Decompositions of the Boolean lattice into rank-symmetric chains

The Boolean lattice $2^{[n]}$ is the power set of $[n]$ ordered by inclusion. A chain $c_{0}\subset...\subset c_{k}$ in $2^{[n]}$ is rank-symmetric, if $|c_{i}|+|c_{k-i}|=n$ for $i=0,...,k$; and it is symmetric, if $|c_{i}|=(n-k)/2+i$. We show that there exist a bijection $$p: [n]^{(\geq n/2)}\rightarrow [n]^{(\leq n/2)}$$ and a partial ordering $<$ on $[n]^{(\geq n/2)}$ satisfying the following properties: (i) $\subset$ is an extension of $<$ on $[n]^{(\geq n/2)}$; (ii) if $C\subset [n]^{(\geq n/2)}$ is a chain with respect to $<$, then $p(C)\cup C$ is a rank-symmetric chain in $2^{[n]}$, where $p(C)=\{p(x): x\in C\}$; (iii) the poset $([n]^{(\geq n/2)},<)$ has the so called normalized matching property. We show two applications of this result. A conjecture of F\"{u}redi asks if $2^{[n]}$ can be partitioned into $\binom{n}{\lfloor n/2\rfloor}$ chains such that the size of any two chains differ by at most 1. We prove an asymptotic version of this conjecture with the additional condition that every chain in the partition is rank-symmetric: $2^{[n]}$ can be partitioned into $\binom{n}{\lfloor n/2\rfloor}$ rank-symmetric chains, each of size $\Theta(\sqrt{n})$.


Introduction
Let us introduce the main definitions and notation used throughout the paper. The notation is mostly standard and can be found in [1], for example.
A chain in a poset is a subset of pairwise comparable elements. A chain C ⊂ 2 [n] with elements c 0 ⊂ ... ⊂ c k is rank-symmetric, if |c i | + |c k−i | = n, and C is skipless, if |c i | = |c 0 | + i for i = 0, ..., k. A chain is symmetric, if it is rank-symmetric and skipless.
A poset P is graded if there exists a partition of its elements into subsets A 0 , A 1 , . . . , A n such that A 0 is the set of minimal elements, and whenever x ∈ A i and x < y with no x < z < y, then y ∈ A i+1 . If there exists such a partition, then it is unique and A 0 , A 1 , . . . , A n are the levels of P . If x ∈ A i , then the rank of x is i and it is denoted by rk(x).
A bipartite graph G = (A, B, E) is a normalized matching graph, if for any where Γ(X) is the set of neighbours of A in B. A graded poset P is a normalized matching poset, or satisfies the normalized matching property, if the bipartite graph induced any two levels A i and A j is a normalized matching graph for i, j ∈ {0, ..., n}, i = j.
It is easy to show that 2 [n] is a rank-symmetric, unimodal normalized matching poset.
By the classical theorem of Sperner [14], the minimum number of chains 2 [n] can be partitioned into is n ⌊n/2⌋ . Also, Brujin et al. [2] showed that 2 [n] admits a chain decomposition into n ⌊n/2⌋ symmetric chains; later Griggs [6] extended their result by proving that every rank-symmetric, unimodal normalized matching poset has a symmetric chain decomposition.
The aim of this paper is to show the existence of a relatively "rich" subposet of (2 [n] , ⊂), where every chain corresponds to a rank-symmetric chain in 2 [n] , and every chain partition corresponds to a chain partition of 2 [n] into rank-symmetric chains. Here, rich means that the poset has the normalized matching property. Our main result is the following. (iii) the poset ([n] (≥n/2) , <) satisfies the normalized matching property.
We also show two applications of this theorem. The first application is the following rank-symmetric variant of a problem of Füredi. Note that in symmetric chain decomposition of 2 [n] , we must have n i − n i−1 chains of size n + 1 − 2i for i = 0, ..., ⌊n/2⌋. In other words, this chain decomposition contains small an large chains as well. In 1985, Füredi [5] proposed the following conjecture.
Conjecture 2. Let n be a positive integer. The Boolean lattice 2 [n] can be partitioned into n ⌊n/2⌋ chains such that the size of any two chains differ by at most 1.
If the conjecture is true, it means that 2 [n] admits a chain partition into n ⌊n/2⌋ chains such that the size of each chain is ≈ π/2 √ n. While this conjecture is still open, there are a few partial results. Hsu, Logan, Shahriari and Towse [8] proved that there exists a chain partition into n ⌊n/2⌋ skipless chains such that the size of each chain is at least √ n/2 + O(1). Also, the author of this paper [15] proved that there is a chain partition of 2 [n] into n ⌊n/2⌋ chains such that the size of each chain is between 0.8 √ n and 13 √ n. We also proposed the following rank-symmetric version of Conjecture 2.
Conjecture 3. Let n be a positive integer. The Boolean lattice 2 [n] can be partitioned into n ⌊n/2⌋ rank-symmetric chains such that the size of any two chains differ in at most 2.
To demonstrate the difficulty of the problem, we challenge the reader to think about the following much weaker result: if h is fixed and n is sufficiently large, then 2 [n] can be partitioned into rank-symmetric chains, each of size at least h. While this problem is not too hard in the case we do not demand our chains to be rank-symmetric, we have to overcome extra obstacles in the rank-symmetric case.
In this paper, we prove the following result concerning Conjecture 3.
For any ǫ > 0 there exists an N ǫ such that if n > N ǫ , then 2 [n] can be partitioned into n ⌊n/2⌋ rank-symmetric chains, all of them of size between (α − ǫ) √ n and O( √ n/ǫ).
In particular, if n is sufficiently large, then there is a chain partition of 2 [n] into n ⌊n/2⌋ rank-symmetric chains such that the size of each chain is between 0.8 √ n and 13 √ n.
We note that this is exactly the same result as Theorem 1.5 in [15], but with the additional condition that our chains are rank-symmetric.
With our second application, we show that 2 [n] has a lot of symmetric chain decompositions. We prove the following theorem. In the proof, we also establish a nontrivial lower bound for the number of matchings in an arbitrary normalized matching graph.

The proof of Theorem 1
In this section, we prove Theorem 1. The proof relies on the idea of the classical symmetric chain decomposition of 2 [n] introduced by De Bruijn, Tengbergen, Kruyswijk [2]. We shall briefly define this chain partition.
First, we introduce some notation. If v is an element of a cartesian product with d terms and i ∈ [d], then v i denotes the ith coordinate of v.
As usual, let [i, j] = {i, i + 1, ..., j} for i, j integers with i < j. Also, for The signature of x, denoted by sg(x), is an element of {0, 1, * } n defined as follows.
If i ∈ x and there exists an integer j with i < j ≤ n such that c x (i, j) = 0, then sg(x) i = 1. The smallest such j is the pair of i in x and is denoted by pr x (i). If there is no such j, then sg(x) i = * .
If i ∈ x and there exists an integer j with 1 ≤ j < i such that c x (j, i) = 0, then sg(x) i = 0. The largest such j is the pair of i in x and is denoted by pr x (i). If there is no such j, then sg(x) i = * .
For example, if x = {2, 3, 4, 6, 7, 10, 11} ∈ 2 [12] , then The signature of x can be interpreted in another way as well. Using our earlier example, write x as )((()(())((), where "(" represents the elements in x and ")" represents the elements not in x. Then, we have sg(x) i = * if the bracket representing i is invalid, in other words does not have a pair. Now we list some of the most important properties of the signature. We shall avoid their proof, as they can be found in [2], for example. Also, the proofs of these properties are very similar to the ones we give in Proposition 9. (i) If i ∈ [n] and j = pr x (i) exists, then pr x (j) exists as well and pr x (j) = i.
(iii) Any two intervals in I(x) are either disjoint, or one is contained in the other. Also, the intervals are disjoint from * (x).
Define the equivalence relation ∼ on 2 [n] such that x ∼ y if sg(x) = sg(y). We show that a complete equivalence class of ∼ is a symmetric chain in 2 [n] . Let x ∈ 2 [n] and let C be the ∼ equivalence class of x. Let 1 ≤ i 1 < i 2 < ... < i t ≤ n be the elements of * (x). Then one can easily show that the elements of C are the subsets of [n] having the form and C is truly a symmetric chain. Thus, the equivalence classes of ∼ form a symmetric chain decomposition of 2 [n] . Now we shall define our bijection p. For simplicity, write Q = [n] (≥n/2) . Our bijection p : Q → [n] (≤n/2) is defined with respect to the symmetric chain decomposition described above: for x ∈ Q, let p(x) be the unique element in the ∼ equivalence class of x such that |x| + |p(x)| = n.
Define the relation < on Q as follows. Let y < x if As ⊂ is a partial order, < is also a partial order. Hence, (Q, <) is a partially ordered set. Furthermore, Q has the following important property Hence, our task is reduced to prove the following theorem. The next proposition shall give us a more useable description of <. Proof. Let C be the ∼ equivalence class of x. The elements of C not larger than x have the form . Now we only need to show that in this case, p(x) ⊂ p(y) holds as well. If b = i 1 , then p(y), p(x), x, y are all elements of C, hence we are done.
But then c y (i u−1 , i u ) = 0, so sg(y) iu−1 = 1 and sg(y) iu = 0. Also, sg(x) and sg(y) agrees on every coordinate other than i u−1 and i u . Thus, This shows that p(x) ⊂ p(y).
By Proposition 8, the bipartite comparability graph of (Q, <) induced on the levels [n] (k−1) and [n] (k) is 2k − n regular from [n] (k) . However, it is not regular from [n] (k−1) . Also, while this proposition lets us identify the neighbours of any However, we shall overcome this obstacle by slightly modifying the definition of the signature. We introduce the circular signature of an element x ∈ 2 [n] . The circular signature of x is denoted by csg(x), and is an element of {0, 1, * } n defined as follows.
View x as a subset of Z n , the ring of integers modulo n.
If i ∈ x and there exists j ∈ Z n such that c ′ x (i, j) = 0, then csg(x) i = 1 and the circular pair of i in x is the j with this property for which |[i, j] n | is the smallest and is denoted by pr ′ x (i). If there is no such j, then csg(x) i = * . If i ∈ x and there exists j ∈ Z n such that c x (j, i) = 0, then csg(x) i = 0 and the circular pair of i in x is the j with this property for which |[j, i] n | is the smallest and is denoted by We note that the idea of circular signature can be also found in [10]. Taking our earlier example x = {2, 3, 4, 6, 7, 10, 11} ∈ 2 [12] , now we have csg(x) = (0, * , * , 1, 0, 1, 1, 0, 0, 1, 1, 0). The following proposition lists the properties of the circular signature and helps us compare it with the signature.
(iii) Any two intervals in I ′ (x) are either disjoint or one is contained in the other. Also, the intervals are disjoint from * ′ (x).
Proof. (i) Look at case i ∈ x, the proof in the other case is similar. The function c ′ x (i, y) changes by 1 as y changes by 1 mod n, and c ′ , so this part trivially follows.
x (a, y) changes by 1 as y changes by , then the j for which |[i r , j] n | is minimal, must be equal to one of j 1 , ..., j t−2k+n . Suppose j = j r ′ . As j r ′ < i r , We can show similarly that c (ii) Also, if y ∈ [n] (k−1) and [j 1 , pr y (j 1 )] n , ..., [j s , pr y (j s )] n are the maximal intervals in I ′ (y), then there are s elements in [n] (k) that are <-larger than y, namely y ∪ pr ′ y (j r ) for r = 1, ..., s.
We finished analyzing the poset (Q, <). But before we can start the proof of Theorem 7, we still need the following well known properties of normalized matching graphs and posets. Proposition 11. Let (P, <) be a graded poset with levels A 0 , ..., A n . For k = 0, ..., n − 1, let G k = (A k , A k+1 , E k ) be the bipartite graph, where x ∈ A k and y ∈ A k+1 are joined by an edge if x < y. If G k is a normalized matching graph for k = 0, ..., n − 1, then (P, <) is a normalized matching poset.
Proof. We need to show that for any positive integers i and j with 0 ≤ i, j ≤ n, i = j, the bipartite subgraph induced on A i ∪ A j is a normalized matching graph. Suppose that i < j, the other case being similar. Let X 0 ⊂ A i and for l = 1, ..., j − i, define X l to be the set of elements of A i+l which are larger than some element of X l−1 . We need to show that But this is obvious as we have The following well known result can be found in various sources [3,12], we state it without proof. Now we are ready to prove Theorem 7, namely that (Q, <) is a normalized matching poset.
Fix k with n/2 + 1 ≤ k ≤ n, and let G = ([n] (k−1) , [n] (k) , E) be the bipartite graph, where x ∈ [n] (k−1) and y ∈ [n] (k) are joined by an edge if x < y. We show that with the choice a = 2n − 2k + 2 and b = 2k there is a weight function w : E → R + satisfying the conditions of Proposition 12.
Define w : E → R + as follows. Suppose that x ∈ [n] (k−1) and y ∈ [n] (k) such that x < y and let the elements of * ′ (y) be i 1 < ... < i 2k−n . Then We show that w suffices. Fix x ∈ [n] (k−1) and let [j 1 , pr ′ x (j 1 )] n , ..., [j t , pr ′ x (j t )] n be the maximal intervals in I ′ (x). The neighbours of x in [n] (k) are Hence, we have x∈e e∈E

Partitioning the Boolean lattice into rank-symmetric chains of uniform size
In this section, we prove Theorem 4. In the proof, we apply the following two theorems from [15].
Theorem 13. Let P be an unimodal normalized matching poset of width w.
Then the poset P can be partitioned into w chains of size at most 2|P | w + 5. Theorem 14. Let P be a normalized matching poset with levels A 0 , A 1 , . . . , A n and let a i = |A i | for i = 0, . . . , n. Suppose that w = a 0 ≥ a 1 ≥ . . . ≥ a n . Let f : {0, . . . , n} → N ∪ {∞} be defined by The poset P can be partitioned into w chains of size at least d + 1.
The proof of Theorem 4 is almost the same as the proof of Theorem 1.5 in [15]. For completeness, we provide the proof here as well, but we shall copy most of it word by word.
Proof of Theorem 4. Again, let Q = [n] (≥n/2) and choose a bijection p : Q → [n] (≤n/2) and partial ordering < such that they satisfy the conditions of Theorem 1. The width of (Q, <) is w = n ⌊n/2⌋ as the symmetric chain decomposition of 2 [n] also defines a chain decomposition of (Q, <) into w chains.
(It also follows from the normalized matching property.) For i = 0, ..., ⌊n/2⌋, let A i = [n] (⌈n/2⌉+i) . First, we need some approximation of the size of the level Using Stirling's approximation one can easily get the estimation that for any t ∈ R we have as n → ∞. For k = 1, 2, . . ., let T k > 0 be the smallest integer such that n ⌈n/2⌉+T k < w/k. Then, by (1) we have Hence, the tail of the sum This implies K = O(1/ǫ).
Let P be the subposet of (Q, <) induced by the levels A 0 , . . . , A TK . Define f and d as in Theorem 14. For k = 2, 3, . . . , K, if T k−1 < j ≤ T k − k, then f (j) = j + k. Hence, Thus, by Theorem 14 there exists a chain partition of P into w chains, each of size at least (1/2 + o(1)) √ n(α − ǫ/2) and at most Let {C x } x∈B ⌈n/2⌉ be such a partition with x ∈ C x for all x ∈ B ⌈n/2⌉ . Let P ′ be the subposet of 2 [n] induced by the levels A TK , . . . , A ⌊n/2⌋ . Then Q is also a unimodal normalized matching poset with width Thus, by Theorem 13, P ′ has a partition into w ′ chains, each of size at most Let {D y } y∈B ⌈n/2⌉+T K be such a partition with y ∈ D y for all y ∈ B ⌈n/2⌉+TK .
Setting ǫ = 0.04, one can get the exact bounds 0.8 √ n and 13 √ n for sufficiently large n. We shall avoid doing these calculations.

The number of symmetric chain partitions of the Boolean lattice
In this section, we present bounds on the number of symmetric chain decompositions of 2 [n] and we prove Theorem 5. First, we show a short proof of a slightly worse bound than the one we have in Theorem 5, without using Theorem 1. This simple proof uses the elegant idea of Kleitman [11], who builds a symmetric chain partition of 2 [n] by induction.
For each sequence (i 1 , ..., i l ) ∈ {0, 1} l , we define the following symmetric chain partition of 2 [n] . It is enough to show that M (G) ≥ n k . We use the following observation: let G ′ = (A, B, E ′ ) be another normalized matching graph on the same vertex set. Then M (G) ∩ M (G ′ ) = ∅. Create the following graded poset (P, <): let the levels of P be A, B, A ′ , where A ′ is a disjoint copy of A; let the partial order < be induced by the following relations: for a ∈ A, a ′ ∈ A ′ and b ∈ B let a < b if ab ∈ E, and b < a ′ if a ′ b ∈ E ′ . Then (P, <) is a unimodal, rank-symmetric normalized matching poset, so by the theorem of Griggs [6] it has a symmetric chain decomposition. This chain decomposition contains chains of size 1 and 3. Let V ⊂ B be the set of elements which are the middle elements of chains of size 3. Then V ∈ M (G) ∩ M (G ′ ).
Let S B denote the set of all permutations of B. For π ∈ S B define the bipartite graph G π = (A, B, E π ) such that a ∈ A and b ∈ B are joined by an edge if aπ(b) ∈ E.
Clearly, G π is a normalized matching graph as G π is isomorphic to G. Hence, we have that M (G) ∩ M (G π ) = ∅. Note that Remark. The result of Theorem 16 is not sharp for any 1 ≤ k ≤ n − 1. However, we cannot get any lower bound better than n k . This is true as every minimal normalized matching graph is a forest (see [3]), and in this case the number of complete matchings from A to B is equal to |M (G)| ≤ n k .
Proof of Theorem 5. Write a k for n k . We have already established that 2 [n] has at least |T M |...|T n−1 | symmetric chain decompositions. Also, by Theorem 16, we have |T k | ≥ a k a k+1 for k = M, ..., n − 1. We shall estimate a k a k−1 for n/2+ √ n < k < n/2+2 √ n. In this case, we have a k = Θ(2 n / √ n) and a k − a k+1 = Θ(2 n /n). Hence, using the well known lower bound a b ≥ (a/b) b for binomial coefficients, we get Thus, we have We note that 2 [n] has at most 2 O(2 n log n) symmetric chain decompositions, so there is only a √ n factor gap in the exponent between the lower and upper bound. We can get the upper bound n 2 n by the following simple observation: every symmetric chain partition corresponds to a unique sequence (U 0 , ..., U n−1 ), where U k is a complete matching from [n] (k) to [n] (k+1) for k < M , and U k is a complete matching from U k+1 to U k for k ≥ M . But for each k, the number of such matchings is less than n ( n k ) , as the comparability subgraph of 2 [n] induced on the levels [n] (k) ∪[n] (k+1) has maximum degree at most n. Hence, the number of symmetric chain decompositions of 2 [n] is less than n k=0 n ( n k ) = n 2 n .

Open problems
In this section, we propose some open problems. Let us extend some of our earlier definitions to rank-symmetric posets. Let (P, <) be a rank-symmetric poset with levels A 0 , ..., A n . A chain C ⊂ P is rank-symmetric, if |C ∩ A i | = |C ∩ A n−i | for i = 0, ..., n; the chain C is skipless, if the set {j : |A j ∩ C| = 1} is an interval. The chain C is symmetric, if C is skipless and rank-symmetric.
After the method we used to prove Theorem 4, it is natural to ask the following question. Let P be a rank-symmetric, unimodal normalized matching poset of width w. Can we partition P into w rank-symmetric chains such that the sizes of the chains are as close to each other as possible?
As we do not know the answer in the simpler case when our chains are not necessarily rank-symmetric, we ask a less ambitious question.
Question 17. Define the function f : Q + → N as follows: for r ∈ Q + , f (r) is the maximal positive integer such that any rank-symmetric, unimodal normalized matching poset P of width |P |/r can be partitioned into rank-symmetric chains of size at least f (r). Is it true that lim r∈Q f (r) = ∞?
Another closely related question is motivated by the following theorem of Lonc [13]. He proved that if given a positive integer h and n is sufficiently large, then 2 [n] has a chain partition, where all but at most one of the chains have size h. The author of this paper [16] showed that the smallest such n is O(h 2 ) and this bound is best possible up to a constant. We propose the following rank-symmetric version of this problem.
Conjecture 18. For every positive integer h there exists a positive integer N (h) such that if n > N (h) and n is odd, then 2 [n] has a partition into rank-symmetric chains, where all but at most one of the chains have size 2h.
We note the following connection between the results of this paper and Conjecture 18. Suppose that the following conjecture is true.
Then Conjecture 18 follows from Theorem 4. Surprisingly, we have not been able to prove the much weaker statement that the bipartite subgraph of (Q, <) induced on [n] (k) and [n] (k+1) is connected for k = n/2 + Θ( √ n). We note that if n is even, then the bipartite subgraph of (Q, <) induced on the levels [n] (n/2) and [n] (n/2+1) is not even connected, it is the union of trees of size n + 1.