Notes on Tolerance Relations of Lattices: a Conjecture of R.N

Journal of Pure and Applied Algebra 68 (1990) 127-134 127 North-Holland

NOTES ON TOLERANCE RELATIONS OF LATTICES: A CONJECTURE OF R.N. McKENZIE*

E. FRIED and G. GRATZER Department ofMathematics, University ofManitoba, Winnipeg, Manitoba, Canada R3T2N2

Received 20 May 1986 Revised 1 February 1987

1. Introduction

A tolerance relation T on a lattice L is defined as a reflexive and symmetric binary relation having the substitution property. A maximal T-connected subset of L is a T-block. The quotient lattice LITconsists of the T-blocks with the natural ordering. If V and Ware lattice varieties, their product VO W consists of all lattices L for which there is a congruence relation 8 satisfying: (i) all 8-classes of L are in V; (ii) LI8 is in W. In general, VoW is not a variety; however, H(VoW) (the class of all homomorphic images of members of VoW) always is. If L is in VO W (established by 8), then LIt/> is a typical member of H(Vo W). On LIt/>, 81t/> is a tolerance relation. The following theorem was conjectured by R.N. McKenzie: a lattice K belongs to the variety generated by VoW iff there is a tolerance relation Ton K satysfying: (i) all J:classes of L are in V; (ii) LITis in W. In this paper we disprove this conjecture:

Theorem. The lattice F ojFig. 1 is in H(M3 0 D). However, there is no A in M 3 0 D with congruence 8 establishing this such that F can be represented as Alt/> and T= 81t/> satisfies (i) all T-classes ofA are in M 3; (ii) AIT is in D.

In this theorem, M 3 is the variety generated by the modular lattice M 3 and D is the variety of all distributive lattices. The Theorem holds for some varieties other than M 3 , see Section 5. In Section 2 we introduce a new lattice construction, called hinged-product which

* The research of both authors was supported by the National Scientific and Research Council of Canada.

Fig. I. we shall utilize to construct an interesting lattice. Since we hope that this construc tion will find other applications as well, we develop it in some detail. In Section 3 we introduce the lattice F of Fig. 1, and using a hinged-product (see

Figs 2 and 3), we show that F is in M 3 0 D. In Section 4 we investigate the tolerance relations of F; they form a lattice shown in Fig. 6. Finally, in Section 5, we prove the Theorem. For the basic concepts and unexplained notations, the reader is referred to [1].

2. Hinged-products

We start with a definition:

Definition 1. We are given a family, L i , iEI, of lattices; in each lattice L i , we are given three elements, the hinge: lis,mis,ui' The hinged-product H=H(Li,li,mi,ui) consists of the following subsets of the direct product of the L i , i E I: (i) the ideal I(H), the direct product of (I;], i E I; (ii) the dual ideal u(H), the direct product of [Ui), i E I; (iii) for every i E I, the ith frame, fi(H), consisting of all elements whose jth coordinate is mj for all j*i. H is partially ordered componentwise.

Observe that these sets may not be disjoint: for instance, if mi = Ii' then I(H) n fi(H) is non-empty; if mi = Ui' then u(H) nfi(H) is non-empty. The element f.1. whose ith coordinate is Ui for all i E I belongs to all fj(H), j E I. Notes on tolerance relations of lattices 129

Lemma 2. H is a lattice.

Proof. Let a,peH. Then the componentwise join, aVP, is always in H (hence it is also the join in H) with one possible exception: a efi(H) and P e}j(H), i *-j. Let a=(ak), P=(bk); now if aivmi>mi and bjvmj>mj, then avp has no upper bound in I(H), in any fk(H), and it has a least upper bound, namely, aVpv/-l, in u(H). Hence, a and P have a least upper bound in H, namely, aVpv/-l. We can argue the meets dually. This completes the proof of Lemma 2. 0

We shall continue to denote componentwise join and meet with V and /\ ; the join and meet in H will be denoted by Y.. and 6, respectively. It is not very easy to visualize H. The ith frame, fi(H) is isomorphic to Li. The fi(H) are glued together at the hinges: Ii, mi' Ui' The glued frames are completed into a lattice by I(H) and u(H). The example of Section 3 may help illuminate this point. If we have homomorphisms ¢i: L i -.L;, then under certain conditions these homomorphisms have a joint extension from the hinged-product H to the hinged product H':

Lemma 3. Let H =H(Li, Ii, mi' Ui) and H'=H(L;, I:, mj, uj) be hinged-products with the same index set I. Let ¢i: Li -.L; be homomorphisms for i e I. Let us assume that ¢i(li) = I:, ¢i(mi) = mj, and ¢i(Ui) = uj. If (i) or (ii) below holds, then the restriction ¢ of the product of the homomorphisms ¢i' i e I, to H is a homo morphism of H into H: (i) For all i e I, Xi V mi > mi implies that ¢(Xi) V mj> mj; and the dual condition for /\. (ii) For all i e I, ¢(Ii) = ¢(Ui)'

Proof. Under the first condition, whenever avp*-ay"p, then ¢(a)v¢(p)* ¢(a)y"¢(p); therefore, ay"p=avpv/-l and ¢(a)y"¢(p)=¢(a)v¢(p)V¢(/-l). It now follows that ¢(a)y"¢(P) = ¢(ay"p). The argument for the meet is dual.

u,m

Fig. 2. 130 E. Fried, G. Gratzer

Under the second condition, t/J(mj) = ¢leu;), so t/J(a) y. t/J(!3) = t/J(a y"P) is obvious.

Note that a necessary and sufficient condition for t/J to be a homomorphism can be easily formulated. We only need the sufficient conditions of Lemma 3 in this paper.

3. Fis in H(M) 0 D)

We start with the lattice of Fig. 2. We take three copies of L, L 1, L 2, and L3• The elements will be denoted accordingly: Ul' U2' and so on. Let A denote the hinged·product (power) of L ll L 2 , and L 3 • The diagram of A is given in Fig. 3; I(H) = {o}, u(H) = [p). Now consider the congruence relation e(O,p) of L, where 0=(11,/2 ,/3) and p= (m1' m2' m3) (U1' U2, U3)' The natural homomorphism of L onto Lle(O,p) ob· viously satisfies (ii) of Lemma 3, hence A has a natural homomorphism onto the appropriate hinged-product of three copies of Lle(O,p).; this new lattice is isomor phic to (el )3. There are eight e·classes: four are isomorphic to M 3, three to (C2)2, and one to (ez)3. Since the congruence classes of e are either distributive or isomorphic to M 3 , and Ale is distributive, we conclude that A belongs to M 3 0 D.

Fig. 3. Notes on tolerance relations of lattices 131

u,m

Fig. 4.

Next consider the congruence e(J.1,k) on L, where k=(kl,k2,k3 ). Lle(/l,k) is the lattice of Fig. 4. The hinged cube of that lattice is isomorphic to the lattice F of Fig. 1. Hence FE H (M 3 0 D). This proves the first sentence of the Theorem.

4. The tolerance relations of F

Fig. 5 represents F with the tolerance relation T= ele(O,/l). T is a natural tolerance on F; unfortunately, FIT is isomorphic to M 3 , and it is not distributive. This makes the Theorem plausible. If there are A, e, tP such that A E M3 0 D is established bye, and T= e/tP satisfies that (i) all T-classes of A are in M3 ; (ii) AIT is in D, then A must be the lattice of Fig. 3 (or a fatter version with larger distributive classes), and e and tP must be as in Section 3. In Section 5 we shall prove this. As a first step, we have to describe all the tolerance relations on F.

with tolerance To

Fig. 5. 132 E. Fried, G. Griitzer

Fig. 6.

Lemma 4. The lattice F has nine tolerance relations; they form a lattice as shown in Fig. 6. The tolerance To is depicted in Fig. 5, and the tolerance ~ in Fig. 7.

Proof. Since F is a finite, simple, modular lattice, any two prime intervals of Fare projective. Therefore, F has a unique minimal proper tolerance relation, To, gener ated by any prime interval. Moreover, if any two distinct elements of a sublattice isomorphic to M 3 are collapsed by a tolerance relation, then the whole sublattice is collapsed. It follows immediately that To is as described in Fig. 5.

Next consider the tolerance relation T1 of Fig. 7. Symmetrically, we can define T2 , and T3• We shall prove that any proper tolerance relation is either a T; or of the form 1jVT;, i*j.

F with tolerance T 1

Fig. 7. Notes on tolerance relations oj lattices 133

So let T be a tolerance relationt T> To. Then there are xtyeF; x

Claim. 0=d; (T) for some i.

Proof. Up to symmetrYt there are three cases to consider: Case 1. al :5x

bz = bzA(d1Vdz) = bzAX = 0 (T). Similarlyt Cz =0 (T)t so dz=0 (T)t as claimed. Case 2. al =X and d3:5Y. Then

0= Ovo = (xAb3)V(XAC3) = (yAb3)v(yAC3) = b3VC3 = d3 (T) as claimed. Case 3. x=o and b3:5Y. Then 0= OvO =xVX= (xAb3)V«(xAa3)vaz)Aal)

= (yAb3)V«(yAa3)vaz)Aal) = b3val = d3 (T) as claimed. This completes the proof of the Claim. D

Now we complete the proof of Lemma 4. The relation d1=0 obviously generates the tolerance 11. Similarly we get Tzt and T3• FinallYt let T be a tolerance relation satisfying T> 11. Then we must have x=y (T)t x

0= ovO = (Cl AaZ)V(C3Aa2) = (CIA l)v(c3A 1) = d1Vd3 (T).

Case 2. The block [OtdtJ. X:50 and d1:5y imply that x=O andt saYt d1vd3:5y. Thus O=d1vd3 (T). Hencet

az = b2 ACZ= (OV b2)/\ (OV cz) = «d1Vd3)vbz)/\«d1Vd3)vcz) = 1 (T).

Case 3. The block [aZtdi vd3]. x:5az and dl vd3:5Yt but not both x=az and d1vd3=y. Hence either x=O and d l Vd3:5Yt in which case we proceed as in Case 2t or X:5 az and y = 1t and we proceed as in Case 1. Case 4. The block [a3t d1Vdz]. We proceedt by symmetrYt as in Case 3. Thus in all four cases we have O=d1Vd3 (T) and az= 1 (T) or O=d1Vdz (T) and a3= 1 (T). Theset obviouslYt describe the tolerances 11 VT3 and T1VTZt respec tively. Since T; V1}t i"*j are maximal tolerancest the proof of Lemma 4 is complete. D 134 E. Fried, G. Gratzer

5. The proof of the Theorem

Let us assume that there is an A in M3 0 D with congruence B establishing this such that F can be represented as AlifJ and T= B/ifJ satisfies (i) all T-classes of A in M 3 ; (ii) A/T is in D. It is easy to see that the lattice AlBA ifJ and the congruences ifJ/BA ifJ and BIBAt:P satisfy the same conditions, and the new congruences are disjoint. In other words, we can assume that BAifJ=w. An element x of F is represented as a t:P congruence class, C(x).

AIT is distributive; since AlTo is isomorphic to M 3 , T> To. Thus by the Claim in the proof of Lemma 4, we can assume that 0==d3 (T). Hence there are elements 0' E C(O) and d; E C(d3} satisfying 0' == d; (6». We can obviously assume that 0'< d;'. Substituting an arbitrary a~ E C(a3) by (O'Y a~) Ad;, we obtain a~ E C(a3) satisfying 0'< a3 < d;'. Similarly, we can choose u' E C(u), bi E C(b3), and ci E C(C3) satisfying u', bi, ci E [ai, d;] and elements ai E C(a!), a~ E C(a2) in [0', u']. Now it is easy to see that the elements 0', ai, a2' a~, u', bi, ci, and d;' form a sublattice of A isomorphic to the interval [0, d3] of F. Indeed, the map (j) : x -+ x' is obviously one-to-one. Since 0' == d; (B), all these elements belong to the same e class. We have to show that the V and Awork properly. As an example, let us show that ai Ya2 u'. Indeed, ai vai == u' (B) since all these elements are in the sameB class. On the other hand, aiva2==u' (ifJ) since both aivai and u' map onto u. Therefore, aiva2,==u' (BAifJ). Since el\ifJ co, we conclude that aivai==u' (co), that is, ai Y u', as claimed. Since every B class is in M3, we get that the interval [0,d3] of F is in M3 , an obvious contradiction which proves the Theorem. It is obvious from the proof, that the Theorem holds for any lattice variety V in place of M3 that does not contain the interval [0, d3] of F. The most general such variety is Mro , the lattice variety generated by M ro , the modular lattice of length two with countably infinite atoms.

References

[1] G. Gratzer, General Lattice Theory (Birkhauser, Basel, 1978). [2] G. Gratzer and G.H. Wenzel, Notes on tolerance relations of lattices, Acta Sci. Math. (Szeged), to appear.