Boolean dimension of a Boolean lattice

For every integer $n$ with $n \geq 6$, we prove that the Boolean dimension of a poset consisting of all the subsets of $\{1,\dots,n\}$ equipped with the inclusion relation is strictly less than $n$.


Introduction
The most widely studied measure of complexity of partially ordered sets (posets for short) is their dimension, introduced by Dushnik and Miller [4] in 1941.Low-dimensional posets admit a compact scheme for handling comparability queries of the form "is x ⩽ y?".In the 1980s, Gambosi, Nešetřil, Pudlák, and Talamo [8,16] introduced the notion of Boolean dimension, which generalizes the notion of dimension with emphasis on the existence of the mentioned compact schemes.See Section 2 for definitions of dimension and Boolean dimension.For a poset P , we write dim(P ) for the dimension of P and bdim(P ) for the Boolean dimension of P .The most prominent (and beautiful) open problem on Boolean dimension comes from the initial paper by Nešetřil and Pudlák [16]: Do posets with planar cover graphs have bounded Boolean dimension?For recent progress towards resolving this problem see [6,2].
In this paper, we consider the following question: What is the Boolean dimension of a Boolean lattice?For a positive integer n, the Boolean lattice of order n, denoted by B n , is the poset on all the subsets of [n] = {1, . . ., n} equipped with the inclusion relation.Although it is wellknown and easy to see that dim(B n ) = n, the problem of determining the dimension of the union of two levels of B n had been heavily studied in the 1990s -see e.g.[3,7,11,14,12,10].Recently, this area of study was revisited due to an increasing interest in the divisibility orders -see [9,15,17].
In the founding paper of the poset dimension theory, Dushnik and Miller introduced the family of posets {S n : n ⩾ 2}, later referred to as standard examples [4,Theorem 4.1].The poset S n is isomorphic to the subposet of B n induced by all singletons and all co-singletons -see Figure 1.The punch line of the family of standard examples is that dim(S n ) = n.On the other hand, the Boolean dimension of every standard example is at most 4. Since bdim(P ) ⩽ dim(P ) for all posets P , we have bdim(B n ) ⩽ n for all positive integers n.The family of Boolean lattices had appeared as a natural candidate to be the canonical example of a family with the property that bdim(B n ) = n.The question, of whether this is true was circulating in the community since Order & Geometry Workshop 2016 held in Poland.In writing, it appeared e.g. in [1,Section 3.4] and [13, page 6].We answer the question in the negative, that is, we prove the following.
More precisely, we prove that bdim(B n ) ⩽ 5  6 n for every positive integer n.The best lower bound we could prove is bdim(B n ) ⩾ n/ log(n + 1) -see Corollary 8. Actually, we tend to believe that the right order of magnitude is o(n).
Next, we give an example of a family of posets with the above-mentioned property for the Boolean dimension.This family is a natural generalization of the family of Boolean lattices.For every positive integer, we define the poset M n , as the poset on the family of all multisets containing elements in [n] equipped with the inclusion relation.Interestingly, the posets M n appeared to be particularly useful in studying the dimension of divisibility posets.It is not hard to see that M n is a product of n infinite chains, and so, bdim(M n ) ⩽ dim(M n ) ⩽ n (see Section 3 for more details).We prove that this bound is tight.
This result can be derived using the Product Ramsey Theorem as mentioned in [1,Section 3.4] (for more on this version of Ramsey Theorem see [1, Section 3.1], [5,Section 4], and [19]).We provide a simple elementary proof, which also implies the mentioned lower bound: bdim(B n ) ⩾ n/ log(n + 1).To be more precise, we prove that for every positive integer n, there exists a positive integer m such that the Boolean dimension of a subposet of M n , induced by all multisets with all elements having multiplicities less than m, is at least n.We give an explicit upper bound on m, namely, m ⩽ n n−1 .
The rest of this paper is organized as follows.In Section 2, we provide some essential definitions and notations used throughout the rest of the paper.In Section 3, we discuss the dimension and the Boolean dimension of products of posets.In Section 4, we prove Theorem 1.In Section 5, we prove Theorem 2. Finally, in Section 6, we present some related open problems.

Preliminaries
The set of the first n positive integers is denoted by [n].By log x we denote the logarithm of x with base 2.
A partially ordered set, or poset for short, is an ordered pair P = (X, ⩽), where X is a nonempty set of elements called the ground set of P , and ⩽ is a binary relation on X (called the order relation in P ), which is reflexive, antisymmetric and transitive.We do not require ground sets to be finite.Sometimes, we abbreviate the phrase x ⩽ y in P to x ⩽ P y.For two posets P and Q, we say that Q is a subposet of P (denoted by Q ⊆ P ) if the ground set of Q is a subset of the ground set of P and the order relation of Q is a restriction of the order relation in P to the ground set of Q.We say that two elements x, y in a poset P are comparable if x ⩽ P y or y ⩽ P x.A poset, where all pairs of elements are comparable is called a linear order.A poset P is a linear extension of P if P is a linear order on the ground set of P such that x ⩽ y in P whenever x ⩽ y in P for every two elements x, y in P .
Let P and Q be two posets with ground sets X and Y respectively.The product of P and Q, denoted by P × Q, is a poset with the ground set X × Y , where for any two pairs (x 1 , y 1 ), (x 2 , y 2 ) ∈ X × Y , we have (x 1 , y 1 ) ⩽ (x 2 , y 2 ) in P × Q if and only if x 1 ⩽ x 2 in P and y 1 ⩽ y 2 in Q.Let n be a positive integer.The n-th power of P is the product of n copies of P , denoted by P n .The posets P and Q are isomorphic if there exists a bijection g : X → Y such that for every two elements x, y in P , we have x ⩽ y in P if and only if g(x) ⩽ g(y) in Q.
For a linear order L and its two elements x, y we define [x ⩽ L y] ∈ {0, 1} to be 1 if x ⩽ y in L and 0 otherwise.For a sequence of linear orders L 1 , . . ., L n on the same ground set and two elements x, y in the ground set, we abbreviate Let P be a poset with at least two elements, and let d be a positive integer.Let L 1 , . . ., L d be linear orders on the ground set of P , and let ϕ : {0, 1} d → {0, 1}.The pair ((L 1 , . . ., L d ), ϕ) is a Boolean realizer of P if for every pair of elements x, y in P , The size of a Boolean realizer is the number of linear orders in the realizer.The Boolean dimension of P is 0 if P consists of only one element; otherwise, the Boolean dimension of P is equal to the minimum size of a Boolean realizer of P .The dimension of P can be defined as the minimum size of a Boolean realizer of P , where the formula ϕ is fixed to be ϕ(ε 1 , . . ., ε d ) = ε 1 • . . .• ε d .This is not the usual way to define the dimension of a poset (see e.g.[18] for the classical definition and basic facts on the dimension).However, this immediately yields bdim(P ) ⩽ dim(P ) for every poset P .It is not hard to see that both Boolean dimension and dimension are monotone under taking subposets.

Boolean dimension and dimension of products of posets
One of the exercises one can solve to familiarize oneself with the notion of the dimension of a poset is to show that for every two posets P and Q, we have dim(P ×Q) ⩽ dim(P )+dim(Q).In particular, this implies that for every poset P and every positive integer n, we have dim(P n ) ⩽ n • dim(P ).Therefore, to upper bound the dimension of a poset, one can express this poset in terms of products of some other posets.It turns out that both B n and M n have very natural representation as a product of n chains (two element chains in the former case and infinite chains in the latter case).Clearly, B n is a subposet of M n .
Let n be a positive integer.Consider a bijection between all elements of M n and {0, 1, . . .} n defined as follows.To a given multiset A of elements in [n], assign v ∈ {0, 1, . . .} n , where for each i ∈ [n], the value v i is the multiplicity of i in A. The bijection transforms the inclusion relation into the coordinate-wise order relation, in other words, the product relation.We obtain that M n is isomorphic to On the other hand, as was already mentioned, B n contains the standard example of order n as a subposet, which yields n ⩽ dim(B n ).Proposition 3.For every positive integer n, We can also prove the additive property of the Boolean dimension for product of posets.We will use this property in the proof of Theorem 1. Lemma 4. For every two posets P and Q, we have bdim(P × Q) ⩽ bdim(P ) + bdim(Q).
Proof.Let P and Q be two posets.If P has exactly one element, then P × Q is isomorphic to Q, and so, bdim(P × Q) = bdim(Q) ⩽ bdim(P ) + bdim(Q).Symmetrically, the assertion follows in the case where Q has exactly one element.Now, we assume that both posets have at least two elements.Let s = bdim(P ) and t = bdim(Q).Let ((L 1 , . . ., L s ), ϕ P ) be a Boolean realizer of P and ((K 1 , . . ., K t ), ϕ Q ) be a Boolean realizer of Q.
The goal is to define a Boolean realized of P ×Q of size s+t.We start with ϕ : {0, 1} s+t → {0, 1} defined as ϕ(δ 1 , . . ., δ s , ε 1 , . . ., ε t ) = ϕ P (δ 1 , . . ., δ s ) • ϕ Q (ε 1 , . . ., ε t ).Fix P and Q arbitrary linear extensions of P and Q respectively.We define two families of linear orders on the ground set of P × Q.First, for each i ∈ [s], we construct a linear order M i .Let (p 1 , q 1 ), (p 2 , q 2 ) be two elements of P × Q.If p 1 ̸ = p 2 , then we order the elements according to L i , that is, (p 1 , q 1 ) ⩽ (p 2 , q 2 ) in M i if and only if p 1 ⩽ p 2 in L i .In the case where p 1 = p 2 , we order the elements according to Q. Next, for each j ∈ [t], we construct a linear order N j similarly, that is, for all (p 1 , q 1 ), (p 2 , q 2 ) elements of P × Q, if q 1 ̸ = q 2 , then we order the elements in N j as in K j and if q 1 = q 2 , then we order the elements in N j as in P .

Boolean dimension of the Boolean lattice
In this section, we prove the following result, which immediately implies Theorem 1.

Posets of multisets
In this section, we prove Theorem 2, that is, for every positive integer n, bdim We define S to be the set of all multisets in M n,m consisting of exactly one element with a positive multiplicity.For every multiset A in M n,m and for every i ∈ [d], we define s i (A) to be the number of elements in S that are less than A in L i .Clearly, 0 ⩽ s i (A) ⩽ |S|.Note that |S| = n(m − 1).For every element A in M n,m , let s(A) = (s 1 (A), . . ., s d (A)) be its signature.
We claim that all elements of M n,m have distinct signatures.See Figure 2 for an illustration of the following argument.
Suppose, contrary to our claim that there exist distinct A, B in M n,m with the same signatures, that is, s(A) = s(B).Fix some i ∈ [d].Since s i (x) = s i (y), both A and B are greater than Table 1.Linear orders L 1 , L 2 , L 3 , L 4 , L 5 on all subsets of [6] forming the Boolean realizer of B 6 .Each column corresponds to one linear order.The greatest element in an order is the top one.
the exact same set of elements from S in L i .In particular, for every S ∈ S, Since A and B are distinct, there is x ∈ [n] that occurs in A and B with different multiplicities.Without loss of generality, assume that the multiplicity of x in A is greater than the multiplicity of x in B. Let T be the multiset consisting of exactly x with the multiplicity equal to the multiplicity of x in A. Note that T ∈ S, and therefore, [T The singletons S = {{1}, {2}, {3}} are highlighted with purple color.We fix A = {2, 3} (in blue) and B = {1, 2, 3} (in red).We have s 1 (A) = 2, since A is greater than {2} and {3} and less than {1}.Similarly, one can check that s 2 (A) = 1 and s 3 (A) = 3.It follows that the signature of A is equal to (2, 1, 3).The signature of B turns out to be the same.We claim that this prevents L 1 , L 2 , L 3 from being a Boolean realizer of B 3 regardless of ϕ : {0, 1} For every positive integer n, the limit lim m→∞ n log m log(nm−n+1) is equal to n.It follows that bdim(M n,m ) = n for a large enough m, and so, bdim(M n ) = n, which concludes the proof of Theorem 2. To be more precise, one can compute how large m should be.Proposition 9.For every integer n with n ⩾ 2, we have bdim(M n,n n−1 ) ⩾ n.

Open problems
To sum up, we list a few related open problems.By Corollary 8 and by Theorem 1, for every positive integer n large enough, we have n log(n+1) ⩽ bdim(B n ) ⩽ 5  6 n .This is not tight, and therefore, we state the following question.
Question 1.What is the order of magnitude of bdim(B n )?
A proper examination of the proof of Lemma 7 in the case of m = 2 (recall that M n,2 is isomorphic to B n ) shows that we never use the fact that a fixed Boolean realizer detects comparabilities between A, B ⊆ [n] with |A|, |B| > 1.In fact, to show the lower bound, we utilize only the detection of comparabilities between singletons and other subsets of [n].Moreover, the singletons play a special role since they form the smallest distinguishing set in B n .This motivates us to consider a relaxed version of the Boolean dimension problem for Boolean lattices.Namely, let bdim(1, B n ) be the minimum positive integer d such that there exists a sequence of linear orders (L 1 , . . ., L d ) of elements of B n and ϕ : {0, 1} d → {0, 1} such that for every x ∈ [n] and A ∈ B n , we have {x} ⩽ A in B n if and only if ϕ [{x} ⩽ L i A] d i=1 = 1.Note that such relaxation was considered for dimension -see e.g.[3,11,12].Clearly, bdim(1, B n ) ⩽ bdim(B n ) and due to the discussion above, n log(n+1) ⩽ bdim(1, B n ).However, even though bounding bdim(1, B n ) is simpler than bounding bdim(B n ), there is no known better upper bound.
Question 2. What is the order of magnitude of bdim(1, B n )?Is it the same as the order of magnitude of bdim(B n )?
The last question is related to the finite subposets of M n .This question was already stated in [1,Section 3.4].Question 3.For a positive integer n, let f (n) be the least positive integer m with bdim(M n,m ) = n.What is the order of magnitude of f (n)?
By Proposition 9 and Theorem 1, for an integer n large enough, we have 3 ⩽ f (n) ⩽ n n−1 .This leaves a substantial gap for further research.

Figure 1 .
Figure 1.For every positive integer n ⩾ 2, the standard example of order n, denoted by S n , is isomorphic to the subposet of B n induced by the singletons and co-singletons.On the right, we show a poset diagram of S 4 and on the right, we show S 4 as a subposet of B 4 .

Lemma 7 .
and by Proposition 3, dim(M n ) = n.Therefore, in order to prove Theorem 2, it suffices to show that n ⩽ bdim(M n ).To this end, we analyze a certain class of subposets of M n .For all positive integers n and m, we define M n,m to be the subposet of M n induced by all multisets such that every element of a multiset has multiplicity less than m.The number of elements in M n,m is equal to m n .Moreover, M n,2 is isomorphic to B n .For all positive integers n, m, we have bdim(M n,m ) ⩾ n log m log(nm−n+1) .Proof.Let n, m be positive integers and assume that bdim(M n,m ) = d for some positive integer d.It follows that there exist linear orders L 1 , . . ., L d and ϕ : {0, 1} d → {0, 1} such that ((L 1 , . . ., L d ), ϕ) is a Boolean realizer of M n,m .
3→ {0, 1}.Indeed, 1 ∈ B\A, however, the relations between {1} and A are the same as the relations between {1} and B in L 1 , L 2 , L 3 .This is a contradiction since {1} ⩽ B and {1} ̸ ⩽ A. The set S in the proof above is distinguishing for the poset M n,m and it is actually the only property that we use.Therefore, Lemma 7 can be easily generalized to the following statement: if a poset P contains a distinguishing set D, then bdim(P ) ⩾ log(|P |) log(|D|+1) .Lemma 7 applied with m = 2 gives a lower bound on bdim(B n ).For every positive integer n, we have bdim(B n ) ⩾ Let us remark on a possible generalization of Lemma 7. A subset D of elements of a poset P is distinguishing for P if for every two distinct elements x, y in P there is z ∈ D such that the relation between z and x is different from the relation between z and y.