Ockham Algebras

°-preserving closure morphism such that the sublattice

p. Define a unary operation

0

:

L

(Vx E L)

-7

-7

L be a

1m

L by the prescription X

O

=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 -7 L will be called a dualMS-algebra if (LOP, +) is an MS-algebra, where LOP denotes the order dual of the lattice 1. It is clear that we can construct dual MS-algebras by using the dual of Theorem 15, which involves a I-preserving dual closure morphism. Example 1.14 Let IN o 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
~"(x) = { gCd~,X}

if x = 00; if x E IN o; if x = 0.

Examples of Ockham algebras; the Berman classes Then

C{Jn

19

is a {O, 00 }-preselVing dual closure morphism with ImC{Jn

= [O,n]u{oo}.

A polarity on 1m C{Jn is given by 0 H 00, 00

x

= IT pfi

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

H

i = IT

H

iE]

pfi-f3t •

iEIU]

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

X

+

=

n gcd{n,x}

-

IT iEIUK

O!i-min{O!, ,'Y;}

Pi

2 Congruence relations Let (L;/) be an Ockham algebra. Then an Ockham algebra congruence (or, briefly, a congruence) on L is an equivalence relation that has the substitution property for both the lattice operations and for the unary operation j. 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) = ;\{
E

Con L I (a, b)

E
In other words, it is the smallest congruence that identifies a and b. Similarly, the principal lattice congruence generated by a, b is

191at(a, b) = ;\{
E

ConlatL I (a, b) E
Note that we then have

191at(a, b):::;; 19(a, b). We recall that, in a distributive lattice,

(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,f) be an Ockham algebra. If a:::;; b in L then 19(a, b) =

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

Congruence relations

21

= V 19lat(rn(a),r(b)).

We show as follows that cp E Con L. n;;,O Suppose that (x,y) E cp, so that there are integers i 1 , ... , i m and elements ZO,Zl,'" ,Zm such that

Proof Let cp

X

where CPik and so

= Z0

¥'il

= 19lat(rik (a),jik (b)).

Zk-l /\lk(a)

¥'im

¥'i2

== Z 1 == Z2 == ... == Zm-l == Zm =y, Now ifi k is even we havejik(a)~fik(b)

= Zk /\lk(a),

Zk-l

V lk(b)

= Zk V fik(b).

Applying f to each of these equalities, we obtain

(r(Zk-l),j(Zk))

(*)

E

19lat(rik+l(a),lk+l(b)) = CPik+ 1 '

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

(a, b) E 19(a, b)

=}

(\In) vn(a),jn(b))

E

19(a, b),

from which we deduce that, for each n,

19(rn(a),r(b)) ~ 19(a, b) and hence that

Thus cp

= 19(a, b) as asserted. 0 then 19(a, b) =

2p+q-l

V 19lat(rn(a),jn(b)). 0 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 equaCorollary

If (L,j)

E Kp,q

tional 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)

= 19B (a, b)lA'

Ockham algebras

22

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

Theorem 2.2 The 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-Ln lAo Now since A is a lattice congruence it has an extension I E Conlat B such that IIA = A. It follows that I;;:: f-Ln for all n and so

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, (1) {) 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 = j2(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, 1) 1\ {)lat(O,j(a») = {)lat(a I\f(a),j(a»). That [a]{)a = [a, a V f (a)] follows immediately from the definition of {)a ; and that {)a = W a ;;::f(a) is immediate from the above. (2) If a ~f2(a) and (x,y) E {)a then ({(x),j(y») E {)a; for, from x I\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) I\f2(a), hence f(x) 1\ a = f(y) 1\ a. (3) If a = f2(a) and a Ilf(a) then by (1) we have {) f(a) = {)lat(a I\f(a), a) and consequently {) a1\ {)f(a) = w. It also follows by (1) that both {)a and {)f(a) are non-trivial. <>

*

Congruence relations

23

Theorem 2.4 Foran Ockham algebra (L;f), (1) {a ELI a ~fZ(a)} is a sublattice of L which contains 0 and 1; (2)a~b

=}

'I9a~1h;

(3)'I9al\'I9b='I9avb'

Proof (1) follows from the fact thatfZ(al\b) = f Z(a)I\j2(b) andj2(av b) = j2(a) V j2(b).

(2) follows from the fact that x 1\ b = Y 1\ b implies x 1\ a x V f(b) = y V f(b) implies x V f(a) = y V f(a).

=y

1\ a, and

(3) If (x,y) E 'I9 a 1\ 'I9 b then xl\a=yl\a, xl\b=Yl\b, xVf(a)=yvf(a), xVf(b)=yvf(b),

which implies that x 1\ (a V b)

=Y 1\ (a V b),

XV f(a V b)

= y V f(a V b).

Consequently, '19 a 1\ '19 b ~ '19 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;f) consider now, for every n E IN, the relation
=r(y).

It is clear that
r(L)

= {fn(x) I x

E

L}

is a subalgebra of L. We now consider some basic results concerning these congruences and subalgebras. Of especial importance in this is the congruence
V
i;;;'O

a practical description of which is as follows. 0 then (x,y) E
Theorem 2.5 If (L;f) Proof If (x,y)

E
x

E

then there exist to,

= to
=r(y)for

, t k and
24

Ockham algebras

Denote the greatest of these
t: fn(1)

=

for all n gives
For every non-trivial Ockham algebra (L;1) it is clear that we have

w =
'"

(
and, with ( meaning 'is a subalgebra of',

{a, I} ( '" (p+l(L) (fi(L) ( ... (f(L) (fO(L) = L. It is readily seen that [x]

Theorem 2.6 If (L;1)

E

Kp,q then,jor n (q,

L/
Theorem 2.7 If (L;1)

E

"-'

f1l(L)

E

Kp,q-n-

~

Kp,q then

w =
Proof Observe that (x,y)

E


We thus have
(x,y)

=
E


If now (x,y)

E


(r- 1(x),r- 1(y)) E
= f q+r- 1(y),

Le. (x,y)

E



V
(1)
(2) (Vi ~ 1)
(3) f is injective.

= w;

=

25

Congrnence relations

Moreover, if (L;/) belongs to the Berman class Kp,q then each ofthe above is equivalent to (L;/) E Kp,o. Proof The 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 <1>1 =f w. Then there exist x,y E L such that x =f y and f(x) = f(y). This implies that f2 P(x) = f2P (y) and hence the contradiction x = y. <> Note that the hypothesis that L belong to some Berman class is necessary in the above. As the following example shows, it is possible for f to be injective with L ¢ }(,z,o for any n.

Example 2.1 On the chain L = {-oo} E9 7L. E9 {oo} define f by

x : -00 ... -2 -1 f(x): 00.. . 3 2

° °

1

2...

00

-2 -3 ...

-00

Then (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 n-class can contain more than one element of fn(L), or even none, as the following example shows.

Example 2.2 Let L be the chain

°< a <

b

< c < 1.

Define f by

x:Oabc1 f(x) : 1 1 c a

°

Then it is readily seen that (L;/) E 0 and that the smallest Berman class that it belongs to is K 1,3' Now f2(L) = {a, a, I}, and

[0]<1>2 [b]2

= {O,a} = {b}

with [0]<1>2 nf2(L) = {O,a}; with [b]2 nf2(L) = 0.

Of some interest is the case where every n-class of L contains exactly one element of 1'l(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,I' Note that in K~o the mapping ep : X f-7 f2 P(X) is a closure operator, since

ep(x) ~ x;

ep2(X) = ep(x);

X~Y

=?

ep(x) ~ ep(y).

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

26

Ockham algebras

[97], it follows that rp(L) is a semiconvex subalgebra of L, Le. if x /\ 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 fq (L). If L E Kffo then every zp -class has a greatest element, this being the only element ofthe class that belongs to fZP (L). Proof Since L E Kp q, for every x E L we have (x, fZP(x)) E q, and therefore (tZP(x),f4 P(x)) E q and so (x,fZkp(X)) E q, Le. (x,jq(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) = fq(y). Now let L E K~o' Then L E Kp,l ~ Kp,zp. Since fZP is a closure operator, the greatest element of [x]zp is the only one that belongs to fZP(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 following statements are equivalent: (1) VB(L)

= Kp,q; = fq+l(L) ="', and VB (tq(L)) = Kp,o, : Fromfq(x) = j2p+q(x) = fq+llfzP-l(X)] Efq+l(L)

(2) L :::> f(L) :::> ... :::> fq(L)

Proof (1)

=} (2) it follows that fq(L) ~ fq+l(L), whence fq(L) = fq+l(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) = fq(L). But, by (1), L ¢ Kp,q_l and so there exists x E L with fq-l(X) t j2p+q-l(X). But fq-l(x) Efq-l(L) = fq(L) and so fq-l(X) = fq(y) for some y E L. Hence fq(y) t fzp+q(y), and this contradicts the fact that L E Kp,q' This then establishes the chain

L :::> f(L) :::> fZ(L) :::> '" :::> fq(L) = r+1(L) = .... Now since fq = j2p+q we see that j2P acts as the identity on fq(L). Hence fq(L) E Kp,o, ThatVB(tq(L)) = Kp,o follows from the fact that iffq(L) E Kt,o C Kp,0 then necessarily tIP and, since fZt is then the identity on (L), we have fq = fZt+ q so that L E Kt,q, which contradicts VB(L) = Kp,q'

r

(2) =} (1) : If VB (tq(L)) = Kp,o then clearly fq = j2p+q and so L E Kp,q' If VB (L) = Kp/ ,q' then from (1) =} (2) it follows that ql = q and pi =p. ~

Corollary 1 IfVB(L)

= Kp,q

then,for 1 ~ i ~ q, VB(ti(L))

= Kp,q_i' ~

27

Congruence relations

Corollary 2 ljVB(L)

= Kp,q

then

w =<1>0 < <1>1 < <1>2 < ... < q =

q+l

= ... ,

and Con L has length at least q + 1 . 0

In a bounded distributive lattice every ideal I is the kernel of at least one congruence, i.e. there is a lattice congruence {J such that [O]{J = I. We also write I = Ker{J. Dually, every filter F is the cokernel of a congruence, i.e. there is a lattice congruence 1f; such that [1]1f; = 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 {J implies (a, 1) E {J, whence the contradiction (0, 1) E {J. 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 I 2n

= (t21l(1))!,

I 21l +1 = (t2n+l(I))i.

Then clearly I 2n is an ideal and pHI is a filter. Finally, let I 00

= V I 2n, n;;,O

10

= V I 2n +1 . 1l;;'0

We shall now investigate, for a given ideal I, the smallest congruence 8 (1) on L that identifies the elements of I. We recall that, in a distributive lattice, 8 (1) is characterised as follows : (x,y)E8(1) ~ (:liE!) 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,

8(1) = 81at (JO) V 81at (100)' Proof This follows immediately from Theorem 2.1 on noting that for every subset X of L we have 8 (X) = V{ {J(a, b) I a, b EX}. 0 Theorem 2.11 For every ideal I of (L;/)

E

(x,y)E8(I) ~ (:liEIoo)(:l}EJO)

0, (xVi)I\}=(yVi)I\}.

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} E JO with (x V i) I\} = (y V i) I\}, whence x I\} = x 1\ (x V i) I\} = x 1\ (y V i) I\}

28

Ockham algebras

and hence (x, x 1\ (y V i)) E 9Iat (JO). Now we also have (x 1\ (y V i)) Vi = (x V i) 1\ (y V i) = (x I\y) V i

and so (x 1\ (y V i), x I\Y) E 9Iat (100)' Thus we see that (x, x 1\ y) E 9Iat (JO) V 9Iat (100)'

Similarly, we can show that (y, x I\Y) E 9Iat (r) V 9Iat (100).

Thus, by Theorem 2.10, (x,y) E 9(1). Conversely, suppose that (x, y) E 9 (1). Then, again by Theorem 2.10, we have the finite sequence x=PI=PZ="'=Pn=y

where each

=is either 9I (JO) or 9I (100)' Suppose that we have at

at

x 9lat (1oJ PI 9Iat (r) pz 9Iat (100) ... 9Iat (r) y. Then we have finitely many equalities XViI

=PI ViI, PIl\jl =pzl\h, pz Viz =P3 Viz, P31\jz =P41\jz,

...

where each ikE 100 and each j k E 1 • Since 100 is an ideal and 1 is a filter, we have Vi k E 100 and !\h E 10 • Moreover, 0

0

(xVVi k ) 1\ !\h = (PI VVi k )l\!\h

= (PI 1\ 1\)k) V (V i k 1\ !\jk) = (Pz 1\ !\h) V (V i k 1\ !\h) = (Pz VVi k ) I\!\h = (y V Vi k ) 1\ !\h, from which we see that (x,y)

E

'P and hence that'P coincides with 9(1).0

The following observation is immediate. Lemma 2.1

If the ideal I

is such that fZ(1) ~ I then,forevery n,

fZn(1) ~ I,

fZn+I(1) ~ (t(1))i,

from which itfollows that

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

29

Congruence relations

(O!) f2(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 that j2 (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(I))i)

(xVi)l\j=il\}.

We have i I\} 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 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;f) 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 a L 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;f). Since Con L is a sublattice of ConlatL, which is known to be distributive, Con L is also distributive. It is also algebraic, in the sense that it is complete and compactly generated (every element of Con L is the supremum of a set of compact elements, a compact element being an element 1J such that if 1J ::;;: sup X for some X ~ Con L then 1J ::;;: 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 Con L is boolean. We also point out that <1>1 is dually dense in Con L, in the sense that if 1J V<1>1 = ~ then 1J = ~. In fact, for any 1J E Con L we have (0,1) E 1J V<1>1 if and only if (0,1) E 1J; for

°

1J Xl <1>1 X2 1J X3 <1>1 ... <1>1 Xn-2 1J Xn-1 <1>1 1

Ockham algebras

30 implies that 1

-a J(xd

=J(xz) -a J(x3) = ... =J(xn-z )

-a J(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 we of Con L, especially when J has a fixed point. In order to illustrate this, consider the following notions. an Let A and B be algebras of the same type. Then we say that A is on extension of B if B is a subalgebra of A. We say that A is a strong extensi which of B if every congruence on B has at most one extension to A, in perfect a is A Finally, A. of case B is said to be a strongly large subalgebra to A, extension of B if every congruence on B has exactly one extension which case B is a perfect subalgebra of A. When this happens, we have in

Con A ~ ConB. In [100] it is shown that, for any element a of a modular lattice L, the ce. cone Ca generated by a, Le. the set at U a r, is a strongly large sublatti This leads to the following property. Theorem 2.13 Let (L;j) E 0 and let a be a fixed point oj L. Then Con L

~

Con Ca'

Proof Clearly, C a is a subalgebra of L and this subalgebra is strongly large. subIn fact, since 0 has the congruence extension property, C a is a perfect <> algebra. Hence we have the isomorphism stated. The usefulness of Theorem 2.13 lies in the fact that it enables us to work with C a instead of L, and we can benefit from this in two ways. Firstly, in the size of C a can be conSiderably less than that of L. Secondly, Ca will e, instanc for L; of general belong to a subvariety of 0 that is smaller than that if it is clear that (Ca;j) always satisfies the axiom x AJ(X) ~y V J(y) even (L;j) does not. Example 2.3 On the lattice L with Hasse diagram

e

define J by x:Oa bcde J1 J(x) : 1 d J cae b 0

31

Congruence relations

Then (L;f) is an Ockham algebra that belongs to K 1,0 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;f) it also belongs to K1 0 and satisfies the supplementary relation x I\f(x) ~ Y V f(y), which L itself'does not (consider the elements e and c). It is easily seen that Con C e 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, b 1, and 11.

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

.0

and made into an Ockham algebra by definingf(x j ) = Xi+l for each i. Note that f is injective so, by the Corollary of Theorem 2.7, W =
< ... -< 19 (Xi+l , Xi+z) -< 19(Xi+l' Xj) -< ... < 'P -<

£

where 'P 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;f) we also have the following results.

Theorem 2.14 If a, bEL with a complement in [w,
~

band f(a)

= f(b)

then 19(a, b) has a

32

Ockham algebras

Proof By Theorem 2.1 we have 'l9(a, b)

= V 'l91at (jn(a),r(b)). 11;;:0

Since f(a) = f(b) by hypothesis, it follows that 'l9(a, b) = 'l91at (a, b) E ConL with, clearly, 19 (a , b) :s:;; <1>1' Write 19 (a , b) = a and observe that since a is a principal lattice congruence it has a complement in Conlat{L), namely

(3 = 'l91~t(O, a) V 'l91at (b, 1). Consider the lattice congruence a' = (3 /\ <1>1' Since every lattice congruence contained in <1>1 is a congruence, we have a' E ConI. Now a V a'

= 'l9(a, b) V ((3/\ <1>1) = ('191at (a, b) V (3) /\ ('191at (a, b) V <1>1) = £ /\ <1>1 = <1>1,

and

a /\ a'

= 'l91at (a, b) /\ (3/\ <1>1 = w /\ <1>1 = w.

It follows that a' is the complement of a in [w, d.

<>

If a is a congruence on an Ockham algebra L then an a-class [ala will be called locallyfinite if, whenever x, y E [a]a with x :s:;; 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-
(a,b)¢n,

(a,b)En+l'

Then n V 'l9(a, b) is an atom of [n, n+d. Moreover, if every n+l -class is locallyfinite then every atom of [n, n+ d is obtained in this way.

Proof Since fn+l(a) = fn+l(b) we have (j(a),j(b)) E no It follows that 'l9(j(a),j(b)):S:;; n and consequently, by Theorem 2.1, (1)

n V 'l9(a, b) = n V 'l91 at (a, b) V 'l9(j(a),j(b)) = n V 'l91 at (a, b).

Clearly, we have n < n V 'l9(a, b):S:;; n+l' Suppose that

n:S:;;

n V 'l9(a, b). Then we note that

(2)

= w.

In fact if (x,y) E

(*)

x /\ a

=y

/\ a,

x V b = y V b,

(x,y) E <po

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

Congruence relations

33

s t- 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) 1\ b = (y V a) 1\ b. Butfrom (*) 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

=

and therefore

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,
E

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


=

V

V{ 19(p, q) I (P, q) ~
Now for suchp, q we have, by Theorem 2.15, that
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 :

loa

x f(x) . 1

1

b 1

c 1

del b b 0

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

Ockham algebras

34

d·

• • • • • • • • • • • • • • • • • • • • • • • • • • • • •

c·

•

•

1•

e·

•

b• a.

I

• •

• •

I

•

t

<1>1

<1>2

"1

"2

"3

I •

•

•

I

•

• • •

• • •

o• W

III

II

"4

"5

"6

II

•

•

•

• •

•

"7 "8 "9 "10 "11

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

The lattice of congruences of L is

17

15

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 < ... < a < ... < Y2 < Yl < 1 made into an Ockham algebra by defining

= 1, f{l} = 0, {Vi} f{Xi} =f(Yi} =f{a} = a. Here we have w = <1>1 with classes {O}, {I}, and C \ {O, I}. The w-class f{O}

C \ {O, I} is not locally finite. Consider now the partition {{O}, {xili~l}, {a}, {Yili~l}, {I}}.

Congruence relations

35

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

Theorem 2.17 If L E Kn,o is finite then Con L is boolean. Moreover, if the length of L is m then Con L has at most m atoms, and hasprecisely m atoms if and only if L itself is boolean. Proof We first show that if a -< b then 'l9(a, b) is an atom of Con L. Suppose that 'l9 E ConL is such that w ~ 'l9 < 'l9(a, b). By Theorem 2.1 we have 'l9(a, b) =

2n-l

V 'l9lat(tk(a)J

k

(b)).

k=O

Since in this case f is a dual automorphism and a -< b, if k is even we have jk(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 'l9 /\ 'l9 lat (tk(a)J k(b)) = w. In fact, if (x,y) E 'l9 /\ 'l9lat(tk(a)Jk(b)) then

= y /\jk(a); xV fk(b) = Y Vjk(b);

(1) x /\fk(a)

(2)

(3) (x,y)

E

'l9.

Consider the elements

u

= (x V fk(a))

/\fk(b),

v = (y

V

fk(a)) /\fk(b).

We have u, v E {fk(a)Jk(b}} and (u, v) E 'l9. Suppose that u =f v. Then one of these elements is jk(a), the other is fk(b), and (tk(a)Jk(b)) E 'l9. Since 'l9 is a congruence it follows that

(a, b) = (t2n-k[Jk(a)], f2n-k[Jk(b}])

E

'l9,

whence 'l9(a, b) ~ 'l9, 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 (1) gives likewise x = y. Using this observation, we now have

'l9 = 'l9/\ 'l9(a, b) =

2n-l

V

('l9 /\ 'l9lat(tk(a),jk(b)))

k=O

=w

and consequently 'l9(a, b) is an atom of Con L. Now take a maximal chain 0=

Zo

-< Zl -< '" -< zm = 1

Ockham algebras

36 in L. Then we have ~

m-l

= V t9(Zi' Zi+l) i=O

a join and so ConL is a bounde d distributive lattice whose greatest element is of at most m distinct atoms, whence it is boolean. is If now L is boolean with m atoms then it is well known that ConL length also boolean with m atoms. Conversely, suppose that L E Kn,o is of that if first e Observ atoms. m and that ConL is boolean with precisely m ra conside x E L \ {O, I} then x Ilf(x). In fact, suppose that x < f(x) and maximal chain passing through both, of the form 1. 0= Zo -< ... -< x = Zi+l -< ... -< f(x) = Zt -< f(Zi) = Zt+l -< ... -< Zm = disSince ConL has precisely m atoms, all the congruences t9(zj,zj +d are we so ConL, of atom an is x) t9(Zi, tinct. But (j(Zi), f(X)) E t9(Zi' x) and show nts argume have the contradiction t9(Zi' x) = t9(j(X),f(Zi))' Similar 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 obobtain servation gives x V f(x) = 1; and from x I\f(x)~f(x I\f(x)) we n. <> boolea is L e whenc y, x I\f(x) = O. Thus x andf(x ) are complementar

Theorem 2.18 Let L be a finite Ockham algebra for which VB(L) Then each summa nd of

= Kp,q'

[w, d E9 [<1>1, <1>2] E9 ... E9 [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

in whence, by Theorem 2.17, Con L/q is boolean; and by a standard result 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; Le. a congruence Oi such that {) ~ Oi for all {) E Con L with {) ::f w. Such a congruence Oi 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,l}U{ajl-n~i~n},

totally ordered by

o < a- n < a-n+I < ... < a-I < ao < al < ... < an-I < an < 1. Define f: Ln Ln by f(O) = 1,1(1) = O,j(ao) = ao, and --t

('rim ~ 1)

f(a m) = a-m+I' f(a- m) = am'

It is readily seen that (L n J) is an Ockham algebra and that it belongs to the

Berman class K I ,2n- From the definition of f it follows that any congruence {) on Ln that identifies an adjacent pair aj, aj+I (Le. any congruence different from w) must identify ao, al' Consequently, Con Ln has a smallest non-trivial element, namely the principal congruence {)(ao, a I ). Hence Ln is subdirectly irreducible. The algebra L o is simple.

Example 3.2 (Tbe centripetal see-saw) With L'l andf as in Example 3.1, let Loo

= U L n = {O, I} U {a

j

liE Z}.

n;;'O

Then (LooJ) is an infinite totally ordered subdirectly irreducible Ockham algebra, with monolith {)(ao, ad.

Ockham algebras

38

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 < az < a4 < ... < 1 and define f

:L

----+ L

f(O)

by

= 1,

f(l)

= 0,

(i;;:: 0) f(a j )

= aj+l'

Then (L;/) is an Ockham algebra. We leave as an exercise for the reader the task of showing that Con L is the lattice t9(ao,aJl t9(ao,l) t9(aO,a2)

t9(O,aJl t9(aj,a3)

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

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

Example 3.4 The pineapple (Example 2.4) is finitely subdirectlyirreducible but not subdirectly irreducible. Given an Ockham algebra (L;/), consider now for each

Tt(L)

i;;:: 1 the subset

= {x ELI /(x) = x}.

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

Subdirectly irreducible algebras

39

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

T 2i (L) ~. T 4i (L) ~ '" ~ T 2"j(L) ~ T 2'l+li(L) ~ .... Consider now the subset

T(L)

=U

T 2n (L),

n;;;:1

Le. the set of x E L for which there exists an even positive integer m x such that fmx(x) = x. Given x,y E T(L), let m = lcm{m x , my}. Then, m being even, we have

fm(x

V

y)

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

V y,

and similarly fm(x Ay) = x Ay. 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)

= {O, I} UTI (L).

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;1) 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 AbE K(L). Since distinct fixed points cannot be comparable, it follows that we must have a V b = 1 and a A b = O. 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) ~ T 2 (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) = {O, I}, T2 (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 afinitely subdirectly irreducible Ockham algebra. If a, bEL are such that f(a) = band f(b) = a then either a = b or {a, b} = {O, I}.

Proof If {a, b} :j:. {O, I} then we have 0 < a < 1 or 0 < b < 1. If a A b = 0 (in which case 1 = f(a) V f(b) = b V a) then, by Theorem 2.1, 'l9(0, a) A'l9(0, b) = ['l91at (0, a) V 'l91at (b, 1)] A['l91at (0, b) V 'l91at (a, 1)]. Since ConlatL is distributive, we can expand the right hand side. Observing that 'l91at (0, a) A 'l91 at (0, b) = w, that 'l91at (0, a) A'l91at (a, 1) = 'l91at (a, a) = w, and

Ockham algebras

40

using the hypothesis that a 1\ b = 0, we see that the right hand side reduces ible. to w. This contradicts the hypothesis that L is finitely subdirectly irreduc 2.1, Consequently we must have a 1\ b t= in which case, by Theorem

°

'l9(0, a

1\

b) 1\ 'l9(a 1\ b, a v b) = ['l91at (0, a 1\ b) V 'l91at (a V b, 1)] 1\ 'l91at (a 1\ b, a V b) = 'l91at (a 1\ b, a 1\ b) V 'l91at (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 Ij L is finitely subdirectly irreducible then K(L) = T 2 (L). either Proof If a E T 2 (L) thenf[ f(a)] = a. Taking b =f(a) we deduce that a = f(a) or {a,j(a )} = {O, I}, i.e. a E T 1(L) U {O, I} = K(L). <> two Corollary 2 Ij(L;f) is finitely subdirectly irreducible then f has at most fixed points. 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, I} U T1(L) most where, by Theorem 3,1, T 1 (L) is either empty or is an antichain of at first was which result, two elements. We can therefore state the following obtained by Kalman [72]. three Theor em 303 In the class M of de Morgan algebras there are only the namely simple, (finitely) subdirectly irreducible algebras, each ofwhich is algebras

I~

B

K

Proof By the above, the algebras shown are the only candidates. It is readily seen that for these we have Con B = Con K = Con M ~ 2, <> so these algebras are indeed subdirectly irreducible; in fact, they are simple.

41

Subdirrxtly irreducible algebras

Theorem 3.4 If LEO is such that K(L) that a < b then '19 (a , b) = L

= Tz(L) and if a, b E T(L) are such

Proof For every x E T(L) let m x be the least even positive integer such that fmx{x) = x. With n x = ~mx, consid ertheel ements Ilx- I

a(x)

= 1\

fZi(x),

i=O

= Observe thatfZ[ a(x)] = a(x) andj2L 8(x)] = f3(x), so a(x),f3 (x) E Tz(L) that M K(L). Now let a, bE T(L) be such that a < b.Cons ider the sublattice is generated by {jZi(a) ,jZj (b) I 0 <, i <, n a , 0 <,}<, nb}' Let Clearly, M is finite withsmallestcelement a(a) and greatest element f3(b). Since M. in (P)] la(aLf3 = B l p be an atom of M and consider the interva p) is every atom of M is of the form 1\ j2i(a) fOf some j, it follows thatj2( ifj tive also an atom of M. Consequently, B is boolean; for it is a finite distribu lattice whose greatest element is a 'join of atoms. Observe that a(a)
(a /\ f3(P), b Nf3(P)) E'I9(a, b). If a /\ ,f3(P) < b /\ f3(P), let c be an atom of B withe 1; a /\f3(P) and c <, b /\ f3(P). Then we have (a(a), c) = (a /\ f3(P) /\ c, b/\ f3(P) /\ c) E 'I9(a, b). < It follows that (a(a), f3(c)) E '19 (a , b). Since a(a), f3(c) E K(L) with a(a) L = b) , (a 9 1 ' re therefo f3(c), we deduce that (0,1) E '19 (a , b) and that If now a /\ f3(P) = b /\ f3(P) let al = a V f3(P) < b vf3(P) = b l . Then a (a 1, bd E '19 (a, b). Moreover, we cannot have f3(P) = 1, so f3(P) must be iction contrad the gives (P) f3 = (b) f3 ise fixed point. Then f3 (b) ;= 1; for otherw a

= a /\ f3(b) = a /\ f3(P) = b /\ f3(P) = b /\ f3(b);= b.

that Considering therefore the interval [f3(P), 1] in M and a coatom q such ~ L. = b) , (a 9 1 ' that r manne dual q ~ a V f3(P) and q j b V f3(P), we see in a

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

Ockham algebras

42 (3) all de Morgan subalgebras of L are simple.

(2) : If (1) holds then by Theore m 3.4 every non-trivial prinis the cipal congru ence on T(L) coincides with t. Since every congru ence T(L) that follows it s, contain it suprem um of the principal congru ences that is simple. (3) : Tz (L) is the largest de Morgan subalgebra of L. (2) by (3) =? (1) : If (3) holds then Tz(L) is simple. But Tz(L) E K 1,a = M; and s, algebra Morgan de Theore m 3.3 there are only three non-isomorphic simple (1) core, in each of which K(L) = Tz(L). Since Tz(L) and L have the same follows. <>

Proof (l)

=?

*

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

= T(L) if and only if

de Proof If K(L) = T(L) then K(L) = Tz(L) = T(L) and so T(L) is simple and Morgan. Conversely, if T(L) is simple de Morgan then K(L) = Tz(L); <> 1. of bra subalge T(L) = Tz(L) since Tz(L) is the largest de Morgan OckIn order to examine more closely the (finitely) subdirectly irreducible n Berma the all s contain that ham algebras, we shall concentrate on a class classes, namely the subclass Kw of 0 defined by

(L;/)

E

Kw

*=?

(Vx

E

L)(:Jm, n

E IN)(m

t 0)

fm+n(x)

=r(x).

even; for Without loss of generality, we may assume that both m and n are fm+n(x) =r(x) implies fZm+Zn(x) = fm+nlrn(x)] = J'llfm+n(x)] = fZn(x). orphic By definition, Kw is closed under the formation of subalgebras, homom y direct images, and arbitrary direct powers. But it is not closed under arbitrar 0 and q) each for Kp,q E L products, as can be seen by taking an algebra q Fang Jie to due result, considering La XLI xL z x .... However, the following and ts, [67], enables us to claim that Kw is closed under finite direct produc therefore forms a generalised variety in the sense of Ash [23].

and Theor em 3.6 Let (L 1;/) and (L z ;/) be Ockham algebras. If Xl ELI with nz mz, and n1 Xz E L z are such that there exist natural numbers m1, fm!+1l! (Xl) = fn! (Xl), fm 2 +n2 (XZ) = fn 2 (xz), then there are natural numbers m, n such that, in the algebra L1 x Lz, f m+n(x 1,xz)

=r(X1, x z)·

Subdirectly irreducible algebras

43

Proof Observe first that jP+q(x) = fq(x) implies jkp+q(x) = fq(x) for all k E IN o. Now let m = lcm{ml,mZ} and n = lcm{nl,nZ}' Then m = mlr, n=nlsand

f m+ll (Xl) =fm l r+1l1 +(S-1)1l1 (Xl) =fill +S1l1-1I1 (x 1)

=f'!(Xl)' Similarly, fm+1l(xz) = f'!(xz). The result now follows. <> A class C of algebras is said to be locallyfinite 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 Tbegeneralised variety Kw is locallyfinite. Proof Suppose that L E Kw is O-generated by {Xl, ... , Xk }. Then there are natural numbers m i ,n i with m i t= 0 such that fm;+1l 1 (Xi) = fill (Xi) 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

fm+ll(x i ) = fll(Xi) [= fZm+ll(Xi)]'

Km,ll

and is DOl-generated by

{jJ(x i )10<'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 Ifan Ockham algebra L is finitely subdirectly irreducible then every «1>1 -class in L contains at most two elements. Proof Suppose that a «1>1 -class contains at least three elements. Then it contains a 3-element chain X < y < Z with f(x) = f(y) = f(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) /\ 19(y,z) = w. <> Theorem 3.9 If L E Kw then the following statements are equivalent: (1) L is finitely subdirectly irreducible; (2) L is subdirectly irreducible. (2) : Since L E Kw, for every x E L we have fm+ll(x) = f'l(X) for some m, n E IN with m t= 0 and even. If «1>1 = w then f is injective and x = fm(x), whence x E T(L). Thus L = T(L) and it follows by Corollary 1 of

Proof (1)

=?-

44

Ockham algebras

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

i

by the hypothesis that L is finitely subdirectly irreducible that 19 (a, b) /\ a "f w. Since 19(a, b) is an atom in Con L it follows that 19(a, b) /\ a = 19(a, b) and hence 19 (a, b) ~ a. Thus 1l( 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 isfinitely subdirectly irreducible but not subdirectly irreducible then necessarily L ~. Kw and f is injective,

Proof ThatL

~

Kw follows from the above.

Suppose thatf were notinjective. Then by Theorem 3.8 the interval [w,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/w ~ T(L).

w' Then x = jP(x), y = fq(y) for some p, q; andr(x) =r(y) for some n. Let r = !cm{p, q, n}; then we have x = r(x) = r(y) = y.

Proof Observe first that

Now in the quotient algebra (L/w;}) we have ]([x]w) = [J(x)]w' Consider the morphism 19 : T(L) -7 L/w given by 19(x) = [x]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 f'l(y) = fm+l1(y). It follows that there exists p ~ n such that

jP+Il(y) = F!(Y) = fZP+l1(y). The first of these equalities gives (y,jP(y») E 11 ~ w; and the second gives\ E Tzp(L) ~ T(L) from which it follows that jP(y) E T(L) since n ~p. Thus we see that

fl1(y)

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

Subdirectly irreducible algebras

45

Theorem 3.11 If L E Kn,o then L isa strong extension ofT2(L). Proof Given -e E Con T 2 (L), let -e l , -e 2 be extensions of -e to L. We may assume without loss of generality that -e l ~ -e 2 . Observe first that if a E T 2(L) then [a]-e l = [a]-e 2. In fact, for every x E L define

n-l

x* =

V f2k(X),

n-l

1\ f2k(X).

x* =

k=O

k=O

If (x, a) E -e 2 then clearly (x*, a) E -e 2 and (x*, a) E -e 2. Since x*, x* E T2(L) and -e l I T2 (L) = -e = -e 2I T 2(L) we have (x*, a) E -e l and (x*, a) E -e l . Since x*~x~x* we deduce that (x,a)E-e l . Thus [aW2 ~ [a]-e l whence we have equality. Suppose now that L is finite and (x,y) E -e 2. To show that (x,y) E -e l it suffices, since L /-e l is a finite distributive lattice, to show that [x]-e l and [y]-e l contain the same set of V-irreducible elements. Suppose then that [c]-e l is Virreducible in L Nl and such that [c]-e l ~ [x]-e l , Le. such that (c A x, c) E -e l . Let m be the smallest positive integer such that [c]-e l = [f2m(c)]-e l , noting that m ~ n. Define

e= { Then [c*]-e l

f2 (c) V ... V f 2m - 2(c)

o

= [c Ve]-e l ; and (c, c A x) E -e l

~ m > 1; ifm=1.

gives

(eve, (cAx)ve)E-el~-e2' But ((c A x) V e, (c Ay) V e) E -e 2 and so (c ve, (c Ay) V e) E -e 2. Since (c*, c V e) E -e l ~ -e 2 we have that (c*, (c A y) ve) E -e 2. Since c* E T2(L) it follows by the observation above that (c*, (c A y) Ve) E -e 1 and therefore (c ve, (c Ay) ve) E -e l . Thus(c, (c Ay) V (c A e)) E -e l and so [c]-e l

= [c Ay]-e 1 V [c Ae]-e 1 .

If m = 1 then e = 0 and we obtain [c]-e l ~ [YWI' If m

> 1 then

[c]-e 1 = [c Ay]-e 1 V [c Af2(C)]-e 1 v··· V [c A f 2m - 2(c)]-e 1 and so, since [c]-e 1 is V-irreducible in LNl' either[c]-e 1 ~ [y]-e 1 or there exists k with 1 ~ k ~ m -1 such that [c]-e 1 ~ [f2k(c)]-e 1 • Since f is injective, the latter gives the contradiction [c]-e 1 = [f2k(c)]-e 1 with 1 ~ k ~ m -1. Hence we have [c]-e 1 ~ [YWI' Similarly, every V-irreducible contained in [y]-e 1 is contained in [x]-e 1 whence [x]-e 1 = [y]-e 1 and consequently -e 1 = -e 2. Suppose now that L is arbitrary and that (x,y) E -e 2 . Let A be the subalgebra of L generated by {x, y }. Since Kn 0 is locally finite by Theorem 3.7, A is finite with T2 (A) = A n T2 (L). By the above, we deduce from

46

Ockham algebras

'l9 l IT2 (A) = 'l9 ZIT2 (A) that 'l9 l l A = 'l9 z lA- Since x,y E A we then have (x,y) E 'l9 l . Hence 'l9 l = 'l9 z as required. 0

Theorem 3.12 If LEO then T(L) is a strong extension of Tz(L). Proof Let i(J E Con Tz(L) and let i(Jl, i(Jz E Con T(L) be extensions of i(J. Suppose that x,y E T(L) are such that (x,y) E i(Jl' If A is the subalgebra of T(L) generated by {x,y} then A E Kp,o for some p. Since i(JlIT2(L) = i(J = i(JZIT2(L) and Tz(A) ~ Tz(L) we have i(JlIT2(A) = i(Jzlrz(A)' By Theorem 3.11, A is a strong extension of Tz(A). Hence i(JlIA = i(JZIA and consequently (x,y) E i(Jz. Thus we have i(Jl ~ i(Jz. Similarly, i(Jz ~ i(Jl and so i(Jl = i(Jz as required. 0 Corollary Con T(L)

~

Con Tz(L).

Proof Immediate from the above and the congruence extension property. 0 We can now use the above results to determine precisely when [
Theorem 3.13 If L E Kw then [
Con L/
The result follows now from the above result of Sankappanavar. 0

Definition We shall say that Con L has comonolith Con L

= [w, CI!]

CI!

whenever

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; (l) K(L) = Tz(L); (2) Con L has comonolith
°

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

19(c, d) =

L

k=O

Suppose now that ep

E

Con L is such that ep =f

L

Since ep

= V

19(a, b)

(a,b) E 'P

we have (a,b)Eep

=}

19(a,b)=f£.

=}

(a,b)EcI>w·

But by the above observation 19(a,b)=f~

Hence ep ~ cI>w and consequently ConL

= [w,cI>w]EB {~}.

(2) =} (3) : This is clear. (3) =} (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) L/cI>w 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 Con L = {w, £} and so cI>w = w since (0, 1) ¢ cI>w. Consequently, cI>1 = w whence f is injective; and cI>w is maximal, so K(L) = T 2 (L). (2) =} (1) : If (2) holds then clearly cI>w = wand, by the above, Con L reduces to {w, ~}. 0 Corollary 3 The simple algebras in Kp,oforwhich K(L) = T 2 (L). 0

Kp,q

are precisely those algebras L in

Note that Corollary 2 above does not extend beyond Kw. This fact is illustrated by the pineapple (Example 2.4).

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

Theorem 3.16 If LEO is subdirectly irreducible and

if f

is not injective

then the monolith of L is
Proof If f is not injective then
f3 in ConlatL. Now f3 1\
= f3 1\
Theorem 3.17 L E Kw is subdirectly irreducible ifand only if Con L reduces to the chain

More precisely, (1) if L belongs to a Berman class and VB(L) = Kp,q thenL 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, «pw] E13{~}.

We now show that the subalgebra f(L) is also subdirectly irreducible. In fact, suppose first that «PI = w. Then we have

Conf(L)

~

Con L/«P 1 = Con L,

whence f(L) is subdirectly irreducible. Suppose now that «PI be the restriction of «PI to f(L). Since (t(x),f(Y)) E «Pt ~ f2(X)

= f2(y)

f

wand let «Pt

~ (x,y) E «P2,

it follows that

«Pt = w ~ «PI

= «P2 = ... = «Pw '

Thus, if «Pt = w then Con L reduces to the three-element chain

w -< «PI

= ... = «Pw -< L

It follows by the congruence extension property thatf(L) is also subdirectly irreducible. On the other hand, if «Pt f w then exactly the same conclusion follows by Theorem 3.15, the monolith of Con f(L) ~ Con L/ «PI being «Pt by Theorem 3.16.

In conclusion, f(L) is subdirectly irreducible, whence so are all r(L) since for every i we have thatp(L) = f[P-1(L)]. We thus have Con L = {w} EI1 [«PI' d with

[«PI, d ~ Con L/«P 1 ~ Conf(L).

Similarly, Con L = {w} EI1 {«PI} EI1 [«P 2, ~] with

[«p 2, ~] ~ Con L/ «P 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

= Kp,q

w = «Po -< «PI -< «P2 -< ... -< «Pq -< L If on the other hand L belongs properly to Kw then since there are infinitely many «Pi' with «Pi+1 covering «Pi and the supremum of the «Pi being «Pw , we conclude that Con L is the infinite chain w = «Po -< «PI -< «P 2 -< ... *= : This is clear.

< «Pw -<

~.

<>

Corollary 1 If L E Kw is subdirectly irreducible then so is every subalgebra of L. <>

50

Ockham algebras

Kw then L is simple if and only if L is subdirectly irreducible and 1 is injective. <>

Corollary 2 If L E

Example 3.5 (The 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,

(Vi

~

1) I(xj) = Xj-1

and extending to the whole of L. It is readily seen that L E Kw. Moreover, every congruence that identifies any adjacent pair also identifies x 0 and x 1 . Consequently the smallest non-trivial congruence is <1>1 = 19(X1, xo), and L is subdirectly irreducible. The congruence w has classes {O}, {I },L \ {O, I}. The reader can easily verify that ConL is the infinite chain

w ~ <1>1

~

<1>2

~

...

< w

~ t

where <1>. I

={

19(xo, Xj) if i is even; 19(xj,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 L

=

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 b j, define

f(O) = 1, f(l) = 0, f(CY.) = CY.; f(a j) = b j, f(b j) = a j- 1 , f(bd = 0; { f(at) = bt, f(bt) = at-I, f(b!) = CY.. 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 f to a dual endomorphism on I. Observe thatf2j(b j) = 1 sof2i+l(a j) = 1; andj2j(bt) = CY. so j2j+l(at) = CY.. It follows that L E Kw. It is readily seen that any congruence on L that identifies a pair x,y E P 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 1 , 1. Hence Lis subdirectly irreducible with monolith <1>1 = 'I9(b 1 , 1). As for w' we have L/w ~ {O, CY., 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. af 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 (Xi 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) => (m) => (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, I-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 ~ A nQ = BnQ

is a congruence. Note that A n Q = B n Q is equivalent to A~ B ~ Qf where means 'symmetric difference' and Qf 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 ofclosed 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);f)

is an Ockham

algebra where (VA E O(X)) Conversely, where

if (L;f)

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

= {a E L If(a) ~ x}.

Moreover, these constrnctions give a dual eqUivalence. <>

Ockham algebras

54

. We now list some further concep ts and terminology that we shall require iant, A subset Q of an Ockham space (X; g) is called a g -subset if it is g-invar in the sense that XEQ =? g(X)E Q. s Q, For every Q ~ X we denote by gW(Q) the smallest g-subse t that contain namely et Q The lattice of all closed g-subsets of X will be denote d by G(X). A g-subs be will {x}, gW of the form gW ({x} ), which we shall write in the simpler form educible called monogenic. Clearly, the monog enic g-subsets are the join-irr elemen ts of G(X).

then Theor em 4.2 Let (X;g) be the dual space of (L;j) EO. If Q E G(X) the eqUivalence relation {}Q defined on O(X) by (A,B)E {}Q

<=?

AnQ= BnQ

is a congruence. Moreover, the lattice Con L is dually isomorphic to G(X). 0 ence For each Q E G(X) the congru ence {}Q will be called the congru . boolean is G(X) if only and if n boolea is associated with Q. Clearly, Con L Q \ X if only and if mented More generally, the congru ence {} Q is comple We thus belong s to G(X), Le. X \ Q is a closed g-subset, whenc e Q is open. have the following

only CoroIIai-y If Q E G(X) then the congruence {} Q is complemented ifand 0 clopen. be to Q s if X \ Q E G(X), a condition that reqUire shall Now in this respect the finite case is worthy of consideration, so we is purely illustrate matters by revisiting Theore m 2.17, the proof of which genera l more is that nt stateme a algebraic, and proVide a very short proof of r Chapte ating than the theorem in question. For this purpos e, and anticip hat 5, we introduce a classification of the subvarieties of 0 which is somew closely sharpe r than that used by Berman. It is also due to Urquhart and is related to the dual space. those For m > n ;;;: 0 denote by P m,1l the subclass of 0 that is formed by r Chapte algebras whose dual space satisfies gm = gil. It will be established in (in other 5 that when m - n is even we have L E P m,1l if and only if fm = and only if m,n P E L have words, PZp+Il,1l = Kp,n); and when m - n is odd we if fm(a) and r(a) are complementary for every a E L.

r

55

Duality theory

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

(1) L E Pn 0 for somen; (2) g is stlrjective; (3) g is injective; (4) Con L is boolean with k atoms, k being the number ofmonogenic gsubsets of X, which attains its maximum value precisely when L is boolean.

Proof (1) =? (2): If L E Pn,o then gn = idx · (3) : Since L is finite, so is X. (2) (3) =? (4) : Let x 1 be an arbitrary element of X. Since X is finite and g is injective, there exists n 1 such that gn l (x 1) = X l' If gW {x d 'f X we can choose Xz EX \gW{xd; then there exists nz such that gn 2 (xz) = xz, 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,

Qz

= gW{xz},

... ,

Qk

= gW{Xk}

of respective cardinalities n1, nz, ... , 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{n1"'" nk} then we see that gn(x) = x for every x E X. Consequently, L E Pn,o, <>

qN

S

p

r

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

pqrst

g(x) s

r q

p t

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

= {P,s},

Qz

= {q,r},

Q3

= {t}.

Ockham algebras

56

Hence Con L ~ 2 3 . The algebra (L;f) is described as follows :

k

a 0 abc d e h j k I 1 f(a) 1 k I h i j e d e abO

h c

a

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

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

I-::p_-,q,--_r_s_ _t__ u_ g(x) u u s 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 +7 1) 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 - t 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 contain p. It follows lhat G(X) ~ 1 EB 2n - 1 . If (L;1) is the dual algebra then Can L ~ 2 n - 1 EB 1, the coatom being the congruence associated with {P}. Since gZ (L;1) E PZ,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 pi. Then it follows that we have (Va. E d i ) f(a.)

= 0;

In particular, if p ~ x for every x E X then f(a.) = 0 for every a. 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 thel) f(a.) = 1 for every x E L \ {l}, whence 1 is join-irreducible and L is a dual Stone algebra.

Ockham algebras

58

In Chapter 2 we introduced the congruences cI>n defined for every n by

(x,y)

E

cI>n

~

PZ(x)

E IN

=pz(y).

In the dual space (X; g) we can consider for every n E IN the g -subset gn(X) = {g1l(x) I x EX}. We can also consider the g-subset

gW(X) =

n g1l(X).

11;;.0

rff Clearly, all gn are continuous maps and, since X is a compact Hausdo the enjoys gW(X) t g-subse The space, allgn(X) are compact, hence closed. same property. Obviously, these closed g-subsets are related by X:2 g(X):2 g2(X):2 ... :2 gW(X):2 {a, I}. The following result [55] elucidates the correspondence betwee n the congruences cI>n and the closed g-subsets gn(x).

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

°

= X. As for cI>l' we have the associated with g(X) : is cI>l that following equivalences which show

Proof Clearly, cI>0 is associated with gO (X) (a, b) E cI>l

f(a) = f(b) ~ X\g-l (A)=X \g-l(B ) ~ g-l(A) = g-l(B) ~ A ng(X) = B ng(x) ~ (A,B) E 19 9 (x}.

~

), Taking into account the fact that, for every n,pz+l(L) is a subalgebra ofPZ(L with ted associa is n cI> that show can and using an easy inductive argument, we g1l(X). Finally, cI>w =

V cI>n is associated with n;;.O

n gn(x) = gW(X). 0 n;;.O

It Example 4.4 Let IN"" consist of INo with an additional point 00 adjoined. one if is known (see [9] for the details) that IN"" becomes a Priestley space

takes as a sub-basis for the topology the subsets U such that U ~ 00 or (U:3 00 and U' is finite).

Duality theory

59

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

V 3 00

The clopen subsets of INoo 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 INoo is discrete, g is trivially order-reversing. It is easily verified that g is continuous. It follows that (IN oo ; g) is an Ockham space which we can represent as follows : e----+) e----+) e----+) e----+l ... e O

I

3

2

4

00

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

and {n,n+l,n+2, ... ,00}=g'Z-1(INoo }'

The closed g-subsets are 0, gn(IN oo } for all n

~

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

gO(IN oo } = INoo ::J g(lN oo } ::J g2(IN oo }::J '" ::J gW(IN oo } ::J 0, and so, if (L;/) is the dual algebra, Con L reduces to the chain

w --( <1>1 --( <1>2 --( '"

< w

--(

~

whence L is subdirectly irreducible. Now (again see [9]) it is known that INoo 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;/) does not belong to any Berman class, but does belong to Kw. For this purpose, let U be a finite subset of INoo that does not contain 00 and let z be its numerically greatest element. Since

/2n(U} = {x E INoo I g2n(x} E U}, we see that /2n (U) is the subset of INoo obtained from U by translating U to the left through an amplitude of 2n and deleting the resulting elements that fall outside INoo' From this observation, we have that if z = 2n then pn(u} = 0 and therefore /21l+2(U} = /2n(u}; and that if z = 2n + I then /2n(u} = {I} and/ 21l +1(U} = INoo, and therefore/ 21l +3(U} =pn+1(U}. The above example can be modified slightly. Consider the effect of taking g to be the constant map given by ('
g(n} = g(oo} = 00.

60

Ockham algebras

Here the closed g-subsets are 0 and all the subsets of IN oo that contain Consequently we have Con L ~ 21N EEl {~ }, the comonolith being irreducible.

<1>1'

00.

In this case, (L;]) E PZ,l and is not subdirectly

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 {~},

the comonolith being <1>1' Since for every a follows that (L;]) is a Stone algebra.

E

L \ {I} we have f(a)

= 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 from R ~ X if Q n R and Tn R::f 0 for every T E G(X) with T:J Q.

=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 t9(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 t9(a, b), we have Q ~ (AilB)'

= (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 t9(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 €(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} to] = L \ {I}. We shall denote by 12n (resp. 12n - 1) the prime ideal that separates X2n and X2n+l (resp. X2n-l and X2n)' The dual space X is then the following ordered set. J

~ IS

1-2

This becomes a Priestley space if we take as a sub-basis for the topology T the subsets II! \ {j}, If! \ {{On, I~m U I~n+l' and I~m U I~n+l where m, nEll. The clopen down-sets of X are 0, X, and UI~ for k E 271 and € E 271+ 1. Then we define g as follows :

Ii

(Vn E 7l) gUn) = In-I, g(J) = {O}, g({O}) =j.

10

(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 {In I nEll}. 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(x2n-l, X2n) is represented by the closed g-subset that is maximally disjoint from {I2n-l}, Le. the closed g-subset generated by 1 2n - 2. Similarly, we see that 19(X2n+l, X2n) is represented by the closed g-subset generated by 12n - 1. The only non-principal congruence 'P corresponds to {{O},]}, and Con L is the chain

'V

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 :

e 1 '--r-'Y'~--i=f 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

{a6,b6}

A

{a6,d6}

F

{aj,dj}

A'

{b6,C6}

F'

{bj,cj}

g(X)

{C6,d6}

B

{a6,bj}

G

{a6}

B'

{b6,aj}

G'

{b6}

g2(X)

{aLbj}

C

{C6,bj}

H

{C6}

C'

{d6,aj}

H'

{d6}

[

{aj}

D'

{b6,cj}

['

{bj}

J

{cj}

E'

{d6,cj}

J'

{dl}

g3(X)

{cLdj}

D

{a6,dj}

g4(X)

{e! }

E

{C6,dj}

The Hasse diagram for Con L is the same as that above with X replaced by K replaced by the corresponding congruence 19K , w; 0 replaced by ~; each gj (X) replaced by j; and each g-subset

Using Theorem 4.5, we can verify that there are 8 non-principal congruences, namely 19 c ,19 0 , 19 F , 19 p , 19H , 19H " 19J , 19],. Moreover, we have

= 19(0, al /\ bd <1>4 = 19(a4 V b4, 1) <1>1

19B = 19(b z , b1 ) 19 D , = 19(az,a 3) 19 E = 19(O,al/\ b3) 191' = 19(a4 V b z , 1)

z

= 19(az V bz , 1)

19A = 19(bo, bd 19B , = 19(az,a 1 ) 19 G = 19(b 4, b3) 19E' = 19(0, a3/\ b1 )

<1>3

= 19(0, a3 /\ b3)

19A , = 19(ao,ad 19D = 19(b z , b3) 19 G, = 19(a4,a3) 19[ = 19(az V b4, 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 i ) = ai-I, f(b i ) = bi- I ,

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

r=

U i~

neN

and s

= U k~. neN

They form two chains eland C 2 as in the following diagraJll. 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 i, s i, r!, s! are not open.

65

Duality theory ·L

Jo ·L

11 ·L

h

The clopen down-sets of X are 0, X, OIL, {3L, and OIL U {3L where OlE C 1 \ {r} and {3 E C z \ {s}. The action of g on X is described by

i5 i5 d ii i! k5 15 ki Ii k! r.

-t

-t

-t

-t

-t ...

(1)

-t

-t

-t

-t

- t . ..

(2)

rrts

.s

The g-subsets of X are easily determined: they are 0; those generated by any element "y of the sequence (1); those generated by any element 8 of the sequence (2); those generated by any pair {"y, 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

The diagram for Con L is the same with the label change X gW(X) - t
gW(X)

-t

w, gi(X)

-t


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

gives the comonolith
66

Ockham 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

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

n 0 abc del

x ConL It is easily seen that <1>1 = 19(e, 1) corresponds to {p,q,r,s}, 19 1 = 19(O,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}, {e, I}} 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].

4.1 Let (X;g) be an Ockham space. Tben a subset Y of X is a gsubset if and only if Y ~ g-1 (Y).

Lemma

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. 0

67

Duality theory Lemma 4.2

If Y is a g -subset oj X and Y

Proof Y ~ Q gives g-l (Y)

~

Q then Y

~ g-l (Q).

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

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

0

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)}

is a sub-base of X, the closure .

Y = n{A' UB I A,B E O(X), Y ~ A' UB}. Now for A' U BEY we have g-l (A' U B) = g-l (AI) U g-l (B)

=J(A) U If(B)l'

and, by Lemma 4.2, Y ~ g-l (A' U B). It follows that g-l (AI U B) E Y. Thus g(Y) ~ Y and so Y is g-closed. 0

Corollary gW (Y) is the least closed g -subset that contains Y. 0 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 ifand 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.O

Corollary 1 If (L;J) E 0 isfinite then (L;J) is subdirectly irreducible ifand 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; Le. if and only if there exists x

E

X such that gW{x}

= X. 0

68

Ockham algebras

Corollary 2 If (L;/) E 0 is finite then (L;/) is simple ifand only ifgW {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. <>

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) :

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 mil is denoted by Lm,n' Clearly, the end of X is unique except in the case where X reduces to a loop; i.e., 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 L m,1l is an (n + 2)-element chain.

<>

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 m n onto X and, by the duality, corresponds to an injective Ockham algebra mor,phism. We therefore have : Corollary 4 Afinite Ockham algebra is subdirectly irreducible ifand only if it is isomorphic to a subalgebra of L m,1l for some m and n. <>

69

Duality theory

We shall now apply the preceding results to describe the subdirectly irreducible algebras in the class P 3 ,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 subdirectly irreducible algebras in P 3 ,1 are the subalgebras of L 3,1, Le. 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~l a

h c

o with the operation f defined by

x f(x)

abc d e h 1 lIb e b e 0 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 PI ,0, P z,0, P z,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 subdirectly 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 f -fixed points in L. By Corollary 3 of Theorem 4.6, we know that Con L has 2 elements if L E P 1,o or L E Pz,o, and 3 elements if L E P Z,l or L E P 3,1' In the latter case we indicate the classes of the only non-trivial congruence $1'

Ockham algebras

70

Ockham algebras

Ockham spaces Pl,D

I~

@

r P 2 ,D q_~_r

hOe 0

H

tn~ P 2,1

~1~

I~

·Oh

p_-@r

0

P 3,1

rlt

t~

~1:

f~

B

M

K

5

51

Kl

K2

Duality theory

71

P3,1

continued

P?q r

q.~I:

b

h«e :$!e h d1

L

&h

N

c

0

PV q

r

p.-ln:

a

p

q

.. r

Ockham spaces

M1

b

0

.-.~

K3

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 L and a'L inheritfrom L the structure ofOckham algebras.

a L define fa (x) and fa (a) = O. If u, vEaL then

Proof For every x

E

= f(x) 1\ a.

= a' then

Then we have fa (0)

=a

fa (u 1\ v) = (t(u) V f(v)) 1\ a = fa (u) V fa (v),

and dually fa (u V v) = fa (u) 1\ fa (v). Hence (aL;fa) is an Ockham algebra. Since f(a) I\f2(a) = 0 and f(a) V f2(a) = 1, we have that f(a') = f2(a) = (t(a))' = a" = a. It follows by the first part that a'L is also an Ockham algebra. <:;

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'. ~ : Suppose that there exists a E Z(L) \ {O, I} such that f(a) = a'. By Theorem 4.7, a L and a'L are Ockham algebras. Define h : L ----7 a L x aiL by h(x) = (x 1\ a, x 1\ a').

Proof

It is well known that h is a lattice isomorphism, and since

f(h(x))

= (t(x 1\ a) 1\ a, f(x 1\ a') 1\ a')

=([J(x) V f(a)] 1\ a,

[J(x) V f(a')] 1\ a')

=(t(x) 1\ a, f(x) 1\ a') =h(t(x)), it follows that h is an Ockham isomorphism.

=> : Suppose that L = U x V with U, V t= 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, 0) = (t u(u), fv(O)) = (0, v), the result follows. <:; The 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' of L are represented in the dual space X by two clopen down-sets A, B such that A nB = 0 and A UB = x. If f(a) = a', Le. f(A) = B, it follows that X \g-l(A) = B, hence g-l(A) = A. Since f(a) = a' 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 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 Xl q P r s X2 q q r s X'2 P P r s X3 q P s r X4 q q s r X'4 P P s r

q X'5 P X6 q X7 q X'7 q Xa q Xs

X~

q P P P P q p P

X9 r r X'9 r r X 10 r r

q q Xi1 P P

X ll

0 1 a b c d e f g h i i h g f e d c b i h h f e e c b i h f f e c c b i e d c h g f b i e e c h h f b 1 0 i i e c c h f f b 1 0 1 i 1 1 i b b 0 b 1 0 i i 1 i i b 0 0 b 1 0 j i 1 j i b a 0 b 1 0 j 1 0 1 1 0 i 1 0 j 1 0 1

r r r r r r q s 1 0 j P s 1 0 d q s 1 0 1 P s 1 0 c p s 1 0 e q s 1 0 1 r s 1 0 1 p s 1 0 e q s 1 0 i

c c c c

h g 0 1 h a 0 1 h h 0 1 h 0 0 1 e h b b i e h h b i 1 h h h c

j a L1 b L2 0 a b 0 b 0 a

j c h g d c h a 1 c h h c c h 0

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

f f

0 f 0 0 h b h f

L~

L3 L4 L'4 Ls L'5 L6 L7 L'7 La L'a L9 L'9 LlO Lll L'l1

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

74

Ockham algebras

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 1\ B of two subvarieties A, B is their intersection, and the join A V B is the equational closure of their union, Le. the smallest subvariety of V that contains AU B. A celebrated theorem of B. Jonsson [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 Af(O) the set of subvarieties of 0 that are generated by finite algebras. In fact, Af(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; M1 = MS of MS-algebras, and M1 = MS of dual MS-algebras. Here and throughout what follows we ignore the trivial subvariety. As a framework for Af(O) we may choose either the Berman classes Kp,q 1, q ~ 0) or the classes Pm,n (m > 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 = Pzp+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

~

Pm,n~Pml,nl

-¢===l>

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,n then the framework at the bottom of Af (0) can be depicted as follows :

Here the label m, n means Pm,n, and the solid circles indicate those that are Berman classes. The Pm,n 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 rva 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 /\, V and rv. A term is atomic if it is of the form rv 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 /\ rv 2 b ::;; rva V rv 3b V c is a basic inequality, as is also rva /\ rv2a

= O.

77

!be lattice ofsubvarieties

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 I\.....,b 1\ .....,zd::;;; .....,za V rvC V rvzC 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

1\

b) = rva

V

rvb,

rv(a

V

b) = rva

1\

rvb

we can convert A and B into equivalent terms A' 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. 0

Example 5.1 The inequality rvZ(a V rvb)::;;; (a 1\ rv(b 1\ c)) V rva

can be written rvZa V rv 3b::;;; (a V .....,a) 1\ (rva V rvb V rv c)

and is equivalent to the conjunction of the four basic inequalities rvza::;;; a V rva, .....,3b::;;;avrva,

rvza::;;; rva V .....,b V rvc, rv3b::;;;rvav.....,bvrvc.

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, ql), (Pz, qz), ... , (Pn, qn)},

Ockham algebras

78

and let

A = {,Jtx i IPi even} U {rvqtXi I qi odd}; B= {,JtXi IPi odd}u{rvqixi I qi even},

where Xi and x j are distinct variablesfor (Pi,qi) Consider the basic inequality

t= (Pj,qj)'

and the universally quantified disjunction (VX)

(**)

V{gPi(X)~gqi(X)

I (Pi,qi)EE},

i

Then (L;rv)EOsatisfies (*) ifandonlyif(X;g) E X satisfies (**). Proof The inequality (*) fails in L if and only if there are clopen down-sets Ai such that ,JiAi n rv qi Ai C1 U ,JiAi U U rv qi Ai' Pi even qi odd Pi odd qi even

n

n

which is the case if and only if there exists

x x

E E

,JiAi rvqiA i

~ ~

X E

X such that

Pi is even, qi is odd.

Since we have that

.-'----'-. { grri.(X) E Ai 1.'f ri 1.'S even; .....--r g '(x) ~ Ai If ri IS odd, the above conditions are equivalent to the existence of down-sets Ai and an element x such that

x

E

rv ri Ai

This in turn is equivalent to the existence of an element x with the property that gPi(X) 'j gqi(X) for every i, which is the case if and only if (**) fails in

X,<> We note that if A = 0 then /\ A = 1, and if B = 0 then VB = 0. 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 ~ and (L; rv) (1) when m - n is odd we have

L E Prn,n

~

E O.

Then

(Va E L) rvrna 1\ rvna = 0, rvrna V rvna = 1;

The lattice of subvarieties

79

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

~

(Va

E

L) rvma

= rv'la.

Proof (1) Consider the equalities rv ma 1\ rv na = 0 and rv ma V rv na = 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 ~ gn, and the second is equivalent to gn ~ gm. Their conjunction is equivalent to gm = gn, Le. to L E Pm,n' (2) The equality rvma = rvna is the conjunction of the inequalities rvma ,,; rvna and rv1Za,,; rvma. For m and n of the same parity, these are equivalent by Urquhart's theorem to gm ~ gn and g1Z ~ gm, Le. to gm = glZ. <> 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), (1, 2), (2, O)}. We have A = {rvoal,rvza3}, B = {rvzal, rvzaz, rvoa3, rv1az}.

Writing a, b, c for a 1 , az, a3 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, I)}. Then A

= {rvza, rva} and B = 0. We have

which is equivalent to

(Vx) gZ(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\ rvza,,; b V rvb V rvzb.

80

Oekham algebras

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

a /\ ",a /\ ",e /\ ",2 e :::;; b V ",b V ",d V ",2d

which is simple and corresponds to the set

= {(O, 1), (1,0), (2, 1), (1, 2)}. it is clear on taking e = a and d = b that (K') implies (K). E

In fact, To obtain the reverse implication, write a V e for a and b /\ d for b in (K) to obtain (a V e) /\ ",a /\ ",e /\ (",2 a V ",2 e):::;; (b /\ d) V ",b V ",d V (",2b /\ ",2d),

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 Mg(x) implies g(x) Mg2(X), the equivalent of (K) is (\ix)

g(x) Mg2(X).

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

(K")

",a /\ ",2 a :::;; ",b V ",2b.

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 ",b V ",2b

= ",2 a V ",b V ",2b

is the conjunction of the inequalities a V ",b V ",2b:::;; ",2 a V ",b V ",2b

¢=>

a:::;; ",2 a V ",b V ",2b

(1)

¢=>

rv2a:::;; a V ",b V ",2b

(2)

and a V ",b V rv2b;;;:: rv 2a V ",b V ",2b

Both (1) and (2) are simple and are respectively equivalent to (\ix) x;;;:: g2(X) (\ix) g2(X);;;:: x

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

The conjunction of these two disjunctions is (Vx)

x

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

The lattice ofsubvarieties

81

Remark For both visual and typographical reasons, we shall henceforth commit an abuse of notation. Specifically, instead of writing the universally quantified disjunction (Vx) V{gPi(X);;::gqi(X) I (Pj,qj)EE} j

we shall write simply

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

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

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

V g;;:: g2; 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)

/\ (g2 ;;:: gO

V gO;;:: g),

and gives the simple inequalities a /\ b /\ rvb ~ rv 2a ~ a /\ b /\ rvb ~ rv2a /\ b /\ rvb, rv 2a /\ b /\ rvb ~ a ~ rv2a /\ b /\ rvb ~ a /\ b /\ rvb.

The conjunction of these gives the equality a /\ b /\ rvb

= rv2a /\ 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 even P's of A, the odd q's of A, the odd P'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 Ci.ij' By way of illustration, corresponding to Examples 5.2-5.6 above we have the following tabulations :

Let T be a tabulation with n rows. If we delete r of these rows (with

o~ 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 ofsubvarieties

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 ~ gll-2k and gll-2k-l ~ gm-2k-l implies gm ~ gil. A tabulation is non-trivial if it is such that

(Vi) Cl!il ::f: Cl!i4, Cl!i2::f: Cl!i3' Otherwise, the corresponding inequality would be trivial, in the sense that it would contain a term of the form rv'l a on each side. The dual (nd) of an inequality (n) is obtained by replacing /\, V, ~ ,0,1 respectively by V, /\, ~ , 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

2 2 1

0

o

2

and represents the inequality

rv 2 a /\ rvb /\ rv 2 b /\ 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 5.4 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 = ... =Pro

,JJl a l

The simple inequality I is then of the form

/\ ",ql al /\ ,JJl a2 /\ ... /\ ,JJl 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 rvqral VB.

In fact, if in (1) we replace a2, ... , a r by al then we obtain (2). Conversely, if in (2) we write a 1 /\ a2 /\ ... /\ a r for al then we obtain ,.Jl al ~

,, I 'jl

/\,.Jl a2 /\ ... /\ ,.Jl a r /\ (",ql al V ",ql a2 V ... V ",ql ar) /\ A

(",qZa 1

/\

rvqZa2 /\ ... /\ rvqZa r ) V··· V (rvqral /\ ",qra2 /\ ... /\ ",qra 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 r()~ remaining unaltered, we thus see that (2) implies (1). .. The same procedure can be applied if an entry appears r times ill we third column. As for columns 2 and 4, the only change required in the apove proof is the substitution a 1 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

1 2 1

0 0 -

-

2 1 3 5 1 -

and the simple inequality

a /\ ",a /\ ",2 b /\ ",b /\ c /\ ",d /\ ",5 e ~ ",2 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 /\ rv 2 b /\ ",b /\ ",5 e ~

",2 a V

rv

3b V

~

rv

3

= Q!42

011

",e,

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

Q!22

Q!12

a V '" 2c V ",e,

= Q!22 = Q!42'

85

Tbe lattice ojsubvarieties 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

(O, 1), (I, 0), (O, 2), (2, 0), (I, 2), (2, 1). The biggest tabulation possible in this case is therefore

o

1

1 0

o

2

2 1 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 -

-

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

I- ""a

V ",,2 a

1

2

I

= 1.

• There are 15 subtabulations consisting of two rows, but we must exclude two of them, namely those that contain (O, 1) and (2, 1), or (I, 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 (I) and (l d ), is added for sake of completeness.

86

Ockham algebras

Tbe lattice ofsubvarieties

87

88

Ockham algebras

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 /\ (rv2 a V rvb) /\ (rv2 a V rv3b) ~

(rv3 a /\ rv2b) va V rvb V (rva /\ rv2b) = B.

Since A ~ rva /\ rv2a /\ band B ~ a V rvb V rv2b, (1]) follows. In contrast, a glance at the dual equivalents of () and (1]) yields a straightforward proof: the tabulation for (() is a subtabulation of that for (1]); equivalently, since g is order-reversing, g 1f gO implies g 1f 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 (a) which is the conjunction of (1) and (1 d)' We write this conjunction in the form a = (1, 1d)' Other conjunctions worthy of note are the following : (1, /3) (3,4)

= (2 d ), (l d , /3) = (2); = (1), (3d,4 d) = (l d);

(3,4 d )=(15) : aV rv a= rv2 av rv a; (3d,4) = (15 d) : a/\ rv a= rv2 a/\ rv a; (3, 6 d ) = (16) : a V rvb V rv 2b = rv 2a V rvb V rv2b; (3d, 6) = (16 d) : a /\ rvb /\ rv 2b = rv2a /\ rvb /\ rv 2b.

89

The lattice ofsubvarieties

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 1\ rva) V b V rvb

=(rv2 a 1\ rva) V b V rvb.

To establish this, observe that dearly (17) implies (9), and also implies a V b V rvb ~ rv 2a 1\ rva, which is readily seen to be equivalent to (12 d ). Conversely, (9) gives

(a

1\

rva) V b V rvb ~ rv2a V b V rvb,

(a

1\

rva) V b V rvb ~ rva V b V rvb,

and since dearly

we have, by distributivity,

(*)

(a

1\

rva) V b V rvb ~ (rv2 a 1\ rva) V b V rvb.

Now (12 d ) yields and since clearly

we obtain

(**) (rv2 a 1\ rva) V b V rvb ~ (a 1\ rva) V b V rvb. Finally, (*) and (**) together give (17). It follows that the dual equivalent of (17) is

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

~

g)

This shows that (17) is self-dual and is implied by each of the axioms (01) and (ry). 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 1\ rva) V rvb V rv 2b = (rv2 a

1\

rva) V rvb V rv2b,

and hence also (12 d , 11) = (18 d )

:

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

The dual equivalents of (18) and (18 d) are respectively

=gO V g ~ g2 V gO ~ g, g2 = gO V g2 ~ g V g ~ gO. g2

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 every 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(P3,1)'

Theorem 5.4 Let K = Var S be a congruence-distributive variety generated by a finite set S offinite algebras, and order the set Si(K) ofsubdirectly irreducible algebras in K by A

~

B

-<===}

A is a homomorphic image ofa subalgebra of B.

Then A(K) is a finite distributive lattice and is isomorphic to 0 (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 every 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 / ep where ep 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

91

!be lattice ofsubvarieties

B

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

Sl S

(2) (a,1) (a) ((3) (2 d )

S

K\

N

(1,4 d ,1) (1,3d,1) (I, 5, 6d ) (3d,4,1) = (15 d,1) (3d, 5d, 6) = (5 d, 16 d)

M1

(I)

M1

K1

K2 K3 L

K2

K3 [ N

(2) (ld,4,1) (1d,3,1) (l d , 5d , 6) (3,4 d ,1) = (15,1) (3, 5, 6d ) = (5, 16) (l d )

The task of determining A{P3,1) 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. !ben, for every x E F,

92

Ockham 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.

93

The lattice ofsubvarieties

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 81 , 82 , ... ,Sn then an equational basis of 8 1 V 8 2 V ... V 8n is obtained by taking the intersection of the sets of axioms satisfied by 8 1 , 8 2 , ... ,8n and then deleting those axioms that are consequences of others. We illustrate this procedure with some examples. Consider for instance the subvariety MVK2 VK3 . We know that equational bases for these three subvarieties are, respectively,

(a),

(ty,I,3 d ),

(I,5,6 d ).

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

B=')'iUliU3~,

C=IiU5iU6~

where n i denotes the up-set generated by axiom (n) in the ordered set of implications. Now since AnB = (aih')'i)u(ainli)u(ain3~)

= 9iU9~UliU3~ an equational basis of M V K 2 is (1, 3d, 9, 9d)' But, using the g-inequalities, we see that (1 ,3d) implies both (9) and (9 d)' So an equational basis of MVK2 is (I,3d)' Now we consider (1 i U 3~) n (1 i U 5 i U 6~) = 1i U (3~ n 5 i ) U (3~ n 6~)

= 1i 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 ain(Iiu5iu6~) = (ainli)u(ain5i)u(ain6~) =IiU6~,

from which we see that an equational basis for M V K3 is (1, 6d)' Using these observations, we can now establish an equational basis for M V8

Clearly, this is (1, 3d, 6d)·

= (M V K 2) 1\ (M V K 3).

Ockham algebras

94

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 ",a 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

f3 3d 4d

lId 8

6d 12 d

rva /\ rv 2a = 0 rva /\ rv 2a ::;; a rv2a::;; 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)= (1,5), (1,f3)= (2 d), (1,12 d,8)= (I,')'), (1,3d,8)= (1, 3d,')'}, (l,4 d,8) = (1,4 d,')'), (a,')'} = (a,8), (a,f3) = (2).

The lattice ofsubvarieties

95

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

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

Adam~Priestley

M1

(I)

MVKz VK3

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

MVK3 MVK1 VKz K z VK3 MVKz MVK1 S SVMVK1 K3 K 1 VKz MVS SVK1 Kz M K1 SVK K B

Blyth-Varlet (I)

(l,lld) (1,6 d) (1,12 d ) (1,5) {I, 3d)

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

(l,6 d ,')') (l,3 d ,')') (a) (l,4 d ,')') (l,3d,6 d ,')') (a,')')

(2)

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

B

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 ofsubvarieties

97

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

L

(ry)

L VK1 VKz LVKz LVSVK1 LVK1

(ry,8 d) (ry,3d) (ry,6) (ry,4) (f3,1)

S

[VK1 VKz [VKz [V8VK1 [VK1

(ry,8) (ry,3) (ry,6 d) (ry,4 d)

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 K3,

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 81 , 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 8 1 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

(8)

NV[VK3 NV[

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

NVKz vK3 NVKz NVK3 81

LVK3 LVK3

(8,6)

(f3) (8,() (8,4)

NVLVK3 NvL NVKz VK3 NVKz NVK3

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

98

Ockham algebras

The lattice ofsubvarieties

99

100

Ockham algebras

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

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 MI and MI (with MI directly above M V K z 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(0). 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 (a), ((3), ... , (1)) and (n) for n = 1, ... , 14. The dual equivalents of the (n) are (a)

= (1) (2) (3) (4) (5) (6)

(7) (8) (9) (E) = (10) (11) (12) (13)

(14)

= gZ = gO gO = gZ V g = gZ V gO ~ gZ ~ g V g ~ g2 ~ gO gO =gZ V gZ ~ gO ~ g V g ~ gO ~ gZ g = gZ V g ff gO gO = gZ V g = gZ V ~ °~ gZ V gZ ~ ~ ° gO = gZ V gO ~ gZ ~ g V g ~ gZ ~ gO g = gZ V gZ ff gO gO = gZ V g ff gO gO

g

g2 ff gO gO

=gZ V gZ ffg

V g=gZ V gffgO g = gZ V gZ ff gO V g ff gO gO = gZ V g = gZ V g ff gO V ~ °~ gZ V gZ ~ ~ ° gO=gZ

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

ties Ubvarie s / 0 e c i t 'The lat

;(° f~ ~~ld;6fthO~U(iiI!

e o " 'l lI g ' fc 11 .

eith n

~(~,)9130~)=h0:: )(1~4) h, 3)

~

"'i

MOreo O) JJs ,(14). ti a c li 2 p 1 ( ll s an)d, iland qualitie 1 4 )

f:;'l!h~ (143,)_7 h (a<) (3, 4) 9) ~ (1) , 5 1 ~ ) ! , (

(11

1"0"1 , " 'e ha ly r a il ll il the 0). S li e s (1 o verify 4 ) nOrl i d ) il ll p " " r, i t is e a s Y t

~

(e)

(6, 12)

(e 12) ~ (

((,6) (c, 6) =:: £) === h , ) (a<, 5) 8 , h 6 h , < a ( (c,12) (a<, 1 ) (c, 9) == (/3, e) ) (/3, I ) =:: e , ) h h ,3) =:: (7 (e,5 (3",9)) == ((ee) h, 6 ' J"' y f '"d ) 5 . Ula ( ) ' " A.(I>3 ,' '" 0 f 16, 3) o s e ti b""rie dual sU al) seJf-

~

) 5 ~ ~ ~ ~ ~ ~ ~

~

n-td.i

rarn.

o ce o f (n ."ing dia8 'ThehJaasWthe f01Io 7 and

4

N

~ (59)

~

Ockham algebras

102

Equational bases for the A-irreducible elements of this lattice are as follows. M: 1 VN

(1/)

MVN LV M: 1 V 8 1 LVM: 1 M: 1 V 8 1 LVMV8 1

(11) (13)

K3 VMV8 1 (6) Kz VMV81 (3) (4) K1 vM

() (8)

N 81

(8) ((3)

(12)

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

L M: 1 B

(01) (ry) (E) (2)

LV81

Kz V M LVM K 3 VL V M V 8 1

(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

1

0 0 0 0 0 0 0 O· 0 0 0 a 0

8

S K1

K1 Kz Kz K

S P3,Z P 3,z

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;'

103

Tbe lattice ofsubvarieties into an algebra in P3,1 in two ways, namely

(C;"'l) : "'l(X,Y ,Z) (C; "'2) : "'2(X,Y ,Z) As far as "'I

= (l-z, l-y,l -z);

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

",I(x,y , z) 1\ "'1 (x,y, z) ~

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

Thus (C; "'d satisfies (8) and so belongs to N. b= Now (13) fails in (C; "'I), as can be seen by taking a = (f, t, Hence (C; "'I) ~ NV LV K3 · Likewise, (13d) fails, c = (t, c = (f, t)· Thus as can be seen by taking a = (t, f), b = (C; "'I) ~ N V LV K 3 . It follows that V(C; "'1) = N. to As for (C; "'2), we have that V(C; "'2) = B1. To see this, it suffices For show that the axiom ('7/), which is an equational basis for:M 1 V N, fails. and b = this purpos e, take a = (t, We refer the interested reader to [46] for further properties.

n

(t, t, t),

t, f)·

(t, t, t),

t,

t,

(f, t, t)·

t, f)

Example 5.10 The algebra of Example 1.10 satisfies the axiom b); for X 1\ ",x = x 1\ (b V x') 1\ a = x 1\ b 1\ a = x 1\ b; { y V ",y =y V b V (y' 1\ a) = y V b V a =y V a. It also satisfies (4) and (4 d); for

{ ) (b Vxl\a 2 ",x=

~ x 1\ a ~ x 1\ b = x 1\ "'x;

= x V a = x V "'x. (1) Hence we have that L E (LVK1) 1\ (LVK1) = K 1 VK1 = K1 . Since neither . K = V(L) re 1 nor (l d ) is satisfied, L ~ K 1 and L ~ K1 . Therefo ::;; b V x V a

Example 5.11 Referring to Example 1.11, consider first the Ockham alge:M , an bra L 1 that corresponds to the polarity p. This does not belong to 1 equatio nal basis for which is (E), since a2n 1\ ",2 b 2n { ",2 a2n V b2n

= a 2n 1\ a 2n+ 1 = a2n; = a 2n - 1 V b2n = b 2n .

for Likewise, it does not belong to N, an equational basis for which is (8); ",2 ao 1\ ",ao = a o II bo = ",2 bo V ",b o. that if However, the relation ('7/) holds. This follows from the observation b } a ~ {ao, b o} then ",a 1\ ",2 a is of the form ak with k < 0, and if b ~ {ao, o :MI' V N to s belong L Hence O. > k 1 then ",b V ",2 b is of the form ak with

Ockham algebras

104

As for the Ockham algebra L z that corresponds to the polarity p', it can be seen that this belongs to L.

Example 5.12 In Example 1.12 we have rvzX =

{xu

{a} if b E X; Xn{a}' ifb¢X,

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

Example 5.13 In Example 1.13 we have rv(Xl, xz, ... ,x n ) = (x~, xi,· .. ,xi); { rvZ(Xl'XZ,··· ,xn) = (Xl,X n"" ,xn)·

Taking into account that Xl ~ x n implies Xl 1\ x~ = 0 and xi V x n = 1 we readily see that axioms (3), .(5) and (6 d ) are satisfied, so that LEN. For instance, taking a = (xl,xz, ... ,xn) and b = (Yl,YZ,'" ,yn) we have rvza ~ a V rvb V rvzb, Le. axiom (6 d ), since (Xl,X n "" ,xn ) ~ (Xl vy~ V Yl, Xz V yi V Yn,"" Xn Vyi V Yn) = (Xl Vy~VYl' 1, ... , 1).

We even have V(L) = N sinceN covers only K 3 v[VS l which satisfies (13d)' an axiom that generally does not hold in (L; rv).

Example 5.14 In Example 1.14 let n

= IT pr;,

a

= IT pf;,

Then a 1\ a+

= IT

p;Uin{{3;,Cl<;-min{Cl
b V b+

b

= IT p7;·

iE]

iEI

= IT

iEK

p;uaxb /Cl<;-min{Cl
iEIUK

iEIU]

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

= min{{3,Oi -min{Oi,{3}},

B

= max{ry,Oi -min{Oi,ry}}.

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

al\a+=gCd{a'gCd{:,a}}

~gcd{a,n}=a++

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

= Kl .

6 Fixed points In an Ockham algebra (L; 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 (X; g) 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 ~ bthen 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].

Theorem 6.1 Let (X;g) be the dual space of (L;1) E O. Then a E L is a fixed point off if and only if the graph of g is bipartite with one ofthe blocks in the partition the down-set A corresponding to a. Proof Given a E L let A E O(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; g) has a fixed point then (L; f) is fixed pointfree; Proof This follows from the fact that a distinguished down-set cannot contain an element of X together with its image under g. <>

106

Ockham algebras

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 troth 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 samefixed points. 0

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

11

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

x
is a falsity ideal of L. 0

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 ~ (:3P ~ (:3P

E E E

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

0

If (L;1) isfixedpointfree then (X; g) has afixed

Proof If (L;1) is fixed point free then, by Lemma 6.1, so is ([(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. 0 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 P m,1l and let (X; g) be its dual space. If the cardinality m - n of the

107

Fixed points

then L loop of X is even then L has at most two fixed points; and if it is odd is fixed pointfree. s Proof Suppose that X contains a distinguished down-set A. This require must that if A contain sgi(x) then A contains neither gi-l(x) nor gi+l(X), but contain gi+Z(x). In other words, the bipartition of the graph of g is {{x,gZ (x),g4( x), ...}, {g(x),g 3(x),g5 (x), ...}}. lity This bipartition is possible (and then is unique ) if and only if the cardina also be of the loop of X is even. In this case, the comple ment of A in X can <> points. fixed two a down-set, which means that L has at most The possible cardinalities of Fix L for an algebra (L; f) E 0 depend heavily ordere d on the subvariety of 0 to which L belongs. Clearly, every totally the Ockha m algebra has at most one fixed point. Some classes of 0 enjoy be will red conside are that s same property. In what follows the algebra t). elemen assume d to be non-trivial (i.e. not to reduce to a single

Theorem 6.5 All algebras in subvarieties of N have at most onefix ed point, and all algebras in subvarieties of Sl are fixed pointfree.

Proof If LEO has at least two fixed points then the axiom "",a 1\ "",za ~ "",b V "",zb,

(8)

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

((3) which characterises the subvariety Sl' cannot be satisfied. <> For the momen t we shall focus our attention on the class M of de Morgan s, and algebras. It has the (non-trivial) proper subclasses B of boolea n algebra where cases The 0. = L K of Kleene algebras. Obviously, if L E B then Fix being much L E K \ Band L E M \ K have to be treated separately, the latter more complicated.

Theorem 6.6 If L E K \ B then IFix LI

E

{O, I}.

Proof By Theore m 65, L has at most one fixed point. Both the possibilities

For of being fixed point free, and of having precisely one fixed point, occur. point; example, the subdirectly irreducible algebra K E K has a single fixed free and the 4-elem ent chain 0 < a < b < 1 can be made into a fixed point <> o. = a, Kleene algebra by defining 0 = 1, a = b, b =

r

Ockham algebras

108

We recall that (L;I) E M if and only if 1 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;1)1 : 1 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 with (L;1) min IFix L I ={ 1

o

max IFix LI

IT n~hJ .0

= { i=l

1

if r i

E M.

Tben

if all n i are odd, otherwise; is even whenever n i is even,

otherwise.

Proof The dual space X of L consists of the disjoint union of r 1 chains of nl - 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 ni 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; 1) has exactly one fixed point. If, on the contrary, some n i 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;1) is fixed point free. Suppose now that, for some i, n i is even and r i 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 n i even always implies r i 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 ri chains D.j - 1 to the number of distinguished down-sets is n}hJ, and the result follows. <> Corollary If L = n 2 and (L;1) E M then max IFix L I = n. <>

Fixed points

109

Example 6.1 The following table illustrates Theorem 6.7 : L

min IFix LI

33

1 1 1 0 0

4

3 2 3 X5 42 X 52 2 5 x 43

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 = {a ELI a ~ rva} = {a /\ rva I a E L}; LV = {a ELI a ~ rva} = {a V rva I a E L}. Clearly, LA is a down-set and LV is an up-set of 1. Lemma 6.3 Let (L; f) EO. Tben the fixed points of f are minimal elements

of LV and maximal elements of LA. If L E M and Fix L t= 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 nL A • Moreover, if b < a < c then ",b ~ ",a = a > b gives bELA \ LV, and ",c ~ ",a = 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 = Ti V (b /\ a) belongs to Fix L since c= (Tivb)/\(Tiva)= b/\(Tiva)=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 subdirectly irreducible algebra K 3' Lemma 6.4 Let L E P 3,1' Tben 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 /\ ",a)

V

(b /\ rvb)

~

",[(a /\ ",a) V (b /\ ",b)] = (",a V rv2a) /\ (rvb V rv2b).

This holds if and only if

(Va, bEL)

a /\ rva ~ rvb v",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 Kz VK3 (resp. L E K). Proof It suffices to note that M1 1\ (N'v L)

= K z VK3 and M 1\ (N'v L) = K. ~

Dually, we have the following results. Lemma 6.5

If L E P 3,1

then LV is a filter of L if and only if LEN V L. ~

By considering (N' V L) 1\ (N V L) we can deduce from Lemmas 6.4 and 6.5 the following : Lemma 6.6 If L E P 3,1 then L/\ is an ideal of Land L v is a filter of L ifand only if L E Sl V L. Proof We have

(N' VL) 1\ (N VL) = (N' 1\ N) V (L 1\ N) V (N' 1\ L) V (L 1\ L) =Sl VKVLVL = Sl V L vL.

~

Theorem 6.8 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

= ,,-,yo

Moreover, L/ <1>1 is dually isomorphic to "-'L . Lemma 6.7

If L E P 3,1

then

(1) V(L)E[B,SdifandonlYifV(,,-,L)=B; (2) V(L) E N\ Sl ifand only ifV(,,-,L) = K; (3) V(L) E P3 ,1 \ N ifand only ifV(,,-,L) = M. Proof (1) : L E Sl if and only if the axiom ([3) is satisfied; i.e. L has a boolean skeleton.

111

Fixed points

(2) : LEN if and only if the axiom (8) is satisfied; Le. L has a Kleene skeleton. (3) : L E P 3,1 if and only if ",3 a = a for every a E L; Le. L has a de Morgan skeleton. <> Since Land ",L 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 \ SI then IFix LIE {O, 1}; (3) if L is countable and V(L) E P3,1 \ N then IFix LI

E

IN \ {1}.

<>

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

Proof We have SI /\ M1 = Sand

N/\ M 1 = (N V N) /\ M 1 = (N /\ Md V (N /\ M 1 ) = K z 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' IfV(L) satisfies axioms (5, 3d, 6) but not axiom (4) then L is fixed pointfree. Proof Observe that (5, 3d , 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 c N 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/\; Le. L/\ = e!. Since L satisfies (16 d ), a::;; e implies that a = ",za. It follows that for every bEL we have b /\ ",b

= ",Z(b /\ ",b) = ",zb /\ ",b,

which means that (15 d ) = (3 d , 4) is satisfied, contrary to the hypothesis.

<>

Corollary If V(L) is any ofthe subvarieties LVS, SVK1 VKz, SVK1 , SVK1 , SVKz, SVK, SVK, LVS, SVK1 VKz, SVK1 , SVK1 , SVKz, SVK,

then L is fixed pointfree.

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

L h,3 d,6) h,3d,4) K VK h,3d,6,8) h,3d,4,8) 1 2 SVK1 h, 3d, 6, 6d) h,3d,4,6 d) K1 h,l d,6) h,l d,4) K2 h,3,3 d,6) h,3,3d,4)= h,I,3d) K h,l d,3,6) h,l d,3,4)= h,I,l d)= (a,ry) h,3,3d,6,6 d) h,3,3d,4,6 d)= h,I,3d,6 d) SvK

By duality, the same conclusion holds for the other subvarieties. 0 For L E M1 a more sophisticated procedure can be used to decide when L is fixed point free.

Theorem 6.11 [35] ffV(L)

= S V M then L is fixed pointfree.

Proof L satisfies the axioms (1, 3d, 6d)' 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 ~ ",L. It follows that every a ~ e is such that a Moreover, (1) implies (3) and L satisfies (16) = (3, 6d ), that is

a V ",b V ",2b

= ",2 a .

= ",2 a V ",b V ",2b.

Hence every a ~ e is such that a = ",2 a . Finally, if a II e then a 1\ (: and a Ve both belong to ",L and, owing to the serniconvexity of ",L in any MS-algebra, we have a E ",L. Thus ",L = L and so L E M, a contradiction. 0

Theorem 6.12 [35] ffV(L)

= S V M V K 1 then L is fixed pointfree.

Proof L satisfies (1, 6 d, 12 d ). Suppose, by way of obtaining a contradiction, that L has a fixed point e. Then the axiom (16) = (3, 6d ), namely a V ",b V ",2b

= rv 2a V rvb V rv2b,

is satisfied and, for every a E L, we have a VeE rvL. Hence we have also

(*)

(VaEL)

aVrvaVeErvL.

113

Fixed points Now L also satisfies the axiom (18 d)

(a

V

",a) 1\ ",b 1\ ",2b

= (11, 12 d ), namely

= (",2 a V ",a) 1\ ",b 1\ ",2b,

which gives

(Va, bEL) In particular, we have

(**)

(Va E L)

(a V ",a) 1\ e E ",L.

By (*), (**), and the convexity of ",L, it follows that a V ",a E ",L for every a E L, which is (15) = (3, 4d ), namely .

a

V

",a

= '" 2a V ",a,

and gives the contradiction L E M V K 1 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) = M1 then IFix LI E IN \ {I} (refer to the Corollary to Theorem 6.9). First, an example of L E M1 that is fixed point free. Consider the lattice L = 2 2 X 3 made into an MS-algebra as follows :

j d

g

x 0 abc d e f g h t J 1 d i jab g jOb g 0

XO 1

a

The thick lines indicate the cI>-classes. To see that V(L) the axiom (lId)' namely

",a

1\ ",2 a ~

a

V

= M1 we observe that

",b V ",2b,

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

c V ",a V ",2 a

= 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) = M1 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 4> -classes.

b'2 b~_4

b~-3

bn - I

bn - 2 bn - 3 b n -4

o 1bere are n - 1 fixed points, marked with circles, namely b n- I ,ji J2 ,... In-4 In-3, an-I'

That V(L) = M I follows from the observation that it contains copies of the subdirectly irreducible algebra M I , namely the intervals [b n - 3, b~-3] and [a n-3, a~-3].

7 Fixed poin t separating congruences we If (L;]) is an Ockham algebra and x,y are distinct elements of L then fixed a By 9. 1 ' ¢ shall say that a congruence '19 on L separates x and y if (x, y) point separating congruence on L we shall mean a congruence that separates point every pair of fixed points of L. We shall denote by F(L) the set of fixed fixed of set the be I} E i ; separating congruences on L. We let Fix L = {O!j ~ 2. LI Fix I that points of (L;]), and we shall assume throughout this chapter All of the results that follow appear in [56}.

a Theorem 7.1 Let (L; -) E M. If 01,13 are distinct fixed points of Land '19 is w. fixed point separating congruence on L then 'I91[0<1\,8,aN,81 = Proof We prove first that '19 I[o<,aN.81 = w. For this purpose, suppos e that O!~x~y~O!vj3

with (x,y) E '19. Define x,8

= x /\ (xv (1).

Then we have ~=XV~/\(1)=~V~/\~v(1)=x/\~v(1)=~

and so x,8 is a fixed point. Likewise, so is y,8

= Y /\ (yV (3).

Y,8 /\ (3 Since (x,y) E '19 we have (x,8,y,8) E '19 whenc e x,8 = Y,8' Then x,8 /\13 = gives x /\ (3 =Y /\ 13, whenc e

x = x /\ (01 V 13) =

01

V (x

/\ 13) =

01

V (y /\ j3) =

Y /\ (01 V (3) = y.

Similarly, we can show that '191[0<1\,8,0<1 = w. Suppose now that 01 /\ 13 ~ x ~ y ~ 01 V 13 with (x,y) E '19. Then

Thus 'I91[0<,o
= w.

O!~O! vx~O! vy~O! v(3 with (O!VX,O!VY)E'I9; O!/\j3~O!/\x~O!/\y~O!

It follows that

01

Vx =

01

V Y and

01 /\

with (O!/\X,O!/\y)E'I9.

x=

01

/\y, whenc e x = y.

The above result can be extend ed to Kw as follows.

<>

116

Ockham algebras

Theorem 7.2 Let (L;1) E Kw and let 01., (3 be distinct fixed points of L. If rJ E :F(L) then every x E (01., 01. V (3] that is separated from 01. by «Pw is also separatedfrom 01. by rJ. Proof Suppose that x E (01., OI.v(3] with (x, 01.) ¢ «pw,i.e.jn(x) rf 01. for every n. Since L E Kw there exist positive integers m, n such that fZm+Zn(x) =pn(x), and clearly Consider the elements

m-l

S

= /\ fZn+Zi+l (x),

t

m-l

= V fZn+Zi(x). i=O

i=O

We have 01. /\(3:( s < 01. < t:( 01. V (3 withf(s) = t andf(t) = s. Consequently, the subalgebra M = (01., (3, s, t) is de Morgan. Since rJ 1M separates fixed points, it follows by Theorem 7.1 that (s, t) ¢ rJ. We now show that rJ separates 01. and x. In fact, if we had (01., x) E rJ then on the one hand (OI.JZk(x)) E rJ would give (01., t) E rJ, and on the other hand (01. JZk+ 1 (x)) E rJ would give (01., s) E rJ, whence we would have the contradiction (s, t) E rJ. <>

Theorem 7.3 If (L;1)

E

Kw

then :F(L) forms a complete ideal of Con L:

Proof It is clear that :F(L) is a down-set of Con L, so it suffices to prove that if A = (rJi)iEI is a family of fixed point separating congruences then V rJ i iEI is fixed point separating. Suppose, by way of obtaining a contradiction, that 01. , (3 are distinct fixed points such that (01., (3) E V rJ i' Then (01., 01. V(3) E V rJ i iEI iEI and there exist Zo,' .. ,Zn ELand rJ 1 , ••. ,rJn E A such that 191

192

19n

193

= Zo < Zl < Zz < ... < Zn = 01. V (3. By Theorem 7.2 there exists k 1 such that fk (zd = 01.. ApplyingfZk1 01.

1

to the

above chain we obtain, for some p and q, _

OI.-f

Zk l

_

_

·Zkl

(zl)-· .. - f

19p

(zp-l)
Zkl

_

(zp)="'-f

Zkl

(Zq_l)

19 q

Applying Theorem 7.2 again, there exists k z such that f Zk2 (tZk 1 (z p)) = 01.. Continuing this argument, we arrive at the existence of t such that

f2t (01. V (3) Le.

01. V

= 01.,

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

n

< ... 19< 01. V (3.

Fixed point separating congruences

117

In general, for L ~ Kw the down-set :F(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

Ci

made into an Ockham algebra by defining

j(O) = 1, j(l) = 0, j(a) = a, (Vi ~ 0) j(Xj) = Xi+I,

j([3) = [3,

and extending to the whole of L. Then we have (Vi ~ 0) j(a j) = aj+I,

Clearly, (L;1) ~ Kw and
{a},

j(b j) = bi+I,

j(yj) = Yi+I,

j(Zj) = Zi+I'

= 19(xo, (3) is fixed point separating with classes [ao,1],

[0, bo],

[zo,Yo].

Now the relation
A=[O,(3),

B={[3},

C=([3,l],

L\{AUBUC}

is also a fixed point separating congruence. Since clearly (a,xo) E
= ~ ~ :F(L).

Ockham algebras

118

Definition We shall say that an Ockham algebra (L; j) is fixedpoint complete if Oi* = V Oij and Oi* = /\ Oij exist with j(Oi*) = Oi* and j(Oi*) = Oi*. jE]

jE]

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

made into an Ockham algebra by defining

j(xo) = Oil,

(Vi) j(Oij) = Oij,

j(z) = 0,

and extending to the whole of L. Here we have

j(

A Oij) = j(O) = 1 =f z = V Oij.

j;;' 1

j;;' 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

Oi*

= /\ Oij > O.

Oi*

= V Oij < 1. jE]

Denote by A the subalgebra generated by

jE]

{Oi*, Oi*} U Fix L. The congruence ep on A whose classes are {O, Oi*}, {Oi*, 1}, and singletons otherwise is fixed point separating with ep > w, and extends to a fixed point separating congruence cp on L. Then 'I' ~ cp and therefore

'l'IA ~ cplA = ep > w.

119

Fixed point separating congruences

But we have
= 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 Tbe angelfish. Consider the lattice

;!I

made into a fixed point complete de Morgan algebra in the obvious way with fixed points Clij. Here we have V Clij = 1 and
'P -classes are shown by thick lines; for example,

(x, x) E 'P.

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

V Clij, jEI

X 1\

V

/\ Clij, (x jEI jEI

1\ Clij)

V Clij = V (x 1\ Clij),

jEI

jEI

E

L,

and /\ (x V Clij) exist we have jEI

X V /\ Clij jEI

= /\ (x V Clij). jEI

When L is fixed point complete and fixed point distributive we have the following analogue of Theorem 7.1.

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

= jEI /\ Clij ~ X ~ Y ~ jEI V Clij = 01*

120

Ockham algebras

with (x,y)

Then for all

E {J.

OIi 1\ OIj

i,l E I we have

~ (x V OIi) 1\ OIj ~

V OIi) 1\ OIj ~ OIj

(y

with ((x V OIi) 1\ OIj' (y V OIi) 1\ OIj) E {J. It follows by Theorem 7.1 that

(x V OIi) 1\ OIj

= (y V OIi) 1\ OIj

whence we have x V

OIi

= (x V OIi) 1\ V OIj JEI

= V[(x V OIi) 1\ OIj] JEI = V [(y V OIi) 1\ OIj] JEI = (y V OIi) 1\ V OIj JEI

=Y V OIi and consequently x

=x

V /\ OIi iEI

= /\ (x V OIi) = /\ (y V OIi) = y iEI

iEI

V /\ OIi iEI

= y. <>

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

e

(x,y) E e ~ (x V 01*) 1\ 01* = (y V 01*) 1\ 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 andfixed point distributive. For i ,1 E I define the relation (x,y) E {Jij ~ (x V OIi) 1\ OIj Then

= (y V OIi) 1\ OIj'

e =/\ {Jij' iJ

Proof That e

~

/\ {Jij follows from the observation that iJ

{Jij

by

121

Fixed point separating congrnences

and that 1\ -aj} jJ

::;;;

8 follows from the observation that (x V 01*) /\ 01*

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

= I\(x V OIj) /\ 01* i = f\[ (x V OIj) /\ 01*] =1\[(XVOIj)/\ VOl}] j

j

}

= 1\ V[(x V OIj) /\ OI}]. I

I:)

J

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

Theorem 7.7 Jj (L; j) E Kw is fixed point complete andfixed point distributive then

Proof Suppose first that L E Kp,q' lben we have jq(L) E Kp,o, Recalling the notation used in Chapter 3, let M = T 2 (r (L)) be the biggest de Morgan subalgebra of r(L). Suppose that x,y EM are such that (x,Y) E 'PIM' lben, for all i,j E I, «(x V 01;) /\ OI}, (y V OIj)

/\ OI})

E 'PIM'

It follows by Theorem 75 that

(Vi,j E I)

(x V OIj) /\ OI}

= (y V OIj) /\ OI},

and therefore, by lbeorem 7.6, (x,y) E 81M' lbus '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 'Plj*) = 8Ij*) and hence 'Pljq(L)

= qlj*) V 'Pljq(L) = qljq(L) V 8Ijq(L) :::;;(q V 8)iJq(L)'

The reverse inequality being trivial, we have that'Pljq(l) Since Conjq(L)·~ [q, d, we deduce that'P = q V 8.

= (q

V 8)ljq(L)'

Suppose now that L E Kw. Let (X,Y)E 'P and let A be the subalgebra generated by {X,Y,Oi*, OI*} U Fix L. lben by lbeorem 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 = qlA V 81A, . whence (x,y) E w V 8. Thus 'P ~ w v8 whence we have equality. I:)

Ockham algebras

122

Note, in fact, that in lbeorems 7.5, 7.6, 7.7 only the existence of a* and = a* and f(a*) = a*.

a* is used; we do not require the properties f(a*)

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

(1) 'P=4>w; (2) Va;=I; ;EI

(3) e=w.

Proof (1) (2)

=}

=} (2) follows from lbeorem 7.4. (3) : If (2) holds then A a; = 0 and clearly e

;EI

(3)

=}

= w.

(1) follows from Theorem 7.7.0

It is of course possible to have 'P = e. By Theorem 7.7, this occurs precisely when 4>w ~ e. 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(a;)

= a;,

f(a*)

= a*,

f(a*)

= a*

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

123

Fixed point separating congruences

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

r = 19(01*,01*). If L is fixed point complete then by Theorem 2.1 we have

r = 191at(OI*, 01*). Theorem 7.9 If L E Kw is fixed point complete then

e

r

is the complement of

in ConL.

Proof Since (0,01*) E e, (01*,01*) E r, and (01*,1) E e, we have

evr=t. Now for all x,y

EL

we have

191at(X,y) 1\ r = 191at(X,y) 1\ 191at(OI*, 01*) = 191at((X V01*) I\y 1\ 01*, Y It follows immediately that (x,y) E e

Since

e

'*

191at(X,y) 1\ r

1\ 01*).

= w.

is a congruence it follows by Theorem 2.1 that (x,y) E e

'*

19(x,y) 1\ r

=w

and therefore that

e 1\ r = V

19(x,y) 1\

(x,y)EE>

r= V

(19(x,y) 1\ r)

= w.

(x,y)EE>

r is the complement of e. 0 Corollary e = 19(0,01*) = 19(01*,1). Proof The principal congruence r = 191at(OI*, 01*) is complemented. plement is 191at(O, 01*) V191at(OI*, 1) = 19(0,01*) = 19(01*,1).0 Consequently,

Its com-

Theorem 7.10 If L E Kw isfixed point complete andfixed point distributive then Con L contains the sublattice

Ockham algebras

124

Proof By Theorems 7.7 and 7.9 we have that

'II 1\ r = (w

V

e) 1\ r

= w 1\ r,

whence it follows that

= (r 1\ 'II) V (w 1\ 'II)

(r V w) 1\ 'II

= (r 1\ w) V (w 1\ 'II) = w 1\ (r V'll) = w 1\ £ = w, and that (w 1\ r) V (w 1\ e)

= w 1\ (r V e) = w 1\ £ = 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 interval [w, £] is boolean ifand only if both the intervals [w, 'II] and ['II, £] are boolean. <> Definition We shall say that L is fixedpoint compact if it is fixed point complete and there is a finite subset 1* of I such that Q!*

Q!*

= A Q!t

(equivalently,

jEl*

= V Q!j)' jEl*

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

x

1\

V Q!j = x 1\ V Q!j = V (x 1\ Q!j) ~ V (x 1\ Q!t),

jEl

jEl*

jEl*

jEl

and the reverse inequality is trivial.

Theorem 7.11 If (L;1) ep E ['II, d is such that

E

Kw is fixed point compact then every congruence

ep='P

V

V

'l9(Q!j,Q!j)' (OIj,OIj)E
Proof Suppose first that L E Kp,q, so thatfq(L) E Kp,o. Consider the de Morgan algebra M = T 2 (tq(L)). For x E M and i,j E I define X jj

= (x V Q!j) 1\ (t(x) V Q!j) 1\ (Q!j V Q!j)'

Fixed point separating congruences

125

Then a simple calculation reveals that Xjj is a fixed point of M. cp E ['P, d, let

V

cp* =

Given

'l9(OIj,OI).

(O<{ ,O<}) E cp

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

E

cp then we have (Xjj,yij)

E

cp. It follows

that we therefore have

((x V OIj) 1\ OIj' (y V OIj) 1\ OIj) E 'l9(x jj ,yjj) ~ cp*. Since L is fixed point compact we have

(x V 01*) 1\ 01* =

V 1\ [(x V OIj) 1\ OIj] JEI* jEI*

and so it follows from the above that

((x V 01*) 1\ 01*, (y V 01*) 1\ 01*) E cp*. Since (x, (x V 01*) 1\ 01*) E e we deduce from this that

(x,y) E e V cp* ~ 'P V cp* and hence that

V

CPIM =

'l9(X,Y)IM ~ ('P V CP*)IM'

(x.Y)EcpIM

The reverse inequality being trivial we therefore have, using Theorem 7.7,

CPIM = ('P V cp*)IM = (q V e V cp*)IM' Arguing as in the proof of Theorem 7.7, we deduce that

cp

= q ve V cp* ~ 'P V cp*,

whence we have equality and so the result holds for L E K p ,q' Suppose now that L E Kw. Let (x, y) E cp ~ 'P and consider the subalgebra B generated by {x, y, 01*, OI*} U Fix 1. We have B E Kp,q for some p, q so, by the above,

(x,y) E cplB

='PIB V CP*IB ~ ('P V CP*)IB

whence (x,y) E 'P V cp* and hence cp ~ 'P 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 'II = wand the congruence V {)(ai, aj) has three classes, namely {OJ, {1}, and L\{O, 1}. iJ

The equality of Theorem 7.11 therefore fails for

I{J

= to

We now give an example of a fixed point compact de Morgan algebra in which the interval [
Example 7.6 Consider the lattice

ao

made into a fixed point compact de Morgan algebra in the obvious way with fixed points ao," . , a oo ' In this, the congruence I{J > 'II whose classes are A=[O,a oo ),

B={a oo },

C=(a oo ,1],

D=L\{AUBUC}

has no complement in ['II, ~]. In contrast, the interval [w

=
Theorem 7.12 Let L E Kw be fixed point complete andfixed point distributive. Then ['II,~] ~

Con T 2 (A)/e

where A is thesubalgebra {OJ EB [a*,a*] EB {I}.

Fixed point separating congruences

127

Proof Suppose first that L E M so that 'I' =

e and T z(A) = A.

Let

A* = ([O!*, O!*];J) and note that, by the Corollary to Theorem 7.9, we have Aje !J. : ['I', £]

-7

~

A*. Define

Con A *

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

= Xo == Xl == ... == Xn = b where each == is either CPI or cpz. Defining Yi = (Xi V O!*) 1\ O!* we have a =Yo == YI == ... ==Yn = b a

with each Yi

E

A*. Consequently, (a, b) E CPIIA*

V

CPZIA*' Thus

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

= CPzIA*;

(2) (x V O!*,y V O!*) E 19(0!*, 1) = e ~ 'I' ~ cpz;

(3) (x 1\ O!*,y 1\ O!*) E 19(0, O!*) = e ~ 'I' ~ cpz. By (1) and (2) we have (x V O!*,Y V O!*) 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. is surjective, let 19 E Con A*. Let 78 E Con A be such that 78IA* = 19; Le. let

[0]78 = [0!*]19 U {O} and [1]78 = [0!*]19 U {I}. By the congruence extension property there exists cP

E

Con L such that cP IA

78. Then !J.(cp V '1') = CPIA* V 'l'IA*

= 19

since, by Theorem 7.5, 'l'IA* = w. Thus !J. 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 ep ~ w the correspondence ep ~ epIT2(L) is therefore a bijection. Thus we see that

Now since the result holds for Tz(L)

E

M we have

['PIT 2(L), £I T2(L)1 ~ Con T z(C)/8 where C

= Tz(L) nA.

CorolIary

Since Tz(C)

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

With L as above, the folloWing statements are equivalent:

(1) ['P, £] is boolean; (2) Tz(A) isfinite; (3) L hasfinitely manyfixed points. (2) : If (1) holds then Tz(A) has finitely many 8-classes and it follows by the definition of 8 that Tz(A) must be finite. (2) =} (3) : This is clear. (3) =} (1) : Observe that for fixed points 01 j , 01j we have Proof (1)

=}

19(OIj,OIj)

= 19(OIj 1\00j,OIj VOIj) = 19Iat(OIj 1\00j,OIj VOIj)'

Since then 19(OIj 1\ OIj' OIj V OIj) 1\ 19(OIj V OIj' 1)

= 19Iat(OIj 1\ OIj' OIj V OIj) 1\ [19Iat(OIj V OIj' 1) V 19Iat(O, OIj 1\ OIj)] =w, 19(OIj 1\ OIj' OIj V OIj) V 19(OIj V OIj' 1) = 19Iat(OIj 1\ OIj' OIj V OIj) V 19Iat(OIj V OIj' 1) V 19Iat(O, OIj

1\ OIj)

=£ we have that

19(OIj,OIj)

is complemented in Con L with complement

19(OIj, OIj)'

= 19(OIj V OIj' 1) = 19(0, OIj 1\ OIj)'

If L has finitely many fixed points it therefore follows by Theorem 7.11 that every congruence ep ~ 'P has a complement in ['P, £]. Hence ['P, £] is boolean. 0

Example 7.7 Consider the lattice

129

Fixed point separating congrnences .1

.0

made into a de Morgan algebra with fixed points aI, a2, a 3' Here
=w

= E> = 'I9(O,a*) = 'I9(a*, 1). By Theorem 7.12, ['11, 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 Jf L isfixed point compact then r

=V'19 (Ci.i' a j)' iJ

Proof By the hypothesis there exists a finite subset 1* of I such that

r = 'I9(a*, a*) = 'I9{ /\ ai, V aJ iEI*

Now, given any j E I*, we have

(Vi and therefore

It follows that

E

I*)

iEI*

Ockham algebras

130 and hence that

r = 19(

V Cl!j):( V19(Cl!j, Cl!j)'

/\ Cl!j, jEI*

jEI*

jJ

1he reverse inequality being trivial, we deduce that

r = V19(Cl!j,Cl!j)' 0 jJ

The following example shows that 1heorem 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 Cl!n = (n, -n) for every n E 7L. 1hen L is fixed point complete and fixed point distributive, but not fixed point compact. Since Cl!* = 1, we have e = wand r = to In contrast, the congruence V19(Cl!j, Cl!j) has three classes, namely {OJ, 7L x 7L, and {I}. jJ

We now consider the question of precisely when the interval [«pw , '1'] 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 andfixed point distributive. Then

[«pw , '1']

~

Con Tz(B)/r

where B is the subalgebra [0, Cl!*] EB [Cl!*, 1].

Proof Suppose first that L EM, so that «Pw = w, 'I' = e, and B = Tz(B). Consider the mapping A : [w, '1'] --+ Con B given by A(!p) = !PIB' Clearly, A is a morphism for 1\. To see that it is also a morphism for v, let (a, b) E B be such that a:( b and (a, b) E (!PI V!PZ)IB' Since !PI,!PZ:( 'I' we have that !PI V !pz :( 'I' so !PI V !pz is fixed point separating. It follows by Theorem 7.5 that (Cl!*,Cl!*) ¢!PI V!pz· Consequently, either a, bE [0,Cl!*] or a, b E [Cl!*, 1]. Now there exist xo,'" ,xn E L such that a

= Xo == Xl

== ... == X n

=b

= (Xj va) 1\ b we have a =Yo == YI == ... == Yn = b

where each == is either !PI or !pz. Defining Yj

and, from the above observation, every Yj E [0, Cl!*] or every Yj E [Cl!*, 1], i.e. every Yj E B. Consequently, (a, b) E !PIIB V !PZIB and so (!PI V !PZ)IB:( !PdB V !PZIB'

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

Fixed point separating congruences

131

To see that A is injective, suppose that rp I, rpz E [w, '1'] are such that rpIIB rpZIB' If (x,y) E rpI then since rpI ~ 'I' = 8 we have

= (y V c¥*) 1\ c¥*; (x 1\ c¥*,y 1\ c¥*) E rpIIB = rpZIB;

(1) (x (2)

=

V c¥*) 1\ c¥*

(3) (x V c¥*,y V c¥*) E rpIIB

= rpZIB'

By (1) and (3) we have (x V c¥*,y V c¥*) E rpz which, together with (2) gives (x,y) E rpz. Hence rpt ~ rpz· Similarly, rpz ~ rpt and so rpt = rpz· Observing that ImA ~ [wIB, 8IB], suppose now that {J E [WIB' 8IB]' Then there exists rp E Con L with ep IB = {J. Let"J = rp 1\ '1'. Then we have A("J)

="JIB = rplB 1\ 'l'IB = {J 1\ 81B = {J.

We thus see that A induces an isomorphism [«pw , '1'] rem 7.9 gives

~ [WIB, 81B].

But Theo-

Hence the result holds for L E M. Suppose now that L E Kw. We have

[«pw , £] ~ Con Lj«Pw ~ Con T(L) ~ Con Tz(L) For rp

~

«Pw the correspondence rp

+-7

= [«pwITz(L), £ITz(L)]'

rpITz(L) is therefore a bijection. Hence

[«pw , '1'] ~ [«pwITz(L), 'l'ITz(L)]' Now since the result holds for Tz(L) E M we have [«pwITz(L), '1'1 Tz(L)] ~ Con Tz(C)jr

where C = Tz(L) n B. Since Tz(C) follows. 0 Corollary [«pw , '1'] is boolean

= Tz(B),

the result for L E Kw now

if and only if Tz(B)

isfinite.

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

Finally, we give an example of a Kw-algebra that is fixed point compact, with the congruences «Pw, '1', 8, r distinct and the interval [«pw , £] boolean.

Example 7.9 Consider the lattice

132

Ockham algebras

o
(3

o

made into a Kw -algebra by defining

!(o) =!({3) = 1, !(1)=0, !(xo)=xo, (Vi ~ 1) !(xzi ) = xzi-z, !(XZi+l) = XZi-1,

!(Xl) = Xl,

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
8 Congruences on

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

(191at(a, b) V cp)/cp

= 191at([a]cp, [b]cp)

(see, for example, [12, page 137]) and consequently

(x,y) E 191at(a, b) V cp

¢==>

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

in L/cp.

Theorem 8.1 If (L;~) E K1,1 and a, b E Lwith a ~ b then (x,y) if and only if (1) x 1\ a 1\ rv 2a 1\ rvb =y 1\ a 1\ rv 2a 1\ rvb; (2) (x 1\ a 1\ rvb) V rv 2b = (y 1\ a 1\ rvb) V rv2b; (3) [(x 1\ a) V rva] 1\ rv 2a = [(y 1\ a) V rva] 1\ rv2a; (4) (x 1\ a) V rva V rv 2b = (y 1\ a) V rva V rv2b; (5) (x V b) 1\ rvb 1\ rv 2a = (y V b) 1\ rvb 1\ rv2a; (6) [(x V b) 1\ rvb] V rv 2b = [(y V b) 1\ rvb] V rv2b; (7) (x V b V rva) 1\ rv 2a = (y V b V rva) 1\ rv2a; (8) x V rva V b V rv 2b = y V rva V b V rv 2b.

E

19(a, b)

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

= 191at(a, b) V191at(rvb, rva) V191at(rv2a, rv2b).

Denote by 1/J the lattice congruence 191at(rvb, rva) V191at(rv2a, rv2b). Then

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

¢==> ¢==> ¢==>

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

in LN

Now we have

(x

1\

a,y 1\ a) E 1/J

¢==> ¢==>

¢==> ¢==>

a], [y 1\ a]) E 191at(rvb, rva) in L/191at(rv2a, rv2b) {[X 1\ a] 1\ [rvb] = [y 1\ a] 1\ [rvb], [x 1\ a] V [rva] = [y 1\ a] V [rva] (x 1\ a 1\ rvb,y 1\ a 1\ rvb) E 191at(rv2a, rv2b), { ((x 1\ a) V rva, (y 1\ a) V rva) E 191at(rv2a, rv2b) (1) --7 (4) hold in L. ([x

1\

Ockham algebras

134 Similarly, we have (X V b ,y V b) E

./,

{((X V b) /\ ,-vb, (y V b) /\ ,-vb) E 19lat(,-v2a, ,-v2b),

'f/ ¢==;>

(x V b V ,-va,y V b V ,-va) E 19lat(,-v2a, ,-v 2b) (5) --? (8) hold in L.

¢==;>

Thus (x,y)

E

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] Jj(L; 0) E MS anda, bEL with a ~ b then (x,y) if and only if (I') x /\ a /\ bO =Y /\ a /\ bO; (3') (x /\ a) V (ao /\ aOo) = (y /\ a) V (aO /\ aOo); (5') (x V b) /\ bO /\ aOO = (y V b) /\ bo /\ aOo; (6') (x /\ bO) V boO = (y /\ bO) V bOo; (7') (x V b V aO) /\ aOO = (y V b V ao) /\ aOo; (8') x V aO V bOO =y V aO V bOo. 0

E

19(a, b)

Since (5') and (7') hold trivially if (L; -) E M, we have:

Corollary 2 [84] Jj (L; -) E M and a, bEL with a ~ b then (x,y) E 19(a, b) if and only if (1/1) x /\ a /\ b =Y /\ a /\ b; (3/1) (x va) /\ a = (y va) /\ a; (6/1) (x /\ b) V b = (y /\ b) V b; (8/1) xvaVb=yvavb. 0 Corollary 3 [74] Jj (L; *) if and only if

E

(i) x /\ a = y /\ a; (ii) (x V b) /\ (a** V b*)

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

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

Proof Writing * 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 /\ b* = Y /\ 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 Corollary 4 Jj (L;') E B and a, bEL with a ~ b then 19(a, b) = 19lat(a, b). Proof Writing' for * in (ii) of Corollary 3 and using distributivity, we obtain xV b =yV b. 0

135

Congruences on KI,I-algebras

Theorem 8.2 Jj(L; rv) E KI,1 and a, bEL with a:::;; b and a /\rva then

= b/\rvb

Proof Note that a/\rva = b/\rvb gives rv2a/\rva = rv 2b/\rvb and rvaVrv2a = rvbV rv 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 Vrvb, b V rva):::;; 19(a, b). But since b /\ rvb :::;; a :::;; b we have

a /\ (a V rvb) = a = (b /\ a) V (b /\ rvb) = b /\ (a V rvb); a V b V rva = b V rva = b V b V ",a, which shows that (a, b) E 191at(a V rvb, b V rva). Since a /\ 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 everyprincipal 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, ",a = b, rvb = a, rvl = O. Here

19(0, a) == {{O,a},{b,l}} 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 :::;; band c:::;; d. Tben 19(a, b) /\ 19(c, d)

= 19(a V c, a V c V (b /\ d)) V 19(a V d, a V d V (b /\ c)).

Proof Using the formula 191at(a, b) /\ 191at(c, d)

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

Ockham algebras

136 and the fact that 19lat(X 1\ y, x)

= 19lat(y, x Vy) we have, by Theorem 2.1,

19(a, b) 1\ 19(c,d) = [19lat(a , b) V191at(b, a)] 1\ [19lat(c, d) V19lat(d, c)] =[19lat(a, b) 1\ 191at(c,d)] V [191at(a, b) 1\ 191at(d,c)] V [19lalb, a) 1\ 191at(c, d)] V[191at(b, a) 1\ 19lat(d, c)] = 191at((a V c) 1\ b 1\ d, b 1\ d) V19lat((a Vd) 1\ b 1\ C, b 1\ c) V 191at((lj" V c) 1\ a 1\ d, a 1\ d) V19lat((li Vd) 1\ a 1\ c, a 1\ c) = 191at(a V c,a V c V (b I\d)) V19lat(a vd,a vdv (b 1\ c)) V 19lat((b V c) 1\ a 1\ d, a 1\ d) V19lat((b V d) 1\ a 1\ c, a 1\ c) =19(a V c,a V c V (b 1\ d)) V 19(a vd,a vdv (b 1\ c)). 0 Theorem 8.5 Every principal congruence on (L, -) E M is complemented. For a, bEL with a ~ b we have

19(a, b)'

= 19 (a vb, 1) V 19(b 1\ b, b) V19 (a, a va).

Proof Consider the congruence

cp That 19(a, b) V cp

= 19(a Vb, 1) V 19(b 1\ b, b) V 19(a, a va).

=~

follows from the observations

(0, b 1\ a) E 19(a vb, 1), (b 1\ a, b 1\ b) E 19(a, b), (b 1\ b, b) E 19(b 1\ b, b), (b, a) E 19(a, b), (a, a va) E 19 (a, a va), (a Va, a V b) E 19(a, b), (av b, 1) E 19 (a V b, 1). That 19(a, b)l\cp = w follows from a routine application of Theorem 8.4 which we leave to the reader. Hence we have 19(a, b)' = cp. 0 The above results provide the following characterisation of the class M of de Morgan algebras.

Theorem 8.6 !be class M is the largest subvariety of K 1,1 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 1 , K 1 in which the property fails. For this purpose, consider the subdirectly irreducible algebras S, 5, K 1, K l' We have Con S ~ Con 5 ~ Con K 1 ~ Con K 1 ~ 3, the non-complemented element in each case being the principal congruence <1>1' 0

Congruences on K I ,I -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. Theorem 8.7 Jj 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 0= Xo 19 Xl 19' X2 19 ... 19' Xn = 1. This implies that

0= Xo 19 rv2XI 19' rv2X2 19 ... 19' rv 2Xn whence 191~2L

V 19'1~2L

= 1,

= ~~2L' ~

Theorem 8.8 Let (L; rv) E KI I . Tben we have (rv 2X, rv 2y) E 19(a, b) if and only if (rv 2X, rv 2y) E 19( rv2 a, rv 2b). 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 2X, rv 2y) E 19(rv2a, rv2b). The converse follows from the fact that (rv2 a, rv2b) E 19(a, b) and therefore 19(rv2a, rv2b) ~ 19(a, b). ~ In what follows we shall use the fact that if (L, rv) E KI,1 then rv 2L = rvL. Theorem 8.9 Let (L, rv)

E

KI,1 and let a, bEL be such that a ~ b. Define

ep(a, b) = 19(0, rvb 1\ rv2a) V 19( rv b 1\ rv2b, rvb) V 19( rv a, rva V rv2a). 1

Tben we have 19(a, b) V ep(a, b) 19(a, by = ep(a, b).

= ~;

and

if 19(a, b) is complemented then

necessari~y

Proof Consider the following chain of elements in L : 1

y s s

e g

Ockham algebras

138

Each pair of elements that are joined by a vertical line segment are congruent modulo 'l9(a, b) and, by Theorem 2.1, 'l9(0, rvb I\ rv2 a) = 'l9( rv a Vrv2b, 1). The above chain therefore shows that (0,1) E 'l9(a, b) V rp(a, b) and so 'l9(a, b) V rp(a, b) = to

Suppose now that 'l9(a, b) is complemented. Then from what we have just proved it follows that 'l9(a, b)' ~ rp(a, b).

To obtain the reverse inequality, we observe by Theorem 8.7 that 'l9(a, b)I~L is 'l9( rv2 a, rv2b) on rvL, and therefore by Theorem 85 that the complement of'l9(a,b)I~L isrp(a,b)I~L' Thus we have that 'l9(a,b)'l~L

= ('l9(a,b)I~L)' = rp(a,b)I~L'

and so 'l9(a,b)' is an extension to L of rp(a,b)I~L' Now it is clear that rp(a, b)I~L identifies each of the pairs (0, rvb 1\ rv2a), (rvb 1\ rv2b, rvb), (rva, rva V rv2a)

of elements of rvL and that so does any extension to L of rp(a, b)I~L' By its definition, therefore, rp (a, b) is the smallest extension to L of rp (a, b) I~L . Consequently, we have that rp(a, b) ~ 'l9(a, b)'

as required. 0 The above knowledge of what 'l9(a, b)' is (whenever it exists) allows us to determine precisely under what conditions it does exist.

Theorem 8.10 (1) (2) (3) (4)

If (L; rv) E K1,1

then 'l9(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 'l9(a, b) is complemented if and only if 'l9(a, b) 1\ rp(a, b) = w. Now, by Theorem 2.1 and the fact that 'l91at (X I\y,x) = 'l91at (Y,X Vy), we have 'l9(0, rvb 1\ rv2a) = 'l91at (0, rvb 1\ rv2a) V 'l91at ( rv2 b V rva, 1); 'l9( rv b 1\ rv2b, rvb) = 'l91at ( rv b 1\ rv2b, rvb); 'l9(rva, rva V rv2a) = 'l91at (rv a , rva V rv2a). It is now a routine matter to show that

Congruences on

K1,1

139

-algebras

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

If (L; rv) E MS then 19(a, b) is complemented if and only if

(I') a /\ rvb = rv 2a /\ b /\ rvb; (2') b /\ (a V rva) = b /\ (rv2 a V rva). Proof If (L; rv) E MS then conditions (2) and (3) of lbeorem 8.10 hold, whereas conditions (1) and (4) are equivalent to (I') and (2') respectively. <>

Recalling from Chapter 2 the principal congruence 19a and the fact that

19 a

= 19lat(a /\ rva, rva),

we deduce the following Corollary 2

If (L; rv) E MS then for 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 lbeorem 8.9 we have

19(0, b)' = tp(O, b) = 19lat(rv2b /\ rvb, rvb)

= 19~2b' <>

We can now apply the above to determine precisely the complemented principal congruences in a Stone algebra. Theorem 8.11

If (L; *) E S then 19(a, b) is complemented if and only if a is

complemented in [0, b]. Moreover, when it exists, 19(a, by = 19lat(a* /\ b**, 1). Proof By Corollary 1 of Theorem 8.10, 19(a, b) is complemented if and only

if

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

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

°

b = b /\ (a

V

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

Ockham algebras

140

and consequently

b 1\ a** = (a

1\

°

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

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

°

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

= 1Jiat(O, x) V 1Jiat(Y, 1),

we have

1J(a, b)' = 19(0, b*) V 1J(a*, 1) = 1Jiat(O, b*) V 1Jiat(b**, 1) V 1Jiat(a*, 1) V 1Jiat(O, a**) = 1Jiat(b*, a*)C V 1Jiat(a**, b**)C =[1Jiat(b*, a*) 1\ 1Jiat(a**, b**)]C = [1Jiat(a* 1\ b** 1\ (b* V a**), a* 1\ b**W = [1Jiat(O, a* 1\ b**)]C = 1Jiat(a* 1\ b**, 1). <>

Corollary 1 (Va

E

(L, *)) 1J(a,a V a*) = 1J(a**, 1).

Proof From the above, 1J(a,a V a*)' = 1Jiat(a*, 1) = 1J(a**, 1)'. <> Corollary 2 If L E B then allprincipal congruences are complemented. <> We shall now consider more closely, for L E K 1,1' 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,l-algebra L we clearly have
Congrnences on KI,I -algebras

141

is a congruence; so, as observed in Chapter 2, when L is finite the interval [w, <1>] of Con L is boolean. Furthermore, [, L] ~ Con L/ ~ Con ",L

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

Theorem 8.14 Let L E KI ,1' Tben is the greatest dually dense element of ConL.

Proof For every rJ

E

Con L we have that

(O,l)ErJV It follows that rJ V

**

(O,l)ErJ.

= L implies rJ = L, and so

is dually dense. Since the dually dense elements of a bounded lattice form an ideal, it remains to prove that if rJ > then rJ is not dually dense. Since the interval [ , L] ~ Con '"L is an algebraic lattice, rJ is a join of compact elements. One of these, say . By Theorem 8.13,

, L], with complement
Theorem 8.15 If L E

KI,I

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 I 1 is such that Con L is boolean then necessarily L E M. But if L ~ M then c:P > wand, by Theorem 8.14, is dually dense. Hence has no complement, a contradiction. 0 We shall now proceed to consider the following question. Given a K I 1algebra L and a subvariety R of K I ,1, what is the least congruence rJ for whi~h L/ rJ 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 rJ R .

Theorem 8.16 rJ M

= <1>.

Proof Clearly, L/ EM. Conversely, if L/fJ E M then (a, b) E => [a]rJ = ",Z([a]rJ) = [",Za]rJ = [",Zb]rJ = ",Z([b]rJ) = [b]rJ

whence

~ rJ.

0

If we denote by rJ~ any congruence on L such that L /fJ~ E R with L /fJ~ subdirectly irreducible then, as proved in [99], we have

rJ R

= /\{rJ~}.

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 ~ Rj then n

{)R

= 1\ {)Ri' 1=1

Denoting by B, K, 5, M the subdirectly irreducible algebras in B, K, S, M respectively, we therefore deduce the following result.

Theorem 8.17 ForeveryL EMS, (1) (2) (3) (4)

= I\{{)B}; {)K = 1\ {{)B, {)d; {)s = 1\{{)B,{)s}; {)M = I\{{)B,{)K,{)M}' 0 {)B

Corollary The coatoms of Con L are of the form {) B' {)K' {)M' and their intersection is
*

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 I. In fact, if (0, x) E sup F then there exist ao, . .. ,an ELand 'l9 1 , ... ,'l9 n+1 E F such that 'l9 1 a1 'l9 2 a2 'l9 3 .,. 'l9 n- 1 a n-1 'l9 n an 19'1+1 x.

°

°

°

Since'l9 1 has a trivial kernel, a1 = and so, since 'l9 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 \ {£}. Then by the above we have that sup F =j:: £ and hence, by its definition, sup F is the unique coatom of ConI. (5) # (6): Observe first that (5) is equivalent to the condition

°

x =j::

° =>

°

'l9(0, x)

= to

Now for x =j:: we have 'l9(O,x) = £ if and only if (0,1) E 'l9(0, x) and by Theorem 8.1 this is the case if and only if

(x /\ ",x) V ",2 X

= ",x V ",2 X ,

Le. if and only if

",x V ",2 X

~ X V ",2 x.

Applying", to this, we see that it is equivalent to ",x ~ ",2 X . 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 £.

Corollary

Proof If Con L has a non-trivial node tp then the centre of Con L reduces

to {w, £}. Since, by Corollary 2 of Theorem 8.10, every principal congruence 'l9(O,x) is complemented, we must have 'l9(O,x) = £ for all x =j:: 0. It follows that every congruence tp =j:: £ has a trivial kernel; for otherwise we have (0, x) E tp for some x =j:: 0, whence the contradiction £ = 'l9(0, x) ~ tp. Thus condition (5) above holds. Consequently, is a non-trivial node of Con L which is covered by to Suppose that tp < <1>. Every lattice congruence contained in is a congruence, so t is a principal ideal of Conlat L. Since this lattice is algebraic, we can find compact congruences 'l9 1 , 'l9 2 such that

w < 'l9 1 ~ tp ~ 19 2 ~ . But for any distributive lattice it is known [15] that the compact elements form a relatively complemented sublattice. Since clearly 'l9 1 cannot have a complement in [w, 'l9 2 ], it follows that we must have tp = <1>. <>

144

Ockham algebras

It goes without saying that ",L is the most significant subalgebra of any K1,1-algebra. Since the class K1,1 enjoys the congruence extension property, it is quite natural to consider on the one hand the restriction 19kL to ",L of any congruence 19 E Con L, and on the other the smallest extension V5 to Con L of ep E Con ",L.

Theorem 8.19 Let L E K1,1 and 19 E Con L. Then (",2 X ,",2 y ) E 19

~

(x,y)

E

19 Vep.

Proof If (",2 X ,",2 y ) E 19 then (",2 X ,",2 y ) E 19 Vep. As (x, ",2 x) E ep for every x E L, it follows that (x, y) E 19 V ep . Conversely, if (x,y) E 19 Vep then (",2 X,",2y ) E 19 vep whence

~

19 Vep

",2 X = Xo 19 Xl ep X2 19 ... 19 Xn -1 ep Xn = ",2y , hence Le.

Thus we have

",2 x

19 ",2 y. 0

The following property is now immediate :

Theorem 8.20 If L E K1,1 and 19 1,19 2 E Con L then thefollowing statements are equivalent: (1) 1911~L

= 1921~L;

(2) 19 1 isanextensionof1921~L; (2') 19 2 is an extension of 1911~L; (3) ",2 a 19 1 ",2b ~ ",2 a 19 2 ",2b; (4) 19 1 vep=19 2 vep.o Theorem 8.21 Let L E K1,1 and let 19 E Con L with 19 ~ ep. Then is an extension of 191~L if and only if Q! V ep = 19.

Q!

E

Con L

Proof This is immediate from Theorem 8.20.0 Corollary Let 19 E Con L with 19 ~ ep. Then the dual pseudocomplement of ep in 19! 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 Q! such that Q! V ep = 19. 0

Congruences on

Kl,l -algebras

145

Theorem 8.22 If L E Kl,l and cp E Con rvL then the smallest lattice congruence on L that extends cp is given by cp

=

V

2

t91at{rv2x, rv y).

(~2x,~2Y)E\O

Moreover, cp E Con L. Proof It is clear that cp E Conlat L and extends cp. If now 19 E Conlat L extends cp then (rv 2x, rv 2y) E cp gives t9 1at {rv 2x, rv 2y) ~ 19 and so cp ~ 19. To prove that cp E Con L, let (a, b) E rp. Then there exist Xl, ... 'X n E L such that

where each serving that

=i

is of the form t91at{rv2x, rv 2y) for some ( rv2 x, rv 2y) E cpo Ob-

(P,q)Et9lat{rv2X,rv2y) =? (rvP,rvq)Et9 1at {rvy, rv x) and that ( rv2 x, rv 2y) E cp gives (rvy, rvx) E cp, we deduce from in which, if =i is t9 lat {rv 2X, rv 2y) then =1 is t9 1at {rvy, rvx), that (rva, rvb) E cpo Thus we see that rp E Con L. <>

Theorem 8.23 If L E Kl,l then the mapping f: Con rvL by f (ip) = cp is a lattice morphism.

---+

Con L described

Proof Clearly, on the one hand, we have rpv 1J ~ ip V 'l/J. On the other hand, (cp V 1J) I~L ;;>- ip, 'l/J and so (rp V 1J) I~L ; >- ip V 'l/J whence rp V 1J;;>- ip V 'l/J. Thus f

is a V-morphism. To show that f is also a I\-morphism, observe that clearly ip 1\ 'l/J ~ cp 1\ 1J. Now, by Theorem 8.22, we have

cpl\'l/J=

V

t9lat{rv2x,rv2y) 1\

(~2x,~2Y)E\O

V

t9lat{rv2a,rv2b)

(~2a,~2b)E!/J

and so, since Con L is a complete distributive lattice in which the infinite distributive law 19 1\ V(i = V{t9 1\ (i) holds, in order to obtain the reverse i

i

inequality ip 1\ 'l/J ; >- cp 1\ 1J it suffices to prove that, for all (rv 2x, rv 2y) E ip and all ( rv2 a, rv2b) E 'l/J, t9 1at {rv 2x, rv 2y) 1\ t9 lat {rv 2a, rv 2b) ~ ip 1\ 'l/J.

Here of course we suppose that rv2x ~ rv 2y and rv2a ~ rv 2b. Now t91at{rv2x, rv 2y)l\t9lat{rv2a, rv 2 b)

= t9 1at {{rv 2XI\rv 2b)v{rv 2y I\rv2a),

rv 2y I\rv 2b),

Ockham algebras

146

and rv 2X ep rv 2y, rv 2a 'lj; rv 2b give respectively (rv2x 1\ rv2b) V (rv 2y 1\ rv2a) ep (rv 2y 1\ rv2b) V (rv 2y 1\ rv2a) (rv2x 1\ rv2b) V (rv 2y 1\ rv2a) 'lj; (rv2 x 1\ rv2b) V (rv 2y 1\ rv2b)

= rv 2y 1\ rv2b, = rv 2y 1\ rv 2b.

It follows that

((rv 2x 1\ rv2b) V (rv 2y 1\ rv2a), rv 2y 1\ rv2b) E ep 1\ 'lj; and so 'I91at(rv2x, rv 2y) 1\ 'I9lat(rv2a, rv2b) ~ ep 1\ 'lj; as required. <> Theorem 8.24 If L E K1 ,1 then the relation ('I9 1 ,'I9 2)Ee

~ '19 1

edefined on Can L by

V
is a lattice congruence and (Can L)/e ~ [
Every e-class is ofthe form [ep I~L , ep] for a unique congruence ep E [ Corollary The mapping ep

f--7

epl~L

is a residuated dual closure on Can L.

Proof This follows immediately by [3, Theorem 15.1]. <> Theorem 8.25 For every L E K1,1 the lattices Can L and Can rvL have isomorphic centres. Proof Consider the mapping f : Can rvL ----t Can L given by f(ep) = Tj5. 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 ep E Can rvL we have ep = ep I~L and therefore ep = Tj5 = ](ep), it follows that] is injective. To show that it is also surjective, let '19 E Z(Con L) and let O! be its complement. By Theorem 8.7, 'I91~L and O!I~L belong to Z(Con rvL) and we have 'I91~L V O!I~L =

£,

'I91~L 1\ O!I~L = w.

Since 'I91~L ~ '19, O!I~L ~ O! and complements are unique, we deduce that '19

= 'I91~L =]('I9I~L)' <>

147

Congruences on K 1,1 -algebras

Theorem 8.26 If L E K1 ,1 isfmite 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 C 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

C ~ 19t = 19!

where + denotes dual pseudocomplements. Now for every 19 E Con L the smallest element of [19]c 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

~_O__l_a__b__c_d__e_=--f---,g:::....-_h_

~

1

°

h

h

g

h

h

g

f

e

d

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

A B C D E F G H I

= {O, 1,a, b, c,d,e,f,g,h}, = {{I, c,f,h}, {O,a, b.d,e,gn, = {{l,g,h},{O,a,b,c,d,e,fn,

= {{I, h}, {c,f}, {g}, {O, a, b, d,en, = {{l},{a,b,c,d,e,f,g,h},{On, = {{I}, {b, c,e,f,g,h}, {a,d}, {On, = {{I}, {c,f,h}, {a, b,d,e,g}, {On, = {{I}, {c,f,h}, {b,e,g}, {a,d}, {On, = {{I}, {d,e,f,g,h}, {a, b,C}, {On, = {{I}, {e,f,g,h}, {b, c}, {d}, {a}, {On,

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

Ockbam a,lgebras

148 K L M

N

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

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

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

o = {{ I}, {b }, { c ,j}, {g}, {b, e}, {a, d}, {O n, P = {{I}, {g,b}, {d,e,j}, {a,b, c}, {{On, Q = {{I}, {g,b}, {e,j}, {b, c}, {d}, {a}, {On, R = {{ 1}, {b }, {f}, {c}, {g}, {d, e}, {a , b}, {O}}, w = {{I}, {b}, {f}, {c}, {g}, {e}, {b}, {d}, {a}, {On· The congruence

B

c

There are eight ~-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 th~ following table, where 'I9(x,y) is at the intersection of column x and row y: 0 w 1 ~ w

a C

I

b C ~ c B A

W

R w

P Q w 0 N L w N 0 M R w ~ f B A L M 0 P Q W g A B F G E J If I w b ~ C D E q fl I K Q w 0 1 a b (:-'71 e f g b

d C e C

~

'>'-

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 Crespo de Morgan space) is a Priestley space X on which there is defined a continuous antitone mapping g such that g2 ~ idx Crespo g2 = idx ). Clearly, an MS-space Crespo a de Morgan space) is the dual space of an MS-algebra Crespo a de Morgan algebra); for g2 ~ idx and g2 = idx are the dual equivalents of axioms (1) and (a) 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 ofan MS-space; (2) there is a continuous dual closure map f) : X ---t X such that 1m f) admits a continuous polarity. Proof (1) =} (2) : If (X;g) is an MS-space, consider the mapping f) = g2. Clearly, f) 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 f). (2) =} (1) : Let f) : X ---t X be a continuous dual closure and suppose that 1m f) admits a continuous polarity a. Define g : X ---t X by the prescription (\Ix E X)

g(x)

= a[f)(x)].

Then g is continuous and antitone. Since f) fixes the elements of 1m f), we have (\Ix EX) whence g2

= f) ~ idx . <>

150

Ockham algebras

Note that the existence of a dual closure f) : X ---+ X such that 1m f) admits a polarity is equivalent to the existence of a subset XI of X which admits a polarity, and a decreasing isotone retraction 7f : X ---+ X l' In fact, it is clear that XI = 1m f) and that 7f is induced by f). Observe also that, by the nature of 7f, 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 f) 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 (tn )n~o has 10 (.en)n~O has .eo = 2, .e 1 = 1.

= 0, fi. = 1 and the Lucas sequence

Definition By an n fence we shall mean an ordered set F 2n of the form

TblA b zA b3

•••

ibn

~ ~ ~

it being assumed that n

~

an

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

< bn+1 ; and

Ffn+l

= F 2n U {ao}

with the single extra relation ao < b 1 • As the notation suggests, the ordered set F1n+l is the dual of F2n+l' In what follows we shall have no interest in F1n+l' for the following reason. If X ~ F1n+l 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 ----t X such that g2 ~ idx , namely that given by

g(a i ) = bn- i +1 , g(b i ) = an-i+l' 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 F 2n 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 For k

~ 2,

IL(Fk)l

= fk+2'

Proof Applying Theorem 5.5 to F 2n with x IL(F 2n )1

= an, we obtain

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

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

and then to F 2n - 1 with x

Writing 0'. k

Now 0'.2

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

= IL(F 2)1 = 3 =14

and 0'.3

= IL(F 3 )1 = 5 = fs·

Example 9.1 Consider the fence F 4

Hence O'.k

= fk+2' 0

:

The mapping g is given by

x al a2 b I b2 g(x) b 2 b1 a 2 al By Theorem 9.2, L(F4 ) has 16

= 8 elements.

Its underlying lattice is

f b

c

152

Ockham algebras

where a = aL b = bL c = at 1 = bt from which the remaining elements are easily identified. Using the fact that JO = X \ g-l (I), 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(F 2n ). 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 =1n+1' Proof Consider the subsets F 2n - 2 = F 2n \ {b 1,a n },

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 2n-4, and conversely. We shall establish the recurrence relation

IFixL(F2n )1

= IFixL(F 2n _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 1 it is clear that I contains either b 1 or anIf b 1 E I (so that an ¢ 1) then 1\ {b 1} is a distinguished down-set of F 2n-2 which contains a 1. If an E I (so that b 1 ¢ 1) and a 1 ¢ 1 then 1\ { an} is a distinguished downset of F 2n - 2 which contains bn . If {aI, an} ~ 1(so that {b 1, bn}nI = 0) thenI\{a1, 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 F2n-2 or F2n-4' Clearly, this correspondence is bijective and the required relation follows. To complete the proof, it suffices to observe from Example 9.1 that IFixL(F 4 )!

= 2 =h.

<)

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

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

b 1 an a n-1 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 +I )) = 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 ),

b n+1

> an = g2(b n +I )·

Example 9.2 Consider the fence F s :

The mapping g is given by x

al

a2 b l

g(x) b2 b l

By Theorem 9.2, L(F s) hasf7

a2

b2 b 3 al

= 13 elements.

bl Its underlying lattice is

f

where c = aI, a = at f = bI, b = bL i = b~, from which the remaining elements are easily identified. The operation ° on L(F s) is given by

j k 1 x 0 abc d e f g h X 1 j j h e e f c c b bOO O

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

Theorem 9.4IFixL(F2n+I)!

= IFixL(F2n )1 = fn+I'

154

Ockham algebras

Proof Suppose that I is a distinguished down-set of F2n+ l ' Then g(I) ~ F 2n +I \ I, g(F 2n +I\ I) ~ I. If bn+ 1 E I then I \ {b n+ d is a distinguished down-set of F2n that does not contain g(b n+I ) = b I ; whereas if b n+I ¢ 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(F 2n ). ~ 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

it being assumed that n If X

~ C 2n

~

2 and all the elements are distinct.

then clearly Xl

= X.

The mapping g described by

g(a j) = b n - 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) = bH!(n+I), k(b j) = aj+!(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 )I

= £2n'

Proof Consider first the 2n-fence F 2n as depicted above. If we insert a line from an to b I 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 F2n that contain b I but not an, that is all the down-sets of F2n \ {b n , a I } that contain b I , hence equivalently all the down-sets of F 2n \ {b I , aI, b n > an}. It therefore follows by Theorem 9.2 that IL(C 2n )I

= IL(F2n)I-!L(F2n-4)1 = f2n+2 = £2n'

- 12n-2 ~

155

MS-spaces; jences, crowns, ...

Example 9.3 The crown C6 is

and the mappings g, k are given by x

al

a 2 a3 b l

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

aL

bt

g

a aL

at

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

1 n pol k m

= b~.

jklmnopl

h g jed j

a c b 0

and on L(C6; k) it is given by xOabcdejghijklmnopl XO

1 n

0

p kIm g h i j d e jab 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 Zn \ {b l , an} is isomorphic to F Zn-Z and that g acts inside it. Suppose that I is a distinguished down-set of C Zn' 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 Zn-Z' Conversely, let] be a distinguished down-set of F zn - Z ' Then]u {an} is a distinguished down-set of C Z1l' This correspondence between the distinguished down-sets of C Zn and those of F Zn-Z is clearly bijective, so

IFixL(Czn;g)1 = IFixL(F zn - z )!· The result now follows from Theorem 9.3. <;

Theorem 9.7 For n odd, IFixL(C zn ;k)!

= R.n.

Proof Recall that, for n odd, k is given by k(a j) = bi+1(n+I), k(b j) = a j + 1(n-I), all subscripts being reduced modulo n. Observe that a distinguished down-set I cannot contain a pair of either of the forms {b j , bj-l+l(n+I)}, .2

In fact, if b j

E

{b j , bj+l(n+I)}' 2

I then {ai-l, a i } ~ I, whence bj-I+t(n+l)

= k(aj_l) ¢ I

and similarly

bi +1(n+l)

= k(a j) ¢ I.

Hence there are n forbidden pairs of the bi' and these can be enumerated cyclically as follows ;

{b l , b 1(n+3)}, {bt(n+3)' bz }, ... , {bn, b 1(n+I)}, {b 1(n+I), bd· Moreover, a distinguished down-set has cardinality n. It contains at most ~ (n - 1) of the elements b i and among them there is of course no forbidden pair. When such elements are chosen, the elements aj are determined since b j and k(b j ) are incomparable. Hence we see that the number of fixed points of L(C Zn; k) is equal to the number of subsets of the set {b l , ... , b,J that contain no forbidden pair. Our problem can therefore be restated as follows : if S11 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 t 1l

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+I possibilities for A, namely those subsets of {2, ... , n} that do not contain two consecutive integers. We deduce from this that

t n =fn-I + fn+I

= 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 {In)n~O by

io

=iI = 1,

("In ~ 2) in

= 2in-I + in-2'

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

Theorem 9.8 Lii i=O

= !-(In + in+I)'

n

Proof Let x n

= L i i and observe that, since i 0 =ir , i=O

3x n

= 2x n + x n

= 2io + (2ir + 10) + ... + (2in + in-I) + in + h + ... + in+I + in = x n + in+I + in· = 2io

It follows that x n = !-(In + in+I)'

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 i ) = bi' gdb i ) = a i ; g2(a i ) = b n- i+1, g2(b i ) = a n-i+1' Since gf = g~ = id, the MS-algebras L(DF 2n ;gl) and L(DF 2n ;g2) are de Morgan algebras. In fact, we can be more explicit: since x Mgl (x) we have that L(DF2n; g 1) is a Kleene algebra whereas, for n ~ 3, V(L (DF2n; g 2)) = M since g2(a1) = bnll a1' 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, Ii) the number of down-sets of X that contain a but not b.

Theorem 9.9 IL(DF 2n )1

=in+1'

Proof Consider the element b n of DF 21!' On the one hand, we have #

(DF 2n ; bn ) = # (DF 2n - 2; bn-d + # (DF 2n - 4 ),

and on the other hand #

(DF 2n ;lin)

= #(DF 2n ;an, lin) + # (DF 2n ;an, lin) = IL(DF 2n-2)1 + [IL(DF2n-2)1 - # (DF2n-2; bn- 1)] = 2IL(DF2n-2)1-#(DF2n-2; bn-d·

It follows that IL(DF 2n )1

= #(DF 2n ;bn) + # (DF 2n ;lin) = 2I L(DF 2n - 2)1 + IL(DF 2n - 4 )1·

Writing OI n = IL(DF 2n )1 we therefore have Since 011 = IL(DF 2)1 deduce that

= 3 = iz

and

012

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

IL (DF 2,J I = OI n =in+1'

Corollary 1

#

<>

(DF 2n ; b n) = !Un-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- 1)·

we

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

159

n-2

I: O'.i = #(DF 2n ; bn) -#(DF4 ; b2)· j=1

Note that # (DF 4 ; b 2 ) = 2 =io + h. Thus we see that n-2 # (DF 2n ;b n ) = I: O'.i + io + h j=1 n-2 = j=1 I:ii+l + io + il n-l = I:ii i=O

= !Un-l + in)

by Theorem 9.8.

<>

Corollary 2 # (DF 2n ;an, b,J = IL(DF 2n - 2)I = O'.n-l =in' <> As for fixed points, we observe that under the mapping gl the only distinguished down-set of DF(2n) is I = {al,"" an}. Consequently, we have IFixL(DF 2n ;gdl = 1. The situation concerning g2 is much more complicated.

Theorem 9.10 IFix L(DF 2n ;g2)1

i 1n 2 J !(n+l)

={ .

ifn is even; ifn is odd.

Proof Consider first the case where n is even. Let A c:::: DF n be the subset (also a double fence) consisting of the elements aI, ... ,a In, b 1 , •.• , b In' For 2 2 every down-set! of A let!* = A'\g2(I) where A' denotes the complement of A in DF211' Then every distinguished down-set of DF2n is of the form I U I * where I is a down-set of A such that al2 n E I and h 2 n ¢ I. By Corollary 2 of Theorem 9.9, the number of such down-sets is i in' Suppose now that n is odd. In this case we consider the subset B consisting of the elements aI, ... ,a!(n+l)' b 1 , ... ,b i(n-l)' Clearly, B is a distinguished down-set of DF 2n ; and every distinguished down-set of DF2n is of the form lUI* where I is a down-set of B that contains a Hn+ 1)' Clearly, this is # (DF n+1 ; at(n+l)' b t(n+l)) which, by Corollary 2 of Theorem 9.9, is i t(n+l)' <> Definition By a double crown we shall mean an ordered set DC2n of the form

of course, DC 2 c:::: 2 and DC4 c:::: DF4 .

160

Ockham algebras

On DC2n there is only one dual closure f with a self-dual image, namely f = id. All antitone maps g on DC2n 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+1' 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 n , k(b i ) = ai+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(Z2n)\ Proof Writing #(Z2n)

= Hfn+2 -

1).

= zn we have

= #(Z2n;aO)+#(Z2n;aO) = #(Z2n\{aO}) + # (Z2n-2) = # (DF 2n+2;bn+1) + Zn-1 = Hin+in+1)+Zn-1,

Zn

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

Zn -Zo Since Zo

= ! i=l IJli + ii+1)'

= 1 = !UO + id we obtain,

using Theorem 9.8,

n

zn

= ! LUi + ii+1) i=O

= tUn + in+1) + tUn+1 + in+2) -! = !Un+2 -1). <> Theorem 9.12 \L(DC 2n )1

= 2in + 1.

161

MS-spaces; fences, crowns, '"

Proof We obtain DC Zn from DF zn by linking an with b l , and al with b w Consider first the effect of adding to DFZn the link an-bl' Clearly, this reduces the number of down-sets. More precisely, in so doing we suppress the down-sets of DF zn that contain b l but not an, i.e. we suppress the downsets of DF Zn \ {b n- l ,an, bn} that contain b l ; equivalently, the down-sets of DFzn\{al, az, b l , bn- l ,an, b,J; equivalently, the down-sets of ZZn-3' A similar reduction in number occurs when we add the link al-b w It therefore follows by Theorems 9.9 and 9.11 that IL(DCzn)1

= IL(DFzn)l- 2 IZ Z(n-3) I =jn+l-jn-I+ 1 = 2jn + 1. 0

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

= 1.

The situation concerning Bz is as follows.

Theorem 9.13 IFix L(DCzn;Bz)1

= {~Hn-Z)

ifn is even; ifn is odd.

J Hn-l)

Proof Observe first that DC\ {aI, an, b l , bn} is isomorphic to DF Zn -4 and that Bz acts inside it. . If I is a distinguished ideal of DC Zn then I must contain a I and an but neither b l nor bw 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, a,J is a distinguished down-set of DCZn' This correspondence of distinguished down-sets is a bijection, so we deduce by Theorem 9.10 that IFixL(DCzn;Bz)1

= IFixL(DFzn-z;Bz)1 = {

j

1(

.2

n-Z

if n is even;

)

if n is odd.

J !(n-l)

0

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

br\)K A

b"

b"+l

~"'<M/ al

Theorem 9.14

#

az

a3

(W Zn +Z) = tUn+3 + 1).

a"

162

Ockham algebras

= #(W2n ) and (3n = #(W2n ; bn ). Then we have Oin+1 = # (W 2n +2 ) = # (W2n +2 ; b n +l ) + #(W2n +2 ; bn +l ) = # (W2n ) + (3n+1 = Oi n + (3n+1 .

Proof For every n let Oi n

We also have

Consequently, Oin+1

= Oi n + II U' n+1 + J.n+2) .

We deduce from this that I " . I . = 0i2 + 113 + J4 + '" + In+1 + 1Jn+2' Since 0i2 =#(W4 ) = 9 = ~ + io + it + h + ~h, we then have

Oin+1

n+1

Oin+1

= ~ + 'LJi + !In+2 i=O = ~ + ~Un+1 + in+2) + !In+2

= ~(l + in+3)'

0

Corollary #(DF 2n ; b l , b n ) = Hjn-I + 1). Proof For n have

= 1, 2 the result follows by direct computation.

Theorem 9.15 For n even, IFix L(DC 2n ; k)1

For n

~

3 we

=i in -1. 2

Proof Consider the subset A = {al, ... , ai2 n , b ll ... , 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] =>

atn+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 b1n> ahd 1* does not 2 contain both b !n+1 and btl' The latter condition is equivalent to al E I and a t n E I. It follows that the number of fixed points of (DC 2n; k) is t

= # (DF n ) -#(DFn ; bl , bl n ) -#(DFn ; al,"ih n ). 2

2

163

MS-spaces; fences, crowns, '"

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

t

= }!1I+1 -

Hi!1I-1

=}1

1 -2

2 11

+1 -

(j12 n -1

+ 1) -#(W~(!1I_2)+2) + 1) - -21 (j12 n +1 + 1)

= }In -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 complex 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 hatracks) are considered and corresponding but quite different results are obtained, some of the algebras in question belonging to MS-subvarieties other than M and M1 . 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 IJI. 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 21l' 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

LtJ

(n _m)2

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

L!I=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

... ,

= 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 , ... , b n+k , each a i being incomparable with b i , ... , b i +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 C!

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 ojordern 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 i < bj if and only if j - i E {O, 1, ... , k -I} modulo n. All vertices of C k (n) have degree k. All C k (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 ofa 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 bi' In fact, Cn(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 - I (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 = ni and IMax CI = nz then the number of edges is nik = nzk whence ni = nz = n. The following examples show that the conditions (1), (2), (3) are independent :

I

I

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

<> A

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 : b

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 of order n in which

(Oiij]

Oiij

I = { 0

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

bI b 2 b 3 b 4 bs b 6 al az

1 0 1 1 0 1 1 0 1 1

a3

as

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

a6

0

a4

1 1 0 1

1 0

1

Any interchange of rows or columns yields another Cn;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 bz b 3 b4 b s b6

al

0

1 1 0 1

1

az

1 0 0 1 1

1

a3

1 1

1 1 1 0 1 1. 1 0

0 0

1 0 0 1 1 0 1 1 0 1

1 1

a4

as a6

to which corresponds the generalised crown bl

b2

b3

b4

bs

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

The dual space ofa 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 matrix 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;f) E 0 is finite with dual space (X; g) then (L;f) is simple if and only if gW(x) = X for 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 gn for every n. (2) If IX/ = 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 P m,n 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, Le. is an N-cycle. (3) If N is odd then the order on X is discrete, Le. 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 Si E IN such that Xi = gSi(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 gSi 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 :

Ockham algebras

168

Theorem 10.1 Every connected component ofthe dual space ofafinite 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:::S. 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( 0) 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 ~ y. If t is the number of connected components of X then for o~ i ~ t -1 we denote by C i the component that contains i. The connected component that contains 0, namely Co = {x E X I x ~ O} has the following important property. If r is the smallest non-zero element ofr(O) then, r being odd, we have 0 -< r, r>- 2r, 2r -< 3r, ... , so Co is the subgroup (r) of the group 71.. 211 , the remaining connected components being the cosets of Co' The following two properties appear in [69].

Theorem 10.2 Co is self-dual. Proof Let p be a fixed element of r (0) and define rpp : Co rpp(q) Ifq

E

Thus Imrpp

= 2n - q + q ~ 2n - q + 2n - p = 2n - p

~ Co.

If now r

E Co

rpp(2n - r - p)

X by

= 2n -p -q.

Co then, since 0 -

0= 2n

----7

then 2n - r -p

= 2n - p

~

- q

E Co

and

- 2n + r + p

= r.

2n -p so

= rpp(q).

The dual space ofa finite simple Ockham algebra

Thus 1m (jJp

= Co'

Since p is odd, we also have

(jJp(r)-j(jJp(s)

Hence

(jJp

169

<¢==?

2n-r-p-j2n-s-p

is a dual order embedding and Co

= 1m (jJp

<¢==?

rts.

is self-dual. <>

If X is not connected then we have 1 = g(O) ¢ Co; for 1 E Co gives o !Xl 1 whence 1 = g(O)!Xlg(l) = 2, and soon, whence X would be connected.

Theorem 10.3 Two connected components of X are either isomorphic or dually isomorphic. More precisely, d

Co'::::'. C 1

d '::::'.

C2

d

d

'::::'. ... '::::'.

Ct - 1

d '::::'.

Co

and the number t ofconnected components is necessarily odd.

Proof It suffices to prove that Cot. C 1 . For this purpose, consider the map 1jJ : Co ----t X defined by the prescription 1jJ{q) = q + 1. Since 0 !Xl q implies 1 !Xl q + 1 we have 1m 1jJ .~ C 1 . But if r E C 1 then r !Xl 1 gives r-1 = r+2n -1 !Xl 0, so that r-1 E Co with 1jJ(r-1) = r. Hence Im1jJ = C 1 . Since 1jJ(r) -j 1jJ(s)

<¢==?

r + 1 -j s + 1

<¢==?

r

= r + 1 + 2n -1 t

s + 1 + 2n -1

=s

it follows that 1jJ is a dual order-embedding and Cot. C l' 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(O) is a singleton. Theorem 10.4 .[/T(O) = {r} then X has t nents each of which is an !f -crown.

= gcd{n, r}

connected compo-

Proof In the cyclic group 7L2n 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 7L2n it follows that Co = {O, r, 2r, ... , e;l -l)r} and we have o-( r, r)- 2r, ... , e;Z -l)r)- O. Consequently we see that Co is an !f-crown. Now each connected component of X is a coset of Co' The number of connected components is therefore 17L2n l/ICol = t. <>

Corollary X is an n -crown ifand 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 connected components are the following 5-crowns : 9

27

15

3

=3

and the

21

~

Theorem 10.4 can be generalised as follows.

Theorem 10.5 Let r(O) = {rl'"'' rd. 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 ~2n' 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 ~2n, namely t = 1~2nl/l(u)1 = gcd{2n, u} = gcd{n, rl,"" rk}' <>

Corollary If r(O) = {rl"'" rd then X is connected integers n, r 1 , .•• , rk are coprime. <> Example 10.2 Let n 3

9

15

21

= 15 27

and r(O) 1

7

= {3, 9}. 13

19

Then t 25

5

if and only if the

= 3 and X is 11

17

23

~2~~

29

Theorem 10.6 Let X be ofcardinality 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 ofeach 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 2jr(O)1 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 ofa finite simple Ockham algebra

171

Corollary There are generalised crowns that cannot be made into the dual space ofa 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 [Q!i) with the property that, modulo n, Q!ij = Q!i+I,j+I' In a circulant (0, 1}-matrix we shall say that the i-th row is symmetric if Q!ij

=1

~ Q!i,n-j

= 1.

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, kzt, ... , 2n -kzt, 2n -kIt,

tn.

all k i being odd, and MinCo = {0,2t, ... , 2(n 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 Cn;k' so it can be represented by a square (0, 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, kzt, ... , 2n -kzt, 2n -kIt

172

Ockham algebras

then 2t is covered by (k 1 +2)t, (k 2 +2)t, ... , 2n-(k 2 -2)t, 2n-(k l -2)t,

Le. 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, I)-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 0 A=

[i

0 0

1 1

1 1

0

~]

1 and

B=

0

[f

0

1

1 1 0 0

1]

with which are associated the graphs 1

3

5

7

o

2

4

6

GA:~

and

GB

1

3

5

7

o

2

4

6

:)%Z

In GA we have x --( g(x) or x)- g(x), whereas in GB 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, I)-matrix of order n that represents a finite simple Ockham algebra and in which the first row is doubly symmetric in the sense that O!lj = 1 "'* O!l,n-j = 1 and (Vj ::;;; n') O!lj = 1 "'* O!l,nl-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,

Tbe dual space ofa finite simple Ockham algebra

173

whereas 4 and 10 are covered by 3, 5, 9, 11 . The corresponding matrix is 10110

1

110 1 1 o 1 101

0 1

101 110

o

1

101 1 1 0

101

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 = [Cliij] be a square (0, I)-matrix of order n. For every pair (i,l) denote by Pi,j(M) the number of values of k for which Cliik = Clijk = 1. In particular, Pi,i+l (M) does not depend on i if M is circulant. Let also Pi,j(M) = {pi,j(M) ; i,l E {I, ... , n Clearly, Pi,j(M) is invariant under arbitrary permutations of the columns and the rows. The symbols Pi,i+l,i+2,..,(M) and Pi,j,k,..,(M) have obvious meanings.

n.

Example 10.4 The C 5:3 whose matrix M is 1

o 1

o 1

100 1 1 1 1 0 0 101 1 1 1 0 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) = {I, 2} whereas Pi,j(M) = {I, 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{l,O, 1, 1,0, I}.

Here we have p.t,}.(A) = p.t,}.(B) = {2 , 3} and p.t,}.(C) = {2 , 4}. Since p.t,}.(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

° M=

1 1 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 CY. n ofnon-isomorphic finite simple Ockham algebras that belong properly to P2n ,o is given by 2 -21n CY.

n

={

2 !(n+l)

if n if n l

is even; is odd.

Ibis includes {)n algebras whose dual spaces are not connected, where {)n is the number ofsubsets {rl,"" rk} of r(o) for which gcd{n, rl,"" rd t= 1.

Proof The Ockham space X is completely determined by its cardinality 2n and the subset r(O). If is even then W(O)I ::; and the number of

n

°::;

!n

Tbe dual space ofa finite simple Ockham algebra

possibilities is

175

(ton) + (tt) + ... + 0:) =

2t

n

.

When n is odd we have 0 ~ 1r(0)1 ~ t(n + 1) and the number of possibilities becomes 2t(n+l). Moreover, if IXI = IX'I with r(0) = {rl'"'' rk} in X and r(O') = {ri, ... , riJ in X' then we have X~X' ~ {rl, ... ,rk}={ri, ... ,rk}.

The final assertion is an immediate consequence of the Corollary to Theorem 10.5. <>

Example 10.7 Let n = 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 (in which it suffices to describe in each case only the component Co and to indicate the number t of components). For r(O)

= 0 we have Co = {O} and t = 30.

For 1r(0)1 = 1 the possibilities are 3

9

15

21

27

9

~ [;kl o

r(0)={3 } 1=3

9

15

15

3

21

mfi I o

10 20 r(0)={5} 1=5

r(0)={9} 1=3

r(0)={15 } 1=15

= 2 the possibilities are

For 1r(0)1 3

27

21

27

3

9

15

21

27

5

15

25

9

27

15

3

21

~~~~ r(0)={3,9} 1=3

r(0)={3 ,15} 1=3

r(0)={5 ,15} 1=5

r(0)={9 ,15} 1=3

For 1r(0)1 = 3 we have

~

o

6 12 18 r(0)={3,9 ,15} 1=3

24

Remark The following table gives the first 35 values of 19 no

176

Ockham algebras n {)n 1 1 2 1 3 2 4 1 5 2 6 2 7 2

n {)n 8 1 9 4 10 2 11 2 12 4 13 2 14 2

n 15 16 17 18 19 20 21

n 10 22 1 23 2 24 8 25 2 26 4 27 18 28

{)n

n 2 29 2 30 16 31 8 32 2 33 32 34 4 35

{)n

{)n

2 38 2 1 66 2 22

There appears to be no obvious rule of formation here. However, the following observations are useful : (1) if n = 2a then

{)n =

1;

(2) if n = pr withp an odd prime then {)n = 2 Hpr-l+1). In particular, if n = p then {) 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 !(pr-l + 1), and {)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

if and only if IXI is odd; L has two fixed points if and only if X is an antichain ofeven cardi-

(1) L is fixed point free (2) nality;

(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. ~ If ::::;; and ::::;; 1 are orders on the same set X and if ::::;; 1 is an extension of ::::;; then a mapping g : X - t X can be order-reversing on (X; ::::;;) but not

The dual space ofa 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.

o•

•1

I

z•

o

(Xl; ~)

~

z•

o

(Xl; ~d

~

z

o

(X z; ~)

z

(X z; ~d

g is order-reversing on (Xl; ~) but not on (Xl; ~d, and is not order-

reversing on (X z; ~) but is on (X z; ~d. The following result, although trivial, is of interest in this context. Lemma Let (X; ~ ,g) be a finite Ockham space.

a way that g remains order-reversing on (X;

~ 1)

If

~1

extends

~

in such

then

(1) (X; ~1 ,g) is also an Ockham space; (2) the dual algebra (L 1;j) of (X; ~ 1 , g) is a subalgebra of the dual algebra (L;j) of (x; ~ ,g); (3) the numberoffixedpoints of (L 1,!) is at most equal to that of (L;j).

Proof It suffices to observe that the operation Jon O(X; ~1 ,g) is the restriction of Jon O(X; ~ ,g), that every down-set of (X; ~d is a down-set of (X; ~), and that every fixed point of (O(X; ~1 ,g);j) is a fixed point of (O(X; ~ ,g);j). <> 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 ofthefinite 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 L o = (2 6;j) whose dual space is the antichain {O, 1, ... , 5} and the subalgebras L 1, Lz, L3 whose dual spaces are

* o

z

Xl

~ i! I

40

Z

Xz

40

Z

4

X3

In the diagrams that follow we omit, for visual reasons, a great many of the lines. As a lattice, L o ~ 26 . The algebra L o has two fixed points, corresponding to the down-sets {O,2,4} and {1,3,5}. Each of L 1,L z ,L 3 has a single fixed point. As a lattice, L 1 ~ 23 EB 23 . The elements of L z are

178

Ockham algebras

those of L 1 together with the three elements that correspond to the down-sets {O, 1, 2}, {2, 3,4}, {O, 4, 5}. The elements of L 3 are those of L 1 together with twelve others. Amongst these are the mid-level elements that correspond to the down-sets {O, 2, 3}, {O, 3, 4}, {O, 1, 4}, {1 , 2, 4}, {2, 4, 5}, {O, 2, 5}. By Theorem 10.11, L 1 is a subalgebra of both L2 and L 3 •

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; rv). If the subvariety V of 0 to which (L; rv) 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

a

b

cannot be made into an M 1-algebra since otherwise we would have 1 = rvO

= rv(a /\ b) = rva V rvb whence either rva = 1 or rvb = 1, so that either Za = 0 or Zb = 0 from which it follows by axiom (1) that either a = 0 or b = 0, a contradiction. rv

rv

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 PZ,l' 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 ~ X by g(x) = y for every x E X. Clearly, g is orderreversing and gZ =g. Thus X has been made into a PZ,l-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(0). Then everyfinite distributive lattice is a reduct ofa V-algebra if and only if V :2 PZ,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 very 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 every 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 ~ by V~W ~

Vr=Wr .

The ~-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 ----> P such that a(Q) is a convex subset of P. For example, the ordered set

N

N

as a subposet but not as a convex subposet. contains A class c of ordered sets is said to be convex-closed if, whenever PEe, every 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;/) E 0 and let (X;g) be its dual space. !ben

L E B ifand only if g = gO; L E K ifand only if gO = g2 and gO Kg; L E S ifand only if gO ~ g; L E S ifand only if gO ~ g; L E S ifand only if g = g2 and gO Kg. 0

For every (finite) LED we denote by XL the dual space of X, and for every (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 (1) Let L be ajinite distributive lattice and let M a closed interval of L. !ben X M is a convex subposet of XL'

= [a, b] be

181

Relative Ockham algebras

(2) Let Q be a convex subset of the finite ordered set P. Then there is a closed interval M of 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 f(p) = P U A. Clearly, f(P) is a down-set and A ~ f(P) ~ B, whence f(P) E M. Moreover, f is injective, f(CJ(B \ A)} = M, and f is a lattice morphism. (2) : Let Q be a convex subset of P. Then

= QL n Qi

Q

= 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. is order-isomorphic to X M where M = [A,B]. <>

Hence Q

Corollary Afinite distributive lattice L is a relative V -algebra if and only if every convex subposet of 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 of ordered sets which have no Pi as a convex poset. • If every member of can be made into a V-space and if cis convexclosed, then the finite relative V-algebras are exactly those L for which XL E C. Moreover, the lattices for which XL 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.

c

c

c

Lemma 11.2 Let P be afinite ordered set. Then thefollowing 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 ofallfinite 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 thefollowing are equivalent: (1) P is a disjoint union oftrees;

A as a subposet; (3) P does not contain A as a convex subposet. (2) P does not contain

Tbe class ofall 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 x' be a minimal element of the set xl nuT n v T. Then x' > u, x' > v, and there is no y such that x' > y > u and x' > y > v. Take u ' to be a maximal element of (X 'l \ {x'} )nuTand Vi to be a maximal element of (x 'l \ {x'}) n v T. Then Q = {x', u ' , v'} is convex by the definition of u ' and Vi. Moreover, u ' II Vi since, for example, u' ~ Vi would give u' E x'l nuT n v T, which is impossible by the minimality of x'. 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 thefollowing are equivalent: (1) P is a disjoint union ofchains;

A nor V as a subposet; (3) P contains neither A nor V as a convex subposet. (2) P contains neither

Tbe class ofall disjoint unions offinite chains is convex-closed.

<>

Theorem 11.4 Tbe P::3-classes ofB, K, S, S are asfollows: (1) [B] = {V E A(O) I V dB, V;;Q K, V;;Q S; V;;Q S}; (2) [K] = {V E A(O) I V d K, V;;Q S; V;;Q S}; (3) [S] = {V E A(O) I V d S; V;;Q S}; (4) [8] = {V E A(O) I V d S; V;;Q S}. Proof We restrict ourselves to establishing (1) and illustrate the 'strategy' In fact, there are described above. The family {P j} j E I is formed by the only three distinct g-maps on the two-element chain x < y, namely those described as follows :

I.

x gl (a) Y g2(a) x g3(a) y a

y

x

x y

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. 0 The next result gives an algebraic description of V r 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. !ben (Q!) if V E [B] then L E V r if and only if L is boolean; ((3) if V E [K] then L E V r if and only if it satisfies any of the following eqUivalent conditions: (1) L is a direct product ofchains; (2) L contains neither 22 EB 1 nor 1 EB 22 as an interval; (3) L has neither 2 2 EB 1 nor 1 EB 22 as a homomorphic image. b) JfV E [S] (resp. if V E [S]) then V r is defined in the following way: (a) the trivial algebra and the 2-element chain are in V r ; (b) anyfinite direct product ofelements of Vr is in Vr; (c) if L E V r then 1 EB L E V r (resp. L EB 1 E V r ); (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. 0 If we apply Theorem 11.5 to the subvariety M1 (see the diagram on page 92), we obtain [B] = {B}, [K] = {K, M, K1 , M V KIl, and [S] consists of all the other subvarieties of MI' So parts ((3) and b) of Theorem ll.s 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-emptyfinite 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

CXi' Define g by g(x)

=

Xj

whenever

l';;;j ';;;k

x MXj' Then g is order-reversing, g2 = g, and g MgO; so (p;g) is an S-space.

Ockham algebras

184

'* : Let (F;g) be the dual space of an S-algebra.

Since gZ = g, we have = g(F) is an antichain; in fact, if g(YI) = gZ(YI) ~ gZ(yz) = g(yz) then gZ(YI) ~ gZ(yz) and so g(YI) = g(yz). Since g ff gO we have F = ,))L U ,))i. 0 that,))

Lemma 11.6 Let (X; g) be the dual space ofa finite 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

< v, u" v, u II y, x II 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 II b, a II d, c" b. u

< x,

Y

< x,

Y

First, consider

XI

= u i n yin (X \ v i).

This set contains x but no element t below v (otherwise u ~ t < v which contradicts u " v). Let x I be a minimal element of X I' Then {u, X I, Y , v} is isomorphic to

N.

Next, consider Xz =

xi 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 "v; for UI ~ v gives the contradiction u ~ v, and UI ~ v gives the contradiction x I ~ v. Moreover, U I We now show that UI -< Xl' Suppose in fact that there existed Uz such that UI < Uz < Xl' Then Uz ~ u; and Uz j v since otherwise we have the contradiction Xl ~ V. By the minimality of Xl we then have Uz j y. We also have Uz 1, y, for otherwise we have the contradiction U I ~ y. Consequently, Uz E X z. But this contradicts the maximality of UI' Hence we have that

"y.

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 gZ(d) = a and we have the contradiction gZ(d) 'f 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 " c. 0

!!elative Ockham algebras The lattice dual of the fence

185

N is the lattice L

8

with Hasse diagram

It i~ ~p.own that the finite distributive lattices which do not have L 8 as a hompfIl.Qrphic image are precisely those whose duals do not contain the

fence N-as a subposet. ","

Ji,:

The !qtter class of ordered sets coincides with the class of series-parallel posets ~58,71}, defined as follows. A (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 jnStlnce, all trees are series-parallel. By Lemma 11.6, the dual spaces of ~r~~j~e~ra~ ~re series-parallel. The. four-element crown ;'!".'"

nnite

cM d

a~b is series-p,arallel, being the linear sum of the antichains {a, b} and {c, d}. By Lemrn.'a' 11.6 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 punt up from 2-elementchains using direct product and vertical sum. Lemmas 11.5 and 11.6 lead mther easily to the following

Theorem 11.6 Let P be a finite series-parallelposet. Then thefollowing statements are equivalent: (1) every connected component of P has a node; (2) no convex.subsetof P' is

(3) in the construction of P from singletons there is a restriction on linear namely that PI E9 P 2 is permitted only when PI has a biggest element P 2 has a smallest element (or both). 0

~t,.tms,

qr

Ix!;

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 the following statements are equivalent:

(1)

LESr ;

(2) L is a member ofthe class C defined as follows: (a) the trivial algebra and the 2 -element chain are in C; (b) the directproduct oftwo members of C is in C; (c) if L 1 and L2 are in C U {0} then L1 61161 L2 is in C; (d) every member of C is obtained in a finite number ofsteps using the properties (a), (b), (c). (3) L does not have L s as a homomorphic image and L contains no interval isomorphic to 2 2 612 2 . 0

12 Double MS-algebras The notion of a double 5tone algebra is well known. This is an algebra (L; V, 1\, *, +,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 M5-algebra is an algebra (L; V, 1\, 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) (D2) (Va E L)

a+O aO+

= a++;

= aCe.

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(21 = {(a, b) E B 2 I a ~ b} then B(21 is a double Stone algebra in which (a, b)o = (b', b') and (a, b)+ = (a',a').

Example 12.2 Every de Morgan algebra (M; -) is a double MS-algebra; it suffices to define aO = a+ =a for every a EM. Example 12.3 If (M; -) is a de Morgan algebra and (5; *, +) is a double Stone algebra then (Mx 5; 0, Ell) is a double MS-algebra where (a, b)o = (a, b*) and (a, b)Ell = (a, b+). Immediate consequences of (Dl) and (D2) are the following. Theorem 12.1 if (L; 0, +) (1) (Va E L)

(2) LOO

is a double M5-algebra then

aO ~ a+;

= r+ .

Proof (1) : Writing aO for a in (D2), we obtain aOo+ = aOOO = aO; and from a ~ aOO we have, since a H a+ is antitone, aOo+ ~ a+. (2) : (Dl) gives L++ ~ LOo; and (D2) gives the reverse inclusion. <> The skeleton 5 (L) of a double MS-algebra (L; °, +) is defined in the same way as for any Ockham algebra, namely by 5(L)

= LO = {XC I x E L}.

Ockham algebras

188 By Theorem 12.1 we also have

5(L)=L+={x+ IXEL} and clearly a E 5(L) if and only if a = a OO = a++. Every double MS-algebra 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 mappingf : E ----t F is said to be residuated if the pre-image under f of every principal down-set yl of F is a principal down-set of E. By [3, Theorem 2.5], f : E ----t F is residuated if and only if it is isotone and there is an isotone mapping g : F ----t E such that fog ~ idF and g 0 f ~ idE' Such a mapping g is necessarily unique, is called the residual of f, and is written as f+. It is readily seen that

(Vy

E

F)

f+(y)

= max{x E E

If(x) ~y}.

For a residuated mapping f we have

f

0

f+

0

f

=f

and f+

0

f

0

f+

= f+ .

Moreover, by [3, Theorem 2.10], the follOWing statements are equivalent:

f is a closure; f+ is a dual closure; f = f+

0

f;

f+ = f

0

f+.

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 ~ a Oo g2 ~ idx

a O+ = a Oo g oh = g2

a~a++

h 2 ~ idx

By way of example, for each x

E Ip (L)

aO~

h(x)={ala+~x}

gh(x) = {a I a O~ h(x)} = {a I a O+ EX} g2(X) = {a I a O~ g(x)} = {a I a Oo EX},

whence a o+ = a OO gives g 0 h

a

~

h(x)

=}

= g2; and

a+

EX=}

a OEX=} a

~

a+

g~h

we have

g(x)={alaO~x},

and so

a+ o = a++ hog = h 2

g(x)

189

Double MS-algebras

so g(x) ~ 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 ofa 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 a. Then {) ={) 0 {)+ and {)+ = {)+ 0 {) with {)+ a closure on X. Define g, h : X ~ X by (Vx E X)

g(x) = a{)(x),

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

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

{)+ ;;;>-

idx . Also,

= a{){)+a{)(x) = a{)a{)(x) =g2(X), = {)+a{)a{)(x) = h 2(x), h = g2 and hog = h 2. ~ (g

0

h)(x)

(h og)(x)

whence g

0

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' ~ Yi

Ockham algebras

190

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

E

E) zez' ~ x;

and strongly regular if it is both.

Theorem 12.3 Let X be an ordered set and let e be an equivalence relation on X. Then the following statements are equivalent: (1) there is a dual closure '13 : X ---t X with Ker '13 = e; (2) e is strongly lower regular and every E>-class is bounded below. Proof (1) => (2) : If (1) holds then clearly every Eklass is bounded below, the smallest element of [x]e being '13 (x). To see that e is strongly lower regular, suppose that z ~ xey. Then 'I3(z) ~ 'I3(x) = 'I3(y) ~ y and so ze'l3(z)~y.

(2) => (1) : Suppose now that (2) holds. For every xE X define

= min [x]e. and Ker e = '13. Since e 'I3(x)

Then clearly {) = {)2 ~ idx is strongly lower regular it follows from z ~ ye'l3(y) that there exists z' E X such that zez' ~ 'I3(y) whence 'I3(z) = 'I3(z') ~ 'l3 2 (y) = 'I3(y). Consequently {) is also isotone and hence is a dual closure. ~

Theorem 12.4 Let X be an orderedset and let '13 : X ---t X be a dual closure. Then thefollowing statements are equivalent: (1) '13 is residuated; (2) e = Ker {) is strongly upper regular and every E>-class is bounded above. Proof (1) => (2) : If the dual closure '13 is residuated then {)+ is a closure on X. Moreover, from the relations '13 = '13+ 0'13 and '13+ ={) 0 {)+ we deduce that Ker '13+ = Ker '13. The dual of Theorem 12.3 now gives (2). (2) => (1) : Suppose now that (2) holds and define