Ockham Algebras (Oxford Science Publications)

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Author: T. S. Blyth | J. C. Varlet

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© T. S Blythandl C Varlet, 1994 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press. Within the UK, exceptions are allowed in respect ofany fair dealing for the purpose of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act, 1988, or in the case of repro graphic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency Enquiries concerning reproduction outside those terms and in other countries should be sent to the Rights Department, Oxford University Press, at the address above. This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, re-sold, hired out, or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser. A catalogue recordfor this book is available from the British Library

Library of Congress Cataloging in Publication Data (Data available) ISBND 19 859938 2 Typeset by T.S Busing MacTeX Printed in Great Britain by Bookcraft (Bath) Ltd Midsomer Norton, Avon

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 <=} {X,YEPwithx(yinP}or{x,YEQwithx(yinQ}. The linear sum P EEl Q consists of P U Q with the order x (y

<=}

{x,y E P with x (y in P) or {x,y or {x E P and x E Q}.

E

Q with x (y in Q)

Ockham algebras

2

Finally, the vertical sum PEBQ is defined only when P has a biggest element a and Q has a smallest element b, and is obtained from P EB Q by identifying a and b. If A, B are ordered sets then a mapping j : A ~ B is said to be orderpreserving or isotone if it is such that (Vx,y

E

A)

x:::;;y =*j(x):::;;j(y),

and order-reversing or antitone if it is such that (Vx,y

E

A)

x:::;;y =*j(x)~j(y).

A and Bare (order-) isomorphic if there is a surjection j : A ~ B such that

(Vx,y E A)

x:::;;y {=?j(x):::;;j(y),

and dually ( order-) isomorphic if there is a surjection j : A (Vx,y E A)

x:::;; y

{=?

j(x)

~

~

B such that

j(y).

An ordered set that is dually isomorphic to itself is said to be self dual. A mapping j : A ~ A such that j2 = idA is called an involution (of period two). An order-reversing involution is called a polarity.

Very often (and this is so for the three examples mentioned above) any pair of elements X,y of an ordered set have a greatest lower bound (or meet, or infimum) which is denoted by x /\y; and a least upper bound (or join, or supremum) which is denoted by x V y. Such ordered sets are called lattices. For example, (IR; :::;;) is a lattice in which x /\y = min{x,y},

xV y = max{x,y};

(IP(E); ~) is a lattice in which X /\ Y

= X n Y,

X VY

= XU Y;

mVn

= lcm{m, n}.

and (INo; I) is a lattice in which m /\ n

= gcd{m, n},

Clearly, any finite subset of a lattice L has a meet and a join. If every subset of L has a meet (resp. join) then L is said to be meet-complete (resp. joincomplete). By a complete lattice we mean a lattice which is both meetcomplete and jOin-complete. The concept of a lattice was introduced at the end of the last century by c. S. Peirce and E. Schroder, but the study of lattices became really systematic with G. Birkhoffs first paper [30] in 1933 and his book [2], the first edition of which appeared in 1940 and was for several decades the bible of lattice theoretists. In recent years, lattice theory has grown considerably. Lattices

Ordered sets, lattices, and universal algebra

3

ap{ear ~n all branches of mathematics: for any algebra the subalgebras, the eq~~::II~e relations, the congruence relations form lattices; in a topological space the open sets, the closed sets, the clopen (Le. closed and open) sets form lattices; the convex subsets of a vector space form a lattice; in classical logic the propositions form a lattice in a way that we shall make precise below. But what are lattices from the point of view of universal algebra? As is well known, the aim of universal algebra is to highlight the properties that various algebraic systems (e.g. groups, rings, fields, modules, lattices, ... ) have in common. If we leave aside some early papers of Whitehead, we might say that the first pioneer of universal algebra was also G. Birkhoff. Fundamental to universal algebra is the notion of an operation. If n is a non-negative integer then an n -ary operation on a set A is a mapping / : All ---7 A. The integer n is called the arity of the operation. We shall be mainly concerned with the cases where n = 0, 1, 2 which give respectively a nullary operation (this simply picks out an element of A), a unary operation, and a binary operation. An algebra o/type (nl, ... , no<) is a pair (A, F) where A is a non-empty set and F is an a-tuple Ui, ... ,io<) such that, for each i with 1 ~ i ~ 01, /; is an nj-ary operation on A. Thus, for example, a lattice is an algebra of type (2,2), the two binary operations being meet and join, and satisfying the folloWing identities : x Ay = Y AX; X

Ax = x;

X A (y AZ) = (X Ay) Az; X A (X V

y) = x;

xvy=yVx; XVx= x; X V

(y

V z)

= (X V y) V z;

XV(XAy)=x.

If a lattice is bounded, Le. if it has a least element 0 and a greatest element 1, then it can be considered as an algebra of type (2,2,0,0). If A and B are algebras of the same type (nl,' .. , no<) then a mapping cP : A ---7 B is a morphism if, for each i such that 1 ~ i ~ 01,

whenever (a 1 , ... , alii) E A IIi . If, in addition, the mapping cp is surjective then cp is said to be an epimorphism with B an epimorphic image of A; if it is injective then it is a monomorphism; and if it is both then it is an isomorphism. A morphism / : A ---7 A is called an endomorphism on A; and an isomorphism / : A ---7 A is called an automorphism on A. Note that if Land M are bounded lattices then any morphism cp : L ---7 M has to satisfy CP(OL) = OM and cp(lL) = 1M,

4

Ockham algebras

As we have mentioned above, lattice theory began properly with the work of George Boole [4] in formal deductive logic with an attempt to codify the laws of thought. In fact, Boole considered very special (but also very important) lattices in which, originally, the meet and the join were the binary connectives called conjunction ('and') and disjunction ('or') respectively, with an additional unary operation called negation ('not'). These so-called boolean lattices have turned out to be very useful in many areas of science and mathematics: in electrical engineering, in computer science, in axiomatic set theory, in model theory, and so on. Precisely, a boolean lattice L has three characteristics :

(1 ) it is bounded; (2) it is distributive in the sense that (Vx,y,z E L)

x /\ (y V z) = (x /\y) V (x /\ z).

It is quite remarkable that this equality is equivalent to its dual x V (y /\z) = (x vy) /\ (x V z).

(Vx,y,z E L)

(3) it is complemented in the sense that for every a E L there exists a' E L (called the complement of a) such that a /\ a' = 0 and a va' = 1; in other words, the centre Z(L) of L, i.e. the set of complemented elements, is L itself. The property of distributivity is shared by many lattices. For instance, each of the three examples given above is distributive. Since all the lattices that we shall deal with will be distributive, we shall say nothing about the various forms of weak or restricted distributivity. On the contrary, the notion of complement is very strong and many weakened forms of it have been considered. Note that in a boolean lattice L the operation x f-7 x' of complementation is a polarity and satisfies the so-called de Morgan laws (Vx,y E L)

(x /\y)' = x' V y',

(x V y)' = x' /\y'.

As observed by H. B. Curry [8], 'the term is customary despite its historical inaccuracy. According to Bochenski, the formulas were known in the Middle Ages'. If, in a bounded distributive lattice, we can define a polarity that satisfies the above de Morgan laws then we obtain what is called a de Morgan algebra. More precisely, this is an algebra (L; /\, V,J, 0,1) of type (2,2,1,0,0) where (L; /\, V, 0, 1) is a bounded distributive lattice and f : L ~ L is a unary operation that satisfies the identities

f(x /\y) = f(x)

V

f(y),

f(x

V y) =

f(x) /\f(y),

f2(X) = X.

Ordered sets, lattices, and universal algebra

5

From these identities it follows that f(O) = 1 and f(l) == O. De Morgan algebras were introduced by G. C. Moisil [75] and investigated by A. Monteiro [76] and his school. A Kleene algebra is a de Morgan algebra satisfying the inequality x I\f(x) ~ Y V f(y)· Another way of generalising the notion of complementation is to retain the identity a 1\ a' == 0 and to drop the other. In this manner we define a semicomplementation. A lattice L that is bounded below is said to be semicomplemented if every a E L has a semicomplement, i.e. a non-zero element that is disjOint from a. Here, of course, the second lattice operation V plays no part, so that the notion of semicomplementation can be defined on a meet semilattice. Of considerable interest are those lattices (or meet semilattices) in which, for any element a, the subset of elements disjOint from a has a greatest element. This is called the pseudocomplement of a and is denoted bya*. Thus a* == max{x ELI a 1\ x = O}. Pseudocomplemented lattices are necessarily bounded. Examples of these are : the lattice of open subsets of a topological space, the pseudocomplement of an open set being the interior of its complement; and the lattice of ideals of a distributive lattice that is bounded below. Of course, if we require that a V a* = 1 for every a E L, then a* becomes the complement of a and, when L is distributive, L is then a boolean lattice. Guided by what occurs in many examples, Stone [92] suggested a restriction of the identity a V a* to those elements a that are pseudocomplements, i.e. that it would be fruitful to consider the identity a* V a** = 1 for all a E L. Distributive pseudocomplemented lattices that satisfy this identity are therefore called Stone lattices. When the unary operation a ~ a* is considered as a fundamental operation of the algebraic system, we shall use the term Stone algebra. Note, therefore, that whereas a Stone lattice is of type (2, 2, 0, 0), a Stone algebra is of type (2, 2, 1 , 0, 0). This distinction is essential not only with regard to morphisms, but also with regard to subalgebras and congruences. A subalgebra B of an algebra A is a non-empty subset of A which is closed under all the operations of A. The operations on B are then those of A restricted to B. An algebra A and its subalgebras are of the same type. For example, a subalgebra of a Stone lattice is just a sublattice , whereas a subalgebra of a Stone algebra is a sublattice that is closed under a ~ a*. A congruence relation on an algebra A of type (nl' ... ,no<) is an equivalence relation -6 on A which satisfies the substitution property : for each i E {I, ... ,a}, (a j ,bj )E-6(j=l, ... ,nj) => Ui(al, ... ,a nl ),};(b 1 , .. ·,b n ))E-6.

6

Ockham algebras

For example, in a Stone lattice L an equivalence relation -e is a congruence if (a,b) E -e and (c,d) E -e imply (a /\ c, b /\ d) E -e

and (a V c, b V d) E -e.

Note that, by the commutativity of /\ and V, this can be simplified to (a,b)E-e => (VCEL) (a/\c,b/\c)E-e, (aVc,bVc)E-e.

In a Stone algebra, however, there is the supplementary requirement (a,b)E-e => (a*,b*)E19.

The set of congruences on an algebra A, ordered by set inclusion, is a lattice with smallest element w = ({x,x) I x E A} and biggest element ~ = A x A. This lattice is called the congrnence lattice of A and is denoted by Con A. If, on a bounded distributive lattice L, there is defined a unary operation f which satisfies the de Morgan laws and is such thatf(l) = 0 andf(O) = 1 (Le. if we drop the assumption that f2 = idL, so that f becomes a duallattice endomorphism, but not necessarily a dual lattice automorphism) then we obtain what is called an Ockham algebra. This idea goes back essentially to 1977 in a short but very deep paper by J. Berman [28]. Two years later, A. Urquhart [94] developed a topological duality theory for this type of algebra, gave a logical motivation for his study, and introduced the name Ockham lattices with the justification : 'the term Ockham lattice was chosen because the so-called de Morgan laws are due (at least in the case of propositional logic) to William of Ockham'. The name Ockham algebra has since become classical and was used in the thorough doctoral thesis of M. Goldberg [68] and in a subsequent paper [69]. Since 1981 many papers have been published on Ockham algebras. The objective of this book is therefore to develop the general properties of this class of algebras and to consider more particularly some important subclasses which are interesting not only in the framework of universal algebra but also for their Significance in the algebra of logic. At this pOint, it is not superfluous to recall that de Morgan algebras arose in the researches on the algebraic treatment of constructive logic with strong negation. The operation f that is involved in an Ockham algebra can also be interpreted as a negation (and for this reasonf(a) is often written as rva), though this does not in general satisfy the law of double negation. If we impose on f the restriction that fn = idL for some n E IN\{l, 2} then we obtain a new logic whose interpretation, as far as we know, has still to be made explicit. In this connection, interesting work has been done by D. Schweigert and M. Szymanska [88] on those Ockham algebras that belong to the class Pn,o (n odd) described in Chapter 4. The

Ordered sets, lattices, and universal algebra

7

class of algebras they deal with is shown to be the semantic for a propositional calculus called correlation logic. The reader will see from the examples of Ockham algebras given in Chapter 1 that the study of Ockham algebras is far from being gratuitous. We close this brief introduction by observing that all the classes of algebra that we have mentioned, and indeed all that we shall consider later, are equational, in the sense that they can be defined by a set of identities. It is a celebrated theorem of Birkhoff that the equational classes of algebras are precisely those that are closed under the formation of subalgebras, epimorphic images, and direct products, Le. are varieties.

1 Examples of Ockham algebras; the Berman classes Recall that a distributive Ockham algebra is an algebra (L; 1\, V,j, 0, 1) of type (2, 2, 1, 0, 0) in which (L; 1\, V, 0, 1) is a bounded distributive lattice and x f-t f(x) is a unary operation such thatf(O) = 1, f(l) = 0 and

(Vx,y E L)

f(x I\y)

= f(x) V f(Y),

f(x V y) = f(x) I\f(y).

Without explicit mention to the contrary, all the Ockham algebras that we shall deal with will be distributive as lattices so we shall agree to drop the adjective 'distributive' and talk of an Ockham algebra. We shall often also denote this by the simpler notation (L;/). The class of Ockham algebras is equational (in other words, a variety), and will be denoted by o. As mentioned in Chapter 0, the concept of an Ockham algebra arose from successive attempts to generalise the notion of a boolean algebra. Important steps in this long history are the de Morgan algebras and the Stone algebras, these forming important subvarieties of the variety O. In 1979, A. Urquhart [94] observed that 'an outstanding open problem is that of determining all equational subclasses of the class of Ockham lattices'. To this day, the problem remains unsolved. However, very important subclasses of 0 were introduced by J. Berman [28] and we shall call them the Berman classes. These are obtained by placing restrictions on the dual endomorphism f. Precisely, if we let jO = id and define r recursively by r (x) = f[r- 1 (x)] for n ~ 1, then for p, q E IN with P ~ 1 and q ~ 0 we define the Berman class Kp:q to be the subclass of 0 obtained by adjoining the equation f

2p +q

=fq ·

The Berman classes are related as follows : Kp,q ~ Kp',q'

-¢==}

PIP'

and q:::;;; q'.

We shall give a simple proof of this later. It follows that the smallest Berman class is the class Kl ,0, which is determined by the equation P = id. This is none other than the class M of de Morgan algebras. The importance of the Berman classes is partly justified by the following property.

Theorem 1.1 Every finite Ockham algebra belongs to a Berman class

Examples of Ockham algebras, the Berman classes

9

Proof Let (L;1) be a finite Ockham algebra, and consider the sets

{f,j3 ,15 , ... },

{f0,j2,j4,j6, ... }

of, respectively, dual endomorphisms and endomorphisms on L. Since both of these must be finite, we have that f q = f 2p +q for some p, q. <> Note that Theorem 1.1 is no longer true if the algebra in question is infinite, as the following example shows.

Example 1.1 Let L = {O, 1} U {a j liE 7l} u {b j liE 7l}, ordered linearly by a j < aj for i < j; b j < bj for i < j; 0 < a j < bj < 1 for aU i,j. Define f: L -tL by

f(O)=l,

f(l)=O,

f(aj)=b_ j, f(bj)=a_i-l'

Then (L; f) is an Ockham algebra. We can depict the effect of f as follows :

t

b2

I bl

I bo

I

D ao

I

For every i E 7l and n E IN, we have the chains

.. < f2n(a i ) < ... < f2(aJ < a < f(a j) < ... < f2n+l(a j) < ... j

... < f2n+l(b j) < ... < f(bJ < bi < f2(b j) < .. < f2n(b j) < .,. It follows that L does not belong to any Berman class. Note also that in this

example f is a bijection.

Ockham algebras

10

The following three examples are by no means surprising.

Example 1.2 In a de Morgan algebra it is traditional to write the unary operation as x 1-7 x. Every de Morgan algebra (L; -) belongs to the Berman class KI,o, In fact, since automorphism.

x = x for every x E L, the operation x

1-7

X is a dual

Example 1.3 In a boolean algebra it is traditional to write the unary operation as x 1-7 Xl. Every boolean algebra belongs to the subclass of K1 obtained by adjoining the equation x I\f(x) = O. In fact, from this it follo~s

°

that f(x) V f2(x) of x.

= 1, i.e. f(x) V x = 1, so that f(x) = Xl is the complement

Example 1.4 In a Stone algebra it is traditional to write the unary operation as x 1-7 x*. Every Stone algebra (L; *) belongs to the subclass ofK11 obtained by adjoining the equation x 1\ f(x) = O. In fact, by the properties of the pseudocomplementation x 1-7 x* we have (x 1\ y)* = x* V y*, (x V y)* = x* 1\ y* with 0* = 1 and 1* = 0, so that (L; *) E O. Since moreover x* = x*** it follows that (L; *) E K I ,l' Finally, x 1\ x* = 0 by the definition of the pseudocomplement. Less trivial are the following examples.

Example 1.5 Let (5; *) be a Stone algebra, let A be a distributive lattice that is bounded below, and let B be a distributive lattice that is bounded above with a dual isomorphism {} : A -7 B. On the linear sum L = A EEl 5 EEl B define a unary operation f by

f(x) = {

{}(x) x* {}-I (x)

if x E A; if x E 5; if x E B.

Then (L; f) is an Ockham algebra that belongs to K 1,1'

Example 1.6 Consider the set F of mappings p : IR+

-7

[0,1] under the

usual order, namely that given by

p ~ q ~ (Vx

E IR+)

p(x) ~ q(x).

It is clear that F is a bounded lattice; the smallest element is the constant map 0 : X 1-7 0, the greatest element is the constant map 1 : X 1-7 I, and for p, q E F their infimum and supremum are respectively the lower and upper

envelopes p 1\ q and p V q given by the prescriptions (p 1\ q)(x) = min{p(x) , q(x)} ,

(p V q)(x)

= max{p(x), q(x)}.

Examples of Ockham algebras; the Berman classes

11

Moreover, this lattice F is distributive. Now let a

E

IR+ be fixed and for every p E F define f(P) by setting

(\:Ix

E

(t(P)}(x) = 1- p(x + a).

IR+)

Clearly, we have f(O) = 1 and f(1) = O. Also, since 1 - min{p(x + a), q(x +

an = max{l - p(x + a), 1 - q(x + an

we see thatf(p !\q) = f(P) V f(q), and likewisef(p V q) = f(P) !\f(q). Thus (F;f) is an Ockham algebra. When a = 0 we have (t(P)){x) = 1 -p(x) and (t2(P)){X) = p(x). It follows that in this case (F;f) E K1,o' Moreover, since min{l, 1 -p} ~ ~ ~ max{q, 1 -q} we have p !\f(P) ~ q V f(q) and (F;f) is a Kleene algebra.

Example 1.7 Consider the bounded distributive lattice consisting of the interval I = [0, 1] of real numbers under the usual order. Every x E I with x :f 1 has a unique decimal representation

X=O.XIXZ x 3··· where each Xj E {O, 1, ... , 9}. For our convenience here, we shall write this as x = (Xj)i;;d' Let a be a fixed positive integer and for each such x define

f(x) = (9 -

Xi+ak;d,

with f( 1) = O. Then (J;f) is an Ockham algebra. In every Ockham algebra (L;f) the subset

S(L)

= {f(x) I x E L}

is clearly a subalgebra of L which we shall call the skeleton of L. The skeleton of L is a de Morgan algebra precisely when f3(x) = f(x) for every x E L, Le. precisely when L belongs to the Berman class K1,1' When this is the case, we shall say that L has a de Morgan skeleton. Note that in this case we also have S(L) = {fZ(x) I x E L}. Every Ockham algebra (L;f) contains a subalgebra that has a de Morgan skeleton. The greatest such subalgebra is

M(L)

= {x ELI f3(x) == f(xn.

For the Ockham algebra (F;f) of Example 1.6, M(F) is the set of those mappings p such that p(x + a) =p(x + 3a) for all x E IR+, Le. those that are

12

Ockham algebras

of period 2a after the point x = a. The skeleton of M(F) consists of those mappings that are of period 2a. In the Ockham algebra (J;J) of Example 1.7, M(J) is the set of real numbers x = (XJi;;'l in fa, 1] with Xi+p = x i +3P' i.e. those that are 2p-repeating after the p-th digit. The skeleton of M(J) consists of those x that are 2prepeating. We now describe a useful way of making bounded distributive lattices into Ockham algebras with de Morgan skeletons. First observe that if (L;J) is an Ockham algebra then the mapping f2 : L ----7 L is a lattice morphism; and if L has a de Morgan skeleton then P is idempotent. These observations yield the following result which, as we shall see, is very useful in constructing examples.

Theorem 1.2 Let L be a bounded distributive lattice and let tp : L ----7 L be an idempotent {O, 1} -lattice morphism such that the sublattice 1m tp admits a polarity p Define a unary operation f : L ----7 L by the prescription (Vx

E

L)

f(x)

=p[tp(x)J

Then (L;J) is an Ockham algebra with a de Morgan skeleton. Moreover, every such Ockham algebra arises in this way

Proof Since tp preserves 0 and 1, we havef(O) = 1 andf(l} = O. Now f(x I\y)

=p[tp(x I\y)] = p[tp(x) 1\ tp(y)] =p[tp(x)] V p[tp(y)] = f(x) V f(y),

and similarly f(x V y) =f(x) I\f(y), so (L;J) is an Ockham algebra. Since tp is idempotent we have that tp acts as the identity on Imtp = Imp. It follows from this that tpptp(x) = ptp(x) for every x E L, so that

f3(x)

= ptpptpptp(x) = ptpp2tp(X) = ptp2(X) = ptp(x) = f(x)

Thus (L;J) has a de Morgan skeleton. Conversely, if (L;J) is an Ockham algebra with a de Morgan skeleton then the mapping tp : X ~ f2(x) describes a {O, l}-lattice morphism on L, and from f3 = f we deduce that tp2 = tp and that f is a polarity on Imtp = {f2(X) I x E L}. 0 As an application of Theorem 1.2, we shall obtain an affirmative answer to the quite natural question of whether, given a bounded distributive lattice L, it is always possible to make L into an Ockham algebra with a de Morgan skeleton. For this purpose, we recall some definitions.

Examples of Ockham algebras; the Berman classes

13

A subset Q of an ordered set P is said to be a down-set if it is decreasing, in the sense that i E Q and j ~ i imply that j E Q. The down-set generated by a subset X of P is defined by

XL

= {y E P I (:Jx E X) Y ~ x}.

In particular, when X = {x} we write XL as xL. An ideal of a lattice L is a sublattice I of L which is also a down-set; and an ideal of the form xL is called a principal ideal. Dually, a subset Q of an ordered set P is said to be an up-set if it is increasing, in the sense that i E Q and j ~ i imply that j E Q. The up-set generated by a subset X of P is defined by

XT = {y E P I (:Jx E X) Y ~ x}. In particular, when X = {x} we write XT as x T. A filter of a lattice L is a sublattice F of L which is also an up-set; and a filter of the form x r is called a principal filter. Ideals and filters are convex, in the sense that if a, bEl Crespo F) and a ~ c ~ b then c E I Crespo F). An ideal or a filter is said to be be proper if it is not the whole lattice. A proper ideal I of L is said to be prime if a, bEL and a 1\ bEl imply that a E lor bEl. The notion of a prime filter is defined dually. The set-theoretic complement of a prime ideal is a prime filter. In a distributive lattice L there are sufficiently many prime ideals, so that any two elements can be 'separated' by a prime ideal, in the sense that if a, bEL with a b then there is a prime ideal I of L that contains one of these elements and not the other.

t-

Theorem 1.3 Every bounded distributive lattice can be made into an Ockham algebra with a de Morgan skeleton.

Proof Let L be a bounded distributive lattice. Then we can find in L a finite chain of prime ideals and a chain C of elements 0= a o < al

< .. < an < 1

such that a i Eli \ I i-I for 1 ~ i ~ n. Consider the mapping cp : L ~ L defined by ifxEIo; if x Eli \ Ii-I; if x ¢ InClearly, cp is an idempotent {a, l}-lattice morphism with Imcp = C. Now on the chain C a polarity p is uniquely defined. By Theorem 1.2, therefore, with

14

Ockham algebras

J(x) = p[cp(x)] for every x E L, we see that (L;1) is an Ockham algebra with

a de Morgan skeleton. <; The preceding proof shows in particular that many non-isomorphic Ockham algebras can be defined on the same distributive lattice; for we can attribute different values to n, and even for a fixed value of n there can exist different chains of prime ideals.

Example 1.8 Let n denote the n-element chain 0= ao

< al < .. < an-l

= 1.

Every antitone mapping cp : n ---7 n such that J(O) = 1 and J(l) = 0 determines an Ockham algebra. Thus the number of non-isomorphic Ockham algebras definable on n is equal to the number of antitone mappings from n - 2 to n, which is known to be Q!n

= (2n -

3) .

n-2

For small values of n, this number is given as follows:

n=234567 Q!n

=

1 3 10 35 126 462

Example 1.9 Consider the unit cube

c = {(x,y,z) E IR3 I x,y,z E [0, In. With respect to the cartesian order, C is a bounded distributive lattice. The mapping cp : C ---7 C given by cp(x,y,z):: (z,y,z)

is a {O, 1}-lattice morphism and is idempotent. Moreover, 1m cp admits the polarity p given by p(x,y,z)=(l-z,l-y,l-z).

By Theorem 1.2 we can therefore make C into an Ockham algebra with a de Morgan skeleton by defining J(x,y,z) = (l-z, 1- y, 1-z).

The skeleton of C is the set of elements (x,y,z) for which x = z, i.e. a diagonal plane. Now there is another polarity p' that can be defined on Imcp, namely that given by p/(X,y,Z) = (1- y, 1- z, 1- y)

15

Examples of Ockham algebras; the Berman classes In this case we obtain

!'(x,y,z) = (1- y, l-z, 1- y), which makes C into a different Ockham algebra with the same de Morgan skeleton.

Example 1.10 Let B be a boolean lattice and let L = {O} EB B EB {1}. Let a, b E B with b < a and define cp : L -'> 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}. ~

E

[b, a], and so

= Im
In Example 1.11, we have

MS(L) = {O, 1, ao} U {a2n+l, b zn I n

E Z}.

In Example 1.12, we have

MS(IP(E)) = [0, {a}'] U [{b},E]. In Example 1.13, we have

MS(B:)

={(Xl,· .. , Xn) I (Vi) Xi ~ X n }·

1.5 Let L be a bounded distributive lattice and let

1m

0

:

(Vx E L)

-t

by

XO

= p[
!ben (L; 0) is an MS-algebra. Moreover, every MS-algebra arises in this way.

Proof Every closure map must also be I-preserving and so by Theorem 1.2 we see that (L; 0) belongs to KI I . Since X OO =
Definition A bounded distributive lattice L together with a unary operation + : L - t L will be called a dualMS-algebra if (LOP, +) is an MS-algebra, where LOP denotes the order dual of the lattice L. It is clear that we can construct dual MS-algebras by using the dual of Theorem 1.5, which involves a I-preserving dual closure morphism.

Example 1.14 Let INo be ordered by the relation of divisibility and let L = {O} EB INo EB {oo}. Let n = IT pfi be the decomposition into prime factors of iEi

a fixed positive integer n. Define
-t

L by

if x = 00; if x E INo; if x = o.

Examples of Ockham algebras; the Berman classes Then

C(Jn

19

is a {O,oo}-preselVing dual closure morphism with ImC(J'1

A polarity on 1m C(Jn is given by 0

x

= IT pft

= [O,n]U{oo}. 00, 00

H

0 and, for x E [1, n],

H

i = IT

H

tE!

pfd3t.

tEIU!

It follows that (L; +) is a dual MS-algebra in which 0+ = 00, 00+ = 0 and, for x = IT p7 t E IN o, iEK

x+

=

n

-

gcd{n,x} -

IT

O
iEIUK Pt

2 Congruence relations Let (L;1) be an Ockham algebra. Then an Ockham algebra congruence Cor, briefly, a congruence) on L is an equivalence relation that has the substitution property for both the lattice operations and for the unary operation f. It follows that every congruence is in particular a lattice congruence and it is essential to distinguish these two types. In order to do so, we shall use the subscript 'lat' to denote a lattice congruence. If a, bEL and a ~ b then the principal congruence 19(a, b) generated by a, b is defined by

19(a, b) = !\{

(X,Y)E191at(a,b) ~ Xl\a=yl\a and xvb=yVb, and that the intersection of two principal lattice congruences is again a principallattice congruence; in fact, if a ~ band c ~ d then 191at(a, b) 1\ 191at(c, d) = 191at((a V c) 1\ b 1\ d, b 1\ d). A fundamental result concerning congruences that we shall require is that if L is an algebra and 19 is a congruence on L then for any congruence

([x]19, [y]19) E
~

(x,y)

E
is a congruence on L /19; and every congruence on L /19 can be uniquely represented as

Theorem 2.1 Let (L,j) be an Ockham algebra. If a ~ b in L then 19(a, b) =

V 191at(r(a),jn(b)). n;;;'O

Congruence relations

21

V 19Jat(r(a),r(b)). We show as follows that cp E Can L. n;;,O Suppose that (x,y) E cp, so that there are integers i 1 , ... , im and elements ZO,Zl,'" ,Zm such that

Proof Let cp ==

X

== Zo

l"i1

==

I"lm

1"12

Zl

== Z2 == ... == Zm-l == Zm == y,

where CPik = 19Jat(jik(a),Jik(b)). Now ifi k is even we have jik(a) ~:Jik(b) and so Zk-l /\/k(a) == Zk /\/k(a),

Zk-l V /k(b) == Zk V jik(b}.

Applying j to each of these equalities, we obtain (j(Zk-l)'/(Zk))

(*)

E

19Jat(ji k+l(a),/k+ 1 (b))

= CPik+ 1 '

If, on the other hand, i k is odd then Ji k ( b) ::;;; Ji k (a), and by a similar argument (*) also holds in this case. The integers i 1 + 1, ... ,im + 1 and the elements j(zo), ... ,/(zm) together with (*) now give (j(x),/(y)) E cpo Consequently, cP E Con L. Clearly, 19Jat(a, b)::;;; cP and hence (a, b) E cpo Since 19(a, b) is, by definition, the smallest congruence to identify a and b, it follows that 19(a, b)::;;; cpo Finally, since 19(a, b) is a congruence we have that (a, b) E 19(a, b)

=9

('in) (r(a),/n(b))

E

19(a, b),

from which we deduce that, for each n, 19(r(a),r(b))::;;; 19(a, b) and hence that cp Thus cp

= V 19Jat(jn(a),/n(b))::;;; V 19(jn(a),r(b))::;;; 19(a, b). n;;'O

n;;,O

= 19(a, b) as asserted. <> 2p+q-l

Corollary

If (L,/) E Kp,q then 19(a, b) == V 19Jat(jn(a),/n(b)). <> n=O

A class K of algebras is said to enjoy the (principal) congruence extension property if, for all A, B E K with A a subalgebra of B, every (principal) congruence 19 on A is the restriction of some congruence cp on B (this being denoted by CPIA = 19). In fact, as was shown by A. Day [66], in an equational class of algebras these properties are equivalent; indeed, they are each equivalent to the condition for all subalgebras A of B and all a, bE A, 19A (a, b) = 19 B (a, b)lA'

Ockham algebras

22

Using this fact, J. Berman [28] established the following result.

Theorem 2.2 Tbe class 0 enjoys the congruence extension property. Proof Let A, BE 0 with A a subalgebra of B. For a, bE A let A == {)A (a, b) and f..L = {)B(a, b). Then, by the above result of Day, it suffices to prove that A == f..L IA' Now by Theorem 2.1 we have A=

V {)lat(r(a),r(b») E Conlat A , n;;:,O

f..L

= V {)lat(r(a),r(b») E ConlatB. n~O

Denote the lattice congruences that appear in the right hand side of these equalities by An and f..Ln respectively. Clearly, A is a sublattice of B and so, since the class of distributive lattices has the congruence extension property, we have An = f..LnlA' Now since A is a lattice congruence it has an extension E Conlat B such that = A. It follows that I ~ f..Ln for all n and so

r

riA

A = IIA ~ whence A = f..LIA'

( V f..Ln) IA ~ A, n;;:,O

<>

Consider now, for any a E L, the relation {)a defined by

(x,y)

E

{)a

¢=>

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
Example 2.1 On the chain L

= {-oo} EEl 7L. EEl {oo} define f

x : -00 '" -2 -1 0 1 2... f(x): 00.. . 3 2 0 -2 -3 '"

by

00 -00

1ben (L; f) is an Ockham algebra with f injective. But for m =f n we have fm(x) =f r(x), so fm =f rand L does not belong to any Berman class. In general, a
Example 2.2 Let L be the chain 0 < a

< b < c < 1. Define f by

x:Oabc1 f(x) : 1 1 c a 0 1ben it is readily seen that (L; j) E 0 and that the smallest Berman class that it belongs to is K 1,3' Now f2(L) = {O, a, I}, and [O]

n-class of L contains exactly one element of 1'1 (L ). Here we shall deal with the subvarieties of 0 defined by x I\f2P(X) = x (see Chapter 1) and denoted by Ramalho and Sequeira [82] by K~o. Clearly, Kp,o C K~o C Kp,l' Note that in Kffo the mapping cp : X f-+ f2P(X) is a closure operator, since

cp(x) ~ x;

cp2(X) = cp(x);

X~Y

'*

cp(x) ~ cp(y).

This closure is additive in the sense that cp(x V y) = cp(x) V cp(y), and also multiplicative in the sense that cp{x 1\ y) = cp(x) 1\ cp(y). By 1beorem 5 of

26

Ockham algebras

[97], it follows that rp(L} is a semiconvex subalgebra of L, Le. if x 1\ y and x V y both belong to rp(L} then both x and y belong to rp(L}. As a direct consequence, rp(L} contains Z(L}, the centre of L (Le. the boolean sublattice of L formed by the complemented elements).

Theorem 2.8 If L E Kp,q and q = 2kp then every q-class contains exactly one element of f q(L). If L E Kff',o then every 2p -class has a greatest element, this being the only element of the class that belongs to j2P (L) . Proof Since L E Kp q' for every x E L we have (x, j2P(x}) E q, and therefore ([ZP(x},j4 P(x}) E q and so (X,j2kP(X}) E q, i.e. (x,fq(x}) E q. It follows that [x]q contains exactly one element of fq(L} since (x,y) E q is equivalent to r(x} = jq(y}. Now let L E K~o. Then L E Kp,l ~ Kp,2p' Since j2P is a closure operator, the greatest element of [X]2p is the only one that belongs to j2P(L}. ¢ If an Ockham algebra (L; f) belongs to a Berman class then there is a smallest Berman class to which it belongs; we denote this by VB (L).

Theorem 2.9 For an Ockham algebra (L; f) the jollowing statements are equivalent: (I) VB(L}

= Kp,q; =jq+l(L} = ''', and VB (r(L)) = Kp,o' : Fromjq(x} = j2p+q(x} = fq+l[j2P-l(X)] Efq+l(L} it fol-

(2) L ~ f(L} ~ ... -:J jq(L}

Proof (I) => (2) lows that fq(L} ~ jq+l(L}, whence fq(L}

= fq+1(L} = .... Suppose now, by way of obtaining a contradiction, that for some n with n ::;; q we have fn-l(L} =r(L}. Then clearly fq-l(L} = jq(L}. But, by (I), L ¢ Kp,q_l and so there exists x E L with fq-1(x} 'f j2p+q-l(X}. But fq-l(x} Ejq-l(L} = r(L} and so jq-l (x) = fq(y} for some y E L. Hence fq(y} 'f f 2p +q(y}, and this contradicts the fact that L E Kp,q' This then establishes the chain L ~ f(L} -:J f2(L}

~

..

~

fq(L} == r+l(L}

= ....

Now since jq =j2p+q we see that f2P acts as the identity on fq(L}. Hence fq(L} E Kp,o' ThatVB(r(L}) = Kp,o follows from the fact that iffq(L} E Kt,o C Kp,o then necessarily tIP and, sincej2t is then the identity onjq(L}, we have f q = j2t+ q so that L E Kt,q' which contradicts VB(L} = Kp,q' (2) => (I) : If VB (tq(L}) = Kp,o then clearly = j2p+q and so L E Kp,q'

r

If VB (L)

= Kp' ,q' then from (I) => (2) it follows that q' = q and p' =p. ¢

Corollary 1 IfVB(L) == Kp,q then,for 1 ~ i::;; q, VB(ti(L}) = Kp,q_i' ¢

Congruence relations

27

Corollary 2 IjVB(L) =

Kp,q then

w = <1>0 < <1>1 < <1>2 < . < q and Con L has length at least q + 1 . 0

= q+1 = ... ,

In a bounded distributive lattice every ideal I is the kernel of at least one congruence, i.e. there is a lattice congruence f} such that [O]f} = I. We also write I = Kerf}. Dually, every filter F is the cokernel of a congruence, i.e. there is a lattice congruence 'If; such that [l]'If; = F. These properties do not carry over to Ockham algebras. For instance, in Example 2.2 the principal ideal c! is not the kernel of any congruence; for (0, c) E f} implies (a, 1) E f}, whence the contradiction (0, 1) E f}. So it is of interest to characterise those ideals of an Ockham algebra that are congruence kernels. Let I be an ideal of the Ockham algebra (L;/). For each n ) 0, define 1211

= (r21l(I))!,

121l+1

= (r21l+1(I))i.

Then clearly I 2n is an ideal and 1 211 +1 is a filter. Finally, let

r = V 12n+1. 1l~0

We shall now investigate, for a given ideal I, the smallest congruence e (I) on L that identifies the elements of I. We recall that, in a distributive lattice, e (I) is characterised as follows : (X,y)Ee(I) ~ (:liEI) xVi=yVi.

With the obvious subscript 'lat' to denote the corresponding smallest lattice congruence, we have the following result.

Theorem 2.10 For every ideal I of (L;/) E 0, e(I)

=elat(r) V elat(Ioo)'

Proof This follows immediately from Theorem 2.1 on noting that for every subset X of L we have

e (X) = V{ f}(a, b) I a, b EX}. 0

Theorem 2.11 For every ideal I of (L;/)

E

(X,Y)Ee(I) ~ (:liEloo)(:ljEr)

0,

(xVi)/\j=(yVi)/\j.

Proof For convenience, denote by 'II the relation defined on L by the above condition. Since L is distributive, 'II is a lattice congruence. Suppose that (x,y) E 'II. Then there exist i E 100 and j E fO with (x V i) /\j = (y V i) /\j, whence x /\j =x /\ (x V i) /\j = x /\ (y V i) /\j

28

Ockham algebras

and hence {x, x /\ (y V i)) E 8Iat (/0). Now we also have {x /\ (y V i)) Vi = (x V i) /\ (y V i) = (x /\y) V i and so

{x /\ (y V i), x /\y) E 81at (/00)' Thus we see that (x, x

I\y) E 8Iat (r) V 81at (/00)'

Similarly, we can show that (y, x /\y)

E

8Iat (/O) V 81at (/00)'

Thus, by Theorem 2.10, (x,y) E 8(1). Conversely, suppose that (x,y) E 8(/). Then, again by Theorem 2.10, we have the finite sequence

x=. P1 =. pz =. ... =. Pn =. y where each =. is either 81at (fO) or 81at (/00)' Suppose that we have

x 81at (/00) P1 8Iat (r) pz 8Iat (/oo) ... 81at (r) y. Then we have finitely many equalities

Pl /\h =P2I\h, pz Viz = P3 Viz, P3 /\iz =P4 /\iz, ... where each i k E 100 and each ik E 1°. Since 100 is an ideal and 10 is a filter, we have Vi k E 100 and I\ik E 10. Moreover, xVi 1 = Pl Vi 1 ,

(x V Vi k) 1\ I\Jk = (P1 V Vi k) /\ I\ik =(P1 /\ I\) k) V {V i k /\ I\} k) = (Pz /\ I\ik) V (V i k /\ I\ik) =(P2 V V i k) /\ I\ik

= (y V Vi k) /\ I\}k, from which we see that (x ,y) E 'P and hence that'P coincides with 8 (/). <> The following observation is immediate. Lemma 2.1

If the ideal I

is such that fZ(/) ~ I then,jorevery n,

fZn(/) ~ I,

f21l+1(/) ~

(t(/))i,

from which it follows that

Theorem 2.12 An ideal I ofan Ockham algebra (L;J) is a congruence kernel if and only if

29

Congruence relations

(O!) P(1) ~ I; ({3) (\Ix

E L)(\lj E

(t(1))i)

Xl\jEI

'* XEI.

Proof If I is the kernel of a congruence 8 on L then for every i E I we have (t2(i),0) E 8, so thatf2(i) E I and we have property (O!). Ifj E L is such that j ~ f(i) for some i E I then (i, 0) E 8 gives (j, 1) E 8, whence (x I\j,x) E 8 for every x E L. Property ({3) is now immediate. Conversely, suppose that (O!) and ({3) hold. By Theorem 2.11, Lemma 2.1, and condition (O!), (x,0)E8(1)

-¢=}

(3iEI)(3jE(t(1))i)

(xVi)l\j=il\j.

We have i I\j E I and so, by ((3), x ViE I and hence x E I. It follows that Ker 8 (1) ~ I. Since I ::::: Ker 8Iat(1) ~ Ker 8 (1), it follows that Ker 8 (1) ::::: I. <>

Corollary 1 If a proper ideal I is a congruence kernel then I nf(1) ::::: 0.

<>

An element of an ordered set P is said to be a node if it is comparable with all the elements of P. A node is non-trivial if it differs from 0 and 1. An element a of an Ockham algebra (L;f) is a fixed point of f if a = f(a). With these notions, we have :

Corollary 2 Let (L;J) E 0 and let a be a non-trivial node of L that is also a (necessarily unique) fixed point of f. Then the congruence kernels of L are the proper ideals of at that satisfy condition (O!) above, and L itself. <> We now proceed with some considerations concerning the structure of the congruence lattice Con L of an Ockham algebra (L;J). Since Con L is a sublattice of ConlatL, which is known to be distributive, Can L is also distributive. It is also algebraic, in the sense that it is complete and compactly generated (every element of Can L is the supremum of a set of compact elements, a compact element being an element 1) such that if 1) ~ sup X for some X ~ Con L then 1) ~ sup Xl for some finite X 1 ~ Con L). As far as the structure of Con L is concerned, it is important to observe that every lattice congruence that is contained in <1>1 is a congruence. It follows that if L is finite then the interval [w, d of Can L is boolean. We also point out that <1>1 is dually dense in Con L, in the sense that if 1) V <1>1 ~ then 1) = ~. In fact, for any 1) E Can L we have (0,1) E 1) V <1>1 if and only if (0, 1) E 19; for

=

o 19 Xl

<1>1 X2 19 X3 <1>1 ... <1>1 Xn-2 19 Xn-1 <1>1 1

Ockham algebras

30 implies that 1

f}

f(xd

= f(X2)

f(x3)

f}

= ...

== f(X n-2)

f}

f(Xn-l)

= O.

We have already seen that the class 0 of Ockham algebras enjoys the congruence extension property. This fact can be useful in the determination of Can L, especially when f has a fixed point. In order to illustrate this, we consider the following notions. Let A and B be algebras of the same type. Then we say that A is an extension of B if B is a subalgebra of A. We say that A is a strong extension of B if every congruence on B has at most one extension to A, in which case B is said to be a strongly large subalgebra of A. Finally, A is a perfect extension of B if every congruence on B has exactly one extension to A, in which case B is a perfect subalgebra of A. When this happens, we have Can A ~ ConB. In [100] it is shown that, for any element a of a modular lattice L, the cone C a generated by a, i.e. the set at U a r, is a strongly large sublattice. This leads to the following property.

Theorem 2.13 Let (L; f)

E0

and let a be a fixed point of L. Tben

Can L ~ Can Ca.

Proof Clearly, C a is a subalgebra of L and this subalgebra is strongly large. In fact, since 0 has the congruence extension property, Ca is a perfect subalgebra. Hence we have the isomorphism stated. <> The usefulness of Theorem 2.13 lies in the fact that it enables us to work with Ca instead of L, and we can benefit from this in two ways. Firstly, the size of C a can be considerably less than that of L. Secondly, C a will in general belong to a subvariety of 0 that is smaller than that of L; for instance, it is clear that (Ca;J) always satisfies the axiom x Af(x} ~y V f(y) even if (L;J) does not.

Example 2.3 On the lattice L with Hasse diagram

e

definef by x:Oabcdef1 f(x) : 1 d f cae b 0

31

Congruence relations

Then (L;/) is an Ockham algebra that belongs to K1,o and has two fixed points, namely c and e. The cone Ce generated by e is a five-element chain. As a subalgebra of (L;/) it also belongs to K1 0 and satisfies the supplementary relation x I\f(x) ~ Y V f(Y), which L itselr'does not (consider the elements e and c). It is easily seen that Con Ce is a four-element boolean lattice and therefore so also is Con L. Note also that by Theorem 2.12 the only ideals of L that are congruence kernels are 0 1, a 1, b1, and 11.

Example 2.4 (!be pineapple) Consider the ordered set L with Hasse diagram e1

eO

and made into an Ockham algebra by definingf{xi) = Xi+1 for each i. Note that f is injective so, by the Corollary of Theorem 2.7, w = <1>1 = ... = w' We leave to the reader the task of verifying that Con L is the chain w

< .. -< 19 {Xi+1 , Xj+z) -< 19{Xi+1, Xi} -< ... < \}I -<

£

where \}I has classes {O}, {I }, L \ {O, 1} . We shall return to this particular example later. Concerning the basic congruences in a general Ockham algebra (L;/) we also have the following results.

Theorem 2.14 If a, bEL with a ~ band f{a} = f{b} then 19{a, b} has a complement in [w, d.

32

Ockham algebras

Proof By Theorem 2.1 we have 19{a, b)

= V 191at (fll{a),r{b)). 11;;:0

Since f{a) = f{b) by hypothesis, it follows that 19{a, b) = 191at{a, b) E ConL with, clearly, 19{a, b):::;;
Consider the lattice congruence Q!I = (3 /\
= 19{a, b) V {(3 /\ 1) = £ /\
and Q! /\ Q!I = 191at{a, b) /\ (3 /\
= W /\
It follows that Q!I is the complement of Q! in [w, If Q! is a congruence on an Ockham algebra L then an Q!-class [a]Q! will be called locally finite if, whenever x, y E [a]Q! with x :::;; y, the interval [x, y] is finite. The following two results are due to Jie Fang [67].

Theorem 2.15 Let a, bEL be such that a -< b,

(a, b) ¢
(a, b)

E

Then
Proof Since fn+l{a) = fn+l{b) we have (f{a),J{b)) E

=
Clearly, we have
(2)

E

ConL is such

= w.

In fact if (x,y) E

(*)

x/\a=y/\a,

xvb=yvb,

{X,Y)E
Writing s = (x V a) /\ band t = (y V a) /\ b we see that (s, t) E

Congruence relations

33

s::f t then one of s, t must be a and the other must be b, whence (a, b) E <po This gives the contradiction 19{a, b) ~ <po Hence we must have s = t, i.e. (x V a) /\ b = (y V a) /\ b. But from (*) we have x Va V b =Y Va V b and so, by the distributivity of L, x V a = y Va. Again by (*) and the distributivity of L, we obtain x = y and hence

n V 19{a, b)) =

n V 191at{a, b)) =

n

and therefore

n> whence

/Z' Hence n V 19{a, b) is an atom of [n, n+l]. Finally, let 01 be an atom of [n, n+ d. Then there exist a, bEL such that a < b, (a, b) ¢ n' and (a, b) E 01 ~ n+l' If every n+l-class is locally finite there existp, q E [a, b] such thatp -< q, (p, q) ¢ n, (p, q) E 01. For suchp, q we have n < n V 19(p, q) ~ 01, whence 01 = n V 19(p, q). <>

Corollary If a, bEL are such that a -< band f{a) an atom of Con L.

=f{b) then 19{a, b) is

Proof Take n = 0 in the above. Then 19{a, b) is an atom of [w, d and hence an atom of Con L. <> Theorem 2.16 Ifthe congruence w has locallyfinite classes then every nontrivial interval [n, n+d of ConL is a complete atomic boolean lattice. Proof For every

n ~

= V{ 19{a, b) I (a, b) E
Thus, if

11 V V{19(a,b) I {a,b)¢n' {a,b)E
E [<1>11'

n+d then, since the n+l-classes are locally finite, we have

= n

V V{19(P,q) I (p,q)

¢ <1>11' (p,q) E
Now for suchp, q we have, by Theorem 2.15, that n V19(p,q) is an atom of [<1>11' n+l]. Consequently, [n, 1l+d is a complete atomistic lattice in which the infinite distributive law x /\ VYi = V{x /\Yi) holds, and so is boolean. <> i

i

Example 2.5 Let L be the chain 0 < a < b < c < d < e < 1 made into an Ockham algebra by defining f as follows :

x f{x)

loa 1

1

b 1

c 1

del b b 0

There are 21 congruences on (L;J), depicted as follows:

Ockham algebras

34 1•

• • • • • • • • • • • • • •

•

e.

• • •

d·

• • •

II

• •

•

c.

•

b• a.

•

o• , w

I

• •

I

•

•

• • • <1>1

<1>2

"1

"2

"3

I

•

• •

•

• •

III

•

•

I

• • •

'"

"5

•

• • •

•

• •

•

II

•

"6 "7 "8 "9 "10 "11

"12 "13 "14 "15 "16 "17

The lattice of congruences of L is

17

15

w

Note that [w, d ~ 24 and [<1>1, <1>2] ~ 2. The condition that every w-class be locally finite cannot be removed from Theorem 2.16, as the following example shows.

Example 2.6 Consider the infinite chain C given by

o < Xl < X2 < ... < 01 < ... < Y2 < Yl < 1 made into an Ockham algebra by defining

j{O} = 1, j{l}=O, {Vi} j{Xj} =j{yj} =j{O!} =

01.

Here we have w = <1>1 with classes {O}, {I}, and C \ {O, I}. The w-class C \ {O, I} is not locally finite. Consider now the partition {{O}, {xjli~l}, {O!}, {yjli~l}, {I}}.

Congruence relations

35

This defines a congruence in [w, «I»d which has no complement in [w, «I»d. So in this case [w, «I»d is not boolean. The following interesting result was obtained by J. Vaz de Carvalho [105].

Theorem 2.17 If L E Kn 0 is finite then Con L is boolean Moreover, if the length of L is m then Con L has at most m atoms, and has precisely m atoms if and only if L itself is boolean. I

Proof We first show that if a -( b then 1?(a, b) is an atom of Con L. Suppose that 1? E ConL is such that w ~ 1? < 1?(a, b). By Theorem 2.1 we have 21Z-1

1?(a, b) =

V

1?lat(rk(a)Jk(b»

k=O

Since in this case f is a dual automorphism and a -( b, if k is even we have fk(a) -( fk(b), and if k is odd thenfk(b} -( fk(a). In what follows we shall suppose that k is even; a similar argument holds when k is odd. We now show that

'I? 1\ 'l?lat(rk(a}Jk(b})

=w.

In fact, if (x,y) E 'I? 1\ 191at(rk(a)Jk(b» then

(1) x I\fk(a} =y I\Jk(a)j

(2) xV fk(b} = Y VJk(b)j (3) (x,y) E 1? Consider the elements

u

= (x V jk(a}) I\jk(b},

v = (y V jk(a» I\jk(b).

We have u, v E (fk(a),jk(b)} and (u, v) E 1? Suppose that u ':f v. Then one of these elements is jk(a}, the other is Jk(b), and (rk(a),jk(b}) E 1? Since 1? is a congruence it follows that

(a, b) = (r21l-k [fk (a)], f211-k[fk(b}]} E 19,

=

whence 1?(a, b} ~ -a, a contradiction. Thus we have u v. This, together with condition (2) and the distributivity of L, gives x V Jk(a} y V fk(a}; and this together with condition (I) gives likewise x = y. Using this observation, we now have

19 = 191\ -a(a, b) =

21Z-1

V

k=O

=

(1? 1\ 1?lat(rk(a},jk(b))} =w

and consequently 19(a, b) is an atom of Con L. Now take a maximal chain 0=

Zo

-< Zl

-(

.. -(

zm

=1

36

Ockham algebras

in L. Then we have ~=

m-1

V 19(Zi,ZI+1)

1=0

and so ConL is a bounded distributive lattice whose greatest element is a join of at most m distinct atoms, whence it is boolean. If now L is boolean with m atoms then it is well known that ConL is also boolean with m atoms. Conversely, suppose that L E Kn,o is of length m and that ConL is boolean with precisely m atoms. Observe first that if x E L \ {O, I} then x Ilf(x). In fact, suppose that x < f(x) and consider a maximal chain passing through both, of the form

0= Zo -< '" -< x

= Zj+1 -< ... -< f(x) = Zt -< f(zj) = Zt+1 -< ... -< Zm = 1.

Since ConL has precisely m atoms, all the congruences 19(Zj,Zj+1) are distinct. But (t(Zi),j(X)) E 19(zj,x) and 19(zi,x) is an atom of ConL, so we have the contradiction 19(zj,x) = 19(t(x),j(Zj))' Similar arguments show that x> f(x) and x = f(x) give contradictions. Since for every x E L we have f{x V f(x)) ~ x V f(x), the above observation gives x V f(x) == 1; and from x Af(x)~f(x Af(x)) we obtain x Af(x) = O. Thus x andf(x) are complementary, whence L is boolean. <>

Theorem 2.18 Let L be ajinite Ockham algebra for which VB(L) Then each summand of

= Kp,q'

[w, d El1 [<1>1, <1>2] El1 ... El1 [q, ~]

is boolean. Proof By Theorem 2.7 we have w L/q

rv

=q and so, by Theorem 2.6, fq(L)

E

Kp,o

whence, by Theorem 2.17, Con L/q is boolean; and by a standard result in universal algebra [13], Con L/q ~ [q,~]. The result now follows by Theorem 2.16. <>

3 Subdirectly irreducible algebras An algebra L is said to be subdirectly irreducible if it has a smallest nontrivial congruence; i.e. a congruence O! such that {) ~ O! for all {) E Con L with

{) "f w. Such a congruence O! is called the monolith of Con L. The importance of such algebras is shown in a classic theorem of Birkhoff [2] which states that in an equational class of algebras every algebra can be embedded in a direct product of subdirectly irreducible algebras. An immediate consequence of the above definition is that if L is subdirectly irreducible then in Con L the trivial congruence w is A-irreducible. A particularly important case of a subdirectly irreducible algebra is a simple algebra, namely one for which the lattice of congruences is the two-element chain {w, ~ }. The existence of infinite subdirectly irreducible Ockham algebras can be seen from the following two examples. These are due to Berman, who also showed that every class Kp,q has only finitely many subdirectly irreducible algebras all of which are finite [28, Theorems 7,8]. Example 3.1 Let n E IN and consider the set Ln = {O, I} U {a j

I - n ~ i ~ n},

totally ordered by

o < a_ n < a-n+l < ... < a_I < ao < al < ... < a n-l < an < 1. Define I: Ln ---+ Ln by 1(0) = 1,/(1) = 0, I(ao) = ao, and I(a m) = a_ m+1 , I(a- m) ::: am· It is readily seen that (Ln;/) is an Ockham algebra and that it belongs to the Berman class K 1 ,2n- From the definition of 1 it follows that any congruence {) on Ln that identifies an adjacent pair aj, a j+1 (Le. any congruence different from w) must identify ao, a 1. Consequently, Con Ln has a smallest non-trivial element, namely the principal congruence {)(ao, al)· Hence Ln is subdirectly irreducible. The algebra La is simple. ('rim ~ 1)

Example 3.2 (Tbe centripetal see-saw) With LlI and 1 as in Example 3.1, let L",,::: ULn={O,l}U{a j liEZ}. n~O

Then (L",,;/) is an infinite totally ordered subdirectly irreducible Ockham algebra, with monolith {)(ao, ad.

38

Ockham algebras

Associated with the notion of a subdirectly irreducible algebra is that of a finitely subdirectly irreducible algebra, this being defined as an algebra in which the intersection of two non-trivial principal congruences is nontrivial. Clearly, every subdirectly irreducible algebra is finitely subdirectly irreducible. The following is an example of a finitely subdirectly irreducible Ockham algebra that is not subdirectly irreducible.

Example 3.3 (The centrifugal see-saw) Let L be the chain

o < ... < a3 < al < ao < a2 < a4 < ... < 1 and define f

:L

----7

L by

f(O} = 1,

f(l} = 0,

(i ~ O) f(a j } = a j + 1 .

Then (L;f) is an Ockham algebra. We leave as an exercise for the reader the task of showing that Con L is the lattice t9(ao,ad

t9(ao,l) t9(aO,a2)

t9(O,ad t9(tlj,tl3)

t9(a2,l) t9(a2,a4)

W.

Clearly, L is finitely subdirectly irreducible but not subdirectly irreducible.

Example 3.4 The pineapple (Example 2.4) is finitely subdirectly irreducible but not subdirectly irreducible. Given an Ockham algebra (L;f), consider now for each i ~ 1 the subset Tj(L}

= {x E L Il(x} = x}.

In particular, T 1 (L) is the set of fixed points off. Of course, it can happen that Tl (L) is empty. More generally, every subset T 2n +1 (L) is either empty or is an antichain; for if x,y E T2n+dL) with x ~y then x =f2n+l(x) ~pn+l(y) =y so x = y. In contrast, T 2 (L} is never empty, for it clearly contains 0 and 1. It

Subdirectly irreducible algebras

39

is readily seen that every subset T 211 (L) is a subalgebra of L; in fact, T 2n (L) is the largest ~l,o-subalgebra of L. Clearly, we have the chains

T2j (L) ~ T4j(L) ~ .. ~ T 21l/(L) ~ T 2'HIAL) ~ Consider now the subset

T(L)

= U T 2n (L), n;<;1

i.e. the set of x E L for which there exists an even positive integer mx such thatrx(x) = x. Given x,y E T(L), let m = lcm{mx,my }' Then, m being even, we have

fm(x

V y)

=fm(x) V fm(y) = X

V y,

and similarly r(x /\y) = x /\y. Since x E T(L) clearly implies f(x) E T(L), it follows that T(L) is also a subalgebra of L. Note that T(L) ~ S(L). Consider also the subset K(L) defined by

K(L)

={a, I} U TdL).

We shall call this the core of L. This is not in general a subalgebra of L.

Theorem 3.1 If the core of (L;J) is a subalgebra then f has at most two fixed points. Proof Suppose that K(L) is a subalgebra and that f has at least two fixed points a, b. Then a V b E K(L) and a /\ b E K(L). Since distinct fixed points cannot be comparable, it follows that we must have a V b = 1 and a /\ b = 0. Thus any two fixed points are complementary. The distributivity of L now shows that there can be at most two fixed points. () It is clear that K(L) ~ T2(L). In general, we have strict inclUSion; for example, the 4-element chain 4 made into a de Morgan algebra is fixed point free and K(4) = {a, I}, T 2 (4) = 4. It is natural to consider the particular case where K(L) = T 2 (L). An important situation in which this happens arises from the follOWing.

Theorem 3.2 Let (L; f) be a finitely subdirectly irreducible Ockham algebra. If a, bEL are such that f(a) = band f( b) = a then either a = b or {a, b} = {a, I}.

°

°

Proof If {a, b} f: {O, I} then we have < a < 1 or < b < 1. If a /\ b = (in which case 1 = f(a) V f(b) = b V a) then, by Theorem 2.1,

°

'19(0, a) /\ '19(0, b) = ['I91at (0, a) V 'I9lat(b, 1)] /\ ['I91at(O, b) V 'I9lat(a, 1)]. Since Conlat L is distributive, we can expand the right hand side. Observing that 191at(0, a) /\ 191at(0, b) = w, that 191at(0, a) /\ 191at(a, 1) = 191at(a, a) = w, and

40

Ockham algebras

using the hypothesis that a 1\ b = 0, we see that the right hand side reduces to w. 1his contradicts the hypothesis that L is finitely subdirectly irreducible. Consequently we must have a 1\ b::f 0 in which case, by Theorem 2.1, 19(0, a 1\ b) 1\ 19(a 1\ b, a V b)

= [191at(0,a 1\ b) V 191at(a V b, 1)] 1\ 191at(a 1\ b,a V b) = 191at(a 1\ b, a 1\ b) V 191at(a V b, a V b) = w. It follows that 'l9(a 1\ b, a V b) = w, whence a 1\ b = a V b and hence a = b.

Corollary 1 If L is finitely subdirectly irreducible then K(L)

<>

= T 2 (L) .

T 2 (L) thenf[f(a)] = a. Taking b =f(a) we deduce that either a = f(a) or {a,j(a)} = {O, I}, i.e. a E Tl (L) U {O,l} = K(L). <>

Proof If a

E

Corollary 2 fixed points.

If (L; f) is finitely subdirectly irreducible then f has at most two

Proof This follows from Corollary 1 and Theorem 3.1. <> Consider now the particular case where L is a finitely subdirectly irreducible de Morgan algebra. By the above, we have L = T 2 (L) = K(L) = {O,l} U T1(L)

where, by Theorem 3.1, TdL) is either empty or is an antichain of at most two elements. We can therefore state the following result, which was first obtained by Kalman [72].

Theorem 3.3 In the class M of de Morgan algebras there are only three (finitely) subdirectly irreducible algebras, each of which is Simple, namely the algebras

I~

B

K

Proof By the above, the algebras shown are the only candidates. It is readily seen that for these we have Can B = Can K = Can M

~

2,

so these algebras are indeed sub directly irreducible; in fact, they are simple. <>

41

Subdirectly irreducible algebras

Theorem 3.4 If LEO is such that K(L) that a

< b then fJ(a, b) =

Proof For every x

= T 2(L) andija, b E T(L) are such

L

T(L) let mx be the least even positive integer such that fmx{x) = x. With nx = ~mx' consider the elements E

Ilx- 1

Ilx- 1

O!(x) =

1\

f2i(X),

i=O

(3(x)

= V f2i(X}. i=O

Observe thatj2[O!(x)] = O!(x) andj2[(3(x)] = (3(x), so O!{x), (3(x) E T 2(L) = K(L). Now let a, bE T{L) be such that a < b. Consider the sublattice M that is generated by

{j2i(a),j2j(b) I 0 ~ i ~ n a , 0 ~j ~ nb}' Clearly, M is finite with smallest element O!(a) and greatest element (3(b). Let p be an atom of M and consider the interval B = {O! (a), (3(P)] in M. Since every atom of M is Gf the form 1\ j2i (a) for some j, it follows that j2 (P) is ifj

also an atom of M. Consequently, B is boolean; for it is a finite distributive lattice whose greatest element is a join of atoms. Observe that O!(a) < (3(P) and so, since both belong to K(L), we have that O!(a) is either 0 or a fixed point, and (3(P) is either 1 or a fixed point. Clearly, a II (3(P) and b II (3(P) belong to B, and

(a II (3(P), b II (3(P») E fJ(a, b). c

~

If a II (3(P) < b II (3(P), let c be an atom of B with c 1; a II (3(P) and b II (3(P). Then we have

(O!(a), c)

= (a II (3(P) II c, b II (3(P) II c) E fJ(a, b).

It follows that (O!(a), (3(c» E 'I9(a, b). Since O!(a), (3(c) E K(L) with O!(a) < (3(c), we deduce that (0,1) E 'I9(a, b) and that therefore 1J(a, b) = t. If now a II (3(P) = b II (3(P) let a 1 = a V (3(P) < b V (3(P) = b l . Then (al, hd E '19 (a , b). Moreover, we cannot have (3(P) = 1, so (3(P) must be a fixed point. Then (3{ b) = 1; for otherwise (3( b) = (3(P) gives the contradiction

a = a II (3(b) = a II (3(P) = b II (3(P) = b II (3(b) = b. Considering therefore the interval [(3(P), 1] in M and a coatom q such that q ~ a V (3(P) and q j b V (3(P), we see in a dual manner that fJ(a, b) = t. 1:/

Theorem 3.5 For an Ockham algebra L the following are equivalent: (1) K(L) = T 2 (L); (2) the subalgebra T(L) is simple;

Ockham algebras

42 (3) all de Morgan subalgebras oj L are simple. Proof (1)

(2) : If (1) holds then by Theorem 3.4 every non-trivial principal congruence on T(L) coincides with £. Since every congruence is the supremum of the principal congruences that it contains, it follows that T(L) =?

is simple. (2) # (3) : T2 (L) is the largest de Morgan subalgebra of 1. (3) =? (1) : If (3) holds then T2 (L) is simple. But T 2 (L) E Kl,o = M; and by Theorem 3.3 there are only three non-isomorphic simple de Morgan algebras, in each of which K(L) = T2(L). Since T2(L) and L have the same core, (1) follows. ~

Corollary For an Ockham algebra L we have K(L) T(L) is a simple de Morgan algebra.

= T(L) if and only if

= T(L) then K(L) = T2(L) = T(L) and so T(L) is simple de Morgan. Conversely, if T(L) is simple de Morgan then K(L) = T2(L)j and T(L) = T2(L) since T 2(L) is the largest de Morgan subalgebra of 1. ~

Proof If K(L)

In order to examine more closely the (finitely) subdirectly irreducible Ockham algebras, we shall concentrate on a class that contains all the Berman classes, namely the subclass Kw of 0 defined by

(L;/)

E

Kw

*'*

(\:Ix

E

L)(:Jm, n

E

IN)(m

t 0)

jm+n(x) =r(x).

Without loss of generality, we may assume that both m and n are even; for

jm+n(x) =r(x) implies

j2m+2n(x) = jm+nlr(x)] =rvm+n(x)] = j2n(x). By definition, Kw is closed under the formation of subalgebras, homomorphic images, and arbitrary direct powers. But it is not closed under arbitrary direct products, as can be seen by taking an algebra Lq E Kp,q for each q ~ 0 and considering Lo XLI xL 2 X .... However, the following result, due to Jie Fang [67], enables us to claim that Kw is closed under finite direct products, and therefore forms a generalised variety in the sense of Ash [23].

Theorem 3.6 Let (L1;/) and (L 2;/) be Ockham algebras. If Xl ELI and X2 E L2 are such that there exist natural numbers ml, nl and m2, n2 with jml+111 (Xl)

=jnl (Xl),

jm 2 +1l 2 (X2)

=jn

2

(X2),

then there are natural numbers m, n such that, in the algebra L1 x L2, jm+n(Xl' X2)

=r(Xl, Xl)'

43

Subdirectly irreducible algebras

Proof Observe first that fP+q(x) = jq(x) implies jkp+q(x) = jq(x) for all k E IN o. Now let m = lcm{m l ,m2} and n = lcm{nl,n2}' Then m = mlr, n = nls and jm+n(XI) =jm l r+1I 1 +(S-I)1I 1 (Xl) =fill +S1I1-1I1 (Xl)

=J'l(XI)' Similarly, jm+lI(X2)

=J'l(X2)'

The result now follows. <>

A class C of algebras is said to be locally finite if every finitely generated member of C is finite. In particular, it is well known that the class DOl of bounded distributive lattices is locally finite.

Theorem 3.7 The generalised variety Kw is locally finite. Proof Suppose that L E Kw is O-generated by {x I , ... ,xk}' Then there are natural numbers mj, nj with mj t- 0 such that jm;+n/(xj) = jn/(Xj) for i = 1, ... , k. By Theorem 3.6 and induction, there exist m, n (m t- 0) such that (i = 1, ... , k) It follows that L belongs to

jm+n(Xj) = jn(Xj) [= j2m+n(Xj)].

Km,1l and is DOl-generated by

{fj(Xj)IO~j<2m+n, 1~i~k}.

The result now follows from the fact that DOl is locally finite. <> Our objective now is to show that for every algebra in Kw (and therefore every algebra that belongs to a Berman class) the properties of being finitely subdirectly irreducible and subdirectly irreducible are equivalent.

Theorem 3.8 If an Ockham algebra L is finitely subdirectly irreducible then every CPI -class in L contains at most two elements.

Proof Suppose that a CPt -class contains at least three elements. Then it contains a 3-element chain X < y < z with j(x) = j(y) = j(z). Then, by Theorem 2.1, we have 19(x,y) = 191at(X,y) and 19(y,z) = 191at(Y,Z), whence we have the contradiction 19(x,y) 1\ 19(y,z) = w. <>

Theorem 3.9 If L E Kw then the jollowing statements are equivalent: (1) L is finitely subdirectly irreducible; (2) L is subdirectly irreducible. Proof (1) => (2) : Since L E Kw, for every X E L we have jm+n(x) == J'l(x) for some m, n E IN with m t- 0 and even. If CPt == w then j is injective and X = jm(x), whence X E T(L). Thus L = T(L) and it follows by Corollary 1 of

44

Ockham algebras

Theorem 3.2 and Theorem 3.5 that L is simple, hence subdirectly irreducible. On the other hand, if cI>1 :f w then by Theorem 3.8 there is a two-element cI>l-class {a, b} and, by the Corollary of Theorem 2.15, 19{a, b) is an atom in the interval [w, cI>d of Con L. If now Q! E Con L with Q!:f w then, since Q! is the supremum of the non-trivial principal congruences which it contains and since Con L satisfies the infinite distributive law {3/\ V"I i = V({3/\"I j), it follows j

j

by the hypothesis that L is finitely sub directly irreducible that 19 (a, b) /\ Q! :f w. Since 19{a, b) is an atom in Con L it follows that 19{a, b) /\ Q! = 19{a, b) and hence 19{a, b) ~ Q!. Thus 'IJ{a, b) is the smallest non-trivial congruence on L, so L is subdirectly irreducible. (2) =} (1) : This is clear. 0

Corollary If an Ockham algebra L is finitely subdirectly irreducible but not subdirectly irreducible then necessarily L ¢ Kw and f is injective. Proof ThatL ¢ Kw follows from the above. Suppose thatj were not injective. Then by Theorem 3.8 the interval [w, cI>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 [
Theorem 3.13 If L

E

Kw

then [
subalgebra T 2(L) isfinite.

Proof By Theorem 3.10 and the Corollary to Theorem 3.12 we have [
Con L/
The result follows now from the above result of Sankappanavar. <>

Definition We shall say that Con L has comonolith

CI!

whenever

ConL = [w,CI!] EB {~}.

The comonolith is therefore the only coatom in Con L, when such exists.

Theorem 3.14 If L E Kw then the follOWing statements are equivalent: (1) K(L) == T 2 (L); (2) Con L has comonolith (2) : Since (0,1) ¢
°

fm+n(a)

=r(a),

fm+n(b)

=r(b).

47

Subdirectly irreducible algebras

Let c = r(a) and d = r(b). Then we have c,d E T(L) with c < d or c > d. It follows by (1) and Theorem 3.4 that 19(c, d) = L Consequently, by Theorem 2.1, n-l

19(a, b) = V 19lat(tk(a)Jk(b))

V

k=O

Suppose now that cp

E

Con L is such that cP f

19(c, d) = L L

Since cp

= V

19(a, b)

(a,b) E 'P

we have

(a,b)Ecp

=}

19(a,b)f£.

=}

(a,b)Ectlw •

But by the above observation

19(a,b)f£ Hence cp ~ ctlw and consequently Con L (2)

=}

(3)

=}

= [w, ctlw] E9 {£}.

(3) : This is clear.

(1) : If (3) holds then, by Theorem 3.10, T(L) is simple, whence

(1) follows by Theorem 3.5. 0

Corollary 1 If L E Kw then the follOwing statements are equivalent: (1) K(L) = T(L); (2) Ljctlw is a simple de Morgan algebra Proof This is immediate from the above and the Corollary to Theorem 3.5. 0 Corollary 2 If L E Kw then the following statements are equivalent: (1) L is simple; (2) K(L) = T 2 (L) and f is injective. Proof (1) =} (2) : If L is simple then ConL = {w,£} and so ctlw = w since (0,1) ¢ ctlw• Consequently, ctl1 = w whencef is injective; and ctlw is maximal, so K(L) = T2(L). (2) =} (1) : If (2) holds then clearly ctlw = wand, by the above, Con L reduces to {w,£}. 0 Corollary 3 The simple algebras in Kp,q are precisely those algebras L in Kp,oforwhich K(L) = T2(L). 0 Note that Corollary 2 above does not extend beyond illustrated by the pineapple (Example 2.4).

Kw.

This fact is

48

Ockham algebras

We now proceed to characterise the (finitely) subdirecdy irreducible algebras in Kw. For this purpose, we require the following results.

Theorem 3.15 Let A be an algebra that belongs to a class that has the congruence extension property If A is subdirectly irreducible with monolith

Proof Every congruence {}* on B with {}* =f w extends to a congruence {} on A such that {} =f w, and therefore {} ~
Theorem 3.16 If LEO is subdirectly irreducible and

0

if f

is not injective

then the monolith of L is
Proof If f is not injective then
= (3 1\
Q!

Q!

Q! Q! give £ = (3 V
Theorem 3.17 L E Kw is subdirectly irreducible if and only if Con L reduces to the chain

More precisely, (1) if L belongs to a Berman class and VB (L) = Kp,q then L is subdirectly irreducible if and only if Con L reduces to the finite chain w

=
~

...

~

=
(2) if L belongs properly to Kw then L is subdirectly irreducible only if Con L reduces to the infinite chain

w =
~

~

...

<
~ £.

if and

Subdirectly irreducible algebras

49

Proof => : Suppose that L is subdirectly irreducible. By Corollary 1 of Theorem 3.2 and Theorem 3.14, we have Con L =: [w, cI>w] E9

{~}.

We now show that the subalgebra f(L} is also sub directly irreducible. In fact, suppose first that cI>1 =: w. Then we have

Conf(L)

~

Con L/CP1

= Con L,

whence f(L} is subdirectly irreducible. Suppose now that cI>1 be the restriction of cI>1 to f(L}. Since

** f2(X)

(t(x),f(y}) E cI>t it follows that

cI>t = w

**

cI>1

=:

f2(y}

**

:f wand let cI>t

(x,y) E cI>2,

= cI>2 = ... =: cI>w·

Thus, if cI>t = w then Con L reduces to the three-element chain

w -< cI>1

=: ...

= cI>w

-<

~.

It follows by the congruence extension property thatf(L} is also subdirectly

irreducible. On the other hand, if cI>t :f w then exactly the same conclusion follows by Theorem 3.15, the monolith of Conf(L) ~ ConL/cI>l being cI>t by Theorem 3.16. In conclusion, f(L} is subdirectly irreducible, whence so are all r(L} since for every i we have that Ji(L} = f[Ji-1(L}]. We thus have Con L =: {w} E9 [cI>1' d with

[CP1, d ~ Con L/CP1 ~ Conf(L}.

Similarly, Con L = {w} E9 {cI>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 -<

= Kp,q

~.

If on the other hand L belongs properly to Kw then since there are infinitely many cI>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 <¢= :

-< cI>1 -< cI>2 -< ... < cI>w -<

~

This is clear. 0

Corollary 1 If L E Kw is subdirectly irreducible then so is every subalgebra ofL.O

50

Ockham algebras

Corollary 2

If L E Kw then L is simple if and only if L is subdirectly irre-

ducible and 1 is injective. <>

Example 3.5 (Tbe sink) Consider the lattice L with Hasse diagram .1

.0

made into an Ockham algebra by defining

1(0) = 1, 1(1) = 0, I(xo)

= xo,

= Xj-1 It is readily seen that L E Kw. Moreover, (Vi ~ 1) I(xj)

and extending to the whole of L. every congruence that identifies any adjacent pair also identifies x 0 and Xl. Consequently the smallest non-trivial congruence is
-<
t

where
={

19(xo, Xj) if i is even; 19(x j , xo) if i is odd.

Every congruence on L is determined by the class containing xo. Note that Examples 3.2 and 3.5 show that, unlike Kp,q' the generalised variety Kw has infinite subdirectly irreducible algebras.

Subdirectly irreducible algebras

51

Example 3.6 Let P be the finite/cofinite algebra on IN o. In the lattice I

=

P EB pop depicted as follows

b~ I

a~ I

where at is the complement in P of the atom a j and bt is the complement in pop of the coatom bi, define

j(O) =1, j(l) =0, j(CI.) = CI.; { j(a j ) = bi' j(b j) = ai-I, j(bd = 0; j(at) = bt, j(bt) = at-I, j(b!) = CI.. Since every element of L can be expressed either as a finite join of the a j or the bt, or a finite meet of the at or the bj, we can extend j to a dual endomorphism oni. Observe thatj2j(b j ) = 1 soj2i+I(a j ) = 1; andj2j(bt) = CI. so j2i+I(at) = CI.. It follows that L E Kw. It is readily seen that any congruence on I that identifies a pair x, YEP identifies 0, a j for some i, whence it identifies 0, a 1 and hence b 1 , 1; and any congruence on L that identifies a pair x,Y E pop identifies bj, 1 for some i, whence it identifies b I , 1. Hence I is subdirectly irreducible with monolith 4>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

AnQ=BnQ

n Q = B n Q is equivalent to Ad B

~

Q' where

d means 'symmetric difference' and Q' is the set-theoretic complement of Q in X. A direct consequence of this is that, for LED,

Con L is dually isomorphic to the lattice of closed subsets of X. At this point we would draw the reader's attention to the 'D- P dictionary' which ends Priestley's survey paper [79] and which summarises some of the most commonly used dual equivalents. Since Ockham algebras are bounded distributive lattices they are dually eqUivalent to a suitable subcategory of P. By an Ockham space we shall mean a Priestley space endowed with a continuous order-reversing map g. Let Q be the category whose objects are the Ockham spaces and whose morphisms are those order-preserving maps that commute with g. Then the category 0 of Ockham algebras is dually equivalent to the category Q. We shall usually abbreviate (X; T, g) to Simply (X; g). Whenever possible, we shall use letters a, b, c, ... for elements of the Ockham algebra and letters p, q, r, ... for elements of the Ockham space. The following important results were first established by Urquhart [94].

Theorem 4.1 If (X;g) is an Ockham space then (O(X);J) is an Ockham algebra where (VA Conversely, where

if (L;J)

E

O(X))

is an Ockham algebra then (Ip(L); g) is an Ockham space

g(x)

= {a ELI f(a) ¢ x}.

Moreover, these constructions give a dual equivalence. 0

Ockham algebras

54

We now list some further concepts and terminology that we shall require. A subset Q of an Ockham space (X; g) is called a g -subset if it is g -invariant, in the sense that

XEQ

=?

g(X)EQ.

For every Q ~ X we denote by gW(Q) the smallest g-subset that contains Q, namely gW(Q) = {gn(x) I n ~ 0, x E Q}. The lattice of all dosed g-subsets of X will be denoted by G(X). A g-subset Q of the form gW ({x}), which we shall write in the simpler form gW {x}, will be called monogenic. Clearly, the monogenic g-subsets are the join-irreducible elements of G(X).

Theorem 4.2 Let (X;g) be the dual space of (L;1) EO. If Q E G(X) then the equivalence relation {}Q defined on CJ(X) by

(A,B)E{}Q

~

AnQ=BnQ

is a congruence.

Moreover, the lattice Con L is dually isomorphic to G(X). <> For each Q E G(X) the congruence {}Q will be called the congruence associated with Q. Clearly, Con L is boolean if and only if G(X) is boolean. More generally, the congruence {} Q is complemented if and only if X \ Q belongs to G(X), i.e. X \ Q is a closed g-subset, whence Q is open. We thus have the following

Corollary If Q E G(X) then the congruence {} Q is complemented ifand only if X \ Q E G(X), a condition that requires Q to be clopen. <> Now in this respect the finite case is worthy of conSideration, so we shall illustrate matters by revisiting Theorem 2.17, the proof of which is purely algebraic, and provide a very short proof of a statement that is more general than the theorem in question. For this purpose, and anticipating Chapter 5, we introduce a classification of the subvarieties of 0 which is somewhat sharper than that used by Berman. It is also due to Urquhart and is closely related to the dual space. For m > 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 <1>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 <1>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 <1>11 is associated with the closed g-subset gn(x), and the congruence w is associated with gW (X).

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

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

~

whence L is subdirectly irreducible. Now (again see [9]) it is known that INco is the I-point compactification of a countable discrete space. As such, it is a boolean space and is (homeomorphic to) the prime ideal space of the finite-cofinite algebra FC(IN). Thus L is an atomic boolean lattice that is not complete. We shall now show that (L;J) does not belong to any Berman class, but does belong to Kw. For this purpose, let U be a finite subset of INco that does not contain 00 and let z be its numerically greatest element. Since

j21l(U) = {x

E

INco I g2n(x) E U},

we see that j2n(u) is the subset of INco obtained from U by translating U to the left through an amplitude of 2n and deleting the resulting elements that fall outside IN"". From this observation, we have that if z = 2n then pn(u) = 0 and therefore j211+2(U) == pn(u); and that if z = 2n + I then j2n(u) = {I} andj211+1(U) = IN co ' and thereforej21l+3(U) ==j2n+l(u}. The above example can be modified slightly. Consider the effect of taking g to be the constant map given by (Vn E IN)

g(n) = g(oo) = 00.

60

Ockham algebras

Here the closed g-subsets are 0 and all the subsets of INoo that contain Consequently we have Con L ::::: 21N EEl {t },

00.

the comonolith being cPl' In this case, (L;J) E P2,1 and is not subdirectly irreducible. Another variation on the same theme consists of ordering INoo as the chain IN with 00 as a greatest element, and taking g to be the same constant map as in the above. Here the clopen down-sets are 00, IN oo ' and n l for every n. Consequently, L is isomorphic to IN EEl 1. The closed g-subsets are 0 and all the subsets of INoo that contain 00, so that Con L :::::

iN EEl {t },

the comonolith being cPl' Since for every a E L \ {I} we have f( a) follows that (L;J) is a Stone algebra.

= 1, it

Our objective now is to characterise those closed g-subsets that correspond to principal congruences under the duality. The following theorem was inspired by Lemmas 2 and 3 of [17]. We shall say that Q E G(X) is maximally disjoint fromR and Tn R :f 0 for every T E G(X) with T:J Q.

~

X if QnR

=0

Theorem 4.5 A closed g -subset Q represents a principal congruence if and only if there is a convex clopen subset R from which Q is maximally disjoint. Proof =* : Let Q E G(X) represent the principal congruence 19(a, b). If the elements a and b correspond to the clopen decreasing subsets A and B of X (with A ~ B) then, since (a, b) E 19(a, b), we have Q ~ (AAB)' = (B \A)'.

Even more, Q is the greatest closed g-subset contained in (B \ A)'. Clearly, = B \ A is a convex clopen subset, and Q is maximally disjoint from R.

R

~ : Conversely, suppose that Q is maximally disjoint from the convex clopen subset R. Since R is closed, Rl is a closed down-set; and since R is open, Rl \ R is also a closed down-set. Since X is a Priestley space, there is a clopen down-set A such that Rl \ R ~ A and RnA = 0. Consider the set B = A U R. Since R is convex and clopen, B is also a clopen down-set. Let a and b be the elements of L that are represented in X by A and B respectively, and let P E G(X) represent the principal congruence 19(a, b). By the first part, P is maximally disjoint from B \ A = R. The fact that P and Q are both closed now gives P = Q, whence Q represents a principal congruence. <>

61

Duality theory Remarks Note that, in the above, (1) the convex clopen subset R is not necessarily unique;

(2) the principal congruence 19(a, b) is represented by the closed g-subset that is maximally disjoint from B \ A; (3) if the complement of a g-closed subset Q is convex and clopen then Q represents a principal congruence. It follows that if L is finite and .e(X) ~ 1 then all congruences of L are principal. (4) if L is finite and the g-subset Q is covered by a unique g-subset P then Q represents a principal congruence (for the convex subset R take {x} in P \ Q).

Example 4.5 Consider once more the pineapple lattice made into an Ockham algebra as in Example 2.4. In this, all prime ideals of L, except {O}, are non-principal. They form two chains connecting {O} toJ= L \ {I}. We shall denote by 1211 (resp. I 2n - 1 ) the prime ideal that separates X21l and X2n+1 (resp. X2n-1 and X21l)' The dual space X is then the following ordered set. J

~

This becomes a Priestley space if we take as a sub-basis for the topology T the subsets III \ {J}, 1;1 \ {{O}}, I~m U I~Il+1' and I~m U I~n+1 where m, n E Z. The clopen down-sets of X are 0, X, and UI~ for k E 2Z and.e E 2Z+ 1. Then we define g as follows :

Ii

(Vn 10

E

Z) g(In) = In-I, g(J) = {O}, g({O}) =].

(X; g, T) is an Ockham space and it is easy to verify that the dual algebra is that described in Example 2.4. The open g-subsets other than 0 and X are generated by any element of {Ill I n E Z}. The closed g-subsets are obtained by adding to the preceding sets the set {{O},]}. Ordered by inclusion, this collection forms an {O} infinite chain with {{O},]} covering 0. The congruence 19(X21l-1, X21l) is represented by the closed g-subset that is maximally disjoint from {I 21l - 1 }, i.e. the closed g-subset generated by I 2n - 2 • Similarly, we see that 19(X2n+1, X2n) is represented by the closed g-subset generated by I 2n - 1 • The only non-principal congruence 'I' corresponds to {{O},]}, and Con L is the chain

Ockham algebras

62

Example 4.6 (Tbe diamond) (1) Tbe finite case. Consider the lattice L, which is the direct product of two 6-element chains, described by the Hasse diagram

and made into an Ockham algebra by defining

then extending to the whole of L. The dual space X is as follows, in which the arrows indicate the action of g :

e1 "'--r-~~--i=4I Jl

at a5 The lattice G(X) of g-subsets has cardinality 26. The Hasse diagram of its dual (G(X))op and a table of generators for each of the g-subsets are as follows:

63

Duality theory

o

G

G'

x

{at,bt}

A

{at,dt}

F

{aj,di}

A'

{b5,C5}

F'

{bj,Cl}

g(X)

{ct,d~}

B

{a5,bj}

G

{at}

B'

{bt,ai}

G'

{btl

g2(X)

{aLb;}

C

{ct,bi}

H

{ct}

c'

{d6,ai}

H'

{dt}

g3(X)

{ci,d;}

D

{at,dD

I

raj}

D'

{bt,cil

I'

{bi}

g4(X)

{e! }

E

{ct,dl}

j

{ell

E'

{d6,ej}

J'

{d;}

The Hasse diagram for Con L is the same as that above with X replaced by K replaced

w; 0 replaced by ~; each gi (X) replaced by «Pi; and each g-subset by the corresponding congruence {JK'

Using Theorem 45, we can verify that there are 8 non-principal congruences, namely {Jc, {Jer, {JF, {Jp, {JH, {JH', {Jj, {J]'. Moreover, we have «PI «P4

= {J(O, al /\ bd = {J(a4 V b 4, 1)

{JA

= {J(b o,b 1 )

{JA'

{JB

= {J(b 2 , bl )

{JB' == {J(a2,al)

{JD

{JD'

= {J(a 2 ,a 3)

{JG

{JE

= {J(O,al /\ b 3)

{Jp

= {J(a4 V b 2 , 1)

«P2 == {J(a2 V b 2, 1)

=:.

{J(b 4 , b 3)

{JE' == {J(O, a3 /\ b 1 )

«P3 == {J(O, a3 /\ b 3)

= {J(ao,ad

= {J(b 2, b 3) {JG' = {J(a4,a3) {JI

=:.

{J(a2 V b 4 , 1)

64

Ockham algebras

(2) The infinite case. Consider the lattice L which is the direct product of two chains isomorphic to IN EB IN°P, described by the Hasse diagram

and made into an Ockham algebra by defining

f(ao) = ao, f(b o) = bo,

(Vi

~ 1)

f(a j ) =

aj-I,

f(b j ) = bj-l,

then extending to the whole of L. All prime ideals of L are principal except for r

=neN U ih

and s

=neN U kh·

They form two chains eland C 2 as in the following diagraJU. The set X = C I U C 2 becomes a Priestley space if we take as a sub-basis for the topology T the subsets {x Ix > p} and {x Ix < p} for every p E X. Note that the sets r r, s r, rl, sl are not open.

Duality theory

65 ·L

16

Jo

lLI

·L 11 ·L Jz

l~

The elopen down-sets of X are 0, X, OIL, {3L, and OIL U {3L where OlE C I \ {r} and {3 E C 2 \ {s}. The action of g on X is described by ·L to

-'> 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
L

$h

N

b

0

.-....

P

q

r

Ockham spaces

d

h

a

c

Ockham algebras

Ockham algebras

72

We close this chapter with some observations on the decomposability of an Ockham algebra [41]. For this purpose, we recall that an algebra is said to be directly decomposable if it is isomorphic to a direct product of two non-trivial algebras.

Theorem 4.7 Let L be an Ockham algebra. If a E Z{L) with f{a) = a ' then a L and aiL inherit from L the structure ofOckham algebras. Proof For every x E a l define fa (x) and fa (a) = O. If u, vEal then

= f{x) A a.

fa (u A v) = (t{u) V f(v») A a

1hen we have fa (O)

=a

=fa (u) V fa (v),

and dually fa (u V v) =fa (u) Afa (v). Hence (al;fa) is an Ockham algebra. Since f(a) Af2(a) = 0 and f(a) V f2(a) = 1, we have that f(a ' ) = j2(a) = (t{a»)' = a" = a. It follows by the first part that a'l is also an Ockham algebra. 0

Theorem 4.8 An Ockham algebra L is directly decomposable if and only if there exists a E Z{L) \ {O, I} such that f{a) = a' . Proof <= : Suppose that there exists a E Z(L) \ {O, 1} such that f(a) = a ' . By 1heorem 4.7, a l and a'l are Ockham algebras. Define h : L ----7 a! x a'l by h(x) = (x A a, x A a'l. It is well known that h is a lattice isomorphism, and since

f(h(x»)

= (t(x A a) A a, f(x A a'l A a ' ) = ([[(x) V f(a)] A a, [[(x) V f(a' )] A a') = (t(x) A a, f(x) A a') =h(t(x»),

it follows that h is an Ockham isomorphism. ~ : Suppose thatL = U x V with U, Vt=L. Then (see, for example, [1]) there exists u,v E Z{L) \ {O, I} with U = u l and V = v l . Since f(u,O) = (t u(u), fv(O») = (0, v), the result follows. 0

1he direct decomposability of an Ockham algebra is sharply reflected in its dual space. We already know that a direct product L x M of distributive lattices corresponds to a disjoint union A l:J B in the dual space.

Theorem 4.9 Let (L;/) be an Ockham algebra and let (X;g) be its dual space. Then L is directly decomposable if and only if X is a disjoint union of two g -subsets.

73

Duality theory

Proof Two complementary elements a and a l of L are represented in the dual space X by two dopen down-sets A, B such that A nB = 0 and A UB = X. If f(a) = aI, i.e. f(A) = B, it follows that X \g-l(A) = B, hence g-l(A) = A. Since f(a) = a l implies f(a /) = a, we also have that B = g-l (B). Hence, by Lemma 4.1, A and Bare g-subsets. <>

Example 4.9 If we are given the Priestley space

!I

• r

•s

we can make it into the Ockham space of a directly decomposable Ockham algebra by defining g in one of the following ways :

p q r s 0 1 a b c d e f g h i j Xl q P r s 1 0 j i h g f e d c b a L1 X2 q q r s 1 0 1 i h h X~ P P r s 1 0 i i h f X3 q P s r 1 0 j i e d X4 q q s r 1 0 1 i e e XI4 P P s r 1 0 i i e c X5 q q r r 1 0 1 i 1 1 Xl5 P P r r 1 0 i i 1 i X6 q P r r 1 0 j i 1 j X7 q P q s 1 0 j c h g XI7 q P P s 1 0 d c h a Xa q q q s 1 0 1 c h h X~ p P P s 1 0 c c h 0 X9 r r p s 1 0 e e h b Xl9 r r q s 1 0 1 e h h X 10 r r r s 1 0 1 1 h h X 11 q q p s 1 0 e c h b X l1 P P q s 1 0 i c h f

e e h h h b b i b 0 1 0 1 0 1 0 1 b i b i h c 0 1 0 1 f f c c c i i

e c g h f b 0 a j d 1

c c i c e i

c c f f f 0 0 0 c c c c c c c c c

b b b b b b b b h h h h f f 0 h h

b 0 a b 0 b 0 a g a h 0 0 f 0 b f

L2 L~

L3 L4 LI4 L5 LI

5

L6 L7 LI7 La LIa L9 LI9 LlO

L11 LI11

This yields 18 non-isomorphic directly decomposable Ockham algebras, each of which has the Hasse diagram

74

Ockham alg

h

c

with the corresponding Ockham operation defined as in the table.

5 The lattice of subvarieties A subclass of a variety V which is also a variety is called a subvariety of V. The subvarieties of V form a lattice which we denote by A(V). In this, the meet A A B of two subvarieties A, B is their intersection, and the join A V B is the equational closure of their union, i.e. the smallest subvariety of V that contains AU B. A celebrated theorem of B. J6nsson [15] states that if V is a variety every algebra of which has a distributive congruence lattice then A(V) is distributive. Applying this to the variety 0 of Ockham algebras, we thus have that A(O) is distributive. The following three remarkable properties of A(O) were established by A. Urquhart [95] : (1) A(O) is uncountable; (2) every subvariety of 0 is generated by its finite members; (3) a subvariety of 0 has finite height in A(O) if and only if it is generated by a finite algebra.

We shall denote by Aj(O) the set of subvarieties of 0 that are generated by finite algebras. In fact, Aj(O) is the interesting part of A(O) since it contains 'at the bottom' the well-known subvarieties B of boolean algebras; K of Kleene algebras; M of de Morgan algebras; S of Stone algebras, and S of dual Stone algebras; P 3 ,1 = K1,1 of Ockham algebras with de Morgan skeletons; Ml = MS of MS-algebras, and Ml = MS of dual MS-algebras. Here and throughout what follows we ignore the trivial subvariety. As a framework for Aj(O) we may choose either the Berman classes Kp,q > 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 -<==? m~m', n~n',

m-nlm'-n'.

76

Ockham algebras

In particular, it follows immediately from this that Kp,q ~ Kp',q' ~

q ~ q', P Ip',

a fact that was mentioned in Chapter 2. If we opt for the Pm ,ll then the framework at the bottom of Af(O} can be depicted as follows :

·4,3

4,1

Here the label m, n means Pm,n, and the solid circles indicate those that are Berman classes. The Pm,ll that are not Berman classes are those for which m -n is odd. All the classes that we shall consider are defined by identities. For this purpose, we shall generally adopt Urquhart's terminology. Note that we shall very often write ",a for f(a}, particularly where axioms are concerned. A term is a polynomial built from variables a, b, c, ... and the constants 0, 1 by means of the operations 1\, V and",. A term is atomic if it is of the form ",n a where a is a variable. An atomic term is even or odd according to whether the exponent n is even or odd. An inequality is an expression of the form A ~ B where A and B are terms. Two inequalities are equivalent if they determine the same equational class in O. An inequality is basic if it is of the form A ~ B with A a meet of atomic terms and B a join of atomic terms. For example, a 1\ ",2 b ~ ",a V ",3 b V c is a basic inequality, as is also

",a

1\ ",2a

= O.

!be lattice of subvarieties

77

The occurrence of a variable a in an inequality is positive if a occurs in an even term of A or in an odd term of B; and the occurrence of a is negative if a occurs in an odd term of A or in an even term of B. A basic inequality is simple if each variable has precisely one positive and one negative occurrence. Whereas, for example, the inequality a 1\ b 1\ rvb 1\ rv2d::;;; rv 2a V rvC V rv2c V d is simple, the inequality is not. Clearly, for an inequality to be simple it is necessary that it contain an even number of atomic terms. The importance of the basic inequalities is emphasised by the following result.

Theorem 5.1 In 0 every inequality is equivalent to a conjunction of basic inequalities. Proof Let A ::;;; B be an inequality. By using the equalities rv(a

b) = rva

1\

V

rvb,

rv(a

V

b) = rva

1\

rvb

we can convert A and B into equivalent terms A I and B' in which no lattice operation occurs within the action of any rv. By the distributive laws, A' can be put in disjunctive normal form, and likewise B' can be put in conjunctive normal form. The new inequality obtained in this way is equivalent to a conjunction of basic inequalities. <>

Example 5.1 The inequality rv 2 (a

V

",b)::;;; (a

1\

rv(b 1\ c)) V rva

can be written

rv 2a

V

rv3b::;;; (a

V

rva) 1\ (rva

V

rvb V rv c)

and is equivalent to the conjunction of the four basic inequalities

",2a::;;; a V rva, rv3b::;;; a V rva,

rv2a::;;; rva V rvb V rvC, rv3b::;;; rva V rvb V rvC.

The special role played by the simple inequalities is illustrated by the following nice property.

Theorem 5.2 [Urquhart] Let E be a finite subset of w x w, say E = {(PI, qd, (P2, q2), ... , (Pn, qn)},

78

Ockham algebras

and let A

= {,JIX j Ipi

even}

u {"AiX i I qj odd};

B= {,JiXj IPj odd}u{"AiXj I qj even}, where Xj and Xj are distinct variablesjor (pj,qj)::f (Pj,qj)' Consider the basic inequality

and the universally quantified disjunction (VX) V{gPI(X)~gql(X) I (Pj,qj)EE}.

(**)

j

Then (L; "') E 0 satisfies (*) ifand only if (X;g) E X satisfies (**).

Proof The inequality (*) fails in L if and only if there are clopen down-sets Aj such that

n

JIAj

n

",qiA j C1 U JiAI U U ",qiA j, ql odd Pi odd q{ even

n

Pi even

which is the case if and only if there exists x E X such that x E ,JiAi ~ pj is even, E ",qiA j ~ qj is odd.

x Since we have that r'A

XE",I

I

~

{ gri (x) E Aj gri (x) ~ Aj

if rj is even; if rj is odd,

the above conditions are equivalent to the existence of down-sets Aj and an element x such that This in turn is equivalent to the existence of an element x with the property that gPi(X) -;} gqi(X) for every i, which is the case if and only if (**) fails in

X·O We note that if A = 0 then AA = 1, and if B = 0 then VB = O. Clearly, a basic inequality is simple if and only if it is of the form (*) in Theorem 5.2. An important consequence of this is the following, which we have mentioned without proof in Chapter 4.

Corollary Let m > 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 <, rv 2a V b V rvb

(8)

a /\ rv 2b <, rva V rv 2a V b g2 Kg O V g ~ g2

(9)

a /\ rva <, rv 2a V b V rvb

g ~gO

gKg O V gO ~g2

0 1 0 2

2 0 1 0

0

2

2 1

(10)

a /\ rv 2b <, rv 2 a V b V rvb

g2 Kg O V g ~gO

(11)

a /\ rvb /\ rv 2 b <, rva V rv2a

g2 Kg

(12)

a /\ rva <, rv 2a V rvb V rv 2b

gO

1 2 0

2 1 2

0 1 0

0 1 2

g VgO

~

g2 Vg ~ g2

2

0 1 2

~

V gO ~g2

1 2 0

0 1 2

(13) a /\ rva /\ rv 2 C gO Kg2 VgO ~ g Vg ~ g2 <, rv 2a V rvb V rv 2b V c

(14) a /\ rva /\ rv 2b <, b V rvC V rv2c

I tabulation I axiom

gO

~

g Vg2

~

gO Vg ~ g2

I dual equivalent

These relations, together with the duals (ld),"" (14 d) can be ordered by logical implication. However, to determine all the implications using only the axioms is rather difficult. For example, though it is clear that (8) implies (71) it is not at all clear that (0 implies (71). Once more, resort to the dual space (either the g-inequalities or the tabulations) greatly helps.

Ockham algebras

88

To see for example that «() implies ('1/), we can proceed as follows: in «() write rva /\ b for a, and a V rvb for b. We obtain

A

= rva /\ b /\ (rv2a V rvb) /\ (rv2a V rv3b) ~

(rv3a /\ rv2b) Va V rvb V (rva /\ rv2b) = B.

Since A ~ rva /\ rv2a /\ band B ~ a V rvb V rv2b, ('1/) follows. In contrast, a and ('1/) yields a straightforward proof: the glance at the dual eqUivalents of tabulation for «() is a subtabulation of that for ('1/)j equivalently, since g is order-reversing, g .Jf gO implies g .Jf g2 .

«)

Ordered by implication, these relations and their duals give the Hasse diagram

6

Recall that we have added to the above list the equality (01) which is the conjunction of (1) and (l d ). We write this conjunction in the form 01 = (1, 1d ). Other conjunctions worthy of note are the following : (1,.8) = (2 d ), (l d ,.B) = (2); (3,4) = (1), (3d,4 d) = (l d); (3,4d) = (15) : aV rv a= rv2 av rv aj (3d, 4) = (15 d) : a /\ rva = rv 2a /\ rvaj (3, 6d ) = (16) : a V rvb V rv 2b = rv 2a V ",b V ",2b; (3d, 6) = (16 d) : a /\ rvb /\ rv 2b =rv2a /\ rvb /\ rv 2 b.

89

The lattice of subvarieties

The dual equivalents of (15), (15 d ), (16), (16 d ) are respectively

gO gO gO gO

=g2 V g ~ gO;

=g2 V gO ~g; =g2 V g ~ g2; =g2 V g2 ~ g.

Less immediate is the following equality :

(9, 12 d ) = (17) : (a" (Va) V b V rvb

=(rv2 a" (Va) V b V rvb.

To establish this, observe that clearly (17) implies (9), and also implies

a V b V (Vb ~ (V2a" (Va, which is readily seen to be equivalent to (12 d ). Conversely, (9) gives

(a" rva) V b V ",b ~ (V2a V b V rvb, and since clearly

(a" rva) V b V rvb ~ (Va V b V (Vb, we have, by distributivity,

(*)

(a" rva) V b V (Vb ~ (rv2 a" (Va) V b V (Vb.

Now (12 d ) yields and since clearly

we obtain

(**) (rv2 a" (Va) V b V (Vb ~ (a" rva) V b V rvb. Finally, (*) and (**) together give (17). It follows that the dual equivalent of (17) is

(g MgO

V gO ~ g2) 1\ (g ~ gO V g2 ~ gO V g2 ~ g)

=g ~ gO V [(g ~ gO V gO ~ g2) 1\ (g2 ~ gO V g2 ~ g)] = g ~ gO V g ~ gO V [gO ~ g2 1\ (g2 ~ gO = g MgO V gO = g2 V gO ~ g2 ~ g =g MgO V gO = g2.

V g2 ~ g)]

(g ~ gO ~ g2 ~ g)

This shows that (17) is self-dual and is implied by each of the axioms (a) and h). In fact, it can be shown that (17) = (15, 15 d ).

90

Ockham algebras In an exactly similar way we can prove that (12, lId) = (18) : (a /\ rva) V rvb V ".}b = (rv2 a /\ rva) V rvb V rv2b,

and hence also (12d' 11) = (18 d ) : (a V rva) /\ rvb /\ rv 2b = (rv2a V rva) /\ rvb /\ rv 2b. The dual equivalents of (18) and (18 d) are respectively

g2 g2

=gO V g ';3 g2 V gO ';3 g, =gO V g2 ';3 g V g ';3 gO.

In Chapter 4 we determined the 19 non-trivial subdirectly irreducible algebras of the class P3,1, as well as their dual spaces. Using the preceding list of inequalities (and mainly their dual equivalents) we can easily give an equational basis for each of the classes generated by these 19 algebras. For notational convenience, we shall use the symbol (n) to denote the conjunction of an inequality (n) with its dual (nd)' Similarly, for evety subvariety A we shall use the notation A for A V A. Since P3,1 (and, more generally, 0) is congruence distributive, we can apply the following fundamental theorem of B. A. Davey [62] and, at least in theory, determine A(P3,d.

Theorem 5.4 Let K = Var 5 be a congruence-distributive variety generated by a finite set 5 offinite algebras, and order the set Si(K) of subdirectly irreducible algebras in K by A ::;:;: B

~

A is a homomorphic image of a subalgebra of B.

Then A(K) is afinite distributive lattice and is isomorphic to O(Si(K)). Moreover, A is a join-irreducible element of A(K) if and only if A = Var A for some A E Si(K). 0 In order to apply this result to the variety P3,1 = Var B 1 , we observe that if B E Si(P3,1) then evety homomorphic image of every subalgebra of B is isomorphic to a subalgebra of B. Indeed, if C is a subalgebra of BE Si(P3,d then by Corollary 1 of Theorem 3.14 we have C E Si(P3,d and so, if A is a homomorphic image of C then by Theorem 3.14 we have A ~ C / cp where cp E {w, ~,£}. Consequently, A ~ C or A ~ f(C) or A is trivial; i.e. A is isomorphic to a subalgebra of C. It follows that we can order Si(P3,1) by A ::;:;: B

~

A is isomorphic to a subalgebra of B,

and thus obtain the Hasse diagram

The lattice of subvarieties

91

B Equational bases for the corresponding varieties are the following : B

(2)

K M Sl S Kl K2 K3

(or,"Y) (or) ((3) (2 d) (1,4 d,"Y) (1,3 d,"Y) (l,5,6 d) (3d,4,"Y) = (15 d,"Y) (3 d, 5d, 6) = (5 d,16 d) (i)

L

N

Ml

S Kl K2 K3 [

N Ml

(2) (ld,4,"Y) (ld,3,"Y) (ld, 5d, 6) (3,4 d,"Y) = (15,"Y) (3,5,6 d)= (5,16) (l d)

The task of determining A{P3,d is rather a daunting one. Even the cardinality of this lattice is not easy to find, though in principle it can be determined by use of the following ingenious result of Berman and Kohler [29].

Theorem 5.5 Let F be a finite ordered set and let 0 (F) be the lattice ofdownsets of F. Then, for every x E F, \O{F)\

= \O{F \ {x})\ + \O{F \ e x )\' <>

92

Ockbam algebras

Although less daunting in nature, this problem requires several applications of the above result, so much so that paper and pencil calculations become unwieldy. Deference to the computer program mentioned in [29] would be in order. This, of course, would tell us only the size of the subvariety lattice and not what it looks like. In what follows we shall determine both by using judiciously chosen subsets of the ordered set Si(P3,1)'

First step We begin by determining A(MS). Applying Davey's theorem to the down-set M i we obtain the lattice

To obtain equational bases for each of the subvarieties of MS, we first determine those of the A-irreducible elements of A(MS) by applying the following obvious principle : (al,a2,'" ,a r ) is an equational basis of VI VV2 V··· VVs if and only if only those subdirectly irreducible algebras contained in VI or V2 or ... or Vs satisfy aI, a2, ... ,a r ·

We can then use these equational bases to obtain those of all the subvarieties in the lattice.

The lattice of subvarieties

93

Favourably, we can exploit the following idea. Since we have the complete list of implications between all the axioms in P3,1 (up to equivalence), if we are given the equational bases of subvarieties S1, S2, ... , Sn then an equational basis of S1 V S2 V ... V Sn is obtained by taking the intersection of the sets of axioms satisfied by S1 , S2, ... ,Sn and then deleting those axioms that are consequences of others. We illustrate this procedure with some examples. Consider for instance the subvariety MVK2VK3. We know that equational bases for these three subvarieties are, respectively,

(a),

(ry,I,3 d ),

(I,5,6 d ).

It follows that they satisfy the following sets of axioms : A=a i ,

B='YiUITU3~,

C=IiU5TU6~

where ni denotes the up-set generated by axiom (n) in the ordered set of implications. Now since AnB = (aTn'YT)u(ainIT)u(aTn3~)

= 9TU9~UliU3~ an equational basis of M V K2 is (1, 3d, 9, 9 d)' But, using the g-inequalities, we see that (1, 3d) implies both (9) and (9d)' So an equational basis of MVK2 is (I,3d)' Now we consider (1 TU 3~) n (1 i U 5TU 6~) = 1TU (3~ n 5T) U (3~ n 6~) = 1 T U 11~. It follows that an equational basis of M V K2 V K3 is (1, lId)'

Similarly, we can compute an equational basis for M V K 3 . In fact, we know that equational bases for M and K3 are, respectively,

(a),

(1,5, 6d ).

Applying the above procedure, we consider aTn(ITu5Tu6~) = (aTnIT)u(aTn5T)u(aTn6~)

=ITU6~,

from which we see that an equational basis for M V K3 is (1, 6 d)' Using these observations, we can now establish an equational basis for M V S::::: (M V K 2) 1\ (M V K3)'

Clearly, this is (1, 3d, 6d).

94

Ockham algebras

In listing the equational bases for the varieties of MS we make a slight digression and compare our results with those obtained by M. E. Adams and H. A. Priestley in [22]. These authors present a 'purely algorithmic approach to the generation of equational bases' for certain subvarieties of distributive lattice ordered algebras, especially Ockham algebras. The identities they obtain by this algorithmic method are of a canonical form, being basic inequalities that involve the minimum possible number of variables and 'can be read off from the dual spaces of the free algebras'. The computation of the free algebras themselves is not necessary. Unfortunately, the theory involves not only the distributive lattice duality that we used intensively in Chapter 4, but also natural dualities based on schizophrenic objects. It is therefore impossible to give here an account of all the deep ideas contained in this long paper which we commend to the interested reader. Adams and Priestley apply their method to the class MS, regarding all varieties as lying in P3,1' For this they use the 7 axioms listed on the left of the following table (the numbering is as in [22]). On the right of the table are our equivalents. Here the reader will observe that an inequality can appear in many forms, sometimes misleading.

1 2 4 12 13 14 24

a V rva V rv 2a = 1 rva /\ rv 2a :( a rv2a :( a V rva rvb /\ rv2b:( a V rva V rv 2a V b b /\ rvb /\ rv 2b :( a V rva V rv 2a rv2b:( a V rva V rv 2a V b V rvb rva /\ rv 2a /\ rv2b:( a V b V rvb

{3

3d 4d lId 8

6d 12d

rva /\ rv 2a = 0 rva /\ rv 2a :( a rv 2a :( a V rva rva /\ rv2a:( a V rvb V rv 2b rva /\ rv2a:( rvb V rv 2b rv2a:( a V rvb V rv 2b rva /\ rv 2a /\ rv2b:( b V rvb

The table that follows compares the equational bases obtained on the one hand by Adams and Priestley and on the other by ourselves. That these axiomatics are equivalent can be shown as follows. We know already the implications

Moreover, we have the following equalities : (1,8)= {l,S), (1,,B) = (2 d), {l,12 d ,8)= (1,')'), (1,3d,8)= {l,3d,')'), {l,4d,8) = {l,4d,')'), (01,')') = (01,8), (OI,{3) = (2).

The lattice of subvarieties

95

By way of illustration, we show that (1, 12 d , 8) = (1, ')'). For this purpose, we obselVe that since b) implies both (12 d ) and (8) it suffices to prove that (1, 12 d, 8) implies b). By (12 d) we have to consider three cases:

(a) g ~ gO : clearly, b) is verified; (b) g2 ~ gO : since by (1) we have gO ~ g2 the equality gO = g2 holds and then (8) yields b); (c) g2 ~ g : using (1) we obtain gO ~ g2 ~ g and b) is satisfied. Using these obselVations we can establish the equivalence of the sets of axioms given in the following table. MS-subvariety

Adam~Priestley

Blyth-Varlet

Ml

(1 ) (1, lId) (1,6 d) (1, 12d) (1,5)

K

(1 ) (l,lld) (1, lId, 6d) (1, lId, 12 d) (1, lId, 8) (1, 3d) (1,4d) (1,,8,3 d) (1, 11 d, 6d, 12 d) (1,lld,8,6 d) (l,lld,8, 12d) (1,3d,6 d) (1, 11 d, 8, 6d, 12 d) (1,3d,8) (1, 3d, 4d) (1,4 d,8) (1, 3d, 8, 6d) (1, 3d, 4d, 8)

(Cl!) (1,4d,')') (1,3d,6 d,')') (Cl!,,),)

B

(1,,8,3d,4d)

(2)

MVK2 VK3 MVK3 MVK1 VK2 K2 VK3 MVK2 MVK1 S SVMVK1 K3 Kl VK2 MVS SVK1 K2 M Kl SVK

(1, 3d) (1,4d) (2 d) (1, 6d, 12 d) (1,5,6 d) (1, ')') (1, 3d, 6d) (1,6 d,,),) (1, 3d,')')

We may conclude that the two axiomatics are equivalent. Of course, the Adams-Priestley axioms are neither independent nor minimal, but their objective is different from ours in that it is more general and practical and in particular enables computer assistance.

96

Ockham algebras

Second step We determine A{L}. The ordered set of subdirectly irreducibles in L is

By Theorem 5.5 the number of down-sets of this ordered set is 53, which is the size of A{L}. Using Davey's theorem we can construct this lattice:

Observe that the south-east and south-west faces consist of MS-algebras and dual MS-algebras respectively.

The lattice of subvarieties

97

Equational bases for the 10 A-irreducible elements of A(L) can be found easily, and are as follows. L LVK1 VK2 LVK2 LVSVK1 LVK1

S

b) b,8 d) b,3d) b,6) b,4) ((3,'1)

[VK1 VK2 [VK2 [VSVK1 [VK1

b,8) b,3) b,6 d) b,4d)

Using the fact that every element of A(L) is a meet of A-irreducible elements, equational bases for the remaining subvarieties in this lattice can now be obtained from the above. Third step We can use the lattice A(L) to construct the lattice A(N). In fact, to obtain the latter we have to add (as v-irreducible elements) to the former the subvarieties generated by K 3, K 3, S 1, N, N. This we do in two stages. First we add the subvarieties K 3 , K3

to obtain the 102-element lattice A(L V K3) which is shown on page 98. We then extend this lattice by adding (as V-irreducibles) the subvarieties SI, N, N. This is achieved by adding the 82-element lattice which is shown on page 99. By pasting the latter as a second layer projecting down onto the former, with SI directly above S, we obtain the lattice A(N). It therefore has 184 elements. Equational bases for the 15 A-irreducible elements of A(N) are as follows. N NV[VK3 NV[ NVK2 VK3 NVK2 NVK3 SI LVK3 LVK3

(8) (8, 13d) (5 d) (8,8 d) (8, 3d) (8,6) ({3) (8,() (8,4)

NVLVK3 NvL NVK2 VK3 NVK2 NVK3

(8,13 ) (5) (8,8) (8,3) (8,6 d )

Ockham algebras

LvK,

The lattice of subvarieties

99

Nv~

Ockham algebras

100

Fourth step Finally, we can obtain the lattice A(P3,1) = A(B1) from the lattice A(N) by adding to the latter (as v-irreducible elements) the varieties M, M1, M1, B1.

We first add M (directly above K), in so doing obtaining a copy of the entire double layer lying above K. This adds another 180 elements. Next, on the new top level, we add Ml and Ml (with Ml directly above M V K2 V K 3). This produces another 97 elements. Finally, we top everything off with B1 . We conclude that, excluding as usual the trivial subvariety, IA(P3,dl = 462. (This corrects an error in [48].) Now we consider a type of subvariety that is particularly interesting in the context of Ockham algebras. A subvariety will be called self-dual if it contains with each of its algebras the order-duals of these algebras. Thus, for example, B, K, M, S are self-dual, as is every subvariety of the form A = A V A for A E A(O). The self-dual subvarieties form the spine of the lattice of subvarieties, in the sense that they constitute the axis of symmetry of this lattice. Moreover, the self-dual subvarieties form a sublattice of A(O). In P 3 ,1 the self-dual subvarieties have equational bases that are conjunctions of the self-dual axioms (Q!), ((3), ... , ('1)) and (1i) for n = 1, ... , 14. The dual equivalents of the (1i) are

(Q!) = (1)

(2)

(3) (4) (5) (6) (7) (8) (9) (E)=(10) (11) (12) (13 ) (14 )

gO = g2 g=gO gO = g2 V g = g2 V gO ~ g2 ~ g V g ~ g2 gO = g2 V g2 ~ gO ~ g V g ~ gO ~ g2 g = g2 V gXgO

~

gO

= g2 V g = g2 V ~ °~ g2 V g2 ~ ~ ° gO = g2 V gO ~ g2 ~ g V g ~ g2 ~ gO g = g2 V g2 Xgo gO = g2 V gHg O gO

g2 Xgo gO = g2 V g2 Hg gO = g2 V g = g2 V g HgO g = g2 V g2 XgO V g HgO gO

= g2 V g = g2 V

g HgO

V ~ °~ g2 V g2 ~ ~ °

If (n) implies (r) or (rd) then (1i) implies (r); but this condition is not necessary for (n) to imply (r), as is shown by the implication (4) =?- (E) = (10)

ice q 'f b e latt

ie (sUbvar

ties

101

'6)

( (c, 6) == « c) h , 8 ) == h',,12) c (c, 9) == ( (,8, c) (,8,7) == :::: ('7) ,9) == (c,3)

(3

Ockham algebras

102

Equational bases for the A-irreducible elements of this lattice are as follows.

(17)

M:l VN MVN

K3 VMVS 1 K2 VMVS 1 Kl vM N

(11)

LV M:l V SI LVM:l M:l V SI LVMVSl

(13)

() (8)

SI

(6) (3) (4) (8) (f3)

(12)

Those that involve only one axiom and do not already appear above are M

(01)

L

b)

M:l

(I:) (2)

B

LV SI K2 VM LVM K3 VLVMVSl

(5) (7) (9) (14)

In the same spirit as in Chapter 2, we shall use the notation V(L) to denote the smallest subvariety of 0 to which L belongs. We close this chapter with some illustrative examples.

< a < b < 1 we can define 10 non-isomorphic Ockham algebras (see Example 1.8). The various operations rv and the corresponding classes are summarised as follows :

Example 5.8 On the 4-element chain 0

0 1 1 1 1 1 1 1 1 1 1

a 0 1 b a a 1 b 1 b 1

b 0 1 b a 0 b a 0 0 a

1 0 0 0 0 0 0 0 0 0 0

S

S Kl Kl K2 K2 K

S P3,2 P 3,2

The first eight subvarieties above represent V(L).

Example 5.9 In Example 1.9 we showed that the unit cube C can be made'

Tbe lattice of subvarieties

103

into an algebra in P3,l in two ways, namely

= (1 -z, 1 - y, 1 -z); = (1 - y, 1 -z, 1 - y). is concerned, we have ",i{x,y,z) = (z,y,z) and so

(C; "'1) : "'1 (x,y,z) (C; "'2) : "'2(X,y,Z) As far as

"'1

",i{x,y, z) A "'1 (x,y, z) ~

(t, t, t) ~ "'t(a, b, c) V "'1 (a, b, c).

Thus (C; "'d satisfies (8) and so belongs to N. Now (13) fails in (C; "'d, as can be seen by taking a == (~, t), b ::: c::: (t,t,~)· Hence (C;",d¢N'VLVK3 · Likewise, (13d) fails, Thus as can be seen by taking a = (t, t, ~), b::: (t,}, }), c = (~, (C; "'1) ¢ N V I: V K3 · It follows that V(C; "'1) = N. As for (C; "'2), we have that V(C, "'2) = B1. To see this, it suffices to show that the axiom (rJ), which is an equational basis for MI V N, fails. For this purpose, take a = -},~) and b = (~, We refer the interested reader to [46] for further properties.

t,

(t,t,t),

t, t).

(t,

t, t)·

Example 5.10 The algebra of Example 1.10 satisfies the axiom b); for {

X A "'x = X A (b V X') A a = X A b A a = X A b; y V "'y =y V b V (y' A a) = y V b V a =y V a.

It also satisfies (4) and (4 d ); for 2 "'X=

(b VxAa ) {

~ X A a ~ X A b ::: X A "'x; ~

b V X Va::: X Va::: X V "'x.

Hence we have that L E (LVK1 ) A (I:VKI ) = Kl VK1 ::: K1 . Since neither (1) nor (l d ) is satisfied, L ¢ KI and L ¢ K1 . Therefore V(L) ::: KI .

Example 5.11 Referring to Example 1.11, consider first the Ockham algebra Ll that corresponds to the polarity p. This does not belong to MI , an equational basis for which is (E), since {

a2n A ",2b zn = a2n A a2n+l ::: a2n; ",2 a2n V b2n ::: a2n-l V b 2n ::: b 2n ·

Likewise, it does not belong to N, an equational basis for which is (8); for

",2 ao

A

",ao

= a o II bo ::: ",2 bo V ",b o.

However, the relation (rJ) holds. This follows from the observation that if a ¢ {ao, b o} then ",a A ",2a is of the form ak with k < 0, and if b ¢ {ao, b o} then rvb V ",2b is of the form ak with k > O. Hence Ll belongs to N V MI.

104

Ockham algebras

As for the Ockham algebra L2 that corresponds to the polarity pi, it can be seen that this belongs to L.

Example 5.12 In Example 1.12 we have ",2 X

= {X U {a}

if b EX;

Xn{a}'

ifb¢X,

hence ",2 X n ",X == 0 for every X. Thus (f3) is satisfied and the algebra belongs to Sl. Now Sl covers only S which is characterised by ({3) and (ry). Since {a} 1\ "'{ a} = {a} and {b} V "'{ b} = {a}' we see that (ry) fails. Hence V(L) = Sl.

Example 5.13 In Example 1.13 we have {

X2,··· ,Xn) = (x~, xi,···, xi); 2(Xl,X2,··· ,xn) = (xl,xn,··· ,xn)·

"'(Xl, rv

Taking into account that Xl ~ xn implies Xl 1\ x~ = 0 and xi V xn = 1 we readily see that axioms (3), (5) and (6 d ) are satisfied, so that LEN. For instance, taking a = (Xl,X2,· .. ,xn) and b = (Yl,Y2, ... ,Yn) we have ",2a ~ a V
(Xl,X n ,··· ,xn) ~ ==

(Xl

Vy~ V Yl, X2 V Yi V Yn,···, Xn Vyi V Yn)

(Xl V y~ V Yl, 1, ... ,

1).

We even have V(L) == N since N covers only K3 VLVS l which satisfies (13d), an axiom that generally does not hold in (L; "').

Example 5.14 In Example 1.14 let n == ITpr i , a == ITpfl, b == iEI

iE]

IT p7 1 • iEK

Then a 1\ a+ ==

IT

p~in{{3i'O/i-rnin{O/I,{3i}},

b V b+ ==

iE!U]

IT

p~ax{-yI'O/i-rnin{O/I''YI}}.

iEIUK

Let Pi be any prime factor of a, b, or n and, deleting subscripts, consider A

= min{{3,OI-min{OI,{3}},

B == max{1,0I-min{0I,1}}.

Considering separately the cases (1) 01 ~ {3, (2) {3 ~ 01 ~ 1, (3) (3 ~ 01 and 1 ~ 01, we see easily that A ~ B. Consequently a 1\ a+ ~ b V b+ and axiom (ry) holds. Moreover, a 1\ a+ == gcd { a, gcd{:, a} }

~ gcd{a, n} == a++

whence axiom (4) holds. We conclude from this that V(L; +) == K l .

6 Fixed points In an Ockham algebra (L j f) the notion of a fixed point of f was introduced in Chapter 2 as an element a E L such thatf(a} = a. Dually, we say that an element x of an Ockham space (Xjg) is a fixed pOint of g if g(x} = x. We shall denote by Fix L [resp. Fix X] the subset of L [resp. Xl formed by the fixed points of f [resp. g]. For every LEO, Fix L is an antichain (possibly empty). In fact, if a, bare fixed points of f with a ~ b then we also have b = f(b} ~ f(a} = a. We shall now show that a fixed point of f corresponds to a special downset of the dual space X, and that a fixed point of g corresponds to a particular prime ideal of L that has been already considered in the framework of de Morgan algebras [103].

11leorem. 6.1 Let (Xjg) be the dual space of (L;/) E O. Then a E L is a .fixed point of f if and only if the graph of g is bipartite with one of the blocks in the partition the down-set A corresponding to a. Proof Given a E L let A E 0 (X) be the down-set that corresponds to a. Then we have

a =f(a}

~ ~ ~

~

A =f(A} =X \ g-l{A} {A,g-l(A}} is a bipartition of X g(X \A} ~ A and g(A} ~ X \A the graph of g is bipartite and A is a block.

<>

Definition A down-set A of X that satisfies the conditions g(A} ~ X \A,

g(X \A} ~ A

will be called a distinguished down-set of X. Observe that if g is surjective (which is the case in particular for algebras in P n,O), then a distinguished down-set A of X satisfies the equalities

g(A}=X\A,

g{X\A}=A.

If, moreover, X is finite then the existence of a distinguished down-set requires IXI to be even. Corollary

If (X j g) has a fixed point then (L;/) is fixed point free.

Proof This follows from the fact that a distinguished down-set cannot contain an element of X together with its image under g. <>

Ockham algebras

106

The converse of the above corollary is in general false, but as we shall see below it is true for L E P3,1' For this purpose, we extend to the class 0 the following notion, which was considered in de Morgan algebras by Belnap and Spencer [27].

Definition A filter F of LEO is a truth filter if, for every a E L, F contains exactly one of the elements a and f(a}. In [27] it is shown that a de Morgan algebra is fixed point free if and only if it has a truth filter. Clearly, a truth filter is a prime filter, and so its complement is a prime ideal which enjoys the same defining property. We shall therefore call such an ideal a falsity ideal. Every falsity ideal is of course closed under the operation x ~ f2(x}. Lemma

6.1 If LEO then Land f(L} have the same fixed points. <>

If LEO and if I is afalsity ideal of L then I nf(L} is afalsity ideal of f(L}. Conversely, if I is afalsity ideal of f(L} then

Lemma 6.2

Il

={x EL I (:3y E 1)

x:::;;y}

is a falsity ideal of L. <>

Theorem 6.2 Let (X;g) be the dual space of (L;1) E O. Then x E X is a fixed point of g if and only if the prime ideal P that corresponds to x is a falsity ideal Proof We have (X;g) has a fixed point

~ ~ ~

Theorem 6.3 Let L E P3 ,1' point.

(:3P E Ip(L)) g(P) = P (:3P E Ip(L)) a E P # f(a} ¢ P (:3P E Ip(L)) P is a falsity ideal of L.

<>

If (L;1) isfixed pOintfree then (X; g) has afixed

Proof If (L;1) is fixed point free then, by Lemma 6.1, so is (f(L);1). Since (L;1) E P 3,1 we have that f(L} is a de Morgan algebra. It follows that f(L} has a falsity ideal Q. By Lemma 6.2, L also has a falsity ideal P to which there corresponds in (X; g) a fixed point. <> We now apply these results to the finite subdirectly irreducible Ockham algebras. Theorem 6.4 Let (L; f) be a subdirectly irreducible Ockham algebra belonging to Pm IZ and let (X; g) be its dual space. If the cardinality m - n of the

107

Fixed points

loop of X is even then 1 has at most two fixed pOints; and if it is odd then 1 is fixed point free.

Proof Suppose that X contains a distinguished down-set A. This requires that if A contains gi(X) then A contains neither gi-l (x) nor gi+l(X), but must contain gi+2(X). In other words, the bipartition of the graph of g is

{{x,g2(x),l(x), ... }, {g(x),g3(x),g5(x), ... }}. This bipartition is possible (and then is unique) if and only if the cardinality of the loop of X is even. In this case, the complement of A in X can also be a down-set, which means that L has at most two fixed points. <> The possible cardinalities of Fix 1 for an algebra (1; f) E 0 depend heavily on the subvariety of 0 to which 1 belongs. Clearly, every totally ordered Ockham algebra has at most one fixed point. Some classes of 0 enjoy the same property. In what follows the algebras that are considered will be assumed to be non-trivial (i.e. not to reduce to a single element).

Theorem 6.5 All algebras in subvarieties of N have at most one fixed point, and all algebras in subvarieties of Sl are fixed point free.

Proof If 1

E

0 has at least two fixed points then the axiom

rva 1\ rv2a ~ rvb V rv2b,

(8)

which characterises the subvariety N, cannot be satisfied. If now 1 has at least one fixed point then the axiom

({3)

rva 1\ rv2a

= 0,

which characterises the subvariety Sl, cannot be satisfied.

<>

For the moment we shall focus our attention on the class M of de Morgan algebras. It has the (non-trivial) proper subclasses B of boolean algebras, and K of Kleene algebras. Obviously, if 1 E B then Fix 1 = 0. The cases where 1 E K \ Band 1 E M \ K have to be treated separately, the latter being much more complicated.

Theorem 6.6 If 1 E K \ B then IFix 11 E {O, 1}. Proof By Theorem 6.5, 1 has at most one fixed point. Both the possibilities of being fixed point free, and of having precisely one fixed point, occur. For example, the subdirectly irreducible algebra K E K has a single fixed point; and the 4-element chain < a < b < 1 can be made into a fixed point free Kleene algebra by defining '0 = 1, a = b, b = a, r = 0. <>

°

Ockham algebras

108

We recall that (L;f) E M if and only iff is a polarity, in which case g is also a polarity in the dual space (X;g). Defining P(L) to be the set of all polarities on L, let min IFix LI

= min{IFix(L;J)1 : f

E

P(L}},

and let max IFix L I be defined similarly. In what follows we shall employ by abuse of language the expression 'odd chain' (resp. 'even chain') to mean a chain with an odd (resp. even) number of elements. Also, for every positive real number x the notation LxJ will denote the greatest integer not exceeding x.

Theorem 6.7 [103] Let L =

n~l x ... x n~k

1 min lFix L I = { 0

max IFix LI =

{ IT n~hJ ~

with (L;J)

E

M. Tben

if all nj are odd, otherwise;

if rj is even whenever nj is even,

j=l

otherwise.

Proof The dual space X of L consists of the disjoint union of r 1 chains of n1 - 1 elements, r2 chains of n2 - 1 elements, ... , rk chains of nk - 1 elements. Every polarity g of X maps each of these chains either onto itself or onto another chain of the same cardinality. If now all the nj are odd then each component of X is an even chain and so, whatever the mapping g may be, X has a distinguished down-set. It follows that (L; f) has exactly one fixed point. If, on the contrary, some nj is even then X has an odd chain C as a component, and g can be defined in such a way that g( C) = C. Then g has a fixed point and, by the Corollary to Theorem 6.1, (L;J) is fixed point free. Suppose now that, for some i, nj is even and rj is odd. Then every polarity g on X maps at least one odd chain of X onto itself; so X has a fixed point, and L is fixed point free. If however nj even always implies rj even then, for IFix LI to be maximum, g has to link the odd chains of X in pairs. As to the even chains, g has to link these in pairs also, with the possible exception of one 'isolated' chain. Since a pair of g-linked chains n contributes a factor n + 1 to the number of distinguished down-sets of X it follows that the contribution from the rj chains ~ - 1 to the number of distinguished down-sets is n}hJ, and the result follows. <> Corollary If L = n 2 and (L;J) E M then max IFix LI

= n. <>

Fixed points

109

Example 6.1 The following table illustrates Theorem 6.7 : L

min IFix LI

33 34 32 X 5 42 X 52

1 1 1 0

43

0

25

x

max IFix LI 3 9

3 20 0

To complete the information about IFix L I when L E M \ K, we require some results concerning the subsets LA and LV defined as follows for every

LEO: LA LV

= {a ELI a ~ rva} = {a 1\ rva I a E L}; = {a ELI a ~ rva} = {a V rva I a E L}.

Clearly, LA is a down-set and LV is an up-set of L. Lemma 6.3 Let (L; f) EO. Then the fixed points of f are minimal elements

of LV and maximal elements of LA. If L E M and Fix L 1- 0 then LV and LA are respectively the up-set and the down-set of L generated by Fix L. Proof If a E Fix L then a E LV nLA. Moreover, if b < a < c then rvb ~ rva = a > b gives bELA \ LV, and rvC ~ rva = a < c gives c E LV \ LA. If now L E M and a E Fix L consider an element bE LV such that b II a. The element c

= b V (b 1\ a) belongs to Fix L since c= (bVb)l\(bva)= bl\(bva)=c.

It now suffices to observe that b ~ c. 0

Note that the second part of Lemma 6.3 is not true for arbitrary Ockham algebras, as is shown for instance by the sub directly irreducible algebra K 3' Lemma 6.4 Let L E P 3,1' Then LA is an ideal of L

if and only if LEN V L.

Proof LA is an ideal if and only if, for all a, bEL, (a

1\

rva)

V

(b

1\

rvb)

~ rv[(a 1\ rva) V (b 1\ rvb)] = (rva V rv2a) 1\ (rvb V rv2b).

This holds if and only if

(Va, bEL)

a 1\ rva ~ rvb V rv 2b

which is the basic inequality (5) that characterises N V L. 0

Ockham algebras

110

Corollary If L E M1 (resp. L EM) then L" is an ideal of L if and only if L E K2 VK3 (resp. L E K) Proof It suffices to note that M1 1\ (N V L) = K2 VK3 and M 1\ (N VL) = K.

<>

Dually, we have the following results.

If L E P3,1

Lemma 6.5

then LV is afilter of L if and only if LEN V L.

<>

By considering (N' V L) 1\ (N V I) we can deduce from Lemmas 6.4 and 6.5 the following : Lemma 6.6 If L E P3,1 then L" is an ideal of Land L v is a filter of L if and only if L E Sl V 1:.

Proof We have

(N' V L) 1\ (N V I) = (N' 1\ N) V (L 1\ N) V (N' 1\ I) V (L 1\ I) =Sl VKVLvI = Sl V LvI. <> Theorem

6.s

For every n E IN \ {I} there exists L E M \ K with IFix L I = n.

Proof First, let n = O. We know that there are fixed point free algebras that belong to M \ K (for example, the subdirectly irreducible algebra M). Now let n = 1. In this case, if L E M and Fix L = {a} then by Lemma 6.3, L" = at and so, by the Corollary to Lemma 6.4, we must have L E K.

Finally, let n ~ 2. By the Corollary to Theorem 6.7 we can obtain a de Morgan algebra with n fixed points, and this algebra does not belong to K because of Theorem 6.6. <> We shall now extend the results obtained for M to a larger class, namely P 3,1' It is almost obvious (and will be shown in Theorem 8.16) that for every L E P 3 ,1 the smallest congruence for which the quotient algebra belongs to

M is <1>1, i.e. the congruence given by (x,y) E <1>1 -<==> "-'x

="-'y.

Moreover, L/l is dually isomorphic to ,,-,L. Lemma 6.7

(1) V(L) (2) V(L) (3) V(L)

If L E P3,1

then

[B,Sd ifand only ifV(,,-,L) = B; E N\ Sl ifand only ifV(,,-,L) = K; E P3 ,1 \ N if and only ifV(,,-,L) = M. E

Proof (1) : L E Sl if and only if the axiom (f3) is satisfied; i.e. L has a boolean

skeleton.

111

Fixed pOints

(2) : LEN if and only if the axiom (8) is satisfied; i.e. L has a Kleene skeleton. (3) : L E P 3 ,1 if and only if ",3 a = a for every a E L; i.e. L has a de Morgan skeleton. <> Since Land rvL have the same fixed points, and since we know the possible cardinalities of Fix L when L E M, the following result is straightforward.

Theorem 6.9 Let L E P3,1' Tben (1) ifV(L) E [B, sd then Fix L = 0; (2) ifV(L) E N \ S1 then IFix LIE {O, I}; (3) if L is countable and V(L) E P3 ,1 \ N then IFix LI E IN \ {I}.

<>

Corollary Let L E M1 Tben (1) ifV(L) E [B, S] then Fix L = 0; (2) ifV(L) E [K, K2 V K3 ] then IFix LIE {O, I}; (3) if L is countable and V(L) E [M, Md then IFix LIE IN \ {I}.

Proof We have S1

N 1\ M1

1\ M1 =

Sand

= (N V N) 1\ M1 = (N 1\ Md V (N 1\ Md = K2 V K 3 . <>

Our purpose now is to sharpen the results in Theorem 6.9(2),(3) for some subvarieties of P 3 ,1'

Theorem 6.10 Let L E P3,1' JjV(L) satisfies axioms (5, 3d, 6) but not axiom (4) then L isfixedpointfree.

Proof Observe that (5,3 d , 6) = (5, 16 d ). Since L satisfies (5), L" is an ideal by Lemma 6.4. Assume, by way of obtaining a contradiction, that L has a fixed point e. The axiom (5) characterises LV N eN and therefore, by Theorem 6.5, this fixed point is unique. By Lemma 6.3, e is a maximal element of the ideal L", hence is the generating element of L"; i.e. L" = e!. Since L satisfies (16 d ), a ~ e implies that a = rv2a. It follows that for every bEL we have b 1\ ",b = r,}(b 1\ ",b) = rv 2b 1\ rvb, which means that (15 d ) = (3 d ,4) is satisfied, contrary to the hypothesis.

<>

Corollary JjV(L) is any of the subvarieties LVS, SVK1 VK2, SVK1, SVK1, SVK2, SVK, SVK, [VS, SVK1 VK2, SVK1, SVK1, SVK2, SVK,

then L isfixedpointfree.

Proof Consider the first seven subvarieties listed. Each of these satisfies the axioms (5, 3d, 6). To show that axiom (4) is not satisfied, it suffices to replace

Ockham algebras

112

axiom (6) by axiom (4) in the equational basis of the subvariety and observe that in so doing we obtain a subvariety that is smaller than (in fact, covered by) the original. This procedure is summarised in the following table. subvariety equational basis LVS SVK1 VK2 SVK1 SVK1 SVK2 SVK SVK

changed to

giving

(ry,3 d,6) L (ry,3d,4) (ry,3d,6,8) (ry,3d,4,8) Kl VK2 (ry, 3d, 6, 6d) SVK1 (ry,3d,4,6 d) (ry,ld,6) (ry,ld,4) Kl (ry,3,3 d,6) (ry,3,3d,4) = (ry,1,3d) K2 K (ry,ld,3,6) (ry,l d,3,4)= (ry,I,ld)= (O!,"Y) (ry,3,3d,6,6 d) (ry,3,3d,4,6 d)= (ry,1,3d,6 d) SVK

By duality, the same conclusion holds for the other subvarieties.

<>

For L E Ml a more sophisticated procedure can be used to decide when L is fixed point free.

Theorem 6.11 [35] {fV(L)

= S V M then L is fixed pointfree.

Proof L satisfies the axioms (1, 3d, 6 d)' Suppose, by way of obtaining a contradiction, that L has a fixed point e. By Lemma 6.3, we have e! ~ LA. Since (1) implies (4), L satisfies (15 d) = (3 d , 4), which says that and means that LA ~ rvL. It follows that every a ~ e is such that a Moreover, (1) implies (3) and L satisfies (16) = (3, 6d ), that is a V rvb V rv 2b = rv 2a V ",b V ",2b.

= rv 2a.

Hence every a ~ e is such that a = rv2a. Finally, if a II e then a 1\ (: and a Ve both belong to rvL and, owing to the serniconvexity of rvL in any MS-algebra, we have a E rvL. Thus rvL = L and so L E M, a contradiction. <>

Theorem 6.12 [35] {fV(L) = S V M V Kl then L isfixedpointfree. Proof L satisfies (1, 6 d, 12d)' Suppose, by way of obtaining a contradiction, that L has a fixed point e. Then the axiom (16) = (3, 6d ), namely a V rvb V rv 2b = rv 2a V rvb V rv2b, is satisfied and, for every a E L, we have a VeE ",L. Hence we have also

(*)

(Va E L) a V rva VeE rvL.

Fixed pOints

113

Now L also satisfies the axiom (18 d ) = (11, I2 d ), namely

(a

V

rva) 1\ rvb 1\ rv 2b = (rv2 a

V

rva) 1\ rvb 1\ rv2b,

which gives

(Va, bEL) In particular, we have

(**)

(a V rva) 1\ e E rvL.

(Va E L)

By (*), (**), and the convexity of rvL, it follows that a V rva E rvL for every a E L, which is (1S) = (3, 4d), namely . and gives the contradiction L E M V Kl C S V M V K 1 . <>

Example 6.2 Here we shall illustrate the fact that if L is countable and such that V(L)

= Ml

then IFix LI E IN \ {I} (refer to the Corollary to Theorem

6.9). First, an example of L E Ml that is fixed point free. Consider the lattice L = 22

X

d

3 made into an MS-algebra as follows :

x 0 abc d e f g h XO 1 d i

Z J

1

jab g jOb g 0

a

The thick lines indicate the
rva 1\ rv 2a ~ a

V

= Ml we observe that

rvb V rv2b,

which appears in the equational basis for M V K2 V K 3 , fails to hold. In fact we have whereas

c V rva V rv2a

= C V d V a =h.

Next we observe that the subdirectly irreducible MS-algebra M 1 itself has two fixed points. For examples of algebras L such that V(L) = Ml and having r fixed points for r ~ 3, we consider the lattice L = n x n with n ~ 4,

114

Ockham algebras

made into an MS-algebra as indicated in the following diagram, the thick lines indicating the cI> -classes.

There are n - 1 fixed points, marked with circles, namely

bn- 1 Ji J2 , ... In-4 In-3 , an-l' That V(L) = Ml follows from the observation that it contains copies of the subdirectly irreducible algebra M 1 , namely the intervals [b n - 3 , b~-3] and [a n-3, a~-3].

7 Fixed point separating congruences If (L; f) is an Ockham algebra and x, yare distinct elements of L then we shall say that a congruence ~ on L separates x and y if (x, y) ¢ ~. By a fixed point separating congruence on L we shall mean a congruence that separates every pair of fixed points of L. We shall denote by :F(L) the set of fixed point separating congruences on L. We let Fix L = {O!i ; i E I} be the set of fixed points of (L;J), and we shall assume throughout this chapter that IFix LI ~ 2. All of the results that follow appear in [56).

Theorem 7.1 Let (L; -) E M. If 01, f3 are distinctfixed points of L and fixed point separating congrnence on L then ~hCl'II,B,Cl'V,Bl = w. Proof We prove first that

~1[Cl',Cl'v,8)

=w.

~

is a

For this purpose, suppose that

O!~x~y~O!vf3

with (x,y)

E~.

Define x,B

= x 1\ (xV (3).

Then we have ~=xV~l\m=~V~I\~Vm=xl\~Vm=~

and so x,B is a fixed point. Likewise, so is y,B

= Y 1\ (yV (3).

Since (x,y) E ~ we have (x,8,y,B) E ~ whence x,B =y,B' Then x,Bl\f3 =y,B 1\ (3 gives x 1\ f3 =Y 1\ f3, whence

= x 1\ (01 V (3) = 01 V (x 1\ (3) = 01 V (y 1\ (3) = y 1\ (01 V (3) = y. Thus ~1[Cl',Cl'v,8J = w. Similarly, we can show that ~1[Cl'II,B,Cl'J = w. x

Suppose now that 01 1\ f3 ~ x ~ y ~ 01 V f3 with (x, y) E iJ. Then 01

~

01

V X ~ 01 V Y ~ 01 V f3

O!I\f3~O!l\x~O!l\y~O!

It follows that 01 V x =

01

VY

with (01 V x, 01 V y) E iJ; with (O!I\X,O!I\Y)EiJ.

and 01 1\ x =

01

I\y, whence x

The above result can be extended to Kw as follows.

=y. ¢

116

Ockham algebras

Theorem 7.2 Let (L;1) E Kw and let 0'., (3 be distinct fixed points of L. If fJ E .F(L) tben every x E (0'., 0'. V (3] that is separated from 0'. by Cl>w is also separated from 0'. by fJ. E (0'., av(3] with (x,a) ¢ Cl>w, Le.fn(x) =j: 0'. foreveryn. Since L E Kw there exist positive integers m, n such that f 2m +2n (x) = j2n(x), and clearly

Proof Suppose that x

Consider the elements

m-l

= A f2n+2i+l (x),

m-l

V f2n+2i(x). i=O We have 0'.1\(3:( s < 0'. < t:( 0'. V (3 withf(s) = t andf(t) = s. Consequently, the subalgebra M = (0'., (3, s, t) is de Morgan. Since fJ 1M separates fixed points, it follows by Theorem 7.1 that (s, t) ¢ fJ. We now show that fJ separates 0'. and x. In fact, if we had (0'., x) E fJ then on the one hand (a,/2k(x)) E fJ would give (0'., t) E fJ, and on the other hand (a,/2k+l(x)) E fJ would give (a,s) E fJ, whence we would have the contradiction (s, t) E fJ. <> S

t=

i=O

Theorem 7.3 If (L;1)

E

Kw then .F(L) forms a complete ideal of Con 1.

Proof It is clear that .F(L) is a down-set of Con L, so it suffices to prove that if A = (fJi)iEl is a family of fixed point separating congruences then V fJ j iEI

is fixed point separating. Suppose, by way of obtaining a contradiction, that 0'. , (3 are distinct fixed points such that (0'., (3) E V fJ j. Then (0'., 0'. V(3) E V fJ j jEI

iEI

and there e:xist Zo,' .. ,Zn ELand fJ 1 , ••. ,fJn E A such that 19!

0'.

192

19~

19n

= Zo < Zl < Z2 < ... < Zn = 0'. V (3.

By Theorem 7.2 there exists kl such that fk! (Zl) = above chain we obtain, for some p and q, 0'.

0'..

Applying f2k! to the

=f2k! (Zl) = ... =f 2k!(Zp_l) ~ fk! (zp) = ... =f2k! (Zq_l) ~

...

Applying Theorem 7.2 again, there exists k2 such that

f2k2(t2k!(Zp))

= 0'..

Continuing this argument, we arrive at the existence of t such that

f2t(a V (3) i.e.

0'.

V

= 0'.,

(3 = a. This provides the required contradiction. <>

~ 0'. V (3.

Fixed paint separating congruences

117

In general, for L ~ Kw the down-set :;=(L) of fixed point separating congruences on L need not be an ideal of Con L. This can be illustrated in the following way.

Example 7.1 Consider the lattice

.{3

made into an Ockham algebra by defining

f(O) = 1, f(l) = 0, f(a) (Vi ~ 0) f(x/) = X/+1,

= a,

f([3)

= [3,

and extending to the whole of L. Then we have

(Vi ~ 0) f(a/)

= aj+l,

Clearly, (L;I) ~ Kw and

1(J1

{a}, Now the relation

1(J2

f(b j) = bi+1,

f(yj)

=Yi+1,

f(Zt)

= Zj+1'

= {)(xo,[3) is fixed point separating with classes [ao,1],

[0, bo],

[zo,Yo].

whose classes are

A=[O,(3),

B={[3},

C=([3,l],

L\{AUBUC}

is also a fixed point separating congruence. Since clearly (a,xo) E 1(J2 and (x 0, (3) E I(J 1 we have (a, (3) E \0 1 V \0 2 and therefore \01 V

1(J2

= ~ ~ :;=(L).

Ockham algebras

118

Definition We shall say that an Ockham algebra (L; f) is fixed point complete if 01* = V O!i and 01* = /\ O!i exist with f{O!*) = 01* and f{O!*) = 01*. iE/

iE/

It is clear that every finite Ockham algebra is fixed point complete.

Example 7.2 For an example of an Ockham algebra that is not fixed point complete, consider the lattice

Xo

made into an Ockham algebra by defining

f{xo)

= 011,

(\Ii) f{O!i)

= O!i,

f{z)

= 0,

and extending to the whole of L. Here we have

f( /\

O!i)

i;;: 1

=f{O)= 1 =!=z=

V O!i'

i;;: 1

It is immediate from Theorem 7.3 that for L E Kw there exists a maximum fixed point separating congruence on L. We shall denote this by '1'. Since
Theorem 7.4 Let L E

Kw befixedpoint complete. If'l' =
Proof Suppose, by way of obtaining a contradiction, that Then we have

01*

01*

= V O!i < 1. iE/

= /\ O!i > 0.

Denote by A the subalgebra generated by

iE/ {O!*,O!*}

U Fix L. The congruence {O,O!*},

{O!*,

cp on A whose classes are

1}, and singletons otherwise

is fixed point separating with cp > w, and extends to a fixed point separating congruence C{3 on L. Then 'I' ~ C{3 and therefore

'l'IA

~

C{3IA = cp > w.

119

Fixed point separating congruences But we have w IA = wand so, by the hypothesis, 'PIA this contradiction that 01* = 1. <>

= w.

We deduce from

The converse of Theorem 7.4 does not hold in general, as is exhibited by the following example.

Example 7.3 The angelfish. Consider the lattice

made into a fixed point complete de Morgan algebra in the obvious way with fixed points OIj. Here we have V OIj = 1 and w = w. But 'II =f w. The j~l

'I' -classes are shown by thick lines; for example,

(x, x) E 'II.

Definition We shall say that L is fixed point distributive if, for every x whenever

E

L,

V OIj, A OIj, V(x 1\ OIj) and A(x V OIj) exist we have jEi

X 1\

jEi

jEi

V OIj = V(x 1\ OIj),

jEi

jEi

jEi

xV

A OIj = A(x V OIj).

jEi

jEi

When L is fixed point complete and fixed point distributive we have the following analogue of Theorem 7.l.

Theorem 7.5 Let (L; -) E M. Suppose that L isfixedpointcompleteandfixed point distributive. Then every {) E :F(L) is such that {)I[a.*,a.*l = w. Proof Suppose that 01*

= jEi A OIj ~ X ~ Y ~ V OIj = 01* iEi

120

Ockham algebras

with (x,y)

Then for all i,j

E {l.

OIj /\ OIj

~ (x V

E

I we have

OIj) /\ OIj

~

(y

V OIj) /\ OIj ~ OIj

with ((x V OIj)

/\ OIj,

(y V OIj) /\ OIj) E {l.

It follows by Theorem 7.1 that

(x V OIj)

= (y V OIj)

/\ OIj

/\ OIj

whence we have x V OIj = (x V OIj)

/\

V OIj JEI

= V [(x V OIj) /\ OIj] JEI = V [(y V OIj) /\ OIj] JEI = (y V OIj) /\ V OIj JEI

= y V

OIj

and consequently

x

= x V A OIj = A (x V OIj) = A (y V OIj) =Y V A OIj = y. <> jEI

jEI

jEI

jEI

Suppose now that L is fixed point complete and consider the relation defined on L by

(x,y) E e

<¢=}

e

(x V 01*) /\ 01* = (y V 01*) /\ 01*.

It is clear that e is a fixed point separating congruence on L. When L is also fixed point distributive, we have the following situation.

Theorem 7.6 Let (L;f) be an Ockham algebra that is fixed point complete and fixed point distributive. For i,j E I define the relation {l jj by (x,y) E {ljj Then

<¢=}

(x V OIj) /\ OIj = (y V OIj) /\ OIj.

e = A{ljj. jJ

Proof That e

~

A{l jj follows from the observation that jJ

121

Fixed point separating congruences and that

1\ -a

j } :::;;

jJ

8 follows from the observation that

(x V 01*) 1\ 01*

= (x V 1\ OIj) 1\ 01* j

= I\{x V OIj) 1\ 01* = I\[(x V OIj) 1\ 01*] = 1\ [(x V OIj) 1\ VOl}] = 1\ V[{x V OIj) 1\ OI}]. j

j

j

j

}

I:)

}

When L E Kw is fixed point complete and fixed pOint distributive we have the following description of the congruence 'P.

Theorem 7.7 If (L;1)

E

Kw

isfixed point complete andfixed point distribu-

tive then

Proof Suppose first that L E Kp,q' Then we have jq{L) E Kp,o' Recalling the

notation used in Chapter 3, let M = rAjq{L)) be the biggest de Morgan subalgebra of jq{L). Suppose that x,y EM are such that (x,y) E 'PIM' Then, for all i,j E I,

({x

V OIj) 1\ OI},

(y

V OIj) 1\ OI}) E

'PIM'

It follows by Theorem 7.s that (Vi,j E /)

(x V OIj) 1\ OI}

= (y V OIj) 1\ OI},

and therefore, by Theorem 7.6, (x,y) E 81M' Thus 'PIM:::;; 81M and consequently, since 8:::;; 'P, we have 'PIM = 81M' Since, by Theorem 3.11, jq(L) is a strong extension of M it follows that 'P 1/*) = 81/*) and hence 'P1/q(L)

= tI>ql/*) V lJIl/q(L) = tI>ql/q(L) V 81/q(L)::;;; (tI>q V 8)I/q(L)'

The reverse inequality being trivial, we have that lJIl/q(L) Since Conjq(L) ~ [tI>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)

E

Kw is fixed point compact then every congruence

V

19(O!j,O!j).

(llIj,llIj)E
Proof Suppose first that L E Kp,q, so thatfq(L) E Kp,o. Consider the de Morgan algebra M = T 2 (r(L)}. For x E M and i,j E / define Xjj

= (x V O!j) /\ (t(x) V O!j} /\ (O!j V O!j).

125

Fixed point separating congruences

Then a simple calculation reveals that Xjj is a fixed point of M. cp E ['I', d, let cp* = V 'I9( OIi,OI).

Given

(01/ ,00f) E ¥'

If x,y E M are such that (x,y) from the equality

E

cp then we have (Xjj'yjj)

E

cpo It follows

that we therefore have ((x V OIj) /\ OIj, (y V OIi) /\ OIj} E 'I9(Xij,Yij) ~ cp*. Since L is fixed point compact we have

(x V 01*) /\ 01* =

V 1\ [(x V OIi) /\ OIj] JEI' iEI'

and so it follows from the above that ((x V 01*) /\ 01*, (y V 01*) /\ 01*) E cp*. Since (x, (x V 01*) /\ 01*) E e we deduce from this that (x,y) E e V cp* ~ 'I' V cp* and hence that CPIM =

V

'I9(X,Y)IM ~ ('I' V CP*)IM'

(xJ')E¥'IM

The reverse inequality being trivial we therefore have, using Theorem 7.7, CPIM = ('I' V CP*)IM = (cI>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 [
Example 7.6 Consider the lattice

0<0

made into a fixed point compact de Morgan algebra in the obvious way with fixed points 01.0, .. . ,01. 00 • In this, the congruence I(J > 'P whose classes are A=[O,OI. oo },

C=(OI. oo ,l],

B={OI. oo },

D=L\{AUBUC}

has no complement in ['P, t]. In contrast, the interval [w

=
Theorem 7.12 Let L E Kw be fixed point complete and fixed point distributive. Then

['P, t]

~

Con T 2 (A}/e

where A is the subalgebra {OJ E9 [01.*,01.*] E9 {I}.

Fixed point separating congruences

127

Proof Suppose first that L E M so that 'P = e and Tz(A)

= A.

Let

A* = ([OI*,OI*];J)

and note that, by the Corollary to Theorem 7.9, we have Aje J.L : ['P, £]

-t

~

A*. Define

Con A *

by setting (with a slight abuse of notation) J.L(cp) = CPIA* where CPIA* is the congruence induced on A * by cP IA . It is clear that J.L is a morphism for 1\. To see that it is also a morphism for V, let a, bE A* be such that a ~ b and (a, b) E (CPI VCPZ)IA*' Then (a, b) E CPI Vcpz and so there exist xo, ... , Xn E L such that

= Xo == Xl == '" == Xn = b where each == is either CPI or CPz. Defining Yi = (Xi V 01*) 1\ 01* we have a =Yo == YI == ... ==Yn = b a

with each Yi

E

A*. Consequently, (a, b) E CPIIA* V CPZIA*' Thus

(CPI V CPZ)IA* ~ CPIIA* V CPZIA*' Since the reverse inequality is trivial, we have that J.L is a V-morphism. To see that J.L is injective, let CPI, cpz E ['P, £] be such that CPIIA* = CPZIA*' If (x,y) E CPI then we have (1) ((x V 01*) 1\ 01*, (y V 01*) 1\ 01*) E CPIIA* and, by the Corollary to Theorem 7.9,

= CPZIA*;

(2) (x V OI*,y V 01*) E 19(01*,1) = e ~ 'P ~ cpz; (3) (x 1\ OI*,y 1\ 01*) E 19(0,01*)

= e ~ 'P ~ CPz.

By (1) and (2) we have (x V OI*,Y V 01*) E cpz which, together with (3) gives (x,y) E CPz· Hence CPI ~ CPZ· Similarly, cpz ~ CPI and so CPI = CPz· To see that J.L is surjective, let 19 E Con A*. Let 78 E Con A be such that 78IA* = 19; i.e. let [0]78 = [01*]19 U {OJ and [1W =[01*]19 U {1}. By the congruence extension property there exists cP E Con L such that cP IA 78. Then J.L(cp V 'P) = CPIA* V 'PIA* = 19 since, by Theorem 7.5, 'PIA* = w. Thus J.L is an isomorphism and consequently the result holds for L E M. Suppose now that L E Kw. We have [w, £] ~ Con Ljw ~ Con T(L) ~ Con Tz(L) = [wITz(L)' £ITz(L)]'

=

Ockham algebras

128

For cP ~
CP!T2(L)

is therefore a bijection. Thus

Now since the result holds for T2 (L) E M we have ['I'!T 2(L), t!T2(L)] ~

where C

= T2 (L) nA.

Corollary

Since T 2 (C)

Con T 2 (C)/8

= T2 (A) the result for L E Kw follows. 0

With L as above, the following statements are equivalent:

(1) ['I', t] is boolean; (2) T 2 (A) isfinite; (3) L hasfinitely manyfixedpoints.

Proof (1) =} (2) : If (1) holds then T2 (A) has finitely many 8-classes and it follows by the definition of 8 that T2 (A) must be finite. (2) =} (3) : This is clear. (3) =} (1) : Observe that for fixed points 01 j , 01j we have 19(O!i,O!j)

= 19(O!j /\O!j,O!j VO!j) = 19Iat(O!j /\O!j,O!j VO!j).

Since then 19(O!j /\ O!j'

O!j

V O!j)

/\ 19(O!j

V O!j,l)

= 19Iat(O!j /\ O!j' O!j V O!j) /\ [19Iat(O!j V O!j' 1) V 19Iat(O, O!j /\ O!j)]

=w, 19(O!j /\ O!j'

O!j

V O!j) V 19(O!j V O!j, 1)

= 19Iat(O!j /\ O!j' O!j V O!j) V 19Iat(O!j V O!j' 1) V 19Iat(O, O!j /\ O!j) = t

we have that

19(O!j,O!j)

is complemented in Con L with complement

19(O!j, O!j)'

= 19(O!j V O!j,I) = 19(0, O!j /\ O!j).

If L has finitely many fixed points it therefore follows by Theorem 7.11 that every congruence cP ~ 'I' has a complement in ['I', t]. Hence ['I', t] is boolean. 0

Example 7.7 Consider the lattice

Fixed point separating congruences

129

.0

made into a de Morgan algebra with fixed points aI, a2, a 3' Here
= E> = 'I9(O,a*) = 'I9(a*, 1). By Theorem 7.12, ['1', L] is boolean. In contrast, [w =
This follows from Corollary 2 to Theorem 7.10 and the fact that, since L is infinite, Con L = [w =
Theorem 7.13 JfL isjixedpointcompacttben

r= V'I9(aj,aj)' jj

Proof By the hypothesis there exists a finite subset /* of / such that

r = 'I9(a*, 01*) = 'I9{ 1\

iEI*

Now, given any j E J*, we have

(Vi and therefore

It follows that

E J*)

aj, V aJ iEI*

Ockham algebras

130 and hence that

The reverse inequality being trivial, we deduce that

r = V -a(OI.i,OI.j)' 0 iJ

The following example shows that Theorem 7.13 fails when L is not fixed point compact.

Example 7.8 Let L be the lattice obtained by adding to the cartesian ordered set 7L. x 7L. a biggest element 1 and a smallest element 0, and make L into a de Morgan algebra in the obvious way with fixed points OI. n = (n, -n) for every n E 7L.. Then L is fixed point complete and fixed point distributive, but not fixed point compact. Since 01.* = 1, we have e = wand r = t. In contrast, the congruence V19(OI.i, OI.j) has three classes, namely {OJ, 7L. x 7L., and {I}. iJ

We now consider the question of precisely when the interval [ct»w, '£I] is boolean. For this purpose, we establish the following analogue of Theorem 7.12.

Theorem 7.14 Let L E Kw be fixed point complete and fixed point distributive. Then [ct»w, '1'] ~ Con T 2 (B)/r

where B is the subalgebra [0,01.*] EB [01.*,1]. Proof Suppose first that L EM, so that ct»w = w, 'I' = e, and B = T2(B). Consider the mapping A : [w, '1'] --+ Con B given by A(/{J) = /{JIB' Clearly, A is a morphism for A. To see that it is also a morphism for v, let (a, b) E B be such that a ~ b and (a, b) E (/{JI V /{J2)IB' Since /{JI, /{J2 ~ 'I' we have that /{JI V /{J2 ~ 'I' so /{JI V C(J2 is fixed point separating. It follows by Theorem 7.5 that (01.*,01.*) ¢ /{JI V /{J2' Consequently, either a, bE [0,01.*] or a, b E [01.*,1]. Now there exist xo, ... ,XII E L such that

= Xo == Xl == ... == Xn = b where each == is either /{JI or /{J2' Defining Yi = (Xi va) A b we have a =Yo == YI == ... == Yn = b a

and, from the above observation, every Yi E [0,01.*] or every Yi E [01.*,1], i.e. every Yi E B. Consequently, (a, b) E /{JIIB V /{J21B and so (C(JI V

C(J2)IB ~ /{JIiB

V

C(J2IB'

Since the reverse inequality is trivial, we have that A is a v-morphism.

Fixed point separating congruences

131

To see that >- is injective, suppose that 10 1,102 E [w, '1'] are such that IOIIB = 1021B' If (x,y) E 101 then since 101 :::;; 'I' = e we have

= (y V a*) /\ a*; (x /\ a*,y /\ a*) E IOIIB = 1021B;

(1) (x (2)

V

a*) /\ a*

(3) (x V a*,y V a*) E IOIIB = 1021B'

By (1) and (3) we have (x V a*,y V a*) E 102 which, together with (2) gives (x,y) E 102' Hence 101 :::;; 102' Similarly, 102:::;; 101 and so 101 = 102' Observing that Im>- ~ [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
Congruences on KI ,1 -algebras

137

Our objective now is to use the description of 19(a, by in a de Morgan algebra to determine precisely when a principal congruence on a KI,I-algebra is complemented. For this purpose, we require the following two results.

1b.eorem 8.7 If LEO and 19 E Z(Con L) then 191~2L

E

Z(Con rv 2L).

Proof Let 19' be the complement of 19 in Con L. From 19 1\ 19' ::::: w it follows that 191~2L 1\ 19'1~2L = W~2L. Since 19 V 19' ::::: ~ there exist xo, ... ,xn E L such that o = Xo 19 Xl 19' X2 19 ... 19' Xn = 1. This implies that

0= Xo 19 rv 2XI 19' rv 2X2 19 ... 19' rv2xn whence

191~2L V 19'1~2L

= 1,

= ~~2L· ~

1b.eorem 8.8 Let (L; rv) E KI,I. Tben we have (rv 2X, rv 2y) E 19(a, b) only if (rv 2 X, rv 2y) E 19( rv2 a, rv 2 b).

if and

Proof If (rv 2x, rv 2y) E 19(a, b) then rv 2X and rv 2y satisfy the eight equations of Theorem 8.1. Applying rv 2 to each, we obtain (rv 2 X, rv 2y) E 19(rv2a, rv2b). The converse follows from the fact that (rv2 a, rv2b) E 19(a, b) and therefore 19( rv2 a, rv2b) ~ 19(a, b). ~ In what follows we shall use the fact that if (L, rv) E KI,I then rv 2L = rvL. Theorem 8.9 Let (L, rv)

E

KI,I and let a, bEL be such that a ~ b. Define

cp(a, b) = 19(0, rvb 1\ rv2a) V 19(rvb 1\ rv2b, rvb) V 19(rva, rva V rv2a). Tben we have 19(a, b) V cp(a, b) necessari~y 19(a, b)' = cp(a, b).

= ~;

and

if 19(a, b) is complemented then

Proof Consider the following chain of elements in L :

o

Ockham algebras

138

Each pair of elements that are joined by a vertical line segment are congruent modulo 19{a, b) and, by Theorem 2.1, 19{O, rvb I\ rv2 a) = 19{rva V rv2b, 1). The above chain therefore shows that (O, 1) E 19{a, b) V cp{a, b) and so 19{a, b) V cp{a, b) = L

Suppose now that 19{a, b) is complemented. Then from what we have just proved it follows that 19{a, b)' ~ cp{a, b).

To obtain the reverse inequality, we observe by Theorem 8.7 that 19{a, b)I~L is 19{ rv2 a, rv2b) on rvL, and therefore by Theorem 8.5 that the complement of19{a,b)I~L iscp{a,b)I~L' Thus we have that 19{a,b)'I~L

= (19{a,b)I~L)' = cp{a,b)I~L'

and so 19{a,b)' is an extension to L of cp{a,b)I~L' cp{a, b)I~L identifies each of the pairs

Now it is clear that

(O, rvb 1\ rv2a), (rvb 1\ rv2b, rvb), (rva, rva V rv2a) of elements of rvL and that so does any extension to L of cp{a, b)I~L' By its definition, therefore, cp (a, b) is the smallest extension to L of cp (a, b) I~L . Consequently, we have that cp{a, b) ~ 19{a, b)'

as required. <> The above knowledge of what 19{a, b)' is (whenever it exists) allows us to determine precisely under what conditions it does exist.

Theorem 8.10 If (L; rv) (1) (2) (3) (4)

E K1,1 then 19{a, b) is complemented if and only if

bl\rvbl\rv2a~a; b~avrvavrv2b;

b 1\ rvb ~ a V rv2b; b 1\ (rva V rv2a) ~ a V rva.

Proof By Theorem 8.9 we have that 19{a, b) is complemented if and only if 19{a, b) 1\ cp{a, b) = w. Now, by Theorem 2.1 and the fact that 191at{X I\y, x) = 191at{Y,X Vy), we have 19{O, rvb 1\ rv2a) = 191at{O, rvb 1\ rv2a) V 191at{rv2b V rva, 1); 19{rvb 1\ rv2b, rvb) = 191at{rvb 1\ rv2b, rvb); 19{ rv a, rva V rv2a) = 191at {rv a , rva V rv2a). It is now a routine matter to show that

Congruences on

K1,1

-algebras

139

19{a, b) 1\ 19{0, rvb 1\ rv2a) = W ~ (I) and (2) are satisfied; 19{a, b) 1\ 19{rvb 1\ rv2b, rvb) = w ~ (3) is satisfied; 19{a, b) 1\ 19{ rv a, rva V rv2a) = W ~ (4) is satisfied, from which the result follows. 0

If (L; rv) E MS then 19{a, b) is complemented if and only if (I') a 1\ rvb = rv2a 1\ b 1\ rvb; (2') b 1\ (a V rva) = b 1\ ( rv2 a V rva).

Corollary 1

Proof If (L; rv) E MS then conditions (2) and (3) of Theorem 8.10 hold, whereas conditions (I) and (4) are equivalent to (I') and (2') respectively. 0 Recalling from Chapter 2 the principal congruence 19a and the fact that

19 a

= 19lat{a 1\ rva, rva),

we deduce the following Corollary 2 If (L; rv) E MS then Jor every bEL the principal congruence 19{0, b) is complemented with 19{0, b)' = 19~2b' Proof The four conditions of Theorem 8.10 are satisfied, so 19{0, b)' exists. Taking a = in Theorem 8.9 we have

°

19{0, by = tp{O, b) = 19lat{rv2b 1\ rvb, rvb)

= 19~2b' 0

We can now apply the above to determine precisely the complemented principal congruences in a Stone algebra.

If (L; *) E S then 19{a, b) is complemented if and only if a is complemented in [0, b]. Moreover, when it exists,

Theorem 8.11

19{a, by = 19lat{a* 1\ b**, 1). Proof By Corollary 1 of Theorem 8.10, 19(a, b) is complemented if and only

if

(2/1) b:( a Va*. Now (2/1) gives b 1\ a**:( (a V a*) 1\ a** = a whence b 1\ a** = a; and this property implies (2/1). Hence 19{a, b)' exists if and only if b 1\ a** = a. Now if this holds then a* V a = a* V b and consequently

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

1\ a*) V

a.

Since b 1\ a* 1\ a = 0, it follows that b 1\ a* is the complement of a in [0, b]. Conversely, if x is such that a 1\ x = and a V x = b then x:( a* and b :( a V a* so that

°

b = b 1\ (a

V

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

Ockham algebras

140

and consequently

b 1\ a** = (a

1\

a**) V (b

1\

°

= a V = a. that a 1\ b* =

a* 1\ a**)

°

As for the final statement, we first observe and hence a** 1\ b* = 0. We can now compute 19(a, b)' using Theorem 8.9. Denoting complements in Conlat L by and recalling that C

191at(X,y)C = 191at(0,X) V 191at(Y, 1), we have

19(a, b)' = 19(0, b*) V 19(a*, 1) = 191at(0, b*) V 191at(b**, 1) V 191at(a*, 1) V 191at(0,a**) = 191at(b*, a*)C V 191at(a**, b**)C =[191at(b*, a*) 1\ 191at(a**, b**}]C = [191at(a* 1\ b** 1\ (b* V a**), a* 1\ b**}]C = [191at(0, a* 1\ b**)]C = 191at(a* 1\ b**, 1). ¢

Corollary 1 (VaE(L,*))

19(a,ava*)=19(a**,l).

Proof From the above, 19(a,a V a*)'

= 191at(a*, 1) = 19(a**, 1)'. ¢

Corollary 2 If L E B then all principal congruences are complemented. ¢ We shall now consider more closely, for L E K1,1, the structure of Con L and its relationship to Con ",L . We know that the lattice Con L is algebraic and distributive. As for Con ",L, it follows from Theorem 8.6 that if ",L is finite then, every congruence being a finite join of principal congruences, Con ",L is boolean. It is remarkable that the converse of this is also true. As mentioned in Chapter 3, this was shown by Sankappanavar [84] using techniques from universal algebra. We thus have : Theorem 8.12 If L EM then Con L is boolean if and only if L isfinite. ¢ Theorem 8.13 If L E M then Con L is a distributive algebraic lattice whose compact elements form a boolean sublattice. Proof By Theorem 8.4 and the fact that Con L is distributive, it follows that the set of compact elements of Con L is closed under intersection whence it forms a sublattice of Con L. That it is boolean follows from Theorem 8.5. ¢ In a Kl 1-algebra L we clearly have cI>1 =. = cI>w and we shall denote this congru'ence simply by cI>. 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 <1>. <> 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 <1>; (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<,rv 2 X, are equivalent to the above. Proof (1) =? (2) : By the Corollary to Theorem 8.17, if there is a unique coatom then this is necessarily . (2) =? (3) : If is a coatom then it is the only one. (3) =? (1) : This is clear. (4) : This is clear. (2) (3) =? (5) : If (3) holds then (5) follows from the fact that, in MS, has trivial kernel.

*

Congruences on K1,1-algebras

143

(5) => (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 < <1>. 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 ~ <1>. 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 = <1>. ~

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

1heorem 8.19 Let L E Kl,l and 19 E Con L. Then (",2 X, ",2y) E 19

~

(x,y) E 19 vet>.

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

<1>= '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 <1>. It is readily seen that g is an isomorphism. For every 7f E Con L let 7f* = 7f V <1>. 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)'

<:)

CorolIary #(DF2n; bl , bn) = tUn-1 + 1). Proof For n = 1, 2 the result follows by direct computation. For n have

Theorem 9.15 For n even, IFix L(DC 2n ; k)1

~

3 we

=i in -1. 2

Proof Consider the subset A = {al,"" ai2 n , bI!"" bin}. LetA' = DC 2n\A. 2 For every down-set I of A let 1* = A'\k(I). Note that if] is a distinguished down-set of DC 2n then b l E] => 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>
Theorem 10.2 Co is self-dual Proof Let p be a fixed element of r( 0) and define

I(Jp :

Co ---) X by

I(Jp(q) = 2n -p-q If q

E

Co then, since 0 -< p implies 0 -< 2n - p, we have q 0= 2n = 2n - q + q

Thus Iml(Jp

~

I>
I>
2n - p so

2n - q + 2n - p = 2n - p - q = I(Jp(q)

Co. If now r E Co then 2n - r -p

E Co

and

I(Jp(2n - r - p) = 2n - p - 2n + r + p = r.

The dual space of a finite simple Ockham algebra Thus 1m ({Jp

= Co·

Since p is odd, we also have

({Jp(r)-j({Jp(s) Hence

({Jp

169

<===?

2n-r-p-j2n-s-p

is a dual order embedding -and Co

<===?

rts.

= 1m ({Jp is self-dual. <>

If X is not connected then we have 1 = g(O) ¢ Co; for 1 E Co gives 0 1><11 whence 1 = g(O) 1><1 g(1) = 2, and so on, whence X would be connected.

Theorem 10.3 Two connected components of X are either isomorphic or dually isomorphic. More precisely, d

d

d

d

d

Co c::: C 1 c::: C 2 c::: ... c::: C t - 1 c::: Co

and the number t of connected components is necessarily odd.

Proof It suffices to prove that Cot. C 1. For this purpose, consider the map 1j; : Co ---) X defined by the prescription 1j;(q) = q + 1. Since 0 1><1 q implies 1 1><1 q + 1 we have 1m 1j; ~ C 1 . But if r E C 1 then r 1><1 1 gives r-l = r+2n -11><10, so that r-l E Co with 1j;(r-l} = r. Hence Im1j; = C 1 . Since

1j;(r) -j 1j;(s)

<===? <===?

r + 1 -j s + 1 r = r + 1 + 2n -1

t s + 1 + 2n -1 = s

it follows that 1j; is a dual order-embedding and Cot. C 1. Finally, that t must be odd is clear from the sequence of dual isomorphisms. <>

Corollary Every connected component of X is self-dual. <> We now proceed to consider the number of connected components. The simplest case arises when r( 0) is a singleton.

Theorem 10.4 .ljT(O) = {r} then X has t nents each of which is an ¥- -crown.

:=

gcd{n, r} connected compo-

Proof In the cyclic group Z2n the order of r is 2;z where t = gcd{2n, r} (equivalently, for r odd, t = gcd{n, r}). Since the elements of Co constitute the subgroup (r) of Z2n it follows that Co = {O, r, 2r, ... , -l)r} and we have o --< r, r)- 2r, ... , (2;Z -l)r )- 0

e;l

Consequently we see that Co is an ¥--crown. Now each connected component of X is a coset of Co. The number of connected components is therefore IZ2nl/ICol = t. <>

Corollary X is an n -crown if and only if r (0) = {r} with r, n coprime. <>

Ockham algebras

170

Example 10.1 Let n = 15 and r(O) = {9}. Then we have t = 3 and the connected components are the following 5-crowns : 9

27

15

3

21

~

Theorem 10.4 can be generalised as follows.

Theorem 10.5 Let r(O) = {rl,'" ,rk}' Then the number ofconnected components of X is t = gcd{n, rl,"" rk} Proof As in the proof of Theorem 10.4, consider the cyclic group Z2n' Here

h) V

... V

(rk)

= (u)

where u = gcd{rl' ... ,rk}, and the order of (u) is 2nlgcd{2n, u}. It follows that the number of components of X is the number of cosets of (u) in Z2n, namely t = IZ2n 1/1(u) I = gcd{2n, u} = gcd{ n, rl, ... ,rk}' <>

Corollary If r(O) = {rl,"" rk} then X is connected integers n, r 1 , .•• , r k are coprime. <>

if and only if the

Example 10.2 Let n = 15 and r(O) = {3, 9}. Then t = 3 and X is 3 9 15 21 27

~

Theorem 10.6 Let X be of cardinality 2n and have t compoents. Then (1) when n is even the degree of each vertex can take any even value between 2 and nit; (2) when n is odd the degree of each vertex can take any value between 1 and nit.

Proof All vertices of X have the same degree, so it suffices to establish the property for the vertex O. If n is even then Max Co has even cardinality and, since 0 --< j implies o --< 2n - j, the degree of 0 is 21r(O)I with 1 ~ Ir(O)1 ~ nl2t. Ifn is odd then Max Co = {t, 3t, ... , n, ... , 2n-3t, 2n-t} and has odd cardinality, and the degree of 0 is 21r(O)I if n ~ r(O) and 2jr(O)I-1 if n E r(O). It follows that the degree of each vertex can take any value between 1 and nit. It is 1 if n = t (Le. X is the disjoint union of n two-element chains), and it is n if t = 1 (Le. X is connected, complete bipartite). <>

The dual space of a finite simple Ockham algebra

171

CoroIlary There are generalised crowns that cannot be made into the dual space of a finite simple Ockham algebra.

Proof Consider the generalised crown

Here we have n ::;: 4 and all vertices have degree 3. By the above, X cannot be made into the dual space of a finite simple Ockham algebra. ~ We can now characterise those generalised crowns that are the connected components of X. We require the notion of a circulant matrix [10], Le. an n x n matrix [O!ii] with the property that, modulo n, O!ii ::;: O!i+I,j+I' In a circulant {O, 1}-matrix we shall say that the i-th row is symmetric if

Theorem 10.7 A generalised crown C n;k is a connected component of X if and only if it can be represented by a circulant (0, I) -matrix of order n in which the first row is symmetric and has k entries 1.

Proof Let M be a matrix satisfying the conditions. We show that M can represent the connected component Co of X. Since the first row of M is symmetric, the entries 1 of this row yield the vertices that cover O. The number t of connected components of X can be chosen arbitrarily, the only restriction being that t must be odd. Then the vertices that cover 0 are kIt, k 2t, ... , 2n -k2t, 2n -kIt,

all k, being odd, and Min Co::;: {O, 2t, ... , 2{n - tn. Since M is circulant, its second row yields the vertices that cover 2t and, more generally, the m-th row yields the vertices that cover 2{m -1}t. Thus we obtain a Cn-k which is a connected component of several X. Conversely, let X be the dual space of a finite simple Ockham algebra and suppose that X has t connected components. By Theorem 10.1 we know that if Co is not a singleton then it is a generalised crown C n k' so it can be represented by a square {O, 1}-matrix M of order n. The row of M that corresponds to the vertex 0 contains k entries 1 and is necessarily symmetric since 0 -< r implies 0 -< 2n -r. By permuting two rows, this row can become the first row. It remains to show that the resulting matrix is equivalent to a circulant matrix. Observe now that if 0 is covered by kIt, k 2t, ... , 2n -k2t, 2n -kIt

Ockham algebras

172 then 2t is covered by

(k 1 +2)t, (k2+2}t, ... , 2n-(k2-2}t, 2n-(k l -2}t,

i.e. there is in M a row whose elements are identical to those of the first row but are moved one place to the right. This row can then become the second row. By repeating this procedure, we obtain a circulant matrix with the required properties. <> At this point we draw the reader's attention to the fact that two circulant (0, l}-matrices which are equivalent can give rise to non-isomorphic dual spaces, hence to non-isomorphic finite simple Ockham algebras. The simplest example is provided by the matrices

A=

[~

0 0 1 0 1 1 0 1

~I

and

B=

[~

1 1 0 1 0 0 1 0

with which are associated the graphs

GA:Pm 1

o

5

3

2

7

and

4

In G A we have x -< g(x) or x

~ g(x},

II

G:~ 1

3

5

7

0

2

4

6

B

whereas in G B we have x Ilg(x}.

Theorem 10.8 For every even n ;;:: 6 there is at least one finite simple Ockham algebra whose dual space is a generalised crown that is not an S~. Proof Taking r(0} = {rl,"" rk, n - rk,"" n - rd we see that 0 and n are both covered by all the elements of

Since n ;;:: 6, we can choose the index k small enough to obtain for 5 a proper subset of Max X. <> We can express Theorem 10.8 in matrix terms: for every even n ;;:: 6, say n = 2n', there is at least one circulant (0, l}-matrix of order n that represents a finite simple Ockham algebra and in which the first row is doubly symmetric in the sense that alj = 1 al,n-j = 1 and (V) :;; n') alj = 1 al,n'-j = 1.

*

*

Example 10.3 Let n = 6 and take r(0} = {I, 5}. Then 0 and 6 are both covered by 1,5,7, 11. It follows that 2 and 8 are covered by 1, 3,7,9,

The dual space of a finite simple Ockham algebra

173

whereas 4 and 10 are covered by 3, 5, 9, 11 . The corresponding matrix is 1

0 1 1

1 0 1 1

1 0 1 1

0 1

0 1

1

0

1 1 0 1 1 0

1

0 1 1

0 1 1 0 1

0 1 1

Now the question to solve is : how to recognise when a square (0,1)matrix all of whose line sums are equal is equivalent to a circulant matrix the first row of which is symmetric? In order to examine this question we require the following notions. Let M = [O!ij] be a square (0, I)-matrix of order n. For every pair (i,j) denote by Pi,j(M) the number of values of k for which O!ik = O!jk = 1. In particular, Pi,i+I (M) does not depend on i if M is circulant. Let also Pi,j(M) = {Pi,j(M) ; i,j E {I, ... , n}}. Clearly, Pi,j(M) is invariant under arbitrary permutations of the columns and the rows. The symbols PI,i+I,i+2, (M) and Pi,j,k, (M) have obvious meanings.

Example 10.4 The C 5;3 whose matrix M is 1

o 1

o 1

100 1 1 1 0 1 0

1 0

101 1 1 0 0 1 1

cannot be the dual space of a finite simple Ockham algebra. In fact, M has to be equivalent either to A = Circ{O, 1,1,1,0} or to B = Circ{l, 0,1,0,1}. But Pi,j(A) = Pi,j(B) = {1,2} whereas Pi,j(M) = {1,2,3}. The fact that n is odd here proVides another direct proof: in A and B the rows are distinct, whereas in M rows 2 and 4 are identical.

Example 10.5 The C6-4 whose (circulant) matrix Mis 1 1 0 1 0 1

1 1 1 0 1 0

0 1 1 1 0 1

1 0 1 1 1 0

0 1 0 1 1 1

1 0 1 0 1 1

174

Ockham algebras

cannot be the dual space of a finite simple Ockham algebra. Indeed, M has to be equivalent to one of A=Circ{O, 1,1,1, 1,0}, B=Circ{l, 1,0,0, 1, I}, C=Circ{I,O, 1, 1,0, I}.

Here we have Pi,j(A) = Pi,j(B) = {2, 3} and Pi,j(C) = {2, 4}. Since Pi,j(M) = {2, 3} we must go further into the analysis of the first two cases. Therefore we consider Pi,j,k and note that Pi,j,k(M) = {I, 3} and Pi,i+l,i+2(A) = Pi,i+l,i+2(B) = 2.

Example 10.6 Suppose that we are given the matrix

° 1 1

M=

1 1

°

1

1

°° 1 1

1

1

°° 1

1

° 1 1 1 1

°

1 1

1 1

°° °° 1 1

1

1

Since both Circ{O, 1,1,1,1, O} and Circ{l, 1, 0, 0,1,1} have distinct rows, we need only decide whether or not M is equivalent to Circ{ 1 , 1 , 1 , ,I} in which, like M, the rows are identical in pairs. To make the third row of M symmetric, we permute columns 2 and 6. Rearranging the rows, we obtain Circ{ 1 , 1 , 1 , ,I}. Consequently the C6;4 represented by M is the dual space of a finite simple Ockham algebra. We are of course aware of the weakness of the above procedure. Given a Cn;k it is necessary to consider all circulant (0, I)-matrices of order n whose first row is symmetric and has k entries 1. If n is large, such an examination can be long. An algorithmic method would be welcome.

°, °

°, °

We have already seen that if IXI = N is odd then there is only one finite simple Ockham algebra in the class PN,o. The situation is quite different if N is even.

Theorem 10.9 Ibe number OI n of non-isomorphic finite Simple Ockham algebras that belong properly to P 2n ,0 is given by 01

-

n -

{

2!n 2 !(n+l)

ifn is even; if n is odd.

Ibis includes {}n algebras whose dual spaces are not connected, where {}n is the number of subsets {rl, ... , rk} of r(o) for which gcd{ n, rl, ... , rk} t= 1.

Proof The Ockham space X is completely determined by its cardinality 2n and the subset r(O). If n is even then ~ Ir(O)1 ~ !n and the number of

°

Tbe dual space of a finite simple Ockham algebra possibilities is

(ton) +

175

Ct) + .. + 0:) =

2tn.

When n is odd we have 0 ~ W(O)I ~ t(n + 1} and the number of possibilities becomes 2t(n+1). Moreover, if IXI = IX'I with r(O) = {rl"'" rk} in X and r(O/) = {ri, ... , r,J in X' then we have X ~ X' ~ {r 1 , ... , r k}

= {ri, ... , rk}'

The final assertion is an immediate consequence of the Corollary to Theorem 10.5. <>

= 15. Then there are 256 non-isomorphic finite simple Ockham algebras that belong properly to P30,0' Here 19 15 = 10 and the dual spaces that are not connected are the following On which it suffices to describe in each case only the component Co and to indicate the number t of components).

Example 10.7 Let n

= 0 we have Co = {O} and t = 30. For W(O)I = 1 the possibilities are For r(O)

3

15

9

21

27

5

25

15

27

9

15

21

3

15

~ txKI ~ I 0

r(0)={3 } 1=3

10 20 r(0)={5} 1=5

0 r(0)={15 } 1=15

r(0)={9} 1=3

For W(O)I = 2 the possibilities are 3

9

15

21

27

3

15

9

21

27

15

5

25

9

27

15

3

21

~~~~ r(0)={3,9} 1=3

For W(O)I

r(0)={3 ,IS} 1=3

0 10 20 r(0)={5 ,IS} 1=5

r(0)={9 ,IS} 1=3

= 3 we have

~

o

6 12 18 r(0)={3,9 ,IS} 1=3

24

Remark The following table gives the first 35 values of 19 n'

176

Ockham algebras n 'I3 n

n

'I3 n

n

'I3 n

n

'I3 n

n

'I3 n

1 2 3 4 5 6 7

8 1 9 4 10 2 11 2 12 4 13 2 14 2

15 16 17 18 19 20 21

10 1 2 8 2 4 18

22 23 24 25 26 27 28

2 2 16 8 2 32 4

29 30 31 32 33 34 35

2 38 2 1 66 2 22

1 1

2 1 2 2 2

There appears to be no obvious rule of formation here. However, the following observations are useful : (1) if n = 2a then 'I3 n = 1; (2) if n = pr withp an odd prime then 'I3 n = 2!-(pr-l+1). In particular, if n =p then '13 11 = 2. Indeed, the integers not exceeding n that are not prime to n are the elements of S = {p, 3p, 5p, ... , pr}. Now S has cardinality t(pr-l + 1), and 'I9 n is the cardinality of the power set of S.

Theorem. 10.10 Let L be a finite simple Ockham algebra and let X be its dual space. Then (1) L is fixed point free if and only if IXI is odd; (2) L has two fixed points if and only if X is an antichain of even cardinality; (3) L has a single fixed pOint if and only if X is a generalised crown or a disjoint union ofgeneralised crowns.

Proof In each case we apply Theorem 6.1. (1) If IXI is odd then the order on X is discrete and there is no bipartition of the graph of g. (2) If X is the antichain {O, 1, ... , 2n -I} then the partition

{{0,2, ... ,2n-2}, {1,3, ... ,2n-l}} gives two fixed points since both the blocks are down-sets. (3) If X is a generalised crown, or a disjoint union of generalised crowns, then the partition {Min X, Max X} gives the unique fixed point. Since these cases exhaust the possibilities, the result follows. 0 If :::;; and :::;; 1 are orders on the same set X and if :::;; 1 is an extension of :::;; then a mapping g : X ~ X can be order-reversing on (X; :::;;) but not

The dual space of a finite simple Ockham algebra

177

on (X; ~d, and conversely, as shown by the following examples in which g(n) = g(n + 1) modulo IXI. 1

•

0

•1

(Xl; ~)

•

2

I 0

•

2

(Xl; ~d

g is order-reversing on (Xl; ~) but not on (Xl; ~d, and is not orderreversing on (X 2 ; ~) but is on (X 2 ; ~d. The following result, although trivial, is of interest in this context. Let (X; ~ , g) be a finite Ockham space. If ~ 1 extends a way that g remains order-reversing on (X; ~ 1) then

Lemma

~

in such

(1) (X; ~1 ,g) is also an Ockham space; (2) the dual algebra (L 1 ;!) of (X; ~1 ,g) is a subalgebra of the dual algebra (L;f) of (X; ~ ,g); (3) the number offixed points of (L 1,1) is at most equal to that of (L;f).

Proof It suffices to observe that the operation Jon LJ(X; ~1 ,g) is the restriction of Jon LJ(X; ~ ,g), that every down-set of (X; ~d is a down-set of (X; ~), and that every fixed point of (LJ(X; ~1 ,g);f) is a fixed point of (LJ(X; ~ ,g);f). <> In fact, the following result is now clear. Theorem 10.11 Every finite simple Ockham algebra whose dual space has cardinality 2n is a subalgebra ofthe finite Simple Ockham algebra whose dual space is the antichain {O, 1, ... , 2n -I}. <>

We illustrate this property for n = 3. By Theorem 10.9, there are four non-isomorphic finite simple Ockham algebras that belong properly to P60, namely Lo = (2 6;f) whose dual space is the antichain {O, 1, ... , 5} and the subalgebras L 1 , L 2 , L3 whose dual spaces are

~

o

2 X2

4

i!i o

2

4

X3

In the diagrams that follow we omit, for visual reasons, a great many of the lines. As a lattice, Lo ~ 26 . The algebra Lo has two fixed points, corresponding to the down-sets {0,2,4} and {1,3,5}. Each of L 1 ,L 2 ,L 3 has a single fixed point. As a lattice, L1 ~ 23 EB 23 . The elements of L2 are

178 J

s algebra Ockham ""''''ets the dO spond to

fhr, With the r e ith corre th e g e elllentso fth1at tb gether W £ , to f to o , e £ S f n e O o e o th P d ts Se 'b l 4 5 } . T e e e1l1en 'd J 3 a l e Oents that corres By a { } 4 . m {2 3 , " " '" th eveJ ele {D, 2 ,5 } {a, 1, 2}, " gst thes3e 4 De ll li - , 2, 4}, {2, 4, 5}, n o m A others. ts {D 2 3} [D ltb,4l} , { l twel"" O e -s l3. n W 2 and o to the d l1 , a.11 l11 Theore1

~i ; SUbalg'ebra}~fb

11 Relative Ockham algebras In Chapter 1 we gave an affirmative answer to the question of whether every bounded distributive lattice L can be made into an Ockham algebra (L; ",). If the subvariety V of 0 to which (L; "') has to belong is prescribed, then the answer is far from being affirmative, even when L is finite. For example, as we observed in [41], the 5-element distributive lattice with Hasse diagram 1

c a

b

cannot be made into an Ml -algebra since otherwise we would have 1 = ",0

= ",(a 1\ b} = ",a v",b = 1 or ",b = 1, so that either ",2 a = 0 or ",2 b = 0 from

whence either ",a which it follows by axiom (1) that either a = 0 or b = 0, a contradiction. So the following general problem arises: a subvariety V of 0 being given, characterise the (finite) distributive lattices that can be made into an algebra in V. A particularly simple first step in the solution of this problem is given by

Theorem 11.1 Let V E A(O} with V :2 P2 ,1' Then every finite distributive lattice can be made into a V -algebra.

Proof Let L be a finite distributive lattice, let X be its dual space, and let y EX. Define g : X ---7 X by g(x} = y for every x E X. Clearly, g is orderreversing and g2 = g. Thus X has been made into a P 2 ,1-space. <>

In [59] G. Bordalo and H. A. Priestley established the converse of Theorem 11.1, thus proving

Theorem 11.2 Let V E A( O}. Then every finite distributive lattice is a reduct of a V -algebra

if and only if V :J- P2 ,l' <>

In the same spirit, they solved [58] for the five subvarieties B, K, S, S, the following interesting question :

S

A subvariety V of 0 being given, what are the finite distributive lattices all of whose closed intervals can be made into a V-algebra?

180

Ockham algebras

We shall follow their work vety closely, reporting the main results (with or without proof) and show that in fact their solution is far from being limited to the above five subvarieties. We begin with some definitions.

Definition Let V be a subvariety of 0 and let L be a finite distributive lattice. Then L is a relative V-algebra if evety interval [a, b] of L can be given the structure of a V-algebra. We shall denote the class of finite relative V-algebras by Yr' On A{O) we can define an equivalence relation f:::! by Vf:::!W ~ Vr=Wr The f:::!-class of V will be denoted by [V]. We shall use duality throughout, but only in the finite case : all lattices and ordered sets involved will be finite. If P and Q are ordered sets then we shall say that P has Q as a subposet if there is an order-embedding of Q into P; and that P has Q as a convex poset if there is an order embedding a : Q ---4 P such that a{Q) is a convex subset of P. For example, the ordered set

N

N

contains as a subposet but not as a convex subposet. A class c of ordered sets is said to be convex-closed if, whenever PEe, evety convex subposet of P also belongs to c. In Chapter 5 the reader can find equational bases for the five subvarieties we are interested in, as well as the dual equivalents of the axioms involved. The following lemma is therefore straightforward. Lemma 11.1 Let (L;J) E 0 and let (X;g) be its dual space. Then

L E B if and only if g = gO; L E K ifand only if gO = g2 and gO Kg; L E S if and only if gO ~ g; L E S if and only if gO :::; g; L E S if and only if g = g2 and gO Kg· 0 For evety (finite) LED we denote by XL the dual space of X, and for evety (finite) ordered set P we denote by L p the lattice of down-sets of P. The following theorem establishes an interesting correspondence between the intervals of L and the convex subposets of XL'

Theorem 11.3 (I) Let L be ajinite distributive lattice and let M = [a, b] be a closed interval of L Then X M is a convex subposet of XL'

Relative Ockham algebras

181

(2) Let Q be a convex subset oj the finite ordered set P. Then there is a closed interval M oj Lp such that X M is order-isomorphic to Q.

Proof (1) : If the down-sets A, B of XL are such that A ~ B then B \ A is a convex subset of XL' In fact, let p, q E B \ A with P :(; x :(; q. Since B is decreasing and q E B, we have x E B. Since A', the complement of A, is increasing and PEA', we have x E A'. Hence x E B \ A. We now show that CJ(B \A) ~ M = [a, b]. For each P E CJ(B \A) define J(P) = P U A. Clearly, J(P) is a down-set and A ~ J(P) ~ B, whence J(P) E M. Moreover, J is injective, J(CJ(B \ A)) = M, and J is a lattice morphism. (2) : Let Q be a convex subset of P. Then

Q = QL n Qi = QL \ (QL \Q) = Ql \ (QL \ (QL n Qi)) = Ql \ (Ql n (XL \ Qi)). Let QL = Band Ql n (XL \ Qi) = A. Both A and B are down-sets. Hence Q is order-isomorphic to X M where M = [A, B]. <>

Corollary Afinite distributive lattice L is a relative V -algebra if and only if every convex subposet oj XL can be endowed with an order-reversing map g which makes it into a V -space. <> This latter property enables us to define the 'strategy' which will be successful in the case of the subvarieties B, K, S, and S. • Seek a family (Pi)iEI of ordered sets (a family that is as 'small' as possible) such that no Pi can be made into a V-space. • Consider the class E of ordered sets which have no P j as a convex poset. • If every member of E can be made into a V-space and if E is convexclosed, then the finite relative V-algebras are exactly those L for which XL E E. Moreover, the lattices for which XL E E are the finite relative W-algebras for any W:2 V which is such that no Pi can be made into a W-space. Lemma 11.2 Let P be a finite ordered set. Then theJollowing are equivalent:

(1) P is an antichain;

I as a subposet; (3) P does not contain I as a convex subposet.

(2) P does not contain

The class oj all finite antichains is convex-closed. <>

182

Ockham algebras

An ordered set P is a tree if, for every x E P, xl is a chain.

Lemma 11.3 Let P be a finite ordered set. Tben the following are equivalent: (1) P is a disjoint union of trees;

A as a subposet; (3) P does not contain A as a convex subposet. (2) P does not contain

Tbe class of all disjoint unions offinite trees is convex-closed. Proof Only (3) ~ (2) is non-trivial. Suppose that{x, u, v} ~ P with x> u, x > v, and u II v. Let Xl be a minimal element of the set xl nuT n v T. Then Xl> U, Xl> v, and there is no y such that Xl> Y > u and Xl> Y > v. Take u l to be a maximal element of (X ll \ {xl})nu T and Vi to be a maximal element of (xll \ {xl})nvT. Then Q = {Xl, u l , Vi} is convex by the definition of u l and Vi. Moreover, u l II Vi since, for example, u l ;;;:: Vi would give u l E X ll nuT n v T,

which is impossible by the minimality of Xl. Hence Q is isomorphic to

A. <>

Combining Lemma 11.3 with its dual version, we obtain

Lemma 11.4 Let P be a finite ordered set. Tben the following are equivalent: (1) P is a disjoint union of chains;

A nor V as a subposet; (3) P contains neither A nor V as a convex subposet. (2) P contains neither

Tbe class of all disjoint unions offinite chains is convex-closed.

<>

Theorem 11.4 Tbe >:d-classes ofB, K, S, S are asfollows: (1) (2) (3) (4)

[B] = {V E A(O) I V dB, V ~ K, V ~ S; V ~ S}; [K] = {V E A(O) I V d K, V ~ S; V ~ S}; [S] = {V E A(O) I V d S; V ~ S}; [S]={VEA(O)IVdS;V~S}.

Proof We restrict ourselves to establishing (1) and illustrate the 'strategy'

described above. The family {Pj}jEI is formed by the only three distinct g-maps on the two-element chain x described as follows : a

x y

gl (a) Y x g2(a) x x g3(a) y y

I.

In fact, there are

< y, namely those

Relative Ockham algebras

183

I

The maps gl,g2,g3 make into a K-, S-, S-algebra respectively. By Lemma 11.2, the class c is the class of finite antichains. Finally, every antichain is a B-space if it is endowed with the identity map as a g-map. The conclusion follows easily. <) The next result gives an algebraic description of Vr where V is in any of the equivalence classes [B], [K], [S], [S].

Theorem 11.5 Let V E A(O) and let L be afinite distributive lattice. Tben

(Cl!) if V E [B] then L E Vr if and only if L is boolean; ((3) if V E [K] then L E Vr if and only if it satisfies any of the following equivalent conditions: (1) L is a direct product of chains; (2) L contains neither 22 EB 1 nor 1 EB 22 as an interval; (3) L has neither 22 EB 1 nor 1 EB 22 as a homomorphic image. h) If V E [S] (resp. if V E [S]) then Vr is defined in the following way: (a) the trivial algebra and the 2-element chain are in Vr ; (b) any finite direct product of elements of Vr is in Vr ; (c) if L E Vr then 1 EB L E Vr (resp. L EB 1 E Vr ); (d) any element ofVr can be constructed by repeated application ofthe properties (a), (b), (c). Further, LEVr if and only if it satisfies either of the following equivalent conditions: (1) L does not contain 22 EB 1 (resp. 1 EB 22) as an interval; (2) L does not have 22 EB 1 (resp. 1 EB 22) as a homomorphic image. <) If we apply Theorem 11.5 to the subvariety Ml (see the diagram on page 92), we obtain [B] = {B}, [K] = {K, M, K 1 , M V KIl, and [S] consists of all the other subvarieties of MI' So parts ((3) and h) of Theorem 11.5 generalise results previously obtained by Varlet [102] and Bordalo [57] for M and S respectively. Whereas the relative V-algebras for V E {B, K, S, S} can be characterised by the exclusion of a set of intervals, for V = S the procedure fails and a more sophisticated method has to be used.

Lemma 11.5 A non-empty finite ordered set P can be made into an S-space if and only if every connected component of P has a node. Proof *= : Suppose that P =

U l~i~k

x MXi' Then g is order-reversing, g2

CXi' Define g by g(x) =

Xj

whenever

= g, and g MgO; so (p;g) is an S-space.

Ockham algebras

184

Let (P;g) be the dual space of an S-algebra. Since gZ = g, we have that)) = g(P) is an antichain; in fact, if g{YI) = gZ{Yd ~ gZ{yz) = g{yz) then gZ{YI) ~ g2{y2) and so g{YI) = g{Y2)' Since g ff gO we have P = ))L U ))i. <> =i> :

Lemma 11.6 Let (X; g) be the dual space ofa fmite relative S-algebra. Then X has no subposet isomorphic to the fence

N.

Proof Suppose that X contains the subposet {u, v, x, y} with u < x, y < x, y < v, u II v, u I y, x I v. We shall show that X has a convex subposet Q = {a, b} U [d, c] such that

a -< c, d -< b, d < c, a I b, a I d, c I b. First, consider This set contains x but no element t below v (otherwise u ~ t < v which contradicts u I v). Let x I be a minimal element of X I' Then {u, X I, Y , v} is isomorphic to

N.

Next, consider X2 =

x! n (X \yi) n (X \yL) n ui .

This set contains u. Let UI be a maximal element of X z . Then we have that UI II v; for UI ~ v gives the contradiction u ~ v, and UI ~ v gives the contradiction Xl ~ v. Moreover, UI Ily. We now show that UI -< Xl' Suppose in fact that there existed U2 such that UI < U2 < Xl' Then Uz ~ u; and U2 j v since otherwise we have the contradiction Xl ~ V. By the minimality of Xl we then have U2 j y. We also have Uz 1, y, for otherwise we have the contradiction U I ~ y. Consequently, Uz E X 2 · But this contradicts the maximality of UI' Hence we have that

UI -<

Xl'

Similarly, we can show the existence of an element VI that covers y and thus we obtain the subposet Q . Finally, we show that no map g can be defined on Q in such a way that it becomes an S-space. Suppose in fact that such a g existed. Then since g ff gO we have g(a) E {a, c}. Now (1) if g(a) = a then g(c) = a and g(d) = c, But then g2(d) = a and we have the contradiction g2(d) t= g(d); (2) if g(a) = c then g(c) = c and g(d) = c. But then g(b) ~ c and therefore g(b) E {a, c}, which is impossible since b II a and b I c. <>

f?elative Ockham algebras

The lattice dual of the fence

185

N is the lattice L8 with Hasse diagram

It is ~pown that the finite distributive lattices which do not have L8 as

a homomorphic image are precisely those whose duals do not contain the as a subposet. fence

N

The latter class of ordered sets coincides with the class of series-parallel posets ~58,71}, defined as follows. A finite (non-empty) ordered set is series-parallel if it can be constructed from singleton sets using the operations of disjoint union and linear sum. For jns14n~e, all trees are series-parallel. By Lemma 11.6, the dual spaces of Sr-~lgebras ~re series-parallel. The four-element crown

cM d aV'!b

is serie~-Harallel, being the linear sum of the antichains {a, b} and {c, d}. By Lemma f1.6 it cannot be made into an Sr-space. The non-trivial finite distributive lattices whose duals are series-parallel posets are those that can be puilt up from 2-element chains using direct product and vertical sum. Lemmas 115 and 11.6 lead ratqer easily to the following

Theorem 11.6 Let P be a finite series-parallelposet. Then the following statements are equivalent: (1) every connected component of P has a node;

(2) no convex subset of P is

I>
(3) in the construction of P from singletons there is a restriction on linear namely that PI EB P 2 is permitted only when PI has a biggest element qr P 2 has a smallest element (or both). 0 ~ums,

186

Ockham algebras

The translation of Theorem 11.6 into algebraic terms is provided by the following result.

Theorem 11.7 Let L be afinite distributive lattice. Then thefollowing statements are equivalent: (1) LESr ; (2) L is a member of the class C defined as follows: (a) the trivial algebra and the 2 -element chain are in C; (b) the direct product of two members of C is in C; (c) if LI and L2 are in C U {0} then LI E9 1 E9 L2 is in C; (d) every member of C is obtained in a finite number of steps using the properties (a), (b), (c). (3) L does not have L8 as a homomorphic image and L contains no interval isomorphic to 22E922. <>

12 Double MS-algebras The notion of a double Stone algebra is well known. This is an algebra (L; V, /\, *, +,0,1) of type (2, 2, 1,1,0,0) such that both (L; *) and (LOP; +) are Stone algebras with the unary operations a H a* and a H a+ linked by the properties Our objective now is to consider the following natural generalisation.

Definition A double MS-algebra is an algebra (L; V, /\, 0, +,0,1) of type (2, 2, 1 , 1 , 0, 0) such that (L; 0) is an MS-algebra, (L; +) is a dual MS-algebra, and the unary operations are linked by the properties (Dl) (Va E L) a+ O= a++; (D2) (Va E L) a O+ = aOO. A double MS-algebra will generally be denoted by (L; 0, +). The double MS-algebras form a variety that we shall denote by OMS.

Example 12.1 If B is a boolean algebra and

B(2)

then BP) is a double Stone algebra in which {a, b)o

=

I a:;:;; b} and (a, b)+ =

({a, b) E B2

= (b', b')

(a',a').

Example 12.2 Every de Morgan algebra (M; -) is a double MS-algebra; it suffices to define a O= a+

= a for every a EM.

Example 12.3 If (M; -) is a de Morgan algebra and (S; *, +) is a double Stone algebra then (M x S; °, $) is a double MS-algebra where {a, b)o = (a, b*) and (a, b)$ = (a, b+). Immediate consequences of (Dl) and (D2) are the following.

Theorem 12.1

if (L; 0, +) is a double MS-algebra then

(I) (Va E L) a O:;:;; a+; (2) LOO = L++ .

Proof (I) : Writing a Ofor a in (D2), we obtain aOo+ = a OOO

=

aO; and from a :;:;; a OO we have, since a H a+ is antitone, aOo+ :;:;; a+. (2) : (Dl) gives L++ ~ LOo; and (D2) gives the reverse inclusion. 0

The skeleton S{L) of a double MS-algebra (L; 0, +) is defined in the same way as for any Ockham algebra, namely by S{L)

= LO = {XO I x E L}.

Ockham algebras

188 By Theorem 12.1 we also have

= L+ = {x+ I x E L} and clearly a E 5(L) if and only if a = aOO = a++. Every double MS-algebra 5(L)

has a de Morgan skeleton. Fundamental to the study of double MS-algebras is the notion of a residuated mapping. We recall here from [3] the following basic facts concerning such mappings. If E and F are ordered sets then a mappingj : E ---7 F is said to be residuated if the pre-image under j of every principal down-set yl of F is a principal down-set of E. By [3, Theorem 2.5], j : E ---7 F is residuated if and only if it is isotone and there is an isotone mapping g : F ---7 E such that jog ~ idF and g 0 j ~ idE' Such a mapping g is necessarily unique, is called the residual of j, and is written as j+. It is readily seen that

(Vy

E

F)

j+(y)

= max{x E E

Ij(x) ~y}.

For a residua ted mapping j we have j

0

j+ 0 j

=j

and j+ 0 j

0

j+

= j+ .

Moreover, by [3, Theorem 2.10], the following statements are equivalent: j is a closure; j+ is a dual closure; j = j+

0

j;

j+ = j

0

j+.

The importance of residuated mappings in our discussion can be seen in the theory of duality that applies to double MS-algebras, which we now proceed to describe. If we consider a double MS-algebra (L; 0, +) then the Ockham algebras (L; 0) and (L; +) give rise to the Ockham spaces (X;g) and (X; h) respectively, each of the mappings g and h being both antitone and continuous. The following table translates properties in L to properties on X. a ~ aOO

a~a++

a O+ = aOO

g2 ~ idx

h 2 ~ idx

g oh

By way of example, for each x

=g2

E Ip (L)

aO~

h(x}={ala+¢x}

gh(x) = {a I aO¢ h(x)} = {a I aO+ EX} g2(X} = {a lao ¢ g(x)} = {a I aOO EX},

whence a O+ = aDO gives g 0 h = g2; and a

¢ h(x)

=}

a+

EX=}

a OEX=} a

a+

g~h

we have

g(x)={alaO¢x},

and so

a+ O= a++ hog = h 2

¢ g(x)

Double MS-algebras so g(x)

~

189

h(x) whence g ~ h.

These considerations lead naturally to the following notion.

Definition A double MS-space (X; g, h) is a Priestley space X on which there are defined two continuous antitone mappings g, h such that g 0 h = g2 ~ idx ~ h 2 = hog. Theorem 12.2 For a Priestley space X the following statements are equivalent: (1) X is the underlying set of a double MS-space; (2) there is a residuated dual closure map {) : X ~ X such that both {) and its residual {)+ are continuous, and 1m {) admits a continuous polarity. Proof (1) => (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

~

X by

{)(y) ~ x

=> y8.,J(y) ~ x => (3z E X) Y ~ z8x =?

Y ~ max [x]8 =
and on the other, since by definition x and -class,

y ~ .,J(y) ~ .,J
Double MS-algebras

191

These observations show that '13 is residuated with '13+ = cp. <> Suppose now that E is an ordered set and that 8 is an equivalence relation on E. If A ~ E then we shall denote by A El the union of all the eclasses that contain an element of A, so that AEl = U{[a]8 I a E A}.

Now the topology on a Priestley space has as a sub-basis the clop en decreasing sets and the clopen increasing sets. We shall say that 8 is lower saturated on X if U decreasing clopen =? UEl clopen;

upper saturated if U increasing clopen =? UEl clopen;

and saturated if both hold.

Theorem 12.5 Let X be a Priestley space and let 8 be an equivalence relation on X. Suppose that there is a dual closure '13 : X --7 X with Ker '13 = e. Then '13 is continuous if and only if 8 is lower saturated.

ue.

Proof First observe that if U is a decreasing subset then '13- 1 (U) = Indeed, if x E then x8u E U so 'I3(x) = 'I3(u) ~ U E U and therefore x E '13- 1 (U); and, conversely, if x E '13- 1 (U) then 'I3(x) E U and therefore, since x8'13(x), we have x E =? : It is immediate from the above observation that if '13 is continuous then 8 is lower saturated. "*= : Suppose now that 8 is lower saturated. If U is dopen and decreasing then '13- 1 (U) = is clopen; and if U is clopen and increasing then '13- 1 (U) = -(-U)El is also dopen. It follows that '13 is continuous. <> The above results give the following characterisation of Priestley spaces that admit the structure of a double MS-space.

ue

ue.

ue

Theorem 12.6 Let X be a Priestley space. Then X is the underlying set of a double MS-space if and only if there can be defined on X an equivalence relation 8 such that (1) 8 has bounded classes; (2) 8 is strongly regular; (3) X /8 admits a continuous polarity; (4) 8 is saturated. Proof =? : By Theorem 12.2 there is a residuated dual closure '13 on X such that '13 and '13+ are continuous and 1m '13 admits a continuous polarity. By

192

Ockham algebras

Theorems 12.3 and 12.4, 0 = Ker f) is strongly regular and every Eklass is bounded, so (1) and (2) hold. As for (3), this follows from 1m f) ~ X/Ker f)

= X/0.

Finally, (4) follows from Theorem 12.5 and its dual.

*=: Ifthe conditions hold then by (1), (2), and Theorems 12.3, 12.4 there is a residuated dual closure f) on X with Ker f) = 0. By (4), Theorem 125 and its dual, both f) and f)+ are continuous. Finally, by (3) and Theorem 12.2, X is the underlying set of a double MS-space. <> As the following two examples show, the four conditions of Theorem 12.6 are independent. Example 12.4 Consider the ordered set X with Hasse diagram

There are five equivalence relations on X, namely 0 1 03

== {{p,r},{q}},

04

== {{p,q},{r}},

05

= W, O2 = £, and == {{p},{q,r}}.

Since X is finite the discrete topology gives a Priestley space, and in this situation condition (4) of Theorem 12.6 is redundant. Of the remaining properties, it is readily verified that 0 1 satisfies all but (3); O 2 satisfies all but (1); 0 3 satisfies all but (1); 0 4 and 0 5 satisfy all but (2).

Example 12.5 Let X = IN EEl 3 EElIN°P where 3 is the chain p < q < r. Endow X with the interval topology, i.e. that generated by the sets {x I x < a} and {x I x > 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. <::::: : Suppose conversely that {} : a ~ aOo is residuated and that {}+ preserves suprema. Then 1m {} == LO O is bicomplete and (Va

E E)

{}+(a) == max(a!

n LOO).

For every a E L define a+

= [{}+(aW == [max(a! nLooW.

Then we have 0+ == 1, 1 + == 0 and (a V bt == [{}+(a V bW

= [{}+(a) V {}+(bW == [{}+(aW 1\ [{}+(bW == a+ 1\ b+

so that (L, +) is a dual MS-algebra. Moreover, aO+ == [max(ao!

and, since [00 == 1m {} = 1m

n LOO)]O == aOo,

{}+ = L++,

a++ == [max(a+! nLooW == a+ o.

Thus (Dl) and (D2) hold, whence (L, 0, +) is a double MS-algebra.

<>

Corollary 1 If an MS-algebra (Lj 0) can be made into a double MS-algebra (L; 0, +) then this can be done in only one way, namely with

(Va E L) Proof Suppose that (L;

also the residual of a Hence the operations

0, Ell) is also a double MS-algebra. Then a

~ aOO

Ell

a+ == [max(a! nLooW.

and

~ a EllEll is and so a EllEll == a++. By (D1) and (D2) we have

+

coincide.

<>

196

Ockham algebras

Jf an MS-algebra (L; 0) can be made into a double MS-algebra then every a E L 00 that is V -reducible in L must be V -reducible in L 00.

Corollary 2

Proof Suppose that a E LOO is such that a = b V c where b, c < a. Then a = a++ = b++ V c++. Since b++ = a gives the contradiction a ~ b, and likewise for c, it follows that b++, c++ < a. <>

Example 12.7 Consider the MS-algebra K 3 . Here we have that 1 is vreducible in K3 but is not v-reducible in Kr~/. Hence, by Corollary 2 of Theorem 12.7, K3 cannot be made into a double MS-algebra. Although we have considered here only double MS-algebras, it is possible to consider other double Ockham algebras. For example, M. Sequeira [91] defines an Oz-algebra to be an algebra (L; /\, v,f,g, 0,1) of type (2, 2, 1, 1,0,0) such that (L; /\, v,f, 0,1) and (L; /\, v,g, 0, 1) are Ockham algebras. In particular, she considers the notion of a double MSn -algebra, namely an algebra (L;j,g) E O2 such that fg = g2n ~ id~f2n = gj.

Clearly, this generalises the notion of a double MS-algebra.

13 Subdirectly irreducible double MS-algebras In order to determine the subdirectly irreducible double MS-algebras we can extend the discussion given in Chapter 4 for MS-algebras. Corresponding to the notion of a g -subset in the dual space (X; g) of an MS-algebra (L; 0), we define a {g, h} -subset in the dual space (X; g, h) of a double MS-algebra (L; 0, +) to be a subset of X that is both a g-subset and an h-subset. Such a subset, for example, is gW{x} U hW{x}. Working with this we see that, corresponding to Corollary 1 of Theorem 4.6, a finite double MS-algebra (L; 0, +) is subdirectly irreducible if and only if there exists x E X such that X = gW{x} u hW{x}. In examining for a double MS-space (X; g, h) the corresponding situation to that in Example 4.8 we have to consider a 'double noose'

.t-:-.: t

~.-

s

....• -.~. p q r

in which the arrows to the right of p indicate the (partial) effect of g and those to the left of p that of h. Here the inequalities g 0 h = g2 ~ idx ~ h 2 = hog force, for example, g(s) = g[h(P)] = g2(P) = r~p. We are therefore led to consider the double MS-space (X; g, h) described as follows, in which the complete actions of g and h on the chains r < p < t and q < s are as indicated: t .................... ~.··········.: s

. .::·:::·1

p

h

......... -

g

Note that here X = gW {P} u hW{P} so the corresponding double MS-algebra is subdirectly irreducible. Using arguments similar to those in Chapter 4, we can see that the subdirectly irreducible double MS-algebras are precisely the subalgebras of this algebra. Up to isomorphism, there are 21 in all. We arrange them in decreasing order of cardinality and label them SID21 , ... ,SID 1 .

198

Ockham a/geb1

Subdirectly irreducible double MS-algebras

199

These algebras were originally determined in [38] without using duality. In a double MS-algebra (L; 0, +) we define the congruence ~ by

(a, b) E

~ ~

(aO

= bO

and a+

= b+).

From the above description, we then have immediately :

Theorem 13.1 A double MS-algebra L is subdirectly irreducible if and only if Con L reduces to the chain w :j ~ --< L. <; Since in SID 21 the non-trivial ~ -classes are {c,j}, {d, g}, {e, h} we also have:

Theorem 13.2 Ofthe 21 non-isom01phic subdirectly irreducible double MSalgebras 11 are non-simple, namely SID4 , SID8 , SID lO , SID ll , SID 12 , SID 15 , SID 16, SID l7 , SID 18 , SID19 , SID 21 ;

and 10 are simple, namely SID 1 , SID 2 , SID3 , SID s , SID6, SID7 , SID 9 , SID 13 , SID 14 , SID20 . <; We note also that SID 1 , SID 2 , SID3 , SID4 , SID7 , SID 9 , SID 13 , SID 14 , SID19 , SID 20 , SID 21 are self-dual, whereas we have dual isomorphisms d

d

SIDs ~ SID6, SID8 ~ SID 10 , SID 11

d ~

d

SID 12 , SID 1S ~ SID 16, SID17

d ~ SID 18 ·

<;

Theorem 13.3 There are 3 non-isomorphic subdirectly irreducible double Stone algebras, namely SID 1 , SID 2 , SID 4. Moreover only one ofthese, namely SID4 , is non-simple. <; We can order the sub directly irreducible double MS-algebras by writing A :::; B if and only if A is isomorphic to a subalgebra of B. In so doing, we obtain the following Hasse diagram, in which n denotes SIDn :

19

4 14

200

Ockham algebras

The lattice of subvarieties of double MS-algebras can, in theory, be obtained from this by applying the theorem of Davey in precisely the same way as we did in Chapter 5 to obtain the lattice of subvarieties of MS-algebras. However, in the case of double MS-algebras this lattice is rather large and so we shall concentrate on some important ideals of it. Specifically, if SIDn denotes the subvariety generated by SID n , we describe the lattices of subvarieties of SID20 , SID I9 , and SID I7 V SID I8 . We also obtain equational bases for the subvarieties generated by the subdirectly irreducible double MS-algebras. (1) Semisimple double MS-algebras An algebra is semisimple if it is isomorphic to a subdirect product of simple algebras. A variety V is semisimple if every member of V is semisimple. It is well known that a variety V is semisimple if and only if every subdirectly irreducible member of V is simple. Now from Theorem 13.2 there are 10 simple double MS-algebras. As can be seen, they constitute the down-set 20L in the above Hasse diagram. So a double MS-algebra is semisimple if and only if it is a subdirect product of copies of SID20 . The variety SID20 can be characterised as follows.

Theorem 13.4 On a double MS-algebra (L; 0, +) the following conditions are equivalent: (Ao) L is semisimple; (A) (Va, bEL) a /\ bOO (AI) cf>~ = w.

~

a++ vb;

Proof (Ao)

=?- (A) : It suffices to observe that SID 20 satisfies (A). (A) =?- (Ao) : Examination of each of the subdirectly irreducible double MS-algebras reveals that only those that are simple satisfy (A). Thus, if L satisfies (A) then by Birkhoff's theorem L is a subdirect product of copies of SID20 . (A) =?- (AI) : Suppose that L satisfies (A) and that a, bEL are such that aO = bO and a+ = b+. Then (A) gives a = a /\ aDO = a /\ bOO ~ a++ V b = b++ V b = b. Similarly, b ~ a and so a = b. (AI) =?- (A) : Suppose that cf>~ = wand consider the elements

p=a/\boo, We have

It follows that

q=a/\boo/\(a++vb).

qO = a OV bO V (a+ /\ be) = aO V bo = po; q+ = a+ V bo V (a+ /\ b+) = a+ V bO =p+.

q

=p, whence (A) follows. <>

5ubdirectly irreducible double M5-algebras

201

We can construct the lattice of subvarieties of semisirnple double MSalgebras by applying Davey's theorem. Its size can of course be predicted by Theorem 5.5 applied to the down-set SID~o. We leave to the reader the verification that it has 30 elements and, with n = SIDn , is 20

14

(2) Locally convex skeletons The skeleton of a double MS-algebra L is the set 5(L) = {a ELI a++ = a = aCe}. Clearly, 5(L) is a de Morgan algebra. We shall say that L has a locally convex skeleton if every interval of the form rae, a+] belongs to S(L). We can characterise the subvariety SID19 as follows.

Theorem 13.5 On a double MS-algebra (L; 0, +) the following conditions are equivalent: (Bo) L has a locally convex skeleton; (B) ('ria, bEL) aDO /\ b+ ~ a++ V be; (Bd L E SID19 · Proof (Bo) # (B) : It is clear that Z E [bO, b+] if and only if z is of the form (a V be) /\ b+, and that this is in S(L) if and only if (aOO

V

which is equivalent to (E).

be) /\ b+ = (a++

V

be) /\ b+,

202

Ockham algebras

(B) .g,. (B 1 ) : It suffices to check that (B) is satisfied by SID 19 but is not satisfied by either SID9 or SID2 · 0 Using the same technique as before, we can construct the lattice of subvarieties of SID19 • This has 24 elements and is

10

(3) Kleene skeletons

We now turn our attention to the subvariety SID17 V SID18 .

Theorem 13.6 Tbe following inequalities are equivalent: (i)

f~b++vb+,

(ii)f~bvb+,

(iii)

f~boovb+,

(iv)

a++l\aO~g,

Uw)

f~boovbo.

Likewise, so are the inequalities U)

aool\aO~g,

Uj)

al\aO~g,

Ujj)

a++l\a+~g.

Proof Clearly, (i) =;. (ii) =;. (iii). For (iii) =;. (i) and (iv) =;. (i), write b++ for b; and for (i) =;. (iv), write bOO for b. 0 Theorem 13.7 Tbe following inequalities are equivalent (v) b V b ~ f E S(L), O

(w) a 1\ a+ ~ g E S(L),

(Vi) b++ V bO ~ f E S(L); (wj) aO o 1\ a+ ~ g E S(L).

0

Subdirectly irreducible double MS-algebras

203

Theorem 13.8 On a double MS-algebra (L; D, +) the following conditions are equivalent:

(Co) L has a Kleene skeleton; (C) (Va, bEL) a 1\ a < b V b+; O

(c 1) L

E

SID17 V SID18 .

Proof (Co) holds if and only if 1\
aD

O

The lattice of subvarieties of double MS-algebras with a Kleene skeleton can be constructed using the same technique as before. The lattice in question has 98 elements and can be visualised as follows. It consists of the three 'layers' shown below, the second projecting down onto the first with SID2 directly above SID l , and the third projecting down onto the second with SID4 directly above SID2 .

17

12

10

Equational bases for SIDIl are not in general unique. In what follows we shall obtain equational bases each of which involves at most three relations. We first add to the results of Theorems 13.4, 135, 13.8 further single-term equational bases.

204

Ockham algebras

Definition For each axiom (n) we define (n*) to be the axiom obtained from (n) by replacing V, 1\, 0, +, 0, 1, ~ respectively by 1\, V, + , °, 1 , 0, ;;;::. When (n) is equivalent to (n*) we shall say that (n) is self-dual in DMS.

Observe for example that although the axiom (2), namely a V aO = 1, is not self-dual in MS it is self-dual in OMS. In fact, a VaO = 1 gives a+ 1\ aDO = 0, whence a+ 1\ a = O. Conversely, a+ 1\ a = 0 gives a++ V aO = a+o V aO = 1, whence a V aO = 1. The axioms (A), (B), (C) above are self-dual in OMS, as are (E), (F), (G), (2d)' (01), h) in the following result.

Theorem 13.9 The Jollowing subvarieties oj double MS-algebras have the equational bases indicated: SID 18 V SID20

(D)

a 1\ bOO ~ a++ V b V b+

SID 17 V SID20

(D*)

a 1\ aO 1\ bOO

SID 15 VSID l 6 VSID 20 (E)

SID7 SID4 V SID ll V SID 12

h)

a 1\ aO

~

SID4

(2 d ) (2)

a 1\ aO

=0

a V aO

=1

SID7 V SID lO

(15)

a V aO

SI07 V SID8

(15*)

SID l

a++ V b

a 1\ a+ 1\ baa ~ a++ V b V b O a Oo 1\ a+ ~ b++ V bO

(F) (G) (01)

SID ll V SID 12 SID2 V SID ll V SID 12

~

a 1\ a+ a

~

= aDO

b V bO

(= aO = a+) b V bo

= aDO V aO a 1\ a+ = a++ 1\ a+

Proof This uses the same principle that we employed in Chapter 5. (D) is satisfied by SID l8 and SID20 , but not by SID8. (D*) is satisfied by SID 17 and SID 20 , but not by SID lO . (E) is satisfied by SID lS , SID l6 , and 5ID 20 , but not by 5ID4 or SID l9 . (F) is satisfied by 5ID ll and 5ID 12 , but not by 5ID2 or SID7. (G) is satisfied by SID2, 5ID ll , and SID 12 but not by SID4, 5ID 13 , or SID7 . (01) is satisfied by SID7 but not by SID 2, 5IDs, or SID6' h) is satisfied by SID4, SID ll , and SID12 but not by SID7 or SID 13 . (2 d ) is satisfied by SID4 but not by SID3. (2) is satisfied by SID l but not by SID2 or 5ID 3. (15) is satisfied by 5ID7 and SID lO , but not by SID 2 or 5ID s . (15 *) is satisfied by SID7 and 5ID8, but not by SID 2 or SID6. <> The above axioms are linked by the following implications.

Subdirectly irreducible double MS-algebras

205

(15*)

Theorem 13.10 An equational basis/or SID15 V SID16 is

(e, E).

Proof Writing SIDn as n, we have that 20 1\ 17 = 13 = 20 1\ 18 and so

15 V 16

= 15 V 16 V 13 = 15v16v [201\ (17V 18)]

= (15 V 16v 20) 1\ (17V 18) whence the result follows by Theorems 13.8 and 13.9. <> In the following result we shall use our knowledge of the various parts of the lattice A(DMS). Theorem 13.11 Equational bases/or n

20 19 18 17 16

(A) (B) (C,D) (e,D*) (C,D,E)

= SIDn (n = 1, ... ,20) are:

15 (e,D*,E) 10 (F,15) 14 (A,B) 9 (A,F) 8 (F,15*) 13 (A,e) 12 (D,F) 7 (a) 11 (D*,F) 6 (A, F, 15)

5 (A,F,15*) 4 (2 d )

3 b,a) 2 (A,2d) 1 (2)

Proof Theorems 13.4, 13.5, 13.9 yield equational bases for 20, 19, 7, 4, 1. Now

18 = 18 V 13 = 18 V (201\ 17) = (18 V 20) 1\ (18 V 17) whence, by Theorems 13.9 and 13.8, 18 has equational basis (D, e). From the lattice in Theorem 13.8, we have 16 = (15 V 16) 1\ 18 and so, by Theorem 13.10 and the above, an equational basis for 16 is (e,D,E). Since 14 = 201\ 19, an equational basis for 14 is (A, B). Since 13 = 201\ (17 V 18), an equational basis for 13 is (A, e). From the lattice of Theorem 13.8 we have 12 = (11 V 12) 1\ 16 so, by Theorem 13.9, the above, and the fact that (F) =? (e,E), an equational basis for 12 is (D,F).

206

Ockham algebras

Since 12 /\ 7 = 3 = 11 /\ 7, we have 10 = 10 V 3 = 10 V [7 /\ (11 V 12)] = (10 V 7) /\ (11 V 12).

Consequently, by Theorem 13.9, an equational basis for 10 is (F, 15). Since 9 = (11 V 12) /\ 13, an equational basis for 9 is (A, F). Since 6 = 9/\ 10, an equational basis for 6 is (A,F, 15). Since 3 = 7 /\ (4 V 11 V 12), an equational basis for 3 is (01, ')'). Since 2 = 20 /\ 4, an equational basis for 2 is (A, 2d)' Finally, equational bases for the remaining five subvarieties can be deduced from the above using the dual isomorphisms on page 199. <> Of the above classes, some are worthy of especial mention, namely: SID l SID3 SID 7 SID4

the class of boolean algebras; the class of Kleene algebras; the class of de Morgan algebras; the class of double Stone algebras.

A further class worth mentioning is the class SID 2 , characterised by the properties (A) and (2d)' From the above, the members of this class are the double Stone algebras that are semisimple. Such algebras are called trivalent Lukasiewicz algebras. In fact, these algebras can be characterised by (G, 2d ) since, from the lattice of Theorem 13.8, we have 2 = (2 V 11 V 12) /\ 4. For further considerations of trivalent Lukasiewicz algebras we refer the reader to

[96,98].

14 Congruences on double MS-algebras We now turn our attention to congruences on double MS-algebras. Here also an important role is played by the principal congruences 1?(a, b) with a ~ b. It is clear from Theorem 2.1 that if L E DMS and a, bEL are such that a ~ b then we have 1?(a, b)

= 1?lat(a, b) V 1?lat(bO, aO) V 1?lat(aOO, bOO) V 1?lat(b+, a+) V 1?lat(a++, b++).

Arguing precisely as in Theorem 8.1, we can characterise 1?(a, b) as follows.

Theorem 14.1 If L E DMS and a, bEL with a ~ b then (x,y) and only if (1) x 1\ a++ /\ bo = Y 1\ a++ /\ be; (2) (x /\ a 1\ be) V b++ = (y 1\ a 1\ be) V b++; (3) (x V aD) /\ a++ /\ b+ = (y V aD) /\ a++ /\ b+; (4) [(x V aD) /\ a /\ b+] V b++ =[(y V aC) /\ a /\ b+] V b++; (5) (x V a+) /\ a++ = (y V a+) /\ a++; (6) (x /\ a) V a+ V b++ = (y 1\ a) V a+ V b++; (7) (x V b) /\ bO 1\ aDO = (y V b) /\ bo /\ aCe; (8) (x /\ be) V bOO = (y /\ bO) V bOO; (9) (x V b V aO) /\ aDO 1\ b+ = (y V b V aD) /\ aOO /\ b+; (10) (x V b V a+) /\ aDO = (y V b V a+) 1\ aCo; (11) (x V a V bOO) 1\ b+ = (y V aO V bOO) 1\ b+; (12) xva+Vboo=yva+vboo. <>

E 1?(a, b)

if

C

In general, these conditions are difficult to handle, but in the case of double Stone algebras a considerable simplification occurs.

Theorem 14.2 In a double Stone algebra (x,y) E 1?(a, b)

if and only if

(13) x 1\ a /\ (a++ V b+) =Y /\ a 1\ (a++ V b+); (14) (x V b) /\ (aOo V bO) 1\ b+ =(y V b) /\ (aOo V be) /\ b+.

Proof Using the identity a 1\ aO forms

= 0, which has the following five equivalent

Ockham algebras

208

and the fact that a ~ b, we see that the equations (I), (2), (3), (6), (7), (10), (11), (12) are trivial. As for (4), this reduces to (x 1\ a 1\ b+) V b++ = (y 1\ a 1\ b+) V b++.

=0 =Y 1\ a 1\ b+ 1\ b++, it follows by distributivity that

Since x 1\ a 1\ b+ 1\ b++ (4) becomes

(13')

x 1\ a 1\ b+

=Y 1\ a 1\ b+.

Clearly, (5), (8), (9) become respectively a++ =y 1\ a++; (14') x V bOO = y V bOo; (14") (x V b) 1\ aOO 1\ b+ = (y V b) 1\ aOO 1\ b+.

(13") x

1\

Now it is clear that (13') and (13") together are equivalent to (13). Since L is a double Stone algebra, {14') is equivalent to xl\bo=yl\bo.

This, together with (14"), gives (14). Conversely, taking the join of each side of (14) with bOo, we obtain (14'); and taking the meet of each side with aOO we obtain (14"). 0 As with MS-algebras, there is a strong relationship between principal congruences on double MS-algebras and principal lattice congruences.

Theorem 14.3 Let L E DMS and let a, bEL be such that a~b,

al\ao=bl\bo, ava+=bvb+

Tben we have '19 (a , b)

= 'l91at (a, b) V 'l91at (bo, aO) V 'l91at (b+, a+)

='l91at ((a V be) 1\ b+, (b V aO) 1\ a+). Proof As observed in the proof of Theorem 8.2, the equality a 1\ aO = b 1\ bO gives 'l91at (bO,ao) = 'l91at(aoo,bOO). This, together with its dual, gives the first equality. As for the second equality, we note first that '19 (a, b) is a lattice congruence that identifies each of the pairs (a,b), (bO,aO), (b+,a+). It follows that ((a V be) 1\ b+, (b V aO) 1\ a+) E '19 (a , b) and so

'l91at ((a V be) 1\ b+, (b V aO) 1\ a+) ~ '19 (a , b).

Congruences on double MS-algebras

209

As for the reverse inequality, observe that

a 1\ (a

V

be) 1\ b+

= a 1\ b+ = [a V (a 1\ aO)] 1\ b+

=(a 1\ b+) V (a 1\ aO 1\ b+) =(a 1\ b+) V (b 1\ bO 1\ b+) = b 1\ (a V be) 1\ b+

and similarly so that we have

(a, b) E 191at((a V be) 1\ b+, (b V aO) 1\ a+). Likewise, we can show that 191at((a V be) 1\ b+, (b V aO) 1\ a+) identifies (a, b). Then, by the first equality, the reverse inequality follows. <>

Corollary If L is a double Stone algebra then every principal congruence is a principal lattice congruence; specifically, 19(a, b) = 191at((a V be) 1\ b+, (b V aO) 1\ a+). <> We can in fact say more.

Theorem 14.4 The class SID4 of double Stone algebras is the largest subvariety of DMS in which every principal congruence is a principal lattice congruence. Proof By considering the ordered set of subdirectly irreducible double MSalgebras (see page 199), we see that it is enough to show that the stated property fails in the subvariety SID3 of Kleene algebras. For this purpose, consider the 4-element chain < a < b < 1 on which the operation ° is defined by 0° = 1, aO = b, bO = a, 1° = 0.

°

This defines a Kleene algebra on which the principal congruence 19(0, a) has the partition {{O,a},{b,l}} and is not a principal lattice congruence. <> Our objective now is to determine precisely when, for a given L E DMS, the lattice Con L is boolean. For this purpose, we embark on an investigation of those principal congruences that are complemented. Given L E DMS and a, bEL with a ::;;; b, define

'Pda, b) = 19(0, bO 1\ aCe) V 19(bOO 1\ bO, be) V 19(aO, aOO vaO), 'Pz(a, b) = 19(0, b+ 1\ a+) V 19(b++ 1\ b+, b+) V 19(a+, a++ va+).

Ockham algebras

210 If we write cp(a, b) = CPl (a, b) /\ CP2(a, b), then by Theorem 8.9 and its dual we have immediately 13(a, b) V cp(a, b)

= to

Theorem 14.5 If 13(a, b) is complemented then its complement is cp(a, b). Proof Suppose that 13(a, b) is complemented and denote its complement by 13 (a, b)c. Then from the above equality it follows that 13(a, b)C :::;; cp(a, b). To establish the reverse inequality, observe that 13(a, b) = 13 o (a, b) V 13+(a, b) where

13 o (a, b) = 19lat(a, b) V 13lat(bO, aO) V 13lat(aOO, bOO)

is an MS-congruence and 19+(a, b) is the corresponding dual MS-congruence. Now, by applying 00 to the six equations in Corollary 1 of Theorem 8.1, we see that (XOO,yOO) E 19 o (a, b) -<==> (XOO,yOO) E 19 o (aOO, bOO). Using this, and its dual, we deduce that in the skeleton 5(L) we have 19(a, b)ls(L)

=19 o (a, b)ls(L) V 19+(a, b)ls(L) =13 o (aOO, bOO) V 19+(a++, b++) =13(aOO, bOO)ls(L) V 13(a++, b++)ls(L)

and consequently, as in Theorem 8.9, 19(a, bYls(L) = (13(a, b)ls(L))c =19(aOO, bOO)Cls(L) /\ 19+(a++, b++)Cls(L) = CPl (a, b)ls(L) /\ CP2(a, b)ls(L)·

It follows that 13(a, bYls(L) = cp(a, b)ls(L) and so 13(a, by is an extension to L

of cp(a, b)ls(L). Now it is clear that cp(a, b)ls(L) identifies each of the pairs (0, bO /\ aCe), (bOO /\ bO, be), (aO, aOO vaO), (0, b+ /\ a++), (b++ /\ b+, b+), (a+, a++ va+) of elements of 5(L), and hence so does every extension toL of cp(a, b)ls(L). In particular, 19(a, b)C identifies each of these pairs, and consequently we have cp(a, b):::;; 13(a, by as required. 0

Congruences on double MS-algebras

211

This knowledge of 1?(a, by, when it exists, allows us to determine precisely under what conditions it does exist.

Theorem 14.6 1?(a, b) is complemented if and only if (1) b 1\ bO1\ a OO ~ b++ Va; (2) b 1\ a OO ~ b++ Va va+;

(3)

bl\b+l\aoo~b++vavao.

Proof By Theorem 14.5, 1?(a, b) is complemented if and only if 1?(a, b) 1\ cp(a, b) = w.

Now 1?(a, b) 1\ cp(a, b)

= 1?(a, b) 1\ CPI (a, b) 1\ CP2(a, b)

and each of these components can be expressed as a join of lattice congruences. Using the identity

1?lat(a, b) 1\ 1?lat(c,d) = 1?lat(b 1\ d 1\ (a V c), b 1\ d), we can then express 1?(a, b) 1\ cp(a, b) as a join of lattice congruences. We leave to the reader the task of verifying that 1?(a, b)l\cp(a, b) can be expressed as the join of the lattice congruences

1?lat(b 1\ bO1\ a OO 1\ (b++ va), b 1\ bO1\ aOo), 1?lat(b 1\ (a OV aOO) 1\ (a V a+ V b++), b 1\ (a OV a OO )) , 1?lat(b 1\ b+ 1\ (a OV aOo) 1\ (b++ Va va o), b 1\ b+ 1\ (a Ova oo )). Clearly, each of these is w if and only if (1), (2), (3) hold. 0 Again, in the case of a double Stone algebra there is a simplification; and in this case we can identify the complement as a principal lattice congruence.

Theorem 14.7 In a double Stone algebra the following are equivalent: (1) 1? (a, b) is complemented; (2) b 1\ b+ ~ a va o; (3) a 1\ b+ is complemented in [0, b 1\ b+]. Moreover, when it exists, the complement of 1?(a, b) is given by 1?(a, b)C = 1?lat((a O 1\ bOO) V (a+ 1\ b++), 1).

Proof (1)

*

(2) : When L is a double Stone algebra the conditions of Theorem 14.6 reduce to

Ockham algebras

212 which is equivalent to

(*)

b /\ b+ /\ aDO = b /\ b+ /\ aDO /\ (b++ Va V aD) = b /\ b+ /\ a.

This implies that b /\ b+

= b /\ b+ /\ (aO V aCe) = b /\ b+ /\ (aO va):::;;; aO Va,

which is (2). Conversely, if (2) holds then b /\ b+ /\ aDO :::;;; (a

V

aD) /\ aDO

= a,

whence (*) follows. (2) =* (3) : If (2) holds then we have b /\ b+ = b /\ b+ /\ (a

V

aD) = (a /\ b+) V (b /\ b+ /\ aD).

Since also (a /\ b+) /\ (b /\ b+ /\ aD) = 0, it follows that b /\ b+ /\ aO is the complement of a /\ b+ in [0, b /\ b+]. (3) =* (2) : Let x be the complement of a /\ b+ in [0, b /\ b+]. Then we have a/\b+ Ax = and (a/\b+)v x = b /\b+. Since (L; 0) is a Stone algebra, it follows that x : :; ; (a/\b+)o = aovb++ and hence that b/\b+:::;;; (a/\b+)VaoVb++. Consequently,

°

b /\ b+ = b /\ b+ /\ [(a /\ b+) V aO V b++] = (a /\ b+) V (b /\ b+ /\ aD) :::;;; a

VaO.

As for the final statement, suppose that 19(a, b)C exists. Then by Theorem 14.5 we have, primes denoting complements in ConlatL, 19(a, b)C = [19(0, be) V 19(aO, 1)] /\ [19(0, b+) V 19(a+, 1)]

= [191at(0, be) V 191at(bOO, 1) V 191at(aO, 1) V 191at(0,aOO)] /\ [191at(0, b+) V 191at(b++, 1) V 191at(a+, 1) V 191at(0, a++)] = [(191at(bO,aO))' V (191at(aOO,bOO))']

/\ [(191at(b+, a+))' V (19lat(a++, b++))']

= [191at( bO, aD) /\ 191at(aOO, bOO)]' /\ [191at (b+ , a+) /\ 191at(a++ , b++)]' = (191at(0, aO /\ bOO))' /\ (191at (0, a+ /\ b++))' == 191at(aO /\ bOo, 1) /\ 191at(a+ /\ b++, 1 ~

= 191at((ao /\ bOO) V (a+ /\ b++), 1).

<>

Congruences on double MS-algebras

213

These results provide the following characterisation of semisimple double MS-algebras.

Theorem 14.8 The subvariety SID20 ofsemisimple double MS-algebras is the largest subvariety of DMS in which every principal congruence is complemented.

Proof The subvariety SID20 of semisimple double MS-algebras is characterised by the axiom (A)

a /I. bOO ~ a++ V b

The conditions of Theorem 14.6 are therefore satisfied, so every principal congruence on a semisimple double MS-algebra is complemented. Using the ordered set of subdirectly irreducible double MS-algebras, we see that in order to establish the result it suffices to produce examples of algebras that belong to the subvarieties SID4 , SIDs, SID lO in which there is a principal congruence that is not complemented. Suitable examples are provided by the subdirectly irreducible algebras SID4 , SIDs , SID lO that generate the subvarieties in question (see page 198). By Theorem 13.1, in each of these Con L is the three-element chain w

-<

~

-<

t.

Moreover, ~ a principal congruence; in the first it is 'I9(d,g), in the second it is 'I9(e,h), and in the third it is 'I9(c,j). <> It is of course natural to ask when a principal congruence has a complement that is also principal.

Theorem 14.9 The subvariety SID2 of trivalent Lukasiewicz algebras is the largest subvariety ofDMS in which everyprincipal congruence has a principal complement.

Proof A trivalent Lukasiewicz algebra is a semisimple double Stone algebra and, as showD; at the end of Chapter 13, is characterised by the axioms (G, 2d)' It follows by (G) and Theorem 14.7 that every principal congruence in such

an algebra has a principal complement. TQ ~omplete the proof it suffices, again by considering the ordered set of subdifectly irreducible double MS-algebras, to provide examples of algebras in SID3 and SID4 in which the property fails. Now SID3 is the subvariety of Kleene algebras. In this, consider the sixelement chain 0 < a < b < c < d < 1 with

214

Ockham algebras

The congruence 19(a, by is described by the partition {{O,a}, {b,c},{d,l}}

and is not principal. As for SID4 , this is the class of double Stone algebras and here, by Theorem 14.8, a principal congruence need not have a complement. 0 We now turn our attention to the problem of determining which double MS-algebras have a boolean lattice of congruences. For this purpose, we require several analogues of results in Chapter 8. The first is the folloWing characterisation of the least extension to a double MS-algebra L of a congruence on its skeleton S{L).

Theorem 14.10 1/ L E DMS and I{J E Con S(L) then the smallest lattice congruence on L that extends I{J is given by I{J

=

V

19lat(xOO,yOO)

(x oo ,yoo) E 'P

Moreover, 7p E Con L .

Proof This is exactly as in Theorem 8.22. 0 Precisely as in Theorem 8.23, we also have

Theorem 14.11 The mapping f: Con S(L)

-t

Con L given by f(rp)

= 7p is

a lattice morphism. 0

Proceeding in the opposite direction, we clearly have

Theorem 14.12 1/19 E ConL is complemented then so is 19ls(L) E ConS(L).O Precisely as in Theorem 8.25, we also have Theorem 14.13 For every L E DMS the lattices Con L and Con S(L) have isomorphic centres. 0

We can now establish the folloWing result.

Theorem 14.14 1/ L E DMS then Con L is boolean if and only if L is semisimple and finite. In this case, Con L is also finite. Proof =>- : Suppose that Con L is boolean. If 19 E Con S(L) then its extension "j E Con L is complemented and therefore, by Theorem 14.12, so is

19 ="j IS(L) E Con S(L). Hence Con S(L) is boolean. It follows by Theorem 14.13 that Con L ~ Con S(L).

Congruences on double MS-algebras

215

Since S(L) must then be finite by Theorem 8.15, it follows that Con L is also finite, and the intersection of all maximal congruences in Con L is w. But every maximal congruence contains cp-:,. Hence cp-:' = wand therefore, by Theorem 11.4, L is semisimple. Suppose now, by way of obtaining a contradiction, that L is infinite. Since the elements of S(L) are precisely the biggest elements in the cpo-classes, it follows from the fact that S(L) is finite that there must be at least one CPO-class that is infinite. But since cp+ 1\ cpo = cp-:, = w, the relation cP+ must separate any two elements in this infinite cpo_ class. There are, consequently, infinitely many cP+-classes. Since the elements of S(L) are precisely the smallest elements in the cP+ -classes, this contradicts the fact that S(L) is finite. We conclude that L must be finite. ~ : Suppose now that L is semisimple and finite. Then, by Theorems 8.15 and 14.13, the centre of Con L is a finite boolean algebra. By Theorem 14.8, there are therefore only finitely many principal congruences on L. Since every cp E Con L can be written

cp

= V

{)(a i , b i ),

(aj,b;)E'P

such a supremum involves only finitely many principal congruences each of which is complemented. It follows that cp is also complemented and hence that Con L is boolean. <)

15 Singles and Doubles In Chapter 5 we showed that every variety of MS-algebras is characterised by adjoining to the basic axioms of MS at most three from a list of ten identities. If a double MS-algebra (L; 0, +) is such that (L; 0) satisfies some of these identities then it is natural to ask when (LOP; +) satisfies the same identities. More particularly, if (L; 0, +) E DMS and V(L; 0) = R, what are the possible subvarieties V(LOP; +)?

Theorem 15.1 Thefollowing axioms are self-dual in DMS : (1), (01),

b),

(2), (2d), (5), (9), (lId), (12d)'

Proof It is clear that (1) is self-dual in DMS. That the others are also self-dual in DMS is a consequence of the following observations.

If (L; 0) satisfies (01)

then so does (LOP; +). OO If a = a for every a E L then, by (D2), a+ = a Oo+ a++ = a O+ = a OO = a. (a)

(b)

If (L; 0) satisfies b) then so does

= a OOO = a O and hence

(LOP; +).

By b) we have a+ /\ a+ o ~ b V bo hence a++ V a+o+ ~ b+ /\ b O+, and therefore a++ V a+ ~ b+ /\ boo. Since a++ ~ a and b ~ bOo, we obtain a V a+ ~ b+ /\ b. (c)

If (L; 0) satisfies (2) then so does

(LOP; +).

This was shown above. (d)

If (L; 0) satisfies (2 d ) then so does

(LOP; +).

In fact, if a/\a o = 0 for every a E L then a+ /\a+ o = 0 which gives a++Va+o+ = 1 so that, by (D1), a++ V a+ = 1 and hence a V a+ = 1. (e)

If (L; 0) satisfies

(5) then so does (LOP; +).

By (5) we have a+ /\ a+ o ~ b+ oO V b+o, i.e., by (D1), a+ /\ a++ ~ b+ V b++. It follows that a++ V a+ ~ b++ 1\ b+ and hence that a V a+ ~ b++ 1\ b+.

(f) If (L; 0) satisfies (9) then so does (Lop; +). Writing a+ for a in a 1\ a O~ a OO V b V bO, we obtain a+ /\ a++ ~ a+ V b V bO whence, applying + to this, we have a++ 1\ b+ /\ bOO ~ a++ V a+. Consequently a++ 1\ b+ 1\ b ~ a V a+, which is (9*).

If (L; 0) satisfies (lId)

then so does (LOP; +). Observe first that (lId)' namely a O1\ a OO ~ a V bo V bOO, is equivalent to (g)

(lId)

(a 1\ a O) V bOV bOO E S(L).

Singles and Doubles

217

In fact, if {II d) holds then

(a

1\

aO) V bOV boo =(a V bOV bOO) 1\ (ao V bOV bOo) =(a V a OV bOV bOO) 1\ (aOO V bOV boO) 1\ (ao V bo V bOO) =(aoo V bOV bOO) 1\ (aO V bOV bOo) E S(L).

Conversely, if (a

1\ aO) V b OV bOO E

(a

1\

S(L) then

aD) V bOV bOO = (aOO

1\

aO) V bO V bOo.

Since the left-hand side is :::;; a V bO V bOo, and the right-hand side is we obtain (lId)' Writing b++ for b in (lld) we obtain

(a

1\

aO) V b+ V b++

~

a Oo I\a o,

= (aOO 1\ aO) V b+ V b++.

Taking the join of each side with b gives

(A)

(a

1\

aO) V b V b+ = (aOO

1\

aO) V b V b+.

Conversely, writing bOO for b in (A) we obtain (lld)' Hence (lld) and (A) are equivalent. Similarly, we have that (ll~) is equivalent to (ll~')

(a V a+) 1\ b++

1\

b+ E S(L).

Writing bOO for b in this, we obtain

(a V a+) 1\ bOO 1\ bO E S(L), which implies that (B)

(a

V

a+) 1\ bOO 1\ bO = (aOO

V

a+) 1\ bOO 1\ bO.

Conversely, writing b++ for b in this and using a similar argument, we obtain (11~'). Thus {Il~) and (B) are equivalent. It suffices, therefore, to show that if (B) fails then so does (A). Now if (B) fails there exist t ~ t+ and SOO :::;; SO such that We then have v Vt = t, and vOO V t > t (since otherwise vOO V t = t which gives vOo :::;; t whence, since voo:::;; soo, we would have voo:::;; v, which contradicts v < voo). Thus we have v V t < vOo V t. Since v 1\ VO

= t 1\ SOO 1\ (to V SO) = t 1\ SOO = v, we conclude that (A) fails.

(b) If(L;o) satisfies (12 d ) tben so does (LOP;+).

Ockbam algebras

218

It is readily seen that (12d)' namely aOO A bO A bOO ~ a V aO, is equivalent to

(a V aO) A bO A bOO E 5(L)

and therefore gives [(a V aO) A bO A bOO]O == [(a V aO) A bO A bOO]+

Writing b+ for b in this, we obtain

(t)

(aO A aCe) V b+ V b++ == (a+ A aCe) V b+ V b++.

Now (12 d ) implies (lId) which is eqUivalent to (a A aO) V bO V bOO E 5(L),

from which we obtain, writing b+ for b,

(tt)

(aOO A aO) V b+ V b++ == (a++ A aO) V b+ V b++.

It now follows from (t), (tt) and the inequalities

that we have the identity (a++ A a+) V b++ V b+ ::: (a 00 A a+) V b++ V b+,

which gives (a A a+) V b++ V b+ E 5(L)

which means that (LOP; +) satisfies (12d)'

0

It is, of course, possible to establish Theorem 15.1 by means of duality. To show, for example, that (lId) is 'self-dual in DMS, suppose that (X;g,b) is the dual space of (L; 0, +). We have to show that if (X; g) satisfies

g2 Mg

V g2 ?: gO

b2Mb

V bO ?: b2.

then (X; b) satisfies Suppose that this were not so. Since g2 Mg is equivalent to b 2Mb, there must exist P E X such that (a) g2(P) Ilg(P), (b) g2(P)?: P (hence g2(P) == P by axiom (1)), (c) b 2(P) I b(P), (d) p 'j b 2(P) (hence p < b2(P) since in DMS we have x ~ b2(x)). Consider the element q ::: b 2(P). We have g(q) ::: gb 2(P) ::: g(P) and so g2(q) == g2(P). It follows by (a) that g2(q) Mg(q); and by (b) and (d) that g2(q) == g2(P) == P 'j b2(P)::: q.

Singles and Doubles

219

The fact that q does not satisfy the axiom (lld) therefore provides the required contradiction.

Theorem 15.2 Let (L; 0, +) be a double MS-algebra. Then

(a) (b)

if (L; 0) satisfies (6 d) then (LOP; +) satisfies (3d); if (L; 0) satisfies (3d) then (LOP; +) satisfies (6 d).

Proof (a) : It is readily seen that (6 d), namely a Oo ~ a V bOV bOo, is equivalent to a V bOV bOO E S{L). Writing b++ for b in this, we obtain

a V b+ V b++ E S{L) which, on taking b = a, gives

a V a+ E S(L) and therefore a++ V a+ ~ a. It follows that (LOP; +) satisfies (3d)' The proof of (b) is similar. (; Before solving the problem stated at the beginning of this chapter, we consider the following strong concept.

Definition We shall say that a subvariety R of MS is fertile if every algebra (L; 0) E R can be made into a double MS-algebra. Theorem 15.3 The fertile subvarieties 0'£ MS are B, K, and M. Proof That M is fertile follows from Example 12.2. Since Band K are subclasses of M, they are also fertile. To show that these are the only fertile subvarieties of MS, it suffices to show that Sand Kl are not fertile. For the first of these, consider the Stone algebra 1 EB 22; and for the second, consider the Kl -algebra

b

By Corollary 2 of Theorem 12.7, neither of these can be made into a double MS-algebra. (;

Ockham algebras

220

Definition We shall say that a subvariety R of MS is barren if no algebra L that properly belongs to R (Le. V(L; 0) = R) can be made into a double MS-algebra. Theorem 15.4 Tbe subvarieties K3 and M v K3 are barren. Proof Suppose, by way of obtaining a contradiction, that M V K3 were not barren, Le. that V(L; 0) = M V K3 and L can be made into a double MSalgebra. Then (L; 0) satisfies (6 d ) and so, by Theorem 15.2, (rap; +) satisfies (3d), hence (12d)' But, by Theorem 15.1, (12 d) is self-dual. Hence (L; 0) also satisfies (12d), and this contradicts the hypothesis that L ~ S V M V K 1 . Similarly, if V(L; 0) = K3 and L can be made into a double MS-algebra then (L; 0) satisfies (6 d), and so (LOP; +) satisfies (3d) and hence (12 d) which is selfdual. Thus (L;o) satisfies (1,5,6d,12d)' hence also (1,0,6 d ,12d). Since, as was shown in Chapter 5, (1,0, 12 d ) is equivalent to (1, 'Y), it follows that (L; 0) satisfies (1, 6d! 'Y)' This contradicts the hypothesis that L ~ S V K1 . <> Definition A subvariety R of MS will be called bistable if, for every double MS-algebra (L; 0, +), whenever V(L; 0) = R we have V(LOP; +) = R. Theorem 15.5 Tbe Jollowing subvarieties oj MS are bistable: MVS, M, SVK, K, S, B [i.e. those that belong to the ideal A oj A (MS) generated by M V S] and Kl VK21 MVK1 VK2! K2 VK3, MVK2 VK31 Ml [i.e. those that belong to the filter B oj A (MS) generated by Kl V K2 ].

Proof All of the subvarieties listed are characterised by axioms that are selfdual or by both the axioms (3d) and (6 d). By Theorems 15.1 and 15.2, (LOP; +) belongs to the same subvariety as (L; 0). Now, suppose that V{l; 0) = MVS. Then (LOP; +) E MVS. IfV(LOP; +) = M or SVK then (L; 0) E M orSVK, contradicting the fact thatV(L; 0) = MVS. The same argument is valid for all subvarieties in A, and for Ml and M V K2 V K3 since M V K3 is barren. Suppose that V(L; 0) = K2 VK3. If (LOP; +) E Kl VK2 then (L; 0) E Kl VK2, a contradiction. It is not possible to have V(L; 0) = K3 since K3 is barren. Finally, suppose that V(L; 0) = M V Kl V K 2. If (LOP; +) E K2 V M then (L; 0) satisfies (1, 6d) whence we have the contradiction (L; 0) E M V K3. If (LOP; +) E S V M V Kl then (L; 0) satisfies (I! 3d) whence we have the contradiction (L; 0) E K2 V M. <>

Singles and Doubles

221

Theorem 15.6 Tbe subvarieties Kl , K 2, S V K 1 , S V M V K 1 , Kl V M, K2 V M are neither barren nor bistable. More precisely, (1) (2) (3) (4) (5) (6)

ijV(L; 0) = Kl then V(LOP; +) = K 2 ; ijV(L; 0) = K2 then V(LOP; +) E {K1 , Kl V S}; ijV(L; 0) = S V Kl then V(iO P ; +) = K 2 ; ijV(L; 0) = S V M V Kl then V(LOP; +) = K2 V M; ijV(L; 0) = Kl V M then V(LOP; +) = K2 V M; ijV(L; 0) = K2 V M then V(LOP; +) E {Kl V M, S V M V K 1 }.

Proof (1) IfV(L; 0) = Kl then (L; 0) satisfies (1,4 d ,,),), hence also (1, 6d ,')'). Then (LOP; +) satisfies (1, 3d, ')') and so belongs to K 2. Since K2 covers only the bistable subvariety S V K, we have V(LOP; +) = Kz. (2) IfV(L; 0) = Kz then (LOP; +) satisfies (1, 6d , ')'), hence belongs to SVK1 which covers the bistable subvariety S V K. It follows that V(LOP; +) is either Kl or Kl V S. (3) IfV(L;o) = SVK1 then (LOP;+) satisfies (1, 3d,,),) and so belongs to K z . In fact, V(LOP; +) = K2 since K2 covers only S V K, which is bistable. (4) If V(L; 0) = S V M V Kl then (LOP; +) satisfies (1, 3d) and so belongs to K2 V M. The latter covers M V s, which is bistable, and K 2. If (Lop; +) E K2 then (L; 0) E Kl V S, contradicting the fact that V(L; 0) = S V.M V K 1 . (5) IfV(L;o) = MVK1 then (L;o) satisfies (1,4d)' hence (1,6 d), and (L oP; +) satisfies (1, 3d) and belongs to K2 V M. The latter covers M V s, which is bistable, and K 2. If (LOP; +) E K2 then (L; 0) E {K1 , Kl V S} which is impossible by the minimality of Kl V M. (6) IfV(L; 0) = MVK2 then (LOP; +) satisfies (1, 6d ) and belongs to MVK3. But the latter is barren. It follows that (LOP; +) belongs to S V M V K 1 , M V K 1 , or S V K 1 . The last of these is excluded since V(iO P ; +) = S V Kl implies V(L; 0) = K 2 • It remains to show that there exist double MS-algebras that satisfy the preceding conditions. For this purpose, consider first the lattice Ll with Hasse diagram 1

e b

c

Ockbam algebras

222

Of the ways in which L 1 can be made into a double MS-algebra, the following four are relevant :

x (1 )

XO

x+ (2)

XO

x+ (3)

XO

x+

(6)

XO

x+

0 1 1 1 1 1 1 1 1

a e 1 d d e 1 d d

b e 1 c c c 1 b b

c e 1 b b e e

c c

d e 1 a a c e a a

e e e 0 a c

c 0 a

1

0 0 0 0 0 0 0 0

Kl K2 K2 Kl SVK1 K2 MVK2 MVK1

Next, consider the lattice L2 ~ Lt, namely

c

d a

o Of the ways in which L 2 can be made into a double MS-algebra, the following two are relevant :

x (2')

XO

x+ (5)

XO

x+

0 1 1 1 1

a c c e 1

b a c e e

c a a c c

d 0 c d d

e 0 a b b

1

0 0 0 0

K2 SVK1 MVK1 MVK2

Finally, let L~6) be the double MS-algebra consisting of Ll with the operations as described in (6) above, and consider L3 = L~6) x SID 4 .

The double MS-algebra (L3; 0, +) is such that

V(L 3; 0) = K2 V M V S = K2 V M,

V(L3; +)

= Kl V M V S.

Singles and Doubles

223

Similarly, with

we have

We know by Corollary 1 of Theorem 12.7 that if an MS-algebra (L; 0) can be made into a double MS-algebra (L; 0, +) then this can be done in one and only one way. The problem that we shall consider now is that of relating the subvarieties of MS to which (L; 0) belongs to the subvarieties of DMS to which (L; 0, +) belongs. Given a double MS-algebra (D; a, +) we shall denote by V(D) the smallest subvariety of MS to which (D; 0) belongs. Consider the relation 8 defined on A(DMS) by (D1 ,D2 )

E

8 ~ (VDl

E

D 1 )(VD 2 E D2 ) V(D 1 )

= V(D 2 )

~ {every algebra in Dl satisfies the same a

-axioms as every algebra in D2 .

Clearly, 8 is an equivalence relation. It is in fact a lattice congruence on A(DMS). For, if (D 1 , D2 ) E 8 and X E A(DMS) then, on the one hand, the °-axioms satisfied by Dl , X are the same as those of D 2 , X, so that we have (Dl J\X, D2 J\X) E 8; and, on the other hand, any o-axiom that holds in Dl VX holds in both Dl and X, hence in D2 and X, and consequently in D2 V X, so that we also have (Dl V X, D2 V X) E 8. Since A(DMS) is a finite lattice, the 8-classes are intervals and A(DMS)/8 can be ordered by writing [Dd8" [D2]8 ~ (:3X E [Dd8)(:3Y E [D2]8) X" Y. We shall denote by A*(MS) the sublattice of A(MS) obtained by deleting the barren subvarieties, which by Theorem 15.4 are K3 and M V K3. If we ignore as usual the trivial subvariety, this is then a 17-element distributive lattice, and we have a lattice isomorphism A(DMS)/8

~

A*(MS).

Our objective now is to determine, for each variety in A*(MS), the corresponding 8-class, i.e. the corresponding interval of A(DMS)'. We shall then determine the cardinality of each of these intervals and use the results to compute the cardinality of A(DMS).

Ockham algebras

224

Theorem 15.7 The subvarieties in A*(MS) are associated with the following intervals in A(DMS) B+-7

h = {SIDd

K+-7 12 M+-7 13 S+-7 14 SVK +-7 Is Kl+-7 16 S V Kl +-7 17 K2+-7 Is MVK1 +-7 19 S V M +-7 110

= {SID3} = {SID7} = [SID2, SID4] = [SID2 V SID3 , SID4 V SID3] = [SID6, SID lO ] = [SID2 V SID6 , SID4 V SID lO ] = [SID5 , SID4 V SIDs] = [SID6 V SID7 , SIDlO V SID7] = [SID2 V SID7 , SID4 V SID7]

Kl V K2 +-7 III = [SID 5 V SID6, SID4 V SID 11 V SIDnl MVK2 +-7112 S V M V Kl +-7 113 K2 V K3 +-7

h4

= [SID5 VSID7 , SID4 VSID7 VSIDs] = [SID2 V SID6 V SID7 , SID4 V SID lO V SID7] = [SID 13 , SID 17 V SID 1S ]

M V Kl V K2 +-7 115 = [SID5 V SID6 V SID7 , SID4 V SID 11 V SID 12 V SID7] M V K2 V K3 +-7 1 16 = [SID7 V SIDI3 , SID7 V SID 17 V SID 1S ] Ml +-7 117

= [SID I4 , SID2d

Proof In order to obtain the biggest element in each of the intervals, the method is routine so we shall illustrate it by an example. Consider the subvariety M V K2 V K3 which has the equational basis (1, lId)' of the subdirectly irreducible double MS-algebras we observe that SID 14 does not satisfy these a-axioms (since its 'MS-part' is M 1), whereas SID7 , SID I7 , and SID 1S do satisfy these axioms. It follows that the biggest element in the E)-class of MVK2 VK3 is SID7 V SID17 V SID 1S ' To obtain the smallest elements in the intervals, observe that if X +-7 [P, Q] then we have [P, Q] = Ql \ U I k kEK

where (Ikh EK denotes the family of ideals generated by the intervals which are associated with the subvaneties of MS that are covered by x. Note that for every X there are at most three such subvarieties. Consequently, the determination of the smallest elements in the intervals can be done by traversing A(MS) upwards.

Singles and Doubles

225

If Q is the supremum in A(DMS) of subvarieties that are generated by linearly ordered ideals, the task is particularly simple. For example, consider the subvariety X = MVKz. We know that in this case Q = SID4 VSID7 vSIDs ' Since SID~, SID~, and SID~ are chains we can depict QL as follows (in which n denotes SIDn) :

4v7v8 8 ~-:;,--4v7

5 3 1

Now the subvariety M V K z covers K2 and M V S. Hence P is the smallest element of QL \ ((SID4 V SID8)L U (SID4 V SID7 )L). From the diagram, this is SIDs V SID7 . If Q fails to have the above property then the determination of P is somewhat more intricate, but can be done by considering only the ordered set of subdirectly irreducible double MS-algebras. For example, consider X = Kl V K 2 . Here we know that Q = SID4 V SID 11 V SID12 . Now the subvariety Kl V K z covers K z and S V K 1 . Hence P is the smallest element of QL \ ((SID4 V SIDlO)L U (SID4 V SIDs)L).

By considering the diagram 11 8

4

1

we see that P = SIDs V SID6· 0

226

Ockham algebras

The partitioning of A(DMS) into 17 intervals makes possible the determination of its cardinality. This is as follows (again with the trivial subvariety ignored).

Theorem 15.8 IA(DMS)I = 380. z, y V z] ~ [x, y]. It follows by Theorem 15.7 therefore that we have the following isomorphisms:

Proof In a distributive lattice we have [x

V

14 ~ Is ~ 1 10 , 16 ~ 1 9 , 18 ~ 1 12 , 17 ~ 1 13 , 111 ~ hs, h4 ~ 116.

Now there are three singletons, namely II' 12, 13 , To determine the cardinality of the other intervals is routine, and we illustrate the method by considering 111 = [SIDs V SID6, SID4 V SID 11 V SIDd.

We require to compute the number of down-sets of the ordered set (4 1 U 11 1 U 121) \ (51 U 61), i.e. the ordered set

:1

l1M12 8

9

10

Using Theorem 5.5, we can see that the number of down-sets is 39. Thus we have 11111 = 39 = Ihsl· In summary, we have

I interval

I cardinality I

17 ,113

1 2 4

18 ,1 12

6

111 , hs

39 41 187

1 1 ,12 ,13 1 4 , Is, 16, 19 , 110

1 14 , h6 117

and conclude from this that IA(DMS)I = 380. 0 Given a double MS-algebra (L; ° , +) let us now consider the operation x ~ x~ defined on L by setting ('
x~

= (x /\ x+) V XO = (x V XC) /\ x+.

5ingles and Doubles

227

Our objective now is to determine when this operation makes L into an Ockham algebra. First we observe that (Vx

E

L)

Hence we see that x~ is the complement of x in the interval [x 1\ x+, X V xc]. Clearly, o~ = 1 and 1~ = O. We also have {

whereas {

x+~

= (X+ V X+o) 1\ X++ = X++,

X~o = (Xo V X++) 1\ Xoo X~+

= xeD;

xo~ = (xo 1\ xo+) V xoo

= (Xo 1\ XeD) V X++;

= (X+ V X++) 1\ Xoo = (X+ 1\ XeD) V X++.

Theorem 15.9 If L E DMS then,Jorevery x (a) x ~ x~~ ~ x I\xo E 5(L); (b) x ~ x~~ ~ x V x+ E 5(L)

E

L,

Proof Using the above, we have {

= (x 1\ x+) V XO V x++ = (x V XC) 1\ (x+ V x++); x~ 1\ x~+ = (x V xc) 1\ x+ 1\ xOO = (x 1\ x+) V (xo 1\ xeD), x~ V x~o

and consequently x~~ = {(X~

1\ x~+) V x~o

(X~

V X~o) 1\ X~+

= (x 1\ x+) V (XO 1\ xeD) V X++; = (x V XC) 1\ (X+ V X++) 1\ xoo.

It follows that

{

x ~ X~~ ~ x ~ XO 1\ xoo ~ x 1\ XO E 5(L) x ~ x~~ ~ x ~ x+ V x++ ~ X V x+ E 5(L).

<>

Corollary x = x~~ ~ x 1\ xc, x V x+ E 5(L). <> A remarkable fact concerning the operation x ~ x~ is the following. Theorem 15.10 If L E DMS then (Vx

E

L)

Proof Using the above observations, we have x~ 1\ x~o

= (x V xc) 1\ x+ 1\ (xo V x++) 1\ XOO = x+ 1\ (x o V x++) 1\ XOO = (X OO V xc) 1\ x+ 1\ (x o V x++) 1\ XOO = x~oo 1\ x~o .

228

Ockham algebras

Similarly we can show that x~

V

x~+

=

x~++

V

x~+

It now follows by the Corollary to Theorem 15.9 that x~

= x~~~. <>

In view of Theorem 15.10 it is natural to consider the question of precisely when (Lj~) is an Ockham algebra. In order to answer this, we require the following result.

Theorem 15.11 If L E DMS then thejollowing conditions are equivalent: (1) (Vx,y E L) (2) (Vx,y E L)

(x V y)~ = (x

x~ Ay~j

AY)~ = x~ V y~j

(3) (Vx,y E L) x A x+ AY+ ~ Y V yOj (4) (Vx,y E L) Y AY+ ~ x V XO Vyo.

Proof (1)"'* (3) : The equality (x

V yt

= x~ Ay~ holds if and only if

[x V y V (XO A yO)] A x+ A y+ = (x V XO) A x+ A (y V yO) A y+,

which is eqUivalent to the inequality [x V y V (XO AyO)] A x+ Ay+

~

(x V XC) A x+ A (y V yO) AY+,

which is equivalent to (x A x+ AY+) V (y A x+ AY+) V (XO AYO) ~ (x V XC) A x+ A (y V yO) AY+.

Clearly, this holds if and only if (3) holds. Dually, we can show that (2) "'* (4). We now show that (3) "'* (4), i.e. that (3) is self-dual. For this purpose, it suffices to prove only that (3) =} (4), a dual proof providing the converse implication. Suppose then that (3) holds. Applying 0, we obtain yO AyoO ~ XO V x++ V y++

which gives (yO AYOO) V x V XO

~

(yo AY++) V X V XO

whence (yo Ay)VXVXO = (yo AyOO)VXVxO, which is axiom (17) = (9, 12 d) of Chapter 5. Now by Theorem 15.1 the axioms (9) and (12 d ) are selfdual in DMS. Hence so is (17), and therefore we have (y V y+) A x A x+ ::::: (y++ V y+) A x A x+ which gives y+ V y++ ~ Y A x A x+, whence y+ V y++

Applying

+

~

yeo A Xoo A x+ .

to this, we obtain

(*)

y++ Ay+

~yO V

XO V x++.

Singles and Doubles Applying

+

229

also to (3), we obtain

(**)

y+ I\yoo:::;; x+ V x++ V y++

It follows that

Y I\y+ :::;; yOOl\y+ :::;; x+ V x++ V (y++ I\y+) :::;; x+ V x++ V yO

by (**)

by (*).

Consequently, y I\y+ = Y I\y+ 1\ (x+ V x++ V yO) = (yl\y+ I\x+)v(yl\y+ I\x++) V (yl\y+ l\yO)

:::;; XVxoVx++Vyo

by (3)

= x V XO vyo, and we have (4). <>

Theorem 15.12

Jf L E DMS then

(L;~)

is an Ockham algebra ifand only if

L E SID2 V SID7 V SID ll V SID 12 .

In this case we have necessarily (L;~) E M V KI .

Proof In view of Theorem 15.10, we have that (L;~) E 0 if and only if the mapping x 1---7 x~ is a dual lattice endomorphism; and by Theorem 15.11 this is the case if and only if L satisfies the axiom x 1\ x+ 1\ y+ :::;; Y V yO

Now a routine verification shows that this is satisfied by the subdirectly irreducible double MS-algebras SID2 , SID7 , SID ll , and SID12 but not by SID4 , SID I3 , or SID I4 . The first statement now follows. To show that when (L;~) is an Ockham algebra it necessarily belongs to the subvariety M V KI of P 3 ,1, we observe that the mapping x 1---7 x~ satisfies in general the inequalities

These follow from the expressions for x~~ obtained in the proof of Theorem 15.9 and the equalities x 1\ x+ = x 1\ x~ and x V XO = x V x~. Thus (L;~) satisfies (4) which is an equational basis for M V K I . <> The lattice of (non-trivial) subvarieties of M VKI is

230

Ockham algebras

B

Theorem 15.13 Let L E DMS be such that (L; ~) E 0 Then the various subvarieties are related as follows: (L;~)

(L; 0,+ ) belongs to

MS-subvariety

DMS-subvariety

(L; 0) belongs to MS-subvariety

MVK.l

SID2 V SID7 V SID l1 V SID 12

MVK1 VK2

MVK1

SID2 V SID7 V SID lO

SVMVK1

MVK1

SID2 V SID7 V SIDB

MVK2

K.l

SID2 V SID ll V SID 12

Kl VK2 SVK1

belongs to

Kl

SID2 V SIDlO

Kl M

SID2 V SIDB SID2 V SID7

K2 SVM

K

SID2 V SID3

SVK

B

SID 1

B

Proof For example, consider (L; ~) K.l =

E K. 1 • An

equational basis of

i: /\ (K.1 V M)

is (ry,4). Since, as observed above, (4) and (4d) hold for (L;~) we deduce that (L; ~) E K.l ~ x /\ x~ ~ Y V y~ ~ x /\ x+ ~ Y V y+. It is readily seen that the biggest subvariety of SID2 V SID7 V SID 11 V SID12 in which this holds is SID2 VSID 11 VSID 12 . By Theorem 15.7, the corresponding subvariety of MS to which (L; 0) belongs is then Kl V K 2 • <>

Bibliography

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[1] R Balbes and Ph. Dwinger : Distributive Lattices, University of Missouri Press, Missouri, 1974. [2] G. Birkhoff : Lattice Theory, 3rd edition, Coll. Pub., XXV, American Mathematical Society, Providence, RI., 1967. [3] T. S. Blyth and M. F. Janowitz : Residuation Theory, Pergamon Press, 1972. [4] G. Boole : An Investigation into the Laws of Thought, London, 1854. Reprinted by Open Court Publishing Co., Chicago, 1940. [5] S. Burris and H. P. Sankappanavar : A Course in Universal Algebra, Graduate Texts in Mathematics, 78, Springer Verlag, 1981. [6] D. A. Cohen: Basic techniques of combinatorial theory, Wiley, 1978. [7] W. H. Cornish: Antimorphic Action. Categories ofAlgebraic Structures with Involutions or Anti-endomorphisms, Research and Exposition in Mathematics, 12, Heldermann Verlag, Berlin, 1986. [8] H. B. Curry: Foundations of Mathematical Logic, McGraw-Hill series in higher mathematics, New York, 1963. [9] B. A. Davey and H. A. Priestley : Introduction to Lattices and Order, Cambridge University Press, 1990. [10] P. J. Davis: Circulant Matrices, Wiley, 1979. [11] G. Gratzer: Lattice Theory; First concepts and Distributive Lattices, Freeman, 1971. [12] G. Gratzer: General Lattice Theory, Birkhauser, 1978. [13] G. Gratzer: UniversalAlgebra, 2nd edition, Springer verlag, 1979. [14] F. Harary : Graph Theory, Addison-Wesley, 1969. [15] B. J6nsson : Topics in Universal Algebra, Springer Verlag, 1972. [16] R N. McKenzie, G. F. McNulty and W. F. Taylor: Algebras, Lattices and Varieties, vol. I, Wadworth and Brooks, 1987.

(b) Research Papers

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232

Ockham algebras

[19] M. E. Adams and R. Beazer : The intersection of principal congruences on de Morgan algebras, Houston]. Math., 16, 1990,59-70. [20] M. E. Adams and H. A. Priestley: Kleene algebras are almost universal, Bull. Austral Math Soc., 34, 1986, 343-373. [21] M. E. Adams and H. A. Priestley : De Morgan algebras are universal, Discrete Math., 66, 1987, 1-13. [22] M. E. Adams and H. A. Priestley: Equational bases for varieties of Ock[23] [24] [25]

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Notation index

P(E)

I Z+ ~,~ sop

--< PUQ PEI7Q PEI7Q

/\, v INo = {1, 2, 3, ... } a'

Z(L) a*

w,

~

rva

(Lj /\, v,f, 0,1) 0

= (L;/)

Kp,q M

a S(L) M(L) xt, xt, Xi, xi n MS MS(L) 1'J(a, b), f}lat(a, b) 1'J a

VB(L) 12n , 12n+l, 100 , fO Ca

1 1 1 1 1 1 1 1 2 2 2 4 4 5 6 6 8 8 8 8 10 11 11 13 14 17 17 20 20 23 25 26 27 30

Tj(L) T(L) K(L)

Kw (XjT,

~)

O(X) Ip(L) (Ip(L)j T, ~) DOl P

Q

(Xj T,g) = (Xjg) gW(Q), gW{x}

G(X) f}Q

Pm,n B

INco B, M, K, S, SI, K l , K2 K 3, M l , L, N, Bl A (V), A(O), J\r(O) K, S, S, Ml = MS, Ml = MS (nd) (n), A SI, K l ,

Kl , K 2 , K2 , K 3 , K3 ,

L,L,N,N

V(L) Fix L P(L) LxJ L",L v F(L) 01*, 01*

'¥

38 39 39 42 52 52 52 52 53 53 53 53 54 54 54 54 55 55 70 71 75 75 83 90 91 102 105 108 108 109 115 117 118

238 8,8jj

r

19R, 19~ in, Pn F 2n , F2n+1 , F~n+l C 2n DF 2n #(X), #(X; a), #(X; a), #(X; a, b) DC 2n Skn Ck{n) Cn;k r(O) X[XJY

Cj

M = [ajj] pjj{M), Pjj{M) an, 19 n

[V] (L;I\,V,o,+,O,l)= (L;o,+) DMS S{L) (X;g, b) A0 <1>0 +

SIDn SIDn 19{a,b)C X~

120 122 140 141 150 150 154 157 158 159 164 164 165 168 168 169 173 173 174 180 187 187 187 189 191 199 199 200 210 226

Index

additive closure 25 algebra de Morgan 4 de Morgan-Stone (= MS-algebra) 17 directly decomposable 72 double MS- 187 double MS n - 196 dual MS- 18 equational 7 Kleene 5 02- 196 Ockham 6,8 semisimple 200 simple 37 Stone 5 subdirectly irreducible 37 trivalent Lukasiewicz 206 universal 3 algebraic lattice 29 anti-symmetric (binary relation) 1 antitone map 2 arity (of an operation) 3 atomic term 76 automorphism 3 barren subvariety 219 basic inequality 76 Berman class 8 bicomplete subset 194 binary operation 3 bistable subvariety 219 boolean lattice 4 bounded lattice 3 centre 4 circulant matrix 171 clopen set 52 closure 16 cokernel27 comonolith 46 compactly generated lattice 29 comparable elements 1 complement 4 complemented lattice 4 complete lattice 2

cone 30 congruence 5 associated with 54 extension property 21 lattice 20 Ockham algebra 20 principal 20 conjunction 4 convex subset 13 core 39 covering relation 1 crown 154 de Morgan algebra 4 laws 4 skeleton 11 space 149 de Morgan-Stone algebra (= MS-algebra) 17 directly decomposable algebra 72 disjoint union 1 disjunction 4 distinguished down-set 105 distributive lattice 4 double crown 159 fence 157 MS-algebra 187 MSn-algebra 196 MS-space 189 Stone algebra 187 doubly symmetric row 172 down-set 13 dual of an inequality 83 of an ordered set 1 of a tabulation 83 MS-algebra 18 space 52 dually dense 29 dually isomorphic 2 end 68 endomorphism 3 epimorphism 3

240 equational algebra 7 equivalent inequalities 76 matrices 166 tabulations 82 even term 76 extension 30 falsity ideal 106 fence 150 fertile subvariety 218 Fibonacci numbers 150 filter 13 prime 13 principal 13 proper 13 truth 106 finitely subdirectly irreducible 38 fixed point 29 compact 124 complete 117 distributive 119 separating congruence 115 generalised crown 165 variety 42 greatest lower bound 2 g-subset 54 {g, h}-subset 197 Hausdorff space 52 ideal 13 falsity 106 prime 13 principal 13 proper 13 incomparable elements 1 inequality 76 infimum 2 involution 2 irreducible tabulation 82 isomorphism 3 isotone map 2 join 2 join-complete lattice 2 kernel 27 Kleene algebra 5 skeleton 202

Ockhama lattice 2 algebraic 29 boolean 4 bounded 3 compactly generated 29 complemented 4 complete 2 distributive 4 join-complete 2 local 142 meet-complete 2 semicomplemented 5 Stone 5 lattice congruence 20 least upper bound 2 linear sum 1 local lattice 142 locally convex skeleton 201 finite congruence class 32 finite class of algebras 43 loop 68 Lucas numbers 150 maximally disjoint 60 meet 2 meet-complete lattice 2 mid-level element 163 monogenic g-subset 54 monolith 37 monomorphism 3 morphism 3 MS-algebra 17 MS-space 150 multiplicative closure 25 negation 4 negative occurrence 77 node 29 non-trivial tabulation 83 nullary operation 3 02-algebra 196 Ockham algebra 6,8 congruence 20 space 53 odd term 76 operation 3 ordered set 1 topological space 52

Index order-preserving map 2 order-reversing map 2 perfect extension 30 subalgebra 30 polarity 2 positive occurrence 77 Priestley space 52 prime filter, ideal 13 ideal space 52 principal congruence 20 filter, ideal 13 proper filter, ideal 13 pseudocomplement 5 reducible tabulation 82 reflexive relation 1 relative V-algebra 180 residual 188 residuated mapping 188 saturated (lower, upper) 191 self-dual axiom 204 inequality 83 ordered set 2 subvariety 100 tabulation 83 semicomplemented lattice 5 semiconvex subalgebra 26 semisimple algebra 200 simple algebra 37 basic inequality 77 skeleton 11 space de Morgan 149 double MS- 189 dual (= prime ideal space) 52 Hausdorff 52 MS- 150 Ockham 53 ordered topological 52 Priestley 52 totally disconnected 52 totally order-disconnected 52 Stone algebra 5 lattice 5

241 strong extension 30 strongly large subalgebra 30 lower, upper regular 189 regular 190 subalgebra 5 subdirectly irreducible algebra 37 subposet 180 substitution property 5 subtabulation 82 subvariety 75 barren 219 bistable 219 fertile 218 self-dual 100 supremum 2 symmetric row 171 tabulation 82 tail 68 term 76 totally disconnected space 52 order-disconnected space 52 transitive relation 1 tree 182 trivalent Lukasiewicz algebra 206 truth filter 106 type of an algebra 3 unary operation 3 universal algebra 3 up-set 13 variety 7 vertical sum 2

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