Lattices with many congruences are planar

Let $L$ be an $n$-element finite lattice. We prove that if $L$ has strictly more than $2^{n-5}$ congruences, then $L$ is planar. This result is sharp, since for each natural number $n\geq 8$, there exists a non-planar lattice with exactly $2^{n-5}$ congruences.


Aim and introduction
Our goal is to prove the following statement. Theorem 1.1. Let L be an n-element finite lattice. If L has strictly more than 2 n−5 congruences, then it is a planar lattice.
In order to point out that this result is sharp, we will also prove the following easy remark. An n-element finite lattice L is dismantlable if there is a sequence L 1 ⊂ L 2 ⊂ · · · ⊂ L n = L of its sublattices such that |L i | = i for every i ∈ {1, . . . , n}; see Rival [14]. We know from Kelly and Rival [10] that every finite planar lattice is dismantlable. Remark 1.2. For each natural number n ≥ 8, there exists an n-element non-dismantlable lattice L(n) with exactly 2 n−5 congruences; this L(n) is non-planar.
We know from Freese [4] that an n-element lattice L has at most 2 n−1 = 16 · 2 n−5 congruences. In other words, denoting the lattice of congruences of L by Con(L), we have that |Con(L)| ≤ 2 n−1 . For n ≥ 5, the second largest number of the set ConSizes(n) := {|Con(L)| : L is a lattice with |L| = n} is 8 · 2 n−5 by Czédli [2], while Kulin and Mureşan [11] proved that the third, fourth, and fifth largest numbers of ConSizes(n) are 5 · 2 n−5 , 4 · 2 n−5 , and n-element lattice L has many congruences with respect to n, then L is necessarily planar. However, the present paper needs a technique different from Kulin and Mureşan [11], because a [11]-like description of the lattices witnessing the sixth, seventh, eighth, . . . , k-th largest numbers in ConSizes(n) seems to be hard to find and prove; we do not even know how large is k. Fortunately, we can rely on the powerful description of planar lattices given by Kelly and Rival [10].
Note that although an n-element finite lattice with "many" (that is, more than 2 n−5 ) congruences is necessarily planar by Theorem 1.1, an n-element planar lattice may have only very few congruences even for large n. For example, the n-element modular lattice of length 2, denoted usually by M n−2 , has only two congruences if n ≥ 5. On the other hand, we know, say, from Kulin and Mureşan [11] that there are a lot of lattices L with many congruences, whereby a lot of lattices belong to the scope of Theorem 1.1.
Outline and prerequisites. Section 2 recalls some known facts from the literature and, based on these facts, proves Remark 1.2 in three lines. The rest of the paper is devoted to the proof of Theorem 1.1.
Due to Section 2, the reader is assumed to have only little familiarity with lattices. Apart from some figures from Kelly and Rival [10], which should be at hand while reading, the present paper is more or less self-contained modulo the above-mentioned familiarity. Note that [10] is an open access paper at the time of this writing; see http://dx.doi.org/10.4153/CJM-1975-074-0.

Some known facts about lattices and their congruences
In the whole paper, all lattices are assumed to be finite even if this is not repeated all the time. For a finite lattice L, the set of nonzero joinirreducible elements, that of nonunit meet-irreducible elements, and that of doubly irreducible (neither 0, not 1) elements will be denoted by J(L), M(L), and Irr(L) = J(L) ∩ M(L), respectively. For a ∈ J(L) and b ∈ M(L), the unique lower cover of a and the unique (upper) cover of b will be denoted by a − and b + , respectively. For a, b ∈ L, let con(a, b) stand for the smallest congruence of L such that a, b ∈ con(a, b). For x, y ∈ J(L), let x ≡ con y mean that con(x − , x) = con(y − , y). Then ≡ con is an equivalence relation on J(L), and the corresponding quotient set will be denoted by Q(L) := J(L)/≡ con . In order to explain how to extract (2.2) and (2.3) from the literature, we recall some facts. A quasiordered set is a structure A; ≤ , where ≤ is a quasiordering, that is, a reflexive and symmetric relation on A. For example, if we let a ≤ con b mean con(a − , a) ≤ con(b − , b), then J(L); ≤ con is a quasiordered set. A subset X of A; ≤ is hereditary, if (∀x ∈ X)(∀y ∈ A)(y ≤ x ⇒ y ∈ X). The set of all hereditary subsets of A; ≤ with respect to set inclusion forms a lattice Hered( A; ≤ ). Freese, Ježek and Nation [5,Theorem 2.35] can be reworded as Con(L); ⊆ ∼ = Hered( J(L); ≤ con ). To recall this theorem in a form closer to [5,Theorem 2.35], for the ≡ con -blocks of a, b ∈ J(L), we define the meaning of a/≡ con ≤ con a/≡ con as a ≤ con b.
In this way, we obtain a poset Q(L); ≤ con /≡ con . With this notation, the original form of [5,Theorem 2.35] states that Con(L) ∼ = Hered(Q(L); ≤ con /≡ con ). Since ≡ con will play an important role later, recall that for intervals [a, b] This relation between the two intervals will be denoted by It is well known and easy to see that Next, we note that Con(L) in (2.3) is a Boolean lattice. Oddly enough, we could find neither this fact, nor (2.3) in the literature explicitly. Hence, in this paragraph, we outline briefly a possible way of deriving these statements from explicitly available and well-known other facts; the reader may skip over this paragraph. So let L be a finite distributive lattice. Since L is modular, Con(L) is a Boolean lattice by Grätzer [6, Corollary 3.12 in page 41]. Pick a maximal chain 0 ≺ a 1 ≺ · · · ≺ a t = 1 in L. Here t is the length of L, and it is well known that t = |J(L)|; see Grätzer [7, Corollary 112 in page 114]. If con(a i−1 , a i ) = con(a j−1 , a j ), then it follows from Grätzer [9] and distributivity (in fact, modularity) that there is a sequence of prime intervals (edges in the diagram) from [a i−1 , a i ] to [a j−1 , a j ] such that any two neighboring intervals in this sequence are transposed. In the terminology of Adaricheva and Czédli [1], the prime intervals [a i−1 , a i ] and [a j−1 , a j ] belong to the same trajectory. Since no two distinct comparable prime intervals of L can belong to the same trajectory by [1, Proposition 6.1], it follows that i = j. Hence, the congruences con(a i−1 , a i ), i ∈ {1, . . . , t}, are pairwise distinct. They are join-irreducible congruences by Grätzer [7, page 213], whereby they are atoms in Con(L) since Con(L) is Boolean. Clearly, t i=1 con(a i−1 , a i ) is 1 Con(L) , which implies that |Con(L)| = 2 t . This proves (2.3) since t = |J(L)|.
Next, a lattice is called planar if it is finite and has a Hasse-diagram that is a planar representation of a graph in the usual sense that any two edges can intersect only at vertices. Let N 0 and N + denote the set {0, 1, 2, . . . } of nonnegative integers and the set {1, 2, 3, . . . } of positive integers, respectively. In their fundamental paper on planar lattices, Kelly and Rival [10] gave a set of finite lattices such that the following statement holds. Note that the lattices A n are selfdual and Kelly and Rival [10] proved the minimality of L KR , but we do not need these facts.
Next, we prove Remark 1.2. The ordinal sum of lattices L and L is their disjoint union L ∪ L such that for x, y ∈ L ∪ L , we have that x ≤ y if and only if x ≤ L y, or x ≤ L y, or x ∈ L and y ∈ L .
Note that L(n) above occurs also in page 93 of Rival [14].

A lemma on subposets that are lattices
While L KR consists of lattices, they appear in Proposition 2.1 as subposets. This fact causes some difficulties in proving our theorem; this section serves as a preparation to overcome these difficulties. The set of join-reducible elements of a lattice L will be denoted by JRed(L). Note that  (i) If a 1 , . . . , a t ∈ K and t ∈ N + , then Note that, according to (ii) and (iv), the distinctness of joins is generally preserved when passing from K to L, but equalities are preserved only under additional assumptions. The dual of a condition (X) will be denoted by (X) d ; for example, the dual of Lemma 3.1(i) is denoted by Lemma 3.1(i) d or simply by 3.1(i) d .
Proof of Lemma 3.1. Part (i) is a trivial consequence of the concept of joins as least upper bounds.
In order to prove (ii), assume that We have equality here, since the converse inequality follows in the same way. Thus, we conclude (ii) by contraposition.
Next, let {c 1 , . . . , c t } be a repetition-free list of JRed(K). For each i in {1, . . . , t}, pick a i , b i ∈ K such that a i b i and c i = a i ∨ K b i . That is, Since a i b i holds also in L, The elements listed in (3.3) are pairwise distinct by part (ii). Therefore, |JRed(K)| = t ≤ |JRed(L)|, proving part (iii). Finally, to prove part (iv), we assume the premise of part (iv), and we let t := |JRed(K)| = |JRed(L)|. Choose c i , a i , b i ∈ K as in (3.2). Since t = |JRed(L)|, part (ii) and (3.3) give that As a part of the premise of (iv), u 1 u 2 has been assumed. Hence, (3.2) yields a unique subscript i ∈ {1, . . . , t} such that Since c 1 , . . . , c t is a repetition-free list of the elements of JRed(K), we have that u 1 ∨ K u 2 = c j = a j ∨ K b j for every j ∈ {1, . . . , t} \ {i}. So, for all j = i, part (ii) gives that u 1 ∨ L u 2 = a j ∨ L b j . But u 1 ∨ L u 2 ∈ JRed(L), whence (3.4) gives that u 1 ∨ L u 2 = a i ∨ L b i . Since the equality v 1 ∨ L v 2 = a i ∨ L b i follows in the same way, we conclude that u 1 ∨ L u 2 = v 1 ∨ L v 2 , as required. This yields part (iv) and completes the proof of Lemma 3.1.
Note that a i ∨ L b i in the proof above can be distinct from c i ; this will be exemplified by Figures 1 and 2.

The rest of the proof
In this section, to ease our terminology, let us agree on the following convention. We say that a finite lattice L has many congruences if |Con(L)| > 2 |L|−5 . Otherwise, if |Con(L)| ≤ 2 |L|−5 , then we say that L has few congruences. (ii) If |JRed(L)| = 3 and there are p, q ∈ J(L) such that p = q and con(p − , p) = con(q − , q), then L has few congruences.
Proof. Let n := |L|. If |JRed(L)| ≥ 4, then (3.1) leads to |J(L)| ≤ n − 5, and it follows by (2.2) that L has few congruences. By duality, this proves part (i). Under the assumptions of (ii), p ≡ con q, and we obtain from (2.1) that |Q(L)| ≤ |J(L)| − 1 = n − 4 − 1 = n − 5, and (2.2) implies again that L has few congruences. This proves the lemma. Proof. Label the elements of K = F 0 as shown in Figure 1. A possible L is given on the right in Figure 1; the elements of K are black-filled. The diagram of L is understood as follows: for y 1 , y 2 ∈ L, a thick solid edge, a thin solid edge, and a thin dotted edge ascending from y 1 to y 2 mean that, in the general case, we know that y 1 ≺ y 2 , y 1 < y 2 , and y 1 ≤ y 2 , respectively. In a concrete situation, further relations can be fulfilled; for example, a thin dotted edge can happen to denote that y 1 = y 2 . The two dashed edges and the element x as well as similar edges and elements can be present but they can also be missing. Note that y 1 ≤ y 2 is understood as y 1 ≤ L y 2 ; for y 1 , y 2 ∈ K, this is the same as y 1 ≤ K y 2 since K is a subposet of L. Since L in the figure carries a lot of information on the general case, the reader may choose to inspect Figure 1 instead of checking some of our computations that will come later. Note also that the convention above applies only for L; for K, every edge stands for covering. Clearly, |JRed(K)| = |MRed(K)| = 3. Hence, Lemma 3.1(iii) gives that |JRed(L)| ≥ 3 and |MRed(L)| ≥ 3. We can assume that none of |JRed(L)| ≥ 4 and |MRed(L)| ≥ 4 holds, because otherwise Lemma 4.1(i) would immediately complete the proof. Hence, |JRed(L)| = 3 and |MRed(L)| = 3. (4.1) Since |JRed(K)| = |MRed(K)| = 3 holds also for K = E 0 , to be given later in Figure 2, note at this point that (4.1) will be valid in the proof of Lemma 4.3. Let p := b ∨ L c ∈ L, and let u 1 ∈ L be a lower cover of p in the interval [b, p] L . Also, let q := b ∧ L d and let u 2 ∈ [q, d] L be a cover of q. Finally, let r := b ∧ L g, and let u 3 ∈ [r, g] L be a cover of r. Since we have formed the joins and the meets of incomparable elements in L such that the corresponding joins are pairwise distinct in K and the same holds for the meets, (4.1) and Lemma 3. Since u 1 ≺ L p, u 1 = p. If we had that u 1 = e ∨ L d, then Replacing d, a by g, c , we obtain similarly that u 1 = e ∨ L g. Hence, (4.2) gives that u 1 ∈ J(L). If we had that u 2 = p, then b ≤ p = u 2 ≤ d would be a contradiction. Similarly, u 2 = e ∨ L d would lead to e ≤ e ∨ L d = u 2 ≤ d while u 2 = e ∨ L g again to e ≤ e ∨ L g = u 2 ≤ d, which are contradictions. Hence, u 2 / ∈ JRed(L) and so 0 L ≤ q ≺ L u 2 gives that u 2 ∈ J(L). We have that u 3 = p, because otherwise b ≤ p = u 3 ≤ g would be a contradiction. Similarly, u 3 = e ∨ L d and u 3 ≤ e ∨ L g would lead to the contradictions e ≤ e ∨ L d = u 3 ≤ g and e ≤ e ∨ L g = u 3 ≤ g, respectively. So, u 3 / ∈ JRed(L) by (4.2). Since r ≺ L u 3 excludes that u 3 = 0, we obtain that u 3 ∈ J(L). Since u 3 = u 2 would lead to which is a contradiction, we have that u 1 , u 2 , u 3 ∈ J(L) and u 2 = u 3 .  Suppose, for a contradiction, that u 2 ≤ u 1 , and pick a maximal chain in the interval [u 2 , u 1 ]. So we pick a lower cover of u 1 , then a lower cover of the previous lower cover, etc., and it follows from (4.6) that this chain contains b. Hence, u 2 ≤ b, and we obtain that q ≺ L u 2 ≤ b ∧ L d = q, a contradiction. Hence, u 2 u 1 . This means that u 1 ∧ L u 2 < u 2 . But q ≤ b ≤ u 1 , so we have that q ≤ u 1 ∧ u 2 < u 2 . Since q ≺ u 2 , we obtain that u 1 ∧ L u 2 = q. Similarly, u 2 ≤ d ≤ p and u 2 u 1 give that u 1 < u 1 ∨ L u 2 ≤ p, whereby u 1 ≺ L p yields that u 1 ∨ L u 2 = p. The last two equalities imply the first half of (4.5). The second half follows basically in the same way, so we give less details. Based on (4.6), u 3 ≤ u 1 would lead to u 3 ≤ b and r ≺ L u 3 ≤ b ∧ L g = r, whence , we obtain that r ≤ u 1 ∧ L u 3 < u 3 and u 1 < u 1 ∨ L u 3 ≤ p. Hence the covering relations r ≺ L u 3 and u 1 ≺ L p imply the second half of (4.5).
We still need another lemma.  Proof. This proof shows a lot of similarities with the earlier proof. In particular, the same convention applies for the diagram of L in Figure 2 and, again, there can be several elements of L not indicated in the diagram. We have already noted that (4.1) holds in the present situation. Figure 2 shows how to pick u 1 , u 2 ∈ L; they are covers of a ∧ L b in [a ∧ L b, a] and b ∧ L c in [b ∧ L c, c], respectively. As a counterpart of (4.2), now we obtain in the same way from (4.1) and Lemma 3.1 that Not all the equalities above will be used but they justify Figure 2. In particular, even if there can be more elements, the comparabilities and the incomparabilities in the figure are correctly indicated. Using (4.7) in the same way as we used (4.2) in the proof of Lemma 4.2 and the above-mentioned correctness of Figure 2, it follows that . Hence, (2.4) gives that con(u − 1 , u 1 ) = con(b, p) = con(u − 2 , u 2 ). Since u 1 and u 2 are distinct by Figure 2 and they belong to J(L) by (4.8) and the a, u 1c, u 2 -symmetry, (4.1) and Lemma 4.1(ii) imply that L has few congruences. This completes the proof of Lemma 4.3.
Now, we are in the position to prove our theorem.
Proof of Theorem 1.1. Let L be an arbitrary non-planar lattice; it suffices to show that L has few congruences. By Proposition 2.1, there is a lattice K in Kelly and Rival's list L KR such that K is a subposet of L or the dual L dual of L. Since Con(L dual ) = Con(L), we can assume that K is a subposet of L. A quick glance at the lattices of L KR , see their diagrams in Kelly and Rival [10], shows that if K ∈ L KR \ {E 0 , F 0 }, then |J(K)| ≥ 4 or |M(K)| ≥ 4. Hence, if K ∈ L KR \ {E 0 , F 0 }, then Lemma 4.1(i) implies that L has few congruences, as required. If K ∈ {E 0 , F 0 }, then the same conclusion is obtained by Lemmas 4.2 and 4.3. This completes the proof of Theorem 1.1.