NOTES ON PLANAR SEMIMODULAR LATTICES. IV. THE SIZE OF A MINIMAL CONGRUENCE LATTICE REPRESENTATION WITH RECTANGULAR LATTICES

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1 NOTES ON PLANAR SEMIMODULAR LATTICES. IV. THE SIZE OF A MINIMAL CONGRUENCE LATTICE REPRESENTATION WITH RECTANGULAR LATTICES G. GRÄTZER AND E. KNAPP Abstract. Let D be a finite distributive lattice with n join-irreducible elements. In Part III, we proved that D can be represented as the congruence lattice of a special type of planar semimodular lattices of O(n 3 ) elements, we called rectangular. In this paper, we show that this result is best possible. Let D be a finite distributive lattice whose order of join-irreducible elements is a balanced bipartite order on n elements. Then any rectangular lattice L whose congruence lattice is isomorphic to D has at least kn 3 elements, for some constant k > Introduction In [5], Part III of this series of notes on semimodular lattices, we defined a rectangular lattice as a planar semimodular lattice L with exactly two doubly-irreducible elements (u l and u r ) on the boundary of L that are complementary and distinct from 0 and 1. We proved the following result. Theorem 1. Let D be a finite distributive lattice with n join-irreducible elements. Then D can be represented as the congruence lattice of a rectangular lattice L of O(n 3 ) elements. Equivalently, Theorem 2. Let P be a finite order of n elements. Then P can be represented as the order of join-irreducible congruences of a rectangular lattice L of O(n 3 ) elements. In this paper, we show that the construction in Part III is of optimal size. An order P is called bipartite, if every element of P is either minimal or maximal. Let A be the set of minimal elements of P and M the set of maximal elements of P. We call P balanced bipartite, if A = M or if A = M + 1. Theorem 3 (Main Theorem). Let L be a rectangular lattice. Let us assume that the order of join-irreducible congruences of L is a balanced bipartite order on n elements. Then, for some constant k > 0, the lattice L has at least kn 3 elements. This is only the second result in the representation theory of finite distributive lattices as congruence lattices of finite lattices proving that a size estimate for the lattice to be constructed is optimal. The first result of this type was published in Date: July 22, Mathematics Subject Classification. Primary: 06C10; Secondary: 06B10. Key words and phrases. Semimodular lattice, planar, congruence, rectangular. The research of the authors was supported by the NSERC of Canada. 1

2 2 G. GRÄTZER AND E. KNAPP two papers. In G. Grätzer, H. Lakser, and E. T. Schmidt [6] it was proved that any finite distributive lattice D with n join-irreducible elements can be represented as the congruence lattice of a (planar) lattice L of O(n 2 ) elements. In G. Grätzer, I. Rival, and N. Zaguia [8], it was proved that O(n 2 ) is optimal in this result. In [8], optimal is proved by assuming that representation with O(n α ) elements is possible, for some α < 2, and deriving a contradiction. In this paper, we prove that O(n 3 ) is optimal by finding a cubic lower bound, kn 3, in case P is a balanced bipartite order. So, combined with the result of [5], we obtain: Theorem 4. Let L be a rectangular lattice. Let us assume that the order of joinirreducible congruences of L is isomorphic to a balanced bipartite order on n elements. If L is of minimal size, then it satisfies (1) kn 3 L 2 3 n3 + 2n n + 1, for some constant k > 0. In Section 6, we verify (1) with the constant k = This paper belongs to the field: representation theory of finite lattices as congruence lattices of finite lattices. For a general survey of this field, see the book [1]. The present paper is a contribution to Problem 9.1 of [1] asking whether the O(n 3 ) result of G. Grätzer, H. Lakser, and E. T. Schmidt [7] (any finite distributive lattice D with n join-irreducible elements can be represented as the congruence lattice of a (planar) semimodular lattice L of O(n 3 ) elements) is optimal for planar semimodular lattices. There are three obvious problems to raise in connection with Theorem 4. The first two raise questions whether (1) can be improved. Problem 1. Is there a lower bound k 1 n 3 (k 1 > 0) that works for planar semimodular lattices, in general? Problem 2. How close can one bring the constant k in (1) to 2 3? The next problem raises the question whether the O(n 2 ) result can be improved. Problem 3. Let L be a (planar) lattice such that the order of join-irreducible congruences of L is a balanced bipartite order on n elements. Is it true that L has at least k 2 n 2 elements, for some constant k 2 > 0? Outline. This paper relies heavily on Parts I III of this series of papers; see [2] (and [3]), [4], and [5]. In Section 2, we revisit the basic definitions and develop a useful lemma. In Section 3, we investigate slim rectangular lattices and we show how the representation theorem of Part I can be strengthened in this case. We also prove the Decomposition Theorem (Theorem 17), which splits a slim rectangular lattice into four smaller rectangular lattices; this is the basis of the induction in Section 5. In Section 4, we investigate tight S 7 sublattices (introduced in Part II) in rectangular lattices. To prove the Main Theorem, we have to find a lot of elements in L; we find them in the tight S 7 -s. The section culminates in the Partition Theorem, showing that the Decomposition Theorem of Section 3 yields a natural partitioning of the tight S 7 -s. The hard proofs are in Section 5. The Lower Bound Theorem (Theorem 27) constructing a large interval is rather computational. Utilizing the results of Sections 4 and 5, the proof of the Main Theorem is easily derived in Section 6.

3 MINIMAL CONGRUENCE LATTICE REPRESENTATION 3 Notation and terminology. We use the notation of [1]. The Glossary of Notation of [1] is available as a pdf file at In particular, for a meet-irreducible element x in a finite lattice, we denote by x the unique upper cover of x; we define x dually. For congruence-perspectivity, we use [a, b] [c, d] for down (that is, a d = c and b d) and [a, b] [c, d] for up. Acknowledgement. We would like to thank the Lattice Theory seminar, especially, R. W. Quackenbush and B. Wolk, for their patient participation in the lectures on this paper. Their comments were constructive and incisive, resulting in a much improved presentation. 2. Preliminaries Let A be a planar lattice, which is, by definition, finite. A left corner (resp., right corner) of the lattice A is a doubly-irreducible element in A {0, 1} on the left (resp., right) boundary of A. A corner of A is either a left or a right corner. We define a rectangular lattice L as a planar semimodular lattice which has exactly one left corner, u l, and exactly one right corner, u r, and they are complementary that is, u l u r = 1 and u l u r = 0. In Part III, we show many diagrams of rectangular lattices. Let L be a rectangular lattice. We define A L as the set of elements on the left boundary of L greater than or equal to u l and call it the top left boundary. Symmetrically and dually, we define B L, C L, D L, respectively, the top right, bottom left, and bottom right boundary of L. We drop the subscript when L is understood. Let L be a rectangular lattice. A top interval I of L is an interval of the form [x y, 1], where x A {1} and y B {1}. We prove in Corollary 15 that I is a rectangular lattice and x is the left corner and y is the right corner of I. 1 A B u l x C x u r C x D D 0 x D Figure 1. The notation for a rectangular lattice Let A be planar lattice. We define a cell B of P : (1) B = C D, where C and D are chains in P ;

4 4 G. GRÄTZER AND E. KNAPP (2) 0 C = 0 D and 1 C = 1 D ; (3) c d, for all c C {0 C, 1 C } and d D {0 D, 1 D }; (4) there are no elements of A in the interior of B. We call A a 4-cell lattice, if each cell in A contains exactly 4 elements. As we observed in [2], if A is semimodular, then A is a 4-cell lattice. In a planar 4-cell lattice A, the element x is immediately to the left of the element y, if x y and x and y belong to the same 4-cell with the two chains being {x y, x, x y} and {x y, y, x y}. Let L be a planar semimodular lattice. If there is no sublattice in L isomorphic to a covering M 3, then we call L slim. If L is not slim, then we may form the slim lattice L slim L by removing from L the interiors of covering M 3 -s; see [2] for further information. We call L slim the slimming of L. Next we state an interesting property of slim, planar, semimodular lattices. Lemma 5. An element of a slim, planar, semimodular lattice L has at most two covers. Proof. Let a, b, and c be distinct covers of x L such that a is to the left of b and b is to the left of c. Since L is semimodular a, c a c. Since L is planar, it follows {x, a, b, c, a c} = M 3. Therefore L is not slim. 1 S a S b S c S d S e S 0 S Figure 2. The lattice S 7 Let S be a sublattice of the lattice L. If S is isomorphic to the lattice S 7 of Figure 2, then we refer to S as an S 7. If S is an S 7 and all the covers of S are preserved with the possible exceptions 0 S d S and 0 S e S, we call S a tight S 7. In a tight S 7, the element a S is immediately to the left of b S, b S is immediately to the left of c S, and a S = b S = c S = 1 S. We denoting by TS7 L the set of tight S 7 -s in the lattice L. For a sublattice S and a congruence Θ of the lattice L, we denote by Θ S the restriction of Θ to S; of course, Θ S is a congruence of S. We denote by Con J L the order of join-irreducible congruences of L. Next we restate the main result of Part II [4]. Theorem 6. Let L be a finite semimodular lattice. Let α β in Con J L. Then there exists a sublattice S TS7 L such that α = con(b S, 1 S ) and β = con(a S, 1 S ).

5 MINIMAL CONGRUENCE LATTICE REPRESENTATION 5 3. Slim rectangular lattices In Part I, we showed that a slim, planar, semimodular lattice is a cover-preserving join-homomorphic image of a direct product U V of two finite chains. 1 The chains U and V are not internal to the lattice but constructed by induction. In contrast, for a slim rectangular lattice we have the following result. Theorem 7. Let L be a slim rectangular lattice. join-homomorphic image of C D. Then L is a cover-preserving To prove this result, we need some lemmas. In this lemma, and also in the rest of this section, let L be a slim rectangular lattice. Lemma 8. If x L is join-irreducible, then x is on the boundary of L. Proof. Let x be join-irreducible and in the interior of L. Then there exist 4-cells X = {0 X, a X, b X, 1 X } with a X to the left of b X and Y = {0 Y, a Y, b Y, 1 Y } with a Y to the left of b Y such that x = b X = a Y. The element x is join-irreducible, so 0 X = 0 Y. Since 1 X = 1 Y contradicts that L is slim, it follows that 1 X 1 Y. But then X Y generates a cover-preserving sublattice, the dual of S 7, contradicting the semimodularity of L. Since all join-irreducible elements in a slim lattice are on the boundary, they form two chains, described in the next few lemmas. We restate Lemma 3 of Part III. Lemma 9. Every a A {1}, b B {1} is meet-irreducible and every c C {0}, d D {0} is join-irreducible. Lemma 10. In the lattice L, if x L is join-irreducible, then x C D. Proof. By Lemma 8, x A B C D. So this lemma follows from Lemma 9 and the definition of a rectangular lattice. Lemma 11. In the lattice L, the relation x y holds, for all x C {0} and y D {0}. Proof. Indeed, if x y and x C {0}, y D {0}, then x u l u r, contradicting that u l and u r are complementary. Proof of Theorem 7. We prove that the map defined by ϕ: C D L (2) c, d ϕ = c d, for c C and d D, is a cover-preserving join-homomorphism of C D onto L. It is obvious that (2) defines a join-homomorphism. Every element of L is a join of join-irreducible elements. Since all the join-irreducible elements of L are in C D by Lemma 10, it follows that every element of L is of the form c d, for some c C and d D. Therefore, ϕ is onto. To prove that ϕ is cover-preserving, let a, b c, d in C D. By symmetry, we can assume that a c and b = d. By the semimodularity of L, it follows that a b c d in L, so a, b ϕ = a b c d = c, d ϕ. 1 This result has been extended to arbitrary finite semimodular lattices by G. Czédli and E. T. Schmidt.

6 6 G. GRÄTZER AND E. KNAPP We also need Lemma 4 of Part III (we added to it a second statement that is obvious from Lemma 9): Lemma 12. The top boundaries A, B are filters and the bottom boundaries C, D are ideals. In addition, all four are chains. Every x L can be represented as a join of join-irreducible elements. So by Lemmas 9 and 10, we can represent x as c d with c C and d D. Obviously, we obtain the largest such representation with c = x u l and d = x u r. We introduce the notation (see Figure 1) (3) x C = x u l and x D = x u r and obtain: Lemma 13. Let x L. Then there is a largest element x C, x D that ϕ maps onto x. Also (4) x = (x u l ) (x u r ). A planar lattice has many planar diagrams. It follows from the considerations of this section, that for slim rectangular lattices, there are some particularly nice ones. Indeed, for a slim rectangular lattice L, form C D, a direct product of two chains. We regard C D as a planar lattice, with C on the left and D on the right lower boundary. Then x (x u l ) (x u r ) is a meet-embedding of L into C D, which also preserves 0 and 1. We regard the image of L in C D the natural planar diagram representing L. For instance, Figure 2 shows the natural planar diagram representing S 7. From now on, we assume that we consider the natural planar diagram for a slim rectangular lattice. From a slim rectangular lattice, we can construct many others. Lemma 14. Let L be a slim rectangular lattice. Let y B {1, u r }. Then [u l y, 1] and [0, y] are slim rectangular lattices. Proof. The interval [u l y, 1] is a slim, planar, semimodular lattice which has exactly one left corner, u l. We have to show that it has exactly one right corner, y. First, we show that y is a right corner of [u l y, 1]. It is, indeed, on the right boundary of [u l y, 1]. Since y B L, it is meet-irreducible in L by Lemma 9, and therefore, in [u l y, 1]. To show that y is join-irreducible in N, assume that it is not, so there are a b [u l y, 1] with a y and b y. Let y u l c, where c is on the left boundary of [u l y, 1]. Obviously, c y and so c a b. By semimodularity, a b (a b) c; furthermore, (a b) c a and (a b) c b. So a b is covered by three distinct elements, contradicting Lemma 5. This proves that y is a right corner. Second, assume that z y is also a right corner of [u l y, 1]. Obviously z < y, otherwise z would also be a right corner of L, contradicting that L is rectangular. The element z is in a 4-cell in [u l y, 1]. Since z is doubly-irreducible, the 4-cell consists of two chains o z i and o a i, where o z i is a chain on the right boundary of [u l y, 1]. Take y u l c, where c is on the left boundary of [u l y, 1]. By semimodularity, o o c; furthermore, o c z and o c a. So o is covered by three distinct elements, contradicting Lemma 5. This shows that [u l y, 1] has exactly one right corner, y. Since u l and y are obviously complementary, [u l y, 1] is rectangular.

7 MINIMAL CONGRUENCE LATTICE REPRESENTATION 7 Now we consider the interval [0, y] of L; it is a slim, planar, semimodular lattice. Since [u l y, 1] is a rectangular lattice, by Lemma 12, the interval [u l y, y] is a chain. So u l y is a corner of [0, y]. So is u r. Now let s be a doubly irreducible element of [0, y]. By (4), every element of [u l y, y] {u l y} is join-reducible in [0, y] and also in L, so s / [u l y, y] {u l y}. By Lemma 10, the element s is on the bottom left or bottom right boundary of L. This implies that s is one of the two corners u l y and u r. By (4), the two corners are complementary. Therefore, the interval [0, y] is a rectangular lattice. In particular, Corollary 15. Let L be a slim rectangular lattice. Then any top interval of L is also a slim rectangular lattice. For a slim rectangular lattice L and x A {1, u l }, y B {1, u r }, we introduce some notations (see Figure 3): (5) (6) (7) (8) L top (x, y) = [x y, 1], L left (x, y) = [u l y, x], L right (x, y) = [x u r, y], L bottom (x, y) = [0, (u l y) (x u r )]. If x and y are understood, we write L top for L top (x, y), and the same for the three other intervals. The unit of L bottom is (u l y) (x u r ); the zero of L bottom is x y. Note that by (4), (u l y) (x u r ) = x y. 1 A x L top y B u l u r L left L right C L bottom D 0 Figure 3. Decomposing a slim rectangular lattice The next statement follows by repeated applications of Lemma 14: Lemma 16. Let L be a slim rectangular lattice, and let x A {1, u l }, y B {1, u r }. Then the intervals L top, L left, L right, L bottom (see (5) (8)) are slim rectangular lattices. By three applications of Lemma 12, we can glue L bottom and L left, also L right and L top, and we can glue the results together to obtain L. We summarize this as follows:

8 8 G. GRÄTZER AND E. KNAPP Theorem 17 (Decomposition Theorem). Let L be a slim rectangular lattice, and let x A {1, u l }, y B {1, u r }. Then L can be decomposed into four slim rectangular lattices L top, L left, L right, L bottom (see (5) (8)), and the lattice L can be reconstructed from these by repeated gluing. 4. Tight S 7 -s To prove Theorem 3, we have to find lots of elements in L. We construct them from tight S 7 -s whose existence was provided by the results of Part II; see Theorem 6. This section contains some more detailed considerations of tight S 7 -s. Recall that by Lemma 5, in a slim rectangular lattice L, an element has at most two covers. For a tight S 7, we use the notation introduced in Figure 2. Note that a 4-cell of a tight S 7 in L, is also a a 4-cell in L. We start out by investigating tight S 7 -s in slim rectangular lattices. The next lemma states that distinct tight S 7 -s have distinct middle elements and these elements have distinct projections on C and D. Lemma 18. Let L be a slim rectangular lattice. Let U V TS7 L. Then b U b V, b C U bc V, and bd U bd V. Proof. We assume that U V TS7 L and that b U = b V. Since b U = b V is meetirreducible in L, it follows that 1 U = (b U ) = (b V ) = 1 V. Since U is a tight S 7, the interval [a U b U, 1 U ] is of length 2. Therefore, {1 U = 1 V, a U, b U = b V, a U b U } is a 4-cell. It does not contain a V because L is slim. So a V is outside this cell, contradicting that a V is immediately to the left of b V. This proves that b U b V. We next prove that b C U bc V and bd U bd V. Assume, to the contrary, that this fails. If b C U = bc V and bd U = bd V, then by Lemma 13, b U = b V, a contradiction. So we can assume, by symmetry, that b C U bc V and b D U = bd V ; and again, by symmetry, we can assume that b C U < b C V and b D U = b D V. So b U < b V and c U < 1 U = b U b V. Since c U b V, we have that c D U = c U u r b V u r b D V. Since c U is to the right of d U, it follows that b D U < cd U and so b D U < cd U bd V, contradicting that bd U = bd V. The next lemma will produce more elements for the proof of Theorem 3. In a slim rectangular lattice, we associate with each pair of tight S 7 s U, V such that b U b V, a single element, b U b V, that determines the pair U, V. Lemma 19. Let L be a slim rectangular lattice. Let U 1, V 1, U 2, V 2 TS7 L and let b U1 b V1 and b U2 b V2. If b U1 b V1 = b U2 b V2 then {U 1, V 1 } = {U 2, V 2 }. Proof. Let x = b U1 b V1 = b U2 b V2. Let {U 1, V 1 } {U 2, V 2 }. Since x is meetreducible L, it follows from Lemma 5 that x has exactly two covers, y and z. By symmetry, we can assume that y b U1, y b U2, z b V1, z b V2, and U 1 U 2. Case 1: b U1 and b U2 are comparable. By symmetry, let b U1 b U2. Since U 1 U 2, by Lemma 18, b U1 < b U2 and so 1 U1 b U2. By semimodularity, b U1 z b U1 and therefore b U1 has two covers, contradicting that it is meet-irreducible. Case 2. b U2 b U1. Let t = b U2 b U1. Since L is semimodular and t z = x, we conclude that t z t. The element t has at most two covers, so either z t b U1 or z t b U2. By symmetry, we can assume that z b U2. So z is a lower bound of b U2 and b V2, contradicting that x = b U2 b V2.

9 MINIMAL CONGRUENCE LATTICE REPRESENTATION 9 Now we claim that the element determining a pair of tight S 7 s cannot be on the boundary of L. Lemma 20. Let L be a slim rectangular lattice. Let U, V TS7 L. Then b U b V is not on the boundary of L. Proof. Since b U and b V are interior elements, we can assume that b U b V. By symmetry, we can further assume that b U is to the left of b V. Set x = b U b V. If x 0 U, then x = 0 U b V. Choose the element y such that x y b V. By semimodularity, 0 U 0 U y. The element 0 U has exactly two covers, namely, d U and e U. Therefore 0 U y d U or 0 U y e U. In either case, y is a lower bound of b U and b V, contradicting that x = b U b V. Thus we can assume that x 0 U < x and, symmetrically, that x 0 V < x. Note that x 0 U = x 0 V would be an element with three covers, contradicting Lemma 5. So x 0 U x 0 V, Therefore, 0 U is to the left of x and 0 V is to the right of x, so x is in the interior of L. In the next three lemmas, we proceed to investigate tight S 7 -s in rectangular lattices in general; we no longer assume that L be slim. Recall that for a lattice L, we denote by L slim the slimming of L. Observe that if L is a rectangular lattice, S TS7 L, and S L slim, then S TS7 L slim also. So for a tight S 7 of L, the statements: S is a tight S 7 of L slim and S L slim are equivalent. Lemma 21. Let L be a rectangular lattice, and let U TS7 L. Then there exists a sublattice V in L slim such that the following conditions hold: (i) V TS7 L slim ; (ii) x U = x V, for x {0, d, e, b, 1}; (iii) the following two congruences hold: con(a V, 1 V ) = con(a U, 1 U ), con(c V, 1 V ) = con(c U, 1 U ). Proof. If U L slim, then define V = U and we are done. So let U L slim. Let t U L slim. Then t is doubly-irreducible, so t = a U or t = c U. By symmetry, let t = a U. So a U is in the interior of some 4-cell in L slim. Let x a U be on the left boundary of this 4-cell; observe that b U is on the right boundary of this 4-cell. Set V = (U {a U }) {x}. It is easy to see that V TS7 L. Since {d V, a V, a U, b V, 1 V } = M 3, we conclude that con(a V, 1 V ) = con(a U, 1 U ). The second congruence is obvious. If V L slim, then we are done. If not, that is, if c U = c V / L slim, then symmetrically, we apply the previous procedure, to obtain a tight S 7 that satisfies the conditions of the lemma. Lemma 22. Let L be a rectangular lattice. Let U TS7 L satisfy U C. Then {0 U, d U } C. Moreover, if a U L slim, then {0 U, d U, a U } C. Proof. Let 0 U C. By Lemma 21, we can replace U by V, a tight S 7 in L slim, satisfying 0 U = 0 V C = C L slim. If d U / C, then 0 U has at least three covers in L slim, contradicting Lemma 5. So d U C. Similarly, if d U C and L is slim, then a U C. Let d U C. By Lemma 12, 0 U C, and we apply the previous case. Finally, let a U C. By Lemma 12, d U C, and we apply the previous case.

10 10 G. GRÄTZER AND E. KNAPP Lemma 23. Let L be a rectangular lattice. Let U, V TS7 L and let U satisfy U C. If 1 V 1 U, then con(a V, 1 V ) con(a U, 1 U ). Proof. We can assume that U V. By Lemmas 12 and 22, a U b V C. Since d V is the leftmost element covered by b V and a U is to the left of b V, it follows that a U b V d V. Now observe that [a U, 1 U ] [a U b V, b V ] [a V, 1 V ]; therefore, con(a V, 1 V ) con(a U, 1 U ). 1 U a U 1 V d U Let 0 U a V b V C d V a U b V Figure 4. Illustrating the proof of Lemma 23 Finally, we switch back to slim rectangular lattices to ascertain the existence of certain left-maximal and right-maximal tight S 7 -s. Let L be a rectangular lattice. Let U TS7 L slim. We call U left-maximal, if the following two conditions hold: (i) U C ; (ii) a W a U holds, for any W TS7 L slim satisfying W C. Note that by Lemma 22, condition (i) is equivalent to {0 U, d U, a U } C. By symmetry, we introduce right-maximal tight S 7 -s. Lemma 24. Let L be a slim rectangular lattice. Let x = u l. If the interval I = [0, x] contains a tight S 7, then there exists a left-maximal tight S 7. Symmetrically, if y = u r and the interval J = [0, y] contains a tight S 7, then there exists a rightmaximal tight S 7. Proof. Clearly, x u r 0, because otherwise x would be a corner of L. So u l and u r x are corners of I, in view of u l x. By (4), every element in L is the join of an element of C and an element of D, the upper right boundary B I of I, contains only one corner. We conclude that I is a slim rectangular lattice.

11 MINIMAL CONGRUENCE LATTICE REPRESENTATION 11 We will show that either I is the direct product of two chains or there is a tight S 7 intersecting C = C I. If 0 x u r, then it follows from the semimodularity of L that I = C m C 2, for some integer m. x u l z s r v C y u w t x u r 0 Figure 5. Illustrating the proof of Lemma 24 Otherwise, 0 is not covered by x u r. Recall that u l x. By Lemma 12, C is a chain, so there is a smallest element y C (follow the notation in Figure 5) such that y z, for some element z [x u r, x]. Let u y in C and v z in [x u r, x]. By Lemma 12, y v = w C. By the minimality of y, there exists an element t L such that w t < v. Set s = u t. By semimodularity, u s. Also by semimodularity, every cell is a 4-cell, so s z. Since s is not on the right boundary, there is an element r z immediately to the right of s. Then s r s, r by the selection of r. Therefore, U = {y r, u, s r, y, s, r, z}, the black-filled elements in Figure 5, is a tight S 7 intersecting the left boundary of L. By the minimality of y, the interval [y, x] is isomorphic to C m C 2, for some integer m. Therefore, U is a left-maximal tight S 7. Applying the proof symmetrically, the result holds for J. Where are the tight S 7 -s in a decomposition of a slim rectangular lattice? We start with the following statement: Lemma 25. For a slim rectangular lattice L and x A {1, u l }, let S TS7 L. Then S [u r x, 1] or S [0, x]. Proof. Assume to the contrary that 0 S [0, x] and 1 S [x u r, 1], that is, 0 S x and x u r 1 S. Then S is split into two parts: S [0, x] and S [x u r, 1]. If the two parts are disjoint, then they are a prime ideal and a prime filter in S, so we can assume that S [0, x] = {0 S, d S, a S }, which is in conflict with a S 1 S. If the

12 12 G. GRÄTZER AND E. KNAPP two parts are not disjoint, then S is the gluing of the two parts, but S is not glue decomposable. Lemma 25 can be rewritten as (9) TS7 L = TS7[u r x, 1] TS7[0, x], where, in fact, the union is disjoint union. This is the simplest form of the Partition Theorem, which in its full generality we now state: Theorem 26 (Partition Theorem). Let L be a slim rectangular lattice, let x A {1, u l }, y B {1, u r }. We decompose L into the lattices L top, L left, L right, L bottom (see (5) (8)). This defines a partitioning of TS7 L: (10) TS7 L = TS7 L top TS7 L left TS7 L right TS7 L bottom. 5. The Lower Bound Theorem Let L be a rectangular lattice. By Theorem 6 and Lemma 21, we may associate each covering pair in Con J L with some tight S 7 in L slim. Form a collection U L of tight S 7 -s in L slim such that each element in U L is associated with a covering pair in Con J L. Furthermore, we take U L to be minimal in size. We refer to U L as a minimal collection of L. Observe that 2 U L is an upper bound for the number of covering pairs in Con J L. We introduce a notation. Let U L be a minimal collection. Let N be a sublattice of L. We denote by U L N the collection of those members of U L that lie in N, that is, U L N = U TS7 N. We start with this crucial result: Theorem 27 (Lower Bound Theorem). Let L be a rectangular lattice with a minimal collection Û. Let N be a top interval of Lslim and let U = Û N. Let n and i be positive integers. Let us make the following assumptions on L, n, and i: (i) Con J L is a suborder of a balanced bipartite order on 2n elements; (ii) U ni. Then one of the following two conclusions hold: (α) N i2 n 2 ; (β) there exists a positive integer j < i and a top interval M of N satisfying the following two conditions: (a) N M + (i j)in 2 ; (b) U M nj. Proof. We work in N, so N as a subscript will be omitted; for instance, we write C for C N and 0 for 0 N, and the corners are u l and u r. Each U U is associated with an element b C U C and an element bd U D see (3). By Lemma 18, either one of these elements determines U. Therefore, by condition (ii), (11) C ni and D ni. Define u = u l u r and W = N top (u l, u r); see Figure 6. We claim that C W =. Indeed, if a C W, then a u and by Lemma 12, u C. Therefore u = u l u r = u l y. Choose an element b with u b u r. Then b u l, because b u l would imply that b = u l u r, a contradiction. So by

13 MINIMAL CONGRUENCE LATTICE REPRESENTATION 13 1 A B u l W u r u l u r u C D 0 Figure 6. Some notation for N semimodularity, u l b u l, therefore, u l = b u l, contradicting that u = u l u r, verifying the claim. Similarly, D W =. Intersecting both sides of (10) with Û, we obtain: (12) U = U W U N left U N right U N bottom. To prove the Lower Bound Theorem, we distinguish three cases. Case I. Û (N left N bottom ) < 2n. We distinguish two subcases. Case I.1. i 2. Then, by (11), N C D {1} 2ni n 2 i2 n 2, and so conclusion (α) holds. Case I.2. i > 2. Set M = [u l u r, 1] and j = i 2; see Figure 7. 1 A B u l u l M u r C u D 0 y = u r Figure 7. Some notation for N in Case I.2

14 14 G. GRÄTZER AND E. KNAPP By (12), U = U M + U [0, x]. It follows that U M U U [0, x] ni 2n nj, utilizing the assumptions for Case I and Case I.2, verifying (b). U M nj, verifying (b). Finally, since C M =, it follows that (i j)in N M + C M + ni M +, 2 verifying (a). This verifies conclusion (β) and completes the proof for Case I.2 and for Case I. Case II. U (N right N bottom ) < 2n. This is symmetric to Case I. Case III. U (N left N bottom ) 2n and U (N right N bottom ) 2n. Set M = W. By Lemma 24, there exists a left-maximal tight S 7, which we denote by U left, and a right-maximal tight S 7, which we denote by U right. We wish to show that U N bottom contains at most n elements. If V U N bottom, then from Lemma 23, we conclude that con(a V, 1 V ) con(a Uleft, 1 Uleft ), con(c V, 1 V ) con(c Uright, 1 Uright ). Also, con(b V, 1 V ) con(a V, 1 V ), con(c V, 1 V ), so (13) (14) con(b V, 1 V ) con(a V, 1 V ) con(a Uleft, 1 Uleft ), con(b V, 1 V ) con(c V, 1 V ) con(c Uright, 1 Uright ). Since a chain in Con J L has at most two elements, it follows that (15) (16) con(b V, 1 V ) con(a Uleft, 1 Uleft ), con(b V, 1 V ) con(c Uright, 1 Uright ). If equality holds, say, in (15), that is, con(b V, 1 V ) = con(a Uleft, 1 Uleft ), then by (13), we conclude that (17) con(b V, 1 V ) = con(a V, 1 V ). Now V U means that V is associated with a covering pair in Con J L, namely, with con(b V, 1 V ) con(a V, 1 V ) or con(b V, 1 V ) con(c V, 1 V ). So (17) implies that con(b V, 1 V ) con(c V, 1 V ) holds. So by (14), equality cannot hold in (16). We conclude that equality holds for at most one of (15) and (16). Assume that it holds for (15). Then, for V U N bottom, the congruence con(b V, 1 V ) and the congruence con(c V, 1 V ) = con(c Uright, 1 Uright ) determine V. The minimality of Û now gives us that U N bottom n. Define U Nleft k 1 =, n U Nright k 2 =. n Since in this case, we assumed that U (N left N bottom ) 2n and U (N right N bottom ) 2n, and in the previous paragraph we proved that U N bottom n, it follows that k 1, k 2 1. Form the set S = { b U b V U U N left, V U N right }

15 MINIMAL CONGRUENCE LATTICE REPRESENTATION 15 By Lemma 19, the element b U b V uniquely determines the pair U, V, so S n 2 k 1 k 2. By Lemma 20, the element b U b V is not in C or D. So S (C {0}) D is a disjoint union in N M. We can easily add an element x to make S (C {0}) D one bigger in N M; for instance, any x U left C. The inequality S n 2 k 1 k 2 combined with (11) yields. (18) S C D {x} n 2 k 1 k 2 + 2ni. There are two subcases to compute. Case III.1. i k 1 + k Observe that by assumption (i), Con J L has at most n 2 covering pairs, so U n 2. Comparing this with assumption (ii), we conclude that i n. Utilizing this and (18): N S C D {x} n 2 k 1 k 2 + 2ni = n 2 (2nk 1k 2 + 4i) n 2 (2ik 1k 2 + 4i) n 2 (ik 1 + ik 2 + 4i) n 2 i(k 1 + k 2 + 3) i2 n 2, and so conclusion (α) holds. Case III.2. i > k 1 + k Then setting j = i (k 1 + k 2 + 3), we obtain that N M S C D {x} n 2 k 1 k 2 + 2ni n2 2 (k 1 + k 2 ) + 2ni (k 1 + k 2 + 3)ni (i j)in =, 2 2 verifying (a). Since we conclude from (12) that U N left n(k 1 + 1), U N right n(k 2 + 1), U N bottom n, U M n(i k 1 k 2 3) = nj, therefore (b) and conclusion (β) are verified. The Lower Bound Theorem does indeed give a lower bound for the size of some intervals of L. Corollary 28. Let L be a rectangular lattice with a minimal collection Û. Let N be a top interval of L slim and let U = Û N. Let n and i be positive integers. Let us make the following assumptions on L, n, and i: (i) Con J L is a suborder of a balanced bipartite order on 2n elements; (ii) U ni. Then (19) N 1 4 i2 n in. Proof. We assume that for all l < i and for all top intervals I = [0 I, 1 L ] of N satisfying U I nl, the inequality (20) I 1 4 l2 n ln holds. By Theorem 27, condition (α) or condition (β) holds for N.

16 16 G. GRÄTZER AND E. KNAPP If (α) holds for N, then N i2 n i2 n in, as required. If (β) holds for N, then there exists a positive integer j < i and a top interval M of N such that U M nj, and N M + Applying (20) with I = M, we obtain that Therefore, by (a), N M + (i j)in 2 (i j)in. 2 M 1 4 j2 n jn. 1 4 j2 n + 1 (i j)in jn = 1 4 j2 n + 1 (i j)in jn + ( i2 n in) + (1 4 i2 n in) = n 4 (j2 + j + 2i 2 2ij i 2 i) + ( 1 4 i2 n in) = n 4 ((i j)2 (i j)) + ( 1 4 i2 n in) 1 4 i2 n in, proving the corollary. We rephrase Corollary 28, so that it be more straightforward to apply. Corollary 29. Let L be a rectangular lattice with a minimal collection Û. Let N be a top interval of L slim and let U = Û N. Let p and i be positive integers. Let us make the following assumptions on L, p, and i: (i) Con J L is a suborder of a balanced bipartite order on p elements; (ii) U pi. Then (21) N 1 8 i2 p ip. Proof. We may assume that p > 1. Let n = p+1 2. Since p 2n, condition (i) of Corollary 28 follows from the present condition (i). Since p n, condition (ii) of Corollary 28 follows from the present condition (ii), so we can apply Corollary 28, to obtain that N 1 4 i2 n in 1 8 i2 p ip. 6. Proof of the Main Theorem Let L be a rectangular lattice such that Con J L is a balanced bipartite order on n elements. We define k = 2 10 and prove that L has at least kn 3 elements. If n 11, then kn 3 2, so the inequality L > kn 3 is trivial. So we can assume that n 12. Let Û be a minimal collection in L. Since Con J L is a balanced bipartite order on n elements, there are at least n covering pairs

17 MINIMAL CONGRUENCE LATTICE REPRESENTATION 17 in Con J L. Each covering pair in Con J L is associated with a tight S 7 in Û and each element of Û is associated with at most two covering pairs in Con J L. It follows that n2 (22) Û n 1 = (n + 1)n n We apply Corollary 29 to the lattice L, the minimal collection Û in L, the top interval N = L slim of L slim, and the integers p = n and i = n 1 8. Let U = Û N. Note that, by definition, Û = U. By (21) and (22), we obtain that L L slim = N 1 n 1 2n 1 n 1 + n = n n 1 ( n 1 ) Since n 8 1, n 16 n 1 8, for n 12, we obtain that L n 8 n 16 n 8 = n = kn3, completing the proof of the Main Theorem. References [1] G. Grätzer, The Congruences of a Finite Lattice, A Proof-by-Picture Approach. Birkhäuser Boston, xxiii+281 pp. ISBN: [2] G. Grätzer and E. Knapp, Notes on planar semimodular lattices. I. Construction, Acta Sci. Math. (Szeged) 73 (2007), [3], A note on planar semimodular lattices, Algebra Universalis 58 (2008), [4], Notes on planar semimodular lattices. II. Congruences. Acta Sci. Math. (Szeged) 74 (2008), [5], Notes on planar semimodular lattices. III. Rectangular lattices Acta Sci. Math. (Szeged) 74 (2008). [6] G. Grätzer, H. Lakser, and E. T. Schmidt, Congruence lattices of small planar lattices, Proc. Amer. Math. Soc. 123 (1995), [7], Congruence lattices of finite semimodular lattices, Canad. Math. Bull. 41 (1998), [8] G. Grätzer, I. Rival, and N. Zaguia, Small representations of finite distributive lattices as congruence lattices, Proc. Amer. Math. Soc. 123 (1995), Correction: 126 (1998), Department of Mathematics, University of Manitoba, Winnipeg, MB R3T 2N2, Canada address, G. Grätzer: gratzer@me.com URL, G. Grätzer: University of Manitoba, Winnipeg, MB R3T 2N2, Canada address, E. Knapp: edward.m.knapp@gmail.com