THE CONE OF FUNCTIONALS ON THE CUNTZ SEMIGROUP
LEONEL ROBERT∗
Abstract
The functionals on an ordered semigroupSin the categoryCu– a category to which the Cuntz semigroup of a C*-algebra naturally belongs – are investigated. After appending a new axiom to the categoryCu, it is shown that the “realification”SRofShas the same functionals asSand, moreover, is recovered functorially from the cone of functionals ofS. Furthermore, ifShas a weak Riesz decomposition property, thenSRhas refinement and interpolation properties which imply that the cone of functionals onSis a complete distributive lattice. These results apply to the Cuntz semigroup of a C*-algebra. At the level of C*-algebras, the operation of realification is matched by tensoring with a certain stably projectionless C*-algebra.
1. Introduction
From its introduction in [5], the Cuntz semigroup of a C*-algebra has been understood as a natural carrier of the dimension functions of the C*-algebra:
they correspond to functionals on the Cuntz semigroup. In [3], Coward, Elliott and Ivanescu define the categoryCu and show that the Cuntz semigroup of a C*-algebra is an object in this category. The idea comes to mind to study functionals on ordered semigroups in the axiomatic setting ofCuand attempt to recover (and push further!) known results in the C*-algebraic context. Such a study was done partly in [6] and [2] and is continued here.
Our starting point is an ordered semigroupSin the categoryCu. However, in order to make progress on questions regarding the functionals on S, we need to assume thatSalso has the almost algebraic order property (see axiom O5 in Subsection 2.1 below). For the Cuntz semigroup of a C*-algebra, this property was proven in [13, Lemma 7.2] and it was also used repeatedly in the arguments of [11]. The results of this paper stress further its importance (see Remark 2.2.4 below).
Assume thatSis in the categoryCuand has almost algebraic order. Denote by F(S)the cone of functionals onS(topologized as in [6]). Eachs ∈Sinduces
∗I was supported by the Danish National Research Foundation (DNRF) through the Center for Symmetry and Deformation while conducting this research.
Received 14 April 2011.
a function on F(S):s(λ)ˆ :=λ(s)for allλ∈F(S). Two natural questions that can be asked are
(i) what can we say abouts, t ∈Sifsˆ = ˆt?
(ii) what can we say about the range of the maps → ˆs?
The first question is answered in Proposition 2.2.6 below. Regarding the second question, we consider a set larger than the range of the maps → ˆs; namely, the closure (under sequential suprema) of theR+-linear span of the range ofs→ ˆs.
This set, denoted bySR, may also be characterized as the “realification” ofS and is the main focus of the results of this paper. It will be shown thatSRcan be recovered functorially from F(S)as a suitable dual of F(S). If we assume further thatShas a weak decomposition property (à la Riesz), thenSRsatisfies a refinement property which in turn implies that F(S)is a complete lattice.
Our results are applicable to C*-algebras. At the level of C*-algebras, the operation of “realification” is matched by tensoring with the stably projec- tionless C*-algebra R studied in [8] and [10]. That is, Cu(A)R ∼= Cu(A⊗ R), where Cu(A)denotes the Cuntz semigroup of the C*-algebraA. Since F(Cu(A))is in bijection with the lower semicontinuous 2-quasitraces onA, it follows that the Cuntz semigroup of anR-absorbing C*-algebra is determ- ined by its cone of lower semicontinuous 2-quasitraces. Cu(A)has the weak Riesz decomposition property mentioned above. Thus, the lower semicontinu- ous 2-quasitraces onAform a complete lattice. This extends Blackadar and Handelmann’s [1, Theorem II.4.4] that the bounded 2-quasitraces of a unital C*-algebraAform a lattice.
In Section 2 we prove some preliminary results on ordered semigroups and we answer question (i) above. In Section 3 we defineSR and show that it is recovered functorially as a dual space of F(S). In Section 4 we prove refinement and interpolation properties forSRand derive from these that F(S)is a complete lattice. The last section contains the results relating to the Cuntz semigroups of C*-algebras. In the last paragraphs we give further evidence of the relevance of the properties of almost algebraic order and weak Riesz decomposition by showing that Glimm’s halving property for non-type I simple C*-algebras is recovered, in the context of ordered semigroups, using these properties.
Acknowledgements. This research was conducted while I was a member of the Center for Symmetry and Deformation at the University of Copenhagen.
I am grateful to the Center, and in particular to Mikael Rørdam, for their hospitality and support. The case thatAis commutative of the isomorphism Cu(A)R ∼=Cu(A⊗R)can be derived using the methods of [14]. I am grateful to Aaron Tikuisis for pointing this out as evidence of the validity of the general result.
2. Preliminaries on ordered semigroups
We call ordered semigroup a monoid endowed with a translation invariant order relation. We always assume that the semigroup is abelian and positive, i.e., 0 is the smallest element of the ordered semigroup. By ordered semigroup map we understand one that preserves the order, the addition operation, and the 0 element.
2.1. The categoryCu
Given elements in an ordered setsandt, we say thatsis sequentially compactly contained int, and denote it bys t, if for any increasing sequence(tn)such thatt supntn we haves tn0 for somen0 ∈ N. (We will often drop the reference to sequences and simply say thats is compactly contained int.)
The objects of the categoryCu– introduced in [3] – are ordered semigroups satisfying a number of axioms. The ordered semigroupSis an object ofCuif
O1 Every increasing sequence has a supremum.
O2 For everys ∈Sthere exists a sequence(sn)such thatsnsn+1for all nands =supnsn.
O3 Ifsi ti, fori=1,2, thens1+s2t1+t2.
O4 If(sn)and(tn)are increasing sequences then supn(sn+tn)=supnsn+ supntn.
The primary example of an ordered semigroup in the categoryCuis the Cuntz semigroup of a C*-algebra. That such an object satisfies the axioms O1–O4 is proven in [3, Theorem 1].
We will also consider the property of almost algebraic order:
O5 Ifs s t then there existsrsuch thats +r t s+r.
It is proven in [13, Lemma 7.2] that the Cuntz semigroup of a C*-algebra satisfies O5.
A sequence(sn)such thatsn sn+1for allnis called rapidly increasing.
Thus, O2 may be restated as saying that every element is the supremum of a rapidly increasing sequence.
A subsetS ⊆ Sis called dense if every element ofSis the supremum of a rapidly increasing sequence of elements inS. If a C*-algebra is separable, then its Cuntz semigroup has a countable dense subset (see Proposition 5.1.1 below).
2.2. Functionals
We call an ordered semigroup map λ:S → [0,∞] a functional on S if it preserves the suprema of increasing sequences. The collection of all functionals onSforms a cone that we denote by F(S)(addition and scalar multiplication are defined pointwise).
Lemma2.2.1.LetSbe an ordered semigroup in the categoryCu. Letλ:S→ [0,∞] be additive and order preserving. Then λ(s)˜ := supssλ(s) is a functional onS. (We callλ˜ the supremum preserving regularization ofλ.)
Remark 2.2.2. The above lemma is proven in [6, Lemma 4.7]. Notice, however, that the hypothesis thatλis order preserving is not included in the statement of [6, Lemma 4.7], although it is tacitly assumed in the proof.
Let us now show that the pointwise order in F(S)coincides with the algeb- raic order ifSis in the categoryCuand has almost algebraic order.
Proposition2.2.3.LetS be an ordered semigroup satisfying the axioms O1–O5. Letαandβ be functionals onS. Thenα(s) β(s)for alls ∈ Sif and only if there exists a functionalγ such thatα+γ =β.
Proof. Defineγ:S→[0,∞] by γ (s)=
β(s)−α(s) ifβ(s) <∞,
∞ otherwise.
It is easy to check thatγ is additive. Let us show that it is also order preserving.
Let s, t ∈ S be such that s t. If β(t ) = ∞ thenγ (t ) = ∞and clearly γ (s) γ (t ). Assume thatβ(t ) <∞. Since supssβ(s) = β(s) <∞, for any given >0 there existss ssuch thatβ(s)β(s )+. By O5 there existsr ∈Ssuch thats +r t s+r. Then,
γ (t )=β(t )−α(t )β(s +r)−α(s+r)β(s)−α(s)γ (s)−. Sincecan be arbitrarily small we get thatγ (t )γ (s).
We haveα+γ = β. Passing to the supremum preserving regularizations we getα+ ˜γ =β.
Remark 2.2.4. It is remarked without proof in [6] – after the proof of [6, Lemma 4.7] – that the above proposition is true for ordered semigroups in the categoryCu. It is not presently clear to me whether this is the case.
Observe that in the above proof we have made use of the axiom O5 (i.e., the property of almost algebraic order). Since [6, Theorem 4.8] relies on this fact, the hypothesis that the ordered semigroups have almost algebraic order must be appended to the statement of [6, Theorem 4.8].
For the remainder of this sectionSdenotes an ordered semigroup satisfying the axioms O1–O5 (i.e., in the categoryCuand with almost algebraic order).
The cone F(S)is endowed with the topology such that a net(λi)converges toλif and only if
(2.1) lim sup
i
λi(s)λ(s)lim inf
i λi(s)
for alls, s ∈ Ssuch thats s. The addition and the scalar multiplication by positive real numbers are jointly continuous operations (see [6, Proposi- tion 3.6]). By [6, Theorem 4.8], F(S)is a compact Hausdorff space. IfSis the Cuntz semigroup of a C*-algebra, then F(S)is isomorphic, as a topological cone, to the cone of lower semicontinuous 2-quasitraces on the C*-algebra (see [6, Theorem 4.4]).
Let us denote by Lsc(F(S)) the set of functionsf: F(S) → [0,∞] that are linear and lower semicontinuous. Lsc(F(S))is endowed with the order of pointwise comparison and the operations of pointwise addition and pointwise scalar multiplication by positive (non-zero) real numbers. Each elements ∈S induces a functionsˆ∈Lsc(F(S))defined bys(λ)ˆ =λ(s)for allλ∈F(S). The maps → ˆsis additive and preserves sequential suprema (because functionals are additive and preserve sequential suprema) but may not preserve the relation of compact containment. However, we do have the following lemma.
Lemma2.2.5.Ifs t ∈Sandα < β ∈(0,∞]thenαsˆ βtˆ(here the relationis taken inLsc(F(S))).
Proof. Suppose that(λi)is a net in F(S)such thatλi →λandλi(s) > α1 for alli. Then
λ(t )lim sup
i
λi(s) 1 α > 1
β. This shows that we have the inclusion
{λ∈F(S)|αs(λ) >ˆ 1} ⊆ {λ|βt (λ) >ˆ 1}.
By [6, Proposition 5.1], this inclusion implies thatαsˆβtˆin Lsc(F(S)).
The following proposition gives an algebraic characterization of the com- parison of elements ofSby functionals (thus answering question (i) from the introduction).
Proposition2.2.6.Let S be an ordered semigroup that satisfies O1–O5 and lets, t ∈ S. Thensˆ tˆif and only if for every >0ands s there existM, N∈Nsuch that M
N >1−andMs N t.
Proof. IfMs N t, withM/N >1−, then(1−)s tˆ. Passing to the supremum over all >0 ands swe get thatsˆ tˆ.
Suppose thatsˆ tˆand lets s and > 0. Comparings andt on the functionalλ:S → [0,∞] such thatλ(x) = 0 ifx ∞ ·t andλ(x) = ∞ otherwise, we conclude thats∞ ·t, and sos Ct for some finiteC >0.
ChooseP , Q∈Nsuch that 1− < P /Q <1. ThenP λ(s) < Qλ(t )for every λ ∈ F(S)such that λ(t ) = 1. Letα:S → [0,∞] be an ordered semigroup map such that α(t ) = 1. Let α˜ be the supremum preserving regularization of α (defined as in Lemma 2.2.1). If α(t )˜ = 0 then P α(s) Pα(s) <˜ Qα(t )˜ Qα(t ). Ifα(t )˜ = 0 thenP α(s )= 0 < Q= Qα(t ). In summary, P α(s) < Qα(t )for any ordered semigroup map α:S → [0,∞] such that α(t )= 1. By [9, Proposition 2.1], this implies that(k+1)P s kQt for all k ∈Nlarge enough. Since we can choosek such that (k+kQ1)P >1−, we are done.
3. The ordered semigroup SR
3.1. Definition and properties ofSR
LetSbe a positive ordered semigroup satisfying axioms O1–O5 (i.e., in the categoryCuand with the almost algebraic order property). We denote bySR the subset of Lsc(F(S))of functions expressible as the pointwise supremum of an increasing sequence(hn), where eachhnbelongs to theQ+-linear span of the image ofSin Lsc(F(S)). That is,f ∈SRif there existsi ∈Sandni ∈N, withi =1,2, . . ., such that the sequence(nsˆi
i)i is increasing and f (λ)=sup
i
ˆ si(λ)
ni for all λ∈F(S).
Proposition3.1.1.LetS be an ordered semigroup satisfying the axioms O1–O5. ThenSRalso satisfiesO1–O5andF(S)∼=F(SR)as topological cones.
Proof. Lets ∈ S and let(si) be a rapidly increasing sequence with su- premums. By Lemma 2.2.5, we have
1− 1i ˆ si
1− i+11 ˆ
si+1, where the relationis taken in Lsc(F(S)). It follows that this relation of compact con- tainment also holds inSR ⊆Lsc(F(S)). Thus,sˆis the supremum of a rapidly increasing sequence. This automatically holds also for nˆs for everyn∈N. Us- ing a standard diagonalization argument (see the proofs of [3, Theorem 1 (i)]
and [6, Proposition 5.1 (iii)]) we can then show that SR is closed under the suprema of increasing sequences (as a subset of Lsc(F(S))), and that every element ofSR is the supremum of a rapidly increasing sequence inSR. Since the supremum of a sequence in Lsc(F(S)) is the pointwise supremum, it is clear thatSRsatisfies O4.
Let us show thatSR satisfies axiom O3. Letfi, gi ∈SR,i = 1,2, be such thatfi gi. In order to prove O3, we may assume thatg1andg2belong to a dense subset. Thus, we may assume that they have the formαtˆ, witht ∈Sand α ∈ Q+. Moreover, multiplying by a suitable integer, we reduce proving O3 to the case thatgi = ˆti,i=1,2. Let us find >0 andti ti, withi=1,2, such thatfi (1−)ti ti. Thenf1+f2(1−)(t1+t2) ˆt1+ ˆt2. This proves O3.
We postpone the proof of O5 to Proposition 3.3.1, where a stronger version of the almost algebraic order property is obtained.
The mapλ→(f →f (λ)), from F(S)to F(SR)is linear and continuous.
It is also bijective, since any functional onSR is uniquely determined by its restriction to the image ofSinSR, and thus gives rise to a unique functional onS.
Since both F(S)and F(SR)are compact Hausdorff spaces,λ→(f →f (λ)) is a homeomorphism.
The ordered semigroup SR can be characterized by a universal property using the property of real multiplication.
Definition 3.1.2. We say that the ordered semigroupO has real multi- plication if there exists a map(0,∞]×O→O
(t, s)→t·s
that is additive on both variables, order preserving on both variables, supremum (of sequences) preserving on both variables, and such that 1·s=s.
SRclearly has real multiplication. An ordered semigroup with real multi- plication is unperforated by definition, i.e.,nxnyimpliesx y. Although SRis not necessarily cancellative, it has the following form of cancellation (a direct consequence of unperforation):
f +hg+h
h∝g ⇒f g.
Hereh∝gmeans thathngfor somen∈N.
The following proposition implies that having real multiplication is a prop- erty rather than additional structure (thus, the scalar multiplication can be uniquely defined, if at all).
Proposition3.1.3.LetS andS be a ordered semigroups satisfyingO1–
O5and suppose thatS has real multiplication. Letα:S→S be an ordered semigroup map that preserves the suprema of increasing sequences. Then there exists a unique ordered semigroup mapα:SR →S that preserves the suprema
of increasing sequences and such that the following diagram commutes:
S α S
α
SR
Proof. Let us show the uniqueness ofαfirst. Suppose thatα1, α2:SR →S satisfy thatα1(s)ˆ =α2(s)ˆ for alls. Thenα1andα2also agree on elements of the forms/nˆ and on the suprema of increasing sequences of such elements.
Thus,α1=α2.
Letα:S→S be given as in the statement of the proposition. Lets1, s2∈S be such thatsˆ1sˆ2. Using Proposition 2.2.6, we can see that(1−)α(s1) α(s2)for all >0 ands1s1. Passing to the supremum over all suchand s1we obtain thatα(s1)α(s2).
Letf ∈ SR. Let(sˆi/ni)and(ˆti/mi)be rapidly increasing sequences with supremumf. Then these sequences intertwine: for everyithere existsj such thatsˆi/ni tˆj/mj and tˆi/mi sˆj/nj. Thus, the sequences(α(si)/ni)and (α(ti)/mi)are also intertwined, and so they have the same supremum. We can thus define
α(f ):=sup
i
α(si) ni .
A straightforward, but tedious, analysis show that this map is additive, order preserving, and supremum preserving.
Corollary3.1.4.LetSbe an ordered semigroup satisfyingO1–O5. Then (SR)R ∼=SR.
Remark 3.1.5. The case can be made that SR is nothing but the tensor product S⊗[0,∞] in the category of ordered semigroups that satisfy the axioms O1–O5. However, tensor products in this category remain a subject yet to be investigated. So we will not pursue this point of view here.
Let us introduce a strengthening of the compact containment relation among the elements ofSR. Letf, g ∈ SR. Let us writef g iff (1−)gfor some >0 andf is continuous at eachλ∈F(S)for whichg(λ)is finite. We will make repeated use of this relation in the coming sections. We remark that
(i) f ghimpliesf h.
(ii) f gimplies thatf g, where the relationis taken in Lsc(F(S)).
This is proven in [6, Proposition 5.1].
(iii) f gandf g implyf +f g+g.
Proposition3.1.6.For each f ∈ SR there exists a sequenceh1 h2 h3. . .inSRwith supremumf.
Proof. It suffices to show that iff f then there exists l such that f l f. Let us choose, recursively, elementsfk
2n ∈ SR indexed by the dyadic rationals in [0,1] in the following manner: f0 = f , f1 f, and fk
2n fk
2n
if 2kn < k
2n . Finally, for eachn∈Nlet ln = 1
2n
2n−1
k=0
fk
2n, ln = 1
2n
2n
k=1
fk
2n.
Then(ln)is increasing,(ln)is decreasing, and f ln ln f for alln.
Letl = supnln. Let us show thatl is continuous at eachλwheref is finite.
Suppose that f (λ) < ∞ and let λi → λ. Sincel is lower semicontinuous, l(λ)lim infil(λi). On the other hand, for everynwe have
lnl lnln+ f 2n. Thus,
lim sup
i
l(λi)lim sup
i
ln(λi)+ 1
2n ·lim sup
i
f1(λi)l(λ)+ f (λ) 2n . Sincenis arbitrary andf (λ) <∞, we have lim supil(λi)l(λ). Thus,lis continuous onλ. In order to arrange thatl f, we first find >0 such that f (1−)f. We then findlsuch thatf l (1−)f andlis continuous on eachλwheref is finite.
Lemma 3.1.7.Let f g and let (fn)n be an increasing sequence with supremumf and such thatfn f for alln. The for every >0there exists Nsuch thatf fn+gfor allnN.
Proof. This follows from the fact thatfnconverges uniformly tof on the set{λ∈F(S):g(λ)1}(by Dini’s theorem).
3.2. SR as dual of F(S)
In this subsectionScontinues to denote an ordered semigroup satisfying ax- ioms O1–O5. Here we show howSR may be recovered solely from the topo- logical cone F(S). Indeed,SRcoincides with the ordered semigroup L(F(S)) introduced in [6].
By L(F(S))we denote the subset of Lsc(F(S))of functionsf expressible as the supremum of an increasing sequence(fn), withfn ∈ Lsc(F(S)) and
fn fn+1for alln. Proposition 3.1.6 implies thatSRis contained in L(F(S)).
Following the same approach used to prove [6, Theorem 5.7], we can show that they are in fact equal:
Theorem3.2.1. Let Sbe an ordered semigroup satisfying O1–O5. Then SR =L(F(S)).
Before proving this theorem, we need some preliminary results.
Lemma3.2.2.LetSbe an ordered semigroup satisfyingO1–O5. Letf, g∈ Lsc(F(S))be such thatf g. Then there existss ∈S such thatf ˆs
∞ ·g.
Proof. Consider the set{λ ∈F(S) |g(λ)= 0}. This set is closed under addition (whence upward directed) and under upward directed suprema (since g is lower semicontinuous). Therefore, it contains a maximum element λ0. The set of functions{ˆs |λ0(s)=0}is closed under addition, whence upward directed. Moreover, the pointwise supremum of these functions is equal to
∞ ·g (ifg(γ ) =0 for some functionalγ thenγ (s) > 0= λ0(s)for some s ∈ S and so∞ · ˆs(γ ) = ∞ ·g(γ )). Sincef ∞ ·g, the function f is compactly contained in∞·g, and so there existssˆ ∈ {ˆs |λ0(s)=0}such that f ˆs ∞ ·g. Hence, there existss s such thatf ˆs2sˆ ∞ ·g.
This proves the lemma.
The following proposition and lemma are analogs of [6, Proposition 5.5]
and [6, Lemma 5.6] (which are stated in the C*-algebraic context). In proving them we will follow the proofs of those results closely.
LetI ⊆Sbe an ideal ofS, i.e., a hereditary subsemigroup closed under the supremum of increasing sequences. LetλI:S→[0,∞] denote the functional such thatλI(s)= 0 ifs ∈ I andλI(s)= ∞otherwise. Finally, let FI(S) ⊆ F(S)be the subcone defined by
(3.1) FI(S):=λI+ {λ∈F(S)|λ(s) <∞for alls s for somes ∈I}. Notice that FI(S)is a cancellative cone: ifλ1+λ=λ2+λ, withλ1, λ2, λ∈ FI(S) thenλ1(s)= λ2(s)for alls such thats s ∈ I for somes. Hence, λ1(s)=λ2(s)for alls ∈Iand soλ1=λ2(since both functionals are infinite outsideI).
Proposition3.2.3.LetV(FI(S)) denote the ordered vector space of lin- ear, real-valued, continuous functions onFI(S). Let: V(FI(S)) →Rbe a positive linear functional onV(FI(S)). Then there existsλ∈FI(S)such that (f )=f (λ)for allf ∈V(FI(S)).
Proof. We will show that the relative topology on FI(S) induced by the topology of F(S)is the weak topology σ (FI(S),V(FI(S))). This will imply
that FI(S)is a weakly complete cancellative cone in the classS of Choquet (see [4, page 194]). The proposition will then follow from [4, Proposition 30.7].
It suffices to show that the relative topology on FI(S)agrees with the topo- logy of pointwise convergence on the functions
(3.2) PI := {f ∈SR|f f ˆs for somef ∈SR, s∈I}.
First observe thatf ˆs, withs ∈ I, implies thatf is finite on FI(S).
Thus, iff f ˆsthenf is continuous on FI(S).
Assume, on the other hand, that(λi)is a net in FI(S)and thatf (λi)→f (λ) for every f ∈ PI. Lets, s ∈ S be such that s s and let us show that the inequalities (2.1) defining the topology of F(S) hold true. If s /∈ I then λi(s)=λ(s)= ∞for alliand so the inequalities (2.1) hold trivially. Suppose thats∈I. Lets be such thats s sand let >0. Since(1−)s ˆs, there existf1, f2 ∈ SR such that (1− )s f1 f2 ˆs. Notice that f1∈PI. So
(1−)lim sups(λi)lim sup
i
f1(λi)=f1(λ)s(λ).
Passing to the supremum over all >0 establishes one half of (2.1). Also, (1−)s (λ)f1(λ)=lim inf
i f1(λi)lim infs(λi).
Passing to the supremum over alls s and >0 we get the other half of (2.1).
Lemma3.2.4.Leth1, h2, h3∈Lsc(F(S))be such thath1h2h3. Then for everyδ >0there isf ∈SRsuch thatf h3andh1δh3+f.
Proof. LetI := {s ∈ S | ˆs ∞ ·h3}. Observe thatI is an ideal of S, i.e., it is a hereditary subsemigroup closed under the suprema of increasing sequences. Consider the compact subsetK⊆F(S)defined by
K:= {λ∈F(S)|h3(λ)1} +λI.
Observe thatKis contained in FI(S). Indeed, ifλ∈Kands s ∈ Ithen ˆ
s∝h3, whenceλ(s) <∞.
The function h1 is continuous on K by hypothesis. Since K ⊆ FI(S), the functions in the set PI (as defined in (3.2)) are also continuous onK.
Let us show that h1 can be uniformly approximated on K by functions in PI. Suppose the contrary. Then there is a real measure m on K such that f dm= 0 for allf ∈PI and
h1dm=1. Letm= m+−m−denote the Jordan decomposition ofm. Then
f dm+ =
f dm− for allf ∈ PI and
h1dm+ =
h1dm−+1. SinceK ⊆ FI(S), we can define positive linear functionals+, −: V(FI(S)→Rby
+(g):=
K
g dm+ and −(g):=
K
g dm−.
By Proposition 3.2.3,+and−are given by the evaluation on functionals λ+andλ−belonging to FI(S). Thus,f (λ+)=f (λ−)for allf ∈ PI. Every ˆ
s withs ∈ I is the supremum of an increasing sequence of elements ofPI. (To see this, find a sequence(fn)inSRsuch thatfn fn+1for allnand with supremums. Thenˆ fn∈PIfor alln.) Thus,λ+(s)=λ−(s)for alls ∈I. Since λ+andλ−are in FI(S), they are both infinite outside ofI. Thus,λ+=λ−.
By Lemma 3.2.2, there exists, s ∈ I such thath2 sˆ, ands s. It follows thath2is finite on FI(S). Soh1is continuous on FI(S). In particular, the restriction ofh1to FI(S)belongs to V(FI(S)). Buth1(λ+)=h1(λ−)+1.
This contradicts the earlier conclusionλ+ = λ−. Therefore, the restriction ofh1 toK must belong to the norm closure of the functions inPI. That is, for everyδ > 0 there existsf ∈PI such thath1−fK < δ. Equivalently, h1f+δh3andf h1+δh3onK. It is easily shown that these inequalities also hold on all F(S). Changingf tof/(1+δ)we can arrange thatf h3.
Proof of Theorem 3.2.1. The inclusion SR ⊆ L(F(S)) follows from Proposition 3.1.6. Let us prove the opposite inclusion. Let(hn)be a sequence in Lsc(F(S))with supremumhand satisfyinghn hn+1for alln. Letμn>0 be such that thathn (1−μn)hn+1for alln. By Lemma 3.2.2, there exists t ∈Ssuch thath3 ˆt ∞ ·h4. Let us chooseM > 0 such thattˆ Mh4
and thenδ > 0 such thatδM < μ3. Finally, using Lemma 3.2.4, let us find g∈SRsuch thatg h3andh1δh3+g.
Letg1=g+δtˆ. Theng1∈SRand
g1=g+δtˆ(1−μ3+δM)h4h4. Also
g1=g+δtˆg+δh3h1.
Soh1 g1 h4. In the same way we may find g2 ∈ SR such thath4 g2 h7. Continuing in this way we get a sequence(gn), withgn ∈ SR and h=supngn. Thus,h∈SR.
A question left unanswered in these paragraphs is what axioms are needed on a topological coneCso that the ordered semigroup L(C)satisfies axioms O1–O5. Furthermore, one can ask if in such a caseCis recovered by passing to the cone of functionals F(L(C)).
Problem 3.2.5. Describe the category of non-cancellative cones dual to the category of ordered semigroups that satisfy axioms O1–O5 and have real multiplication.
3.3. Almost algebraic order ofSR
Here we show thatSRhas almost algebraic order (thus completing the proof of Proposition 3.1.1). We will show that, in fact,SR has the following strength- ening of the almost algebraic order property:
Proposition3.3.1.LetSbe an ordered semigroup satisfying axiomsO1–
O5. Letf , f, g ∈ SRbe such thatf f g. Then there existh, h ∈ SR such thatf hf andh+h =g.
This proposition is an immediate consequence of Proposition 3.1.6 com- bined with the following lemma:
Lemma3.3.2.Letf, g ∈ SRbe such thatf g g for someg ∈ SR. Then there existsh∈SR such thatf +h= g. The elementhmay be chosen such thatf ∝h.
Proof. Let(gn)be a sequence inSRsuch thatg =supngnandgn gn+1
for all n. We may assume that g g1, and so f g1. Let us define the functionshn: F(S)→[0,∞] by
hn(λ):=
gn(λ)−f (λ) ifgn(λ) <∞,
∞ otherwise.
It is easily verified thathnis linear. Let us show that it is also lower semicontinu- ous. Let(λi)be a net converging to a functionalλ. Suppose thatgn(λ) <∞. Thenf is continuous atλ. So,
lim inf
i (gn(λi)−f (λi))=lim inf
i gn(λi)−f (λ)gn(λ)−f (λ).
Thus, hn is lower semicontinuous at λ. Suppose that gn(λ) = ∞. Since f (1−n)gn for some n > 0, we have gn(λi)−f (λi) ngn(λi)if gn(λi)is finite. This implies thathn(λi) ngn(λi), whethergn(λi)is finite or not. Passing to the limit with respect toi we get that lim infihn(λi) lim infingn(λi)= ∞. Thus,hnis lower semicontinuous atλ.
Let us now show thathnhn+1for alln. Ifhn+1(λ) <∞thengn+1(λ) <
∞, and so gn andf are both finite and continuous atλ. It follows from the definition ofhnthat it is also continuous atλ. Also, fromgn(1−n)gn+1for somen>0 and the definition ofhnwe easily deduce thathn (1−n)hn+1. It follows thathnhn+1.
Leth=supnhn. Thenh∈L(F(S))=SR. Sincegn=f+hnfor alln, we conclude thatg = f +h. Finally, in order to arrange forf ∝ h, find >0 such thatf g (1−)g. Find thenh such thatf +h =(1−)gand seth=h +g. This concludes the proof of the lemma.
4. Refinement and interpolation properties
Let S be an ordered semigroup. In this section, in addition to the axioms O1–O5, we assume thatSsatisfies the following axiom:
O6 Ifs, t, r∈Sare such thats r+t, then for everys sthere existr andt such that
s r +t , r r, s and t t, s.
Notation convention. In order to state multiple inequalities more com- pactly, we will often use the notationa, b, c, . . . x, y, z, . . .to mean that every element listed on the left side is less than or equal to every element listed on the right side.
Lemma4.0.1. Let S be an ordered semigroup satisfying axioms O1–O6.
ThenSRalso satisfiesO1–O6.
Proof. We have already shown in Proposition 3.1.1 thatSRsatisfies O1–O5.
Letf, g, h∈SRbe such thatf g+h. In order to prove axiom O6, it suffices to verify that it holds forf,g, andhbelonging to a dense subsemigroup ofSR. So we may assume that they all belong to theQ+-linear span of the image of SinSR. Moreover, multiplying by a sufficiently large integer, we may assume thatf,gandhbelong to the image ofSinSR. So let us suppose thatsˆrˆ+ ˆt. Lets s s. By Proposition 2.2.6, given > 0 there existM, N ∈ N such thatM/N >1−andMs N r+N t. Thus, by axiom O6 applied to Sthere existr andt such that
Ms r +t , r N r, Ms and t N t, Ms. Thus, settingNrˆ =gand Ntˆ =h, we get that
(1−)sˆ g+h, gr,ˆ sˆ and ht ,ˆ s.ˆ
Since the elements of the form(1−)sˆ , with >0 ands s, are compactly contained insˆand have supremums, the proof is complete.ˆ
In what follows SR denotes the realification of an ordered semigroup S that satisfies axioms O1–O6. SinceSRsatisfies the same axioms (by Propos- ition 3.3.1 and Lemma 4.0.1), and(SR)R ∼= SR, we may alternatively regard
SR as an arbitrary ordered semigroup with real multiplication and satisfying axioms O1–O6.
4.1. Refinement
The following form of refinement property holds inSRand suffices to conclude that F(S)is a lattice.
Theorem4.1.1.LetSbe an ordered semigroup satisfying axiomsO1–O6.
Let(fi)ni=1and(gj)jm=1be elements ofSRsuch that n
i=1
fi m
j=1
gj.
Let(fi)ni=1be such thatfi fi for alli. Then there exist elementshij, with i=1,2, . . . , nandj =1,2, . . . , m, such that
fi m
j=1
hi,j fi for alli, (4.1)
n
i=1
hi,j gj for allj.
(4.2)
Proof. Notice that it suffices to prove the theorem with the inequality relationin place of the compact containment relationin (4.1). Once the inequalities are obtained, the compact containment is easily arranged by finding interpolating elementsfi fi fi fi and applying the theorem, with inequality relations, for the pairsfi fi .
Let us first prove the theorem forn=1 andm=2. Letf , f, g∈SRbe such thatf f g1+g2. Let us assume thatf ∝g2. By Proposition 3.1.6, there existslsuch thatf lf. Let >0 be such thatl(1−)f. Sincel f, for anyδ >0 we can apply Lemma 3.1.7 to getl such thatf l l and ll +δf. Sincef ∝g2, we can chooseδsmall enough so thatl l +g2. In summary, we first findl and >0 such thatf l (1−)f and then findl such thatf l landl l +g2.
By axiom O6 applied to
l l (1−)g1+(1−)g2, there existg1andg2such thatl g1+g2and
g1(1−)g1, l, g2(1−)g2, l.
Let us chooseh1g1such thatl h1+g2. Sinceg1l, by Proposition 3.3.1 we may chooseh1that is algebraically complemented inl, i.e., such that there existsh2such thatl=h1+h2. Then
h1+h2=ll +g2h1+g2+g2h1+g2.
Sinceh1f ∝g2, we can cancelh1to obtainh2g2. This proves the case n=1,m=2 of the theorem under the assumption thatf ∝g2.
It follows by induction that iff f ni=1gi, andf ∝ gnthen there exist(hi)ni=1such thatf ni=1hi f andhi gi for alli.
Let us now go back to the casen=1 andm=2 and remove the assumption f ∝g2. Suppose again thatf f g1+g2. Letland >0 be such that f l(1−)f. By axiom O6 there existg1andg2such thatl g1+g2
and g1(1−)g1, (1−)f,
g2(1−)g2, (1−)f.
Then we trivially havelg1+g2+2(g1+g2)andl ∝ 2(g1+g2). So, there existh1,h2andh3such that
f h1+h2+h3l, h1g1, h2g2 and h3
2(g1+g2).
Set h1+ 2g1 = h1 and h2 + 2g2 = h2. Then h1 g1, h2 g2, and f h1+h2. Also,
h1+h2=h1+h2+
2(g1+g2)l+
2(g1+g2).
Butl (1−)f and 2(g1+g2) f. So,h1+h2 f. This proves the theorem forn=1 andm=2.
The reader may verify that the casen=1 and arbitrarymnow follows by induction, building on the case that was just established.
Finally, let us consider the general case of the theorem. Let us assume that the theorem has been proved for certainnandmand then show that it is also valid forn+1 and m. Suppose that ni=+11fi jm=1gj and letf1 f1. Then there exist(hj)jm=1such thatf1 jm=1hj f1andhj gj for allj. For eachj let us findhj hj andgj such thathj+gj gj hj+gj and f1 jm=1hj. Then
f1+
n+1
i=2
fi m
j=1
gj m
j=1
hj+ m
j=1
gj f1+ m
j=1
gj.
Thus,
(4.3) f1+
n+1
i=2
fi f1+ m
j=1
gj
By Lemma 3.3.2, the elementsgj may be chosen such thatgj ∝ gj. So f1∝ jm=1gjand we can cancelf1on both sides of (4.3). By induction, there exist(hi,j),i= 2, . . . , n+1,j =1, . . . , m, such thatfi jm=1hi,j fi fori = 2,3, . . . , n+1 and ni=1hi,j gj for allj. We now seth1,j = hj. The elementshi,j have the desired properties. This completes the induction.
Theorem 4.1.2. Let S be an ordered semigroup satisfying axioms O1–
O6. Then F(S) is a complete distributive lattice. Furthermore, addition is distributive with respect to∧and∨:
(λ1∨λ2)+λ3=(λ1+λ3)∨(λ2+λ3), (4.4)
(λ1∧λ2)+λ3=(λ1+λ3)∧(λ2+λ3) (4.5)
Proof. Since F(S) ∼= F(SR), it suffices to prove the same properties for F(SR)(or alternatively, to assume thatShas real multiplication). The pointwise supremum of an upward directed set of functionals is also a functional, and so the supremum of the set. Thus, in order to show that F(SR)is a complete lattice, it suffices to show that any two functionals have a least upper bound.
Letλ1andλ2be in F(SR). Let us defineλ:SR→[0,∞] by (4.6) λ(f ):=sup{λ1(f1)+λ2(f2)|f1+f2f}.
That λ is sub-additive follows from general considerations. The inequality λ(f )+λ(g) λ(f +g)follows from the refinement property obtained in Theorem 4.1.1. Thus,λis additive. It is clear thatλis the least upper bound ofλ1 andλ2among all the ordered semigroup maps fromSRto [0,∞]. Letλ˜denote the supremum preserving regularization ofλ. That is,λ(f )˜ :=supffλ(f ).
Thenλis a functional onSR(see [6, Lemma 4.7]) andλ1∨λ2= ˜λ.
The identity (4.4) follows from the fact thatλ1∨λ2is the lower semicon- tinuous regularization of the functional given by (4.6). The reader is referred to the proof of [6, Theorem 3.3] for the details of this argument. Similarly, in order to prove (4.5) we need a Kantorovich-type formula forλ1∧λ2. Consider the mapλ:SR →[0,∞] defined by
λ(f ):=inf{λ1(f1)+λ2(f2)|f f1+f2}.
Thatλis sub-additive follows again from general considerations. The refine- ment property of Theorem 4.1.1 can then be used to show that
λ(f )+λ(g)λ(f +g),
for allf f andg g. It follows thatλ(f )˜ :=supff λ(f )is additive.
Moreover, proceeding as in the proof of [6, Lemma 4.7] we get that λ˜ is a functional onSR. Ifγ ∈ F(SR)is such thatγ λ1, λ2 then clearlyγ λ.
Sinceγ (f )=supffγ (f ), we also have thatγ λ. Therefore,˜ λ˜ =λ1∧λ2. Identity(4.5) can now be derived proceeding as in the proof of [6, Theorem 3.3].
Finally, the identities (4.4) and (4.5) imply that F(SR)is a distributive lattice (by [6, Proposition 3.4]).
4.2. Interpolation
Here we show that if S satisfies axioms O1–O6 and has a countable dense subset then there exists a greatest lower boundf ∧g for any two elements f, g∈SR.
Lemma4.2.1.Letf, g ∈ SR withf ∝g. Then the set of elementsh∈ SR such thathh f, gfor someh, is an upward directed set.
Proof. Let p and q be elements of SR such that p p f, g and q q f, g. Writingp as the supremum of a rapidly increasing sequence as in Proposition 3.1.6, we can findp1andp2such thatpp1p2p. Similarly, we findq1andq2such thatqq1q2q. In order to prove the lemma, it suffices to findr1 ∈SRsuch thatp1, q1 r1f, g, for then there existsrsuch thatp, q rr1f, g.
Let us prove the existence ofr1satisfying thatp1, q1r1f, g. In what follows, the relevant properties ofp1andq1are that
(i) there exists >0 such thatp1, q1(1−)f, (1−)g, and
(ii) p1andq1have algebraic complements in both(1−)f and(1−)g (this follows from Lemma 3.3.2).
Let us choosepf,qf, andqgsuch that
p1+pf =q1+qf =(1−)f, q1+qg =(1−)g.
Then q1+qf =(1−)f =p1+pf
(1−)g+pf
=q1+qg+pf.
