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Characterization of
Extremal KMS States on
Groupoid C*-Algebras
Rafael Pereira Lima
Dissertação apresentada
ao Instituto de Matemática e Estatística
da
Universidade de São Paulo
para
obtenção do título
de
Mestre em Ciências
Programa: Mestrado em Matemática Aplicada
Orientador: Prof. Dr. Rodrigo Bissacot
Durante o desenvolvimento deste trabalho o autor recebeu auxílio �nanceiro do CNPq
São Paulo, julho de 2019
Characterization of
Extremal KMS States on
Groupoid C*-Algebras
Esta dissertação contém as correções e alterações
sugeridas pela Comissão Julgadora durante a defesa
realizada por Rafael Pereira Lima em 1/7/2019.
O original encontra-se disponível no Instituto de
Matemática e Estatística da Universidade de São Paulo.
Comissão Julgadora:
• Prof. Dr. Cristian Ortiz (presidente) - IME-USP
• Prof. Dr. Alcides Buss - UFSC
• Prof. Dr. Alexandre Baraviera - UFRGS
Agradecimentos
Consegui escrever esta dissertação por causa da ajuda de algumas pessoas. Gostaria de
agradecer aos meus pais, minha irmã e minha madrinha pelo incentivo que sempre tive.
Sou grato pela ótima orientação do professor Rodrigo Bissacot, que me preparou para
a pesquisa e se preocupa com a carreira dos alunos, além de ser um exemplo para mim.
Aos colegas do grupo pela ajuda durante o desenvolvimento da dissertação, principalmente
Lucas, João, Rodrigo e Thiago, que revisaram o trabalho durante vários seminários. Em
particular, gostaria de agradecer ao Lucas, por me recomendar para o professor Rodrigo.
Agradeço ao professor Severino Toscano pela ajuda durante o mestrado. Especialmente,
sou muito grato pela ajuda e apoio do professor Paulo Cordaro, o que foi fundamental para
eu conseguir terminar o mestrado.
1
Resumo
Nesta dissertação de mestrado, estudamos um teorema de Neshveyev [17] que descreve todos
os estados KMS em uma C*-álgebra de um grupóide étale localmente compacto Hausdor�
satisfazendo o segundo axioma de enumerabilidade. Depois estudamos um resultado provado
por Thomsen [26] que caracteriza os estados KMS extremais nessa C*-álgebra para um
grupóide de Renault-Deaconu.
Palavras-chave: C*-álgebras, estados KMS, medidas conformes, grupóides.
2
Abstract
In this master's thesis we study a theorem due to Neshveyev [17] which describes all KMS
states on the groupoid C*-algebra for a locally compact Hausdor� second countable étale
groupoid. Then we study a result due to Thomsen [26] which characterizes the extremal
KMS states on this C*-algebra for a Renault-Deaconu groupoid.
Keyworkds: C*-algebras, KMS states, conformal measures, groupoids.
3
Contents
1 Introduction 6
2 Measure Theory 12
2.1 Radon-Nikodym Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Pushforward Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.3 Purely Atomic and Non-Atomic Measures . . . . . . . . . . . . . . . . . . . 17
2.4 Measures on Locally Compact Spaces . . . . . . . . . . . . . . . . . . . . . . 18
2.5 µ-Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.6 Vector-Valued Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3 Groupoids 41
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.2 Topological Groupoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
3.3 Groupoid C*-Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4 Renault's Disintegration Theorem 74
4.1 Haar Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
4.2 Borel Hilbert Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
4.3 Renault's Disintegration Theorem . . . . . . . . . . . . . . . . . . . . . . . . 92
4
5 Neshveyev's Theorems 95
5.1 KMS States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
5.2 First Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
5.3 Second Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
6 Renault-Deaconu Groupoid 159
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
6.2 Full orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
6.3 Conformal Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
6.4 KMS States on the Renault-Deaconu Groupoid . . . . . . . . . . . . . . . . 201
7 Concluding Remarks 214
5
Chapter 1
Introduction
The purpose of this thesis is to �nd all KMS states on groupoid C*-algebras when the
groupoid satis�es certain topological conditions. This result was proved by Neshveyev in
[17]. Later we study a theorem due to Thomsen [26] which applies Neshveyev's theorem to
a Renault-Deaconu groupoid to characterize its extremal KMS states.
Groupoids are a generalization of groups where not every pair of elements can be multiplied
but each element has an inverse. This structure can be seen as a collection of arrows attached
to points on a plane, as shown in Figure 1.1. Such arrows can be composed if the end (called
range) of the �rst arrow is the source of the second. The inverse is obtained by reversing the
direction of the arrow and each point is identi�ed with an element of the groupoid assuming
its corresponding vector is the null vector.
s(g)
r(g)
s(h)
r(h)
gh
(a)
r(g)
s(g) = r(h)
s(h)g
h
gh
(b)
s(g)
r(g)
s(g)
r(g)
s(g) = g−1g
r(g) = gg−1
g g−1
(c)
Figure 1.1: Groupoids can be seen as arrows on a plane. s(g) and r(g) denote the sourceand range of g. (a) g and h are not composable, since s(g) 6= r(h); (b) The composition ofg and h is gh; (c) g−1 is the inverse of g. Note that g−1g = s(g) and gg−1 = r(g).
6
Given a groupoid G, G(2) is the set of composable elements. It consists of all pairs of
elements in G which can be multiplied. G(0) ⊂ G is the set of units. G is endowed with the
multiplication (also called composition) and inversion operations. r, s : G → G(0) are the
range and source maps. Later we will de�ne formally the notion of groupoids. The results
on groupoids in this thesis can be found in Rodrigo Frausino's thesis [9]. In fact, this thesis
can be seen as a sequel of his work because he also describes groupoid C*-algebras and the
Renault-Deaconu groupoid. In addition, many results here are based on his work.
Under certain conditions, we can equip the groupoid with a topology in such a way that
r, s are local homeomorphisms and the sets Gxx = s−1(x)∩ r−1(x) are discrete and countable
groups, and we assume this topology satis�es other conditions. In this case, we can equip
the space of continuous and compactly supported functions on G, denoted by Cc(G), with an
involution and a convolution which is not the pointwise multiplication. Then Cc(G) becomes
a ∗-algebra, not necessarily commutative.
In order to de�ne the groupoid C*-algebra C*(G), we equip Cc(G) with a norm which
depends on the ∗-representations of Cc(G). Then C*(G) is de�ned as the completion of
Cc(G) with respect to this norm.
Let c be a continuous R-valued 1-cocycle, that is, a continuous function c : G → R such
that c(g1g2) = c(g1) + c(g2) for (g1, g2) ∈ G(2). Then we �x a dynamics on C*(G) de�ned by
τt(f)(g) = eitc(g)f(g) for every f ∈ Cc(G), g ∈ G and t ∈ R. For f ∈ Cc(G), we can extend
the de�nition of τ to complex parameters, that is, τz(f) is well-de�ned. Given β ∈ R, we say
that a state ϕ on C*(G) is a KMS state if ϕ(f1τiβ(f2)) = ϕ(f2f1) for every f1, f2 ∈ Cc(G).
KMS states characterizes the equilibrium states in quantum statistical mechanics. A
theorem due to Neshveyev describes every KMS state ϕ on C*(G) by an explicit formula.
In fact, there is a correspondence between ϕ and a pair (µ, {ϕx}x∈G(0)) satisfying some
conditions, such that µ is a probability measure on G(0) and each ϕx is a state on C*(Gxx).
An important step in the proof of this theorem is the Renault's disintegration theorem [15],
which will be used to obtain {ϕx}x∈G(0) and µ when a KMS state ϕ on C*(G) is given.
In the �nal part of the thesis, we de�ne the Renault-Deaconu groupoid and prove Thom-
7
sen's theorem.
Let X be a locally compact, second countable, locally Hausdor� space. Given σ : X → X
a local homeomorphism, the Renault-Deaconu groupoid is de�ned by
G = {(x, k, y) : k = n−m,σn(x) = σm(y)},
with composition (x, k, y)(y, l, z) = (x, k + l, z) and inversion (x, k, y)−1 = (y,−k, x).
Although the de�nition of G is abstract, it is useful to have an intuition about this
structure. Note that the sequence {σn(x)}n∈N can be seen as a trajectory starting at x.
Given y ∈ X, (x, k, y) ∈ G means that the trajectories of x and y eventually meet. k can be
interpreted as the delay of one trajectory with respect to the other. Figure 1.2 shows this
idea.
y
x · σ(y) ·
σ(x) σ2(y)
σ2(x) = σ3(y)
Figure 1.2: If (x, k, y) ∈ G then the trajectories {σl(x)}l∈N and {σl(y)}l∈N eventually meet.k can be seen as the delay of one trajectory with respect to the other. In this �gure, k = −1,since σ2(x) = σ3(y).
Given a continuous function F : X → R, we can de�ne a continuous R-valued 1-cocycle
cF by
cF (x, k, y) =n−1∑j=0
F (σj(x))−m−1∑j=0
F (σj(y))
for n,m ∈ N such that k = n − m and σn(x) = σm(y). In fact, there exists a bijection
between R-valued 1-cocycles on G and continuous real-valued functions on X. Then we
de�ne the dynamics on C*(G) by τt(f)(g) = eitcF (g)f(g). We want to describe the KMS
8
states on C*(G) with respect to this dynamics.
Since extremal KMS states are su�cient to describe all KMS states on a C*-algebra,
Thomsen's theorem characterizes only the extremal KMS states on the full C*-algebra of
this groupoid. In this case, we show that the probability measures corresponding to the
KMS states are eβF -conformal measures on X.
The orbit O(x) of x denotes the set of points y ∈ X such that (x, k, y) ∈ G for some k.
There is a bijection between orbits in X and the set of extremal atomic eβF -conformal prob-
ability measures on X. Thomsen's theorem divides extremal KMS-states ϕ corresponding
to measures m in three cases:
• when m is continuous;
• when m purely atomic and corresponds to a periodic orbit;
• when m purely atomic and corresponds to an aperiodic orbit.
In each case the theorem gives a formula for ϕ.
This thesis is structured in the following way:
Chapter 2: we recall some concepts of measure theory. This chapter is important to
understand the properties of the measures corresponding to KMS states on grupoid C*-
algebras. We also de�ne the integral of vector-valued functions on a Banach space.
Chapter 3: we de�ne groupoids and topological groupoids. Then we de�ne the groupoid
C*-algebra and prove some properties of this C*-algebra.
Chapter 4: we de�ne concepts necessary to understand Renault's disintegration theorem
and we state this theorem. However, we do not prove this result.
Chapter 5: we de�ne KMS states on arbitrary C*-algebras and prove some properties.
Then we prove two theorems due to Neshveyev, used to describe KMS states on some
groupoid C*-algebras. We state these theorems below and we refer to them as Neshveyev's
�rst theorem and Neshveyev's second theorem, respectively.
9
Theorem. [17, Theorem 1.1] Let G be a locally compact Hausdor� second countable étale
groupoid. There is a one-to-one correspondence between states on C*(G) with centralizer
containing C0(G(0)) and pairs (µ, {ϕx}x) consisting of a probability measure µ on G(0) and
a µ-measurable �eld of states ϕx on C*(Gxx). Namely, the state corresponding to (µ, {ϕx}x)
is given by
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x) for f ∈ Cc(G).
Theorem. [17, Theorem 1.3] Let G be a locally compact second countable Hausdor� étale
groupoid. Let c be a continuous R-valued 1-cocycle on G and τ be the dynamics on C*(G)
de�ned by τt(f)(g) = eitc(g)f(g) for f ∈ Cc(G), g ∈ G. Fix β ∈ R. Then there exists a one-
to-one correspondence between KMSβ-states on C*(G) and pairs (µ, {ϕx}x∈G(0)) consisting
of a probability measure µ on G(0) and a µ-measurable �eld of states ϕx on C*(Gxx) such
that:
(i) µ is quasi-invariant with Radon-Nikodym derivative e−βc;
(ii) ϕx(ug) = ϕr(h)(uhgh−1) for every g ∈ Gxx and h ∈ Gx, for µ-a.e. x; in particular, ϕx is
tracial for µ-a.e. x;
(iii) ϕx(ug) = 0 for all g ∈ Gxx \ c−1(0), for µ-a.e. x.
Chapter 6: we de�ne the Renault-Deaconu groupoid, describe some of its properties, then
we characterize the extremal KMS-states proving the following theorem due to Thomsen:
Theorem. [26, Theorem 2.2] Let β ∈ R \ {0}. Assume that the periodic points of σ are
countable. The extremal KMSβ-states for τ are
1. States φm, where m is an extremal and continuous (non-atomic) eβF -conformal Borel
probability measure on X;
10
2. The states φλx, where λ ∈ C, |λ| = 1 and x is periodic with minimum period p, such that
p−1∑j=0
F (σj(x)) = 0 and∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)<∞; (1.1)
3. The states φmz where z is aperiodic and β-summable.
11
Chapter 2
Measure Theory
The main theorems in this thesis, described in Chapters 5 and 6, shows that there is a relation
between a KMS state on a particular groupoid C*-algebra and a probability measure on a
subset of this groupoid. In order to understand these theorems, we should recall some results
from measure theory. We also generalize the notion of integral to functions from a measurable
space to a Banach space.
2.1 Radon-Nikodym Theorem
The Radon-Nikodym theorem proves that, under certain conditions, two measures ν, µ are
related by a non-negative measurable function f , denoted the Radon-Nikodym derivative.
In this case, ν can be interpreted as the integral of f with respect to µ. The results in this
section can be found in [14].
De�nition 2.1.1. Let X be a measurable space and let µ, ν be measures on X. We say ν
is absolutely continuous with respect to µ if
µ(A) = 0 implies ν(A) = 0, A measurable.
We denote ν � µ.
12
Note that � de�nes a partial order on the set of measures on X (assuming the σ-algebra
is �xed.)
Theorem 2.1.2. (Radon-Nikodym Theorem) Let X be a measurable space and ν, µ be
σ-�nite measures on X. If ν � µ then there exists a measurable nonnegative function f on
X such that f is �nite µ-a.e. and
ν(A) =
∫A
fdµ, A ⊂ X measurable.
Moreover, ν is �nite if and only if f is integrable.
The function f in Theorem 2.1.2 is called the Radon-Nikodym derivative of ν with respect
to µ and is denoted by
f =dν
dµ. (2.1)
Although we write 2.1 as an equality, the function f is not unique. If there exists a function
g satisfying 2.1, then f = g µ-a.e. We assume equality since we can neglect values of f on a
null set.
Remark 2.1.3. If the measure space X is locally compact Hausdor�, the Radon-Nikodym
derivative is a local property. That is, if we want to �nd the Radon-Nikodym derivative
dνdµ
(x) on a neighborhood of a point x, it is su�cient to study the relation between ν, µ on
this neighborhood.
In fact, let U be an open neighborhood of x and assume there exists a measurable function
∆ on U such that
∫U
f(y)dν(y) =
∫U
f(y)∆(y)dµ(y),
13
for every f ∈ Cc(U). Then using the de�nition of dν/dµ, we have
∫U
f(y)dν
dµ(y)dµ(y) =
∫U
f(y)∆(y)dµ(y).
Since f is arbitrary, we have
dν
dµ(y) = ∆(y), for µ-a.e. y ∈ U .
Example 2.1.4. Let µ be the Lebesgue measure on R. De�ne the measure ν on R by
ν([a, b]) = a3 − b3, for every closed interval [a, b]. Then
ν([a, b]) =
∫ b
a
3x2µ(x).
Then we have, by Remark 2.1.3,
dν
dµ(x) = 3x2.
Now we state some results on the Radon-Nikodym derivative which will be used through-
out the thesis.
Proposition 2.1.5. Let µ, ν be σ-�nite measures on X such that ν � µ. Then, for every
integrable function g with respect to ν we have
∫X
gdν =
∫X
gdν
dµdµ.
Proposition 2.1.6. If µ, ν are σ-�nite measures on X such that ν � µ and dν/dµ 6= 0
µ-a.e., then µ� ν and
dµ
dν=
(dν
dµ
)−1
.
14
Proposition 2.1.7. (Chain rule) If µ, ν, η are measures on X satisfying η � ν � µ, then
dη
dµ=dη
dν
dν
dµ.
2.2 Pushforward Measure
Given a measurable function T : X → Y between two measurable spaces, assume X is en-
dowed with a measure µ. Then we can de�ne a measure on Y , referred to as the pushforward
measure. This notion is de�ned in [24].
This notion will be used to prove Theorem 6.3.21 on page 194:
Theorem. Let β ∈ R. A measure µ on G(0) is eβF -conformal if, and only if, µ is quasi-
invariant with Radon-Nikodym derivative e−βcF .
This theorem shows that one of the hypothesis of Neshveyev's second theorem holds for
every eβF -conformal measure on the unit space of the Renault-Deaconu groupoid. This will
be used to prove Thomsen's theorem.
De�nition 2.2.1. Let X, Y be measurable spaces. Let µ be a measure on X. Given a
measurable function σ : X → Y , we de�ne the pushforward measure σ∗µ on Y by
∫Y
fd(σ∗µ) =
∫X
f ◦ σdµ. (2.2)
Lemma 2.2.2. Equation (2.2) is equivalent to
σ∗µ(A) = µ(σ−1(A)), for every A ⊂ Y measurable. (2.3)
Proof. Assume (2.2) holds. Let A ⊂ Y be measurable. Then χA is a measurable function
15
on Y . σ is measurable, then χA ◦ σ is measurable on X. Note that, for x ∈ A,
χA ◦ σ(x) =
1 if σ(x) ∈ A
0 otherwise=
1 if x ∈ σ−1(A)
0 otherwise= χσ−1(A).
Hence,
σ∗µ(A) =
∫Y
χA(y)d(σ∗µ) =
∫X
χA ◦ σ(x)dµ(x) =
∫X
χσ−1(A)dµ(x) = µ(σ−1(A)).
Conversely, suppose (2.3) holds. Let ϕ be a simple nonnegative measurable function on
Y . There exist a1, . . . , an ≥ 0, A1, . . . , An measurable on Y such that ϕ =∑n
i=1 aiχAi . Then∫Y
ϕdσ∗µ =n∑i=1
aiσ∗µ(Ai) =n∑i=1
aiµ(σ−1(Ai))
=n∑i=1
ai
∫X
χσ−1(Ai)(x)dµ(x)
=n∑i=1
ai
∫X
χAi ◦ σ(x)dµ(x)
=
∫X
ϕ ◦ σ(x)dµ(x).
Let f be a measurable function on Y . Assume f is nonnegative. Then there exists a sequence
of simple nonnegative functions bounded by f and converging to f . Hence (2.2) holds for f .
Therefore (2.2) holds for every measurable function on Y .
Lemma 2.2.3. Let µ be a measure on X. Let σ2 : X → Y , σ1 : Y → Z be measurable.
Then σ1∗σ2∗µ = (σ1 ◦ σ2)∗µ.
Proof. Let A ⊂ X be measurable. Then,
σ1∗σ2∗µ(A) = σ1∗(σ2∗µ)(A) = σ2∗µ(σ−11 (A))
= µ(σ−12 (σ−1
1 (A))) = µ((σ1 ◦ σ2)−1(A))
16
= (σ1 ◦ σ2)∗(A).
Lemma 2.2.4. Let µ, ν be measures on X such that ν � µ. Let σ : X → Y be a measurable
bijection such that σ−1 is measurable. Then σ∗ν � σ∗µ and
dσ∗ν
dσ∗µ(y) =
dν
dµ(σ−1(y)) y ∈ Y .
Proof. Let f be a σ∗ν-integrable function on Y . Then
∫Y
f(y)dσ∗ν(y) =
∫X
f ◦ σ(x)dν(x)
=
∫X
f ◦ σ(x)dν
dµ(x)dµ(x)
=
∫Y
f(y)dν
dµ(σ−1(y))dσ∗µ(y)
Then σ∗ν � σ∗µ and
dσ∗ν
dσ∗µ(y) =
dν
dµ(σ−1(y)) y ∈ Y .
2.3 Purely Atomic and Non-Atomic Measures
In this section we recall that every �nite Borel measure can be decomposed uniquely as a
sum of two measures, one being purely atomic and the other one being continuous. As a
consequence, an extremal probability measure is either purely atomic or continuous.
In order to prove Thomsen's theorem, we will show in Chapter 6 that every extremal KMS
17
state corresponds to an extremal probability m, then m has one of the properties de�ned
below. The results in this section can be found in [12] and [25].
De�nition 2.3.1. A �nite Borel measure m on the topological space X is non-atomic or
continuous when m({x}) = 0 for every x ∈ X and purely atomic if there is a Borel set
A ⊂ X such that m(A) = m(X) and m({a}) > 0 for all a ∈ A.
Given a measure µ, we write µ(x) = µ({x}).
Proposition 2.3.2. If µ is a σ-�nite measure on a σ-algebra then there exist unique measures
µa and µc such that µ = µa + µc and such that µa is purely atomic and µc is non-atomic.
2.4 Measures on Locally Compact Spaces
The results in this section are presented in [6]. Here we introduce the notion of Radon
measures and we conclude that, if the topological space satis�es certain conditions, every
probability measure is Radon.
First we prove some properties of Hausdor� spaces.
Proposition 2.4.1. Let X be a Hausdor� space, and let K and L be disjoint compact
subsets of X. Then there are disjoint open subsets U and V of X such that K ⊂ U and
L ⊂ V .
Proof. We can assume that K and L are both non-empty (otherwise we could use ∅ as one
of our open sets and X as the other). Let us begin with the case where K contains exactly
one point, say x. We show that there are open disjoint sets Ux, Vx such that x ∈ Ux and
L ⊂ Vx.
Since X is Hausdor�, for each y ∈ L there is a pair Uy, Vy of disjoint open sets such that
x ∈ Uy and y ∈ Vy. Since L is compact, there is a �nite family y1, . . . , yn such that the sets
Vy1 , . . . , Vyn cover L. The sets Ux and Vx de�ned by Ux = ∩ni=1Uyi , Vx = ∪ni=1Vyi are then the
required sets.
18
Next consider the case where K has more than one element. We have just shown that for
each x ∈ K there are disjoint open sets Ux and Vx such that x ∈ Ux and L ⊂ Vx. Since K
is compact, there is a �nite family x1, . . . , xn such that Ux1 , . . . , Uxn cover K. The proof is
complete if we de�ne U = ∪ni=1Uxi , V = ∩ni=1Vxi .
Proposition 2.4.2. Let X be a locally compact Hausdor� space, x a point in X, and U an
open neighborhood of x. Then x has an open neighborhood whose closure is compact and
included in U .
Proof. Since X is locally compact, there is an open neighborhood of x, sayW , whose closure
is compact. By replacing W with W ∩U , we assume that W is included in U . The di�culty
is that W may extend outside U .
Use Proposition 2.4.1 to choose disjoint open sets V1 and V2 that separate the compact
sets {x} andW \W . Note that the closure of V1∩W is included inW . In fact, suppose there
exists y ∈ V1 ∩W such that y /∈ W . Then y ∈ W \W . By de�nition, V2 is a neighborhood
of y. Since y ∈ V1 ∩W , there exists x1 ∈ V1 ∩ W such that x1 ∈ V2. This leads to a
contradiction because V1 ∩ V2 = ∅.
Then V1 ∩W is compact and included in W , and hence in U ; thus V1 ∩W is the required
open neighborhood of x.
A subset of a topological space X is a Gδ if it is the intersection of a sequence of open
subsets of X, and Fσ if it is the union of a sequence of closed subsets of X.
Proposition 2.4.3. Let X be a locally compact Hausdor� space, let K be a compact subset
of X, and let U be an open subset of X that includes K. Then there is an open set V of X
that has a compact closure and satis�es K ⊂ V ⊂ V ⊂ U .
Proof. Proposition 2.4.2 implies that each point in K has an open neighborhood whose
closure is compact and included in U . Since K is compact, some �nite collection of these
neighborhoods covers K. Let V be the union of these sets in such a �nite collection; then V
is the required set.
19
Proposition 2.4.4. Let X be a locally compact Hausdor� second countable space. Then
each open subset of X is an Fσ, and is in fact the union of a sequence of compact sets.
Likewise, each closed subset is a Gδ.
Proof. Suppose that U is a countable basis for the topology of X. Let U be an open set in
X. Given x ∈ U , it follows from Proposition 2.4.2 that there exists an open neighborhood
Wx of x such that Wx is compact and Wx ⊂ U . Since U is the basis for the topology of X,
there exists Vx ∈ U such that x ∈ Vx ⊂ Wx. Then Vx is compact and Vx ⊂ U . Thus,
U =⋃x∈U
Vx.
Since each Vx ∈ U and U is countable, then U is a countable union of compact sets. Therefore
U is Fσ.
Let A ⊂ X be a closed set. Then Ac is open, and Ac is the union of a sequence {Fn}
consisting of closed sets. Hence,
A = (Ac)c =
(∞⋃n=1
Fn
)c
=∞⋂n=1
F cn.
Therefore A is Gδ.
Lemma 2.4.5. Let X be a locally compact Hausdor� second countable space. Given an
open subset U ⊂ X, there exists a sequence {Kn} of compact subsets such that Kn ⊂ Kn+1
for every n, and U = ∪∞n=1Kn.
Proof. Let U ⊂ X be an open set. It follows from Proposition 2.4.4 that there is a sequence
of compact sets {Fn} such that U = ∪nFn. De�ne, for each n ≥ 1, Kn = ∪ni=1Fi. Clearly
each Kn is compact and Kn ⊂ Kn+1. Moreover, U = ∪∞n=1Kn.
Let X be a Hausdor� topological space. Then B(X), the Borel σ-algebra on X, is the
σ-algebra generated by the open subsets of X; the Borel subsets of X are those that belong
to B(X).
20
We turn to terminology for measures. Let X be a Hausdor� topological space. A Borel
measure on X is a measure whose domain is B(X). Suppose that A is a σ-algebra on X
such that B(X) ⊂ A. A positive measure µ on A is Radon if
(a) each compact subset K of X satis�es µ(K) <∞,
(b) each set A in A satis�es
µ(A) = inf{µ(U) : A ⊂ U and U is open}, and
(c) each open set U of X satis�es
µ(U) = sup{µ(K) : K ⊂ U and K is compact}.
A Radon Borel measure on X is a Radon measure whose domain is B(X). A measure that
satis�es condition (b) is often called outer regular , and a measure that satis�es condition
(c), inner regular .
Now we de�ne the support of a Radon Borel measure. The following theorem is necessary
to show that the support is well-de�ned.
Proposition 2.4.6. Let X be a locally compact Hausdor� space, let µ be a Radon Borel
measure on X. Then the union of all open subsets of X that have measure zero under µ is
itself an open set that has measure zero under µ.
Proof. Let U be the collection of all open subsets of X that have measure zero under µ, and
let U be the union of the sets in U . Then U is open. If K is a compact subset of U , then K
can be covered by a �nite collection U1, . . . , Un of sets that belong to U , and so we have
µ(K) ≤n∑i=1
µ(Ui) = 0.
Since K is arbitrary, it follows from the de�nition of Radon measure that µ(U) = 0.
21
It follows from Proposition 2.4.6 that, for X, µ, there is the largest open subset U ⊂ X
with µ(U) = 0. Then we de�ne the support as follows.
De�nition 2.4.7. LetX be a locally compact Hausdor� space, and µ a Radon Borel measure
on X. We de�ne the support of µ as the complement of the largest open subset of X with
measure zero. We denote the support of X by supp(µ).
Note that supp(µ) is closed. Now we prove some properties of Radon measures.
Lemma 2.4.8. Let X be a Hausdor� space in which each open set is an Fσ, and let µ be a
�nite Borel measure on X. Then each Borel subset A of X satis�es
µ(A) = inf{µ(U) : A ⊂ U and U is open}, (2.4)
µ(A) = sup{µ(F ) : F ⊂ A and F is closed}. (2.5)
In particular, µ is Radon.
Proof. Let R denote the set of Borel sets A ⊂ X that satisfy conditions (2.4) and (2.5). We
prove that R contains all open subsets of X. Let U ⊂ X open. Clearly U satis�es (2.4). By
hypothesis there exists a sequence {Fn} of closed sets such that U = ∪nFn. We can assume
that Fn ⊂ Fn+1 for each n without loss of generality. Then µ(U) = limn µ(Fn). Therefore
(2.5) holds.
Now we show that conditions (2.4) and (2.5) hold for an arbitrary Borel set A if, and only
if, for every ε > 0 there are U open, F closed, such that
F ⊂ A ⊂ U and µ(U \ F ) < ε. (2.6)
In fact, assume (2.4) and (2.5) hold. Let A be measurable. Given ε > 0, by (2.4) there exists
U open such that A ⊂ U and
µ(U) < µ(A) + ε/2. (2.7)
22
Aplying (2.5), there exists F ⊂ A closed satisfying
µ(F ) > µ(A)− ε/2. (2.8)
Then, by (2.7) and (2.8), we have
µ(U \ F ) = µ(U)− µ(F ) <(µ(A) +
ε
2
)−(µ(A)− ε
2
)= ε.
Then (2.6) holds.
Conversely, assume (2.6) holds. Given ε > 0, there are U open, F closed, such that
F ⊂ A ⊂ U and µ(U \ F ) < ε. Hence,
µ(A) ≤ µ(U) = µ(A) + µ(U \ A)
≤ µ(A) + µ(U \ F ) , since F ⊂ A,
< µ(A) + ε,
and
µ(F ) ≤ µ(A) = µ(F ) + µ(A \ F )
≤ µ(F ) + µ(U \ F ) , since A ⊂ U ,
< µ(F ) + ε.
Since ε > 0 is arbitrary, it follows that conditions (2.4) and (2.5) are satis�ed.
We can now show that R is a σ-algebra. Clearly ∅ ∈ R, since ∅ is open. Given A ∈ R,
ε > 0, there are U open, F closed such that F ⊂ A ⊂ U and µ(U \ F ) < ε. Then
F c is open, U c is closed, and U c ⊂ Ac ⊂ F c. Since F c \ U c = U \ F , it follows that
µ(F c \ U c) = µ(U \ F ) < ε. Therefore Ac ∈ R.
Let {An} be a sequence of sets in R. Then, for every n ≥ 1, there exists Un open, Fn
23
closed such that
Fn ⊂ An ⊂ Un and µ(Un \ Fn) <ε
2n+1.
Let U = ∪nUn and F = ∪nFn. Then U , F satisfy the relations F ⊂ ∪nAn ⊂ U and
µ(U \ F ) ≤ µ
(∞⋃n=1
(Un \ Fn)
)≤
∞∑n=1
µ(Un \ Fn) <∞∑n=1
ε
2n+1=ε
2. (2.9)
The set U is open, but the set F can fail to be closed. However for each N the set ∪Nn=1Fn
is closed, and since
µ(U \ F ) = µ(U)− µ(F ) = µ(U)− limN→∞
µ
(N⋃n=1
Fn
),
we can choose N such that
µ
(U \
N⋃n=1
Fn
)< ε.
Thus U and ∪Nn=1Fk are the sets required in (2.6), and R is closed under countable unions.
We have now shown that R is a σ-algebra on X that contains the open sets. Since B(X) is
the smallest σ-algebra on X that contains the open sets, it follows that B(X) ⊂ R. Therefore
this lemma is proved.
Remark 2.4.9. Let X be a locally compact Hausdor� second countable space, then every
probability measure on X is Radon. In fact, it follows from Proposition 2.4.4 that every
open set is Gδ. Then, given a probability measure µ on X, it follows from Lemma 2.4.8 that
µ is Radon.
We assume in Lemma 2.4.8 that the measure µ is �nite, but this result can be generalized
to σ-�nite measures that are �nite on compact sets.
24
Proposition 2.4.10. Let X be a locally compact Hausdor� space that has a countable
basis, and let µ be a Borel measure on X that is �nite on compact sets. Then µ is Radon.
Proof. First consider the inner regularity of µ. Let U be an open subset of X. Lemma 2.4.5
implies that U is the union of a sequence {Kj} of compact subsets, then
µ(U) = limn→∞
µ
(n⋃j=1
Kj
).
The inner regularity follows.
Let {Un} be a sequence of open sets such that X = ∪nUn and such that µ(Un) <∞ holds
for each n (for instance, take a countable basis U for X, and arrange in a sequence those
sets U in U for which U is compact).
For each n de�ne a Borel measure µn on X by µn(A) = µ(A ∩ Un). The measures µn are
�nite, and so Lemma 2.4.8 implies that they are outer regular. Hence if A belongs to B(X)
and if ε is a positive number, then for each n there is an open set Vn that includes A and
satis�es µn(Vn) < µn(A) + ε/2n. Consequently,
µ((Un ∩ Vn) \ A) < ε/2n.
Then set V de�ned by V = ∪n(Un ∩ Vn) is open, includes A and satis�es
µ(V \ A) ≤∑n
µ((Un ∩ Vn) \ A) < ε.
Hence µ(V ) ≤ µ(A) + ε, and the outer regularity of µ follows.
Assume X is a locally compact second countable Hausdor� space. By de�nition, a Radon
measure µ on X is �nite on compact subsets of X. It follows from Proposition 2.4.10 that
a Borel measure on X is Radon if, and only if, it is �nite on the compact subsets of X.
Proposition 2.4.11. [6, Proposition 7.2.6] Let X be a Hausdor� space, let A be a σ-algebra
25
on X that includes B(X), and let µ be a Radon measure on A. If A belongs to A and is
σ-�nite under µ, then
µ(A) = sup{µ(K) : K ⊂ A and K is compact }. (2.10)
Remark 2.4.12. Let X be a locally compact second countable Hausdor� space, and µ a
Borel measure which is �nite on compact subsets of X. It follows from Propositions 2.4.10
and 2.4.11 that, for every A ⊂ X Borel, we have
µ(A) = inf{µ(U) : A ⊂ U and U is open},
µ(A) = sup{µ(K) : K ⊂ A and K is compact}.
Lemma 2.4.13. Let X be a locally compact Hausdor� second countable space, µ be a
Radon Borel measure on X, B ⊂ X a Borel set, U ⊂ X an open set satisfying B ⊂ U .
Given a continuous non-negative function f on X, we have
∫B
f(x)dµ(x) = infB⊂V⊂UV open
∫V
f(x)dµ(x).
Proof. Since f is continuous and non-negative, we can de�ne the Borel measure ν on X by
ν(A) =
∫A
f(x)dµ(x), A Borel set.
The function f is continuous and µ is �nite on compact subsets, then ν is �nite on compact
subsets of X. Hence, from Proposition 2.4.10, ν is Radon. Therefore, for every B ⊂ X
Borel,
∫B
f(x)dµ(x) = ν(B) = infB⊂VV open
ν(V ) = infB⊂VV open
∫V
f(x)dµ(x).
For every open set V such that B ⊂ V , it follows that µ(B) ≤ µ(V ∩ U) ≤ µ(U) and
26
V ∩U is open. Hence, we can take the in�mum over the open sets V such that B ⊂ V ⊂ U .
Therefore,
∫B
f(x)dµ(x) = ν(B) = infB⊂V⊂UV open
ν(V ) = infB⊂V⊂UV open
∫V
f(x)dµ(x).
Lemma 2.4.14. Let X be a locally compact Hausdor� second countable space and µ a
Radon measure on X. Given an open subset U ⊂ X, we have
µ(U) = supf∈Cc(X)0≤f≤χU
∫X
f(x)dµ(x).
Proof. • Assume µ(U) <∞.
Note that for every f ∈ Cc(X) such that 0 ≤ f ≤ χU , we have∫X
f(x)dµ(x) ≤ µ(U).
Given ε > 0, there is a compact set K ⊂ U satisfying µ(U \K) < ε by Remark 2.4.12.
Since X is locally compact Hausdor�, there is an open set V such that V is compact
and K ⊂ V ⊂ V ⊂ U by Proposition 2.4.3.
By Urysohn's lemma, there exists a continuous function f assuming values in the
interval [0, 1] such that f equals one on K and vanishes outside V ⊂ U . Then f ∈
Cc(X) and 0 ≤ f ≤ χU .
Using the fact that µ(U)− µ(K) ≤ ε, we have µ(K) ≥ µ(U)− ε. Hence,
∫X
f(x)dµ(x) = µ(K) +
∫U\K
f(x)dµ(x)
≥ µ(K)
≥ µ(U)− ε.
27
Since ε is arbitrary, we have
µ(U) = supf∈Cc(X)0≤f≤χU
∫X
f(x)dµ(x).
• Suppose µ(U) =∞.
Let n be a natural number. By Remark 2.4.12, there exists a compact set Kn ⊂ U
such that µ(Kn) ≥ n. From Urysohn's lemma, we can choose a continuous compactly
supported function fn assuming values in the interval [0, 1] such that fn(x) = 1 for
every x ∈ Kn and fn vanishes of U . Hence,
∫X
fn(x)dµ(x) ≥∫Kn
fn(x)dµ(x) = µ(Kn) ≥ n.
Therefore,
supf∈Cc(X)0≤f≤χU
∫X
f(x)dµ(x) ≥ supn∈N
∫X
fn(x)dµ(x) =∞.
Hence the result follows.
2.5 µ-Measurable Functions
In this section we de�ne the µ-completion of a σ-algebra. The de�nition here can be found
in [6]. This σ-algebra will be necessary to understand one of the conditions in Neshveyev's
�rst theorem.
De�nition 2.5.1. Let (X,A) be a measurable space and let µ be a measure on A. The
completion of A under µ is the collection Aµ of subsets A of X for which there are sets E
and F in A such that
28
E ⊂ A ⊂ F and µ(F \ E) = 0.
A set that belongs to Aµ is said to be µ-measurable.
In fact, Aµ is a σ-algebra on X. We say that a function f on X is µ-measurable if it is
measurable with respect to the σ-algebra Aµ. Note that if f is measurable with respect to
the σ-algebra A, then f is µ-measurable.
Lemma 2.5.2. LetX be a topological space and µ a Borel purely atomic probability measure
on X. Every complex-valued function on X is µ-measurable.
Proof. Let f be a complex valued function on X. Let I be the set of points x ∈ X such that
µ({x}) > 0, then I is countable µ(X \ I) = 0. Given V ⊂ C measurable, let A = f−1(V )
and J = I \ A. Then I, J are measurable since both are countable.
Note that I ∩ A ⊂ A ⊂ X \ J . Since (X \ J) \ I ⊂ X \ I, it follows that
µ((X \ J) \ I) ≤ µ(X \ I) = 0.
Then A is µ-measurable and, therefore, f is µ-measurable.
2.6 Vector-Valued Integration
Now we introduce the concept of vector-valued integral, that is, the integral of functions
f : R → B where B is a complex Banach space. This section is based on [21]. We will
need this notion to prove Proposition 5.1.19 on page 105, and then de�ne KMS states on a
arbitrary C*-algebra.
Recall that one of the main steps in the construction of the Lebesgue integral is the notion
of simple functions. A function ϕ : R→ R is simple if there are A1, · · · , An measurable sets
29
and a1, . . . , an real numbers such that
ϕ =n∑i=1
aiχAi .
Then we de�ne its integral by
∫Rϕ(t)dµ(t) =
n∑i=1
aiµ(Ai).
Then, under certain conditions, the integral of a measurable function can be approximated
by the integral of simple functions. We will try to de�ne the integral of vector-valued
functions similarly.
De�nition 2.6.1. Let µ be a Borel measure on R and B a Banach space. A function
ϕ : R→ B is simple if there are a1, . . . , an ∈ B, and A1, . . . , An Borel subsets of B such that
ϕ =n∑i=1
aiχAi , (2.11)
and each χAi : R→ R is de�ned by
χAi(x) =
1, if x ∈ Ai,
0, if x /∈ Ai,
for x ∈ R. We call (2.11) a representation of ϕ.
In the next example, we show that the representation (2.11) is not necessarily unique.
Example 2.6.2. Let ϕ : R→ R3 be de�ned by
ϕ(x) =
(1, 1, 1), if 0 < x ≤ 1,
(1, 2, 2), if 1 < x ≤ 2,
(1, 2, 3), if 2 < x ≤ 3.
30
Then,
ϕ = (1, 1, 1)χ(0,1] + (1, 2, 2)χ(1,2] + (1, 2, 3)χ(2,3]
= (1, 1, 1)χ(0,3] + (0, 1, 1)χ(1,3] + (0, 0, 1)χ(2,3].
Therefore ϕ is simple and can be written in at least two di�erent representations.
De�nition 2.6.3. Let µ be a Borel measure on R and B a Banach space. Given a simple
function ϕ with representation (2.11), we de�ne its integral by
∫Rϕ(x)dµ(x) =
n∑i=1
aiµ(Ai).
Lemma 2.6.4. The integral in De�nition (2.6.3) is well-de�ned, that is, it does not depend
on the representation.
Proof. Let B be a Banach space. Given a simple function ϕ : R → B, let a1, . . . , an ∈ B,
b1, . . . , bm ∈ B, and let A1, . . . , An, B1, . . . , Bm be Borel sets such that
ϕ =n∑i=1
aiχAi =m∑j=1
biχAi .
Let
x =n∑i=1
aiµ(Ai) and y =m∑j=1
bjµ(Bj).
Note that x, y ∈ B. Choose an arbitrary Λ ∈ B*. Then Λ ◦ ϕ : R→ C is a simple function
with
Λ ◦ ϕ =n∑i=1
Λ(ai)χAi =m∑j=1
Λ(bi)χAi .
Since the integral of complex-valued functions does not depend on the representation, we
31
have
∫R(Λ ◦ ϕ)(t)dµ(t) =
n∑i=1
Λ(ai)µAi =m∑j=1
Λ(bi)µBi
= Λ
(n∑i=1
aiµAi
)= Λ
(m∑j=1
bjµBj
)
= Λ(x) = Λ(y)
Since Λ is arbitrary and B* separates points in B, it follows that x = y.
Remark 2.6.5. Given a simple function ϕ : R → B, Λ ∈ B*, Λ ◦ ϕ : R → C is a simple
function. Note that we used the property
Λ
(∫Rϕ(t)dµ(t)
)=
∫R
Λ(ϕ(t))dµ(t)
in Lemma 2.6.4 to show that the integral is well-de�ned. Similarly, we will de�ne the integral
in such a way that this property holds when we replace ϕ by a Borel function f : R → B.
Given Λ ∈ B*, we denote Λf = Λ ◦ f . Note that both Λf and Λϕ are measurable functions.
De�nition 2.6.6. Given a Banach space B, a function f : R → B is weakly measurable if
Λf is measurable for every Λ ∈ X*.
Remark 2.6.7. Note that every Borel function f : R → B is weakly measurable. In
particular, every continuous function from R to B is weakly measurable.
De�nition 2.6.8. Let µ be a Borel measure on R. Given a Banach space B, let f : R→ B
be weakly measurable. If there exists y ∈ B such that for every Λ ∈ B*,
Λy =
∫R
Λf(t)dµ(t),
then we de�ne the integral of f by
∫Rf(t)dµ(t) = y. (2.12)
32
Remark 2.6.9. Note that there exists at most one y such that (2.12) holds. This follows
from the fact that B* separates points in B.
De�nition 2.6.10. Given a Banach space B, Cc(R, B) denotes the space of compactly
supported functions f : R → B which are continuous. Recall that the support of f is the
closure of the set {t ∈ R : f(t) 6= 0}.
Note that every function f ∈ Cc(R, B) is weakly measurable.
Given a Banach space B, we de�ne the norm on Cc(R, B) by
‖f‖∞ = supt∈R‖f(t)‖.
Lemma 2.6.11. Let µ be a Borel measure on R, B a Banach space and f ∈ Cc(R, B) such
that there exists y =∫R f(t)dµ(t). Then
‖y‖ ≤∫R‖f(t)‖dµ(t).
Proof. Since B is a Banach space, we have
‖y‖ = supΛ∈B*‖Λ‖≤1
|Λy|.
However, for every Λ ∈ B* such that ‖Λ‖ ≤ 1, we have
|Λy| =∣∣∣∣∫
RΛf(t)dµ(t)
∣∣∣∣ ≤ ∫R|Λf(t)|dµ(t) ≤
∫R‖Λ‖‖f(t)‖dµ(t) ≤
∫R‖f(t)‖dµ(t).
Lemma 2.6.12. Let µ be a Radon measure on R. Let f ∈ Cc(R, B). Then, for every ε > 0,
there are A1, . . . An disjoint Borel sets, t1, . . . , tn ∈ R, such that the function
ϕ =n∑i=1
f(ti)χAi
33
satis�es the following property: ‖ϕ− f‖∞ ≤ ε.
Proof. Let ε > 0. Since f is continuous, for every t ∈ R, there exists an open set Ut such
that for every s ∈ Ut, ‖f(s)− f(t)‖ < ε.
Let K be the support of f . Then there are t1, . . . tn ∈ K such that Ut1 , . . . , Utn is an open
cover for K. Let, for i = 1, . . . , n,
Ai =
Ut1 if i = 1
Uti \ Ai−1 if i = 2, . . . , n.
Then each Ai is Borel, and ∪ni=1Ai = ∪ni=1Uti . De�ne ϕ by
ϕ =n∑i=1
f(ti)χAi .
Now we prove that ‖ϕ − f‖∞ ≤ ε. Let t ∈ R. If t /∈ ∪ni=1Uti , then ϕ(t) = 0 by de�nition.
Moreover, t /∈ K, since Ut1 , . . . , Utn cover K. Then f(t) = 0 and ‖f(t)− ϕ(t)‖ = 0 ≤ ε.
Assume t ∈ ∪ni=1Uti . By de�nition of A1, . . . , An, there exists a unique i such that t ∈ Ai.
Hence ϕ(t) = f(ti). Since Ai ⊂ Uti , we have
‖ϕ(t)− f(t)‖ = ‖f(ti)− f(t)‖ < ε.
Therefore,
‖ϕ− f‖∞ = supt∈R‖ϕ(t)− f(t)‖ ≤ ε.
In order to prove the existence of the integral of functions in Cc(R, B), we will state
Theorem 2.6.14, which is an application of a theorem proved in [22].
De�nition 2.6.13. Let X be a normed vector space. Given a subset S of X, we de�ne
34
its convex hull as the smallest convex set containing S. We denote the convex hull of S by
co(S).
Theorem 2.6.14. [22, Theorem 3.25] Suppose H is the convex hull of a compact set K in
a Banach space B. Then H is compact.
Theorem 2.6.15. Let µ be a Radon measure on R. Let B be a Banach space. Given
f ∈ Cc(R, B), the integral of f :
y =
∫Rf(t)dµ(t)
exists.
Proof. Assume µ is a probability measure.
Let K = supp(f). Let L = f(K) ∪ {0}. This set is compact because f is continuous and
K is compact. De�ne H to be the closure of co(L). Then H is compact by Theorem 2.6.14.
Given k ∈ N with k ≥ 1, it follows from Lemma 2.6.12 that there is a simple function
ϕ(k) : R → B such that there are disjoint Borel sets A(k)1 , . . . , A
(k)nk , and t
(k)1 , . . . , t
(k)nk ∈ R
satisfying
ϕ(k) =
nk∑i=1
f(t(k)i )χ
A(k)i
and ‖ϕ(k) − f‖∞ <1
k.
Let
yk =
∫Rϕ(k)dµ(t) =
nk∑i=1
f(t(k)i )µ(A
(k)i ). (2.13)
Since the sets A(k)1 , . . . , A
(k)nk are disjoint and µ is a probability measure, we have
nk∑i=1
µ(A(k)i ) ≤ 1 and ‖yk‖ ≤ ‖f‖∞. (2.14)
35
Moreover, since each f(t(k)i ) ∈ H and 0 ∈ H, it follows from (2.13) and (2.14) that yk ∈ H.
H is compact, then {yk}k∈N has a subsequence {ykj}j∈N converging to some y ∈ H.
Let Λ ∈ B*. Assume Λ 6= 0 without loss of generality. Since Λ is continuous, we have
Λykj → Λy. However, by Remark 2.6.5,
Λykj = Λ
(∫Rϕ(kj)(t)dµ(t)
)=
∫R
Λϕ(kj)(t)dµ(t). (2.15)
By de�nition, each ϕ(kj) satis�es ‖ϕ(kj)(t)‖ ≤ ‖f‖∞. Hence, for t ∈ R,
|Λϕ(kj)(t)| ≤ ‖Λ‖‖ϕ(kj)(t)‖ ≤ ‖Λ‖‖f‖∞.
Moreover, Λϕ(kj) converges to Λf pointwise. Since µ is a probability measure, we can apply
the dominated convergence theorem, obtaining
∫R
Λf(t)dµ(t) = limj→∞
∫R
Λϕ(kj)(t)dµ(t) = limj→∞
Λykj = Λy.
Since Λ is arbitrary, we have
y =
∫Rf(t)dµ(t).
Now assume µ is an arbitrary Radon measure on R. Let K be the support of f .
Suppose µ(K) = 0, then for each Λ ∈ B*,
∫R
Λ(f(t))dµ(t) =
∫K
Λ(f(t))dµ(t) = 0.
Therefore∫R f(t)dµ(t) = 0.
Now suppose µ(K) > 0. De�ne the measure µ on R by
µ(I) =µ(I ∩K)
µ(K),
36
for every Borel set I ⊂ R. Then µ is a probability Borel measure on R.
Let y =∫R f(t)dµ(t). Then, for every Λ ∈ B*,
Λy =
∫R
Λ(f(t))dµ(t)
=
∫K
Λ(f(t))dµ(t)
=1
µ(K)
∫K
Λ(f(t))dµ(t)
=1
µ(K)
∫R
Λ(f(t))dµ(t).
Therefore,
µ(K)y =
∫Rf(t)dµ(t).
Proposition 2.6.16. Let µ be a Borel measure on R and let B be a Banach space. Let
f : R → B be a continuous function such that∫∞−∞ ‖f(t)‖dµ(t) < ∞. Then the integral∫
R f(t)dµ(t) exists.
Proof. Let, for each n, hn be a continuous function such that hn equals 1 in the closed
interval [−n, n] and vanishes outside ]− n− 1, n + 1[. For each n, fhn ∈ Cc(R, B). De�ne,
for every n,
yn =
∫Rhn(t)f(t)dµ(t).
In order to show that the sequence of yn converges, we only need to prove that {yn} is a
Cauchy sequence because B is complete. Given ε > 0, let n0 be such that
∫|t|≥n0
‖f(t)‖ < ε
2.
37
Given n,m ≥ n0, assume n ≥ m without loss of generality. By de�nition, we have
hn(t) = hm(t) = 1 for t satisfying |t| < n0. (2.16)
In this case, |hn(t)− hm(t)| = 0. Then,
‖yn − ym‖ =
∥∥∥∥∫Rhn(t)f(t)dµ(t)−
∫Rhm(t)f(t)dµ(t)
∥∥∥∥=
∥∥∥∥∫R(hn(t)− hm(t))f(t)dµ(t)
∥∥∥∥≤∫R|hn(t)− hm(t)|‖f(t)‖dµ(t)
=
∫|t|≥n0
|hn(t)− hm(t)|‖f(t)‖dµ(t), by (2.16),
≤∫|t|≥n0
(|hn(t)|+ |hm(t)|)‖f(t)‖dµ(t)
≤ 2
∫|t|≥n0
‖f(t)‖dµ(t), because hn, hm assume values in [0, 1],
≤ 2ε
2= ε.
Therefore yn → y for some y ∈ B.
Now we prove that∫R Λ(hn(t)f(t))dµ(t)→
∫R Λ(f(t))dµ(t). For every t, we have
|Λ(hn(t)f(t))| ≤ ‖λ‖‖f(t)‖.
By assumption, the function t 7→ ‖f(t)‖ is integrable. Then, the dominated convergence
theorem implies,
limn→∞
∫R
Λ(hn(t)f(t))dµ(t) =
∫R
(limn→∞
Λ(hn(t)f(t)))dµ(t) =
∫R
Λ(f(t))dµ(t).
38
Note that, for every n,
Λ
(∫Rhn(t)f(t)dµ(t)
)=
∫R
Λ(hn(t)f(t))dµ(t).
The left-hand side equals to Λyn and thus converges to Λy. As we already proved, the
right-hand side converges to∫R Λ(f(t))dµ(t). Therefore,
Λy =
∫Rλ(f(t))dµ(t).
Λ is arbitrary, then the integral
y =
∫Rf(t)dµ(t)
exists.
Corollary 2.6.17. Let B be a Banach space and µ a Borel measure on X. Let f : R→ B
be a continuous function such that t 7→ ‖f(t)‖ is integrable. Given a linear and bounded
operator L : B → B1 such that B1 is a Banach space, then
L
(∫Rf(t)dµ(t)
)=
∫RL(f(t))dµ(t).
Proof. Let y =∫R f(t)dµ(t). The function Lf : R → B1 is continous, moreover, t 7→
‖L(f(t))‖ is integrable, since
∫R‖L(f(t))‖dµ(t) ≤ ‖L‖
∫R‖f(t)‖dµ(t) <∞.
Let z =∫R L(f(t))dµ(t). Given Λ ∈ B1*, ΛL ∈ B*. Hence,
Λz = Λ
∫RL(f(t))dµ(t)
=
∫R
ΛL(f(t))dµ(t)
39
=
∫R(ΛL)(f(t))dµ(t)
= (ΛL)
∫R(f(t))dµ(t)
= (ΛL)y
= Λ(Ly).
Since Λ is arbitrary, it follows that z = Ly. Thus,
L
(∫Rf(t)dµ(t)
)=
∫RL(f(t))dµ(t).
40
Chapter 3
Groupoids
Groupoids can be understood as a generalization of groups where the unit is not unique
and not every pair of elements can be multiplied. Each groupoid G is endowed with two
functions r and s from G to the subset G(0) of units. We equip the groupoid with a topology
such that r, s are continuous.
If G has some nice topological properties, we can de�ne Cc(G), the space of continuous
and compactly supported complex functions on G. Moreover, we can endow this space with
an involution and a product which is not necessarily commutative. Then we de�ne a norm on
Cc(G) which depends on the ∗-representations of Cc(G). Then the full groupoid C*-algebra,
denoted C*(G) is the completion of Cc(G) with respect to this norm.
Most de�nitions and results in this chapter can be found in [9].
3.1 Introduction
In this section we de�ne groupoids and give some examples. The results in this section are
taken from [9] and [20].
De�nition 3.1.1. A groupoid is a set G together with a subset G(0) (called units , unit space
or objects), two surjective maps r, s : G → G(0) (called range and source, respectively) and
41
a law of composition
(g, h) ∈ G(2) 7→ gh = g · h ∈ G,
where G(2) = {(g, h) ∈ G × G : s(g) = r(h)} is called the set of composable elements or
composable pairs .
A groupoid satis�es the following properties for g, h, k ∈ G:
(i) s(gh) = s(h) and r(gh) = r(g) if (g, h) ∈ G(2);
(ii) r(x) = s(x) if x ∈ G(0);
(iii) gs(g) = g and r(g)g = g;
(iv) (gh)k = g(hk) if (g, h), (h, k) ∈ G(2);
(v) g has a two-sided inverse g−1 such that gg−1 = r(g) and g−1g = s(g).
The maps (g, h) ∈ G(2) 7→ gh and g 7→ g−1 are called product and inverse, respectively.
We can interpret groupoids as a collection of arrows attached to points on a plane. Two
arrows can be composed only if the end of the �rst arrow meets the start of the second.
Units are points with the null vector and the inverse of an element is obtained by reversing
the direction of the arrow. Figure 3.1 shows this idea.
s(g)
r(g)
s(h)
r(h)
gh
(a)
r(g)
s(g) = r(h)
s(h)g
h
gh
(b)
s(g)
r(g)
s(g)
r(g)
s(g) = g−1g
r(g) = gg−1
g g−1
(c)
Figure 3.1: Groupoids can be seen as arrows on a plane. s(g) and r(g) denote the sourceand range of g. (a) g and h are not composable, since s(g) 6= r(h); (b) The composition ofg and h is gh; (c) g−1 is the inverse of g. Note that g−1g = s(g) and gg−1 = r(g).
42
Example 3.1.2. Every group is a groupoid. Let G be a group with unit e. Let G(0) = {e},
G(2) = G×G and de�ne the range and source maps by r(g) = s(g) = e.
Since the range and source of each element is e and G is associative, one can easily show
that properties (i)-(v) are satis�ed.
Example 3.1.3. We show that a group action de�nes a groupoid. Let G be a group with
identity e and X a set. Recall that a group action [19] is a map G × X → X denoted by
(g, x) 7→ gx, satisfying the following properties:
(i) g(hx) = (gh)x for g, h ∈ G, x ∈ X,
(ii) ex = x for x ∈ X.
If G is an action, we say that G acts on X.
The cartesian productH = G×X has a groupoid structure with unit spaceH(0) = {e}×X.
The range and source maps are s(g, x) = (e, x) and r(g, x) = (e, gx) and the operations are
de�ned by
(g, hx)(h, x) = (gh, x) and (g, x)−1 = (g−1, gx).
H is a groupoid, called the action groupoid (or transformation groupoid [23]). In fact,
(i) s((g, x)(h, y)) = s(g, y) = (e, y) = s(h, y) for (g, x), (h, y) composable;
(ii) r(e, x) = (e, ex) = (e, x) = s(e, x) for x ∈ X;
(iii) Given (g, x) ∈ H,
(g, x)s(g, x) = (g, x)(e, x) = (ge, x) = (g, x)
r(g, x)(g, x) = (e, gx)(g, x) = (eg, x) = (g, x);
(iv) Let (g, x), (h, y), (k, z) ∈ H such that ((g, x), (h, y)), ((h, y), (k, z)) ∈ H(2). By hypoth-
43
esis, x = hy and y = kz. Hence, x = hkz. Then
[(g, x)(h, y)] (k, z) = (gh, y)(k, z)
= (ghk, z)
= (g, x)(hk, z)
= (g, x) [(h, y)(k, z)] .
(v) Given (g, x) ∈ H,
(g, x)(g, x)−1 = (g, x)(g−1, gx) = (gg−1, gx) = (e, gx) = r(g, x)
(g, x)−1(g, x) = (g−1, gx)(g, x) = (g−1g, x) = (e, x) = s(g, x).
Example 3.1.4. Let ∼ be an equivalence relation on a set X. Let
G = {(x, y) ∈ X ×X : x ∼ y},
G(0) = {(x, x) : x ∈ X}, and
G(2) = {((x, y), (y, z)) : x ∼ y, y ∼ z}.
De�ne the range and source maps by r(x, y) = (x, x) and s(x, y) = (y, y). Let (x, y)−1 =
(y, x) and (x, y)(y, z) = (x, z). The inverse and multiplication maps are well-de�ned by the
re�exivity and transitivity of ∼. Hence G is a groupoid.
Remark 3.1.5. Note that [9] and [20] de�ne groupoids di�erently. On the one hand, [9]
introduces a groupoid as in De�nition 3.1.1. On the other hand, Renuault [20] describes
groupoids as follows:
A groupoid is a set G endowed with a product map (g, h) 7→ gh : G(2) → G, where G(2) is
a subset of G×G called the set of composable pairs, and an inverse map g 7→ g−1 : G→ G
such that the following relations are satis�ed:
(i') (g−1)−1 = g;
44
(ii') If (g, h), (h, k) ∈ G(2), then (gh, k), (g, hk) ∈ G(2) and (gh)k = g(hk);
(iii') (g−1, g) ∈ G(2) and if (g, h) ∈ G(2), then g−1(gh) = h;
(iv') (g, g−1) ∈ G(2) and if (h, g) ∈ G(2), then (hg)g−1 = h.
Given g ∈ G, we de�ne r(g) = gg−1 and s(g) = g−1g. The unit space is de�ned by
G(0) = s(G) = r(G).
These de�nitions are equivalent.
First, suppose G is a groupoid as in De�nition 3.1.1. Note that r(x) = s(x) = x for each
x ∈ G(0). In fact, r(x) = s(x) by property (ii). Since s : G→ G(0) is surjective, there exists
g ∈ G such that x = s(g) = g−1g. Hence,
s(x) = s(s(g)) = s(g−1g) = s(g) = x.
Now we prove properties (i')�(iv').
(i') (g−1)−1 = g
Since s(g) = g−1g and s(s(g)) = r(s(g)), we have
s(s(g)) = s(g) = g−1g,
r(s(g)) = r(g−1g) = r(g−1) = g−1(g−1)−1.
Then g−1g = g−1(g−1)−1 and therefore g = (g−1)−1.
(ii') (gh)k = g(hk)
This holds by property (iv).
(iii') g−1(gh) = h
Note that s(g) = r(h). Then
g−1(gh) = (g−1g)h = s(g)h = r(h)h = h.
45
(iv') (hg)g−1 = h
Note that s(h) = r(g). Then
(hg)g−1 = h(gg−1) = hr(g) = hs(h) = h.
Then G is a groupoid as de�ned in [20].
Conversely, assume that G is a groupoid as in [20].
First we show that G(2) = {(g, h) : s(g) = r(h)}. Suppose (g, h) ∈ G(2). Then,
g−1(gh) = h by (iii'),
[g−1(gh)]h−1 = hh−1 by (iv'),
[(g−1g)h]h−1 = hh−1
g−1g = hh−1
s(g) = r(h).
Suppose g, h ∈ G are such that s(g) = r(g). Then
(h, h−1), (h−1, h) ∈ G(2) by (iii'), (iv'),
⇒ (hh−1, h) ∈ G(2)
⇒ (r(h), h) ∈ G(2)
⇒ (s(g), h) ∈ G(2)
⇒ (g−1g, h) ∈ G(2)
⇒ (g−1g, h) ∈ G(2).
46
On the other hand,
(g, g−1), (g−1, g) ∈ G(2) by (iii'), (iv'),
⇒ (g, g−1g) ∈ G(2).
Then (g, g−1g), (g−1g, h) ∈ G(2). Therefore (g, h) ∈ G(2).
Now we prove properties (i)�(v).
(i) s(gh) = s(h), r(gh) = r(g)
By assumption, (g, h) ∈ G(2). Also, (h, h−1) ∈ G(2) by (iii). Then s(h) = r(h−1) and
(gh, h−1) ∈ G(2). This implies s(gh) = r(h−1) = s(h).
The proof of r(gh) = r(g) is analogous.
(ii) r(x) = s(x) if x ∈ G(0)
Given x ∈ G(0), there exists g ∈ G such that x = r(g) = gg−1. Then
s(x) = s(gg−1) = s(g−1) = r(g) = x.
Note that s(gg−1) = s(g−1) by (i).
(iii) gs(g) = g and r(g)g = g
gs(g) = g(g−1g) = (g−1)−1(g−1g) = g,
r(g)g = (gg−1)g = (gg−1)(g−1)−1 = g.
(iv) (gh)k = g(hk)
This is equivalent to (ii')
(v) r(g) = gg−1, g−1g = s(g)
This follows from the de�nition of r, s.
47
Therefore the de�nitions are equivalent.
Remark 3.1.6. Given A,B subsets of a groupoid G, one may form the following subsets of
G:
A−1 = {g ∈ G : g−1 ∈ A}, AB = {gh ∈ G : g ∈ A, h ∈ B}.
Given x, y ∈ G(0):
Gx = r−1(x), Gy = s−1(y), and Gxy = Gx ∩Gy.
Gx (resp. Gy) is called the r-�ber of G over x (resp. s-�ber of G over y) as in [7].
Note that Gxx is a group. It is called the isotropy group at x. In fact,
(i) gh ∈ Gxx for g, h ∈ Gx
x;
(ii) gg−1 = g−1g = x for g ∈ Gxx. Hence x is the unity of Gx
x;
(iii) the product in Gxx associative.
Notation 3.1.7. Unless otherwise speci�ed, we will use the following notation in this thesis:
G denotes a groupoid; its units are denoted by the letters x, y, z; g, h are elements in G.
Subsets of G may be written as the uppercase letters U, V . The letters may be indexed or
marked with an accent or symbol.
3.2 Topological Groupoids
If G is a groupoid endowed with a topology, it is useful that its operations have interesting
topological properties. We de�ne the notion of topological groupoid, where its operations
are continuous. We also de�ne étale groupoids, where the range and source maps are local
homeomorphisms. This section is based on [9] and [20].
48
De�nition 3.2.1. A topological groupoid is a groupoid G with a topology such that G(2) has
the induced topology from G×G, and both the product and inverse maps are continuous.
Remark 3.2.2. Let G be a topological groupoid. Since r and s are de�ned by r(g) = gg−1
and s(g) = g−1g, it follows that these functions are continuous.
Now we de�ne the notion of étale groupoid. The main results in this thesis assume the
groupoid has this property.
De�nition 3.2.3. A topological groupoid is étale if the maps r and s are local homeomor-
phisms.
Example 3.2.4. Every discrete groupoid G is étale. In fact, for every g ∈ G, the subsets
{g}, {r(g)}, {s(g)} are open in G. Moreover, the maps r|{g} : {g} → {r(g)}, s|{g} : {g} →
{s(g)} are homeomorphimsms. In particular, every discrete group is étale.
Example 3.2.5. Let X be a topological space and let r, s : X → X be identity maps.
Moreover, de�ned for each x ∈ X, xx = x and x−1 = x. Then X is an étale groupoid
because r, s are homeomorphimsms.
Another example of étale groupoid is the transformation groupoid G×X when the group
G is discrete. We prove this in the following lemma:
Lemma 3.2.6. Let G be a group endowed with a topology. Let X be a topological space
and �x a continuous group action G×X → X. Suppose G×X is the action groupoid as in
Example 3.1.3 and equip this space with the product topology. Then G×X is étale if, and
only if, G is discrete.
Proof. • Suppose G is not discrete.
There exists g ∈ G such that for each neighborhood U of g, U \ {g} 6= ∅.
Fix x ∈ X and let V be an arbitrary open neighborhood of x. Let U be an open
neighborhood of g. Then there exists h 6= g such that h ∈ U . Then (g, x), (h, x) ∈ U×V
49
and s(g, x) = s(h, x) = (e, x). Since U, V are arbitrary, it follows that s is not a local
homeomorphimsm. Therefore, G is not étale.
• Now suppose that G is discrete
Then the product and inverse maps on the group G are continuous. Note that product
and inverse maps on the groupoid are continuous because they are compositions of
continuous functions. Then G×X is a topological groupoid.
For each g ∈ G the map X 7→ X de�ned by x 7→ gx is a homeomorphimsm with
inverse x 7→ g−1x. Then, for every open set U ⊂ X, the set gU = {gx : x ∈ U} is open
in X.
Now we show that G ×X is étale. Let (g, x) ∈ G ×X, U a neighborhood of x ∈ X.
Then {g} × U is an open neighborhood of (g, x). Then
r({g} × U) = {(e, gx) : x ∈ U} = {e} × gU,
s({g} × U) = {(e, x) : x ∈ U} = {e} × U.
Then r({g} × U), s({g} × U) are open sets in G×X.
The function s|{g}×U is injective. The function x ∈ U 7→ gx is injective, then r
is injective on {g} × U . Since g and U are arbitrary, it follows that r, s are open
bisections. Therefore, G×X is étale.
De�nition 3.2.7. An open subset U of an étale groupoid is an open bisection of G if
r(U), s(U) are open in G(0), and r|U : U → r(U) and s|U : U → s(U) are homeomorphisms.
Notation 3.2.8. We will usually denote an open bisection by the cursive letter U . This
letter may be indexed or marked with an accent or symbol.
Remark 3.2.9. Many times throughout the thesis, we will evaluate sums which take into
account values f(g) such that g ranges over Gx or Gx, assuming f ∈ Cc(G). However, if this
50
function is supported on an open bisection, we can consider only one term in the sum. This
element is usually denoted hx (resp. hx) and hx ∈ Gx ∩ U (resp. hx ∈ Gx ∩ U).
Later we prove that every f ∈ Cc(G) can be written as a �nite sum of continuous functions
supported on open bisections. Hence, in many cases, we can assume f is supported on an
open bisection without loss of generality.
Proposition 3.2.10. Let G be an étale groupoid. The set of open bisections of G forms an
open base for the topology of G.
Proof. Let U be an open set of G. We will show that for every g ∈ U there exists an open
bisection Ug such that g ∈ Ug ⊂ U . In fact, let g ∈ U . Since G is étale, r, s are local
homeomorphimsms. Then there exist Rg, Sg open neighborhoods of g such that r(Rg) and
s(Sg) are open in G(0), and r|Rg : Rg → r(Rg), s|Sg : Sg → s(Sg) are homeomorphisms.
Let Ug = Rg ∩ Sg ∩ U . r(Ug) is open in r(Rg), then r(Ug) is open in G(0). Hence,
r|Ug : Ug → r(Ug) is a homeomorphism. Analogously s|Ug : Ug → s(Ug) is a homeomorphism.
Therefore, for every open set U , we have U =⋃g∈U Ug.
Proposition 3.2.11. If G is an étale groupoid, then the subspace topology of Gx and Gx is
equivalent to the discrete topology for all x ∈ G(0). Furthermore, if G is second countable,
then Gx and Gx have a countable number of elements.
Proof. Let g ∈ Gx. There exists an open bisection Ug containing g. We show that Ug ∩Gx =
{g}. Suppose there exists h 6= g such that h ∈ Ug ∩Gx. Then r(h) = x. Contradiction, since
r is injective on Ug. Hence {g} is open in Gx. Therefore Gx is endowed with the discrete
topology.
Assume G is second countable. Then Gx is second countable. Since the sets {g}, g ∈ Gx,
form a family of disjoint open sets in Gx, it follows that Gx is countable. The proof for Gx
is analogous.
Proposition 3.2.12. If G is a locally compact Hausdor� étale groupoid, then G(0) is a
clopen subset of G. We assume G(0) is endowed with the subspace topology.
51
Proof. We divide the proof in two parts.
• G(0) is closed
Let xi be a net in G(0) converging to x ∈ G. The function r is continuous, then
r(xi)→ r(x). As xi ∈ G(0), we have r(xi) = xi. Hence x = r(x) ∈ G(0). Therefore G(0)
is closed.
• G(0) is open
Let x ∈ G(0). Let U ⊂ G be an open bisection containing x. Let V = G(0) ∩ U . Then
V is an open neighborhood in G(0) of x. Moreover, V ⊂ r(U), since r(y) = y for every
y ∈ V .
Since r|U : U → r(U) is a homeomorphimsm, r|−1U (V ) = V . Then V is open in G.
Therefore G(0) is open.
Let G′ = ∪x∈G(0)Gxx, called the isotropy bundle. The following lemma shows that G′ is
closed.
Lemma 3.2.13. Let G be a locally compact Hausdor� second countable étale groupoid.
Given g ∈ G such that r(g) 6= s(g), there exists an open bisection U including g such that
r(U) ∩ s(U) = ∅. Moreover, G′ ∩ U = ∅. In particular, G′ is closed.
Proof. Suppose this lemma is false. Then there exists g ∈ G \ G′ such that for every open
bisection U including g, we have
r(U) ∩ s(U) 6= ∅.
Since G is second countable and étale, we can choose a countable family {Un} of open
bisections containing g such that every neighborhood of g contains at least one Un. Hence,
52
for every n there are gn, hn ∈ Un satisfying
r(gn) = s(hn). (3.1)
By de�nition, both sequences {gn}n∈N, {hn}n∈N converge to g as n→∞. Then, by continuity
of r, s, we have r(gn)→ r(g) and s(hn)→ s(g). However, from (3.1), we have s(hn)→ r(g).
Hence r(g) = s(g). This leads to a contradiction because we assumed g /∈ G′.
Therefore we can choose U satisfying r(U) ∩ s(U) = ∅. Moreover, G′ ∩ U = ∅. Since
g ∈ G \G′ is arbitrary, it follows that G′ is closed.
Remark 3.2.14. Let G be a groupoid and V ⊂ G(0). We de�ne G|V = G∩r−1(V )∩s−1(V ).
Note that G|V is a groupoid. If G is a topological groupoid, then G|V is also a topological
groupoid. Analogously, if G is étale, so is G|V .
3.3 Groupoid C*-Algebras
Now we de�ne the full groupoid C*-algebra and prove some properties of this C*-algebra.
The results in this section can be found in [5], [9] and [23].
Let G be a locally compact second countable Hausdor� étale groupoid. Denote Cc(G) by
Cc(G) = {f : G→ C : f is continuous and supp(f) is compact}.
Recall that the support of f is de�ned by supp(f) = {g ∈ G : f(g) 6= 0}. We de�ne the
convolution and involution operations on Cc(G) by
(f1 · f2)(g) =∑g1g2=g
f1(g1)f2(g2) and f*(g) = f(g−1). (3.2)
Example 3.3.1. Let n be a positive integer and de�ne the groupoid G = {(i, j) : i, j =
53
1, . . . , n} such that
G(0) = {(i, i) : i = 1, . . . , n}
G(2) = {((i, k), (k, j)) : i, j, k = 1, . . . , n},
and de�ne the operations
(i, k)(k, j) = (i, j) and (i, j)−1 = (j, i).
Equip G with the discrete topology. Then G is locally compact Hausdor�. Moreover, G is
étale by Example 3.2.4.
Note that there is a bijection from Cc(G) toMn(C) given by f 7→ F such that Fi,j = f(i, j).
Moreover, we can identify Cc(G) with Mn(C). In fact, let f (1), f (2) ∈ Cc(G). Assume A is a
matrix corresponding to f (1) · f (2). Then, for every i, j = 1, . . . , n,
Ai,j = (f (1) · f (2))(i, j) =n∑k=1
f (1)(i, k)f (2)(k, j) =n∑k=1
F(1)i,k F
(2)k,j = (F (1)F (2))i,j,
then A = F (1)F (2).
Now assume F ∈Mn(C) corresponds to f ∈ Cc(G). Then F* corresponds to f*. In fact,
for i, j = 1, . . . , n,
F *i,j = Fj,i = f(j, i) = f((i, j)−1) = f*(i, j).
Therefore we can identify Mn(C) with Cc(G). Moreover, Cc(G) is not commutative.
Notation 3.3.2. We usually denote a function in Cc(G) by f . Note that, for an open
subset U ⊂ G, every function in Cc(U) can be extended uniquely to a function in Cc(G)
whose support lies in U . Thus, for every open set U ⊂ G, we will denote without loss of
54
generality,
Cc(U) = {f ∈ Cc(G) : supp(f) ⊂ U}.
Thus Cc(U) is a subspace of Cc(G).
The letter h sometimes denotes elements in Cc(G(0)). However, h is also used to indicate
elements in G.
Given f1, f2 ∈ Cc(G), f1f2 denotes the pointwise product of these functions, while f1 · f2
denotes the convolution product.
Lemma 3.3.3. Let G be a locally compact second countable Hausdor� étale groupoid.
Given f1, f2 ∈ Cc(G), g ∈ G,
(f1 · f2)(g) =∑
h∈Gs(g)
f1(gh−1)f2(h) (3.3)
=∑
h∈Gr(g)f1(h)f2(h−1g). (3.4)
Proof. Let g1g2 ∈ G such that g1g2 = g. This is equivalent to g1 = gg−12 . This equation
holds only if g2 ∈ Gs(g). Therefore, for every h ∈ Gs(g) we can choose g2 = h and g1 = gh−1.
Then,
(f1 · f2)(g) =∑
h∈Gr(g)f1(gh−1)f2(h).
This sum is �nite for every g, since Gr(g) is countable and f2 is compactly supported, then the
set of elements h ∈ Gr(g) such that f2(h) 6= 0 is �nite. The proof for (3.4) is analogous.
Lemma 3.3.4. Let f ∈ Cc(G), h ∈ Cc(G(0)), g ∈ G. Then
(h · f)(g) = h(r(g))f(g), and (f · h)(g) = f(g)h(s(g)).
55
Proof. It follows from Lemma 3.3.3 that
(h · f)(g) =∑
k∈Gr(g)h(k)f(k−1g).
Suppose h(k) 6= 0 for some k ∈ Gr(g). Then k ∈ G(0) and r(k) = r(g). Hence k = r(g) and
therefore,
(h · f)(g) = h(r(g))f(r(g)−1g) = h(r(g))f(r(g)g) = h(r(g))f(g).
The proof for f · h is analogous.
Now we show that every function in Cc(G) can be decomposed as a sum of continuous
compactly supported functions whose support are included in open bisections. This result
will be used many times in the thesis because many results are easier to prove when the
function is supported on an open bisection.
Lemma 3.3.5. Let G be a locally compact second countable étale Hausdor� groupoid.
Given f ∈ Cc(G), there are U1, . . . ,Un open bisections and f1, . . . , fn functions such that
f = f1 + . . . + fn and each fi ∈ Cc(Ui). Moreover, if f is non-negative, we can choose each
fi to be non-negative.
Proof. Let f ∈ Cc(G) with support K. From Proposition 3.2.10 the set of open bisections
forms an open base for G. Then there exists a �nite cover U1, . . .Un of K such that each Uiis an open bisection.
Let Un+1 = G \K. Then {Ui}n+1i=1 is an open cover of G. Let {αi}n+1
i=1 be the partition of
unit subordinate to the the open cover {Ui}n+1i=1 . Note that fαn+1 = 0 since αn+1 is supported
on Un+1 \K. Then,
f =n+1∑i=1
fαi =n∑i=1
fαi.
56
De�ne fi = fαi for i = 1, . . . , n. By de�nition of αi, each fi ∈ Cc(Ui). Moreover,
since αi assumes values in the interval [0, 1], if f is non-negative, it follows that each fi is
non-negative.
Lemma 3.3.6. Let G be a locally compact Hausdor� second countable étale groupoid. If
U ,V ⊂ G are open bisections, then
UV = {gh : g ∈ U , h ∈ V , (g, h) ∈ G(2)}
is an open bisection.
Proof. Before we prove UV is an open bisection, we will show that we can assume s(U) = r(V)
without loss of generality. Let W = s(U) ∩ r(V). Then W is an open set in G(0) because U ,
V are open bisections.
Let U0 = s|−1U (W ) and V0 = r|−1
V (W ). Both U0,V0 are open bisections, since they are
open subsets of open bisections. Moreover, we have
s(U0) = s ◦ s|−1U (W ) = W = r ◦ r|−1
V (W ) = r(V0).
Now we show that U×V∩G(2) = U0×V0∩G(2). In fact, given (g, h) ∈ U×V∩G(2), we have
g ∈ U , h ∈ V and s(g) = r(h). If we de�ne x = s(g), then x ∈ W . Moreover, g = s|−1U (x),
which implies g ∈ U0. Analogously, h ∈ V0. Then (g, h) ∈ U0 × V0 ∩ G(2). Therefore
U×V∩G(2) ⊂ U0×V0∩G(2). Since U0 ⊂ U and V0 ⊂ V , we have U×V∩G(2) = U0×V0∩G(2).
By de�nition UV , we have
UV = {gh : g ∈ U , h ∈ V , (g, h) ∈ G(2)}
= {gh : (g, h) ∈ U × V ∩G(2)}
= {gh : (g, h) ∈ U0 × V0 ∩G(2)}
= U0V0.
57
Therefore we can assume s(U) = r(V) without loss of generality.
Now we prove UV is an open bisection. Assume s(U) = r(V). Let φ : U → V be the
homeomorphism de�ned by φ = r|−1V ◦ s|U .
De�ne the map f from U to U × V by f(g) = (g, φ(g)). The image f(U) is included in
U × V ∩G(2), since
r(φ(g)) = r(r|−1V ◦ s(g)) = s(g).
We claim f(U) = U × V ∩G(2). Suppose (g, h) ∈ U × V ∩G(2). Then s(g) = r(g), g ∈ U ,
h ∈ V . Hence
h = r|−1V ◦ s|U(g) = φ(g).
Thus (g, h) = (g, φ(g)) = f(g). Therefore f(U) = U × V ∩G(2).
By de�nition of f , we have that f is injective, thus. We will show that f is a homeomor-
phism. Let π : U ×V ∩G(2) → U be the projection onto the �rst coordinate. π is continuous
by de�nition. So we will show that π is the inverse of f .
Given g ∈ U ,
π ◦ f(g) = π(g, φ(g)) = g.
Given (g, h) ∈ U × V ∩G(2), we have (g, h) = (g, φ(g)) = f(g) since f is a bijection. Then
(f ◦ π)(g, h) = f(g) = (g, φ(g)) = (g, h).
Therefore π is the inverse of f and f is a homeomorphism. Hence the set U × V ∩ G(2) is
open in U × V .
58
Now we can consider the product p : U × V ∩G(2) → UV and observe that
r|UV ◦ p = r|U ◦ π. (3.5)
In fact, given (g, h) ∈ U × V ∩G(2),
r|UV ◦ p(g, h) = r(gh) = r(g) = r|U ◦ π(g, h).
Equation (3.5) shows that r|UV ◦ p is a homeomorphism. Moreover, we conclude that p is
surjective and r|UV is injective.
In addition, p is injective because if p(g1, h1) = p(g2h2), we have by (3.5) the following
result,
r|U ◦ π(g1, h1) = r|U ◦ π(g2, h2)
r|U(g1) = r|U(g1)
g1 = g2, since U is an open bisection,
r(h1) = r(h2) because (g1, h1), (g2, h2) ∈ U × V ∩G(2),
h1 = h2 since V is an open bisection.
Therefore p is injective. Hence r|UV , p are continuous bijections such that their composi-
tion is a homeomorphism. Therefore p : U × V ∩G(2) → UV is a homeomorphism.
Since U ×V ∩G(2) is an open set, so is UV . We already proved that r|UV is injective. The
proof for s|UV is analogous. Therefore, UV is an open bisection.
Lemma 3.3.7. Let G be a locally compact Hausdor� second countable étale groupoid. If
U ⊂ G is an open bisection, then U−1 = {g−1 : g ∈ U} is an open bisection.
Proof. Let ι : G → G be the inverse map. ι is continuous and ι ◦ ι is the identity. Then
U−1 = ι(U) is open.
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Let g1, g2 ∈ U−1 such that r(g1) = r(g2). There exist h1, h2 ∈ U such that gi = h−1i ,
i = 1, 2. Then
s(h1) = r(g1) = r(g2) = s(h2).
Since U is an open bisection, we have h1 = h2. Then g1 = g2. The proof for s is analogous.
Therefore U−1 is an open bisection.
Lemma 3.3.8. Let G be a locally compact Hausdor� second countable étale groupoid.
(i) Given U1,U2 open bisections, f1 ∈ Cc(U1), f2 ∈ Cc(U2), then f1 · f2 ∈ Cc(U1U2).
(ii) If U is an open bisection and f ∈ Cc(U), we have f* ∈ Cc(U−1).
Proof. (i) Note that U1U2 is an open bisection by Lemma 3.3.6.
Let g /∈ U1U2. Then f1 · f2(g) = 0 since f(g) 6= 0 implies that there are g1 ∈ U1, g2 ∈ U2
satisfying g1g2 = g. Therefore the support of f1 · f2 lies in U1U2.
Since U1U2 is an open bisection, the maps u1, u2 are homeomorphisms where u1 :
U1U2 → U1 is de�ned by u1 = r|−1U1 ◦ r and u2 : U1U2 → U2 is de�ned by u2 = s|−1
U2 ◦ s.
Given g ∈ U1U2, g1 = u1(g), g2 = u2(g) are the only elements satisfying g1 ∈ U1, g2 ∈
U2, g = g1g2. In fact, suppose there are (h1, h2) ∈ U1 × U2 ∩ G(2) such that g = h1h2.
Then r(h1) = r(g). Since U1 is an open bisection, we have h1 = r|−1U1 ◦ r(g) = g1.
Analogously h2 = g2.
Therefore, for every g ∈ U1U2,
(f1 · f2)(g) = f1(u1(g))f2(u2(g)).
For i = 1, 2, ui : U1U2 → Ui is continuous and fi : Ui → C is continuous. Hence f1 · f2
is continuous on U1U2. Since f1 · f2 vanishes outside U1U2, we have f1 · f2 ∈ Cc(U1U2).
60
(ii) Let ι : G → G be the inverse map. ι is continuous. Since f* is de�ned by f* = f ◦ ι,
then f* is continuous.
Let K = supp(f) and L = supp(f*). Then
L = {g ∈ G : f*(g) 6= 0}
= {g ∈ G : f(g−1) 6= 0}
= {g ∈ G : f(g−1) 6= 0}
= {g ∈ G : f(g) 6= 0}−1
= ({g ∈ G : f(g) 6= 0})−1
= ({g ∈ G : f(g) 6= 0})−1, since the inversion is continuous,
= K.
The inversion on G is continuous, thus L is compact. Moreover, L ⊂ U−1. Thus
f* ∈ Cc(U−1).
Theorem 3.3.9. Let G be a locally compact Hausdor� second countable étale groupoid.
Cc(G) with the operations (3.2) is a ∗-algebra.
Proof. Clearly Cc(G) is a vector space.
• The product is bilinear.
Let f1, f2, f ∈ Cc(G), λ ∈ C, g ∈ G. Then,
[(f1 + λf2) · f ](g) =∑g1g2=g
(f1 + λf2)(g1)f(g2)
=∑g1g2=g
f1(g1)f(g2) + λ∑g1g2=g
f2(g1)f(g2)
= [f1 · f ](g) + λ[f2 · f ](g).
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The proof for f · (f1 + λf2) is analogous.
• The product is associative.
Let f1, f2, f3 ∈ Cc(G). Given g ∈ G,
[f1 · (f2 · f3)](g) =∑g1h=g
f1(g1)(f2 · f3)(h)
=∑g1h=g
∑g2g3=h
f1(g1)f2(g2)f3(g3)
=∑
g1g2g3=g
f1(g1)f2(g2)f3(g3)
=∑hg3=g
∑g1g2=h
f1(g1)f2(g2)f3(g3)
=∑hg3=g
( ∑g1g2=h
f1(g1)f2(g2)
)f3(g3)
=∑hg3=g
(f1 · f2)(h)f3(g3)
= [(f1 · f2) · f3](g).
• f1 · f2 ∈ Cc(G) if f1, f2 ∈ Cc(G).
Since the product is bilinear and, from Lemma 3.3.5, every function in Cc(G) can be
written as a �nite sum of continuous functions supported on open bisections, it su�ces
to show that f1 · f2 ∈ Cc(U1U2) for f1 ∈ Cc(U1), f2 ∈ Cc(U2) where U1,U2 are open
bisections. Note that U1U2 is an open bisection by Lemma 3.3.6. However, we already
proved f1 · f2 ∈ Cc(U1U2) in Lemma 3.3.8.
• For f ∈ Cc(G), f** = f .
Let g ∈ G, then
f**(g) = f*(g−1) = f((g−1)−1) = f(g).
• For f1, f2 ∈ Cc(G), (f1 · f2)* = f2* · f1*.
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Let g ∈ G. Then,
(f1 · f2)*(g) = (f1 · f2)(g−1)
=∑
g1g2=g−1
f1(g1)f2(g2)
=∑
g1g2=g−1
f2*(g−12 )f1*(g−1
1 )
=∑
g−12 g−1
1 =g
f2*(g−12 )f1*(g−1
1 ),
making the change of variables h1 = g−12 , h2 = g−1
1 ,
=∑
h1h2=g
f2*(h1)f1*(h2)
= (f2* · f1*)(g).
• The involution is conjugate-linear.
Let f1, f2 ∈ Cc(G), λ ∈ C, g ∈ G. Then,
(f1 + λf2)*(g) = (f1 + λf2)(g−1) = f1(g−1) + λf2(g−1) = f1*(g) + λf2*(g).
• If f ∈ Cc(G), then f* ∈ Cc(G).
Since the involution is conjugate-linear, we can assume f ∈ Cc(U) for an open bisection.
It follows from Lemma 3.3.8 that f* ∈ Cc(U).
In Theorem 3.3.9 we proved Cc(G) is a ∗-algebra. Now we will equip this space with a
norm such that its completion is a C*-algebra.
Lemma 3.3.10. Let G be a locally compact Hausdor� second countable étale groupoid.
Cc(G(0)) is a sub-∗-algebra of Cc(G). Moreover, Cc(G(0)) is commutative with product given
by the pointwise multiplication and involution de�ned by f*(x) = f(x).
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Proof. From Proposition 3.2.12 it follows that G(0) is open. Then Cc(G(0)) is a subspace of
Cc(G) as described in Notation 3.3.2.
• f* ∈ Cc(G(0)) if f ∈ Cc(G(0))
Let f ∈ Cc(G(0)). Then f* ∈ Cc(G) by Theorem 3.3.9. Let g ∈ G such that f(g) 6= 0.
Then f*(g) = f(g−1) 6= 0. Thus g−1 ∈ G(0) by assumption. Then g = g−1 ∈ G(0).
Therefore f* is also supported on G(0).
• f1 · f2 = f1f2 ∈ Cc(G(0))
Now let f1, f2 ∈ Cc(G(0)). Then f1 · f2 ∈ Cc(G) by by Theorem 3.3.9. Given g ∈ G, we
have
(f1 · f2)(g) =∑g1g2=g
f1(g1)f2(g2).
Suppose there are g1, g2 such that g = g1g2 and f1(g1)f2(g2) 6= 0. Then g1, g2 ∈ G(0).
Hence g ∈ G(0) and g = g1 = g2. Therefore, (f1 · f2)(g) = f1(g)f2(g).
Therefore Cc(G(0)) is commutative.
Proposition 3.3.11. Let G be a locally compact Hausdor� second countable étale groupoid.
For each f ∈ Cc(G), there is a constant Kf ≥ 0 such that ‖π(f)‖ ≤ Kf for every ∗-
representation π : Cc(G) → B(H) of Cc(G) on a Hilbert space H. If f is supported on an
open bisection, we can take Kf = ‖f‖∞.
Proof. Suppose π is a ∗-representation. Then π|Cc(G(0)) is a ∗-representation of the commu-
tative ∗-algebra Cc(G(0)), and so ‖π(h)‖ ≤ ‖h‖∞ for every h ∈ Cc(G(0)).
Let f ∈ Cc(G). There are f1, . . . , fn with f =∑n
i=1 fi such that each fi ∈ Cc(Ui) and Uiis an open bisection. Fix i, hence fi* ∈ Cc(U−1) and therefore fi* · fi ∈ Cc(U−1U). However,
64
UU−1 = s(U). In fact,
U−1U = {gh : g ∈ U−1, h ∈ U , s(g) = r(h)}
= {g−11 g2 : g1, g2 ∈ U , r(g1) = r(g2)}
= {g−1g : g ∈ U} since U is an open bisection
= s(U).
Thus fi*fi ∈ Cc(s(U)). So
‖π(fi)‖2 = ‖π(fi* · fi)‖ ≤ ‖fi* · fi‖∞ = ‖fi‖2.
Let Kf =∑n
i=1 ‖fi‖. Applying triangle inequality, we have ‖π(f)‖ ≤ Kf .
Proposition 3.3.12. Let f ∈ Cc(G) such that f 6= 0. There exists a ∗-representation of
Cc(G) such that π(f) 6= 0.
Proof. Let x ∈ G(0) such that f(hx) 6= 0 for some hx ∈ Gx. Since Gx is countable,
`2(Gx) =
{{ξg}g∈Gx : ξg ∈ C,
∑g∈Gx
|ξg|2 <∞
}
is a Hilbert space with inner product given by
〈ξ, ζ〉 =∑g∈Gx
ξgζg.
De�ne πx : Cc(G)→ B(`2(Gx)) by
(πx(f1)ξ)g =∑
h1h2=g
f1(h1)ξh2 .
65
Note that
(πx(f1)ξ)g =∑
h∈Gs(g)
f1(gh−1)ξh =∑
h∈Gr(g)f1(h)ξh−1g,
making the change of variables h2 = h, h1 = gh−1 in the �rst sum, and h1 = h, h2 = h−1g
in the second sum. Note that πx is linear.
• πx is well-de�ned.
We will prove that the image of πx is in B(`2(Gx)). Since πx is linear, it is su�cient
to show that πx(f1) ∈ B(`2(Gx)) for every f ∈ Cc(U) such that U ⊂ G is an open
bisection.
Let U ⊂ G be an open bisection, f1 ∈ Cc(U). Let L denote the set of g ∈ Gx such that
there exists h ∈ Gr(g) satisfying f(h) 6= 0. h is unique for every g ∈ L and it will be
denoted by hr(g). Then, for every ξ ∈ `2(Gx), g ∈ L,
(πx(f1)ξ)g = f1(hr(g))ξ(hr(g))−1g.
Note that (πx(f1)ξ)g = 0 if g ∈ Gx \ L.
Suppose there are g1, g2 ∈ L such that (hr(g1))−1g1 = (hr(g2))−1g2. Then s(hr(g1)) =
s(hr(g2)). Since hr(g1), hr(g2) ∈ U , we have hr(g1) = hr(g2). Then g1 = g2. Since hr(g)g ∈
Gx, the family {ξhr(g)g}g∈L has distinct elements. Hence,
‖πx(f1)ξ‖2 =∑g∈Gx
|(πx(f1)ξ)g|2
=∑g∈L
|(πx(f1)ξ)g|2
=∑g∈L
|f1(hr(g))ξ(hr(g))−1g|2
≤ ‖f1‖2∞
∑g∈L
|ξ(hr(g))−1g|2
66
≤ ‖f1‖2∞
∑h∈Gx
|ξh|2 , since {ξhr(g)g}g∈L has distinct elements,
≤ ‖f1‖2∞‖ξ‖2.
Therefore ‖πx(f1)‖ ≤ ‖f1‖∞ <∞.
• πx(f) 6= 0.
Let ξ ∈ `2(Gx) such that ξx = 1 and ξl = 0 if l 6= x. Let g ∈ Gx such that f(g) 6= 0.
Then, by de�nition of ξ,
(πx(f)ξ)g =∑
h∈Gr(g)f(h)ξh−1g = f(g)ξg−1g = f(g)ξx = f(g) 6= 0.
• πx(f1 · f2) = πx(f1)πx(f2)
Let f1, f2 ∈ Cc(G). Then,
[πx(f1)(πx(f2)ξ)]g =∑g1h=g
f1(g1)(πx(f2)ξ)h
=∑g1h=g
f1(g1)∑g2g3=h
f2(g2)ξg3
=∑
g1g2g3=g
f1(g1)f2(g2)ξg3
=∑hg3=g
( ∑g1g2=h
f1(g1)f2(g2)
)ξg3
=∑hg3=g
(f1 · f2)(h)ξg3
= [πx(f1 · f2)ξ]g .
• πx(f1*) = πx(f1)*
67
Let f1 ∈ Cc(G), ξ ∈ `2(Gx). Then
〈ξ, πx(f1*)ξ〉 =∑g∈Gx
ξg[πx(f1*)]g
=∑g∈Gx
ξg∑h∈Gx
f1*(gh−1)ξh
=∑g∈Gx
ξg∑h∈Gx
f1(hg−1)ξh
=∑h∈Gx
(∑g∈Gx
f1(hg−1)ξg
)ξh
=∑h∈Gx
[πx(f1)ξ]h ξh
= 〈πx(f1)ξ, ξ〉.
Therefore πx(f1*) = πx(f1)*.
It follows that πx is a ∗-representation of Cc(G) such that πx(f) 6= 0.
Theorem 3.3.13. AssumeG is a locally compact Hausdor� second countable étale groupoid.
There exists a C*-algebra C*(G) such that Cc(G) is dense in C*(G) and the norm on C*(G)
satis�es
‖f‖ = sup{‖π(f)‖: π : Cc(G)→ B(Hπ) is a ∗-representation of Cc(G)},
for every f ∈ Cc(G).
Proof. For every f ∈ Cc(G), Proposition 3.3.11 shows that the set
{‖π(f)‖ : π is a ∗-representation of Cc(G)}
is bounded above, and it is nonempty because of the zero representation. So we can de�ne
68
ρ : Cc(G)→ [0,∞) by
ρ(f) = sup{‖π(f)‖ : π : Cc(G)→ B(Hπ) is a ∗-representation}.
ρ is a norm on Cc(G). In fact, given λ ∈ C, f ∈ Cc(G),
ρ(λf) = supπ‖π(λf)‖ = |λ| sup
π‖π(f)‖ = |λ|ρ(f).
Given f1, f2 ∈ Cc(G),
ρ(f1 + f2) = supπ‖π(f1 + f2)‖ ≤ sup
π‖π(f1)‖+ sup
π‖π(f2)‖ = ρ(f1) + ρ(f2).
Given f ∈ Cc(G) such that f 6= 0, ρ(f) > 0 by Proposition 3.3.12.
The norm is submultiplicative. Given f1, f2 ∈ Cc(G),
ρ(f1 · f2) = supπ‖π(f1 · f2)‖ = sup
π‖π(f1)π(f2)‖ ≤ sup
π‖π(f1)‖ sup
π‖π(f2)‖ = ρ(f1)ρ(f2).
Given f ∈ Cc(G),
ρ(f*) = supπ‖π(f*)‖ = sup
π‖π(f)*‖ = sup
π‖π(f)‖ = ρ(f).
Moreover, ρ satis�es the C*-identity. Indeed, given f ∈ Cc(G),
ρ(f*f) = supπ‖π(f*f)‖ = sup
π‖π(f)*π(f)‖ = sup
π‖π(f)‖2 = sup
π‖π(f)‖ = ρ(f)2.
So we de�ne C*(G) to be the completion of Cc(G) with respect to the norm ρ. C*(G) is a
C*-algebra.
De�nition 3.3.14. Given a locally compact Hausdor� second countable étale groupoid G,
C*(G) is called the full C*-algebra of G.
Remark 3.3.15. In this thesis we also say C*(G) is the groupoid C*-algebra for G. However,
69
this is not the unique C*-algebra de�ned as the completion of Cc(G). For example, in
[9] the reduced C*-algebra is de�ned as the closure of Cc(G) with respect to the norm
‖f‖ = ‖πλ(f)‖, where πλ is a ∗-representation of Cc(G) called the regular representation of
Cc(G).
Lemma 3.3.16. Let G be a locally compact second countable Hausdor� étale groupoid.
Then C0(G(0)) is a sub-C*-algebra of C*(G) and the norm on C0(G(0)) is the uniform norm.
Moreover, Cc(G(0)) is dense in C0(G(0)).
Proof. From Proposition 3.2.12, we have that G(0) is clopen in G. Moreover, G(0) is an open
bisection because r and s are injective on G(0). From Lemma 3.3.10 we have that Cc(G(0))
is a sub-∗-algebra with product given by pointwise multiplication and involution de�ned by
f*(x) = f(x).
Let h ∈ Cc(G(0)). It follows from Proposition 3.3.11 that ‖π(h)‖ ≤ ‖h‖∞ for every
representation π of Cc(G). Then ‖h‖ ≤ ‖h‖∞.
We will show that ‖h‖ = ‖h‖∞ for every h ∈ Cc(G(0)). Given x ∈ G(0), let πx : Cc(G)→
B(`2(Gx)) be the ∗-representation as in the proof of Proposition 3.3.12. Then, for every
h ∈ Cc(G(0)), ξ ∈ `2(Gx), we have
[πx(h)ξ]g =∑g1g2=g
h(g1)ξg2 = h(r(g))ξg,
because h vanishes outside G(0). Let ζ ∈ `2(Gx) such that ζg = 0 if g 6= x and ζx = 1. Then
‖ζ‖ = 1 and
‖πx(h)‖2 ≥ ‖πx(h)ζ‖2 =∑g∈Gx
|πx(h)ζg|2 = |πx(h)|2 = |h(r(x))|2 = |h(x)|2.
Then,
‖h‖ = sup{‖π(h)‖ : π : Cc(G)→ B(Hπ) is a ∗-representation}
70
≥ supx∈G(0)
|πx(h)| = supx∈G(0)
|h(x)| = ‖h‖∞.
Then ‖h‖ = ‖h‖∞. Recall that C0(G(0)) is the closure of Cc(G(0)) with respect to the
norm ‖ · ‖∞. Therefore, C0(G(0)) is a sub-C*-algebra of C*(G).
Example 3.3.17. Let X be the groupoid of Example 3.2.5 and assume that X is locally
compact Hausdor� second countable. Then the operations on Cc(X) are the pointwise
multiplication and the complex conjugate by Lemma 3.3.10. It follows from Lemma 3.3.16
that C0(X) = C*(X) and the norm on this C*-algebra is the uniform norm.
Now we de�ne the inductive limit topology. Later we show that convergence with respect
to the inductive limit topology on Cc(G) implies convergence in the norm of C*(G). This
de�nition can be found in [10]. Then we will prove that C*(G) is separable.
De�nition 3.3.18. SupposeX is a locally compact Hausdor� second countable space. Given
a sequence {fn}n∈N on Cc(X) and f ∈ Cc(X), we say that fn → f with respect to the
inductive limit topology if, and only if, fn → f uniformly and there exists a compact set K
in X such that, eventually, all the fn and f vanish outside K. Given a topological space Y ,
we will say that a function F : Cc(X) → Y is continuous in the inductive limit topology if
F (fn)→ F (f) whenever fn → f with respect to the inductive limit topology.
Lemma 3.3.19. Let G be a locally compact Hausdor� second countable Hausdor� étale
groupoid. Let {fi}i∈N be a sequence in Cc(G) such that fi → f with respect to the inductive
limit topology. Then fi → f in C*(G).
Proof. Let K be a compact set such that fi eventually vanishes outside K. Let U1, . . . ,Unbe open bisections which cover K. Let p1, . . . , pn be a partition of unit subordinate to the
open cover.
Let π : Cc(G)→ B(Hπ) be a ∗-representation of Cc(G). Fix j = 1, . . . , n. Then pjf and
each pjfi are supported on the open bisection Uj. Hence, by Proposition 3.3.11, we have for
71
every i,
‖π(pjfi)− π(pjf)‖ ≤ ‖pjfi − pjf‖∞ ≤ ‖fi − f‖∞.
By taking the supremum on π, we have ‖pjfi − pjf‖ ≤ ‖fi − f‖∞. Since fi converges to f
uniformly, it follows that pjfi → pjf in C*(G). Since {pj} is a partition of unit subordinate
to the open cover of K, and the sequence fi is eventually supported on K, it follows that
fi → f in C*(G).
We state the Stone-Weierstrass theorem below, which can be found in [11] . This theorem
will be used to prove that C*(G) is separable.
Theorem 3.3.20. (Stone-Weierstrass theorem for complex-valued functions) Let K be a
compact space, A a subalgebra of C(X) which separates points in K, that is, for every
x1, x2 ∈ K, there is f ∈ A such that f(x1) 6= f(x2). Assume f ∈ A for every f ∈ A.
Moreover, suppose that for every x ∈ K there exists f ∈ A with f(x) 6= 0. Therefore A is
dense in C(K).
Lemma 3.3.21. Let X be a locally compact Hausdor� second countable space. Let U be
an open subset of X with compact closure. Then Cc(U) is separable with respect to the
supremum norm.
Proof. Let F = {Un} be a countable family of open sets such that Un is compact and
U = ∪nUn. Given n,m such that Un ∩ Um = ∅, let fn,m ∈ Cc(U) such that fn,m|Un = 1 and
fn,m|Um = 0.
Let A be the algebra generated by fn,m, and A0 be the set generated by sums and products
of fn,m, and also by multiplication of scalars in Q+ iQ. Note that A0 is countable and dense
in A. Moreover, if f ∈ A, then f ∈ A.
Note that A separates points in U . Let x1, x2 ∈ U . It follows from Propositions 2.4.1 and
2.4.2 that there are U1, U2 ∈ F such that U1 ∩ U2 = ∅, x1 ∈ U1, x2 ∈ U2. Then f1,2(x1) = 1
and f1,2(x2) = 0.
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Therefore, by the Stone-Weierstrass theorem, A0 is dense in Cc(U).
Proposition 3.3.22. Let G be a locally compact Hausdor� second countable étale groupoid.
Then C*(G) is separable.
Proof. Let I be a countable family of open bisections with compact support that covers G.
From Lemma 3.3.21, there exists a countable subset AU of Cc(U) such that AU is dense in
Cc(U) with respect to the supremum norm. Let A0 be the set generated by �nite sums of
elements in ∪U∈IAU . Then A0 is countable.
Let f ∈ Cc(G). There is a �nite family U1, . . . ,Un ∈ I that covers the support of f . Let
p1, . . . , pn be a partition of unit subordinate to U1, . . . ,Un.
Let ε > 0. Given i = 1, . . . , n, pif ∈ Cc(Ui). Then there exists Fi ∈ AUi satisfying
‖Fi − pifi‖∞ < ε/n. It follows from Proposition 3.3.11 that ‖π(Fi) − π(pif)‖ < ε/n, for
every ∗-representation of Cc(G). Then ‖Fi − pif‖ < ε/n.
Let F =∑n
i=1 Fi. Then F ∈ A0. Moreover,
‖F − f‖ =
∥∥∥∥∥n∑i=1
Fi −n∑i=1
pifi
∥∥∥∥∥ ≤n∑i=1
‖Fi − pifi‖ < nε
n= ε.
Therefore A0 is dense in Cc(G). Since Cc(G) is dense in C*(G), it follows that C*(G) is
separable.
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Chapter 4
Renault's Disintegration Theorem
Neshveyev's theorems describe KMS states ϕ on a groupoid C*-algebra. It is possible to
write ϕ(f) as an integral on G(0) with respect to a probability measure µ such that, for each
x, there is a state ϕx on C*(Gxx). Moreover, the family of states ϕx depends on µ.
Recall from the theory of C*-algebras [2], [16] that we can write ϕ(f) = 〈π(f)ξ, ξ〉 for a
representation π : C*(G)→ H, and ξ ∈ H. Analogously, we can write ϕx(f) = 〈πx(f)ξx, ξx〉
for a representation π : C*(Gxx)→ Hx, and ξx ∈ H
A fundamental step in the proof of Neshveyev's theorem is the Renault's disintegration
theorem which shows a relation between π and the family of πx. Moreover, it proves that
we can assume ξ = {ξx}x∈G(0) . In this case we say ξ ∈∫ ⊕G(0)Hxdµ(x).
The results in this chapter can be found in [4] and [10].
4.1 Haar Systems
Given a measure µ on G(0), it is possible to de�ne a measure ν on G if we have a family
of measures λx supported on Gx. The family {λx}x∈G(0) is called a Haar system. A Haar
system is a generalization of the notion of Haar measures on groups.
If we �x the Haar system, we can de�ne a family of measures λx supported on Gx and,
74
with this family, we can construct another measure ν−1 on G. When ν, ν−1 are equivalent,
we say µ is quasi-invariant.
Given a state ϕ on C*(G), it follows from Neshveyev's �rst theorem that there is corre-
sponding probability measure µ on G(0). In addition, if this state is KMS, then this measure
is necessarily quasi-invariant with respect to the Haar system given by counting measures
λx on Gx.
The results in this section are based on [4] and [10].
De�nition 4.1.1. Let G be a locally compact Hausdor� groupoid, a (left) Haar system
{λx}x∈G(0) for G is a family of Radon measures on G, such that the following conditions
hold:
(i) supp(λx) = Gx for every x ∈ G(0);
(ii) (continuously varying) for f ∈ Cc(G), the function
x 7→∫G
f(g)dλx(g)
is in Cc(G(0));
(iii) (left invariance) for f ∈ Cc(G), h ∈ G,
∫G
f(hg)dλs(h)(g) =
∫G
f(g)dλr(h)(g).
Now we prove that Haar systems are a generalization of Haar measures, as de�ned in
[6]. A topological group is a group G endowed with a topology such that its operations are
continuous functions. Given a locally compact Hausdor� group G, a (left) Haar measure is
a non-zero Radon Borel measure µ on G satisfying
µ(gA) = µ(A) for every g ∈ G and A ⊂ G measurable.
75
Lemma 4.1.2. Let G be a locally compact Hausdor� topological group. Let µ be a Radon
measure on G. Then µ is a Haar measure if, and only if, {λx}x∈G(0) is a Haar system assuming
λ1 = µ.
Proof. Note that a group G is a groupoid such that G(0) = {1}.
Assume µ is a Haar measure. First we show that supp(µ) = G1 = G. We prove that the
properties of Haar system hold for {λx}x∈G(0)
(i) supp(λ1) = G1 = G
Suppose there exists an open non-empty set U ⊂ G such that µ(U) = 0. Let h ∈ U .
Then h−1U is an open neighborhood of 1 and µ(h−1U) = µ(U) = 0. Let V = h−1U .
LetK be a compact set. For every g ∈ K, gV is an open neighborhood of g. Then there
are g1, . . . , gn ∈ K such that g1V, . . . , gnV is an open cover of K. However, µ(g1V ) = 0
for every i = 1, . . . , n. Hence µ(K) = 0. Then, by de�nition of Radon measure, we
have
µ(G) = sup{K ⊂ G : K is compact} = 0,
which leads to a contradiction. Therefore, µ is positive on all open subsets of G.
Now let F be the support of µ. Then, by de�nition, F is closed and µ(G \ F ) = 0.
Since G \ F is open, we have G \ F = ∅, that is, G = F .
(ii) Property (ii) in De�nition 4.1.1 since G(0) is singleton.
(iii) Left invariance in De�nition 4.1.1
Let f ∈ Cc(G). We can assume f ≥ 0 without loss of generality. Let A ⊂ G measurable
and a ≥ 0 such that ϕ = aχA ≤ f .
Given h ∈ G, the function g 7→ ϕ(hg) satis�es ϕ(hg) = χh−1A(g) ≤ f(hg) for every
76
g ∈ G. Hence,
∫G
ϕ(g)dµ(g) = aµ(A) = aµ(h−1A) =
∫G
ϕ(hg)dµ(g). (4.1)
Equation (4.1) holds for every simple function ϕ ≤ f by linearity of the integral. Then,
by taking the supremum over ϕ ≤ f ,
∫G
f(g)dµ(g) ≤∫G
f(hg)dµ(g).
The other inequality is proven analogously.
Conversely, assume {λx}x∈G(0) is a Haar system with λ1 = µ.
Let U ⊂ G be an open set and let h ∈ G. Note that there is a correspondence between
measurable functions 0 ≤ f ≤ χU and 0 ≤ f ≤ χh−1U given by the relation
f(g) = f(h−1g). (4.2)
In fact, given f satisfying f ≤ χU , g ∈ G such that f(g) 6= 0, then f(h−1g) 6= 0. Hence
h−1g ∈ U . Then g ∈ hU . Therefore 0 ≤ f ≤ χhU . Analogously, given 0 ≤ f ≤ χhU , there is
a unique 0 ≤ f ≤ χU such that (4.2) holds.
Hence, by Lemma 2.4.14,
µ(U) = supf∈Cc(G)0≤f≤χU
∫G
f(g)dµ(g) = supf∈Cc(G)0≤f≤χU
∫G
f(h−1g)dµ(g) = supf∈Cc(G)
0≤f≤χU
∫G
f(g)dµ(g) = µ(hU).
Let A be a Borel set. Since the map g 7→ hg is a homeomorphism, for every open set V ⊃ hA,
there exists a unique open set A ⊂ such that V = hU . Then,
µ(A) = infA⊂UU open
µ(U) = infA⊂UU open
µ(hU) = infhA⊂VV open
µ(V ) = µ(hA).
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Therefore µ is a Haar measure.
Now we prove that the family λx of counting measures on Gx is a Haar system. In fact,
this will be the Haar system used in Chapter 5.
Proposition 4.1.3. Let G be a locally compact étale groupoid. If λx is the counting measure
on Gx, then {λx}x∈G(0) is a left Haar system.
Proof. Each Gx is countable, then the counting measure λx is well de�ned. We show the
properties of De�nition 4.1.1 hold.
(i) Given x ∈ G(0), Gx is closed in G. Moreover, by Proposition 3.2.11, Gx is countable.
Then Gx is the support of λx.
Since G is étale, for every g ∈ Gx there exists an open bisection Ug such that Ug∩Gx =
{g}. Let K ⊂ G be compact. Then K ′ = K ∩ Gx is compact and λx(K) = λx(K ′).
Moreover, there are g1, . . . gn ∈ Gx such that K ′ ⊂ Ug1 ∪ . . . ∪ Ugn . Hence,
λx(K) = λx(K ′) ≤ λx(Ug1 ∪ . . . ∪ Ugn) ≤n∑i=1
λx(Ugi) =n∑i=1
λx(gi) = n.
Then λx is �nite on compact subsets. Therefore, λx is Radon by Proposition 2.4.10.
(ii) Let f ∈ Cc(G). First assume f ∈ Cc(U), where U is an open bisection. Let K =
supp(f). De�ne the open set V = r(U) and the compact set L = r(K). Hence, we can
de�ne the function f ∈ Cc(r(V )) by
f(y) =
f(r|−1U (y)), if y ∈ L,
0, otherwise.
Note that, for every y ∈ G(0),
f(y) =
∫G
f(g)dλy(g).
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Now let f ∈ Cc(G) arbitrary with support K = supp(f). Since K is compact and G
is étale, K has a �nite open cover U1, . . . ,Un of open bisections. By Proposition 3.3.5,
there are f1, . . . , fn such that f = f1 + . . .+fn and each fi ∈ Cc(Ui). As we have shown,
for every i = 1, . . . , n, the function
y 7→∫G
fi(g)dλy(g)
is continuous and compactly supported. Then the function
y 7→∫G
f(g)dλy(g)
is continuous and compactly supported.
(iii) Given f ∈ Cc(G), h ∈ G,
∫G
f(hg)dλs(h)(g) =∑
g∈Gs(h)f(hg).
We can use the change of variables g = hg because the function from Gs(h) to Gr(h)
de�ned by g 7→ hg is injective. Then we have g ∈ Gr(h) and
∑g∈Gs(h)
f(hg) =∑
g∈Gr(h)f(g) =
∫G
f(g)dλr(h)(g).
Given a Haar system {λx}, we de�ne for x ∈ G(0) the measure λx by λx(E) = λx(E−1),
for every E ⊂ G measureable.
Lemma 4.1.4. If G is a locally compact Hausdor� étale groupoid, and λx is the counting
measure on Gx, then λx is the counting measure on Gx.
Proof. Given a set A, |A| denotes the number of elements in this set. Let E ⊂ G measurable,
79
then
λx(E) = λx(E−1)
= |{g : g ∈ E−1 ∩Gx}|
= |{h−1 : h ∈ E ∩Gx}|, by the change of variables h−1 = g,
= |{h : h ∈ E ∩Gx}|,
since h 7→ h−1 is a bijection. Then λx is the counting measure on Gx.
De�nition 4.1.5. Let G be a locally compact Hausdor� groupoid with a Haar system
{λx}x∈G(0) . Given a Radon measure µ on G(0), we de�ne the induced measures on G by
ν(E) =
∫G(0)
λx(E)dµ(x), ν−1(E) =
∫G(0)
λx(E)dµ(x),
for every Borel set E. Or equivalently, for every f ∈ Cc(G),
∫f(g)dν(g) =
∫G(0)
∫Gxf(g)dλx(g)dµ(x),
∫f(g)dν−1(g) =
∫G(0)
∫Gx
f(g)dλx(g)dµ(x).
We denote ν =∫G(0) λ
xdµ(x), ν−1 =∫G(0) λxdµ(x).
If G a is locally compact Hausdor� étale groupoid and endowed with a Haar system {λx}
such that each λx is a counting measure on Gx, then we denote the induced measures ν and
ν−1 by µr and µs. In this case,
∫f(g)dµr(g) =
∫G(0)
∑g∈Gx
f(g)dµ(x),∫f(g)dµs(g) =
∫G(0)
∑g∈Gx
f(g)dµ(x).
De�nition 4.1.6. Suppose µ is a measure on G(0). We say µ is quasi-invariant if ν and ν−1
are equivalent measures. In this case we take ∆ : G → (0,∞) to be the Radon-Nikodym
derivative dν/dν−1.
Remark 4.1.7. It follows from [10, Theorem 3.72] that it is possible to choose ∆ to be a
80
homomorphism from G to (0,∞), where (0,∞) is a group with respect to the product.
Proposition 4.1.8. 1 Let G be a locally compact Hausdor� étale groupoid with Haar system
{λx} given by counting measures λx on Gx. If µ is a quasi-invariant measure on G(0), then
for µ-a.e. x and all g ∈ Gxx, we have ∆(g) = 1.
Proof. Let G′ be the isotropy bundle de�ned by G′ = ∪x∈G(0)Gxx. Note that G
′ is closed in G.
In fact, if {gi} is a sequence in G′ converging to some g in G, it follows from the continuity
of r and s that r(gi) → r(g) and s(gi) → s(g). Since r(gi) = s(gi) for every i, we have
r(g) = s(g), that is, g ∈ G′. Therefore G′ is measurable.
Let f be a positive measurable function whose support lies in G′. Then, by assumption,
∫G
f(g)dµr(g) =
∫G
f(g)∆(g)dµs(g). (4.3)
It follows from the de�nition of µr that∫G
f(g)dµr(g) =
∫G(0)
∑g∈Gx
f(g)dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)dµ(x), (4.4)
since f is supported of G′. Note that f∆ is also supported on G′. Then, by de�nition of µs,
we have
∫G
f(g)∆(g)dµs(g) =
∫G(0)
∑g∈Gx
f(g)∆(g)dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)∆(g)dµ(x). (4.5)
1Thanks Frausino!
81
Applying (4.4) and (4.5) in (4.3), we have
∫G(0)
∑g∈Gxx
(∆(g)− 1)f(g)dµ(x) = 0
De�ne B+ = {g ∈ G′ : ∆(g) ≥ 1} and B− = {g ∈ G′ : ∆(g) ≤ 1}. If we choose f = χB+ ,
we have
∫G(0)
∑g∈Gxx∩B+
(∆(g)− 1)dµ(x) = 0.
Since the function x 7→∑
g∈Gxx∩B+(∆(g)− 1) is non-negative, it follows that for µ-a.e. x,
∑g∈Gxx∩B+
(∆(g)− 1) = 0.
From the de�nition of B+, for µ-a.e. x and all g ∈ Gxx, if ∆(g) ≥ 1, it follows that ∆(g) = 1.
Analogously for B−, for µ-a.e. x and all g ∈ Gxx, if ∆(g) ≤ 1, it follows that ∆(g) = 1.
Therefore, for µ-a.e. x and all g ∈ Gxx, we have ∆(g) = 1.
Remark 4.1.9. When we study Neshveyev's second theorem, we will consider measures
µ such that ∆(g) = ec for a continuous R-valued 1-cocycle c : G → R. It follows from
Proposition 4.1.8 that for µ-a.e. x ∈ G(0), all g ∈ Gxx, we have c(g) = 0. In Remark 6.3.22
on page 200, we prove this result for extremal eβF -measures using the properties of the
Renault-Deaconu groupoid.
4.2 Borel Hilbert Bundles
Given a group G, [18] de�nes a unitary representation of G as a pair (L,H), where H is a
complex Hilbert space and L is homomorphism from G to the group of unitary operators on
H, usually written U(H), with product as group operation. We will denote Lg as the image
of g in G under the map L.
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We want to extend this notion to groupoids. By de�nition of groupoids, two elements are
not necessarily composable. In order to de�ne a unitary representation of groupoids, we will
consider unitary operators Lg : Hs(g) → Hr(g), where {Hx}x∈G(0) is a family of Hilbert spaces
indexed by G(0) and satisfying certain conditions.
Given this family of Hilbert spaces, we de�ne a Hilbert space L2(X ∗ H, µ), also denoted
by∫ ⊕XHxdµ(x). By Renault's Disintegration Theorem, a representation π : Cc(G) → H
corresponds to a groupoid representation. Moreover, H can be identi�ed with∫ ⊕XHxdµ(x).
In this section we use results from [27], [10], [4], and [15].
Let H = {Hx}x∈X be a collection of separable (nonzero) complex Hilbert spaces indexed
by a locally compact Hausdor� second countable space X. Then the total space is de�ned
by
X ∗ H = {(x, h) : h ∈ Hx},
and we let p : X ∗ H→ X be de�ned by p(x, h) = x.
Remark 4.2.1. The total space X ∗ H is de�ned in [27] assuming X is analytic. However,
we will always assume that X is locally compact Hausdor� second countable without loss of
generality. Theorem 5.3 in [13] shows that every locally compact Hausdor� second countable
space is analytic. More details about analytic and Polish spaces can be found in [6].
A section of X ∗H is a function f : X → X ∗H such that (p ◦ f)(x) = x for each x ∈ X.
The total space can be seen as the union of Hilbert spaces such that each Hx is glued to a
point x ∈ X. A section maps x to some point in {x} ×Hx, as shown in Figure 4.1.
Example 4.2.2. Let Hx = C for every x ∈ X. Then X ∗ H = X × C. In this case, each
section f corresponds to a complex-valued function f such that f(x) = (x, f(x)). Then we
can identify f with f without loss of generality.
Given a Borel measure µ on X, we want to de�ne a space of square-integrable sections,
which we denote by L2(X ∗ H, µ), such that it is a Hilbert space and every section f ∈
83
x
f(x)
Hx
y
f(y)
Hy
z
f(z)
Hz
X
Figure 4.1: X ∗H can be seen as a union of Hilbert spaces Hx such that every Hx correspondsto a point x in X. A section is a function that maps x to an element of {x} ×Hx.
L2(X ∗ H, µ) satis�es
∫X
‖f(x)‖2dµ(x) <∞.
Here we identify functions that are equal µ-a.e.
If we endow X ∗ H with a Borel structure, we would naturally be interested in a subset
of measurable sections, denoted by Sec(X ∗ H) such that
x 7→ 〈f1(x), f2(x)〉 is measurable for every f1, f2 ∈ Sec(X ∗ H), (4.6)
since we want to de�ne the inner product on L2(X ∗ H) as
〈f1, f2〉 =
∫X
〈f1(x), f2(x)〉dµ(x).
We will replace condition (4.6) by simpler conditions which consider a countable family of
sections. Later we will see that for every pair of measurable functions, (4.6) holds.
De�nition 4.2.3. Let H = {Hx}x∈X be a family of separable Hilbert spaces indexed by a
locally compact Hausdor� second countable space X. Then (X ∗ H, p) is an analytic Borel
84
Hilbert bundle if X ∗ H has a Borel structure such that
(a) For every E ⊂ X, p−1(E) is a Borel in X ∗ H if, and only if, E is Borel in X;
(b) There is a sequence of sections {fn} such that
(i) the map fn : X ∗ H→ C de�ned by fn(x, h) = 〈fn(x), h〉Hx is Borel for each n,
(ii) for each n and m, x 7→ 〈fn(x), fm(x)〉Hx is Borel, and
(iii) the functions {fn} and p separate points of X ∗ H.
The sequence {fn} is called a fundamental sequence for (X ∗ H, p).
Remark 4.2.4. Given an analytic Borel Hilbert bundle (X, p), we let Sec(X ∗ H) denote
the set of sections f : X → X ∗ H such that
x 7→ 〈f(x), fn(x)〉
is Borel for every fn. It follows from [27, Remark F.3] that a section f of X ∗ H is in
Sec(X ∗ H) if, and only if, it is Borel.
Remark 4.2.5. Note that given a section f of X ∗ H, f(x) is an element of {x} × Hx.
Although this is an abuse of notation, we will identify {x} × Hx with Hx. Hence, the inner
product 〈f(x), h〉 is well-de�ned for h ∈ Hx. We can also write 〈f1(x), f2(x)〉 for sections
f1, f2 without further comments.
Example 4.2.6. Let X be a locally compact Hausdor� second countable space. Let H be
a separable Hilbert space. De�ne Hx = H for each x ∈ X. Then X ∗ H = X ×H. Endow
X ×H with the product topology. Given E ⊂ X, p−1(E) = E ×H. Thus p−1(E) is Borel
if, and only if, E is Borel. Therefore property (a) of De�nition 4.2.3 holds.
Assume {en} is an orthonormal basis for H. De�ne a sequence of sections of X ∗ H by
fn(x) = (x, en) ∈ {x} ×Hx. We will prove that {fn} is a fundamental sequence:
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(i) fn : X ×H → C is de�ned by fn(x, h) = 〈fn(x), h〉 = 〈en, h〉. fn is continuous, then it
is Borel.
(ii) Given n,m, the function x 7→ 〈fn(x), fm(x)〉 = 〈en, em〉 is constant, therefore this
function is Borel.
(iii) Let (x, h), (y, k) ∈ X ×H. If x 6= y, then p(x, h) 6= p(y, k) by de�nition of p. Assume
x = y and h 6= k. There exists en such that 〈en, h〉 6= 〈en, k〉. Thus, fn(x, h) 6= fn(y, k).
Therefore (X ×H, p) is an analytic Borel Hilbert bundle.
Example 4.2.7. Given a locally compact Hausdor� second countable space X, let X =
X∞ ∪ X1 ∪ X2 ∪ . . . be a Borel partition of X, i.e., every Xd is Borel and the collection of
Xd is disjoint.
For every d = ∞, 1, 2, . . ., let H(d) be a Hilbert space of dimension d and basis {edn}n=dn=1.
Then
X ∗ H =d=∞⋃d=1
Xd ×H(d).
EndowX∗H with a Borel structure such that E ⊂ X∗H is Borel if, and only if, E∩(Xd×H(d))
is Borel for all d.
De�ne, for every n =∞, 1, 2, . . ., the section fn such that
fn(x) =
edn, if x ∈ Xd and 1 ≤ n ≤ d,
0, otherwise.
We prove that (X ∗ H, p) is an analytic Borel Hilbert bundle. Recall from Example 4.2.6
that (Xd×H(d), p|Xd) is an analytic Borel Hilbert bundle. Moreover, {fn|Xd} is a fundamental
sequence. Here we assume fn|Xd : Xd → Xd ×H(d) and p|Xd×H(d) : Xd ×H(d) → Xd without
loss of generality.
Now we show that all properties in De�nition 4.2.3 hold for (X ∗ H, p).
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(a) Let E ⊂ X. Then
p−1(E) =d=∞⋃d=1
(E ∩Xd)×H(d). (4.7)
Assume E is Borel and �x d =∞, 1, 2, . . .. Then E ∩Xd is a Borel subset of Xd. Since
(Xd ×H(d), p|Xd×H(d)) is a Borel Hilbert bundle, it follows that
p|−1Xd×H(d)(E ∩Xd) = (E ∩Xd)×H(d) is Borel.
Note that d is arbitrary. From (4.7) we have that p−1(E) is Borel.
Conversely, assume p−1(E) is Borel. Fix d =∞, 1, 2, . . .. Then p−1(E) ∩ (Xd ×H(d)) is
Borel in Xd×H(d). However, from (4.7), p−1(E)∩ (Xd×H(d)) = (E∩Xd)×H(d). Hence
E ∩Xd is Borel in X. Since d is arbitrary, it follows that E is Borel.
(b) (i) Fix n. fn|Xd is Borel for every n, d from Example 4.2.6.
Since fn = fn|X∞ + fn|X1 + fn|X2 + . . ., it follows that fn is Borel.
(ii) Fix d and let n,m ∈ {∞, 1, 2, . . .}. {fn|Xd} is a fundamental sequence for the Borel
Hilbert bundle (Xd ×H(d), p|Xd×H(d)). Then the function from Xd to C de�ned by
x 7→ 〈fn(x), fm(x)〉 is Borel.
Since d is arbitrary, the function from X to C de�ned by x 7→ 〈fn(x), fm(x)〉 is
Borel.
(iii) Let (x, h), (y, k) ∈ X ∗ H. If x 6= y, p(x, h) = p(y, k).
Assume x = y. Let d be such that x ∈ Xd. Since (Xd × H(d), p|Xd×H(d)) is an
analytic Borel Hilbert bundle with fundamental sequence {fn|Xd×H(d)}n=∞n=1 , there
exists n such that fn|Xd×H(d)(x, h) 6= fn|Xd×H(d)(y, h). Thus fn(x, h) 6= fn(x, k).
Therefore (X ∗ H, p) is an analytic Borel Hilbert bundle.
Given f1, f2 ∈ Sec(X ∗H), [27, Proposition F.6] shows that the function x 7→ 〈f1(x), f2(x)〉
is Borel. If µ is a Borel measure on X, we can de�ne the normed vector space L2(X ∗ H, µ)
87
formed by the quotient of
L2(X ∗ H, µ) = {f ∈ Sec(X ∗ H) : x→ ‖f(x)‖2 is µ-integrable},
where functions agreeing µ-almost everywhere are identi�ed. L2(X ∗H, µ) is a Hilbert space
with inner product de�ned by
〈f1, f2〉 =
∫X
〈f1(x), f2(x)〉dµ(x).
Here we also denote L2(X ∗ H) by∫ ⊕XHxdµ(x).
Given a Borel Hilbert bundle X ∗ H, we can de�ne a groupoid, called the isomorphism
bundle, consisting of unitary operators between Hilbert spaces in the family {Hx}.
De�nition 4.2.8. GivenX a locally compact Hausdor� second countable space, let (X∗H, p)
be an analytic Borel Hilbert bundle. The isomorphism bundle of X ∗ H is the set
Iso(X ∗ H) = {(x, L, y) such that L : Hy → Hx is unitary}
endowed with the smallest Borel structure such that, for all f1, f2 ∈ Sec(X ∗ H),
φf1,f2(x, L, y) = 〈Lf1(y), f2(x)〉 (4.8)
de�ne Borel functions from Iso(X ∗ H) to C.
Lemma 4.2.9. Iso(X∗H) is a groupoid whose units are de�ned by (x, idHx , x). Two elements
(x, L, y) and (y′,M, z) are composable if y = y′. In this case, we de�ne their product by
(x, L, y)(y,M, z) = (x, LM, z). The inverse in Iso(X∗H) is de�ned by (x, L, y)−1 = (y, L*, x).
The range and source maps are de�ned by r(x, L, y) = (x, idHx , x) and s(x, L, y) = (y, idHy , y)
Proof. Given x ∈ X, idHx is unitary. Hence (x, idHx , x) ∈ Iso(X∗H). Then r, s are surjective.
Note that we can identify (x, idHx , x) with x without loss of generality. Thus we will assume
88
X = Iso(X ∗ H)(0).
Now we show that the product in Iso(X ∗ H) is well-de�ned. In fact, given (x, L, y),
(y,M, z) ∈ Iso(X ∗ H), then LM : Hz → Hx is a unitary operator. Thus (x, LM, z) ∈
Iso(X ∗H). The inverse is well-de�ned as well, since (x, L, y)−1 = (y, L*, x) by de�nition and
L* : Hx → Hy is unitary.
Now we prove the conditions in De�nition 3.1.1. Let g = (x, L, y), h = (y,M, z) and
k = (z,N,w) be elements of Iso(X ∗ H). Then
(i) s(gh) = s((x, L, y)(y,M, z)) = s(x, LM, z) = z = s(y,M, z) = s(h), and
r(gh) = r((x, L, y)(y,M, z)) = r(x, LM, z) = x = r(x, L, y) = r(g).
(ii) x = r(x, idHx , x) = r(x), and x = s(x, idHx , x) = s(x).
(iii) gs(g) = gy = (x, L, y)(y, idHy , y) = (x, LidHy , y) = (x, L, y) = g, and
r(g)g = xg = (x, idHx , x)(x, L, y) = (x, idHxL, y) = (x, L, y) = g.
(iv) (gh)k = [(x, L, y)(y,M, z)] (z,N,w) = (x, LM, z)(z,N,w) = (x, LMN,w), and
g(hk) = (x, L, y) [(y,M, z)(z,N,w)] = (x, L, y)(y,MN,w) = (x, LMN,w).
Then (gh)k = g(hk).
(v) gg−1 = (x, L, y)(y, L*, x) = (x, LL*, x) = (x, idHx , x) = x = r(g), and
g−1g = (y, L*, x)(x, L, y) = (y, L*L, y) = (y, idHy , y) = y = s(g).
Therefore Iso(X ∗ H) is a groupoid.
De�nition 4.2.10. Let G be a locally compact Hausdor� second countable étale groupoid
with Haar system {λx}x∈G(0) . A unitary representation of G is a triple (µ,G(0) ∗ H, L) such
that µ is a quasi-invariant measure on G(0), L = {Lg}g∈G is a family of unitary operators
Lg : Hs(g) → Hr(g), and L : G→ Iso(G(0) ∗H) is a Borel homomorphism such that g 7→ Lg =
(r(g), Lg, s(g)).
89
Now we de�ne a norm on Cc(G) which will be an upper bound for the norm of a rep-
resentation obtained with the Renault's disintegration theorem. This norm is de�ned in
[20].
De�nition 4.2.11. Let G be a locally compact Hausdor� second countable étale groupoid
with Haar system {λx}x∈G(0) . We de�ne the I-norm on Cc(G) by
‖f‖I = max
{supx∈G(0)
∫G
|f(g)|dλx(g), supx∈G(0)
∫G
|f(g)|dλx(g)
}.
If the Haar system is given by counting measures λx on Gx, then
‖f‖I = max
{supx∈G(0)
∑g∈Gx|f(g)|, sup
x∈G(0)
∑g∈Gx
|f(g)|
}.
Lemma 4.2.12. Let G be a locally compact Hausdor� second countable étale groupoid with
Haar system given by counting measures λx on Gx. Then ‖f‖I < ∞ for every f ∈ Cc(G).
Moreover, ‖ · ‖I de�nes a norm on Cc(G).
Proof. Let f1, f2, f ∈ Cc(G), λ ∈ C.
• ‖f‖I = 0 implies f = 0.
Assume ‖f‖I = 0. Let g ∈ G and x = r(g). Then,
|f(g)| ≤∑h∈Gx|f(h)| ≤ ‖f‖I = 0.
Therefore f = 0.
• ‖λf‖ = |λ|‖f‖.
Let x ∈ G(0). Then
∑g∈Gx|λf(g)| = |λ|
∑g∈Gx|f(g)|.
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Thus,
supx∈G(0)
∑g∈Gx|λf(g)| = |λ| sup
x∈G(0)
∑g∈Gx|f(g)|.
Analogously,
supx∈G(0)
∑g∈Gx
|λf(g)| = |λ| supx∈G(0)
∑g∈Gx
|f(g)|.
Therefore ‖λf‖I = |λ|‖f‖I ..
• ‖f1 + f2‖I ≤ ‖f1‖I + ‖f2‖I .
‖f1 + f2‖I = max
{supx∈G(0)
∑g∈Gx|f1(g) + f2(g)|, sup
x∈G(0)
∑g∈Gx
|f1(g) + f2(g)|
}
≤ max
{supx∈G(0)
∑g∈Gx
(|f1(g)|+ |f2(g)|), supx∈G(0)
∑g∈Gx
(|f1(g)|+ |f2(g)|)
}
≤ max
{supx∈G(0)
∑g∈Gx|f1(g)|+ ‖f2‖I , sup
x∈G(0)
∑g∈Gx
|f1(g)|+ ‖f2‖I
}
= ‖f1‖I + ‖f2‖I .
Now we show that ‖f‖I <∞ for every f ∈ Cc(G).
Let U be an open bisection of G and let f ∈ Cc(U). For every x ∈ r(U), there is a unique
hx ∈ U satisfying r(hx) = x. Note that x = r|−1U (x). Hence,
∑g∈Gx|f(g)| = |f(hx)|.
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Since r|U is a homeomorphism, we have
supx∈r(U)
∑g∈Gx|f(g)| = sup
x∈r(U)
|f(hx)| = supx∈r(U)
|f ◦ r|−1U (x)| = sup
g∈U|f(g)| = ‖f‖∞.
If x /∈ r(U), then f(g) = 0 for every g ∈ Gx. Thus,
supx∈G(0)
∑g∈Gx|f(g)| = ‖f‖∞.
Analogously, we can prove that
supx∈G(0)
∑g∈Gx
|f(g)| = ‖f‖∞.
Therefore ‖f‖I = ‖f‖∞ for f ∈ Cc(U).
Now let f ∈ Cc(G). It follows from Proposition 3.3.5 that there exist f1, . . . fn continuous
functions supported on open bisections such that f = f1 + . . . + fn. Then ‖f‖I ≤ ‖f1‖ +
. . .+ ‖fn‖ <∞. Therefore ‖ · ‖I is a norm on Cc(G).
4.3 Renault's Disintegration Theorem
The results in this section can be found in [15]. The Renault's Disintegration Theorem can
also be found in [4] and [10]. However, both use the abstract notion of upper semi-continuous
C*-bundle and the theorem is stated in a more general case than the results presented here.
Proposition 4.3.1. Let G be a locally compact Hausdor� étale groupoid with Haar system
{λx}x∈G(0) . If (µ,G(0) ∗ H, L) is a unitary representation of G, then we obtain a ‖ · ‖I-
norm bounded representation π of Cc(G) on∫ ⊕G(0)Hxdµ(x), called the integrated form of
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(µ,G(0) ∗ H, L), determined by
〈π(f)h, k〉 =
∫G
f(g)〈Lghs(g), kr(g)〉∆(g)−12dν(g), (4.9)
where ν =∫G(0) dλ
xdµ(x), ν−1 =∫G(0) dλxdµ(x) and ∆ = dν/dν−1 as described in De�nition
4.1.5.
Renault's disintegration theorem shows that a representation π : Cc(G) → H can be
identi�ed with a groupoid representation (µ,G(0) ∗ H, L) such that L(G) is measurable in
Iso(G(0) ∗ H). Condition (4.8) on page 88 will be necessary in order to write (4.9).
De�nition 4.3.2. Given a vector space H0, Lin(H0) denotes the set of linear operators
T : H0 → H0. Note that Lin(H0) is an algebra whose product is de�ned by composition of
operators.
Theorem 4.3.3. (Renault's disintegration theorem) Let G be a locally compact Hausdor�
étale groupoid. Suppose that H0 is a dense subspace of a complex Hilbert space H. Let
π : Cc(G)→ Lin(H0) be a homomorphism such that:
(a) {π(f)h : f ∈ Cc(G) and h ∈ H0} is dense in H;
(b) For each h, k ∈ H0,
f 7→ 〈π(f)h, k〉
is continuous in the inductive limit topology on Cc(G);
(c) For f ∈ Cc(G) and h, k ∈ H0, we have
〈π(f)h, k〉 = 〈h, π(f*)k〉.
Then each π(f) is bounded and extends to an operator π(f) on H of norm at most ‖f‖I .
Furthermore, π is a representation of Cc(G) on H and there is a unitary representation
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(µ,G(0) ∗H, L) of G such that H ∼ L2(G(0) ∗H, µ) and π is equivalent to the integrated form
of (µ,G(0) ∗ H, L).
Remark 4.3.4. Recall that the inductive limit topology is introduced in De�nition 3.3.18
on page 71.
Lemma 4.3.5. Let G be a locally compact Hausdor� second countable étale groupoid. Let
ϕ be a state on C*(G) with GNS-triple (H, π, ξ). Then π satis�es the conditions of Renault's
disintegration theorem.
Proof. Assume ϕ is a state on C*(G). Let (H, π, ξ) be the corresponding GNS-triple.
(a) {π(f)h : f ∈ Cc(G) and h ∈ H} is dense in H
Since ξ is a cyclic vector, it follows that π(C*(G))ξ is dense in H. Recall that Cc(G) is
dense in C*(G), then π(Cc(G))ξ is dense in π(C*(G))ξ. Therefore, the property holds.
(b) For h, k ∈ H, f 7→ 〈π(f)h, k〉 is continuous in the inductive limit topology on Cc(G)
Let h, k ∈ H. Given f ∈ Cc(G), let {fi}i∈N be a sequence in Cc(G) such that fi → f
with respect to the inductive limit topology. It follows from Lemma 3.3.19 that fi → f
in C*(G). Then ‖π(fi)− π(f)‖ → 0 and therefore 〈(π(fi)− π(f))k, h〉 → 0.
(c) 〈π(f)h, k〉 = 〈h, π(f*)〉
Let h, k ∈ H, f ∈ Cc(G). Since π is an ∗-representation of C*(G), we have π(f*) =
π(f)*. Therefore the property holds.
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Chapter 5
Neshveyev's Theorems
In this chapter we describe the KMS states of a groupoid C*-algebra. Neshveyev proves
this result in [17] in two theorems: the �rst theorem describes all states satisfying a certain
condition, and the second theorem describes all KMS states and it is a corollary of the �rst
result. These theorems show a correspondence between a KMS state ϕ on C*(G) and a pair
(µ, {ϕx}x∈G(0)) consisting of a probability measure µ on G(0) and a family of states ϕx on
C*(Gxx) satisfying a certain condition.
Before proving Neshveyev's theorems, we de�ne KMS states on a C*-algebra as described
in [2], [3] and [9].
5.1 KMS States
In this section we de�ne the notion of KMS states and prove some of their main properties.
The results in this section can be found in [2], [3] and [9]. KMS states characterizes the
equilibrium states in quantum statistical mechanics.
Before we de�ne KMS states, let us recall the de�nition of approximate unit and prove
some properties of states on a C*-algebra.
Let A be a C*-algebra. An approximate unit for A is an increasing net {uλ}λ∈Λ of positive
95
elements in the closed unit ball of A such that a = limλ auλ for all a ∈ A. Equivalently,
a = limλ uλa. From [16], it follows that every C*-algebra A contains an approximate unit.
Moreover, if A is separable, then it admits an approximate unit which is a sequence.
Recall the following theorems on positive linear functionals from [16]:
Theorem 5.1.1. If ϕ is a positive linear functional on a C*-algebra A, then it is bounded.
Theorem 5.1.2. Let ϕ be a bounded linear functional on a C*-algebra A. The following
conditions are equivalent:
(i) ϕ is positive.
(ii) For each approximate unit {uλ}λ∈Λ, ‖ϕ‖ = limλ ϕ(uλ).
(iii) For some approximate unit {uλ}λ∈Λ, ‖ϕ‖ = limλ ϕ(uλ).
Proof. Assume ϕ 6= 0. First we show the implication (i)⇒ (ii) holds.
Note that the map A2 → C de�ned by (a, b) 7→ ϕ(b*a) is a positive sesquilinear form on
A. Hence ϕ(b*a) = ϕ(a*b) and the Cauchy-Schwarz inequality |ϕ(b*a)| ≤ ϕ(b*b)1/2ϕ(a*a)1/2
holds.
Let (uλ)λ∈Λ be an approximate unit of A and let a ∈ A with a 6= 0. Then, for every λ ∈ Λ,
|ϕ(auλ)|2 ≤ ϕ(a*a)ϕ(uλ*uλ) = ϕ(a*a)ϕ(u2λ), (5.1)
by the Cauchy-Schwarz inequality. Since the net {uλ}λ∈Λ is increasing and its elements are in
the unit ball, then the net {u2λ}λ∈Λ is also increasing and included in the unit ball. Because
ϕ is positive and bounded, we have supλ∈Λ ϕ(u2λ) = limλ∈Λ ϕ(u2
λ). Then, for every i ∈ Λ, we
have by (5.1),
|ϕ(aui)|2 ≤ ϕ(a*a) limλ∈Λ
ϕ(u2λ) ≤ ‖ϕ‖‖a*a‖ lim
λ∈Λϕ(u2
λ) = ‖ϕ‖‖a‖2 limλ∈Λ
ϕ(u2λ).
96
Therefore, using the continuity of ϕ,
limλ∈Λ|ϕ(aui)|2 ≤ ‖ϕ‖‖a‖2 lim
λ∈Λϕ(u2
λ)
|ϕ(a)|2 ≤ ‖ϕ‖‖a‖2 limλ∈Λ
ϕ(u2λ)
|ϕ(a)|2
‖a‖2≤ ‖ϕ‖ lim
λ∈Λϕ(u2
λ).
Since a is arbitrary, ‖ϕ‖ ≤ limλ∈Λ ϕ(u2λ). On the other hand, for each λ ∈ Λ,
ϕ(u2λ) ≤ ‖ϕ‖‖u2
λ‖ ≤ ‖ϕ‖.
Then ‖ϕ‖ = limλ∈Λ ϕ(u2λ). Note that u
2λ−uλ ≤ 0 for every λ, thus ϕ(u2
λ) ≤ ϕ(uλ). Therefore,
‖ϕ‖ = limλ∈Λ
ϕ(u2λ) ≤ lim
λ∈Λϕ(uλ) ≤ ‖ϕ‖.
It is obvious that (ii)⇒ (iii).
Now we show that (iii) ⇒ (i). Suppose that {un}n∈N is an approximate unit such that
1 = limn→∞ τ(un). Let a be a selft-adjoint element of A such that ‖a‖ ≤ 1 and write
τ(a) = α + iβ where α, β are real numbers. To show that τ(a) ∈ R, we may suppose that
β ≤ 0. If k is a positive integer, then
‖a− iuλ‖2 = ‖(a+ ikun)(a− ikun)‖
= ‖a2 + k2u2n − ik(aun − una)‖
≤ 1 + k2 + k‖aun − una‖,
so |τ(a− ikun)|2 ≤ 1 + k2 + k‖aun − una‖ because ‖τ‖ = 1.
However, limk→∞ τ(a− ikun) = τ(a)− ik by hypothesis, and limk→∞(aun − una) = 0, so
97
in the limit as n→∞, we get
|α + iβ − ik|2 = |τ(a)− ik|2 = |α + iβ − ik|2 ≤ 1 + k2,
then
α2 + β2 − 2kβ + k2 ≤ 1 + n2
⇒− 2kβ ≤ 1− β2 − α2.
Since β is not positive and this inequality holds for all positive integers n, β must be zero.
Therefore, τ(a) is real if a is hermitian.
Now suppose that a is positive and ‖a‖ ≤ 1. Then un− a is hermitian and ‖un− a‖ ≤ 1,
so τ(un− a) ≤ 1. But then 1− τ(a) = limn→∞ τ(un− a) ≤ 1, and therefore τ(a) ≥ 0. Thus,
τ is positive and we have shown (iii)⇒ (i).
We want to show that, if the C*-algebra A is commutative and unital, then the extremal
states are precisely the characters on A. We need some results on convex spaces before
studying this.
De�nition 5.1.3. Given a commutative C*-algebra A, a character is a non-zero homomor-
phism ϕ : A→ C.
De�nition 5.1.4. Let X be a normed vector space. A functional η in a convex subset
C ⊂ X* is an extreme point in C (or extremal) if the condition η = tη1 + (1 − t)η2, where
η1, η2 ∈ C, 0 < t < 1, implies that η = η1 = η2.
De�nition 5.1.5. We say that a state ϕ on A is pure if it has the property that whenever
ρ is a positive linear functional on A such that ρ ≤ ϕ, necessarily there is a number t ∈ [0, 1]
such that ρ = tϕ.
Theorem 5.1.6. [3, Theorem 2.3.15] Let A be a unital C*-algebra. The set of states on A
is convex and its extremal points are the pure states.
98
Theorem 5.1.7. [16, Theorem 5.1.6] Let ϕ be a state on a commutative C*-algebra A.
Then ϕ is pure if, and only if, it is a character on A.
Lemma 5.1.8. Let X be a Borel space endowed with a probability measure µ. Let A be a
separable C*-algebra and A0 a dense subset of A. Let ψx be a family of states on A de�ned
for µ-a.e. x ∈ X such that for every a ∈ A0 the map x 7→ ψx(a) is µ-measurable. De�ne ϕ
by
ϕ(a) =
∫X
ψx(a)dµ(x).
Then ϕ is a state on A.
Proof. Let V ⊂ X be a conull set such that ψx is de�ned for every x ∈ V . De�ne for every
a ∈ A the function Fa : X → C by
Fa(x) =
ψx(a) if x ∈ V
0 otherwise.
By assumption, Fa is µ-measurable for every a ∈ A0.
Let a ∈ A. Since A0 is dense in A, there exists a sequence {an}n∈N in A0 converging to a.
Then, using the continuity of ψx, we have for µ-a.e. x,
Fa(x) = ψx(a) = limn→∞
ψx(an) = limn→∞
Fan(x).
Therefore Fa is measurable. In other words, the function x 7→ ψx(a) is µ-measurable for
every a ∈ A.
In order to show that ϕ is a state, we begin by proving that ϕ is a positive linear functional.
Since each ψx linear, it follows from the de�nition of ϕ that ϕ is also linear. Let a be a positive
99
element in A. Then ψx(a) ≥ 0 for µ-a.e. x. Thus,
ϕ(a) =
∫X
ψx(a)dµ(x) ≥ 0.
By Theorem 5.1.1, ϕ is bounded. It follows from Theorem 5.1.2 that, for some approximate
unit {un}n∈N of A,
‖ϕ‖ = limn→∞
ϕ(un) = limn→∞
∫X
ψx(un)dµ(x).
However,
|ψx(un)| ≤ ‖ψx‖‖un‖ = ‖un‖ ≤ 1,
for every n and for µ-a.e. x ∈ X because each ψx is a state and every un satis�es ‖un‖ ≤ 1.
Therefore we can apply the dominated convergence theorem, then
‖ϕ‖ = limn→∞
∫X
ψx(un)dµ(x)
=
∫X
(limn→∞
ψx(un))dµ(x)
=
∫X
1dµ(x), since each ψx is a state,
= 1.
Now we de�ne the notion of dynamical system on a C*-algebra and prove some of its
properties.
De�nition 5.1.9. We say that τ = {τt}t∈R is a one-parameter group of ∗-automorphisms of
a C*-algebra A if τt : A→ A is an ∗-automorphism and
(i) τt+s = τt ◦ τs for all t, s ∈ R;
100
(ii) τ0 = id.
Example 5.1.10. Consider the algebra of square matrices Mn(C) for some n. Let H ∈
Mn(C), and de�ne each τt : Mn(C) → Mn(C) by τt(A) = eitHAe−itH . Note that τt is linear
and τ0 is the identity. Each τt is a homomorphism because
τt(AB) = eitHABe−itH = eitHAe−itHeitHBe−itH = τt(A)τt(B).
Also, the equality τt+s = τt ◦ τs holds, since
τt+s(A) = ei(t+s)HAe−i(t+s)H = eitHeisHAe−isHe−itH = eitHτs(A)e−itH = τt(τs(A)).
De�nition 5.1.11. A C*-dynamical system is a pair (A, τ) where A is a C*-algebra and
τ = {τt}t∈R is a one-parameter group of ∗-automorphisms strongly continuous , i.e., t 7→ τt(a)
is continuous in the norm for all a ∈ A. If we �x τ , we say that τ is the dynamics on A.
Now we prove that, under certain conditions, if we de�ne a one-parameter group of ∗-
automorphisms τ on a dense ∗-algebra of a C*-algebra A, then we can extend the operators
uniquely on A. Moreover, τ de�nes a dynamics on A.
Lemma 5.1.12. Let A0 be a dense ∗-subalgebra of a C*-algebra A. Let τ = {τt}t∈R be a
family of ∗-automorphisms τt : A0 → A0 such that
(i) τt+s = τt ◦ τs for t, s ∈ R,
(ii) τ0 = id,
(iii) t 7→ τt(a) is continuous in the norm for each a ∈ A0.
Moreover, assume
‖τt(a)‖ ≤ ‖a‖, for every t ∈ R, a ∈ A0. (5.2)
101
Then τ can be extended uniquely to a dynamics on A, which we also denote by τ = {τt}t∈Rwithout loss of generality.
Proof. First we show that each τt can be extended uniquely to A. Let a ∈ A. Since A0 is
dense in A, there exists a sequence {an}n∈N converging to a.
Using the fact that A0 is a vector space and the equation (5.2), we have that {τt(an)}n∈Nis a Cauchy sequence, then we can de�ne τt(a) = limn→∞ τt(an).
Note that τt(a) is well-de�ned. In fact, let {bn}n∈N be an arbitrary sequence converging
to a, and let x = limn→∞ τt(bn). Then
‖τt(a)− x‖ = limn→∞
‖τt(an)− τt(bn)‖ ≤ limn→∞
‖an − bn‖ = 0.
It follows from the de�ntion of τt on A that ‖τt(a)‖ ≤ ‖a‖ for every a ∈ A.
Note that the extension τ also satis�es τ0 = id and τt+s = τt ◦ τs. In fact, given a ∈ A, let
{an}n∈N be a sequence in A0 converging to a. Then
τ0(a) = limn→∞
τ0(an) = limn→∞
an = a
τt+s(a) = limn→∞
τt+s(an) = limn→∞
τt ◦ τs(an) = τt ◦ τs(a).
Then τt is invertible with inverse τ−t.
Now we show that τt : A → A is an automorphism. Let a, b ∈ A and {an}n∈N, {bn}n∈Nsequences in A0 converging to a, b respectively. Then
τt(ab) = limn→∞
τt(anbn) = limn→∞
τt(an)τt(bn) = τt(a)τt(b),
τt(a*) = limn→∞
τt(an*) = limn→∞
τt(an)* = τt(a)*.
Finally, we show that τ is strongly continuous. Let a ∈ A and �x t0. Given ε, there exists
a0 ∈ A0 such that ‖a − a0‖ < ε/3. Let δ > 0 such that ‖τt(a0) − τs(a0)‖ < ε/3 for every
t ∈ R such that |t− t0| < δ.
102
Given t with |t− t0| < δ, we have
‖τt(a)− τt0(a)‖ ≤ ‖τt(a)− τt(a0)‖+ ‖τt(a0)− τt0(a0)‖+ ‖τt0(a0)− τt0(a)‖
≤ 2‖a− a0‖+ ‖τt(a0)− τt0(a0)‖
≤ 2ε
3+ε
3
= ε.
Let X be a complex Banach space and X* its dual. Let σ(X,X*) denote the topology
on X induced by the functionals on X. This topology is denoted the weak topology on X.
Remark 5.1.13. We say that a function f : R→ X is σ(X,X*)-continuous if η ◦f : R→ C
is continuous for every η ∈ X*.
De�nition 5.1.14. Given a Banach space X, let {τt}t∈R be a family of linear and bounded
operators τt : X → X. This family is called a one-parameter σ(X,X*)-continuous group of
isometries if
(i) τt+s = τt ◦ τs for all t, s ∈ R;
(ii) τ0 = id;
(iii) ‖τt‖ = 1 for all t ∈ R;
(iv) t 7→ τt(a) is σ(X,X*)-continuous for all a ∈ X.
Lemma 5.1.15. Let A be a C*-algebra and τ its dynamics. Then τ is a one-parameter
σ(A,A*)-continuous group of isometries.
Proof. Note that properties (i) and (ii) in De�nitions 5.1.9 and 5.1.14 are equal. In both,
every τ is a linear operator on A. So we need to show properties (iii) and (iv) of De�nition
5.1.14.
103
(iii) ‖τt‖ = 1
This follows from the fact that every ∗-automorphism on a C*-algebra is an isometry.
(iv) t 7→ τt(a) is σ(A,A*)-continuous for all a ∈ A
Let a ∈ A and t0 ∈ R. τ is strongly continuous by assumption. Then τt(a) → τt0(a)
in the norm as t → t0. Thus, for every η ∈ A*, η(τt(a)) → η(τt0(a)) as t → t0. Thus
property (iv) holds.
Therefore τ is a one-parameter σ(A,A*)-continuous group of isometries.
Note that, for every a ∈ X, the function t 7→ η ◦ τt(a) is continuous for all η ∈ X*. If we
can extend this function to an analytic function on a strip in C which we will de�ne later,
we say that a is analytic. We will prove that the set of analytic elements is dense in X.
De�nition 5.1.16. Let τ be a one-parameter σ(X,X*)-continuous group of isometries. An
element a ∈ X is analytic for τ if there exists λ > 0 and a function f : Iλ → X, where
Iλ = {z ∈ C : |Im(z)| < λ}, such that
(i) f(t) = τt(a) for all t ∈ R;
(ii) The function z 7→ η(f(z)) is analytic in Iλ for all η ∈ X*.
Under these conditions we write
τz(a) = f(z), for z ∈ Iλ.
If λ =∞, we say that a is entire analytic for τ .
Remark 5.1.17. Suppose a1, a2 ∈ X are entire analytic, and α ∈ C. Let f1, f2 be the
corresponding functions as in De�nition 5.1.16. Then, for every t ∈ R,
f1(t) + αf2(t) = τt(a1) + ατt(a2) = τt(a1 + αa2). (5.3)
104
Moreover, the function z 7→ η(f1(z) +αf2(z)) is analytic for every η ∈ X*. Thus a1 +αa2 is
entire analytic. Therefore the set of entire analytic elements in X, denoted Xτ , is a subspace
of X. For a1, a2 ∈ Xτ , α, z ∈ C, it follows from (5.3) that τz(a1 + αa2) = τz(a1) + τz(a2).
Later we will prove that Xτ is also dense in X, both in the σ(X,X*)-topology (see
Proposition 5.1.19) and with respect to the norm (see Corollary 5.1.24).
When τ de�nes a dynamics on a C*-algebra A, Aτ is not only a vector space, but a
∗-subalgebra. We will prove this result in Lemma 5.1.22 on page 110.
Lemma 5.1.18. Let n be a positive integer, and δ > 0. Then√n
π
∫ ∞−∞
e−nt2
dt = 1, (5.4)√n
π
∫|t|≥δ
e−nt2
dt =1√π
∫|t|≥√nδ
e−t2
dt. (5.5)
Moreover, the second integral converges to zero as n→∞.
Proof. Let δ ≥ 0. Since the function t 7→ e−nt2is even, we have√
n
π
∫|t|≥δ
e−nt2
dt = 2
√n
π
∫t≥δ
e−nt2
dt
=2√π
∫s≥√nδ
e−s2
ds, by the change of variables s =√nt,
=1√π
∫|s|≥√nδ
e−s2
ds.
Thus the equality (5.5) follows. If δ = 0, we have√n
π
∫ ∞−∞
e−nt2
dt =
√n
π
∫|t|≥0
e−nt2
dt =1√π
∫|s|≥0
e−s2
ds =1√π
∫ ∞−∞
e−s2
ds = 1.
Therefore the integral√
nπ
∫|t|≥δ e
−nt2dt converges to zero as n→∞.
Proposition 5.1.19. If τ is a one-parameter σ(X,X*)-continuous group of isometries, and
105
a ∈ X, de�ne for n = 1, 2, . . .,
an =
√n
π
∫ ∞−∞
τt(a)e−nt2
dt.
Then, for each n, an is an entire analytic element for τ and ‖an‖ ≤ ‖a‖. In addition,
τz(an) =
√n
π
∫ ∞−∞
τt(a)e−n(t−z)2dt,
for every z ∈ C. Moreover, an → a on the σ(X,X*) topology as n→∞. In particular, Xτ ,
is a σ(X,X*)-dense subspace of X.
Proof. Let a ∈ X. De�ne for each n = 1, 2, . . . the function fn : C→ X by
fn(z) =
√n
π
∫ ∞−∞
τt(a)e−n(t−z)2dt.
This function is well-de�ned since t 7→ e−n(t−z)2 is an integrable function, ‖τt(a)‖ ≤ ‖a‖ for
every t, and t 7→ e−n(t−z)2τt(a) is continuous, then we apply Proposition 2.6. Note that for
each s ∈ R, we have
fn(s) =
√n
π
∫ ∞−∞
τt(a)e−n(t−s)2dt
=
√n
π
∫ ∞−∞
τt+s(a)e−nt2
dt
=
√n
π
∫ ∞−∞
τs ◦ τt(a)e−nt2
dt
= τs
(√n
π
∫ ∞−∞
τt(a)e−nt2
dt
), by Corollary 2.6.17,
= τs(an).
Let η ∈ X*. Let z, z0 ∈ C such that z 6= z0. By de�nition of fn, it follows that∣∣∣∣η(fn(z))− η(fn(z0))
z − z0
−√n
π
∫ ∞−∞
2n(t− z)e−n(t−z)2η(τt(a))dt
∣∣∣∣106
equals to∣∣∣∣∣√n
π
∫ ∞−∞
η(τt(a))
(e−n(t−z)2 − e−n(t−z0)2
z − z0
)dt−
√n
π
∫ ∞−∞
2n(t− z)e−n(t−z)2η(τt(a))dt
∣∣∣∣∣=
√n
π
∣∣∣∣∣∫ ∞−∞
(e−n(t−z)2 − e−n(t−z0)2
z − z0
− 2n(t− z)e−n(t−z)2)η(τt(a))dt
∣∣∣∣∣≤√n
π
∫ ∞−∞
∣∣∣∣∣e−n(t−z)2 − e−n(t−z0)2
z − z0
− 2n(t− z)e−n(t−z)2∣∣∣∣∣ |η(τt(a))|dt
≤‖η‖‖a‖√n
π
∫ ∞−∞
∣∣∣∣∣e−n(t−z)2 − e−n(t−z0)2
z − z0
− 2n(t− z)e−n(t−z)2∣∣∣∣∣ dt,
since |η(τt(a))| ≤ ‖η‖‖τt(a)‖ ≤ ‖η‖‖a‖ for each t. This integral goes to zero when z → z0
and the entire analyticity follows. Also, fn(z) = τz(an) for every z ∈ C.
In addition, we have ‖an‖ ≤ ‖a‖. In fact,
‖an‖ =
∥∥∥∥√n
π
∫ ∞−∞
τt(a)e−nt2
dt
∥∥∥∥≤√n
π
∫ ∞−∞‖τt(a)‖e−nt2dt, from Lemma 2.6.11,
≤ ‖a‖√n
π
∫ ∞−∞
e−nt2
dt
= ‖a‖, from Lemma 5.1.18.
Now we prove that an converges to a on the σ(X,X*) topology. Let η ∈ X*. If η = 0
or a = 0, it follows that η(an) = η(a) = 0 for every n. Then we assume η 6= 0 and a 6= 0
without loss of generality.
Note that
η(an − a) = η(an)− η(a)
=
√n
π
∫ ∞−∞
η(τt(a))e−nt2
dt− η(a), by de�nition of an,
107
=
√n
π
∫ ∞−∞
(η(τt(a))− η(a))e−nt2
dt,
since√
nπ
∫∞−∞ e
−nt2dt = 1 by Lemma 5.1.18.
Let ε > 0. Since the map t 7→ τt(a) is continuous and η ∈ X*, there exists δ > 0 such
that
|t| < δ implies |η(τt(a))− η(a)| < ε/2. (5.6)
Moreover, from Lemma 5.1.18, there exists N > 0 such that, for every n ≥ N , we have√n
π
∫|t|≥δ
e−nt2
dt <ε
4‖η‖‖a‖. (5.7)
Then, for each n ≥ N ,
|η(an − a)| =∣∣∣∣√n
π
∫ ∞−∞
e−nt2
(η(τt(a))− η(a))dt
∣∣∣∣≤∣∣∣∣√n
π
∫|t|≥δ
e−nt2
(η(τt(a))− η(a))dt
∣∣∣∣+
∣∣∣∣√n
π
∫|t|<δ
e−nt2
(η(τt(a))− η(a))dt
∣∣∣∣≤√n
π
∫|t|≥δ
e−nt2|η(τt(a))− η(a))|dt+
√n
π
∫|t|<δ
e−nt2|η(τt(a))− η(a))|dt
≤√n
π
∫|t|≥δ
e−nt2|η(τt(a))− η(a))|dt+
ε
2
√n
π
∫|t|<δ
e−nt2
dt, from (5.6),
≤√n
π
∫|t|≥δ
e−nt2‖η‖(‖τt(a)‖+ ‖a‖)dt+
ε
2
√n
π
∫|t|<δ
e−nt2
dt
≤√n
π
∫|t|≥δ
e−nt2‖η‖(‖a‖+ ‖a‖)dt+
ε
2
√n
π
∫|t|<δ
e−nt2
dt, since ‖τt‖ = 1,
=2‖η‖‖a‖√n
π
∫|t|≥δ
e−nt2
dt+ε
2
√n
π
∫|t|<δ
e−nt2
dt
≤2‖η‖‖a‖ ε
4‖η‖‖a‖+ε
2
√n
π
∫|t|<δ
e−nt2
dt, from (5.7),
≤2‖η‖‖a‖ ε
4‖η‖‖a‖+ε
2
√n
π
∫ ∞−∞
e−nt2
dt,
108
=ε
2+ε
2, from Lemma 5.1.18,
=ε.
Hence an → a in the topology σ(X,X*).
Remark 5.1.20. Let τ be a one-parameter σ(X,X*)-continuous group of isometries. Given
λ, we say that a is analytic on the strip Iλ if the conditions in De�nition 5.1.16 hold for a
and λ.
We say that a ∈ X is strongly analytic on the strip Iλ if there exists f : Iλ → X such that
condition (i) in De�nition 5.1.16 is satis�ed and the limit
limh→0
f(z + h)− f(z)
h
exists for every z ∈ Iλ. We now show that these two notions are equivalent.
Lemma 5.1.21. Let τ be a one-parameter σ(X,X*)-continuous group of isometries, and
a ∈ Xτ . Given w ∈ C, τw(a) ∈ Xτ . Moreover, for every z ∈ C we have
τz+w(a) = τz ◦ τw(a).
Proof. Let a ∈ Xτ . Then there exists a function f : C→ X such that f(t) = τt(a) for every
t ∈ R, and η ◦ f is entire analytic for η ∈ X*.
Fix s ∈ R. Let fs : C → X be the function de�ned by fs(w) = f(s + w). Then η ◦ fs is
entire analytic for all η ∈ X*.
De�ne g : C→ X by g(w) = τs ◦τw(a). Note that η ◦g is entire analytic for every η ∈ X*.
In fact, given w ∈ C,
η ◦ g(w) = η ◦ τs ◦ τw(a) = (η ◦ τs) ◦ f(z).
Since η ◦ τs ∈ X* and a is entire analytic, it follows that (η ◦ τs) ◦ f is entire analytic. Thus
109
η ◦ g is entire analytic. Moreover, for t ∈ R, we have
η ◦ g(t) = η ◦ τs ◦ τt(a) = η ◦ τs+t(a) = η ◦ f(t+ s) = η ◦ fs(t).
Thus η ◦ g and η ◦ fs are entire analytic functions which agree on the real line. Thus, by
the uniqueness theorem for analytic functions [1, Theorem 6.9] we have η ◦ g = η ◦ fs. η is
arbitrary and X* separates poins in X, thus g = fs. Therefore, for every w ∈ C,
τs ◦ τw(a) = τs+w(a). (5.8)
Now we �x w ∈ C and assume s is a real variable. De�ne fw : C→ X by fw(z) = f(w+z).
Since a is entire analytic, the function η ◦fw is entire analytic for every η ∈ X*. By equation
(5.8), we have for all t ∈ R the equality
fw(t) = f(w + t) = τt+w(a) = τt ◦ τw(a).
Therefore τw(a) is entire analytic and
τz ◦ τw(a) = fw(z) = f(w + z) = τz+w(a).
Let τ de�ne a dynamics on a C*-algebra. We say that a subset A1 is τ -invariant if
τt(a) ∈ A1 for every t ∈ R, a ∈ A1.
Lemma 5.1.22. Let (A, τ) be a C*-dynamical system. Then Aτ is a ∗-subalgebra which is
τ -invariant.
Proof. Aτ is a vector space by Remark 5.1.17. Moreover, Aτ is τ -invariant by Lemma 5.1.21.
Given a, b ∈ Aτ , let f1, f2 : C → A be the functions such that η ◦ f1, η ◦ f2 are analytic
for all η ∈ A*, and such that f1(t) = τt(a), f2(t) = τt(b) for every t ∈ R.
110
Let f = f1f2. Then η ◦ f is analytic for all η ∈ A*. Moreover, given t ∈ R,
f(t) = f1(t)f2(t) = τt(a)τt(b) = τt(ab).
Therefore ab is entire analytic.
De�ne f1* by f1*(z) = f1(z)*. Then, for t ∈ R,
f1*(t) = f1(t)* = τt(a)* = τt(a*).
Given η ∈ A*, let η1 ∈ A* be de�ned by η1(b) = η(b*) for b ∈ A. Then,
η ◦ f1*(z) = η(f1(z)*) = η1 ◦ f(z).
Note that the function z 7→ η1 ◦ f1(z) is analytic. In fact,
limh→0
η1 ◦ f1(z + h)− η1 ◦ f1(z)
h=
(limh→0
η1 ◦ f1(z + h)− η1 ◦ f1(z)
h
).
This limit exists since η1 ◦ f1 is analytic. Then η ◦ f1* is analytic, and therefore a* is entire
analytic.
Proposition 5.1.23. Let τ be a one-parameter σ(X,X*)-continuous group of isometries.
Given λ > 0, a ∈ X is analytic on the strip Iλ if, and only if, a is strongly analytic on Iλ.
Proof. Let η ∈ X*. Let z0 ∈ Iλ. There exists r > 0 such that D(z0, r/2) ⊂ D(z0, r) ⊂ Iλ.
Let h, k ∈ D(0, r/2). Then z0 + h, z0 + k ∈ D(z0, r/2). Using Cauchy's integral formula
and assuming C = ∂D(z0, r), we have
η(f(z0)) =1
2πi
∮C
η(f(z))
z − z0
dz,
η(f(z0 + h)) =1
2πi
∮C
η(f(z))
z − z0 − hdz,
η(f(z0 + k)) =1
2πi
∮C
η(f(z))
z − z0 − kdz.
111
Then
η
(f(z0 + h)− f(z0)
h− f(z0 + k)− f(z0)
k
)=
1
2πi
∮C
η(f(z))
[1
h
(1
z − z0 − h− 1
z − z0
)− 1
k
(1
z − z0 − k− 1
z − z0
)]dz (5.9)
Note that 1a−b −
1a
= ba(a−b) for a, b complex numbers such that a 6= b and a 6= 0. Applying
this, we have that (5.9) equals to
1
2πi
∮C
η(f(z))
[1
h
h
(z − z0)(z − z0 − h)− 1
k
k
(z − z0)(z − z0 − k)
]dz
=1
2πi
∮C
η(f(z))
[1
(z − z0)(z − z0 − h)− 1
(z − z0)(z − z0 − k)
]dz
=1
2πi
∮C
η(f(z))
z − z0
[1
z − z0 − h− 1
z − z0 − k
]dz
=1
2πi
∮C
η(f(z))
z − z0
h− k(z − z0 − h)(z − z0 − k)
dz
Note that for every z such that |z − z0| = r, we have
|z − z0 − h| ≥ |z − z0| − |h| ≥ r − r
2=r
2
Thus
1
|z − z0 − h|≤ 2
h. (5.10)
Analogously, 1|z−z0−k| ≤
2k. Recall that C denotes the set of points z satisfying |z − z0| = r.
Therefore,
∣∣∣∣η(f(z0 + h)− f(z0)
h− f(z0 + k)− f(z0)
k
)∣∣∣∣ =
∣∣∣∣ 1
2πi
∮C
η(f(z))
z − z0
h− k(z − z0 − h)(z − z0 − k)
dz
∣∣∣∣≤ rmax
z∈C
|η(f(z))||z − z0|
|h− k||z − z0 − h||z − z0 − k|
112
≤ r4
r2maxz∈C
|η(f(z))||z − z0|
|h− k|, by (5.10),
=4
r2maxz∈C|η(f(z))||h− k|
≤ 4
r2‖η‖max
z∈C‖f(z)‖|h− k|.
Using the fact that ‖x‖ = sup‖η‖=1 ‖η(x)‖ in a Banach space, we obtain
∣∣∣∣f(z0 + h)− f(z0)
h− f(z0 + k)− f(z0)
k
∣∣∣∣ ≤ 4
r2maxz∈C|f(z)||h− k|.
Since X is complete, it follows that the limit
limh→0
f(z0 + h)− f(z0)
h
exists. In other words, a is strongly analytic on Iλ.
Conversely, suppose that a is strongly analytic on the strip Iλ. Let z ∈ Iλ. By hypothesis,
there exists x ∈ X such that
x = limh→0
f(z + h)− f(z)
h.
Let η ∈ X*. Then
η(x) = η
(limh→0
f(z + h)− f(z)
h
)= lim
h→0
η(f(z + h))− η(f(z))
h.
Hence η ◦ f is analytic at z. Since z and η are arbitrary, it follows that a is analytic on
Iλ.
Corollary 5.1.24. If τ = {τt}t∈R is a one-parameter σ(X,X*)-continuous group of isome-
tries, then τ is strongly continuous and Xτ is norm-dense in X.
Proof. Note that by strongly continuous, we mean that the map t 7→ τt(a) is continuous with
respect to the norm in X for every a ∈ X.
113
Suppose Xτ is not dense in X. Let H be the norm closure of Xτ . By hypothesis, H is a
proper subspace of X. Let y ∈ X \H. By the Hahn-Banach for the sets H and {y}, there
exists a ϕ ∈ X* such that
Re(ϕ(y)) < Re(ϕ(y)), h ∈ H. (5.11)
Since H is a proper subspace of X and ϕ is linear, Re(ϕ(H)) must be either {0} or R. By
equation (5.11), Re(ϕ(H)) = {0}.
Note that Im(ϕ(H)) = 0. In fact, let h ∈ H. Then there is λ ∈ R such that ϕ(h) = λi.
Since H is a complex vector space and ϕ is linear, ih ∈ H and ϕ(ih) = λ. It follows that
λ = 0. The choice of h is arbitrary, thus Im(ϕ(H)) = {0}. Hence ϕ is zero on H and
ϕ(y) 6= 0.
From Proposition 5.1.19, Xτ is dense in X with respect to the σ(X,X*)-topology. Since
H contains Xτ , H is also dense in X in this topology. Thus there exists a sequence {yn} in
H converging to y in the σ(X,X*)-topology. Since ϕ ∈ X*, we have ϕ(yn) → ϕ(y). This
leads to a contradiction because ϕ(y) 6= 0 and ϕ(yn) = 0 for every n. Therefore H = X,
that is, Xτ is norm dense in X.
Now we show that τ is strongly continuous. Given a ∈ Xτ , z ∈ C, it follows from
Proposition 5.1.23 that the limit
limh→0
∥∥∥∥τz+h(a)− τz(a)
h
∥∥∥∥exists. Thus ‖ τz+h(a)−τz(a)
h‖ converges to zero as h → 0, i.e., the function z 7→ τz(a) is
continuous with respect to the norm.
Now choose a ∈ X arbitrary, �x t ∈ R and let ε > 0. Since Xτ is norm dense in X, there
exists aε ∈ Xτ such that ‖a− aε‖ < ε/3.
Since aε ∈ Xτ , there exists δ > 0 such that for every h ∈ R with |h| < δ, we have
114
‖τt+h(aε)− τt(aε)‖ < ε/3. Then
‖τt+h(a)− τt(a)‖ ≤ ‖τt+h(a)− τt+h(aε)‖+ ‖τt+h(aε)− τt(aε)‖+ ‖τt(aε)− τt(a)‖
≤ ‖τt+h‖‖a− aε‖+ ‖τt+h(aε)− τt(aε)‖+ ‖τt‖‖aε − a‖
= ‖a− aε‖+ ‖τt+h(aε)− τt(aε)‖+ ‖aε − a‖
<ε
3+ ‖τt+h(aε)− τt(aε)‖+
ε
3
=2
3ε+ ‖τt+h(aε)− τt(aε)‖
≤ 2
3ε+
ε
3
= ε.
Then the map t 7→ τt(a) is continuous. Since a is arbitrary, τ is strongly continuous.
De�nition 5.1.25. Let (A, τ) be C*-dynamical system, ϕ a state on A and β ∈ R. We say
ϕ is a τ -KMSβ-state if
ϕ(aτiβ(b)) = ϕ(ba)
for all a, b in a ∗-subalgebra A0 composed of entire analytic elements such that A0 is norm-
dense and τ -invariant.
When τ is implicit, we just say that ϕ is a KMSβ-state in order to assert that ϕ is a
τ -KMSβ-state. Moreover, if β is �xed, we just say ϕ is a KMS state.
Remark 5.1.26. If ϕ is a KMS-state and β = 0, we have
ϕ(ab) = ϕ(ba),
for every a, b ∈ A0. From the continuity of ϕ, this equality holds for every a, b ∈ A. In this
case, we say the state ϕ is tracial .
De�nition 5.1.27. Let (A, τ) be a C*-dynamical system. We say that a ∈ A is τ -invariant
115
if τt(a) = a for every a ∈ A.
De�nition 5.1.28. Let A be a C*-algebra with dynamics τ . We say that a state ϕ on A is
τ -invariant if ϕ(τt(a)) = ϕ(a) for every a ∈ A and t ∈ R.
In this section we de�ned KMS-states as described in [3]. However, Neshveyev's theorem
assumes a diferent de�nition of KMS-states in [17]. The de�nition used is the item (ii) of
the next theorem. Now we show an equivalence in the de�nition of KMS states when β 6= 0.
Proposition 5.1.29. Let A be a C*-algebra with dynamics given by τ and let β ∈ R \ {0}.
Given a state ϕ on A, the following are equivalent:
(i) ϕ is a KMSβ-state;
(ii) ϕ is τ -invariant and
ϕ(aτiβ(b)) = ϕ(ba) (5.12)
for a dense set of analytic elements a, b ∈ A.
(iii) Equation (5.12) holds for every a, b ∈ Aτ .
Proof. Assume β > 0 without loss of generality.
(i)⇒ (ii) Let ϕ be a KMSβ-state. By de�nition, there exists a subalgebra A0 composed of
entire analytic elements such that A0 is norm-dense, τ -invariant and such that (5.12) holds
for every a, b ∈ A0. Thus, we only need to show that ϕ is τ -invariant.
Let a ∈ A0. De�ne f : C→ C by f(z) = ϕ(τz(a)). Since a is entire analytic and ϕ ∈ A*,
it follows that f is entire analytic.
Let {uλ}λ∈Λ be an approximate unit for A. Fix λ ∈ Λ. Since A0 is norm dense in A, for
every ε > 0 there exists u(ε)λ ∈ A0 satisfying ‖u(ε)
λ − uλ‖ < ε. Both u(ε)λ , a ∈ A0 and ϕ is a
KMSβ-state, then
ϕ(u(ε)λ τiβ(a)) = ϕ(au
(ε)λ ).
116
Using the continuity of ϕ, we have
ϕ(uλτiβ(a)) = ϕ(auλ),
for every λ ∈ Λ. Again, using the continuity of ϕ and the fact that {uλ}λ∈Λ is an approximate
unit, we have
ϕ(τiβ(a)) = limλϕ(uλτiβ(a)) = lim
λϕ(auλ) = ϕ(a).
De�ne g : C→ C by g(z) = ϕ(τz+iβ(a))−ϕ(τz(a)). Since a is entire analytic and ϕ ∈ A*,
it follows that g is entire analytic. Note that g is zero on the real line. In fact, given t ∈ R,
we have by Lemma 5.1.21
g(t) = ϕ(τt+iβ(a))− ϕ(τt(a)) = ϕ(τiβ(τt(a)))− ϕ(τt(a)) = 0.
Hence g(z) = 0 for every z ∈ C, that is,
ϕ(τz+iβ(a)) = ϕ(τz(a)). (5.13)
Let f : C → C be the analytic function de�ned by f(z) = ϕ(τz(a)). From (5.13) it follows
that f is periodic, that is, for every z ∈ C there exists s ∈ R and 0 < t ≤ β such that
f(z) = f(s+ it). Then, for every z ∈ C, we have
|f(z)| ≤ supw∈C|f(w)|
≤ sups∈R
0<t≤β
|f(s+ it)|
= sups∈R
0<t≤β
|ϕ(τs+it(a))|
≤ sups∈R
0<t≤β
‖τs+it(a)‖
117
= sups∈R
0<t≤β
‖τs(τit(a))‖, by Lemma 5.1.21,
≤ sup0<t≤β
‖τit(a)‖, by de�nition of τs for s ∈ R,
<∞,
since f is continuous. Then f is entire analytic and bounded. Therefore f is constant. Thus,
for every z ∈ C, we have
ϕ(a) = f(0) = f(z) = ϕ(τz(a)).
In particular, for every a ∈ A0 and t ∈ R, we have
ϕ(τt(a)) = ϕ(a). (5.14)
Since A0 is dense in A, it follows that (5.14) holds for every t ∈ R, a ∈ A. Therefore ϕ is
τ -invariant.
(ii) ⇒ (iii) Let ϕ be a τ -invariant state and suppose that there exists a dense set A1 of
analytic elements such that (5.12) holds for every a, b ∈ A1. Let
Dβ = {z ∈ C : 0 ≤ Im(z) ≤ β}.
Let a, b ∈ Aτ . De�ne η ∈ A* by η(x) = ϕ(ax). Then we can de�ne the entire analytic
function f : C→ C by f(z) = ϕ(aτz(b)) because f(z) = η(τz(a)).
Since A1 is dense in A, there are sequences {an}n∈N, {bn}n∈N in A1 such that an → a and
bn → b. Let a′ = supn ‖an‖, b′ = supn ‖bn‖. Both a′, b′ are �nite.
For every n, there exists λn > β such that bn is analytic on the strip Iλn . So we can de�ne
the analytic function fn by fn(z) = ϕ(anτz(bn)) for z ∈ Iλn . Let fn = fn|Dβ . We will show
that {fn}n∈N is a Cauchy sequence with respect to the supremum norm.
118
In fact, given n,m, de�ne λn,m = min{λn, λm}. Then fn|Iλn,m − fm|Iλn,m is analytic on
Iλn,m . Then the supremum of |fn − fm| is attained on the boundary of Dβ. Hence, for each
z ∈ Dβ, we have
|fn(z)− fm(z)| ≤ max
{supt∈R|fn(t)− fm(t)|, sup
t∈R|fn(t+ iβ)− fm(t+ iβ)|
}.
However, for every t ∈ R,
|fn(t)− fm(t)| = |ϕ(anτt(bn))− ϕ(amτt(bm))|
≤ |ϕ(anτt(bn))− ϕ(anτt(bm))|+ |ϕ(anτt(bm))− ϕ(amτt(bm))|
≤ ‖anτt(bn)− anτt(bm)‖+ ‖anτt(bm)− amτt(bm)‖
≤ ‖an‖‖τt(bn − bm)‖+ ‖an − am‖‖τt(bm)‖
≤ ‖an‖‖bn − bm‖+ ‖an − am‖‖bm‖
≤ a′‖bn − bm‖+ ‖an − am‖b′.
On the other hand,
|fn(t+ iβ)− fm(t+ iβ)| = |ϕ(anτt+iβ(bn))− ϕ(amτt+iβ(bm))|
= |ϕ(τt(bn)an)− ϕ(τt(bm)am)|
= |ϕ(τt(bn)an)− ϕ(τt(bn)am)|+ |ϕ(τt(bn)am)− ϕ(τt(bm)am)|
≤ ‖τt(bn)an − τt(bn)am‖+ ‖τt(bn)am − τt(bm)am‖
≤ ‖τt(bn)‖‖an − am‖+ ‖τt(bn)− τt(bm)‖‖am‖
≤ ‖bn‖‖an − am‖+ ‖bn − bm‖‖am‖
≤ b′‖an − am‖+ ‖bn − bm‖a′.
Therefore ‖fn− fm‖ ≤ b′‖an− am‖+ ‖bn− bm‖a′. Hence {fn}n∈N is a Cauchy sequence and
then this sequence converges to a continuous function f : Dβ → C which is analytic on the
119
interior of Dβ. Note that for every t ∈ R,
f(t) = ϕ(aτt(b)) = limn→∞
ϕ(anτt(bn)) = f(t),
by the continuity of ϕ and τt. Therefore f |Dβ = f by the uniqueness theorem for analytic
functions. Moreover,
ϕ(aτiβ(b)) = f(iβ)
= limn→∞
fn(iβ)
= limn→∞
ϕ(anτiβ(bn))
= limn→∞
ϕ(bnan)
= ϕ(ba),
for every a, b ∈ Aτ . Therefore ϕ is a KMSβ-state.
(iii) ⇒ (i) From Lemma 5.1.22, Aτ is a ∗-subalgebra and is τ -invariant. By Corollary
5.1.24, Aτ is dense in A. Hence ϕ is a KMSβ-state.
Remark 5.1.30. Let ϕ be a τ -invariant state on a C*-algebra A. In order to prove that ϕ is
a KMSβ-state, it su�ces to show that ϕ(aτiβ(b)) = ϕ(ba) in an arbitrary dense ∗-subalgebra
of A. In fact, if β 6= 0, this follows directly from Proposition 5.1.29. If β = 0, this follows
from the continuity of ϕ.
De�nition 5.1.31. Given a C*-algebra A and a state ϕ on A, we de�ne the centralizer of
ϕ as the set
{a ∈ A : ϕ(ab) = ϕ(ba) for every b ∈ A}.
Lemma 5.1.32. Let (A, τ) be a C*-dynamical system, β > 0, and ϕ a τ -KMSβ-state. Then,
for every a ∈ A such that a is τ -invariant, it follows that a is in the centralizer of ϕ.
120
Proof. Let a ∈ A be τ -invariant. Then τt(a) = a for every t. Then we can de�ne the function
f : C → A by f(z) = a. Then we have f(t) = τt(a) for every t ∈ R. The function f is
constant, then f is strongly analytic on C. Then, by Proposition 5.1.23, a is entire analytic.
Let b ∈ Aτ . Then,
ϕ(ba) = ϕ(bτiβ(a)) = ϕ(ab),
because ϕ is a τ -KMSβ-state.
Now we show that the set of KMS states, for a �xed β > 0 is convex. Moreover, in order
to describe all KMS states on a C*-algebra, it is su�cient to �nd only its extremal KMS
states.
Lemma 5.1.33. Let A be a C*-algebra with dynamics τ . Let β ∈ R. The set of KMSβ-states
is convex.
Proof. Let A be a C*-algebra with dynamics τ . Fix β ∈ R. Given ϕ1, ϕ2 KMS states,
0 < t < 1, de�ne ϕ = tϕ1 + (1− t)ϕ2. Now we show that ϕ is a KMS state.
(i) ϕ is a state.
Note that ϕ is positive. In fact, given a positive element a ∈ A,
ϕ(a) = tϕ1(a) + (1− t)ϕ2(a) ≥ 0,
since both ϕ1(a), ϕ2(a) ≥ 0. Moreover, we prove that ϕ has norm 1. Applying Lemma
5.1.2, we have for every approximate unit {uλ}λ∈Λ of A,
‖ϕ‖ = limλϕ(uλ)
= limλ
(tϕ1(uλ) + (1− t)ϕ2(uλ))
= t limλ
(ϕ1(uλ)) + (1− t) limλ
(ϕ2(uλ))
121
= t‖ϕ1‖+ (1− t)‖ϕ2‖
= t+ (1− t)
= 1.
(ii) ϕ is KMS
Given a, b ∈ Aτ ,
ϕ(aτiβ(b)) = tϕ1(aτiβ(b)) + (1− t)ϕ2(aτiβ(b)) = tϕ1(ba) + (1− t)ϕ2(ba) = ϕ(ba).
In order to prove that the extremal KMS-states are su�cient to describe all KMS-states,
we are going to use a the Krein-Milman theorem. But, before using this theorem, we show
that the set of KMSβ-states is compact with respect to a topology we de�ne below, the
weak* topology. Results used in this part can be found on [16] and [22].
Let X be a normed vector space. The weak*-topology on X* is generated by the family
of seminorms px : X*→ R such that η 7→ px(η) = |η(x)|.
Theorem 5.1.34. (The Banach-Alaoglu theorem) If V is a neighborhood of 0 in a normed
vector space X and if
K = {η ∈ X* : |η(x)| ≤ 1 for every x ∈ V },
then K is compact in the weak*-topology.
In this case, we can also say that K is weak*-compact .
Lemma 5.1.35. Let A be a C*-algebra with dynamics τ . Let β ∈ R. The set of KMSβ-states
is compact in the weak*-topology.
122
Proof. Let A be the unitization of A. Let
K = {η ∈ A* : |η(a)| ≤ 1 for a ∈ A with ‖a‖ ≤ 1}
= {η ∈ A* : ‖η‖ ≤ 1}.
It follows from the Banach-Alaoglu theorem that K is weak*-compact.
Let {ϕλ}λ∈Λ be a net of KMSβ-states on A. De�ne for each λ, ϕλ the extension of ϕλ on
A. Since each ϕλ ∈ K and K is compact, there exists a subnet {ϕλj}j∈J converging to some
ϕ ∈ K in the weak*-topology. Hence, for every a ∈ A,
ϕ(a) = limj∈J
ϕλj(a).
Also, since each ϕλj is a state
ϕ(1) = limj∈J
ϕλj(1) = 1.
It follows from Theorem 5.1.2 that ϕ is a state. Hence ϕ is a state. We will prove that ϕ is
a KMSβ-state.
Let a, b ∈ Aτ . Then
ϕ(aτiβ(b)) = limj∈J
ϕj(aτiβ(b)) = limj∈J
ϕj(ba) = ϕ(ba).
Therefore the set of KMSβ-states is weak*-compact.
We will use the fact that the set of KMS states is compact and convex to prove that the
extremal KMS states are su�cient to describe all KMS states. Before proving this, we de�ne
some concepts about convex sets and state then state the Krein-Milman theorem. Here all
results are de�ned for subsets of X*. For more general results, see [16].
De�nition 5.1.36. Let X be a normed vector space. Given a subset S of X*, co(S) denotes
the closure of co(S) with respect to the weak*-topology.
123
Now we state the Krein-Milman theorem for X*.
Theorem 5.1.37. Let X be a normed vector space. Let C ⊂ X* be a non-empty convex
weak*-compact subset. Then the set E of extreme points of C is non-empty and
C = co(E).
Moreover, if S is a closed subset of C with respect to the weak*-topology such that co(S) = C,
then S contains E.
Now we apply the Krein-Milman theorem to show that the extremal KMS states are
su�cient to describe all KMS states on A.
Corollary 5.1.38. Let A be a C*-algebra and τ its dynamics. Fix β > 0. Let C denote the
set of KMSβ-states on A and assume C 6= ∅. Let E be the set of extremal KMSβ-states.
Then C = co(E) and E 6= ∅.
Proof. It follows from Lemmas 5.1.33 and 5.1.35 that C is convex and compact. By hypoth-
esis, C 6= ∅. Then we can apply the Krein-Milman theorem and the result follows.
5.2 First Theorem
Let G be a locally compact Hausdor� second countable étale groupoid. Neshveyev's �rst
theorem shows that for every state ϕ on C*(G) with centralizer containing C0(G(0)) there
is a corresponding pair (µ, {ϕx}x), where µ is a probability measure on G(0), and {ϕx}x is a
µ-measurable �eld of states ϕx on C*(Gxx) for x ∈ G(0). The results in this section can be
found in [17].
In Section 5.3, we will de�ne a dynamics τ on C*(G) such that every function in C0(G(0))
is τ -invariant. Thus, by Lemma 5.1.32, the centralizer of every KMS state contains C0(G(0)).
Therefore, we can apply Neshveyev's �rst theorem to KMS states.
Now we describe µ-measurable �eld of states as described in [5].
124
De�nition 5.2.1. Let G be a locally compact Hausdor� second countable étale groupoid,
and let µ be a Radon Borel measure on G(0). For each x ∈ G(0), g ∈ Gxx, we let ug denote the
canonical unitary generators of C*(Gxx), i.e., ug ∈ Cc(Gx
x) is a function de�ned by ug(g) = 1
and ug(h) = 0 if h 6= g.
We call a collection {ϕx}x∈G(0) a µ-measurable �eld of states if each ϕx is a state on
C*(Gxx) and the function:
G(0) 3 x 7→∑g∈Gxx
f(g)ϕx(ug)
is µ-measurable for each f ∈ Cc(G).
Remark 5.2.2. Fix a probability measure µ on G(0). Given a conull set V ⊂ G(0), let
{ϕx}x∈V be a family of states ϕx on C*(Gxx) such that for every f ∈ Cc(G) the function
x 7→ χV (x)∑g∈Gxx
f(g)ϕx(ug) (5.15)
is µ-measurable. Then there exists a µ-measurable �eld of states {ϕx}x∈G(0) such that ϕx =
ϕx for every x ∈ V . In fact, any C*-algebra C*(Gxx) has at least on state, then you can just
choose any family of states ϕx on C*(Gxx) such that ϕx = ϕx for x ∈ V . Since the function
de�ned by (5.15) is µ-measurable, then {ϕx} is a µ-measurable �eld of states.
Later we will prove in Neshveyev's �rst theorem that if {ϕx}x∈G(0) , {ϕx}x∈G(0) are two
µ-measurable �eld of states whose states are equal on a conull subset of G(0), then (µ, {ϕx}x)
and (µ, {ϕx}x) de�ne the same state.
Remark 5.2.3. When there is no risk of confusion, we denote the µ-measurable �eld of
states {ϕx}x∈G(0) by {ϕx}x.
Lemma 5.2.4. Let G be a locally compact Hausdor� second countable étale groupoid and
let x ∈ G(0). Given g1, g2, g ∈ Gxx, ug1 · ug2 = ug1g2 and ug* = ug−1 .
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Proof. Let h ∈ Gxx. Then
ug1 · ug2(h) =∑h1h2
ug1(h1)ug2(h2).
Note that for every h1, h2 such that h1 6= g1 or h2 6= g2, we have ug1 · ug2(h) = 0. Thus,
ug1 · ug2(h) =
1 if h = g1g2,
0 otherwise= ug1g2(h).
We also have
ug*(h) = ug(h−1)
= ug(h−1), since ug is real-valued,
=
1 if h−1 = g,
0 otherwise,
=
1 if h = g−1,
0 otherwise,
= ug−1(h).
Therefore the result follows.
Remark 5.2.5. If µ is purely atomic, it follows from Lemma 2.5.2 that that every family
{ϕx}x∈G(0) of states ϕx on C*(Gxx) is a µ-measurable �eld of states.
Given a state ϕ on C*(G), we obtain from Renault's disintegration theorem a unitary
representation of G. The following lemma uses this representation to �nd a representation
πx on C*(Gxx) for each x ∈ G(0). In the proof of Neshveyev's �rst theorem, we will use πx to
de�ne a state ϕx on C*(Gxx).
126
Lemma 5.2.6. Let G be a locally compact Hausdor� second countable étale groupoid. Let
(µ,G(0)∗H, L) be a unitary representation of G. Let x ∈ G(0). The linear map πx : C*(Gxx)→
B(Hx) de�ned by πx(ug) = Lg is a representation of C*(Gxx).
Proof. Let x ∈ G(0). Let g ∈ Gxx, then
πx(ug*) = πx(ug−1) = Lg−1 = Lg* = πx(ug)*.
Given g1, g2 ∈ Gxx,
πx(ug1ug2) = πx(ug1g2) = Lg1g2 = Lg1Lg2 = πx(ug1)πx(ug2).
Given a state ϕx on C*(Gxx), it corresponds to a representation πx : C*(Gx
x) → B(Hx).
We can use πx to de�ne a unitary map Lg : Hx → Hx by Lg = πx(ug), for every g ∈ Gxx.
Using this fact, we can apply Lemma 5.2.7 and obtain a Hilbert space Kx. We use this result
in the proof of Neshveyev's �rst theorem to show that there exists a state ψx corresponding
to Kx. Then we use Lemma 5.1.8 to �nd the state ϕ on C*(G).
Lemma 5.2.7. Let G be a locally compact Hausdor� second countable étale groupoid with
Haar system de�ned by the counting measures on Gx. Let {Lh}h∈G′ be a family of unitary
operators Lh : Hx → Hx on a Hilbert space Hx de�ned for all h ∈ Gxx and x ∈ G(0). Given
x ∈ G(0), let Kx be the family of functions v : Gx → Hx such that
v(gh) = Lh*v(g) for g ∈ Gx and h ∈ Gxx (5.16)
and
∑g∈Gx/Gxx
‖v(g)‖2 <∞. (5.17)
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Then Kx is a Hilbert space.
Proof. Note that Gx/Gxx is de�ned by the following equivalence relation on Gx:
g1 ∼ g2 if g1 = g2h for some h ∈ Gxx.
In order to prove that (5.17) is well-de�ned, we will show that ‖v(g1)‖ = ‖v(g2)‖ in this case.
In fact, by de�nition of v,
‖v(g1)‖ = ‖v(g2h)‖ = ‖Lh*v(g2)‖ = ‖v(g2)‖,
since Lh is unitary.
Kx is a vector space. In fact, let u, v ∈ Kx, λ ∈ C. Then, for g ∈ Gx, h ∈ Gxx,
(u+ λv)(gh) = u(gh) + λv(gh)
= Lh*u(g) + Lh*λv(g)
= Lh* [(u+ λv)(g)] .
Now we can de�ne the inner product on Kx by
〈u, v〉 =∑
g∈Gx/Gxx
〈u(g), v(g)〉. (5.18)
Note that (5.18) de�nes an inner product on Kx. In fact, given g1, g2 ∈ Gx, h ∈ Gxx such
that g1 = g2h, we have for u, v ∈ Kx,
〈u(g1), v(g1)〉 = 〈u(g2h), v(g2h)〉 = 〈Lh*u(g2), Lh*v(g2)〉 = 〈u(g2), v(g2)〉.
It is easy to show this operation satis�es
〈u1 + λu2, v〉 = 〈u1, v〉+ λ〈u2, v〉 , 〈u, u〉 ≥ 0 and 〈u, v〉 = 〈v, u〉,
128
for u, u1, u2, v ∈ Kx, λ ∈ C.
Suppose 〈v, v〉 = 0. Let g ∈ Gx, then 〈v(g), v(g)〉 = 0 by de�nition. Therefore v = 0. We
denote the norm de�ned in (5.17) by ‖ · ‖.
Now we prove that Kx is Banach. Let {vn} be a Cauchy sequence on Kx. Given g ∈ Gx,
{vn(g)} is a Cauchy sequence on Hx, hence vn(g)→ v(g), where v : Gx → Hx is a function.
We show that v satis�es (5.16). Let g ∈ Gx, h ∈ Gxx. Then vn(gh)→ v(gh). But for each
n, vn(gh) = Lh*vn(g). Since Lh is unitary, it follows that
v(gh) = limn→∞
vn(gh) = limn→∞
Lh*vn(g) = Lh*v(g).
Now we prove that vn converges to v with respect to the norm. Since G is second countable
and étale, it follows from Proposition 3.2.11 that Gx is countable. Then Gx/Gxx is countable.
If Gx/Gxx is �nite, then pointwise convergence implies convergence in the norm. So we assume
Gx/Gxx is in�nite and denote its elements by a sequence {gk}. Given ε > 0, let n0 ≥ 0 such
that for every n,m ≥ n0, we have ‖vn − vm‖ < ε.
Fix k1. If n ≥ n0, we have
k1∑k=1
‖vn(gk)− v(gk)‖2 = limm→∞
k1∑k=1
‖vn(gk)− vm(gk)‖2 ≤ ε2.
Since k1 is arbitrary, we have
‖vn − v‖2 = limk1→∞
k1∑k=1
‖vn(gk)− v(gk)‖2 ≤ ε2.
Hence ‖vn − v‖ → 0. Therefore vn converges to v in the norm. Moreover, it is easy to see
that ‖v‖ < ∞. In fact, choose n such that ‖vn − v‖ < 1. Hence ‖v‖ ≤ ‖vn − v‖ + ‖vn‖ <
1 + ‖vn‖ <∞.
Proposition 5.2.8. Let G be a locally compact Hausdor� second countable étale groupoid.
Let f ∈ Cc(G) and assume there exists an open set U including the support of f such that
129
U ∩G′ = ∅. Then we can write f = f1 + . . . + fn where for every i = 1, . . . , n, fi ∈ Cc(Ui)
and Ui is an open bisection satisfying r(Ui) ∩ s(Ui) = ∅.
Proof. Let f ∈ Cc(G) and U an open set U containing its support such that U ∩ G′ = ∅.
Since K is compact, there exists an open set V whose closure is compact and K ⊂ V ⊂ U .
Then, by Lemma 3.2.13, for every g ∈ V there exists an open bisection Ug containing g
such that r(Ug) ∩ s(Ug) = ∅. However, V is compact, then there are g1, . . . , gn such that
Ug1 , . . . ,Ugn cover V . Denote Ui = Ugi for i = 1, . . . , n.
Let hii = 1n be the partition of unity subordinate to the open cover {Ui}ni=1. For each
i = 1, . . . , n, let fi = hif . Then fi ∈ Cc(Ui) and f = f1 + . . .+ fn.
Theorem 5.2.9. (Neshveyev) Let G be a locally compact Hausdor� second countable étale
groupoid. There is a one-to-one correspondence between states on C*(G) with centralizer
containing C0(G(0)) and pairs (µ, {ϕx}x) consisting of a probability measure µ on G(0) and
a µ-measurable �eld of states ϕx on C*(Gxx). Namely, the state corresponding to (µ, {ϕx}x)
is given by
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x) for f ∈ Cc(G).
Proof. Endow G with the Haar system given by counting measures λx on Gx.
Assume ϕ is a state on C*(G) with centralizer containing C0(G(0)). Let (H, π, ξ) be the
corresponding GNS-triple. It follows from Lemma 4.3.5 that π satis�es the conditions of
Renault's disintegration theorem. Therefore there is a unitary representation (µ,G(0) ∗H, L)
of G such that H is isomorphic to L2(G(0) ∗H, µ) and π is equivalent to the integrated form
of (µ,G(0) ∗H, L). Here we identify H with L2(G(0) ∗H, µ) without loss of generality. Hence,
ϕ(f) = 〈π(f)ξ, ξ〉
=
∫G
f(g)〈Lgξs(g), ξr(g)〉Hr(g)∆(g)−12dµr(g)
130
=
∫G(0)
∑g∈Gx
f(g)〈Lgξs(g), ξx〉Hx∆(g)−12dµ(x). (5.19)
Now we prove that for every f ∈ Cc(G) and for µ-a.e. x ∈ G(0) we have
∑g∈Gx\Gxx
f(g)〈Lgξs(g), ξx〉Hx∆(g)−12 = 0. (5.20)
Using the lineariy of (5.20) with respect to f we can assume it is supported on an open
bisection U . Since the isotropy bundle G′ is closed from Lemma 3.2.13 and the sum in (5.20)
does not take into account elements in G′, we can assume U ∩G′ = ∅. Then, by Proposition
5.2.8, we can assume r(U) ∩ s(U) = ∅ without loss of generality.
Let h ∈ Cc(r(U)). Then f ·h = 0. In fact, given g ∈ G, (f ·h)(g) = f(g)h(s(g)). Suppose
f(g) 6= 0. Then g ∈ U . Thus s(g) ∈ s(U). Since s(U) ∩ r(U) = ∅ by hypothesis, it follows
that s(g) /∈ r(U). Thus h(s(g)) = 0. Therefore f · h = 0.
Since ϕ has centralizer containing C0(G(0)), we have ϕ(h · f) = ϕ(f · h) = 0. Applying
(5.19) for h · f , it follows that
0 = ϕ(h · f)
=
∫G(0)
∑g∈Gx
(h · f)(g)〈Lgξs(g), ξx〉∆(g)−12dµ(x)
=
∫G(0)
∑g∈Gx
h(r(g))f(g)〈Lgξs(g), ξx〉∆(g)−12dµ(x), by Lemma 3.3.4,
=
∫G(0)
h(x)∑g∈Gx
f(g)〈Lgξs(g), ξx〉∆(g)−12dµ(x)
=
∫r(U)
h(x)∑g∈Gx
f(g)〈Lgξs(g), ξx〉∆(g)−12dµ(x), since h ∈ Cc(r(U)),
=
∫r(U)
h(x)∑
g∈Gx\Gxx
f(g)〈Lgξs(g), ξx〉∆(g)−12dµ(x), since U ∩G′ = ∅.
Since h ∈ Cc(r(U)) is arbitrary, it follows that for µ-a.e x ∈ r(U), (5.20) holds. Since
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f ∈ Cc(U), (5.20) holds for every x /∈ r(U). Indeed, given x /∈ r(U), g ∈ Gx, then g /∈ U .
Hence f(g) = 0. Therefore (5.20) holds for all f ∈ Cc(U), for µ-a.e. x. Therefore, (5.20) is
valid for every f ∈ Cc(G).
Let f ∈ Cc(G). Applying (5.20) on (5.19), we have
ϕ(f) =
∫G(0)
∑g∈Gx
f(g)〈Lgξs(g), ξx〉Hx∆(g)−12dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)〈Lgξs(g), ξx〉Hx∆(g)−12dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)〈Lgξx, ξx〉Hx∆(g)−12dµ(x).
From Lemma 5.2.6, we can de�ne for every x ∈ G(0) a representation πx : C*(Gxx) →
B(Hx) such that πx(ug) = Lg for all g ∈ Gxx. Then
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hx∆(g)−12dµ(x).
By Proposition 4.1.8, ∆(g) = 1 for all g ∈ Gxx for µ-a.e. x ∈ G(0). Hence,
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hxdµ(x).
Since ξ ∈ L2(X ∗ H, µ), the set V ⊂ G(0) of elements x such that ξx 6= 0, is measurable.
Let µ be a measure on G(0) such that dµ/dµ(x) = ‖ξx‖2. Note that µ is supported on V .
Moreover, µ is a probability measure. In fact,
∫G(0)
dµ(x) =
∫G(0)
‖ξx‖2dµ(x) = ‖ξ‖2 = 1.
Let ξ be de�ned by ξx = 0 if x /∈ V , and ξx = ξx/‖ξx‖ if x ∈ V . Then ξ ∈ L2(X ∗ H, µ) and
132
‖ξ‖ = 1. Moreover,
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hxdµ(x)
=
∫V
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hxdµ(x)
=
∫V
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hx
‖ξx‖2‖ξx‖2dµ(x)
=
∫V
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hx‖ξx‖2dµ(x)
=
∫V
∑g∈Gxx
f(g)〈πx(ug)ξx, ξx〉Hxdµ(x)
For every x ∈ V , there is a state ϕx on C*(Gxx) such that
ϕx(ug) = 〈πx(ug)ξx, ξx〉Hx , for every g ∈ Gxx. (5.21)
From Remark 5.2.2, we can choose a µ-measurable �eld of states {ϕx} such that ϕx is de�ned
by (5.21) for every x ∈ V . Therefore,
ϕ(f) =
∫V
∑g∈Gxx
f(g)ϕx(ug)dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x).
Conversely, let µ be a probability measure on G(0) and let a µ-measurable �eld of states
ϕx on C*(Gxx).
Given x, let (Hx, πx, ξx) be the GNS-triple for ϕx. De�ne for every h ∈ Gxx, Lh = πx(uh).
Each Lh is unitary, since
Lh* = πx(uh)* = πx(uh*) = πx(uh−1) = πx(u−1h ) = πx(uh)
−1 = L−1h .
133
Let Kx be the Hilbert space as in Lemma 5.2.7. Let ϑx : Cc(G)→ B(Kx) be de�ned by
(ϑx(f)v)(g) =∑
h∈Gr(g)f(h)v(h−1g) for f ∈ Cc(G), g ∈ Gx.
First we show that each ϑx(f) is in B(Kx). In fact, ϑx(f) is linear on Kx by de�nition.
Let v ∈ Kx, g ∈ Gx and k ∈ Gxx. Then
(ϑx(f)v)(gk) =∑
h∈Gr(g)f(h)v(h−1gk)
=∑
h∈Gr(g)f(h)Lk*v(h−1g)
= Lk*
∑h∈Gr(g)
f(h)v(h−1g)
= Lk* [(ϑx(f)v)(g)] .
Also, we need to show that ‖ϑx(f)‖2 < ∞ for every f ∈ Cc(G). Since ϑx is linear, we can
assume f ∈ Cc(U), where U ⊂ G is an open bisection.
Let g ∈ Gx. Assume r(g) /∈ r(U). Then, for every h ∈ Gr(g), it follows that h /∈ U , which
implies f(h) = 0. Hence (ϑx(f)v)(g) = 0. Now assume r(g) ∈ r(U). Since U is an open
bisection, there exists a unique hr(g) in U such that hr(g) ∈ Gr(g). Then
(ϑx(f)v)(g) =∑
h∈Gr(g)f(h)v(h−1g) = f(hr(g))v((hr(g))−1g).
Note that if g1, g2 ∈ Gx are equivalent, then r(g1) = r(g2). Indeed, there is k ∈ Gxx such
that g1 = g2k. Hence r(g1) = r(g2k) = r(g2). Moreover, if r(g1) ∈ r(U), it follows that
‖v((hr(g1))−1g1)‖ = ‖v((hr(g1))−1g2k)‖
= ‖v((hr(g2))−1g2k)‖, since r(g1) = r(g2),
= ‖Lk*v((hr(g2))−1g2)‖
134
= ‖v((hr(g2))−1g2)‖.
Therefore ‖v(hr(g)g)‖ is well-de�ned for g ∈ Gx/Gxx with r(g) ∈ r(U). Then,
‖ϑx(f)v‖2 =∑
g∈Gx/Gxxr(g)∈r(U)
‖(ϑx(f)v)(g)‖2
=∑
g∈Gx/Gxxr(g)∈r(U)
|f(hr(g))|2‖v((hr(g))−1g)‖2
≤ ‖f‖2∞
∑g∈Gx/Gxxr(g)∈r(U)
‖v((hr(g))−1g)‖2. (5.22)
Note that for g1, g2 with r(g1) ∈ U , g1, g2 are equivalent if, and only if (hr(g1))−1g1 and
(hr(g2))−1g2 are equivalent. In fact, suppose g1 and g2 are equivalent. Then g1 = g2k for
some k ∈ Gxx. Recall that r(g1) = r(g2). Then hr(g1) = hr(g2) and (hr(g1))−1g1 = (hr(g2))−1g2k.
Therefore (hr(g1))g1 and (hr(g2))g2 are equivalent.
Conversely, assume (hr(g1))−1g1 and (hr(g2))−1g2 are equivalent. Then there is k ∈ Gxx such
that (hr(g1))−1g1 = (hr(g2))−1g2k. Note that
s(hr(g1)) = r((hr(g1))−1) = r((hr(g1))−1g1) = r((hr(g2))−1g2) = r((hr(g2))−1) = s(hr(g2)).
Since hr(g1), hr(g2) ∈ U and U is an open bisection, it follows that hr(g1) = hr(g2). Then
g1 = g2k. Therefore g1 and g2 are equivalent.
Then the set of elements (hr(g))−1g for g ∈ Gx/Gxx with r(g) ∈ r(U) is a subset of the set
{g : g ∈ Gx/Gxx}. Hence, by (5.22),
‖ϑx(f)v‖2 ≤ ‖f‖2∞
∑g∈Gx/Gxxr(g)∈r(U)
‖v((hr(g))−1g)‖2
≤ ‖f‖2∞
∑g∈Gx/Gxx
‖v(g)‖2
135
= ‖f‖2∞‖v‖2
<∞.
Therefore, ‖ϑx(f)‖ <∞ for every f ∈ Cc(G).
Now we prove that ϑx is a representation of Cc(G) on Kx.
• ϑx(f1 · f2) = ϑx(f1)ϑx(f2).
Given f1, f2 ∈ Cc(G), v ∈ Kx, g ∈ Gx,
(ϑx(f1 · f2)v)(g) =∑
h∈Gr(g)(f1 · f2)(h)v(h−1g)
=∑
h∈Gr(g)
∑k∈Gr(h)
f1(k)f2(k−1h)
v(h−1g), by (3.4) on page 55,
=∑
h∈Gr(g)
∑k∈Gr(g)
f1(k)f2(k−1h)v(h−1g)
=∑
k∈Gr(g)f1(k)
∑h∈Gr(g)
f2(k−1h)v(h−1g)
=∑
k∈Gr(g)f1(k)
∑h∈Gr(g)
f2(k−1h)v(h−1kk−1g)
=∑
k∈Gr(g)f1(k)
∑h∈Gr(g)
f2(k−1h)v((k−1h)−1k−1g).
Making the change of variables h = k−1h, then h ∈ Gr(k−1) = Gr(k−1g). Then,
(ϑx(f1 · f2)v)(g) =∑
k∈Gr(g)f1(k)
∑h∈Gr(k−1g)
f2(h)v(h−1k−1g)
=∑
k∈Gr(g)f1(k)(ϑx(f2)v)(k−1g)
= ϑx(f1)(ϑx(f2)v)(g).
• ϑx(f*) = ϑx(f)*.
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Given v, w ∈ Kx,
〈ϑx(f)v, w〉 =∑
g∈Gx/Gxx
〈(ϑx(f)v)(g), w(g)〉
=∑
g∈Gx/Gxx
∑h∈Gr(g)
f(h)〈v(h−1g), w(g)〉.
On the other hand,
〈v, ϑx(f*)w〉 =∑
g∈Gx/Gxx
〈v(g), (ϑx(f*)w)(g)〉
=∑
g∈Gx/Gxx
∑h∈Gr(g)
f*(h)〈v(g), w(h−1g)〉
=∑
g∈Gx/Gxx
∑h∈Gr(g)
f(h−1)〈v(g), w(h−1g)〉 (5.23)
Note that for every g ∈ Gx/Gxx, h ∈ Gr(g), there exist unique g ∈ Gx/G
xx, h ∈ Gr(g)
satisfying h = h−1
g = h−1g.
Then we can make the change of variables in (5.23), obtaining
〈v, ϑx(f*)〉 =∑
g∈Gx/Gxx
∑h∈Gr(g)
f(h)〈v(h−1g), w(g)〉
= 〈ϑx(f)v, w〉.
Thus ϑx(f*) = ϑ(f)*.
Let ζx be de�ned by ζx(g) = πx(ug*)ξx if g ∈ Gxx, and ζx(g) = 0 if g ∈ Gx \Gx
x. Note that
137
ζx is in Kx. Given h ∈ Gxx, g ∈ Gx, ζx(gh) = 0 if g /∈ Gx
x. If g ∈ Gxx, by de�nition of Kx,
ζx(gh) = πx(ugh*)ξx = πx(uh*ug*)ξx = πx(uh*)π(ug*)ξx = πx(uh*)ζx(g).
Moreover, ‖ζx‖ = ‖ζx(x)‖ = ‖ξx‖ = 1. Let ψx be the state on C*(G) de�ned by ψx(f) =
〈ϑx(f)ζx, ζx〉. Then, for f ∈ Cc(G),
ψx(f) =∑
g∈Gx/Gxx
〈ϑx(f)ζx(g), ζx(g)〉
= 〈(ϑx(f)ζx)(x), ζx(x)〉
=∑h∈Gx
f(h)〈ζx(h−1x), ζx(x)〉
=∑h∈Gx
f(h)〈ζx(h−1), ζx(x)〉
=∑g∈Gxx
f(g)〈πx(ug−1*)ξx, πx(ux*)ξx〉, by de�nition of ζx,
=∑g∈Gxx
f(g)〈πx(ug**)ξx, πx(ux*)ξx〉
=∑g∈Gxx
f(g)〈πx(ug)ξx, πx(ux)ξx〉
=∑g∈Gxx
f(g)ϕx(ug). (5.24)
Note that C0(G(0)) is in the centralizer of ψx. From Lemma 3.3.16, we have that Cc(G(0))
is dense in C0(G(0)). Using the continuity of ψx, it is su�cient to show that Cc(G(0)) is in
the centralizer of ϕ. Given f ∈ Cc(G) and h ∈ Cc(G(0)), we have
ψx(f · h) =∑g∈Gxx
(f · h)(g)ϕx(ug) =∑g∈Gxx
f(g)h(x)ϕx(ug) =∑g∈Gxx
(h · f)(g)ϕx(ug) = ψx(h · f).
By assumption, the map x 7→ ψx(f) is µ-measurable for every f ∈ Cc(G). By Lemma 5.1.8,
we can de�ne a state ϕ on C*(G) by ϕ(f) =∫G(0) ψx(f)dµ(x).
138
Finally we show that if (µ, {ϕx}x) and (µ, {ϕx}x) de�ne the same state ϕ, then µ = µ and
ϕx = ϕx for µ-a.e. x.
Recall from Proposition 3.2.12 that G(0) is clopen in G. Let f ∈ Cc(G(0)). Then,
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x) =
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x)∫G(0)
f(x)ϕx(ux)dµ(x) =
∫G(0)
f(x)ϕx(ux)dµ(x), since f ∈ Cc(G(0)),∫G(0)
f(x)dµ(x) =
∫G(0)
f(x)dµ(x), ux is the unity in C*(Gxx).
Since f is arbitrary, we have µ = µ.
We will prove that ϕx = ϕx for µ-a.e. x. Let W ⊂ G be the set of g ∈ G′ such that
x = r(g) and ϕx(ug) 6= ϕx(ug). Let V ⊂ G(0) be the set of x ∈ G(0) such that ϕx 6= ϕx. Note
that V = r(W ).
Given g ∈ W , let Ug be an open bisection containing g. Using the topological properties
of G, we can assume the family {Ug}g∈W is countable without loss of generality.
Given f ∈ Cc(r(Ug)), there exists F ∈ Cc(Ug) such that f = F ◦ r|−1Ug . We denote by
hx = r|−1Ug (x), hence F (hx) = f(x) for every x ∈ r(Ug). Hence,
∫r(Ug∩G′)
f(x) [ϕx(hx)− ϕx(hx)] dµ(x)
=
∫r(Ug∩G′)
F (hx) [ϕx(hx)− ϕx(hx)] dµ(x)
=
∫r(Ug∩G′)
∑g∈Gxx
F (g) [ϕx(hx)− ϕx(hx)] dµ(x), since hx ∈ G′,
=
∫r(Ug)
∑g∈Gxx
F (g) [ϕx(hx)− ϕx(hx)] dµ(x)
=ϕ(F )− ϕ(F )
=0.
139
Since f is arbitrary, it follows that ϕx(uhx) = ϕx(uhx) for µ-a.e. x ∈ r(Ug ∩ G′). Then
µ(r(W∩Ug)) = 0. The family of Ug indexed by g ∈ W is countable, and V = ∪g∈W r(W∩Ug).
Then µ(V ) = 0.
Remark 5.2.10. If we de�ne a dynamics τ on C*(G) such that every function in C0(G(0)) is
τ -invariant, it follows from Lemma 5.1.32, that every KMS state ϕ on C*(G) has centralizer
containing C0(G(0)). In this case, we can apply the �rst Neshveyev's theorem.
5.3 Second Theorem
The second theorem shows the conditions that the pair (µ, {ϕx}x) satis�es if its corresponding
state is KMS. We begin by de�ning a dynamics τ on C*(G).
De�nition 5.3.1. Let G be a topological groupoid. A continuous R-valued 1-cocyle on G
is a continuous function c : G→ R such that c(gh) = c(g) + c(h) for every (g, h) ∈ G(2).
Now we will prove a lemma which will help us to show that the dynamics de�ned by the
cocycle is well-de�ned.
Lemma 5.3.2. Let G be a locally compact Hausdor� second countable étale groupoid. Let
K be R or C. Let F : K×G→ C be a continuous function. Let U ⊂ G be an open bisection
and suppose there exists a compact set K ⊂ U such that F (z, g) = 0 for every z ∈ K, g /∈ K.
De�ne for every z ∈ K the function Fz : G→ C by Fz(g) = F (z, g). Then the map from
K to Cc(G) de�ned by z 7→ Fz is continuous with respect to the norm of C*(G).
Proof. We can assume K = C without loss of generality. Note that each Fz is continuous
and its support is included in the compact set K.
Fix z0 ∈ C. Note that F is continuous at (z0, g) for every g ∈ K. Thus, for each g ∈ K
there exists δg > 0 and an open neighborhood Ug of g such that
|F (z, h)− F (z0, g)| ≤ ε
2,
140
for every (z, h) satisfying z ∈ Bδg(z0) = {w ∈ C : |w − z0| < δg}, h ∈ Ug.
The family {Bδg(z0)× Ug}g∈K forms an open cover for the compact set {z0} ×K. Then
there are g1, . . . , gn ∈ K such that {Bδgi(z0)× Ugi}ni=1 covers {z0} ×K. Let
δ =1
2min{δg1 , . . . , δgn}.
Now let z ∈ C such that |z − z0| < δ. Let g ∈ K. There exists i ∈ {1, . . . , n} such that
(z0, g) ∈ Bδgi(z0)× Ugi .
Since δ < δgi , we have (z, g) ∈ Bδgi(z0)× Ugi . Thus,
|F (z, g)− F (z0, g)| ≤ |F (z, g)− F (z0, gi)|+ |F (z0, gi)− F (z0, g)|
≤ ε
2+ε
2
≤ ε,
because (z, g), (z0, g) ∈ Bδgi× Ugi . Since g ∈ K is arbitrary, we have
‖Fz − Fz0‖ ≤ ‖Fz − Fz0‖∞, from Proposition 3.3.11,
= supg∈K|Fz(g)− Fz0(g)|
= supg∈K|F (z, g)− F (z0, g)|
≤ ε.
Therefore the map z 7→ Fz is continuous.
The next lemma is useful to prove that the elements of Cc(G) are entire analytic for our
dynamics.
Corollary 5.3.3. Using the same conditions of Lemma 5.3.2 for K = C, suppose that for
141
every g ∈ C the function z 7→ F (z, g) is di�erentiable. De�ne F ′ : C×G→ C by
F ′(z, g) =∂F
∂z(z, g) = lim
h→0
F (z + h, g)− F (z, g)
h.
Suppose F ′ is continuous. For every z ∈ C, de�ne F ′z : G → C by Fz(g) = F ′(z, g). Then
we have the limit
limh→0
∥∥∥∥Fz+h − Fzh− F ′z
∥∥∥∥ = 0,
for every z ∈ C.
Proof. Fix z ∈ C. By de�nition, F ′z is supported on K. De�ne the function H : C×G→ C
by
H(h, g) =
Fz+h(g)− Fz(g)
h− F ′z(g) if h 6= 0,
0 if h = 0.
This function continuous. Moreover, using the notation from of 5.3.2, we have that Hh is
supported on K for every h. Then by Lemma 5.3.2,
0 = limh→0‖Hh −H0‖
= limh→0‖Hh‖
= limh→0
∥∥∥∥Fz+h − Fzh− F ′z
∥∥∥∥ .
Now we de�ne a dynamics on C*(G) for a continuous R-valued cocycle. Throughout this
section, the dynamics is �xed.
Lemma 5.3.4. Let G be a locally compact Hausdor� second countable étale groupoid and
c : G → R a continuous R-valued cocycle. De�ne τ = {τt}t∈R by τt(f)(g) = eitc(g)f(g) for
142
every g ∈ G, f ∈ Cc(G). Then (C*(G), τ) is a C*-dynamical system.
Proof. Fix t ∈ R. Note that τt(f) ∈ Cc(G) for every f ∈ Cc(G). By de�nition, τt is a linear
map on Cc(G).
• τt is a ∗-homomorphism.
Given f1, f2 ∈ Cc(G), g ∈ G,
τt(f1 · f2)(g) = eitc(g)(f1 · f2)(g)
= eitc(g)∑g1g2=g
f1(g1)f2(g2)
=∑g1g2=g
eitc(g1g2)f1(g1)f2(g2)
=∑g1g2=g
(eitc(g1)f1(g1))(eitc(g2)f2(g2))
=∑g1g2=g
τt(f1)(g1)τt(f2)(g2)
= (τt(f1) · τt(f2))(g).
Given f ∈ Cc(G), g ∈ G, we have
τt(f*)(g) = eitc(g)f*(g) = e−itc(g)f(g−1) = eitc(g−1)f(g−1) = τt(f)(g−1) = τt(f)*(g).
Thus τt is a ∗-homomorphism.
• τt ◦ τs = τt+s and τ0 is the identity
Given f ∈ Cc(G), g ∈ G, t, s ∈ R,
τt ◦ τs(f)(g) = τt(τs(f))(g)
= eitc(g)τs(f)(g)
= eitc(g)eisc(g)f(g)
143
= ei(t+s)c(g)f(g)
= τt+s(f)(g).
By de�nition, τ0 is the identity.
Each τt is invertible. Therefore every τt is a ∗-automorphism.
• τ is strongly continuous
Let f ∈ Cc(G). Let K be the support of f . Assume there exists and open bisection U
such that K ⊂ U .
De�ne F : R×G→ C by F (t, g) = τt(f)(g) = eitc(g)f(g). Both c and f are continuous
functions, then F is continuous. Let K be the compact support of f . Then F (t, g) = 0
for every t ∈ R, g /∈ K. Using the notation of Lemma 5.3.2, we have τt(f) = Ft.
Therefore, by Lemma 5.3.2, the function t 7→ τt(f) is continuous.
Now let f ∈ Cc(G) be arbitrary. There are open bisections U1, . . . ,Un and functions
f1, . . . , fn such that each fk ∈ Cc(Uk) and f = f1 + . . . + fn. Since the function
t 7→ τt(fk) is continuous for every k and each τt is linear, it follows that the function
t 7→ τt(f) is continuous.
Note that, for every f ∈ Cc(G), ‖τ(f)‖ ≤ ‖f‖. In fact, let π be a ∗-representation of
Cc(G). Then π ◦ τ is an ∗-representation of Cc(G). Then. by de�ntion of the norm on
Cc(G), we have
‖π(τt(f))‖ = ‖π ◦ τt(f)‖f‖.
Since π is arbitrary, using the de�ntion of ‖π(τt(f))‖, we have
‖τt(f)‖ ≤ ‖f‖.
Therefore, from Lemma 5.1.12, τ de�nes a dynamics on C*(G).
144
Lemma 5.3.5. Let G be a locally compact second countable étale groupoid. Fix β > 0.
Assume ϕ is a KMSβ-state. Then the centralizer of ϕ contains C0(G(0)).
Proof. Let h ∈ C0(G(0)). Note that for every x ∈ G(0), c(x) = c(xx) = 2c(x) = 0. Then
c(x) = 0. Since h is supported on G(0), we have for every t ∈ R,
τt(h)(g) = eitc(g)h(g) = h(g) for g ∈ G.
Then τt(h) = h. Therefore, by Lemma 5.1.32, h is in the centralizer of ϕ.
Lemma 5.3.5 shows that we can apply Theorem 5.2.9 for every KMS state ϕ on C*(G).
Then ϕ corresponds to a pair (µ, {ϕy}y∈G(0)) as shown in that theorem.
Lemma 5.3.6. Let G be a locally compact Hausdor� second countable étale groupoid. Then
every f ∈ Cc(G) is entire analytic for τ .
Proof. Let f ∈ Cc(G). Since the set of entire analytic elements in C*(G) forms a vector space
and each function in Cc(G) can be decomposed as a sum of functions in Cc(G) supported on
open bisections, we can assume f is supported on an open bisection without loss of generality.
Let K = supp(f).
De�ne F : C × G → C by F (z, g) = eizc(g)f(g). Then F is continuous. De�ne for every
z ∈ C, Fz : G→ C by
Fz(g) = F (z, g) = eizc(g)f(g).
Note that Ft(f) = τt(f) for every t ∈ R. For every z ∈ C, Fz is continuous and supported
on K. Analogously, de�ne F ′ : C×G→ by
F ′(z, g) =∂F
∂z(z, g) = ic(g)eizc(g)f(g).
145
Then F ′ is continuous. From Corollary 5.3.3 we have the limit
limh→0
∥∥∥∥Fz+h − Fzh− F ′z
∥∥∥∥ = 0.
Therefore f is entire analytic. In this case, we can write τz(f) = Fz for z ∈ C.
The following lemma proves some properties of compactly supported functions. These
properties will be used in the proof of Neshneyev's theorem.
Lemma 5.3.7. Let G be a locally compact Hausdor� second countable étale groupoid. Let
U be a an open bisection and let f1 ∈ Cc(U), f2 ∈ Cc(G). Then, given g ∈ G,
(f1 · f2)(g) =
f1(hx)f2((hx)−1g), for x ∈ r(U), g ∈ Gx,
0, if r(g) /∈ r(U).(5.25)
(f2 · f1)(g) =
f1(hx)f2(g(hx)−1), for x ∈ s(U), g ∈ Gx,
0, if s(g) /∈ s(U).(5.26)
For x ∈ r(U), hx denotes the unique element in U ∩Gx. Analogously, for x ∈ s(U), hx is the
unique element in U ∩Gx.
Proof. Let f1 ∈ Cc(U), f2 ∈ Cc(G).
Equation (5.25): Let x ∈ G(0), g ∈ Gx. Note that Gx ∩ U = ∅ if x /∈ r(U). From Lemma
3.3.3, we have
(f1 · f2)(g) =∑h∈Gx
f1(h)f2(h−1g)
=∑
h∈Gx∩U
f1(h)f2(h−1g), since f1 ∈ Cc(U),
=
f1(hx)f2((hx)−1g), for x ∈ r(U), g ∈ Gx,
0, if r(g) /∈ r(U).
146
Equation (5.26): Let x ∈ G(0), g ∈ Gx. Note that Gx ∩ U = ∅ if x /∈ s(U). From Lemma
3.3.3, we have
(f2 · f1)(g) =∑h∈Gx
f2(gh−1)f1(h)
=∑
h∈Gx∩U
f2(gh−1)f1(h), since f1 ∈ Cc(U),
=
f2(g(hx)−1)f1(hx), if x ∈ s(U),
0, otherwise.
Lemma 5.3.8. Let G be a locally compact Hausdor� second countable étale groupoid. Let
c be an R-valued 1-cocyle. A measure µ on G(0) is quasi-invariant with Radon-Nikodym
derivative ec if, and only if, for every open bisection U ⊂ G, we have
dT∗µ
dµ(x) = ec(hx), (5.27)
for x ∈ s(U), where hx ∈ U is the unique element such that s(hx) = x and T : r(U)→ s(U)
is the homeomorphism de�ned by T (r(hx)) = x. In particular, T = s|U ◦ r|−1U .
Proof. Let U be an open bisection in G. Then hx = s|−1U (x) for every x ∈ s(U). Let
T : r(U)→ s(U) such that T (r(hx)) = x for every x ∈ s(U). Then, for every x,
x = T (r(hx)) = T (r|U(hx)) = T (r|U(s|−1U (x))).
Therefore T = (r|U ◦ s|−1U )−1 = s|U ◦ r|−1
U .
First we show a formula which holds if, and only if, condition (5.27) is satis�ed. Note
that equation (5.27) holds if, and only if, for every f ∈ Cc(s(U)),
∫s(U)
f(x)d(T∗µ)(x) =
∫s(U)
ec(hx)f(x)dµ(x). (5.28)
147
Recall that a measure µ on G(0) is quasi-invariant with Radon-Nikodym derivative ec if,
and only if, for every f ∈ Cc(G),
∫G
f(g)dµr(g) =
∫G
ec(g)f(g)dµs(g).
From the de�nition of µr, µs, this is equivalent to∫r(U)
∑g∈Gx
f(g)dµ(x) =
∫s(U)
∑g∈Gx
f(g)dµ(x). (5.29)
Since f ∈ Cc(U), we can consider only g ∈ U in the integrals. Recall that U ∩ Gx = {hx},
if x ∈ r(U), where hx = r|−1U (x). Analogously, U ∩ Gx = {hx} if x ∈ s(U). Then we can
rewrite (5.29) as
∫r(U)
f(hx)dµ(x) =
∫s(U)
ec(hx)f(hx)dµ(x). (5.30)
Therefore µ is quasi-invariant with Radon-Nikodym derivative ec if, and only if, (5.30) holds
for every open bisection U , f ∈ Cc(U).
Note that for every x ∈ r(U), hx = r|−1U (x) = s|−1
U ◦ s|U ◦ r|−1U (x) = s|−1
U (Tx) = hTx. Then
(5.30) is equivalent to
∫r(U)
f(hTx)dµ(x) =
∫s(U)
ec(hx)f(hx)dµ(x). (5.31)
There is a bijection from Cc(s(U)) to Cc(U) given by f 7→ f = f ◦s|U . This holds because
s|U : U → s(U) is a homeomorphism. Therefore, for every x ∈ s(U), f(x) = f(s(hx)) =
f(hx). In the rest of this proof, given f ∈ Cc(s(U)), we denote by f its corresponding function
in Cc(U). Analogously, given f ∈ Cc(U), f is the corresponding function in Cc(s(U)).
Assume µ is quasi-invariant with Radon-Nikodym derivative ec. Let U be an open bisec-
148
tion of G. Let f ∈ Cc(s(U)). Then
∫s(U)
f(x)d(T∗µ)(x) =
∫s(U)
f(hx)d(T∗µ)(x)
=
∫r(U)
f(hTx)dµ(x), from the de�nition of T∗µ,
=
∫s(U)
ec(hx)f(hx)dµ(x), by (5.31)
=
∫s(U)
ec(hx)f(x)dµ(x).
Hence, (5.28) holds, then (5.27) holds.
Conversely, suppose (5.28) holds. Given an open bisection U , f ∈ Cc(U), we have
∫r(U)
f(hTx)dµ(x) =
∫r(U)
f(Tx)dµ(x)
=
∫s(U)
f(x)d(T∗µ)(x)
=
∫s(U)
ec(hx)f(x)dµ(x)
=
∫s(U)
ec(hx)f(hx)dµ(x).
Then (5.31) holds for f . Therefore, µ is quasi-invariant with Radon-Nikodym derivative
ec.
Lemma 5.3.9. Let ϕ be a state on C*(G) with centralizer containing C0(G(0)). Assume ϕ
corresponds to the pair (µ, {ϕx}x). Then ϕ is τ -invariant if, and only if,
ϕx(ug) = 0 for every g ∈ Gxx \ c−1(0), µ-a.e. x. (5.32)
Proof. Assume ϕ is τ -invariant. It follows from the continuity of c that c−1(0) is closed. Let
g ∈ G such that c(g) 6= 0. Let t ∈ R such that tc(g) ∈ (0, 2π). Then 1 − eitc(g) 6= 0. There
exists an open bisection U containing g such that 1− eitc(g) 6= 0 for every g ∈ U .
149
Let f ∈ Cc(s(U)) and de�ne f = f ◦ s|U ∈ Cc(U). Let x ∈ G(0). Suppose g ∈ Gxx ∩ U .
Then x ∈ s(U) and g = hx, where hx = s|−1U (x) denotes the unique element in U ∩ Gx.
Hence, we can write
ϕ(f) =
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x)
=
∫G(0)
∑g∈Gxx∩U
f(g)ϕx(ug)dµ(x)
=
∫s(U)
∑g∈Gxx∩U
f(g)ϕx(ug)dµ(x)
=
∫s(U)
∑g∈Gx∩U
χG′(g)f(g)ϕx(ug)dµ(x)
=
∫s(U)
χG′(hx)f(hx)ϕx(uhx)dµ(x)
=
∫s(U)
χG′(hx)ϕx(uhx)f(x)dµ(x), by de�nition of f . (5.33)
Analogously,
ϕ(τt(f)) =
∫s(U)
eitc(hx)χG′(hx)ϕx(uhx)f(x)dµ(x). (5.34)
Since ϕ(τt(f)) = ϕ(f) by hypothesis, then from equations (5.33) and (5.34) we have
∫s(U)
[1− eitc(hx)]χG′(hx)ϕx(uhx)f(x)dµ(x) = 0.
The function f ∈ Cc(s(U)) is arbitrary and each 1− eitc(hx) 6= 0. Then, for µ-a.e. x ∈ s(U),
χG′(hx)ϕx(uhx) = 0,
or equivalently, for µ-a.e. x ∈ s(U), g ∈ U ∩Gx
χG′(g)ϕx(ug) = 0. (5.35)
150
Note that G \ c−1(0) can be covered by a countable family of open bisections {Un}n∈Nsuch that (5.35) holds for each Un. Therefore, for µ-a.e.x ∈ G(0), for all g ∈ Gx \ c−1(0),
χG′(g)ϕx(ug) = 0.
Thus, for µ-a.e. x ∈ G(0), for all g ∈ Gxx, ϕx(ug) = 0.
Conversely, assume (5.32) holds. Let f ∈ Cc(G). Then, for µ-a.e. x ∈ G(0) and all g ∈ Gxx,
ϕ(ug) 6= 0 implies c(g) = 0. Then, for µ-a.e. x ∈ G(0),
∑g∈Gxx
f(g)ϕx(ug) =∑g∈Gxx
eitc(g)f(g)ϕx(ug). (5.36)
Therefore,
ϕ(τt(f)) =
∫G(0)
∑g∈Gxx
eitc(g)f(g)ϕx(ug)dµ(x)
=
∫G(0)
∑g∈Gxx
f(g)ϕx(ug)dµ(x)
= ϕ(f).
Since ϕ and ϕ ◦ τt are continuous functions and Cc(G) is dense in C*(G), it follows that
ϕ(τt(a)) = ϕ(a) for every a ∈ C*(G). In other words, ϕ is τ -invariant.
Now we prove Neshveyev's second theorem. Note that in this theorem we assume a
di�erent de�nition for KMS-states. Given β ∈ R, a state ϕ on a C*-algebra A is a KMSβ-
state if ϕ is τ -invariant and ϕ(aτiβ(b)) = ϕ(ba) for a dense subset of analytic elements
a, b ∈ A. This de�nition corresponds to item (ii) in Proposition 5.1.29, so it is equivalent to
the de�nition introduced in Section 5.1 when β 6= 0.
Theorem 5.3.10. (Neshveyev) Let G be a locally compact second countable Hausdor� étale
groupoid. Let c be a continuous R-valued 1-cocycle on G and τ be the dynamics on C*(G)
de�ned by τt(f)(g) = eitc(g)f(g) for f ∈ Cc(G), g ∈ G. Fix β ∈ R. Then there exists a one-
151
to-one correspondence between KMSβ-states on C*(G) and pairs (µ, {ϕx}x∈G(0)) consisting
of a probability measure µ on G(0) and a µ-measurable �eld of states ϕx on C*(Gxx) such
that:
(i) µ is quasi-invariant with Radon-Nikodym derivative e−βc;
(ii) ϕx(ug) = ϕr(h)(uhgh−1) for every g ∈ Gxx and h ∈ Gx, for µ-a.e. x; in particular, ϕx is
tracial for µ-a.e. x;
(iii) ϕx(ug) = 0 for all g ∈ Gxx \ c−1(0), for µ-a.e. x.
Proof. From Lemma 5.3.5, the centralizer of any τ -KMS-state ϕ contains C0(G(0)). By
Theorem 5.2.9, ϕ is de�ned by a pair (µ, {ϕx}x) consisting of a probability measure µ on
G(0) and a µ-measurable �eld of states ϕx on C*(Gxx). It follows from Lemma 5.3.9 that
property (iii) is satis�ed if and only if ϕ is τ -invariant.
Therefore we have to prove that properties (i) and (ii) are satis�ed if, and only if,
ϕ(f1 · f2) = ϕ(f2 · τiβ(f1)), for every f1, f2 ∈ Cc(G). (5.37)
Each function in Cc(G) can be decomposed as a �nite sum of continuous functions sup-
ported on open bisections from Lemma 3.3.5. Every function in Cc(G) is entire analytic by
by Lemma 5.3.6. Note that τiβ is linear on Cc(G) by de�nition. Therefore (5.37) holds if,
and only if, for every open bisection U we have
ϕ(f1 · f2) = ϕ(f2 · τiβ(f1)), for every f1 ∈ Cc(U), f2 ∈ Cc(G). (5.38)
We will show (5.38) is equivalent to another equation and we will use this to prove the
equivalence between (5.38) and conditions (i) and (ii). Given an open bisection U , de�ne
hx = r|−1U (x) for every x ∈ r(U). Analogously, de�ne hx = s|−1
U (x) for x ∈ s(U). Then, for
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every x ∈ r(U),
Tx = s|U ◦ r|−1U (x) = s(hx), and
hTx = s|−1U (Tx) = s|−1
U (s|U ◦ r|−1U (x)) = r|−1
U (x) = hx.
Suppose (5.37) holds. Let f1 ∈ Cc(U), f2 ∈ Cc(G). It follows that
ϕ(f1 · f2) =
∫G(0)
∑g∈Gxx
(f1 · f2)(g)ϕx(ug)dµ(x)
=
∫r(U)
∑g∈Gxx
f1(hx)f2((hx)−1g)ϕx(ug)dµ(x), from Lemma 5.3.7,
=
∫r(U)
f1(hx)∑g∈Gxx
f2((hx)−1g)ϕx(ug)dµ(x).
Since τiβ(f1) ∈ Cc(U), we can apply Lemma 5.3.7 and obtain
ϕ(f2 · τiβ(f1)) =
∫G(0)
∑g∈Gxx
(f2 · τiβ(f1))(g)ϕx(ug)dµ(x)
=
∫s(U)
∑g∈Gxx
τiβ(f1)(hx)f2(g(hx)−1)ϕx(ug)dµ(x), from Lemma 5.3.7,
=
∫s(U)
∑g∈Gxx
e−βc(hx)f1(hx)f2(g(hx)−1)ϕx(ug)dµ(x)
=
∫s(U)
e−βc(hx)f1(hx)∑g∈Gxx
f2(g(hx)−1)ϕx(ug)dµ(x).
Therefore (5.38) is equivalent to the following equation for f1 ∈ Cc(U), f2 ∈ Cc(G), U open
bisection.
∫r(U)
f1(hx)∑g∈Gxx
f2((hx)−1g)ϕx(ug)dµ(x) =
∫s(U)
e−βc(hx)f1(hx)∑g∈Gxx
f2(g(hx)−1)ϕx(ug)dµ(x).
(5.39)
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Suppose (5.39) holds. Let f ∈ Cc(s(U)), then we can de�ne f1 ∈ Cc(U) such that
f |U = f ◦ s|U . Let f2 = f1*. Then f2 ∈ Cc(U−1) by Lemma 3.3.8.
Let x ∈ r(U), g ∈ Gxx such that f2((hx)−1g) 6= 0. Then (hx)−1g ∈ U−1 which implies
g−1hx ∈ U . Recall that U is an open bisection and hx ∈ U . Moreover, s(g−1hx) = s(hx),
then g−1hx = hx, hence g = x. Therefore, for all x ∈ r(U), we have
∑g∈Gxx
f2((hx)−1g)ϕx(ug) = f2((hx)−1x)ϕx(ux) = f2((hx)−1).
Therefore, for f1, f2, we can rewrite (5.39) as
∫s(U)
e−βc(hx)f1(hx)f2((hx)−1)dµ(x) =
∫r(U)
f1(hx)f2((hx)−1)dµ(x)
=
∫r(U)
f1(hTx)f2((hTx)−1)dµ(x), (5.40)
because hTx = hx for x ∈ r(U). Using the de�nition of f1 and f2, we have for every x ∈ s(U),
f1(hx) = f ◦ s|U(hx) = f ◦ s|U(s|−1U (x)) = f(x),
f2((hx)−1) = f1*((hx)
−1) = f1(hx) = f(x).
Then we can replace the values in the integrals in (5.40) and obtain
∫s(U)
e−βc(hx)|f(x)|2dµ(x) =
∫r(U)
|f(Tx)|2dµ(x).
Since f is arbitrary, it follows that for every f ∈ Cc(s(U)),
∫s(U)
e−βc(hx)f(x)dµ(x) =
∫r(U)
f(Tx)dµ(x).
HencedT∗µ
dµ(x) = e−βc(hx). It follows from Lemma 5.3.8 that property (i) holds.
Now we show that property (ii) is satis�ed. Let U be an open bisection and f ∈ Cc(r(U)),
154
de�ne f1 ∈ Cc(U) such that f1|U = f ◦r|U . Then f1(hx) = f(x). Given f2 ∈ Cc(G) arbitrary,
de�ne the function F : G(0) → C by
F (x) =∑g∈Gxx
(f2 · f1)(g)ϕx(ug).
This function is µ-measurable because f2 · f1 ∈ Cc(G) and {ϕx}x is a µ-measurable �eld of
states. Moreover, by Lemma 5.3.7,
F (x) =
f1(hx)∑
g∈Gxxf2(g(hx)
−1)ϕx(ug), if x ∈ s(U),
0, otherwise.
Then
∫r(U)
f(x)∑g∈GTxTx
f2(g(hx)−1)ϕTx(ug)dµ(x) =
∫r(U)
f1(hx)∑g∈GTxTx
f2(g(hx)−1)ϕTx(ug)dµ(x)
=
∫r(U)
f1(hTx)∑g∈GTxTx
f2(g(hTx)−1)ϕTx(ug)dµ(x)
=
∫r(U)
F (Tx)dµ(x)
=
∫s(U)
F (x)dT∗µ(x)
=
∫s(U)
e−βc(hx)F (x)dµ(x),
by property (i) that we already proved. Then,
=
∫s(U)
e−βc(hx)f1(hx)∑g∈Gxx
f2(g(hx)−1)ϕx(ug)dµ(x),
=
∫r(U)
f1(hx)∑g∈Gxx
f2((hx)−1g)ϕx(ug)dµ(x),
155
by (5.39). Then,
=
∫r(U)
f(x)∑g∈Gxx
f2((hx)−1g)ϕx(ug)dµ(x),
if we make the change of variables g = hxg(hx)−1, we get
=
∫r(U)
f(x)∑g∈GTxTx
f2(g(hx)−1)ϕx(u(hx)−1ghx)dµ(x).
Assume U has compact closure. Let V be an open bisection with compact closure such
that s(V) ⊂ s(U). We can assume s(V) = s(U) without loss of generality. In fact, if this is
not the case, just replace U by s|−1U ◦ s(V). VU−1 is an open bisection from Lemmas 3.3.6
and 3.3.7.
Assume f2 ∈ Cc(G) is such that f2 is positive on VU−1 and vanishes outside this set.
Since f ∈ Cc(r(U)) is arbitrary, we have for µ-a.e. x,
∑g∈GTxTx
f2(g(hx)−1)[ϕx(u(hx)−1ghx)− ϕTx(ug)] = 0. (5.41)
Let x ∈ r(U) be such that (5.41) holds. Let g ∈ V ∩ GTx. Then g(hx)−1 is the unique
element in VU−1. Thus we can write (5.41) as
ϕx(u(hx)−1ghx) = ϕTx(ug) for g ∈ GTxTx ∩ V , µ-a.e. x ∈ r(U). (5.42)
Since the preimage of s(U) under s can be covered by a countable family of open bisections
V with compact closure such that s(V) ⊂ s(U), it follows that for µ-a.e. x ∈ r(U), g ∈ GTxTx,
we have
ϕx(u(hx)−1ghx) = ϕTx(ug) = ϕs(hx)(ug).
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Note that the set
{x ∈ G(0) : ϕx(ug) 6= ϕr(h)(uhgh−1) for some g ∈ Gxx, h ∈ Gx}
={x ∈ G(0) : ϕx(ug) 6= ϕr(h−1)(uh−1gh) for some g ∈ Gxx, h ∈ Gx}
={x ∈ G(0) : ϕx(ug) 6= ϕs(h)(uh−1gh) for some g ∈ Gxx, h ∈ Gx}
=⋃U
{x ∈ G(0) : ϕx(ug) 6= ϕs(h)(uh−1gh) for some g ∈ Gxx, h ∈ Gx ∩ U}
=⋃U
{x ∈ r(U) : ϕx(ug) 6= ϕs(h)(uh−1gh) for some g ∈ Gxx, h ∈ Gx ∩ U}
=⋃U
{x ∈ r(U) : ϕx(ug) 6= ϕs(hx)(u(hx)−1ghx) for some g ∈ Gxx}
=⋃U
{x ∈ r(U) : ϕx(ug) 6= ϕTx(u(hx)−1ghx) for some g ∈ Gxx}
has measure zero. Here U ranges over a countable open cover of G such that r, s are injective
on U . Therefore property (ii) holds.
Conversely, assume properties (i), and (ii) are satis�ed. Given an open bisection U , let
f1 ∈ Cc(U), f2 ∈ Cc(G). Then
∫r(U)
f1(hx)∑g∈Gxx
f2((hx)−1g)ϕx(ug)dµ(x)
=
∫r(U)
f1(hx)∑g∈GTxTx
f2(g(hx)−1)ϕx(uhxg(hx)−1)dµ(x) , making g = (hx)−1ghx,
=
∫r(U)
f1(hx)∑g∈GTxTx
f2(g(hx)−1)ϕr(hx)(uhxg(hx)−1)dµ(x)
=
∫r(U)
f1(hTx)∑g∈GTxTx
f2(g(hTx)−1)ϕr(hTx)(uhTxg(hTx)−1)dµ(x)
=
∫r(U)
f1(hTx)∑g∈GTxTx
f2(g(hTx)−1)ϕTx(ug)dµ(x), from property (ii),
=
∫s(U)
e−βc(hx)f1(hx)∑g∈Gxx
f2(g(hx)−1)ϕx(ug)dµ(x), from (i) and Lemma 5.3.8.
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Then (5.39) holds. However, we already proved this is equivalent to equation (5.37).
158
Chapter 6
Renault-Deaconu Groupoid
In this chapter we prove a theorem due to Thomsen [26] which characterizes the extremal
KMS states on the full C*-algebra of the Renault-Deaconu groupoid. The de�nition of this
groupoid depends on a local homeomorphism σ : X → X such that X has some topological
properties. We can identify the subset of units with X.
We can apply Neshveyev's theorems to this groupoid C*-algebra in order to describe its
KMS states. In this chapter we show that, on this groupoid, quasi-invariant measures are
the same as conformal measures. Moreover, the corresponding measure of an extremal KMS
state is either continuous or supported on an orbit.
Under certain conditions, Thomsen's theorem gives an explicit formula for the extremal
KMS states. The results in this chapter are based on [9] and [26].
6.1 Introduction
Now we de�ne the Renault-Deaconu groupoid and prove some of its topological properties.
De�nition 6.1.1. Let X be a locally compact second countable Hausdor� space. Let σ :
159
X → X be a local homeomorphism. The Renault-Deaconu groupoid is the groupoid
G = {(x, k, y) ∈ X × Z×X : ∃n,m ∈ N, k = n−m,σn(x) = σm(y)},
such that
G(2) = {((x1, k1, y1), (x2, k2, y2)) ∈ G × G : y1 = x2},
with the following multiplication and inversion laws
(x, k1, y)(y, k2, z) = (x, k1 + k2, z) and (x, k, y)−1 = (y,−k, x),
and unit space de�ned by G(0) = {(x, 0, x) : x ∈ X}.
The range and source maps are de�ned by r(x, k, y) = (x, 0, x) and s(x, k, y) = (y, 0, y).
Since the map x 7→ (x, 0, x) is a bijection from X to G(0), we identify X with G(0).
Remark 6.1.2. We assume, by convention, that σ0 is the identity.
The groupoid G can be understood intuitively as follows: given x ∈ X, we can interpret
the sequence {σn(x)}n∈N as a trajectory starting at x, as shown in Figure 6.1.
x σ(x) σ1(x) σ2(x) σ3(x) . . .
Figure 6.1: The sequence {σn(x)}n∈N, x ∈ X, can be interpreted as the trajectory of x.
If, for some y ∈ X, there is some n such that σn(y) is an element of the trajectory of
x, we can say the trajectories eventually meet. In other words, there exists m such that
σm(x) = σn(y). k = m − n is the delay of one trajectory with respect to the other. Hence
160
(x, k, y) ∈ G if the trajectories of x and y eventually meet. This idea is shown in Figure 6.2.
y
x · σ(y) ·
σ(x) σ2(y)
σ2(x) = σ3(y)
Figure 6.2: If (x, k, y) ∈ G then the trajectories {σl(x)}l∈N and {σl(y)}l∈N eventually meet.k can be seen as the delay of one trajectory with respect to the other. In this �gure, k = −1,since σ2(x) = σ3(y).
Proposition 6.1.3. The Renault-Deaconu groupoid is a groupoid.
Proof. Let G be the Renault-Deaconu groupoid. Clearly the maps r and s are surjective.
The product is well-de�ned: In fact, let g, h ∈ G be composable. Then g = (x, k, y),
h = (y, l, z) for some x, y, z ∈ X, k, l ∈ Z.
By de�nition of G, there exist m,n ∈ N such that k = m− n and σm(x) = σn(y). There
are p, q ∈ N satisfying l = p− q and σp(y) = σq(z). Hence,
σm+p(x) = σp(σm(x)) = σp(σn(y)) = σp+n(y) = σn(σp(y)) = σn(σq(z)) = σn+q(z).
The inverse is well-de�ned: Given (x, k, y) ∈ G, there exist m,n ∈ N such that
k = m−n and σm(x) = σn(y). Then −k = n−m and σn(y) = σm(x). Hence (y,−k, x) ∈ G.
Now we show G satis�es properties (i)-(v) of De�nition 3.1.1:
(i) Let g = (x, k, y) ∈ G, h = (y, l, z) ∈ G, then
s(gh) = s(x, k + l, z)= z = s(h)
r(gh) = r(x, k + l, z)= x = r(g).
(ii) Given x ∈ X, r(x, 0, x) = (x, 0, x) = s(x, 0, x).
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(iii) Given g = (x, k, y) ∈ G,
gs(g) = (x, k, y)(y, 0, y)= (x, k, y) = g
r(g)g = (x, 0, x)(x, k, y)= (x, k, y) = g.
(iv) Let (g1, g2), (g2, g3) ∈ G(2). Then g1 = (x, k1, y), g2 = (y, k2, z), g3 = (z, k3, w). Hence,
(g1g2)g3 = (x, k1 + k2, z)(z, k3, w) = (x, k1 + k2 + k3, w)
g1(g2g3) = (x, k1, y)(y, k2 + k3, w)= (x, k1 + k2 + k3, w).
(v) Given g = (x, k, y) ∈ G,
gg−1 = (x, k, y)(y,−k, x)= (x, 0, x) = r(g)
g−1g = (y,−k, x)(x, k, y)= (y, 0, y) = s(g).
Example 6.1.4. Let X = {x = {xn}n∈N : xn ∈ {0, 1}}. Endow X with the metric
d(x, y) = 2−min {n∈N : xn 6=yn}.
Moreover, σ is a local homeomorphism. Indeed, let x ∈ X. De�ne the map ρ : X → X by
ρ(y)n =
x0, if n = 0
yn−1 if n ≥ 1.
Then ρ is continuous, since for y, z ∈ X,
d(σ(y), σ(z)) = 2−min{n≥1:yn−1 6=zn−1} = 2−1d(y, z)
162
Moreover, for every y ∈ B(x, 1), y0 = x0. Hence y = ρ ◦ σ(y) = σ ◦ ρ(y). Therefore σ is
invertible on B(x, 1) with inverse ρ. That is, ρ is a local homeomorphism.
Then (x, k, y) ∈ G if there are n,m ∈ N such that k = n −m and σn(x) = σm(y). For
instance, if x = (0, 1, 0, 0, . . .) and y = (1, 1, 1, 1, 0, 0, . . .) then (x,−2, y) ∈ G, as shown in
Figure 6.3.
x = (0, 1, 0, 0, . . .)
y = (1, 1, 1, 1, 0, 0, . . .)
σ(x)
σ3(y)
Figure 6.3: It follows from the equality σ(x) = σ3(y) = (1, 0, 0, . . .) that (x,−2, y) ∈ G.
Now we de�ne a topology on G which makes the Renault-Deaconu groupoid an étale
groupoid.
Given A,B open subsets of X, m,n ∈ N, let
Un,mA,B = {(x, n−m, y) ∈ G : σn(x) = σm(y), x ∈ A, y ∈ B}.
These sets form a basis of the topology on G.
Proposition 6.1.5. The family of sets Un,mA,B , for A,B open subsets of X and n,m ∈ N,
generates a topology on G. Moreover, G is second countable.
Proof. First we show that G is the union of these sets. Let (x, k, y) ∈ G. Then there exist
n,m such that σn(x) = σm(y). Hence (x, n−m, y) ∈ Un,mX,X .
Now we prove that for every (x, k, y) ∈ Un1,m1
A1,B1∩ Un2,m2
A2,B2there exists Un,mA,B such that
(x, k, y) ∈ Un,mA,B ⊂ Un1,m1
A1,B1∩ Un2,m2
A2,B2.
Let (x, k, y) ∈ Un1,m1
A1,B1∩ Un2,m2
A2,B2. Then,
x ∈ A1 ∩ A2 , y ∈ B1 ∩B2 , σn1(x) = σm1(y) , σn2(x) = σm2(y).
163
Let p1 = n2, p2 = n1. Note that m1 − n1 = m2 − n2 implies m1 + p1 = m2 + p2. From the
de�nition of p1, p2, we have n1 + p1 = n2 + p2.
For i = 1, 2, let Ui be an open neighborhood of σni(x) such that σpi is injective on Ui. Let
n = n1 + p1 = n2 + p2
m = m1 + p1 = m2 + p2
A = A1 ∩ A2 ∩ σ−n1(U1) ∩ σ−n2(U2)
B = B1 ∩B2 ∩ σ−m1(U1) ∩ σ−m2(U2).
From continuity of σni , σmi , it follows that A and B are open sets.
We show that (x, k, y) ∈ Un,mA,B . Clearly n−m = n1 +p1− (m1 +p1) = n1−m1 = k. Then,
σn(x) = σp1+n1(x) = σp1(σn1(x)) = σp1(σm1(y)) = σp1+m1(y) = σm(y).
For i = 1, 2, σni(x) ∈ Ui, σmi(y) = σni(x) ∈ Ui. Then x ∈ A, y ∈ B. Therefore (x, k, y) ∈
Un,mA,B .
Now we show that Un,mA,B ⊂ Un1,m1
A1,B1∩ Un2,m2
A2,B2. Let (u, k, v) ∈ Un,mA,B .
For i = 1, 2, we have
σn(u) = σm(v)
σni+pi(u) = σmi+pi(v)
σpi(σni(u)) = σpi(σmi(u)).
Since σni(u), σmi(v) ∈ U and σpi is injective on this set, it follows that σni(u) = σmi(v).
Hence (u, k, v) ∈ Uni,miAi,Bi. Therefore (u, k, v) ∈ Un1,m1
A1,B1∩ Un2,m2
A2,B2.
Now we show that G is second countable.
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Let B be a countable base of X. Then the family of sets of the form
Un,mA,B such that n,m ∈ N and A,B ∈ B
is also countable. We show this family form a base for G. Let Un,mA,B with A,B arbitrary open
sets in X and n,m ∈ N. Let (x, n−m, y) ∈ Un,mA,B . There exists A, B such that x ∈ A ⊂ A,
y ∈ B ⊂ B. Then (x, n−m, y) ∈ Un,mA,B⊂ Un,mA,B .
Lemma 6.1.6. Let {(xi, ki, yi)}i∈N be a sequence in G converging to (x, k, y). Then xi → x,
yi → y and there exists i0 such that ki = k for every i ≥ i0. Hence we can assume, without
loss of generality, that ki is constant.
Proof. Let n,m ∈ N such that σn(x) = σm(y) and k = n −m. Let A,B be neighborhoods
of x, y, respectively. Then there exists i0 such that for every i ≥ i0, (xi, ki, yi) ∈ Un,mA,B , then,
x ∈ A, y ∈ B, ki = n−m = k.
Therefore ki is eventually constant, xi → x and yi → y.
Lemma 6.1.7. Fix n0,m0 ∈ N. Given a sequence {(xi, k, yi)}i∈N a net assume that for all
neighborhoods A of x and B of y there exists i0 such that (xi, k, yi) ∈ Un0,m0
A,B for i ≥ i0.
Then for every A,B open neighborhoods of x, y, respectively, n,m such that (x, k, y) ∈
Un,mA,B . There exists i0 such that
(xi, k, yi) ∈ Un,mA,B for i ≥ i0.
Then (xi, k, yi)→ (x, k, y).
Proof. Note that k = n0−m0. Let n,m ∈ N such that σn(x) = σm(y) and k = n−m. Then
n0 +m = n+m0.
Let V be an open neighborhood of σn(x) = σm(y) where σn0 and is injective. Since xi → x
and yi → y and σ is continuous, there exists i0 such that for every i ≥ i0, σn(xi), σ
m(yi) ∈ V .
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Then, given i ≥ i0,
σn+n0(xi) = σn(σn0(xi)) = σn(σm0(yi)) = σn+m0(yi) = σn0+m(yi).
Then σn0(σn(xi)) = σn0(σm(yi)). Since σn(xi), σm(yi) ∈ V and σn0 is injective on V , it
follows that σn(xi) = σm(yi) for every i ≥ i0.
Note that (x, k, y) /∈ Un,mA,B if n−m 6= k.
Let A,B be open neighborhoods of x, y respectively. Let m,n such that m−n = k. Since
xi → x and yi → y, there exists i0 such that if i ≥ i0,
xi ∈ A, yi ∈ B, σn(xi) = σm(yi),
or equivalently, (xi, k, yi) ∈ Un,mA,B for every i ≥ i0.
Corollary 6.1.8. Let (x, n −m, y) ∈ G. Let {xi}i∈N, {yi}i∈N be sequences in X such that
xi → x and yi → y. If σn(xi) = σm(yi) for each i, then (xi, n−m, yi)→ (x, n−m, y) in G.
Proof. Let A be an open neighborhood of x, B an open neighborhood of y. There exists
i0 such that, for every i ≥ i0, xi ∈ A, yi ∈ B. By hypothesis, σn(xi) = σm(yi). Then
(xi, n−m, yi) ∈ Un,mA,B . From Lemma 6.1.7, it follows that (xi, n−m, yi) → (x, n−m, y) in
G.
Theorem 6.1.9. The Renault-Deaconu groupoid G, with topology generated by Un,mA,B is a
topological groupoid, locally compact Hausdor�, second countable and étale.
Proof. • G is a topological groupoid
(i) G(2) is closed in G × G.
Assume {(gi, hi)}i∈N is a sequence in G(2) converging to (g, h) ∈ G × G. Then
gi = (xi, ki, yi), hi = (yi, li, zi) for each i. Assume g = (x, k, y) and h = (y, l, z).
It follows from Lemma 6.1.6 that yi → y and yi → y in X. Since X is Hausdor�,
we have y = y. Therefore (g, h) ∈ G(2).
166
(ii) The inverse is continuous
Denote the inverse map by ι. Let Um,nB,A be an open set in the topological base of
G. Then,
ι−1(Um,nB,A) = {(x, k, y) : (y,−k, x) ∈ Um,nB,A}
= {(x, k, y) : y ∈ B, x ∈ A, σm(y) = σn(x), k = n−m}
= Un,mA,B .
Therefore the inverse map is continuous.
(iii) The product is continuous
Let (gi, hi)→ (g, h) in G(2). We can assume g = (x, k, y), h = (y, l, z) and for each
i, gi = (xi, k, yi) and hi = (yi, l, zi).
Let n1,m1, n2,m2 ∈ N such that
k = n1 −m1 , σn1(x) = σm1(y)
l = n2 −m2 , σn2(y) = σm2(z).
Since (xi, k, yi) → (x, k, y) and (yi, l, zi) → (y, l, z), it follows that for A, B
neighborhoods of x and z, there exists i0 such that
(xi, k, yi) ∈ Un1,m1
A,X and (yi, k, zi) ∈ Un1,m1
X,B for i ≥ i0.
Hence, if i ≥ i0,
σn1+n2(xi) = σn2(σn1(xi)) = σn2(σm1(yi)) = σn2+m1(yi)
= σm1(σn2(yi))= σm1(σm2(zi)) = σm1+m2(zi).
167
Then (xi, k + l, zi) ∈ Un1+n2,m1+m2
A,B . It follows from Lemma 6.1.7 that
(xi, k + l, zi)→ (x, k + l, z).
• G is Hausdor�
Let gi = (xi, ni −mi, yi) ∈ G such that σni(xi) = σmi(yi), i = 1, 2. Assume g1 6= g2.
(i) If n1 −m1 6= n2 −m2,
then g1 ∈ Un1,m1
X,X , g2 ∈ Un2,m2
X,X and Un1,m1
X,X ∩ Un2,m2
X,X = ∅.
(ii) If n1 −m1 = n2 −m2,
Then x1 6= x2 or y1 6= y2. Assume x1 6= x2. Since X is Hausdor�, we can choose
A1, A2 open neighborhoods of x1, x2 respectively, such that A1 ∩ A2 = ∅. Then
gi ∈ Uni,miAi,X, i = 1, 2, and Un1,m1
A1,X∩ Un2,m2
A2,X= ∅.
The proof for y1 6= y2 is analogous.
• G is locally compact.
Let A,B be open sets of X such that A,B are compact, and let n,m ∈ N. Then
Un,mA,B ⊂ Un,m
A,B, where
UA,B = {(x, n−m, y) ∈ G : σn(x) = σm(y), x ∈ A, y ∈ B}.
Let {(xi, n−m, yi)}i∈N be a sequence in UA,B. Then {(xi, yi)}i∈N is a sequence in the
compact set A×B. Then there exists a subsequence {(xij , yij)}j∈N such that xij → x
for some x ∈ A and yij → y for some y ∈ B. By continuity of σ, σn(x) = σm(y).
Therefore UA,B is compact.
• G is étale.
Let (x, n − m, y) ∈ G such that σn(x) = σm(y). Since σ is a local homeomorphism,
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there are A,B open neighborhoods of x, y, respectively, satisfying
σn(A) is open and σn|A : A→ σn(A) is a homeomorphism
σm(B) is open and σm|B : B → σm(B) is a homeomorphism.
Then (x, n−m, y) ∈ Un,mA,B . Since G is a topological groupoid, r is continuous. In order
to prove r is a local homeomorphism, we will show r is injective on Un,mA,B , r(Un,mA,B ) is
open and r|−1Un,mA,B
is continuous.
(i) r is injective.
Suppose there exist x1, x2 ∈ A, y1, y2 ∈ B such that r(x1, n −m, y1) = r(x2, n −
m, y2). Then y1 = y2. Moreover,
x1 = σ−n|A(σm(y1)) = σ−n|A(σm(y2)) = x2.
Therefore (x1, n−m, y1) = (x2, n−m, y2).
(ii) r(Un,mA,B ) is open.
r(Un,mA,B ) = {(y, 0, y) ∈ G : (x, n−m, y) ∈ Un,mA,B}
= {(y, 0, y) ∈ G : x ∈ A, y ∈ B, σn(x) = σm(y)}
= {(y, 0, y) ∈ G : y ∈ B, x = σ|−nA (σm(y)) ∈ A}
= {(y, 0, y) ∈ G : y ∈ B, σm(y) ∈ σn(A)} note that σn(A) is open
= {(y, 0, y) ∈ G : y ∈ B, y ∈ σ|−mB (σn(A))}
= U0,0C,C ,
where C = B ∩ σ|−mB (σn(A)).
(iii) r|−1Un,mA,B
is continuous.
Let {(yi, 0, yi)}i∈N be a sequence in r(Un,mA,B ) converging to some (y, 0, y) in r(Un,mA,B ).
Then yi → y ∈ B. De�ne the sequence xi = σ|−nA (σm(yi)). Then xi → x =
169
σ|−nA (σm(y)).
Note that xi is the only element in A satisfying σn(xi) = σm(yi), hence,
(xi, n−m, yi) = r|−1Un,mA,B
(yi, 0, yi).
Analogously, (x, n−m, y) = r|−1Un,mA,B
(y, 0, y). Then, it follows from Corollary 6.1.8
that (xi, n−m, yi)→ (x, n−m, y).
Analogously, we can show s is a local homeomorphism.
The next lemma shows that we can identify X wth the unit space of G. In this chapter,
we �x X and σ, and we assume G is the Renault-Deaconu groupoid.
Lemma 6.1.10. We can identify the unit space G(0) with the set X. In fact, both have the
same topology.
Proof. Clearly the map ι : G(0) → X de�ned by (x, 0, x)→ x is a bijection. Denote this map
by ι. We will show ι is a homeomorphism.
Let n,m ∈ N, A,B ⊂ X open sets. If Un,mA,B ∩ G(0) 6= ∅, then n = m. In this case,
Un,mA,B ⊂ G(0). Hence,
ι(Un,mA,B ) = ι({(x, 0, x) : x ∈ A ∩B}) = A ∩B
is open in X. On the other hand, for any open set A ⊂ X,
ι−1(A) = {(x, 0, x) : x ∈ A} = U0,0A,A,
Therefore ι is a homeomorphism.
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6.2 Full orbits
Given x ∈ X, the full orbit of x denotes the set of elements inX whose trajectories eventually
meet the trajectory of x. There are two types of orbits, periodic and aperiodic. In this section
we will study their properties.
De�nition 6.2.1. Let x ∈ X, the full orbit of x is the set
O(x) = {y ∈ X : there exists k ∈ Z such that (x, k, y) ∈ G}.
Lemma 6.2.2. Given x, y ∈ X, O(x) = O(y) if, and only if, y ∈ O(x).
Proof. Let y ∈ O(x). There exists (x, k, y) ∈ G.
• O(y) ⊂ O(y) Let z ∈ O(y). There exists (y, l, z) ∈ G. Hence (x, k + l, z) ∈ G. Then
z ∈ O(x). Therefore O(y) ⊂ O(x).
• O(x) ⊂ O(y) Let z ∈ O(x). There exists (x, l, z) ∈ G. Then (z, k − l, y) ∈ G. Thus
z ∈ O(y). Then O(x) ⊂ O(y).
Therefore O(x) = O(y).
Conversely, assume O(x) = O(y). By de�nition, (y, 0, y) ∈ G. Then y ∈ O(y) = O(x).
Remark 6.2.3. Lemma 6.2.2 is a fact of general groupoids. Given a groupoid G, if we de�ne
the set O(x) = r(Gx) for every x ∈ G(0), then the Lemma 6.2.2 holds.
Lemma 6.2.4. Given x ∈ X, O(x) is countable.
Proof.
O(x) = {y ∈ X : there exists k ∈ Z such that (x, k, y) ∈ G}
= {y ∈ X : Gyx 6= ∅}
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= {r(g) : g ∈ Gx 6= ∅} identifying X with G(0)
= r(Gx).
G is second countable étale from Theorem 6.1.9. It follows from Proposition 3.2.11 that Gxis countable. Therefore O(x) is countable.
There are two types of full orbits: periodic and aperiodic orbits. This di�erence will be
fundamental when we de�ne the extremal conformal measures later.
De�nition 6.2.5. Given x ∈ X, we say it is periodic or σ-periodic if there is a positive
integer p such that
σp(x) = x. (6.1)
The minimum positive natural number such that (6.1) holds is called the minimal period of
x.
De�nition 6.2.6. A point z ∈ X is called aperiodic if O(z) does not contain periodic points.
De�nition 6.2.7. Let y ∈ X. IfO(y) has periodic points, O(y) is called periodic. Otherwise,
O(y) is aperiodic.
When represented graphically, periodic and aperiodic orbits look di�erent. An aperiodic
orbits look like a tree while a periodic orbits has a single cycle, the trajectory of a periodic
172
element. Figure 6.4 shows these two types of orbits.
x
(a)
z
. . .
(b)
Figure 6.4: Periodic and aperiodic orbits look di�erent. The periodic orbit (a) has a cycle,while the aperiodic orbit looks like a tree. Circles in green represent the trajectories of xand z.
Given y in the orbit of x, the trajectory of y eventually meets the trajectory of x. If x is
periodic, we can see in Figure 6.4 that y eventually meets the point x, that is, there exists
an n such that σn(y) = x. Now we will prove this result.
Lemma 6.2.8. Let x ∈ X be a periodic point with minimum period p. Given y ∈ O(x),
there exists n ∈ N such that σn(y) = x.
Proof. Let y ∈ O(x). Let n1,m1 ∈ N such that σn1(x) = σm1(x). Let N be a natural number
such that N +m1 ∈ pN. De�ne n = N + n1. Then
σn(y) = σN+n1(y) = σN+m1(x) = x.
Lemma 6.2.9. Assume the set of periodic points in X is countable. Let I denote the set of
aperiodic points. Then I is countable
Proof. Let N be the set of points with periodic orbits. Then I = X \N and we can write
N =⋃x∈X
x periodic
O(x).
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O(x) is countable for every x ∈ X by Lemma 6.2.4. Moreover, the set of periodic points is
countable by hypothesis. Then N is countable and µ(N) = 0.
Lemma 6.2.10. Let z ∈ X be aperiodic. Then Gzz = {(z, 0, z)}. If y ∈ X is such that O(y)
has a periodic point with minimum period p, then Gyy = {(y, kp, y) : k ∈ Z}.
Proof. Let z ∈ X be aperiodic. Let k ∈ Z such that (z, k, z) ∈ G. Then there exists n,m ∈ N
such that σn(z) = σm(z), k = n−m.
Assume without loss of generality that n > m. Denote x = σm(z) ∈ O(z). Then x is
periodic, since
σn−m(x) = σn−m(σm(z)) = σn(z) = σm(z) = x.
Contradiction. Then k = 0 and therefore Gzz = {(z, 0, z)}.
Let y ∈ X such that O(y) is periodic. Then there exist x ∈ X periodic with minimum
period p, and l ∈ Z such that (x, l, y) ∈ G. Then (y,−l, x)(x, p, x)(x, l, p) = (y, p, y) ∈ G.
Therefore, by induction, (y, kp, y) ∈ G for every k ∈ Z.
Suppose there exists k /∈ pZ such that (y, k, y) ∈ Gyy . Let k1, k2 ∈ N such that k = k1+k2p,
0 ≤ k1 < p. Note that k1 6= 0 by hypothesis.
Then (x, k, x) ∈ G since (x, k, x) = (y,−l, x)(y, k, y)(y, l, x). Hence there exist n,m ∈ N
such that n−m = k and σn(x) = σm(x).
Assume n > m without loss of generality. Let N be an integer such that N + m ∈ pN.
Then,
x = σm+N(x) = σn+N(x) = σn−m(σm+N(x)) = σn−m(x)
= σk(x) = σk1+k2p(x) = σk1(σk2p(x)) = σk1(x).
Contradiction, since x has minimum period p. Therefore Gyy = {(y, kp, y) : k ∈ Z}.
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6.3 Conformal Measures
We will show explicitly all the extremal atomic eβF -conformal probability measures on G.
for β 6= 0.
Now we de�ne conformal measures as described in [8].
De�nition 6.3.1. Consider a measurable function T : X → X on a measurable space
(X,F) and a measurable nonnegative function f on X. A measure µ on (X,F) is called
f -conformal if
µ(T (A)) =
∫A
f(x)dµ(x),
whenever A ⊂ X is a measurable set, for which T (A) is measurable and T : A → T (A) is
invertible.
A set A as in De�nition 6.3.1 is called special .
The set of f -conformal probability measures µ forms a convex set. We say µ is extremal
if µ is an extremal point in this set.
De�nition 6.3.2. Let µ be a f -conformal probability measure. We say that µ is extremal
if for all µ1, µ2, t ∈ (0, 1) such that µ = tµ1 + (1− t)µ2, it follows that µ1 = µ2 = µ.
Lemma 6.3.3. Let µ be a �nite measure on the topological space X with µ = µa + µc, µa
purely atomic and µc non-atomic. Then µ is f -conformal if and only if µa, µc are f -conformal.
Proof. Since µ = µa + µc, both µa, µc are �nite measures.
Assume µ is f -conformal. LetXa be the Borel set such that µa(Xa) = µa(X) and µ(x) > 0
for every x ∈ Xa. Since µa is �nite, then Xa is countable and µc(Xa) = 0.
Let A be a special set. Since Xa is countable, T (A∩Xa) is also countable, then T (A∩Xa)
is measurable. Note that T (A) = T (A∩Xa)∪ T (A \Xa) and T (A∩Xa)∩ T (A \Xa) = ∅.
Hence T (A \Xa) is measurable.
µc(T (A ∩Xa)) = 0 since T (A ∩Xa) is countable. Now we show µa(T (A \Xa)) = 0. Let
y ∈ T (A \ Xa). There exists a ∈ A \ Xa such that y = T (a). Using that µ is f -conformal
175
and a /∈ Xa, we have
µa(T (a)) = µ(T (a)) = f(a)µ(a) = f(a)µa(a) = 0.
Since y is arbitrary and µa is atomic, we have µa(T (A \Xa)) = 0. Then
µa(T (A)) = µa(T (A ∩Xa)) + µa(T (A \Xa))
= µa(T (A ∩Xa))
= µa(T (A ∩Xa)) + µc(T (A ∩Xa))
= µ(T (A ∩Xa))
=
∫A∩Xa
f(x)dµ(x)
=
∫A
f(x)dµa(x).
Since µ = µa on Xa and µa = 0 outside Xa, then µa is f -conformal. The proof for µc is
analogous.
Conversely, assume µa, µc are f -conformal. Let A be a Borel set such that T : A→ T (A)
is invertible. Then
µ(T (A)) = µa(T (A)) + µc(T (A))
=
∫A
f(x)dµa(x) +
∫A
f(x)dµc(x)
=
∫A
f(x)dµ(x).
Then µ is f -conformal.
Remark 6.3.4. Fix a continuous function F : X → R and assume β 6= 0. It follows
from Lemma 6.3.3 that every extremal eβF -conformal measure is either purely atomic or
non-atomic. In fact, let µ be an extremal eβF -conformal probability measure and assume is
neither purely atomic nor non-atomic. By Lemma 6.3.3, µ = µa + µc, µa is purely atomic
176
and µc is non-atomic.
Let t = µa(X). Then t > 0 (otherwise µ = µc) and t < 1 (otherwise µ = µa). De�ne
µ1 = t−1µa, µ2 = (1− t)−1µc. Note that µ1, µ2 are probability measures. In fact,
µ1(X) = t−1µa(X) = t−1t = 1
and
µ2(X) = (1− t)−1µc(X)
= (1− t)−1(µ(X)− µa(X)) , since µ = µa + µc,
= (1− t)−1(1− t) = 1.
Then µ1, µ2 are probability measures. Now let A be a special set. Then
µ1(T (A)) = t−1µa(T (A)) = t−1
∫A
eβF (x)dµa(x) =
∫A
eβF (x)dµ1(x)
and
µ2(T (A)) = (1− t)−1µc(T (A)) = (1− t)−1
∫A
eβF (x)dµc(x) =
∫A
eβF (x)dµ2(x).
Therefore µ1, µ2 are eβF -conformal.
In this section we want to �nd all extremal eβF -conformal purely atomic probability
measures on X. Given µ eβF -conformal, then µ(σ(x)) = eβF (x)µ(x).
Lemma 6.3.5. Let β ∈ R and µ an eβF -conformal measure on X, n a positive natural
number. Given y ∈ X, we have
µ(σn(y)) = exp
(β
n−1∑k=0
F (σk(y))
)µ(y). (6.2)
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Proof. We prove this by induction. Assume 6.2 holds for n. Then
µ(σn+1(y)) = µ(σ(σn(y))) = eβF (σn(y))µ(σn(y)) , since µ is eβF -conformal,
= eβF (σn(y)) exp
(β
n−1∑k=0
F (σk(y))
)µ(y) , by hypothesis,
= exp
(β
n∑k=0
F (σk(y))
)µ(y)
= exp
β (n+1)−1∑k=0
F (σk(y))
µ(y).
Then (6.2) holds for n+ 1. Let n = 1. Using the fact that µ is eβF -conformal, we have
µ(σn(y)) = µ(σ(y)) = eβF (y)µ(y) = exp
(βn−1∑k=0
F (σk(y))
)µ(y).
Proposition 6.3.6. Given β ∈ R, let µ be an eβF -conformal measure on X, x ∈ X. Then,
for every y ∈ O(x),
µ(y) = exp
(−β
(m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(x))
))µ(x),
where σn(x) = σm(y).
Proof. Let y ∈ O(x). There exist m,n > 0 such that σn(x) = σm(y). Then
µ(σn(x)) = exp
(β
n−1∑j=0
F (σj(x))
)µ(x) and µ(σm(y)) = exp
(β
m−1∑j=0
F (σj(y))
)µ(y).
Since σn(x) = σm(y), it follows that
exp
(β
m−1∑j=0
F (σj(y))
)µ(y) = exp
(β
n−1∑j=0
F (σj(x))
)µ(x)
178
µ(y) = exp
(βn−1∑j=0
F (σj(x))
)µ(x) exp
(−β
m−1∑j=0
F (σj(y))
)
µ(y) = exp
(−β
(m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(x))
))µ(x).
It follows from Proposition 6.3.6 that if two eβF -conformal measures µ1, µ2 are equal on
a point x ∈ X, then µ1, µ2 are equal on O(x). Moreover, if µ is an eβF -conformal measure
such that µ(x) > 0, then µ(y) > 0 for every y ∈ O(x).
Corollary 6.3.7. Let x ∈ X and β ∈ R. There exists at most one eβF -conformal probability
measure that vanishes outside O(x). In particular, if µ is an eβF -conformal probability
measure that vanishes outside O(x), then µ is extremal.
Proof. Let µ1, µ2 be a eβF -conformal probability measures vanishing outside O(x). It follows
from Proposition 6.3.6 that µ1(x), µ2(x) > 0. In fact, let i = 1, 2. Since O(x) is countable
and µi is a probability measure whose support lies in O(x), there exists y ∈ O(x) such that
µi(y) > 0. Then µi(x) > 0 by Proposition 6.3.6.
Suppose that µ1(x) < µ2(x). Then µ1(y) < µ2(y) for every y ∈ O(x) by Proposition 6.3.6.
Therefore
1 = µ1(X) = µ1(O(x)) = µ2(O(x)) = µ2(X) = 1,
which is a contradiction. is not a probability measure. The proof is analogous for µ1(x) >
µ2(x). Then µ1(x) = µ2(x). Then µ1 and µ2 are equal on O(x) by Proposition 6.3.6.
Therefore µ1 = µ2.
Now we show µ is extremal. Let µ1, µ2 be two eβF -conformal probability measures such
that µ = tµ1 + (1 − t)µ2, 0 < t < 1. Let A = X \ O(x). A is measurable. Since µ(A) = 0
and µ1(A), µ2(A) ≥ 0, we have µ1(A) = µ2(A) = 0. By previous arguments, µ = µ1 = µ2.
Therefore µ is extremal.
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Corollary 6.3.8. Assume β ∈ R \ {0}. Let µ be an eβF -conformal measure. Let x ∈ X
periodic with minimum period p such that µ(x) > 0. It follows that
p−1∑j=0
F (σj(x)) = 0.
Proof. σp(x) = x. Then,
µ(x) = µ(σp(x)) = exp
(β
p−1∑k=0
F (σk(x))
)µ(x).
µ(x) > 0, hence
1 = exp
(β
p−1∑k=0
F (σk(x))
).
Note that β 6= 0. Thereforep−1∑k=0
F (σk(x)) = 0.
Lemma 6.3.9. Let µ be a purely atomic extremal eβF -conformal probability measure, β ∈ R.
Then there exists x ∈ X such that for y ∈ X, µ(y) > 0 if, and only if, y ∈ O(x).
Proof. Let x ∈ X such that µ(x) > 0.
Suppose there exists y ∈ X \ O(x) such that µ(y) > 0. Since µ is a probability measure,
µ(x) > 0 and x /∈ O(y), we have 0 < µ(O(y)) < 1. Let t = µ(O(y)). De�ne µ1 by
µ1(z) =
t−1µ(z) if z ∈ O(y)
0 otherwise.
Then µ1 is a probability measure. Note that σ(z) ∈ O(y) if, and only if, z ∈ O(y). In fact,
O(σ(z)) = O(z) from Lemma 6.2.2. Then
µ1(σ(z)) = t−1µ(σ(z)) = eβF (z)t−1µ(z) = eβF (z)µ1(z).
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Therefore µ1 is eβF -conformal. De�ne µ2 by
µ2(z) =
(1− t)−1µ(z) if z /∈ O(y)
0 if z ∈ O(y)
.
µ2 is an eβF -conformal probability measure as well. In fact,
µ2(X) = µ2(X \ O(y))
= (1− t)−1µ(X \ O(y))
= (1− t)−1 [µ(X)− µ(O(y))]
= (1− t)−1 [1− t] , since t = µ(O(x)),
= 1.
Then µ2 is a probability measure. Given z /∈ O(y), σ(z) /∈ O(y). In fact, suppose σ(z) ∈
O(y). Then O(σ(z)) = O(y) by Lemma 6.2.2, but O(z) = O(σ(z)). Then O(σ(z)) = O(y).
Therefore z ∈ O(y). Contradiction. Then,
µ2(σ(z)) = (1− t)−1µ(σ(z))
= eβF (z)(1− t)−1µ(z) , since µ is eβF -conformal
= eβF (z)µ2(z).
Therefore µ2 is eβF -conformal.
Moreover, µ = tµ1 +(1−t)µ2. Therefore µ is not extremal. Contradiction. Then µ(y) = 0
if y /∈ O(x).
Let y ∈ O(x). It follows from Proposition 6.3.6 that µ(y) > 0.
For β 6= 0, each extremal atomic eβF -conformal probability measure corresponds to an
orbit O(x). However, an orbit O(x) does not necessarily have a correspondent extremal
atomic eβF -conformal probability measure.
181
Proposition 6.3.10. Let β ∈ R \ {0}. Let x ∈ X be a periodic point with minimum period
p. There exists an extremal atomic eβF -conformal probability measure with support O(x)
if, and only if,
p−1∑j=0
F (σj(x)) = 0, (6.3)
M =∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)<∞. (6.4)
where Yn = σ−n(x) \⋃n−1j=0 σ
−j(x) for n ≥ 1, and Y0 = {x}.
In this case, the measure is denoted by
mx = (1 +M)−1
[δx +
∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)δy
]. (6.5)
Proof. Let µ be an extremal atomic eβF -conformal probability measure with support O(x).
It follows from Corollary 6.3.8 that (6.3) holds.
Let y ∈ O(x), by Lemma 6.2.8 there exists a minimum natural number n such that
σn(y) = x. Hence y ∈ Yn. Therefore
O(x) =∞⋃n=0
Yn and µ(O(x)) =∞∑n=0
µ(Yn).
Let y ∈ Yn. From Proposition 6.3.6,
µ(y) = exp
(−β
n−1∑j=0
F (σj(y))
)µ(x).
182
O(x) is countable by Lemma 6.2.4. Thus,
µ(Yn) =∑y∈Yn
µ(y) =∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)µ(x).
Hence,
1 = µ(O(x)) =∞∑n=0
µ(Yn)
= µ(x) +∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)µ(x)
= µ(x)
[1 +
∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)]
= µ(x)(1 +M).
Then M <∞ and µ(x) = (1 +M)−1. Given y ∈ Yn, n ≥ 1,
µ(y) = (1 +M)−1 exp
(−β
n−1∑j=0
F (σj(y))
).
Therefore (6.5) holds.
Conversely, assume (6.3), (6.4) hold. Given n > 1,
mx(Yn) = (1 +M)−1∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
).
Hence,
mx(O(x)) = mx(x) +∞∑n=1
mx(Yn)
= (1 +M)−1
[1 +
∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)]
183
= (1 +M)−1(1 +M) = 1.
Then mx is a probability measure. Now we prove mx is eβF -conformal.
Let y ∈ Yn, n ≥ 2. Then σ(y) ∈ Yn−1. In fact, σn−1(σ(y)) = σn(y) = x. Let 0 ≤ l < n−1,
then σl(σ(y)) = σl+1(y) 6= x, since 1 ≤ l < n. Let y′ = σ(y), then
mx(σ(y)) = (1 +M)−1 exp
(−β
n−2∑j=0
F (σ(y′))
)
= (1 +M)−1 exp
(−β
n−1∑j=1
F (σ(y))
)
= eβF (y)(1 +M)−1 exp
(−β
n−1∑j=0
F (σ(y))
)
= eβF (y)mx(y).
Let y ∈ Yn, n = 1. Then σ(y) = x. Hence,
mx(σ(y)) = mx(x) = (1 +M)−1
= eβF (y)(1 +M)−1 exp (−βF (y))
= eβF (y)(1 +M)−1 exp
(−β
n−1∑j=0
F (σj(y))
)
= eβF (y)mx(y).
Note that σ(x) ∈ Yp−1. In fact, σp−1(σ(x)) = σp(x) = x. Given 0 ≤ l < p − 1, σl(σ(x)) =
σl+1(x) 6= x since 1 ≤ l + 1 < p. Then, if x′ = σ(x), we have
mx(σ(x)) = (1 +M)−1 exp
(−β
p−2∑j=0
F (σj(x′))
)
= (1 +M)−1 exp
(−β
p−1∑j=1
F (σj(x))
)
184
= eβF (x)(1 +M)−1 exp
(−β
p−1∑j=0
F (σj(x))
)
= eβF (x)(1 +M)−1 from (6.3)
= eβF (x)mx(x).
Therefore mx is eβF -conformal. mx has support O(x) by de�nition. From Corollary 6.3.7,
mx is extremal.
De�nition 6.3.11. Given a continuous function F : X → R we de�ne cF : G → R by
cF (x, k, y) =n−1∑j=0
F (σj(x))−m−1∑j=0
F (σj(y)),
for n,m ∈ N such that k = n−m and σn(x) = σm(y).
Lemma 6.3.12. cF is well-de�ned.
Proof. Let (x, k, y) ∈ G. For i = 1, 2, let ni,mi ∈ N such that σni(x) = σmi(y), k = ni −mi.
Then n2 − n1 = m2 −m1. Assume n2 > n1 without loss of generality. Then m2 > m1 and
n2−1∑j=0
F (σj(x))−m2−1∑j=0
F (σj(y))
=
(n1−1∑j=0
F (σj(x)) +
n2−1∑j=n1
F (σj(x))
)−
(m1−1∑j=0
F (σj(y)) +
m2−1∑j=m1
F (σj(y))
)
=
(n1−1∑j=0
F (σj(x)) +
n2−n1−1∑j=0
F (σj+n1(x))
)−
(m1−1∑j=0
F (σj(y)) +
m2−m1−1∑j=0
F (σj+m1(y))
)
=
(n1−1∑j=0
F (σj(x))−m1−1∑j=0
F (σj(y))
)+
(n2−n1−1∑j=0
F (σj+n1(x))−m2−m1−1∑
j=0
F (σj+m1(y))
)
=
(n1−1∑j=0
F (σj(x))−m1−1∑j=0
F (σj(y))
)+
(n2−n1−1∑j=0
F (σj(σn1(x)))−n2−n1−1∑j=0
F (σj(σm1(y)))
)
=
(n1−1∑j=0
F (σj(x))−m1−1∑j=0
F (σj(y))
)+
(n2−n1−1∑j=0
F (σj(σm1(y)))−n2−n1−1∑j=0
F (σj(σm1(y)))
)
185
=
(n1−1∑j=0
F (σj(x))−m1−1∑j=0
F (σj(y))
).
Therefore cF (x, k, y) does not depend on the choice of n,m satisfying k = n−m.
Proposition 6.3.13. cF is a continuous R-valued 1-cocycle on G.
Proof. Let (x, k, y), (y, l, z) ∈ G. There exist m,n, p, q ∈ N such that k = m− n, l = p− q,
σm(x) = σn(y) and σp(y) = σq(z).
Then m+ p− n− q = k + l and σm+p(x) = σn+p(y) = σn+q(z). Hence,
cF (x, k + l, z) =
m+p−1∑j=0
F (σj(x))−n+q−1∑j=0
F (σj(z))
=m−1∑j=0
F (σj(x)) +
m+p−1∑j=m
F (σj(x))−q−1∑j=0
F (σj(z))−n+q−1∑j=q
F (σj(z))
=m−1∑j=0
F (σj(x)) +
p−1∑j=0
F (σj+m(x))−q−1∑j=0
F (σj(z))−n−1∑j=0
F (σj+q(z))
=m−1∑j=0
F (σj(x)) +
p−1∑j=0
F (σj+n(y))−q−1∑j=0
F (σj(z))−n−1∑j=0
F (σj+p(y))
=m−1∑j=0
F (σj(x)) +
p+n−1∑j=n
F (σj(y))−q−1∑j=0
F (σj(z))−p+n−1∑j=p
F (σj(y))
=m−1∑j=0
F (σj(x))−n−1∑j=0
F (σj(y)) +
p+n−1∑j=0
F (σj(y))
−q−1∑j=0
F (σj(z)) +
p−1∑j=0
F (σj(y))−p+n−1∑j=0
F (σj(y))
= cF (x, k, y) + cF (y, l, z).
Now we prove the continuity of cF . Let (xi, ki, yi) → (x, k, y). Then xi → x and yi → y.
There exists m,n ∈ N with m−n = k and i0 such that σm(xi) = σn(yi) for every i ≥ i0. By
186
continuity of σ, σm(x) = σn(y).
Hence, for every i ≥ i0,
cF (xi, ki, yi) =m−1∑j=0
F (σj(xi))−n−1∑j=0
F (σj(yi)).
Since F and σ are continuous on X, we have cF (xi, ki, yi)→ cF (x, k, y).
The following lemma will be used to prove that every continuous R-valued 1-cocycle
corresponds to a unique cF .
Lemma 6.3.14. Let (x, n −m, y) ∈ G such that σn(x) = σm(y). Suppose that n,m ≥ 1.
Then
(x, n−m, y) =(x, 1, σ(x))(σ(x), 1, σ2(x)) · · · (σn−1(x), 1, σn(x))
(σm(y),−1, σm−1(y))(σm−1(y),−1, σm−2(y)) · · · (σ(y),−1, y). (6.6)
Proof. First we show that for every natural number N ≥ 1,
(x,N, σN(x)) = (x, 1, σ(x)) · · · (σN−1(x), 1, σN(x)). (6.7)
Clearly (6.7) holds for N = 1. Suppose that (6.7) is satis�ed for an arbitrary N . Then
(x,N + 1, σN+1(x)) = (x,N, σN(x))(σN(x), 1, σN+1(x))
= (x, 1, σ(x)) · · · (σN(x), 1, σN+1(x)).
Hence, (6.7) is satis�ed for every N . In particular, this equality holds for N = n. By the
same argument,
(y,m, σm(y)) = (y, 1, σ(y)) · · · (σm−1(y), 1, σm(y)).
187
Then, applying the inverse on both sides, we have
(σm(y),−m, y) = (σm(y),−1, σm−1(y)) · · · (σ(y),−1, y).
Since σn(x) = σm(y), we have
(x, n−m, y) =(x, n, σn(x))(σm(y),−m, y)
=(x, 1, σ(x))(σ(x), 1, σ2(x)) · · · (σn−1(x), 1, σn(x))
(σm(y),−1, σm−1(y))(σm−1(y),−1, σm−2(y)) · · · (σ(y),−1, y).
Now we will prove that there exists a bijection between continuous R-valued 1-cocycles
on G and the continuous functions from X to R. Recall from Proposition 6.3.13 that every
cF is a continuous R-valued 1-cocycle.
Proposition 6.3.15. For every continuous R-valued 1-cocycle c on G, there exists a unique
continuous function F : X → R such that c = cF .
Proof. Let c : G → R be a continuous R-valued 1-cocycle. De�ne the function F : X → G
by F (x) = c(x, 1, σ(x)).
The function F is continuous. In fact, let {xn}n∈N be a sequence inX converging to x ∈ X.
By de�nition of G, (x, 1, σ(x)) ∈ G and each (xn, 1, σ(xn)) ∈ G. Since σ is continuous, we
have σ(xn) → σ(x). It follows from Corollary 6.1.8 that (xn, 1, σ(xn)) → (x, 1, σ(x)). Since
c is continuous, we have F (xn)→ F (x). Therefore F is continuous.
Let (x, n − m, y) ∈ G such that σn(x) = σm(y). Assume that n,m ≥ 1 without loss of
generality. Then, by Lemma 6.3.14,
(x, n−m, y) =(x, 1, σ(x))(σ(x), 1, σ2(x)) · · · (σn−1(x), 1, σn(x))
=(σm(y),−1, σm−1(y))(σm−1(y),−1, σm−2(y)) · · · (σ(y),−1, y).
188
Then
c(x, n−m, y) =c(x, 1, σ(x)) + · · ·+ c(σn−1(x), 1, σn(x))
+ c(σm(y),−1, σm−1(y)) + · · ·+ c(σ(y),−1, y).
=n−1∑i=0
c(σi(x), 1, σi+1(x)) +m−1∑i=0
c(σi+1(y),−1, σi(y))
=n−1∑i=0
c(σi(x), 1, σi+1(x))−m−1∑i=0
c(σi(y), 1, σi+1(y))
=n−1∑i=0
F (σi(x))−m−1∑i=0
F (σi(y))
=cF (x, n−m, y).
Finally we prove that cF is unique. Suppose that there exists a continuous function
H : X → R such that cH = cF . Then, for every x ∈ X,
H(x) = cH(x, 1, σ(x)) = cF (x, 1, σ(x)) = F (x).
Therefore H = F .
The following lemma shows a equality for cF on Gyy where O(y) is periodic.
Lemma 6.3.16. Let x ∈ X be a periodic point with minimum period p. Then
cF (y, kp, y) = k
p−1∑j=0
F (σj(x)),
for every y ∈ O(x), k ∈ Z.
Proof. Since σp(x) = x, we have
cF (x, p, x) =
p−1∑j=0
F (σj(x)).
189
Let y ∈ O(x). Then, from Lemma 6.2.8, there exists n such that σn(y) = x. Thus
(y, n, x) ∈ G. Then
cF (y, kp, y) = kcF (y, p, y)
= kcF ((y, n, x)(x, p, x)(x,−n, y))
= k[cF (y, n, x) + cF (x, p, x) + cF (x,−n, y)]
= k[cF (y, n, x) + cF (x, p, x)− cF (x,−n, y)]
= kcF (x, p, x)
= k
p−1∑j=0
F (σj(x)).
Proposition 6.3.17. Let β 6= 0. Let z ∈ X be aperiodic. There exists an extremal atomic
eβF -conformal probability measure with support O(z) if, and only if,
M =∑y∈O(z)
e−βF(y) <∞, (6.8)
where
F(y) =m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(z)), (6.9)
with σm(y) = σn(z). In this case the measure is de�ned by
mz = M−1∑y∈O(z)
e−βF(y)δy. (6.10)
In particular, if z satis�es condition (6.8), we say z is β-summable.
Proof. First we prove F is well-de�ned. Let y ∈ O(z). There exists a unique k ∈ Z such
that (z, k, y) ∈ G. In fact, suppose there are k1, k2 ∈ Z satisfying (z, k1, y), (z, k2, y) ∈ G.
190
Then (z, k2 − k1, z) ∈ G. k2 − k1 = 0 from Lemma 6.2.10.
Let m,n ∈ N such that σm(y) = σn(z). Then (z, n−m, y) ∈ G. Hence k = n−m and
m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(z)) = −cF (z, k, y).
Therefore F(y) does not depend on the choice of m,n.
Let µ be an extremal atomic eβF -conformal probability measure with support O(z). Given
y ∈ O(z) there exist m,n ∈ N such that σm(y) = σn(z). Then by Proposition 6.3.6,
µ(y) = exp
(−β
(m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(z))
))µ(z) = e−βF(y)µ(z).
Since µ is a probability measure, it follows that
1 = µ(O(z)) =∑y∈O(z)
µ(y) =∑y∈O(z)
e−βF(y)µ(z).
Hence,
µ(z) =
∑y∈O(z)
e−βF(y)
−1
= M−1.
Therefore M <∞ and µ(y) = M−1e−βF(y) for y ∈ O(z).
Conversely, assume (6.8) holds. We prove that mz is an extremal eβF -conformal proba-
bility measure.
mz(O(z)) = M−1∑y∈O(z)
e−βF(y) = M−1M = 1,
thenmz is a probability measure. Now we showmz is eβF -conformal. Let y ∈ O(z), n,m ∈ N
191
such that σm(y) = σn(z), m ≥ 2. Then σm−1(σ(y)) = σn(z) and
F(σ(y)) =m−2∑j=0
F (σj(σ(y)))−n−1∑j=0
F (σj(z)) =m−1∑j=1
F (σj(y))−n−1∑j=0
F (σj(z))
= −F (y) +m−1∑j=0
F (σj(y))−n−1∑j=0
F (σj(z))
= −F (y) + F(y).
Then,
mz(σ(y)) = M−1e−βF(σ(y)) = M−1e−β[−F (y)+F(y)] = M−1eβF (y)e−βF(y) = eβF (y)mz(y).
Therefore mz is eβF -conformal. By de�nition of mz, its support is O(z). From Corollary
6.3.7, mz is extremal.
Lemma 6.3.18. Let X be a locally compact second countable Hausdor� topological space,
µ a Borel measure which is �nite on compact subsets of X. Given a local homeomorphism
σ : X → X, and a non-negative function f on X, µ is f -conformal if, and only if,
µ(σ(A)) =
∫A
f(x)dµ(x), (6.11)
for every open set A such that σ|A is injective.
Proof. The measure µ is Radon by Proposition 2.4.10. Assume µ is f -conformal, then (6.11)
holds by de�nition.
Conversely, suppose (6.11) holds. Let A be a measurable subset such that σ is injective
on A. First we assume there exists an open set U such that A ⊂ U and σ is injective on U .
Given an open set W including σ(A), there exists an open set W ′ = W ∩ σ(U) such that
σ(A) ⊂ W ′ ⊂ σ(U) and µ(W ′) ≤ µ(W ). Since µ is a Radon measure and W is arbitrary, we
192
have
µ(σ(A)) = infσ(A)⊂WW open
µ(W ) = infσ(A)⊂W ′⊂σ(U)
W ′ open
µ(W ′).
The function σ is injective on U , then for every open set W satisfying σ(A) ⊂ W ⊂ σ(U),
there is a unique open set V such that A ⊂ V ⊂ U and W = σ(V ). Clearly σ is injective on
each V . Then,
µ(σ(A)) = infA⊂V⊂UV open
µ(σ(V ))
= infA⊂V⊂UV open
∫V
f(x)dµ(x), by hypothesis,
=
∫A
f(x)dµ(x), by Lemma 2.4.13.
Now let A be an arbitrary measurable set A such that σ|A is injective. Since σ is a local
homeomorphism and X is second countable, there exists a countable open cover {Un}n∈N of
A such that σ is injective on each Un. De�ne A1 = A ∩ U1 and, for every n,
An+1 = A ∩ Un+1 \n⋃j=1
Aj.
Then A = ∪∞n=1An and the family {An}n∈N is disjoint. Moreover, An ⊂ Un for every n.
Then,
µ(σ(An)) =
∫An
f(x)dµ(x).
Since σ is injective on A, we have,
µ(σ(A)) =∞∑n=1
µ(σ(An)) =∞∑n=1
∫An
f(x)dµ(x) =
∫A
f(x)dµ(x).
193
Therefore µ is f -conformal.
Lemma 6.3.19. Let σ : X → X be a local homeomorphism. Let n ∈ N*, x ∈ X. Then
there exists an open neighborhood U of x such that for j = 1, . . . , n,
σ|σj−1(U) : σj−1(U)→ σj(U) is a homeomorphism and σj(U) is open. (6.12)
Proof. We prove this by induction. Let n = 1. There exists an open neighborhood U of x
such that σ(U) is open and σ|U : U → σ(U) is a homeomorphism. Then the result holds for
n = 1.
Now assume the result holds for n ≥ 1. Since σ is a local homeomorphism, there exists
an open neighborhood W of σn(x) in σn(U) such that σ(W ) is open and σ|W : W → σ(W )
is a homeomorphism.
Let V = σ|−nU (W ). V is an open neigborhood of x in U . Let j = 1, . . . , n. Note that
σ|jU : U → σj(U) is a homeomorphism. Since V ⊂ U , then σj(V ) is open and σj(V ) ⊂ σj(U).
Hence σ|σj−1(V ) : σj−1(V )→ σj(V ) is a homeomorphism for i = 1, . . . , n.
Note that σn+1(V ) = σ(σ|nU ◦ σ|−nU (W )) = σ(W ). Then σn+1(V ) is open. It follows that
σ|σn(V ) : σn(V )→ σn+1(V ) is a homeomorphism, therefore the result holds for n+ 1.
Remark 6.3.20. Let x, y ∈ X such that σn(x) = σm(y) for n,m ∈ N. Assume A is an
open neigborhood of x such that (6.12) holds for j = 1, . . . , n, replacing U by A. By the
same argument, suppose there exists B, an open neigborhood of y such that (6.12) holds for
j = 1, . . . ,m. Then we can assume without loss of generality that σn(A) = σm(B).
In fact, let V = σn(A) ∩ σm(B). V is an open neigborhood of σn(x). Let A0 = σ|−nA (V ),
B0 = σ|−mB (V ). Clearly A0 and B0 are open neigborhoods of x, y, respectively. Then, for
j = 1, . . . , n, (6.12) holds for A0, since σj−1(A0) ⊂ σj−1(A) and σj(A0) ⊂ σj(A) are open
sets. Analogously (6.12) holds for B0.
Moreover,
σn(A0) = σn(σ|−nA (V )) = V = σm(σ|−mB (V )) = σm(B0).
194
Theorem 6.3.21. Let β ∈ R. A measure µ on G(0) is eβF -conformal if, and only if, µ is
quasi-invariant with Radon-Nikodym derivative e−βcF .
Proof. Assume µ is eβF -conformal.
Let U ⊂ G be an open bisection. Let T : r(U)→ s(U) be the homeomorphism de�ned by
T = s|U ◦ r|−1U . Given y ∈ s(U), let hy = s|−1
U (y) = (x, k, y). There are A,B ⊂ X open sets,
n,m ∈ N such that hy ∈ Un,mA,B ⊂ U .
By Lemma 6.3.19 there exists an open neigborhood of x, A0 ⊂ A, such that for j =
1, . . . , n,
σ|σj−1(A0) : σj−1(A0)→ σj(A0) is a homeomorphism and σj(A0) is open.
By the same argument, there exists an open neigborhood B0 of y, with B0 ⊂ B, such that
for j = 1, . . . ,m,
σ|σj−1(B0) : σj−1(B0)→ σj(B0) is a homeomorphism and σj(B0) is open.
We can assume σn(A0) = σm(B0) without loss of generality. Then hy ∈ Un,mA0,B0⊂ Un,mA,B .
Note that s(Un,mA0,B0) = B0 and r(Un,mA0,B0
) = A0. In fact, since σ|nA0, σ|mB0
are homeomor-
phisms, it follows that
Un,mA0,B0= {(x′, n−m, y′) : x′ ∈ A0, y
′ ∈ B0, σn(x′) = σm(y′)}
= {(σ|−nA0(σ|mB0
(y′)), n−m, y′) : y′ ∈ B0}
= {(x′, n−m,σ|−mB0(σ|nA0
(x′))) : x′ ∈ A0}.
Moreover, T |A0 = σ|−mB0◦ σ|nA0
. Note that
σ|−mB0◦ σ|nA0
= σ|−1B0◦ σ|−1
σ(B0) ◦ . . . ◦ σ|−1σm−1(B0) ◦ σ|
nA0
= σ|−mB0◦ σ|σn−1(A0) ◦ σ|σn−2(A0) ◦ . . . σ|σ1(A0) ◦ σ|A0 .
195
Given j = 0, . . . , n+m, let
Tj =
σ|−mB0◦ σ|n−j
σj(A0), if 0 ≤ j ≤ n,
σ|−m−n+jB0
, if n ≤ j ≤ n+m.
Then T0 = T |A0 and Tn+m = id|B0 . Note that Tj : σj(A0) → B0 if j ≤ n and Tj :
σn+m−j(B0)→ B0 if j ≥ n. Moreover,
• if 0 ≤ j ≤ n− 1,
Tj = σ|−mB0◦ σ|n−j
σj(A0)= σ|−mB0
◦ σ|n−j−1σj+1(A0)
◦ σ|σj(A0) = Tj+1 ◦ σ|σj(A0).
Then Tj+1 = Tj ◦ σ|−1σj(A0)
. Also,
T−1j y = σ|j−n
σj(A0)◦ σ|mB0
(y)
= (σ|n−jσj(A0)
)−1 ◦ σ|nA0(x)
= (σ|nA0◦ σ|−jA0
)−1 ◦ σ|nA0(x)
= σ|jA0◦ σ|−nA0
◦ σ|nA0(x)
= σ|jA0(x).
Let B1 ⊂ B0 be measurable. Then,
Tj+1∗µ(B1) = Tj∗σ|−1σj(A0)∗
µ(B1)
= σ|−1σj(A0)∗
µ(Tj−1(B1))
= µ(σ(Tj−1(B1))).
Tj−1(B1) is measurable, since Tj is continuous. Since Tj
−1(B1) ⊂ σj(A0), σ is injective
196
on σj(A0) and µ is eβF -conformal, we have
µ(σ(Tj−1(B1))) =
∫T−1j (B0)
eβF (u)dµ(u)
=
∫T−1j (B0)
eβF (T−1j Tju)dµ(u)
=
∫B0
eβF (T−1j u)dTj∗µ(u) by (2.2) on page 15
ThenTj+1∗µ
Tj∗µ(y) = eβF (T−1
j y) = eβF (σj(x)).
Therefore,
Tj∗µ
Tj+1∗µ(y) = e−βF (σj(x)). (6.13)
• if n ≤ j ≤ n+m− 1,
Tj = σ|−m−n+jB0
= (σ|m+n−jB0
)−1
= (σ|σm+n−j−1(B0) ◦ σ|m+n−j−1B0
)−1
= (σ|m+n−j−1B0
)−1 ◦ σ|−1σm+n−j−1(B0)
= (σ|−m−n+j+1B0
)−1 ◦ σ|−1σm+n−j−1(B0)
= Tj+1 ◦ σ|−1σm+n−j−1(B0)
.
Moreover,
T−1j y = σ|m+n−j
B0(y) = σm+n−j(y).
197
Let B1 ⊂ B0 measurable. Then,
Tj∗µ(B1) = Tj+1∗σ|−1σm+n−j−1(B0)∗
µ(B1)
= σ|−1σm+n−j−1(B0)∗
µ(T−1j+1(B1))
= µ(σ(T−1j+1(B1))).
Note that T−1j+1(B1) is measurable by continuity of Tj+1. Since
Tj+1−1(B1) ⊂ σm+n−j−1(B0),
σ is injective on σm+n−j−1(B0) and µ is eβF -conformal, we have
µ(σ(T−1j+1(B1))) =
∫T−1j+1(B1)
eβF (u)dµ(u)
=
∫T−1j+1(B1)
eβF (T−1j+1Tj+1u)dµ(u)
=
∫B1
eβF (T−1j+1u)dTj+1∗µ(u) by (2.2) on page 15.
Then,
dTj∗µ
dTj+1∗µ(y) = eβF (T−1
j+1(y)) = eσm+n−j−1(y). (6.14)
Therefore,
dT∗µ
dµ(y) =
dT0∗µ
dTm+n∗µ(y)
=dT0∗µ
dT1∗µ(y)
dT1∗µ
dT2∗µ(y) . . .
dTn+m−1∗µ
dTn+m∗µ(y)
=
(n−1∏j=0
dTj∗µ
dTj+1∗µ(y)
)(n+m−1∏j=n
dTj∗µ
dTj+1∗µ(y)
)
198
=
(n−1∏j=0
e−βF (σj(x))
)(n+m−1∏j=n
eβF (σm+n−j−1(y))
)from (6.13) and (6.14),
=
(n−1∏j=0
e−βF (σj(x))
)(m−1∏j=0
eβF (σm−j−1(y))
).
Making the change of variables j 7→ m− j − 1 in the second product, we have
=
(n−1∏j=0
e−βF (σj(x))
)(m−1∏j=0
eβF (σj(y))
)
= exp
(−β
n−1∑j=0
F (σj(x)) + β
m−1∑j=0
F (σj(y))
)
= exp
(−β
(n−1∑j=0
F (σj(x))−m−1∑j=0
F (σj(y))
))
= e−βcF (x,k,y)
= e−βcF (hy).
Since y ∈ s(U) is arbitrary, the equality holds for every y ∈ s(U). U is any open bisection.
Therefore, by Lemma 5.3.8, µ is quasi-invariant with Radon-Nikodym derivative e−βcF .
Conversely, assume µ is quasi-invariant with Radon-Nikodym derivative e−βcF .
Let A ⊂ X be an open set such that σ|A is invertible. U0,1σ(A),A is an open bisection. In
fact,
U0,1σ(A),A = {(x,−1, y) : x ∈ σ(A), y ∈ A, x = σ(y)}
= {(σ(y),−1, y) : y ∈ A}.
Since σ is injective on A, it follows that r, s are injective on U0,1σ(A),A. Moreover, A = s(U0,1
σ(A),A),
σ(A) = r(U0,1σ(A),A), and T : σ(A)→ A is given by T = σ|−1
A . From Lemma 5.3.8,
dT∗µ
dµ(y) = e−βcF (hy)
= exp (−βcF (σ(y),−1, y))
199
= exp
(−β
(0∑j=0
F (σj(σ(y))−1∑j=0
F (σj(y))
))
= exp (−β [F (σ(y))− F (y)− F (σ(y))])
= eβF (y).
Hence,
µ(σ(A)) = µ(σ|A(A)) = µ(T−1(A)) = T∗µ(A) =
∫A
eβF (y)dµ(y).
It follows from Lemma 6.3.18 that µ is eβF -conformal.
Remark 6.3.22. Let β 6= 0. Given a eβF -conformal measure, it follows from Theorem 6.3.21
that µ is quasi-invariant with Radon-Nikodym derivative e−βcF . From Remark 4.1.9 we have
−βcF (g) = 0 for µ-a.e. x ∈ G(0) and all g ∈ Gxx, (6.15)
then
cF (g) = 0 for µ-a.e. x ∈ G(0) and all g ∈ Gxx, (6.16)
We can show this fact for extremal eβF -conformal probability measures using the properties
of G when the set of periodic points in X is countable.
Let µ be an extremal eβF -conformal probability measure. From Lemma 6.3.3 and Lemma
6.3.9, the measure falls in one of these cases: µ is continuous; µ is purely atomic and
supported on a periodic orbit; or µ is purely atomic and supported on an aperiodic orbit.
Suppose µ is continuous. Since the set of periodic points is countable, it follows that for
µ-a.e. y, Gyy = {(y, 0, y)}. However, cF (y, 0, y) = 0. Then the result holds for µ continuous.
If µ is purely atomic and supported on an aperiodic orbit O(z), then Gyy = {(y, 0, y)} for
200
every y ∈ O(z). Then (6.16) holds.
If µ is purely atomic and supported on a periodic orbit O(x) for a periodic point x with
minimum period p, it follows from 6.3.8 that
p−1∑i=0
F (σi(x)) = 0.
Given y ∈ O(x), Gyy = {(y, kp, y) : k ∈ Z} from Lemma 6.2.10. From Lemma 6.3.16, we have
cF (y, kp, y) = 0 for every k ∈ Z. Then (6.16) holds.
Hence, the result follows for every extremal eβF -conformal probability measure.
6.4 KMS States on the Renault-Deaconu Groupoid
In this section we �nd all extremal KMS states on C*(G) using Neshveyev's Theorems.
Since the continuous R-valued 1-cocycle on G is given by cF as in De�nition 6.3.11, we �x
the dynamics τ on C*(G), given by τt(f)(g) = eitcF (g)f(g).
We will show that all extremal KMS states are φm, φmz , φλx, de�ned below.
Let m be an extremal eβF -conformal non-atomic probability measure on X. We de�ne
φm by
φm(f) =
∫X
f(y, 0, y)dm(y),
where f ∈ Cc(G).
Given an aperiodic β-summable point z, by Proposition 6.3.17 mz is an extremal eβF -
conformal probability measure. We de�ne φmz by
φmz(f) =
∫X
f(y, 0, y)dmz(y),
with f ∈ Cc(G).
201
Given a periodic point x with period p with x satisfying the conditions of Proposition
6.3.10, λ ∈ C with |λ| = 1, we de�ne φλx by
φλx(f) =
∫X
∑k∈Z
λkf(y, kp, y)dmx(y),
for f ∈ Cc(G).
Lemma 6.4.1. Let x ∈ G(0). Given g1, g2 ∈ Gxx , ug1 · ug2 = ug1g2 . As a consequence, C*(Gxx)
is a commutative C*-algebra with identity ux.
Proof. Let g ∈ G, then
(ug1 · ug2)(g) =∑ab=g
ug1(a)ug2(b)
=
1 if g1g2 = g
0 otherwise
= ug1g2(g).
Lemma 6.4.2. Let x ∈ X, y ∈ O(x), h ∈ Gxy . There exists an ∗-isomorphism P : C*(Gyy )→
C*(Gxx) given by P (ug) = uhgh−1 . Moreover, P is an isometry.
Proof. Let P : C*(Gyy ) → C*(Gxx) be the linear map de�ned by P (ug) = uhgh−1 . Note that
P is invertible with inverse given by P−1 : C*(Gxx)→ C*(Gyy ), de�ned by P−1(ug) = uh−1gh.
First we show that P is a homomorphism. Given g1, g2, g ∈ Gyy ,
P (ug1 · ug2) = P (ug1g2) = uhg1g2h−1 = uhg1h−1hg2h−1 = uhg1h−1 · uhg2h−1 = P (ug1) · P (ug2),
P (ug*) = P (ug−1) = uhg−1h−1 = u(hgh−1)−1 = u(hgh−1)* = P (ug)*.
202
Since P is linear, we have that P is a ∗-homomorphism. So is P−1 by the same arguments.
Then P is an ∗-isomorphism between C*-algebras and therefore P is an isometry.
Remark 6.4.3. Lemma 6.4.2 can be generalized for a locally compact Hausdor� second
countable étale groupoid G. Given x, y ∈ G(0) such that Gxy 6= ∅, there exists an ∗-
isomorphism P :: C*(Gyy)→ C(Gx
x) given by P (ug) = uhgh−1 .
Proposition 6.4.4. Suppose the set of periodic points in X is countable. Let β ∈ R \ {0}.
Let φ be the KMSβ-state corresponding to the pair (µ, {ϕy}y). Then φ is extremal if and
only if µ is an extremal eβF -conformal measure and ϕy is a character for µ-a.e. y.
Proof. It follows from Theorem 5.3.10 that µ is quasi-invariant with Radon-Nikodym deriva-
tive e−βcF . It follows from Theorem 6.3.21 that µ is eβF -conformal.
Assume φ is extremal.
• Suppose µ is not extremal, then µ = tµ1 + (1 − t)µ2 where µ1, µ2 are eβF -conformal
probability measures, 0 < t < 1 and µ1 6= µ2.
Then µ1 and µ2 are quasi-invariant with Radon-Nikodym derivative e−βcF . Since
µ1, µ2 � µ, conditions (ii) and (iii) of Theorem 5.3.10 are satis�ed. Then (µ1, {ϕy}y)),
(µ2, {ϕy}y)) correspond to the KMS states φ1, φ2, respectively. Note that φ1 6= φ2.
Then
tφ1(f) + (1− t)φ2(f) = t
∫X
∑g∈Gxx
f(g)ϕx(ug)dµ1(x) + (1− t)∫X
∑g∈Gxx
f(g)ϕx(ug)dµ2(x)
=
∫X
∑g∈Gxx
f(g)ϕx(ug)d(tµ1 + (1− t)µ2)(x)
=
∫X
∑g∈Gxx
f(g)ϕx(ug)dµ(x)
= φ(f).
Then φ is not extremal. Contradiction.
203
• Suppose µ is extremal and non-atomic. Let I be the set aperiodic points in X. Then
µ(I) = 1 by Lemma 6.2.9. Given y ∈ I, Gyy = {y} by Lemma 6.2.10. Then every
element in C*(Gyy ) in the form auy, where a ∈ C. In addition ϕy(uy) = 1 because ϕ is
a state and uy is the unit of C*(Gyy ). Therefore ϕy is a character.
• Assume µ is extremal and atomic. Then there exists an orbit I with µ(I) = 1. Suppose
there is x ∈ I such that ϕx is not a character. Then, by Lemma 6.4.1, there are states
ϕ(1)x , ϕ
(2)x , t ∈ (0, 1) such that ϕ(1)
x 6= ϕ(2)x and ϕx = tϕ
(1)x + (1− t)ϕ(2)
x .
De�ne the �eld of states {ϕ(1)y }y, {ϕ(2)
y }y by
ϕ(i)y (ug) =
ϕy(ug) if y /∈ I
ϕ(i)x (uhgh−1) if y ∈ I, and h ∈ Gx
y arbitrary,
for i = 1, 2. Note that hgh−1 does not depend on the choice of g. In fact, given
y ∈ O(x), g ∈ Gyy , h ∈ Gxy , there exists k, l ∈ N such that g = (y, k, y) and h = (x, l, y).
Then hgh−1 = (x, k, x).
It is clear that ϕ(i)y is a state for y /∈ I. We will show that ϕ(i)
y is also a state when
y ∈ I. Given h ∈ Gxy , let P : C*(Gyy )→ C*(Gxx) be de�ned by P (ug) = uhgh−1 for every
g ∈ Gyy . It follows from Lemma 6.4.2 that P is an isometry. Moreover, by de�nition of
ϕ(i)y , we have ϕ(i)
y = ϕ(i)x ◦ P . ϕ(i)
y is linear and bounded because
‖ϕ(i)y ‖ ≤ ‖ϕ(i)
x ‖‖P‖ = ‖ϕ(i)x ‖.
Since ϕ(i)x = ϕ
(i)y ◦ P−1, we can show analogously that ‖ϕ(i)
x ‖ ≤ ‖ϕ(i)y ‖. Then ‖ϕ(i)
y ‖ =
‖ϕ(i)x ‖ = 1. Moreover,
ϕ(i)y (uy) = ϕ(i)
x (uhyh−1) = ϕ(i)x (uhh−1) = ϕ(i)
x (ux) = 1.
Note that uy is the unit of C*(Gyy ). It follows from Theorem 5.1.2 that ϕ(i)y is a state.
204
Since µ is atomic, {ϕ(i)y }y de�nes a µ-measurable �eld of states by Remark 5.2.5. For
i = 1, 2, let φ(i) be the state de�ned by (µ, {ϕ(i)y }y). We will show that φ(i) is a KMS
state. In order to prove this, we have show that properties (i)-(iii) in Theorem 5.3.10
hold.
We already know that µ is quasi-invariant with Radon-Nikodym derivative e−βcF , so
property (i) holds.
We will prove property (ii). Let y ∈ I, then there exists k ∈ Gxy . Let h ∈ Gy. Then
r(k) ∈ I and kh−1 ∈ Gxr(h). For every g ∈ Gyy , we have
ϕ(i)r(h)(uhgh−1) = ϕ(i)
x (ukh−1(hgh−1)hk−1), by de�nition of ϕ(i)r(h),
= ϕ(i)x (ukgk−1)
= ϕ(i)y (uy), by de�nition of ϕ(i)
y .
Then property (ii) holds.
For µ-a.e. y ∈ X, all g ∈ Gyy , we have cF (g) = 0 by Remark 6.3.22. Then property (iii)
holds.
By de�nition of ϕ(i)y , ϕy(ug) = tϕ
(1)y (ug) + (1− t)ϕ(2)
y (ug) if y /∈ I. If y ∈ I, there exists
h ∈ Gxy . Then,
ϕy(ug) = ϕr(h)(uhgh−1)
= ϕx(uhgh−1)
= tϕ(1)x (uhgh−1) + (1− t)ϕ(2)
x (uhgh−1)
= tϕ(1)y (ug) + (1− t)ϕ(2)
y (ug).
Hence ϕy = tϕ(1)y + (1− t)ϕ(2)
y and ϕ(1)y 6= ϕ
(2)y for every y ∈ I. By de�nition of {ϕ(i)
y }y,
(µ, {ϕ(i)y }y) de�nes a state φ(i), i = 1, 2. Note that φ(1) 6= φ(2).
205
Then, for f ∈ Cc(G),
φ(f) =
∫X
∑g∈Gyy
f(g)ϕy(ug)dµ(y)
=
∫X
∑g∈Gyy
f(g)[tϕ(1)
y (ug) + (1− t)ϕ(2)y (ug)
]dµ(y)
= t
∫X
∑g∈Gyy
f(g)ϕ(1)y dµ(y) + (1− t)
∫X
∑g∈Gyy
f(g)ϕ(2)y dµ(y)
= tφ(1)(f) + (1− t)φ(2)(f).
Then φ is not extremal. Contradiction. Therefore µ is extremal and ϕy is a character
for µ-a.e. y.
Conversely, suppose µ is extremal and ϕy is a character for µ-a.e. y. Suppose there exist
KMS states φ(1), φ(2), t ∈ (0, 1) such that φ = tφ(1) + (1− t)φ(2) and each φ(i) corresponds to
the pair (µi, {ϕ(i)y }y). Since X is clopen G, we have for every f ∈ Cc(X),
φ(f) = tφ(1)(f) + (1− t)φ(2)(f)∫X
f(x)dµ(x) =
∫X
f(x)dµ1(x) + (1− t)∫X
f(x)dµ2(x).
Then µ = tµ1 + (1 − t)µ2. Since µ is extremal, we have µ1 = µ2 = µ. Then, for every
f ∈ Cc(G),
φ(f) = tφ(1)(f) + (1− t)φ(2)(f)
= t
∫X
∑g∈Gxx
f(g)ϕ(1)x (ug)dµ(x) + (1− t)
∫X
∑g∈Gxx
f(g)ϕ(2)x (ug)dµ(x)
=
∫X
∑g∈Gxx
f(g)[tϕ(1)
x (ug) + (1− t)ϕ(2)x (ug)
]dµ(x).
Each tϕ(1)x +(1−t)ϕ(2)
x is a state on C*(Gxx). Moreover, {tϕ(1)x +(1−t)ϕ(2)
x }x is a µ-measurable
�eld of states. Then the pair (µ, {tϕ(1)x +(1−t)ϕ(2)
x }x) also de�nes φ. It follows from Theorem
206
5.2.9 that
ϕx = tϕ(1)x + (1− t)ϕ(2)
x , for µ-a.e. x
Since ϕx is a character µ-a.e., it follows that ϕx = ϕ(1)x = ϕ
(2)x for µ-a.e. x. Then, by Theorem
5.2.9, φ = φ(1) = φ(2). Therefore φ is extremal.
Lemma 6.4.5. Let y ∈ X be aperiodic. There is a unique state ϕy on C*(Gyy ). In particular,
ϕy is a character.
Proof. Since Gyy = {y}, it follows that C*(Gyy ) is isomorphic to C. Hence there is a unique
state on C*(Gyy ), which is a character.
Lemma 6.4.6. Let y ∈ X such that O(y) is periodic with period p. Let λ ∈ C such that
|λ| = 1. De�ne the linear functional ϕλy on C*(Gyy ) by ϕλy(u(y,kp,y)) = λk. Then ϕλy is a
character on C*(Gyy ). In fact, ϕλy are the only characters de�ned on C*(Gyy ).
Proof. Note that we can identify Gyy with Z by the isomorphism (y, kp, y) 7→ k. Moreover,
C*(Z) is isomorphic to C(S1), the set of continuous functions on the complex unit circle. In
fact, for k ∈ Z, let uk : Z→ C be de�ned by
uk(l) =
1 if l = k
0 otherwise,
and let pk : S1 → C be de�ned by pk(z) = zk.
There exists an isomorphism from C*(Z) to C(S1) given by uk 7→ pk. Since {pk}k∈Zgenerates the commutative C*-algebra C(S1), it follows that all characters of C(S1) corre-
spond to elements on the unit circle S1. Hence, each character ϕ on C*(Gyy ) corresponds to
a character λ on C(S1) such that
ϕ(u(y,kp,y)) = pk(λ) = λk.
207
Now we describe all extremal KMSβ-states on C*(G). Recall that the dynamics on C*(G)
is given by τt(f)(g) = eitcF (g)f(g).
Theorem 6.4.7. [26, Theorem 2.2] Let β ∈ R \ {0}. Assume that the periodic points of σ
are countable. The extremal KMSβ-states for τ are
1. States φm, where m is an extremal and continuous (non-atomic) eβF -conformal Borel
probability measure on X;
2. The states φλx, where λ ∈ C, |λ| = 1 and x is periodic with minimum period p, such that
p−1∑j=0
F (σj(x)) = 0 and∞∑n=1
∑y∈Yn
exp
(−β
n−1∑j=0
F (σj(y))
)<∞; (6.17)
3. The states φmz where z is aperiodic and β-summable.
Proof. Let φ be an extremal KMSβ-state corresponding to the pair (µ, {ϕy}y). From Propo-
sition 6.4.4, µ is an extremal eβF -conformal probability measure, and ϕy is a character for
µ-a.e. y.
It follows from Theorem 6.3.21 that µ is quasi-invariant with Radon-Nikodym derivative
e−βcF . Since µ is extremal, µ is either atomic or non-atomic.
(i) Suppose µ = m is non-atomic or µ = mz for z aperiodic.
If µ = mz, then z is β-summable by Proposition 6.3.17.
Let φ be an extremal KMSβ-state corresponding to the pair (m, {ϕy}y). De�ne I by
• the set of aperiodic points, i.e. points X whose orbits are aperiodic, if µ is non-
atomic. From Lemma 6.2.9 we have m(I) = 1.
• I = O(z) if µ = mz.
208
Then µ(I) = 1 and y ∈ I is aperiodic for every y ∈ I. Then by Lemma 6.4.5, ϕy
de�ned by ϕy(uy) = 1 is the unique character de�ned on C*(Gyy ).
Hence φ is de�ned by
φ(f) =
∫X
∑g∈Gyy
f(g)dµ(y) =
∫I
f(y, 0, y)dµ(y) =
∫X
f(y, 0, y)dµ(y).
Then φ = φm if µ = m and φ = φmz if µ = mz.
(ii) Suppose µ = mx with x periodic with minimum period p.
From Proposition 6.3.10 conditions (6.17) must hold.
Let I = O(x). Let φ be an extremal KMSβ-state with corresponding pair (mx, {ϕy}y).
It follows from Proposition 6.4.4 and Lemma 6.4.6 that ϕx = ϕλx for some λ ∈ S1.
Let y ∈ O(x), then there exists h = (x, l, y) ∈ G for some l ∈ Z. Let g ∈ Gyy ,
then g = (y, kp, y) for some k ∈ Z by Lemma 6.2.10. Then applying property (iii) of
Theorem 5.3.10,
ϕy(ug) = ϕr(h)(uhgh−1)
= ϕx(u(x,l,y)(y,kp,y)(y,−l,x))
= ϕx(u(x,kp,x))
= ϕλx(u(x,kp,x))
= λk
= ϕλy(u(y,kp,y))
= ϕλy(ug).
Then ϕy = ϕλy for every y ∈ I. Then for f ∈ Cc(G),
φ(f) =
∫X
∑g∈Gyy
f(g)ϕy(ug)dmx(y)
209
=
∫I
∑g∈Gyy
f(g)ϕy(ug)dmx(y)
=
∫I
∑g∈Gyy
f(g)ϕλy(ug)dmx(y)
=
∫I
∑k∈Z
f(y, kp, y)ϕλy(uy,kp,y)dmx(y)
=
∫I
∑k∈Z
f(y, kp, y)λkdmx(y)
=
∫X
∑k∈Z
f(y, kp, y)λkdmx(y)
= φλx(f).
Hence every extremal KMS state has the form φm, φmz or φλx.
Now we prove that φm, φmz , φλx satisfying the conditions of the theorem always de�ne
extremal KMS states.
Note that, for an extremal eβF -conformal probability measure µ on X, it follows from
Remark 6.3.22 that for µ-a.e. x ∈ X, all g ∈ Gxx , we have cF (g) = 0. Then in order to
prove that a state de�ned by (µ, {ϕy}y) is KMSβ, we only need to show that property
(ii) of Theorem 5.3.10 holds.
• φm
Let m be an extremal and continuous eβF -conformal Borel probability measure
on X. Let {ϕy}y∈X be a family of states ϕy on C*(Gyy ).
Let I be the set of aperiodic points. m(X \I) = 0 by Lemma 6.2.9. Let f ∈ Cc(G).
Given y ∈ I, Gyy = {y}. Then
∑g∈Gyy
f(g)ϕ(ug) = f(y, 0, y).
Thus {ϕy}y is a m-measurable �eld of states. Moreover, it follows from Lemma
6.4.5 that ϕy is a character for every y ∈ I.
210
Now we show that (m, {ϕy}y) de�nes a KMS state. Let y ∈ I, then its orbit is
aperiodic. Let h ∈ Gy. g = y is the unique element in Gyy . Then
ϕy(ug) = ϕy(uy) = 1, and
ϕr(h)(uhgh−1) = ϕr(h)(uhyh−1) = ϕr(h)(uhh−1) = ϕr(h)(ur(h)) = 1.
Then property (ii) of Theorem 5.3.10 and, therefore, φm is a KMS state. Since m
is extremal and ϕy is extremal for m-a.e. y, it follows from Proposition 6.4.4 that
φm is extremal.
• φmz
This case is analogous to the proof for m continuous if we de�ne I = O(z). Note
that mz is de�ned only if z is β-summable by Proposition 6.3.17.
• φλx
Let x ∈ X be periodic with minimum period p such that (6.17) holds. Then
the extremal eβF -conformal probability measure mx supported on O(x) exists by
Proposition 6.3.10. Let {ϕy}y∈X be a family os states such that ϕy = ϕλy for every
y ∈ O(x). Then {ϕy}y∈X de�nes a mx-measurable �eld of states. Moreover, ϕy is
a character for µ-a.e. y by Lemma 6.4.6.
Now we show that (mx, {ϕy}y) de�nes a KMS state, so we will show that property
(ii) of Theorem 5.3.10 holds. Let h ∈ Gx. There is l ∈ Z, such that h = (r(h), l, x).
Let g ∈ Gxx . Then there exists k ∈ Z such that g = (x, kp, x). Hence,
ϕx(ug) = ϕx(u(x,kp,x)) = λk, and
ϕr(h)(uhgh−1) = ϕr(h)(u(r(h),l,x)(x,kp,x)(x,−l,r(h))) = ϕr(h)(u(r(h),kp,r(h))) = λk.
Then (mx, {ϕy}y) de�nes a KMS state. Moreover, this state is extremal by Propo-
211
sition 6.4.4. Given f ∈ Cc(G), this state is de�ned by
∫X
∑g∈Gyy
f(g)ϕx(ug)dmx =
∫I
∑g∈Gyy
f(g)ϕλx(ug)dmx
=
∫I
∑k∈Z
f(y, kp, y)ϕλx(u(y,kp,y))dmx
=
∫I
∑k∈Z
f(y, kp, y)λkdmx
=
∫X
∑k∈Z
f(y, kp, y)λkdmx
= φλx(f).
Therefore the extremal KMS states on C*(G) are precisely φm, φmz and φλx.
Corollary 6.4.8. Let β ∈ R \ {0}. Assume F is positive on X. Then there exists a
correspondence between the extremal KMSβ-states and the extremal and continuous eβF -
conformal probability measures on X.
Proof. Let m be an extremal and continuous eβF -conformal probability measure on X. Then
φm is a KMSβ-state on C*(G) by Theorem 6.4.7.
Conversely, suppose φ is an extremal KMSβ-state. We will see that φ is not in cases 2
and 3 of theorem Theorem 6.4.7:
2. Suppose φ = φλx for some λ ∈ C with |λ| = 1 and x ∈ X periodic with minimum period
p > 0. Then
p−1∑j=0
F (σj(x)) = 0
Contradiction, since F assumes positive values.
3. Suppose φ = φmz for some z ∈ X aperiodic.
212
De�ne the sequence {yn}n∈N by yn = σn(z). Then
F(yn) = −n−1∑j=0
F (σj(z)) = −n−1∑j=0
F (yj).
Then, for each n, F(yn+1) < F(yn). Hence, e−βF(yn+1) > e−βF(yn). So,
∞∑n=0
e−βF(yn) =∞.
Then
∑y∈O(z)
e−βF(y) =∞.
Therefore z is not β-summable. Contradiction, since we assumed φ = φmz .
Therefore φ = φm for an extremal and continuous eβF -conformal Borel probability measure
on X.
213
Chapter 7
Concluding Remarks
In this thesis we described KMS states on groupoid C*-algebras for locally compact Hausdor�
second countable étale groupoids using Neshveyev's theorems. Then we studied a theorem
due to Thomsen which characterizes the extremal KMS states for the Renault-Deaconu
groupoid.
Neshveyev's theorem proved to be a useful tool to �nd an explicit formula for all KMS
states. However, the proof of this theorem depends on the fact that the groupoid is étale.
When the groupoid is not étale, it is possible to de�ne groupoid C*-algebras which are similar
to the C*-algebras studied in this thesis. For instance, [10] de�nes the crossed product of
a C*-algebra by a groupoid G where this groupoid is locally compact Hausdor� and is
endowed with a Haar system. This space is a closure of a space of continuous and compactly
supported functions f on X such that, for every x ∈ X, f(x) is an element of a C*-algebra.
The operations in this space are analogous to the operations in a full C*-algebra. So, one
challenge is to extend Neshveyev's theorem to non-étale groupoids.
Christensen [5] generalized the theorem of Neshveyev to describe KMS weights. This
result can be applied to describe KMS weights for di�erent groupoids.
214
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