Ockham Algebras (Oxford Science Publications)

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Preface

An Ockham algebra is a bounded distributive lattice with a dual endomor-

phism, the nomenclature being chosen since the notion of de Morgan negation has been attributed to the logician William of Ockham (c1290-c1349). The class of such algebras is vast, containing in particular the well-known classes of boolean algebras, de Morgan algebras, Kleene algebras, and Stone algebras. Pioneering work by Berman in 1977 has shown the importance of Ockham algebras in general, and has since stimulated much research in this area, notably by Urquhart, Goldberg, Adams, Priestley, and Davey. Here our objective is to provide a reasonably self-contained and readable account of some of this research. Our collaboration began in 1982 in the consideration of a common abstraction of de Morgan algebras and Stone algebras which we called MS-algebras. This class of Ockham algebras is characterised by the fact that the dual endomorphism] satisfies]O ::;;; j2, which implies that ] =]3. The subvariety M of de Morgan algebras is characterised by]O =j2. In general, it seems an impossible task to describe all the subvarieties of Ockham algebras. The subvarieties of paramount importance are those in which]q =]2p+q for some p, q; these are denoted by Kp,q and are called the Berman varieties. Of these, the most Significant seems to be Kl 1 in which each algebra L is such that ](L) E M. Here we concentrate particularly on K1 ,1, its subvarieties, subdirectly irreducibles, and congruences. No study of Ockham algebras can be considered complete without mention of the theory of duality, in which the work of Priestley is fundamental. We make full use of Priestley duality in considering the subvariety K 1,1' Chapters 0-11 deal entirely with Ockham algebras whereas Chapters 1215 are devoted to a brief study of double algebras. More precisely, we consider algebras (L; 0, +) for which (L; o) is an MS-algebra and (L; +) is a dual MSalgebra with the unary operations ° and + linked by the identities aO+ = aDO and a+ o = a++. Particular subvarieties of double MS-algebras are those of double Stone algebras and trivalent Lukasiewicz algebras. As we have written this text with beginning graduate students in mind, we have included many illustrative examples and diagrams, as well as several useful tabulations. We would add that we make no claim that the list of references is complete, nor that what is due to C~sar has not been attributed to Brutus. T.S.B., J.c.v.

Contents

O. ordered sets, lattices, and universal algebra

1

1. Examples of Ockham algebras; the Berman classes

8

2. Congruence relations

20

3. Subdirectly irreducible algebras

37

4. Duality theory

52

5. The lattice of subvarieties

75

6. Fixed points

105

7. Fixed point separating congruences

115

8. Congruences on K 1 ,1 -algebras

133

9. MS-spaces; fences, crowns, ...

149

10. The dual space of a finite simple Ockham algebra

164

11. Relative Ockham algebras

179

12. Double MS-algebras

187

13. Subdirectly irreducible double MS-algebras

197

14. Congruences on double MS-algebras

207

15. Singles and doubles

216

Bibliography

231

Notation index

237

Index

239

o

Ordered sets, lattices, and universal algebra

It is of course impossible to give a full account of ordered sets, lattices, and universal algebra in a few pages, so we refer the reader to the various books cited in the bibliography. Nevertheless, in order to make this monograph reasonably self-contained, we shall summarise in this introductory chapter the fundamental notions that we shall use throughout. More specific concepts that we shall require will be defined as necessary. As far as notation is concerned, we provide an index of the various symbols that are used throughout. The concept of order plays in mathematics a very prominent role, probably as important as that of size, though its importance has only rather recently been recognised. It is probably the success of the work of George Boole in the first half of the last century that has acted as a catalyst in producing a new area of research, namely that of ordered sets and, more particularly, lattices. An ordered set (or partially ordered set or poset) is a set S on which there is defined a binary relation R which is reflexive (aRa for all a E S), transitive (for all a, bE S the relations aRb and bRc imply aRc), and anti-symmetric (for all a, bE S the relations aRb and bRa imply a = b). Mathematics is replete with examples of such order relations; for example, the relation of magnitude on the set of real numbers, the relation ~ of inclusion on the power set IP{E} of any set E, the relation I of divisibility on the set INo of strictly positive integers, etc.. Usually, an order relation is denoted by ( and its converse by ~. Two elements x, y of an ordered set are said to be comparable (in symbols, x My) if x (y or y ( x, and incomparable (in symbols, x I y) if x 1, y and y 1, x. The (order) dual Sop of an ordered set S is the same set equipped with the converse order. We write x --< y if x ( y and {z I x < z < y} = 0. If x --< y then we say that x is covered by y, or that y covers x. The relation j is clearly an order. There are several useful ways in which disjoint ordered sets P, Q can be combined to produce a third ordered set. In particular, the disjoint union P U Q consists of the set P U Q with the order defined by

x(y L by cp(O} = 1, cp(l} = 0 and

(Vx E B)

cp(x} = (b V x) /\ a = b V (x /\ a).

Clearly, cp is an idempotent {O, 1}-lattice morphism. We can define a polarity p on Imcp = {O} U [b,a] U {I} by p(O} = 1, p(1} = 0 and

(Vx

E

[b, aD

p(x) = cp(x'}

where x' is the complement of x in B. It is easy to verify that p(x} is the relative complement of x in [b,a]. It follows from Theorem 1.2, withf(x} = p[cp(x)] for every x E L, that (L;1) is an Ockham algebra with a de Morgan skeleton. Here we havef(O} = 1,J(I} = 0 and, for x E B,

f(x}

=p[cp(x)] = cp([cp(x}]') = cp[(b' /\ x'} va'] = b V (a /\ x') = cp(x'}.

Example 1.11 Let L consist of the (possibly infinite) lazy tongs lattice with a new smallest element 0 and a new greatest element 1 adjoined. Consider the mapping cp : L -'> L given by

x

ifxE{0,I,ao,bo}U{a2n+dnE~;

cp(x} = { a2n-l

if x

= a2rl

a2n+l

if x

= b21l (n :f O).

.1

(n :f O); b2

In the Hasse diagram opposite, the arrowheads indicate the effect of cp. It is readily seen that cp is an idempotent {O, 1 }-lattice morphism. There are two polarities on Imcp, namely p,p' given by

p(o}

bo

=p'(O} = 1; = 0;

p(l} = p'(I}

p(a 2n +1} = p'(a2n+d = a-2n-1; p(ao} = ao, p(b o} = bo; p'(ao} = bo, p'(b o} = a o·

.0

These polarities give rise to different Ockham algebras with the same de Morgan skeleton.

16

Ockham algebras

Example 1.12 Let E be a set and let a, b be distinct elements of E. Consider the mapping ip : IP(E}

-7

IP(E} given by X _

ip(

}-

{xu

{a}

Xn{a}'

if bE X; if b

¢ x.

Roughly speaking, ip adds a if X already contains b, and removes a if X does not contain b. Clearly, ip(0} = 0 and ip(E} = E. It is readily seen that ip is both a u-morphism and an n-morphism. Moreover, ip is idempotent. Now 1m ip

= [0, {a, b}'] u [{a, b}, E]

and so admits the polarity p provided by complementation. We can therefore make IP(E} into an Ockham algebra with a de Morgan skeleton by defining

j(X}=[ (X)]' ip

= {xln{a}1

ifbEX; X' U {a} if b ¢ X.

Example 1.13 Let B be a boolean lattice. Given n B~ = {(Xl,"', x,J E BII

I Xl

~

3, define

:,;; xn}'

Then B~ is a sublattice of BIl, with (0, ... , O) as smallest element and (1 , ... ,I) as greatest element. Define ip : B~ -7 B~ by

ip(XI,X2,'" ,xn} = (XI,X n "" ,xn )· Clearly, ip is an idempotent {O, l}-lattice morphism. A polarity p on Imip is

We can therefore make B~ into an Ockham algebra with a de Morgan skeleton by defining

We have seen abuve that if (L;/) E KI,1 then the mapping X 1--7 j2(x} is an idempotent lattice morphism. We shall now investigate an important special case of this, namely when the mapping X 1--7 j2(X} is a closure. We recall that a closure on an ordered set E is an isotone mapping j : E -7 E such thatj = j2 ~ idE' Thus X 1--7 j2(X} is a closure precisely when

(Vx,YEL) x:';;Y => j2(X}:';;j2(y}; (Vx E L) X :,;;j2(X}; (Vx E L) j2(X} = j4(x}. Note that the first of these properties is satisfied by all LEO, and that the third is satisfied by all L E KI,I'

Examples of Ockham algebras; the Berman classes

17

Definition By a de Morgan-Stone algebra, or an MS-algebra, we mean an algebra (L; A, V, 0, 0, 1) of type (2,2,1,0,0) such that (L; A, V, 0,1) is a bounded distributive lattice and x 1---7 XO is a unary operation on L such that (MSl) (MS2) (MS3)

1° = 0; (Vx,y E L) (x AY)O (Vx E L) x ~ xeD.

= XO V yO;

Clearly, de Morgan algebras and Stone algebras are MS-algebras; hence the terminology. In fact, when we introduced the notion of an MS-algebra in 1983 [33] our objective was to stress the numerous similarities between these two classes of algebras. In a de Morgan algebra, x 1---7 X is a dual automorphism and x 1---7 X is the identity. In a Stone algebra, x 1---7 x* is a dual endomorphism and x 1---7 x** is a closure. So the notion of an MSalgebra arises quite naturally by retaining the properties that are common to these two classes of algebras. The class MS of MS-algebras is equational; it is the subclass of KII obtained by adjoining the equation x Af2(X) = x. In this connection, w~ note that M. Ramalho and M. Sequeira [82] have considered more generally the subvarieties of 0 defined by x Af21l(X) =x. The relation of MS-algebras to Ockham algebras is as follows.

Theorem 1.4 Every MS-algebra is an Ockham algebra with a de Morgan skeleton. An Ockham algebra (L;I) is an MS-algebra ifand only if x ~f2(X) for every x E L. Proof Let (L; 0) be an MS-algebra. Then, by (MSl) and (MS3), we have 0° = 100 ~ 1 and so 0° = 1. By (MS2), the mapping x 1---7 XO is antitone. So (x

V y)O ~

XO

A

yO ~ (XO

A yO)OO

= (x 00 V yOO)O .

Since clearly x V y ~ xOo V yOO, which implies that (x V y)o ~ (x 00 V yOO)O, we deduce that (x V y)o = XO Ayo. It follows that (L; 0) is an Ockham algebra. Now by (MS3) we have x ~ XOO and so XO ~ xooo. But, again by (MS3), XO ~ XOoo. Consequently, we have XO = XOOO and so (L; 0) has a de Morgan skeleton. The second statement is immediate from the definitions. In an Ockham algebra (L; f) the biggest MS-subalgebra is MS(L) = {x ELI X ~f2(X)}.

In the case where (L;I) belongs to K I ,I and is obtained as in Theorem 1.2, MS(L) = {x ELI x ~ cp(x)}.

Ockham algebras

18

Thus, in Example 1.9 we have (for both of the algebras described)

MS(C) In Example 1.10, we have x

= {(x,y,z) E C I x ~ z}. ~



x 1\ a = y 1\ a and xV f(a}

= y V f(a}.

Theorem 2.3 For every a E L, (I) {) a is a principal lattice congruence; moreover, [a]{) a = [a, a V f (a)] and{)a == W a ~f(a). (2) If a ~ f2 (a) then {) a is a principal congruence. (3) If a = f2(a) and a Ilf(a) then {)a and {)f(a) are non-trivial congruences with {) a 1\ {)f{a) = w.

*

Proof (1) By its very definition, {)a = {)lat(a, I} 1\ 191at(O,f(a») = {)lat(a I\f(a),f(a». That [a]{) a = [a, a V f (a)] follows immediately from the definition of 19a ; and that {) a = W a ~ f(a) is immediate from the above. (2) If a ~f2(a) and (x,y) E {)a then (t(x),f(y}) E f)a; for, from x 1\ a = y 1\ a we obtainf(x) V f(a) == f(y) V f(a) and from x V f(a) == y V f(a) we obtain f(x} 1\j2(a) == f(y) 1\j2(a), hence f(x) 1\ a =f(y) 1\ a. (3) If a = j2(a) and a Ilf(a) then by (1) we have {) f{a) = {)lat(a I\f(a}, a) and consequently -a a 1\ 19f{a) = w. It also follows by (1) that both f) a and f) f(a) are non-trivial.

*

Congruence relations

23

Theorem 2.4 Foran Ockham algebra (L;1), (1) {a ELI a ~f2(a)} is a sublattice of L which contains 0 and 1;

(2) a ~ b

=}

'I9 a ~ 'I9 b;

(3)'I9a/\'I9b='I9avb'

Proof (1) follows from thefact thatf2(a/\b) j2(a) V j2(b).

=f2(a)/\j2(b) andj2(av b) =

(2) follows from the fact that x /\ b = Y /\ b implies x /\ a x V f(b) = y V f(b) implies x V f(a) = y V f(a).

= y /\ a,

and

(3) If (x,y) E 'I9 a /\ 'I9 b then x/\a=y/\a, x/\b=y/\b, xVf(a)=yvf(a), xVf(b)=yvf(b), which implies that x /\ (a V b)

=y /\ (a V b),

xV f(a V

b) = y V f(a V b).

Consequently, 'I9 a /\ 'I9 b ~ 'I9 avb ' The converse inequality follows from (2). Theorems 2.3 and 2.4 show that in MS the meet of two principal congruences each of the form '19 a is again a principal congruence. Unfortunately, this property does not hold for arbitrary principal congruences, even when the Ockham algebra belongs to a 'small' class such as that of de Morgan algebras. The interested reader may consult M. E. Adams [17]. For an Ockham algebra (L;1) consider now, for every n E IN, the relation «Pn on L defined by (x,y) E «PII *=> PZ(x) = PZ(y)· It is clear that «Pn is a congruence on L. Moreover, the subset

is a subalgebra of L. We now consider some basic results concerning these congruences and subalgebras. Of especial importance in this is the congruence «Pw =

V «Pi'

i~O

a practical description of which is as follows.

Theorem 2.5 If (L;1)

then (x,y) E «Pw ifand only if r(x) some n. Moreover, if L is non-triVial then «Pw < ~.

Proof If (x,y)

E

E0

=r(y)for

«Pw then there exist to, ... , tk and «Pi!,"" «Pik such that

x

= to cIl ! t1 «Pi2 t2 j

... t k- 1 «Pik tk = y.

24

Ockham algebras

Denote the greatest of these q>j by q>n. Then we have x q>nY, i.e. fn(x) = fn(y). Conversely, if r(x) =r(Y) then clearly (x,y) E q>w' Finally, if L is non-trivial, r(O)

t: fn(l)

for all n gives q>w

t: L ~

For every non-trivial Ockham algebra (L;I) it is clear that we have W

= q>o :::;;; q>1 :::;;; q>2:::;;;

.. :::;;; q>j :::;;; q>i+1 :::;;; ... :::;;; q>w

< t,

and, with :::;;; meaning 'is a subalgebra of',

{O,l}:::;;; ... :::;;;/+l(L) :::;;;/(L):::;;; ... :::;;;f(L):::;;;fO(L)=L. It is readily seen that [x]q>j

/(x) describes an Ockham algebra isomorphism when i is even and a dual isomorphism when i is odd, a situation which we shall denote by writing L/ q> j ' " / (L). The following result is therefore clear.

Theorem 2.6 If (L i f)

E Kp ,q

f--t

then, for n :::;;; q,

L/q>n '" fn(L)

Theorem 2.7 If (L;I) W

E Kp,q

E K p,q_1l'

~

then

== q>o :::;;; q>1:::;;; ... :::;;; q>q

= q>q+1 = ... = q>w'

Proof Observe that (x,y)

E

q>q+l

We thus have q>q gives

f q+ r (y)

~

f q+1(x) =r+l(y) fq(x) = f2P-l[jq+l(X)]

~

(x,y)

~

= q>q+1'

E

q>q.

If now (x,y) E q>q+r with r ~ 1 then fq+r(x) =

(tr-l (x),r- 1(y)) and so f q+r- 1(x) = f q+r-l(y), i.e. (x,y) q>q =

and therefore q>w

= pp-l [jq+1 (y)] = fq(y)

E E

q>q+l

= q>q

q>q+r-1' Consequently,

q>q+l = q>q+2 = ... ,

= V q>j == q>q.

~

j;;;:O

Corollary !be following statements are equivalent: (1) q>w == Wi

(2) (Vi ~ 1) q>j

(3) f is injective.

= Wi

25

Congrnence relations

Moreover, if (L;/) belongs to the Berman class Kp,q then each of the above is equivalent to (L;/) E Kp,o. Proof 1be equivalence of (1), (2), (3) is clear. As for the final statement, if for

every x E L we havej2p+q(x) = fq(x) then, by (3), we havej2P(x) = x and then (L;/) E Kp,o' Conversely, suppose that (L;/) E Kp,o and that d contains an atom 19{a, b). As shown above, this implies that L is subdirectly irreducible, a contradiction. 0 The Kw-analogue of Theorem 2.6 is the following.

Theorem 3.10 If L E Kw then L/cI>w

~

T{L).

Proof Observe first that cI>wIT(L) = w. In fact, let x,y E T{L) be such that (x,y) E cI>w. Then x = fP{x), y = jq(y) for some p, q; andr{x) =r(Y) for some n. Let r = lcm{p,q, n}; then we have x = r(x) = r(Y) = y. Now in the quotient algebra (L/cI>w;]) we have ]([x]cI>w) = [j{x)]cI>w. Consider the morphism 19 : T{L) ~ L/cI>w given by 19(x) = [x]cI>w. By the above observation, 19 is injective. To see that 19 is also surjective, observe that, since L E Kw by hypothesis, for every y E L there exist m, n such that pi (y) = f m+n (y). It follows that there exists p ;;;: n such that fP+1I(y)

= f'!(y) = j2p+n(y).

The first of these equalities gives (y,JP(y» E cI>1I ~ cI>w; and the second gives\ ~ T{L) from which it follows that fP(y) E T(L) since n ~p. Thus we see that

fn(y) E T 2p {L)

and so 19 is also surjective. 0 Our objective now is to determine precisely when, for L E Kw, the interval of Con L is boolean. For this purpose, we require the following result of J. Vaz de Carvalho [106]. [cI>w,~]

Subdirectly irreducible algebras

45

Theorem 3.11 If L E Kn,o then L is a strong extension of T 2(L). Proof Given

1'J E Con T2 (L), let 1'J 1, 1'J 2 be extensions of 1'J to 1. We may assume without loss of generality that 1'J 1 ~ 1'J 2 . Observe first that if a E T 2(L) then [a]1'J 1 = [a]1'J 2. In fact, for every x E L define

If (x, a) E 1'J 2 then clearly (x*, a) E 1'J 2 and (x*, a) E 1'J 2 • Since x*, x* E T2(L) and 1'J 1 I T 2(L) = 1'J = 1'J 2I T 2(L) we have (x*, a) E 1'J 1 and (x*, a) E 1'J 1. Since x* ~ x ~ x* we deduce that (x, a) E 1'J 1. Thus [a]1'J 2 ~ [a]19 1 whence we have equality. Suppose now that L is finite and (x,y) E 19 2 , To show that (x,y) E 19 1 it suffices, since L /191 is a finite distributive lattice, to show that [x]19 1 and [Y]191 contain the same set of V-irreducible elements. Suppose then that [c]19 1 is Virreducible in L /191 and such that [c ]19 1 ~ [x]19 1 , i.e. such that (c /\ x, c) E 19 1, Let m be the smallest positive integer such that [c]19 1 = (f2m(c)]19 1 , noting that m ~ n. Define

e= {

f2(C) V .. ~V f2m-2(c)

if m > 1; ifm=1.

Then [c*]19 1 = [c V e}19 1 ; and (c, c /\ x) E 19 1 gives (Cve, (c/\x)ve)E191~192' But (c /\ x) ve, (c /\ y) V c) E 19 2 and so (c Ve, (c /\ y) V c) E 19 2, Since (c*, c V c) E 19 1 ~ 19 2 we have that (c*, (c /\ y) V c) E 19 2, Since c* E T 2(L) it follows by the observation above that (c*, (c /\y) ve) E 19 1 and therefore (cve, (c/\y)Ve)E19 1. Thus(c, (c/\y)v(c/\e)) E19 1 and so

[c]19 1 = [c /\y]19 1 V [c /\ e]19 1 . If m

= 1 then e = 0 and we obtain [c]19 1 ~ [Y]191'

If m

> 1 then

[c]19 1 = [c /\y]19 1 V [c /\f2(c)]19 1 v··· V [c /\f2m-2(c)]19 1 and so, since [c ]19 1 is V-irreducible in L/191, either[c]19 1 ~ [y]19 1 or there exists k with 1 ~ k ~ m -1 such that [c]19 1 ~ (f2k(c)]19 1 • Since f is injective, the latter gives the contradiction [c]19 1 = (f2k(c)]19 1 with 1 ~ k ~ m -1. Hence we have [c]19 1 ~ [y]19 1 . Similarly, every V-irreducible contained in [y]19 1 is contained in [x]19 1 whence [x]19 1 = [y]19 1 and consequently 19 1 = 19 2 , Suppose now that L is arbitrary and that (x,y) E 19 2, Let A be the subalgebra of L generated by {x,y}. Since Kn,o is locally finite by Theorem 3.7, A is finite with T 2 (A) = A n T2(L). By the above, we deduce from

46

Ockham algebras

'!91I T2(A) == '!92I T2(A) that '!9 11A == '!9 2 1A- Since x,y E A we then have (x,y) E '!91. Hence '!9 1 == '!9 2 as required.

Theorem 3.12 If LEO then T(L) is a strong extension of T2(L). Proof Let i(J E Con T 2(L) and let i(J1, i(J2 E Con T(L) be extensions of i(J. Suppose that X,Y E T(L) are such that (x,y) E i(J1. If A is the subalgebra of T(L) generated by {x,y} then A E Kp,o for somep. Since i(J1I T2(L) = i(J == i(J2I T2(L) and T 2(A) ~ T 2(L) we have i(J1I T2(A) = i(J2IT2(A). By Theorem 3.11, A is a strong extension of T2(A). Hence i(J11A = i(J21A and consequently (x,y) E i(J2. Thus we have i(J1 ~ i(J2. Similarly, i(J2 ~ i(J1 and so i(J1 = i(J2 as required. Corollary Con T(L)

~

Con T2(L).

Proof Immediate from the above and the congruence extension property. We can now use the above results to determine precisely when [1} E9 [cI>z, ~] with

[cI>2 , ~] ~ Con L/ cI>2 ~ Con f2 (L).

We conclude from this that if L belongs to a Berman class and VB(L} then, by Theorem 2.7, Con L is the finite chain w = cI>o

-< cI>1 -< cI>2 -< . . . -< cI>q -
j, with cI>i+1 covering cI>j and the supremum of the cI>j being cI>w, we conclude that Con L is the infinite chain w = cI>o 1 -< cI>2 -< ... < cI>w -
1 = 19(b I , 1). As for 4>w, we have L/4>w ~ {O,CI., I} = T(L).

4 Duality theory Duality theory in the context of distributive lattice-ordered algebras has rather a long history. It was in 1933 that G. Birkhoff established his famous representation theorem for finite distributive lattices and, about three years later, M. H. Stone developed a representation theory for arbitrary boolean algebras, using topological methods. In the early seventies, H. A. Priestley provided an ingenious common generalisation of both these theories, thus allowing questions of a lattice-theoretic nature 'to be translated into the language of ordered topological spaces' and usually be resolved more easily, the dual space generally being simpler and more tractable than the algebra itself. Of course, it is impossible to give here a complete survey of the theory of duality; for this we refer the reader to [78], [79], and [63]. Here we shall simply introduce the concepts and results that we shall require. A set X which carries a topology T and an order relation ~ is called an ordered topological space. Such a space (X; T, ~ ) is said to be totally order-disconnected if (TaD) given x, y E X with x i y there exists a clopen (= closed and open) down-set U such that y E U and x ¢ U. Clearly, every totally order-disconnected X is totally disconnected in the sense that (TD) given x, y E X with x =f y there exists a clopen subset U such that x E U andy ¢ U. Every totally disconnected space X is Hausdorff in the sense that (H) given x, y E X with x =f y there exist open sets UI , U2 such that x E U1> Y E U2 , and UI n U2 = 0. In summary, therefore, we have (TaD) =? (TD) =? (H). A compact totally order-disconnected space is called a Priestley space. The family of clopen down-sets of a Priestley space X will be denoted by O(X). Let L be a distributive lattice and let Ip(L) be the set of prime ideals of L. The dual space or prime ideal space of L is (X; T,~) where X = Ip(L) and T has as a base the sets {x E Ip(L) I x 3 a} and {x E Ip(L) I x ~ a} for every a E L. The fundamental result concerning this is the following :

Duality theory

53

(X; T,~) is a Priestley space and L ~ O(X) via a ~ {x E X I x ~ a}. Conversely, if P is a Priestley space then O(p) is a distributive lattice and P ~ (Ip(O(P)); T, ~ ).

In the language of category theory, if we denote by DOl the category of bounded distributive lattices and 0, 1-preserving lattice homomorphisms, and by P the category of Priestley spaces and continuous order-preserving maps, then the above isomorphisms give a dual equivalence between DOl and P. The power of duality theory is particularly evident in the study of congruence relations. If LED and X = (Ip(O(P)); T, ~) is the dual space of L then for every closed subset Q of X the relation -8 Q defined on O(X) by (A,B)E-8 Q

is a congruence. Note that A

n ~ 0 denote by P m,1l the subclass of 0 that is formed by those algebras whose dual space satisfies gm =gil. It will be established in Chapter 5 that when m - n is even we have L E P m,1I if and only if fm = fn (in other words, P2p +n ,n = Kp •n); and when m - n is odd we have L E P m.n if and only if fm (a) and fn (a) are complementary for every a E L.

55

Duality theory

Theorem 4.3 Let (L;/) be a finite Ockham algebra and let (X; g) be its dual space. Then the jollowing statements are equivalent:

(1) L E Pnojorsomen; (2) g is sU'rjective; (3) g is injective; (4) Con L is boolean with k atoms, k being the number oj monogenic gsubsets oj X, which attains its maximum value precisely when L is boolean.

Proof (1)

=? (2): If L E PIZ,o then glZ = idx . (3) : Since L is finite, so is x. (2) (3) =? (4) : Let Xl be an arbitrary element of X. Since X is finite and g is injective, there exists n 1 such that g'll (x 1) = X l' If gW {x d =f X we can choose X2 EX \gW{xd; then there exists n2 such that gn2(X2) = x 2 , and so on. Since X is finite, this procedure terminates after finitely many steps. In this way we obtain k monogenic g-subsets

*

Q1

= gW{xd,

Q2

=gW{X2},

... ,

Qk

= gW{xk}

of respective cardinalities nl, n2, "', nk' Moreover, since g is injective, these g-subsets are pairwise disjoint. It follows that G(X) ~ 2k and so Con L is a boolean lattice with k atoms. Note that in this case the maximum value of k is obtained when every monogenic g-subset reduces to a singleton. This is so precisely when g(x) = x for every x E X, in which case L E P 1 ,o = B, the class of boolean algebras. (4) =? (2) : If Con L is boolean then so is G(X). In this case g(X) and its complement are both g-subsets. This implies that X \ g(X) = 0, whence X = g(X) and so g is surjective. (3) =? (1) : If g is injective then, as we have shown above, G(X) is the disjoint union of k monogenic g-subsets Q1,' .. ,Qk of respective cardinalities n1,"" nk' If we consider n = lcm{nl,"" nk} then we see that gn(x) = x for every x E X. Consequently, L E Pn,o'

qN

Example 4.1 Let (X;g) be the Ockham space

p

S

x pqrst g(x) s r q p t

r

The dual algebra (L;/) belongs to P 2,o and so is a de Morgan algebra. There are 3 monogenic g-subsets, namely Q1

= {P,s},

Q2 = {q,r},

Q3

= {t}.

56 Hence Con L

Ockham algebras ~

23 . The algebra (L;J) is described as follows :

~ f(~)

°

abc d e h j k I 1 1 k I h i j e d e abO

The three coatoms of Con L are 1J QI ' 1J Q2' and 1J Q3' descriptions of which are obtained by Theorem 4.2.

The other three non-trivial congruences are easily described. Note also that 1J Q1 = 1J(a,k), 1J Q2 = 1J(b,l), and 1J Q3 = 1J(O,i).

Example 4.2 Consider again Example 2.5. The dual space X is the 6element chain p < q < r < s < t < u with g defined by q

u

r s

stu s s s

Here the monogenic g-subsets are {s}, {s, u}, {s, t}, {r,s}, {p,s, u}, {q,s, u} The Hasse diagram of (G(X))oP is depicted below and, compared with the diagram for Con L given in Chapter 2, shows the correspondence Q ~ 1J Q obtained in Theorem 4.2. To simplify the notation in the diagram, w~ write {q,s, u} as qsu, etc ..

Duality theory

57

o

rs

rst

pqsu

pqstu

x Example 4.3 Let X be an ordered set (the order may be discrete) of cardinality n. Let p be an element of X and let g : X -7 X be the constant map given by g(x) = p for every x E X. Clearly, (X;g) is an Ockham space. The non-empty g-subsets of X are all the subsets of X that containp. It follows that c(X) ~ 1 EB 2n - 1 . If (L;1) is the dual algebra then Can L ~ 2n - 1 EB 1, the coatom being the congruence associated with {P}. Since g2 (L;1) E P2,l'

= g we have

LetA be a down-set of X. Ifp E A then we haveg-1(A) = X andf(A) = 0. If P ¢ A then g-l(A) = 0 and f(A) = X. Let d be the element of L that represents p!. Then it follows that we have (VOl. E d i ) f(OI)

= 0;

(VOl. ¢ d i ) f(OI.)

= 1.

In particular, if p ~ x for every x E X then f(OI.) = 0 for every 01. E L \ {O}, which implies that 0 is meet-irreducible. In this case, L is a Stone algebra. If, on the contrary, we have p ~ x for every x E X then f(OI.) = 1 for every x E L \ {I}, whence 1 is join-irreducible and L is a dual Stone algebra.

Ockham algebras

58

In Chapter 2 we introduced the congruences 12 defined for every n E IN by In the dual space (X;g) we can consider for every n E IN the g-subset

gll(X) = {gil (x) I x EX}. We can also consider the g-subset

gW(X) =

n g12(X).

1I~0

Clearly, all gil are continuous maps and, since X is a compact Hausdorff space, all g12 (X) are compact, hence closed. The g-subset gW (X) enjoys the same property. Obviously, these closed g-subsets are related by X

:2 g(X) :2 g2(X):2 .. :2 gW(X) :2 {O, I}.

The following result [55] elucidates the correspondence between the congruences 11 and the closed g-subsets gil (X).

Theorem 4.4 Let (L;1) E 0 and let (X; g) be its dual space. Tben for every n ~ 0 the congruence 11 is associated with the closed g-subset gn(x), and the congruence w is associated with gW (X).

= X. As for 1, we have the following equivalences which show that 1 is associated with g (X) :

Proof Clearly, 0 is associated with gO (X) (a, b) E 1

f(a) = f(b) ~ X\g-l(A)=X\g-I(B) ~ g-l(A) = g-l(B) ~ A ng(X) =B ng(x) ~ (A,B) E 199 (x). ~

Taking into account the fact that, for every n,r+l(L) is a subalgebra ofr(L), and using an easy inductive argument, we can show that n is associated with

gn(x). Finally, w

= V n is associated with n~O

n gn(x) = gW(X). n~O

Example 4.4 Let INoo consist of INo with an additional point 00 adjoined. It is known (see [9] for the details) that INoo becomes a Priestley space if one takes as a sub-basis for the topology the subsets U such that U~

00

or (U

300

and U' is finite).

Duality theory

59

Then a subset V is dosed if or (V ~ 00 and V is finite).

V 3 00

The dopen subsets of INco are therefore the finite sets that do not contain 00, together with their complements. Now define g by setting g(n) = n + I for every n E IN, and g(oo) = 00. Since the order on INco is discrete, g is trivially order-reversing. It is easily verified that g is continuous. It follows that (INco;g) is an Ockham space which we can represent as follows : .----~).----~).----~).----~)

I

3

2

4

....

~

00

The g-subsets are 0, {oo} and, for every n, {n,n+l,n+2, ... }

and

{n,n+I,n+2, ... ,00}=gll-1(IN co ).

The dosed g -subsets are 0, gn (IN co ) for all n

~

0, and gW (IN co ) = {oo}. Now

g°(lN co ) = INco :) g(lN co ) :) g2(IN co ):) ... :) gW(INco) :) 0, and so, if (L; j) is the dual algebra, Con L reduces to the chain w

-< cI>1 -< cI>2 -< .. < cI>w -
J·L0 -'> Xl·L -'>11·L·L -'> X2 -'> ... (1) k6 -'> 15 -'> ki -'> Ii -'> k~ -'> . .. (2)

r.

·L t2

kL2

·L tl

ki

·L to

kL0

CI

r+-ts

.s

The g-subsets of X are easily determined: they are 0; those generated by any element "( of the sequence (1); those generated by any element 8 of the sequence (2); those generated by any pair {"(, 8}; and finally all of the previous with {r, s} adjoined. Only and the last mentioned g-subsets are closed, so (G(X))oP is given by

o

C2

The diagram for Con L is the same with the label change X -'> w, gi(X) -'> 4>i' gW(X) -'> 4>w, and 0 -'> to Note that

gW(X)

= n gi(X) = {r,s} i~O

gives the comonolith 4>w which is not principal, and that L/4>w is isomorphic to the simple 4-element de Morgan algebra.

66

Ockbam algebras

Example 4.7 Here is an example of a small Ockham algebra with only one non-principal congruence. 1

e

q

c

b

r

x L

n 0 abc d e 1 f(n) 1 d b c a 0 0

x p q r s t g(x) s

r q p p

{q, r}

19 1

{p,s,t}

ConL It is easily seen that cI>1 = 19(e, 1) corresponds to {p, q, r, s}, 19 1 = 19(0, a) corresponds to {q, r}, and 19 2 = 19 (a, d) corresponds to {p, s, t}. Only

19 3 = 19{p,s} == {{O}, {a, b,c,d}, ie, 1}} is not principal. In fact, the greatest convex sets disjoint from {p, s} are {q, r} and {t}; and {p, s} is not maximally disjoint from either. We now consider subdirectly irreducible and simple algebras from the duality point of view. For this purpose, we require the following fundamental results of Urquhart [94]. Lemma 4.1 Let (X;g) be an Ockbam space. Tben a subset Y of X is a gsubset if and only if Y ~ g-1 (Y).

Proof If Y is a g-subset then g(Y) ~ Y gives Y ~ g-1 [g(Y)] ~ g-1 (Y). Conversely, Y ~ g-1 (Y) gives g(Y) ~ g[g-1 (Y)] ~ Y.

67

Duality theory Lemma 4.2

Proof Y

~

If Y is a g -subset oj X and Y

~

Q then Y

~ g-l (Q).

Q gives g-l(y) ~ g-l(Q) and the result follows by Lemma 4.1.

Lemma 4.3 Let (X; g) be an Ockham space.

If Y is a g -subset oj X then so

is its closure Y.

Proof Since {A I A E O(X)}U{A' I A Y of any g-subset Y is

E O(X)}

Y = n{A' UB I A,B E O(X),

is a sub-base of X, the closure Y ~ A' UB}

Now for A' U BEY we have g-l (A' U B) = g-l (A') U g-l (B)

=J(A) U [f(B)l'

and, by Lemma 4.2, Y ~ g-l (A' U B). It follows that g-l (A' U B) E 17. Thus g(Y) ~ Y and so Y is g-closed.

Corollary gw (Y) is the least closed g -subset that contains Y. Theorem 4.6 Let (X; g) be the dual space oj (L ;j) E 0 and let Q

= {x E X I gW {x} ~ X}.

Tben (L; J) is subdirectly irreducible if and only if Q ~ X.

Proof *' : Suppose that Q ~ X and let X be such that gW {x} ~ X. Since gW{g(x)} ~ gW{x}, we have gW{g(x)} ~ gW{x} C X and, by Lemma 4.3, Q E G(X). Let R E G(X) \ {X}. If x E R then gW{x} ~ R and x E Q. Thus R ~ Q and Q is maximal in G(X) \ {X}. Consequently, by Theorem 4.2, L is subdirectly irreducible. ==} : Suppose now that L is subdirectly irreducible and denote by Y the maximum element of G(X) \ {X}. If x E Y then gW{x} ~ Y and gw{x} ~ X. Conversely, if gW{x} ~ X then gW{x} ~ Y, which gives x E Y. We thus have Y=Q.

Corollary 1 If (L; J) E 0 is finite then (L; J) is subdirectly irreducible if and only if gW{x} = X Jor some x E x. Proof The topology on X being discrete, L is subdirectly irreducible if and only if {x I gW{x} ~X} ~X; i.e. if and only if there exists x

E

X such that gW{x} = X.

68

Ockham algebras

Corollary 2 If (L; f) E 0 is finite then (L; f) is simple if and only if gW {x} X for every x E X.

=

Proof For every x E X we have gW{x} E G(X) \ {0}, and (L;/) is simple if and only if G(X) has only two elements, namely 0 and X. 0

By Corollary 1 above, the dual space (X; g) of a finite subdirectly irreducible algebra LEO can be represented as follows (in which the order on X is ignored and the action of g is indicated by the arrows) :

o

1

2

e--+-e--+-e-

We suppose that gm(o) = gll(O). The element 0 then generates X under g; such an element is called an end of X. The subsets {O, 1 , ... , n - I} and {n, n + 1 , ... , m - I} are called the tail and the loop of X respectively. Such an Ockham space with the discrete order is usually denoted by m n . The dual algebra of mn is denoted by Lm,n' Clearly, the end of X is unique except in the case where X reduces to a loop; Le., by Corollary 2 above, precisely when L is simple. It is clear that the non-empty g-subsets of X are the sets

{k,k+1, ... ,m-1}

(O~k~n)

The following is now a consequence of Theorem 4.2. Corollary 3 Con Lm,n is an (n + 2)-element chain. 0

Any order on {O, 1 , ... , m - I} with respect to which g is order-reversing yields the dual space of a subdirectly irreducible Ockham algebra. Conversely, all dual spaces of finite subdirectly irreducible Ockham algebras arise in this way. Now the identity map on {O, 1, ... , m -I} is an Ockham space morphism of mn onto X and, by the duality, corresponds to an injective Ockham algebra morphism. We therefore have : Corollary 4 Afinite Ockham algebra is subdirectly irreducible if and only if it is isomorphic to a subalgebra of L m,1l for some m and n. 0

Duality theory

69

We shall now apply the preceding results to describe the subdirecdy irreducible algebras in the class P3,1 = K1,1' These were determined in 1980 by Sankappanavar but not published until 1985 [85]. During this time they were determined independently by Beazer [24] using both algebraic and topological approaches. Of course, all MS-algebras belong to P 3,1; the subdirectly irreducible MS-algebras were determined by the authors in 1983 [33].

Example 4.8 The subdirecdy irreducible algebras in P 3 ,1 are the subalgebras of L 3,1, i.e. the algebra whose dual space X = {p, q, r} has discrete order and on which g acts as follows :

p• -

q• :;:::=:::!: r•

The Hasse diagram of L 3,1 is the boolean lattice

d@lh a c

o with the operation J defined by

x 0 abc d e h 1 J{x) l I b e b e 0 0 We could of course seek to find all the subalgebras of L 3 ,1' Instead, we shall adopt the tactic of determining them from their dual spaces. For this purpose we have to determine all the orders on {p, q, r} with respect to which g is order-reversing. The enumeration is made simpler by the observation that for such an order we must have p < q ~ r < q and dually, whereas the relations p < r, q < r and their duals have no consequences. There are in all 19 non-trivial subdirectly irreducible algebras in P3 ,1 (including its subclasses P 1,Q,P2,Q,P2,1)' These are as shown on pages 70 and 71, together with their duals. Note that B,M,K,Sl,B 1 are the only subdirecdy irreducible algebras in P 3 ,1 that are self-dual. In these diagrams we use the notation @ to denote g-fixed points in X and J-fixed points in L. By Corollary 3 of Theorem 4.6, we know that Con L has 2 elements if L E P 1,Q or L E P 2,Q, and 3 elements if L E P 2,1 or L E P 3,1' In the latter case we indicate the classes of the only non-trivial congruence $1'

Ockham algebl

Ockham algebras

Ockham spaces Pl,D

I~

® r

P2,D q_~_r

hOe 0

H

tn~

B

M

K

P 2,1

~1~

H

·Ob

p_-®r

0

P 3 ,1

rlt

t~

~1~

f~

S

Sl

Kl

K2

)uality theory

71

P3,1

continued

P?q

h«e

r

K3

q.~I:

b

~e c

Ml

0

PV q

r

p.-1D:

a

d n ~ 0). Every finite Ockham algebra is in Kp,q for some p,q and in Pm,n for some m,n. Every Kp,q is a Pm,n (precisely, Kp,q = P 2p +q ,q) but not conversely. Since every P m,n is generated by a single subdirectly irreducible algebra, it is join-irreducible in A(O). By comparing the tails and loops in the dual spaces of subdirectly irreducible algebras it is clear that we have

(p

~ 1, q ~ 0) or the classes Pm,n (m

Pm,n~Pml,nl - n

~

0 and (L; "') EO. Then

(1) when m - n is odd we have LEPm,n ~ (VaEL) ""m a A",,1!a=O, ""m av ""tl a =l;

79

The lattice of subvarieties

(2) when m - n is even we have L E Pm,n

¢=>

(Va E L) rvma = rv!la.

Proof (1) Consider the equalities rvma 1\ rv ll a = 0 and rvma V rv ll a = 1. Without loss of generality we may assume that m is even and n is odd. By Urquhart's theorem the first of these is equivalent to gm ~ gll, and the second is equivalent to gn ~ gm. Their conjunction is equivalent to gm = gn, i.e. to L E P m ,ll'

(2) The equality rvma = rv ll a is the conjunction of the inequalities rvma ~ rv ll a and rvlla ~ rvma. For m and n of the same parity, these are equivalent by Urquhart's theorem to gm ~ gil and gil ~ gm, i.e. to gm = gil. Urquhart's theorem is extremely powerful. In order to make its impact more transparent, we consider some examples.

Example 5.2 Let E = {( 0, 2), (I, 2), (2, 0

n. We have

A = {rvoal, rv2a3}' B = {",2 al , rv2a2, rvoa3, rvla2}'

Writing a, b, c for a 1, a2, a 3 respectively, the corresponding simple inequality in the language of 0 is

and is equivalent in X to the universally quantified disjunction

(**) Example 5.3 Let E

= {(2, In.

Then A

= {",2 a , ",a} and B = 0. We have

which is equivalent to (Vx)

(**)

g2(X) ~ g(x).

Urquhart's theorem applies only to simple inequalities. Fortunately, however, as shown by Urquhart [95, Theorem 2.4], every basic inequality is equivalent to a simple inequality. Hence, if we are given a basic inequality we are able to find its equivalent in the dual space.

Example 5.4 Consider the generalised Kleene identity (K)

a

1\

rva 1\ ",2a ~ b V rvb V ",2b.

80

Ockham algebras

This is not simple, but is easily seen to be equivalent to (K')

a 1\ rva 1\ rvC 1\ rv2c::,;; b V rvb V rvd V rv 2d

which is simple and corresponds to the set E =

{(O, 1), (1,0), (2, 1), (1, 2)}.

In fact, it is clear on taking c =a and d = b that (K') implies (K). To obtain the reverse implication, write a V c for a and b 1\ d for b in (K) to obtain (a V c) 1\ rva 1\ rvC 1\ ( rv2 a V rv2c)::,;; (b 1\ d) V rvb V rvd V (rv2b 1\ rv2d),

which clearly implies (K'). The corresponding universally quantified disjunction in X is (\Ix)

x;.;:: g(x)

V g(x);';:: x V g2(x);.;:: g(x) V g(x);.;:: g2(X).

Since x Hg(x) implies g(x) Hg2(X), the equivalent of (K) is (\Ix)

g(x) Hg2(X).

It follows from this that (K) can be simplified, and indeed be replaced by

rva 1\ rv2a::,;; rvb V rv 2b.

(K")

Every equality can be considered as a conjunction of two inequalities, so the procedure can be applied to them also.

Example 5.5 The equality a V rvb V rv 2b = rv 2a V rvb V rv 2b

is the conjunction of the inequalities a V rvb V rv2b::';; rv2a V rvb V A}b ~ a ~ rv2a V rvb V rv 2b

(1)

and a V rvb V A}b;.;:: rv 2a V rvb V rv 2b ~ rv2a ~ a V rvb V rv 2b

Both (1) and (2) are simple and are respectively equivalent to (\Ix) (\Ix)

x;.;:: g2(X) g2(X);';:: X

V g(x);.;:: g2(X)j V g(x);.;:: g2(X).

The conjunction of these two disjunctions is (\Ix)

x

= g2(X) V g(x);.;:: g2(X).

(2)

The lattice of subvarieties

81

Remark For both visual and typographical reasons, we shall henceforth commit an abuse of notation. Specifically, instead of writing the universally quantified disjunction

V{gPt (x) ;:: gqj (x) I (pj, qj) E E}

(Vx)

j

we shall write simply

V{gPi;:: gqi I (pj,qj) E E}. j

Thus, for instance, in Example 5.5 we can write x;:: g2(X) (Vx) g2(X);:: X (Vx)

V g(x);:: g2(X)j V g(x);:: g2(X).

in the simpler form gO;:: g2 g2 ;:: gO

V g;:: g2j V g;:: g2 ,

the conjunction of which is gO

=g2 V g;:: g2 .

In what follows, the reader should bear in mind this abuse of notation. The 'translation' given by Urquhart's theorem can be done in the opposite direction, in the sense that if we are given the g-inequalities we can find the corresponding rv-inequality.

Example 5.6 Consider (with the above caveat on notation) the disjunction gO

= g2 V gO ;:: g.

This can be written in the form (go;:: g2

V gO;:: g) 1\ (g2 ;:: gO V gO;:: g),

and gives the simple inequalities

a /\ b /\ rvb ~ rv 2a ~ a /\ b /\ rvb ~ rv2a /\ b /\ rvb, rv2a /\ b /\ rvb ~ a ~ rv 2a /\ b /\ rvb ~ a /\ b /\ rvb. The conjunction of these gives the equality

a /\ b /\ rvb

= rv 2a /\ b /\ rvb.

With every simple inequality A ~ B having 2n atomic terms we shall associate a rectangular array having n rows and 4 columns. Reading from left to right, these columns contain the evenp's of A, the odd q's of A, the oddp's of B, the even q's of B.

Ockham algebras

82

Each row of the array has only two entries, those of the i-th row being Pi and q i, and in that order except when both are odd, in which case the order is reversed. We shall call such a rectangular array a tabulation and denote its elements by CY.ij' By way of illustration, corresponding to Examples 5.2-5.6 above we have the following tabulations :

o

2 1 2

2

o

2 1

-I

o

1

1

0

2 1 1 2

o

2

and

1 2

T56

:

I~

2

and

1

2 -

- 0 1 2

2 -

-

o

0

1

Let T be a tabulation with n rows. If we delete r of these rows (with 0:;;; r:;;; n -1) we obtain a new tabulation which we call a subtabulation of T. In this way we can form 2n -1 subtabulations of T. If T' is a subtabulation of T then the inequality associated with T' implies that associated with T. We shall say that two tabulations are equivalent if they correspond to equivalent inequalities. For example, if we permute two rows of a tabulation T then we obtain an eqUivalent tabulation. A tabulation will be called reducible if it is equivalent to another tabulation with fewer rows. For example, the tabulation T 5 4 is reducible; in fact, because gO Mg implies g2 Mg, it is equivalent to the tabulation

I= ~

~

;I

A non-reducible tabulation will be called irreducible. Note that if an irre-

ducible tabulation contains the ordered pair (m, n) as one of its rows then it

The lattice of subvarieties

83

cannot contain either (m - 2 k , n - 2 k) for any k ~ 1 or (n - 2k -1, m - 2k -1) for any k ~ O. This is a direct consequence of Urquhart's theorem since each of gm-2k ~ gn-2k and gll-2k-l ~ gm-2k-l implies gm ~ gil. A tabulation is non-trivial if it is such that

(Vi) Cl!il t= Cl!i4, Cl!i2 f. Cl!i3 OthelWise, the corresponding inequality would be trivial, in the sense that it would contain a term of the form ",Ila on each side. The dual (nd) of an inequality (n) is obtained by replacing 1\, V, ~ ,0,1 respectively by V, 1\, ~ , 1 , O. The dual of a tabulation T is the tabulation T' obtained from T by rewriting T from right to left; e.g. the dual of T 5 2 is

and represents the inequality ",2a 1\ rvb

2 2 1

0

o

2

1\ ",2b 1\

c~a

V

rv 2

c.

If a tabulation represents a particular inequality then it is clear that the dual tabulation represents the dual inequality. The notions of self-dual tabulations and inequalities are obvious. An example is provided by T 54 and the equality it represents. A simple inequality has the advantage of being suitable for a direct application of Urquhart's theorem. But it has the disadvantage of involving too many atomic terms and variables. The aim of the following result is to remedy this situation.

Theorem 5.3 Let T be the tabulation corresponding to a simple inequality I. if T has n rows and if the same entry appears r times (r ~ 2) in the same column then I can be reduced to an equivalent inequality having n - r + 1 variables and 2n - r + 1 atomic terms. Proof Suppose that the same entry appears r times in the first column of T. Without loss of generality, we may assume that this occurs in the first r rows of T, so that T has the form

84

Ockham algebras

with PI

(1)

=P2 = ... =Pr'

,JJI al

The simple inequality I is then of the form

/\ ",ql al /\ ,JJI a2 /\ .. /\ ,JJI a r

/\

A

~

",qZa2 V ... V ",qr a r VB.

We claim that this is equivalent to (2)

,JJ1al/\ ", q1 a l /\A

~ ",qZal

v· . V ",qral VB.

In fact, if in (1) we replace a2, ... , a r by al then we obtain (2). Conversely, if in (2) we write a l /\ a2 /\ ... /\ a r for al then we obtain J1al/\ J1a2/\" /\ J1ar /\ (", q1 a l V ", q1 a2 V··· V ", q1 a r ) /\A ~ (",qZal /\ ",qZa2 /\ ... /\ ",qZa r ) V··· V (",qral /\ rvq, a2 /\ ... /\ rvqra r ) VB.

Since the first member of the latter inequality is greater than or equal to the first member of (1) and the second member of the latter inequality is les$ than or equal to the second member of (1), the part of T under the r-th row remaining unaltered, we thus see that (2) implies (1). The same procedure can be applied if an entry appears r times in flle third column. As for columns 2 and 4, the only change required in the above proof is the substitution al V a2 V··· V a r for al'

Example 5.7 Corresponding to the set

E = {(O,1),(2,1),(O,2),(3,1),(1,5)} we have the tabulation 0 2 0

1 1

-

1 5

-

2 3 1 -

and the simple inequality a /\ rva /\ ",2b /\ rvb /\ c /\ ",d /\ ",5 e ~ r,}C V ",3d V ",e.

Applying Theorem 5.3 to Q!ll = Q!3I on the one hand, and to the other, we obtain the simplified inequality a /\ "'a /\ ",2 b /\ ",b /\ ",5 e ~

",2 a V ",3 b V

",3 a V ",2 C V

= Q!42

on

",e,

which has 3 variables and 8 atomic terms. Alternatively, we could take advantage of the fact that This yields the inequality a /\ "'a /\ ",2 a /\ c /\ ",5 e ~

Q!22

Q!l2

",e,

= Q!22 = Q!42'

85

Tbe lattice oj subvarieties which has the same degree of complexity as the previous simplification.

From now on we shall focus our attention on the class P3,1 and describe all of its subclasses. In P 3,1 there are only 6 possible ordered pairs (p,q) involved in the set E, namely (0,1), (1,0), (0, 2), (2, 0), (1,2), (2, 1). The biggest tabulation possible in this case is therefore

o

1

1

o 2 1 2

0 2 0 2

1

We observe immediately that this tabulation is reducible (since gO Hg implies g2 Hg). Its irreducible subtabulations will yield all of the desired inequalities.

To obtain these, we use the following observations. • There are 6 subtabulations consisting of a single row, but the tabulations

I2

I

and

/\ ",a = 0

*'*

1 -

-

I- -

1

2

I

give in P3 ,1 the same axiom: ",2 a

",a V

",2 a

= 1.

• There are 15 subtabulations consisting of two rows, but we must exclude two of them, namely those that contain (0,1) and (2,1), or (1,0) and (1,2), which are clearly reducible. • There are 20 subtabulations consisting of three rows, 8 of which are reducible. • There are 15 subtabulations consisting of four rows, 11 of which are reducible. • All subtabulations having more than four rows are clearly reducible. It follows from these observations that in P 3,1 we can define 34 nonequivalent axioms, and that these axioms involve at most 3 variables. A complete list of these is given below, including the corresponding tabulations and the dual space eqUivalents. We begin with those that are self-dual, for which we use the Greek letters (a), . .. ,(7)). The 14 axioms (i) that follow, together with their duals (i d ), complete the list. Note that the equality (a), which is the conjunction of (1) and (l d ), is added for sake of completeness.

86

Ockbam algebras

I tabulation -

I dual equivalent

I axiom (O!) a = ",,2a

gO = g2

-I

(f') ""a 1\ ",,2a = 0

g = g2 gHgO

12

1

I~

1 - 1

~I

h)

1

;1

(8) ""a 1\ ",,2a

~I

(E) a 1\ ",,2b ~ ",,2a vb

1=

1 -

I~ 0

~ ""b V ",,2b

g2Hg

g2H g O

1 1

0 2 0 2 1

2

a 1\ ""a ~ b V ""b

0 2 0

gHgO

V g2 HgO

2 0 2

(71) ",a 1\ ",,2a 1\ b ~ a V ""b V ",2b g2 Hg

V g2 HgO

21

(I) a

(()

a 1\ ""a 1\ ",,2b ~ ",2a V b V ",b

1

10

~ ",2a

=1

gO ~g2 g~gO

1- -

1

01

(2) a V ""a

I~

-

1

~I

(3)

a ~ ""a V",,2a

gO ~g2

V g ~g2

I~

1

-

=1

(4)

a 1\ ""a ~ ",,2a

gO ~g2

V gO ~g

I~

1 -

1

;1

(5)

al\""a~""bV",,2b

gO ~g

I~

1

-

=1

(6)

a 1\ ""b 1\ ",,2b ~ ",,2a

gO ~g2

V g ~g2 V g2 ~g

The lattice of subvarieties

I~

-

1

~I

0 2

2 0 1 2

0

2 1 0

87

V gO ~g2

(7)

a q, t], we deduce that'P = tI>q V 8.

= (tI>q

V 8)I/q(L)'

Suppose now that L E Kw. Let (x,y) E 'P and let A be the subalgebra generated by {x,y, 01*, OI*} U Fix L. Then by Theorem 3.6 applied first to the elements x, y we see that A E Kp,q for some p, q. It follows by the above that (x,y) E 'PIA = tI>qlA V 81A, whence (x,y) E tI>w V 8. Thus 'P:::;; tI>w V 8 whence we have equality. I:)

Ockham algebras

122

Note, in fact, that in lbeorems 7.5, 7.6, 7.7 only the existence of Q!* and Q!* is used; we do not require the properties f(Q!*) = Q!* and f(Q!*) = Q!*.

Theorem 7.8 If (L;J) E Kw is fixed point complete and fixed point distributive then the following statements are equivalent:

(1) 'P=cI>w; (2) VQ!j=l; jEI

(3) 0=w.

Proof (1) =* (2) follows from lbeorem 7.4. (2) =* (3) : If (2) holds then A Q!j = 0 and clearly 0 = w. jEI

(3) =* (1) follows from Theorem 7.7·0 It is of course possible to have 'P = 0. By Theorem 7.7, this occurs precisely when cI>w ~ 0. An example of this situation is the following.

Example 7.4 Consider the lattice

y

made into an Ockham algebra by defining

f(x)=f(Y)=z,

f(z)=y,

(Vi) f(Q!j)=Q!j,

f(Q!*)=Q!*,

f(Q!*)=Q!*

and extending to the whole of L. Then L is fixed point complete and fixed point distributive. Here we have 'P = 0, the classes being indicated by thick

Fixed point separating congroences

123

lines. The cpw-classes are the interval [x,y], the intervals 'parallel' to it, and singletons otherwise. Consider now the congruence

r = 19{0I*,0I*). If L is fixed point complete then by Theorem 2.1 we have r = 191at{0I*,0I*).

Theorem 7.9 If L E Kw is fixed point complete then r is the complement of

e

in ConL.

Proof Since (O,OI*) E e, (OI*,OI*) E r, and (OI*, 1) E e, we have evr=L. Now for all x,y

EL

we have

191at{X,y) A r = 191at{X,y) A 191at{0I*, 01*) = 191at{{X V 01*) Ay A 01*, Y A OI*}. It follows immediately that (x,y) E e => 191at{X,y) A r

Since

e

= w.

is a congruence it follows by Theorem 2.1 that (x,y) E e => 19{x,y) A r

=w

and therefore that

eAr =

V

19{x,y) A r

(x,y}EE>

= V

(x,y}EE>

(19{x,y) A r}

= w.

r is the complement of e. Corollary e = 19{0, 01*) = 19{0I*, 1). Proof The principal congruence r = 191at{0I*, 01*) is complemented. plement is 191at{0,0I*) V 191at{0I*, 1) = 19{0,0I*) = 19{0I*, 1). Consequently,

Its com-

Theorem 7.10 If L E Kw is fixed point complete and fixed point distributive then Con L contains the sublattice

Ockham algebras

124 Proof By Theorems 7.7 and 7.9 we have that

'II /\ r

= (w V e) /\ r = w /\ r,

whence it follows that

(r V w) /\ 'II = (r /\ 'II) V (w /\ 'P) = (r /\ w) V (w /\ 'II) = w /\ (r V'll)

= w /\ £ = w, and that (w /\ r) V (w /\ e)

= w /\ (r V e) = w /\ £ = w.

We then have the sublattice illustrated.



Corollary 1 [w, £] ~ [w, 'II] x ['II, £]. Proof'll is a complemented element of [w, £]. Corollary 2 The interoal [w, £] is boolean [w, 'P] and ['II, £] are boolean.



if and only if both the interoals

Definition We shall say that L is fixed point compact if it is fixed point complete and there is a finite subset /* of / such that O!* O!* =

= A O!t

(equivalently,

tEl*

V O!t)· tEl*

If L is fixed point compact then necessarily L is fixed point distributive. In fact,

x /\

V O!t = x /\ V O!t = V (x /\ O!t) ~ V (x /\ O!t),

tEl

tEl*

tEl*

lEI

and the reverse inequality is trivial.

Theorem 7.11 If (L;1)

q

ve V CP*)IM'

Arguing as in the proof of Theorem 7.7, we deduce that cp = cI>q

ve V cp* ~ 'I' V cp*,

whence we have equality and so the result holds for L E Kp,q' Suppose now that L E Kw. Let (x, y) E cp ~ 'I' and consider the subalgebra B generated by {x, y, 01*, OI*} U Fix L. We have B E Kp,q for some p, q so, by the above, (x,y) E CPIB = 'l'IB V CP*IB ~ ('I' V CP*)IB whence (x,y) E 'I' V cp* and hence cp ~ 'I' V cp*. The reverse inequality is trivial. As the following example shows, Theorem 7.11 does not hold in general if L is not fixed point compact.

Example 7.5 Consider the Ockham algebra obtained by adding to the angel fish of Example 7.3 a fixed point 010 as a complement of 011 in the interval

126

Ockham algebras

[x,X]. We obtain an Ockham algebra that is fixed point complete and fixed point distributive, but not fixed point compact. In this we have 'P = wand the congruence V {)(OI.i' OI.j) has three classes, namely {OJ, {I}, and L\ {O, I}. iJ

The equality of Theorem 7.11 therefore fails for

I(J

= t.

We now give an example of a fixed point compact de Morgan algebra in which the interval [- ~ [wIB, eIB], suppose now that {) E [wIB, elB]. Then there exists 10 E Con L with cp IB = {). Let""J = 10 /\ '1'. Then we have >- (""J) = ""JIB

= 10 IB /\ 'I'IB = {) /\ e IB = {).

We thus see that >- induces an isomorphism [cI>w, '1'] rem 7.9 gives

~

[wIB, eIB]' But Theo-

Hence the result holds for L E M. Suppose now that L E Kw. We have [cI>w, t] ~ Con LjcI>w ~ Con T(L) ~ Con T 2 (L)

For 10

~

= [cI>wIT2(l), tIT2(L)]'

cI>w the correspondence 10 ~ IOI T2(L) is therefore a bijection. Hence [cI>w, '1'] ~ [cI>w!rz(L), 'I'!rz(L)1.

Now since the result holds for T 2 (L) E M we have [cI>wIT2(L), 'I'I T2(L)] ~ Con T 2 (C)jr

where C = T 2 (L) n B. Since T 2 (C) follows. 0 Corollary [cI>w, '1'] is boolean

= T2 (B),

the result for L E Kw now

if and only if T 2 (B) isjinite.

Proof From the above, [cI>w, '1'] is boolean if and only if T 2 (B) has finitely many r-classes. It is clear that this is so if and only if T 2 (B) is finite. 0

Finally, we give an example of a Kw-algebra that is fixed point compact, with the congruences cI>w, '1', e, r distinct and the interval [cI>w, t] boolean.

Example 7.9 Consider the lattice

132

Ockham algebras

OIO=XO

o made into a Kw -algebra by defining

!(o) =!(f3) = 1, !(1)=0, !(xo)=xo, (Vi ~ 1) !(xu)

= XZi-2,

!(XZi+l)

!(Xl) = Xl,

= XZi-l,

and extending to the whole of L. Then L, though not a complete lattice, is fixed point compact. It can readily be seen that here the congruences w, 'P, a, r are all distinct. In fact, if we denote the four parts of [O!*, O!*] by the cardinal points W, E, N, S then the w-classes are W,E,N,S,{O,,8},{l}; the 'P-classes are W, E, N U {I}, S U {a, ,8}; the a-classes are {O!*, I}, {0,,8, O!*}, singletons otherwise; the r -classes are [O!*, O!*], singletons otherwise. Here [w, L] is the four-element boolean lattice.

8 Congruences on Kl,l-algebras

We now take a close look at congruences on a KI,1 -algebra L. We begin by characterising the principal congruences. Observe that in any lattice every congruence cp satisfies

(19lat(a, b) V cp)/cp = 19lat([a]cp, [b]cp) (see, for example, [12, page 137]) and consequently (x,y) E 191at(a, b) V cp

Theorem 8.1 If (L;~)

E KI,1

¢=>

([x], [y]) E 191at([a], [b])

in L/cp.

and a, bEL with a ~ b then (x,y) E 19(a, b)

if and only if (1) x /\ a /\ rv 2a /\ rvb =y /\ a /\ rv 2a /\ rvb; (2) (x /\ a /\ rvb) V rv 2b = (y /\ a /\ rvb) V rv2b; (3) [(x /\ a) V rua] /\ ru2a = [(y /\ a) V rva] /\ rv2a; (4) (x /\ a) V rva V rv 2b = (y /\ a) V rva V rv2b; (5) (x V b) /\ rvb /\ rv 2a = (y V b) /\ rvb /\ rv2a; (6) [(x V b) /\ rub] V rv 2b = [(y V b) /\ rub] V rv2b; (7) (x V b V rua) /\ rv 2a = (y V b V rva) /\ rv2a; (8) x V rva V b V rv 2b = y V rva V b V rv 2b. Proof By Theorem 2.1 we have 19(a, b)

= 19lat(a, b) V 19lat(rvb, rva) V 191at(rv2a, rv2b).

Denote by 1/; the lattice congruence 19lat(rvb, rva) V 191at(rv2a, rv2b). Then

(x,y) E 19(a, b)

¢=> ¢=> ¢=>

([x], [y]) E 191at([a], [b]) in LN [x] /\ [a] = [y] /\ [a], [x] V [b] = [y] V [b] (x /\ a,y /\ a) E 1/;, (x V b,y V b) E 1/;.

in LN

Now we have

(x /\ a,y /\ a) E 1/;

¢=> ¢=>

¢=> ¢=>

([x /\ a], [y /\ a]) E 191at(rvb, rva) in LNlat(rv2a, rv2b) {[X /\ a] /\ [rvb] = [y /\ a] /\ [rvb], [x /\ a] V [rva] = [y /\ a] V [rva] (x /\ a /\ rvb,y /\ a /\ rvb) E 19lat(rv2a, rv2b), { ((x /\ a) V rva, (y /\ a) V rva) E 191at(rv2a, rv2b) (1) ---+ (4) hold in L.

Ockham algebras

134 Similarly, we have (X V b ,y V b) E ./, 'f/

-{===}

-{===}

Thus (x,y)

E

{((X V b) 1\ rvb, (y V b) 1\ rvb) E 191at (rv2 a, rv2b), (x V b V rva,y V b V rva) E 191at(rv2a, rv2b) (5) ~ (8) hold in L.

19(a, b) if and only (1) ~ (8) hold. 0

Since (2) and (4) hold trivially if (L; 0) EMS, we have: Corollary 1 [36] Jf(L; 0) E MS anda, bEL with a:;;;;; b then (x,y) E 19(a, b) if and only if

(I') (3') (5') (6') (7') (8')

x 1\ a 1\ bO = Y 1\ a 1\ be; (x 1\ a) V (ao 1\ aCe) = (y 1\ a) V (aO 1\ aCe); (x V b) 1\ bO 1\ aOO = (y V b) 1\ bo 1\ aCe; (x 1\ be) V boO = (y 1\ be) V bOo; (x V b V aO) 1\ aOO = (y V b V aO) 1\ aCe; x V aO V bOO =y V aO V bOo. 0

Since (5') and (7') hold trivially if (L; -) E M, we have: Corollary 2 [84] Jf (L; -) E M and a, bEL with a:;;;;; b then (x,y) E 19(a, b) if and only if

(I") x 1\ a 1\ b = Y 1\ a 1\ b; (3") (x va) 1\ a = (y va) 1\ a; (6") (x 1\ b) V b = (y 1\ b) V b; (8") xvavb=yvavb. 0 Corollary 3 [74] Jf (L; *) E S and a, bEL with a:;;;;; b then (x,y) E 19(a, b) if and only if (i) x 1\ a = y 1\ a; (ii) (x V b) 1\ (a** V b*)

= (y V b) 1\ (a** V b*).

* for ° in the equalities of Corollary 1, we see that (I') and (5') hold trivially, (3') gives (i), and (6') implies (8'). As (6') implies that x 1\ b* = Y 1\ b*, (ii) follows from (6') and (7'). Conversely, taking the supremum with b** and the infimum with a** of both sides of (ii), we obtain (6') and (7') respectively. 0

Proof Writing

Corollary 4 Jf (L;') E B and a, bEL with a:;;;;; b then 19(a, b) = 191at(a, b). Proof Writing' for * in (ii) of Corollary 3 and using distributivity, we obtain xV b =yV b. 0

135

Congruences on K1,1-algebras

Theorem 8.2 Jf(L; rv) E K1,1 and a, bEL with a ~ b and a I\rva then

= bl\rvb

Proof Note that al\rva = bl\rvb gives rv2al\rva == rv 2bl\rvb and rvaV rv2 a = rvbVrv 2b. Hence 191at(rvb, rva) = 191at(rv2a, rv 2b) and the first equality follows. Now (a, b) E 19(a, b) and (rvb, rva) E 19(a, b), so (avrvb, bVrva) E 19(a, b) and consequently 191at(a V rvb, b V rva) ~ 19(a, b).

But since b 1\ rvb ~ a ~ b we have

a 1\ (a V rvb) = a = (b 1\ a) V (b 1\ rvb) == b 1\ (a V rvb); a V b V rva == b V rva == b V b V rva, which shows that (a, b) E 191at(a V rvb, b V rva). Since a 1\ rva ~ rvb ~ rva we have likewise that (rvb, rva) E 191at(a V rvb, b V rva). Hence 19(a, b) == 191at(a, b) V 191at(rvb, rva) ~ 191at(a V rvb, b V rva).

~

Theorem 8.3 Tbe class S is the largest subvariety ofMS in which every principal congruence is a principal lattice congruence. Proof It follows immediately from Theorem 8.2 that if (L; *) E S then 19(a, b) == 191at(a V b*, b V a*), a fact that was first observed in [74]. To complete the proof, we need therefore only exhibit an algebra in K in which not every principal congruence is a principal lattice congruence. For this purpose, consider the four-element chain 0< a < b < 1 with rvO == 1, rva = b, rvb = a, rvl = O. Here

19(0, a) == {{O, a}, {b, I}} is not a principal lattice congruence. ~ We shall now consider the question of when a principal congruence 19(a, b) is complemented in Con L. For this purpose, we first concentrate on the case where L E M. Here the situation is described by the following results of Sankappanavar [84].

Theorem 8.4 Let (L, -) E M and let a, b, c, dEL be such that a c~d. Tben 19(a, b) 1\ 19(c, d) = 19(a V c, a V c V (b

band

1\ d)) V 19(a V d, a V d V (b 1\ c)).

Proof Using the formula 191at(a, b) 1\ 191at(C, d)

~

= 191at((a V c) 1\ b 1\ d, b 1\ d)

Ockham algebras

136

and the fact that '!91at {X /\y, x) = '!91 at {Y, x Vy) we have, by Theorem 2.1,

'!9{a, b) /\ '!9{c,d) = ['!91at {a, b) V '!91at(b, a)] /\ ['!91at {c,d) V '!91at {d, c)] =['!91at {a, b) /\ '!91at {c,d)] V ['!91at {a, b) /\ '!91at {d, c)] V ['!91alb, a) /\ '!91at {c, d)] V ['!91at {b, a) /\ '!91at {d, c)] = '!91at({a V c) /\ b /\ d, b /\ d) V '!91at({a V d) /\ b /\ c, b /\ c) V '!91at((tj" V c) /\ a /\ d, a /\ d) V '!91at((lj" V d) /\ a /\ c, a /\ c) = '!91at(a V c,a V c V (b /\d)) V '!91at (a vd,a vdv (b /\ c)) V '!91at({b V c) /\ a /\ d, a /\ d) V '!91at({b V d) /\ a /\ c, a /\ c) = '!9(a V c,a V c V (b /\d)) V '!9(a vd,a vdv (b /\ c)). 0 Theorem 8.5 Every principal congruence on (L, -) E M is complemented. For a, bEL with a :::;; b we have

'!9{a, b)'

= '!9{a vb, 1) V '!9{b /\ b, b) V '!9{a, a va).

Proof Consider the congruence

cp That '!9{a, b) V cp =

= '!9{a Vb, 1) V '!9{b /\ b, b) V '!9{a, a va). ~

follows from the observations

(O, b /\ a)

'!9{a vb, 1), (b /\ a, b /\ b) E '!9{a, b), (b /\ b, b) E '!9{b /\ b, b), (b, a) E '!9{a, b), (a,av a) E '!9{a,av a), (aV a,av b) E '!9{a, b), (av b, 1) E '!9{av b, 1). E

That '!9{a, b)/\cp = w follows from a routine application of Theorem 8.4 which we leave to the reader. Hence we have '!9{a, b)' = cpo 0 The above results provide the following characterisation of the class M of de Morgan algebras.

Theorem 8.6 !be class :tt,I is the largest subvariety of KI,1 in which every principal congruence is complemented. Proof By Theorem 8.5, every principal congruence on a de Morgan algebra is complemented. To complete the proof it therefore suffices to exhibit algebras in S, S, K I , KI in which the property fails. For this purpose, consider the subdirectly irreducible algebras S, 5, K I, K I. We have Con S ~ Con 5 ~ Con K 1 ~ Con K I ~ 3, the non-complemented element in each case being the principal congruence . Every lattice congruence that is contained in

Congruences on K 1 ,1 -algebras

141

ep is a congruence; so, as observed in Chapter 2, when L is finite the interval [w, ep] of Con L is boolean. Furthermore, [ep, L]

~

Con Ljep

~

Con ",L

so, by Theorem 8.12, [ep, L] is boolean if and only if ",L is finite. A further important property of the congruence ep is the following.

Theorem 8.14 Let L

E K1 ,1'

Tben ep is the greatest dually dense element of

ConL.

Proof For every .,J E Con L we have that (O,l)E.,JVep It follows that .,J V ep =

~

(O,l)E.,J.

implies .,J = L, and so ep is dually dense. Since the dually dense elements of a bounded lattice form an ideal, it remains to prove that if .,J > ep then.,J is not dually dense. Since the interval [ep, L] ~ Con ",L is an algebraic lattice, .,J is a join of compact elements. One of these, say cp, has to be distinct from ep. By Theorem 8.13, cp is complemented in [ep, L], with complement cpl say. Then we have .,J V cpl = L with cpl t- L, whence .,J is not dually dense. L

We can now extend Theorem 8.12 to the whole of K 1,1'

Theorem 8.15 If L E K 1 ,1 then Con L is boolean

if and only if L

is afinite

de Morgan algebra.

Proof It suffices to prove that if L E K 1 ,1 is such that Con L is boolean then necessarily L E M. But if L ~ M then ep > wand, by Theorem 8.14, ep is dually dense. Hence ep has no complement, a contradiction. We shall now proceed to consider the following question. Given a K1,1algebra L and a subvariety R of Kl ,1, what is the least congruence .,J for which L j.,J E R? Put another way, what is the greatest homomorphic image of L that belongs to R? Here we consider this question for RE {B,K,M,S}, the least such congruence being denoted by.,JR'

Theorem 8.16 .,JM = ep Proof Clearly, L/ep EM. Conversely, if LN E M then (a, b) E ep

whence ep

=> [a].,J = ",2([a].,J) = [",2a].,J = [",2b].,J = ",2 ([b].,J) = [b].,J

~.,J.



If we denote by .,J~ any congruence on L such that L N~ E R with L N~ subdirectly irreducible then, as proved in [99], we have .,JR = !\{.,J~}.

142

Ockham algebras

Thus, if R has finitely many subdirectly irreducible algebras, say R 1, ... , R n, and if {}Ri is any congruence such that Lj{)Ri ~ R j then 11

{}R =

1\ {}Ri'

i=1

Denoting by B, K, S, M the subdirectly irreducible algebras in B, K, S, M respectively, we therefore deduce the following result.

Theorem 8.17 ForeveryL EMS, (1) {}B = I\{{}B}; (2) {}K = 1\ {{} B, {} K}; (3) {}s = I\{ {}B, {}s}; (4) {}M = I\{{}B,{}K,{}M}'



Corollary The coatoms of Con L are of the form {}B, {} K, {}M, and their intersection is . We shall now use the above results in considering the possible existence of a non-trivial node in Con L, i.e. a congruence {} ¢ {w, L} that is comparable with all elements of Con L. For this purpose, we shall say that a lattice is local if it contains a unique coatom. This terminology is borrowed from ring theory: a local ring is a commutative ring having a unique maximal ideal. We shall say that a congruence has a trivial kernel if its kernel reduces to {O}. Theorem 8.18 For a K 1 ,1 -algebra (L, rv) the following statements are equivalent: (1) Con L is a local lattice ; (2) is maximal in Con L; (3) Con L has comonolith ; (4) the de Morgan algebra L / is simple. If, moreover, (L, rv) E MS then the conditions (5) every congruence on L, other than L, has a trivial kernel; (6) (\lxEL\{O}) rvx (1) : Observe that if F is a family of congruences on L each of which has a trivial kernel then so does sup F in the complete lattice Con L. In fact, if (0, x) E sup F then there exist a o,' .. ,an ELand 19 1 , ... ,19 11 +1 E F such that 019 1 a1 19 2 a2 19 3 ... 19 11 - 1 a ll -1 19 11 all 19 11 +1 x.

°

°

Since 19 1 has a trivial kernel, a1 = and so, since 19 2 has trivial kernel, a2 = and so on. We deduce in this way that x = and hence that sup F has trivial kernel. If now (5) holds, choose F = Con L \ {t}. Then by the above we have that sup F t= L and hence, by its definition, sup F is the unique coatom of

°

ConL.

(5)

#

(6): Observe first that (5) is equivalent to the condition

x

°

t=

° =>

19(O,x) = L.

Now for x t= we have 19(O,x) = L if and only if (0,1) E 19(O,x) and by Theorem 8.1 this is the case if and only if

(x

1\ "'x) V ",2 X

= ",x V ",2 x,

Le. if and only if

",x V ",2X ~ X

V ",2X.

Applying rv to this, we see that it is equivalent to ",x ~ ",2x. Consequently, (5) and (6) are equivalent. ~

If L E MS then Con L has at most one non-trivial node. When such exists, it is necessarily and is covered by L.

Corollary

Proof If Con L has a non-trivial node tp then the centre of Con L reduces to {w, L}. Since, by Corollary 2 of Theorem 8.10, every principal congruence 19(O,x) is complemented, we must have 19(O,x) = L for all x t= 0. It follows that every congruence tp t= L has a trivial kernel; for otherwise we have (O,x) E tp for some x t= 0, whence the contradiction L = 19(0, x) ~ tp. Thus condition (5) above holds. Consequently, is a non-trivial node of Con L which is covered by L. Suppose that tp < . Every lattice congruence contained in is a congruence, so 1 is a principal ideal of COnlat L. Since this lattice is algebraic, we can find compact congruences 19 1 ,19 2 such that

w < 19 1 ~ tp ~ 19 2 ~ . But for any distributive lattice it is known [15] that the compact elements form a relatively complemented sublattice. Since dearly 19 1 cannot have a complement in [w, 19 2 ], it follows that we must have tp = . ~

144

Ockham algebras

It goes without saying that rvL is the most significant subalgebra of any

Kl,l-algebra. Since the class Kl,l enjoys the congruence extension property, it is quite natural to consider on the one hand the restriction 191~L to rvL of any congruence 19 E Con L, and on the other the smallest extension V5 to Con L of

.

Proof If (rv2 x, rv 2y) E 19 then (rv2 x, rv 2y) E 19 vet>. As (x, rv 2x) E et> ~ 19 Vet> for every x E L, it follows that (x, y) E 19 Vet>. Conversely, if (x,y) E 19 V et> then (rv2 x, rv 2y) E 19 vet> whence ",2X = Xo 19 Xl et> X2 19 ... 19 Xn-l et> Xn = rv 2y, hence i.e.

Thus we have rv 2X 19 ",2y. 0 The following property is now immediate :

1heorem 8.20 If L E Kl,l and 19 1 ,19 2 E Con L then thefollowing statements are equivalent: (1) 19l1~L

= 1921~L;

(2) 19 1 isanextensionof1921~L; (2') 19 2 is an extension of19ll~L; (3) rv 2a 19 1 rv 2b ~ rv 2a 19 2 rv2b; (4) 19 1 Vet> = 19 2 Vet>. 0

1heorem 8.21 Let L E K1 ,1 and let 19 E Con L with 19 ~ et>. Then is an extension of 191~L if and only if CI! V et> = 19.

CI!

E Con L

Proof This is immediate from Theorem 8.20.0 Corollary Let 19 E Con L with 19 ~ et>. Then the dual pseudocomplement of et> in 19 1 exists and is the least extension to L of 191~L' Proof Since the meet of any family of extensions of 191~L to L is an extension of 191~L' the existence of a least extension is clear. By Theorem 8.21, the least extension of 191~L to L is the smallest CI! such that CI! vet> = 19. 0

Congruences on K1,1-algebras

145

Theorem 8.22 If L E K 1,1 and cp E Con ",L then the smallest lattice congruence on L that extends cp is given by cp :::::

191at (",2 x , ",2y).

V (~2x,~2Y)EI"

Moreover, 7(5 E Con L. Proof It is clear that 7(5 E Conlat L and extends cp. If now 19 E Conlat L extends cp then (",2 x, ",2 y) E cp gives 191at (",2 x, ",2y) ~ {) and so 7(5 ~ 19. To prove that 7(5 E Con L, let (a, b) E 7(5. Then there exist Xl, ... ,X n E L such that

where each serving that

=i

is of the form 19 1at (",2 x , ",2y) for some (",Zx, ",Zy) E cpo Ob-

(P,q)E{)lat(",2 X,,,,2y )

'*

(",p,,,,q)E19 1at (,,,y,,,,x)

and that (",2 X, ",2 y) E cp gives (",y, "'x) E cp, we deduce from

in which, if =i is 19 1at (",2x , ",2y) then Thus we see that 7(5 E Con L.

=1 is {)lat("'Y, ",x), that (",a, "'b) E cpo

Theorem 8.23 If L E K 1 ,1 then the mapping f: Con ",L by f(cp)::::: 7(5 is a lattice morphism.

-t

Con L described

Proof Clearly, on the one hand, we have 7(5 V1f ~ cp V 1/J. On the other hand, (7(5 V 1f)I~L ~ cp, 1/J and so (7(5 V 1f)I~L ~ cp V 1/J whence 7(5 V 1f ~ cp V 1/J. Thus f

is a V-morphism. To show that f is also a I\-morphism, observe that clearly cp 1\ 1/J ~ 7(51\ 1f. Now, by Theorem 8.22, we have 7(51\1/J:::::

V

19 1at (",Zx,"'zy) 1\

(~2x,~2Y)EI"

V

19 1at (",2 a , ",Zb)

(~2a,~2b)E.p

and so, since Con L is a complete distributive lattice in which the infinite distributive law 19 1\ V (i ::::: V(19 1\ (i) holds, in order to obtain the reverse i

i

inequality cp 1\ 1/J ~ 7(5 1\ 1f it suffices to prove that, for all (",2 X, ",Zy) E cp and all (",Za, ",Zb) E 1/J, 19 1at (",2 x, ",2 y) 1\ {)lat( ",2 a, ",2 b) ~ cp 1\ 1/J. Here of course we suppose that ",2X ~ ",2y and ",2a ~ ",zb. Now 19 1at (",2 x , ",Zy)I\19 1at (",2 a , ",2b)::::: 191at((",zxl\",Zb)v(",2yl\",2a), ",}yl\",Zb),

Ockham algebras

146 and

rv 2 X

I(J

y,

rv 2

rv 2

a 'lj;

rv 2

b give respectively

(rv2x /\ rv2b) V (rv 2y /\ rv2a) I(J (rv 2y /\ rv2b) V (rv 2y /\ rv2a) = rv 2y /\ rv2b, (rv2x /\ rv2b) V (rv 2y /\ rv2a) 'lj; (rv2x /\ rv2b) V (rv 2y /\ rv2b) = rv 2y /\ ",2b. It follows that

((",2 X and so

'I9lat(rv2x,

/\

",2b) V (rv 2y /\ rv2a), rv 2y /\ rv2b) E I(J /\ 'lj;

rv 2y) /\

'I9lat(rv2a,

rv2b) ~ I(J /\ 'lj; as required.

Theorem 8.24 If L E K1 ,1 then the relation Cdefined on Con L by ('I9 1 ,'I9 2 )EC ~ '19 1 V

= '19 2 V

is a lattice congruence and

(Con L)/c ~ [, £].

Every C-class is of the form

[I(J I~L , I(J]

for a unique congruence

I(J E

[, £].

Proof The relation Cis clearly an equivalence relation which is compatible with V. That it is also compatible with /\ follows from the distributivity of Con L. Consider the map g : (Con L)/c ----) [, £] given by g(['I9]c) = '19 V . It is readily seen that g is an isomorphism. For every 7f E Con L let 7f* = 7f V . By Theorem 8.20, [7f]c is the set of all extensions to L of 7f*I~L' has biggest element 7f* and smallest element 7f*I~L. In particular, [w]c = 1 and, since is dually dense, [£]c = {£}.

Corollary The mapping

I(J

t---7

I(JI~L

is a residuated dual closure on Con L.

Proof This follows immediately by [3, Theorem 15.1]. Theorem 8.25 For every L E K1,1 the lattices Con L and Con rvL have isomorphic centres. Proof Consider the mapping f : Con rvL ----) Con L given by f(l(J) = cpo By Theorem 8.23, this is a lattice morphism. Let] be the restriction of f to Z(Con L). Then] is also a lattice morphism. Since ](w) = wand ](£) = £, we see that in fact] is a mapping into Z(Con L). Since for I(J E Con rvL we have I(J = I(J I~L and therefore I(J = cp = ]( I(J), it follows that] is injective. To show that it is also surjective, let '19 E Z(Con L) and let Q! be its complement. By Theorem 8.7, 'I91~L and Q!I~L belong to Z(Con rvL) and we have 'I91~L V Q!I~L =

Since

'I91~L ~ '19, Q!I~L ~ Q!

£,

'I91~L

/\ Q!I~L = w.

and complements are unique, we deduce that '19

= 'I91~L =]('I9I~L).

Congruences on

K1 ,1 -algebras

147

Theorem 8.26 If L E K 1 ,1 isfinite then Con L is a dual Stone lattice. Proof If L is finite then Con L is a finite distributive lattice and is therefore pseudocomplemented. Since, by Theorem 8.14, is dually dense in Con L, the congruence ~ defined on Con L as in Theorem 8.24 is the dual of the Glivenko congruence and so can be described by (19 1 , 19 2 ) E ~

{:=}

19t = 19!

where + denotes dual pseudocomplements. Now for every 19 E Con L the smallest element of [19]~ is 19++ = 191~L. Since {19++ I 19 E Con L} is then a sublattice of Con L by Theorem 8.23, it follows that Con L is a dual Stone lattice.

Example 8.1 Consider the MS-algebra L described as follows:

g 1

a

bed

o

h

h

g

h

e f g g f

h

h

e

d

Con L has 20 elements, namely those given by the following partitions : £

= {O,I,a,b,e,d,e,/,g,h},

A = {{I, e,/, h}, {O, a, b.d,e,g}}, B = {{1,g,h},{O,a,b,e,d,e,/}}, C = {{I, h}, {e,/}, {g}, {O, a, b, d,e}}, D = {{I},{a,b,c,d,e,/,g,h},{O}}, E = {{I}, {b, c,e,/,g,h}, {a,d}, {O}}, F = {{I}, {e,/,h}, {a, b,d,e,g}, {O}}, G = {{I}, {e,/,h}, {b,e,g}, {a,d}, {O}}, H = {{I}, {d,e,/,g,h}, {a, b,e}, {O}}, I = {{I}, {e,/,g,h}, {b, e}, {d}, {a}, {O}}, ] = {{I}, {f,h}, {e}, {d,e,g}, {a, b}, {O}},

148

Ockham algebras K = {{I}, {f,h}, {c}, {e,g}, {b}, {d}, {a}, {On, L M

= {{I},{g,h},{a,b,c,d,e,j},{On, = {{I},{g,h},{b,c,e,j},{a,d},{On,

N = {{1},{h},{c,j},{g},{a,b,d,e},{On,

° = {{ I}, P

{h }, { c ,j}, {g}, { b, e}, {a, d}, {O

n,

= {{I}, {g,h}, {d,e,j}, {a,b, c}, {{O}},

Q = {{I}, {g,h}, {e,j}, {b, c}, {d}, {a}, {On, R = {{ 1 }, {h }, {f}, { c }, {g}, {d, e}, { a , b}, {O

n,

w = {{I}, {h}, {f}, {c}, {g}, {e}, {b}, {d}, {a}, {On. The congruence ell is N and has 6 classes. The Hasse diagram for Con L is

B

c

~-classes, namely {I,E,H,D}, {K,G,],F}, {Q,M,P,L}, {w,O,R,N}, {A}, {B}, {C}, {~}. The principal congruences are as in the

There are eight

following table, where 19(x,y) is at the intersection of column x and row y :

o

w

1

~

w

a

C

I

W

b C ~ c B A

R

w

P

Q w

d C e C

°°

f

~

~

B A

N

N L M

L

w

M

R

w

°

P

Q w

g A B F G E ] K I w h ~ C D E fJ H I K Q w 0 1 a b ~ (1 e f g h

9 MS-spaces; fences, crowns, ... In this chapter we shall apply both the theory of duality and the results on fixed points to a consideration of finite MS-algebras whose dual space is of a particularly simple nature, namely is connected and of length 1. As we shall see, such ordered sets are amenable to rather interesting combinatorial considerations [50, 53]. In order to proceed, we must first characterise the dual spaces of MS-algebras.

Definition An MS-space (resp. de Morgan space) is a Priestley space X on which there is defined a continuous antitone mapping g such that g2 ~ idx (resp. g2 = idx ). Clearly, an MS-space (resp. a de Morgan space) is the dual space of an MS-algebra (resp. a de Morgan algebra); for g2 ~ idx and g2 = idx are the dual equivalents of axioms (1) and (01) respectively. Note that if (X; g) is an MS-space then g2[g(X)] ~ g(x) gives g3 ~ g. But since g is antitone we also have g .g2 ~ g .idx , Le. g3 ~ g. It therefore follows that g3 = g.

Theorem 9.1 For a Priestley space X the following statements are equivalent: (1) X is the underlying set of an MS-space; (2) there is a continuous dual closure map rJ : X - t X such that 1m rJ admits a continuous polarity. Proof (1) =} (2) : If (X;g) is an MS-space, consider the mapping rJ = g2. Clearly, rJ is a dual closure on X and is continuous. Since g is antitone with g3 = g, it is equally clear that g induces a continuous polarity on 1m rJ. (2) =} (1) : Let rJ : X - t X be a continuous dual closure and suppose that 1m rJ admits a continuous polarity 01. Define g : X - t X by the prescription (Vx E X)

g(x) = OI[rJ(X)].

Then g is continuous and anti tone. Since rJ fixes the elements of 1m rJ, we have (Vx EX) whence g2

= rJ ~ idx .

150

Ockham algebras

Note that the existence of a dual closure -a : X ---7 X such that 1m -a admits a polarity is equivalent to the existence of a subset XI of X which admits a polarity, and a decreasing isotone retraction 7r : X ---7 X l' In fact, it is clear that XI =1m -a and that 7r is induced by -a. Observe also that, by the nature of 7r, the subset XI contains all the minimal elements of X; and that if a, b E XI then every minimal element of the set of upper bounds of {a, b} must also belong to XI' The special case where XI = X is important. Here -a is necessarily idx , so that g2 = idx and X is then a de Morgan space. We now proceed to consider some particularly simple MS-spaces, the underlying sets of which are connected and of length 1. For each of these we shall determine the cardinality of the associated MS-algebra and that of its set of fixed points. Somewhat surprisingly, these involve the Fibonacci numbers and the Lucas numbers. In order to avoid any ambiguity, we record here that for these numbers we adopt the following definitions. The generating recurrence relation in each case being the Fibonacci sequence {fn )n~O has 10 (~n)n~O has ~o = 2, ~1 = 1.

= 0, fi = 1 and the Lucas sequence

Definition By an n fence we shall mean an ordered set F 211 of the form

it being assumed that n

~

1 and all the elements are distinct.

We can define two non-isomorphic fences with the same odd number 2n + 1 of elements as follows. We let

F 2n +1 = F 2n U {bn+d

with the single extra relation an

< bll +1; and

Ffn+l = F21l U {aD}

with the single extra relation ao < b 1 • As the notation suggests, the ordered set Ffn+l is the dual of F 2n +1. In what follows we shall have no interest in Ffll+1' for the following reason. If X ~ Ffll+1 then by the observation following Theorem 9.1 we would require XI = X. But here X is not self-dual, so there is no appropriate g that can be defined on it.

MS-spaces; fences, crowns, ...

151

Consider first X ~ F2n' Here it is clear that there is only one antitone mapping g : X -7 X such that g2 ::;; idx , namely that given by

g(a i ) = bn - i +1 , g(b i ) = an-Hl' In fact this mapping g is such that g2 = idx . We can therefore deduce that

there is a unique MS-algebra L(F 2n ) associated with F2n and that it is a de Morgan algebra. As to the size of L(F 2n ), this was determined by Berman and Kohler [29] as an application of Theorem 5.5.

Theorem 9.2

Fork~2,

IL(F k )!=fk+2'

Proof Applying Theorem 5.5 to F 2n with x = an' we obtain IL(F 2n )1

= IL(F2n \

{an})1 + IL(F 2n - 2)1 = IL(F 2n - 1 )1 + IL(F 2n - 2)1;

= bn, we obtain IL(F2n-dl = IL(F2n-2)1 + IL(F2n- 3 )1·

and then to F2n - 1 with x

Writing Ci k

= IL (F k) I, we therefore have the recurrence relation

Example 9.1 Consider the fence F4 :

The mapping g is given by

x al a2 b1 b 2 g(x) b 2 b1 a2 al By Theorem 9.2, L (F 4) has 16 = 8 elements. Its underlying lattice is

f b

c

152

Ockham algebras

at.

at

bt

where a = b = bi, c = 1 = from which the remaining elements are easily identified. Using the fact that r = X \ g-I (J), we can obtain the corresponding de Morgan negation; it is given by x 0 abc d e 1 1 XO 1 e b 1 d a c 0 We now proceed to consider the number of fixed points of L(F2n)' By the considerations in Chapter 6, this is the number of distinguished down-sets of F 2n' The calculation of this is somewhat more complicated, but the outcome is pleasantly simple.

Theorem 9.3 IFixL(F 2n )1

=111+1 .

Proof Consider the subsets {b I , an}, F 2n-4 = F 2n-2 \ {a 1, bn}· With g as defined above, we note that g acts inside F 2n-2 and F 2n-4' Moreover, a distinguished down-set of F 2n-2 cannot be a distinguished down-set of F 2»-4, and conversely. We shall establish the recurrence relation F 2n-2

= F 2n \

IFixL(F2n )1

= IFixL(F2n - 2 )1 + IFixL(F2n _ 4 )1·

For this purpose, let! be a distinguished down-set of F 2n- Then as g( bd = an and g(a n) = b i it is clear that I contains either b i or anIf b i E I (so that an ~ 1) then 1\ {bIl is a distinguished down-set of F 2n-2 which contains a 1 . If an E I (so that b i ~ 1) and al ~ I then 1\ {an} is a distinguished downset of F 2n - 2 which contains bnIf {aI, an} ~ 1(so that {b l , bn}nI = 0) then 1\ {aI, an} is a distinguished down-set of F 2n-4' Thus, to every distinguished down-set of F 2n there corresponds a distingUished down-set of either F 2n-2 or F 2n-4' Clearly, this correspondence is bijective and the required relation follows. To complete the proof, it suffices to obselVe from Example 9.1 that IFixL(F 4 )1=2=i3.



We now turn our attention to the fence F 2n+ I' It is clear that if X '.:::: F 2n+ 1 then there are precisely two antitone mappings g : X ----+ X such that g2 ~ idx , namely those given by

g(x) bll bn - I h (x) b n + 1 b n

b i an an-I b 2 b n+ 1 an

MS-spaces; fences, crowns, ...

153

It is readily seen by relabelling X that these mappings give rise to isomorphic MS-algebras, so we shall consider only the mapping g. Since g is not surjective the corresponding MS-algebra L(F2n + l ) is not a de Morgan algebra. In fact, V(L(F 2n + l )) = MI if n > 1, L(F3) being the subdirectly irreducible K 3. To prove this, it suffices to show that axiom (11 d) fails; or alternatively that its dual equivalent, namely g2 Mg V gO ~ g2, fails. The latter is easier. Consider b n + l ; we have

g2(b n +l ) = an II b l

= g(b n +I ),

Example 9.2 Consider the fence F 5

bn +1 > an

= g2(b n +1 )·

:

The mapping g is given by

x a l a2 b1 b2 b3 g(x) b 2 b l a2

By Theorem 9.2, L(F5) hasf7

al

= 13 elements.

b1

Its underlying lattice is

f

where c = aI, a = a~, f = bI, b = bL i = b~, from which the remaining elements are easily identified. The operation ° on L (F 5) is given by j k 1 x 0 abc d e f g h 1 j j h e e f c c b bOO

XO

Observe from Example 9.2 that the number of fixed points of L(F 5 ) is 2, which is the same as the number of fixed points of L(F4)' In fact, we have the following result.

Theorem 9.4 IFixL(F 2n + I )1

= IFixL(F2n )1 = fn+l'

154

Ockham algebras

Proof Suppose that I is a distinguished down-set of F 2n+ I' Then g(J) ~ F2n+1 \ I, g(F2n+l\ I) ~ I. If bn+1 E I then 1\ {bn+d is a distinguished down-set of F 2n that does not contain g(b n+I ) == b l ; whereas if bn+1 ¢ I then I is a distinguished down-set of F 2n that contains g(b n+I ) == b l . It follows that the number of distinguished down-sets of L(F 2n +I ) is the same as that of L(F2n)' 0 We now turn our attention to a slightly more complicated ordered set.

Definition By an n -crown we shall mean an ordered set C 2n of the form

~b3 ... Abtl ~ ~ ~an it being assumed that n If X

~ C 2n

~

2 and all the elements are distinct.

then clearly X I == X. The mapping g described by

g(a j) == bn- HI , g(b j) == an-HI is antitone and such that g2 ~ idx . When n is even, this is the only such mapping. However, when n is odd there is another, namely the mapping k given by k(a j) == bHt(n+I), k(b j) == aj+t(n-I), all subSCripts being reduced modulo n. In fact, we have g2 == idx == k 2 and so, whatever the parity of n, L(C 2n ) belongs to the subvariety M. Indeed V(L(C 2n )) == M if n > 2 since, as we shall see below, it has more than one fixed point. Note that V(L(C 4)) == K and L(C 4) has only one fixed point. As to the cardinality of L ( C 2n), this rather nicely involves the Lucas numbers.

Theorem 9.5 IL(C 2n )1 == £2n' Proof Consider first the 2n-fence F 2n as depicted above. If we insert a line from an to b l we obtain C 2n' Now by adding this line we reduce the number of down-sets. More precisely, in so doing we suppress all the down-sets of F 2n that contain b l but not an, that is all the down-sets of F 2n \ {b n , a l } that contain b l , hence equivalently all the down-sets of F 2n \ {bl,al, bn,a n }. It therefore follows by Theorem 9.2 that IL(C 2n )1 == IL(F2n)I-IL(F2n-4)1 == 12n+2 - hn-2 == £2n

0

155

MS-spaces; jences, crowns, ...

Example 9.3 The crown C6 is

and the mappings g, k are given by

x al a2 a3 b i b2 b 3 g(x) b3 b 2 b i a3 a 2 a l k(x) b3 b i b 2 a2 a3 al By Theorem 9.5, L( C 6) has l6 = 18 elements. Its underlying lattice is the free distributive lattice on 3 generators

g

aL

at

in which a = b= c = a~, g = bt i = hi, j On L (C 6; g) the operation ° is given by

= b~.

xOabcdejgh jklmnopl XO 1 n p o l k m h g jed j a c b 0 and on L(C6; k) it is given by

xOabcdejghijklmnopl XO 1 n 0 p k i m g h i j d e j a b c 0 As the following results show, the number of fixed points is governed again in a very agreeable way by the Fibonacci numbers and the Lucas numbers.

Theorem 9.6 For every n, IFixL(C 211 ;g)1

=jll

Proof Recall that g is defined on C 2n by g(a j ) = bn - HI , g(b j ) =

an-HI'

156

Ockham algebras

Observe that the subset C 2n \ {b I , an} is isomorphic to F 2n-2 and that g acts inside it. Suppose that I is a distinguished down-set of C 2n. We may assume without loss of generality that I contains an but not b l . Then 1\ {an} is a distinguished down-set of F 2n-2. Conversely, let J be a distinguished down-set of F 2n - 2. ThenJU {an} is a distinguished down-set of C 21l" This correspondence between the distinguished down-sets of C 2n and those of F 2n-Z is clearly bijective, so

IFixL(C 2n ;g)1 = IFixL(F 2n - 2)1· The result now follows from Theorem 9.3.0

Theorem 9.7 For n odd, IFixL(C zn ; k)1 = €n. Proof Recall that, for n odd, k is given by k(a i) = bH!(n+I),

k(b j ) = ai+t(n-I),

all subscripts being reduced modulo n. Observe that a distinguished down-set I cannot contain a pair of either of the forms {bj , bj-I+!(n+I)}, {b i , bj+t(n+I)}. In fact, if b i E I then {aj-l,a j }

~

I, whence

bi-I+t(n+l) = k(aj_l)

¢I

and similarly

bi+±(n+l) = k(a j ) ¢ I. Hence there are n forbidden pairs of the bj, and these can be enumerated cyclically as follows : {bl, b t (n+3)}' {bt(n+3)' b2}, ... , {b n, bt(n+I)}, {b!(n+I), bd·

Moreover, a distinguished down-set has cardinality n. It contains at most ~(n -1) of the elements bi and among them there is of course no forbidden pair. When such elements are chosen, the elements a j are determined since b j and k(bJ are incomparable. Hence we see that the number of fixed points of L (C 2n; k) is equal to the number of subsets of the set {b l , ... , b,zl that contain no forbidden pair. Our problem can therefore be restated as follows : if Sn is a set of n points on a circle, how many subsets of Sn do not contain two neighbouringpoints? To solve this, we shall use the well-known fact that the number of subsets of {I, ... , n} that do not contain two consecutive integers is fn+z; see, for example, [6]. Let the points of Sn be labelled consecutively 1, ... , n and let tn

157

MS-spaces; fences, crowns, ..

be the number of subsets of Sn that do not contain two neighbouring points. rf A is a subset of Sn that is counted in t n then there are two possibilities : (1 ) 1 E A : in this case 2 ~ A and n ~ A, so by the above fact there are fn -1 possibilities for A, namely {I} U X where X is a subset of {3, ... , n - 1} that does not contain two consecutive integers. (2) 1 ~ A : in this case there are fn+1 possibilities for A, namely those subsets of {2, ... , n} that do not contain two consecutive integers. We deduce from this that

tn

=fn-1

+ fn+1

=Rn,

whence the result follows. 0 We now consider some more complicated ordered sets, also of length 1. For this purpose we define the sequence (jn)n~O by

io =i1

= 1,

("In ~ 2) in

= 2in-1 + in-2

A property of this sequence that we shall require is the following. n

Theorem 9.8 "£ij j=O

= ~(jn + in+1)'

n

Proof Let x 1I = "£ii and observe that, since io j=O

3xn

=ir,

= 2xn + xn

= 2io + (2Jr + 10) +

.. + (2in + in-d + in = 2io + iz + .. + i1l+1 + in = xn + in+1 + in· It follows that xn = ~(jn + in+1)'

0

Definition Bya double fence we shall mean an ordered set DF 2n of the form

Note that n = 1 and n = 2 give respectively 2 and C 4. On DF 2n there is clearly only one dual closure f with a self-dual image, namely f = id. There are two dual isomorphisms on rmf = DF 2n , namely

Ockham algebras

158

a reflection gl in the horizontal, and a rotation g2 through 180°; specifically, for each i, gl (a j) = bj, gdb j) = a i ;

g2(a i ) = b n- i+l , g2(b i ) = an-i+l Since gI = g~ = id, the MS-algebras L(DF2n;gd and L(DF 2n ;g2) are de Morgan algebras. In fact, we can be more explicit: since x Xgl (x) we have thatL(DF 2n ;gl) is a Kleene algebra whereas, for n ~ 3, V(L(DF 2n ;g2)) = M since g2(a l ) = bnll al' In what follows, for every ordered set X we shall denote by # (X) the number of down-sets of X, by #(X; a) the number of down-sets of X that contain the element a of X, by # (X; a) the number of down-sets of X that do not contain a, and by # (X; a, Fi) the number of down-sets of X that contain a but not b.

Theorem 9.9 IL(DF 2n )1 = ill+1 Proof Consider the element b n of DF21Z' On the one hand, we have #

(DF 2n ; b n ) =

#

(DF 2n - 2; b n- l ) + # (DF 21l - 4 ) >

and on the other hand #

(DF 2n ; Fin)

= #(DF2n; an> Fin) + # (DF 2n ;an> bn) = IL(DF 2n - 2)1

+ [lL(DF21l-2)1-#(DF2n-2; bn-dl

= 2IL(DF2n-2)1-#(DF2n-2; bn-d

It follows that

IL(DF 2n )1 = #(DF2n; b n) + # (DF 2n ; Fin) = 2IL(DF 21l _2)1 + IL(DF 2n -4)1· Writing Oi n = IL(DF 2n )1 we therefore have Since Oil = IL(DF 2)1 deduce that

Corollary 1

#

= 3 = 12

and 0i2

= IL(DF4)1 = IL(C4 )1 = 7 = 13, we

(DF 21l ; b ll ) = tUn-1 + in)·

Proof From the first observation above we have Oi n-2

= IL(DF 2n - 4)1 = # (DF 2n ; b n) -#(DF 2n - 2; bn- l )·

MS-spaces; fences, crowns, '"

159

Consequently, 11-2

I: 011 = #(DF 2n ; bn) -#(DF4; b2)· 1=1

Note that # ( DF4; b2) = 2 =io + h. Thus we see that 1/-2

#

(DF2ni bn )

= i=1 LOll + io + h n-2

= ;=1 I:ii+l+io+h 1l-1

= i=O L); = !Vn-l + in)

by Theorem 9.8.

~

(DF 2n ; an, 11n) = IL(DF 2n - 2)I = OI n-1 =in' ~ As for fixed points, we observe that under the mapping g 1 the only distinguished down-set of DF(2n) is 1= {a1' ... ,an}' Consequently, we have

corollary 2

#

IFix L(DF2n ig1)1 = 1.

The situation concerning g2 is much more complicated.

Theorem 9.10 IFix L{DF 2n ;g2)1 = {

i1

ifn is even;

J !(n+1)

ifn is odd.

.!n

~ DFn be the subset (also a double fence) consisting of the elements al>'" ,ai n, b 1 , •.• , bin. For every down-setJ of A letJ* = A'\g2(I) where A' denotes the complement of A in DF 2,1' Then every distinguished down-set of DF2n is of the form I U 1* where I is a down-set of A such that at n E I and b!n ~ I. By Corollary 2 of Theorem 9.9, the number of such down-sets is i}w Suppose now that n is odd. In this case we consider the subset B conSisting of the elements all'" , a!(n+l)' bI!"" b Hn-1)' Clearly, B is a distinguished down-set of DF 2n ; and every distinguished down-set of DF2n is of the form IUI* where I is a down-set of B that contains a!(n+l)' Clearly, this is

Proof Consider first the case where n is even. Let A

#

(DF n+1; at(n+1)' bt(n+1») which, by Corollary 2 of Theorem 9.9, isi t(n+1)' ~

Definition By a double crown we shall mean an ordered set DC 2n of the form

Of course, DC 2 ~ 2 and DC 4 ~ DF4'

160

Ockham algebras

On DC 2n there is only one dual closure f with a self-dual image, namely f = id. All antitone maps g on DC 2n such that g2 ~ id are then such that g2 = id and give rise to de Morgan algebras. For every value of n there are the following : (1) the horizontal reflection gl given by gl (a i ) = bi, g(b i ) = a i ; (2) the rotationg2 given by g2(a i ) = bn-i+1 , g2(b i ) = an-i+l' For odd n these are the only possibilities. For n even, however, there is also (3) the slide-reflection k given by

k(ai) = bi+l2 1Z , k(b j )

= aj+1n, 2

the subscripts being reduced modulo n. In order to determine the cardinality of L(DC 2n) we shall make use of the ordered set Z 2n with Hasse diagram

Theorem 9.11

IL(Z2JI = H.in+2 - 1).

Proof Writing #(Z2n) = Zn we have Zn

= #(Z2n;aO)+#(Z211;aO) = # (Z2n \ {ao}) + # (Z2n-2) = # (DF 2n+2; bn+d + Zn-l = H.in+in+l)+Zn-l,

the final equality following from Corollary 1 of Theorem 9.9. It follows that 11

Zn -Zo =

! i=l IJh + ij+l)'

Since Zo = 1 = !Uo + 11) we obtain, using Theorem 9.8, n

zn =

i i=O LUi + ii+l)

= tUn + in+l) + tUn+l + i12+2) -! = iUn+2 -1). 0 Theorem 9.12 IL(DC 2n )1 = 2in + 1.

161

MS-spaces; fences, crowns, ..

Proof We obtain DC 2n from DF 2n by linking an with b l , and al with b w Consider first the effect of adding to DF 2n the link a n- b l . Clearly, this reduces the number of down-sets. More precisely, in so doing we suppress the down-sets of DF 2n that contain b i but not an' i.e. we suppress the downsets of DF2n\{bn-l,an,bn} that contain b l ; equivalently, the down-sets of DF 2n \{ aI, a2, b l , bn- I , an, b,J; equivalently, the down-sets of Z2n-3' A similar reduction in number occurs when we add the link aI-bIZ' It therefore follows by Theorems 9.9 and 9.11 that IL(DC 2n )1 = IL(DF 2n )l- 2 IZ 2(n-3) I =jn+l-jn-I+ 1 =2jn+1.

As for fixed points, we observe that under the mapping BI the only distinguished down-set of DC(2n) is 1= {aI, ... , an}. Consequently, IFix L(DC 2n ;BI)1

= 1.

The situation concerning B2 is as follows.

Theorem 9.13 IFix L(DC 2n ;B2)1

= {~Hn-2)

ifn is even; ifn is odd.

J Hn-I)

Proof Observe first that DC \ {a I, an' b i , bn } is isomorphic to DF 2n-4 and that B2 acts inside it. If I is a distinguished ideal of DC 2n then I must contain a I and an but neither b i nor bn . Consequently, 1\ {aI, an} is a distinguished down-set of DC\ {aI' an, b l , bn}. Conversely, it is clear that if] is a distinguished downset of DC\ {aI, an, b l , bn} then] u {aI, an} is a distinguished down-set of DC 2n' This correspondence of distinguished down-sets is a bijection, so we deduce by Theorem 9.10 that IFix L(DC 2n; B2)1 = IFix L(DF 2n - 2;B2)1 = {

jl( .7:

n-2

)

J !(n-I)

ifniseven; if n is odd.



In order to determine the number of fixed points of L (DC 2n; k) with n even, we shall make use of the ordered set W 2n +2 with Hasse diagram

~ al

Theorem 9.14

#

Q2

.. <W"TI

a3

(W 2n +2) = tUn+3 + 1).

a"

162

Ockham algebras

Proof For every n let Oi n = #(W2n ) and f3n Oin+1

= #(W2n ; bn)'

Then we have

= # (W2n+2) = # (W2n+2; b n+l ) + #(W2n +2; b n+l ) = # (W2n ) + f3n+1 = Oin + f3n+1 .

We also have f3n

= #(W2n ;bn) = #(Z2n;aO) = # (DF 2n+2; b n+l ) = ~Un

Consequently, Oin+1

+ in+d·

. ). = Oin + 2'I U' n+1 + 1n+2

We deduce from this that I " . I . = 0i2 + 2'13 + J4 + ... + In+1 + 2'111+2' Since 0i2 =#(W4 ) = 9 = t + io + h + h + th, we then have

Oin+1

Oin+1 = ~ +

n+1

"£ h + Vn+2

i=O

= ~ + ~Un+1 + in+2) + ~in+2

= ~(1 + in+3)'

at n+1 = k(b l )

¢] =>

bt n ¢j.

Using the geometric nature of k it can readily be seen that a subset] of DC 2n is a distinguished down-set if and only if it is of the form I U 1* where I is a down-set of A that does not contain both b l and bin, and 1* does not 2 contain both b tn+1 and btl' The latter condition is equivalent to al E I and a tn E I. It follows that the number of fixed points of (DC 2n; k) is t

= # (DFn) -#(DFn; bl , bin) -#(DFn; aI, ai n )· 2

2

MS-spaces; fences, crowns, ...

163

Using Theorem 9.9, the Corollary to Theorem 9.14, and the dual of Theorem 9.14, we deduce that

t

= i!n+l - Hi!n-l + 1) -#(W~(!n-2)+2)

=i!n+l - Hitn-1 + 1) - t(itn+l + 1) =hn -1. 2

The reader will by now have realised that, even in the few cases that we have considered above, there are many problems of a combinatorial nature that arise in connection with Ockham algebras whose dual spaces are finite ordered sets of small length, the solutions to which require conSiderably com.plex arguments. There are of course many other small ordered sets to which similar considerations can be applied. As it is not our intention here to develop a 'cottage industry' in this, we simply refer the reader to [53] where particular ordered sets of length 2 (called batracks) are considered and corresponding but quite different results are obtained, some of the algebras in question belonging to MS-subvarieties other than M and MI' Suppose now that X is a finite ordered set and that I is a down-set of X. Then the length of I in O(X) is III. This is immediate from the observation that if we delete a maximal element of I then we obtain a down-set that is covered by I. In particular, consider the case where X is F 2n or C 2w Here o (X) =L (X) is a de Morgan algebra of length 2n. By a mid-level element we shall mean an element of length n. The question of precisely how many midlevel elements L (X) has is a difficult one and involves further combinatorial arguments. In [50] this is answered for F 2n and C 2n . Specifically, the number of mid-level elements of L(F 2n ) is

Lt (n _m)2 J

m and the number of mid-level elements of L(C 2n ) is m=O

LtI=l)J

n (n -m-1)

2

n -2m m Generating functions for these can also be found in [50]. m=O

10 The dual space of a finite simple Ockham algebra

We recall that an n -crown (with n ~ 2) is an ordered set C with 2n elements Xl, ... , X 2n whose only comparabilities are

... ,

CI = IMin CI = n, every minimal element is covered by two maximal elements, and every maximal element covers two minimal elements; so all vertices of C have degree 2. There have been some attempts to generalise this notion. We mention two of these. In [93], W. T. Trotter, Jr. defines a crown S~ as follows : for n ~ 3 and k ~ 0, S~ is an ordered set of length 1 with n + k minimal elements al, ... ,an+k and n + k maximal elements b l , ... ,bn+k , each a i being incomparable with bi , ... ,bi +k and being covered by the remaining n -1 maximal elements. Here, of course, the subscripts have to be interpreted cyclically. For example, the graph of S~ is as follows: An n-crown C is connected, has length 1, IMax

Clearly, every vertex of S~ has degree n -1 and, for every n ~ 3, an n-crown corresponds to S3-3 . The definition of a k -crown oj order n as given by B. Sands [83] is very close to Trotter's definition. A k-crown of order n (with n ~ k > 0) is an ordered set Ck(n) = A UB where A = {al,"" an} and B = {b l ,··., b n } are antichains and a j < bj if and only if j - i E {O, 1, ... , k -I} modulo n. All vertices of Ck(n) have degree k. All Ck(n) are connected except if k = 1, in which case C 1 (n) is the disjoint union of n two-element chains. For example, C 3 (6) has the following graph, clearly isomorphic to S~ :

The dual space of a finite simple Ockham algebra

165

Whereas in an S~ every a i is incomparable with at least one bi' in Cn(n) all a i are comparable with all b i' In fact, C n (n) is a complete bipartite ordered set. In every C k (n) with k 'f n, for two distinct minimal elements there is a maximal element that covers one of them but not the other. Except in the special case described above, S~ is isomorphic to Cn-l(n + k). For our purposes here, we introduce a more general concept.

Definition A generalised crown C n;k is an ordered set C of cardinality 2n (with n ~ 1) that satisfies the following conditions: (1 ) C is connected; (2) C has length 1; (3) all vertices of C have the same degree k (with 1 ::;; k::;; n). From this definition it follows that every C n,k has n minimal elements and n maximal elements. In fact, if IMin CI = nl and IMax CI = nz then the number of edges is nlk = nzk whence nl = nz = n. The following examples show that the conditions (1), (2), (3) are independent :

I

I

A

satisfies (2) and (3) but not (1);

satisfies (1) and (3) but not (2);

satisfies (1) and (2) but not (3).

The family of generalised crowns is strictly larger than that of the S~. This fact is llustrated by the following C 6;4 :

Here both al and a4 are covered by the same elements, namely b l , b 3 , b 4 , b 6 . Hence this ordered set is not an S~. This example also shows that for some n, k there are non-isomorphic C n,k'

166

Ockham algebras

Every generalised crown C n,k can be represented by a square (0, 1 )-matrix [OIij] of order n in which

if a i -< bj ; otherwise, and in which all line sums (i.e. all row sums and all column sums) are equal to k. Clearly, there is a one-one correspondence between such matrices and generalised crowns Cn;k' For example, to the preceding C 6;4 corresponds the matrix M 1 given by b I b2 b3 b4 b s b6

al a2 a3 a4 as a6

1 1 0 1 1 0

0 1 1 1 0 1 1 1 0 0 1 1 1 0 1 1 1 0

0 1 1 0 1 1 0 1 1 0 1 1

Any interchange of rows or columns yields another C n,k order-isomorphic to the original. For example, interchanging rows 1 and 3, and columns 2 and 6 in M 1 we obtain the matrix M 2 given by b I b2 b3 b4 b s b6

al a2 a3 a4 as a6

0 1 1 1 1 0

1 0 1 1 0 1

1 0 1 1 0 1

0 1 1 1 1 0

1 1

1 1

0 0 0 0 1 1 1 1

to which corresponds the generalised crown

Two matrices such as M 1 and M2 will be called equivalent. More precisely, M 2 is equivalent to M 1 if there are permutation matrices P and Q such that M2 = PMIQ. Clearly, two generalised crowns are order-isomorphic if and only if they have equivalent matrices.

The dual space oj a finite simple Ockham algebra

167

Finally, we note that the 6 x 6 matrix that we associate with C6.4 is in fact a reduced form of the 12 x 12 adjacency matnx of the labeled graph with 12 points as it is usually defined in graph theory [14]. Since MinC6 4 and Max C6;4 are totally unordered, there are only zeros in the NW and SE quarters of the adjacency matrix. Moreover, the latter is symmetric, hence the NE quarter (which is our 6 x 6 matrix) describes the situation unambiguously. Our objective here is to characterise the dual space of a finite simple Ockham algebra and to determine the number of finite simple Ockham algebras in each class Pm,n' For this purpose we begin by showing that every connected component of the dual space of a finite simple Ockham algebra is either a generalised crown or a singleton. In this connection, the basic result on which our investigation rests is Corollary 2 of Theorem 4.6: if (L;j) E 0 is finite with dual space (X; g) then (L; J) is simple if and only if gW (x) = X Jor every x EX. It follows that the dual space (X; g) of a finite simple Ockham algebra satisfies the following properties : (1) g is surjective, hence bijective; likewise so is gil for every n. (2) If IXI = N then for every x E X the elements x,g(x), ... ,gN-l(X) are distinct and gN(X) = x. If we ignore the order relation, this means that X is a 'loop' and the smallest class Pm,ll to which the dual algebra L belongs is PN,o' The mapping g is an order-reversing permutation of the N elements of X with a unique orbit, i.e. is an N-cycle. (3) If N is odd then the order on X is discrete, i.e. X is an antichain. In fact, from x < y we obtain gN(X) > gN(y), giving the contradiction x> y. (4) g maps every maximal element of X onto a minimal element, and conversely. In fact, if N is odd then Max X = Min X and the result is trivial. On the other hand, if N is even, suppose that x E Max X and y < g(x). Then gN-l(y) > gN-l(g(X)) = gN(X) = x, which contradicts the fact that x E Max X. It follows from this that IMax XI = IMin XI. (5) The length €(X) of X is at most one; otherwise g would act inside the set X\(Max X U Min X) in contradiction to property (2). (6) All vertices of X have the same degree. Indeed, let x be an arbitrary vertex of X. Without loss of generality, we may assume that x E Min X. Let q be the degree of x. For every Xi E X there exists s i E IN such that Xi = gS;(x). As observed above, gS; is injective. If Si is even then gS; is order-preserving, Xi E Min X and is covered by q elements; if Si is odd then gS; is order-reversing, Xi E Max X and covers exactly q elements. Hence all the vertices of X have the same degree. Thus we have proved :

168

Ockham algebras

Theorem 10.1 Every connected component of the dual space of ajinite simple Ockham algebra is either a generalised crown or a singleton. In what follows, (X;g), or simply X, will always denote the dual space of a finite simple Ockham algebra. We shall write the elements of X as 0, 1 , ... , N - 1 with 0 E Min X, and g (0) = 1, g( 1) = 2, ... , g (N - 1) = O. With this simplification of notation, gr(o) = r modulo N and, more generally, gr(i) = i + r modulo N. We point out, once and for all, that such equalities have to be understood modulo N. Since we are using integers to denote the elements of X we shall denote the order on X by :j. Note that if i -< j then we have i + r -< j + r if r is even, and i + r )- j + r if r is odd. We have already seen that the case where N is odd is uninteresting. Suppose then that N is even, say N = 2n. Note that in this case, if i -< i + r then i -< i + 2n - r; in fact, since 2n - r must be odd, we have

i -< i + r =? i + 2n - r )- i + r + 2n - r = i + 2n = i. In particular, if 0 -< j then 0 -< 2n - j. Note also that

r -< s

~

2n - r -< 2n -so

We define the subset r(O) by r(O)

= {x E X I 0 -< x,

x ~ n},

noting that all the elements of r(O) are odd. If x,Y E X are connected then we write x I> (2) : If (X;g,h) is a double MS-space, consider the mapping {) = g2. Clearly, {) is a dual closure on X and is continuous. Since g is antitone and g3 = g, it is equally clear that g induces a polarity on 1m {). Now g2 0 h 2 = gog 0 h 0 h = g3 0 h = g 0 h = g2 ~ idx , and similarly h 2 0 g2 = h 2 ;;::: idx . Consequently, {) is residuated with residual {)+ = h 2 which is continuous. (2) => (1) : Let {) be a residuated dual closure on X. Suppose that {) and {)+ are continuous, and that 1m {) has a continuous polarity 01. Then {) = {) 0 {)+ and {)+ = {)+ 0 {) with {)+ a closure on X. Define g, h : X ~ X by (Vx E X)

g(x) = OI{)(X),

h(x) = {)+g(x).

Then g, hare antitone and continuous. Now {) fixes the elements of 1m {), so we have g2(X) = OI{)OI{)(X) = 0I 2{)(X) = {)(x); h 2(x) = {)+0I19{)+0I{)(x) = {)+OI{)OI{)(X) = {)+{)(x) = {)+(x), whence g2 = {) ~ idx and h 2 =

{)+ ;;:::

idx . Also,

(g 0 h)(x) = OI{){)+OI{)(X) = OI{)OI{)(X) = g2(X), (h og)(x) = {)+OI{)OI{)(X) = h 2(x), whence g 0 h = g2 and hog = h 2. Precisely when a Priestley space X is a double MS-space can also be determined using equivalence relations. For this purpose, we recall that an equivalence relation 8 on an ordered set E is strongly lower regular if

z

~

x8y => (:JZI

E

E) z8z' ~ y;

Ockham algebras

190

strongly upper regular if z ~ y8x => (3z'

E

E) z8z' ~ x;

and strongly regular if it is both.

1heorem 12.3 Let X be an ordered set and let 8 be an equivalence relation on X. Then the jollowing statements are equivalent: (1) there is a dual closure .,J : X ~ X with Ker .,J = 8; (2) 8 is strongly lower regular and every 8-class is bounded below. Proof (1) => (2) : If (1) holds then clearly every Eklass is bounded below, the smallest element of [x]8 being .,J(x). To see that 8 is strongly lower regular, suppose that z ~ x8y. Then .,J(z) ~ .,J(x) = .,J(y) ~ y and so z8.,J(z)~y.

(2) => (1) : Suppose now that (2) holds. For every x

E

X define

.,J(x) = min [x]8. Then clearly {) = {)2 ~ idx and Ker 8 = .,J. Since 8 is strongly lower regular it follows from z ~ y8.,J(y) that there exists z' E X such that z8z' ~ .,J(y) whence .,J(z) = .,J(z') ~ .,J2(y) = .,J(y). Consequently {) is also isotone and hence is a dual closure.

1heorem 12.4 Let X be an ordered set and let .,J : X ~ X be a dual closure. Then the jollowing statements are equivalent: (1) .,J is residuated; (2) 8 = Ker {) is strongly upper regular and every 8-class is bounded above. Proof (1) => (2) : If the dual closure .,J is residuated then {)+ is a closure on X. Moreover, from the relations .,J = .,J+ O.,J and.,J+ = {) 0 {)+ we deduce that Ker.,J+ = Ker.,J. The dual of Theorem 12.3 now gives (2). (2) => (1) : Suppose now that (2) holds and define

(3z E X) Y ~ z8x =?

Y ~ max [x]8 = a} for every a EX. Then X is a Priestley space. Consider the dual closure f) : X ----t X given by f)(x) = {px

if xi q; if x = q.

Here 0 = Ker f) clearly satisfies conditions (1), (2), (3) of Theorem 12.6. However, it does not satisfy condition (4). For example, U = ql is clopen but =p 1 is not open. Note that in this example f) is residuated; in fact, we have

ue

f)+(x)

Although f) is continuous,

pl is not.

f)+

= {Xq

if xi p; if x = p.

is not; for example, q 1 is open but

(f)+)-l (q 1)

=

Double MS-algebras

193

For a finite ordered set X, regarded as a Priestley space under the discrete topology, the number of double MS-spaces definable on X depends on the number of polarities on X /9 for each appropriate equivalence relation 9 on X. This is illustrated in the following example.

Example 12.6 If X has Hasse diagram

s

q

r

then on X there are five eqUivalence relations that satisfy the conditions of Theorem 12.6, namely 9 1 = W, 92 = L, 9 3 == {{p, r,s}, {q, t}}, 9 4 == {{r,s,t},{p,q}}, 9 4 == {{r,s},{p},{q},{t}}.

There are therefore five corresponding dual closures ~i such that Ker namely x p q r s t r s t ~l(X) P q ~2(X)

~3(x) ~4(X)

~s(x)

P P P P

~i

=9 i,

P P P P q P P q P r r r q

r

r

t

Since X /9 s is the four-element boolean lattice, which admits two distinct polarities, there are in all six distinct antitone mappings g (these being given as in the proof of Theorem 12.2 by g(x) = Q!~(x) where Q! is a polarity on 1m ~), namely p q r s t x gl(X) t q s r p

g2(X) g3(x) g4(X) gs(x) gs(x)

P p P P p q p q q p r r p p p t q r r p t r q q p

194

Ockham algebras

In an entirely similar manner, we can compute the mappings h. Now L

6 corresponding antitone

= CJ(X) has Hasse diagram

c d

and on L we can define six non-isomorphic double MS-algebras, namely

x

0

a

XO

1

f f

b e e

c

d

d d

c c

e b b

f

1

a a

0 0

x+

1

XO

1

0

0

0

0

0

0

0

x+

1

1

1

1

1

1

1

0

0

0

0

1

0 1

c

0

Ll L2

XO

1

c

c

x+

1

1

1

c c

XO

1

d

0

0

1

1

1

1

d d

0 d

0 d

0 0

L4

x+

f f f f

d

d d

c c

a

0

Ls

c c

d d

a a a a

XO

1

x+

1

XO

1

x+

1

f c

f

c a d

L3

0 0 0

L6

Note that Ll is a Kleene algebra and L2 is a double Stone algebra. We shall now consider the following purely algebraic question : given an MS-algebra (L; 0), precisely when can this be made into a double MS-algebra (L; 0, +)? For this purpose, we recall that a non-empty subset M of an ordered set E is said to be bicomplete if, for every x E E, the set xl n M has a biggest element and the set x TnM has a smallest element. In an MS-algebra (L; 0) the smallest element of x T nLoo is xeD. By [3, Theorem 20.1], there is a bijection between the set of residuated closure maps on E and the set of bicomplete

195

Double MS-algebras

subsets of E; if {} : E ----> E is a residuated closure then 1m {} bicomplete and {}+ is given by (Vx E E)

{}+(x)

= 1m {}+

is

= max(x! nLOO).

Theorem 12.7 An MS-algebra (L; 0) can be made into a double MS-algebra (L; 0, +) if and only if the closure a ~ aOo is residuated and its residual preseroes suprema.

'* :

If (L; 0, +) is a double MS-algebra then for every a by (D1) and (D2),

Proof

=a

a++OO

= a+ ooo = a+o = a++ ~ a.

+++

L we have,

= aO+ = aO >a;

aOO++

O

E

O

Consequently the closure map a ~ aOo is residuated with residual the dual closure a ~ a++, which clearly preserves suprema.