# The cone of lower semicontinuous traces on a C*-algebra

George. A. Elliott and Leonel Robert and Luis Santiago  George A. Elliott,
Department of Mathematics, University of Toronto, Toronto, Canada M5S 2E4
<EMAIL_ADDRESS>Leonel Robert, The Fields Institute, Toronto, Canada
M5T 3J1<EMAIL_ADDRESS>Luis Santiago, Department of Mathematics,
University of Toronto, Toronto, Canada M5S 2E4<EMAIL_ADDRESS>
###### Abstract.

The cone of lower semicontinuous traces is studied with a view to its use as
an invariant. Its properties include compactness, Hausdorffness, and
continuity with respect to inductive limits. A suitable notion of dual cone is
given. The cone of lower semicontinuous 2-quasitraces on a (non-exact)
C*-algebra is considered as well. These results are applied to the study of
the Cuntz semigroup. It is shown that if a C*-algebra absorbs the Jiang-Su
algebra, then the subsemigroup of its Cuntz semigroup consisting of the purely
non-compact elements is isomorphic to the dual cone of the cone of lower
semicontinuous 2-quasitraces. This yields a computation of the Cuntz semigroup
for the following two classes of C*-algebras: C*-algebras that absorb the
Jiang-Su algebra and have no non-zero simple subquotients, and simple
C*-algebras that absorb the Jiang-Su algebra.

##  1\. Introduction

The most standard invariants in the classification of nuclear, simple,
C*-algebras are their K-groups and their traces. The traces are assumed to be
bounded in the unital case, and lower semicontinuous and densely finite in the
non-unital case. If one has in mind the classification of non-simple
C*-algebras, it is clear that these two kinds of traces will not suffice and a
broader class should be considered. In this paper we study the properties of
the cone of all lower semicontinuous traces on a C*-algebra, with the purpose
of applying our results to questions in the classification of non-simple
C*-algebras. We also consider lower semicontinuous 2-quasitraces, since they
appear naturally as functionals on the Cuntz semigroup of the algebra. If the
algebra is exact (this is the case that we are mostly concerned with in the
classification program), then lower semicontinuous traces and 2-quasitraces
coincide. However, some of our considerations apply equally to traces and
2-quasitraces without assuming exactness of the C*-algebra. Thus, we treat
both classes for arbitrary C*-algebras.

Recall that a trace on a C*-algebra $A$ is a linear map $\tau$ on the positive
elements of $A$, with values in $[0,\infty]$, that vanishes at 0 and satisfies
the trace identity $\tau(xx^{*})=\tau(x^{*}x)$ (see [dixmier]). Every trace on
$A$ extends to a trace on $A\otimes\mathcal{K}$. A 2-quasitrace on $A$ is a
map on $(A\otimes\mathcal{K})^{+}$, with values in $[0,\infty]$, that vanishes
at 0, satisfies the trace identity, and is linear on pairs of positive
elements that commute (see [kirchberg-blanchard, Definition 2.22] and
[kirchberg-blanchard, Proposition 2.24]). If a trace or 2-quasitrace is lower
semicontinuous, then it is invariant under approximately inner automorphisms.
This makes the cones of lower semicontinuous traces and 2-quasitraces on a
C*-algebra $A$—let us denote these cones by $\mathrm{T}(A)$ and
$\mathrm{QT}_{2}(A)$—natural classification invariants associated to $A$.

In Section 2 of this paper we study the basic properties of the cone
$\mathrm{T}(A)$. Even though various classes of traces have been studied in
the past (e.g., [dixmier], [pedersen], [perdrizet]), we have found no
bibliographic source for the properties of $\mathrm{T}(A)$. On the other hand,
the subcone of $\mathrm{T}(A)$ consisting of densely finite traces has been
studied more thoroughly (e.g., in [pedersen] and [perdrizet]). We will show
that some well known properties of the cone of densely finite traces persist
as properties of $\mathrm{T}(A)$. Furthermore, some properties appear that are
not present in the cone of densely finite traces, notably, the compactness of
$\mathrm{T}(A)$ (in a suitable topology).

In Section 3 we turn our attention to $\mathrm{QT}_{2}(A)$. By results of
Blanchard and Kirchberg (see [kirchberg-blanchard, Proposition 2.24]), which
in turn extend work by Cuntz, Blackadar, Handelman, and Goodearl, the lower
semicontinuous 2-quasitraces are in bijective correspondence with the
additive, order-preserving, extended positive real-valued maps on the Cuntz
semigroup that vanish at 0 and preserve the suprema of increasing
sequences—which henceforth we shall just call functionals. Thus, we may think
of $\mathrm{QT}_{2}(A)$ as the cone of functionals on the Cuntz semigroup.

Section 4 contains the description of suitable dual cones for
$\mathrm{QT}_{2}(A)$ and $\mathrm{T}(A)$. The main results of this section,
Theorem 5.7 and Theorem 5.12, relate the dual cones of $\mathrm{QT}_{2}(A)$
and $\mathrm{T}(A)$ with the functions that the positive elements of $A$
induce on $\mathrm{QT}_{2}(A)$ and $\mathrm{T}(A)$.

The last section contains applications of our results to understanding the
structure of the Cuntz semigroup of certain C*-algebras, in particular those
C*-algebras that absorb the Jiang-Su algebra. For this class, we identify a
natural subsemigroup of the Cuntz semigroup that is isomorphic to the dual
cone of $\mathrm{QT}_{2}(A)$. The complement of this subsemigroup consists of
the elements that become compact, and not a multiple of infinity, after
passing to the quotient by some closed two-sided ideal. The last result of the
paper is the computation of the Cuntz semigroup for two (disjoint) classes of
C*-algebras: C*-algebras that absorb the Jiang-Su algebra and have no non-zero
simple subquotients, and simple C*-algebras that absorb the Jiang-Su algebra.
The computation of the Cuntz semigroup for the latter class extends a previous
result of Brown, Perera, and Toms (see [brown-perera-toms]); in their
computation they made the additional assumptions that the algebra was unital,
exact, and of stable rank one.

##  2\. Preliminary results

###  2.1. The Cuntz-Pedersen equivalence relation

Let $A$ be a C*-algebra and let $a$ and $b$ be positive elements of $A$. Let
us say that $a$ is Cuntz-Pedersen equivalent to $b$, and write $a\sim b$, if
$a=\sum_{i=1}^{\infty}x_{i}x_{i}^{*}$ and
$b=\sum_{i=1}^{\infty}x_{i}^{*}x_{i}$ for some sequence $x_{i}\in A$,
$i=1,2,\dots$. Let us say that $a$ is Cuntz-Pedersen smaller than $b$ if
$a\sim a^{\prime}$ for some $a^{\prime}\in A^{+}$ with $a^{\prime}\leq b$. In
this case let us write $a\preccurlyeq b$. In [cuntz-pedersen] Cuntz and
Pedersen showed that the relations $\sim$ and $\preccurlyeq$ are transitive.
This is a consequence of the following result of Pedersen ([pedersen,
Proposition 1.1]).

######  Proposition 2.1.

(Riesz-Pedersen decomposition property.) Suppose that $x_{i},y_{i}\in A$,
$i=1,2\dots$, are such that
$\sum_{i=1}^{\infty}x_{i}x_{i}^{*}=\sum_{i=1}^{\infty}y_{i}^{*}y_{i}$. Then
there are elements $z_{i,j}$, $i,j=1,2\dots$, such that
$x_{i}^{*}x_{i}=\sum_{j=1}^{\infty}z_{i,j}^{*}z_{i,j}$ and
$y_{j}y_{j}^{*}=\sum_{i=1}^{\infty}z_{i,j}z_{i,j}^{*}$, for all $i,j\geq 1$.

######  Lemma 2.2.

Let $a,b\in A^{+}$ be such that $\|a-b\|<\epsilon$. Then
$(a-\epsilon)_{+}\preccurlyeq b$.

###### Proof.

By [kirchberg-rordam, Lemma 2.2], there is $d\in A$ with $\|d\|\leq 1$ and
$(a-\epsilon)_{+}=dbd^{*}$. Hence, $(a-\epsilon)_{+}\sim
b^{1/2}d^{*}db^{1/2}\leq b$. ∎

The following proposition is a summary of the properties of the relations
$\preccurlyeq$ and $\sim$ between positive elements of $A$ that will be needed
later.

######  Proposition 2.3.

(i) For every $\epsilon>0$ there is $\epsilon^{\prime}>0$ such that

(2.1) $\displaystyle(a_{1}+a_{2}-\epsilon)_{+}$ $\displaystyle\preccurlyeq$
$\displaystyle(a_{1}-\epsilon^{\prime})_{+}+(a_{2}-\epsilon^{\prime})_{+},$
(2.2) $\displaystyle(a_{1}-\epsilon)_{+}+(a_{2}-\epsilon)_{+}$
$\displaystyle\preccurlyeq$
$\displaystyle(a_{1}+a_{2}-\epsilon^{\prime})_{+}.$

(ii) For every $\epsilon>0$ and $x\in A$ we have
$(xx^{*}-\epsilon)_{+}\sim(x^{*}x-\epsilon)_{+}$.

(iii) (Riesz decomposition property.) Suppose that
$\sum_{i=1}^{\infty}a_{i}\sim\sum_{i=1}^{\infty}b_{i}$. Then there are
elements $w_{i,j}$, $i,j=1,2,\dots$, such that
$a_{i}\sim\sum_{j=1}^{\infty}w_{i,j}$ and
$b_{j}\sim\sum_{i=1}^{\infty}w_{i,j}$.

###### Proof.

(i) Since $(a_{1}-\epsilon^{\prime})_{+}+(a_{2}-\epsilon^{\prime})_{+}$ tends
to $a_{1}+a_{2}$ as $\epsilon^{\prime}$ tends to 0, we get (2.1) from Lemma
2.2.

Let us prove (2.2). Let $(e_{n})$ be an approximate unit for
$\overline{(a_{1}+a_{2})A(a_{1}+a_{2})}$ such that
$e_{n}(a_{1}+a_{2})e_{n}\leq(a_{1}+a_{2}-1/n)_{+}$ (e.g.,
$e_{n}=\phi_{n}(a_{1}+a_{2})$, with $\phi_{n}(t)=\frac{1}{t}(t-1/n)_{+}$).
Since $e_{n}a_{1}e_{n}\to a_{1}$ and $e_{n}a_{2}e_{n}\to a_{2}$, by Lemma 2.2
there exists $n$ such that
$(a_{1}-\epsilon)_{+}+(a_{2}-\epsilon)_{+}\preccurlyeq
e_{n}(a_{1}+a_{2})e_{n}\leq(a_{1}+a_{2}-1/n)_{+}$.

(ii) Let us show that $(x^{*}x-\epsilon)_{+}$ is Murray-von Neumann equivalent
to $(xx^{*}-\epsilon)_{+}$ (i.e., Cuntz-Pedersen equivalent by means of a
single element). Consider the polar decomposition $x=u|x|$ of $x$ in the
bidual of $A$. The element $y=u(x^{*}x-\epsilon)_{+}^{1/2}$ belongs to $A$,
and we have $y^{*}y=(x^{*}x-\epsilon)_{+}$ and $yy^{*}=(xx^{*}-\epsilon)_{+}$.

(iii) Suppose that $\sum_{i}a_{i}=\sum_{j}x_{j}^{*}x_{j}$ and
$\sum_{k}b_{k}=\sum_{j}x_{j}x_{j}^{*}$. By the Riesz-Pedersen decomposition
property, there are $y_{i,j}$s such that $a_{i}=\sum_{j}y_{i,j}^{*}y_{i,j}$
and $x_{j}x_{j}^{*}=\sum_{i}y_{i,j}y_{i,j}^{*}$. We have
$\sum_{i,j}y_{i,j}y_{i,j}^{*}=\sum_{k}b_{k}$. Therefore, there are
$z_{i,j,k}$s such that $b_{k}=\sum_{i,j}z_{i,j,k}^{*}z_{i,j,k}$ and
$y_{i,j}^{*}y_{i,j}=\sum_{k}z_{i,j,k}z_{i,j,k}^{*}$. Set
$\sum_{j}z_{i,j,k}^{*}z_{i,j,k}=w_{i,k}$. Then
$\sum_{k}w_{i,k}=\sum_{j,k}z_{i,j,k}^{*}z_{i,j,k}\sim\sum_{j}y_{i,j}^{*}y_{i,j}=a_{i}$.
Also, $\sum_{i}w_{i,k}\sim\sum_{i,j}z_{i,j,k}z_{i,j,k}^{*}=b_{k}$. ∎

###  2.2. Non-cancellative cones.

Let us introduce the terminology non-cancellative cone for an abelian
semigroup endowed with a scalar multiplication by strictly positive real
numbers. The semigroup may not have cancellation, that is to say,
$\tau+\tau_{1}=\tau+\tau_{2}$ may not imply that $\tau_{1}=\tau_{2}$. However,
we will often refer to non-cancellative cones simply as cones, and refer to
standard cones that embed in a vector space as cancellative cones.

Notice that we have not included scalar multiplication by 0 or $\infty$ in the
definition of non-cancellative cone. For some of the cones that we shall
consider here—of traces and, more generally, of 2-quasitraces—we will be able
to extend the scalar multiplication to include $0$ and $\infty$. However, it
will not necessarily be the case that scalar multiplication by $0$ will result
in the zero element of the cone (see (3.4)).

Non-cancellative cones satisfy the following form of restricted cancellation.

######  Lemma 2.4.

(Cancellation lemma.) Let $S$ be a non-cancellative cone. Suppose that
$x+z=y+z$ for some $z$ such that $z+z_{1}=nx$ and $z+z_{2}=ny$. Then $x=y$.

###### Proof.

By induction we have $nx+z=ny+z$. So let us assume that $n=1$. Then
$x+y=x+z+z_{2}=y+z+z_{2}=2y$. In the same way $x+y=2x$, and so $x=y$. ∎

##  3\. The cone of lower semicontinuous traces

###  3.1. The cone $\mathrm{T}(A)$

Let $A$ be a C*-algebra. Let us say that $\tau\colon A^{+}\to[0,\infty]$ is a
trace on $A$ if $\tau$ is linear (i.e., additive, homogeneous with respect to
strictly positive scalars, and vanishing at 0) and satisfies the trace
identity $\tau(xx^{*})=\tau(x^{*}x)$.

The following lemma is well known (see [kirchberg-blanchard, Remark 2.27
(iv)]).

######  Lemma 3.1.

If $\tau\colon A^{+}\to[0,\infty]$ is a trace then $\widetilde{\tau}$ defined
by $\widetilde{\tau}(a):=\sup_{\epsilon>0}\tau((a-\epsilon)_{+})$ is a lower
semicontinuous trace, and is the largest such trace majorized by $\tau$.

###### Proof.

By parts (i) and (ii) of Proposition 2.3 we have that $\widetilde{\tau}$ is a
trace. We also have that
$\widetilde{\tau}(a)=\sup_{\epsilon>0}\widetilde{\tau}((a-\epsilon)_{+})$. Let
us show that this implies that $\widetilde{\tau}$ is lower semicontinuous.
Suppose that $\widetilde{\tau}(a)>\alpha$ for some $\alpha\geq 0$. Let
$\epsilon>0$ be such that $\widetilde{\tau}((a-\epsilon)_{+})>\alpha$. By
Lemma 2.2, if $\|a^{\prime}-a\|<\epsilon$ then $(a-\epsilon)_{+}\preccurlyeq
a^{\prime}$, whence
$\widetilde{\tau}(a^{\prime})\geq\widetilde{\tau}((a-\epsilon)_{+})>\alpha$.
If $\sigma$ is another lower semicontinuous trace with $\sigma\leq\tau$, then
for any $a\in A^{+}$,

$\sigma(a)=\sup_{\epsilon>0}\sigma((a-\epsilon)_{+})\leq\sup_{\epsilon>0}\tau((a-\epsilon)_{+})=\widetilde{\tau}(a)\qed$

Let us denote by $\mathrm{T}(A)$ the collection of all lower semicontinuous
traces of $A$. This set is a non-cancellative cone endowed with the operations
of pointwise addition and pointwise scalar multiplication by strictly positive
real numbers. (We will later extend the scalar multiplication to include $0$
and $\infty$.) We shall also consider $\mathrm{T}(A)$ endowed with the order
induced by its addition operation. (When we consider the dual cone of
$\mathrm{T}(A)$ in Section 5 below, we shall also need to consider its
pointwise order, but, as we shall now show, for $\mathrm{T}(A)$ itself this is
determined by addition.)

The following proposition is well known for various classes of traces on a
C*-algebra (e.g., see [dixmier, Proposition 6]).

######  Proposition 3.2.

Let $\tau_{1},\tau_{2}\in\mathrm{T}(A)$. Suppose that
$\tau_{1}(x)\leq\tau_{2}(x)$ for all $x\in A^{+}$. Then there exists
$\tau_{3}\in\mathrm{T}(A)$ such that $\tau_{1}+\tau_{3}=\tau_{2}$.

###### Proof.

Define $\tau\colon A^{+}\to[0,\infty]$ as follows:

$\tau(x):=\left\\{\begin{array}[]{cl}\tau_{2}(x)-\tau_{1}(x)&\hbox{ if
}\tau_{2}(x)<\infty,\\\ \infty&\hbox{ otherwise.}\\\ \end{array}\right.$

It is easily verified that $\tau$ is linear, satisfies the trace identity, and
satisfies $\tau_{1}+\tau=\tau_{2}$. Set $\widetilde{\tau}=\tau_{3}$, where
$\widetilde{\tau}$ is the lower semicontinuous regularization of $\tau$
described in Lemma 3.1. Taking the suprema of both sides with respect to
$\epsilon$ in the equation

$\tau_{1}((a-\epsilon)_{+})+\tau((a-\epsilon)_{+})=\tau_{2}((a-\epsilon)_{+}),$

we get that $\tau_{1}+\tau_{3}=\tau_{2}$. ∎

In [pedersen, Theorem 3.1], Pedersen used the Riesz-Pedersen property to show
that the cone of densely finite lower semicontinuous traces is a lattice. We
shall follow a similar method here to show that the whole of $\mathrm{T}(A)$
is a lattice, and is in fact complete.

######  Theorem 3.3.

The cone $\mathrm{T}(A)$ is a complete lattice with respect to the order
determined by addition (equivalently, by Proposition 3.2, the pointwise
order). For all $\tau_{1},\tau_{2},\tau_{3}\in\mathrm{T}(A)$ we have

(3.1)
$\displaystyle\tau_{1}\vee\tau_{2}+\tau_{3}=(\tau_{1}+\tau_{3})\vee(\tau_{2}+\tau_{3}),$
(3.2)
$\displaystyle\tau_{1}\wedge\tau_{2}+\tau_{3}=(\tau_{1}+\tau_{3})\wedge(\tau_{2}+\tau_{3}).$

###### Proof.

The properties of lower semicontinuity and linearity, and also the trace
identity, are preserved under passing to the pointwise supremum of an upward
directed collection of lower semicontinuous traces. Thus, $\mathrm{T}(A)$ is
closed under passage to directed suprema. In order to prove that
$\mathrm{T}(A)$ is a complete lattice it is then enough to show that the
supremum of any two lower semicontinuous traces exists. (The supremum of any
non-empty set will then exist, and the supremum of the empty set is 0. It
follows that the infimum of any set also exists.)

Let $\tau_{1}$ and $\tau_{2}$ be in $\mathrm{T}(A)$. Define $\tau\colon
A^{+}\to[0,\infty]$ by the Riesz-Kantorovich formula:

$\tau(x):=\sup\\{\,\tau_{1}(x_{1})+\tau_{2}(x_{2})\mid x_{1}+x_{2}\sim
x\,\\}.$

We clearly have $\tau(xx^{*})=\tau(x^{*}x)$. The linearity of $\tau$ follows
from the Riesz decomposition property (i.e., Proposition 2.3 (iii)), by a
standard argument that goes back to Riesz (see [riesz, Theorem 1]). It is
clear that $\tau_{1}\leq\tau$, $\tau_{2}\leq\tau$, and that any trace that
majorizes $\tau_{1}$ and $\tau_{2}$ is greater than or equal to $\tau$. With
$\widetilde{\tau}$ the lower semicontinuous regularization of $\tau$ of Lemma
3.1, i.e., $\widetilde{\tau}(a)=\sup_{\epsilon>0}\tau((a-\epsilon)_{+})$, we
have $\widetilde{\tau}\leq\tau$, and, furthermore, both
$\tau_{1}\leq\widetilde{\tau}$ and $\tau_{2}\leq\widetilde{\tau}$ (as
$\tau_{i}=\sup_{\epsilon>0}\tau_{i}((a-\epsilon)_{+})\leq\sup_{\epsilon>0}\tau((a-\epsilon)_{+})=\widetilde{\tau}(a)$).
Therefore, $\tau\leq\widetilde{\tau}$, and so $\tau=\widetilde{\tau}$; in
other words, the supremum of $\tau_{1}$ and $\tau_{2}$ in the cone of all
traces belongs to $\mathrm{T}(A)$.

The identity (3.1) follows from the Riesz-Kantorovich formula for the supremum
of two traces in $\mathrm{T}(A)$. (Note that we have shown that this formula
does describe the supremum in $\mathrm{T}(A)$.)

Consider now $\tau\colon A^{+}\to[0,\infty]$ defined by

$\tau(x):=\inf\\{\,\tau_{1}(x_{1})+\tau_{2}(x_{2})\mid x_{1}+x_{2}\sim
x\,\\}.$

By the Riesz decomposition property $\tau$ is a trace. Very much as shown
above for the supremum, $\tau$ is seen to be lower semicontinuous and to be
the infimum of $\tau_{1}$ and $\tau_{2}$ in $\mathrm{T}(A)$. The identity
(3.2) now follows from the Riesz-Kantorovich formula that we have just
established for the infimum of $\tau_{1}$ and $\tau_{2}$ in $\mathrm{T}(A)$. ∎

Vector lattices, i.e., ordered vector spaces that are lattices with respect to
their order, have a number of properties that are implied by their lattice
structure. For example, a vector lattice is always distributive and satisfies
the identities (3.1) and (3.2) (see [vector-lattices]). The cone
$\mathrm{T}(A)$ cannot be embedded in a vector space since it is not
cancellative. For instance, if $I$ denotes a closed two-sided ideal of $A$
then $\tau_{I}$ defined by

$\tau_{I}(x):=\left\\{\begin{array}[]{lc}0&x\in I^{+},\\\ \infty&x\notin
I^{+},\end{array}\right.$

is a lower semicontinuous trace and satisfies $\tau_{I}+\tau_{I}=\tau_{I}$.
Indeed, the lower semicontinuous traces with the only possible values $0$ and
$\infty$—i.e., that satisfy $\tau+\tau=\tau$—are, as is easily seen, in order
reversing bijection with the closed two-sided ideals of $A$ by the map
$I\mapsto\tau_{I}$.

Making use of equations (3.1) and (3.2), and the restricted cancellation of
Lemma 2.4, we can show that $\mathrm{T}(A)$ has some of the properties of a
vector lattice.

######  Proposition 3.4.

(i) We have $\tau_{1}\vee\tau_{2}+\tau_{1}\wedge\tau_{2}=\tau_{1}+\tau_{2}$
for all $\tau_{1}$ and $\tau_{2}$ in $\mathrm{T}(A)$.

(ii) $\mathrm{T}(A)$ is a distributive lattice.

###### Proof.

(i) Taking $\tau_{3}=\tau_{1}\wedge\tau_{2}$ in (3.1) yields

$\tau_{1}\vee\tau_{2}+\tau_{1}\wedge\tau_{2}=(\tau_{1}+\tau_{1}\wedge\tau_{2})\vee(\tau_{2}+\tau_{1}\wedge\tau_{2})\leq\tau_{1}+\tau_{2}.$

Taking $\tau_{3}=\tau_{1}\vee\tau_{2}$ in (3.2) yields

$\tau_{1}\wedge\tau_{2}+\tau_{1}\vee\tau_{2}=(\tau_{1}+\tau_{1}\vee\tau_{2})\wedge(\tau_{2}+\tau_{1}\vee\tau_{2})\geq\tau_{1}+\tau_{2}.$

(ii) Let us prove that
$(\tau_{1}\vee\tau_{2})\wedge\tau_{3}=(\tau_{1}\wedge\tau_{3})\vee(\tau_{2}\wedge\tau_{3})$.
It is enough to prove this equality after adding
$\tau_{1}\wedge\tau_{2}\wedge\tau_{3}$ to both sides, since this term may be
cancelled by Lemma 2.4. Considering the right-hand side, we have

$\displaystyle(\tau_{1}\wedge\tau_{3})\vee(\tau_{2}\wedge\tau_{3})+\tau_{1}\wedge\tau_{2}\wedge\tau_{3}$
$\displaystyle=\tau_{1}\wedge\tau_{3}+\tau_{2}\wedge\tau_{3}$
$\displaystyle=(\tau_{1}+\tau_{2}\wedge\tau_{3})\wedge(\tau_{3}+\tau_{2}\wedge\tau_{3})$
$\displaystyle=(\tau_{1}+\tau_{2})\wedge(\tau_{1}+\tau_{3})\wedge(\tau_{2}+\tau_{3})\wedge
2\tau_{3}.$

Considering the left-hand side, we obtain the same quantity:

$\displaystyle(\tau_{1}\vee\tau_{2})\wedge\tau_{3}+\tau_{1}\wedge\tau_{2}\wedge\tau_{3}$
$\displaystyle=(\tau_{1}\vee\tau_{2}+\tau_{1}\wedge\tau_{2}\wedge\tau_{3})\wedge(\tau_{3}+\tau_{1}\wedge\tau_{2}\wedge\tau_{3})$
$\displaystyle=(\tau_{1}+\tau_{2})\wedge(\tau_{1}\vee\tau_{2}+\tau_{3})\wedge(\tau_{3}+\tau_{1}\wedge\tau_{2}\wedge\tau_{3})$
$\displaystyle=(\tau_{1}+\tau_{2})\wedge(\tau_{3}+\tau_{1}\wedge\tau_{2}\wedge\tau_{3})$
$\displaystyle=(\tau_{1}+\tau_{2})\wedge(\tau_{1}+\tau_{3})\wedge(\tau_{2}+\tau_{3})\wedge
2\tau_{3}.\qed$

###  3.2. The topology on $\mathrm{T}(A)$

Let us endow the cone $\mathrm{T}(A)$ with the topology in which the net
$(\tau_{i})$ converges to $\tau$ if

(3.3)
$\displaystyle\limsup\tau_{i}((a-\epsilon)_{+})\leq\tau(a)\leq\liminf\tau_{i}(a)$

for any $a\in A^{+}$ and $\epsilon>0$. Equivalently (by Lemma 3.1—both
parts—combined with compactness of the infinite product of copies of
$[0,\infty]$—one for each $a\in A^{+}$), $\tau_{i}\to\tau$ if, whenever a
subnet of $(\tau_{i})$ converges pointwise to a trace $\sigma$, the
regularization $\widetilde{\sigma}$ of $\sigma$ given by Lemma 3.1 is equal to
$\tau$ (cf. proof of Theorem 3.7, below). A sub-basis of neighbourhoods for
the trace $\tau$ is given by the sets

$\displaystyle U(\tau;a,\epsilon)$
$\displaystyle:=\\{\,\tau^{\prime}\in\mathrm{T}(A)\mid\tau((a-\epsilon)_{+})\leq\tau^{\prime}(a)+\epsilon\hbox{
or }\tau^{\prime}(a)>\frac{1}{\epsilon}\,\\},$ $\displaystyle
V(\tau;a,\epsilon)$
$\displaystyle:=\\{\,\tau^{\prime}\in\mathrm{T}(A)\mid\tau^{\prime}((a-\epsilon)_{+})\leq\tau(a)+\epsilon\,\\}.$

######  Remark 3.5.

In order to define the topology of $\mathrm{T}(A)$ the element $a$ can be
restricted to vary in a dense subset $S$ of $A^{+}$ such that $a\in S$ implies
that $(a-1/n)_{+}\in S$ for all $n\geq 1$. Let us verify this. Let $S$ be such
a set. Let $a\in A^{+}$ and $\epsilon>0$. Choose $a^{\prime}\in S$ and
$n\in\mathbb{N}$ such that $\|a-(a^{\prime}-2/n)_{+}\|<\epsilon$ and
$\|a-a^{\prime}\|<1/n$. By Lemma 2.2 (applied twice), we have

$(a-\epsilon)_{+}\preccurlyeq(a^{\prime}-\frac{2}{n})_{+}\preccurlyeq(a^{\prime}-\frac{1}{n})_{+}\preccurlyeq
a.$

Hence, $U(\tau;(a^{\prime}-\frac{1}{n})_{+},\frac{1}{n})\subseteq
U(\tau;a,\epsilon)$ and
$V(\tau;(a^{\prime}-\frac{1}{n})_{+},\frac{1}{n})\subseteq
V(\tau;a,\epsilon)$.

One can verify using (3.3) that $\alpha\tau\to\tau_{\ker\tau}$ when
$\alpha\to\infty$. One also verifies that
$\alpha\tau\to\tau_{\operatorname{fin}\tau}$ when $\alpha\to 0$, where
$\operatorname{fin}\tau$ is the closed two-sided ideal spanned by $\\{\,x\in
A^{+}\mid\tau(x)<\infty\,\\}$. (In the terminology of [dixmier],
$\operatorname{fin}\tau$ is the closure of the ideal of definition of $\tau$;
we shall refer to this ideal as the (closed) ideal of finiteness of $\tau$.)
In view of these computations, we may extend by continuity the scalar
multiplication in order to include the scalars 0 and $\infty$:

(3.4)
$0\cdot\tau=\tau_{\operatorname{fin}\tau},\quad\infty\cdot\tau=\tau_{\ker\tau}.$

######  Proposition 3.6.

Addition and (extended) scalar multiplication in $\mathrm{T}(A)$ are jointly
continuous.

###### Proof.

We will prove here that if $\tau_{i}\to\tau$ then $0\cdot\tau_{i}\to
0\cdot\tau$. The other parts of the proposition are easily verified from the
definition of the topology of $\mathrm{T}(A)$ by the inequalities (3.3).

Let us show that
$\limsup(0\cdot\tau_{i})((a-\epsilon)_{+})\leq(0\cdot\tau)(a)$ for all $a\in
A^{+}$ and $\epsilon>0$. If $(0\cdot\tau)(a)=\infty$ this is obvious. Suppose
that $(0\cdot\tau)(a)=0$. Then $a\in\operatorname{fin}\tau$, and so
$\tau((a-\epsilon)_{+})<\infty$ for all $\epsilon>0$. Since $\tau_{i}\to\tau$,
we have
$\limsup\tau_{i}((a-\epsilon)_{+})\leq\tau((a-\epsilon/2)_{+})<\infty$. Hence
$(0\cdot\tau_{i})((a-\epsilon)_{+})=0$ for all $i\geq i_{0}$ for some $i_{0}$,
and from this the desired inequality follows.

Let us show now that $(0\cdot\tau)(a)\leq\liminf(0\cdot\tau_{i})(a)$ for all
$a\in A^{+}$. If $\tau((a-\epsilon)_{+})<\infty$ for all $\epsilon>0$ then
$(0\cdot\tau)(a)=0$, and so the desired inequality clearly holds. Suppose that
$\tau((a-\epsilon)_{+})=\infty$ for some $\epsilon>0$. Then
$\tau_{i}((a-\epsilon)_{+})=\infty$ for all $i\geq i_{0}$, for some $i_{0}$.
Hence $(0\cdot\tau_{i})(a)=\infty$ for all $i\geq i_{0}$, and from this the
desired inequality follows. ∎

######  Theorem 3.7.

$\mathrm{T}(A)$ is a compact Hausdorff space. If $A$ is separable then
$\mathrm{T}(A)$ has a countable basis.

###### Proof.

Let us show that $\mathrm{T}(A)$ is Hausdorff. Let $\tau_{1}$ and $\tau_{2}$
be distinct points in $\mathrm{T}(A)$. Since either $\tau_{1}\nleq\tau_{2}$ or
$\tau_{2}\nleq\tau_{1}$, we may suppose that we are in the first case. Then
there are $a\in A^{+}$ and $\epsilon>0$ such that
$\tau_{1}((a-\epsilon)_{+})\nleq\tau_{2}(a)+\epsilon$. Let us choose
$\epsilon>0$ such that $\tau_{2}(a)<2/\epsilon-\epsilon/2$ (this is possible
since, necessarily, $\tau_{2}(a)<\infty$). Then the sets
$U(\tau_{1};(a-\epsilon/2)_{+},\epsilon/2)$ and $V(\tau_{2};a,\epsilon/2)$ are
disjoint neighbourhoods of $\tau_{1}$ and $\tau_{2}$ respectively. For suppose
that $\tau$ belongs to their intersection. Then, either

$\tau_{1}((a-\epsilon)_{+})\leq\tau((a-\epsilon/2)_{+})+\epsilon/2\leq\tau_{2}(a)+\epsilon$

or

$\frac{2}{\epsilon}<\tau((a-\epsilon/2)_{+})\leq\tau_{2}(a)+\frac{\epsilon}{2}.$

In either case, this is a contradiction.

The following simple proof of the compactness of $\mathrm{T}(A)$ was suggested
to us by E. Kirchberg (our original proof was much longer).

Let $(\tau_{i})_{i\in\Lambda}$ be a net of traces in $\mathrm{T}(A)$. By
Tychonoff’s theorem (using the compactness of $[0,\infty]$), we can choose a
subnet $(\tau_{i})_{i\in\Lambda^{\prime}}$ converging pointwise to $\sigma$.
The function $\sigma\colon A^{+}\to[0,\infty]$ is linear and satisfies the
trace identity. With $\widetilde{\sigma}$ the lower semicontinuous trace of
Lemma 3.1, i.e.,
$\widetilde{\sigma}(a)=\sup_{\epsilon>0}\sigma((a-\epsilon)_{+})$, let us show
that $(\tau_{i})_{i\in\Lambda^{\prime}}$ converges to $\widetilde{\sigma}$ in
$\mathrm{T}(A)$ (i.e., the inequalities (3.3) are satisfied for all $a\in
A^{+}$ and $\epsilon>0$). Let $a\in A^{+}$ and let $\epsilon>0$. Then

$\limsup_{i\in\Lambda^{\prime}}\tau_{i}((a-\epsilon)_{+})=\sigma((a-\epsilon)_{+})\leq\widetilde{\sigma}(a)\leq\sigma(a)=\liminf_{i\in\Lambda^{\prime}}\tau_{i}(a).$

(This convergence is also immediate from the alternative form of the
definition.) This shows that $\mathrm{T}(A)$ is compact.

Now suppose that $A$ is separable. It follows from the remark made after the
definition of the topology of $\mathrm{T}(A)$ that if $A$ is separable then
$\mathrm{T}(A)$ is first countable, and in fact there is a countable basis of
symmetric entourages for a uniform structure giving rise to the topology of
$\mathrm{T}(A)$; let us choose such a basis. Inspection of the entourages
described shows that not only are they symmetric but also the corresponding
neighbourhoods in the topology are open; we shall assume therefore that our
countable basis consists of such entourages.

Separability of $A$ also implies that there is a countable dense subset of
$\mathrm{T}(A)$: as a set of maps from $A^{+}$ to $[0,\infty]$, we may
naturally identify $\mathrm{T}(A)$ with a subset of $\Pi_{a\in
A^{+}}[0,\infty]$, and (since the maps in $\mathrm{T}(A)$ are lower
semicontinuous) in fact, as we shall now show, with a subset of $\Pi_{a\in
S}[0,\infty]$ where $S$ is a suitable countable dense subset of $A^{+}$. Since
lower semicontinuous functions are not determined on just any dense subset, we
must choose $S$ to consist of a countable dense subset of $A^{+}$ (any such
subset) together with, for each $a\in S$, the set of all elements
$(a-\epsilon)^{+}$ with $\epsilon=1/n$, $n=1,2,3,\ldots$. (It follows from
Lemma 2.2 that $S$ separates elements of $\mathrm{T}(A)$: if
$\tau,\tau^{\prime}\in\mathrm{T}(A)$ and $\tau$ and $\tau^{\prime}$ agree on
$S$, then for any $a\in A^{+}$, with $b\in S$ such that $\|a-b\|<\epsilon$, we
have $(b-\epsilon)_{+}\preceq a$, in the sense of Cuntz and Pedersen, and so

$\tau(b)=\sup_{n\geq 1}\tau((b-1/n)_{+})=\sup_{n\geq
1}\tau^{\prime}((b-1/n)_{+})\leq\tau^{\prime}(b).$

In other words, $\tau\leq\tau^{\prime}$, and so by symmetry
$\tau=\tau^{\prime}$.)

Not only is the embedding of $\mathrm{T}(A)$ in $\Pi_{a\in S}[0,\infty]$
injective, but also by the alternative definition of the topology on
$\mathrm{T}(A)$ the inverse of this map, from the image with the coordinate-
wise topology, is continuous. In other words, as we shall now show, if
$\tau_{i}(a)\to\tau(a)$ for all $a\in S$, with $\tau_{i}$ and $\tau$ in
$\mathrm{T}(A)$, then $\tau_{i}\to\tau$ in $\mathrm{T}(A)$. It is enough to
show that if $\tau^{\prime}$ is a trace and $\tau_{i}\to\tau^{\prime}$
pointwise on $A^{+}$ then $(\tau^{\prime})^{\sim}=\tau$. By hypothesis,
$\tau^{\prime}$ agrees with $\tau$ on $S$. By the choice of $S$,
$(\tau^{\prime})^{\sim}$ also coincides with $\tau$ on $S$, and therefore by
injectivity $(\tau^{\prime})^{\sim}=\tau$ in $\mathrm{T}(A)$. Hence, one
obtains a countable dense subset of $\mathrm{T}(A)$ as the image under the
inverse map of a countable dense subset of its domain—which exists as the
countable Cartesian product is a metrizable compact space.

It follows that $\mathrm{T}(A)$ has a countable basis for the topology under
consideration, namely, the collection of all neighbourhoods of a fixed dense
sequence $\tau_{1},\tau_{2},\ldots$ in $\mathrm{T}(A)$ corresponding to the
countable basis of symmetric entourages for the uniform structure referred to
above. (The proof of this is just as if the entourages were determined by a
metric, as the degrees of closeness corresponding to a sequence of distances
converging to zero. Let $\tau$ be a point in $\mathrm{T}(A)$, and let $W$ be
an arbitrary open neighbourhood of a symmetric entourage $U$ such that if
$(\tau,s)\in U$ then $s\in W$. Choose an entourage $V$ such that if
$(\tau,\tau^{\prime})\in V$ and $(\tau^{\prime},\sigma)\in V$ then
$(\tau,\sigma)\in U$. We may choose $V$ to be one of the countable basis of
symmetric entourages chosen above and in particular such that the
neighbourhood of any point determined by $V$ is open. Choose $n$ such that
$(\tau,\tau_{n})\in V$. The neighbourhood of $\tau_{n}$ determined by the
symmetric entourage $V$ then both includes the point $\tau$ and is included in
the neighbourhood of $\tau$ determined by $U$, and therefore also in the given
open neighbourhood $W$ of $\tau$. Since this neighbourhood is open by the
choice of $V$, we have identified a countable basis of open sets (the
neighbourhoods of $\tau_{1},\tau_{2},\cdots$ determined by the chosen
countable basis of entourages).) ∎

######  Proposition 3.8.

(i) The order relation in $\mathrm{T}(A)$ is continuous (i.e., the set
$\\{(\tau_{1},\tau_{2})\mid\tau_{1}\leq\tau_{2}\\}$ is closed in
$\mathrm{T}(A)\times\mathrm{T}(A)$).

(ii) An upward directed subset of $\mathrm{T}(A)$ converges to its supremum
(when indexed by itself), and a downward directed subset converges to its
infimum.

(iii) The complete, distributive, lattice $\mathrm{T}(A)$ is join continuous;
that is, for any subset $S$ of $\mathrm{T}(A)$, and for any
$\tau\in\mathrm{T}(A)$,

$(\bigwedge_{\sigma\in S}\sigma)\vee\tau=\bigwedge_{\sigma\in
S}(\sigma\vee\tau).$

###### Proof.

(i) Let $((\tau_{i},\mu_{i}))$ be a net converging to $(\tau,\mu)$ and suppose
that $\tau_{i}\leq\mu_{i}$ for all $i$. Then
$\mu_{i}=\tau_{i}+\mu_{i}^{\prime}$ for some $\mu_{i}^{\prime}$. Passing to a
convergent subnet of $\mu_{i}^{\prime}$ (by compactness) and then passing to
the limit we get that $\mu=\tau+\mu^{\prime}$, whence $\tau\leq\mu$.

(ii) Let $(\tau_{i})_{i\in\Lambda}$ be a decreasing net with infimum $\tau$.
It is enough by compactness to show that every convergent subnet of
$(\tau_{i})_{i\in\Lambda}$ converges to $\tau$, and so we may assume without
loss of generality that $(\tau_{i})_{i\in\Lambda}$ converges to
$\tau^{\prime}$. For every $i$ we have $\tau\leq\tau_{i}$. Thus, passing to
the limit and using part (i) of this proposition we conclude that
$\tau\leq\tau^{\prime}$. On the other hand, for every $i$ and $j$ with $i\leq
j$ we have $\tau_{j}\leq\tau_{i}$. Fixing $i$ and passing to the limit in $j$
we obtain $\tau^{\prime}\leq\tau_{i}$. Since this holds for all $i$ we
conclude that $\tau^{\prime}\leq\tau$.

One may proceed in a similar way for upward directed subsets of
$\mathrm{T}(A)$.

(iii) By the distributivity of $\mathrm{T}(A)$ we have

$(\bigwedge_{i\in F}\tau_{i})\vee\tau=\bigwedge_{i\in F}(\tau_{i}\vee\tau)$

for every finite subset $F$ of $S$. Let us consider both sides as downward
directed families of traces indexed by the finite subsets of $S$. The infimum
of the right side is $\bigwedge_{i\in\Lambda}(\tau_{i}\vee\tau)$. Set
$\bigwedge_{i\in F}\tau_{i}=\mu_{F}$. It is enough to prove that if the
downward directed subset $\\{\mu_{F}\\}$ has infimum $\mu$, then the infimum
of $\\{\mu_{F}\vee\tau\\}$ is $\mu\vee\tau$. By Proposition 3.4 (i), we have
$\mu_{F}\vee\tau+\mu_{F}\wedge\tau=\mu_{F}+\tau$. By (ii) together with
Proposition 3.6, for any two downward directed sets $S_{1}$ and $S_{2}$,
$\bigwedge(S_{1}+S_{2})=\bigwedge S_{1}+\bigwedge S_{2}$. Hence, taking infima
on both sides of $\mu_{F}\vee\tau+\mu_{F}\wedge\tau=\mu_{F}+\tau$ we get

$\bigwedge_{F}(\mu_{F}\vee\tau)+\bigwedge_{F}(\mu_{F}\wedge\tau)=\bigwedge_{F}\mu_{F}+\tau=(\bigwedge_{F}\mu_{F})\wedge\tau+(\bigwedge_{F}\mu_{F})\vee\tau.$

Cancelling $(\bigwedge_{F}\mu_{F})\wedge\tau$ (using Lemma 2.4 with $n=1$) we
obtain $\bigwedge_{F}(\mu_{F}\vee\tau)=(\bigwedge_{F}\mu_{F})\vee\tau$, as
desired. ∎

######  Remark 3.9.

Proposition 3.8 (i) and (ii) may be proved directly from the definition of the
topology of $\mathrm{T}(A)$. The proof given above, however, applies to an
arbitrary topological cone that is a complete lattice, and is compact and
Hausdorff. The infinite distributivity of Proposition 3.8 (iii) implies that
the lattice obtained by reversing the order of $\mathrm{T}(A)$ is a continuous
lattice (in the sense of [compendium]; see [compendium, Theorem I-2.7]). Since
the map $I\mapsto\tau_{I}$ is an order reversing embedding of
$\mathrm{Lat}(A)$ as a subcomplete sublattice of $\mathrm{T}(A)$, we deduce
from Proposition 3.8 (iii) the well known fact that $\mathrm{Lat}(A)$ is a
continuous lattice.

_Question._ Is the map $\mu\mapsto\mu\vee\tau$ continuous in the topology of
$\mathrm{T}(A)$? Is the topology of $\mathrm{T}(A)$ the Fell-Lawson topology
of the complete lattice obtained by reversing the order of $\mathrm{T}(A)$?

Let us write $\mathrm{T}_{I}(A)$ for the subcone of $\mathrm{T}(A)$ of traces
with ideal of finiteness $I$ (i.e., $\tau$ with $\operatorname{fin}\tau=I$).
The subcone $\mathrm{T}_{I}(A)$ is in bijective correspondence with the
densely finite lower semicontinuous traces on the ideal $I$, because for every
such trace we get a trace in $\mathrm{T}_{I}(A)$ by setting it equal to
$\infty$ outside $I^{+}$.

######  Proposition 3.10.

(i) For each ideal $I$ of $A$, the relative topology on the subcone
$\mathrm{T}_{I}(A)$ of $\mathrm{T}(A)$ is the topology of pointwise
convergence on the positive elements of the Pedersen ideal of $I$.

(ii) The relative topology on the subset $\mathrm{Lat}(A)$ of
$\mathrm{T}(A)$—the image of the embedding $I\mapsto\tau_{I}$—is the Fell-
Lawson topology.

###### Proof.

(i) Let $\tau\in\mathrm{T}_{I}(A)$ and let $(\tau_{i})$ be a net in
$\mathrm{T}_{I}(A)$ converging to $\tau$ in the topology of $\mathrm{T}(A)$.
Let us show that $\tau_{i}(a)\to\tau(a)$ for all $a\in\mathrm{Ped}(I)^{+}$. By
the alternative definition of limit in $\mathrm{T}(A)$ as the regularization
of every pointwise convergent subnet, it is sufficient to show that an
arbitrary densely finite trace $\sigma$ on $I^{+}$ satisfies
$\sigma=\sup_{\epsilon>0}\sigma((a-\epsilon)_{+})$ for each
$a\in\mathrm{Ped}(I)^{+}$. This holds by [pedersen2, Corollary 3.2].

Now suppose that we have a net $(\tau_{i})$ of traces in $\mathrm{T}_{I}(A)$
converging pointwise on $\mathrm{Ped}(I)^{+}$ to a trace $\tau$, also in
$\mathrm{T}_{I}(A)$. Let $a\in A^{+}$ and $\epsilon>0$. We need to show that
the inequalities (3.3) hold. If $a\notin I^{+}$ then this is true, since
$\tau_{i}(a)=\tau(a)=\infty$ for all $i$. Suppose that $a\in I^{+}$. Then
$(a-\epsilon)_{+}\in\mathrm{Ped}(I)^{+}$ for all $\epsilon>0$. So

$\limsup\tau_{i}((a-\epsilon)_{+})=\tau((a-\epsilon)_{+})\leq\tau(a),$

and

$\tau((a-\epsilon)_{+})=\liminf\tau_{i}((a-\epsilon)_{+})\leq\liminf\tau_{i}(a),$

for all $\epsilon>0$.

(ii) The traces that are a multiple of 0 form a closed subset of
$\mathrm{T}(A)$. Hence $\mathrm{Lat}(A)$ is compact and Hausdorff in the
relative topology inherited from $\mathrm{T}(A)$. Let us show that this
topology is finer than the Fell-Lawson topology. This will give the desired
result, since $\mathrm{Lat}(A)$ is compact and Hausdorff in both topologies.

Recall that the Fell-Lawson topology has the sub-basis of open sets
$U_{I}=\\{J\in\mathrm{Lat}(A)\mid I\nleq J\\}$, and
$V_{I}=\\{J\in\mathrm{Lat}(A)\mid I\ll J\\}$, where $I$ ranges in
$\mathrm{Lat}(A)$. Here we have denoted by $\ll$ the (countable) far below
relation in the ordered set $\mathrm{Lat}(A)$; see Section 4.2 below (cf. also
[compendium], where uncountable increasing nets are allowed). Suppose that
$(J_{i})_{i\in\Lambda}$ is a net converging to $J$ in the relative topology,
and $J\in U_{I}$. If we have $I\leq J_{i}$ for a subnet
$(J_{i})_{i\in\Lambda^{\prime}}$, then $\tau_{J_{i}}\leq\tau_{I}$ for all
$i\in\Lambda^{\prime}$, whence $\tau_{J}\leq\tau_{I}$. This contradicts the
relation $I\nleq J$. Therefore, there exists $i_{0}$ such that $I\nleq J_{i}$
for all $i\geq i_{0}$. This shows that the set $U_{I}$ is open in the relative
topology.

Let $J\in V_{I}$. For every $b\in J^{+}$ and $\epsilon>0$ consider the ideal
$J_{b,\epsilon}=\mathrm{Ideal}((b-\epsilon)_{+})$. By Proposition 2.3 (i), the
ideals $J_{b,\epsilon}$ form an upward directed subset of $\mathrm{Lat}(A)$.
Since they have supremum $J$, we must have
$I\subseteq\mathrm{Ideal}((b-\epsilon)_{+})$ for some $b\in J^{+}$ and
$\epsilon>0$. If $(J_{i})$ is a net such that $J_{i}\to J$ in the relative
topology, then $\limsup\tau_{J_{i}}((b-\epsilon/2)_{+})\leq\tau_{J}(b)=0$.
Thus $(b-\epsilon/2)_{+}\in J_{i}$ for all $i\geq i_{0}$, for some $i_{0}$. In
other words,
$I\subseteq\mathrm{Ideal}((b-\epsilon)_{+})\ll\mathrm{Ideal}((b-\epsilon/2)_{+})\subseteq
J_{i}$ for all $i\geq i_{0}$. This shows that the set $V_{I}$ is open in the
relative topology. ∎

###  3.3. The functor $\mathrm{T}(\cdot)$

Homomorphisms between C*-algebras induce morphisms in the opposite direction
between their cones of traces; given $\phi\colon A\to B$ the map
$\mathrm{T}(\phi)\colon\mathrm{T}(B)\to\mathrm{T}(A)$ is defined by
$\mathrm{T}(\phi)(\tau)=\tau\circ\phi$. It is easily verified that
$\mathrm{T}(\phi)$ is linear and continuous.

Let us denote by $\mathcal{C}$ the category of compact Hausdorff non-
cancellative cones with jointly continuous addition and jointly continuous
scalar multiplication by $[0,\infty]$, with, as morphisms, continuous linear
maps between cones. (Here, linear means additive, homogeneous with respect to
scalars in $[0,\infty]$, and takes $0$ into $0$.) By Proposition 3.6 and
Theorem 3.7, the cone $\mathrm{T}(A)$ is in the category $\mathcal{C}$.

######  Theorem 3.11.

$\mathrm{T}(\cdot)$ is a continuous contravariant functor from the category of
C*-algebras to the category $\mathcal{C}$.

###### Proof.

It is straightforward that $\mathrm{T}(\cdot)$ is a functor. (If $\phi:A\to
B)$ is a homomorphism of C*-algebras and $\tau_{i}\to\tau$ in $\mathrm{T}(B)$,
then to show that $\tau_{i}\phi\to\tau\phi$ in $\mathrm{T}(A)$, passing to a
subnet with $\tau_{i}\phi$ converging pointwise to $\sigma$ we must show that
$\widetilde{\sigma}=\tau\phi$. We may suppose that
$\tau_{i}\to\sigma^{\prime}$ pointwise, so $\sigma=\sigma^{\prime}\phi$, and
since $(\sigma^{\prime})^{\sim}=\tau$, we have
$(\sigma^{\prime}\phi)^{\sim}=(\sigma^{\prime})^{\sim}\phi=\tau\phi$, as
desired.) Let $A=\varinjlim(A_{i},\phi_{i,j})$ be an inductive limit of
C*-algebras. N.B: we will not assume that the index set in this inductive
system is countable. Let $C$ denote the subset of the Cartesian product
$\prod_{i}\mathrm{T}(A_{i})$ of vectors $(\tau_{i})$ compatible with the
projective system $(\mathrm{T}(A_{i}),\mathrm{T}(\phi_{i,j}))$; that is,
$\tau_{i}=\mathrm{T}(\phi_{i,j})(\tau_{j})$ for all $i<j$. Denote by
$\mu_{i}\colon C\to\mathrm{T}(A_{i})$ the projection onto the $i$th
coordinate. It is well known that $C$ is the projective limit of
$(\mathrm{T}(A_{i}),\mathrm{T}(\phi_{i,j}))$ in the category of compact
Hausdorff spaces. It is easily verified that $C$ is a cone when endowed with
the operations of coordinate-wise addition and scalar multiplication, that $C$
belongs to the category $\mathcal{C}$, and that $(C,\mu_{i})$ is in fact the
projective limit of the system of cones
$(\mathrm{T}(A_{i}),\mathrm{T}(\phi_{i,j}))$ in the category $\mathcal{C}$.

Let $m\colon\mathrm{T}(A)\to C$ denote the map given by
$m(\tau):=(\mathrm{T}(\phi_{i,\infty}(\tau)))$. In order to show that
$\mathrm{T}(A)$ and $C$ are isomorphic, it is enough to prove that $m$ is
bijective, since a continuous bijection between compact Hausdorff spaces has
continuous inverse.

Suppose that $\tau_{1}$ and $\tau_{2}$ are traces in $\mathrm{T}(A)$ such that
$m(\tau_{1})=m(\tau_{2})$, i.e.,
$\tau_{1}\circ\phi_{i,\infty}=\tau_{2}\circ\phi_{i,\infty}$ for all $i$. Then
$\tau_{1}$ and $\tau_{2}$ agree on the set
$\bigcup_{i}\phi_{i,\infty}(A_{i}^{+})$ of positive elements coming from the
algebras $A_{i}$. Let us call this set $B$. We have that $B$ is dense in
$A^{+}$ and is such that if $a\in B$ then $(a-\epsilon)_{+}\in B$ for all
$\epsilon>0$. It follows, by Remark 3.5, that $\tau_{1}$ and $\tau_{2}$ cannot
be separated in the topology of $\mathrm{T}(A)$. Since $\mathrm{T}(A)$ is
Hausdorff (by Theorem 3.7), this shows that $\tau_{1}=\tau_{2}$.

Let $(\tau_{i})$ be a vector in $C$. Let us find a trace $\tau$ such that
$m(\tau)=(\tau_{i})$. For $a\in A_{i}^{+}$ write
$\tau_{i}(a)=\tau(\phi_{i,\infty}(a))$. Let us show that $\tau$ is well
defined on the set $B$. Suppose that $\phi_{i,\infty}(a)=\phi_{i,\infty}(b)$.
For every $\epsilon>0$ there exists $j$ such that
$\|\phi_{i,j}(a)-\phi_{i,j}(b)\|<\epsilon$. Using Lemma 2.2 we get

$\tau_{i}((a-\epsilon)_{+})=\tau_{j}(\phi_{i,j}((a-\epsilon)_{+}))\leq\tau_{j}(\phi_{i,j}(b))=\tau_{i}(b).$

In the limit as $\epsilon\to 0$ we obtain $\tau_{i}(a)\leq\tau_{i}(b)$. Hence
by symmetry, $\tau_{i}(a)=\tau_{i}(b)$. So $\tau$ is well defined on $B$.

Let us extend $\tau$ from $B$ to $A^{+}$ as follows. Define
$\widetilde{\tau}\colon A^{+}\to[0,\infty]$ by

$\widetilde{\tau}(a):=\sup\\{\,\tau(a^{\prime})\mid a^{\prime}\in
B,a^{\prime}\preccurlyeq(a-\epsilon)_{+}\hbox{ for some }\epsilon>0\,\\}.$

Let us show that $\widetilde{\tau}$ is a trace; clearly, it extends $\tau$.
For every $a^{\prime}\in B$ and $\epsilon>0$ such that
$a^{\prime}\preccurlyeq(a-\epsilon)_{+}$ we have
$\widetilde{\tau}(a^{\prime})\leq\widetilde{\tau}((a-\epsilon/2)_{+})$. It
follows from this that
$\widetilde{\tau}(a)=\sup_{\epsilon>0}\widetilde{\tau}((a-\epsilon)_{+})$.
Since $(xx^{*}-\epsilon)_{+}\sim(x^{*}x-\epsilon)_{+}$ for all $\epsilon>0$
and $x\in A$, we have $\widetilde{\tau}(xx^{*})=\widetilde{\tau}(x^{*}x)$.
Also, it can be shown using (2.1) (see Proposition 2.3 (i)) that
$\widetilde{\tau}$ is superadditive, i.e.,
$\widetilde{\tau}(a)+\widetilde{\tau}(b)\leq\widetilde{\tau}(a+b)$. In
particular $\widetilde{\tau}$ is increasing.

It remains to show that $\tau$ is subadditive. (Homogeneity is clear.) Before
proceeding with the proof of the subadditivity of $\tau$ let us prove a
preliminary fact. Suppose that $c,c^{\prime}\in A^{+}$ are such that
$\|c-c^{\prime}\|<\epsilon$. By [kirchberg-rordam, Lemma 2.2] there exists a
contraction $d\in A$ such that $(c-\epsilon)_{+}=dc^{\prime}d^{*}$. Since
$\tau$ is increasing and satisfies the trace identity, it follows that
$\widetilde{\tau}((c-\epsilon)_{+})\leq\widetilde{\tau}(c^{\prime})$ whenever
$\|c-c^{\prime}\|<\epsilon$.

Let us show that $\tau$ is subadditive. Let $a,b\in A^{+}$ and $\epsilon>0$
and assume, as we may without loss of generality, that $\|a\|,\|b\|\leq 1$.
Choose $a^{\prime},b^{\prime}\in B$ such that $\|a-a^{\prime}\|<\epsilon/4$
and $\|b-b^{\prime}\|<\epsilon/4$. Then

$\displaystyle\widetilde{\tau}((a+b-\epsilon)_{+})$
$\displaystyle\leq\widetilde{\tau}((a^{\prime}-\frac{\epsilon}{2})_{+}+(b^{\prime}-\frac{\epsilon}{2})_{+})=$
$\displaystyle=\widetilde{\tau}((a^{\prime}-\frac{\epsilon}{2})_{+})+\widetilde{\tau}((b^{\prime}-\frac{\epsilon}{2})_{+})\leq\widetilde{\tau}(a)+\widetilde{\tau}(b).$

Passing to the supremum on the left side with respect to $\epsilon$, we deduce
that $\widetilde{\tau}$ is subadditive. This shows that $\widetilde{\tau}$ is
a trace. That $\widetilde{\tau}$ is lower semicontinuous follows from Lemma
3.1 and the equation
$\widetilde{\tau}(a)=\sup_{\epsilon>0}\widetilde{\tau}((a-\epsilon)_{+})$. ∎

######  Remark 3.12.

In addition to being an object in the category $\mathcal{C}$, we have seen
that the cone $\mathrm{T}(A)$ is a complete lattice and satisfies the
identities (3.1) and (3.2). Inspection of the proofs of Propositions 3.4 and
3.8 shows that they also hold on replacing $\mathrm{T}(A)$ by any topological
cone in $\mathcal{C}$ that is a complete lattice and satisfies the identities
(3.1) and (3.2). Is Proposition 3.10 (i) true for such cones too?

##  4\. Functionals on the Cuntz Semigroup

###  4.1. Quasitraces and functionals

Let $\mathrm{Cu}(A)$ denote the stabilized Cuntz semigroup of $A$, i.e., the
ordered semigroup of Cuntz equivalence classes of positive elements in
$A\otimes\mathcal{K}$ (see [coward-elliott-ivanescu] and [rordam]). Given
$\tau\in\mathrm{T}(A\otimes\mathcal{K})$ let us define the function
$\lambda_{\tau}([a]):=\sup_{n}\tau(a^{1/n})$, where $[a]\in\mathrm{Cu}(A)$
denotes the equivalence class of $a\in(A\otimes\mathcal{K})^{+}$. This is
known to be a well defined function on $\mathrm{Cu}(A)$ with values in
$[0,\infty]$ and with the following properties:

(1) $\lambda_{\tau}$ is additive and order preserving, and
$\lambda_{\tau}(0)=0$,

(2) $\lambda_{\tau}$ preserves the suprema of increasing sequences.

Let us refer to a function $\lambda\colon\mathrm{Cu}(A)\to[0,\infty]$ with the
properties (1) and (2) as a functional on $\mathrm{Cu}(A)$.

######  Lemma 4.1.

If $\tau:A^{+}\to[0,\infty]$ is a quasitrace then $\widetilde{\tau}$ defined
by $\widetilde{\tau}(a)=\sup_{\epsilon>0}\tau((a-\epsilon)_{+})$ is a lower
semicontinuous quasitrace on $A$, and is the largest such quasitrace majorized
by $\tau$.

###### Proof.

If $B$ is a commutative sub-C*-algebra of $A$, then the restriction $\tau|B$
of $\tau$ to $B$ is a trace on $B$, and the restriction of $\widetilde{\tau}$
to $B$ is the lower semicontinuous regularization $(\tau|B)^{\sim}$ of Lemma
3.1. This shows that $\widetilde{\tau}$ is additive on elements that commute.
Since for every $x\in A$ and $\epsilon>0$ there is $y\in A$ such that
$(x^{*}x-\epsilon)_{+}=y^{*}y$ and $(xx^{*}-\epsilon)_{+}=yy^{*}$ (see the
proof of Proposition 2.3 (ii)), we have
$\widetilde{\tau}(x^{*}x)=\widetilde{\tau}(xx^{*})$ for all $x\in A$. So
$\widetilde{\tau}$ is a quasitrace. This, together with the defining equation
$\widetilde{\tau}(a)=\sup_{\epsilon>0}\tau((a-\epsilon)_{+})$, implies that
$\widetilde{\tau}$ is lower semicontinuous (see the last remark in [kirchberg-
blanchard, Definition 2.2]). If $\sigma$ is another lower semicontinuous
quasitrace with $\sigma\leq\tau$, then (as in the proof of Lemma 3.1), for any
$a\in A^{+}$,

$\sigma(a)=\sup_{\epsilon>0}\sigma((a-\epsilon)_{+})\leq\sup_{\epsilon>0}\tau(a-\epsilon)_{+})=\tau(a).\qed$

To repeat, we shall denote by $\mathrm{QT}_{2}(A)$ the cone of lower
semicontinuous quasitraces of $A\otimes\mathcal{K}$. The notation
$\mathrm{QT}_{2}(A)$ is explained by the result of Blanchard and Kirchberg
that every lower semicontinuous 2-quasitrace of $A$ extends to a lower
semicontinuous quasitrace of $A\otimes\mathcal{K}$; see [kirchberg-blanchard,
Remark 2.27 (viii)]). Let $\mathrm{F}(\mathrm{Cu}(A))$ denote the cone of
functionals on $\mathrm{Cu}(A)$, as defined above.

######  Proposition 4.2.

Given $\tau\in\mathrm{QT}_{2}(A)$ the function
$\lambda_{\tau}([a]):=\sup_{n}\tau(a^{1/n})$ is well defined and gives a
functional on $\mathrm{Cu}(A)$, i.e., an element of
$\mathrm{F}(\mathrm{Cu}(A))$. Given $\lambda\in\mathrm{F}(\mathrm{Cu}(A))$ the
function $\tau_{\lambda}(a)=\int^{\infty}_{0}\lambda([(a-t)_{+}])\,dt$ is a
lower semi-continuous quasitrace on $A\otimes\mathcal{K}$, i.e., an element of
$\mathrm{QT}_{2}(A)$. The maps $\tau\mapsto\lambda_{\tau}$ and
$\lambda\mapsto\tau_{\lambda}$ are the inverses of each other.

###### Proof.

Let $\tau\in\mathrm{QT}_{2}(A)$ and set $\sup_{n}\tau(a^{1/n})=D(a)$, for
$a\in(A\otimes\mathcal{K})^{+}$. The restriction of $\tau$ to a commutative
sub-C*-algebra $B$ of $A\otimes\mathcal{K}$ consists of integration by some
measure on the spectrum of $B$. If $a\in B^{+}$ then $D(a)$ is the measure of
the open subset of the spectrum of $B$ consisting of points where $a$ does not
vanish. From this observation it follows that (1) $D(a+b)=D(a)+D(b)$ if $a$
and $b$ are orthogonal, (2) $D(a)\leq D(b)$ if $a$ and $b$ commute and $a\leq
Mb$ for some $M>0$, and (3) $D(a)=\sup_{\epsilon>0}D((a-\epsilon)_{+})$.

Let us show that $D$ is constant on the Cuntz equivalence classes of positive
elements. First, note that for any $x\in A\otimes\mathcal{K}$ and $g\in
C_{0}(\mathbb{R}^{+})$ there is $y$ such that $g(x^{*}x)=y^{*}y$ and
$g(xx^{*})=yy^{*}$. Therefore, $D(x^{*}x)=D(xx^{*})$. Suppose now that $a$ is
Cuntz smaller than $b$, i.e., $d^{*}_{n}bd_{n}\to a$ for some sequence
$(d_{n})$. Let $\epsilon>0$. Choose $d\in A\otimes\mathcal{K}$ and $\delta>0$
such that $\|a-d^{*}(b-\delta)_{+}d\|<\epsilon$. By the proof of Lemma 2.2,
there is $y$ such that $(a-\epsilon)_{+}=y^{*}y$ and
$yy^{*}\leq(b-\delta)_{+}$. Choose a continuous function $g$ with $g(0)=0$
such that $g(b)\leq Mb$ for some $M>0$ and
$g(b)(b-\delta)_{+}=(b-\delta)_{+}$. We have $g(b)yy^{*}=yy^{*}g(b)=yy^{*}$.
In particular, $yy^{*}$ and $g(b)$ commute and $yy^{*}\leq\|y\|^{2}g(b)$.
Hence $D(yy^{*})\leq D(g(b))\leq D(b)$, by (2) of the previous paragraph.
Thus, $D((a-\epsilon)_{+})=D(yy^{*})\leq D(b)$. Letting $\epsilon$ tend to $0$
and applying (3) of the previous paragraph we obtain $D(a)\leq D(b)$.

It follows from the previous discussion that $\lambda$ defined by
$\lambda([a]):=D(a)$ is well defined on $\mathrm{Cu}(A)$, additive, order
preserving, takes $0$ into $0$, and satisfies
$\lambda([a])=\sup_{\epsilon>0}\lambda([(a-\epsilon)_{+}])$. In order to show
that $\lambda$ is a functional it remains to show that it preserves the
suprema of arbitrary increasing sequences. Let $([a_{n}])$ be increasing with
supremum $[a]$. It is known that $[(a-\epsilon)_{+}]\ll[a]$ (see the first
paragraph of the next subsection for the definition of the relation $\ll$—and
see [coward-elliott-ivanescu] for the statement $[(a-\epsilon)_{+}]\ll[a]$).
This implies that $[(a-\epsilon)_{+}]\leq[a_{n}]$ for some $n$. Thus,
$\lambda([a-\epsilon)_{+}])\leq\sup_{n}\lambda[a_{n}]),$ and letting
$\epsilon$ go to $0$ we get $\lambda([a])\leq\sup_{n}\lambda([a_{n}])$. The
reverse inequality is clearly true, since $\lambda$ is order preserving.

Let us now start with a functional $\lambda$ and let $\tau_{\lambda}$ be
defined as in the statement of the proposition. If $B$ is commutative
sub-C*-algebra of $A\otimes\mathcal{K}$, and $a\in B^{+}$, then $\lambda([a])$
depends only on the set of points in the spectrum of $B$ where $a$ does not
vanish. Moreover, $\lambda$ defines a Borel measure on the spectrum of $B$ in
this way. By Fubini’s theorem, $\tau_{\lambda}(a)$ is the integral of $a$ with
respect to that measure. Therefore, $\tau_{\lambda}$ is additive on $B$ and
$\tau_{\lambda}(a)=\sup_{\epsilon>0}\tau_{\lambda}((a-\epsilon)_{+})$.

For every $x\in A\otimes\mathcal{K}$ and $\epsilon>0$ there exists $y$ such
that $(x^{*}x-\epsilon)_{+}=y^{*}y$ and $(xx^{*}-\epsilon)_{+}=yy^{*}$. Hence
$\lambda([x^{*}x-\epsilon)_{+}])=\lambda([xx^{*}-\epsilon)_{+}])$, and so
$\tau_{\lambda}(x^{*}x)=\tau_{\lambda}(xx^{*})$. It follows that
$\tau_{\lambda}$ is a quasitrace. We also know that
$\tau_{\lambda}(a)=\sup_{\epsilon>0}\tau_{\lambda}((a-\epsilon)_{+})$. This
implies that $\tau_{\lambda}$ is lower semicontinuous (see [kirchberg-
blanchard, Definition 2.22]).

Finally, we need to show that the maps $\tau\mapsto\tau_{\lambda}$ and
$\lambda\mapsto\lambda_{\tau}$ are inverse to each other. It is immediate from
the definitions of these two maps that it is enough to prove this on the
commutative sub-C*-algebra generated by a positive element. In this case the
result follows from standard results in the theory of integration. ∎

######  Remark 4.3.

A theorem of Haagerup says that if $A$ is exact and unital then every bounded
2-quasitrace on $A$ is a trace. It was observed in [kirchberg-blanchard,
Remark 2.29 (i)] that after a number of elementary reductions this theorem can
be extended to obtain that every lower semicontinuous 2-quasitrace on an exact
C*-algebra must be a trace. It follows that if $A$ is exact then every
functional on $\mathrm{Cu}(A)$ arises from a lower semicontinuous trace.

Let us endow $\mathrm{QT}_{2}(A)$ with the topology in which the net
$(\tau_{i})$ converges to $\tau$ if

(4.1) $\limsup\tau_{i}((a-\epsilon)_{+})\leq\tau(a)\leq\liminf\tau_{i}(a)$

for all $a\in(A\otimes\mathcal{K})^{+}$ and $\epsilon>0$. Alternatively (as
for $\mathrm{T}(A)$), $\tau_{i}\to\tau$ if, whenever a subnet of $(\tau_{i})$
converges pointwise to a function
$\sigma\colon(A\otimes\mathcal{K})^{+}\to[0,\infty]$, the regularization
$\widetilde{\sigma}$ of $\sigma$ given by Lemma 4.1 above is equal to $\tau$.
A neighbourhood basis of a point of $\mathrm{QT}_{2}(A)$ (or for a uniform
structure determining the topology) can be described just as before for
$\mathrm{T}(A)$. Notice that the relative topology of
$\mathrm{T}(A\otimes\mathcal{K})$, as a subset of $\mathrm{QT}_{2}(A)$, is the
same topology that we assigned to $\mathrm{T}(A\otimes\mathcal{K})$ in the
previous section.

Let us say that the net $(\lambda_{i})$ converges to $\lambda$ on the cone
$\mathrm{F}(\mathrm{Cu}(A))$ of functionals on $\mathrm{Cu}(A)$ if

$\limsup\lambda_{i}([(a-\epsilon)_{+}])\leq\lambda([a])\leq\liminf\lambda_{i}([a])$

for all $[a]\in\mathrm{Cu}(A)$ and $\epsilon>0$. Alternatively (as for
$\mathrm{T}(A)$), $\lambda_{i}\to\lambda$ if
$\lambda([a])=\sup_{\epsilon>0}\lambda^{\prime}([(a-\epsilon)_{+}])$ for all
$a\in(A\otimes\mathcal{K})^{+}$, whenever a subnet of $(\lambda_{i})$
converges pointwise to a function
$\lambda^{\prime}\colon\mathrm{Cu}(A)\to[0,\infty]$.

######  Theorem 4.4.

The cones $\mathrm{QT}_{2}(A)$ and $\mathrm{F}(\mathrm{Cu}(A))$ are compact
and Hausdorff, and the map $\tau\mapsto\lambda_{\tau}$ is a a homeomorphism
between them.

###### Proof.

The proof that $\mathrm{QT}_{2}(A)$ is compact and Hausdorff is similar to the
proof given above for $\mathrm{T}(A)$ (Theorem 3.7). This is also the case for
the proof that $\mathrm{F}(\mathrm{Cu}(A))$ is compact and Hausdorff (see
Theorem 4.8 below for a generalization of this). In order to show that
$\tau\mapsto\lambda_{\tau}$ is a homeomorphism it is enough to show that it is
continuous. Let $(\tau_{i})$ be a net in $\mathrm{QT}_{2}(A)$ such that
$\tau_{i}\to\tau$ in $\mathrm{QT}_{2}(A)$. Let $[a]\in\mathrm{Cu}(A)$. We have

$\tau(a^{1/n})\leq\liminf\tau_{i}(a^{1/n})\leq\liminf\lambda_{\tau_{i}}([a]),$

for all $n$. Therefore,
$\lambda_{\tau}([a])\leq\liminf\lambda_{\tau_{i}}([a])$.

Let $\epsilon>0$. Choose $f\in C_{0}(\mathbb{R}^{+})$ such that $0\leq f\leq
1$ and $f(t)=1$ for $t\in[\epsilon,\|a\|]$. Set $f(a)=a^{\prime}$. We have

$(1-\epsilon)\limsup\lambda_{\tau_{i}}([(a-\epsilon)_{+}])\leq\limsup\tau_{i}(a^{\prime}-\epsilon)_{+}\leq\tau(a^{\prime})\leq\lambda_{\tau}([a]),$

for all $\epsilon>0$. This implies that
$\limsup\lambda_{\tau_{i}}([(a-\epsilon)_{+}])\leq\lambda_{\tau}([a])$. ∎

######  Remark 4.5.

If $A$ is exact then $\mathrm{T}(A\otimes\mathcal{K})=\mathrm{QT}_{2}(A)$ (see
Remark 4.3). Since it is always the case that
$\mathrm{T}(A)\cong\mathrm{T}(A\otimes\mathcal{K})$, we have that if $A$ is
exact then
$\mathrm{T}(A)\cong\mathrm{T}(A\otimes\mathcal{K})=\mathrm{QT}_{2}(A)\cong\mathrm{F}(\mathrm{Cu}(A))$.

###  4.2. The category $\mathcal{C}u$

In [coward-elliott-ivanescu], Coward, Elliott, and Ivanescu showed that
$\mathrm{Cu}(A)$ belongs to a particular category of ordered semigroups
denoted by $\mathcal{C}u$. Let us recall the definition of this category here.

For elements $a$ and $b$ of an ordered set, let us say that $a$ is far below
$b$, and write $a\ll b$, if for any increasing sequence $(b_{n})$ with
supremum greater than or equal to $b$ there exists $n$ such that $a\leq
b_{n}$. (Then in particular $a\leq b$.) The category $\mathcal{C}u$ has for
objects the ordered semigroups $S$ with 0 such that

(1) increasing sequences in $S$ have a supremum,

(2) for every $a\in S$ there is a sequence $a_{1},a_{2},\dots$ with supremum
$a$ such that $a_{n}\ll a_{n+1}$ for all $n$,

(3) if $a_{1}\ll b_{1}$ and $a_{2}\ll b_{2}$ then $a_{1}+a_{2}\ll
b_{1}+b_{2}$, and

(4) if $(a_{n})$ and $(b_{n})$ are increasing sequences then
$\sup(a_{n}+b_{n})=\sup a_{n}+\sup b_{n}$.

The morphisms of the category $\mathcal{C}u$ are the ordered semigroup
morphisms (i.e., the additive and order preserving maps) that preserve suprema
of increasing sequences and the far below relation.

######  Remark 4.6.

The far below relation (also referred to as the way below relation, or, more
formally, compact containment), is usually defined with respect to increasing
nets $(b_{i})$ instead of increasing (countable) sequences $(b_{n})$.
Nevertheless, it is (countable) increasing sequences that we wish to consider
here. To avoid confusion we might say countable compact containment.

Let $S$ be a semigroup in the category $\mathcal{C}u$. Let us call functionals
on $S$ those additive and order preserving functions from $S$ to $[0,\infty]$
that take 0 into 0 and preserve the suprema of increasing sequences. Let us
denote by $\mathrm{F}(S)$ the cone of functionals on $S$ endowed with
pointwise addition and scalar multiplication by strictly positive real
numbers. Let us consider $\mathrm{F}(S)$ with the topology in which a net
$(\lambda_{i})$ converges to a point $\lambda$ if

(4.2)
$\displaystyle\limsup\lambda_{i}(x)\leq\lambda(y)\leq\liminf\lambda_{i}(y)$

for all $x,y\in S$ such that $x\ll y$. If $\phi\colon S\to T$ is a morphism in
the category $\mathcal{C}u$ then
$\mathrm{F}(\phi)(\lambda):=\lambda\circ\phi$, $\lambda\in\mathrm{F}(T)$, is a
continuous linear map from $\mathrm{F}(T)$ to $\mathrm{F}(S)$.

######  Lemma 4.7.

Let $S$ be a semigroup in the category $\mathcal{C}u$ and let $\lambda\colon
S\to[0,\infty]$ be an additive map on $S$. Then $\widetilde{\lambda}$ defined
by $\widetilde{\lambda}(x):=\sup\\{\,\lambda(z)\mid z\ll x\,\\}$ is a
functional on $S$, and is the largest functional majorized by $\lambda$.

###### Proof.

If $x\leq y$ and $x^{\prime}\ll x$ then $x^{\prime}\ll y$. This allows us to
conclude that $\widetilde{\lambda}(x)\leq\widetilde{\lambda}(y)$. If
$x^{\prime}\ll x$ and $y^{\prime}\ll y$ then $x^{\prime}+y^{\prime}\ll x+y$,
and so
$\widetilde{\lambda}(x)+\widetilde{\lambda}(y)\leq\widetilde{\lambda}(x+y)$.
If $z\ll x+y$ then there are $x^{\prime}\ll x$ and $y^{\prime}\ll y$ such that
$z\ll x^{\prime}+y^{\prime}$. It follows that
$\widetilde{\lambda}(x+y)\leq\widetilde{\lambda}(x)+\widetilde{\lambda}(y)$.
Finally, let $(x_{n})$ be an increasing sequence with supremum $x$. Since
$\widetilde{\lambda}$ is order preserving we have
$\sup\widetilde{\lambda}(x_{n})\leq\widetilde{\lambda}(x)$. On the other hand,
for all $x^{\prime}\ll x$ we have $x^{\prime}\ll x_{n}\leq x$ for some $n$.
Therefore, $\widetilde{\lambda}(x)\leq\sup\widetilde{\lambda}(x_{n})$. The
last statement is proved as in 3.1 and 4.1. ∎

The order of pointwise comparison of functionals in $\mathrm{F}(S)$ is the
same as the order arising from the semigroup structure. The proof of this is
identical to the proof for $\mathrm{T}(A)$ (see Proposition 3.2), provided
that Lemma 4.7 is used instead of Lemma 3.1.

Just as with $\mathrm{T}(A)$, we may extend the scalar multiplication of
$\mathrm{F}(S)$ to include 0 and $\infty$:

(4.3) $\displaystyle(\infty\cdot\lambda)(x)$ $\displaystyle:=0\hbox{ if
}\lambda(x)=0,\hbox{ and }(\infty\cdot\lambda)(x):=\infty\hbox{ otherwise},$
(4.4) $\displaystyle(0\cdot\lambda)(x)$ $\displaystyle:=0\hbox{ if
}\lambda(z)<\infty,\forall z\ll x,\hbox{ and }(0\cdot\lambda)(x)=\infty\hbox{
otherwise.}$

Notice that $0\cdot\lambda$ is the regularization (as in the statement of
Lemma 4.7) of the additive map $x\mapsto 0\cdot\lambda(x)$, where
$0\cdot\infty$ is taken to be $\infty$. Notice also that $\alpha\lambda\to
0\cdot\lambda$ and $\frac{1}{\alpha}\lambda\to\infty\cdot\lambda$ as
$\alpha\to 0$, and, indeed, extended scalar multiplication is (jointly)
continuous overall.

######  Theorem 4.8.

$\mathrm{F}(\cdot)$ is a sequentially continuous contravariant functor from
the category $\mathcal{C}u$ to the category of topological cones
$\mathcal{C}$.

###### Proof.

The proof that $\mathrm{F}(S)$ is compact and Hausdorff is similar to the
proof for $\mathrm{T}(A)$ (cf. also Theorem 4.4). We use Lemma 4.7 instead of
Lemma 3.1.

Let us show that $\mathrm{F}(\cdot)$ is a sequentially continuous functor. As
can be seen from the construction given in [coward-elliott-ivanescu],
inductive limits in the category $\mathcal{C}u$ are characterized as follows:
$S$ is the inductive limit of $(S_{i},\phi_{i,j})$ if

(1) every element of $S$ is supremum of an increasing sequence of elements
coming from the $S_{i}$s,

(2) if $x,y\in S_{i}$ are such that $\phi_{i,\infty}(x)\leq\phi_{i,\infty}(y)$
in the limit, then for all $z\ll x$ in $S_{i}$ there is $n\geq i$ such that
$\phi_{i,n}(z)\leq\phi_{i,n}(y)$ in $S_{n}$.

Let $C$ denote the projective limit of
$(\mathrm{F}(S_{i}),\mathrm{F}(\phi_{i,j}))$ in the category $\mathcal{C}$
(cf. proof of Theorem 3.11). If $\lambda_{1}$ and $\lambda_{2}$ are two
functionals on $S$ that agree on the elements coming from finite stages, then
$\lambda_{1}$ and $\lambda_{2}$ are equal by the property (1) above of
inductive limits in $\mathcal{C}u$. Thus, the map from $C$ to $\mathrm{F}(S)$
is in injective. In order to see that this map is surjective we need to show
that for any sequence of functionals $\lambda_{i}\in\mathrm{F}(S_{i})$
compatible with the inductive limit, there is $\lambda\in\mathrm{F}(S)$ such
that $\mathrm{F}(\phi_{i,\infty})(\lambda)=\lambda_{i}$. Let us define
$\lambda$ on the subsemigroup $\bigcup_{i}\phi_{i,\infty}(S_{i})$ of $S$ by
$\lambda(\phi_{i,\infty}(x))=\lambda_{i}(x)$. Let us check that this map is
well defined. Suppose that $\phi_{i,\infty}(x)=\phi_{i,\infty}(y)$. Then by
the property (2) of inductive limits in the category $\mathcal{C}u$, for every
$z\ll x$ there is $n$ such that $\phi_{i,n}(x)\leq\phi_{i,n}(y)$. So
$\lambda_{i}(z)=\lambda_{n}(\phi_{i,n}(z))\leq\lambda_{n}(\phi_{i,n}(y))=\lambda_{i}(y)$.
Since this holds for all $z\ll x$ we have $\lambda_{i}(x)\leq\lambda_{i}(y)$,
whence, by symmetry, $\lambda_{i}(x)=\lambda_{i}(y)$.

Let us write $T=\bigcup_{i}\phi_{i,\infty}(S_{i})$. Let us extend $\lambda$
from $T$ to all of $S$ as follows:

$\widetilde{\lambda}(x)=\sup\\{\,\lambda(x^{\prime})\mid x^{\prime}\ll
x,x^{\prime}\in T\,\\}.$

One can now show that $\widetilde{\lambda}$ is a functional on $S$ that
extends $\lambda$. We will only show here that $\widetilde{\lambda}$ is
additive. Let $x,y\in S$. Let $x^{\prime}\ll x$, $y^{\prime}\ll y$ and
$x^{\prime},y^{\prime}\in T$. Then $x^{\prime}+y^{\prime}\ll x+y$ and
$x^{\prime}+y^{\prime}\in T$. This implies that $\widetilde{\lambda}$ is
superadditive. On the other hand, if $z^{\prime}\ll x+y$ , $z^{\prime}\in T$,
then there are $x^{\prime},y^{\prime}\in T$ such that $z\leq
x^{\prime}+y^{\prime}\ll x+y$ and $x^{\prime}\ll x$, $y^{\prime}\ll y$. From
this we conclude that $\widetilde{\lambda}$ is subadditive. ∎

######  Remark 4.9.

It was shown in [coward-elliott-ivanescu] that $\mathrm{Cu}(\cdot)$ is a
sequentially continuous covariant functor from the category of C*-algebras to
the category $\mathcal{C}u$. Therefore, by Theorem 4.8,
$\mathrm{F}(\mathrm{Cu}(\cdot))$ is a sequentially continuous contravariant
functor from the category of C*-algebras to the category $\mathcal{C}$.

##  5\. Dual cones for $\mathrm{F}(\mathrm{Cu}(A))$ and $\mathrm{T}(A)$

###  5.1. The space $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$

Before discussing the dual cone of the cone $\mathrm{F}(\mathrm{Cu}(A))$ let
us begin with some general considerations concerning the cones in the category
$\mathcal{C}$.

Let $C$ be a cone in the category $\mathcal{C}$. Let $\mathrm{Lsc}(C)$ denote
the set of lower semicontinuous functions on $C$ with values in $[0,\infty]$
that are additive, homogeneous (with respect to the scalar multiplication by
$\mathbb{R}^{+}$), and take 0 into 0. We shall regard $\mathrm{Lsc}(C)$ as a
non-cancellative cone endowed with the operations of pointwise addition and
pointwise multiplication by strictly positive scalars. We shall also consider
$\mathrm{Lsc}(C)$ as ordered by the order of pointwise comparison of
functions. Notice that $\mathrm{Lsc}(C)$ is closed under passage to suprema of
upward directed sets. Thus, we can extend the scalar multiplication to include
$\infty$ by setting $\sup_{n}n\cdot f=\infty\cdot f$. For functions $f$ and
$g$ in $\mathrm{Lsc}(C)$ we shall write $f\ll g$ if for every increasing
sequence $(g_{n})$ such that $g\leq\sup g_{n}$ there is $g_{n_{0}}$ such that
$f\leq g_{n_{0}}$.

Let us denote by $\mathrm{L}(C)$ the subset of $\mathrm{Lsc}(C)$ composed of
those functions $f$ for which there is an increasing sequence $(h_{n})$,
$h_{n}\in\mathrm{Lsc}(C)$, with the following two properties:

(I) the supremum of the $h_{n}$s is $f$,

(II) $h_{n}$ is continuous at each point where $h_{n+1}$ is finite.

The definition of $\mathrm{L}(C)$ is motivated by Proposition 5.3 below. It is
shown there that the functions arising from the positive elements of a
C*-algebra $A$ (and of $A\otimes\mathcal{K}$) on the cones $\mathrm{T}(A)$ and
$\mathrm{F}(\mathrm{Cu}(A))$ satisfy (I) and (II) (for suitable increasing
sequences).

For $f\in\mathrm{Lsc}(C)$ let us write $\mathrm{Set}(f)=\\{\,\lambda\in C\mid
f(\lambda)>1\,\\}$. Notice that $f\leq g$ if and only if
$\mathrm{Set}(f)\subseteq\mathrm{Set}(g)$.

######  Proposition 5.1.

Let $f$ and $g$ be in $\mathrm{Lsc}(C)$.

(i) If $f\leq(1-\mu)g$ for some $\mu>0$, and $f$ is continuous at each point
where $g$ is finite, then
$\overline{\mathrm{Set}(f)}\subseteq\mathrm{Set}(g)$.

(ii) If $\overline{\mathrm{Set}(f)}\subseteq\mathrm{Set}(g)$ then $f\ll g$ in
$\mathrm{Lsc}(C)$.

(iii) $\mathrm{L}(C)$ is a subcone of
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ closed under passage to suprema of
increasing sequences.

###### Proof.

(i) Let $(\lambda_{i})$ be a net in $C$ such that $\lambda_{i}\to\lambda$ and
$\lambda_{i}\in\mathrm{Set}(f)$. If $g(\lambda)<\infty$ then $f(\lambda)=\lim
f(\lambda_{i})\geq 1$, and so $g(\lambda)>1$.

(ii) Let $(g_{n})$ be an increasing sequence of functions in $\mathrm{Lsc}(C)$
with pointwise supremum greater than or equal to $g$. Then
$\mathrm{Set}(g)\subseteq\bigcup_{n}\mathrm{Set}(g_{n})$. Since
$\overline{\mathrm{Set}(f)}$ is compact, we must have that
$\overline{\mathrm{Set}(f)}\subseteq\mathrm{Set}(g_{n_{0}})$ for some $n_{0}$.
Therefore, $f\leq g_{n_{0}}$.

(iii) It follows easily from its definition that $\mathrm{L}(C)$ is closed
under addition and multiplication by strictly positive scalars.

Let $(f_{n})$ be an increasing sequence of functions in $\mathrm{L}(C)$ with
supremum $f$. For every $f_{n}$ let $(h_{k}^{n})_{k=1}^{\infty}$ be a choice
of the corresponding sequence satisfying (I) and (II). We may assume without
loss of generality that $h_{k}^{n}\leq(1-\mu_{k}^{n})h_{k+1}^{n}$ for some
$\mu_{k}^{n}>0$. Let $k_{1}=1$. Since $h_{k_{1}+1}^{1}\ll f_{2}$ on
$\mathrm{Lsc}(C)$ (by (i) and (ii)), there is $h_{k_{2}}^{2}$ such that
$h_{k_{1}+1}^{1}\leq h_{k_{2}}^{2}$. In the same way we may find
$h_{k_{3}}^{3}$ such that $h_{k_{2}+1}^{2}\leq h_{k_{3}}^{3}$. Continue in
this way to obtain a sequence $(h_{k_{n}}^{n})$ such that $h_{k_{n}+1}^{i}\leq
h_{k_{n+1}}^{n+1}$. By proceeding as in the proof of [coward-elliott-ivanescu,
Theorem 1 (i)], we can choose this sequence so that its supremum is $f$. If
$h_{k_{n+1}}^{n+1}(\lambda)<\infty$ then $h_{k_{n}+1}^{n}(\lambda)<\infty$,
and it follows that $h_{k_{n}}^{n}$ is continuous at $\lambda$. This shows
that $f$ belongs to $\mathrm{L}(C)$. ∎

######  Remark 5.2.

The cone $\mathrm{L}(C)$ is not to be confused with a cone belonging to the
category $\mathcal{C}$. In particular, no topology will be defined on
$\mathrm{L}(C)$. Instead, we shall consider $\mathrm{L}(C)$ as a non-
cancellative cone endowed with an order—that of pointwise comparison of
functions—which may not coincide with the order arising from the addition
operation of $\mathrm{L}(C)$. Also, we shall not define a scalar
multiplication by $0$ in $\mathrm{L}(C)$.

Let us now specialize the study of $\mathrm{L}(C)$ to the case that
$C=\mathrm{F}(\mathrm{Cu}(A))$ for some C*-algebra $A$. Our main result is
Theorem 5.7. As applications of this theorem we will obtain that
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is an ordered semigroup in the
category $\mathcal{C}u$ and that $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\cdot)))$
is a sequentially continuous functor from the category of C*-algebras to the
category $\mathcal{C}u$. We will also make use of this theorem in the next
section when we look at the structure of the Cuntz semigroup for certain
C*-algebras.

Let $A$ be a C*-algebra and $a$ be a positive element of
$A\otimes\mathcal{K}$. The Cuntz semigroup element $[a]$ and the positive
element $a$ give rise to functions on $\mathrm{F}(\mathrm{Cu}(A))$:

(5.1) $\displaystyle\widehat{[a]}(\lambda)$ $\displaystyle:=\lambda([a]),$
(5.2) $\displaystyle\widehat{a}(\lambda)$
$\displaystyle:=\tau_{\lambda}(a)=\int_{0}^{\infty}\lambda([(a-t)_{+}])\,dt,$

where $\lambda\in\mathrm{F}(\mathrm{Cu}(A))$ and $\tau_{\lambda}$ is the
quasitrace associated to $\lambda$ by Proposition 4.2. The function
$\widehat{[a]}$ belongs to $\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ by the
inequalities (4.2) that define the topology on $\mathrm{F}(\mathrm{Cu}(A))$).
The function $\widehat{a}$ belongs to
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ by the inequalities that define the
topology on $\mathrm{QT}_{2}(A)$ and the isomorphism between
$\mathrm{F}(\mathrm{Cu}(A))$ and $\mathrm{QT}_{2}(A)$.

For the rest of this section if $a$ is a positive element of a C*-algebra we
will use the notation $a_{\epsilon}$ to mean the positive element
$(a-\epsilon)_{+}$.

######  Proposition 5.3.

For all $a\in A^{+}$ and $\epsilon>0$ the function $\widehat{a_{\epsilon}}$ is
continuous at each point where $\widehat{a}$ is finite.

###### Proof.

Let $\lambda$ be such that $\widehat{a}(\lambda)<\infty$ and let
$(\lambda_{i})$ be a net in $\mathrm{F}(\mathrm{Cu}(A))$ that converges to
$\lambda$. We have
$\tau_{\lambda}(a_{\epsilon})\leq\liminf\tau_{\lambda_{i}}(a_{\epsilon})$. Let
$\mu>0$ and set $a^{\prime}=a_{\epsilon}+\mu a$. There is
$\epsilon^{\prime}>0$ such that $a_{\epsilon}\leq
a_{\epsilon^{\prime}}^{\prime}$ (this is easily verified in $C^{*}(a)$).
Therefore,
$\limsup\tau_{\lambda_{i}}(a_{\epsilon})\leq\tau_{\lambda}(a^{\prime})=\tau_{\lambda}(a_{\epsilon})+\mu\tau_{\lambda}(a)$.
This is true for all $\mu>0$. Since $\tau_{\lambda}(a)$ is finite we conclude
that
$\limsup\tau_{\lambda_{i}}(a_{\epsilon})\leq\tau_{\lambda}(a_{\epsilon})$. ∎

Proposition 5.3 implies that $\widehat{a}$ is in
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ for every
$a\in(A\otimes\mathcal{K})^{+}$. Since
$\widehat{[a]}=\sup_{n}\widehat{(a^{1/n})}$ and
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is closed under passage to suprema of
increasing sequences, $\widehat{[a]}$ is in
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ too. Notice also that since
$\widehat{a_{\epsilon}}\leq(1-\epsilon)\widehat{a}$, we have that
$\widehat{a_{\epsilon}}\ll\widehat{a}$, by Propositions 5.1 and 5.3.

Let $I$ be a closed two-sided ideal of $A\otimes\mathcal{K}$. Let
$f_{I}\colon\mathrm{F}(\mathrm{Cu}(A))\to[0,\infty]$ denote the function given
by

(5.5) $\displaystyle f_{I}(\lambda)=\left\\{\begin{array}[]{ll}0&\hbox{if
}\lambda([a])=0\hbox{ for all }a\in I^{+},\\\
\infty&\hbox{otherwise}.\end{array}\right.$

It can be verified that $f_{I}$ is in
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$. Moreover, every function in
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ with the only possible values $0$
and $\infty$ has the form $f_{I}$ for some ideal $I$. For
$f\in\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ let us write
$\mathrm{Ideal}(f)$ for the ideal of $A\otimes\mathcal{K}$ such that
$\infty\cdot f=f_{\mathrm{Ideal}(f)}$. If $a$ is a positive element then
$\mathrm{Ideal}(\widehat{a})$ is the closed two-sided ideal generated by $a$
(i.e., $\mathrm{Ideal}(a)$).

######  Lemma 5.4.

Let $f,g\in\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$. If
$\overline{\mathrm{Set}(f)}\subseteq\mathrm{Set}(g)$ then
$\mathrm{Ideal}(f)\ll\mathrm{Ideal}(g)$ (in $\mathrm{Lat}(A)$).

###### Proof.

Let $(I_{i})$ be an upward directed collection of ideals with supremum
$\mathrm{Ideal}(g)$. The functions $f_{I_{i}}$ form an upward directed subset
of $\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ with supremum $\infty\cdot g$,
and so $\mathrm{Set}(\infty\cdot g)=\bigcup_{i}\mathrm{Set}(f_{I_{i}})$. Since
$\overline{\mathrm{Set}(f)}$ is compact and
$\overline{\mathrm{Set}(f)}\subseteq\mathrm{Set}(g)\subseteq\mathrm{Set}(\infty\cdot
g)$ (as $g\leq\infty\cdot g$), we have
$\mathrm{Set}(f)\subseteq\mathrm{Set}(f_{I_{i_{0}}})$ for some $i_{0}$. From
this we get $\infty\cdot f\leq f_{I_{i_{0}}}$ that is to say,
$\mathrm{Ideal}(f)\subseteq I_{i_{0}}$. ∎

The following proposition relies on a result in the duality theory of
topological vector spaces.

######  Proposition 5.5.

Let $\mathrm{F}_{A}(\mathrm{Cu}(A))$ denote the subcone of
$\mathrm{F}(\mathrm{Cu}(A))$ of functionals such that $0\cdot\lambda=0$. Let
$\mathrm{V}(\mathrm{F}_{A}(\mathrm{Cu}(A)))$ denote the ordered vector space
of linear, real-valued, continuous functions on
$\mathrm{F}_{A}(\mathrm{Cu}(A))$. Then for every positive linear functional
$\Lambda\colon\mathrm{V}(\mathrm{F}_{A}(\mathrm{Cu}(A)))\to\mathbb{R}$ there
is $\lambda\in\mathrm{F}_{A}(\mathrm{Cu}(A))$ such that
$\Lambda(f)=f(\lambda)$.

###### Proof.

By (4.4), $0\cdot\lambda=0$ if and only if $\lambda([a_{\epsilon}])<\infty$
for all $a\in(A\otimes\mathcal{K})^{+}$ and $\epsilon>0$. Let us identify—via
Proposition 4.2—the subcone $\mathrm{F}_{A}(\mathrm{Cu}(A))$ with the
quasitraces of $A\otimes\mathcal{K}$ that are densely finite. Notice that
$\mathrm{F}_{A}(\mathrm{Cu}(A))$ is a cancellative cone, since the quasitraces
in it are densely finite.

Let us show that the relative topology on $\mathrm{F}_{A}(\mathrm{Cu}(A))$
induced by the topology of $\mathrm{F}(\mathrm{Cu}(A))$ is the topology of
pointwise convergence on the set $\\{\,a_{\epsilon}\mid
a\in(A\otimes\mathcal{K})^{+},\epsilon>0\,\\}$. Let $(\lambda_{i})$ be a net
in $\mathrm{F}_{A}(\mathrm{Cu}(A))$ and
$\lambda\in\mathrm{F}_{A}(\mathrm{Cu}(A))$. Suppose that
$\lambda_{i}\to\lambda$ in the topology of $\mathrm{F}(\mathrm{Cu}(A))$. Since
$\widehat{a_{\epsilon/2}}(\lambda)<\infty$ for all
$a\in(A\otimes\mathcal{K})^{+}$ and $\epsilon>0$, by Proposition 5.3
$\widehat{a_{\epsilon}}(\lambda_{i})\to\widehat{a_{\epsilon}}(\lambda)$.
Suppose on the other hand that
$\widehat{a_{\epsilon}}(\lambda_{i})\to\widehat{a_{\epsilon}}(\lambda)$ for
all $a\in A^{+}$ and $\epsilon>0$. Then we may proceed as in the proof of
Proposition 3.10 (i) to conclude that $\lambda_{i}\to\lambda$ in the relative
topology of $\mathrm{F}_{A}(\mathrm{Cu}(A))$.

We conclude that the relative topology on $\mathrm{F}_{A}(\mathrm{Cu}(A))$ is
the weak*-topology of pointwise convergence on the set $\\{\,a_{\epsilon}\mid
a\in(A\otimes\mathcal{K})^{+}\,\\}$. Therefore,
$\mathrm{F}_{A}(\mathrm{Cu}(A))$ is a weakly complete cancellative cone in the
class $\mathcal{S}$ of Choquet (see [choquet, page 194]). Now the theorem
follows from [choquet, Proposition 30.7]. ∎

######  Lemma 5.6.

Let $h_{1},h_{2},h_{3}\in\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ be such
that $h_{i}\leq(1-\mu_{i})h_{i+1}$ and $h_{i}$ is continuous at each point
where $h_{i+1}$ is finite, for $i=1,2$ and some $\mu_{1},\mu_{2}>0$. Then for
every $\delta>0$ there is $a\in(A\otimes\mathcal{K})^{+}$ such that
$\widehat{a}\leq h_{3}$ and $h_{1}\leq\delta h_{3}+\widehat{a}$.

###### Proof.

We may assume without loss of generality that
$\mathrm{Ideal}(h_{3})=A\otimes\mathcal{K}$, i.e., $h_{3}(\lambda)>0$ unless
$\lambda=0$. Then we have $\widehat{b}\leq\infty\cdot h_{3}$ for every
$b\in(A\otimes\mathcal{K})^{+}$. Since $\widehat{b_{\epsilon}}\ll\widehat{b}$
for every $\epsilon>0$, we obtain that $\widehat{b_{\epsilon}}\leq Mh_{3}$ for
some finite $M>0$. Set $K=\\{\,\lambda\in\mathrm{F}(\mathrm{Cu}(A))\mid
h_{3}(\lambda)\leq 1\,\\}$. By Proposition 5.3, the function
$\widehat{b_{\epsilon}}$ is continuous on $K$ for all
$b\in(A\otimes\mathcal{K})^{+}$ and all $\epsilon>0$. Also, by hypothesis,
$h_{1}$ is continuous on $K$.

Let us show that $h_{1}$ can be uniformly approximated on $K$ by convex
combinations of functions of the form $\widehat{b_{\epsilon}}$. Suppose the
contrary. Then there is a real measure $m$ on $K$ such that
$\int\widehat{b_{\epsilon}}\,dm=0$ for all $b\in(A\otimes\mathcal{K})^{+}$ and
$\epsilon>0$, and $\int h_{1}\,dm=1$. Let $m=m_{+}-m_{-}$ denote the Jordan
decomposition of $m$. Then

$\int h_{1}\,dm_{+}=\int h_{1}\,dm_{-}+1\hbox{\quad and
\quad}\int\widehat{b_{\epsilon}}\,dm_{+}=\int\widehat{b_{\epsilon}}\,dm_{-},\hbox{\quad
for all }b,\epsilon.$

For $\lambda\in K$ we have $h_{3}(0\cdot\lambda)=0$, and since
$\mathrm{Ideal}(h_{3})=A\otimes\mathcal{K}$, we conclude that
$0\cdot\lambda=0$. Therefore, $K$ is contained in
$\mathrm{F}_{A}(\mathrm{Cu}(A))$, with $\mathrm{F}_{A}(\mathrm{Cu}(A))$ as
defined in Proposition 5.5. So we can define positive linear functionals
$\Lambda_{+}$, $\Lambda_{-}$ on the vector space
$\mathrm{V}(\mathrm{F}_{A}(\mathrm{Cu}(A)))$ by $\Lambda_{+}(f)=\int
fd\,m_{+}$ and $\Lambda_{-}=\int f\,dm_{-}$. By Proposition 5.5, $\Lambda_{+}$
and $\Lambda_{-}$ are given by evaluation on functionals $\lambda_{+}$ and
$\lambda_{-}$ belonging to $\mathrm{F}_{A}(\mathrm{Cu}(A))$. For all
$b\in(A\otimes\mathcal{K})^{+}$ and $\epsilon>0$ we have
$b_{\epsilon}\in\mathrm{V}(\mathrm{F}_{A}(\mathrm{Cu}(A)))$. So
$\widehat{b_{\epsilon}}(\lambda_{+})=\widehat{b_{\epsilon}}(\lambda_{-})$ for
all $b$ and $\epsilon$. This implies that $\lambda_{+}=\lambda_{-}$.

Let us show that the restriction of $h_{1}$ to
$\mathrm{F}_{A}(\mathrm{Cu}(A))$ is also in
$\mathrm{V}(\mathrm{F}_{A}(\mathrm{Cu}(A)))$. Let
$\lambda\in\mathrm{F}_{A}(\mathrm{Cu}(A))$, i.e., $0\cdot\lambda=0$. If
$h_{2}(\lambda)=\infty$ then $h_{2}(\lambda/n)=\infty$ for all $n$. Since
$\overline{\mathrm{Set}(h_{2})}\subseteq\mathrm{Set}(h_{3})$ (by Proposition
5.1 (i)), $0\cdot\lambda\in\mathrm{Set}(h_{3})$. This implies that
$h_{3}(0\cdot\lambda)=\infty$, which contradicts the equation
$0\cdot\lambda=0$. We conclude that $h_{2}(\lambda)<\infty$ for all
$\lambda\in\mathrm{F}_{A}(\mathrm{Cu}(A))$, and so $h_{1}$ is continuous on
$\mathrm{F}_{A}(\mathrm{Cu}(A))$.

We now have $h_{1}(\lambda_{+})=h_{1}(\lambda_{-})+1$. This contradicts the
earlier conclusion $\lambda_{+}=\lambda_{-}$. Therefore, the restriction of
$h_{1}$ to $K$ belongs to the closure of the convex set spanned by the
functions $\widehat{b_{\epsilon}}$. Hence, for every $\delta>0$ there exists a
positive element $a$ such that $\|h_{1}-a\|_{K}<\delta$. Equivalently,
$h_{1}\leq\widehat{a}+\delta h_{3}$ and $\widehat{a}\leq h_{1}+\delta h_{3}$
on $K$. It is easily shown that these inequalities also hold on all
$\mathrm{F}(\mathrm{Cu}(A))$. Changing $\widehat{a}$ to
$\widehat{a}/(1+\delta)$ we can arrange that $\widehat{a}\leq h_{3}$. ∎

######  Theorem 5.7.

Let $f$ be in $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. Then $f$ is the
supremum of an increasing sequence of $\widehat{a}$s
($a\in(A\otimes\mathcal{K})^{+}$). Such a sequence may even be chosen to be
rapidly increasing: $a_{1}\ll a_{2}\ll\cdots$.

###### Proof.

Let $(h_{n})$ be a an increasing sequence satisfying (I) and (II). We may
assume without loss of generality that $h_{n}\leq(1-\mu_{n})h_{n+1}$ for some
$\mu_{n}>0$, for all $n$. By Proposition 5.1,
$\overline{\mathrm{Set}(h_{n})}\subseteq\mathrm{Set}(h_{n+1})$ and $h_{n}\ll
h_{n+1}$ for all $n$. Hence, by Lemma 5.4,
$\mathrm{Ideal}(h_{n})\ll\mathrm{Ideal}(h_{n+1})$ in $\mathrm{Lat}(A)$ for all
$n$. Let us choose $b\in(A\otimes\mathcal{K})^{+}$ and $\epsilon_{0}>0$ such
that $\mathrm{Ideal}(h_{4})\subseteq\mathrm{Ideal}(b_{\epsilon_{0}})$ and
$\mathrm{Ideal}(b)\subseteq\mathrm{Ideal}(h_{5})$. We have that
$h_{4}\leq\infty\cdot\widehat{b_{\epsilon_{0}}}$ and
$\widehat{b}\leq\infty\cdot h_{5}$. Therefore, there is a constant $M>0$ such
that $h_{2}\leq M\widehat{b_{\epsilon_{0}}}$ and
$\widehat{b_{\epsilon_{0}}}\leq Mh_{5}$. Let us choose $\delta$ such that
$\delta M<\mu_{3}$. Finally, using Lemma 5.6, let us find $a$ in
$(A\otimes\mathcal{K})^{+}$ such that $\widehat{a}\leq h_{3}$ and
$h_{1}\leq(\delta/M)h_{3}+\widehat{a}$.

By the stability of $A\otimes\mathcal{K}$, we may assume that the positive
elements $a$ and $b$ that we found in the previous paragraph are orthogonal to
each other. (If they are not, we may replace them by Murray-von Neumann
equivalent elements that are orthogonal.) Let $a_{1}=a+\delta
b_{\epsilon_{0}}$. Then

$\widehat{a}_{1}=\widehat{a}+\delta\widehat{b_{\epsilon_{0}}}\leq(1-\mu_{3}+\delta
M)h_{5}\leq h_{5}.$

Also

$\widehat{a}_{1}=\widehat{a}+\delta\widehat{b_{\epsilon_{0}}}\geq\widehat{a}+\frac{\delta}{M}h_{3}\geq
h_{1}.$

So $h_{1}\leq\widehat{a}_{1}\leq h_{5}$. In the same way we may find
$\widehat{a}_{2}$ such that $h_{5}\leq\widehat{a}_{2}\leq h_{9}$. Continuing
in this way we get the desired sequence—by Proposition 5.1 (i) and (ii) even
rapidly increasing. ∎

######  Corollary 5.8.

Every function in $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is the supremum of
an increasing (even rapidly increasing) sequence of functions of the form
$\sum_{i=1}^{n}\alpha_{i}\widehat{[c_{i}]}$, where
$\alpha_{i}\in\mathbb{R}^{+}$ and $[c_{i}]\in\mathrm{Cu}(A)$ for
$i=1,\dots,n$.

###### Proof.

By (5.2)—cf. Proposition 4.2—, the function $\widehat{a}$ is the supremum of
the increasing sequence of Riemann sums
$\sum_{i=1}^{2^{n}}1/2^{n}\widehat{[(a-i/2^{n})]}$. Let $f$ be an arbitrary
function in $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. Then by Theorem 5.7
there is a rapidly increasing sequence $(\widehat{a_{i}})$, with
$a_{i}\in(A\otimes\mathcal{K})^{+}$, such that $f=\sup\widehat{a_{i}}$. We may
now interpolate between $\widehat{a_{i}}$ and $\widehat{a_{i+i}}$ an element
of the form $\sum_{i=1}^{n}\alpha_{i}\widehat{[c_{i}]}$. This proves the
result. ∎

######  Corollary 5.9.

For every functional
$\Lambda\colon\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))\to[0,\infty]$ (i.e,
additive, order-preserving map, taking 0 into 0, and preserving suprema of
increasing sequences) there is $\lambda\in\mathrm{F}(\mathrm{Cu}(A))$ such
that $\Lambda(f)=f(\lambda)$.

###### Proof.

Set $\Lambda(\widehat{[a]})=\lambda([a])$. Then $\Lambda(f)=f(\lambda)$ for
every $f$ of the form $\widehat{[a]}$. By the previous corollary this equality
also holds all $f\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. ∎

Let $\phi\colon A\to B$ be homomorphism of C*-algebras. Recall that
$\mathrm{F}(\phi)$ is a continuous linear map from
$\mathrm{F}(\mathrm{Cu}(B))$ to $\mathrm{F}(\mathrm{Cu}(A))$. It follows that
$f\mapsto f\circ\mathrm{F}(\phi)$ maps
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(A)))$ to
$\mathrm{Lsc}(\mathrm{F}(\mathrm{Cu}(B)))$. Moreover, in this way
$\widehat{a}$ is mapped to $\widehat{\phi(a)}$. Thus, by Theorem 5.7,
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\phi)))(f):=f\circ\mathrm{F}(\mathrm{Cu}(\phi))$
is a map from $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ to
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$.

######  Theorem 5.10.

$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\cdot)))$ is a sequentially continuous
covariant functor from the category of C*-algebras to the category
$\mathcal{C}u$.

###### Proof.

Let us first show that $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is an ordered
semigroup in the category $\mathcal{C}u$. We have already seen that the
supremum of an increasing sequence in
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$, with respect to the pointwise order
under consideration, exists, and is equal to the pointwise supremum of the
sequence. By Theorem 5.7 every element is the supremum of a rapidly increasing
sequence (i.e., a sequence satisfying the axiom (2) of the category
$\mathcal{C}u$) of functions that also belong to
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. We clearly have the axiom (3) of the
category $\mathcal{C}u$ too, since the supremum of an increasing sequence of
functions in $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is the pointwise
supremum of the sequence. Suppose that $f_{1}\ll g_{1}$ and $f_{2}\ll g_{2}$
in $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. Let $h_{1}$ be such that
$f_{1}\leq h_{1}\leq(1-\mu)g_{1}$, and $h_{1}$ is continuous at each point
where $g_{1}$ is finite. Suppose that $h_{2}$ is in the same relationship with
respect to $f_{2}$ and $g_{2}$. Then $f_{1}+f_{2}\leq
h_{1}+h_{2}\leq(1-\mu)(g_{1}+g_{2})$, and $h_{1}+h_{2}$ is continuous at the
points where $g_{1}+g_{2}$ is finite. Hence, $f_{1}+f_{2}\ll g_{1}+g_{2}$.
This shows that $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is in $\mathcal{C}u$.

If $\phi\colon A\to B$ is a homomorphism of C*-algebras then
$\mathrm{F}(\mathrm{Cu}(\phi))$ is continuous and linear. Keeping this in
mind, it is easy to show that $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\phi)))$
preserves suprema of increasing sequences and the relation $\ll$.

Let $A=\varinjlim(A_{i},\phi_{i,j},i,j\in\mathbb{N})$ be a sequential
inductive limit of C*-algebras. Recall from the proof of Theorem 4.8 the two
conditions (1) and (2) that characterize inductive limits in the category
$\mathcal{C}u$. In order to show that $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$
is the inductive limit of the $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A_{i})))$s it
is enough to show that these conditions are satisfied with respect to
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ and the inductive system
$(\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A_{i}))),\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\phi_{i,j})))$.

Let us show that the condition (1) is satisfied. It was shown in [coward-
elliott-ivanescu, Theorem 2] that
$\mathrm{Cu}(A)=\varinjlim(\mathrm{Cu}(A_{i}),\mathrm{Cu}(\phi_{i,j}),i,j\in\mathbb{N})$
in the category $\mathcal{C}u$. Thus, for every $[c]$ in $\mathrm{Cu}(A)$
there is an increasing sequence $([a_{n}])$ of elements coming from the finite
stages of the limit and with supremum $[c]$. It follows that the same is true
for every element of $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ of the form
$\sum_{i=1}^{m}\alpha_{i}\widehat{[c_{i}]}$. Finally, since, by Corollary 5.8,
every $f\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is the supremum of an
increasing—even rapidly increasing—sequence of functions of the form
$\sum_{i=1}^{m}\alpha_{i}\widehat{[c_{i}]}$, we deduce by a standard argument
that $f$ is also the supremum of an increasing sequence of elements coming
from finite stages.

Let us now show that the condition (2) of the proof of Theorem 4.8 is
satisfied. For $h\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A_{1})))$ let us denote
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\phi_{1,i})))(h)$ by $h_{i}$, for
$i=2,\dots,\infty$. Let
$f^{\prime},f,g\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A_{1})))$ be such that
$f^{\prime}\ll f$ and $f_{\infty}\leq g_{\infty}$. Then the compact sets
$\overline{\mathrm{Set}(f_{i}^{\prime})}\cap\mathrm{Set}(g_{i})^{c}$ have as
projective limit the set
$\overline{\mathrm{Set}(f_{\infty}^{\prime})}\cap\mathrm{Set}(g_{\infty})^{c}$.
This last set is empty, since $f_{\infty}^{\prime}\ll g_{\infty}$. Therefore,
for some $i$ we must have
$\overline{\mathrm{Set}(f_{i}^{\prime})}\subseteq\mathrm{Set}(g_{i})$, and so
$f_{i}^{\prime}\leq g_{i}$. ∎

The following lemma will be used in the next section.

######  Lemma 5.11.

If $[a]\ll[b]$ then $\widehat{[a]}\ll(1+\delta)\widehat{[b]}$ for all
$\delta>0$.

###### Proof.

By Proposition 5.1 (ii), it is sufficient to show that
$\overline{\mathrm{Set}(\widehat{[a]})}\subseteq\mathrm{Set}((1+\delta)\widehat{[b]})$.
Let $(\lambda_{i})$ be a net in $\mathrm{Set}(\widehat{[a]})$ converging to
$\lambda$. By the definition of the topology of $\mathrm{F}(\mathrm{Cu}(A))$
we have $1\leq\limsup\lambda_{i}[a]\leq\lambda([b])$. So
$\lambda\in\mathrm{Set}((1+\delta)\widehat{[b]})$ for any $\delta>0$. ∎

###  5.2. The space $\mathrm{L}(\mathrm{T}(A))$

Here we briefly review the properties of the space
$\mathrm{L}(\mathrm{T}(A))$.

If $a\in A^{+}$ then $\bar{a}(\tau)=\tau(a)$ defines a lower semicontinuous
function in $\mathrm{L}(\mathrm{T}(A))$. All the propositions and lemmas that
were proved before for the ordered cone
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ have obvious counterparts for
$\mathrm{L}(\mathrm{T}(A))$. The proofs of these results are entirely
analogous to the ones that we have seen above. We therefore have the following
theorem:

######  Theorem 5.12.

Let $f$ be in $\mathrm{L}(\mathrm{T}(A))$. Then $f$ is the supremum of an
increasing sequence $(\bar{a}_{n})$, with $a_{n}\in A^{+}$.

######  Remark 5.13.

Notice that the positive elements $a_{n}$ are now chosen in the C*-algebra $A$
and not in $A\otimes\mathcal{K}$ (unlike in Theorem 5.7). The stability of
$A\otimes\mathcal{K}$ was used in the proof of Theorem 5.7 to find orthogonal
elements $a$ and $b$ that were Murray-von Neumann equivalent to two given
elements. This step is not needed in proving Theorem 5.12, since the
additivity on pairs of orthogonal elements of quasitraces is now replaced by
the full additivity of traces.

######  Remark 5.14.

As was done before for $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$, Theorem 5.12
may be used to show that $\mathrm{L}(\mathrm{T}(A))$ is an ordered semigroup
in the category $\mathcal{C}u$ and $\mathrm{L}(\mathrm{T}(\cdot))$ is a
continuous functor from the category of C*-algebras to the category
$\mathcal{C}u$. Theorem 5.12 is also used, and at the same time improved, in
[tracecomp, Theorem 2]. It is shown there that if $A$ is stable then
$f\in\mathrm{L}(\mathrm{T}(A))$ if and only if $f=\bar{a}$ for some $a\in
A^{+}$.

##  6\. The structure of the Cuntz semigroups of certain C*-algebras

Let us apply the results of the previous sections to study the structure of
the Cuntz semigroup, in certain well-behaved cases.

######  Proposition 6.1.

Suppose that $A$ is an AH C*-algebra with no dimension growth. Then there is a
constant $M$ such that for all $[a]$ and $[b]$ in $\mathrm{Cu}(A)$ we have
$\widehat{[a]}\leq\widehat{[b]}$ if and only if $k[a]\leq(k+M)[b]$ for all
$k\in\mathbb{N}$.

###### Proof.

First suppose that $A$ is a direct sum of homogeneous algebras. For
$a\in(A\otimes\mathcal{K})^{+}$ let $\operatorname{rank}([a])$ denote the
lower semicontinuous function on the spectrum of $A$ given by
$\operatorname{rank}([a])(x):=\operatorname{rank}(a(x))$. In this case we
shall obtain the desired result as a corollary of the following theorem of
Toms (see [toms, Theorem 3.15]):

_There is a constant $K$ such that for every finite dimensional compact
Hausdorff space $X$, if $[a],[b]\in\mathrm{Cu}(C_{0}(X))$ satisfy
$\operatorname{rank}[a]+K\dim X\leq\operatorname{rank}[b]$, then
$[a]\leq[b]$._

Let us see how. Suppose that $\widehat{[a]}\leq\widehat{[b]}$. This implies
that $\operatorname{rank}[a]\leq\operatorname{rank}[b]$. So
$k\cdot\operatorname{rank}[a]+K\dim X\leq k\cdot\operatorname{rank}[b]+K\dim
X$ for all $k\in\mathbb{N}$. We may assume without loss of generality that
$b(x)\neq 0$ for all $x$. So $\operatorname{rank}b\geq 1$ and
$k\cdot\operatorname{rank}[a]+K\dim X\leq(k+K\dim
X)\cdot\operatorname{rank}[b]$. We conclude that $k[a]\leq(k+M)[b]$ for
$M=K\dim X$ and all $k\in\mathbb{N}$.

Let $A$ be an AH algebra with no dimension growth. Suppose that
$A=\varinjlim(A_{i},\phi_{i,j},i,j\in\mathbb{N})$ where the $A_{i}$s are
homogeneous algebras with spectra of bounded dimension. Since
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(\cdot)))$ is a sequentially continuous
functor, $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is the limit of the
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A_{i})))$s in the category $\mathcal{C}u$.
First let us suppose that $[a]$ and $[b]$ come from finite stages of the
sequence $(A_{i})$, say, $[a]=\phi_{1,\infty}([a_{1}])$ and
$[b]=\phi_{1,\infty}([b_{1}])$. Let us write $a_{n}=\phi_{1,n}(a_{1})$ and
$b_{n}=\phi_{1,n}(b_{1})$. For every $\epsilon>0$ and $k>0$ we have
$\widehat{[(a_{1}-\epsilon)_{+}]}\ll(1+1/k)\widehat{[a_{1}]}$ by Lemma 5.11.
By the condition (2) for inductive limits in the category $\mathcal{C}u$ (see
proof of Theorem 4.8), there is $n$ such that
$\widehat{[(a_{n}-\epsilon)_{+}]}\leq(1+1/k)\widehat{[b_{n}]}$. Therefore,
$k\widehat{[(a_{n}-\epsilon)_{+}]}\leq(k+1)\widehat{[b_{n}]}$. We have already
established that as this is at a finite stage of the sequence $(A_{i})$ this
implies that $k[(a_{n}-\epsilon)_{+}]\leq(k+M)[b_{n}]$, where $M$ depends only
on the bound on the dimensions of the spectra of this $A_{i}s$—and, as we may
take the best choice for different inductive limit decompositions, therefore
depends only on $A$. Recalling that $a_{n}$ and $b_{n}$ map into $[a]$ and
$[b]$ in $A$, we have $k[(a-\epsilon)_{+}]\leq(k+M)[b]$.

Let us consider the case that $[a]$ and $[b]$ do not necessarily come from
finite stages. Let $[f_{i}]$ and $[g_{i}]$, $i=1,2,\dots$, be increasing
sequences of elements coming from finite stages and with suprema $[a]$ and
$[b]$ respectively. Since
$\widehat{[f_{i}]}\ll\widehat{[a]}\leq\widehat{[b]}$, there is $[g_{j}]$ such
that $\widehat{[f_{i}]}\leq\widehat{[g_{j}]}$. Hence as before,
$k[f_{i}]\leq(k+C)[g_{j}]\leq(k+M)[b]$. Passing to the supremum with respect
to $i$ we get that $k[a]\leq(k+M)[b]$. ∎

We now turn to the setting of arbitrary C*-algebras with almost unperforated
Cuntz semigroup. Recall from [rordam] that an ordered semigroup is said to be
almost unperforated if the inequality $(k+1)x\leq ky$ for some
$k\in\mathbb{N}$ implies that $x\leq y$.

The following proposition is an improvement of [rordam, Proposition 3.2] for
ordered semigroups in the category $\mathcal{C}u$.

######  Proposition 6.2.

Let $S$ be an ordered semigroup in the category $\mathcal{C}u$. Then $S$ is
almost unperforated if and only if the following condition is fulfilled: for
all $x$ and $y$ in $S$ with $x\leq\infty\cdot y$ and $\lambda(x)<\lambda(y)$
for any functional on $S$ such that $\lambda(y)=1$, one has $x\leq y$.

###### Proof.

Suppose that $S$ satisfies the condition of comparison of elements by
functionals described in the statement of the proposition (this condition is
often referred to as “strict comparison”). If $(k+1)x\leq ky$ then
$\lambda(x)\leq k/(k+1)<\lambda(y)$ for any $\lambda$ such that
$\lambda(y)=1$. Since $x\leq ky\leq\infty\cdot y$, we conclude that $x\leq y$,
as desired.

Suppose that $S$ is almost unperforated. Let $x,y\in S$ be such that
$x\leq\infty\cdot y$ and $\lambda(x)<\lambda(y)$ for all $\lambda$ such that
$0<\lambda(y)<\infty$. Let $z\ll x$. Then $z\leq ky$ for some $k$. We shall
prove that for every additive, order preserving, function $D$ on $S$—not
necessarily preserving suprema of increasing sequences—such that $D(y)=1$, we
have $D(z)<D(y)$. By [rordam, Proposition 3.2], this will imply that $z\leq
y$, from which the desired result will follow on taking the supremum over all
$z$ that are far below $x$.

Let $D\colon S\to[0,\infty]$ be additive, order preserving, and such that
$D(y)=1$. Set $\widetilde{D}w:=\sup\\{\,Dw^{\prime}\mid w^{\prime}\ll w\,\\}$,
for each $w\in S$. By Lemma 4.7, $\widetilde{D}$ is a functional on $S$, and
it is clear that $\widetilde{D}(y)\leq 1<\infty$.

Case 1. Suppose that $\widetilde{D}y\neq 0$. Then

$D(z)\leq\widetilde{D}(x)<\widetilde{D}(y)\leq D(y).$

Case 2. Suppose that $\widetilde{D}y=0$. Then $\widetilde{D}x=0$ (because
$x\leq\infty\cdot y$), and so

$D(z)\leq\widetilde{D}(x)=0<D(y).\qed$

######  Corollary 6.3.

Let $S$ be an almost unperforated ordered semigroup in the category
$\mathcal{C}u$. Let $x,y\in S$. Then $\lambda(x)\leq\lambda(y)$ for every
functional $\lambda$ on $S$ if and only if $kx\leq(k+1)y$ for all
$k\in\mathbb{N}$.

###### Proof.

The implication that $\lambda(x)\leq\lambda(y)$ for every functional $\lambda$
if $kx\leq(k+1)y$ for all $k\in\mathbb{N}$ is obvious. Let us prove the
converse.

By considering functionals with the only possible values $0$ or $\infty$ we
conclude that $x\leq\infty y$. We may now apply the previous proposition with
$kx$ and $(k+1)y$ in place of $x$ and $y$, respectively. ∎

By a theorem of Rørdam (see [rordam, Theorem 4.5]), the Cuntz semigroup of a
C*-algebra that absorbs the Jiang-Su algebra is almost unperforated. It
follows by Corollary 6.3 that if $A$ absorbs the Jiang-Su algebra then
$\widehat{[a]}\leq\widehat{[b]}$ if and only if $k[a]\leq(k+1)[b]$ for all
$k\in\mathbb{N}$. In the sequel we shall denote the Jiang-Su algebra by the
letter $Z$.

Let us now try to identify $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ with a
subsemigroup of $\mathrm{Cu}(A)$. Notice that if
$f\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ is compact (i.e., $f\ll f$ in
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$) then $(1-\epsilon)f=f$ for some
$0\not=\epsilon\not=1$ and so $f$ takes 0 and $\infty$ as its only possible
values. This observation suggests the following definition.

Let us say that $[a]\in\mathrm{Cu}(A)$ is purely non-compact if it has the
property that if the image of $[a]$ in the quotient by some ideal $I$—let us
denote this by $[a_{I}]$—is compact (i.e., $[a_{I}]\ll[a_{I}]$), then
$[a_{I}]$ is a multiple of infinity (i.e., $2[a_{I}]=[a_{I}]$).

######  Proposition 6.4.

(i) The purely non-compact elements of $\mathrm{Cu}(A)$ form a subobject of
$\mathrm{Cu}(A)$, i.e., a subsemigroup closed under the passage to suprema of
increasing sequences.

(ii) If $[a]$ is purely non-compact then for all $[b]$, with $[b]\ll[a]$, and
for all $M\in\mathbb{N}$, we have $(k+M)[b]\leq k[a]$ for all sufficiently
large $k$.

(iii) If $\mathrm{Cu}(A)$ is almost unperforated the converse of (ii) is true.

(iv) If $A$ is simple then every element of $\mathrm{Cu}(A)$ is purely non-
compact except for the element $[p]$ where $p$ is any non-zero finite
projection.

(v) If $[a]=\sum_{i=1}^{\infty}[c_{i}]$ and
$\mathrm{Ideal}(c_{i})=\mathrm{Ideal}(a)$, then $[a]$ is purely non-compact.

###### Proof.

Before proving the proposition we need some preliminary formulas.

Suppose that $[a]$ is purely non-compact. Let $\epsilon>0$ and choose a
positive function $c_{\epsilon}(t)$ different from 0 precisely on the interval
$(0,\epsilon)$. Then $(a-\epsilon)_{+}$ and $c_{\epsilon}(a)$ are orthogonal
and $(a-\epsilon)_{+}+c_{\epsilon}(a)\leq C_{1}a\leq
C_{2}((a-\epsilon/2)_{+}+c_{\epsilon}(a))$ for some positive scalars $C_{1}$
and $C_{2}$. Hence,

(6.1)
$[(a-\epsilon)_{+}]+[c_{\epsilon}(a)]\leq[a]\leq[(a-\epsilon/2)_{+}]+[c_{\epsilon}(a)].$

These inequalities imply that $[a]$ is compact after passing to the quotient
by the ideal $\mathrm{Ideal}(c_{\epsilon}(a))$. Let us call this ideal $I$ and
let us denote with the subscript $I$ the images of elements of $A$ in $A/I$.
We have that $2[a_{I}]=[((a-\epsilon)_{+})_{I}]$. By [crs, Theorem 1], this
means that $2[a]\leq[(a-\epsilon)_{+}]+[g]$ for some $[g]$ such that
$\mathrm{Ideal}(g)=I$. Since $[(a-\epsilon)_{+}]\ll[a]$ and $[g]$ is the
supremum of an increasing sequence $[g_{i}]$ with $[g_{i}]\ll[g]$, so that
also $2[(a-\epsilon)_{+}]\ll
2[a]\leq[(a-\epsilon)_{+}]+[g]=\sup_{i}([(a-\epsilon)_{+}]+[g_{i}])$, there is
$[g^{\prime}]\ll[g]$ such that
$2[(a-\epsilon)_{+}]\leq[(a-\epsilon)_{+}]+[g^{\prime}]$. Since
$[g^{\prime}]\ll\infty[c_{\epsilon}(a)]$, there is $k\in\mathbb{N}$ such that
$[g^{\prime}]\leq k[c_{\epsilon}(a)]$. Thus,

(6.2) $2[(a-\epsilon)_{+}]\leq[(a-\epsilon)_{+}]+k[c_{\epsilon}(a)]$

for sufficiently large $k$.

(i) Suppose that $[a_{i}]$ is an increasing sequence of purely non-compact
elements with supremum $[a]$, and that $[a]$ is compact in $\mathrm{Cu}(A/I)$
for some ideal $I$. Without loss of generality, in order to prove that
$[a_{I}]$ is a multiple of infinity, we may assume that $I=0$. Thus, $[a]$ is
compact, and so $[a]=[a_{i_{0}}]$ for some $i_{0}$. Since $[a_{i_{0}}]$ is
purely non-compact, $[a_{i_{0}}]$ is a multiple of infinity, and therefore so
also is $[a]$, as desired.

Let $[a]$ and $[b]$ be purely non-compact and suppose that $[a]+[b]$ is
compact in $\mathrm{Cu}(A/I)$ for some ideal $I$. Again we may assume that
$I=0$. Thus, $[a]+[b]$ is compact, and so

$[(a-\epsilon)_{+}]+[(b-\epsilon)_{+}]+[c_{\epsilon}(a)]+[c_{\epsilon}(b)]\leq[a]+[b]=[(a-\epsilon)_{+}]+[(b-\epsilon)_{+}]$

for some $\epsilon$. Let $k$ be such that (6.2) holds both for $[a]$ and for
$[b]$. Then we have

$[a]+[b]=[(a-\epsilon)_{+}]+[(b-\epsilon)_{+}]+k([c_{\epsilon}(a)]+[c_{\epsilon}(b)])\geq
2([a]+[b]).$

Therefore, $[a]+[b]$ is a multiple of infinity, as desired.

(ii) It is enough to assume that $[b]=[(a-\epsilon)_{+}]$. From (6.2) we get
by induction that for every $M\in\mathbb{N}$ we have
$M[(a-\epsilon)_{+}]\leq[(a-\epsilon)_{+}]+k[c_{\epsilon}(a)]$ for
sufficiently large $k$. On the other hand we deduce from (6.1) that
$k[(a-\epsilon)_{+}]+k[c_{\epsilon}(a)]\leq k[a]$. Combining these two
equations we get the desired result.

(iii) Suppose that $\mathrm{Cu}(A)$ is almost unperforated and let $[a]$ be
such that for every $\epsilon>0$ we have $k[(a-\epsilon)_{+}]\leq(k+1)[a]$ for
$k$ large enough. Suppose that $[a]$ is compact. Then for some $\epsilon>0$ we
have $[(a-\epsilon)_{+}]=[a]$. So $(k+1)[a]=k[a]$, and this implies that
$2(k+1)[a]=k[a]$. Since $\mathrm{Cu}(A)$ is almost unperforated it follows
that $2[a]=[a]$. The same implication holds in any quotient of $A$.

(iv) Suppose that $A$ is simple. Then for any
$0\not=a\in(A\otimes\mathcal{K})^{+}$ the element $\infty\cdot[a]$ of
$\mathrm{Cu}(A)$ is the same, and is the largest element of $\mathrm{Cu}(A)$.
(In general, the Cuntz semigroup may not have a largest element.)

Fix a $\in(A\otimes\mathcal{K})^{+}$. The element $[a]$ of $\mathrm{Cu}(A)$ is
purely non-compact unless $[a]$ is compact and not equal either to 0 or to the
largest element of $\mathrm{Cu}(A)$. Suppose that the latter is true, i.e.,
that $[a]$ is compact and different from both zero and the largest element of
$\mathrm{Cu}(A)$. For some $\epsilon>0$ we have $[(a-\epsilon)_{+}]=[a]$. By
(6.1) we have $[a]+[c_{\epsilon}(a)]=[a]$. If $c_{\epsilon}(a)$ were non-zero
then, again as $A$ is simple, $\infty\cdot[c_{\epsilon}(a)]$ would be the
largest element of $\mathrm{Cu}(A)$, and hence $[a]$ would be the largest
element of $\mathrm{Cu}(A)$. Therefore, $c_{\epsilon}(a)=0$, and so $[a]=[p]$
for some non-zero projection $p$. The projection $p$ must be finite, as
otherwise $[p]$ would be the largest element of $\mathrm{Cu}(A)$.

(v) Suppose that $[a]$ is compact. Then $[a]=[s]+[r]=[s]$ for some
$[r],[s]\in\mathrm{Cu}(A)$ such that
$\mathrm{Ideal}(r)=\mathrm{Ideal}(s)=\mathrm{Ideal}(a)$ ($[s]$ may be chosen
as a partial sum in the given representation of $[a]$ as an infinite series
and $[r]$ as the remainder term). Since $\mathrm{Ideal}(r)=\mathrm{Ideal}(a)$
we have $[a]\leq\infty[r]$. Since $[a]$ is compact we therefore have $[a]\leq
k[r]$ for some $k\in\mathbb{N}$. We now have $[a]=[s]+k[r]\geq 2[a]$. The same
implication holds in any quotient of $A$. ∎

######  Lemma 6.5.

If $A$ absorbs the Jiang-Su algebra then for every
$a\in(A\otimes\mathcal{K})^{+}$ there is $[c]\in\mathrm{Cu}(A)$ such that
$\widehat{a}=\widehat{[c]}$. One can always choose $[c]$ to be purely non-
compact.

###### Proof.

First let us prove that if the existence of $[c]$ is guaranteed (in general),
then we can always choose it so that it is purely non-compact. Let $[c_{i}]$,
$i=1,2,\dots$, be such that $\widehat{a}/2^{i}=\widehat{[c_{i}]}$ for all $i$.
Then $[c]=\sum_{i=1}^{\infty}[c_{i}]$ is purely non-compact by Proposition 6.4
(v), and $\widehat{[c]}=\sum_{i=1}^{\infty}\widehat{[c_{i}]}=\widehat{a}$.

Let us now prove that $[c]$ exists with $\widehat{[c]}=\widehat{a}$. Every
positive element of $A\otimes\mathcal{K}\otimes Z$ is approximately unitarily
equivalent to one of the form $b\otimes 1$ with
$b\in(A\otimes\mathcal{K})^{+}$ (see the proof of [brown-perera-toms, Theorem
5.5]). Therefore, we may assume that the given positive element $a$ has the
form $b\otimes 1\in(A\otimes\mathcal{K})\otimes Z$. Recall that

$\widehat{(b\otimes 1)}(\lambda)=\int_{0}^{\infty}\lambda([(b-t)_{+}\otimes
1])\,dt.$

Since $t\mapsto\lambda([(b-t)_{+}])$ is a decreasing function of $t$, the
Riemann sums

$\sum_{i=1}^{2^{n}}\frac{1}{2^{n}}\lambda([(b-\frac{i}{2^{n}})_{+}\otimes 1])$

converge to the integral. Choose a positive element $e_{1/2^{n}}$ of the
Jiang-Su algebra with rank $1/2^{n}$. (That such an element exists follows,
for instance, by the computation of the Cuntz semigroup of the Jiang-Su
algebra obtained in [brown-perera-toms].) Then the Riemann sum above is equal
to

$\lambda\left(\sum_{i=1}^{2^{n}}[(b-\frac{i}{2^{n}})_{+}\otimes
e_{1/2^{n}}]\right).$

Let us show that the Cuntz semigroup elements

(6.3) $\displaystyle\sum_{i=1}^{2^{n}}[(b-\frac{i}{2^{n}})_{+}\otimes
e_{1/2^{n}}]$

form an increasing sequence. Comparing two consecutive terms of this sequence
we see that it is enough to show that $[(b-i/2^{n})_{+}\otimes
e_{1/2^{n}}]=2[(b-i/2^{n})_{+}\otimes e_{1/2^{n+1}}]$. This is true, since
$[e_{1/2^{n}}]=2[e_{1/2^{n+1}}]$, as follows from the computation of
$\mathrm{Cu}(Z)$ in [brown-perera-toms]. We have $\widehat{b\otimes
1}=\widehat{[c]}$, with $[c]$ the supremum of the sequence (6.3). ∎

######  Theorem 6.6.

Let $A$ be a C*-algebra. Suppose that $\mathrm{Cu}(A)$ is almost unperforated.
If $[a]$ and $[b]$ are in $\mathrm{Cu}(A)$ and $[a]$ is purely non-compact,
then $[a]\leq[b]$ if and only if $\widehat{[a]}\leq\widehat{[b]}$.

Suppose further that $A$ absorbs the Jiang-Su algebra. Then the map
$[a]\mapsto\widehat{[a]}$ is an isomorphism of the ordered semigroups of
purely non-compact elements of $\mathrm{Cu}(A)$ and of
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$.

###### Proof.

Suppose that $[a]$ is purely non-compact, $[b]\in\mathrm{Cu}(A)$, and
$\widehat{[a]}\leq\widehat{[b]}$. By Corollary 6.3, for every $k\in\mathbb{N}$
we have $k[a]\leq(k+1)[b]$. On the other hand, by Proposition 6.4 (ii), for
every $\epsilon>0$ there exists $k$ such that $(k+2)[(a-\epsilon)_{+}]\leq
k[a]\leq(k+1)[b]$. Since $\mathrm{Cu}(A)$ is almost unperforated, it follows
that $[(a-\epsilon)_{+}]\leq[b]$ for all $\epsilon>0$. Hence, $[a]\leq[b]$.

Let us now prove that the map $[a]\mapsto\widehat{[a]}$ is a surjection from
the purely non-compact elements to $\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$.
Let $f\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. By Theorem 5.7 there exists
an increasing sequence $(\widehat{a_{i}})$ with supremum $f$, where
$a_{i}\in(A\otimes\mathcal{K})^{+}$. By Lemma 6.5, for each $i=1,2,\dots$
there exists a purely non-compact element $[c_{i}]\in\mathrm{Cu}(A)$ such that
$\widehat{a_{i}}=\widehat{[c_{i}]}$. We have
$\widehat{[c_{i}]}\leq\widehat{[c_{i+1}]}$, and so as shown above
$[c_{i}]\leq[c_{i+1}]$; that is, $[c_{i}]$, $i=1,\dots$, is an increasing
sequence. Set $\sup[c_{i}]=[c]$. Since the map $[a]\mapsto\widehat{[a]}$
preserves suprema of increasing sequences, we have $f=\widehat{[c]}$. In order
to ensure that $[c]$ is purely non-compact let us choose $[c_{i}^{\prime}]$
such that $1/2^{i}f=\widehat{[c_{i}^{\prime}]}$ for all $i=1,2,\dots$. Then
$[c^{\prime}]=\sum_{i=1}^{\infty}[c_{i}^{\prime}]$ is purely non-compact by
Proposition 6.4 (i), and $f=\widehat{[c^{\prime}]}$. ∎

######  Corollary 6.7.

Let $A$ be a C*-algebra that absorbs the Jiang-Su algebra. If $A$ has no non-
zero simple subquotients then
$\mathrm{Cu}(A)\cong\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. (Here by $\cong$
is meant a natural isomorphism of functors into the category $\mathcal{C}u$.)

###### Proof.

If $A$ has no non-zero simple subquotients then no non-zero element of
$\mathrm{Cu}(A)$, or of $\mathrm{Cu}(A/I)$ for an ideal $I$, is compact (and
in particular, every element of $\mathrm{Cu}(A)$ is purely non-compact, as
desired). For if $[a]$ is compact and non-zero then $[a]=[(a-\epsilon)_{+}]$
for some $\epsilon>0$, whence
$\mathrm{Ideal}(a)=\mathrm{Ideal}((a-\epsilon)_{+})$. It follows, for instance
from Lemma 2.2, that $\mathrm{Ideal}(a)$ is a compact ideal of $A$ (a compact
element of $\mathrm{Lat}(A)$). (Here we do not mean only countable
compactness; however, that would be sufficient for our purposes since
$\mathrm{Ideal}(a)$ is singly generated.) Hence, $\mathrm{Ideal}(a)/J$ is
simple for any maximal proper ideal $J$ of $\mathrm{Ideal}(a)$ ($J$ exists, by
Zorn’s Lemma, since $\mathrm{Ideal}(a)$ is compact and non-zero). ∎

As another corollary of Theorem 6.6, let us give a computation of the Cuntz
semigroup of a simple C*-algebra absorbing the Jiang-Su algebra (this
computation was previously obtained in [brown-perera-toms] with the additional
assumptions that the algebra was unital, exact, and of stable rank one). Let
$A$ be a simple C*-algebra absorbing the Jiang-Su algebra. Let $\mathrm{V}(A)$
denote the semigroup of Murray-von Neumann equivalence classes of projections
of $A\otimes\mathcal{K}$. Let us define on the abstract disjoint union
$(\mathrm{V}(A)\backslash\\{0\\})\sqcup\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$
an order and an addition operation, making it what might be called the
lexicographic ordered semigroup disjoint union. Inside the two subsets
$(\mathrm{V}(A)\backslash\\{0\\})$ and
$\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ let us retain the order and addition
with which these sets are endowed. Let $[p]\in\mathrm{V}(A)\backslash\\{0\\}$
and $f\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$. Let us define
$f+[p]\in\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$ as the function
$f+\widehat{[p]}$ if $f\not=0$, and as the class $[p]$ if $f=0$. Let us say
that $f\leq[p]$ if $f\leq\widehat{[p]}$, and that $[p]\leq f$ if
$\widehat{[p]}+g=f$ for some $0\not=g\in L(F(\mathrm{Cu}(A)))$. It is not
difficult to verify that
$(\mathrm{V}(A)\backslash\\{0\\})\sqcup\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$
is an object in $\mathcal{C}u$.

######  Corollary 6.8.

Let $A$ be a simple C*-algebra absorbing the Jiang-Su algebra. Then either $A$
is purely infinite and $\mathrm{Cu}(A)\cong\\{0,\infty\\}$, or every
projection in $A\otimes\mathcal{K}$ is finite and
$\mathrm{Cu}(A)\cong(\mathrm{V}(A)\backslash\\{0\\})\sqcup\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$.
(Here by $\cong$ is meant a natural isomorphism of functors into the category
$\mathcal{C}u$.)

###### Proof.

If $\mathrm{F}(\mathrm{Cu}(A))\cong\\{0,\infty\\}$ then by Proposition 6.2,
$\mathrm{Cu}(A)\cong\\{0,\infty\\}$, in other words, $A$ is purely infinite
(this is also obtained in [rordam, Corollary 5.1]). If, on the other hand,
$\mathrm{F}(\mathrm{Cu}(A))$ contains a non-trivial functional, i.e., a
functional with a non-zero finite value, then every projection in
$A\otimes\mathcal{K}$ must be finite. Indeed, if $\lambda$ is a non-trivial
functional, and $p$ is a projection in $A\otimes\mathcal{K}$, then, by
[coward-elliott-ivanescu], the class $[p]$ is compact in $\mathrm{Cu}(A)$, and
so by simplicity is majorized by a finite multiple of any non-zero element,
and hence is finite on $\lambda$. Again by simplicity, Ker $\lambda=0$, and it
follows that if $[p]+[a]=[p]$ in $\mathrm{Cu}(A)$ then $[a]=0$, and in
particular the projection $p$ is finite.

Suppose that every projection of $A\otimes\mathcal{K}$ is finite. By
Proposition 6.4 (iv), the complement of the set of purely non-compact elements
of $\mathrm{Cu}(A)$, in the case of a simple C*-algebra, is the set of
elements $[p]$ such that $p$ is a non-zero finite projection. It is easy to
show (and well known) that among finite projections Cuntz equivalence amounts
to Murray-von Neumann equivalence. Therefore, by Theorem 6.6, the map from
$\mathrm{Cu}(A)$ to
$(\mathrm{V}(A)\backslash\\{0\\})\sqcup\mathrm{L}(\mathrm{F}(\mathrm{Cu}(A)))$
given by $[p]\mapsto[p]$ if $p$ is a projection, and $[a]\mapsto\widehat{[a]}$
if $[a]$ is purely non-compact, is a bijection.

To prove that that this (natural) map is an isomorphism of ordered semigroups,
let $0\not=p\in A\otimes\mathcal{K}$ be a projection and let
$0\not=a\in(A\times\mathcal{K})^{+}$ be such that $[a]$ is purely non-compact
in $\mathrm{Cu}(A)$. The sum $[p]+[a]$ in $\mathrm{Cu}(A)$ is then also purely
non-compact. (By 6.4 (iv) it is enough to show that $[p]+[a]$ is not the class
of a non-zero finite projection, if $[a]$ itself is not the class of a non-
zero finite projection. We may assume that $pa=0$. If $[p]+[a]=[p+a]$ is the
class of a non-zero finite projection, say $q$, then, by the formulation of
Cuntz equivalence given in [coward-elliott-ivanescu], and since, also by
[coward-elliott-ivanescu], $[q]$ is compact, and
$\sup_{\epsilon>0}[(a-\epsilon)_{+}]=[a]$, whence
$\sup_{\epsilon>0}[p+(a-\epsilon)_{+}]=[p+a]$, the Hilbert $A$-module $qA$ is
(by compactness) isomorphic to a sub Hilbert module $X$ of
$((p+(a-\epsilon)_{+})A)^{-}$ for some $\epsilon>0$, and (for any
$\epsilon>0$) $((p+(a-\epsilon)_{+})A)^{-}$ is isomorphic to a sub Hilbert
module $Y$ of $qA$. Hence, if $X^{\prime}$ denotes the isomorphic copy of $X$
contained in $((p+(a-\epsilon)_{+})A)^{-}$, and $X^{\prime\prime}$ and
$Y^{\prime}$ the resulting isomorphic images of $X^{\prime}$ and $Y$ in $qA$,
so that $X^{\prime\prime}\subseteq Y^{\prime}\subseteq qA$ and
$X^{\prime\prime}$ is isomorphic to $qA$, by finiteness of $q$ it follows that
$X^{\prime\prime}=qA$—so that in particular $Y=qA$, and so
$((p+(a-\epsilon)_{+})A)^{-}=q^{\prime}A$ for some projection
$q^{\prime}=q^{\prime}_{\epsilon}$. Then necessarily
$p+(a-\epsilon)_{+}=q^{\prime}$, i.e., $(a-\epsilon)_{+}$ is a projection
(namely, $q^{\prime}-p$). Since $\epsilon>0$ may be arbitrarily small, it
follows that $a$ itself is a projection, equal to $(a-\epsilon)_{+}$ for some
$\epsilon>0$, and therefore $\leq q^{\prime}_{\epsilon}$ for that $\epsilon$;
and therefore finite. By hypothesis, $[a]\not=0$; we have therefore proved the
contrapositive.)

It follows that the image of $[p]+[a]$ in the disjoint union is
$([p]+[a])^{\wedge}$. On the other hand, the image of $[p]$ is $[p]$ and the
image of $[a]$ is $\widehat{[a]}$, and the sum of $[p]$ and $\widehat{[a]}$ in
the (lexicographic) disjoint union is by definition
$\widehat{[p]}+\widehat{[a]}$, which is equal to $([p]+[a])^{\wedge}$. The one
case remaining in which to check additivity, that the first element is $[p]$
and the second is the purely non-compact element $0\in\mathrm{Cu}A$, is
trivial: the sum of these elements is $[p]$, and the sum of the images,
$\widehat{[p]}$ and $0$, is $\widehat{[p]}$, the image of the sum.

It remains to verify that the relation $[a]\leq[p]$ or $[p]\leq[a]$ in
$\mathrm{Cu}(A)$ implies the relation $\widehat{[a]}\leq[p]$ or
$[p]\leq\widehat{[a]}$ for the images in the disjoint union, and conversely
(with $p$ and $a$ still as above). If $[a]\leq[p]$ in $\mathrm{Cu}(A)$ then
$\widehat{[a]}\leq\widehat{[p]}$ in $L(F(\mathrm{Cu}(A)))$, which is just the
definition of $\widehat{[a]}\leq[p]$ in the disjoint union. Conversely, if
$\widehat{[a]}\leq[p]$ in the disjoint union, i.e., if
$\widehat{[a]}\leq\widehat{[p]}$ in $L(F(\mathrm{Cu}(A))$, then, as $a$ is
purely non-compact, by Theorem 6.6, $[a]\leq[p]$.

If $[p]\leq[a]$ in $\mathrm{Cu}(A)$, then the compact Hilbert module $pA$ is
isomorphic (by [coward-elliott-ivanescu]) to a sub Hilbert module of
$(aA)^{-}$, necessarily complemented: $(aA)^{-}\cong pA\oplus(bA)^{-}$ (with,
e.g., $b=(1-p)a)$. Since $a$ is non-zero and purely non-compact, $b\not=0$.
Hence, $\widehat{[a]}=\widehat{[p]}+g$ with $0\not=g=\widehat{\,[b]\,}\in
L(F(\mathrm{Cu}(A)))$, i.e., $[p]\leq\widehat{[a]}$ in the disjoint union.
Conversely, if $\widehat{[a]}=\widehat{[p]}+g$ with $0\not=g\in
L(F(\mathrm{Cu}(A)))$, then by Theorem 6.6 there exists a purely non-compact
element $0\not=[b]\in\mathrm{Cu}(A)$ such that $\widehat{\,[b]\,}=g$. Then, as
shown above, also $[p]+[b]$ is purely non-compact, and since
$\widehat{[a]}=\widehat{[p]}+{\widehat{\,[b]\,}}=([p]+[b])^{\wedge}$, and also
$[a]$ is non-compact, by Theorem 6.6, $[a]=[p]+[q]$, and in particular
$[p]\leq[a]$ in $\mathrm{Cu}(A)$. ∎

######  Remark 6.9.

Certain C*-algebras with no non-zero simple subquotients were considered by
the second author in [robert]—these were closed two-sided ideals of AI
algebras—and were classified by means of tracial data. (The techniques of
[robert] apply in fact to arbitrary ideals of AI algebras with no non-zero
simple subquotients—equivalently, with the K1-group of every ideal equal to
zero.) For these algebras, the results of [robert] may therefore be viewed as
a determination of the Cuntz semigroup in terms of tracial data, although the
way this ordered semigroup is determined is only implicit. Note that, in
[ciuperca-elliott], the Cuntz semigroup, together with the special element
consisting of the class of the strictly positive elements, was shown to be a
complete invariant for arbitrary AI algebras, or ideals of AI algebras. The
results of [robert] could be deduced from this together with Corollary 6.7.
The problem of describing the Cuntz semigroup in terms of K-theoretical and
tracial data in this more general setting would seem to be very interesting.

## References