# Cyclic dualizing elements in Girard quantales.

[section]1. Preliminaries

Quantales were introduced by C.J. Mulvey in [1] with the purpose of studying the spectrum of C*-algebras and the foundations of quantum mechanics. The study of such partially ordered algebraic structures goes back to a series of papers by Ward and Dilworth [2, 3] in the 1930s. It has become a useful tool in studying noncommutative topology, linear logic and C*-algebra theory [4-6]. Following Mulvey, various types and aspects of quantales have been considered by many researchers [7-9]. The importance of quantales for linear logic is revealed in Yetter's work [10]. Yetter has clarified the use of quantales in linear logic and he has introduced the term "Girard quantale" . In [11], J. Paseka and D. Kruml have shown that any quantale can be embedded into a unital quantale. In [12], K.I. Rosenthal has proved that every quantale can be embedded into a Girard quantale. Thus, it is important to study Girard quantale. This is the motivation for us to investigate Girard quantale. In the note, we shall study the interior structures of Girard quantale and the cyclic dualizing element in Girard quantales.

We use 1 to denote the top element and 0 the bottom element in a complete lattice. For notions and concepts, but not explained, please to refer to [12].

Definition 1.1. A quantale is a complete lattice Q with an associative binary operation "&" satisfying:

a&([??][b.sub.[alpha]]) = [??] (a&[b.sub.[alpha]]) and ([??][b.sub.[alpha]])&a = [??]([b.sub.[alpha]]&a)

for all a [member of] Q,{[b.sub.[alpha]} [??] Q.

An element e [member of] Q is called a unit if a&e = e&a = a for all a [member of] Q. Q is called unital if Q has the unit e.

Since a&_ and _&a preserve arbitrary sups for all a [member of] Q, they have right adjoints and we shall denote them by a [[right arrow].sub.r] and a [[right arrow].sub.l] - respectively.

Proposition 1.2. Let Q be a quantale, a, b, c [member of] Q. Then

(1) a&(a [[right arrow].sub.r] b) [less than or equal to] b;

(2) a [[right arrow].sub.r] (b [[right arrow].sub.r] c) = b&a [[right arrow].sub.r] c;

Again, analogous results hold upon replacing [[right arrow].sub.r] by [[right arrow].sub.l].

Definition 1.3. Let Q be a quantale, An element c of Q is called cyclic, if a [[right arrow].sub.r] c = a [[right arrow].sub.l] c for all a [member of] Q. d [member of] Q is called a dualizing element, if a = (a [[right arrow].sub.l] d) [[right arrow].sub.r] d = (a [[right arrow].sub.r] d) [[right arrow].sub.l] d for all a [member of] Q.

Definition 1.4. A quantale Q is called a Girard quantale if it has a cyclic dualizing element d.

Let Q be a Girard quantale with cyclic dualizing element d and a, b [member of] Q, define the binary operation "[parallel]" by a[parallel]b = ([a.sup.[perpendicular to]]&[b.sup.[perpendicular to]])1, then we can prove that a[parallel]_ and _[parallel]a preserve arbitrary infs for all a [member of] Q, hence they have left adjoints and we shall denote them by a [[??].sub.r] _ and a [[??].sub.r] _ respectively. If a [[right arrow].sub.r] d = a [[right arrow].sub.l]d, we shall denote it by a [right arrow] d, or more frequently by [a.sup.[perpendicular]] if d is a cyclic dualizing element.

[section]2. The equivalent descriptions for Girard quantale

In this section, we shall study the interior structures of Girard quantale and give some equivalent descriptions for Girard quantale. According to the above, we know that there are six binary operations on a Girard quantale such as _&_, _ [[right arrow].sub.r] _, _ [[right arrow].sub.l] _, _[parallel]_, _ [[??].sub.r]_, _ [[??].sub.l] _, we shall respectively call them multiplying, right implication, left implication, Par operation, dual right implication and dual left implication for convenience.

Theorem 2.1. Let Q be a unital quantale, [perpendicular] : Q [right arrow] Q an unary operation on Q. Then Q is a Girard quantale if and only if

(1) a [[right arrow].sub.l] b = [(a&[b.sup.[perpendicular]]).sup.[perpendicular]]; (2) a [[right arrow].sub.r] b = [(b&[a.sup.[perpendicular]]).sup.[perpendicular]].

Proof. The necessity is obvious. Sufficiency: suppose (1) and (2) hold, a [member of] Q. Denote the unit element by e on Q, then a = e [[right arrow].sub.l] a = [(e&[a.sup.[perpendicular]]).sup.[perpendicular]] = [([a.sup.[perpendicular]]).sup.[perpendicular]]. Thus a = [a.sup.[perpendicular][perpendicular]], hence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Similarly we get [a.sup.[perpendicular]] = a [[right arrow].sub.r] [e.sup.[perpendicular]]. Take d = [e.sup.[perpendicular]], thus d is a cyclic element of Q. Again [for all]a [member of] Q, (a [right arrow] [e.sup.[perpendicular]]) [right arrow] [e.sup.[perpendicular]] = [a.sup.[perpendicular]] [right arrow] [e.sup.[perpendicular]] = [([a.sup.[perpendicular]]).sup.[perpendicular]] = a. This proves d = [e.sup.[perpendicular]] is a dualizing element on Q. Thus the proof is completed.

Theorem 2.2. Let Q be a complete lattice. _ [[right arrow].sub.r] _ : Q x Q [right arrow] Q is a binary operation on Q, a [[right arrow].sub.r] _ : Q [arrow] Q and _ [[right arrow].sub.r] a : Q [right arrow] [Q.sup.op] preserve arbitrary sups for all a [member of] Q. [perpendicular] : Q [right arrow] Q is a unary operation on Q, e [member of] Q. For all a, b, c [member of] Q,

(1) e [[right arrow].sub.r] a = a; a [[right arrow].sub.r] [e.sup.[perpendicular]] = [a.sup.[perpendicular]];

(2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then Q is a Girard quantale and _ [[right arrow].sub.r] _ is the right implication operation.

Proof. Define the binary operation a&b = [(b [[right arrow].sub.r] [a.sup.[perpendicular]]).sup.[perpendicular]] for all a, b [member of] Q. & satisfies associative law: In fact, for all a, b, c [member of] Q,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Using the condition (3), we have (c [[right arrow].sub.r] [(b [[right arrow].sub.r], [a.sup.[perpendicular]])).sup.[perpendicular]] = [((c [[right arrow].sub.r] [b.sup.[perpendicular]]).sup.[perpendicular]] [[right arrow].sub.r] [a.sup.[perpendicular]].sup.[perpendicular]. By the definition of the binary operation & we can get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So (a&b)&c = a&(b&c).

Using the condition (4), we have a&b [less than or equal to] c [??] [(b [[right arrow].sub.r], [a.sup.[perpendicular]]).sup.[perpendicular]] [less than or equal to] c [??] [c.sup.[perpendicular]] [less than or equal to] b [[right arrow].sub.r] [a.sup.[perpendicular]] [??] b [less than or equal to] a [[right arrow].sub.r] c for all a, b, c [member of] Q.

For any a [member of] Q, [{[b.sub.i]}.sub.i[member of]I] [??] Q. If I = 0, then a&0 = [(a [[right arrow].sub.r] [0.sup.[perpendicular]]).sup.[perpendicular]] = [(a [[right arrow].sub.r] 1).sup.[perpendicular]], since again [(a [[right arrow].sub.r] 1).sup.[perpendicular]] [less than or equal to] 0 [??] 1 [less than or equal to] a [[right arrow].sub.r] 1 [??] a&1 [less than or equal to] 1, the last inequality obviously holds. So a&0 = 0. Thus a&_ preserves empty-sups. If I = 0, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [([??].sub.i[member]I] ([b.sub.i] [[right arrow.sub.r] [a.sup.[perpendicular]])).sup.[perpendicular]]

= [[??].sub.i[member]I] ([b.sub.i] [[right arrow.sub.r] [a.sup.[perpendicular]]).sup.[perpendicular]]

= [[??].sub.i[member]I] (a&[b.sub.i]).

Hence a&_ preserves arbitrary sups for all a [member of] Q. Similarly, we can prove &a preserves arbitrary sups for all a [member of] Q. Thus (Q, &) is a quantale.

In accordance with the condition (1), we know e is the unit element corresponding to & on Q and a [[right arrow].sub.r] [e.sup.[perpendicular]] = [a.sup.[perpendicular]]. Denote by a [[right arrow].sub.l] _ the right adjoint of _&a. Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [??] {x [member of] Q|x&a [less than or equal] [e.sup.[perpendicular]]}

= [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [??] {x [member of] Q|e [less than or equal to] a [[right arrow].sub.r] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|a&e [less than or equal to] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|a [less than or equal to] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|x [less than or equal to] [a.sup.[perpendicular]]}

= [a.sup.[perpendicular]].

This show [e.sup.[perpendicular]] is a cyclic element in Q. Using conditions (1) and (2) we know [e.sup.[perpendicular]] is also a dualizing element on Q. Hence (Q, &, [perpendicular]) is a Girard quantale. We can easily prove _ [[right arrow].sub.r] _ is the right implication operation on Q by the above consideration.

Theorem 2.3. Let Q be a complete lattice. _ [[??].sub.r], _ : Q x Q [right arrow] Q is a binary operation in Q, a [[??].sub.r] _ : Q [right arrow] Q and _ [[??].sub.r] a : [Q.sup.op] [right arrow] Q preserve arbitrary sups for all a [member of] Q. 1 : Q [right arrow] Q is an unary operation in Q, d [member of] Q. For all a, b, c [member of] Q,

(1) d [[??].sub.r] a = a; a [[??].sub.r] [d.sup.[perpendicular]] = [a.sup.[perpendicular]];

(2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(3) [(a [[??].sub.r] b).sup.[perpendicular]] [[??].sub.r] c = a [[??].sub.r] ([b.sup.[perpendicular]] [[??].sub.r] c);

(4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then Q is a Girard quantale and _ [[??].sub.r] _ is the dual right implication operation.

Proof. Define binary operation a&b = ([b.sup.[perpendicular]] [[??].sub.r] a) for all a, b [member of] Q, (i) The binary operation & is associative: Since [for all]a, b, c [member of] Q,

(a&b)&c = ([b.sup.[perpendicular]] [[??].sub.r] a)&c

= [c.sup.[perpendicular]] [[??].sub.r] ([b.sup.[perpendicular]] [[??].sub.r] a)

= [[c.sup.[perpendicular]] [[??].sub.r] b).sup.[perpendicular]] [[??].sub.r] a

= a&([c.sup.[perpendicular]] [[??].sub.r] b)

= a&(b&c).

(ii) Using the condition (2), we can prove

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

for any set I and [{[a.sub.i]}.sub.i[member of]I] [??] Q.

(iii) For all a [member of] Q, [{[b.sub.i]}.sub.i[member of]I] [??] Q, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Similarly we have ([??].sub.i[member of]I] [b.sub.i])&a = ([??].sub.i[member of]I] ([b.sub.i]&a). Hence (Q, &)is a&[d.sup.[perpendicular]] = ([d.sup.[perpendicular]].sup.[perpendicular]] [[??].sub.r] a = d [[??].sub.r] a = a; [d.sup.[perpendicular]]&b = [b.sup.[perpendicular]] [[??].sub.r] [d.sup.[perpendicular]] = b, thus (Q,&) is a unit quantale with unit element [d.sup.[perpendicular]].

(iv) If a [member of] Q, we have

a [[right arrow].sub.l] d = [??]{x [member of] Q|x [less than or equal to] a [[right arrow].sub.l] d}

= [??]{x [member of] Q|x&a [less than or equal to] d}

= [??]{x [member of] Q|[a.sup.[perpendicular]] [[??].sub.r] x [less than or equal to] d}

= [??]{x [member of] Q|d [[??].sub.r] a [less than or equal to] [x.sup.[perpendicular]]}

= [??]{x [member of] Q|x [less than or equal to] [a.sup.[perpendicular]]}

= [a.sup.[perpendicular]].

Similarly, a [[right arrow].sub.r] d = [a.sup.[perpendicular]], hence d is a cyclic element in Q. d is also a dual element in Q by condition (2). Thus (Q, &) is a Girard quantale with cyclic dual element d. We easily know _ [[??].sub.r] _ is the dual right implication operation in Q by the definition of &.

Obviously, Theorem 2.2 and Theorem 2.3 also hold if [[right arrow].sub.r] and [[??].sub.r] are substituted by [[right arrow].sub.l] and [[??].sub.r] respectively, "right" and "left" replace each other.

Theorem 2.4. Let Q be a unital quantale with a unary operation [perpendicular] satisfying the condition

CN : [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

for all a, b [member of] Q. Then Q is a Girard quantale.

[section]3. The cyclic dualizing element of Girard quantale

According to the definition of Girard quantale, we know that the cyclic dualizing element plays an important role in Girard quantale, so we shall discuss the cyclic dualizing element in this section. We shall account for whether the cyclic dualizing element is unique in a Girard quantale; when it is unique; whether these Girard quintiles determined by different cyclic dualizing elements are different. Let us see the following example

Example 3.1. Let Q = {0, a, b, c, 1}, the partial order on Q be defined as Fig 1, the operator & on Q be defined by Table 1. Then we can prove that Q is a commutative Girard quantale. And we can prove that a, b and c are cyclic dualizing elements of Q.

[FIGURE 1 OMITTED]

Proposition 3.2. Let Q be a unital quantale with the unit element e. [sup.[perpendicular]1] and [sup.[perpendicular]2] satisfy the condition CN in Theorem 2.4. Then [e.sup.[perpendicular]1] = [e.sup.[perpendicular]2] if and only if [sup.[perpendicular]1] = [sup.[perpendicular]2].

Proposition 3.3. Let Q be a quantale, [d.sub.1], [d.sub.2] are cyclic dualizing elements of Q, [sup.[perpendicular]1] [sup.[perpendicular]2] are unary operations on Q induced by [d.sub.1], [d.sub.2] respectively. Then [d.sub.1] = [d.sub.2] if and only if [sup.[perpendicular]1] = [sup.[perpendicular]2].

Theorem 3.4. Let Q be a Girard quantale. Then there is a one-to-one correspondence between the set of cyclic dualizing elements in Q and the set of unary operations satisfying the condition CN in Theorem 2.4.

Proposition 3.5. Let Q be a Girard quantale. If 0 is a cyclic dualizing element of Q, then Q is strictly two-sided.

Proof. Assume 0 is a cyclic dualizing element in Q. Then [0.sup.[perpendicular]] = 0 [right arrow] 0 = 1 is the unit of Q, hence [for all]a [member of] Q, a&1 = 1&a = a, this finished the proof.

Proposition 3.6. If Q is a two-sided Girard quantale, then the unique cyclic dualizing element is the least element 0.

Proof. If Q is a two-sided Girard quantale, then we have a = a&e [less than or equal to] a&1 [less than or equal to] a for all a [member of] Q. Similarly, we have 1&a = a. Thus Q is strictly two-sided. Suppose d is a cyclic dual element in Q, [sup.[perpendicular]] is the unary operation induced by d, then we have d = 1 [right arrow] d = [1.sup.[perpendicular]] = 0. the proof is finished.

Corollary 3.7. Let Q be a Girard quantale with cyclic dualizing element 0. Then the cyclic dualizing element of Q is unique.

Theorem 3.8. Any complete lattice implication algebra is a Girard quantale with unique cyclic dualizing element 0.

According the above conclusions, we have a question : Whether the cyclic dualizing element must be the least element 0 if a Girard quantale has an unique cyclic dualizing element. Theanswer is negative. Let us see the following example.

Example 3.9. Let Q = {0, e, 1}, the partial order on Q be defined by 0 < e < 1, the binary operation & be defined by Table 2

It is immediate to verify Q being a Girard quantale with the unique cyclic dualizing element e.

Question 3.10. What is the necessary condition when the cyclic dualizing element of Guard quantale is unique?

References

[1] C. J. Mulvey, Suppl. Rend. Circ. Mat. Palermo Ser. II, 12(1986), 99-104.

[2] M. Ward, Structure residation, Ann. Math., 39(1938), 558-569.

[3] R. P. Dilworth, Noncommutative residuated lattices, Trans. Amer. Math. Soc., 46(1939), 426-444.

[4] B. Banaschewski and M. Erne, On Krull's separation lemma, Order, 10(1993), 253-260.

[5] J. Y. Guard, Linear logic, Theoret. Comp. Sci., 50(1987), 1-102.

[6] F. Borceux and G. Van den Bossche, Quantales and their sheaves, Order, 3(1986), 61-87.

[7] D. Kruml, Spatial quantales, Appl. Categ. Structures, 10(2002), 49-62.

[8] S. Q. Wang, B. Zhao, Prequantale congruence and its properties, Advances in Mathematics(In Chinese), 34(2005), 746-752.

[9] S. W. Han, B. Zhao, The quantale completion of ordered semigroup, Acta Mathematics Sinica(In Chinese), 51(2008), No.6, 1081-1088.

[10] D. Yetter, Quantales and (noncommutative) linear logic, J. Symbolic Logic, 55(1990), 41-64.

[11] J. Paseka, D. Kruml, Embeddings of quantales into simple quantales, J. Pure Appl. Algebra, 148(2000), 209-216.

[12] K. I. Rosenthal, Quantales and their applications, Longman Scientific and Technical, London, 1990.

Bin Zhao ([dagger]) and Shunqin Wang ([double dagger])

([dagger]) College of Mathematics and Information Science, Shaanxi Normal University, Shaanxi, Xi'an 710062, P.R. China

([double dagger]) School of Mathematics and Statistics, Nanyang Normal University, Henan, Nanyang 473061, P.R. China

zhaobin@snnu.edu.cn;math.wangsq@163.com

(1) This work was supported by the National Natural Science Foundation of China(Grant No.10871121) and the Research Award for Teachers in Nangyang Normal University, China (nynu200749)

Quantales were introduced by C.J. Mulvey in [1] with the purpose of studying the spectrum of C*-algebras and the foundations of quantum mechanics. The study of such partially ordered algebraic structures goes back to a series of papers by Ward and Dilworth [2, 3] in the 1930s. It has become a useful tool in studying noncommutative topology, linear logic and C*-algebra theory [4-6]. Following Mulvey, various types and aspects of quantales have been considered by many researchers [7-9]. The importance of quantales for linear logic is revealed in Yetter's work [10]. Yetter has clarified the use of quantales in linear logic and he has introduced the term "Girard quantale" . In [11], J. Paseka and D. Kruml have shown that any quantale can be embedded into a unital quantale. In [12], K.I. Rosenthal has proved that every quantale can be embedded into a Girard quantale. Thus, it is important to study Girard quantale. This is the motivation for us to investigate Girard quantale. In the note, we shall study the interior structures of Girard quantale and the cyclic dualizing element in Girard quantales.

We use 1 to denote the top element and 0 the bottom element in a complete lattice. For notions and concepts, but not explained, please to refer to [12].

Definition 1.1. A quantale is a complete lattice Q with an associative binary operation "&" satisfying:

a&([??][b.sub.[alpha]]) = [??] (a&[b.sub.[alpha]]) and ([??][b.sub.[alpha]])&a = [??]([b.sub.[alpha]]&a)

for all a [member of] Q,{[b.sub.[alpha]} [??] Q.

An element e [member of] Q is called a unit if a&e = e&a = a for all a [member of] Q. Q is called unital if Q has the unit e.

Since a&_ and _&a preserve arbitrary sups for all a [member of] Q, they have right adjoints and we shall denote them by a [[right arrow].sub.r] and a [[right arrow].sub.l] - respectively.

Proposition 1.2. Let Q be a quantale, a, b, c [member of] Q. Then

(1) a&(a [[right arrow].sub.r] b) [less than or equal to] b;

(2) a [[right arrow].sub.r] (b [[right arrow].sub.r] c) = b&a [[right arrow].sub.r] c;

Again, analogous results hold upon replacing [[right arrow].sub.r] by [[right arrow].sub.l].

Definition 1.3. Let Q be a quantale, An element c of Q is called cyclic, if a [[right arrow].sub.r] c = a [[right arrow].sub.l] c for all a [member of] Q. d [member of] Q is called a dualizing element, if a = (a [[right arrow].sub.l] d) [[right arrow].sub.r] d = (a [[right arrow].sub.r] d) [[right arrow].sub.l] d for all a [member of] Q.

Definition 1.4. A quantale Q is called a Girard quantale if it has a cyclic dualizing element d.

Let Q be a Girard quantale with cyclic dualizing element d and a, b [member of] Q, define the binary operation "[parallel]" by a[parallel]b = ([a.sup.[perpendicular to]]&[b.sup.[perpendicular to]])1, then we can prove that a[parallel]_ and _[parallel]a preserve arbitrary infs for all a [member of] Q, hence they have left adjoints and we shall denote them by a [[??].sub.r] _ and a [[??].sub.r] _ respectively. If a [[right arrow].sub.r] d = a [[right arrow].sub.l]d, we shall denote it by a [right arrow] d, or more frequently by [a.sup.[perpendicular]] if d is a cyclic dualizing element.

[section]2. The equivalent descriptions for Girard quantale

In this section, we shall study the interior structures of Girard quantale and give some equivalent descriptions for Girard quantale. According to the above, we know that there are six binary operations on a Girard quantale such as _&_, _ [[right arrow].sub.r] _, _ [[right arrow].sub.l] _, _[parallel]_, _ [[??].sub.r]_, _ [[??].sub.l] _, we shall respectively call them multiplying, right implication, left implication, Par operation, dual right implication and dual left implication for convenience.

Theorem 2.1. Let Q be a unital quantale, [perpendicular] : Q [right arrow] Q an unary operation on Q. Then Q is a Girard quantale if and only if

(1) a [[right arrow].sub.l] b = [(a&[b.sup.[perpendicular]]).sup.[perpendicular]]; (2) a [[right arrow].sub.r] b = [(b&[a.sup.[perpendicular]]).sup.[perpendicular]].

Proof. The necessity is obvious. Sufficiency: suppose (1) and (2) hold, a [member of] Q. Denote the unit element by e on Q, then a = e [[right arrow].sub.l] a = [(e&[a.sup.[perpendicular]]).sup.[perpendicular]] = [([a.sup.[perpendicular]]).sup.[perpendicular]]. Thus a = [a.sup.[perpendicular][perpendicular]], hence

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Similarly we get [a.sup.[perpendicular]] = a [[right arrow].sub.r] [e.sup.[perpendicular]]. Take d = [e.sup.[perpendicular]], thus d is a cyclic element of Q. Again [for all]a [member of] Q, (a [right arrow] [e.sup.[perpendicular]]) [right arrow] [e.sup.[perpendicular]] = [a.sup.[perpendicular]] [right arrow] [e.sup.[perpendicular]] = [([a.sup.[perpendicular]]).sup.[perpendicular]] = a. This proves d = [e.sup.[perpendicular]] is a dualizing element on Q. Thus the proof is completed.

Theorem 2.2. Let Q be a complete lattice. _ [[right arrow].sub.r] _ : Q x Q [right arrow] Q is a binary operation on Q, a [[right arrow].sub.r] _ : Q [arrow] Q and _ [[right arrow].sub.r] a : Q [right arrow] [Q.sup.op] preserve arbitrary sups for all a [member of] Q. [perpendicular] : Q [right arrow] Q is a unary operation on Q, e [member of] Q. For all a, b, c [member of] Q,

(1) e [[right arrow].sub.r] a = a; a [[right arrow].sub.r] [e.sup.[perpendicular]] = [a.sup.[perpendicular]];

(2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(3) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then Q is a Girard quantale and _ [[right arrow].sub.r] _ is the right implication operation.

Proof. Define the binary operation a&b = [(b [[right arrow].sub.r] [a.sup.[perpendicular]]).sup.[perpendicular]] for all a, b [member of] Q. & satisfies associative law: In fact, for all a, b, c [member of] Q,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Using the condition (3), we have (c [[right arrow].sub.r] [(b [[right arrow].sub.r], [a.sup.[perpendicular]])).sup.[perpendicular]] = [((c [[right arrow].sub.r] [b.sup.[perpendicular]]).sup.[perpendicular]] [[right arrow].sub.r] [a.sup.[perpendicular]].sup.[perpendicular]. By the definition of the binary operation & we can get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

So (a&b)&c = a&(b&c).

Using the condition (4), we have a&b [less than or equal to] c [??] [(b [[right arrow].sub.r], [a.sup.[perpendicular]]).sup.[perpendicular]] [less than or equal to] c [??] [c.sup.[perpendicular]] [less than or equal to] b [[right arrow].sub.r] [a.sup.[perpendicular]] [??] b [less than or equal to] a [[right arrow].sub.r] c for all a, b, c [member of] Q.

For any a [member of] Q, [{[b.sub.i]}.sub.i[member of]I] [??] Q. If I = 0, then a&0 = [(a [[right arrow].sub.r] [0.sup.[perpendicular]]).sup.[perpendicular]] = [(a [[right arrow].sub.r] 1).sup.[perpendicular]], since again [(a [[right arrow].sub.r] 1).sup.[perpendicular]] [less than or equal to] 0 [??] 1 [less than or equal to] a [[right arrow].sub.r] 1 [??] a&1 [less than or equal to] 1, the last inequality obviously holds. So a&0 = 0. Thus a&_ preserves empty-sups. If I = 0, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [([??].sub.i[member]I] ([b.sub.i] [[right arrow.sub.r] [a.sup.[perpendicular]])).sup.[perpendicular]]

= [[??].sub.i[member]I] ([b.sub.i] [[right arrow.sub.r] [a.sup.[perpendicular]]).sup.[perpendicular]]

= [[??].sub.i[member]I] (a&[b.sub.i]).

Hence a&_ preserves arbitrary sups for all a [member of] Q. Similarly, we can prove &a preserves arbitrary sups for all a [member of] Q. Thus (Q, &) is a quantale.

In accordance with the condition (1), we know e is the unit element corresponding to & on Q and a [[right arrow].sub.r] [e.sup.[perpendicular]] = [a.sup.[perpendicular]]. Denote by a [[right arrow].sub.l] _ the right adjoint of _&a. Then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [??] {x [member of] Q|x&a [less than or equal] [e.sup.[perpendicular]]}

= [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

= [??] {x [member of] Q|e [less than or equal to] a [[right arrow].sub.r] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|a&e [less than or equal to] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|a [less than or equal to] [x.sup.[perpendicular]]}

= [??] {x [member of] Q|x [less than or equal to] [a.sup.[perpendicular]]}

= [a.sup.[perpendicular]].

This show [e.sup.[perpendicular]] is a cyclic element in Q. Using conditions (1) and (2) we know [e.sup.[perpendicular]] is also a dualizing element on Q. Hence (Q, &, [perpendicular]) is a Girard quantale. We can easily prove _ [[right arrow].sub.r] _ is the right implication operation on Q by the above consideration.

Theorem 2.3. Let Q be a complete lattice. _ [[??].sub.r], _ : Q x Q [right arrow] Q is a binary operation in Q, a [[??].sub.r] _ : Q [right arrow] Q and _ [[??].sub.r] a : [Q.sup.op] [right arrow] Q preserve arbitrary sups for all a [member of] Q. 1 : Q [right arrow] Q is an unary operation in Q, d [member of] Q. For all a, b, c [member of] Q,

(1) d [[??].sub.r] a = a; a [[??].sub.r] [d.sup.[perpendicular]] = [a.sup.[perpendicular]];

(2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

(3) [(a [[??].sub.r] b).sup.[perpendicular]] [[??].sub.r] c = a [[??].sub.r] ([b.sup.[perpendicular]] [[??].sub.r] c);

(4) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then Q is a Girard quantale and _ [[??].sub.r] _ is the dual right implication operation.

Proof. Define binary operation a&b = ([b.sup.[perpendicular]] [[??].sub.r] a) for all a, b [member of] Q, (i) The binary operation & is associative: Since [for all]a, b, c [member of] Q,

(a&b)&c = ([b.sup.[perpendicular]] [[??].sub.r] a)&c

= [c.sup.[perpendicular]] [[??].sub.r] ([b.sup.[perpendicular]] [[??].sub.r] a)

= [[c.sup.[perpendicular]] [[??].sub.r] b).sup.[perpendicular]] [[??].sub.r] a

= a&([c.sup.[perpendicular]] [[??].sub.r] b)

= a&(b&c).

(ii) Using the condition (2), we can prove

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

for any set I and [{[a.sub.i]}.sub.i[member of]I] [??] Q.

(iii) For all a [member of] Q, [{[b.sub.i]}.sub.i[member of]I] [??] Q, we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Similarly we have ([??].sub.i[member of]I] [b.sub.i])&a = ([??].sub.i[member of]I] ([b.sub.i]&a). Hence (Q, &)is a&[d.sup.[perpendicular]] = ([d.sup.[perpendicular]].sup.[perpendicular]] [[??].sub.r] a = d [[??].sub.r] a = a; [d.sup.[perpendicular]]&b = [b.sup.[perpendicular]] [[??].sub.r] [d.sup.[perpendicular]] = b, thus (Q,&) is a unit quantale with unit element [d.sup.[perpendicular]].

(iv) If a [member of] Q, we have

a [[right arrow].sub.l] d = [??]{x [member of] Q|x [less than or equal to] a [[right arrow].sub.l] d}

= [??]{x [member of] Q|x&a [less than or equal to] d}

= [??]{x [member of] Q|[a.sup.[perpendicular]] [[??].sub.r] x [less than or equal to] d}

= [??]{x [member of] Q|d [[??].sub.r] a [less than or equal to] [x.sup.[perpendicular]]}

= [??]{x [member of] Q|x [less than or equal to] [a.sup.[perpendicular]]}

= [a.sup.[perpendicular]].

Similarly, a [[right arrow].sub.r] d = [a.sup.[perpendicular]], hence d is a cyclic element in Q. d is also a dual element in Q by condition (2). Thus (Q, &) is a Girard quantale with cyclic dual element d. We easily know _ [[??].sub.r] _ is the dual right implication operation in Q by the definition of &.

Obviously, Theorem 2.2 and Theorem 2.3 also hold if [[right arrow].sub.r] and [[??].sub.r] are substituted by [[right arrow].sub.l] and [[??].sub.r] respectively, "right" and "left" replace each other.

Theorem 2.4. Let Q be a unital quantale with a unary operation [perpendicular] satisfying the condition

CN : [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

for all a, b [member of] Q. Then Q is a Girard quantale.

[section]3. The cyclic dualizing element of Girard quantale

According to the definition of Girard quantale, we know that the cyclic dualizing element plays an important role in Girard quantale, so we shall discuss the cyclic dualizing element in this section. We shall account for whether the cyclic dualizing element is unique in a Girard quantale; when it is unique; whether these Girard quintiles determined by different cyclic dualizing elements are different. Let us see the following example

Example 3.1. Let Q = {0, a, b, c, 1}, the partial order on Q be defined as Fig 1, the operator & on Q be defined by Table 1. Then we can prove that Q is a commutative Girard quantale. And we can prove that a, b and c are cyclic dualizing elements of Q.

[FIGURE 1 OMITTED]

Table 1. & 0 a b c 1 0 0 0 0 0 0 a 0 b c a 1 b 0 c a b 1 c 0 a b c 1 1 0 1 1 1 1

Proposition 3.2. Let Q be a unital quantale with the unit element e. [sup.[perpendicular]1] and [sup.[perpendicular]2] satisfy the condition CN in Theorem 2.4. Then [e.sup.[perpendicular]1] = [e.sup.[perpendicular]2] if and only if [sup.[perpendicular]1] = [sup.[perpendicular]2].

Proposition 3.3. Let Q be a quantale, [d.sub.1], [d.sub.2] are cyclic dualizing elements of Q, [sup.[perpendicular]1] [sup.[perpendicular]2] are unary operations on Q induced by [d.sub.1], [d.sub.2] respectively. Then [d.sub.1] = [d.sub.2] if and only if [sup.[perpendicular]1] = [sup.[perpendicular]2].

Theorem 3.4. Let Q be a Girard quantale. Then there is a one-to-one correspondence between the set of cyclic dualizing elements in Q and the set of unary operations satisfying the condition CN in Theorem 2.4.

Proposition 3.5. Let Q be a Girard quantale. If 0 is a cyclic dualizing element of Q, then Q is strictly two-sided.

Proof. Assume 0 is a cyclic dualizing element in Q. Then [0.sup.[perpendicular]] = 0 [right arrow] 0 = 1 is the unit of Q, hence [for all]a [member of] Q, a&1 = 1&a = a, this finished the proof.

Proposition 3.6. If Q is a two-sided Girard quantale, then the unique cyclic dualizing element is the least element 0.

Proof. If Q is a two-sided Girard quantale, then we have a = a&e [less than or equal to] a&1 [less than or equal to] a for all a [member of] Q. Similarly, we have 1&a = a. Thus Q is strictly two-sided. Suppose d is a cyclic dual element in Q, [sup.[perpendicular]] is the unary operation induced by d, then we have d = 1 [right arrow] d = [1.sup.[perpendicular]] = 0. the proof is finished.

Corollary 3.7. Let Q be a Girard quantale with cyclic dualizing element 0. Then the cyclic dualizing element of Q is unique.

Theorem 3.8. Any complete lattice implication algebra is a Girard quantale with unique cyclic dualizing element 0.

According the above conclusions, we have a question : Whether the cyclic dualizing element must be the least element 0 if a Girard quantale has an unique cyclic dualizing element. Theanswer is negative. Let us see the following example.

Example 3.9. Let Q = {0, e, 1}, the partial order on Q be defined by 0 < e < 1, the binary operation & be defined by Table 2

Table 2. & 0 e 1 0 0 0 0 e 0 e 1 1 0 1 1

It is immediate to verify Q being a Girard quantale with the unique cyclic dualizing element e.

Question 3.10. What is the necessary condition when the cyclic dualizing element of Guard quantale is unique?

References

[1] C. J. Mulvey, Suppl. Rend. Circ. Mat. Palermo Ser. II, 12(1986), 99-104.

[2] M. Ward, Structure residation, Ann. Math., 39(1938), 558-569.

[3] R. P. Dilworth, Noncommutative residuated lattices, Trans. Amer. Math. Soc., 46(1939), 426-444.

[4] B. Banaschewski and M. Erne, On Krull's separation lemma, Order, 10(1993), 253-260.

[5] J. Y. Guard, Linear logic, Theoret. Comp. Sci., 50(1987), 1-102.

[6] F. Borceux and G. Van den Bossche, Quantales and their sheaves, Order, 3(1986), 61-87.

[7] D. Kruml, Spatial quantales, Appl. Categ. Structures, 10(2002), 49-62.

[8] S. Q. Wang, B. Zhao, Prequantale congruence and its properties, Advances in Mathematics(In Chinese), 34(2005), 746-752.

[9] S. W. Han, B. Zhao, The quantale completion of ordered semigroup, Acta Mathematics Sinica(In Chinese), 51(2008), No.6, 1081-1088.

[10] D. Yetter, Quantales and (noncommutative) linear logic, J. Symbolic Logic, 55(1990), 41-64.

[11] J. Paseka, D. Kruml, Embeddings of quantales into simple quantales, J. Pure Appl. Algebra, 148(2000), 209-216.

[12] K. I. Rosenthal, Quantales and their applications, Longman Scientific and Technical, London, 1990.

Bin Zhao ([dagger]) and Shunqin Wang ([double dagger])

([dagger]) College of Mathematics and Information Science, Shaanxi Normal University, Shaanxi, Xi'an 710062, P.R. China

([double dagger]) School of Mathematics and Statistics, Nanyang Normal University, Henan, Nanyang 473061, P.R. China

zhaobin@snnu.edu.cn;math.wangsq@163.com

(1) This work was supported by the National Natural Science Foundation of China(Grant No.10871121) and the Research Award for Teachers in Nangyang Normal University, China (nynu200749)

Printer friendly Cite/link Email Feedback | |

Author: | Zhao, Bin; Wang, Shunqin |
---|---|

Publication: | Scientia Magna |

Article Type: | Report |

Geographic Code: | 9CHIN |

Date: | Jan 1, 2009 |

Words: | 2953 |

Previous Article: | On the instantaneous screw axes of two parameter motions in Lorentzian space. |

Next Article: | [nu]-compact spaces. |

Topics: |