Sup-lattice 2-forms and quantales

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Journal of Algebra 276 (2004) 143–167

www.elsevier.com/locate/jalgebr

Sup-lattice 2-forms and quantales✩

Pedro Resende

Departamento de Matemática, Instituto Superior Técnico, Av. Rovisco Pais, 1049-001 Lisboa, Portug

Received 20 November 2002

Communicated by Michel Broué

Abstract

A 2-form between two sup-latticesL andR is defined to be a sup-lattice bimorphismL × R → 2.Such 2-forms are equivalent to Galois connections, and we study them and their relation to quinvolutive quantales and quantale modules. As examples we describe applications to C*-alge 2004 Elsevier Inc. All rights reserved.

Keywords:Sup-lattice; Galois connection; Quantale; Involutive quantale; Quantale module; C*-algebra

1. Introduction

Let L andR be sup-lattices. A Galois connection betweenL andR is a pair of antitonemaps(−)⊥ :L → R and⊥(−) :R → L such thatx � ⊥(x⊥) andy � (⊥y)⊥ for all x ∈ L

andy ∈ R. In fact all the information present in the Galois connection is already avaiin each of the maps(−)⊥ and⊥(−), due to completeness of the lattices, or equivalentlthe mapϕ :L × R → 2 given by

ϕ(x, y) = 0 ⇔ x � ⊥y(⇔ y � x⊥)

,

which is a bimorphism of sup-lattices. In this paper we study such bimorphisms, anthem (sup-lattice) 2-forms. The purpose is to provide a useful framework within whistudy various aspects of quantales and their modules, including involutive quantaltheir applications to C*-algebras.

✩ Research supported in part by FEDER and FCT/POCTI/POSI through the Research Units Funding Programand grant POCTI/1999/MAT/33018.

E-mail address:[email protected].

0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2004.01.020

144 P. Resende / Journal of Algebra 276 (2004) 143–167

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ersction.

We study two notions of map between 2-forms: the orthomorphisms, in Section 3,are analogous to isometries, and the continuous maps, in Section 4, so-named becausegeneralize the continuous maps of topological spaces. In particular, the set of continuouendomaps of a 2-form has the structure of a quantale, we show that under mild restrthe 2-form can be recovered from it, and we obtain generalizations of well known facts [concerning the right and left sides of quantales of sup-lattice endomorphisms, and aconcerning involutive quantales. In Section 5 we deal with principal quantale mo(i.e., modules with a single generator), and in Section 6 we relate them to 2-forms. Fin Section 7 we address the particular case of symmetric 2-forms and involutive quaand we discuss applications to C*-algebras.

We are indebted to the work of Mulvey and Pelletier [7], which was one of the msources of inspiration for our paper. They implicitly use parts of the theory of 2-formsthis is reflected in the fact that we obtain, in Section 7, a much shorter proof of otheir main theorems [7, Theorem 9.1], which concerns the relation between quantalesC*-algebras. We hope in this way to bring out more explicitly some of the principles thalie behind that relation.

2. Background

In this section we present some basic facts, terminology and notation concerninlattices, quantales and quantale modules, however without attempting to be comFurther basic reading about sup-lattices andquantales can be found in the first chaptof the book by Rosenthal [13], and further references will be cited throughout this se

By a sup-latticeis meant a partially ordered setS each of whose subsetsX ⊆ S has ajoin (supremum)

∨X in S (hence, a sup-lattice is a complete lattice). By ahomomorphism

of sup-latticesf :S → T is meant a map that preserves arbitrary joins:

f(∨

X)

=∨{

f (x) | x ∈ X}, for all X ⊆ S.

The greatest element∨

S of a sup-latticeS (the top) is denoted by 1S , or 1, and theleast element

∨∅ (the bottom) by 0S , or 0. The two-element sup-lattice{0,1} is denotedby 2. The order-dual of a sup-latticeS, i.e.,S with the order reversed, is denoted bySop.A homomorphism of sup-latticesf :S → T is said to bestrongif f (1) = 1, anddenseifthe conditionf (x) = 0 impliesx = 0 for all x ∈ S.

Any sup-lattice homomorphismf :S → T has a right adjointf∗ :T → S, whichpreserves all the meets (infima) inT and is defined by

f∗(y) =∨{

x ∈ S | f (x) � y}.

Equivalently,f∗ is the unique monotone map that satisfies the condition

f (x) � y ⇔ x � f∗(y)

for all x ∈ S andy ∈ T .

P. Resende / Journal of Algebra 276 (2004) 143–167 145

m.

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,

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rs with

If S is a sup-lattice, andj :S → S is a closure operator onS, the set of fixed-points ofj ,Sj = {x ∈ S | j (x) = x}, is a sup-lattice whose joins are given by

∨jX = j (

∨X), and the

mapj :S → Sj that sends eachx ∈ S to j (x) is a surjective sup-lattice homomorphisAny quotient of a sup-lattice arises like this, up to isomorphism, for iff :S → T is asurjective sup-lattice homomorphism thenj = f∗ ◦ f is a closure operator onS, andT ∼= Sj [1,13].

The categorySL of sup-lattices is monoidal [1], and a semigroup in it is aquantale,unital if the semigroup is a monoid,involutiveif the semigroup has an involution. A le(respectively right)moduleover a quantaleQ is a left (respectively right) action inSL. Themultiplication of two elementsa andb in a quantaleQ is denoted bya · b; if the quantaleis unital, its multiplicative unit is denoted byeQ, or simplye; if the quantale is involutivethe involution assigns to eacha ∈ Q an element that is denoted bya∗. The action of anelementa ∈ Q on x ∈ M, whereM is a left Q-module, is denoted byax (or xa for aright Q-module), and the module isunital if ex = x for all x ∈ M (respectivelyxe = x

for a right module). An elementa of a quantale isleft-sided(respectivelyright-sided) if1 · a � a (respectivelya · 1 � a). An element which is both left- and right-sided istwo-sided. The set of left-sided elements of a quantaleQ is denoted by L(Q) (this is a rightQ-module under multiplication), and the set (a leftQ-module) of right-sided elementsdenoted by R(Q). A factor is a quantaleQ whose set of two-sided elements is{0,1}.

For any sup-latticeS the setQ(S) of sup-lattice endomorphisms ofS is a unital quantaleunder the pointwise ordering, with multiplication given by composition,f · g = g ◦ f , andwe have L(Q(S)) ∼= S and R(Q(S)) ∼= Sop [6]. Explicitly, for a unital quantaleQ we haveL(Q) = 1 · Q and R(Q) = Q · 1, and thus, forQ(S), a left-sided element is the same a“constant” map for somes ∈ S,

cs(x) ={

s if x = 0,

0 if x = 0,

and a right-sided element is an annihilator of somes ∈ S,

as(x) ={

1 if x � s,

0 if x � s.

It also follows from this thatQ(S) is a factor.Another example of unital quantale, for any monoidM, is the powerset 2M under

pointwise multiplication:

X · Y = {xy | x ∈ X,y ∈ Y }.

This construction is universal in the sense that for any unital quantaleQ and anyhomomorphism of monoidsh :M → Q there is a unique homomorphism of uniquantales̄h : 2M → Q such that̄h({x}) = h(x) for all x ∈ M. Hence, any unital quantalea quotient of one of the form 2M , for some monoidM.

Quotients of quantales and modules can be described in terms of closure operatoadditional properties: aquantic nucleus[13], or simply anucleus, on a unital quantaleQ


talof

e

p

,

l

dum”he

is a closure operatorj on Q such that for alla, b ∈ Q we havej (a) · j (b) � j (a · b);and anucleuson a leftQ-moduleM is a closure operatork on M such that for alla ∈ Q

andx ∈ M we haveak(x) � k(ax) (see [10] or [12, Section 2.5]). Given nucleij andk asabove,Qj is a unital quantale with multiplication(a, b) �→ a∗b defined bya∗b = j (a ·b),and Mk is a left Q-module with action(a, x) �→ a • x defined bya • x = k(ax). Thesurjective mapsj :Q → Qj andk :M → Mk are respectively a homomorphism of uniquantales and a homomorphism of leftQ-modules, and any quotient of quantales orleft Q-modules arises like this, up to isomorphism. Ifj andk are further related by thconditionj (a)x � k(ax), for all a ∈ Q andx ∈ M, thenMk is also a leftQj -module.

If R is a ring with unit then the map that sends each subset ofR to the additive subgrouit generates is a nucleus on the unital quantale 2R, and thus Sub(R), the set of additivesubgroups ofR, is a unital quantale with multiplication defined by

a · b = {r1s1 + · · · + rnsn | ri ∈ a, si ∈ b}.

The left-sided elements of Sub(R) are then the left ideals ofR.More generally, in the case of a unitalk-algebraA with k an arbitrary commutative ring

the set Subk(A) of all thek-submodules ofA is a unital quantale, and ifA is a topologicalk-algebra then the setSubk(A) of all the closedk-submodules ofA is a unital quantalewith multiplication defined by

a · b = {r1s1 + · · · + rnsn | ri ∈ a, si ∈ b},

where(−) denotes topological closure.We can obtain examples of modules in a similar way. IfR is a ring andM is a left

R-module then the set Sub(M) of additive subgroups ofM is a left module over Sub(R),with action defined by

ax = {r1m1 + · · · + rnmn | ri ∈ a, mi ∈ x}.

A similar expression, but including closure for the topology, gives us a leftSubk(A)-moduleSubk(M), consisting of all the closedk-submodules ofM, from any topologicaleft A-moduleM over a topologicalk-algebraA.

If A is a complex C*-algebra with unit, the unital quantaleSubC(A) is involutive, withinvolution obtained pointwise from the involution ofA. This involutive quantale is denoteby MaxA in [4,5,7,8], where it plays the role of the “noncommutative maximal spectrof A. If H is a Hilbert space, its norm-closed linearsubspaces can be identified with tprojections onH , and we denote the sup-latticeSubC(H) by P(H). Any C*-algebrarepresentationπ :A → B(H) of A on H makesH a topological leftA-module, thusmakingP(H) a left MaxA-module.

Let L, R, andM be sup-lattices, and(−) ∗ (−) :L× R → M a sup-latticebimorphism,i.e., a map that preserves joins in each variable (e.g., the multiplicationQ × Q → Q of aquantaleQ, or the actionQ × M → M of Q on a left moduleM):

(∨X

)∗ y =

∨{x ∗ y | x ∈ X},


ae

usge

y

ft.

hismre-

ht.

x ∗(∨

Y)

=∨

{x ∗ y | y ∈ Y }.

We will consistently use the following notation for the residuations associated to∗ (i.e.,the right adjoints to the homomorphisms(−) ∗ y andx ∗ (−)), for eachx ∈ L, y ∈ R, andz ∈ M:

z/y =∨

{x ∈ L | x ∗ y � z},x\z =

∨{y ∈ R | x ∗ y � z}.

Also, we define the following annihilators: ann(x) = x\0, and ann(y) = 0/y. Hence, wehave

y � x\z ⇔ x ∗ y � z ⇔ x � z/y,

y � ann(x) ⇔ x ∗ y = 0 ⇔ x � ann(y),

and also the following (in)equalities:(z/y) ∗ y � z, ann(y) ∗ y = 0, x ∗ (x\z) � z,x ∗ann(x) = 0,x � (x ∗y)/y, y � x\(x ∗y), ((x ∗y)/y)∗y = x ∗y, x ∗(x\(x ∗y)) = x ∗y,((x/y) ∗ y)/y = x/y, x\(x ∗ (x\y)) = x\y.

3. 2-forms and orthomorphisms

Let S be a sup-lattice. Since its dual,Sop, is order isomorphic to hom(S,2) [1],any Galois connection between two sup-latticesL and R is uniquely determined bysup-lattice homomorphismL → hom(R,2), which in turn is equivalent to a sup-latticbimorphismL × R → 2 (because we have an order isomorphism hom(L ⊗ R,M) ∼=hom(L,hom(R,M)) for any sup-latticesL,R,M [1]). Such bimorphisms are analogoto bilinear forms on ring modules, and provide us with a convenient alternative languafor describing Galois connections.

Definition 3.1. Let L andR be sup-lattices. A mapϕ :L × R → 2 that preserves arbitrarjoins in each variable is called a 2-form betweenL andR, and we usually write〈x, y〉 or〈x, y〉ϕ instead ofϕ(x, y). Two elementsx ∈ L andy ∈ R areorthogonalif 〈x, y〉 = 0, inwhich case we writex ⊥ y. The form isdense on the rightif 1 ⊥ y impliesy = 0, denseon the leftif x ⊥ 1 impliesx = 0, anddenseif it is both dense on the right and on the leThe form isfaithful on the rightif x = y whenever〈z, x〉 = 〈z, y〉 for all z ∈ L, faithfulon the leftif x = y whenever〈x, z〉 = 〈y, z〉 for all z ∈ R, andfaithful, or non-singular,or a duality, if it is both faithful on the right and on the left. A 2-formϕ :S × S → 2 issymmetricif 〈x, y〉 = 〈y, x〉 for all x, y ∈ S.

Density on the right is equivalent to requiring the sup-lattice homomorp〈1,−〉 :L → 2 to be dense, which justifies our terminology. It is also equivalent toquiring y = 0 wheneverz ⊥ y for all z ∈ L, i.e., whenever〈z, y〉 = 〈z,0〉 for all z ∈ L,which shows that faithfulness on the right isa stronger condition than density on the rig


ringce, forll needforn the

lar torily

ion is

rized as

siony

Of course, these two conditions would be equivalent if we were dealing with forms onmodules, and equivalent to saying that a form is non-degenerate on the right. Hensup-lattices there are two natural notions of non-degeneracy on the right. We shaboth of them, so we have decided to use a different word for each, and non-degeneracynone in order to avoid ambiguity. Similar remarks apply to density and faithfulness oleft.

In view of these remarks it may seem surprising that we have defined non-singumean the same as faithful, since for ring modules a non-degenerate form is not necessanon-singular, but 3.4 below shows that in the case of sup-lattices this identificatappropriate.

Definition 3.2. Let ϕ :L × R → 2 be a 2-form,x ∈ L andy ∈ R. The (right) orthogonalimageof x is the elementx⊥ ∈ R defined by

x⊥ =∨

{y ∈ R | x ⊥ y}.

Similarly, the (left) orthogonal imageof y is given by

⊥y =∨

{x ∈ L | x ⊥ y}.

The correspondence between Galois connections and 2-forms can be summafollows:

Proposition 3.3. For any 2-form between sup-latticesL and R, the orthogonal image(−)⊥ :L → R and ⊥(−) :R → L form a Galois connection, and any Galois connectbetweenL andR is uniquely determined in this way by the2-form whose orthogonalitrelation is given by

x ⊥ y ⇔ x � ⊥y(⇔ y � x⊥)

.

Furthermore, we have:

(1) The following are equivalent:(a) ϕ is dense on the right;(b) 1⊥ = 0;(c) 0 is the unique elementy ∈ R such that⊥y = 1.

(2) The following are equivalent:(a) ϕ is dense on the left;(b) ⊥1= 0;(c) 0 is the unique elementx ∈ L such thatx⊥ = 1.

(3) The following are equivalent:(a) ϕ is faithful on the right;(b) (−)⊥ is surjective;(c) ⊥(−) is injective;(d) (⊥y)

⊥ = y for all y ∈ R.


ms.

sure-

d ite

s

ght)t).

(4) The following are equivalent:(a) ϕ is faithful on the left;(b) (−)⊥ is injective;(c) ⊥(−) is surjective;(d) ⊥(x⊥) = x for all x ∈ L.

Corollary 3.4. A 2-form between sup-latticesL and R is faithful if and only if (−)⊥(equivalently,⊥(−)) is an antitone order isomorphism.

Let us see some explicit examples of Galois connections in the language of 2-for

Example 3.5. Let X be a topological space, with topologyτX . Then we define a 2-formbetween 2X andτX by

S ⊥ U ⇔ S ∩ U = ∅,

which is faithful on the right and dense on the left. More generally, given any clooperatorj :L → L on a sup-latticeL, the assignmentx �→ j (x) defines a surjective suplattice homomorphismj :L → Lj (cf. Section 2), and we may define a 2-form onL×L

opj

by x ⊥ y ⇔ x � y in L. Furthermore, this form is necessarily faithful on the right, anis dense on the left if and only ifj (0) = 0 (i.e., the closure is dense). In particular, if wtakeL to be 2X andLj to be the lattice of closed sets ofX thenL

opj

∼= τX and we obtainthe same as before.

Example 3.6. Let ρ be a binary relation between two setsS andT . Then we have a 2-formbetween 2S and 2T given by

X ⊥ Y ⇔ xρy for all x ∈ X, y ∈ Y.

For instance, ifS = T and we takexρy to bex = y, we obtain

X ⊥ Y ⇔ X ∩ Y = ∅,

as in the topological example above.

Example 3.7. Let R be a commutative ring,M andN R-modules, andf :M × N → R

a bilinear form. Thenf induces an orthogonalityrelation betweenM andN , with respectto which we can define a sup-lattice 2-formϕ : SubR(M)×SubR(N) → 2 as in the previouexample: for all submodulesX ⊆ M andY ⊆ N , putX ⊥ Y if and only if f (x, y) = 0 forall x ∈ X andy ∈ Y . This sup-lattice 2-form is dense on the left (respectively on the riif and only if the bilinear formf is non-degenerate on the left (respectively on the righ


yf

for

Definition 3.8. Let ϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be 2-forms. Anorthomorphismϕ → ϕ′ is a pair of sup-lattice homomorphismsf :L → L′ andg :R → R′ such that forall x ∈ L andy ∈ R we have

⟨f (x), g(y)

⟩ = 〈x, y〉.

If both f and g are surjective the orthomorphism(f, g) is said to be aquotientorthomorphism. In that caseϕ′ is anorthoquotient, or simplyquotient, of ϕ.

Proposition 3.9. Let ϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be2-forms, and(f, g) :ϕ → ϕ′an orthomorphism. Ifg is surjective then(f, g) commutes with(−)⊥ in the sense that

g(x⊥) = f (x)⊥

for all x ∈ L. If furthermoref is strong then(f, g) preserves density on the right(i.e.,ϕ′is dense on the right ifϕ is), and, ifg is also dense, thenϕ is dense on the right if and onlif ϕ′ is dense on the right. Obvious dual statements apply to⊥(−) and density on the left if andg are interchanged.

Proof. Assume thatg is surjective, and letx ∈ L. Then,

g(x⊥) = g

(∨{y | x ⊥ y}

)=

∨{g(y) | x ⊥ y

}

=∨{

g(y) | f (x) ⊥ g(y)} =

∨{z ∈ R′ | f (x) ⊥ z

} = f (x)⊥.

If furthermoref is strong andϕ is dense on the right we have

1L′⊥ = f (1L)⊥ = g(1L

⊥) = g(0) = 0,

i.e., ϕ′ is dense on the right; and ifg is also dense we have 1L⊥ = 0 if and only if

g(1L⊥) = 0, and thusϕ is dense on the right if and only ifϕ′ is. The dual facts, with

f andg interchanged, are proved in a similar way.�Proposition 3.10. Letϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be2-forms, and(f, g) :ϕ → ϕ′an orthomorphism. Ifg is surjective andϕ is faithful on the left thenf is an orderembedding, and iff is surjective andϕ is faithful on the right theng is an orderembedding.

Proof. Assume thatg is surjective andϕ is faithful on the left, i.e.,⊥(z⊥) = z for all z ∈ L

(cf. 3.3). Then(f, g) preserves(−)⊥, by the previous proposition, and thus we have,all x, z ∈ L,

f (x) � f (z) ⇒ f (x) � ⊥(f (z)⊥

) ⇔ f (x) � ⊥(g(z⊥)) ⇔ f (x) ⊥ g

(z⊥)

⇔ x ⊥ z⊥ ⇔ x � ⊥(z⊥) = z,


f

tientrm wect thatlosed

sly

d

i.e., f is an order embedding. Forf surjective andϕ faithful on the right everything issimilar. �Corollary 3.11. Letϕ andϕ′ be2-forms, and(f, g) :ϕ → ϕ′ a quotient orthomorphism. Iϕ is faithful then bothf andg are order isomorphisms.

Hence, the faithful 2-forms are “simple” in the sense that their only quoorthomorphisms are isomorphisms. The next proposition states that from any 2-focan always obtain a faithful one by means of a quotient, and corresponds to the faany Galois connection restricts to a dual isomorphism between the lattices of “celements”.

Proposition 3.12. Letϕ :L × R → 2 be a2-form, and letf :L → L andg :R → R be theclosure operators defined byx �→ ⊥(x⊥) andy �→ (⊥y)⊥. LetL′ = ⊥R = {⊥(x⊥) | x ∈ L}andR′ = L⊥ = {(⊥y)⊥ | y ∈ R} be the corresponding quotients ofL andR, and define amapϕ′ :L′ × R′ → 2 to be the restriction ofϕ to L′ × R′. Thenϕ′ is a faithful2-form andthe pair(f, g) defines a(quotient) orthomorphism fromϕ to ϕ′.

Proof. First we remark that for allx ∈ L andy ∈ R we have

x ⊥ y ⇔ y � x⊥ = (⊥(x⊥))⊥ ⇔ ⊥(

x⊥) ⊥ y.

Hence, for each subsetX ⊆ L′ and eachy ∈ R we have

⊥((∨X

)⊥)⊥ y ⇔

∨X ⊥ y ⇔ x ⊥ y for all x ∈ X,

which shows thatϕ′ preserves joins in the left variable, because the join ofX in L′ is⊥((

∨X)

⊥). Similarly, ϕ′ preserves joins on the right and is thus a 2-form, obviou

faithful, see 3.3. Finally, we also obtain, for allx ∈ L andy ∈ R,

x ⊥ y ⇔ ⊥(x⊥) ⊥ y ⇔ ⊥(

x⊥) ⊥ (⊥y)⊥

,

which means(f, g) is an orthomorphism. �Definition 3.13. We refer to the faithful 2-formϕ′ of the previousproposition as theorthogonal quotientof ϕ.

We conclude this section with the following proposition, which will not be needeelsewhere in this paper, but which can be regarded as the “soft” version of 3.10:

Proposition 3.14. Letϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be2-forms, and(f, g) :ϕ → ϕ′an orthomorphism. Ifg is strong andϕ is dense on the left thenf is dense. Iff is strongandϕ is dense on the right theng is dense.


, after

s),

al

tors

,

Proof. Assume thatϕ is dense on the left. Ifg is strong then we have

f (x) = 0 ⇒ f (x) ⊥ 1 ⇔ f (x) ⊥ g(1) ⇔ x ⊥ 1 ⇔ x = 0,

i.e.,f is dense. The second part of the proof is similar.�

4. Quantales and 2-forms

The multiplication of a quantaleQ has the property that, for allX ⊆ Q anda ∈ Q,(∨

X) · a = 0 if and only ifx · a = 0 for all x ∈ X, anda · ∨X = 0 if and only ifa · x = 0for all x ∈ X. Hence, we obtain a 2-form from any quantale, as follows:

Definition 4.1. For any quantaleQ, we define a 2-formΦ(Q) between L(Q) and R(Q) byputting, for eacha ∈ L(Q) andb ∈ R(Q),

a ⊥ b ⇔ a · b = 0.

Now we study a converse to this, i.e., a way of obtaining a quantale from a 2-formwhich we relate the two constructions.

Definition 4.2. Letϕ :L×R → 2 andϕ′ :L′ ×R′ → 2 be 2-forms. Acontinuous mapfromϕ to ϕ′ is a pair(f, g) of contravariant sup-lattice homomorphisms, wheref :L → L′ andg :R′ → R, such that the followingcontinuity condition is satisfied: for allx ∈ L andy ∈ R′,

⟨f (x), y

⟩ = ⟨x,g(y)

⟩.

Example 4.3. The above terminology is justified as follows. LetX andY be topologicalspaces, with topologiesτX andτY , let f :X → Y be a map (not necessarily continuouand let g : τY → τX be a sup-lattice homomorphism. SeeingX and Y as 2-forms asin 3.5, the pair(f̃ , g), wheref̃ is the direct image map off , f̃ (S) = {f (x) | x ∈ S}, isa continuous map of 2-forms if and only iff :X → Y is a continuous map of topologicspaces andg = f −1.

A generalization of this situation can be obtained from any pair of closure operajandj ′ on sup-latticesL andL′, respectively. From these we obtain 2-forms onL × L

opj and

L′ × L′opj ′ , as in the second part of 3.5, and iff :L → L′ is a sup-lattice homomorphism

then (f, g) is a continuous map of 2-forms if and only iff satisfies the conditionf ◦ j � j ′ ◦ f (i.e., f is continuous with respect to the closure operators) andg is therestriction toLj ′ of the right adjointf∗—see 4.6 below.

Proposition 4.4. The continuity condition is equivalent to each of the following:

(1) g∗(x⊥) = f (x)⊥ for all x ∈ L,(2) f∗(⊥y) = ⊥g(y) for all y ∈ R,


s

y

hichcient

(3) f (x) ⊥ g∗(x⊥) andx ⊥ g(f (x)⊥) for all x ∈ L,(4) f∗(⊥y) ⊥ g(y) andf (⊥g(y)) ⊥ y for all y ∈ R.

Proof. (1) Continuity can be rewritten as

f (x) ⊥ y ⇔ x ⊥ g(y),

which in turn is equivalent to the condition

y � f (x)⊥ ⇔ g(y) � x⊥,

whose right-hand side is equivalent toy � g∗(x⊥).(2) This is similar to the previous case, once we rewrite the continuity condition a

f (x) � ⊥y ⇔ x � ⊥g(y),

since now the left-hand side is equivalent tox � f∗(⊥y).(3) From the first condition, continuity is equivalent to the conjunction

g∗(x⊥)

� f (x)⊥ and g∗(x⊥)

� f (x)⊥.

The inequalityg∗(x⊥) � f (x)⊥ is equivalent tof (x) ⊥ g∗(x⊥), and the other inequalitis equivalent tox⊥ � g(f (x)⊥), i.e., tox ⊥ g(f (x)⊥).

(4) Similar to the previous case, now using the second condition.�Corollary 4.5. Letϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be2-forms.

(1) If ϕ is faithful on the right then for each sup-lattice homomorphismf :L → L′ thereis at most one sup-lattice homomorphismg :R′ → R such that(f, g) is a continuousmap of2-formsϕ → ϕ′.

(2) If ϕ′ is faithful on the left then for each sup-lattice homomorphismg :R′ → R thereis at most one sup-lattice homomorphismf :L → L′ such that(f, g) is a continuousmap of2-formsϕ → ϕ′.

Proof. (1) ϕ is faithful on the right if and only if the map(−)⊥ :L → R is surjective.Hence, the first condition of the proposition,g∗(x⊥) = f (x)⊥ for all x ∈ L, completelydetermines the right adjointg∗, and thus it determinesg.

(2) Similar, taking into account the second condition of the proposition.�Notice that we do not state, e.g., in the first part of this corollary, that for everyf there

is ag such that(f, g) is continuous. The second part of 4.3 provides an example in wfor only somef this holds, and the following proposition gives a necessary and sufficondition for suchg to exist.

Proposition 4.6. Let ϕ :L × R → 2 andϕ′ :L′ × R′ → 2 be2-forms, both faithful on theright. Let alsof :L → L′ be a homomorphism. Then the following are equivalent:


re

en

e

r, the

own

(1) There is a homomorphismg :R′ → R such that(f, g) is continuous.(2) f (⊥(x⊥)) � ⊥(f (x)⊥) for all x ∈ L (i.e.,f is continuous with respect to the closu

operators⊥((−)⊥)).

If in addition ϕ is faithful on the left, then for eachf :L → L′ there is exactly oneg :R′ → R such that(f, g) is continuous.

Proof. Assume that (1) holds. Then we havef (x) ⊥ f (x)⊥, and

f (x) ⊥ f (x)⊥ ⇔ x ⊥ g(f (x)⊥

) ⇔ ⊥(x⊥) ⊥ g

(f (x)⊥

)⇔ f

(⊥(x⊥)) ⊥ f (x)⊥ ⇔ f

(⊥(x⊥))

� ⊥(f (x)⊥

).

Now assume that (2) holds. Writej for the closure operator⊥((−)⊥) on L, andk for thesimilar closure onL′. Then (2) is the conditionf ◦j � k ◦f . We shall prove that the imagof the restriction off∗ to L′

k is contained inLj . Indeed, this is equivalent to the conditiothatj ◦ f∗ ◦ k � f∗ ◦ k, which holds because

f ◦ j � k ◦ f ⇔ j � f∗ ◦ k ◦ f ⇒ j ◦ f∗ ◦ k � f∗ ◦ k ◦ f ◦ f∗ ◦ k � f∗ ◦ k,

where the latter inequality follows from the fact thatf ◦ f∗ � idL′ andk ◦ k = k. Hence,f∗ defines a meet preserving mapL′

k → Lj . Due to right faithfulness ofϕ and ϕ′ wehave order isomorphismsLj

∼= Rop andL′k

∼= R′op, and thusg :R′ → R can be definedby composingf∗ with the isomorphisms. Finally, ifϕ is also faithful on the left we hav⊥(x⊥) = x for all x ∈ L, and thusf is trivially continuous with respect to⊥((−)⊥). �

Clearly, continuous maps are closed under composition, and thuswe obtain anothecategory of 2-forms, which furthermore is sup-lattice enriched. In particular, thencontinuous endomaps of any 2-form form a unital quantale:

Definition 4.7. Let ϕ :L × R → 2 be a 2-form. Thequantale ofϕ, denoted byQ(ϕ),is the quantale of continuous endomaps ofϕ, with (f, g) � (f ′, g′) if and only iff (x) � f ′(x) andg(y) � g′(y) for all x ∈ L andy ∈ R, and with multiplication givenby (f, g) · (f ′, g′) = (f ′ ◦ f,g ◦ g′).

In the case of a symmetric 2-formϕ we have(f, g) ∈ Q(ϕ) if and only if (g, f ) ∈ Q(ϕ),and at once we remark:

Proposition 4.8. Let ϕ :L × L → 2 be a symmetric2-form. Then the quantaleQ(ϕ) isinvolutive, with the involution given by(f, g)∗ = (g, f ). Conversely, ifQ is an involutivequantale thenΦ(Q) is isomorphic to a symmetric2-form, andQ(Φ(Q)) is involutive, withthe involution given by(f, g)∗ = (g′, f ′), wheref ′ : R(Q) → R(Q) andg′ : L(Q) → L(Q)

are defined byf ′(y) = f (y∗)∗ andg′(x) = g(x∗)∗.

Example 4.9. Let us relate the quantales of endomorphisms of 2-forms to the well knendomorphism quantales of sup-lattices.


ents

ce,

t,

(1) Let ϕ :L × R → 2 be a faithful 2-form. From 4.6 it follows that the quantalesQ(ϕ)

andQ(L) are isomorphic.(2) Let L be a sup-lattice, and define a 2-formϕ :L × Lop → 2 by x ⊥ y if and only

if x � y in L (in other words, consider the Galois connection betweenL and Lop

defined by the identity(−)⊥ = idL :L → (Lop)op). This 2-form is faithful, and thusthe quantalesQ(ϕ) andQ(L) are isomorphic.

(3) Letϕ :L × L → 2 be both symmetric and faithful. ThenQ(ϕ) is isomorphic toQ(L),which is thus involutive. The involution is defined onQ(L) in the usual way forquantales of endomorphisms on self-dual sup-lattices [6]:

f ∗(y) =(∨{

x | f (x) � y⊥})⊥.

Example 4.10. Kruml [3] defines aGalois quantaleto be a quantaleQ(G) = {(f, g) ∈Q(S) ×Q(T ) | g ◦ G = G ◦ f } for some sup-lattice homomorphismG :S → T . From 4.4it follows that Galois quantales are the same as quantales of 2-forms:Q(G) is isomorphicto Q(ϕ) for the 2-formϕ :S × T op → 2 such that(−)⊥ = G.

Lemma 4.11. Letϕ :L × R → 2 be a dense two-form. ThenL(Q(ϕ)) is order isomorphicto L, andR(Q(ϕ)) is order isomorphic toR. FurthermoreQ(ϕ) is a factor quantale.

Proof. First we remark thatQ(ϕ) is a subquantale ofQ(L) × Q∗(R), whereQ∗(R) isthe quantaleQ(R) with reversed multiplication, i.e., withf · g = f ◦ g. Also, the topof Q(L) × Q∗(R) belongs toQ(ϕ) becauseϕ is dense: for allx ∈ L and y ∈ R, ifeither x = 0 or y = 0 then both conditions 1Q(L)(x) ⊥ y and x ⊥ 1Q∗(R)(y) are true,whereas ifx = 0 andy = 0 then both conditions are false. Hence, the left-sided elemof Q(ϕ) are precisely those which are left-sided as elements ofQ(L) × Q∗(R), i.e., theyare the continuous maps of the form(cl, ar) for somel ∈ L andr ∈ R, wherecl andar

are respectively a “constant” map and an annihilator, as described in Section 2. Hencontinuity means that for any pair of elementsx ∈ L andy ∈ R we must have

⟨cl(x), y

⟩ = ⟨x, ar(y)

⟩.

Takingx = 1 yields〈l, y〉 = 〈1, ar(y)〉, and thus

l ⊥ y ⇔ 1 ⊥ ar(y)

⇔ ar(y) = 0 (due to density)

⇔ y � r.

Therefore a necessary condition for continuity isr = l⊥. The condition is also sufficiensince:

• if x = 0 then, trivially,〈cl(x), y〉 = 〈x, ar (y)〉 = 0;


,pch

d

ll

ma

to

• if x = 0 thencl(x) = l and we obtain

⟨cl(x), y

⟩ = 0 ⇔ l ⊥ y ⇔ y � r

⇔ ar(y) = 0 ⇔ ⟨x, ar(y)

⟩ = 0,

where the latter step follows from density on the left and the fact thatar(y) equals either 0or 1. Hence, the generic form of a left-sided element ofQ(ϕ) is (cl, al⊥), which means wehave an assignmentl �→ (cl, al⊥) that defines a surjective mapL → L(Q(ϕ)). Furthermorewe havel � k if and only if cl � ck , andl � k impliesal⊥ � ak⊥ , which makes the maL → L(Q(ϕ)) an order-isomorphism.For right-sided elements the proof is analogous: earight-sided element ofQ(ϕ) must be of the form(al, cr) for somel ∈ L andr ∈ R, andcontinuity is the condition

⟨al(x), y

⟩ = ⟨x, cr(y)

⟩.

Taking again density ofϕ into account, we conclude thatl = ⊥r, and thus the right-sideelements must be of the form(a⊥r , cr ). Hence,R is isomorphic to R(Q(ϕ)). Finally, theonly elements that are simultaneously left- and right-sided are those for which(cl, al⊥) =(a⊥r , cr ), with l ∈ L and r ∈ R. The only solutions are(1,1) and (0,0), correspondingrespectively tol = r = 1 andl = r = 0, i.e.,Q(ϕ) is a factor. �Example 4.12. Let L be a sup-lattice, andϕ the 2-form onL × Lop of 4.9(2) (i.e., withx ⊥ y ⇔ x � y). From the isomorphismQ(L) ∼= Q(ϕ) we immediately obtain the weknown isomorphisms L(Q(L)) ∼= L and R(Q(L)) ∼= Lop [6].

Theorem 4.13. Letϕ :L×R → 2 be a dense2-form. Thenϕ andΦ(Q(ϕ)) are isomorphic2-forms.

Proof. All that we have to do is show that the isomorphisms of the previous lemcommute with the forms, i.e., that for alll ∈ L and r ∈ R we havel ⊥ r if and only if,in Q(ϕ), the following condition holds,

(cl, al⊥) · (a⊥r , cr ) = (0,0),

or, equivalently, if and only if the two following conditions hold: (i)a⊥r ◦ cl = 0, and(ii) al⊥ ◦ cr = 0. Since we havecl(1) = l, condition (i) holds if and only ifa⊥r (l) =a⊥r (cl(1)) = 0, which is equivalent tol � ⊥r. Similarly, condition (ii) holds if and onlyif al⊥(r) = 0, which is equivalent tor � l⊥. Hence, both (i) and (ii) are equivalentl ⊥ r. �

It is not in general true that for a quantaleQ we haveQ ∼=Q(Φ(Q)), but there is alwaysa comparison homomorphismκ :Q →Q(Φ(Q)):

Proposition 4.14. Let Q be a quantale, and leta ∈ Q. The right action ofa on L(Q)

and the left action ofa on R(Q) jointly define a continuous endomap((−) · a, a · (−))


.

tion

t

the

of the2-form Φ(Q). The comparison homomorphismκ :Q → Q(Φ(Q)) defined bya �→((−) · a, a · (−)) is a homomorphism of quantales, unital ifQ is unital. If furthermoreQ isinvolutive(in which caseQ(Φ(Q)) is involutive, see4.8) thenκ preserves the involution

Proof. The first part is immediate from the associativity of multiplication inQ, for:

〈x · a, y〉 = 0 ⇔ (x · a) · y = 0 ⇔ x · (a · y) = 0 ⇔ 〈x, a · y〉 = 0.

Now let us see thatκ is a homomorphism of quantales. First, it preserves multiplicabecause composition of continuous maps of 2-forms is defined by(f ′, g′) ◦ (f, g) =(f ′ ◦ f,g ◦ g′), and thus for alla, b ∈ Q the productκ(a) · κ(b) equals

κ(b) ◦ κ(a) = (((−) · a) · b, a · (b · (−)

)) = ((−) · (a · b), (a · b) · (−)

) = κ(a · b).

If Q is unital thenκ(e) is the unit of Q(Φ(Q)), and if Q is involutive we haveκ(a∗)(x, y) = (x · a∗, a∗ · y) = ((a · x∗)∗, (y∗ · a)∗) = κ(a)∗(x, y), see 4.8. �

The comparison homomorphism is injective if and only if, for alla, b ∈ Q, if x ·a = x ·banda · y = b · y for all x ∈ L(Q) andy ∈ R(Q) then we havea = b. A quantale satisfyingthis condition is usually said to befaithful [2,9,11].

5. Principal quantale modules

Definition 5.1. Let Q be a quantale. A leftQ-moduleM is principal if it has agenerator,i.e., an elementx ∈ M such thatQx = {ax | a ∈ Q} = M. Similar definitions apply to righmodules.

Some basic obvious properties of principal modules are the following:

Proposition 5.2. LetQ be a quantale.

(1) Any leftQ-module quotient of a principal leftQ-module is principal.(2) If M is a principal leftQ-module then it is a leftQ-module quotient ofQ.(3) If Q is unital thenM is a principal Q-module if and only if it is a leftQ-module

quotient ofQ.

Proof. (1) If f :M → N is a surjective homomorphism of leftQ-modules andM has ageneratorx thenf (x) is a generator ofN .

(2) If M is a leftQ-module with a generatorx then the mapQ → M defined bya �→ ax

is a surjective homomorphism of leftQ-modules.(3) If Q is unital thene is a generator of itself as a module. The rest follows from

previous two. �Definition 5.3. LetQ be a quantale, andM a leftQ-module. An elementx ∈ M is invariantif ax � x for all a ∈ Q (equivalently, if 1Qx � x).


the

rle

d

ation

ones,

Hence, the left-sided elements of a quantaleQ are the invariant elements ofQ whenQ

is seen as a left module over itself.

Proposition 5.4. Let Q be a quantale,M a left Q-module, andm an element ofM. Thefollowing are equivalent:

(1) ↑m is a left Q-module quotient ofM, with action defined by(a, x) �→ ax ∨ m andquotient projectionQ → ↑m given byx �→ x ∨ m.

(2) ↓m is a leftQ-submodule ofM.(3) m is an invariant element ofM.

Proof. (1) ⇔ (3). Condition (1) holds if and only if the map(−)∨m :M → M is a nucleusof left Q-modules. So assume thatm is invariant. Then, for alla ∈ Q andx ∈ M we have

a(x ∨ m) = ax ∨ am � ax ∨ m,

i.e., (−) ∨ m is a nucleus. Now assume that(−) ∨ m is a nucleus. Then for alla ∈ Q wehaveam = a(0∨ m) � a0∨ m = m, i.e.,m is invariant.

(2) ⇔ (3). ↓m is a sub-sup-lattice, so it is a submodule if and only if it is closed foraction. Let thenm be invariant,x ∈ ↓m, anda ∈ Q. Thenax � am � m. Let now↓m be asubmodule. Thenam ∈ ↓m for all a ∈ Q, i.e.,m is invariant. �Example 5.5. Let R be a ring, andM a leftR-module. Then Sub(M) is a left module ovethe quantale Sub(R), and an invariant elementN ∈ Sub(M) is the same as a submoduof M. Given such a submoduleN , it follows that Sub(N) is a left Sub(R)-submodule ofSub(M) and coincides with↓N , whereas↑N , which is order-isomorphic to Sub(M/N),is also isomorphic as a left Sub(R)-module when↑N is given the action of 5.4 anSub(M/N) is given the action induced by the leftR-module structure ofM/N .

Proposition 5.6. Let Q be a quantale,M a left Q-module, andx ∈ M. The map(−)x :Q → M sends left-sided elements to invariant elements, and the residu(−)/x :M → Q sends invariant elements to left-sided elements. Furthermore, ifx is agenerator then the residuation also reflects left-sided elements back into invarianti.e.,m ∈ M is invariant if and only ifm/x is left-sided.

Proof. If a ∈ Q is left-sided then 1(ax) = (1a)x � ax, i.e.,ax is invariant. Now assumemis invariant. We always have(m/x)x � m, and thus 1(m/x)x � 1m. Sincem is invariant wealso have 1(m/x)x � m, which is equivalent to 1(m/x) � m/x, i.e.,m/x is left-sided. Nowassume thatx is a generator. Then the map(−)x is surjective, the inequality(m/x)x � m

becomes the equalitym = (m/x)x, and thusm is the invariant element to which(−)x

maps any left-sided element of the formm/x. �Corollary 5.7. Let Q be a quantale,M a left Q-module, andx ∈ M. Thenann(x) is left-sided.


reastionselating

o

.

ereuled

d

Proof. ann(x) = 0/x is left-sided because 0 is invariant.�Proposition 5.8. Let Q be a quantale,M a left Q-module, andx ∈ M a generator. Thenthe quotient(−)x :Q → M factors through the quotient(−) ∨ ann(x) :Q �→ ↑ann(x)

and a dense homomorphismϕ :↑ann(x) → M. Furthermore,ϕ restricts to an orderisomorphismM/x = {m/x | m ∈ M} ∼= M.

Proof. For eacha ∈ Q we haveax = ax ∨ 0 = ax ∨ ann(x)x = (a ∨ ann(x))x. Hence,we have(a ∨ ann(x))x � ax, which is equivalent toa ∨ ann(x) � (ax)/x, i.e., the closureoperatora �→ (ax)/x on Q is greater or equal toa �→ a ∨ ann(x), and thusϕ is just therestriction of(−)x to ↑ann(x). It is dense becauseax = 0 is equivalent toa � ann(x), andM is isomorphic toM/x becauseϕ is surjective. �Corollary 5.9. x is a generator ofM if and only if(↑ann(x))x = M.

In [7] a notion ofpoint of a quantale is based on having “enough generators,” whein other papers it is related only to a kind of irreducibility [2,9,11]. Since these nohave some importance, we devote the rest of this section to some simple results rthe two, although they are not needed in the rest of the paper.

Definition 5.10. Let Q be a quantale, andM a leftQ-module.M is said to beirreducibleif it has no invariant elements besides 0 and 1, andeverywhere principalif for all non-zerom ∈ M there is a generatorx � m.

Theorem 5.11. LetQ be a quantale, andM a leftQ-module. If either of the following twconditions holds, thenM is irreducible:

(1) M is everywhere principal.(2) M has a generatorx such thatann(x) is a maximal left-sided element ofQ.

Proof. (1) For this it suffices to see that ifx ∈ M is any generator then 1M is the onlyinvariant abovex. So assume thatm is an invariant such thatx � m, wherex is a generatorThen 1M = ∨

M = ∨Qx = 1Qx � 1Qm � m.

(2) Let m ∈ M be invariant. Thenm/x is left-sided, by 5.6, and thus eitherm/x =ann(x) or m/x = 1Q. But m = (m/x)x becausex is a generator, and thusm = ann(x)x =0M or m = 1Qx = ∨

Qx = ∨M = 1M , i.e.,M is irreducible. �

Remark 5.12. In [7] the pointsof a(n involutive) quantaleQ are certain rightQ-moduleswhich are atomic as sup-lattices and whose atoms are generators, being thus everywhprincipal (see also Section 7). In certainplaces in [7] the hypothesis that the modsatisfies an additional condition known asnon-triviality is assumed. Although formulatedifferently, this is equivalent to the requirement that ann(x) be a maximal right-sideelement for some atomx, which itself implies irreducibility, by the previous theorem.


fs

ed in

fromIn

he

sent

ct

6. Quantale modules and 2-forms

Let Q be a quantale,ϕ :L × R → 2 a 2-form, andh :Q → Q(ϕ) a homomorphism oquantales. For eacha ∈ Q, the continuous endomaph(a) is a contravariant pair of mapthat defines a right action ofQ onL and a left action ofQ onR:

h(a) = ((−)a, a(−)

).

Examples of such homomorphisms are the comparison homomorphisms definSection 4, for which we haveϕ = Φ(Q). By definition of continuity the actions ofQon these modules satisfy, for allx ∈ L, y ∈ R, anda ∈ Q, the following “middle-linearity”condition:

〈xa, y〉 = 〈x, ay〉.

(In other words, the 2-form can be identified with a sup-lattice homomorphismL ⊗Q R, rather than justL ⊗ R, to 2—for tensor products of sup-lattices, see [1].)this section we shall study such pairs of modules:

Definition 6.1. Let Q be a quantale, andϕ :L × R → 2 a 2-form. Anactionof Q on ϕ

consists of a right action ofQ on L and a left action ofQ on R,1 such that for allx ∈ L,y ∈ R, anda ∈ Q we have〈xa, y〉 = 〈x, ay〉. When the latter condition holds we say t2-form isbalanced(with respect to theQ-modulesL andR), or that it is a 2-form overQ.If Q is unital, the 2-form isunital if both L andR are unital modules.

Proposition 6.2. Let Q be a quantale,L a right Q-module,R a left Q-module, andϕ :L × R → 2 a sup-lattice2-form. The following are equivalent:

(1) a\(x⊥) = (xa)⊥ for all x ∈ L anda ∈ Q,(2) (⊥y)/a = ⊥(ay) for all y ∈ R anda ∈ Q,(3) xa ⊥ a\(x⊥) andx ⊥ a((xa)⊥) for all x ∈ L anda ∈ Q,(4) (⊥y)/a ⊥ ay and(⊥(ay))a ⊥ y for all y ∈ R anda ∈ Q,(5) ϕ is a 2-form overQ.

Proof. This is an immediate consequence of 4.4, for being a 2-form overQ is the sameas the map((−)a, a(−)) being continuous, and(−)/a anda\(−) are the right adjoints to(−)a anda(−), respectively. �

Now we introduce the notion of orthomorphism that is appropriate in the precontext.

1 Our notation was suggested by the fact thatL is the “left part” of the 2-form, andR is the “right part.” Whileunfortunately this has led toL being aright module andR a left module, the notation is consistent with the fathat oftenL is L(Q) andR is R(Q) for some quantaleQ.


, is

ft

Definition 6.3. Let ϕ andϕ′ be 2-forms over a quantaleQ. An orthomorphism overQ, orsimply aQ-orthomorphism, is an orthomorphism(f, g) :ϕ → ϕ′ such thatf is a homo-morphism of rightQ-modules andg is a homomorphism of leftQ-modules.

Balance is preserved by surjections, as follows:

Lemma 6.4. Let Q be a quantale, andϕ :L × R → 2 a 2-form over Q. Let alsoϕ′ :L′×R′ → 2 be any2-form, such thatL′ is a rightQ-module, andR′ is a leftQ-module.Let (f, g) :ϕ → ϕ′ be an orthomorphism such that bothf andg are surjectiveQ-modulehomomorphisms(respectively right and left). Thenϕ′ is balanced.

Proof. Let x ′ ∈ L′, y ′ ∈ R′, and a ∈ Q. Due to surjectivity, there isx ∈ L such thatx ′ = f (x), andy ∈ R such thaty ′ = g(y). Hence,

ϕ′(x ′a, y ′) = ϕ′(f (x)a, g(y)) = ϕ′(f (xa), g(y)

) = ϕ(xa, y)

= ϕ(x, ay) = · · · = ϕ′(x ′, ay ′). �Lemma 6.5. Let Q be a quantale, andϕ :L × R → 2 a 2-form overQ. Then the closureoperators onL andR defined byx �→ ⊥(x⊥) andy �→ (⊥y)⊥ are nuclei ofQ-modules.

Proof. We prove this only for⊥((−)⊥), as the other case is similar. Letx ∈ L anda ∈ Q.The conditionxa ⊥ (xa)⊥ is always true and equivalent tox ⊥ a((xa)⊥), which in turnis equivalent toa((xa)⊥) � x⊥ = (⊥(x⊥))⊥, i.e., to ⊥(x⊥) ⊥ a((xa)⊥). Finally, thisis equivalent to(⊥(x⊥))a ⊥ (xa)⊥, i.e., (⊥(x⊥))a � ⊥((xa)⊥), which is precisely thestatement that(⊥(−))⊥ is a nucleus of rightQ-modules. �Theorem 6.6. Let Q be a quantale,ϕ :L × R → 2 a 2-form overQ, andf :L → L andg :R → R the closure operators defined byx �→ ⊥(x⊥) andy �→ (⊥y)⊥. LetL′ = ⊥R ={ ⊥(x⊥) | x ∈ L} and R′ = L⊥ = {(⊥y)⊥ | y ∈ R} be the corresponding quotients ofL

andR, and let the2-form ϕ′ :L′ × R′ → 2 be the restriction ofϕ to L′ × R′. Thenϕ′ is a2-form overQ.

(In other words, ifϕ is a 2-form overQ then its simple quotient, as defined in 3.12also a 2-form overQ.)

Proof. Corollary of the previous lemmas and 3.12.�Lemma 6.7. LetQ be a quantale, andn ∈ Q. Define a mapϕn :Q × Q → 2 by

ϕn(x, y) = 0 ⇔ x · y � n.

Thenϕn is a 2-form, and it is balanced with respect toQ, seen both as a right and a lemodule over itself.


tedn

s

--

nital

Proof. We haveϕn(∨

i xi, y) = 0 if and only if∨

i xi · y � n, which holds if and only ifxi · y � n for all i, i.e.,ϕn(xi, y) = 0 for all i. A similar fact holds for joins on the righvariable, and thusϕn is a 2-form onQ×Q. Furthermore, this 2-form is obviously balancwith respect to the actions ofQ on itself, due to the associativity of the multiplicatioin Q. �Definition 6.8. Let Q be a quantale. A 2-formϕ :L× R → 2 overQ is principal if both L

andR are principalQ-modules.

Hence, ifQ is unital andn ∈ Q thenϕn is principal, and so is any of its quotients.

Definition 6.9. Let Q be a quantale, andϕ :L × R → 2 a principal 2-form with generatorx ∈ L andy ∈ R. Theorthogonalizerof x andy is defined to be

orth(x, y) =∨

{a ∈ Q | x ⊥ ay}.

Notice that the following equivalences hold,

a � orth(x, y) ⇔ x ⊥ ay ⇔ ay � x⊥ ⇔ a �(x⊥)

/y,

and thus orth(x, y) = (x⊥)/y. Also, we have

x ⊥ ay ⇔ xa ⊥ y ⇔ xa � ⊥y,

whence orth(x, y) = x\(⊥y).

Theorem 6.10. Let Q be a quantale,ϕ :L × R → 2 a principal 2-form overQ, andn = orth(x, y) for a pair of generatorsx ∈ L andy ∈ R. Then there is aQ-orthoquotient(f, g) :ϕn → ϕ.

Proof. The mapf :Q → L defined bya �→ xa is a surjective rightQ-module homomorphism, and the mapg :Q → R defined byb �→ by is a surjective leftQ-module homomorphism. Finally, for alla, b ∈ Q we have

ϕn(a, b) = 0 ⇔ a · b � n ⇔ x ⊥ aby ⇔ xa ⊥ by ⇔ f (a) ⊥ g(b),

which shows that(f, g) is aQ-orthomorphism. �In particular, this gives us a classification of the principal 2-forms over a u

quantaleQ:

Corollary 6.11. Let Q be a unital quantale. Thenϕ is a principal2-form overQ if andonly if it is aQ-orthoquotient ofϕn for somen ∈ Q.

If furthermore the 2-forms are required to be faithful we obtain:


les of

-

s

tt

tse

r.

Corollary 6.12. Let Q be a unital quantale. Thenϕ is a faithful principal2-form overQif and only if it is isomorphic to the orthogonal quotient(necessarily aQ-orthoquotient,by6.6)of ϕn for somen ∈ Q.

We conclude this section relating these 2-forms with the upsegment moduSection 5.

Lemma 6.13. Let Q be a quantale,n ∈ Q, r ∈ R(Q), and l ∈ L(Q), such thatr ∨ l � n,and letψ :↑r × ↑l → 2 be the restriction ofϕn to ↑r × ↑l. Then:

(1) with the quotient module structures of↑r and↑l, ψ is a2-form overQ which further-more is a quotient ofϕn, and the orthogonal quotient ofϕn factors through it;

(2) if ψ is dense on the right(respectively left) thenl (respectivelyr) is the greatest leftsided(respectively right-sided) element belown;

(3) if Q is unital then ψ is dense on the right(respectively left) if and only if l

(respectivelyr) is the greatest left-sided(respectively right-sided) element belown.

Proof. (1) First we remark that the joins in↑r are precisely the same as inQ, except forthe join of the empty set, which in↑r is r. Similarly, the joins in↑l are those ofQ butwith the empty join beingl. Hence, forψ to be a 2-form it suffices to verify that it satisfieψ(r, y) = ψ(x, l) = 0 for all y ∈ ↑l andx ∈ ↑r. But we haveψ(r, y) = ϕn(r, y) = 0 if andonly if r ·y � n, which is true becauser is right-sided:r ·y � r � n. Similarly,ψ(x, l) = 0becausel is left-sided andl � n, and we conclude thatψ is a 2-form. Since it is a quotienof ϕn, which is a 2-form overQ, ψ is also a 2-form overQ. Finally, the orthogonal quotienof ϕn is the least quotient ofϕn and thus factors throughψ .

(2) Now assume thatψ is dense on the right, and leta � n be a left-sided elemenof Q. Thena ∨ l � n, and 1· (a ∨ l) � n, i.e.,ψ(1, a ∨ l) = 0 (this makes sense becaua ∨ l ∈ ↑l). Hence, sinceψ is dense it follows thata ∨ l = l, i.e.,a � l, which shows thatlis the greatest left-sided element belown. The situation with density on the left is simila

(3) Assume thatQ is unital and thatl is the greatest left-sided element belown. Letx ∈ ↑l such that 1⊥ x, i.e., such that 1· x � n. Then 1· x � l because 1· x is left-sided,and thusx � l, i.e., x = l, becauseQ is unital and thusx � 1 · x. This shows thatψ isdense on the right. Density on the left is handled similarly.�Theorem 6.14. Let Q be a quantale, andϕ :L × R → 2 a principal 2-form overQ withgeneratorsx ∈ L andy ∈ R. Then:

(1) if ϕ is dense on the right(respectively left) thenann(y) (respectivelyann(x)) is thegreatest left-sided(respectively right-sided) element beloworth(x, y);

(2) if Q is unital thenϕ is dense on the right(respectively left) if and only if ann(y)

(respectivelyann(x)) is the greatest left-sided(respectively right-sided) element beloworth(x, y).

Proof. Let n = orth(x, y). From 6.10 it follows thatϕ is a Q-orthoquotient ofϕn,and 5.8 implies that this quotient factors through(f, g) :ψ → ϕ, where the 2-form


e.

ds

ularave

.8 that

aithfultation

ψ :↑ann(x) × ↑ann(y) → 2 is as in 6.13, and bothf and g are surjective and densHence, by 3.9 we conclude thatϕ is dense on the right if and only ifψ is, and similarly onthe left. The result now follows from 6.13.�

7. Involutive modules

If Q is an involutive quantale and we have a 2-formϕ :L × R → 2 where bothL andR are left Q-modules, it still makes sense to define when it is thatϕ is balanced, forthe involution makesL a right module:xa = a∗x. Hence, being balanced corresponto the condition〈a∗x, y〉 = 〈x, ay〉 for all x ∈ L, y ∈ R, and a ∈ Q, or, equivalently,〈ax, y〉 = 〈x, a∗y〉. We will not pursue this in general, but rather study the particsituation whereL = R and the 2-form is symmetric. Since in this situation we h(−)⊥ = ⊥(−), we shall write(−)⊥ for both.

Definition 7.1. Let Q be an involutive quantale,M a left Q-module, andϕ a symmetric2-form onM. The pair(M,ϕ) (or justM, when no confusion may arise) is aninvolutive(left) Q-moduleif for all a ∈ Q andx, y ∈ M we have

⟨a∗x, y

⟩ = 〈x, ay〉.

A homomorphismof involutive left Q-modules is a homomorphism of leftQ-modulesfsuch that(f,f ) is an orthomorphism:〈f (x), f (y)〉 = 〈x, y〉. An involutive right moduleis defined analogously by the condition

〈xa, y〉 = ⟨x, ya∗⟩.

The fact that we have restricted to symmetric 2-forms allows us to use the fact 4Q(ϕ) is an involutive quantale:

Proposition 7.2. LetQ be an involutive quantale, andϕ :M ×M → 2 a symmetric2-form.There is a bijection between involutive leftQ-module structures on(M,ϕ) and involutionpreserving homomorphisms fromQ to Q(ϕ).

Proof. A quantale homomorphismh :Q → Q(ϕ) is the same as an action ofQ on ϕ,with h(a) = ((−)a, a(−)). Hence,h preserves involution if and only if((−)a∗, a∗(−)) =((−)a, a(−))∗ = (a(−), (−)a), i.e., if and only if(−)a = a∗(−) for all a ∈ Q, i.e., if andonly if (M,ϕ) is an involutive leftQ-module. �

From here and 4.9 we see that in the case when the 2-forms involved are fthe notion of involutive module corresponds precisely to that of involutive represenQ →Q(S) of [7,11]. In other words we have, as a corollary of 6.2:

Proposition 7.3. Let Q be an involutive quantale,M a left Q-module, andϕ a symmetric2-form onM. Then(M,ϕ) is an involutive leftQ-module if and only if(a∗x)⊥ = a\(x⊥)


o

lutive

in the

by

for all a ∈ Q and x ∈ M. In the case whenϕ is faithful this condition is equivalent t

a∗x = (a\(x⊥))⊥

.

All the previous definitions and results can be specialized to the case of invomodules. We highlight just a few facts:

Proposition 7.4. Let Q be an involutive quantale,m a left-sided element, andn a self-adjoint element such thatm � n. Then the leftQ-module↑m is involutive, with thesymmetric2-form being defined bya ⊥ b ⇔ a∗ · b � n.

Proof. Immediate consequence of 6.13, becausem∗ is right-sided andm∗ � n∗ = n. �Proposition 7.5. Let Q be an involutive quantale,M an involutive leftQ-module, andx ∈ M. Thenorth(x, x) is self-adjoint.

Proof. For alla ∈ Q we have:

a � orth(x, x) ⇔ x ⊥ ax ⇔ a∗x ⊥ x ⇔ x ⊥ a∗x

⇔ a∗ � orth(x, x). �Proposition 7.6. LetQ be an involutive quantale, andM an involutive leftQ-module witha generatorx ∈ M. Then↑ann(x) is an involutive leftQ-module with the symmetric2-formdefined bya ⊥x b ⇔ a∗ · b � orth(x, x), and the map↑ann(x) → M defined bya �→ ax

is a surjective homomorphism of involutive leftQ-modules.

Proof. Let us just see that the mapa �→ ax is an orthomorphism:

a ⊥x b ⇔ a∗ · b � orth(x, x) ⇔ x ⊥ a∗bx ⇔ ax ⊥ bx. �Corollary 7.7. LetQ, M, andx be as in the previous proposition. If the symmetric2-formof ↑ann(x) is faithful then↑ann(x) andM are isomorphic as involutive leftQ-modules.

We conclude by pointing out a few immediate consequences of these resultscase of quantales associated to C*-algebras. Recall from Section 2 that ifA is a unitalC*-algebra then the set of all the closed linear subspaces ofA is a unital involutive quantaleMaxA. We then have:

Lemma 7.8. LetA be a unital C*-algebra,m a maximal left-sided element ofMaxA, andn = m ∨ m∗. Then the symmetric2-form on↑m determined byn is faithful.

Proof. From C*-algebra theory we know that there is a unique pure stateϕ :A → C whosekernel isn, and that the quotientH = A/m is a Hilbert space with inner product defined〈a +m,b+m〉 = ϕ(b∗a). Hence, two vectorsa +m,b+m ∈ H are orthogonal if and onlyif b∗a ∈ n, i.e.,a∗b ∈ n, and thus the isomorphismf :↑m → P(H) is an orthomorphism


t

is, as[7],

]. Thehave

Maxincipaln

ratornoof ofthat

c.,

because, for allc, d ∈ ↑m, f (c) is orthogonal tof (d) in the latticeP(H) of closed linearsubspaces ofH if and only if c∗ · d � n. Therefore the 2-form on↑m is faithful becausethe 2-form onP(H) is. �Theorem 7.9. Let A be a unital C*-algebra, andM an involutive leftMaxA-modulewith a generatorx. Assume also thatann(x) is a maximal left-sided element. ThenM

is isomorphic as an involutive leftMaxA-module to↑ann(x).

Proof. Let m = ann(x). The topological leftA-module structure ofA/m makesP(A/m)

a left MaxA-module (cf. Section 2). Furthermore,A/m is involutive in the sense tha〈ax, y〉 = 〈x, a∗y〉, and from here it follows easily thatP(A/m) is involutive as a MaxA-module. Hence, we have a surjective homomorphism↑m → M of involutive left MaxA-modules, which must be an isomorphism because the 2-form on↑m is faithful. �

Following the terminology of [7], let us define aHilbert representationof MaxA to bean involutive left MaxA-module isomorphic to one of the formP(H) determined by arepresentation ofA onH , in the manner of Section 2. Then we obtain:

Corollary 7.10. Let A be a unital C*-algebra, andM an involutive leftMaxA-modulewith a generatorx. Assume also thatann(x) is a maximal left-sided element. ThenM isa Hilbert representation determined by an irreducible representation ofA.

Proof. This follows from the previous results and the fact that for a maximal idealm thequotientA/m defines an irreducible representation.�

The existence of a generator whose annihilator is a maximal left-sided elementwas already mentioned in 5.12, equivalent to the property known as non-triviality inand the above corollary corresponds to one of the implications in [7, Theorem 9.1main difference between the proof in [7] and what we have done above is that weused 2-forms. Also, we have focused less on the properties of those elements ofA

known as “pure states” and instead more on the annihilators of the generators of prMaxA-modules, e.g., formulating non-triviality directly in terms of the annihilators. I[7, Theorem 9.1] it is further assumed that the moduleM is analgebraically irreduciblerepresentation, i.e., thatM is atomic as a sup-lattice and that each atom is a gene(equivalently,M is atomic and everywhere principal).Therefore our present formulatiois more general, even though it is not so in an essential way because in the pr[7, Theorem 9.1] the extra conditions are not used. In [7] it is further conjecturedevery algebraically irreducible representation of MaxA is non-trivial.

References

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[6] C.J. Mulvey, J.W. Pelletier, A quantisation of the calculus of relations, Canad. Math. Soc. Conf. Proc.(1992) 345–360.

[7] C.J. Mulvey, J.W. Pelletier, On the quantisation of points, J. Pure Appl. Algebra 159 (2001) 231–295.[8] C.J. Mulvey, J.W. Pelletier, On the quantisation of spaces, J. Pure Appl. Algebra 175 (2002) 289–32[9] J. Paseka, Simple quantales, in: Proc. 8th Prague Topological Symposium 1996, in: Topology Atlas, 199

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[11] J.W. Pelletier, J. Rosický, Simple involutive quantales, J. Algebra 195 (1997) 367–386.[12] P. Resende, Tropological systems are points of quantales, J. Pure Appl. Algebra 173 (2002) 87–120.[13] K. Rosenthal, Quantales and Their Applications, in: Pitman Res.Notes Math. Ser., vol. 234, Longma

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Sup-lattice 2-forms and quantales

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