that (2.2) holds. Then since b(am−1 b)n−1 = a, replacing (am−1 b)n−1 by
b−1 a and canceling b−1 , we obtain a new expression for w of the form
w = bs0 · · · bsk−n −1 ark +mℓ+1 ,
which contradicts the minimality of the length of the normal form. This closes
the proof.

86

CHAPTER 2. A PLETHORA OF ORDERS

Step II of the proof of Theorem 2.2.47 can be established via several approaches. Here, we chose the dynamical one, based on the fact that Gm,n embeds
g R). To see this, let us first come back to the presentation
into PSL(2,
Gm,n = c, d : cm = dn ,

which exhibits Gm,n as a central extension of the group
Gm,n = c̄, d¯: c̄m = d¯n = id .
A concrete realization of Gm,n inside PSL(2, R) arises when identifying c̄ to the
, and d¯ to an hyperbolic rotation of angle 2π
centered
circle rotation of angle 2π
m
n
at a point different from the origin in such a way that, if we let p0 := p, p1 :=
¯
c̄(p), . . . , pm−1 := c̄m−1 (p) and q0 := p, q1 := d(p),
. . . , qn−1 := d¯n−1 (p) for a certain
p ∈ S1 , we have that all the qi ’s lie between p0 and p1 , and qn−1 = p1 . This
g R) by identifying c ∈ Gm,n to the
realization allows embedding Gm,n into PSL(2,
lifting of c̄ to the real line given by x 7→ x + 2π
, and d to the unique lifting of
m
¯
¯
d to the real line satisfying x ≤ d(x) ≤ x + 2π for all x ∈ R. (Actually, the
arguments given so far only show that the above identifications induce a group
g
homomorphism from Gm,n into PSL(2,
R), and the injectivity follows from the
arguments given below.)
The dynamics of the action of Gm,n on the circle is illustrated in Figure 10.
Passing to the generators a, b, we have that b̄ = c̄−(m−1) d¯ = c̄d¯ is a parabolic
Möbius transformation fixing p1 , and ā = c̄. Using this, we next proceed to
show that no element w in ha, bi+ ⊂ Gm,n represents the identity. By taking
inverses, this will imply that no element in ha−1 , b−1 i+ represents the identity,
thus completing the proof.
c̄ = ā

.........
............ ................
.....
....
...
...
.
. . ......................
..................... ..........
.
.
.
.
.
.
.
.
0
.
.....
.. . ....................................................... ..................
..
.
.
.
.
............
...
...
........ m−1.....
..........
...........
.
.
.
.
.
.
.
...
......
.
.
..
.
1........
.
.
...
.
.....
.
.
.
.
.
.
.
.
.
.
.
..... ....
...
.............. ........ ...........................
.... ...
....
..
...
...
...
..
...
..
.
2..
... m−2
...
...
..
..
..
..
..
..
...
.
..
...
.
.
...
.
.
...
..
...
..
..
3 ......
... m−3
.
...
...
....
....
....
.
.
.
.....
.
.....
....
......
.....
.........
...... m−4
.
.
.
.
.
.
.
.............
.
...........................................................

c̄

p
•

p

p

p=
•p

d¯

c̄

p

•

•

p
•

•

•
p

•
p

Figure 10

87

2.2. THE SPACE OF LEFT-ORDERS
We begin by writing w in the form
w = bs0 ar1 bs1 · · · ark amℓ ,

ℓ ≥ 0,

with the corresponding restrictions on the exponents. Here, we may assume that
no expression (bam−1 )n−1 b appears, otherwise we may replace it by a.
Assume that w is not a power of a, and let us consider its reduction
g 2 (R).
w̄ = b̄s0 ār1 b̄s1 · · · ārk ∈ PSL

Using b̄ = ād¯ and simplifying ām = id, we may rewrite this in the form
′
′
′
′
g 2 (R),
w̄ = d¯s0 ār1 d¯s1 · · · ārk′ ∈ PSL

with similar restrictions on the exponents ri′ , s′i . What is crucial here is that the
fact that no expression (bam−1 )n−1 b appeared in the original form implies that this
new expression is nontrivial, as it can be easily checked. (Indeed, no cancellation
d¯n = id will be performed.)
¯ we may conjugate it into either some w̄ ′ ∈ hā, b̄i+
Unless w̄ is a power of dā,
beginning and finishing by ā and so that all the exponents of ā lie in {1, . . . , m−1},
¯ + beginning and finishing with d¯ with the same restricor to some w̄ ′′ ∈ hā, di
tion on the
 exponents of ā. An easy
 ping-pong type argument then shows that
w̄ ′ ]p0 , p1 [ ⊂]p1 , p0 [ and w̄ ′′ ]p1 , p0 [ ⊂]p0 , p1 [, hence w̄ ′ 6= id and w̄ ′′ 6= id.
Thus, to conclude the proof, we need to check that neither a nor da are torsion
¯ sends [p0 , p1 ] into the
elements. That da has infinite order follows from that dā
¯
strict subinterval [p0 , d(p2 )], hence no iterate of it can equal the identity. Finally,
to see that a also has infinite order, just notice that it identifies to the translation
g R).
in PSL(2,
by 2π
m

Some other examples. Searching for more examples of finitely-generated positive cones in groups with infinitely many left-orders has become a topic of much
activity over the last years. The examples given above as well as the techniques
used in proofs have been pursued in three directions. First, there is the close relation with Dehornoy-like orders in which the previous examples fit, as described in
[151] and later in [104]. (See also Remark 3.2.45.) Second, there is an approach
based on partial cyclic amalgamation, which is fully developed in [103]. This
allows iterative implementation, thus establishing for instance that the groups
m2
mn
1
Gm1 ,m2 ,...,mn := a1 , . . . , an : am
1 = a2 = . . . = an

do admit finitely-generated positive cones. This approach was somewhat complemented in [102]; however, the orders constructed therein are only ensured to

88

CHAPTER 2. A PLETHORA OF ORDERS

be isolated, and knowing whether their positive cones are finitely generated remains as an interesting question. Finally, there is a more combinatorial approach
starting from group presentations introduced in [55]. Roughly, in case these
presentations have a triangular form, finitely-generated positive cones naturally
appear. As a random example, we can mention that the groups
Hm,n := a, b, c : a = ba2 (b2 a2 )m c, b = c(ba2 )n ba
fall in this category.
We do not pursue on this nice subject here; we just refer the reader to the
works mentioned above for the announced results and further developments (see
also Remark 2.2.32). Nevertheless, let us mention that none of these approaches
has provided a new proof of Dehornoy’s theorem concerning the D-order on Bn
for n ≥ 4. This issue seems to be beyond the scope of these methods.

Chapter 3
ORDERABLE GROUPS AS
DYNAMICAL OBJECTS
3.1

Hölder’s Theorem

The results of this section –essentially due to Hölder– are classical and perhaps
correspond to the most beautiful founder pieces of the theory. They characterize group left-orders satisfying an Archimedean type property: the underlying
ordered group must be ordered isomorphic to a subgroup of (R, +).
Definition 3.1.1. A left-order  on a group Γ is said to be Archimedean if
for all g, h in Γ such that g 6= id, there exists n ∈ Z satisfying g n ≻ h.
Theorem 3.1.2. Every group endowed with an Archimedean left-order is orderisomorphic to a subgroup of (R, +).
Hölder proved this theorem under the extra assumption that the group is
Abelian. However, his arguments work verbatim without this hypothesis but assuming that the left-order is bi-invariant. That this hypothesis is also superfluous
was first remarked by Conrad in [51].
Lemma 3.1.3. Every Archimedean left-order on a group is bi-invariant.
Proof. Let  be an Archimedean left-order on a group Γ. We need to show that
its positive cone is a normal semigroup.
Suppose that g ∈ P+ and h ∈ P− are such that hgh−1 ∈
/ P+ . Let n be
the smallest positive integer for which h−1 ≺ g n . Since hgh−1 ≺ id, we have
89

90

CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

h−1 ≺ g −1h−1 ≺ g n−1, which contradicts the definition of n. We thus conclude
that P+ is stable under conjugacy by elements in P− .
Assume now that g, h in P+ verify hgh−1 ∈
/ P+ . In this case, hg −1 h−1 ≻ id,
and since h−1 ∈ P− , the first part of the proof yields h−1 (hg −1 h−1 )h ∈ P+ ,
that is, g −1 ∈ P+ , which is absurd. Hence, P+ is also stable under conjugacy by
elements in P+ , which concludes the proof.

Exercise 3.1.4. Prove the preceding lemma by using dynamical realizations. More
precisely, show that the dynamical realization of every Archimedean left-order on a
countable group is a subgroup of Homeo+ (R) acting freely on the line (c.f. §1.1.3;
compare Example 3.1.6).

Proof of Theorem 3.1.2. Let Γ be a group endowed with an Archimedean leftorder . By Lemma 3.1.3, we know that  is bi-invariant. Let us fix a positive
element f ∈ Γ, and for each g ∈ Γ and each p ∈ N, let us consider the unique
integer q = q(p) such that f q  g p ≺ f q+1 .

Claim (i). The sequence q(p)/p converges to a real number as p goes to infinity.

Indeed, if f q(p1 )  g p1 ≺ f q(p1 )+1 and f q(p2 )  g p2 ≺ f q(p2 )+1 , then the
bi-invariance of  yields
f q(p1 )+q(p2 )  g p1 +p2 ≺ f q(p1 )+q(p2 )+2 .

Therefore, q(p1 ) + q(p2 ) ≤ q(p1 + p2 ) ≤ q(p1 ) + q(p2 ) + 1. The convergence of
the sequence (q(p)/p) then follows from Exercise 3.1.5.
Claim (ii). The map φ : Γ → (R, +) is a group homomorphism.

Indeed, let g1 , g2 be arbitrary elements in Γ. Let us suppose that g1 g2  g2 g1
(the case g2 g1  g1 g2 is analogous). Since  is bi-invariant, if f q1  g1p ≺ f q1 +1
and f q2  g2p ≺ f q2 +1 , then
f q1 +q2  g1p g2p  (g1 g2 )p  g2p g1p ≺ f q1 +q2 +2 .
From these relations one concludes that
q1 + q2
q1 + q2 + 1
≤ φ(g1g2 ) ≤ lim
= φ(g1 ) + φ(g2 ),
p→∞
p→∞
p
p

φ(g1 ) + φ(g2 ) = lim

and therefore φ(g1 g2 ) = φ(g1 ) + φ(g2 ).

3.2. THE CONRAD PROPERTY

91

Claim (iii). The homomorphism φ is one-to-one and order-preserving.
That φ is order-preserving (in the sense that if g1  g2 then φ(g1 ) ≤ φ(g2))
follows from the definition. To show injectivity, first notice that φ(f ) = 1. Let h
be an element in Γ such that φ(h) = 0. Assume that h 6= id. Then there exists
n ∈ Z such that hn  f . Consequently, 0 = nφ(h) = φ(hn ) ≥ φ(f ) = 1, which is
absurd. Therefore, if φ(h) = 0 then h = id.

Exercise 3.1.5. Let (an )n∈Z be a sequence of real numbers. Assume that there exists
a constant C ∈ R such that, for all m, n in Z,
|am+n − am − an | ≤ C.

(3.1)

Show that there exists a unique θ ∈ R such that the sequence |an − nΘ| is bounded.
Check that this number θ is equal to the limit of the sequence (an /n) as n goes to ±∞
(in particular, this limit exists).


Hint. For each n ∈ N let In := (an − C)/n, (aT
n + C)/n . Check that Imn is contained
in In for every m, n in N. Conclude that I := n∈N In is nonempty (any θ in I satisfies
the desired property).

Example 3.1.6. Examples of groups admitting Archimedean left-orders are groups
acting freely on the real line. Indeed, from such an action one may define  on Γ by
letting g ≺ h if g(x) < h(x) for some (equivalently, for all) x ∈ R. This order relation is
total, and using the fact that the action is free, one readily shows that it is Archimedean
(as well as bi-invariant).
Notice that, by the proof of Theorem 3.1.2, the left-order  above induces an
embedding φ of Γ into (R, +). If φ(Γ) is isomorphic to Z, then the action of Γ is
conjugate to the action by integer translations. Otherwise, unless Γ is trivial, φ(Γ) is
dense in (R, +). For each point x in the line we may then define

ϕ(x) = sup φ(h) ∈ R : h(0) ≤ x .

It is easy to see that ϕ : R → R is a non-decreasing map. Moreover, it satisfies
ϕ(h(x)) = ϕ(x) + φ(h) for all x ∈ R and all h ∈ Γ. Finally, ϕ is continuous, as
otherwise the set R \ ϕ(R) would be a nonempty open set invariant by the translations
of φ(Γ), which is impossible. In summary, every free action on the line is (continuously)
semiconjugate to an action by translations.

3.2
3.2.1

The Conrad Property
The classical approach revisited

A left-order  on a group Γ is said to be Conradian (a C-order, for short)
if for all positive elements f, g, there exists n ∈ N such that f g n ≻ g.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Bi-invariant left-orders are Conradian, as n = 1 works in the preceding inequality for bi-orders. In this direction, it is quite remarkable that one may
actually take n = 2 in the general definition above, as the next proposition
shows (the nice proof below, due to Jiménez, is taken from [106]).
Proposition 3.2.1. If  is a Conradian order on a group, then f g 2 ≻ g holds
for all positive elements f, g.
Proof. Suppose that two positive elements f, g for a left-order ′ on a group Γ
are such that f g 2 ′ g. Then (g −1 f g)g ′ id, and since g is a positive element,
this implies that g −1 f g is negative, and therefore f g ≺′ g. Now for the positive
element h := f g and every n ∈ N, one has
f hn = f (f g)n = f (f g)n−2(f g)(f g) ≺′ f (f g)n−2(f g)g
= f (f g)n−2f g 2 ′ f (f g)n−2g = f (f g)n−3f g 2 ′ f (f g)n−3g ′ . . .
′ f (f g)g = f f g 2 ′ f g = h.
This shows that ′ does not satisfy the Conrad property.



The following is an easy (but important) corollary to the previous proposition,
and we leave its proof to the reader. (Compare Exercise 2.2.3.)
Corollary 3.2.2. For every left-orderable group, the subspace CO(Γ) of Conradian orders is closed inside the space of left-orders. Moreover, this subspace is
invariant under the conjugacy action.
Perhaps the most important theorem concerning C-orderable groups is the
next one. The direct implication is due to Conrad [51]; the converse is due to
Brodski [25], yet it was independently rediscovered by Rhemtulla and Rolfsen
[169]. We postpone the proof of the first part, and for the second we offer an elementary one taken from [153]. Recall that a group is said to be locally indicable
if each nontrivial finitely-generated subgroup admits a nontrivial homomorphism
into (R, +).
Theorem 3.2.3. A group Γ is C-orderable if and only if it is locally indicable.
To show that local indicability implies C-orderability (the converse will be
proved in §3.2.3), we will need the following lemma whose proof is left to the
reader. (Compare §1.1.2.)

93

3.2. THE CONRAD PROPERTY

Lemma 3.2.4. A group Γ is C-orderable if and only if for every finite family G
of elements in Γ \ {id}, there exist a choice of exponents η : G → {−1, +1} such
that id does not belong to the smallest subsemigroup hhGii satisfying:
– It contains all the elements g η(g) , with g ∈ G;
– For all f, g in the semigroup, the element g −1 f g 2 also belongs to it.
Local indicability implies C-orderability. We need to check that every
locally indicable group Γ satisfies the condition of the preceding lemma. Let
{g1 , . . . , gk } be a finite family of elements in Γ different from the identity. By
hypothesis, there is a nontrivial homomorphism φ1 : hg1 , . . . , gk i → (R, +). Let
i1 , . . . , ik′ be the indices (if any) such that φ1 (gij ) = 0. Again by hypothesis, there
exists a nontrivial homomorphism φ2 : hgi1 , . . . , gik′ i → (R, +). Letting i′1 , . . . , i′k′′
be the indices in {i1 , . . . , ik′ } for which φ2 (gi′j ) = 0, we may choose a nontrivial
homomorphism φ3 : hgi′1 , . . . , gi′k′′ i → (R, +)... Notice that this process must finish
in a finite number of steps (indeed, it stops in at most k steps). Now, for each
i ∈ {1, . . . , k}, choose the (unique) index j(i) such that φj(i) is defined at gi and
φj(i) (gi ) 6= 0, and let ηi := η(gi ) ∈ {−1, +1} be so that φj(i) (giηi ) > 0. We claim
that this choice of exponents ηi is “compatible”. Indeed, for every index j and
every f, g for which φj are defined, one has φj (f −1 gf 2) = φj (f ) + φj (g). Therefore, φ1 (h) ≥ 0 for every h ∈ hhg1η1 , . . . , gkηk ii. Moreover, if φ1 (h) = 0, then h
ηi
ηi
actually belongs to hhgi1 1 , . . . , gik′k′ ii. In this case, the preceding argument shows
ηi′

ηi′

that φ2 (h) ≥ 0, with equality if and only if h ∈ hhgi1 1 , . . . , gi′ ′′k′′ ii... Continuing
k
in this way, one concludes that φj (h) must be strictly positive for some index j.
Thus, the element h cannot be equal to the identity, and this finishes the proof.

If a group Γ contains a normal subgroup Γ∗ so that both Γ∗ and Γ/Γ∗ are
locally indicable, then Γ itself is locally indicable. Equivalently, the extension of
a C-orderable group by a C-orderable group is C-orderable. In a similar way we
have:
Exercise 3.2.5. Let (Γ, ) be a C-ordered group, and let Γ∗ be a convex subgroup.
Show that for any C-order ∗ of Γ∗ , the extension of ∗ by  is still Conradian. In
particular, every left-order obtained from a C-left-order by flipping a convex subgroup
is Conradian (c.f. §2.1.1).
Example 3.2.6. A remarkable theorem independently obtained by Brodski [25] and
Howie [100] asserts that torsion-free, 1-relator groups are locally indicable. Also, all
knot groups in R3 are locally indicable (see [101, Lemma 2]).

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Examples of left-orderable, non C-orderable groups. Only a few examples are
known. Historically, the first was exhibited (in a slightly different context) by Thurston
[189], and rediscovered some years later by Bergman [12]. It corresponds to the lifting
g
to PSL(2,
R) of the (2, 3, 7)-triangle group, and has the presentation
Γ = f, g, h : f 2 = g3 = h7 = f gh .

g
Left-orderability follows from that PSL(2,
R) is a subgroup of Homeo+ (R). The fact
that Γ is not C-orderable is a consequence of that it has no nontrivial homomorphism
into (R, +), which may be easily deduced from the presentation above. Actually, Γ is
the π1 of an homological sphere, and this was the motivation of Thurston to deal with
this group in his generalization of the famous Reeb stability theorem for codimension-1
foliations. We strongly recommend the lecture of [189] for all of this.
Below we elaborate on a different and quite important example, namely braid groups
e of Thompson’s group G to the real line;
Bn for n ≥ 5. Another example is the lifting G
see [38] for more details.
Example 3.2.7. The braid groups B3 and B4 are locally indicable. For B3 , this may
be easily deduced from the exact sequence
0 −→ [B3 , B3 ] ∼ F2 −→ B3 −→ B3 /[B3 , B3 ] ∼ Z −→ 0,
where the isomorphism [B3 , B3 ] ∼ F2 may be shown by looking the action on the circle
g
of B3 ∼ PSL(2,
Z), and B3 /[B3 , B3 ] ∼ Z appears by taking “total exponents”. For B4 ,
there is an exact sequence
0 −→ F2 −→ B4 −→ B3 −→ 0.

Here, the homomorphism from B4 to B3 is the one that sends σ1 and σ3 to σ1 , and σ2
to σ2 . Its kernel is generated by σ1 σ3−1 and σ2 σ1 σ3−1 σ2−1 . To show that these elements
are free generators, one may consider the homomorphism φ : B4 → Aut(F2 ) defined by
φ(σ1 )(a) := a, φ(σ1 )(b) := ab, φ(σ2 )(a) := b−1 a, φ(σ2 )(b) := b, φ(σ3 )(a) := a, φ(σ3 )(b) :=
ba, and notice that φ(σ1 σ3−1 ) (resp. φ(σ2 σ1 σ3−1 σ2−1 )) is the conjugacy by a (resp.
b−1 a).

Incompatibility between bi-orders on PBn and left-orders on Bn . In
contrast to B3 and B4 , the groups Bn fail to be locally indicable for n ≥ 5. Indeed,
for n ≥ 5, the commutator subgroup B′n is (finitely generated and) perfect (i.e.
it coincides with its own commutator subgroup), as shown below.
Example 3.2.8. As is well-known (and easy to check), the commutator subgroup B′n is
generated by the elements of the form σi,j := σi σj−1 . Also, recall that all the generators

95

3.2. THE CONRAD PROPERTY

σi of Bn are conjugate between them. Indeed, letting ∆ := σ1 σ2 · · · σn−1 , one readily
checks that σi ∆ = ∆ σi−1 . Thus, for all i ∈ {1, . . . , n − 3}, the equality
σi,i+2 = (σi σi+1 )−1 [σi,i+2 , σi+1,i ] (σi σi+1 )
shows that σi,i+2 belongs to B′′n . We will close the proof by showing that, for n ≥ 5, the
normal closure H (in Bn ) of the family of elements σi,i+2 (equivalently, of each σi,i+2 ) is
B′n . To do this, notice that σi,j and σi,j ′ are conjugate whenever {j, j ′ }∩{i−1, i+1} = ∅.
Indeed, one may perform a conjugacy between σj and σj′ as above but inside the
subgroup B′n−2 ⊂ Bn consisting of braids for which the i and i+1 strands remain “fixed”;
such a conjugacy does not change σi . Therefore, σi,j belongs to H for all j ∈
/ {i−1, i+1}.
Moreover, since for all j ∈
/ {i − 1, i, i + 1, i + 2} (resp. j ∈
/ {i − 2, i − 1, i, i + 1}),
σi,i+1 = σi,j σj,i+1

(resp. σi,i−1 = σi,j σj,i−1 ),

the elements si,i+1 and si,i−1 also belong to H. This shows that H coincides with B′n .
We recommend [146] for more details on this example as well as generalizations in
the context of Artin groups.

A nice consequence of the example above is that the bi-orders on PBn do not
extend to left-orders on Bn for any n ≥ 5. (This fact was established, independently, in [67] and [169].) Indeed, we have the following
Proposition 3.2.9. Let Γ0 be a finite-index subgroup of a left-orderable group
Γ. If  is a left-order on Γ whose restriction to Γ0 is Conradian, then  is
Conradian.
Proof. Let f ≻ id and g ≻ id be elements in Γ. One has f m ∈ Γ0 and g n ∈ Γ0
for some positive n, m smaller than or equal to the index of Γ0 in Γ. Hence,
f m g 2n ≻ g n ≻ g. We claim that this implies that either f g ≻ g or f g 2n ≻ g.
Otherwise, g −1 f g ≺ id and g −1 f g 2n ≺ id. Thus,
id ≺ g −1f m g 2n = (g −1f g)m−1 (g −1f g 2n ) ≺ id,
which is a contradiction.



A criterion of non left-orderability. Proposition 3.2.9 allows showing that
certain “small” groups cannot be left-ordered. In concrete terms, we have the
following result due to Rhemtulla [17, Chapter 7]:
Proposition 3.2.10. Let Γ be a finitely-generated group containing a finite-index
subgroup Γ0 all of whose left-orders are Conradian. If Γ has no nontrivial homomorphism into (R, +), then Γ is not left-orderable.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Indeed, if Γ were left-orderable then, by Proposition 3.2.9, every left-order
on it would be Conradian. Since Γ is finitely generated, Theorem 3.2.3 would
provide us with a nontrivial homomorphism into (R, +).
Example 3.2.11. In §1.4.1, we introduced the group
Γ = a, b : a2 ba2 = b, b2 ab2 = a ,
which contains an index-4 Abelian subgroup, namely ha2 , b2 , (ab)2 i ∼ Z3 . From the presentation, it follows that Γ admits no nontrivial homomorphism into (R, +). Since biinvariant left-orders are Conradian, Theorem 3.2.10 implies that Γ is not left-orderable.

3.2.2

An approach via crossings

An alternative –dynamical– approach to the theory of Conradian orders has
been recently developed in [153, 158]. We begin with the definition of the notion
of crossing, which is the most important tool in this approach.1
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u

fNv

w

gM u

v

Figure 11: A reinforced crossing.
Let  be a left-order on a group Γ. Following [158], we say that a 5-uple
(f, g; u, v, w) of elements in Γ is a crossing (resp. reinforced crossing) for
(Γ, ) if the following conditions are satisfied:
– u ≺ w ≺ v;
– g n u ≺ v and f n v ≻ u for every n ∈ N (resp. f u ≻ u and gv ≺ v);
– There exist M, N in N such that f N v ≺ w ≺ g M u.
1

It should be noticed that an equivalent notion –namely that of overlapping elements–
was introduced by Glass in his dynamical study of lattice-orderable groups [85], though no
conexion with the Conrad property is exhibited therein.

3.2. THE CONRAD PROPERTY

97

Clearly, every reinforced crossing is a crossing. Conversely, if (f, g; u, v, w) is a
crossing, then one easily checks that (f N g M , g M f N ; f N w, g M w, w) is a reinforced
crossing.
An equivalent notion to the above ones is that of a resilient pair, namely a
4-uple of group elements (f, g; u, v) satisfying
u ≺ f u ≺ f v ≺ gu ≺ gv ≺ v.
Indeed, if (f, g; u, v, w) is a reinforced crossing, then (f N , g M ; u, v) is a resilient
pair for the corresponding exponents M, N. Conversely, if (f, g; u, v) is a resilient
pair, then (f 2 , g; u, v, f v) is a reinforced crossing.

Figure 12: A resilient pair.
Theorem 3.2.12. The left-order  is Conradian if and only if (Γ, ) admits no
(reinforced) crossing.
Proof. Suppose that  is not Conradian, and let f, g be positive elements such
that f g n ≺ g for every n ∈ N. We claim that (f, g; u, v, w) is a crossing for (Γ, )
for the choice u := 1, v := f −1 g, w := g 2 . Indeed:
– From f g 2 ≺ g one obtains g 2 ≺ f −1 g, and since g ≻ 1, this yields 1 ≺ g 2 ≺
f −1 g, that is, u ≺ w ≺ v;
– From f g n ≺ g it follows that g n ≺ f −1 g, that is, g n u ≺ v (for every n ∈ N);
moreover, since both f, g are positive, we have f n−1 g ≻ 1, and thus f n (f −1 g) ≻ 1,
that is, f n v ≻ u (for every n ∈ N);
– The relation f (f −1g) = g ≺ g 2 may be read as f N v ≺ w for N = 1; finally,
the relation g 2 ≺ g 3 is w ≺ g M u for M = 3.
Conversely, let (f, g; u, v, w) be a crossing for (Γ, ) for which for certain M, N
in N,
f N v ≺ w ≺ g M u.

98

CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

We will prove that  is not Conradian by showing that, for h := g M f N and
h̄ := g M , both elements w −1hw and w −1 h̄w are positive, but
(w −1 hw)(w −1h̄w)n ≺ w −1h̄w,

for all n ∈ N.

To do this, first notice that gw ≻ w, as otherwise
w ≺ g N u ≺ g N w ≺ g N −1 w ≺ . . . ≺ gw ≺ w,
which is absurd. Clearly, the inequality gw ≻ w implies g M w ≻ w, hence
w −1 h̄w = w −1 g M w ≻ 1.

(3.2)

Moreover, hw = g M f N w ≻ g M f N f N v = g M f 2N v ≻ g M u ≻ w, thus
w −1hw ≻ 1.

(3.3)

Now notice that, for every n ∈ N,
hh̄n w = hg M n w ≺ hg M n g M u = hg M n+M u ≺ hv = g M f N v ≺ g M w = h̄w.
After multiplying by the left by w −1, the last inequality becomes
(w −1 hw)(w −1h̄w)n = w −1 hh̄n w ≺ w −1 h̄w,
as we wanted to check. Together with (3.2) and (3.3), this shows that  is not
Conradian.

Exercise 3.2.13. Using the characterization of the Conrad property in terms of resilient pairs, show that the subspace of C-left-orders is closed inside the space of leftorders of a group (c.f. Corollary 3.2.2).
Exercise 3.2.14. Using the notion of crossings, give an alternative proof for Proposition 3.2.9.
Hint. If (f, g; u, v) is a resilient pair, then the same is true for (f n , g n ; u, v), for all n ≥ 1.
Exercise 3.2.15. Proceed similarly with Proposition 3.2.1.
Hint. Show that, if f, g are positive elements for which f g2 ≺ g, then (f, f g; id, f g, g)
is a crossing for M = N = 2 (see Figure 13 below).

99

3.2. THE CONRAD PROPERTY
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g2

Figure 13: The n = 2 condition.
Example 3.2.16. The Dehornoy left-order D on the braid group Bn (where n ≥ 3) is
not Conradian. Indeed, as we next show, (f, g; u, v, w) := (σ2−1 , σ1 , σ2 , σ2 σ1 , σ2−1 σ1 ) is
a crossing for ≺D with M = N = 1 (see [159] for an alternative argument):

– It holds σ2 ≺D σ2−1 σ1 is u ≺D w; moreover, one easily checks that σ2 σ1 ≻D σ1 ≻D
σ2−1 σ1 , hence w ≺D v.

– For all k > 0, we have gk (u) = σ1k (σ2 ) ≺D σ2 σ1 = v, where the middle inequality
follows from σ1−1 σ2−1 σ1k σ2 = σ1−1 σ1 σ2k σ1−1 = σ2k σ1−1 ≺D 1; analogously, for k ∈ N, we
−(k−1)
have f k (v) = σ2−k (σ2 σ1 ) = σ2
σ1 ≺D σ2 σ1 , where the last inequality follows from
σ1−1 σ2k−1 σ2 σ1 = σ2 σ1k σ2−1 ≺D id.
– We have f (v) = σ2−1 (σ2 σ1 ) = σ1 ≻D σ2−1 σ1 = w and g(u) = σ1 (σ2 ) ≻D σ1 ≻D
σ2−1 σ1 = w.
Exercise 3.2.17. Show that the isolated left-order on the group Gm,n constructed in
§2.2.3 is not Conradian for (m, n) 6= (2, 2).
Remark 3.2.18. The dynamical characterization of the Conrad property should serve
as inspiration for introducing other relevant properties for group left-orders. (Compare
[153, Question 3.22].) For instance, one may say that a 6-uple (f, g; u1 , v1 , u2 , v2 ) of
elements in an ordered group (Γ, ) is a double resilient pair if both (f, g; u1 , v1 ) and
(g, f −1 ; u2 , v2 ) are resilient pairs and u1 ≺ u2 ≺ v1 (see Figure 14). Finding a simpler
algebraic counterpart of the property of not having a double crossing for a left-order
seems to be an interesting problem.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

The notion of n-resilient pair can be analogously defined. This corresponds to an
(2n + 2)-uple (f, g; u1 , v1 , u2 , v2 , . . . , un , vn ) such that:
– (f, g; u1 , v1 ), (g, f −1 ; u2 , v2 ), (f −1 , g−1 ; u3 , v3 ), (g−1 , f ; u4 , v4 ), (f, g; u5 , v5 ), etc, are
all resilient pairs,
– ui ≺ ui+1 ≺ vi , for all i ∈ {1, . . . , n − 1}.

An eventual affirmative answer for the question below would have interesting consequences; see Proposition 4.1.9.
Question 3.2.19. Let Γ be a left-orderable group such that no left-order admits an
n-resilient pair for some (large) n ∈ N. Does Γ admit a C-order ?
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g

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Figure 14: A double resilient pair.
Non-Conradian orders yield free subsemigroups. Let  be a non-Conradian
order on a group Γ. Let (f, g; u, v) ∈ Γ4 be a resilient pair for , and denote
A := [u, f v] := {w : u  w  f v},

B := [gu, v].

Then A and B are disjoint, and for all n ∈ N, we have f n (A ∪ B) ⊂ A and
g n (A ∪ B) ⊂ B. A direct application of the Positive Ping-Pong Lemma (see [93])
shows that the semigroup generated by f and g is free.
This shows in particular that all left-orders on torsion-free, virtually-nilpotent
groups are Conradian, a fact first established in [129] by different methods. (This
is no longer true for left-orderable polycyclic groups, even for metabelian ones;

3.2. THE CONRAD PROPERTY

101

see [17, Corollary 7.5.6].) Similarly, the equality LO(Γ) = CO(Γ) holds for
left-orderable groups Γ with subexponential growth, as for example GrigorchukMaki’s group [87, 154] (see Exercise 4.3.7 for a precise definition). As a consequence of Proposition 3.2.53, we obtain:
Theorem 3.2.20. The space of left-orders of a countable, torsion-free, virtuallynilpotent group with infinitely many left-orders is homeomorphic to the Cantor
set. The same holds for countable, left-orderable groups without free subsemigroups and having infinitely many left-orders.
Notice also that all left-orders on Tararin groups (i.e. groups with finitely
many left-orders; see §2.2.1) are Conradian. Indeed, if Γ has finitely many leftorders, then for every g ∈ Γ and every left-order  on Γ, the left-order g−n must
coincide with  for some finite n (actually, for an n smaller than or equal to
the cardinality of LO(Γ)). Thus, f ≻gn id holds for every -positive element f ,
that is, g −n f g n ≻ id. In particular, if g ≻ id, then f g n ≻ g n ≻ g, which shows
that  is Conradian.
Question 3.2.21. Suppose all left-orders on a finitely-generated, left-orderable
group are Conradian. Must the group be residually almost-nilpotent ?
Every non-Conradian order leads to uncountably many left-orders. Using the notion of crossings, we show an straightened version of Theorem 2.2.9 in
presence of non-Conradian orders.
Lemma 3.2.22. If  is a non-Conradian order on a group Γ, then there exists
(f, g, h; u, v) in Γ5 such that
u ≺ f u ≺ f v ≺ hu ≺ hv ≺ gu ≺ gv ≺ v.

Proof. Let (f¯, ḡ; u, v) be a resilient pair for , so that
u ≺ f¯u ≺ f¯v ≺ ḡu ≺ ḡv ≺ v.

Let f := f¯, h := ḡ f¯, g := ḡ 2. Then:
– The inequality f v ≺ hu is f¯v ≺ ḡ f¯u, which follows from
f¯u ≻ u =⇒ ḡ f¯u ≻ ḡu ≻ f¯v;

– The inequality hv ≺ gu is ḡ f¯v ≺ ḡ 2 u, which follows from
f¯v ≺ ḡu =⇒ ḡ f¯v ≺ ḡ 2u.



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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Theorem 3.2.23. If  is a non-Conradian order on a group Γ, then the closure
of the orbit of  in LO(Γ) contains a Cantor set.
Proof. Fix (f, g, h; u, v) as in the previous lemma. Let I denote the closure of
the subset {w : u  w  v} of LO(Γ). (Recall that w is the left-order with
positive cone w −1 P+ w.) Let I + := {′ ∈ I : h ≻′ id} and I + := {′ ∈ I : h ≺′ id}.
We claim that f (I) ⊂ I + and g(I) ⊂ I − . Indeed, to show that f (I) ⊂ I + , we
need to check that h(f w) ≻ f w for all u  w  v. But this follows from
h(f w)  h(f u) ≻ hu ≻ f v  f w.
The proof of the containment g(I) ⊂ I − is analogous.
Denote Λ := {0, 1}N , and let h0 := f and h1 := g. Consider the map

Λ → P orb() ,

ι = (i1 , i2 , . . .) 7→

\

n≥1

hi1 hi2 · · · hin (I) = ι(I).

By the claim above, if ι 6= ι′ , then ι(I) ∩ ι′ (I) = ∅. The theorem follows.

3.2.3



An extension to group actions on ordered spaces

Let Γ be a group acting by order-preserving bijections on a totally ordered
space (Ω, ≤). A crossing for the action of Γ on Ω is a 5-uple (f, g; u, v, w), where
f, g belong to Γ and u, v, w are in Ω, such that:
– It holds u ≺ w ≺ v;
– For every n ∈ N, we have g n u ≺ v and f n v ≻ u;

– There exist M, N in N so that f N v ≺ w ≺ g M u.
Analogous definitions of reinforced crossings and resilient pairs may be given.
Notice that for a left-ordered group (Γ, ), the notions of the preceding section
correspond to the above ones for the action by left-translations on the ordered
space (Γ, ).
For another relevant example, recall from Remark 2.1.6 that, given a -convex
subgroup Γ0 of a left-ordered group (Γ, ), the space of left cosets Ω = Γ/Γ0
carries a natural total order ≤ that is invariant by the left-translations. (Taking
Γ0 as the trivial subgroup, this reduces to the preceding example.) Whenever
this action has no crossing, we will say that Γ is a -Conradian extension of
Γ0 . Of course, this is the case of every convex subgroup Γ0 if  is Conradian.

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103

Remark 3.2.24. Let (Γ, ) be a left-ordered group, and let Γ0 be a -convex subgroup.
Given any left-order ∗ on Γ0 , let ′ be the extension of ∗ by . One readily checks
that Γ is a -Conradian extension of Γ0 if and only if it is a ′ -Conradian extension
of it.
Exercise 3.2.25. Let Γ be a subgroup of Homeo+ (R). Say that an open interval I is
an irreducible component of a nontrivial element g ∈ Γ if it is fixed by g and contains
no fixed point inside. Equivalently, I is a connected component of the complement of
the set of fixed points of g.
(i) Show that if the action of Γ is without crossings, then for any pair of different
irreducible components, either one of them contains the other, or they are disjoint.
(ii) Show that the converse of (i) also holds.

For a general order-preserving action of a group Γ on a totally ordered space
(Ω, ≤), the action of an element f ∈ Γ is said to be cofinal if for all x < y in Ω
there exists n ∈ Z such that f n (x) > y. Equivalently, the action of f is not cofinal
if there exist x < y in Ω such that f n (x) < y for every integer n. If (Γ, ) is a
left-ordered group, then f ∈ Γ is cofinal if it is so for the corresponding left action
of Γ on itself.
Proposition 3.2.26. Let Γ be a group acting by order-preserving bijections on
a totally ordered space (Ω, ≤). If the action has no crossings, then the set of
elements whose action is not cofinal forms a normal subgroup of Γ.
Proof. Let us denote the set of elements whose action is not cofinal by Γ0 . This
set is normal. Indeed, given g ∈ Γ0 , let x < y in Ω be such that g n (x) < y for all
n. For each h ∈ Γ we have g n h−1 (h(x)) < y, hence (hgh−1 )n (h(x)) < h(y) (for
all n ∈ Z). Since h(x) < h(y), this shows that hgh−1 belongs to Γ0 .
It follows immediately from the definition that Γ0 is stable under inversion,
that is, g −1 belongs to Γ0 for all g ∈ Γ0 . The fact that Γ0 is stable under multiplication is more subtle. For the proof, given x ∈ Ω and g ∈ Γ0 , we will denote by
Ig (x) the convex closure of the set {g n (x): n ∈ Z}, that is, the set formed by
the y ∈ Ω for which there exist m, n in Z so that g m (x) ≤ y ≤ g n (x). Notice that
Ig (x) = Ig (x′ ) for all x′ ∈ Ig (x). Moreover, Ig−1 (x) = Ig (x) for all g ∈ Γ0 and all
x ∈ Ω. Finally, if g(x) = x, then Ig (x) = {x}. We claim that if Ig (x) and If (y) are
non-disjoint for some x, y in Ω and f, g in Γ0 , then one of them contains the other.
Indeed, assume that there exist non-disjoint sets If (y) and Ig (x), none of which
contains the other. Without loss of generality, we may assume that Ig (x) contains
points to the left of If (y) (if this is not the case, just interchange the roles of f
and g). Changing f and/or g by their inverses if necessary, we may assume that

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

g(x) > x and f (y) < y, thus g(x′ ) > x′ for all x′ ∈ Ig (x), and f (y ′) < y ′ for all
y ′ ∈ Iy (f ). Take u ∈ Ig (x) \ If (y), w ∈ Ig (x) ∩ If (y), and v ∈ If (y) \ Ig (x). Then
one easily checks that (f, g; u, v, w) is a crossing, which is a contradiction.
Now, let g, h be elements in Γ0 , and let x1 < y1 and x2 < y2 be points in Ω
such that g n (x1 ) < y1 and hn (x2 ) < y2 , for all n ∈ Z. Set x := min{x1 , x2 } and
y := max{y1 , y2 }. Then g n (x) < y and hn (x) < y, for all n ∈ Z; in particular, y
does not belong to neither Ig (x) nor Ih (x). Since x belongs to both sets, we have
either Ig (x) ⊂ Ih (x) or Ih (x) ⊂ Ig (x). Both cases being analogous, let us consider
only the first one. Then for all x′ ∈ Ig (x) we have Ih (x′ ) ⊂ Ig (x′ ) = Ig (x). In
particular, h±1 (x′ ) belongs to Ig (x) for all x′ ∈ Ig (x). Since the same holds for
g ±1 (x′ ), this easily implies that (gh)n (x) ∈ Ig (x), for all n ∈ Z. As a consequence,
(gh)n (x) < y holds for all n ∈ Z, thus showing that gh belongs to Γ0 .

Slightly extending Example 2.1.1, a convex jump of a left-ordered group
(Γ, ) is a pair (G, H) of distinct -convex subgroups such that H is contained
in G, and there is no -convex subgroup between them.
Theorem 3.2.27. Let (Γ, ) be a left-ordered group, and let (G, H) be a convex
jump in Γ. Suppose that G is a Conradian extension of H. Then H is normal in
G, and the left-order induced by  on the quotient G/H is Archimedean.
Proof. Let us consider the action of G on the space of cosets G/H. Each element
of H fixes the coset H, hence its action is not cofinal. If we show that the action of
each element in G \ H is cofinal, then Proposition 3.2.26 will imply the normality
of H in G.
Now given f ∈ G \ H, let Gf be the smallest convex subgroup of G containing
(H and) f . We claim that Gf coincides with the set
Sf := {g ∈ G : f m ≺ g ≺ f n for some m, n in Z}.
Indeed, Sf is clearly a convex subset of G containing H and contained in Gf .
Thus, for showing that Gf = Sf , we need to show that Sf is a subgroup. To do
this, first notice that, with the notation of the proof of Proposition 3.2.26, the
conditions g ∈ Sf and Ig (H) ⊂ If (H) are equivalent. Therefore, for each g ∈ Sf
we have Ig−1 (H) = Ig (H) ⊂ If (H), thus g −1 ∈ Sf . Moreover, if ḡ is another
element in Sf , then ḡgH ∈ ḡ(If (H)) = If (H), hence Iḡg (H) ⊂ If (H). This
means that ḡg belongs to Sf , thus concluding the proof that Sf and Gf coincide.
Each f ∈ G \ H leads to a convex subgroup Gf = Sf strictly containing
H. Since (G, H) is a convex jump, we necessarily have Sf = G. Given g1 ≺ g2
in G, choose m1 , n2 in Z for which f m1 ≺ g1 and g2 ≺ f n2 . Then we have

3.2. THE CONRAD PROPERTY

105

f n2 −m1 g1 ≻ f n2 −m1 f m1 = f n2 ≻ g2 , hence f n2 −m1 (g1 H) ≥ g2 H. This easily
implies that the action of f is cofinal.
We have then showed that H is normal in G. The left-invariant total order
on the space of cosets G/H is therefore a group left-order. Moreover, given f, g
in G, with f ∈
/ H, the previous argument shows that there exists n ∈ Z such that
f n ≻ g, thus f n H  gH. This is nothing but the Archimedean property for the
induced left-order on G/H.

Corollary 3.2.28. Under the hypothesis of Theorem 3.2.27, up to multiplication
by a positive real number, there exists a unique homomorphism τ : G → (R, +)
such that ker(τ ) = H and τ (g) > 0 for every positive element g ∈ G \ H.
The homomorphism τ above will be referred to as the Conrad homomorphism associated to the corresponding Conradian extension (jump).
Exercise 3.2.29. Let Γ be a subgroup of Homeo+ (R). Show that the action of g ∈ Γ
is not cofinal if and only if g has fixed points on the line. If Γ is finitely generated
and acts without crossings, show that the normal subgroup formed by the elements
having fixed points has global fixed points. If the action corresponds to the dynamical
realization of a left-order , show that this subgroup coincides with the kernel of the
Conrad homomorphism associated to the convex jump with respect to the maximal
proper -convex subgroup (c.f. Example 2.1.2).

C-orderability implies local indicability. Let  be a Conradian order on
a group Γ. Let Γ0 be a nontrivial subgroup of Γ generated by finitely many
positive elements f1 ≺ . . . ≺ fk . Let Γf (resp. Γf ) be the largest (resp. smallest)
convex subgroup which does not contain f := fk (resp. which contains f ). By
the corollary above, there exists a nontrivial homomorphism τ : Γf → (R, +) such
that ker(τ ) = Γf . This shows that Γ is locally indicable.
Remark 3.2.30. The homomorphism τ produced above respects orders: if f  g, then
τ (f ) ≤ τ (g). Moreover, it is trivial when restricted to the maximal convex subgroup.
As commutators are mapped into zero by τ , we conclude that every element in [Γ, Γ]
is strictly smaller than any other element f satisfying τ (f ) > 0.

We close this section with the following analogue of Proposition 2.1.7.
Proposition 3.2.31. Let Γ be a C-orderable group, and let {Γλ : λ ∈ Λ} be a
family of subgroups each of which is convex with respect to aTC-left-order λ .
Then there exists a C-left-order on Γ for which the subgroup λ Γλ is convex.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

The proof is based on a result concerning left-orders obtained from actions on
a totally ordered space.
Proposition 3.2.32. Let Γ be a group acting faithfully by order-preserving transformations on a totally ordered space (Ω, ≤). If the action has no crossings, then
all induced left-orders on Γ are Conradian.
Proof. Suppose that the left-order  on Γ induced from the action via a wellorder 6wo on Ω (c.f. §1.1.3) is not Conradian. Then there are -positive elements
f, g in Γ such that f g n ≺ g, for every n ∈ N. This easily implies f ≺ g. Let
w̄ := min6wo {wf , wg }. (Recall that wf := min6wo {w : f (w) 6= w}, and similarly
for wg .) We claim that (f g, f g 2; w̄, g(w̄), f g 2(w̄)) is a crossing for the action.
Indeed:
– From id ≺ f ≺ g we obtain w̄ = wg 6wo wf and g(w̄) > w̄; moreover f (w̄) ≥ w̄,
which together with f g n ≺ g yield
w̄ < f g 2(w̄) < g(w̄).
– The preceding argument actually shows that f g n (w̄) < g(w̄), for all n ∈ N.
As a consequence, f g 2 f g 2(w̄) < f g 3 (w̄) < g(w̄). A straightforward inductive
argument then shows that (f g 2)n (w̄) < g(w̄), for all n ∈ N. Moreover, from
g(w̄) > w̄ and f (w̄) ≥ w̄, we conclude that w̄ < (f g)n (g(w̄)).

– Finally, w̄ < f g 2 (w̄) implies f g 2 (w̄) < f g 2 (f g 2(w̄)) = (f g 2)2 (w̄), whereas
f g 2(w̄) < g(w̄) implies (f g)2(g(w̄)) = f g(f g 2(w̄)) < f g(g(w̄)) = f g 2(w̄).


The proof of Proposition 3.2.31 proceeds
as that of Proposition 2.1.7. We
Q
consider the left action of Γ on Ω := λ∈Λ Γ/Γλ × Γ endowed
T with the lexicographic order. The stabilizer of ([idλ ])λ∈Λ × Γ coincides with λ Γλ , which may
be made convex for an induced left-order  on Γ. Now the main point is that,
as the action of Γ on each Γ/Γλ has no crossings, the same holds for the action
of Γ on Ω. By Proposition 3.2.32, the left-order  is Conradian, thus concluding
the proof.
Exercise 3.2.33. Prove the following converse to Proposition 3.2.32: If (Γ, ) is a
countable C-ordered group, then its dynamical realization is an action on the real line
without crossings (c.f. §1.1.3).

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3.2. THE CONRAD PROPERTY

3.2.4

The Conradian soul of a left-order

A subgroup of a left-ordered group (Γ, ) will be said to be Conradian if the
restriction of  to it is a Conradian order.
Definition 3.2.34. The Conradian soul C (Γ) of (Γ, ) is the (unique) subgroup that is -convex, -Conradian, and that is maximal among subgroups
verifying these two properties simultaneously.
Example 3.2.35. Recall from Example 3.2.7 that the commutator subgroup [B3 , B3 ]
is free in σ1 σ2−1 and σ12 σ2−2 . Denote by  the restriction of the Dehornoy left-order to
[B3 , B3 ]. As we show below,  has no proper convex subgroups.2 Since, as it is easily
shown,  is non-Conradian (compare Example 3.2.16), its Conradian soul is trivial.
Let C ⊂ F2 = [B3 , B3 ] be a nontrivial convex subgroup. Clearly, we may choose a
1-positive σ ∈ F2 . If σ commutes with σ2 , then one may show that σ is of the form
σ = ∆2p σ2q for some integers p, q satisfying 3p = −q > 0, where ∆ = σ1 σ2 σ1 . We thus
have ∆2 ≺ ∆4p σ2−6p = σ 2 . Since ∆2 is cofinal for the Dehornoy left-order and central,
σ is cofinal as well. Since C is a convex subgroup containing C, it must coincide with
F2 .
Suppose now that σ and σ2 do not commute. By the Subword Property (c.f. §1.2.6),
for every k > 0 the braid σσ2k σ −1 is 1-positive, as well as σσ2k σ −1 σ2−k . Next, σ2k σ −1 σ2−k
is 1-negative, so that σσ2k σ −1 σ2−k ≺ σ. By convexity, σσ2k σ −1 σ2−k must lie in C. Since
σ ∈ C, both σ2k σ −1 σ2−k and σ2k σσ2−k belong to C. Now σ may be represented as σ2m σ1 w,
where m is an integer, and w is a 1-positive, 1-neutral, or empty word. Choose k > 0
so that m′ = k + m > 0, and set σ ′ := σ2k σσ2−k . We know that σ ′ lies in C, and
it may be represented by the 1-positive braid word σ2ℓ σ1 wσ2−k . We will now proceed
to show that C must contain both generators of F2 , thus C = F2 . First notice that
σ2 (σ1−1 σ2ℓ σ1 )wσ2−k = σ2 (σ2 σ1ℓ σ2−1 )wσ2−k is 1-positice. Therefore,
id ≺ σ2 σ1−1 σ2ℓ σ1 wσ2−k

=⇒

σ1 σ2−1 ≺ σ2ℓ σ1 wσ2−k = σ ′ ∈ C,

and since id ≺ σ1 σ2−1 , this implies that σ1 σ2−1 ∈ C by convexity. Concerning the second
generator σ12 σ2−2 , observe that
σ22 σ1−1 σ1−1 σ2ℓ σ1 wσ2−k = σ22 σ1−1 σ2 σ1ℓ σ2−1 wσ2−k = σ22 σ2 σ1 σ2−1 σ1ℓ−1 σ2−1 wσ2−k
is 1-positive. Thus,
id ≺ σ22 σ1−2 σ2ℓ σ1 wσ2−k

=⇒

σ12 σ2−2 ≺ σ2ℓ σ1 wσ2−k = β ′ ∈ C,

and since 1 ≺ σ12 σ2−2 , we conclude from the convexity of C that σ12 σ2−2 ∈ C.
This example is due to Clay [45]. However, the existence of a left-order on F2 with no proper
convex subgroups also follows from the work of McCleary [136]. See also [159] for left-orders
on braid groups without proper convex subgroups.
2

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Example 3.2.36. The Conradian soul of D on Bn is the cyclic subgroup generated
by σn−1 . Indeed, this follows from the facts that the only n -convex subgroups of
Bn are {id}, hσn−1 i, hσn−2 , σn−1 i, . . . , hσ2 , . . . , σn−1 i and Bn itself, and that the
restriction of D to hσn−2 , σn−1 i ∼ B3 is not Conradian (c.f. Example 3.2.16). Let
us examine the case of B3 by denoting a = σ1 σ1 and b = σ2−1 . (For a general Bn ,
one uses a similar argument together with Theorem 2.2.44.) Recall that the family of
D -convex subgroups coincides with that of DD -convex ones. Clearly, hbi does not
properly contain any nontrivial convex subgroup. Suppose that there exists a DD convex subgroup B of B3 such that hbi ( B ( B3 . Let ′ , ′′ , and ′′′ , be the
left-orders defined on hbi, B, and B3 , respectively, by:
– ′ is the restriction of DD to hbi;

– ′′ is the extension of ′ by the restriction of DD to B;

– ′′′ is the extension of ′′ by DD .

The left-order ′′′ is different from DD (the DD -negative elements in B \ hbi are
′′′ -positive), but its positive cone still contains the elements a, b. Nevertheless, this is
impossible, since these elements generate the positive cone of DD .

Exercise 3.2.37. Let Γ∗ := C (Γ) be the Conradian soul of a left-ordered group
(Γ, ). Show that, for any Conradian order ∗ on Γ∗ , the extension of ∗ by  has
Conradian soul Γ∗ .

To give a dynamical counterpart of the notion of Conradian soul in terms of
crossings, we consider the set C + formed by the elements h ≻ id such that h  w
for every crossing (f, g; u, v, w) satisfying id  u. Analogously, we let C − be the
set formed by the elements h ≺ id such that w  h for every crossing (f, g; u, v, w)
satisfying v  id. Finally, we let
C := {id} ∪ C + ∪ C − .
A priori, it is not clear that the set C has a nice structure (for instance, it is not
at all evident that it is a subgroup). Nevertheless, we have the following
Theorem 3.2.38. The Conradian soul of (Γ, ) coincides with the set C above.
Before passing to the proof, we give four general lemmas on crossings for
left-orders (notice that the first three lemmas still apply to crossings for actions
on totally ordered spaces). The first one allows us replacing the “comparison
element” w by its “images” under positive iterates of either f or g.

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109

Lemma 3.2.39. If (f, g; u, v, w) is a crossing, then both (f, g; u, v, g nw) and
(f, g; u, v, f nw) are also crossings, for every n ∈ N.
Proof. We only consider the first 5-uple (the other is analogous). Since gw ≻ w,
for every n ∈ N we have u ≺ w ≺ g n w; moreover, v ≻ g M +n u = g n g M u ≻ g n w.
Hence, u ≺ g n w ≺ v. Furthermore, f N v ≺ w ≺ g n w. Finally, from g M u ≻ w
we get g M +n u ≻ g n w.

Our second lemma allows replacing the “limiting” elements u and v by more
appropriate ones.
Lemma 3.2.40. Let (f, g; u, v, w) be a crossing. If f u ≻ u (resp. f u ≺ u) then
(f, g; f n u, v, w) (resp. (f, g; f −n u, v, w)) is also a crossing for every n ≥ 1. Analogously, if gv ≺ v (resp. gv ≻ v), then (f, g; u, g nv, w) (resp. (f, g; u, g −nv, w))
is also crossing for every n ≥ 1.
Proof. Let us only consider the first 5-uple (the second case is analogous).
Suppose that f u ≻ u (the case f u ≺ u may be treated similarly). Then f n u ≻ u,
which yields g M f n u ≻ g M u ≻ w. To show that f n u ≺ w, assume by contradiction
that f n u  w. Then f n u ≻ f N v yields u ≻ f N −n v, which is absurd.

The third lemma relies on the dynamical nature of the crossing condition.
Lemma 3.2.41. If (f, g; u, v, w) is a crossing, then (hf h−1 , hgh−1 ; hu, hv, hw) is
also a crossing, for every h ∈ Γ.
Proof. The three conditions to be checked are nothing but the three conditions
in the definition of crossing multiplied by h on the left.

A direct application of the lemma above shows that, if (f, g; u, v, w) is a crossing, then the 5-uples (f, f n gf −n; f n u, f n v, f n w) and (g n f g −n , g; g nu, g nv, g n w) are
also crossings, for every n ∈ N.
Lemma 3.2.42. If (f, g; u, v, w) is a crossing and id  h1 ≺ h2 are elements in
Γ such that h1 ∈ C and h2 ∈
/ C, then there exists a crossing (f˜, g̃; ũ, ṽ, w̃) such
that h1 ≺ ũ ≺ ṽ ≺ h2 .
Proof. Since id ≺ h2 ∈
/ C, there must be a crossing (f, g; u, v, w) such that
id  u ≺ w ≺ h2 . Fix N ∈ N such that f N v ≺ w, and consider the crossing
(f, ḡ; ū, v̄, w̄) := (f, f N gf −N ; f N u, f N v, f N w).

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Notice that v̄ = f N v ≺ w ≺ h2 . We claim that h1  w̄ = f N w. Indeed, if
f N u ≻ u then f n u ≻ id, and by the definition of C we must have h1  w̄.
If f N u ≺ u, then we must have f u ≺ u, thus by Lemma 3.2.40 we know that
(f, ḡ; u, v̄, w̄) is also a crossing, which still allows concluding that h1  w̄.
Now, for the crossing (f, ḡ; ū, v̄, w̄), there exists M ∈ N such that w̄ ≺ ḡ M ū.
Let us consider the crossing (ḡ M f ḡ −M , ḡ; ḡ M ū, ḡ M v̄, ḡ M w̄). If ḡ M v̄ ≺ v̄, then
ḡ M v̄ ≺ h2 , and we are done. If not, then we must have ḡv̄ ≻ v̄. By Lemma 3.2.40,
(ḡ M f ḡ −M , ḡ; ḡ M ū, ḡ M v̄, w̄) is still a crossing, and since v̄ ≺ h2 , this concludes the
proof.

Proof of Theorem 3.2.38. The proof is divided into several steps.
Claim (i). The set C is convex.
This follows directly from the definition of C.
Claim (ii). If h belongs to C, then h−1 also belongs to C.
Assume that h ∈ C is positive and h−1 does not belong to C. Then there
exists a crossing (f, g; u, v, w) such that h−1 ≺ w ≺ v  id.
We first notice that, if h−1  u, then after conjugating by h as in Lemma
3.2.41, we get a contradiction because (hgh−1 , hf h−1 ; hu, hv, hw) is a crossing
with id  hu and hw ≺ hv  h. To reduce the case h−1 ≻ u to this one, we
first use Lemma 3.2.41 and consider the crossing (g M f g −M , g; g M u, g M v, g M w).
Since h−1 ≺ w ≺ g M u ≺ g M w ≺ g M v, if g M v ≺ v then we are done. If not,
Lemma 3.2.40 shows that (g M f g −M , g; g M u, g M v, w) is also a crossing, which still
allows concluding.
In the case where h ∈ C is negative, we proceed similarly but we conjugate
by f N instead of g M . Alternatively, since id ∈ C and id ≺ h−1 , if we suppose
that h−1 ∈
/ C then Lemma 3.2.42 provides us with a crossing (f, g; u, v, w) such
that id ≺ u ≺ w ≺ v ≺ h−1 , which gives a contradiction after conjugating by h.
Claim (iii). If h and h̄ belong to C, then hh̄ also belongs to C.
First, we show that for every pair of positive elements in C, their product
still belongs to C. (Notice that, by Claim (ii), the same will be true for pairs
of negative elements in C.) Indeed, suppose that h, h̄ are positive elements,
with h ∈ C but hh̄ ∈
/ C. Then, by Lemma 3.2.42, we may produce a crossing
(f, g; u, v, w) such that h ≺ u ≺ v ≺ hh̄. After conjugating by h−1 , we obtain
the crossing (h−1 f h, h−1 gh; h−1 u, h−1 v, h−1 w) satisfying id ≺ h−1 u ≺ h−1 w ≺ h̄,
which shows that h̄ ∈
/ C.
Now, if h ≺ id ≺ h̄, then h ≺ hh̄. Thus, if hh̄ is negative, then the convexity

3.2. THE CONRAD PROPERTY

111

of C yields hh̄ ∈ C. If hh̄ is positive, then h̄−1 h−1 is negative, and since h̄−1 ≺
h̄−1 h−1 , the convexity gives again that h̄−1 h−1 , hence hh̄, belongs to C. The
remaining case h̄ ≺ id ≺ h may be treated similarly.

Claim (iv). The subgroup C is Conradian.
In order to apply Theorem 3.2.12, we need to show that there are no crossings in C. Suppose by contradiction that (f, g; u, v, w) is a crossing such that
f, g, u, v, w all belong to C. If id  w then, by Lemma 3.2.41, we have that
(g n f g −n , g; g nu, g nv, g n w) is a crossing. Taking n = M so that g M u ≻ w, this
contradicts the definition of C, because id  w ≺ g M u ≺ g M w ≺ g M v ∈ C. The
case w  id may be treated analogously by conjugating by powers of f instead
of g.
Claim (v). The subgroup C is maximal among -convex, -Conradian subgroups.

Indeed, if H is a subgroup strictly containing C, then there is a positive
element h ∈ H \ C. By Lemma 3.2.42, there exists a crossing (f, g; u, v, w) such
that id ≺ u ≺ w ≺ v ≺ h. If H is convex, then u, v, w belong to H. To conclude
that H is not Conradian, it suffices to show that f and g belong to H.
On the one hand, since id ≺ u, we have either id ≺ g ≺ gu ≺ v or id ≺
−1
g ≺ g −1u ≺ v. In both cases, the convexity of H implies that g belongs to
H. On the other hand, if f is positive, then from f N ≺ f N v ≺ w we get f ∈ H,
whereas in the case of a negative f , the inequality id ≺ u gives id ≺ f −1 ≺
f −1 u ≺ v, which still shows that f ∈ H.


3.2.5

Approximation of left-orders and the Conradian soul

The notion of Conradian soul was introduced in [153] as a tool for leading with
the problem of approximating a group left-order by its conjugates. We begin with
the case of trivial Conradian soul. (Compare Proposition 3.3.3 and its proof.)
Theorem 3.2.43. If the Conradian soul of an infinite left-ordered group (Γ, )
is trivial, then  may be approximated by its conjugates.
We will give two different proofs for this theorem, each of which gives some
complementary information. The first one, due to Clay [46], shows that every leftorder which is not approximated by its conjugates admits a nontrivial, convex,
bi-ordered subgroup. This may also be obtained by using the method of the
second proof below (which is taken from [158]) under the stronger assumption
that  is isolated in LO(Γ). Nevertheless, though more elaborate than the first

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

(it uses the results of the preceding section), this second proof is suitable for
generalization in the case where the Conradian soul is “almost trivial” (i.e. it is
nontrivial but admits only finitely many left-orders; see Theorem 3.2.46 below).
First proof of Theorem 3.2.43. Suppose that  cannot be approximated by
its conjugates, and let g1 , . . . , gk be finitely many positive elements such that the
only conjugate of  lying in Vg1 ∩ . . . ∩ Vgk is  itself. (Recall that Vg denotes the
set of left-orders making g a positive element.) For each index i ∈ {1, . . . , k}, let

Bi+ = h ∈ Γ : id  h  gin for some n ∈ N ,

Bi = h ∈ Γ : gi−m  h  gin for some m, n in N .

Claim (i). For some j ∈ {1, . . . , k} we have h−1 P+ h = P+ for every h ∈ Bj+ .
If not, then for each i there exists hi ∈ Γ such that id ≺ hi  gini for some
+
+
+
−1 +
ni ∈ N and h−1
i P hi 6= P . Let h := min{h1 , . . . , hk }. Then h P h 6= P .
ni
n
Moreover, h  gi for each i, thus h−1 gi i  id. Since h is necessarily positive,
this yields h−1 gini h ≻ id, which implies h−1 gi h ≻ id, that is gi ∈ h−1 P+ h. Since
this holds for every i, by hypothesis the conjugate left-order h−1 must coincide
with , which is a contradiction.
Claim (ii). All elements in Bj+ stabilize P+ (under conjugacy).

Indeed, from gj−m  h  gin we obtain id  gjm h  gjm+n . Thus, gjm h
belongs to Bj+ . Since gjm also belongs to Bj+ , by Claim (i) above we have
(gjm h)−1 P+ (gjm h) = P+ ,

gj−m P+ gjm = P+ .

This easily yields h−1 P+ h = P+ , which in its turn implies hP+ h−1 = P+ .
Claim (iii). The set Bj is a -convex subgroup of Γ, and the restriction of  to
it is a bi-order (hence a C-order).
The convexity of Bj as a set is obvious. Now, for each h ∈ Bj , the relations
−m
gj  h  gjn and hP+ h−1 = P+ easily yield gjm  h  gjn , thus showing that
h−1 ∈ Bj . Similar arguments show that h1 h2 belongs to Bi+ for all h1 , h2 in Bj+ ,
as well as the bi-invariance of the restriction of  to Bj+ .

Second proof of Theorem 3.2.43. Let f1 ≺ f2 ≺ . . . ≺ fk be finitely many
positive elements of Γ. We need to show that there exists a conjugate of  that
is different from  but for which all the fi ’s are still positive.
Since id ∈ C (Γ) and f1 ∈
/ C (Γ), Theorem 3.2.38 and Lemma 3.2.42 imply
that there is a crossing (f, g; u, v, w) such that id ≺ u ≺ v ≺ f1 . Let M, N in

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3.2. THE CONRAD PROPERTY

N be such that f N v ≺ w ≺ g M u. We claim that id ≺v fi and id ≺w fi hold for
all 1 ≤ i ≤ k, but g M f N ≺v id and g M f N ≻w id. Indeed, since id ≺ v ≺ fi ,
we have v ≺ fi ≺ fi v, thus id ≺ v −1 fi v. By definition, this means that fi ≻v id.
The inequality fi ≻w id is proved similarly. Now notice that g M f N v ≺ g M w ≺ v,
hence g M f N ≺v id. Finally, from g M f N w ≻ g M u ≻ w we deduce g M f N ≻w id.
Now the preceding relations imply that the fi ’s are still positive for both v−1
and w−1 , but at least one of these left-orders is different from . This concludes
the proof.

We next deal with the case where the Conradian soul is nontrivial but admits
finitely many left-orders (i.e. it is a Tararin group; see §2.2.1). It turns out
that, in this case, the left-order may fail to be an accumulation point of its
conjugates. A concrete example is given by the DD-left-order on Bn . Indeed,
its Conradian soul is isomorphic to Z (c.f. Example 3.2.36), though it is an
isolated point of the space of braid left-orders because its positive cone is finitely
generated (c.f. §2.2.3). Now the DD-left-order has the Dehornoy left-order D
as a natural “associate”, in the sense that the latter may be obtained from the
former by successive flipings along convex jumps. For the case of B3 , this reduces
to changing the left-order on the Conradian soul in the unique possible way. As
shown below, D is an accumulation point of its conjugates. Moreover, there is
a sequence of conjugates of DD that converges to D as well.
Example 3.2.44. The sequence of conjugates j of D by σ2j σ1−1 converges to D in
a nontrivial way. Indeed, if w = σ2k for some k > 0, then
σ1−1 σ2j wσ2−j σ1 = σ1−1 σ2k σ1 = σ2 σ1k σ2−1 ≻D id.
k

If, on the other hand, w is a σ1 -positive word, say w = σ2k1 σ1 σ2k2 . . . σ2 ℓ−1 σ1 σ2kℓ , then
k

k

σ1−1 σ2j wσ2−j σ1 = σ1−1 σ2j σ2k1 σ1 σ2k2 . . . σ2 ℓ−1 σ1 σ2kℓ σ2−i σ1 = σ2 σ1j+k1 σ2−1 σ2k2 . . . σ2 ℓ−1 σ1 σ2kℓ σ2−n σ1 .

Thus, σ1 σ2−j wσ2j σ1 is 1-positive for sufficiently large j (namely, for j > −k1 ). This
proves the desired convergence. Finally, j is different from D for each positive
integer j, since its smallest positive element is the conjugate of σ2 by σ1 σ2j , and this is
different from the smallest positive element of D , namely σ2 . We leave to the reader
the task of checking that the sequence of conjugates of DD by σ1−1 σ2j converges to D
as well.
Remark 3.2.45. The B3 -case of the preceding example can be generalized as follows:
For all m, n larger than 1, with (m, n) 6= (2, 2), the left-order  on Gm,n = ha, b :
(bam−1 )n−1 b = ai with positive cone ha, bi+ given by Theorem 2.2.47 has Conradian

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

soul hbi ∼ Z. Flipping this order on the Conradian soul yields a left-order ′ that is
accumulated by its conjugates. Moreover, there is a sequence of conjugates of  that
also converges to ′ . See [151] as well as [102, 103, 104] for more on this and related
examples.

It turns out that the phenomenon described above for braid groups occurs for
general left-ordered groups. To be more precise, let Γ be a group having a leftorder  whose Conradian soul admits finitely many left-orders 1 , 2 , . . . , 2n ,
where 1 is the restriction of  to its Conradian soul. Each j induces a left-order
j on Γ, namely the convex extension of j by . (Notice that 1 coincides with
.) All the left-orders j share the same Conradian soul (c.f. Exercise 3.2.37).
Assume throughout that  is not Conradian, which is equivalent to that Γ is not
a Tararin group.
Theorem 3.2.46. With the notation above, at least one of the left-orders j is
an accumulation point of the set of conjugates of .
Corollary 3.2.47. At least one of the left-orders j is approximated by its conjugates.
Proof. Assuming Theorem 3.2.46, we have that k belongs to the set of accumulation points acc(orb(1 )) of the orbit of 1 for some k in {1, . . . , 2n }. Theorem
3.2.46 applied to this k instead of  shows the existence of k ′ ∈ {1, . . . , 2n } so
′
′
that k ∈ acc(orb(k )), and hence k ∈ acc(orb(1 )). If k ′ equals either 1 or k
then we are done; if not, we continue arguing in this way... In at most 2n steps
we will find an index j such that j ∈ acc(orb(j )).

Theorem 3.2.46 will follow from the next
Proposition 3.2.48. Given an arbitrary finite family G of -positive elements
in Γ, there exists h ∈ Γ and a positive h̄ ∈
/ C (Γ) such that id ≺ h−1 f h ∈
/ C (Γ)
−1
for all f ∈ G \ C (Γ), but id ≻ h h̄h ∈
/ C (Γ).
Proof of Theorem 3.2.46 from Proposition 3.2.48. Let us consider the
directed net formed by the finite sets G of -positive elements. For each such a G,
let hG and h̄G be the elements in Γ provided by Proposition 3.2.48. After passing
to subnets of (hG ) and (h̄G ) if necessary, we may assume that the restrictions of
hG to C (Γ) all coincide with a single j . Now the properties of hG and h̄G
imply:

3.2. THE CONRAD PROPERTY
– f ≻j id and f (≻j )hG id, for all f ∈ G \ C (Γ);
– h̄G ≻ id, but h̄G (≺j )hG ≺ id.
This clearly shows the theorem.

115



For the proof of Proposition 3.2.48 we will use some lemmas.
Lemma 3.2.49. For every id ≺ c ∈
/ C (Γ), there is a crossing (f, g; u, v, w) such
that u, v, w do not belong to C (Γ) and id ≺ u ≺ w ≺ v ≺ c.
Proof. By Theorem 3.2.38 and Lemma 3.2.42, for every id  s ∈ C (Γ) there
exists a crossing (f, g; u, v, w) such that s ≺ u ≺ w ≺ v ≺ c. Clearly, v does not
belong to C (Γ). The element w is also outside C (Γ), as otherwise the element
a := w 2 would satisfy w ≺ a ∈ C (Γ), which is absurd. Taking M > 0 so that
g M u ≻ w, this gives g M u ∈
/ C (Γ), g M w ∈
/ C (Γ), and g M v ∈
/ C (Γ). Consider
M
−M
M
M
M
the crossing (g f g , g; g u, g v, g w). If g M v ≺ v, then we are done. If not,
then gv ≻ v, and Lemma 3.2.40 ensures that (g M f g −M, g; g M u, v, g M w) is also a
crossing, which still allows concluding.

Lemma 3.2.50. Given id ≺ c ∈
/ C (Γ), there exists id ≺ a ∈
/ C (Γ) (with a ≺ c)
−1
such that, for all id  b  a and all c̄  c, one has id ≺ b c̄b ∈
/ C (Γ).
Proof. Let us consider the crossing (f, g; u, v, w) such that id ≺ u ≺ w ≺ v ≺ c
and such that u, v, w do not belong to C (Γ). We affirm that the lemma holds
for a := u. Indeed, if id  b  u, then from b  u ≺ v ≺ c̄ we obtain
id  b−1 u ≺ b−1 v ≺ b−1 c̄, thus the crossing (b−1 f b, b−1 gb; b−1 u, b−1 v, b−1 w) shows
that b−1 c̄ ∈
/ C (Γ). Since id  b, we conclude that id ≺ b−1 c̄  b−1 c̄b, and the
convexity of S implies that b−1 c̄b ∈
/ C (Γ).

Lemma 3.2.51. For every g ∈ Γ, the set g C (Γ) is convex. Moreover, for every
crossing (f, g; u, v, w), one has uC (Γ) < wC (Γ) < vC (Γ), in the sense that
uh1 ≺ wh2 ≺ vh3 for all h1 , h2 , h3 in C (Γ).
Proof. The verification of the convexity of gC (Γ) is straightforward. Suppose
next that uh1 ≻ wh2 for some h1 , h2 in C (Γ). Then, since u ≺ w, the convexity
of both left classes uC (Γ) and wC (Γ) gives the equality between them. In
particular, there exists h ∈ C (Γ) such that uh = w. Notice that such an h must
be positive, hence id ≺ h = u−1 w. But since (u−1 f u, u−1gu; id, u−1v, u−1 w) is
a crossing, this contradicts the definition of C (Γ). The proof of the fact that
wC (Γ) ≺ vC (Γ) is similar.


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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Proof of Proposition 3.2.48. Indexing the elements of G = {f1 , . . . , fr } so that
f1 ≺ . . . ≺ fr , let k be such that fk−1 ∈ C (Γ) but fk ∈
/ C (Γ). Recall that, by
Lemma 3.2.50, there exists id ≺ a ∈
/ C (Γ) such that, for every id  b  a, one
has id ≺ b−1 fk+j b ∈
/ C (Γ) for all j ≥ 0. We fix a crossing (f, g; u, v, w) such that
id ≺ u ≺ v ≺ a and u ∈
/ C (Γ). Notice that the conjugacy by w −1 yields the
−1
−1
crossing (w f w, w gw; w −1u, w −1v, id).
Case I. One has w −1 v  a.
In this case, we claim that the proposition holds for the choice h := w −1v and
h̄ := w −1g M +1 f N w. To show this, first notice that neither w −1 gw nor w −1 f w
belong to C (Γ). Indeed, this follows from the convexity of C (Γ) and the
inequalities w −1 g −M w ≺ w −1 u ∈
/ C (Γ) and w −1f −N w ≻ w −1 v ∈
/ C (Γ).
−1 M N
We also have id ≺ w g f w, hence id ≺ w −1 gw ≺ w −1g M +1 f N w, which
shows that h̄ ∈
/ C (Γ). Moreover, the inequality w −1g M +1 f N w(w −1v) ≺ w −1v
can be written as h−1 h̄h ≺ id. Finally, Lemma 3.2.39 applied to the crossing (w −1 f w, w −1gw; w −1u, w −1v, id) shows that, for every n ∈ N, the 5-uple
(w −1 f w, w −1gw; w −1u, w −1v, w −1 g M +nf N w) is also a crossing. For n ≥ M we
have w −1 g M +1f N w(w −1v) ≺ w −1 g M +n f N w. Since w −1g M +n f N w ≺ w −1v, Lemma
3.2.51 easily implies that w −1g M +1 f N w(w −1v)C (Γ) ≺ w −1 vC (Γ), which yields
/ C (Γ).
h−1 h̄h ∈
Case II. One has a ≺ w −1 v and w −1 g mw  a, for all m > 0.
We claim that, in this case, the proposition holds for the choice h := a
and h̄ := w −1g M +1 f N w. This may be checked in the very same way as in
Case I by noticing that, if a ≺ w −1 v but w −1 g m w  a for all m > 0, then
(w −1 f w, w −1gw; w −1u, a, id) is a crossing.

Case III. One has a ≺ w −1 v and w −1g m w ≻ a for some m > 0. (Notice that the
first condition follows from the second one.)
We claim that, in this case, the proposition holds for the choice h := a and
h̄ := w ∈
/ C (Γ). Indeed, we have g m w ≻ ha (and w ≺ ha). Since g m w ≺ v ≺ a,
we also have wa ≺ a, which means that h−1 h̄h ≺ id. Finally, from Lemmas 3.2.39
and 3.2.51, we obtain
waC (Γ)  g m wC (Γ) ≺ vC (Γ)  aC (Γ).
This implies that a−1 waC (Γ) ≺ C (Γ), which means that h−1 h̄h ∈
/ C (Γ). 
Remark 3.2.52. In the context of Theorem 3.2.46, it is possible that one of the
orders j may be not approximated by its conjugates despite being non-isolated. An
illustrative example of this fact for free groups is the subject of the Appendix of [158].

3.2. THE CONRAD PROPERTY

3.2.6

117

Groups with finitely many Conradian orders

The starting point of this section is the following
Proposition 3.2.53. Let Γ be a C-orderable group. If Γ admits a Conradian
left-order having a countable neighborhood in LO(Γ), then Γ admits finitely many
left-orders.
Before showing this proposition, let us show how it leads to a
Proof of Theorem 2.2.9. We provide three different arguments (see §4.4.3
for still another one that gives supplementary information). First, as we saw
in §2.2.1, the proof is reduced to showing Proposition 2.2.10. So, let (Γ, ) be
a left-ordered group admitting a finite-index subgroup restricted to which  is
bi-invariant. By Proposition 3.2.9, the left-order  is Conradian. By Proposition
3.2.53, if Γ admits infinitely many left-orders, then all neighborhoods of  in
LO(Γ) are uncountable.
An alternative argument proceeds as follows. As was shown in §3.2.2, if a
group admits a non Conradian order, then it has uncountably many left-orders.
Assume that Γ is left-orderable and all of its left-orders are Conradian. By
Proposition 3.2.53, if some of them has a countable neighborhood inside LO(Γ) =
CO(Γ) (in particular, if LO(Γ) is countable), then Γ admits only finitely many
left-orders.
As a final argument, notice that Proposition 3.2.53 together with a convex
extension argument (c.f. Section 2.1.1) show that, if Γ is a left-orderable group
such that LO(Γ) has an isolated point , then the Conradian soul C (Γ) cannot
have infinitely many left-orders. If C (Γ) is trivial (resp. if it is nontrivial
and admits finitely many left-orders), then Proposition 3.2.43 (resp. Proposition
3.2.46) yields the existence of a left-order ∗ on Γ that is accumulated by its
conjugates. As we have already remarked, the closure of the orbit under the
conjugacy action of such a left-order is uncountable.

Proof of Proposition 3.2.53. Let Γ be a group admitting a Conradian order
 having a countable neighborhood in LO(Γ), say

Vf1 ∩ . . . ∩ Vfk = ′: fi ≻′ id for all i ∈ {1, . . . , k} .

Claim (i). The chain of -convex subgroups is finite.
Otherwise, there exists an infinite ascending or descending chain of convex
jumps Γgn ⊳ Γgn so that fm ∈
/ Γgn \ Γgn for every m, n. As in the proof of

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Proposition 2.2.17, for each ι = (i1 , i2 . . .) ∈ {−1, +1}N let us define the left-order
ι on Γ by:


T
T
Γ \ (Γgn \ Γgn ) = P+
– P+ι
Γ \ (Γgn \ Γgn ) , for each n ∈ N;
– Pι ∩ (Γgn \ Γgn ) equals P+ ∩ (Γgn \ Γgn ) (resp. P− ∩ (Γgn \ Γgn )) if in = +1
(resp. in = −1).
This yields a continuous embedding of the Cantor set {−1, +1}N into LO(Γ).
Moreover, since fm ∈
/ Γgn \ Γgn for every m, n, the image of this embedding is
contained in Vf1 ∩ . . . ∩ Vfk . This proves the claim.
Claim (ii). Denote by {id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ the chain of all -convex
subgroups. Then each quotient Γi−1 /Γi is torsion-free, rank-1 Abelian.
If the rank of some Γi−1 /Γi were larger than 1, then the induced left-order on
the quotient would be non-isolated in the space of left-orders of Γi−1 /Γi . This
would allow to produce –by a convex extension type procedure– uncountably
many left-orders on any given neighborhood of , which is contrary to our hypothesis.
Claim (iii). In the series above, the group Γk−2 is not bi-orderable.
First notice that Γk−2 is not Abelian. Otherwise, it would have rank 2. This
would imply that every neighborhood of the restriction of  to Γk−2 is uncountable, which implies –by convex extension– the same property for .
Now as in the case of Proposition 2.2.19, if Γk−2 were bi-orderable, then it
would be contained in the affine group Aff + (R). The space of left-orders of a
non-Abelian countable group inside Aff + (R) was roughly described in §1.2.2: it is
homeomorphic to the Cantor set. (See also §3.3.) In particular, no neighborhood
of the restriction of  to Γk−2 is countable, which implies –by convex extension–
that the same is true for Γ. For sake of completeness, we give an explicit sequence
of approximating left-orders. To do this, notice that, for some q > 0, the group
Γk−2 can be identified with the group whose elements are of the form

 k
q a
,
(k, a) ∼
0 1
where a ∈ Γk−1 and k ∈ Z. Let (k1 , a1 ), . . . , (kn , an ) be an arbitrary family of
-positive elements indexed in such a way that k1 = k2 = . . . = kr = 0 and
kr+1 6= 0, . . . , kn 6= 0 for some r ∈ {1, . . . , n}. Four cases are possible:
(i) a1 > 0, . . . , ar > 0 and kr+1 > 0, . . . , kn > 0;
(ii) a1 < 0, . . . , ar < 0 and kr+1 > 0, . . . , kn > 0;

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3.2. THE CONRAD PROPERTY

(iii) a1 > 0, . . . , ar > 0 and kr+1 < 0, . . . , kn < 0;
(iv) a1 < 0, . . . , ar < 0 and kr+1 < 0, . . . , kn < 0.
As in §1.2.2, for each irrational number ε, let ε be the left-order on ε whose
positive cone is

P+ε = (k, a) : q k + εa > 1 .

In case (i), for ε positive and very small, the left-order ε is different from 
but still makes positive all the elements (ki , ai ). The same is true in case (ii)
for ε negative and near zero. In case (iii), this still holds for the order ¯ε when
ε is negative and near zero. Finally, in case (iv), one needs to consider again
the order ¯ε but for ε positive and small. Now letting ε vary over a Cantor set
formed by irrational numbers3 very close to 0 (and which are positive or negative
according to the case), this shows that the neighborhood of (the restriction to
 of)  consisting of the left-orders on Γk−2 that make positive all the elements
(ki , ai ) contains a homeomorphic copy of the Cantor set.

Claim (iv). The series of Claim (ii) is normal (hence rational) and no quotient
Γi−2 /Γi is bi-orderable.
By Theorem 2.2.13, the group Γk−2 admits a unique rational series, namely
{id}⊳Γk−1 ⊳Γk−2 . Since for every h ∈ Γk−3 the series {id}⊳hΓk−1 h−1 ⊳hΓk−2 h−1
is also rational for Γk−2 , they must coincide. Hence, the rational series
{id} ⊳ Γk−1 ⊳ Γk−2 ⊳ Γk−3
is normal. Moreover, proceeding as in Claim (iii) with the induced left-order on
Γk−3 /Γk−1, one readily checks that this quotient is not bi-orderable. Once again,
Theorem 2.2.13 implies that the rational series of Γk−3 is unique... Continuing
arguing in this way, the claim follows.
We may now conclude the proof of the proposition. Indeed, we have shown
that, if Γ is a group having a C-order with a countable neighborhood in LO(Γ),
then Γ admits a rational series
{id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ
such that no quotient Γi−2 /Γi is bi-orderable. By Theorem 2.2.13, Γ has only
finitely many left-orders.

3

Take for example the set of numbers of the form
P
it by j≥1 42j2 .

P

ik
i≥1 4k ,

where ik ∈ {0, 1}, and translate

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

We now turn to the study of the space of Conradian orders. The next result
from [173] is the analogue of Tararin’s theorem describing left-orderable groups
with finitely many left-orders; see §2.2.1.
Theorem 3.2.54. If a C-orderable group Γ has only finitely many C-orders, then
it has a unique (hence normal) rational series {id} = Γk ⊳ Γk−1 ⊳ . . .⊳ Γ0 = Γ. In
this series, no quotient Γi−2 /Γi is Abelian. Conversely, if Γ is a group admitting
a normal rational series {id} = Γk ⊳ Γk−1 ⊳ . . . ⊳ Γ0 = Γ such that no quotient
Γi−2 /Γi is Abelian, then the number of C-orders on Γ is 2k .
Proof. The proof will be divided into four independent claims.
Claim (i). If Γ is a C-orderable group admitting only finitely many C-orders, then
for every C-order  on Γ, the sequence of -convex subgroups is a rational series.
Indeed, for each convex jump Γg ⊳ Γg , we may flip the left-order on Γg to produce a new left-order (c.f. Example 2.1.4) which is still Conradian (c.f. Exercise
3.2.5). If there were infinitely many -convex subgroups, then this would allow
to produce infinitely many C-orders on Γ, contrary to our hypothesis. Let then
{id} = Γk ( Γk−1 ( . . . ( Γ0 = Γ
be the sequence of all -convex subgroups. As in the proof of Proposition 2.2.17,
Γi is normal in Γi−1 , and Γi−1 /Γi is torsion-free Abelian. The rank of this quotient
must be 1, as otherwise it would admit uncountably many orders, which would
allow to produce –by convex extension– uncountably many C-orders on Γ.
Claim (ii). If a left-orderable group admits only finitely many C-orders, then it
has a unique (hence normal) rational series.
The proof is almost the same as that of Proposition 2.2.18. We just need
to change the word “left-order” by “C-order” along that proof, and replace the
(crucial) use of Proposition 2.1.7 by Proposition 3.2.31.
Claim (iii). If a group Γ with a normal rational series {id} = Γk ⊳Γ⊳. . .⊳Γ0 = Γ
admits only finitely many C-orders, then no quotient Γi−2 /Γi is Abelian.
First notice that every group admitting a rational series is C-orderable. Actually, using the rational series above, one may produce 2k Conradian orders on
Γ. If one of the quotients Γi−2 /Γi were Abelian, then it would have rank 2, hence
it would admit uncountably many left-orders. This would allow to produce –by
convex extension– uncountably many C-orders on Γ.
Claim (iv). If a group Γ has a normal rational series {id} = Γk ⊳ Γ ⊳ . . . ⊳ Γ0 = Γ
such that no quotient Γi−2 /Γi is Abelian, then this series coincides with that

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3.2. THE CONRAD PROPERTY

formed by the -convex subgroups, where  is any C-order on Γ. In particular,
such a series is unique.
As we have already seen, the rational series above leads to 2k Conradian leftorders. We have to prove that these are the only possible C-orders on G. To
show this, let  be a C-order on Γ. By Claim (iii), there exist non-commuting
elements g ∈ Γk−1 and h ∈ Γk−2 \ Γk−1 . Denote the Conrad homomorphism
of the group hg, hi (endowed with the restriction of  ) by τ . Then we have
τ (g) = τ (hgh−1 ) 6= 0. Since Γk−1 is rank-1 Abelian, hgh−1 must be equal to
g s for some rational number s 6= 1. Hence, τ (g) = sτ (g), which implies that
τ (g) = 0. Therefore, g n ≺ |h| for every n ∈ Z, where |h| := max{h−1 , h}. Since
Γk−2 /Γk−1 has rank 1, this actually holds for every h 6= id in Γk−2 \ Γk−1 . Thus,
Γk−1 is -convex in Γk−2 .
Repeating the argument above, though now with Γk−2 /Γk−1 and Γk−3 /Γk−1
instead of Γk−1 and Γk−2 , respectively, we see that the rational series we began
with is no other than the series given by the -convex subgroups. Since each
Γi−1 /Γi is rank-1 Abelian, if we choose gi ∈ Γi−1 \ Γi for each i, then any C-order
on Γ is completely determined by the signs of these elements. This shows the
claim, and concludes the proof of Theorem 3.2.54.

Example 3.2.55. The Baumslag-Solitar group BS(1, ℓ) = ha, b : aba−1 = bℓ i, ℓ ≥ 2,
admits the rational series
1

{id} ⊳ bZ[ ℓ ] := hc : cℓ = b for some integer i > 0i ⊳ BS(1, ℓ),
i

which satisfies the conditions of Theorem 3.2.54. Therefore, it admits four C-orders –all
of which are bi-invariant–, though its space of left-orders is uncountable (c.f. §1.2.2).
See §3.3.1 for more details on this example.
Example 3.2.56. Examples of groups having exactly 2k left-orders (hence 2k Conradian orders) were introduced in Example 2.2.14. Namely, one may consider Kk =
ha1 , . . . , ak | Rk i, where the set of relations Rk is
−1
a−1
i+1 ai ai+1 = ai

if

i < k,

ai aj = aj ai

if

|i − j| ≥ 2.

The existence of groups with 2k Conradian orders but infinitely many (hence uncountably many) left-orders is more subtle. As we have seen in the preceding example, for
n = k this is the case of the Baumslag-Solitar groups BS(1, ℓ) for ℓ ≥ 2. To construct
examples for higher k having BS(1, ℓ) as a quotient by a normal convex subgroup, we
choose an odd integer ℓ ≥ 3 and we let Cn (ℓ) be the group
c, b, a1 , . . . , an | cbc−1 = bℓ , cai = ai c , ban b−1 = a−1
n , bai = ai b if i 6= n , Rn .

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

This corresponds to the set Z × Z[ 31 ] × Zn endowed with the product rule

  m′
 m
c, k , a1 , . . . , an · c′ , k′ , a′1 , . . . , a′n =
ℓ
ℓ


′
m
m
= c + c′ , ℓc k′ + k , (−1)a2 a′1 + a1 , . . . , (−1)an a′n−1 + an−1 , (−1)m a′n + an .
ℓ
ℓ
Notice that this is well-defined, as (−1)m = (−1)m̄ whenever m/ℓk = m̄/ℓk̄ (it is here
where the fact that ℓ is odd becomes important). The group Cn (ℓ) admits the rational
series
1

{id} ⊳ ha1 i ⊳ ha1 , a2 i ⊳ . . . ⊳ ha1 , . . . , an i ⊳ a1 , . . . , an , bZ[ ℓ ] ⊳ Cn (ℓ).
By Theorem 3.2.54, it admits exactly 2n+2 Conradian orders. However, it has BS(1, ℓ)
as quotient by the normal convex subgroup Kn . Since BS(1, ℓ) admits uncountably
many (left) left-orders, the same is true for Cn (ℓ).

We close this section with a result (also taken from [173]) to be compared
with Theorem 2.2.9.
Theorem 3.2.57. Every C-orderable group admits either finitely many or uncountably many C-orders. In the last case, none of these left-orders is isolated in
the space of C-orders.
To prove this theorem, we need the lemmas below.
Lemma 3.2.58. If Γ is C-orderable group such that CO(Γ) has an isolated point
, then the family of -convex subgroups (is finite and) is a rational series such
that no quotient of the form Γi−2 /Γi is Abelian.
Proof. As in Claim (i) of Proposition 3.2.53, the family of -convex subgroups
is finite, say
{id} = Γk ( Γk−1 ( . . . ( Γ0 = Γ.
Since  is Conradian, Γi is normal in Γi−1 for each i. The proofs of that Γi−1 /Γi
has rank-1 and no quotient Γi−2 /Γi is Abelian are similar to those of Theorem
3.2.54, and we leave them to the reader.

Lemma 3.2.59. For any C-orderable group whose space of C-orders has an isolated point , the rational series formed by the -convex subgroups is normal.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

123

Proof. The proof is by induction on the length k of the rational series. For
k = 1, there is nothing to prove; for k = 2, the series is automatically normal.
Assume that the claim of the lemma holds for k, and let
{id} = Γk+1 ⊳ Γk ⊳ . . . ⊳ Γ1 ⊳ Γ0 = Γ

(3.4)

be the rational series of length k + 1 associated to a C-order on a group Γ that
is isolated in CO(Γ). Notice that the truncated chain of length k
{id} = Γk+1 ⊳ Γk ⊳ . . . ⊳ Γ1

(3.5)

is a rational series for Γ1 . Moreover, this series is associated to a C-order on Γ1
(namely the restriction of ) that is isolated in CO(Γ1 ) (otherwise,  would be
non-isolated in CO(Γ)). By the inductive hypothesis, this series is normal. By
the preceding lemma, for each i ∈ {3, . . . , k + 1}, the quotient Γi−2 /Γi is nonAbelian. We are hence under the hypothesis of Theorem 3.2.54, which allows us
to conclude that this is the unique rational series of Γ1 .
Now, since Γ1 is normal in Γ, for each h ∈ Γ, the conjugate series
{id} = hΓk+1 h−1 ⊳ hΓk h−1 ⊳ . . . ⊳ hΓ1 h−1 = Γ1

is also a rational series for Γ1 . By the uniqueness above, this series coincides with
(3.5). Therefore, (3.4) is a normal rational series.

The proof of Theorem 3.2.57 is now at hand. Indeed, the two preceding
lemmas imply that, if a C-orderable group admits an isolated C-order, then it
has a normal rational series satisfying the hypothesis of Theorem 3.2.54, thus it
has finitely many C-orders. If, otherwise, no C-order is isolated in the space of
C-orders, then this is a Hausdorff, totally disconnected, topological space without
isolated points, hence uncountable ([95, Theorem 2-80]).
Exercise 3.2.60. By slightly pursuing on the arguments above, show the following analogue of Proposition 3.2.53: If a C-orderable group admits infinitely many C-orders,
then every neighborhood of such a left-order in the space of Conradian orders is uncountable.

3.3
3.3.1

An Application: Ordering Solvable Groups
The space of left-orders of finite-rank solvable groups

Following [174], we will see in §3.3.2 that the space of left-orders of a countable
left-orderable virtually-solvable group has no isolated point, except for the cases

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

where it is finite that are described in §2.2.1. This result requires both algebraic
and dynamical issues. As a major particular case, here we focus on finite-rank
solvable groups and their finite extensions, for which the result above will follow
from a rough classification of all actions on the line without global fixed points.
As concrete relevant examples, we will treat the cases of the Baumslag-Solitar
groups and the groups Sol at the end of this section.
Recall that a group Γ is said to be virtually finite-rank solvable if it
contains a finite-index subgroup Γ̃ that admits a normal series
en ⊳ Γ
e1 ⊳ . . . ⊳ Γ
e0 = Γ
e
{id} = Γ

ei−1 /Γ
ei is finite-rank Abelian.4 (Notice that such a group
in which every quotient Γ
P
ei−1 /Γ
ei ) is independent from
Γ is necessarily countable.) The number i rank(Γ
both the finite-index subgroup and the normal series chosen. (In particular, we
e as being normal in Γ.) We call this number the rank of
can –and we will– take Γ
Γ. We leave to the reader the task of checking that this number strictly decreases
when passing to either an infinite-index subgroup or to a quotient by an infinite
subgroup. (See [176] in case of problems.)
Exercise 3.3.1. Show that every left-order on a virtually finite-rank solvable group
admits a maximal proper convex subgroup (despite the fact that such a group can be
non-finitely-generated).
Hint. Proceed by induction on the rank, noticing that if G ⊂ H are distinct convex
subgroups, then the rank of G is strictly smaller than that of H.

The main result of this section is
Theorem 3.3.2. The space of left-orders of a virtually finite-rank solvable group
is either finite or a Cantor set.
The first step to show this result deals with left-orders induced from nonAbelian affine actions. (Compare Theorem 3.2.43.)
Proposition 3.3.3. Let Γ be a subgroup of the affine group endowed with a
left-order  induced (in a way) from its affine action on the real line. If Γ is
non-Abelian, then  is an accumulation point of its set of conjugates.
4

In case such a series can be taken so that each Γi−1 /Γi is cyclic, the group is said to be
virtually polycyclic.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

125

Proof. First notice that, as affine homeomorphisms fix at most one point, the
dynamical-lexicographic order  is completely determined by the first two comparison points, that we denote x1 , x2 . (In case of a single point, we let x2 := x1 .)
By assumption, Γ contains both nontrivial homotheties and nontrivial translations. It follows that the translations in Γ form a subgroup with dense orbits,
hence the set of points that are fixed by some nontrivial homothethy in Γ is dense
in R. Therefore, given any two distinct points in R, there is a nontrivial homothethy whose unique fixed point lies between them. As a consequence, for any
pair of comparison points y1 , y2 such that y1 6= x1 , the induced left-order ′ is
different from .
We next show that y1 , y2 may be chosen so that ′ is close to . Given a finite
set G ⊂ Γ of -positive elements, we write it as a disjoint union G = G1 ⊔ G2 ,
where G1 is the subset of elements of G lying in the stabilizer of x1 in Γ. Let I
denote the open interval with endpoints x1 and x2 . On the one hand, since G2 is
finite, there is a small neighborhood U of x1 such that f (x) > x for every x ∈ U
and every f ∈ G2 . On the other hand, for every f ∈ G1 , we have f (x) > x for
every x ∈ I. (Notice that each f ∈ G1 is an homothethy.) Thus, if we choose any
y1 in the nonempty open set I ∩ J (and y2 arbitrary), then the resulting left-order
′ is such that all elements in G are still ′ -positive. Finally, we can choose such
a y1 in the Γ-orbit of x1 , say y1 = h(x1 ). For this choice (and letting y2 := h(x2 )),
we have that ′ is the conjugate of  by h, as desired.

Corollary 3.3.4. Let (Γ, ) be a countable, left-ordered group. Suppose there is
a homomorphism Φ : Γ → Aff + (R) with -convex kernel and non-Abelian image.
Suppose further that the dynamical realization of (Γ, ) is semiconjugate to the
action given by Φ. Then  is non-isolated in LO(Γ).
Proof. By Proposition 2.2.1, it suffices to deal with the case where Φ is injective. Let ϕ denote the semiconjugacy assumed by hypothesis, and let Γ0 be the
stabilizer of ϕ(0) in Φ(Γ). This is an Abelian subgroup of Γ. We claim that it is
-convex. Indeed, if h1 ≺ g ≺ h2 , then h1 (0) ≤ g(0) ≤ h2 (0), thus ϕ(h1 (0)) ≤
ϕ(g(0)) ≤ ϕ(h2 (0)), and hence Φ(h1 )(ϕ(0)) ≤ Φ(g)(ϕ(0)) ≤ Φ(h2 )(ϕ(0)). In
particular, if h1 , h2 lie in Γ0 , then Φ(g)(ϕ(0)) = ϕ(0), that is, g also lies in Γ0 .
If Γ0 is trivial, then Proposition 3.3.3 directly applies, since in this case 
coincides with the left-order induced from ϕ(0) in the action given by Φ. If Γ0
has rank 1, then the restriction of  to Γ0 is completely determined by the sign of
any nontrivial element therein, say Φ(h) ∈ Γ0 , with h ≻ id. As Φ(h) is a nontrivial
homothethy, there exists x ∈ R such that Φ(h)(x) > x. It follows that  coincides

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

with the left-order induced from the action Φ using the comparison points x1 :=
ϕ(0) and x2 := x. Therefore, Proposition 3.3.3 still allows concluding that  is
non-isolated. Finally, the case where Γ0 has rank > 1 is slightly different, as we
cannot argue that  is completely induced from the affine action. However, by
§1.2.1, the restriction of  to Γ0 is non-isolated. Therefore, by convex extension,
 itself is non-isolated, as desired.

To proceed with the proof of Theorem 3.3.2, we need some general results on
the structure of finite-rank solvable groups. If Γ is such a group and is torsion-free,
then it contains a finite-index subgroup Γ̃ whose commutator subgroup [Γ̃, Γ̃] is
e By maximality,
nilpotent [168, 176]. Let R be a maximal nilpotent subgroup of Γ.
e hence normal in Γ; moreover, it is unique (see
R is a characteristic subgroup of Γ,
e
the exercise below). It is sometimes called the nilpotent radical of Γ.

Exercise 3.3.5. Let Γ be a group and G, H two normal nilpotent subgroups. Show
that the set GH := {gh : g ∈ G, h ∈ H} is a nilpotent subgroup.

Theorem 3.2.20 implies Theorem 3.3.2 in the case where R has finite index in
e
Γ (in particular, when the rank of Γ is 1). Hence, in what follows, we assume that
e
Γ/R
is infinite. We proceed by induction, thus we assume that Theorem 3.3.2
holds for every virtually finite-rank solvable group having smaller rank than that
of Γ. Let  be a left-order on Γ. Consider its dynamical realization, and denote
by Γ0 ⊂ R the set of elements in R having fixed points. Since R is normal in Γ,
we have that Γ0 is also normal in Γ. The following lemma implies that Γ0 has a
global fixed point. (Compare Exercise 3.2.29.)
Lemma 3.3.6. Assume that a nilpotent group with finite rank acts by orientationpreserving homeomorphisms of the real line. If every element admits fixed points,
then there is a global fixed point for the action.
Proof. If the nilpotence length of the underlying group G is 0, then the group is
trivial, and there is nothing to prove. We continue by induction on the nilpotent
length, denoting by H the center of G. This is a finite-rank Abelian group, hence
it contains a subgroup H0 isomorphic to a certain Zd such that H/H0 is a torsion
group. It follows that the (closed) set F ix := F ix(H0 ) of fixed points of H0 is
nonempty. Since H/H0 is torsion, F ix coincides
F with the set of fixed points of H.
The complement of F ix is a disjoint union i Ii of open intervals Ii . Moreover,
since H ⊳ G, we have that F ix is G-invariant. In particular, the intervals in
the complement of F ix are permuted by G. Furthermore, since every element
of G has fixed points, we have that every element in G must fix some point in

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

127

F ix. Let us now extend in a piecewise affine way the action of G on F ix to the
complementary intervals. Doing this, we obtain a new action of G on R which
factors throughout G/H. Since this action coincides with the original one on
F ix, every element of G admits fixed points. We can hence apply the induction
hypothesis, thus concluding that G/H has a global fixed point in F ix, hence G
has a global fixed point.

Now, since R is nilpotent, every left-order on it is Conradian (see the discussion before Theorem 3.2.20). Using Corollary 3.2.28 (more precisely, by Exercise
3.2.29), we thus obtain that Γ0 contains [R, R], and R/Γ0 is torsion-free. Moreover, since Γ0 is normal in Γ, we have that F ix(Γ0 ) is Γ-invariant, hence Γ0 admits
a Z-indexed sequence of global fixed points going from −∞ to ∞.
We divide the induction argument into two separated cases.
e on
Case I. Either R/Γ0 is trivial, or it has rank 1 and the conjugacy action of Γ
it is by multiplication by ±1.
In this case, we have

e 0 is Abelian.
Claim (i). The quotient Γ/Γ

e Γ]
e ⊆ R. Otherwise,
Indeed, if R/Γ0 is trivial, then this follows from that [Γ,
e does not centralize Γ̃/Γ0 . As Γ̃ centralizes Γ̃/R,
assume for a contradiction that Γ
this means that Γ̃ does not centralize R/Γ0 . Hence, there are f ∈ R \ Γ0 and
e such that, modulo Γ0 , one has the equality gf g −1 = f −1 . Now, since f
g ∈Γ
acts without fixed points, changing f by f −1 if necessary, we can assume that
f (y) > y for every y ∈ R. Thus, if we let x be in the set of fixed points of Γ0 , we
have that gf g −1(x) = f −1 (x) < x, which implies that f g −1(x) < g −1(x), contrary
to our assumption on f .
Let I be the smallest closed interval containing the origin whose endpoints
are fixed by Γ0 , and let H be its stabilizer in Γ. Since Γ0 is normal in Γ, for every
g ∈ Γ, either g(I) equals I or it is disjoint from it. By Proposition 2.1.3, this
implies that H is a convex subgroup.
e
Claim (ii). The subgroup H has smaller rank than Γ.

e cannot be equal to Γ,
e since the latter does
Indeed, on the one hand, H ∩ Γ
e
not have global fixed points. On the other hand, since Γ0 is contained in H ∩ Γ,
e
e
e
e
Claim (i) above implies that H ∩ Γ is a normal subgroup of Γ and that Γ/(H ∩ Γ)
e
e is left-orderable, it has rank
is Abelian. Therefore, as the quotient Γ/(H
∩ Γ)
> 0, thus showing the claim.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

It follows by induction that the space of left-orders of H is either finite or a
Cantor set. Hence, by Proposition 2.2.1, if  is isolated in LO(Γ), then H is a
Tararin group. However, if H is a Tararin group, then every left-order on H is
e in Γ
e and the fact that
Conradian (c.f. Lemma 2.2.16). By the convexity of H ∩ Γ
e
e is Abelian, we have that the restriction of  to Γ
e is Conradian (c.f.
Γ/(H
∩ Γ)
Exercise 3.2.5). Therefore, by Theorem 3.2.9, we have that  is a Conradian
order of Γ. As a consequence, using Proposition 3.2.53 we conclude that, if  is
isolated in LO(Γ), then Γ must be a Tararin group, as desired.
e which
Case II. Either rank(R/Γ0 ) ≥ 2, or rank(R/Γ0 ) = 1 and there exists g ∈ Γ
does not act on R/Γ0 by multiplication by ±1. (In particular, R/Γ0 cannot be
isomorphic to Z.)
In this case, Proposition 3.5.16 and Remark 3.5.17 provide us with an Rinvariant Radon5 measure ν, associated to which there is a translation number
homomorphism τν : R → (R, +) defined by τν (g) := ν([x, g(x))) (here and
below, we use the convention ν([x, y]) := −ν([y, x]) for y < x). Notice that this
definition does not depend on the choice of the point x.
Exercise 3.3.7. Show that τν coincides (up to a positive multiple) with the Conrad
homomorphism on R associated to the convex jump with respect to the maximal proper
convex subgroup (c.f. §3.2.3)
Exercise 3.3.8. Show that if G is a subgroup of Homeo+ (R) with no global fixed
point and whose action preserves a Radon measure for which the translation-number
homomorphism has image non isomorphic to Z, then its action is semiconjugate to an
action by translations.
Exercise 3.3.9. Let G be a subgroup of Homeo+ (R) preserving a Radon measure ν.
(i) Show that the kernel of τν coincides with the subset G0 made of the elements having
fixed points. Moreover, show that for all x in the support supp(ν), its stabilizer in G
coincides with G0 .
(ii) Conclude that if every element in G has fixed points, then there is a global fixed
point for the action.

The next proposition (which is interesting in its own right) tell us that, up to
multiplication by a positive constant, ν is the unique R-invariant Radon measure.
It is somewhat a dynamical counterpart of Exercise 3.3.7 above.
5

Recall that a Radon measure is a measure giving finite mass to compact sets.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

129

Proposition 3.3.10. Let G be a subgroup of Homeo+ (R) preserving a Radon
measure ν. Then, for any other (nontrivial) G-invariant Radon measure ν ′ , there
is a positive real number κ such that κτν = τν ′ . Moreover, if τν (G) is dense in
(R, +), then κν = ν ′ .
Proof. It easily follows from Exercise 3.3.9 that τν (G) and τν ′ (G) are simultaneously either discrete or dense in R. In the former case, the claim of the proposition
is obvious. Below we deal with the latter case.
Fix g ∈
/ G0 and a point x that is fixed by G0 . Then, as a combination of
Exercises 3.1.5 and 3.3.8, we have that for all f ∈ G,
nq
o
q
p
q+1
τν (f ) = τν (g) lim
: g (x) ≤ f (x) < g (x) ,
p→∞ p
and the same holds changing ν by ν ′ . Therefore, we have τν ′ (g)τν (f ) = τν ′ (f )τν (g)
for every f ∈ G, hence τν ′ equals κτν for a certain positive κ.
Next, we claim that the supports of ν and ν ′ coincide. Indeed, the density of
τν ′ (G) implies that ν ′ has no atoms and the action of G on supp(ν ′ ) is minimal
(i.e. every orbit is dense). It follows that if there is a point x ∈ supp(ν ′ )\supp(ν),
then there exists g ∈ G such that g(x) > x and ν([x, g(x))) = 0, contradicting
the fact that ker(τν ′ ) = ker(τν ). Therefore, supp(ν ′ ) ⊂ supp(ν), and the reverse
inclusion is proven analogously.
Finally, let x < y be two points in the (common) supports of ν and ν ′ , and
let gn ∈ G be such that gn (x) converges to y. Then



ν ′ [x, y] = lim ν ′ [x, gn (x)] = lim τν ′ (gn ) = lim κτν (gn ) = κν [x, y] ,
n→∞

which finishes the proof.

n→∞

n→∞



Using the R-invariant Radon measure, we can describe the action of Γ up to
semiconjugacy. More precisely, we have
Claim (i). There is a homomorphism Φ : Γ → Aff + (R) such that Φ(R) contains
nontrivial translations, and Γ0 coincides with ker(Φ)∩R. Moreover, the dynamical
realization of (Γ, ) is (continuously) semiconjugate to this affine action.
Indeed, let us continue denoting by ν an R-invariant Radon measure. Since
R/Γ0 is not isomorphic to Z, Proposition 3.3.10 (and its proof) implies that ν is
unique up to a scalar multiple, and that Γ0 is the kernel of the translation-number
homomorphism τν . As R is normal in Γ, this implies that for each g ∈ Γ, the
measure g∗ (ν) is also R-invariant. Thus, for every g ∈ Γ, there is λg > 0 such

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

that g∗ (ν) = λg ν. This yields a group homomorphism λ : Γ → R∗ into the group
of positive reals (with multiplication). We then define Φ : Γ → Aff + (R) by
Φ(g)(x) :=

1
x + ν([0, g(0)]).
λg

One can easily check that this is a homomorphism that extends τν .
To show that the dynamical realization of  is semiconjugate to this affine
action, for each x ∈ R we let ϕ(x) := ν([0, x]). Then ϕ is a continuous, nondecreasing surjective map, and a direct computation shows that for all g ∈ Γ and
every x ∈ R,
ϕ(g(x)) = Φ(g)(ϕ(x)),
which shows the desired semiconjugacy.
Next, we let Iν := (a, b), where a := sup{x < 0 : x ∈ supp(ν)} and b :=
inf{x > 0 : x ∈ supp(ν)}. We also let Γν be the stabilizer in Γ of Iν . The
subgroup Γν is easily seen to be convex. Moreover, Γν ∩ R = Γ0 .
Notice that the rank of Γν is smaller than that of Γ. Thus, by the induction
hypothesis, if Γν admits infinitely many left-orders, then no left-order on it is
isolated. By convex extension, we conclude that  is non-isolated in LO(Γ). For
the other case, we have the next
Claim (ii). If Γν is a Tararin group, then ker(Φ) is convex.
Indeed, since Φ(Γν ) does not contain any nontrivial translation, it can only
contain homotheties centered at 0; in particular, it is Abelian. If it is trivial,
then ker(Φ) = Γν , so it is convex, as desired. Assume Φ(Γν ) is nontrivial, and
let {id} = Γn ⊳ Γn−1 ⊳ . . . ⊳ Γ0 = Γν be the series of all convex subgroups of the
Tararin group Γν . (Recall that Γi−1 /Γi has rank 1 and that the action of Γi−1 on
Γi−1 /Γi is by multiplication by some negative number.) By Exercise 2.2.21, Γν
has a unique torsion-free Abelian quotient, namely Γν /Γ1 . As this must coincide
with Φ(Γ), we conclude that ker(Φ) equals Γ1 , hence it is convex.
Knowing that ker(Φ) is convex, we can proceed to show that  is non-isolated.
Indeed, either Γ/ ker(Φ) is Abelian of rank at least 2, or it is a non-Abelian
subgroup of the affine group. In the former case, it has no isolated left-orders
(see §1.2.1), hence -by convex extension- the left-order  is non-isolated in LO(Γ).
In the latter case, we are under the hypothesis of Proposition 3.3.3, which yields
the same conclusion. This finishes the proof of Theorem 3.3.2.
Left-orders on Baumslag-Solitar’s groups. Perhaps the simplest examples

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

131

of finite-rank solvable groups that are non virtually-nilpotent are the BaumslagSolitar groups BS(1, ℓ) := hh, g : hgh−1 = g ℓ i, where ℓ > 1. We have seen that
BS(1, ℓ) admits only four Conradian orders (c.f. Example 3.2.55), yet it also
admits the left-orders induced from the affine faithful actions on the line (c.f.
§1.2.2). Below we follow the lines of the previous proof to show that, actually,
these are the only possible left-orders on B(1, ℓ).
As in §1.2.2, we can see
B(1, ℓ) as a semidirect product Z[ 1ℓ ] ⋊ Z, where the

Z-factor acts on Z[ 1ℓ ] := ℓkm : k, m in Z by multiplication by ℓ. In this way, it
easily follows that the nilpotent radical of B(1, ℓ) is R := Z[ 1ℓ ].
Now, given a left-order on BS(1, ℓ), we consider its dynamical realization.
Since R = Z[ 1ℓ ] has rank one, two cases may arise.
Case I. There is a global fixed point for R.
As R is normal in BS(1, ℓ), the set of R-fixed points is BS(1, ℓ)-invariant;
thus, in this case, it is unbounded in both directions. In particular, the convex
subgroup H stabilizing the interval that contains the origin and is enclosed by
two consecutive R-fixed points must coincide with R. As a consequence,  is
Conradian, and therefore  is non-isolated, since BS(1, ℓ) is not a Tararin group.
Case II. Every nontrivial element of R is fixed-point free.
In this case, the action of R is continuously semiconjugate to that of a dense
group of translations, thus it preserves a Radon measure ν without atoms that
is unique up to a scalar factor. Moreover, h does not preserve ν, otherwise we
would have τν (g ℓ−1 ) = τν (g −1hgh−1 ) = 0, contradicting the fact that g ℓ−1 acts
freely. Thus, the dynamical realization of  is semiconjugate to (the action given
by) a faithful embedding of BS(1, ℓ) into Aff + (R), and the left-order  coincides
with a left-order induced from this affine action. The fact that  is non-isolated
in this case follows from Corollary 3.3.4.
Notice that the affine-like orders of Case II approximate the Conradian ones
of Case I just by letting the first comparison point tending to either −∞ or ∞.
It is worth pointing out that the description above –as well as its proof– applies
not only to dynamical realizations of left-orders, but also to general (faithful)
actions on the line with no global fixed point. Such an action is hence either
without crossings (with R being the subgroup of elements having fixed points) or
semiconjugate to an affine action. (See [173] for more details on this.)
Left-orders on Sol groups. Relevant examples of finite-rank solvable groups
that are non virtually-polycyclic are those of the form Sol := Z2 ⋊A Z, where A is
an hyperbolic automorphism of Z2 (i.e. given by a matrix in SL(2, Z) with trace

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

greater than 2, so that it has two irrational eigenvalues). Below we follow the
lines of the previous proof in this particular case to get an accurate description
of the space of left-orders and its subspaces of bi-invariant and Conradian orders.
Actually, the methods employed yield a complete description of all faithful actions
on the line with no global fixed point.
We denote by R the commutator subgroup of Sol –which coincides with the
2
Z -factor–, and we let f be the element of Z acting on R as A. The subgroup R
is easily seen to coincide with the nilpotent radical of Sol.
Given a left-order  on Sol, let us consider its dynamical realization. Since AT
is Q-irreducible and R is Abelian and finitely generated, the next three properties
are equivalent:
– There is an element in R having a fixed point;
– Every element of R has a fixed point;
– There is a global fixed point for R.
Indeed, having a fixed point for g ∈ R is equivalent to that τν (g) = 0 for an Rinvariant Radon measure ν (see Exercise 3.3.9), and A also acts at the level of
translation numbers, as it is shown in the next
Exercise 3.3.11. Let g1 , g2 be the canonical basis of R = Z2 . Show that




τν (f g1 f −1 )
τν (g1 )
T
=A
.
τν (f g2 f −1 )
τν (g2 )

Thus, the two cases considered in the proof of Theorem 3.3.2 fit with those
considered below.
Case I. The subgroup R has a global fixed point.
Since Sol acts without global fixed points and R is normal in Sol, in this case
the set of R-fixed points is unbounded in both directions (and Γ-invariant). As
for BS(1, ℓ), this implies that  is Conradian. To see that  is non-isolated,
one may argue by convex extension by noticing that R is convex and rank-two
Abelian. Alternatively, Sol is not a Tararin group...
Case II. There is no global fixed point for R.
In this case, R is semiconjugate to a dense group of translations, thus it
preserves a Radon measure without atoms ν that is unique up to a scalar factor.
As f is hyperbolic, it cannot preserve ν: it acts as an homothethy with ratio one
of the eigenvalues of AT . Thus, the dynamical realization of  is semiconjugate
to (the action given by) a faithful embedding of Γ into Aff + (R). The fact that 
is non-isolated in this case follows from Corollary 3.3.4.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

133

Figure 15: Depicting the space of left-orders of Sol.
It follows from the previous analysis that, as it was the case for the BaumslagSolitar groups, there are two types of left-orders on Sol:
Case I. Conradian orders.
These correspond to those left-orders for which the normal subgroup R = Z2
is convex. Therefore, CO(Sol) is made of two copies of the Cantor set LO(Z2 ),
each of which corresponds to a choice of sign for f . (In Figure 15, these are
represented as two “vertical dashed circles”.) Observe that among all bi-orders
on R, those that are invariant under conjugacy by f are those that that correspond
(under Conrad’s homomorphisms) to eigendirections of the matrix AT . Since the
corresponding left-orders on Sol are the bi-invariant ones, we conclude that Sol
supports exactly eight bi-orders.6
Case II. Left-orders coming from affine actions.
These form an open set –which is locally a Cantor set– that complements the
subspace of Conradian orders. These affine-like orders can be described as in
§1.2.2. Notice, however, that Sol admits four embeddings into Aff + (R). These
left-orders hence appear as four “horizontal dotted lines” in Figure 15. These
“lines” accumulate at the eight bi-invariant orders in CO(Sol), in a similar way
to the approximation of the four bi-orders of Baumslag-Solitar’s groups by affinelike orders previously described.
Based on all of this, it is not difficult to describe the dynamics of the conjugacy
action of Sol on its space of left-orders. We leave this as an exercise to the reader.

3.3.2

The general case

In this section, we prove
6

A classification of finitely-generated solvable groups admitting only finitely many bi-orders
can be found in [17, 121].

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Theorem 3.3.12. The space of left-orders of a countable virtually-solvable group
is either finite or homeomorphic to the Cantor set.
In the preceding section, we proved this theorem for virtually finite-rank solvable groups. Our main tool in that case was the fact that these groups are virtually nilpotent-by-Abelian, with finite-rank nilpotent part. As such a nilpotent
group preserves a Radon measure when acting by orientation-preserving homeomorphisms of the real line, we thus conclude that actions of virtually finite-rank
solvable groups on the line quasi-preserve a nontrivial Radon measure7 . Thus, the
action is semiconjugate to an affine action provided the measure is non-atomic.
This dynamical picture no longer arises for general solvable groups (not even for
metabelian groups!), as the next classical example (due to Plante [165]) shows.
L
Example
3.3.13.
The
wreath
product
Z
≀
Z
:=
Z Z ⋊ Z is a metabelian group having
L
H := Z Z as its maximal nilpotent subgroup. We next describe an action of Z ≀ Z
on the real line with the property that for every shift-invariant subgroup H, no global
fixed point arises, although every element therein admits fixed points. This implies in
particular that there is no quasi-invariant measure for Z ≀ Z. Indeed, such a measure
would be invariant by the derived subgroup [Z ≀ Z , Z ≀ Z], which is shift-invariant. In
view of the announced properties, this is in contradiction with Exercise 3.3.9.
For the construction, let f denote the homothethy x 7→ 2x. Let I0 := [−1, 1],
and for i ∈ Z, denote Ii := f i (I0 ). Let g : I0 → I0 be a homeomorphism such that
g(−1/2) = 1/2 and g(x) > x for all x ∈ (−1, 1). We define gi : Ii → Ii by gi := f i gf −i .
Notice that this is equivalent to that f −1 gi (x) = gi−1 f −1 (x) holds for all x ∈ Ii , that
is, gi f (y) = f gi−1 (y) for all y ∈ Ii−1 . Below, we extend the definition of each gi to the
whole line in such a way that f and g0 generate a group isomorphic to Z ≀ Z.
One easily convinces that there is a unique way to extend the maps gi into commuting homeomorphisms of the real line. For instance, to ensure commutativity, we must
necessarily have gi−1 (x) := gim gi−1 gi−m (x) for x ∈ gim (Ii−1 ). The (proof of the uniqueness of the) extension can then be easily achieved inductively. We continue denoting
by gi the resulting homeomorphisms. We claim that f gi f −1 = gi+1 holds. Indeed, this
follows from the definition for x ∈ Ii+1 . Assume inductively that f gi f −1 (x) = gi+1 (x)
holds for all x ∈ Ik for a certain k ≥ i + 1, and let x ∈ Ik+1 . Letting m ∈ Z be such
m (y) for a certain y ∈ I , we have
that x = gk+1
k
m
f gi f −1 (x) = f gi f −1 (gk+1
(y)) = f gi gkm f −1 (y)
m
m
= f gkm gi f −1 (y) = gk+1
f gi f −1 (y) = gk+1
gi+1 (y) = gi+1 (x),
7

In our context, a Radon measure ν is quasi-preserved by a group Γ if for every g ∈ Γ, there
is a positive real number λg such that g∗ (ν) = λg ν, where by definition g∗ (ν)(X) := ν(g −1 (X))
for every measurable set X.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

135

where the second and fourth equalities follow from the definition of gk , the third from
the commutativity between gi and gk , and the fifth from the induction hypothesis.

Figure 16: Plante’s action of Z ≀ Z.

Example 3.3.14. There exists a left-order on Z ≀ Z whose dynamical realization is
semiconjugate to the action constructed in the preceding example. Indeed, using the
exact sequence
0 −→ H −→ Z ≀ Z −→ Z −→ 0,
we can produce the bi-order ′ on Z ≀ Z as the convex extension of the lexicographic
order on H by one of the two possible orders on the cyclic factor Z = hf i. Notice that
given any element h ∈ H, we can find two elements h± in H such that h− ≺′ hn ≺′ h+
holds for all n ∈ Z. Moreover, H is the maximal proper ′ -convex subgroup. Using
Lemma 2.1.13, we can hence produce a left-order  on Z ≀ Z that has f as its smallest
positive element and that coincides with ′ on H. It is not very hard to see that the
dynamical realization of  is semiconjugate to the action constructed in Example 3.3.13
above.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

To deal with this new phenomenon, we will use the machinery developed in
§3.2. The price to pay is that, unlike §3.3.1, here we are not able to give a
classification –up to semiconjugacy– of all actions on the real line of the involved
groups. Rather, we give a rough local description of the dynamics that allows
concluding Theorem 3.3.12. We start with an exercise that follows from an easy
reformulation of part of the proof of Propositions 2.2.17 and 3.2.53.
Exercise 3.3.15. Show that every left-order on a group admitting infinitely many
convex subgroups is non-isolated in the corresponding space of left-orders.

Due to the preceding exercise, in order to prove Theorem 3.3.12, it suffices to
consider left-orders with finitely many convex subgroups. Let  be such an order
on a group Γ, with
{id} = Cn ( Cn−1 ⊂ . . . ( C0 = Γ
being the family of convex subgroups. One of these subgroups must coincide with
the Conradian soul G := C (Γ), that is, with the maximal -convex subgroup
restricted to which  is Conradian (c.f. §3.2.4). If G is not a Tararin group,
then by Proposition 3.2.53, the restriction of  to G is non-isolated in LO(G);
by convex extension,  is not isolated in LO(Γ). Hence, in all what follows, we
assume that G is a Tararin group.
If G = Γ, then we are done: Γ admits only finitely many left-orders. If
G is trivial, then  is non-isolated in LO(Γ), due to Theorem 3.2.43. We hence
suppose that G is a nontrivial, proper subgroup of Γ, say G = Cℓ , with n > ℓ > 0.
We will show that the restriction of  to Cℓ−1 is non-isolated; by convex extension,
this in turns implies that  is non-isolated in LO(Γ), as desired. As the claim to
be shown only involves Cℓ−1 , to simplify will denote this group as Γ; equivalently,
we will assume that ℓ = 1, that is, there is no convex subgroup strictly between
G = C1 and Γ.
We consider the dynamical realization of . Since G is a proper convex
subgroup, it has at least one fixed point on each side of the origin. We let IG be
the smallest open interval fixed by G that contains the origin. By Proposition
2.1.3, the convexity of G immediately implies
Lemma 3.3.16. Every element of Γ either fixes IG or moves it into a disjoint
interval. In particular, the stabilizer of IG coincides with G.
e be a finite-index,
From now on, we assume Γ to be virtually solvable. Let Γ
e0 := Γ
e and Γ
ej := [Γ
ej−1 , Γ
e j−1] be the
normal, solvable subgroup of Γ. We let Γ

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

137

associated derived series:
ek ⊳ Γ
e k−1 ⊳ . . . ⊳ Γ
e1 ⊳ Γ
e0 = Γ
e ⊳ Γ.
{id} = Γ

ej is normal in Γ. We let i be the minimal index such that Γ
ei is
Notice that each Γ
contained in G. Since G is a nontrivial, proper, convex subgroup, we have that
k > i ≥ 1; see Figure 17 below.
Γ
•............

..... ............................
...........
.....
i−1
.....
.
.
.
.
.
...
..
...
.....
.
.
.
.
...
.
.
.
.........
...
....
....
...
....
...
.
....
.
....
...
....
...
....
.... .....
.... ..
.......
...
i−1
...
..
...
...
...
i
...
...
..

G•

•Γ
e

• G∩Γ
e

• Γ
e
•
ek
{id} = Γ

e i and Γ
ei−1 .
Figure 17: The groups G, Γ

ei−1 will be of most importance in our analysis: although it is
The subgroup Γ
ei−1 will be shown to be Conradian.
not always nilpotent, the restriction of  to Γ
Thus, dynamically, it will play the role played by the nilpotent radical in the
finite-rank case. We like to think it as a kind of Conradian sckeleton of (Γ, ).
ei−1 is Conradian.
Lemma 3.3.17. The order  restricted to Γ

ei = [Γ
ei−1 , Γ
e i−1 ] is contained in G. Therefore,
Proof. By definition, the subgroup Γ
ei−1 ∩ G is normal in Γ
ei−1 , as well as convex therein. Moreover, as the quotient
Γ
i−1
i−1
e /Γ
e ∩ G is Abelian, it only admits Conradian orders. Since  restricted to
Γ
i−1
e ∩ G is Conradian, this implies that  restricted to Γ
ei−1 is a convex extension
Γ
of a Conradian order by a Conradian one, hence Conradian (c.f. Exercice 3.2.5).
ei−1 has no global fixed point.
Lemma 3.3.18. The action of Γ

Proof. Let I be the smallest open interval containing the origin that is fixed by
ei−1 . Since Γ
e i−1 is normal in Γ, the interval I is either fixed or moved disjointly
Γ
by each g ∈ Γ. In particular, the stabilizer StabΓ (I) of I is a convex subgroup

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

of Γ (c.f. Proposition 2.1.3). Now, if I was not the whole line, then the maxiei−1 ⊂ G, which is a
mality of G would imply that StabΓ (I) ⊆ G, thus yielding Γ
contradiction.


ei−1 is Conradian, its action on the real line has no crossSince  restricted to Γ
ei−1 having fixed
ings (c.f. Exercise 3.2.33). It follows that the set of elements in Γ
e i−1 (actually, of Γ); see e.g. Proposition 3.2.26.
points is a normal subgroup of Γ
e i−1 does not act freely, then the set of fixed points of g
In particular, if g ∈ Γ
accumulates at both −∞ and +∞. Thus, in order to prove Theorem 3.3.12, we
need to analyze two cases.
ei−1 contains elements without fixed points. (Such a case
Case I. The subgroup Γ
ei−1 has finite rank.)
arises for instance if Γ

ei−1 preserves a nontrivial Radon measure
We first observe that in this case Γ
i−1
e
ν. Indeed, since the order on Γ is Conradian, its action on the real line has no
e i−1 acting freely, Proposition 3.2.26 easily
crossings. Further, since there is g0 ∈ Γ
i−1
e
ei−1 :
implies that Γ
has a maximal proper convex subgroup, namely {g ∈ Γ
g has a fixed point}. It then follows from Proposition 3.5.16 and Remark 3.5.17
ei−1 preserves a non-trivial Radon measure on the line.
that Γ
ei−1 in Γ and Proposition 3.3.10, we have that
Now, from the normality of Γ
there is a homomorphism λ : g 7→ λg from Γ into R∗ satisfying τg∗ ν = λg τν . The
next lemma comes from the work of Plante [165].

Lemma 3.3.19. If the homomorphism λ is trivial, then Γ preserves a Radon
measure on the real line. Otherwise, Γ quasi-preserves a Radon measure which is
ei−1 -invariant.
Γ

ei−1 ) is dense, then ν is quasiProof. Recall that by Proposition 3.3.10, if τν (Γ
invariant. This occurs for instance if λ is nontrivial. Indeed, choosing g ∈ Γ such
ei−1 ,
that λg < 1, we have for every f ∈ Γ


τν g −1 f g = ν [g −1(x), g −1 f (x)) = g∗ ν([x, f (x))) = τg∗ ν (f ) = λg τν (f ). (3.6)

ei−1 ) ≃ Z is not dense. In particular, we assume that
Thus, we assume that τν (Γ
ei−1 ) ≃ Z.
λ is trivial and that τν (Γ
e i−1 | τν (g) = 0}. We have seen that H is made
We let H := ker(τν ) = {g ∈ Γ
ei−1 having fixed points (c.f. Exercise 3.3.9). Therefore, H
of by the elements in Γ
ei−1 , but also in Γ. Moreover, the condition τν (Γ
ei−1 ) ≃ Z
is normal not only in Γ
ei−1 /H ≃ Z. We claim that Γ
ei−1 /H is in the center of Γ/H.
translates into that Γ

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

139

ei−1 be a generator of Γ
e i−1 /H, for each g ∈ Γ we have that
Indeed, letting f ∈ Γ
g −1 f g = f n h holds for certain h ∈ H and n ∈ Z. We need to show that n = 1.
Now, by (3.6), we have
n τν (f ) = τν (g −1 f g) = λg τν (f ) = τν (f ) 6= 0,
which implies that n = 1, as desired.
ei−1 ≃ (Γ/H)/(Γ
ei−1/H) acts on the compact
Finally, the quotient group Γ/Γ
ei−1 and all x ∈ F ix(H).
quotient F ix(H)/ ∼, where x ∼ f (x) for each f ∈ Γ
This space is easily seen to be homeomorphic to the circle. Therefore, as Γ is
amenable, it preserves a probability measure on it (see §4.1). Pulling back this
measure to the real line, we obtain a Γ-invariant Radon measure on R.

We now claim that the dynamical realization of  is semiconjugate to a nonAbelian affine action. As in the preceding section, this will follow once we show
that the homomorphism λ is nontrivial. Assume for a contradiction that λ is
trivial. Then by the preceding Lemma, there is a Γ-invariant Radon measure
ν. Moreover, as the origin is moved by every nontrivial element and Γ contains
elements having fixed points (for instance, those of G), the origin does not belong
to the support of ν. Let Iν be the connected component of the complement of
the support of ν containing the origin. The interval Iν is either fixed or moved
disjointly by each element of Γ, hence its stabilizer StabΓ (Iν ) is a convex subgroup
of Γ. Since this subgroup contains G and since G is the maximal proper convex
subgroup of Γ, we must have StabΓ (Iν ) = G. Further, StabΓ (Iν ) coincides with
the kernel of the translation-number homomorphism τν : Γ → (R, +), thus it is
normal in Γ. We thus conclude that G is normal and co-Abelian in Γ. Therefore,
 is a convex extension of a Conradian order by a Conradian one, hence it is
Conradian (c.f. Exercise 3.2.5). However, this contradicts the fact that G is the
Conradian soul of Γ.
We can finally show that  is non-isolated by invoking Corollary 3.3.4. Indeed,
it easily follows from the construction of the dynamical realization that the kernel
of the induced homomorphism from Γ into Aff + (R) is a -convex subgroup, hence
the hypothesis of the corollary are fulfilled.
ei−1 admits fixed points on both sides of the origin.
Case II. Every element of Γ

In this case, we will prove that the approximation scheme by conjugates developed in §3.2.5 applies. More precisely, starting from the dynamical realization
of , we will induce a new left-order using a comparison point that is outside
but very close to IG . The main issue here is to ensure that this procedure can

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

be made in such a way that the order restricted to G remains untouched (compare Theorem 3.2.46). Along the proof, it will actually arise that the action is
somewhat similar to the one described in Example 3.3.13.
ei−1 , let us denote by Ig the connected comFor each nontrivial element g ∈ Γ
ponent of the complement of its set of fixed points that contains the origin. It
follows from Lemma 3.3.18 that the union of all the Ig ’s is the whole real line.
ei−1 , one of the following possibilities
Lemma 3.3.20. For each f ∈ Γ and g ∈ Γ
occurs:
– f (Ig ) = Ig ;
– f (Ig ) is disjoint from Ig ;
– Up to changing f by its inverse if necessary, we have Ig ⊂ f (Ig ).
e i−1 is Conradian, hence
Proof. By Lemma 3.3.17, the order  restricted to Γ
ei−1 acts without crossings (c.f. Exercise 3.2.33). As Γ
e i−1 is normal in Γ, the
Γ
lemma easily follows.

The next two lemmas are similar to the preceding one. The first follows from
the convexity of G, and the second from the nonexistence of crossings for the
ei−1 and the fact that Γ
e i−1 is normal in Γ. Details are left to the reader.
action of Γ

Lemma 3.3.21. In the preceding lemma, if g does not belong to G, then the
ei−1 \ G
second possibility cannot occur for f ∈ G. In other words, for all g ∈ Γ
−1
and each f ∈ G, either f fixes Ig or (up to changing f by f if necessary), we
have Ig ⊂ f (Ig ).
ei−1 \G.
Lemma 3.3.22. Let I be the intersection of all the intervals Ig , with g ∈ Γ
Then each element f ∈ Γ either moves I into a disjoint interval, or up to replacing
it by its inverse, we have I ⊂ f (I).
The next lemma is a kind of refined version of Theorem 3.2.38 knowing that
ei−1 is
G has finite rank and/or admits only finitely many left-orders, and that Γ
normal and -Conradian.

ei−1 \ G coincides
Lemma 3.3.23. The intersection of all the intervals Ig for g ∈ Γ
with IG .

e i−1 is a -Conradian subgroup (c.f. Lemma 3.3.17), its action has
Proof. Since Γ
ei−1 \ {id},
no crossings, which implies that the family of intervals Ig , with g ∈ Γ

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

141

is totally ordered by inclusion (c.f. Exercise 3.2.25). Moreover, as G is convex,
ei−1 \ G we have that Ig strictly contains IG . Therefore, letting I
for each g ∈ Γ
e i−1 \ G, we have that I is a bounded
be the intersection of all the Ig ’s for g ∈ Γ
interval containing IG .
Assume that no element f ∈ Γ is such that I strictly contains f (I). Then, by
Corollary 3.3.22, the stabilizer of I in Γ is a proper convex subgroup of Γ; moreover, this subgroup contains G. As G is the maximal proper convex subgroup,
this necessarily implies that IG equals I.
Therefore, we are left to prove that no f ∈ Γ satisfies f (I) ( I. Assume
T
e i−1 \ G
otherwise for a certain f . As I = g∈Γei−1 \G Ig , there must exist g ∈ Γ
ei−1 , by the
such that I ∩ f (Ig ) is strictly contained in I. Since f gf −1 belongs to Γ
−1
definition of I we must have that f gf belongs to G. Actually, the same holds
for f n gf −n , for all n ≥ 1.
Notice that f n (Ig ) is an open interval (not necessarily containing the origin)
that is fixed by f n gf −n and has no fixed point inside. Moreover, as f (Ig ) ( Ig ,
we have f n+1 (Ig ) ( f n (Ig ), for all n ≥ 1. Together with the fact that the action
of G has no crossings, this easily implies that f k gf −k belongs to StabG (f n (Ig )),
for all k ≥ n. As a consequence, StabG (f n (Ig )) is an strictly decreasing sequence
of convex subgroups of G for any left-order induced from a sequence starting with
a point in the (nonempty) intersection of the compact intervals f n (Ig ). However,
this contradicts the fact that G is a Tararin group (as well as that G has finite
rank).

The next lemma follows almost directly from Proposition 3.2.48 and its proof.
Actually, it is a kind of restatement of it for dynamical realizations. We leave the
details to the reader.
Lemma 3.3.24. For each x not belonging to IG , every left-order ′ induced from
the dynamical realization of  induced from a sequence starting with x is different
from . Moreover, as x ∈
/ IG converges to any of the endpoints of IG , we have
′
that  converges to  outside G.
For the sake of concreteness, the order ′ above may –and will– be taken as
the one for which the second comparison point is the origin (so that no other
comparison point is necessary). The convergence in the statement means that for
any sequence xn converging to an endpoint of IG from outside, given g ∈ Γ \ G
we have that g ≻ id holds if and only if g ≻′n id holds for all large-enough n
(where ′n denotes the order associated to the point xn and the origin, as before).

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Therefore, to prove that  is non-isolated, we are left to showing that for a wellchosen sequence (xn ) as above, the left-orders ′n coincide with  when restricted
to G for large-enough n. This is achieved by the next lemma, which closes the
proof of Theorem 3.3.12.
Lemma 3.3.25. There exists a sequence of points (xn ) converging to an endpoint
of IG from outside such that the induced left-orders ′n coincide with  on G for
all n.
To show this, we need one more general lemma for Tararin groups.
Lemma 3.3.26. Let T be a Tararin group, with chain of convex subgroups
{id} = T k ⊳ T k−1 ⊳ . . . ⊳ T 1 ⊳ T 0 = T,
and let fT be an element in T \ T 1 acting on T 1 /T 2 as the multiplication by a
negative (rational) number. Suppose T is acting by orientation-preserving homeomorphisms of the line in such a way that the sets of fixed points of nontrivial
elements have empty interior (as it is the case for dynamical realizations). Then,
for every left-order  on T and each y ∈ R not fixed by fT , there exists a point
x between fT−2 (y) and fT2 (y) such that  coincides on T 1 with any left-order ′
induced from a sequence starting with x.
Proof. As the sets of fixed points of nontrivial elements have empty interior,
and since T is countable, there is a point z between fT−1 (y) and y whose orbit
under T is free. Any such point induces –in a dynamical-lexicograhic way– a
left-order ∗ on T . Since T 1 is necessarily convex for this order, there is an open
interval I containing z that is fixed by T 1 and contains no other fixed point of T 1
inside. Moreover, I is moved into a disjoint interval by any nontrivial power of
fT ; in particular, it contains at most one point of the orbit of y under hfT i. As a
consequence, I strictly lies between fT−2 (y) and fT (y). Now, by Proposition 2.2.23,
there exists an element g in either T 1 or fT T 1 such that  and ∗g coincide on
T 1 . As ∗g is the dynamical-lexicographic order with comparison point x := g(z),
this point satisfies the conclusion of the lemma.

e i−1 acts without
Proof of Lemma 3.3.25. Due to Lemma 3.3.23 (and since Γ
ei−1 \G such that Ign converges
crossings), there exists a sequence of elements gn ∈ Γ
to IG . As in the preceding lemma, let fG be an element in G\G1 acting on G1 /G2
as the multiplication by a negative number. According to Lemma 3.3.21, we may
pass to a subsequence for which one of the two possibilities below occur.

3.3. AN APPLICATION: ORDERING SOLVABLE GROUPS

143

Subcase (i). Each interval Ign is fixed by fG .
We first claim that G must fix each interval Ign . Indeed, letting y be an
endpoint of any of the Ign ’s, we may induce a left-order on G from a sequence
having y as its initial point. For such an order, the stabilizer of y in G is a convex
subgroup of G containing fG . It must hence coincide with G, and therefore G
fixes y, as desired.
We may now let xn be any of the endpoints of Ign , say the right one. As G
fixes xn and the second comparison point of ′n is the origin, the restriction of
′n to G coincides with that of . Moreover, since Ign converges to IG , the points
xn converge (from outside) to the right endpoint of IG .
Subcase (ii). For each n, we have either Ign ⊂ fG (Ign ) or Ign ⊂ fG−1 (Ign ).

Up to taking a subsequence, we may assume that for all n, we have Ign ( Ign−1
and, changing fG by its inverse if necessary, Ign ⊂ fG (Ign ). In case where fG ≺ id
(resp. fG ≻ id), let yn be the left (resp. right) endpoint of Ign . Notice that yn
converges to a fixed point of fG (hence of G). Moreover, for all n ≥ 1, if fG ≺ id
(resp. fG ≻ id), then
fG (yn ) < yn (resp. fG (yn ) > yn ).

(3.7)

Now, for each yn , let us consider the point xn provided by Lemma 3.3.26.
Then ′n coincides with  in restriction to the maximal proper convex subgroup
G1 of G. Moreover, as xn lies between fG−2 (yn ) and fG2 (yn ), it has the same limit
point as yn , hence it converges to an endpoint of IG from outside. Furthermore,
(3.7) obviously holds for xn instead of yn . This implies that ′n coincides with 
over the whole of G, as desired.

To close this section, let us mention that it is unclear what is the most general
framework in which Theorem 3.3.12 still holds. This naturally yields to the next
Question 3.3.27. Is the space of left-orders of a countable amenable group
either finite or a Cantor set ? What about groups without free subgroups in two
generators ?
It is very likely that the previous methods can be extended to a wide family of
amenable groups, namely that of elementary amenable ones. Roughly, this is
the smallest family of groups that contains all Abelian groups and that is stable
under taking extensions, direct limits, quotients and subgroups (see [43, 155] for
more on this). A relevant example is considered in the next

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Example 3.3.28. The group Γ := Z ⋉ (...Z ≀ (Z ≀ ...)...))) in which the conjugacy action
of the left factor consists in shifting (the level of) the factors in the right wreath product
is obviously elementary amenable. It is somewhat a variation of Plante’s example (yet
it is non-solvable), and was simultaneously introduced in [23] and [155]. A natural
(faithful) action of this group on the line comes from identifying the generator of the
left factor with te map x 7→ 2x and a generator of the 0th -factor of the wreath product
with a homeomorphism with support contained in [−2, 2] and sending −1 to 1. It is
very likely that its space of left-orders is a Cantor set, and it actually seems reasonable
trying to describe all of its actions on the line.

3.4

Verbal Properties of Left-Orders

Let W be the set of reduced words in two letters a, b. (This naturally identifies
to the free group in two generators.) We distinguish three subsets of W, namely
W + , W − , and W ± , the set of words involving only positive, negative, or mixed
exponents in a and b, respectively. Given elements f, g in a group Γ and W ∈ W,
we let W (f, g) be the element in Γ obtained from the expression of W by replacing
a by f and b by g, respectively.
Definition 3.4.1. A left-order  on a group Γ satisfies the verbal property
W , or it is a W -order, if whenever f and g are -positive, the element W (f, g)
is also -positive.
Notice that this defines a nontrivial property only in the case where W ∈ W ± ,
hence in the sequel we will only consider these words.
Example 3.4.2. For W (a, b) := b−1 ab, one easily checks that the set of W -left-orders
coincides with that of bi-orders.
Example 3.4.3. For W (a, b) := b−1 ab2 , Proposition 3.2.1 tells us that the set of W orders corresponds to that of Conradian ones.

The next two questions become natural in this context.
Question 3.4.4. Does there exist a word W such that the W -orders are those
that satisfy an specific and relevant algebraic property different from bi-orderability
or the Conradian one ?
Question 3.4.5. Is the property of not having a double crossing (c.f. Example
3.2.18) for a left-order equivalent to a verbal property (or to an interesection of
finitely many ones) ?

3.4. VERBAL PROPERTIES OF LEFT-ORDERS

145

As it is easy to check, the subset of W -orders is closed inside LO(Γ), and the
conjugacy action preserves this subset. The next result on free groups is only
stated for two generators, though it can be easily extended to more generators.
Theorem 3.4.6. The free group on two generators admits left-orders satisfying
no verbal property W ∈ W ± . Actually, this is the case of a Gδ -dense subset of
LO(F2 ).
Let us first show that the existence of a single left-order satisfying no verbal
property implies that this is the case for most left-orders. Thies relies on Lemma
2.2.27, as shown by the next
Lemma 3.4.7. Every left-order on F2 having a dense orbit under the conjugacy
action satisfies no verbal property W ∈ W ± .
Proof. Otherwise, as the closure of such an orbit only contains W -orders, we
would be in contradiction with Theorem 3.4.6.

Question 3.4.8. It is a nontrivial fact that the real-analytic homeomorphisms
of the line given by x 7→ x + 1 and x 7→ x3 generate a free group [50]. By
analyticity, a Gδ -dense subset S of points in the line have a free orbit under this
action. Given a point x ∈ S, we may associate to it the left-order on F2 defined
by f ≻ g whenever f (x) > g(x). Is the set of x ∈ S for which the associate order
satisfies no verbal property still a Gδ -dense subset of R ?
We next proceed to the proof of the first claim of Theorem 3.4.6, which is done
via a very simple dynamical argument. Namely, given W ∈ W ± , we will construct
two increasing homeomorphisms of the real line f, g, both moving the origin to the
right, such that in the action of F2 given by a → f , b → g, the homeomorphism
W (f, g) moves the origin to the left. Then, any dynamical-lexicographic left-order
 associated to a sequence starting at the origin will be such that f ≻ id, g ≻ id,
and W (f, g) ≺ id. This is enough for our purposes except for that the action we
will produce will be not necessary faithful. However, this is just a minor detail
that may be solved in many ways. For instance, one can make the action faithful
by perturbing it close to infinity, as in §2.2.2; alternatively, one may consider a
convex-like extension of the order , as in §2.1.1.
The construction of the desired action is done as follows. By interchanging
a and b if necessary, we may assume that the word W = W (a, b) writes in the
form W = W1 a−n W2 , where W2 is either empty or a product of positive powers

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

of a and b, the integer n is positive, and W1 is arbitrary. Let us consider two
local homeomorphisms defined on a right neighborhood of the real line such that
f (0) > 0, g(0) > 0 and W2 (f, g)(0) < f n (0). This can be easily done by taking
f (0) ≫ g(0) and letting g be almost flat on a very large right-neighborhood of
the origin. If W1 is empty, just extend f and g into homeomorphisms of the
real line. Otherwise, write W1 = ank bmk . . . an2 bm2 an1 bm1 , where all mi , ni are
nonzero excepting perhaps nk . The extension of f and g to a left-neighborhood
of the origin depends on the signs of the exponents mi , ni , and is done in a
constructive manner. Namely, first extend f slightly so that f −n W2 (f, g)(0) is
defined and f has a fixed point x1 to the left of the origin. Then extend g to a
left-neighborhood of the origin so that g m1 f −n W2 (f, g)(0) < x1 and g has a fixed
point y1 to the left of x1 . Notice that m1 > 0 forces g to be right-topologicallyattracting towards y1 on an interval containg f −n W2 (f, g)(0), whereas m1 < 0
forces right topological repulsion. Next, extend f to a left neighborhood of x1
so that f n1 g m1 f −n W2 (f, g)(0) < y1 and f has a fixed point x2 to the left of
y1 . Again, if n1 > 0, this forces right-topological-attraction towards x2 , whereas
n1 < 0 implies right-topological-repulsion.
.
..
.. .......................................................
.
..........
.
.....
..
.....
.
.. ..
.
... ....
.
.. ..
..
... ... . .......................................
.... ...............................
...............................................................................................................................................................................................................
.. ... ..
... . ... ..
... .... ....
.
.
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.
.
.
......................................................
.. ..
....
.. ..
..
..
.. ...
.
.
...
.
.
.... ..
..
..
.......................................... ...
................................•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
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.
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.
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.
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f
g

Figure 18 : The case W1 = an2 bm2 an1 bm1 , where m1 > 0, n1 < 0, m2 < 0, n2 > 0.

Continuing the procedure in this manner (see Figure 18 for an illustration),
we get partially-defined homeomorphisms f, g for which
0 > f nk g mk . . . f n2 g m2 f n1 g m1 f −n W2 (f, g)(0) = W (f, g)(0).
Extending f, g arbitrarily into homeomorphisms of the real line, we finally obtain
the desired action.

147

3.5. A NON LEFT-ORDERABLE GROUP, AND MORE

3.5
3.5.1

A Non Left-Orderable Group, and More
No left-order on finite-index subgroups of SL(n, Z)

Proposition 3.2.9 gave us a simple criterium for non left-orderability of certain
groups. In the same spirit, an important result due to Morris-Witte [192] establishes that finite-index subgroups of SL(n, Z) are non left-orderable for n ≥ 3.
(Notice that most of these groups are torsion-free, bacause of the classical Selberg
lemma [179].)
Theorem 3.5.1. If Γ is a finite-index subgroup of SL(n, Z), with n ≥ 3, then Γ
is non left-orderable.
Proof. Since SL(3, Z) injects into SL(n, Z) for every n ≥ 3, it suffices to consider
the case n = 3. Assume for a contradiction that  is a left-order on a finiteindex subgroup Γ of SL(n, Z). Notice that for large-enough k ∈ N, the following
elements must belong to Γ:






1 k 0
1 0 k
1 0 0
g1 =  0 1 0  ,
g2 =  0 1 0  ,
g3 =  0 1 k  ,
0 0 1
0 0 1
0 0 1



1 0 0
g4 =  k 1 0  ,
0 0 1




1 0 0
g5 =  0 1 0  ,
k 0 1




1 0 0
g6 =  0 1 0  .
0 k 1

It is easy to check that for each i ∈ Z/6Z, the following relations hold:
gi gi+1 = gi+1 gi ,

[gi−1 , gi+1 ] = gik .

In particular, the group generated by gi−1 , gi and gi+1 is nilpotent.
For g ∈ Γ, we define |g| := g if g  id, and |g| := g −1 in the other case. We
also write g ≫ h if g ≻ hn for every n ≥ 1. We claim that either |gi−1| ≫ |gi | or
|gi+1 | ≫ |gi |. Indeed, as  restricted to the subgroup hgi−1, gi , gi+1 i is Conradian
(c.f. Theorem 3.2.20) and a power of gi is a commutator, this follows from Remark
3.2.30.
Assume for instance that |g1 | ≪ |g2|, the case where |g2| ≪ |g1 | being analogous. Then we obtain |g1 | ≪ |g2 | ≪ |g3 | ≪ |g4| ≪ |g5 | ≪ |g6 | ≪ |g1 |, which is
absurd.


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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

It follows from an important theorem due to Margulis that for n ≥ 3, every
normal subgroup of a finite-index subgroup of SL(n, Z) either is finite or has
finite index (see [140]). As a corollary, we obtain the following strong version of
Theorem 3.5.1.
Theorem 3.5.2. For n ≥ 3, no torsion-free, finite-index subgroup of SL(n, Z)
admits a total, nontrivial, left-invariant preorder.
Proof. If Γ is such a group and admits a nontrivial, total preorder, then by
Exercise 1.1.6, there is a nontrivial quotient Γ/H that is left-orderable. Since Γ
is torsion-free, it has no nontrivial finite subgroup. Therefore, there are only two
possible cases: either H is trivial, in which case we contradict Theorem 3.5.1,
or Γ/H is finite and nontrivial, which is impossible as no nontrivial finite group
admits a nontrivial, left-invariant preorder. (Indeed, if f ≻ id for such a preorder,
then f n ≻ id for all n ∈ N.)

In terms of semigroups, this translates into the next
Corollary 3.5.3. If n ≥ 3 and Γ is a torsion-free, finite-index subgroup of
SL(n, Z), then there is only one subsemigroup P of Γ satisfying P ∪ P −1 = Γ,
namely P = Γ.
The results above are conjecturally true for all lattices in simple Lie groups
of rank≥ 2. Although this is still open in full generality, Morris-Witte’s theorem
extends to the case of higher Q-rank, as well as other non-cocompact lattices
[122]. However, no proof is available for cocompact lattices.

3.5.2

A canonical decomposition of the space of left-orders

Let (Γ, ) be a finitely-generated, left-ordered group, and let Γ0 be its maximal
-convex subgroup (c.f. Example 2.1.2). There are two possibilities, namely the
action of Γ on Γ/Γ0 is or is not Conradian. In the first case, we will say that 
is of type I. The next proposition generalizes Corollary 3.2.2.
Proposition 3.5.4. The set of left-orders of type I is closed inside LO(Γ).
Proof. Since Γ is finitely generated, Γ/Γ′ may be written as Zk × G, where
k ≥ 1 and G is a finite Abelian group. Let n be a sequence of type-I left-orders
on Γ converging to a left-order . We must show that  is also of type I. To
do this, notice that associated to each n , there is a Conrad’s homomorphism

3.5. A NON LEFT-ORDERABLE GROUP, AND MORE

149

τn , which may be though of as defined on Zk . This homomorphism may be
chosen normalized. More precisely, if we let {g1 , . . . , gk } be a family of elements
whose representatives generate Γ/Γ′ and denote an,i := τn (gi ), then the vector
(an,1 , . . . , an,k ) belongs to the (k − 1)-sphere Sk−1 , for each n.
Claim (i). The points (an,1 , . . . , an,k ) converge to some limit (a1 , . . . , ak ) ∈ Sk−1 .
Otherwise, there are subsequences (τni ) and (τmi ) so that the associated vectors converge to two different points of Sk−1 with orthogonal hyperplanes H1 , H2 ,
respectively. These hyperplanes divide Rk into four regions. Let us pick a point
of integer coordinates on each of these regions, and let h1 , h2 , h3 , h4 be elements of
Γ which project to these points by the quotient Γ → Zk × G. For i large-enough,
the value of both τni (hj ) and τmi (hj ) is nonzero for each j, but the signs of these
numbers must be different for some j. As Conrad’s homomorphisms are nondecreasing, after passing to a subsequence of (ni ) and (mi ) this implies that,
for some j, the element hj will have different sign for ni and mi . However, this
is in contradiction to the convergence of n .
Now, notice that the vector (a1 , . . . , ak ) gives raise to a group homomorphism
τ : Zk → R (which may be though of as defined on Γ), namely, for g ∼ g1n1 · · · gknk
in Γ/Γ′ ,
k
X
ai ni .
τ (g) :=
i=1

Claim (ii). The kernel of τ is a -convex subgroup of Γ.
Indeed, let g ∈ Γ and f ∈ ker(τ ) be such that id  g  f . As Conrad’s
homomorphisms are order preserving, for each n, we have
0 = τn (id) ≤ τn (g) ≤ τn (f ).
As τn pointwise converges to τ and τ (f ) = 0, the inequalities above yield, after
passing to the limit, τ (g) = 0. Thus, g belongs to ker(τ ).
As a consequence of Claim (ii), the maximal -convex subgroup Γ0 contains ker(τ
 ). Also, the action of Γ on Γ/Γ0 is order-isomorphic to that on
Γ/ ker(τ ) Γ0 / ker(τ ). Since the latter is an action by translations, the former
is, in particular, Conradian. Therefore,  is of type I.

The case where the action of Γ on Γ/Γ0 is not Conradian is dynamically more
interesting. We know by definition that there must exist a crossing for the action.
The question is “how large” can be the “domain of crossing”. To formalize this

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

idea, for each h ∈ Γ, let us consider the “interval”

I(h) := h̄ ∈ Γ: there exists a crossing (f, g; u, w, v) such that f v ≺ h ≺ h̄ ≺ gu .

By definition, I(h) is a convex subset of Γ.

Lemma 3.5.5. If the set I(h) is bounded from above for some h ∈ Γ, then it is
bounded from above for all h ∈ Γ.
Proof. As the notion
of crossing is invariant under conjugacy, for all h1 , h2 in Γ

we have h1 I(h2 ) = I(h1 h2 ). The lemma easily follows from this.

If (the action of Γ on Γ/Γ0 is not Conradian and) I(h) is bounded from above
for all h ∈ Γ, we will say that  is of type II. If not,  will be said of type
III. We then have a canonical decomposition of the space of left-orders of Γ into
three disjoint subsets (compare [84, Theorem 7.E]):
LO(Γ) = LOI (Γ) ⊔ LOII (Γ) ⊔ LOIII (Γ).
Example 3.5.6. Every Conradian order is of type I. Therefore, by Theorem 3.2.3,
finitely-generated, locally-indicable groups admit left-orders of type I.
Example 3.5.7. Smirnov’s left-orders ǫ (with ε irrational; see §1.2.2) on subgroups
of the affine group are prototypes of type III left-orders. However, these groups being
bi-orderable, they also admit left-orders of type I, which actually arise as limits os
Smirnov type orders. Moreover, the description given in 1.2.2 shows that these groups
do not admit type II left-orders. As a consequence, LOIII (Γ) is not necessarily closed
inside LO(Γ).
The last remark above may be straightened. Indeed, if we choose inside the
Baumslag-Solitar group BS(1, ℓ) a sequence (gn ) so that gn−1 (ε) tends to +∞, then all
conjugate left-orders (ε )gn are of type III, but the limit left-order ∞ is bi-invariant,
hence of type I. Thus, a limit of type III left-orders in the same orbit of the conjugacy
action may fail to be of type III.
Exercise 3.5.8. Let (xi ) and (yn ) be two sequences of points in ]0, 1[ so that xi
converges to the origin and {yn } is dense. For each i ≥ 1, let (zn,i )n be the sequence
having xi as its first term and the yn ’s as the next ones. Associated to this sequence
there is a dynamical-lexicographic left-order i on Thompson’s group F, namely, f ≻i
id if and only if the smallest n for which f (zn,i ) 6= zn,i is such that f (zn,i ) > zn,i (c.f.
§1.1.3). Show that i is of type III for all i, but any adherence point of the sequence
(i ) in LO(F) is of type I.

3.5. A NON LEFT-ORDERABLE GROUP, AND MORE

151

Remark 3.5.9. The subset of left-orders of type II on the free group F2 is dense in
LO(F2 ). Roughly, the proof proceeds as follows. (Compare §2.2.2.) Start with an
arbitrary left-order  on F2 together with an integer n ∈ N. Consider the dynamical
realization of  as well as a very large compact subinterval I in the real line on which
the dynamics captures all inequalities between elements in the ball Bn (id) or radius n
in F2 . Then consider a new action of F2 which coincides with this dynamical realization
on I and commutes with a translation of the line (of very large amplitude). This new
action induces a (perhaps partial) left-order n , which can be easily turned into a total
one (by convex extension) which is not of type I (by adding crossings). Clearly, the
left-orders n are all of type II and converge to .
Remark 3.5.10. A similar construction allows to produce a sequence of left-orders of
type III on F2 that converges to an order of type II. Roughly, starting with a type II
left-order , we consider its dynamical realization, we keep it untouched on a very large
compact interval I, and outside I we perturb it by including infinitely many crossings
for the generators along larger and larger domains. The new action will then induce a
left-order on F2 which is of type III and very close to . We leave the details to the
reader.

The main problem related to this discussion is
Question 3.5.11. Does there exist a finitely-generated, left-orderable group all
of whose left-orders are of type III ?
We will come back to this question later.
Cofinal elements and the type of left-orders. Recall from §3.2.3 that an
element f of a left-ordered group (Γ, ) is -cofinal if for any g ∈ Γ there exist
integers m, n such that f m ≺ g ≺ f n . In terms of dynamical realizations (c.f.
§1.1.3), for countable groups, this corresponds to that f has no fixed point on
the real line.
Following [47], we say that f is a cofinal element of Γ if it is -cofinal for
every left-order  on Γ. The following should be clear from the discussion above.
Proposition 3.5.12. If a finitely-generated, left-orderable group Γ has a cofinal,
central element, then no left-order on Γ is of type III.
Example 3.5.13. In §3.2, we introduced the group
Γ = f, g, h : f 2 = g3 = h7 = f gh ,

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

which is left-orderable but admits no nontrivial homomorphism into the reals (hence
no left-order of type I). We claim that the central element ∆ := f gh is cofinal. Indeed,
if ∆ = f 2 = g3 = h7 has a fixed point for a dynamical realization, then this is fixed
by f, g, h, hence by the whole group, which is impossible. As a consequence, every
left-order on Γ is of type II.
Example 3.5.14. Recall that the center of the braid group Bn is generated by the
square of the so-called Garside element ∆n . Moreover, one has
∆2n = (σ1 σ2 · · · σn−1 )n = (σ12 σ2 · · · σn−1 )n−1 .
We next reproduce the proof given in [47] of that ∆2n is cofinal in Bn .
Claim (i). If  is a left-order on Bn for which ∆n ≻ id, then for any braid σ that is
conjugate to either αn := σ1 σ2 · · · σn or βn := σ12 σ2 · · · σn we have id ≺ σ ≺ ∆2n .
Indeed, as σ k = ∆2 for k equal to either n or n − 1, we must have
id ≺ σ ≺ σ 2 ≺ σ 3 ≺ . . . ≺ σ k = ∆2n .
2
Claim (ii). If  is as above, then ∆−2
n ≺ σi ≺ ∆n , for all i ∈ {1, . . . , n − 1}.

Indeed, since ∆2n is central, by Claim (i) we have, for all δ ∈ Bn ,
−1 −1
∆−2
≺ id,
n ≺ δαn δ

id ≺ δβn δ−1 ≺ ∆2n .

−2
−1 ≺ ∆2 , and since all the σ ’s are conjugate
Since βn α−1
i
n = σ1 , this yields ∆n ≺ δσ1 δ
n
between them, this shows the claim.

Claim (iii). The element ∆2n is cofinal in Bn .
Let  be a left-order on Bn . Using again the fact that ∆2n is central, the set
2s
{σ ∈ Bn : ∆2r
n ≺ σ ≺ ∆n for some r, s in Z} is easily seen to be a subgroup of Bn . By
Claim (ii), it contains the σi ’s. Thus, it coincides with Bn , which concludes the proof.
Remark 3.5.15. We do not know whether there exist type-III left-orders on the derived
groups B′n for n ≥ 5. Notice that these groups do not admit left-orders of type I since
they admit no nontrivial homomorphism into the reals (c.f. Example 3.2.8). By the
preceding example, the restriction of any left-order on Bn to them is of type II.

A dynamical view. As we showed in §1.1.3 (and used many times), finitelygenerated, left-ordered groups may be realized as groups of orientation-preserving
homeomorphisms of the real line. In what follows, we use this to visualize the
dynamical differences between orders of different type. For example, type-I leftorders are characterized as follows.

3.5. A NON LEFT-ORDERABLE GROUP, AND MORE

153

Proposition 3.5.16. If  is a left-order of type I on a (finitely-generated) group
Γ, then its dynamical realization preserves a Radon8 measure on R. Conversely,
any left-order induced from a faithful action of Γ on the real line that preserves
a (nontrivial) Radon measure is of type I.
Proof. Let  be a left-order of type I on Γ, let Γ0 be its maximal proper convex
subgroup, and consider the dynamical realization of . By convexity, Γ0 fixes the
interval [a, b], where a, b are, respectively, the infimum and the supremum of the
orbit of the origin under Γ0 . Moreover, by Theorem 3.2.27 and Corollary 3.2.28,
we have that Γ0 is normal in Γ, that Γ0 = ker(τ ), where τ : Γ → (R, +) is
the Conrad homomorphism, and that the induced order on Γ/Γ0 is Archimedean.
In particular, the set F ix(Γ0 ) of global fixed points of Γ0 , is Γ-invariant (hence
infinite), and the action of Γ/Γ0 on F ix(Γ0 ) is free.
Now, if F ix(Γ0 ) is P
discrete (equivalently, if τ (Γ) ∼ Z), then for each x ∈
F ix(Γ0 ), the measure h∈Γ/Γ0 δh(x) is a Γ-invariant Radon measure. If F ix(Γ0 )
is non-discrete (equivalently, if τ (Γ) is a dense subgroup of R), we may proceed as
in Example 3.1.6 to show that the action of Γ is continuously semiconjugate to an
action by translations that factors throughout Γ/Γ0 . Pulling back the Lebesgue
measure by this semiconjugacy, we obtain a Γ-invariant Radon measure.
Conversely, assume that an action of Γ by orientation-preserving homeomorphism of the real line preserves a Radon measure ν. Then there is a translationnumber homomorphism τν : Γ → (R, +) defined by τν (g) := ν([y, g(y))). (Recall
that the value is independent of y, due to invariance.) We claim that ker(τν ) is a
convex subgroup for any left-order induced from the action. Indeed, let x ∈ R be
the first reference point for inducing such a left-order on Γ (c.f. §1.1.3). On the
one hand, if x lies in the support of ν, then ker(τν ) coincides with the stabilizer of
x, hence it is a convex subgroup. On the other hand, if x does not belong to the
support of ν, let I be the connected component of the complement of the support
of ν containing x. At least one endpoint of I is finite, which easily allows to show
that for each g ∈ Γ, either g(I) ∩ I is empty or coincides with I. It follows from
this that the stabilizer of I is a convex subgroup of Γ that coincides with ker(τν ).
Notice that for all g, h in Γ, the inequality τν (g) > τν (h) implies g(x) > h(x) for
every x ∈ R. It easily follows from this and the discussion above that ker(τν ) is
the maximal convex subgroup. Finally, the action of Γ on Γ/ ker(τν ) is Conradian,
because it is order-isomorphic to an action by translations.


8

Recall that a Radon measure is a measure giving finite mass to compact sets.

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Remark 3.5.17. In the proof above, the finite-generation hypothesis was only used
in the direct implication to ensure the existence of a maximal proper convex subgroup.
Since this is known to exist in some other situations (see, for instance, Exercise 3.3.1),
the proposition still holds in these cases.

To deal with type-II and type-III left-orders, we closely follow [63] (compare
[133]). We say that the action of a subgroup Γ of Homeo+ (R) is locally contracting if for every x ∈ R there is y > x such that the interval [x, y] can be
contracted into a point by a sequence of elements in Γ. We say that the action is
globally contracting if such a sequence of contractions exists for any compact
^ + (S1 ) the group of homeomorphisms of
subinterval of R. We denote by Homeo
the line that are liftings of orientation-preserving circle homeomorphisms. The
next lemma is to be compared with Example 2.1.2.
Lemma 3.5.18. Every finitely-generated subgroup of Homeo+ (R) preserves a
nonempty minimal closed subset of the line. This set is unique in case no discrete
orbit exists.
Proof. Fix a point x0 and a compact interval I containing x0 as well as all its
images under the (finitely many) generators of the group Γ. By obvious reasons,
every closed set of the line that is invariant under the action must intersect I.
Therefore, the standard argument using Zorn’s lemma to detect (nonempty) minimal sets may be applied by looking at the (compact) “traces” in I of nonempty
invariant closed subsets of the line. We leave the details to the reader (see [150,
Proposition 2.1.12] in case of problems). To prove uniqueness, notice that for a
closed invariant subset K, the set of acummulation points K ′ is also closed and
invariant, hence K ′ = K in case K is not a discrete orbit. Assume K is not
the whole line (otherwise, the uniqueness is obvious). It is then easy to see that
every connected component of the complement of K ′ has a sequence of images
converging to any point in K ′ . In other words, every orbit acummulates at K,
which obviously implies the uniqueness of the nonempty minimal invariant closed
set.

Theorem 3.5.19. Let Γ be a finitely-generated subgroup of Homeo+ (R) whose
action admits no global fixed point. Then one of the following mutually-exclusive
possibilities occur:
(i) Γ is semiconjugate to a group of translations;
^ + (S1 );
(ii) Γ is semiconjugate to a minimal, locally contracting subgroup of Homeo
(iii) Γ is globally contracting.

3.5. A NON LEFT-ORDERABLE GROUP, AND MORE

155

Proof. Assume there is no discrete orbit for the action. By Lemma 3.5.18, there
is a unique minimal nonempty closed Γ-invariant subset K. In case K is not the
whole line, collapse each connected component of the complement of K to a point
in order to continuously semiconjugate Γ to a group Γ whose action is minimal.
If Γ preserves a Radon measure, then after semiconjugacy, this measure becomes
a Γ-invariant Radon measure of total support and no atoms. Therefore, Γ (resp.
Γ) is conjugate (resp. semiconjugate) to a group of translations.
Suppose next that Γ has no invariant Radon measure. Then the action of Γ
cannot be free. Otherwise, Γ would be conjugate to a group of translations (c.f.
Example 3.1.6), and the pull-back of the Lebesgue measure by the semiconjugacy
would be a Γ-invariant Radon measure.
Let ḡ ∈ Γ be a nontrivial element having fixed points, and let x̄0 be a point in
the boundary of Fix(ḡ). Then there is a left or right neighborhood I of x̄0 that is
contracted to x̄0 under iterates of either ḡ or its inverse. By minimality, every x̄
has a neighborhood that can be contracted to a point by elements in Γ. Coming
back to the original action, we conclude that every x ∈ R has a neighborhood that
can be contracted to a point by elements in Γ. Notice that such a limit point can
be chosen arbitrarily in K; in particular, it may be chosen to belong to a compact
interval I that intersects every orbit (as in the proof of Lemma 3.5.18).
For each x ∈ R, let M(x) ∈ R∪{+∞} be the supremum of the y > x such that
the interval (x, y) can be contracted to a point in I by elements of Γ. Then either
M ≡ +∞, in which case the group Γ is globally contracting, or M(x) is finite
for every x ∈ R. In the last case, M induces a non-decreasing map M : R → R
commuting with all the elements in Γ. Since the union of the intervals on which
M is constant is invariant by Γ, the minimality of the action implies that there
is no such interval, that is, M is strictly increasing. Moreover, the interior of
R \ M (R) is also invariant, hence empty because the action is minimal. In other
words, M is continuous. All of this shows that M induces a homeomorphism
of R into its image. Since the image of M is Γ-invariant, it must be the whole
line. Therefore, M is a homeomorphism from the real line to itself. Observe that
M (x) > x for any point x, which implies that M is conjugate to the translation
+
^ (S1 ).
x 7→ x + 1. After this conjugacy, Γ becomes a subgroup of Homeo

The next proposition should now be clear to the reader.
Proposition 3.5.20. Let Γ be a finitely-generated left-orderable group, and let
 be a left-order on it. Then  is of type I (resp. II, resp. III) if and only if its
dynamical realization satisfies property (i) above (resp. (ii), resp. (iii)).

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CHAPTER 3. ORDERABLE GROUPS AS DYNAMICAL OBJECTS

Having this proposition at hand, we come back to Question 3.5.11. We claim
that, in case of a negative answer, no lattice in a higher-rank simple Lie group
can be left-orderable. Indeed, on the one hand, orders of type I lead to nontrivial
homomorphisms into the reals in an obvious way. On the other hand, orders of
type II yield (perhaps unfaithful) actions on the circle (viewed as the space of
orbits of the freely-acting commuting homeomorphism) that have no invariant
probability measure (otherwise, we fall into type I). Both cases are impossible
for higher rank lattices: the former contradicts Kazhdan’s property (T) (see [9]),
and the latter contradicts a theorem of Ghys (see [80]).
Remark 3.5.21. As cited above, lattices in higher-rank simple Lie groups have Kazhdan’s property (T), meaning that every affine isometric action on a Hilbert space has
a fixed point (see [9]). It is unknown whether an infinite Kazhdan group may be leftorderable. A challenging case is that of lattices in Sp(n, 1). Indeed, orderability of such
a lattice Γ would provide an affirmative answer to this and to Question 3.5.11 (due to
Proposition 3.5.20 above), and non-left-orderability would yield new examples (besides
those of [116]) related to Question 1.4.8 provided the injectivity radius of Sp(n, 1)/Γ
is large enough, in virtue of Example 1.3.7 (indeed, the associated symmetric space
carries an invariant negatively-curved metric).

Chapter 4
PROBABILITY AND
LEFT-ORDERABLE GROUPS
4.1

Amenable Left-Orderable Groups

Starting from the work of von Neumann and Day, amenability became one of
the deepest notions in the theory of infinite groups. Von Newmann defined this
concept as follows.
Definition 4.1.1. A countable group Γ is said to be amenable if there exists
an invariant mean on L∞ (Γ), that is, a linear functional M on L∞ (Γ) that
satisfies:
– (Positivity) If φ is non-negative, then M(φ) ≥ 0;
– (Normalization) If 1Γ denotes the constant function equal to 1 along Γ, then
M(1Γ ) = 1;
– (Invariance) For all φ ∈ L∞ (Γ) and all g ∈ Γ, one has M(φ) = M(φ ◦ Rg ),
where R is the right action of Γ on L∞ (Γ), that is, Rg (φ)(h) := φ(hg).

Among the many equivalent definitions, we next state the one that will be
useful in this section: a group is amenable if every action by homeomorphisms of
a compact metric space admits an invariant probability measure.

Exercise 4.1.2. Prove the equivalence between the two definitions above (see [108] or
[195] in case of problems).

Our aim now is to discuss another nice result due to Morris-Witte [145]. The
theorem below was conjectured by Linnell in [125], but it was already suggested
by Thurston (see [189, page 348]).
157

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Theorem 4.1.3. Every amenable, left-orderable group is locally indicable.
For the proof, we will say that a left-order  is right-recurrent if for every
pair of elements f, h in Γ such that f ≻ id, there exists n ∈ N satisfying f hn ≻ hn .
Notice that every right-recurrent order is Conradian. (The converse does not
hold; see Example 4.1.7.) As subgroups of amenable groups are amenable [195],
this implies that Theorem 4.1.3 follows from the next
Proposition 4.1.4. If Γ is a finitely-generated, amenable, left-orderable group,
then Γ admits a right-recurrent order.
To prove this proposition, we will need the following weak form of the Poincaré
Recurrence Theorem. We recall the proof for the reader’s convenience.
Theorem 4.1.5. If S is a measurable map that preserves a probability measure
µ on a space X, then for every measurable subset A of X and µ-almost-every
point x ∈ A, there exists n ∈ N such that S n (x) belongs to A.
Proof. The
B of points in A that do not come back to A under iterates of
S set −n
S is A \ n∈N S (A). One easily checks that the sets S −i (B), with i ≥ 1, are
two-by-two disjoint. Since S preserves µ, these sets have the same measure, and
since the total mass of µ equals 1, the only possibility is that this measure equals
zero. Therefore, µ(B) = 0, that is, µ-almost-every point in A comes back to A
under some iterate of S.

Exercise 4.1.6. In the framework above, show that for µ-almost every point x ∈ X,
the set of positive integers n such that S n (x) ∈ A is unbounded.

Proof of Proposition 4.1.4. It directly follows from on of the definitions
that if a countable left-orderable group Γ is amenable, then its action on (the
compact metric space) LO(Γ) preserves a probability measure µ. We claim that
µ-almost-every point in LO(Γ) is right-recurrent. To show this, for each g ∈ Γ,
let us consider the subset Vg of LO(Γ) formed by the left-orders  on Γ such
that g ≻ id. By
Recurrence Theorem, for each f ∈ Γ, the set
S the Poincaré
−n
Bg (f ) S
:= Vg \ n∈N f (Vg ) has null µ-measure. Therefore, Sthe measure of
Bg := f ∈Γ Bg (f ) is also zero, as well as the measure of B := g∈Γ Bg . Let us
consider an arbitrary element  in the (µ-full measure) set LO(Γ) \ B. Given
g ≻ id and f ∈ Γ, from the inclusion Bg (f ) ⊂ B we deduce that  does not
belong to Bg (f ). Thus, there exists n ∈ N such that  belongs to f −n (Vg ), hence
f n is in Vg . In other words, one has g ≻f n id, that is, gf n ≻ f n . Since g ≻ id
and f ∈ Γ were arbitrary, this shows the right-recurrence of .

4.1. AMENABLE LEFT-ORDERABLE GROUPS

159

Example 4.1.7. Following [145, Example 4.5], we next show that there exist Corderable groups which do not admit right-recurrent orders. This is the case of the
semidirect product Γ = F2 ⋉Z2 , where F2 is any free subgroup of SL(2, Z) acting linearly
on Z2 . (Such a subgroup may be taken of finite-index.) Indeed, that Γ is C-orderable
follows from the local indicability of both F2 and Z2 . Assume throughout that  is a
right-recurrent left-order on Γ. For a matrix f ∈ F2 and a vector v = (m, n) ∈ Z2 , let
us denote by f¯ and v̄ the corresponding elements in Γ, so that f (v) = f¯v̄ f¯−1 . Let τ
be the Conrad’s homomorphism associated to the restriction of  to Z2 , so that we
have v ≻ id whenever τ (v) > 0, and τ (v) ≥ 0 for all v ≻ id (c.f. Corollary 3.2.28).
Let f be a hyperbolic matrix in F2 , with positive eigenvalues α1 , α2 and corresponding
eigenvectors v1 , v2 in R2 . Since v1 and v2 are linearly independent, we may assume
that τ (v1 ) 6= 0. Furthermore, we may assume τ (v1 ) > 0 and α1 > 1 after replacing
v1 with −v1 and/or f with f −1 , if necessary. Let L : R2 → R be the (unique) linear
functional that satisfies L(v1 ) = 1 and L(v2 ) = 0. Given any v ∈ Z2 such that τ (v) > 0,
right-recurrence provides us with an increasing sequence (ni ) such that v̄ f¯−ni ≻ f¯−ni
for every i. This implies that f¯ni v̄ f¯−ni ≻ id, hence τ (f ni (v)) ≥ 0. Since
τ (f ni (v))
=τ
lim
i→∞
αn1 i



f ni (v)
lim
n
i→∞ α1 i



= τ (L(v)v1 ) = L(v)τ (v1 ),

we conclude that L(v) ≥ 0. Since v is an arbitrary element of Z2 satisfying τ (v) > 0,
this necessarily implies that ker(τ ) = ker(L) is an eigenspace of f . But f is an arbitrary
hyperbolic matrix in F2 , and it is easy to show that there are hyperbolic matrices in
F2 with no common eigenspace. This is a contradiction.
Remark 4.1.8. The group Γ = F2 ⋉ Z2 is particularly interesting in our context by
this and many other reasons; see [152] for a discussion of this.

An extension for left-orderable groups without free subgroups ? A
prototype of a non-amenable group is the free group in two generators. Indeed,
this group acts by projective diffeomorphisms of the circle without an invariant
probability measure (one of these actions is the so-called Schottky action, to
which Klein’s ping-pong argument perfectly applies). It is an easy exercise to
show that every subgroup of an amenable group is amenable. Therefore, every
group containing a (non-Abelian) free group is also non-amenable. The converse
is, however, false, even in the framework of bi-orderable, finitely-presented groups
(see §4.2).
It is unknown whether Theorem 4.1.3 extends to groups without free subgroups, that is, whether left-orderable groups not containing F2 are locally indicable. (See [124] for an interesting result pointing in the affirmative direction.)

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

We next show that this would be the case if the answer to Question 3.5.11 is negative. Indeed, as type-I orders yield homomorphisms into the reals, what we need
to show is that type-II orders imply the existence of free subgroups. However,
as we saw at the end of §3.5.2, type-II orders yield group actions on the circle
without invariant probability measures. Now, a theorem of Margulis establishes
that a group admitting such an action necessarily contains a copy of F2 (see [139]
as well as [79, 150]).
A relevant class of groups that do not contain free subgroups consists of
those satisfying a nontrivial law (or identity). This is a reduced word W =
W (x1 , . . . , xk ) in positive and negative powers such that W (g1, . . . , gn ) is trivial
for every g1 , . . . , gn in the group. For instance, Abelian groups satisfy a law,
−1
namely W1 (x1 , x2 ) := x1 x2 x−1
1 x2 . Nilpotent and solvable groups also satisfy
group laws. Another important but less understood family is the one given by
groups satisfying an Engel condition WkE , where

E
−1
WkE (x1 , x2 ) := W1E Wk−1
(x1 , x2 ), x2 ,
W1E (x1 , x2 ) := x1 x2 x−1
1 x2 .
It is an open question whether left-orderable groups satisfying an Engel condition
must be nilpotent. This is known to be true if the group is Conrad-orderable (see
[84, Theorem 6.G]). In other words, if Γ is an Engel group having a left-order
without resilient pairs, then Γ is nilpotent. In this direction, the next proposition
becomes interesting, and shows the pertinence of Question 3.2.19. The (easy)
proof is left to the reader. (See [148] for more on this.)
Proposition 4.1.9. If Γ is a left-orderable group satisfying a law, then there
exists n ∈ N such that no left-order on Γ admits an n-resilient pair.
Locally-invariant orders on amenable groups. The ideas involved in the
proof of Theorem 4.1.3 yield interesting results for other type of orders on amenable
groups. The next result is due to Linnell and Morris-Witte [126].
Theorem 4.1.10. Every amenable group admitting a locally-invariant order is
left-orderable (hence locally indicable).
Proof. As for Theorem 4.1.3, we may assume that Γ is finitely generated.
First, it is not hard to extend the claim of Exercise 1.3.8 to describe the
restriction of a locally-invariant order to any left coset of a cyclic subgroup: For
every f ∈ Γ and g 6= id, either
f g n ≺ f g n+1

for all n ∈ Z,

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4.1. AMENABLE LEFT-ORDERABLE GROUPS
or
f g n ≺ f g n−1

or there exists ℓ ∈ Z such that
f g n ≺ f g n+1

for all n ∈ Z,

for all n ≥ ℓ and f g n ≺ f g n−1

for all n < ℓ.

We next argue that for amenable groups, there is a locally-invariant order for
which the third possibility never arises.
Recall from Exercise 2.2.4 that the space of locally-invariant orders is a compact topological space, which is metrizable whenever Γ is countable. The group Γ
acts on LIO(Γ) by left and right translations. Since Γ is amenable, both actions
preserve probability measures on LIO(Γ). Let µ be a probability measure that
is invariant under the right action. We leave to the reader the task of showing
that a generic locally-invariant order  is strongly right-recurrent. More precisely,
there is a subset A of full µ-measure such that for every  in A, the following
happens: If f ≺ g, then given h ∈ Γ, the set of integers n such that f hn ≺ ghn
is unbounded in both directions. (Compare Exercise 4.1.6) Since in the third
case above this property fails, we conclude that a generic locally-invariant order
is either the canonical one or its reverse whenever restricted to a left-coset of a
cyclic subgroup.
We next show that every  in A is a left-order. Indeed, by the definition
of locally-invariant order, for every g 6= id we have either g ≻ id or g −1 ≻ id.
Both inequalities cannot hold simultaneously, otherwise the restriction of  to
the cyclic subgroup hgi wouldn’t be neither the canonical order nor its reverse.
Therefore, the positive cone P := {g : g ≻ id} is disjoint from its inverse, and
their union covers Γ \ {id}.
It remains to show that P is a semigroup. Assume for a contradiction that
g, h in P are such that gh ∈
/ P . Then gh ≺ id. Thus g ≻ id ≻ gh, hence using the
property of a locally-invariant order, one easily checks that, necessarily, gh2 ≺ gh.
More generally,
g ≻ gh ≻ gh2 ≻ gh3 ≻ . . . .
Therefore, ghn ≺ id, for all n ≥ 1. However, due to the right-recurrence of ,
there is some n ∈ N such that ghn ≻ hn  h ≻ id. This is a contradiction.

The theorem above makes natural the following

Question 4.1.11. Does there exist an amenable U.P.P. group that is not leftorderable ? (See §1.4.3.)

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

4.2

Non-Amenable, Left-Orderable Groups with
no Free Subgroup

The natural problem of finding non-amenable groups without (non-Abelian)
free subgroups goes back to von Neumann. This was first solved by Ol’shanskii
in [160] and soon after by Adian [1], in the beginning of the eighties. Their
examples, however, do not admit a finite presentation, and also contain a lot
of torsion elements. Actually, torsion is fundamental for their constructions; for
instance, Adian proves that the free Burnside group B(2, n) is non-amenable for
large enough n (c.f. Example 1.4.14). The first example of a finitely-presented,
non-amenable group without free subgroups was constructed much later in [161];
however, this group still has a lot of torsion.
Here we present a construction of a bi-orderable (hence torsion-free) nonamenable group having no free subgroups and which, moreover, admits a finite
presentation. This example comes from the beautiful recent work of Lodha and
Moore [128], who were able to isolate a particular finitely-presented group inside
a much bigger family of groups previously studied by Monod [144], motivated by
a prior theorem of Carrière and Ghys [39].

4.2.1

(Non-)amenable actions

In this section, we will show that PP+ (R), the group of orientation-preserving,
piecewise-projective homeomorphisms of the real line (c.f. §1.2.5), contains many
countable subgroups that are non-amenable [144]. The key ingredient is a result of
Carrière and Ghys [39], presented as Theorem 4.2.1 below. For the statement, we
say that an action of a countable group Γ on a measured space X by measurable
maps (a measurable action, for short) is regular (or non-singular) if the
images of zero-measure sets under group elements have zero measure. The orbital
equivalence relation associated to such an action is said to be amenable if for
each x ∈ X there is a linear functional (mean) Mx : L∞ (Γx) → R (where Γx
denotes the orbit of the point x ∈ X) that satisfies:
– (Positivity) If φ is non-negative, then Mx (φ) ≥ 0;
– (Normalization) For almost-every x ∈ X, we have Mx (1Γx ) = 1, where 1Γx
denotes the constant function equal to 1 along Γx;
– (Invariance) For all g ∈ Γ and almost every x ∈ X, one has Mx = Mg(x) ;
– (Measurability) If ψ is a bounded measurable function defined on the graph

(x, y) : there exists g ∈ Γ such that y = g(x)

4.2. NON-AMENABLE, LEFT-ORDERABLE GROUPS WITH NO FREE SUBGROUP163
of the equivalence relation, then x 7→ Mx (ψ(x, ·)) is a measurable function from
X into R.
Theorem 4.2.1. Let Γ ⊂ PGL(2, R) be a countable subgroup. If Γ contains a
non-discrete copy of F2 , then the equivalence relation given by the Γ-orbits for its
projective action on P1 (R) is non-amenable.
Proof. Assume for a contradiction that the relation induced by the Γ-action
on P1 (R) is amenable. Then the induced action of the free subgroup F2 is also
amenable. Indeed, we may define the family M ′ of means along the F2 -orbits by
letting Mx′ (f ) := Mx (f¯), where f¯ coincides with f along the F2 -orbit of x and
equals 1 on Γx \ F2 x. Therefore, we may assume that Γ = F2 .
Let {a, b} be a free generating set of F2 = Γ, and let A (resp. B) be the set of
elements in F2 whose reduced form end with a nontrivial power of a (resp. b). Assuming the existence of the linear functionals Mx as above, let ψ = ψa : Γ → [0, 1]
be defined by ψ(x) := Mx (1Ax ), where 1Ax stands for the characteristic function
of the corresponding set Ax := {h(x) : h ∈ A}. By applying (Measurability)
to the function (x, y) 7→ 1Ax (y), we conclude that ψ is a measurable function.
Moreover, the action of F2 on P1 (R) is almost everywhere free, because every
nontrivial element in PGL(2, R) fixes at most two points. Since the sets Abi are
two-by-two disjoint for i ∈ Z, by (Positivity) and (Normalization) we obtain
X
0≤
Mx (1Abi x ) ≤ 1.
i

Since by (Invariance) we also have
ψ(bi x) = Mbi x (1Abi x ) = Mx (1Abi x ),
we obtain
0≤

X
i

ψ(bi x) ≤ 1.

(4.1)

In particular, if ψ(x) > 1/2, then ψ(bi x) < 1/2 for all i 6= 0.
Using (Invariance) and the fact that the action is almost free, one easily concludes that Mx (1{x} ) = 0 holds for almost every x. Therefore, by (Normalisation),
we have ψb := 1 − ψa . This allows us to conclude in the very same way as above
that if ψ(x) < 1/2, then ψ(ai x) > 1/2 for all i 6= 0.
These properties fit into the framework of the classical Klein’s ping-pong
argument for the measurable subsets P and Q of P1 (R) defined by
P := {x ∈ G | ψ(x) < 1/2} and Q := {x ∈ G | ψ(x) > 1/2}.

164

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Indeed, we have ai P ⊂ Q and bj Q ⊂ P for every nonzero i, j. In particular, for
every element g ∈ F2 that (in reduced form) begins and ends by a power of a, we
have g(P ) ⊂ Q.
By hypothesis, there is a sequence of nontrivial elements gn ∈ F2 converging
to the identity in PGL(2, R). We claim that we may take these elements to begin
and end with nontrivial powers of a (and hence gn (P ) ⊂ Q for every n). Indeed,
If gn begins or finishes by a power of b, then at least one of the following elements
agn a−1 or a−1 gn a begins and finishes by powers of a, and such a conjugate stays
close to the identity.
From the Lebesgue Density Theorem, it follows that

lim µ gn (P ) ∩ P = µ(P ),
n→∞

where µ stands for the Lebesgue measure on P1 (R). Since gn (P ) ⊂ Q and P
and Q are disjoint, we obtain that µ(P ) = 0. The same reasoning shows that
µ(Q) = 0. In particular, almost surely the function ψ takes the value 1/2, but
this contradicts the finiteness of the series (4.1).


Quite naturally, if a group Γ is amenable, then every regular action of Γ on
a measured space X is amenable. Indeed, if M is a mean on Γ, then the family
of means on the Γ-orbits defined by Mx (φ) := M(φ̃x ), where φ̃x (g) := φ(gx),
satisfies the desired properties. The only nontrivial property is (Invariance), but
this follows from the invariance of M under right multiplication together with
the equality φ̃hx = φ̃x ◦ Rh .
The previous discussion is quite useful to detect non-amenable groups. Before
stating the next result, we introduce some notation: We consider the action of
PGL(2, R) on P1 (R) ∼ R ∪ {∞}, where ∞ ∼ [1 : 0]. Given a subgroup Γ of
PGL(2, R) acting on P1 (R), we denote P(Γ) the subgroup of Homeo+ (R) formed
by the homeomorphisms that coincide with the restriction of an element of Γ on
each piece of a division of the real line into finitely many intervals. Observe that
Γ is not assumed to fix ∞ whereas P (Γ) fix it.
Proposition 4.2.2. Let Γ be a subgroup of PGL(2, R). Assume that Γ contains
a nontrivial translation x 7→ x + t. Then the orbit relations induced on R \ Γ∞
by both the actions of Γ and P(Γ) coincide.
Proof. After conjugating by a suitable affine map, we may assume that the
translation x 7→ x + 1 is contained in Γ. Let x, y be any pair of points not lying in
the Γ-orbit of ∞ ∈ P1 (R) but lying in the same Γ-orbit. Let g ∈ Γ, say g(z) = az+b
cz+d

4.2. NON-AMENABLE, LEFT-ORDERABLE GROUPS WITH NO FREE SUBGROUP165
(with ad − bc 6= 0), be such that y = g(x). We need to show that there exists
an element of P(Γ) sending x to y. If c = 0, then g already fixes ∞, and the
restriction of g to R belongs to P(Γ), hence we are done in this case. Assume now
that c 6= 0. For each n ∈ Z of large-enough modulus, the equation g(z) = z − n
has two solutions, namely
s
!
4c(dn + b)
a − d + cn
1± 1+
.
z± =
2c
(a − d + cn)2
These satisfy

d
z− →|n|→∞ − .
c
d
−1
Since x 6= − c = g (∞), by choosing n large enough or small enough according
to whether x > − dc or x < − dc , we see that x lies inside the interval I with
endpoints z− and z+ . Define ĝ(z) := g(z) if z belongs to I, and ĝ(z) := z − n if
not. Then ĝ is an element of P(Γ) which sends x to y, as desired.

z+ ∼|n|→∞ n

and

Corollary 4.2.3. If Γ ⊂ PGL(2, R) is a countable subgroup containing a nondiscrete copy of F2 and a nontrivial translation x 7→ x + t, then the group P(Γ)
is non-amenable and does not contain any copy of F2 .
Proof. Combine Theorem 1.2.13, Proposition 4.2.2 and Theorem 4.2.1.



Remark 4.2.4. Another result of [39] is that any countable dense subgroup of
PGL(2, R) contains a non-discrete copy of F2 . Therefore, the preceding corollary
as well as Theorem 4.2.1 can be extended by only demanding the density of Γ. It
should be noticed that, in [22], it is shown that if Γ is a countable dense subgroup
of a connected, real, semi-simple Lie group G (as for instance PGL(2, R)), then
Γ contains a copy of F2 which is also dense in G.

4.2.2

A finitely-presented version

In this section, we let Γ ⊂ PGL(2, R) be the group generated by

 √




0 1
1 1
2
0√
and c̃ =
, b̃ =
ã =
.
−1 0
0 1
0 1/ 2
Notice that the first two elements generate PSL(2, Z). By Theorem 1.2.13, P(Γ)
has no free subgroup. Using the results of the preceding section, we will show

166

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

that P(Γ) is non-amenable. Besides, following [128], will show that the group
P(Γ) contains a finitely-presented subgroup which is still non-amenable.
The action of P(Γ) on P1 (R) is non-amenable. This will follow from Theorem
4.2.1 provided we check that Γ contains a copy of F2 which is non-discrete in
PGL(2, R). To do this, fix n ≥ 1, and consider


0
1
= c̃2n b̃ã−1 c̃2n ∈ Γ.
g :=
−1 1/2n
Letting δ := 1/2n , we see that the eigenvalues of g are
√
δ i 4 − δ2
.
λ± = ±
2
2
Notice that g is an elliptic element of PGL(2, R) because λ+ λ− = 1 and both
λ− , λ+ have nontrivial imaginary part. Hence the projective action of g is conjugate to a rotation.
We claim that g acts hyperbolically on Q2 × Q2 , where Q2 denotes the 2adic rationals (see [118] for background). Indeed, to check this, we just need to
compute |λ± |2 , the 2-adic norm of λ± , and show that none of them is equal to 1.
To do this, we write
1
λ± = n+1 w± ,
2
√
where w± = 1±i 22n+2 − 1. Thus we need to show that |w± |2 6= 1/2n+1. Looking
for a contradiction, we assume that |w− |2 = |w+ |2 = 1/2n+1 . Then
√
w+ w+
w+
1 + i 22n+2 − 1
1=
=
+ 1 = 2n+1 + 1 .
2n+1
w− w+ 2
2
2
2
2
But for a ultrametric norm (such as the 2-adic norm) we have that |x| < |y| implies
2n+1
w+
w+
that |x − y| = |y|. Since 22n+1
= 22n+1 = 2n , this yields 1 = 22n+1
= 2n , which
2
2
is the desired contradiction.
Thus, g acts hyperbolically on Q2 × Q2 , hence it has infinite order. As a
consequence, g is conjugate to an irrational rotation on P1 (R), otherwise it would
have finite order. (As the angle of rotation θ of an element h ∈ PGL(2, R) satisfies
2 cos(θ) = ±tr(h), this shows that arccos(1/2n+1) is an irrational multiple of π
for each n ≥ 1.)
Now, since Γ is not a solvable group, we can find a conjugate f of g such that
f and g do not share any eigenvector (this is an easy exercise; see [111] in case of

4.2. NON-AMENABLE, LEFT-ORDERABLE GROUPS WITH NO FREE SUBGROUP167
problems). By a ping-pong argument applied to the action on Q2 × Q2 , it follows
that g and f generate a free group. This free group is not discrete in PGL(2, R)
since g is conjugate to an irrational rotation.
A finitely presented subgroup. This is the main novelty of [128]. We first
provide a crucial construction that will allow us to deal with P (Γ) in combinatorial
terms.
Exercise 4.2.5. The Hurwitz application is the map φ : {0, 1}N → [0, ∞] recursively
defined by
φ(0ξ) :=

1
1
1 + φ(ξ)

and

where ξ ∈ {0, 1}N .

φ(1ξ) := 1 + φ(ξ),

(i) Check that, for n1 ≥ 0 and all n2 , n3 , . . . positive integers,
φ(1n1 0n2 1n3 0n4 . . .) = n1 +

1

n2 +

1
n3 + n

.

1
4 +...

(ii) Show that φ is one-to-one, except at points in {0, 1}N that become eventually
constant yet are not constant. Check that these points map under φ to the rational
points in ]0, ∞[, and that the non-injectivity comes from that, for every finite sequence
s of 0’s and 1’s,
φ(s01) = φ(s10),
where 0 (resp. 1) stands for the constant sequence with all entries 0 (resp. 1).
(iii) Show that φ is increasing (resp. continuous) with respect to the lexicographic order
(resp. product topology) on {0, 1}N .
(iv) Let φ0 : {0, 1}N → [0, 1] be defined by φ0 (ξ) := φ(0ξ), and remind the map
φ2 : {0, 1}N → [0, 1] from Exercise 1.2.11 defined by
X ij
φ2 (ξ) =
,
where ξ = (i1 , i2 , . . .), ij ∈ {0, 1}.
2j
j≥1

Show that ψ := φ0 ◦ φ−1
2 : [0, 1] → [0, 1] is a well-defined homeomorphism that sends
bijectively the dyadic numbers into the rationals. More accurately, show that, if we let
0 = p0 /q0 < p1 /q1 < . . . pi /qi < . . . < p2n /q2n = 1 be the nth step Farey sequence of
rationals, then ψ sends each point i/2n into pi /qi .
Hint. Although the claim above can be proven by induction, a dynamical argument
proceeds as follows. Let H2 : [0, 1] → [0, 1] be the map defined by H2 (t) := {2t} for
t < 1 and H(1) = 1. Also, let H0 : [0, 1] → [0, 1] be defined by
 t
if 0 ≤ t < 21 ,
 1−t
H0 (t) :=
 2t−1
if 21 ≤ t ≤ 1.
t

168

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Show that both H0 and H1 are conjugate to the shift σ : {0, 1}N → {0, 1}N defined by
σ(i1 , i2 , i3 , . . .) := (i2 , i3 , . . .). More precisely, show that
H0 ◦ φ0 = φ0 ◦ σ,

H2 ◦ φ2 = φ2 ◦ σ.

(v) For each ξ ∈ {0, 1}N , let ξ¯ be the conjugate of ξ, which results from ξ by changing
all 0’s into 1’s and viceversa. Check that φ(ξ)φ(ξ̄) = 1 holds for all sequences ξ that
are not constant.
(vi) Let Φ : {0, 1}N → R ∩ {∞} = P1 (R) be defined by
Φ(0ξ) := −φ(ξ̄),

Φ(1ξ) := φ(ξ).

Translate the properties of φ above into analog properties of Φ.

by

Remind from Exercise 1.2.11 the homeomorphisms â and b̂ of {0, 1}N given


0η





10η
â(ξ) :=





11η

if ξ = 00η,
if ξ = 01η,

and

b̂(ξ) :=

if ξ = 1η,

Exercise 4.2.6. Define a, b in P (Γ) by letting

a(t) := ã(t) = t + 1,

and

b(t) :=


t







t




ξ








 10η

if ξ = 0η,
if ξ = 100η,



110η







111η

1−t




3 − 1t






t+1

if ξ = 101η,
if ξ = 11η.

if t ≤ 0,
if 0 ≤ t ≤ 21 ,
if

1
2

≤ t ≤ 1,

if t ≥ 1.

Check that Φ(â(ξ)) = a(Φ(ξ)) and Φ(b̂(ξ)) = b(Φ(ξ)) for all ξ ∈ {0, 1}N .

 √


2
0√
Observe that the element c̃ =
above corresponds to multipli0 1/ 2
cation by 2. It is hence crucial to encode the action of the map x → 2x in the
coordinates given by φ.

4.2. NON-AMENABLE, LEFT-ORDERABLE GROUPS WITH NO FREE SUBGROUP169
Exercise 4.2.7. Let ĉ : {0, 1}N → {0, 1}N be recursively defined by

0ĉ(η)
if ξ = 00η,





10ĉ−1 (η)
if ξ = 01η,
ĉ(ξ) :=





11ĉ(η)
if ξ = 1η.

Show that, for all ξ ∈ {0, 1}N ,

φ(ĉ(ξ)) = 2φ(ξ).

Hint. Check the equality above for sequences ξ that become eventually constant (to do
this, use an induction argument on the length of the largest finite subword of ξ before it
becomes constant). Show also that ĉ is a homeomorphism, and conclude by continuity
that the equality above holds for all ξ.

Similarly to Exercise 1.2.12, given a finite binary sequence s, we let âs (resp.
ĉs ) be the map that consists of the action of â (resp. ĉ) localized at the subtree
starting at the terminal vertex of the path s. In precise terms,


if ξ = sη,
if ξ = sη,
 sâ(η)
 sĉ(η)
âs (ξ) :=
ĉs (ξ) :=


ξ
otherwise,
ξ
otherwise.

Notice that b̂ = ĉ1 . Notice also that ĉs is conjugate to ĉ10 for each non-constant,
nonempty, binary sequence s.

Exercise 4.2.8. Let c be the piecewise projective homeomorphism of the real line
defined by
 2t
if t ∈ [0, 1],
 1+t
c(t) :=

t
otherwise.

Check that Φ(ĉ10 (ξ)) = c(Φ(ξ)) holds for all ξ ∈ {0, 1}N .

We denote by G0 the subgroup of P(Γ) generated by the elements a, b, c from
Exercises 4.2.6 and 4.2.8. Then G0 has no free subgroup. Moreover, by Proposition 4.2.2, the orbit equivalence relation of G0 coincides with that of P(Γ), hence
it is non-amenable, and therefore G0 is a non-amenable group. Our task now is
to give a finite presentation for G0 . To do this, we will deal with its isomorphic
version acting on {0, 1}N generated by â, b̂ and ĉ.
We start by noticing that the subgroup of G0 generated by â and b̂ = â1 is
isomorphic to Thompson’s group F (c.f. Exercise 1.2.11).

170

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Remark 4.2.9. The map ψ = φ0 ◦ φ−1
from Exercise 4.2.5 conjugates the standard
2
dyadic action of F into an action by piecewise projective homeomorphisms of [0, 1]. (The
latter was first constructed by Thurston, and it is well described in [38] and [150].) It
hence corresponds to the Ghys-Sergiescu conjugacy between these two actions; see [82].
It may be proved that ψ is totally singular with respect to the Lebesgue measure; see
for instance [62].

Remind from Exercise 1.2.12 that for finite binary sequences s, t, we let ât (s)
be the action of ât on s whenever it is defined. The main result of Lodha and
Moore [128] may be stated as follows.
Theorem 4.2.10. Consider G0 as a group generated by âs and ĉt , where s is
any (perhaps empty) sequence and t any nonempty, non-constant sequence. Then
the family of relations below provide a presentation of G0 with respect to these
generators:
(i) If ât (s) is defined, then ât âs = âât (s) ât ;
(ii) If ât (s) is defined, then ât ĉs = ĉât (s) ât ;
(iii) If s, t are (nonempty, non-constant and) incompatible, then ĉs ĉt = ĉt ĉs ;
(iv) For each s a nonempty, non-constant, binary sequence, ĉs = ĉs11 ĉ−1
s10 ĉs0 âs .
By Exercise 1.2.12, the first group of relations correspond to those of a presentation of F, and by Exercise 1.2.10, these may be summarized into
[â−1 b̂, âb̂â−1 ] = [â−1 b̂, â2 b̂â−2 ] = id,
where b̂ := â1 . Using conjugacy by elements in F, it is not hard to see that the
second group of relations may be reduced to
[â0 , ĉ10 ] = [â01 , ĉ10 ] = [â11 , ĉ10 ] = [â111 , ĉ10 ] = id.
By a similar procedure, the third group reduces to
[ĉ10 , ĉ01 ] = [ĉ10 , ĉ001 ] = id,
and the last group to the single non-commuting relation
ĉ10 = ĉ1011 ĉ−1
1010 ĉ100 â10 ,
thus providing a finite presentation.

4.2. NON-AMENABLE, LEFT-ORDERABLE GROUPS WITH NO FREE SUBGROUP171
Exercise 4.2.11. By rewriting the relations above in terms of a ∼ â, b ∼ â1 and
c ∼ ĉ10 , check that the list of relations above reduces to
[a−1 b, aba−1 ] = [a−1 b, a2 ba−2 ] = id,
[c, a−1 b−1 a2 ] = [c, a−1 b−1 ab−1 a−1 b2 a] = [c, aba−1 ] = [c, a2 ba−2 ] = id,
[a−1 ca, c] = [a−2 ca2 ] = id,
c = b−1 ab−1 aca−1 c−1 ba−1 cab−1 a−1 b2 ,
the last one being a simplification of
c = (b−1 ab−1 aca−1 ba−1 b)(b−1 ab−1 c−1 ba−1 b)(b−1 cb)(b−1 ab−1 a−1 b2 ).

The relations in Theorem 4.2.10 are easy to check. Those of type (i) and (ii)
arise by conjugacy. Commutativity in case (iii) holds because ĉs and ĉt correspond
to maps with disjoint supports whenever s and t are incompatible. Finally, relations of type (iv) reduce by conjugacy to ĉ10 = ĉ1011 ĉ−1
1010 ĉ100 â10 , which is straightforward to check. (Notice that, by the definition of ĉ, we have ĉ = ĉ0 ĉ−1
10 ĉ11 â.) The
goal now is to provide the main ideas of the proof of Theorem 4.2.10. Although
we will leave this as exercises, we will illustrate most of the crucial steps with
examples. We refer to [128] in case of problems with the formal arguments.
We first consider words representing elements in G0 made of letters of the
form âs and ĉt , where s is arbitrary (perhaps empty) and t is nonempty and
non-constant. We say that such a word is standard if, from right to left, it is
the concatenation of a word on the âs and a word on the ĉt , and whenever both
ĉt1 and ĉt2 occur with nontrivial exponents for certain t2 ( t1 , the first of these
appears before the other one. The depth of such a word is the smallest ℓ for which
there is some ĉt appearing in it satisfying length(t) = ℓ. (In case no ĉt appears,
the depth is defined to be infinite.) Two words are said to be equivalent if it is
possible to derive one of them from the other one by applying the relations listed
in Theorem 4.2.10.
Exercise 4.2.12. For most of the claims below, use an inductive procedure for the
proof.
(i) Prove that for every nonempty, non-constant, binary sequence s and each ℓ ≥ 1,
there are standard words W1 , W2 equivalent to ĉs and ĉ−1
s , respectively, such that:
– If ât occurs in Wi , then t entends s;
– If ĉt occurs in Wi , then t extends s, has length ≥ ℓ, and the exponent of ĉt is ±1;
– If ĉt1 and ĉt2 occur in Wi for t1 6= t2 , then t1 and t2 are incompatible.

172

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Hint. Use the relations of type (iv) above and their inverse versions, namely:
−1
−1
−1 −1
−1 −1
ĉ−1
s = âs ĉs0 ĉs10 ĉs11 = ĉs00 ĉs01 ĉs1 âs .

For instance, for s = 10 and ℓ = 5, we have:
ĉ10 = ĉ1011 ĉ−1
1010 ĉ100 â10
−1
−1
−1
−1
= (ĉ101111 ĉ−1
101110 ĉ10110 â1011 )(ĉ101000 ĉ101001 ĉ10101 â1010 )(ĉ10011 ĉ10010 ĉ1000 â100 )â10
−1
−1
−1
−1
= ĉ101111 ĉ−1
101110 ĉ10110 ĉ101000 ĉ101001 ĉ10101 ĉ10011 ĉ10010 ĉ1000 â1011 â1010 â100 â10
−1
−1
−1
−1
−1
= ĉ101111 ĉ−1
101110 ĉ10110 ĉ101000 ĉ101001 ĉ10101 ĉ10011 ĉ10010 ĉ100011 ĉ100010 ĉ10000 â1000 â1011 â1010 â100 â10 .

(ii) Given a word W in the âs , show that there exists ℓ ≥ 1 such that for any standard
word W ′ of depth ≥ ℓ, the word W W ′ is equivalent to a standard word of depth ≥ ℓ−k,
where k is the word-length of W ′ .
Hint. Use relations of type (ii).
(iii) Using (i) and (ii), prove that for every ℓ ≥ 1, each word in the âs , ĉt (with t
nonempty and non-constant) is equivalent to a standard word.

If W is a standard word and ĉs occurs in W , we say that s is exposed if there
is an infinite binary path starting at s along which no yt occurs in W . The word
is sufficiently expanded if, whenever ĉs occurs in W and s is not exposed, we have
that:
– ĉs0 occurs in W if ĉs occurs with a positive exponent;
– ĉs1 occurs in W if ĉs occurs with a negative exponent.
Exercise 4.2.13. Show that every standard word is equivalent to one which is sufficiently expanded.
Hint. Use the relations of type (iv) and their inverse versions. For instance, this yields
−1
2
2
ĉ30 ĉ001 â1 = ĉ20 ĉ00 ĉ−1
010 ĉ011 â0 ĉ001 â1 = ĉ0 ĉ00 ĉ010 ĉ011 ĉ010 â0 â1 = ĉ0 ĉ00 ĉ011 â0 â,

where the left-side expression is not sufficiently expanded yet the right-side one is.

For a word Λ in the alphabet {0, 1, ĉ, ĉ−1 }, we define the process of advancing
the occurrence of a certain ĉ±1 the result of replacing
ĉ00 → 0ĉ,

ĉ01 → 10ĉ−1 ,

ĉ1 → 11ĉ,

ĉ−1 0 → 00ĉ−1 ,

ĉ−1 10 → 01ĉ,

ĉ−1 11 → 1ĉ−1 .

The resulting word Λ′ after advancing this occurrence of ĉ±1 several times is said
to be an advanced version of Λ. A potencial cancellation in Λ is a concatenation
of the form ĉĉ−1 or ĉ−1 ĉ obtained in an advanced version of Λ.

4.3. ALMOST-PERIODICITY

173

Exercise 4.2.14. Let Λ be a word in the alphabet {0, 1, ĉ, ĉ−1 } having no potential
cancellation.
(i) Prove that no advanced version of Λ contains potential cancellations.
(ii) Prove that there exist binary, finite sequences s, t such that Λs can be transformed
into tĉn by advancing all occurrences of ĉ±1 (n is the number of these occurrences).
Exercise 4.2.15. Let W be a nontrivial, sufficiently expanded word on ĉs , where s
ranges along nonempty, non-constant, binary sequences. The goal is to define finite
binary sequences u, v such that for all ξ ∈ {0, 1}N , the image of uξ under W is vĉn (ξ)
for some n > 0.
n
n
(i) Assume the existence of u, v as above. Show that W sends u02 102 1 . . . into
n
n
v012 012 . . . As these two points have non-equivalent tails, conclude that W does
not represent the identity.
(ii) Let yt0 be the last entry (from right to left) in W . In case t0 is exposed, let u
be any path extending t0 that points to infinite without passing through the ti ’s for
which yti appears in W . If t0 is not exposed, extend it as follows: if yt0 appears with
a positive exponent, add a 0 to the right, and a 1 otherwise; continue the procedure
until getting an exposed index, and conclude the construction of a path u0 . Let Λ be
the word on {0, 1, ĉ, ĉ−1 } obtained by inserting ĉn just after s whenever ĉns appears in
W . By Exercise 4.2.14, there exist s, t such that Λs can be transformed into tĉn by the
procedure of advancing. Show that the claim holds for u := u0 s and v := t.
Hint. First notice that Λ contains no potential cancellation, and then check that applying W to uξ corresponds to advancing the word Λu.

The conclusion of the proof of Theorem 4.2.10 is now at hand. Namely,
every word in âs , ĉt can be transformed into a sufficiently expanded standard one
using the relations of type (i), (ii), (iii), (iv) above. If we assume that such a
word represents the identity, then its part in the generators ĉt must be the trivial
word, otherwise by Exercise 4.2.15 there would be a point being sent into another
one with a non-equivalent tail, which is impossible. As a consequence, W is a
word in the generators âs . Now, every such word representing the identity can
be transformed into the trivial word using the relations of type (i), as these give
a presentation of F ∼ hâs i.

4.3

Almost-Periodicity

In this section, we develop the notion of almost-periodicity for group actions
on the real line. A left-orderable group being given, the set of such actions
equipped with the compact-open topology can be used as a substitute to the

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

space of left-orders. As an example, we will show how this yields an alternative
proof of Theorem 4.1.3 which does not rely on the theory of Conradian orders.

4.3.1

Almost-periodic representations

The group of orientation-preserving homeomorphisms of the real line is equipped
with the compact-open topology, which turns it a topological group. A homeomorphism h ∈ Homeo+ (R) is said to be almost-periodic if the set
{τs−1 ◦ h ◦ τs | s ∈ R}
is relatively compact in Homeo+ (R), where τs (t) := s + t. The set of almostperiodic, orientation-preserving homeomorphisms of the line is denoted AP H+ (R).
Example 4.3.1. Certainly, every homeomorphism that is periodic (i.e. that commutes
with a nontrivial translation) is almost-periodic. An example of an almost-periodic
homeomorphism that is non-periodic is given by
ϕ(t) := t +

√ 
1
sin(t) + sin( 2t) .
3

(4.2)

The fact that ϕ is almost-periodic can be checked directly, yet a more conceptual
argument proceeds as follows: Consider the diffeomorphism of the 2-torus defined by
!
√


2
1
sin(x) + sin(y) , y +
sin(x) + sin(y) .
F (x, y) = x +
3
3
This diffeomorphism preserves the orbits
√ ∂ of the (irrational) flow SR = {Ss }s∈R gener∂
ated by the linear vector field ∂x
, 2 ∂y
. Moreover, we have F (St (x0 )) = Sϕ(t) (x0 )
for every t ∈ R, where x0 := (0, 0). From this and Proposition 4.3.3 below, it follows
that ϕ is almost-periodic.

Lemma 4.3.2. The subset AP H+ (R) is a subgroup of Homeo+ (R).
Proof. This is a consequence of the continuity of the composition and inverse
operations on Homeo+ (R) with respect to the compact-open topology.

A group action on the real line whose image is contained in AP H+ (R) will
be said to be almost-periodic. There are several ways to construct faithful
almost-periodic actions of a given left-orderable, countable group Γ on the line.
The simplest one consists in considering a faithful action on the interval and then

4.3. ALMOST-PERIODICITY

175

to extend it to the whole line so that it commutes with the translation t 7→ t + 1.
This somewhat trivial construction shows that AP H+ (R) contains a copy of every
left-orderable, countable group. Nevertheless, at this point we should stress that,
in order to carry out a study that also involves actions that appear as limits of
conjugates of a given one, we are forced to consider actions that may be unfaithful.
This is the reason why we use the notation Φ : Γ → Homeo+ (R) in what follows.
Starting with an almost-periodic action of a group Γ on the real line, we next
provide a compact one-dimensional foliated space together with a Γ-action on it
that preserves the leaves. It is this construction which gives interest to considering
almost-periodic actions.
Proposition 4.3.3. Let Γ be a finitely-generated group and Φ0 : Γ → Homeo+ (R)
an action by orientation-preserving homeomorphisms of the line. Then Φ0 is
almost-periodic if and only if there exists a topological flow SR = {Ss }s∈R acting
freely on a compact space X, an action of Γ on X by homeomorphisms preserving
every S-orbit together with its orientation, and a point x0 ∈ X, such that for every
g ∈ Γ and every t ∈ R,

g St (x0 ) = SΦ0 (g)(t) (x0 ).
(4.3)
Moreover, the flow can be taken so that the S-orbit of x0 is dense in X.

Proof. Let us first show that if there is a compact space X together with a flow
SR and a Γ-action verifying (4.3), then the action Φ0 is almost-periodic. Indeed,
for each x ∈ X, we can lift the Γ-action on X to an action Φx : Γ → Homeo+ (R)
verifying

g St (x) = SΦx (g)(t) (x),

which is well-defined since the flow SR is free. Moreover, as the Γ-action on
X is by homeomorphisms, for every g ∈ Γ, the map x 7→ Φx (g) from X into
Homeo+ (R) is continuous. Hence, the set of elements Φx (g), where x ∈ X, is
compact. Now, for every s, t in R and every x ∈ X, we have


g St (Ss (x)) = g St+s (x) = SΦx (g)(t+s) (x) = SΦx (g)(t+s)−s (Ss (x)),
which yields

ΦSs (x) (g) = τ−s ◦ Φx (g) ◦ τs .
Therefore, for every g ∈ Γ, the conjugates of Φx (g) by the translations τs stay
in a compact set, which proves that Φx is almost-periodic for every x ∈ X. In
particular, for x = x0 , we deduce that Φ0 = Φx0 is almost-periodic.

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Conversely, let us start with an almost-periodic action Φ0 , and let us provide
the compact space X together with the flow S = SR and the Γ-action verifying (4.3). Denote by AP A+ (Γ) the set of almost-periodic actions of Γ on the real
line. This can be seen as a closed subset of AP H+ (R)G , where G is a finite generating set of Γ. Define the translation flow SR of conjugacies by translations
on AP A+ (Γ), namely
Ss (Φ)(g) := τ−s ◦ Φ(g) ◦ τs ,
(4.4)

where Φ ∈ AP A+ (Γ) and g ∈ Γ. This is a topological flow acting on AP A+ (Γ).
Denote by X the closure of the S-orbit of Φ0 . This is a compact S-invariant
subset of AP A+ (Γ), since Φ0 is almost-periodic.
We claim that the formula
g(Φ) := τ−Φ(g)(0) ◦ Φ ◦ τΦ(g)(0)

(4.5)

defines an action of Γ on AP A+ (Γ). One can verify this by a tedious computation,
but a more conceptual argument proceeds as follows. Consider the actions of R
and Γ on the product space AP A+ (Γ) × R given by


s(Φ, t) := Ts (Φ), t − s
and g(Φ, t) = Φ, Φ(g)(t) .

A point in AP A+ (Γ) × R can be thought of as an almost-periodic action of Γ
together with a marker. The action of the reals on AP A+ (Γ) × R corresponds
to translating the marker while conjugating the almost-periodic action by the
same translation. The Γ-action on AP A+ (Γ) × R corresponds to acting on the
marker using the action of the first coordinate while leaving the almost-periodic
representation unchanged. An easy computation shows that these two actions
commute. Hence, there is a natural action of Γ on the quotient of AP A+ (Γ) × R
by R, which naturally identifies with AP A+ (Γ) via the embedding
Φ ∈ AP A+ (Γ) 7→ (Φ, 0) ∈ AP A+ (Γ) × R.
The action of Γ on AP A+ (Γ) induced by this idenfication is given by the formula (4.5).
A priori, there is no reason to expect for the flow SR on X to be free. However,
it is possible to change it into a free one by the following procedure: Let Y be any
compact space endowed with a topological flow SR acting freely. (For instance,
the toral flow of Example 4.3.1; see also Exercise 4.3.4 below.) Consider the space
X̃ := X × Y together with the Γ-action on it defined as

g : (Φ, y) 7→ g(Φ), SΦ(g)(0) (y)

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4.3. ALMOST-PERIODICITY
and the (diagonal) flow

s : (Φ, y) 7→ Ss (Φ), Ss (y) .

Then all the properties above are still satisfied, and moreover the flow on X̃ is
now free. Hence, we can (and we will) assume that the flow SR on X is free.
Equation (4.3) is obvious from the construction, as well as the fact that X
can be taken as being the closure of a single point. This closes the proof.

Exercise 4.3.4. A Delone set D in R is a subset that is discrete and almost dense in
a uniform way. More concretely, there exist positive constants ε, δ such that |x − y| ≥ ε
for all x 6= y in D, and for all z ∈ R there is x ∈ D such that |x − y| ≤ δ. Two Delone
sets are close if they coincide over a very large interval centered at the origin (this
induces a topology that is metrizable). A Delone set is repetitive if for all r > 0 there
is R = R(r) > 0 such that for every pair of intervals I, J of length r, R, respectively,
there is a translated copy of D ∩ I contained in D ∩ J.
Assume that D0 is a repetitive yet non-periodic Delone set in R (it is easy to build
such sets). Show that the natural translation flow St : D 7→ D + t, restricted to the
closure of the orbit of D0 , is a minimal flow.

4.3.2

A bi-Lipschitz conjugacy theorem

We denote by BiLip+ (R) the group of orientation-preserving, bi-Lipschitz
homeomorphisms of the real line. For every h ∈ BiLip+ (R), we let L(h) be its
bi-Lipschitz constant, that is, the minimum of the numbers L ≥ 1 such that
L−1 |y − x| ≤ |h(y) − h(x)| ≤ L|y − x|

for all x, y in R.

(4.6)

We equip BiLip+ (R) with the topology of uniform convergence on compact sets.
Theorem 4.3.5. Every finitely-generated group of homeomorphisms of the real
line is topologically conjugate to a group of bi-Lipschitz homeomorphisms.
Proof. Let ν = λ(t)dt be a probability measure on R with a smooth, positive
density λ such that for |t| large enough, we have λ(t) = 1/t2 . The following
observation will be central in what follows: If for some constant L ≥ 1, a homeomorphism h of the real line satisfies
h∗ (ν) ≤ Lν

and

(h−1 )∗ (ν) ≤ Lν,

(4.7)


then h is Lipschitz. To prove this fact, first notice that ν [t,+∞) = 1/t for
all large-enough positive numbers t (and similarly, ν (−∞, t] = 1/|t| if |t| is

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

large-enough and t is negative). Thus, the left-side inequality in (4.7) shows that
|h(t)| ≤ L|t| holds for |t| large enough. The density of (h−1 )∗ (ν) being
 given by
Dh(t)λ(h(t)), the right-side inequality in (4.7) yields Dh(t) ≤ Lλ(t) λ(h(t)) for
almost-every t. Thus, up to sets of zero Lebesgue measure, the derivative Dh is
bounded on every compact interval, and for |t| large enough, we have
Dh(t) ≤

Lλ(t)
L|h(t)|2
=
≤ L3 .
λ(h(t))
|t|2

This proves that Dh is a.e. bounded, hence h is Lipschitz, with Lipschitz constant
at most L3 .
Next, let Γ be a finitely-generated subgroup of Homeo+ (R), and let G be
a finite, symmetric system of generators of Γ. Let φ ∈ L1 (Γ) be a function
taking positive values such that, for every h ∈ G, there is a constant Lh satisfying
φ(hg) ≤ Lh φ(g) for all g ∈ Γ. For instance, one can take φ(g) = κ||g||, where kgk
is the word-length of g with respect to G and κ a small-enough positive number.
(For κ < 1/|G|, one can ensure that φ belongs toPL1 (Γ); see also Excercise 4.3.7
below.) Let us normalize the function φ so that g∈Γ φ(g) = 1, and let us define
the probability measure ν0 on R by letting
X
ν0 :=
φ(g) g∗(ν).
g∈Γ

Notice that for each h ∈ Γ, we have
X
h∗ (ν0 ) =
φ(g) (hg)∗(ν) ≤ Lh ν0 .
g∈Γ

The measure ν0 has no atoms and is of full support (i.e. the ν0 -measure of
every nonempty open set is positive). Thus, there exists a homeomorphism ϕ of
the real line sending ν0 into ν. For each h ∈ Γ, we have
(ϕ ◦ h ◦ ϕ−1 )∗ (ν) = ϕ∗ h∗ (ν0 ) ≤ Lh ϕ∗ (ν0 ) = Lh ν.

From the discussion at the beginning of the proof, we deduce that the conjugate
of Γ by ϕ is contained in BiLip+ (R).

The proof above was taken from [60]. In §4.4, we will give a more conceptual
(yet quite elaborate) proof based on probabilistic arguments. For analogous results for transverse pseudo-groups of codimension-one foliations or groups acting
on the circle, see [59, Proposition 2.5] and [61, Théorème D]. We point out that
a conjugacy into a group of C 1 diffeomorphisms is, in general, impossible; see
[16, 152].

4.3. ALMOST-PERIODICITY

179

Exercise 4.3.6. Let Γ be a subgroup of BiLip+ ([0, 1]), and let ϕ : [0, 1] → R be an
orientation-preserving homeomorphism such that ϕ(x) = −1/x for x close to zero, and
ϕ(x) = 1/(1 − x) for x close to 1. Check that the conjugate of Γ by ϕ is a subgroup of
BiLip+ (R).
Exercise 4.3.7. Assume that a subgroup Γ of Homeo+ (R) has subexponential growth,
that is, the number of elements in the ball of radius n with respect to a finite generating system growth subexponentially in n. (This condition does not depend on the
generating set.) Show that, for every L > 1, it is possible to simultaneously conjugate
the generators of Γ into L-bi-Lipschitz homeomorphisms.
Hint. In the proof above, the (positive) function φ ∈ L1 (Γ) can be taken so that
φ(hg) ≤ L1/3 φ(g) for every h ∈ G and every g ∈ Γ. See [149] for more on this.

We should stress that there is no analogue of Theorem 4.3.5 in higher dimension, even for actions of (infinite) cyclic groups; see [94].
Exercise 4.3.8. Let D be a Delone subset of R (c.f. Exercise 4.3.4). Show that there
exists a bi-Lipschitz homeomorphism of the real line that sends D onto Z. (Again,
there is no two-dimensional analogue of this fact; see [28, 137] as well as [52].)

4.3.3

Actions almost having fixed points

Let Γ be a finitely-generated group with finite generating set G, and let Φ
be an almost-periodic action of Γ on R. (Recall that we do not assume Φ to be
faithful.) We say that Φ almost has fixed points if
inf sup Φ(g)(t) − t = 0.

t∈R g∈G

An equivalent way to think about this property is that the action of Γ on the
compact space constructed in Proposition 4.3.3 has a global fixed point in the
closure of the S-orbit of Φ0 .
It is not obvious how to construct almost-periodic actions that do not almost
have fixed points. (Consider, for instance, the case of affine groups.) This is the
goal of the next
Theorem 4.3.9. All actions of finitely-generated groups by orientation-preserving
homeomorphisms of the real line are topologically conjugate to almost-periodic actions. Moreover, if the original action has no fixed point, then there is such a
conjugate that does not almost have fixed points.

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

To prove this result, let Γ be a finitely-generated group provided with a finite,
symmetric system of generators G. Given constants L > 1 and D > D ′ > 0, we
denote R = R(Γ, G, L, D, D ′) the set of representations Φ : Γ → BiLip+ (R) such
that every g ∈ G satisfies L(Φ(g)) ≤ L and
t − D ≤ min Φ(g)(t) ≤ t − D ′ ≤ t + D ′ ≤ max Φ(g)(t) ≤ t + D
g∈G

g∈G

(4.8)

for all t ∈ R. This set can be seen as a closed subset of BiLip+ (R)G , and as such
is equipped with the product topology. Relations (4.6) and (4.8) imply that R
is compact, by Arzela-Ascoli’s theorem. Moreover, the same relations show that
the translation flow SR defined by (4.4) preserves R. Hence, every element of R is
an almost-periodic action of Γ, and (4.8) shows that, moreover, such an element
does not almost have fixed points.
Lemma 4.3.10. Let Φ0 : Γ → Homeo+ (R) be a faithful action without global
fixed points of a finitely-generated group Γ. Then there are constants L > 1 and
D > D ′ > 0, as well as a finite, symmetric generating system of Γ, such that the
corresponding set R contains a representation that is conjugate to Φ0 .
Proof. By Theorem 4.3.5, it is enough to prove the statement in the case where
Γ is a subgroup of BiLip+ (R). We see Γ as being contained in Homeo+ (R) via
Φ0 . Let G be a finite, symmetric generating set of Γ, and let L be a constant
such that every g ∈ G is L-bi-Lipschitz. Let (tn )n∈Z be the sequence of points in
R defined by t0 := 0 and tn+1 := maxg∈G g(tn ). Equivalently, tn−1 = ming∈G g(tn ),
as G is symmetric. Since Γ has no fixed point on the real line, we have
lim tn = ±∞.

n→±∞

Let ϕ be the homeomorphism of the real line that sends tn to n and that is affine on
each interval [tn , tn+1 ]. We claim that the action of Γ defined by Φ(g) := ϕ◦g◦ϕ−1
belongs to R(Γ, G, L6 , 1, 4) for the generating set G := G ∪ G 2 .
To prove this, we first notice that the distortion of the sequence tn is uniformly
bounded. In concrete terms, if for each n ∈ Z we denote δn := tn+1 − tn , then
L−1 δn+1 ≤ δn ≤ L δn+1 .

(4.9)

Indeed, let gn ∈ G be such that tn+1 = gn (tn ). By definition, gn (tn+1 ) ≤ tn+2 ,
and since gn is an L-bi-Lipschitz map, we have
tn+2 − tn+1 ≥ gn (tn+1 ) − gn (tn ) ≥ L−1 (tn+1 − tn ),

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4.3. ALMOST-PERIODICITY

which yields the right-side inequality in (4.9). (The left-side inequality is obtained
analogously.) Notice that, by construction, for every all w, z in [tn , tn+1 ],
|z − w|
|z − w|
≤ ϕ(z) − ϕ(w) ≤
.
δn
δn

(4.10)

We next claim that for every g ∈ G, we have L(Φ(g)) ≤ L3 . To show this,
it suffices to prove that each such Φ(g) is Lipschitz on every interval [n, n + 1],
with Lipschitz constant smaller than or equal to L3 . To check this, consider two
arbitrary points x, y in [n, n + 1], and define w := ϕ−1 (x) and z := ϕ−1 (y). Then
w, z both belong to [tn , tn+1 ], which in virtue of (4.9) and (4.10) yields
Φ(g)(y) − Φ(g)(x) = ϕ(g(z)) − ϕ(g(w)) ≤

L2 z − w
L g(z) − g(w)
≤
≤ L3 |y − x|,
δn
δn

as desired.
By construction, for every generator g ∈ G and all x ∈ R,
x − 2 ≤ Φ(g)(x) − x ≤ x + 2.
Indeed, the integer points just after and before x are moved a distance less than
or equal to 1 by Φ(g). Moreover, as for every n ∈ Z we have Φ(gn+1 gn )(n) = n+2,
taking n being the integer part of x, this yields Φ(gn+1 gn )(x) ≥ x + 1. We have
hence proved that Φ belongs to R(Γ, G, L6 , 1, 4).

Theorem 4.3.9 immediately follows from the preceding lemma in the case
where Γ has no global fixed point. In the case where such a point exists, we
replace Γ by the free product Γ ⋆ Z (or a quotient of it) and we extend Φ0 so that
the generator the Z-factor is mapped to a nontrivial translation. This new group
has no fixed point, so that the preceding lemma applies to it, hence to Γ.
A substitute to the space of left-orders. The previous construction allows
replacing the space of left-orders of a given left-orderable group by an object
provided with a flow which is more natural when dealing with dynamical realizations.
Corollary 4.3.11. Let Γ be a finitely-generated, left-orderable group. Then there
exists a compact space X, a free flow SR on X, and an action of Γ on X without
global fixed points which preserves the S-orbits together with their orientations.
Proof. Since Γ is finitely-generated and left-orderable, it admits a faithful action
by orientation-preserving homeomorphisms of the real line without global fixed

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

point (c.f. §1.1.3). By Theorem 4.3.9, this action is conjugate to an almostperiodic action Φ0 that does not almost have fixed points. Consider the space X
constructed in (the proof of) Proposition 4.3.3 together with the free flow S = SR
and the Γ-action on it. Because Φ0 does not almost have fixed points and SR (Φ0 )
is dense in X, there is no fixed point for the Γ-action on X. Moreover, Γ stabilizes
every S-orbit, and preserves the orientation on each of them.


4.3.4

Indicability of amenable left-orderable groups revisited

Based on the previous construction, and following [60], we next give an alternative proof of Theorem 4.1.3. Let Γ be a finitely-generated, left-orderable group,
and let X be a compact space equipped with a free action SR and an action of
Γ, as described by Corollary 4.3.11. If Γ is amenable, then there exists a probability measure µ on X that is invariant by Γ. Consider the conditional measures
of µ along the orbits of the translation flow SR . These are Radon measures on
µ-almost every S-orbit that are well-defined up to multiplication by a positive
constant. We denote by µl this Radon measure on a S-orbit l. More precisely, in
a flow box [0, 1] × Λ where the flow S is given by the formula Ss (t, l) = (t + s, λ),
we disintegrate the measure µ as
µ(dt, dl) = µl (dt) µ(dλ)
where µ is the image of µ under the projection [0, 1] × Λ → Λ into the transversal
Λ and the measures µl are measures on the unit interval; see [177, Section 3].
The measures µl depend on the flow box; however, they are well-defined up to a
positive constant on almost-every orbit.
Because Γ preserves µ and is countable, for µ-almost-every S-orbit l in X, the
measure µl is nonzero, and every g ∈ Γ multiplies it by a certain factor:
g∗ (µl ) = cl (g)µl ,

where cl (g) > 0.

If µl is not preserved by Γ, then the map g 7→ log cl (g) is a nontrivial homomorphism from Γ into (R, +). (See also Remark 4.3.12 below.) Otherwise, µl is
preserved by Γ. If µl has an atom, then its orbit must be discrete, and Γ acts by
translations along this orbit, thus giving rise to a nontrivial homomorphism into
the integers. If µl has no atom, then the Γ-action on l is semiconjugate to an
action by translations, which induces a nontrivial homomorphism into the reals.

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

183

Remark 4.3.12. It seems that the the condition µl = cl (g)µl for a function cl that is
not identically equal to 1 cannot arise in the proof above. Actually, this can be proven
under the extra hypothesis that there is no discrete orbit for the action of Γ on l (which
is identified to the real line). Indeed, in this case, it is not hard (yet not straightforward)
to prove that this action is semiconjugate to that of an affine group. We claim that the
image group must be Abelian for a generic leaf. To see this, assume the image group
is non-Abelian. Then there is a resilient pair u < f (u) < f (v) < g(u) < g(v) < v for the
action on l after this semiconjugacy (where the order is given by the orientation of l).
With no loss of generality, we may assume that the origin 0 belongs to the region of
crossing; let Φ be the associated representation. Besides, there is an element h ∈ Γ
whose inverse (and all its iterates) sends [u, v] into a disjoint interval (hence to a region
where no crossing for f, g arises; c.f. §3.2.2). As a consequence, for all n ∈ N, the
conjugate representations hn (Φ) remain outside a certain neighborhood of Φ. However,
this is in contradiction with Poincaré’s recurrence, hence cannot arise for a µ-generic
leaf l. (We refer to Examples 4.4.16 and 4.4.18 for another application of this idea.)

4.4

Random Walks on Left-Orderable Groups

In what follows, we provide a more conceptual proof of the existence of almostperiodic actions for left-orderable groups based on probabilistic arguments. Along
this section, Γ will denote a finitely-generated group and ρ a probability measure
on Γ whose support is finite, generates Γ, and is symmetric, in the sense that
ρ(g) = ρ(g −1 ) for all g ∈ Γ.
We start with an emphasis on a particular type of actions, namely, those
for which the Lebesgue measure is stationary, i.e. invariant in mean. These
actions are called ρ-harmonic. Their basic properties are listed in the next two
subsections. In particular, it is proven that they are almost-periodic. The main
result is proven in the third and fourth subsections. It states that, under suitable
conjugacies, all actions on R become harmonic, and the conjugacy is unique up
to post-composition with an affine map.

4.4.1

Harmonic actions and Derriennic’s property

Let Γ be a subgroup of Homeo+ (R) having no global fixed point in the line.
The action of Γ is said to be ρ-harmonic (or just harmonic, if the probability ρ
is clear from the context) if the Lebesgue measure is stationary, that is, if for

184

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

every x, y in R,
y−x=

Z

Γ

X


g(y) − g(x) dρ(g) =
g(y) − g(x) ρ(g).

(4.11)

g∈G

Actually, no group action on the line satisfying this property can have a global
fixed point; see Exercise 4.4.8.
Obviously, ρ-harmonic actions include those that satisfy, for every x ∈ R,
Z
x = g(x) dρ(g).
Γ

This will be called the Derriennic property , as it corresponds to a weak form
of a property studied by Derriennic in [64] in the more general context of Markov
processes on the line (not necessarily coming from a group action). Quite suprisingly, all ρ-harmonic actions do satisfy this property.
Proposition 4.4.1. Every ρ-harmonic action has the Derriennic property.
For the proof, we need the following
Lemma 4.4.2. For all h ∈ Homeo+ (R) and each compact interval [a, b], we have
Z b


(h(x) − x) + (h−1 (x) − x) dx = ∆h (b) − ∆h (a),
(4.12)
a

where ∆h (x) is the non-signed area of the region depicted in Figure 19:


 Rc
if h(c) ≥ c,
 h−1 (c) h(s) − c ds,
h
∆ (c) :=

 R c  −1
h
(s)
−
c
ds,
if h(c) ≤ c.
h(c)

Proof. Denoting |A| the Lebesgue measure of a subset A ⊂ R2 , we have that
Rb
(h(x) − x) dx equals
a


(x, y) : a < x < b, x < y < h(x) − (x, y) : a < x < b, h(x) < y < x ,
which may be rewritten as



(x, y) : a < x < b, b < y < h(x) + (x, y) : a < x < b, a < y < b, x < y < h(x)


− (x, y) : a < x < b, h(x) < y < a − (x, y) : a < x < b, a < y < b, h(x) < y < x .

185

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS
A similar equality holds for h−1 . Now, in the sum
Z b
Z b
(h(x) − x) dx +
(h−1 (x) − x) dx,
a

a

the corresponding second and fourth terms above cancel each other. Indeed, these
terms involve all couples (x, y) ∈ [a, b]2 , and we have x < y < h(x) if and only if
h−1 (y) < x < y. Therefore, the second term for h is exactly the negative of the
fourth term for h−1 , and viceversa.
As a consequence, the value of
Z b


(h(x) − x) + (h−1 (x) − x) dx
a

equals

(x, y) : a < x < b, b < y < h(x)

− (x, y) : a < x < b, h(x) < y < a


(x, y) : a < x < b, b < y < h−1 (x)

− (x, y) : a < x < b, h−1 (x) < y < a ,

+

and one can easily check that the expressions above and below are equal to ∆h (b)
and ∆h (a), respectively. This proves the desired equality.

h(c) ... . . . . . . . . . . . . . . ........................
c

....
..... . . . . . .
........ .. .. .. .. .. .. ..
.
.
.
....
..... . . . . . . . . . . .
.... . . . . . . . . . . . . .
.
.
.
.
.
.
.
. . . . . . . . . .....
.
...
..
.... .
....
..... .
.
.
....
.
.
.
.
.
.
.
.
.
.
..
....
.
....
....
.
....
....
....
.
.
.
.
.
.
.
.
.
. .......
....
....
....
....
.....
.
.
.
. .
.
.
.
.
..... ..
....
.
....
.
.
.
.
....
.
....
.
.
.
.
.
.....
.
.
.
.
.
.
.

−→ ∆h (c)

h

h−1 (c)

−→ ∆ (c)

h(c)

c

....
...
.... .. ..
.
.
h
...
... . . . .
..... .. .. .. .. ..
.
.
.
.... . . . . ....... . . . . . ........
...
.
.... .
.. . ....
....... .
.. .. .. .......
........
.
.
.
.
.
.
.
.
.
. . . . . . . . . . ...... . ...... . . ...............
.
..
..
.
−1..... ........ .....................
.
.
.. ..... ...........
............................. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..........
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
............ ............
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. ..
.......
.
.
.
.
.
. ..
.
.
.
. ...
.
.
.
.
.
.
.
..
....
.
.
.
.
.
.
.
.
..
.....
. . . . . . ..
.

h

h

h(c) c

h−1 (c)

Figure 19: The definition of ∆h (c) in the two possible cases.
Proof of Proposition 4.4.1. First notice that, by ρ-harmonicity, the value of
Z

g(x) − x dρ(g)
Γ

186

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

is independent of x. We call it the drift of the action and denote it by Dr(Γ, ρ).
The statement to be proved is hence equivalent to the vanishing of the drift. To
show this, we integrate (4.12) over Γ and use the symmetry of ρ to obtain, for all
a < b,
Z

2(b − a)Dr(Γ, ρ) = (∆g (b) − ∆g (a) dρ(g).
Denoting now ∆ρ (c) :=

R

Γ

Γ

∆g (c) dρ(g), this yields

2(b − a)Dr(Γ, ρ) = ∆ρ (b) − ∆ρ (a).
The last equality shows that ∆ρ is an affine function. On the other hand, ∆ρ is
an average of non-negative functions, thus it is non-negative. Therefore, ∆ρ must
be constant, which implies that Dr(Γ, ρ) = 0, as desired.

The next proposition shows the relevance of the Derriennic property in the
study of the smoothness of a group action.
Proposition 4.4.3. If the Γ-action is ρ-harmonic, then every g ∈ Γ is Lipschitz.
Moreover, the displacement function x 7→ g(x) − x is uniformly bounded in x. In
particular, every ρ-harmonic action is almost-periodic.
Proof. It suffices to prove the lemma for the elements of the support of ρ.
Indeed, this is obvious for the Lipschitz property, and for the boundedness of the
displacements, this is a consequence of the relation


gh(x) − x = gh(x) − h(x) + h(x) − x .
Now, for every element g in the support of ρ and every x < y, we have
Z




ρ(g) g(y) − g(x) ≤
g(y) − g(x) dρ(g) = y − x.
Γ

Hence,

g(y) − g(x) ≤

y−x
,
ρ(g)

proving that g has Lipschitz constant at most 1/ρ(g). To show that the displacements are bounded, we will show that the value of ∆g (x) is comparable, up to a multiplicative constant, to [g(x) − x]2 . In concrete terms, set
L := max{1/ρ(h) : ρ(h) > 0}. Then g −1 has Lipschitz constant at most L.

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

187

Letting x ∈ R be such that g(x) ≥ x, for each y ∈ R such that g −1(x) ≤ y ≤ x,
we have

1
g(y) − x ≥
y − g −1 (x) .
L
Therefore,
Z

x

g −1 (x)




1
g(y) − x dy ≥
L

Z

x

g −1 (x)




2
−1
x
−
g
(x)
y − g −1(x) dy =
.
2L



Since g(x) − x ≤ L x − g −1 (x) , using the notation of the preceding proposition,
this implies that
2

g(x)
−
x
∆g (x) ≥
,
2L3
hence

2
Z
g(x)
−
x
.
∆ρ (x) = ∆g (x) dρ(g) ≥
2L3
Γ
However, by (the proof of) Proposition 4.4.1, the function ∆ρ (·) is constant.
Thus, the value of |g(x) − x| is bounded from above by a constant, as we wanted
to show. Analogous arguments apply in the case where g(x) ≤ x.

Exercise 4.4.4. If Γ is a free Abelian group, show that every ρ-harmonic action of Γ
on the real line is an action by translations.
Hint. Observe that the functions g 7→ g(x) are ρ-harmonic and Lipschitz for every
x ∈ R, and use the fact that Lipschitz harmonic functions
P on an Abelian group are
linear. (Here, φ : Γ → R is said to be harmonic if φ(g) = h∈Γ φ(gh)ρ(h) holds for all
g ∈ Γ.)

4.4.2

Properties of stationary measures

As we have seen, very nice properties hold for actions for which the Lebesgue
measure is stationary. Although for most actions this is certainly not the case, in
§4.4.3, we will see that, under very mild hypothesis, there is always a stationary
measure, which is unique up to multiplication by a positive constant and can be
transformed into the Lebesgue one by a semiconjugacy. We will also show several
general properties of stationary measures that actually will be crucial for the
proof of this fact. Throughout this section, Γ will continue denoting a subgroup
of Homeo+ (R) having no global fixed point.

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CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Following [63], let us introduce the Markov process on the line defined by
Xxn = gn · · · g1 (x),
where (gn ) is a family of independent random variables with law ρ. (For a general
introduction to the theory of Markov processes, we refer the reader to the very
nice book [69].) The transition probabilities of this process are
X
ρX (x, y) :=
ρ(g).
y=g(x)

The associated Markov operator P = PX acting on the space of bounded continous functions Cb (R) is given by
Z
x
P φ(x) = E(φ(X1 )) = φ(gx) dρ(g).
(4.13)
Γ

The iterates of this operator correspond to the operators associated to the convolutions of ρ. More precisely, we have Pρn = Pρ∗n , where the convolution of
two probabilities ρ1 , ρ2 on Γ is
X
ρ1 (f ) ρ2 (g),
ρ1 ∗ ρ2 (h) :=
f g=h

and ρ∗n := ρ ∗ ρ ∗ · · · ∗ ρ (n times).
We will still denote by P the dual action on the space of Radon measures on
the line. Such a measure is said to be stationary if it is P -invariant, that is,
P ν = ν. Equivalently,
X
ν=
g∗ (ν) ρ(g).
g∈Γ

By definition, an action is harmonic if and only if the Lebesgue measure is stationary.

Lemma 4.4.5. Every nonzero stationary Radon measure ν on the real line is
bi-infinite (i.e. ν(x, ∞) = ∞ and ν(−∞, x) = ∞, for all x ∈ R).

Proof. Suppose that there exists x ∈ R such that ν(x, ∞) < ∞. Since we are
assuming that the Γ-action has no global fixed point on R, for every y ∈ R, there
is an element g ∈ Γ such that g(x) < y. As the support of ρ generates Γ, we can
choose n > 0 such that p⋆n (g −1) > 0. Then
ν(y, ∞) ≤ ν(g(x), ∞) ≤

ν(x, ∞)
< ∞.
ρ⋆n (g −1)

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

189

This shows that ν(y, ∞) < ∞ holds for all y ∈ R.
Now let φ : R → (0, ∞) be the function defined by φ(x) := ν(x, ∞). Since
ρ is symmetric, this function is harmonic; in other words, we have P φ = φ,
where P φ is defined by (4.13). (This definition still makes sense yet φ is not
necessarily continuous.) Fix a real number A satisfying 0 < L < ν(−∞, ∞), and
let ψ := max{0, L − φ}. The function ψ is subharmonic, which means that
ψ ≤ P ψ. Moreover, it vanishes on a neighborhood of −∞ and is bounded
on
R
Ra neighborhood of ∞. This implies that ψ is ν-integrable, and since P ψ dν =
ψ dν, the function ψ must be ν-a.e P -invariant. Now a classical lemma from
[77] asserts that a measurable function which is in L1 (ν) and P -invariant must
be a.e. Γ-invariant (see Exercise 4.4.6 for a schema of proof). Thus, ψ is constant
on almost-every orbit. However, this is impossible, since every orbit intersects
every neighborhood of −∞ (where ψ vanishes) and of ∞ (where ψ is positive).
This contradiction establishes the lemma.

Exercise 4.4.6. Let Γ be a countable group and ρ a measure on Γ whose support
generates Γ. Assume that Γ acts on a probability space (X, ν) by measurable maps
and that ν is ρ-stationary, meaning that
Z
g∗ (ν)dρ(g).
ν=
Γ

Prove that any function φ ∈ L1 (X, ν) that is a.e. P -invariant is a.e. Γ-invariant.
Hint. Pick a constant L ∈ R and consider the function ψ := max{φ, L}. Observe that ψ
belongs to L1 (ν) and satisfies ψ ≤ P ψ, and deduce that ψ is ρ-harmonic on almost-every
Γ-orbit. Conclude by ranging L along all rational numbers.
Exercise 4.4.7. Let Γ be a finitely-generated subgroup of Homeo+ (R), and let ρ be a
symmetric probability measure on it with generating support. Prove that for all x ∈ R,
every compact interval I, and almost-every sequence (gn ) ∈ ΓN , the set of integers n
for which Xxn belongs to I has density zero, that is,
1 
n ∈ {1, . . . , k} : Xxn ∈ I
k→∞ k
lim

= 0.

Hint. Let νk be the measure on the line defined by
νk (I) :=

k

1 X ⋆n
ρ {h : h(x) ∈ I} .
k
n=1

Assuming that the zero-density above doesn’t hold, show that, up to a subsequence, νk
converges to a nonzero, finite, stationary measure, thus contradicting Lemma 4.4.5.

190

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Exercise 4.4.8. Prove that no nontrivial group action on the line satisfying (4.11) for
all x, y in R can have a global fixed point.
Hint. By definition, the Lebesgue measure is P -invariant for a ρ-harmonic action. Apply
Lemma 4.4.5 to the restriction of the action to a connected component of the set of
global fixed points.

We next study the case when the Γ-action admits discrete orbits. Obviously,
such an orbit supports an invariant Radon measure, namely the counting measure;
in particular, this measure is P -invariant. The next two lemmas prove that in the
case there are discrete orbits, all P -invariant Radon measures lie in the convex
closure of the set of counting measures along discrete orbits.
Lemma 4.4.9. Let ν be a stationary measure on the line. If there is a discrete
orbit, then ν is supported on the union of discrete orbits and is totally invariant.
Proof. If there is a discrete orbit O, then Γ acts on it by translating its points.
Thus, the normal subgroup Γ∗ formed by the elements acting trivially on O is
recurrent, by Polya’s classical theorem [166] (see also Corollary 4.4.15 for an
alternative proof of this fact). Let ρ∗ be the (symmetric) measure on Γ∗ obtained
by balayage of ρ to Γ∗ ; more precisely, we consider the random walk on Γ
with initial distribution ρ and we stop it at the first moment where it visits
Γ∗ , thus yielding a random variable with values in Γ∗ whose distribution is ρ∗ .
Observe that the restriction of ν to each connected component C of R \ O is
a finite measure that is invariant for the Markov process induced by ρ∗ on C.
It follows from Lemma 4.4.5 that this measure is supported on Fix(Γ∗ ) ∩ C,
the set of global fixed points for the group Γ∗ contained in the closure of C.
(Notice that Lemma4.4.5 dod not use that the support of ρ was finite.) As a
consequence, Γ acts by “integer translations” on the support of ν, which consists
of discrete orbits. To see that ν is invariant, notice that for each atom x ∈ R,
the function g 7→ ν(g(x)) viewed as a function defined on Γ/Γ∗ ∼ Z is harmonic,
hence constant.

Exercise 4.4.10. Assume that the Γ-action is harmonic and admits a discrete orbit.
Prove that Γ is contained in the (cyclic) group generated by a translation of the line.

Lemma 4.4.11. Let ν be a stationary measure on the real line. If the atomic
part of ν is nontrivial, then it is supported on a union of discrete orbits.

191

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

Proof. The proof uses the recurrence property that will be proven in §4.4.3. Let
x ∈ R be a point such that ν(x) > 0. Let O = Ox be the orbit of x endowed with
the discrete topology, and let ν be the measure on O defined by ν(y) := ν(y).
We claim that ν is an invariant measure for the Markov process induced by ρ on
O. To see this, consider an arbitrary function φ : O → R with finite support. Let
φε : R → R be a family of functions with support a union of intervals of length ε
centered at the points of the support of φ. Then,
Z
Z
Z
Z
Z
φ dν = lim φε dν = lim φε dP ν = lim P φε dν = P φ dν,
O

ε→0

R

ε→0

R

ε→0

R

O

which is sufficient to establish our claim.
Let K = [A, B] be an interval such that for any element g of the support of
ρ, we have g(A) < B. (Recall that ρ is assumed to be finitely supported.) Let I
be an arbitrary compact interval containing K, and let OI := I ∩ O. We want
to show that OI is finite. To do this, first observe that OI is a recurrent subset
of O, by Corollary 4.4.15. Let Y be the Markov process on OI defined by the
first return of the original Markov process X to OI . The process Y is symmetric,
because
P X is symmetric. Moreover, the restriction of ν to OI is stationary. Now,
since y∈OI ν(y) < ∞, there must be an atom y ∈ OI for which ν(y) is maximal.
The PY -invariance of ν and the symmetry of the transition probabilities ρY yield
X
X
ρY (y, z)ν(z) =
ρY (z, y)ν(z) = ν(y).
z∈O

z∈OI

The maximum principle now implies that ν(z) = ν(y). Thus, all the atoms of ν
contained in O have the same mass, hence there is only a finite number of them.
As this holds for an arbitrary compact interval I containing K, this shows that
O is discrete.

Next, we consider the case where Γ has no discrete orbits. Recall that in this
situation, there is a unique nonempty minimal invariant closed set for the action,
that we denote M; see Lemma 3.5.18.
Lemma 4.4.12. Assume that Γ does not have any discrete orbit on the real line.
Then any stationary measure is supported on the minimal set M.
Proof. Let ν be a stationary measure on the real line. Then ν is quasi-invariant
by Γ, because for all h in the support of ρ, we have
1 X
ν
h∗ (ν) ≤
g∗ (ν) ρ(g) =
.
ρ(h) g∈Γ
ρ(h)

192

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Therefore, the support of ν is a closed Γ-invariant subset of the line, hence it
contains M. Thus, it suffices to show that ν does not charge any connected
component of the complement Mc .
Assume Mc is nonempty, and collapse each of its connected components to
a point, thus obtaining a topological line carrying a Γ-action for which all orbits
are dense. The stationary measure ν can be pushed to a stationary measure ν
for this new action. If a component of Mc has a positive ν-measure, then ν has
atoms. Lemma 4.4.11 then implies that the new Γ-action cannot be minimal,
which is a contradiction.

Exercise 4.4.13. Prove that if the action of Γ is harmonic, then Γ either is a group
of translations or acts minimally on the real line.

4.4.3

Recurrence

As in previous sections, we continue considering a finitely-supported, symmetric probability measure ρ on a group Γ acting on the real line without global
fixed points. We also assume that the support of ρ generates Γ. We start with an
oscillation result claiming that almost-every random orbit escapes to the infinity
in both directions.
Proposition 4.4.14. For every x ∈ R, almost surely we have
lim sup Xxn = ∞

and

n→∞

lim inf Xxn = −∞.
n→∞

Proof. Denote P := ρN , and given points L and x on the real line, let
h
i
pL (x) := P lim sup Xxn > L .
n→∞

Since Γ acts by orientation-preserving homeomorphisms, for all x ≤ y, we have
o
o n
n
(gn ) ∈ ΓN : lim sup Xxn > L ⊂ (gn ) ∈ ΓN : lim sup Xyn > L .
n→∞

n→∞

In particular, pL (x) ≤ pL (y), that is, pL is non-decreasing. Moreover, since pL is
the probability of the tail event
h
i
lim sup Xxn > L
n→∞

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

193

and X is a Markov chain, pL is a harmonic function, that is, for every x ∈ R and
every integer n ≥ 0,
X


pL (x) =
pL g(x) ρ⋆n (g) = E pL (Xxn ) .
g∈Γ

Now, we would like to see pL as the distribution function of a finite measure
on the line. However, this is only possible when pL is continuous on the right,
which is a priori not necessarily the case. We are hence led to consider the
right-continuous function
pL (x) :=

lim

y→x, y>x

pL (y).

This function is still non-decreasing. Therefore, there exists a finite measure ν
on R such that for all x < y,
ν(x, y] = pL (y) − pL (x).
Since pL is harmonic and Γ acts by homeomorphisms, pL is also harmonic. Since
ρ is symmetric, this yields that ν is P -invariant. Now recall that Lemma 4.4.5
implies that any P -invariant finite measure identically vanishes (see also [61,
Proposition 5.7]). Therefore, pL is constant, hence in particular its value does
not depend on the starting point x. The 0-1 law then allows to conclude that
(for any fixed L) either pL ≡ 0 or pL ≡ 1.
Let us now show that pL identically equals to 1 for each L. To do this, fix any
x0 > L. As for any g ∈ Homeo+ (R), we have either g(x0 ) ≥ x0 or g −1 (x0 ) ≥ x0 ,
the symmetry of ρ yields that Xxn0 ≥ x0 holds with probability at least 1/2, for
all n ∈ N. It is then easy to see that
h
i
pL = pL (x0 ) ≥ P lim sup Xxn0 ≥ x0 ≥ 1/2.
n→∞

As we have already shown that pL equals 0 or 1, this implies that pL is identically
equal to 1.
The latter means that for every x ∈ R,
lim sup Xxn = ∞
n→∞

holds almost surely. Analogously, for every x ∈ R, almost surely we have
lim inf Xxn = −∞.
n→∞

194

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

This completes the proof of the proposition.



We are now ready to prove the main result of this section, namely the recurrence of the Markov process.
Corollary 4.4.15. There exists a compact interval K such that, for every x ∈ R,
almost surely the sequence (Xxn ) intersects K infinite many times.
Proof. Consider a closed interval K as in the proof of Lemma 4.4.11, that is,
K = [A, B], where A < B are such that for every g of the support of ρ, we
have g(A) < B. (Recall that ρ is finitely supported.) By Proposition 4.4.14, for
every x ∈ R, almost surely the sequence (Xxn ) will pass from (−∞, A] to [B, +∞)
infinite many times. Now the desired conclusion follows from the observation
that the choice of A and B imply that every time this happens, (Xxn ) must cross
the interval K.

On left-orders that are generic with respect to a stationary measure.
Given a finitely-supported probability on a left-orderable group Γ with generating
support, we can also consider stationary probability measures for the action of Γ
on its space of left-orders (see Example 4.4.6). By this, we mean a probability
measure µ on LO(Γ) such that
X
µ=
g∗ (µ)ρ(g).
(4.14)
g∈Γ

Since LO(Γ) is compact, such a probability measure µ always exists. (This follows from a direct application of either Kakutani’s fixed point theorem or the
Bogoliubov-Krylov procedure.) It seems quite interesting to study the relation
of µ with the algebraic properties of Γ as well as its dependence on ρ. We give
below two examples on this.
Example 4.4.16. We next give still another proof of Theorem 2.2.9 for finitelygenerated groups. To do this, fix ρ and µ as above. We can assume that µ is ergodic, in
the sense that it cannot be written as a nontrivial convex combination of two different
stationary probability measures. We have two possibilities:
Case (i). The measure µ has an atom.
If  is an atom of maximal µ-measure, then (4.14) easily implies that its orbit
must be finite. (Actually, by ergodicity, this orbit coincides with the support of µ.)
In particular,  is right-reccurrent, hence Conradian. Thus, if Γ has infinitely many
left-orders, then Proposition 3.2.53 implies that LO(Γ) is uncountable, as desired.

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

195

Case (ii). The measure µ is non-atomic.


By ergodicity, for almost-every , (gn ) in LO(Γ)×ΓN (endowed with the measure
µ × ρN ) the sequence (σn (ω) , σ n (ω)) is dense in supp(µ) × ΓN , where σ stands for the
shift σ((gn )) := (gn+1 ). Let us fix such a pair , (gn ) , and let Uk be a sequence of
open subsets of positive ρ-measure in LO(Γ), none of which containing , but which
do converge to . For each k, there exists n(k) ∈ N such that σn(k) (ω) belongs to
Uk . Hence, σn(k) (ω) converges to , with σn(k) (ω) being distinct from  for all k.
Therefore, the closure of the orbit of  under the action of Γ is a totally disconnected
compact metric space with no isolated point, that is, a Cantor set. In particular, LO(Γ)
is uncountable.
Remark 4.4.17. Notice that the approximation by conjugates in the example above
is essentially different from that of §3.2.5. Indeed, the conjugating elements therein are
positive but “small’ (outside the Conradian soul). In the proof above, the conjugating
elements are “random”, and hence by Exercise 4.4.7, they are mostly “near the infinite”
(in any direction), despite the recurrence of the associated random walk on the line (in
the case where ρ is symmetric).
Example 4.4.18. According to Example 3.2.55, for each integer ℓ ≥ 2, the BaumslagSolitar group B(1, ℓ) = ha, b : aba−1 = bℓ i admits four Conradian orders, which are
actually bi-invariant and come from the exact sequence
0 −→ Z

h1i
ℓ

−→ B(1, ℓ) −→ Z −→ 0.

We claim that, although LO(B(1, ℓ)) is a Cantor set (c.f. §3.3.1), for every symmetric
probability distribution ρ on B(1, ℓ) as above, every stationary probability measure µ on
LO(B(1, ℓ)) is supported on these four points. Indeed, we proved in §3.3.1 that for every
left-order  on BS(1, ℓ) that is not bi-invariant, the associated dynamical realization
is semiconjugate to a non-Abelian subgroup of the affine group. In particular, there
exist elements whose sets of fixed points are bounded and for which −∞ and +∞ are
topologically-repelling fixed points. Let g be such an element (actually, such a g can
be taken as the image of a), and denote by F ix(g) its set of fixed points. Let f1 , f2
be in the realization of B(1, ℓ) so that f1 (resp. f2 ) sends the leftmost (resp. the
rightmost) fixed point of g to the right (resp. left) of 0 = t(id). Denote g1 := f1 gf1−1
and g2 := f2 gf2−1 . If we identify elements in BS(1, ℓ) with their realizations, we have
g1 ≻ id and g2 ≺ id. Moreover, hgi h−1 ≺ id holds for both i = 1 and i = 2 provided
h is sufficiently large (say, larger than a certain element h+ ). Similarly, hgi h−1 ≻ id
holds for i = 1 and i = 2 provided h is smaller than a certain element h− ; see Figure
20 below.

196

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS
..
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g1 = f1 gf1−1
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•
•....
•

t(h− )

t(id) = 0

t(h+ )

...
.....
.......
....
............................
.
.
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.............................
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....
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.........
g2 = f2 gf2−1
.....
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....
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Figure 20: The elements g1 , g2 , h− and h+ .
Assume µ is an stationary probability measure on LO(BS(1, ℓ)) that is not fully
supported on the four bi-orders. Then any ergodic component of this measure outside
these bi-orders is still stationary, and supported on the complement of the bi-orders.
For simplicity, we still denote this measure by µ. Let  a point in the support of µ. If
we perform the construction of the elements
g1 , g2 , h− , h+ above, then the measure of

the open neighborhood Vg1 ∩ Vg−1 = ′ : g1 ≻′ id, g2 ≺′ id of  must be positive,
2
say equal to κ > 0. A direct application of the ergodic theorem then shows that, for
a generic random path (hn ) ∈ B(1, ℓ)N , the set of integers n for which X n lies in
Vg1 ∩ Vg−1 , where X n := hn · · · h1 , has density κ. Nevertheless, among these integers
2
n, with density 1 we have either X n ≺ h− or X n ≻ h+ (see Exercise 4.4.7), thus
providing a contradiction.

4.4.4

Existence of stationary measures

Using the recurrence result of the preceding section, we can now establish the
existence of a P -invariant Radon measure via a quite long but standard argument.
Theorem 4.4.19. Let Γ be a finitely-generated subgroup of Homeo+ (R) endowed
with a symmetric probability measure whose support generates Γ. Then there
exists a (nonzero) ρ-stationary measure on the real line.
Proof. Fix a continuous compactly-supported function ξ : R → [0, 1] such that
ξ ≡ 1 on K. For any initial point x, let us stop the process Xxn at a random

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

197

stopping time T chosen in a Markovian way so that, for all n ∈ N,


P T = n + 1 | T ≥ n = ξ(Xxn+1).

(Here, T = T (ω), where ω = (gi )i∈N .) In other words, after each step of the initial
random walk arriving to a point y = Xxn+1 , we stop with probability ξ(y), and
we continue the compositions with probability 1 − ξ(y).
Denote by Yx the random stopping point XxT , and consider its distribution
ρx (notice that T is almost-surely finite since the process Xxn almost surely visits
K and ξ ≡ 1 on K). Due to the continuity of ξ, the measure ρx on R depends
continuously (in the weak topology) on x. Therefore, the corresponding diffusion
operator Pξ defined by
Z

φ(y) dρx(y)
Pξ (φ)(x) = E φ(Yx ) =
R

acts on the space of continuous bounded functions on R, and hence it acts by
duality on the space of probability measures on R. Notice that for any such
probability measure, its image under Pξ is supported on K̂ := supp(ξ). Thus,
by applying the Bogolyubov-Krylov procedure of time averaging (and extracting
a convergent subsequence), we see that there exists a Pξ -invariant probability
measure ν0 .
To construct a Radon measure that is stationary for the initial process, we
proceed as follows. For each point x ∈ R, let us take the sum of the Dirac measures
supported in its random trajectory before the stopping time T . In other words,
we consider the “random measure”
T (w)−1

mx (ω) :=

X

δXxj

j=0

and its expectation



T (w)−1



mx := E mx (ω) = E 

X
j=0



δXxj 

as a measure on R. Finally, we integrate
R mx with respect to the measure ν0 on x,
thus yielding a Radon measure ν := mx dν0 (x) on R. Formally speaking, for
any compactly supported function φ, we have


Z
Z
T (w)−1
X
φ dν =
E
φ(Xxj ) dν0 (x).
(4.15)
R

R

j=0

198

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

Notice that the right-side expression in (4.15) is well-defined and finite. Indeed,
there exist N ∈ N and p0 > 0 such that with probability at least p0 a trajectory
starting at any point of supp(φ) hits K in at most N steps. Therefore, the
distribution of the measure mx (w) on supp(φ) (i.e. the number of steps that are
spent in supp(φ) until the stopping time) has an exponentially decreasing tail.
Thus, its expectation is finite and bounded uniformly on x ∈ supp(φ), which
implies the finiteness of the integral.
Next, let us check that the measure ν is P -invariant. To do this, let us rewrite
the measure ν as follows. First, notice that, by definition, we have
" n
#
n
X X
Y
Y


mx =
ρ(gj )
1 − ξ(gj · · · g1 (x)) δgn ···g1 (x) .
n≥0 g1 ,...,gn ∈G

j=1

j=1

Thus,
P (mx ) =

X

ρ(g) g∗(mx )

g∈G

=

X

ρ(g) g∗

X

X

n≥0 g1 ,...,gn ∈G

g∈G

=

X

X

ρ(g)

n≥0 g1 ,...,gn ,g∈G

=

X

X

n≥0 g1 ,...,gn ,gn+1 ∈G

n
Y

" n
Y
j=1

p(gj )

j=1

n+1
Y
j=1

ρ(gj )

ρ(gj )

!

n
Y

j=1

n
Y
j=1



δgn ···g1 (x)

!



1 − ξ(gj · · · g1 (x)) g∗ δgn ···g1 (x)

! (n+1)−1
Y 
j=1

1 − ξ(gj · · · g1 (x))

#



1 − ξ(gj · · · g1 (x)) δgn+1 gn ···g1 (x) .

As
before, the last expression equals the expectation of the random measure
PT (ω)
j=1 δXxj . In this sum, we are counting the stopping time, but not the initial
one. Therefore,
P mx = mx − δx + E(δYx ).
By integrating with respect to ν0 , this yields
Z
 Z
Pν = P
mx dν0 (x) =
P (mx ) dν0(x) =
R
R
Z
Z
Z
mx dν0 (x) − δx dν0 (x) + E(δY x ) dν0(x) = ν − ν0 + Pξ (ν0 ).
R

R

R

Since ν0 is Pξ -invariant, we finally obtain P ν = ν, as we wanted to show.



4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

199

Theorem 4.4.20. Every minimal action Φ : Γ → Homeo+ (R) is topologically
conjugate to a ρ-harmonic action.
Proof. Take a P -invariant Radon measure ν whose existence follows from Theorem 4.4.19. By Lemma 4.4.5, it is bi-infinite. By Lemma 4.4.11, it has no atoms.
Finally, by Lemma 4.4.12, it has total support. As a consequence, there exists
a homeomorphism ϕ : R → R such that ϕ∗ (ν) is the Lebesgue measure. The
conjugate action ϕ ◦ Φ ◦ ϕ−1 is then ρ-harmonic.

When the action of Γ admits discrete orbits, we know that every stationary Radon measure must be Γ-invariant. However, two such measures may be
supported on different orbits. We next establish the uniqueness (up to a scalar
factor) of the stationary measure in the case there is no discrete orbit. Recall
that, in this case, there exists a unique nonempty, closed, minimal Γ-invariant set
M (see Lemma 3.5.18).
Proposition 4.4.21. Assume that there is no discrete orbit for the Γ-action on
the line. Then the P -invariant Radon measure ν is unique up to a scalar factor.
We begin the proof by some reductions. First, we can assume that the action
is minimal, since stationary measures are supported on M (c.f. Lemma 4.4.12)
and the action semiconjugates into a minimal one. Moreover, after changing coordinates (by a topological conjugacy), we may assume that the Lebesgue measure
is stationary, that is, that the action is ρ-harmonic.
Recall that a P -invariant measure is said to be ergodic if every Γ-invariant
measurable subset of the line either has measure 0 or its complement has measure
0. Every P -invariant measure decomposes as an integral of ergodic measures.
Thus, to prove Theorem 4.4.21, it suffices to show that there exists a unique
ergodic harmonic measure, up to multiplication by a constant.
Lemma 4.4.22. Assume that the action of Γ is minimal and ρ-stationary. Let ν
be an ergodic P -invariant measure. Then for all continuous functions φ, ψ with
compact support, with φ ≥ 0 and φ ≡ 1 on the recurrence interval K given by
Corollary 4.4.15, and for every x ∈ R, it almost surely holds
R
ψ dν
Sk ψ(x, ω)
−→ R
(4.16)
Sk φ(x, ω)
φ dν
as n tends to infinity, where Sk ψ(x, ω) := ψ(Xx0 ) + ψ(Xx1 ) + . . . + ψ(Xxk−1) (and
similarly for Sk φ).

200

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

For the proof, we will apply Hopf’s ratio ergodic theorem [99] (see also [110])
to the system (RN0 , σ, ν̂), where σ is the shift operator σ(X n )n = (X n+1 )n , and ν̂
is the image of the measure ν × ρN under the map

x, ω = (gn ) →
7 (Xx0 = x, Xx1 , . . . , Xxn , . . .).

We leave as an exercise to the reader to verify that ν̂ is invariant by σ. (Actually,
this is nothing but a reformulation of the fact that ν is P -invariant).
We claim that the system (RN0 , σ, ν̂) is ergodic, that is, every measurable
σ-invariant subset A of RN0 has either zero or full ν̂-measure. Indeed, for such an
A, and for a fixed x ∈ R, let pA (x) be the probability that the sequence (Xxn )n≥0
belongs to A. The thus defined function pA : R → [0, 1] is measurable. Since A is
σ-invariant, the property of belonging to A does only depend on the tail of the
sequence, hence the function pA is P -invariant. (This can be easily checked by the
reader.) We claim that this function is indeed constant. To prove this, notice that
we cannot directly apply Exercise 4.4.6, because the function pA has no reason
to belong to L1 (R, ν). To overcome this difficulty, let us consider a compact
interval I containing the recurrence interval K. Given a point x ∈ I, we denote
by Yx1 , . . . , Yxm . . . the points of the sequence Xx1 , . . . , Xxn , . . . that belong to I. As
we are assuming that the Lebesgue measure is P -stationary, the Markov process
Y on I leaves invariant the restriction of the Lebesgue measure on I. Moreover,
the restriction of the function pA is still harmonic for the Markov process Y ,
namely we have pA (x) = E(pA (Yx1 )) for every x ∈ I. The Lebesgue measure of I
being finite, an easy extension of Exercise 4.4.6 for Markov processes shows that
pA is almost-surely constant on I. As this is true for every compact interval I
containing K, we conclude that pA is almost everywhere constant, as was claimed.
Now, the 0 − 1 law shows that this constant is either 0 or 1, thus showing that
A has measure 0 or its complementary has measure 0. This concludes the proof
that the system (Rn , σ, ν̂) is ergodic.
Next, let φ : R → R be a non-negative function with compact support such
b (X n )n≥1 ) := φ(x),
that φ ≡ 1 on the recurrence interval K. Then, letting φ(x,
the function φb belongs to L1 (RN0 , νb), and the recurrence property implies that
for ν̂-almost-every (x, (X n )), we have
X
k≥0


φb σ k (x, (X n )n ) = ∞.

Hopf’s ratio ergodic theorem then states that for every function ψb ∈ L1 (RN0 , νb),

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

201

almost surely we have the convergence
R
b
ψdν̂
ψb + ψb ◦ σ + . . . + ψb ◦ σ k−1
.
−→ R
b
φdν̂
φb + φb ◦ σ + . . . + φb ◦ σ k−1

b (X n )n ) := ψ(x), where ψ : R → R is
Applying this to a function of the form ψ(x,
continuous with compact support, and noticing that
Z
Z

b
φ dν̂ = φ dν
and
φb + φb ◦ σ + . . . + φb ◦ σ k−1 x, (X n )n = Sk φ(x, ω)

(and similarly for ψ), we conclude that (4.16) holds for ν-almost-every x ∈ R.
The difficulty now is to extend (4.16) to every x ∈ R. This will follow from
the following contraction property for ρ-harmonic actions:

Lemma 4.4.23. For any fixed number 0 < p < 1 and all x, y, with probability at
least p we have
|x − y|
lim |Xxn − Xyn | ≤
.
n→∞
1−p
Proof. Assume y < x for simplicity. Since ν is P -invariant, the sequence of
random variables ω 7→ Xxn − Xyn is a positive martingale. In particular, for every
integer n ≥ 1, we have
E(Xxn − Xyn ) = x − y.
By the Martingale Convergence Theorem, the sequence Xxn − Xyn almost surely
converges to a non-negative random variable v(x, y). By Fatou’s inequality, we
have
E(v(x, y)) ≤ lim E(Xxn − Xyn ) = x − y.
n→∞

The lemma then follows from Chebyshev’s inequality.



Let now y ∈ R and the functions φ, ψ as in the hypothesis of Lemma 4.4.22 be
fixed. We claim that, for any m ≥ 1, with probability at least 1 − 1/m we have
R
ψ dν
Sk ψ(y, ω)
1
lim sup
≤ .
−R
(4.17)
Sk φ(y, ω)
m
φ dν
k→∞
Once this established, it will obviously implies that (4.16) holds almost surely at
all points, as desired.

202

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

To show (4.17), notice that since ν has total support, one can find x that is
generic, so that (4.17) holds at x for every m. Moreover, given ε > 0, the point x
may be chosen sufficiently close to y so that |x − y| ≤ ε. By Lemma 4.4.23, with
probability at least 1 − 1/m we have for all k sufficiently large, say k ≥ k0 (ω),
|Xyk − Xxk | ≤ mε.

(4.18)

Next, as we already know that (with probability 1)
R
ψ dν
Sk ψ(x, ω)
,
lim
= R
k→∞ Sk φ(x, ω)
φ dν

instead of estimating the difference in (4.17), it suffices to obtain estimates of the
“relative errors”
lim sup
k→∞

and
lim sup
k→∞

Sk ψ(y, ω) − Sk ψ(x, ω)
≤ δ1 (ε)
Sk φ(x, ω)

(4.19)

Sk φ(y, ω) − Sk φ(x, ω)
≤ δ2 (ε),
Sk φ(x, ω)

(4.20)

in such a way that δ1 (ε) → 0 and δ2 (ε) → 0 as ε → 0.
Since the estimate (4.20) for φ is a particular case of the estimate (4.19),
we will only check (4.19). Now, (4.18) implies that |Sk ψ(y, ω) − Sk ψ(x, ω)| is at
most

mod(mε, ψ) card k0 (ω) ≤ j ≤ k : either Xxj or Xyj is in suppψ + 2k0 (ω) max |ψ|
≤ mod(mε, v) card{j ≤ k | Xxj ∈ Umε (suppψ)} + const(ω).

Here, mod(·, ψ) stands for the modulus of continuity of ψ with respect to the
distance d on the variable, and Umε (suppψ) denotes the mε-neighborhood of the
support of ψ, again with respect to d.
Let ξ be a continuous function satisfying 0 ≤ ξ ≤ 1 and that is equal to 1 on
Umε (suppψ) and to 0 outside U(m+1)ε (suppψ). We have

card j ≤ k : Xxj ∈ Umε (suppψ) ≤ Sk ξ(x, ω).
Thus,

Sk ψ(y, ω) − Sk ψ(x, ω)
Sk φ(x, ω)

const(ω) + mod(mε, ψ) Sk ξ(x, ω)
Sk φ(x, ω)
R
ξ dν
=: δ1 (ε).
−→ mod(mε, ψ) R
φ dν
≤

4.4. RANDOM WALKS ON LEFT-ORDERABLE GROUPS

203

Here, we have applied the fact that, by our choice of x, equality (4.16) holds
with ξ in the numerator and φ in the denominator. Since mod(mε, ψ) tends to 0
as ε → 0 and the quotient
R
ν(U(m+1)ε (suppφ))
ξ dν
R
R
≤
φ dν
φ dν

remains bounded, this yields δ1 (ε) → 0 as ε → 0.



Having Lemma 4.4.22 at hand, it is now easy to finish the proof of Proposition
4.4.21. Indeed, given any two ergodic P -invariant Radon measures ν1 , ν2 , for each
x ∈ M and every compactly supported, real-valued function ψ, almost surely we
have
R
ψ dνi
Sk ψ(x, ω)
−→ R
,
Sk φ(x, ω)
φ dνi
R
R
R
R
where i ∈ {1, 2}. Thus, ψ dν1 = λ ψ dν2 , with λ := φ dν1 / φ dν2 . This
proves that ν1 = λν2 , and concludes the proof of Proposition 4.4.21.
Exercise 4.4.24. Show that the condition on φ in Lemma 4.4.22 can be relaxed to
φ ≥ 0 and φ 6≡ 0.

We close with the next
Theorem 4.4.25. The conjugacy of a minimal action to a ρ-harmonic one is
unique up to post-composition by an affine map.
Proof. Given a minimal action Φ : Γ → Homeo+ (R) and two homeomorphisms
ϕi : R → R such that each ϕi ◦ Φ ◦ ϕ−1
is ρ-harmonic, the images of the Lebesgue
i
−1
measure by ϕ−1
,
ϕ
are
ρ-stationary
for
Φ, hence they differ by multiplication
1
2
−1
by a constant. Therefore, ϕ2 ◦ ϕ1 sends the Lebesgue measure into a multiple
of itself, which means that ϕ2 ◦ ϕ−1

1 is an affine map.

204

CHAPTER 4. PROBABILITY AND LEFT-ORDERABLE GROUPS

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