A Local-Global Principle for Macaulay Posets
A Local-Global Principle for Macaulay Posets Sergei L. Bezrukov∗ Department of Mathematics and Computer Science University of Paderborn F¨ urstenallee 11, 33102 Paderborn, Germany
Xavier Portas†,
Oriol Serra
Departament de Matem`atica Aplicada i Telem`atica Modul C3, Campus Nord, Universitat Polit`ecnica de Catalunya Jordi Girona 1, 08034 Barcelona, Spain
Abstract We consider the shadow minimization problem (SMP) for cartesian powers P n of a Macaulay poset P . Our main result is a local-global principle with respect to the lexicographic order Ln . Namely, we show that under certain conditions the shadow of any initial segment of the order Ln for n ≥ 3 is minimal iff it is so for n = 2. These conditions include such poset properties as additivity, shadow increasing, final shadow increasing and being rank-greedy. We also show that these conditions are essentially necessary for the lexicographic order to provide nestedness in the SMP.
1
Introduction
Let (P, ≤) be a finite poset with a partial order ≤. The poset (P, ≤) is called ranked if there exists a function rP : P 7→ IN such that minx∈P rP (x) = 0 and rP (x) + 1 = rP (y) whenever x < y and there is no z ∈ P yielding x < z < y. We call the numbers rP (x) and r(P ) = maxy∈P r(y) the rank of x ∈ P and of P , respectively. Usually we denote a poset and its element set by the same letter. The set Pt = {x ∈ P | rP (x) = t} is called the tth level of P . For A ⊆ Pt and t ≥ 0 define the shadow of A as ∆(A) = {x ∈ Pt−1 | x ≤ y for some y ∈ A}. ∗ †
Partially supported by the Spanish Research Council under project TIC97-0963. Granted by the Spanish Research Council, PN94 40433776.
1
The shadow minimization problem (SMP for brevity) is one of the most important problems in combinatorics and has a lot of applications: for a ranked poset P and given natural numbers t > 0 and m, 1 ≤ m ≤ |Pt |, find a set A ⊆ Pt such that |A| = m and |∆(A)| ≤ |∆(B)| for any B ⊆ Pt with |B| = m. We call such a set optimal . See [9] for an introduction to the subject. Let us concentrate on posets decomposable under the cartesian product operation. For posets (P 0 , ≤P 0 ) and (P 00 , ≤P 00 ) we define their cartesian product as a poset with the element set P 0 × P 00 and with the partial order ≤× defined as follows: (x1 , y1 ) ≤× (x2 , y2 ) iff x1 ≤P 0 x2 and y1 ≤P 00 y2 . Since the cartesian product is an associative operation, the products of more than two posets are well defined. It can be easily shown that the cartesian product of ranked posets is a ranked poset too. We denote by P n the nth cartesian power of a poset P .
s
s Q
s
s
Q Q Q Qs s
.. .
s
s
s
.. .
.. .
.. .
s
s
s
s
s
s
a.
b.
s
s Q Q Q Q s · · · Qs
c.
···
s
d.
Figure 1: The basic posets The SMP for cartesian products of various posets was considered in the literature. Examples of such posets include the n-cube (the nth cartesian power of a poset in Fig. 1a) [12, 13], the lattice of multisets (the cartesian products of chains, cf. Fig. 1b) [8], and the star poset (the cartesian powers of posets in Fig. 1c) [14]. Recently, the spider poset (the cartesian powers of posets in Fig. 1d) was studied in [4, 6]. This poset includes all the above mentioned posets as special cases. For all these posets, the solution to the SMP problem provides nested families of optimal subsets. An exact formulation of this phenomenon leads to the notion of a Macaulay poset. Let be a total order on P . For z ∈ Pt denote Ft (z) = {x ∈ Pt | x z}. We call a subset A ⊆ Pi initial segment if A = Ft (z) for some z ∈ Pt . A poset P is called Macaulay, if there exists a total order (called Macaulay order ), such that for any t > 0 the following properties hold:
2
N1 (nestedness) : For any z ∈ Pt the initial segment Ft (z) has minimal shadow among all subsets of Pt of the same cardinality; N2 (continuity) : The shadow of an initial segment is an initial segment itself, i.e. for any z ∈ Pt there exists z 0 ∈ Pt−1 such that ∆(Ft (z)) = Ft−1 (z 0 ). Note that a Macaulay order is not claimed to be a linear extension of ≤. We refer to a Macaulay poset as to a triple (P, ≤, ). For a poset P we introduce its dual P ∗ as a poset with the same element set P and with inverse partial order. Proposition 1 (cf. [9]) A poset P is Macaulay iff so is its dual P ∗ . Moreover, if is the Macaulay order on P then the inverse of is the Macaulay order on P ∗ . Therefore, the posets in Fig. 1 and their duals are Macaulay. The above mentioned results imply that the same holds for cartesian powers of these posets. There exists a powerful technique for establishing the Macaulayness of cartesian powers of posets. This technique is based on compression (see below) and usually involves induction on the number n of posets in the product. However, for this technique it is principal that n ≥ 3. The case n = 2 is a special one and must be considered separately. A similar situation also occurs in the edge isoperimetric problems on graphs. In [1, 2] even a more general problem of the minimization of submodular functions on graphs has been studied. For a finite set S a function f : 2S 7→ IR is called submodular if for any A, B ⊆ S f (A) + f (B) ≥ f (A ∪ B) + f (A ∩ B). If S is the vertex set of a graph G = (VG , EG ), then the number of the cut edges separating a set A ⊆ VG from VG \A is an example of a submodular function. Based on a submodular function f defined on 2VG , some special functions f (n) on the nth cartesian power of G are considered in [1]. These functions f (n) are, in a sense, decomposable, i.e. they can be represented as certain sums of functions f (n−1) . Without going into details, consider, for example, the function f (n) defined as the number of cut edges which separate a subset of vertices of the n-cube from its complement. Then f (n) can be represented as the sum over i = 1, . . . , n of the number of cut edges which are parallel to the ith dimension. There are two important results in [1, 2] we want to highlight. Firstly, the submodularity of the function f defined on 2VG provides a sufficient condition for applying the compression for minimization of f (n) defined on the nth cartesian power of the graph G. Secondly, if the lexicographic order (see Section 2) provides nestedness (cf. N1 ) in minimization of f (n) for n = 2, then it is so for any n ≥ 3. Readers are referred to [2] for more details. It turned out that the last result, which is called the local-global principle in [2], is valid also with respect to some other total orders [3]. In what concerns the SMP, the above approach can not be directly applied because of the necessity to maintain the level structure of a poset. Another difficulty in applying the results of [1] is that the function |∆(·)| is not decomposable in the sense above. 3
However, the applicability of compression and some general principles in the proof techniques for establishing the Macaulayness of posets and solving edge-isoperimetric problems (cf. [3]) gave us a starting point to look for a local-global principle with respect to the SMP. Our main result is a local-global principle with respect to the lexicographic order, which is formulated in Theorem 1 in the next section. It turns out that for the validity of this result it is important that the poset satisfies some additional conditions, which have no analogies for graphs yet (cf. Section 2 for precise definitions). In Theorem 2 we show that these conditions are essential for the lexicographic order to provide the Macaulayness of any cartesian power of a poset. The next section of our paper is devoted to the formulation of some poset properties and statements of our main results: Theorem 1 and Theorem 2. In Section 3 we present some auxiliary propositions which will be used throughout the text. Sections 4 and 5 are devoted to the proof of Theorems 1 and 2 respectively. An application of our technique to establishing the Macaulayness of a new class of posets is demonstrated in Section 6. Final remarks in Section 7 conclude the paper.
2
Strongly Macaulay posets
Let (P, ≤, ) be Macaulay. Consider the SMP for the cartesian powers P n . We represent the elements of P n as n-dimensional vectors (x1 , . . . , xn ) and denote them by bold letters. The lexicographic order n on the set P n is defined as follows: (x1 , . . . , xn ) n (y1 , . . . , yn ) iff there exists an i such that xi yi and xj = yj for all j < i. We often omit the superscript n in n if n is uniquely defined by the context. For a subset A ⊆ P n we refer to a minimal or a maximal element of A as the corresponding elements in the order n . We call a set F ⊆ Pi a segment if it consists of elements that are consecutive in Pi in the order . We introduce initial and final segments of Pi defined naturally. For a segment F ⊆ Pi let G be the set of elements in Pi that precede each element of F in order . The new shadow of F is defined as ∆new (F ) = ∆(F ) \ ∆(G). A Macaulay poset (P, ≤, ) is called additive if for any i ≥ 1 |∆new (F1 )| ≥ |∆new (F2 )| ≥ |∆new (F3 )|
(1)
for all segments F1 , F2 , F3 ⊆ Pi with F1 being initial, F3 being final and |F1 | = |F2 | = |F3 |. Proposition 2 (cf. [9]) A Macaulay poset is additive iff so is its dual. It is worth to mention another (equivalent) definition of additivity, which relates to submodularity. For a Macaulay poset (P, ≤, ) and i = 0, . . . , r(P ) define the shadow function sfi : 2Pi 7→ IN by setting, for F ⊆ Pi , sfi (F ) = |∆(F 0 )|, 4
with F 0 ⊆ Pi being an initial segment with |F 0 | = |F |. The shadow function is called little submodular if sfi (F1 ) + sfi (F2 ) ≥ sfi (F1 ∪ F2 ) + sfi (F1 ∩ F2 ) for all F1 , F2 ⊆ Pi
(2)
with F1 ∩ F2 = ∅ or F1 ∪ F2 = Pi .
(3)
The poset (P, ≤, ) is called little submodular if the shadow function sfi is little submodular for any i = 0, . . . , r(P ). It is known that a poset is little submodular iff it is additive (cf. [9]). If we do not claim the condition (3) then the shadow function will be submodular (cf. Section 1). It is, however, not the case for the powers of posets in Fig. 1 (cf. a counterexample on page 349 in [9]). It is worth to mention that the function |∆(·)| is submodular. Furthermore, a Macaulay poset (P, ≤, ) is called shadow increasing if for any i ≥ 1 |∆(F1 )| ≥ |∆(F2 )|
(4)
for any initial segments F1 ⊆ Pi and F2 ⊆ Pi−1 with |F1 | = |F2 |. Cartesian powers of the posets shown in Fig. 1 are additive and shadow increasing [9]. Similarly, the poset P is called final shadow increasing, if for any i ≥ 1 |∆new (F1 )| ≥ |∆new (F2 )|
(5)
for any final segments F1 ⊆ Pi and F2 ⊆ Pi−1 with |F1 | = |F2 |. This notion has been already appeared in [10], where extremal ideals in posets were studied. Obviously, if (P, ≤, ) is shadow increasing then (4) is valid for any initial segments F1 ⊆ Pi and F2 ⊆ Pj with |F1 | = |F2 | and i ≥ j. A similar remark concerns a final shadow increasing poset and (5). Denote by M the class of ranked posets having only one maximum and only one minimum element. The next proposition is a bit weaker form of Propositions 7 and 8 of [10]. Proposition 3 (cf. [10]). Let (P, ≤, ) ∈ M be Macaulay. a. If P is final shadow increasing then the dual P ∗ is shadow increasing; b. If P is additive and shadow increasing, and if the dual P ∗ is shadow increasing, then P is final shadow increasing. We call a Macaulay poset P strongly Macaulay if it is additive, shadow increasing and final shadow increasing. Proposition 4 A poset P ∈ M is strongly Macaulay iff so is its dual P ∗ .
5
Proof. Obviously, P ∗ ∈ M. Then Propositions 1, 2 and 3a respectively imply that P ∗ is Macaulay, additive and shadow increasing. Since (P ∗ )∗ = P , then Proposition 3b applied to P ∗ implies P ∗ is final shadow increasing. Thus, P ∗ is strongly Macaulay. The second part of the assertion follows from (P ∗ )∗ = P . 2 For F ⊆ P and x ∈ P we write F ≺ x if y ≺ x for any y ∈ F . A Macaulay poset (P, ≤, ) is called rank-greedy if the order is a linear extension of ≤, and if ∆(x) ≺ y and rP (x) > rP (y) imply x ≺ y.
(6)
Proposition 5 (cf. [9]). If P is a rank-greedy Macaulay poset, then so is its dual P ∗ . Our main result is the following theorem. Theorem 1 (the local-global principle for SMP). Let (P, ≤, ) ∈ M be strongly Macaulay and rank-greedy. Let the order 2 be Macaulay for P 2 . Then for any n ≥ 2 the order n is a Macaulay order for P n . The following result shows that the assumptions concerning the poset P in Theorem 1 are essential. Theorem 2 Let (P, ≤, ) be a Macaulay poset. Furthermore, let r(P ) ≥ 3 and assume the orders 2 and 3 are Macaulay for P 2 and P 3 , respectively. Then for any n ≥ 1 a. P n ∈ M; b. P n is rank-greedy; c. P n is additive; d. P n is shadow increasing; e. P n is final shadow increasing We prove Theorems 1 and 2 in Sections 4 and 5, respectively, after establishing some properties of the order n in the next section.
3
Some auxiliary results
The following property, which is called consistency, is straightforward. Lemma 1 Let x = (x1 , . . . , xn ) ∈ P n and y = (y1 , . . . , yn ) ∈ P n . Furthermore, let xi = yi for some i. Then x n y iff x0 n−1 y0 for x0 and y0 obtained from x and y, respectively, by omitting their ith entries. 2 6
Let A ⊆ P n . For a fixed i and z ∈ P denote P n (i, z) = {(ξ1 , . . . , ξn ) ∈ P n | ξi = z} A(i, z) = A ∩ P n (i, z). Denote by (P n (i, z), v) the subposet of P n with induced partial order (and the total order induced in it by n ). Obviously, this poset is isomorphic to P n−1 . Moreover, by the consistency of the order n , the natural isomorphism between the two posets sends initial segments to initial segments. Proposition 6 Let P be an additive and Macaulay poset. Then ∆(Pi ) = Pi−1
for i = 1, . . . , r(P ).
(7)
Proof. Indeed, assume it is not so for some i ≥ 1 and denote Z = Pi−1 \ ∆(Pi ). Consider the dual P ∗ . By Propositions 1 and 2, P ∗ is Macaulay and additive. Since Z is a final segment in P , it is an initial segment in P ∗ . Then, ∆(Z) = ∅ in P ∗ contradicts the additivity of P ∗ . 2 Lemma 2 Let (P, ≤, ) be additive, Macaulay and rank-greedy. Then P n satisfies the property N2 with respect to the order n for any n ≥ 1. Proof. We prove the lemma by induction on n. Since it is true for n = 1, we proceed with n ≥ 2 and t ≥ 1. Let A ⊆ Ptn be an initial segment and let x = (x1 , . . . , xn ) be the largest element of A in order n . Then A =
[
A(1, z) ∪ A(1, x1 ),
z≺x1
∆(A) =
[
∆new (A(1, z)) ∪ ∆new (A(1, x1 )),
(8)
z≺x1
where all the unions are disjoint. Note that for any z ≺ x1 one has Ptn ∩ P n (1, z) ⊆ A.
(9)
n Furthermore, note that (7) implies ∆(Pin ) = Pi−1 for any n ≥ 2 and any i = 1, . . . , r(P n ). n n This and (9) imply Pt−1 ∩ P (1, z) ⊆ ∆(A), i.e. n ∆new (A(1, z)) = Pt−1 ∩ P n (1, z).
(10)
Note that A(1, x1 ) is an initial segment in the subposet (P n (1, x1 ), v). Since this subposet is isomorphic to P n−1 and since the natural isomorphism sends initial segments to initial segments, then the induction hypothesis implies that ∆new (A(1, x1 )) is an initial segment in (P n (1, x1 ), v). This, (8) and (10) imply the lemma. 2 This lemma is essential for the proof of Lemma 3. Let A ⊆ Ptn and let i be fixed. We say that A is i-compressed if A(i, z) is an initial segment (with respect to the order n ) in the subposet (P n (i, z), v) for any z ∈ P . The set A is called compressed if it is i-compressed for i = 1, . . . , n. 7
Lemma 3 Let (P, ≤, ) be additive, Macaulay and rank-greedy. Suppose for some n ≥ 2 that P n−1 is Macaulay and that the order n−1 is a Macaulay order. Then for any A ⊆ Ptn and for a fixed i, 1 ≤ i ≤ n, there exists an i-compressed set B ⊆ Ptn such that |B| = |A| and |∆(B)| ≤ |∆(A)|. Proof. For z ∈ P with rP (z) < r(P ) denote ∇(z) = {x ∈ P | z ∈ ∆(x)}. Let A ⊆ Ptn . Then |∆(A)| = |
[ z∈P
(∆(A) ∩ P n (i, z))| ≥
X z∈P
max {|∆(A(i, z)) ∩ P n (i, z)|, |A(i, x)|}.
x∈∇(z)
(11)
Let B be the i-compression of A. Then B(i, z) and ∆(B(i, z)) ∩ P n (i, z) (cf. Lemma 2) are initial segments in the subposet (P n (i, z), v). Therefore, the lower bound (11) for the set B is tight. Since (P n (i, z), v) is isomorphic to the Macaulay poset P n−1 , then |∆(A(i, z)) ∩ P n (i, z)| ≥ |∆(B(i, z)) ∩ P n (i, z)|. This implies |∆(A)| ≥ |∆(B)|. 2 Applying the compression for i = 1, . . . , n sufficiently many times one gets a compressed set C ⊆ Ptn such that |C| = |A| and |∆(C)| ≤ |∆(A)|.
4
Proof of Theorem 1
Obviously, the total order provides a numbering of P n with numbers 1, . . . , |P n |. According to this, for z ∈ Ptn denote by N (z) the number of z and for A ⊆ Ptn let P N (A) = z∈A N (z). Note that with A and B as in Lemma 3 it holds N (B) ≤ N (A). We prove the theorem by induction on n. Since for n = 2 it is true, we proceed for n ≥ 3. Let A ⊆ Ptn be an optimal set. By Lemma 3 we can assume that A is compressed. Let x = (x1 , . . . , xn ) be the largest element of A, and let y = (y1 , . . . , yn ) be the smallest element of Pt \ A. If A is not an initial segment then y ≺ x. Without loss of generality we can assume that xi 6= yi for i = 1, . . . , n, since otherwise y ∈ A follows from Lemma 1 and the fact that A is compressed. In particular, y1 ≺ x1 . Lemma 4 If n ≥ 3 then rP (xn ) ≥ rP (yn ). Proof. Assume rP (xn ) < rP (yn ). We show that this assumption leads to a contradiction. For i = 0, . . . , r(P n ) denote by fin and lni the smallest and largest elements of Pin respectively. Since P ∈ M and P is rank-greedy, then fin = (0, . . . , 0, fg1 , 1, . . . , 1) |
{z h
}
lni = (1, . . . , 1, l1g , 0, . . . , 0), |
{z
}
h
1 where 0 = f01 , 1 = fr(P ) , and g and h are defined by i = r(P ) · h + g, 0 ≤ g < r(P ).
8
Denote t0 = t − rP (x1 ) − rP (xn ) and t00 = t − rP (y1 ) − rP (yn ), and let u = (y1 , ln−2 t00 , yn ),
v = (x1 , ftn−2 , xn ). 0
Case 1. Assume rP (x1 ) ≤ rP (y1 ). Now rP (xn ) < rP (yn ) implies t0 > t00 . Consider n 00 000 0 z = (y1 , ln−2 t000 , xn ) ∈ Pt with t < t ≤ t . Lemma 1 and the fact that P is rank-greedy imply y u ≺ z ≺ v x. (12) Since x and v agree in the first entry, and since v x, then v ∈ A because A is 1-compressed. By a similar argument {z, u, y} ⊆ A, and we have a contradiction. Case 2. Assume rP (y1 ) < rP (x1 ). If rP (x1 ) + rP (xn ) > rP (y1 ) + rP (yn ) then consider an element z1 ∈ P such that rP (z1 ) = rP (x1 )−(rP (yn )−rP (xn )). Obviously, rP (z1 ) < rP (x1 ). On the other hand, rP (z1 ) = (rP (x1 ) + rP (xn )) − rP (yn ) > rP (y1 ). Since P is rank-greedy, then z1 can be chosen so that y1 ≺ z1 ≺ x1 . Therefore, z = (z1 , ftn−2 , yn ) ∈ Ptn . Since n ≥ 3 then (12) holds and we have a contradiction as above. 0 Assume rP (x1 ) + rP (xn ) ≤ rP (y1 ) + rP (yn ). Now if rP (x1 ) > rP (y1 ) + 1 then consider , zn ) ∈ Ptn with some zn and some z1 , such that rP (z1 ) = rP (y1 ) + 1 w = (z1 , ftn−2 0 and y1 ≺ z1 ≺ x1 . Since P is rank-greedy, then such an element z1 does exist. If rP (x1 ) = rP (y1 ) + 1 then set w = v. Now for q = rP (yn ) − rP (zn ) one has 0 < q = (t − rP (y1 ) − t00 ) − rP (zn ) = (t − rP (z1 ) − rP (zn )) + 1 − t00 = t0 − t00 + 1. 00 Thus, if t00 ≥ 1 then q ≤ t0 . In this case for z = (z1 , ftn−2 0 −q , yn ) one has z ≺ w v. If t = 0 then take z = (y1 , fqn−2 , xn ). In both cases (12) holds and leads to a contradiction. 2
For z = (z1 , . . . , zn−1 ) ∈ P n−1 and A ⊆ P n denote P n (z) = {(ξ1 , . . . , ξn ) ∈ P n | ξi = zi , i = 1, . . . , n − 1} A(z) = A ∩ P n (z). Obviously, the subposet of (P n , ≤× ) with the element set P n (z) and the induced partial order is isomorphic to (P, ≤). Furthermore, denote x0 = (x1 , . . . , xn−1 ), y0 = (y1 , . . . , yn−1 ) and let k 0 = rP (xn ) and k 00 = rP (yn ). Then k 0 ≥ k 00 by Lemma 4. Moreover, the sets A(x0 ) and A(y0 ) are initial segments in Pkn0 (x0 ) and Pkn00 (y0 ) respectively (these sets are shown by bold lines in Fig. 2). Our goal is to take some elements from A(x0 ) and add them to A(y0 ). More precisely, denote δ = min{|Pkn00 (y0 ) \ A(y0 )|, |A(x0 )|}. Let Fx ⊆ Pkn0 (x0 ) be the segment consisting of the last δ elements of the segment A(x0 ) and let Fy ⊆ Pkn00 (y0 ) be the segment consisting of δ elements of Pkn00 (y0 ) \ A(y0 ) such that y ∈ Fy . Consider the set B = (A \ Fx ) ∪ Fy . Obviously, B ⊆ Ptn and |B| = |A|. We show |∆(A)| ≥ |∆(B)|. 9
(13)
'$
'$
P n (x0 )
P n (y0 )
s XX XXX XXX XX zy c 00
k0
x
k
'
$
n
s 0
P (n, 0)
s & %
(x1 , . . . , xn−1 , 0)
s & %
(y1 , . . . , yn−1 , 0)
&
%
Figure 2: Usage of the shadow increase property in the proof of Theorem 1 For this note that ∆(B \ A) \ P n (y0 ) ⊆ ∆(A). Indeed, if k 00 = r(P ) then k 0 = r(P ) as well. Since |Pr(P ) | = 1 by the assumption P ∈ M, then the nth entries of y and x agree. This implies y ∈ A, which contradicts the definition of y. Therefore k 00 < r(P ). Now any element z ∈ ∆(B \ A) \ P n (y0 ) is of the form z = (z1 , . . . , zn−1 , yn ) and the vectors (z1 , . . . , zn−1 ) and y0 differ just in one entry. Since z ∈ ∆(u) for u = (z1 , . . . , zn−1 , zn ) with yn ≤ zn , and since u ≺ y, then z ∈ ∆(A). Therefore, to prove (13) it is sufficient to show that |∆(A \ B) ∩ P n (x0 )| ≥ |∆(B \ A) ∩ P n (y0 )|. Equivalently, by the natural identification of P n (x0 ) and P n (y0 ) with P , it suffices to show that, in (P, ≤), |∆new (Fx )| ≥ |∆new (Fy )|. (14) Assume |Pkn00 (y0 ) \ A(y0 )| ≥ |A(x0 )|. Let F with |F | = δ be the initial segment of Pkn00 (y0 ). Since P is additive and shadow increasing, then (1) and (4) imply |∆new (Fx )| = |∆(Fx )| ≥ |∆(F )| = |∆new (F )| ≥ |∆new (Fy )|. Now assume |Pkn00 (y0 ) \ A(y0 )| < |A(x0 )|. Then Fy is a final segment in Pkn00 (y0 ). Denote by F , |F | = δ, the final segment of Pkn0 (x0 ). Since P is additive and final shadow increasing, then (1) and (5) imply |∆new (Fx )| ≥ |∆new (F )| ≥ |∆new (Fy )|. Therefore, (14) and, thus, (13) are established. Obviously, N (B) < N (A). Hence, by successive applications of the above arguments we can transform the set A into an initial segment without increasing the shadow. 2
10
1 s
s3
@ @
2@s
p+1 sa aa QQ Qaaa s Qs p a s s a · · ·Q 1 HH3 4 !s p + 2 !! @ ! HH@ ! H !! H @ s ! 2
a.
b.
Figure 3: Counterexamples to Theorem 2 for n = 2
5
Proof of Theorem 2
By Proposition 1 the inverse of n is a Macaulay order for the dual of P n , n = 1, 2, 3. It is easy to show that the inverse of n is the lexicographic order too. Thus, the assumptions of Theorem 2 are valid for (P ∗ )n for n = 2, 3. In the proof we often refer to this assertion. We use some ideas of [9] in the proof of propositions c) – e) of the theorem. a. It suffices to prove the assertion for n = 1 only. For this denote k = r(P ) and assume |Pk | ≥ 2. Let b be the first element of Pk . Then (b, b) is the first element of 2 P2k . Let c ∈ Pk be the successor of b, and let a ∈ ∆(b). If a ≺ b then (a, c) 6∈ ∆(b, b),
(a, c) ≺ (b, a) ∈ ∆(b, b).
This implies ∆(b, b) is not an initial segment, which contradicts N2 . 3 Assume a b. Similarly to above, (b, b, b) is the smallest element of P3k . One has
(b, c, a) 6∈ ∆(b, b, b),
(b, c, a) ≺ (a, b, b) ∈ ∆(b, b, b).
This implies ∆(b, b, b) is not an initial segment, which again contradicts N2 . Applying the above arguments to the dual P ∗ , one gets that it has an only maximum element. This implies |P0 | = 1. Note that |P0 | = |Pr(P ) | = 1 implies, in particular, that the Hasse diagram of P is connected. For n = 2 Theorem 2a is not true for poset shown in Fig. 3a together with a Macaulay order. b. It also suffices to consider the case n = 1 only. First assume the order is not a linear extension of ≤, i.e. there exist a, b ∈ P such that a ∈ ∆(b) and a b. We call such a pair {a, b} inverted pair and show that the existence of an inverted pair implies P is a chain. Denote by fi (resp. li ) the first (resp. last) element of Pi , i = 1, . . . , r(P ). Fact 1 ∆(z) = Pi−1 for any z ∈ Pi , i = 1, . . . , r(P ). Proof. Let us choose an inverted pair {a, b} satisfying if x ∈ ∆(y) and rP (y) > rP (b) then x ≺ y b is the smallest element of ∇(a). 11
(15) (16)
Assume there exists an element u ∈ Pi−1 \ ∆(fi ) for some i ≥ 1. Let t = rP (b) 2 and denote B = Ft+i (b, fi ) ⊆ Pt+i . Then (b, u) 6∈ ∆(B). Indeed, if (b, u) ∈ ∆(x, y) for some (x, y) ∈ B then either b ∈ ∆(x) or u ∈ ∆(y). If b ∈ ∆(x) then b ≺ x by (15) and, thus, (b, fi ) ≺ (x, y). If u ∈ ∆(y) then fi ≺ y and x = b. This implies (b, fi ) ≺ (x, y) again. Therefore, one has (b, u) 6∈ ∆(B) and (b, u) ≺ (a, fi ) ∈ ∆(B). Since B is an initial segment, then we have a contradiction with N2 . Therefore, the element u mentioned above does not exist, i.e. ∆(fi ) = Pi−1 for i = 1, . . . , r(P ). This and the Macaulayness of P imply the assertion for any z ∈ Pi . 2 Fact 2 Let {a, b} be an inverted pair, and let t = rP (b). Then a = lt−1 and b = ft . Proof. First we show that a = lt−1 . Since this is true if |Pt−1 | = 1, let us assume |Pt−1 | > 1. Let b be the smallest element in ∇(a) such that b ≺ a. Suppose a ≺ lt−1 . Then, by the choice of b, (a, lt−1 ) 6∈ ∆(F2t−1 (b, a)) and (a, lt−1 ) ≺ (lt−1 , a) ∈ ∆(b, a), thus contradicting N2 . Therefore, a = lt−1 . The second part of the statement follows by applying similar arguments to the dual poset (P ∗ )2 . 2 Fact 3 Let {a, b} be an inverted pair, and let t = rP (b). Then |Pt | = |Pt−1 | = 1. Proof. Assume the contrary, i.e. |Pt | > 1. By the part a) of the Theorem, t < r(P ). Fact 1 implies ∇(ft ) = ∇(lt ) = Pt+1 , and it follows from Fact 2 that b = ft ≺ Pt+1 . Therefore, (b, lt , a) 6∈ ∆(F3t (b, b, b)) and (b, lt , a) ≺ (a, b, b) ∈ ∆(b, b, b). This contradicts N2 . Similar arguments in the dual poset (P ∗ )3 imply |Pt−1 | = 1.
2
We are now ready to conclude the proof of the part b). Let {a, b} be an inverted pair 2 and let t = rP (b). Assume t < r(P ) and ft+1 b. Consider B = Ft+1 (b, f1 ) ⊆ Pt+1 . Now ∆new (b, f1 ) = {(b, f0 ), (a, f1 )}. Furthermore, (ft+1 , f0 ) 6∈ B and ∆(ft+1 , f0 ) = {(b, f0 )}. Therefore, |∆((B \ (b, f1 )) ∪ (ft+1 , f0 ))| = |∆(B)| − |∆new (b, f1 )| + 1 < |∆(B)|, which contradicts the optimality of B. Hence, ft+1 ≺ b and, thus, {b, ft+1 } is an inverted pair. Fact 3 implies |Pt+1 | = 1. By iterating the argument with respect to the inverted pair {ft+i , ft+i+1 } for 1 ≤ i ≤ r(P ) − t we get b = ft ft+1 · · · fr(P )
and |Pt | = |Pt+1 | = · · · |Pr(P ) | = 1. 12
Similar arguments in the dual (P ∗ )2 provide f0 f1 · · · ft−1 = a and |P0 | = |P1 | = · · · |Pt−1 | = 1. Therefore, P is a chain and f0 · · · fr(P ) . Denote k = r(P ) and consider 3 B = F2k−1 (fk−1 , fk , f0 ) ⊆ P2k−1 . Then |∆(B)| = 3k − 3, while the shadow of the 3 final segment in P2k−1 of the same size consists of 2k − 1 elements. This contradicts to the optimality of B if k ≥ 3. This proves that, if r(P ) ≥ 3 then the Macaulay order is a linear extension of the partial order ≤. Finally, assume (6) is not fulfilled for P . Thus, there exist a, b ∈ P such that ∆(b) ≺ a, rP (b) > rP (a) and b a. Denote q = rP (b) − rP (a) and consider the set B = FrP (a)+q (a, fq ) ⊆ Pr2P (a)+q . Since (b, f0 ) (a, fq ), then (b, f0 ) 6∈ B. Furthermore, ∆(b) ≺ a implies ∆(b, f0 ) ⊆ B. Since a 6= b and q ≥ 1, then ∆(b, f0 ) ∩ ∆(a, fq ) = ∅. Since the order is a linear extension of ≤, and since q ≥ 1, then (a, fq−1 ) ∈ ∆new (a, fq ). Thus, |∆new (a, fq )| ≥ 1. One has |∆((B \ (a, fq )) ∪ (b, f0 ))| = |∆(B)| − |∆new (a, fq )| < |∆(B)|, which contradicts the optimality of B. This concludes the proof. It is worth noting that Theorem 2b is not necessarily true for n = 2. The poset P shown in Fig. 3b has the Macaulay order presented in the figure which is not rank greedy, and P 2 admits the lexicographic order as Macaulay order for p = 1, 2. However, the lexicographic order is not Macaulay for P n for any n ≥ 3 and p ≥ 1. Moreover, the theorem is not true for r(P ) = 2, i.e. if P is a chain with 2 elements a and b with a ∈ ∆(b). In the case b ≺ a the order n is Macaulay for n = 2, 3, however, not for n ≥ 4. c. Taking into account that the assertion is true for the lattice of multisets [6, 9], let j be minimal index such that |Pj | ≥ 2. Furthermore, let y be the smallest element of Pj and let x ∈ Pj be the successor of y in Pj . Let n ≥ 1 and let F1 , F2 ⊆ Pin be segments with F1 being initial and |F2 | = |F1 |. Denote F10 = {(x, z) | z ∈ F1 } and F20 = {(y, z) | z ∈ F2 }.
(17)
n+1 Then F10 , F20 are segments in Pi+j . One has
∆new (F10 ) = {(x, w) | w ∈ ∆new (F1 )}, ∆new (F20 ) = {(y, w) | w ∈ ∆new (F2 )}. Hence, |∆new (F10 )| = |∆new (F1 )| and |∆new (F20 )| = |∆new (F2 )|.
(18)
n+1 Denote by G2 the set of elements in Pi+j that precede each element of F20 and let n+1 A = G2 ∪ F10 and B = G2 ∪ F20 . Note that B is an initial segment in Pi+j . Since n+1 P is Macaulay (Theorem 1), then
|∆(A)| = |∆(G2 )| + |∆new (F10 )| ≥ |∆(B)| = |∆(G2 )| + |∆new (F20 )|. 13
Thus, |∆new (F10 )| ≥ |∆new (F20 )| and the first inequality in (1) follows by taking (18) into account. To show the second inequality in (1), take segments F1 , F2 ∈ Pin with F2 being final n+1 and |F1 | = |F2 | as above. Construct the segments F10 , F20 ⊆ Pi+j according to (17). For these segments (18) is valid. Denote by G1 the set of elements of {(x, z) | z ∈ P n } that precede each element of F10 and consider the set G2 as above. Furthermore, denote A = G1 ∪ G2 ∪ F10 and n+1 B = G1 ∪ G2 ∪ F20 . Since B is an initial segment in Pi+j then |∆(A)| ≥ |∆(B)|, which is equivalent to |∆(G2 )| + |∆new (G1 )| + |∆new (F10 )| ≥ |∆(G2 )| + |∆new (G1 )| + |∆new (F20 )|, and the second inequality in (1) follows from (18). d. Let y be the only element of P0 and let x be the smallest element of P1 . Since y ≤ x n and P is rank-greedy by b) then y ≺ x. Let n ≥ 1 and let F1 ⊆ Pin and F2 ⊆ Pi−1 be initial segments such that |F1 | = |F2 |. Denote F10 = {(y, z) | z ∈ F1 } and F20 = {(x, z) | z ∈ F2 }.
(19)
Then F10 , F20 are segments in Pin+1 with F10 being initial. One has ∆new (F10 ) = {(y, w) | w ∈ ∆(F1 )}, ∆new (F20 ) = {(x, w) | w ∈ ∆(F2 )}. Hence, |∆new (F10 )| = |∆(F1 )| and |∆new (F20 )| = |∆(F2 )|.
(20)
Now since P n+1 is Macaulay by Theorem 1, then P n+1 is additive by c). Thus, the first inequality in (1) applied to P n+1 implies |∆new (F10 )| ≥ |∆new (F20 )|,
(21)
which, by taking (20) into account, implies (4). e. The proof of this proposition is similar to the last one. Let x be the only largest element of Pr(P ) and let y be the largest element of Pr(P )−1 . Then (7) implies y ≤ x. Moreover, y ≺ x, since P is rank-greedy by b). n Let n ≥ 1 and let F1 ⊆ Pin and F2 ⊆ Pi−1 be final segments such that |F1 | = |F2 |. n+1 0 0 We introduce F1 and F2 according to (19). Then F10 , F20 are segments in Pr(P )−1+i 0 0 0 with F2 being final. For the segments F1 and F2 equality (18) is valid.
Now since P n+1 is Macaulay, then P n+1 is additive by c). Thus, the second inequality in (1) applied to P n+1 implies (21), from where (5) follows. 2
14
a3 s
−T3
A A Asb2 a2 s @ @ a1 s @sb1 A A b0As
a3 s
6 s
−T2
A A Asb2 a2 s
A A As 5 3 s
−T1
a1 s
2 s
−T0
sb1
A A b0As
a.
b.
s4 A A 1 As
c.
Figure 4: Posets (T (3), ≤) (a.), 3 × 2 grid (b.), and torus C(3) (c.)
6
Applications
Firstly, let us mention that Theorem 1 being applied to chains implies the Kruskal-Katona theorem [12, 13] and a particular case of the Clements-Lindstr¨om theorem [8] if all the chains in the product are of the same length. Now consider the following poset (T (k), ≤) ∈ M of rank k. For 1 ≤ i ≤ k − 1 the ith level of T (k) consists of two elements ai and bi . Denote by b0 and ak the elements of T0 and Tk respectively. The partial order is defined as follows: x < y iff r(x) < r(y). The Hasse diagram of (T (3), ≤) is shown in Fig. 4a. We define the total order on T (k) by setting bi−1 ≺ ai for i = 1, . . . , k and ai ≺ bi for i = 1, . . . , k − 1. Obviously, the order is Macaulay on (T (k), ≤). Theorem 3 For any k ≥ 1 and any n ≥ 1 the poset (T n (k), ≤× , n ) is Macaulay. Proof. Denote T = T (k) for brevity. Since for k ≤ 2 the theorem is true, we proceed for k ≥ 3. Obviously, (T (k), ≤) is strongly Macaulay. Therefore, by Theorem 1 it suffices to prove Theorem 3 for n = 2 only. Note that if the properties N1 and N2 are satisfied with respect to a total order on a ∗ ranked poset P for some t > 0, then these properties are also valid for Pr(P )−t+1 and the inverse of (cf. the proof of Proposition 8.1.2. of [9]). Therefore, if P is isomorphic to P ∗ then P is Macaulay iff Pt satisfies the properties N1 and N2 for t ≤ dr(P )/2e. Applying this assertion to T 2 with r(T 2 ) = 2k, it suffices to consider the case t ≤ k only. 2 To avoid trivial cases, we assume that ∆(A) 6= Tt−1 . In particular, t > 1. By Lemma 2, 2 T satisfies the property N2 . Thus, we only have to check the property N1 . Let A ⊆ Tt2 be an optimal set and assume that A is not an initial segment. By Lemma 3 we may assume that A is compressed. In the sequel we construct a set B such that |B| = |A|, For S ⊆ Tt2 denote
|∆(B)| ≤ |∆(A)| and N (B) < N (A). ˜ ∆(S) = ∆(S) \ ∆(A \ S). 15
(22)
Note that if B = (A \ S) ∪ R for S ⊆ A and R ⊆ Tt2 \ A, then ˜ ˜ |∆(B)| = |∆(A)| + |∆(R)| − |∆(S) \ ∆(R)|.
(23)
Let y = (y1 , y2 ) be the smallest element in Tt2 \ A and let x = (x1 , x2 ) ∈ A with y ≺ x. ˜ Such an element exists because A is not an initial segment. We can assume |∆(y)| > 0, since otherwise B = (A \ {x}) ∪ {y} satisfies (22). Since A is compressed, then x1 > y 1 ,
x2 < y 2 .
(24)
˜ Note that if y2 = bj for some j ≥ 0, then |∆(y)| = 0, which contradicts our assumptions. Therefore, y2 = aj for some j ≥ 1. ˜ Note that, since t ≤ k, then |∆(y)| ≤ |∆new (y2 )| ≤ 2. We consider two cases. Case 1. Assume j < k. Since y 6∈ A and A is compressed, then (y1 , bj ) ∈ Tt2 \ A. Moreover, ˜ ˜ ∆({y, (y1 , bj )}) = ∆(y). (25) ˜ Case 1a. Assume |∆(y)| = 2. Then j ≥ 2. Let z ∈ A be the smallest element such that y ≺ z. Then z = (ap , aq ) for some q < j (cf. (24)) and p = t − q. We show j − q ≥ 2. Indeed, if j − q = 1 then p − rT (y1 ) = 1. Thus, (y1 , aq ) ∈ ∆(z) ∩ ∆new (y). This contradicts ˜ |∆(y)| = 2. ˜ Therefore, Z = {(ξ, aq ) | ξ ∈ ∆(ap )} ⊆ ∆(z). Since p ≥ rT (y1 ) + 2 ≥ 2, then, according to (24), |Z| = |∆(ap )| = 2. Consider (
B=
(A \ {z}) ∪ {y}, if (bp , aq ) 6∈ A (A \ {z, (bp , aq )}) ∪ {y, (y1 , bj )}, otherwise.
˜ ˜ In the first case |∆(z)| ≥ |Z| = 2, and in the second one |∆({z, (bp , aq )})| ≥ |Z| = 2. This, (23) and (25) imply the set B satisfies (22). ˜ Case 1b. Assume |∆(y)| = 1. Note that j ≥ 2. Indeed, if j = 1, then rT (y1 ) = t − 1 and, ˜ thus, rT (x1 ) = t according to (24). This contradicts |∆(y)| = 1. Denote i = rT (y1 ). Note that (ai+1 , aj−1 ) ∈ A and A ∩ {(ai+1 , bj−1 ), (bi+1 , bj−1 )} = ∅. Assume y = (ai , aj ). Then let (
B=
(A \ {(ai+1 , aj−1 )}) ∪ {y}, if (bi+1 , aj−1 ) 6∈ A (A \ {(ai+1 , aj−1 ), (bi+1 , aj−1 )}) ∪ {y, (y1 , bj )}, otherwise.
Assume y = (bi , aj ). Now if A ∩ {(ai+2 , bj−2 ), (bi+2 , bj−2 )} = ∅ then we construct B as above. Otherwise, if (ai+2 , bj−2 ) ∈ A and (bi+2 , bj−2 ) 6∈ A then let (
B=
(A \ {(ai+2 , bj−2 )}) ∪ {y}, if (bi+1 , aj−1 ) 6∈ A (A \ {(ai+2 , bj−2 ), (bi+1 , aj−1 )}) ∪ {y, (y1 , bj )}, otherwise. 16
Finally, if S = {(ai+2 , bj−2 ), (bi+2 , bj−2 )} ⊆ A then let (
B=
(A \ S) ∪ {y, (y1 , bj )}, if (bi+1 , aj−1 ) 6∈ A (A \ S \ {(bi+1 , aj−1 )}) ∪ {y, (y1 , bj ), (ai+1 , bj−1 )}, otherwise.
Case 2. Assume y2 = ak . Thus, t = k and y = (b0 , ak ). Let φ be the poset automorphism of T such that φ(ai ) = bi , 0 < i < k and φ(b0 ) = b0 , φ(ak ) = ak . It induces an automorphism ψ in T 2 defined as ψ(α, β) = (φ(β), φ(α)). Let I be the largest initial segment of Tt2 \ A and let F be the largest final segment of Tt2 \ A. Since |∆(A)| = |∆(ψ(A))|, we may assume that |F | ≥ |I| > 0. Denote Zi = {(α, β) ∈ Tk2 | rT (α) = i}, i = 0, . . . , k. Now |F | ≥ |I| > 0 implies (Z0 ∪ Zk ) ∩ A = ∅. Furthermore, if Z1 ∩ A = ∅ then Zk−1 ∩ A = ∅ too. In this case construct the set B by replacing each (αi , βj ) ∈ A with (αi+1 , βj−1 ) for α, β ∈ {a, b}. ˜ ˜ If A∩Z1 6= ∅, then |∆(y)| < 2. Thus, without loss of generality we can assume |∆(y)| = 1. This implies A ∩ {(a1 , bk−1 ), (b1 , bk−1 )} = ∅. Let x be the largest element in A. Now if ˜ x2 = bq then 1 ≤ q < k − 1. Thus, for S = {x, (x1 , aq )} one has: |∆(S)| ≥ 1 and ∆(S) ∩ ∆(y) = ∅. In this case set B = (A \ S) ∪ {y, (a1 , bk−1 )}. ˜ Finally, assume x2 = aj . If q = 1 then |∆(x)| = 1 and ∆(x) ∩ ∆(y) = ∅, because k ≥ 3. ˜ If q > 1 then |∆(x)| = 2 and |∆(x) ∩ ∆(y)| ≤ 1. We set B = (A \ {x}) ∪ {y}. It is easy to check that in all cases the set B satisfies (22). Therefore, by applying similar arguments (and the compression, if necessary) to the set B we transform it into an initial segment without increasing the shadow. 2 It is shown in [5] how to deduce a solution to an edge-isoperimetric problem on a graph G if a certain poset P is Macaulay. In fact, for any graph G one can construct the associated poset P . Without going into details (for which we refer readers to [3, 5]), let us mention that the poset (T (k), ≤) corresponds to the complete bipartite graph Kk,k . Moreover, it is shown in [3] that any cartesian power of the graph G corresponds in the sense above to the same cartesian power of P . Therefore, Theorem 3 implies a solution for an edgeisoperimetric problem on the graph (Kk,k )n . This extends a corresponding result in [2], where these graphs were studied with respect to the edge-isoperimetric problem.
7
Concluding remarks
Our first remarks concern the lexicographic order. Let us modify the partial order in T (k) a bit by making the elements ai and bi+1 incomparable for i = 1, . . . , k − 1. Then the resulting poset Q is isomorphic to the k × 2 grid (cf. Fig. 4b). Due to the ClementsLindstr¨om theorem [8] the cartesian product of any chains Z1 , . . . , Zn of lengths k1 , . . . , kn is Macaulay. Moreover, if k1 ≥ · · · ≥ kn (26)
17
then the lexicographic order over the set of n-dimensional vectors is a Macaulay order. However, there is no Macaulay order on Q such that the lexicographic order n is Macaulay for n ≥ 2. The reason for this is that regardless of the total order , the condition (26) is not satisfied. For n ≥ 3 this condition is essential for the lexicographic order to work for the lattice of multisets and, thus, for the poset Qn too. If we further modify the poset T (k) by making the elements bi and ai+1 incomparable for i = 1, . . . , k − 1, then this results in a cycle C(k) (cf. Fig. 4c). This poset is rank-greedy and the corresponding Macaulay order is essentially unique (it is shown in the figure). However, the lexicographic order does not work for C(k) × C(k) (the element (2, 6) is the smallest one in the fourth level, but its shadow is greater than one of (3, 3)). Nevertheless, any cartesian power of C(k) (which is isomorphic to a torus) is Macaulay. The Macaulay order can be derived from [11, 15], where a vertex-isoperimetric problem for tori was studied, and is not lexicographic. Although Theorem 2 provides some necessary conditions for posets, all whose cartesian powers are Macaulay, there are many infinite series of such posets. We illustrate this by providing a tool for constructing such series. Given a Macaulay poset (P, ≤, ), construct another poset Q = (P, v) as follows. Take an element a ∈ Pi for some i > 1 and consider Fi (a). Then ∆(Fi (a)) = Fi−1 (b) for some b ∈ Pi−1 . Let c ∈ Fi−1 (b) and assume c 6≤ a. Now we extend the partial order ≤ by setting c ≤ a (cf. Fig. 5). 9 s
9 s @ @
@ @
@sa 8@ @ @ @ @s 3 sc 5@ 7@s b @ @ @ @s 2@ 4@s @ 1@s
@ 8s a @ @ @ @ @s s s b 3 c 5@ 7@ @ @ @ @ s 2@ 4@s @ 1@s
a.
b.
6 s
6 s
Figure 5: Posets P (a) and Q (b) Denote ∆P (m, i) = min |∆(A)|, where the minimum runs over all A ⊆ Pi with |A| = m. Since the partial order ≤ is a suborder of v, then ∆P (A) ⊆ ∆Q (A) for any A ⊆ Pi , Thus, ∆P (m, i) ≤ ∆Q (m, i) for all m = 1, . . . , |Pi | and i = 1, . . . , r(P ).
(27)
However, ∆P (Fi (a)) = ∆Q (Fi (a)) for any a ∈ Pi . Therefore, since P is Macaulay, then the lower bound (27) for Q is tight. This implies (P, v, ) is Macaulay too. Moreover, if P is strongly Macaulay and rank-greedy then so is Q. Obviously, applying any number of the above extensions leads to a Macaulay poset.
18
Now consider P 2 . Since ∆P 2 (Fi ((x, y))) = {(x, ξ) | ξ ∈ ∆P (Fi−rP (x) (y))} ∪ {(ξ, y) | ξ ∈ ∆P (Fi−rP (y) (x))}, then ∆P 2 (Fi ((x, y)) = ∆Q2 (Fi ((x, y))). Therefore, if P satisfies the assumptions of Theorem 1, then so is for Q. On the other hand, since the lexicographic order is Macaulay for P 2 , then so is for P 4 , for example. Extending P 2 as shown above results in a new poset, for which Theorem 1 is applicable. In particular, the Macaulayness of the torus poset C n (k) implies a similar result for the powers of k × 2 grids (cf. Fig. 4b,c). It would be interesting to establish local-global principles with respect to some other extremal problems. In our forthcoming paper [7] we present a local-global principle for vertex-isoperimetric problems. Sperner theory [9] and the Erd¨os-Ko-Rado type results are closely related with Macaulay theory, and are further examples were local-global principles may exist.
Acknowledgements The authors would like to thank Konrad Engel (University of Rostock, Germany) for his helpful comments concerning the draft version of the manuscript.
References [1] R. Ahlswede, N. Cai, General edge-isoperimetric inequalities, Part I: Informationtheoretical method, Europ. J. Combin., 18 (1997), 355–372. [2] R. Ahlswede, N. Cai, General edge-isoperimetric inequalities, Part II: A local-global principle for lexicographic solution, Europ. J. Combin., 18 (1997), 479–489. [3] S.L. Bezrukov, Edge isoperimetric problems on graphs, to appear in Graph Theory and Combinatorial Biology, Bolyai Soc. Math. Stud. 7, L. Lov´asz, A. Gyarfas, G.O.H. Katona, A. Recski, L. Sz´ekely eds., Budapest. [4] S.L. Bezrukov, On posets whose products are Macaulay, J. Comb. Theory, A-84 (1998), 157–170. [5] S.L. Bezrukov, On an equivalence in discrete extremal problems, to appear in Discr. Math. [6] S.L. Bezrukov, R. Els¨asser, The spider poset is Macaulay, preprint, 1997. [7] S.L. Bezrukov, X. Portas, O. Serra, A local-global principle for vertex-isoperimetric problems, preprint, 1998. [8] G.F. Clements, B. Lindstr¨om, A generalization of a combinatorial theorem of Macaulay, J. Comb. Th. 7 (1969), No. 2, 230–238. 19
[9] K. Engel, Sperner theory, Cambridge University Press, 1997. [10] K. Engel, U. Leck, Optimal antichains and ideals in Macaulay posets, Preprint 96/21, University of Rostock, to appear in Graph Theory and Combinatorial Biology, Bolyai Soc. Math. Stud. 7, L. Lov´asz, A. Gyarfas, G.O.H. Katona, A. Recski, L. Sz´ekely eds., Budapest. [11] V.M. Karachanjan, A discrete isoperimetric problem on multidimensional torus, (in Russian), Doklady AN Arm. SSR, vol. LXXIV (1982), No. 2, 61–65. [12] G.O.H. Katona, A theorem of finite sets, in: Theory of graphs, Academia Kiado, Budapest, 1968, 187–207. [13] J.B. Kruskal, The optimal number of simplices in a complex, in: Math. Optimization Tech., University of California Press, Berkeley, California, 1963, 251–268. [14] U. Leck, Extremalprobleme f¨ ur den Schatten in Posets, Ph. D. Thesis, FU Berlin, 1995; Shaker-Verlag Aachen, 1995. [15] O. Riordan, An ordering on the discrete even torus, SIAM J. Discr. Math., 11 (1998), No. 1, 110–127.
20
Xavier Portas†,
Oriol Serra
Departament de Matem`atica Aplicada i Telem`atica Modul C3, Campus Nord, Universitat Polit`ecnica de Catalunya Jordi Girona 1, 08034 Barcelona, Spain
Abstract We consider the shadow minimization problem (SMP) for cartesian powers P n of a Macaulay poset P . Our main result is a local-global principle with respect to the lexicographic order Ln . Namely, we show that under certain conditions the shadow of any initial segment of the order Ln for n ≥ 3 is minimal iff it is so for n = 2. These conditions include such poset properties as additivity, shadow increasing, final shadow increasing and being rank-greedy. We also show that these conditions are essentially necessary for the lexicographic order to provide nestedness in the SMP.
1
Introduction
Let (P, ≤) be a finite poset with a partial order ≤. The poset (P, ≤) is called ranked if there exists a function rP : P 7→ IN such that minx∈P rP (x) = 0 and rP (x) + 1 = rP (y) whenever x < y and there is no z ∈ P yielding x < z < y. We call the numbers rP (x) and r(P ) = maxy∈P r(y) the rank of x ∈ P and of P , respectively. Usually we denote a poset and its element set by the same letter. The set Pt = {x ∈ P | rP (x) = t} is called the tth level of P . For A ⊆ Pt and t ≥ 0 define the shadow of A as ∆(A) = {x ∈ Pt−1 | x ≤ y for some y ∈ A}. ∗ †
Partially supported by the Spanish Research Council under project TIC97-0963. Granted by the Spanish Research Council, PN94 40433776.
1
The shadow minimization problem (SMP for brevity) is one of the most important problems in combinatorics and has a lot of applications: for a ranked poset P and given natural numbers t > 0 and m, 1 ≤ m ≤ |Pt |, find a set A ⊆ Pt such that |A| = m and |∆(A)| ≤ |∆(B)| for any B ⊆ Pt with |B| = m. We call such a set optimal . See [9] for an introduction to the subject. Let us concentrate on posets decomposable under the cartesian product operation. For posets (P 0 , ≤P 0 ) and (P 00 , ≤P 00 ) we define their cartesian product as a poset with the element set P 0 × P 00 and with the partial order ≤× defined as follows: (x1 , y1 ) ≤× (x2 , y2 ) iff x1 ≤P 0 x2 and y1 ≤P 00 y2 . Since the cartesian product is an associative operation, the products of more than two posets are well defined. It can be easily shown that the cartesian product of ranked posets is a ranked poset too. We denote by P n the nth cartesian power of a poset P .
s
s Q
s
s
Q Q Q Qs s
.. .
s
s
s
.. .
.. .
.. .
s
s
s
s
s
s
a.
b.
s
s Q Q Q Q s · · · Qs
c.
···
s
d.
Figure 1: The basic posets The SMP for cartesian products of various posets was considered in the literature. Examples of such posets include the n-cube (the nth cartesian power of a poset in Fig. 1a) [12, 13], the lattice of multisets (the cartesian products of chains, cf. Fig. 1b) [8], and the star poset (the cartesian powers of posets in Fig. 1c) [14]. Recently, the spider poset (the cartesian powers of posets in Fig. 1d) was studied in [4, 6]. This poset includes all the above mentioned posets as special cases. For all these posets, the solution to the SMP problem provides nested families of optimal subsets. An exact formulation of this phenomenon leads to the notion of a Macaulay poset. Let be a total order on P . For z ∈ Pt denote Ft (z) = {x ∈ Pt | x z}. We call a subset A ⊆ Pi initial segment if A = Ft (z) for some z ∈ Pt . A poset P is called Macaulay, if there exists a total order (called Macaulay order ), such that for any t > 0 the following properties hold:
2
N1 (nestedness) : For any z ∈ Pt the initial segment Ft (z) has minimal shadow among all subsets of Pt of the same cardinality; N2 (continuity) : The shadow of an initial segment is an initial segment itself, i.e. for any z ∈ Pt there exists z 0 ∈ Pt−1 such that ∆(Ft (z)) = Ft−1 (z 0 ). Note that a Macaulay order is not claimed to be a linear extension of ≤. We refer to a Macaulay poset as to a triple (P, ≤, ). For a poset P we introduce its dual P ∗ as a poset with the same element set P and with inverse partial order. Proposition 1 (cf. [9]) A poset P is Macaulay iff so is its dual P ∗ . Moreover, if is the Macaulay order on P then the inverse of is the Macaulay order on P ∗ . Therefore, the posets in Fig. 1 and their duals are Macaulay. The above mentioned results imply that the same holds for cartesian powers of these posets. There exists a powerful technique for establishing the Macaulayness of cartesian powers of posets. This technique is based on compression (see below) and usually involves induction on the number n of posets in the product. However, for this technique it is principal that n ≥ 3. The case n = 2 is a special one and must be considered separately. A similar situation also occurs in the edge isoperimetric problems on graphs. In [1, 2] even a more general problem of the minimization of submodular functions on graphs has been studied. For a finite set S a function f : 2S 7→ IR is called submodular if for any A, B ⊆ S f (A) + f (B) ≥ f (A ∪ B) + f (A ∩ B). If S is the vertex set of a graph G = (VG , EG ), then the number of the cut edges separating a set A ⊆ VG from VG \A is an example of a submodular function. Based on a submodular function f defined on 2VG , some special functions f (n) on the nth cartesian power of G are considered in [1]. These functions f (n) are, in a sense, decomposable, i.e. they can be represented as certain sums of functions f (n−1) . Without going into details, consider, for example, the function f (n) defined as the number of cut edges which separate a subset of vertices of the n-cube from its complement. Then f (n) can be represented as the sum over i = 1, . . . , n of the number of cut edges which are parallel to the ith dimension. There are two important results in [1, 2] we want to highlight. Firstly, the submodularity of the function f defined on 2VG provides a sufficient condition for applying the compression for minimization of f (n) defined on the nth cartesian power of the graph G. Secondly, if the lexicographic order (see Section 2) provides nestedness (cf. N1 ) in minimization of f (n) for n = 2, then it is so for any n ≥ 3. Readers are referred to [2] for more details. It turned out that the last result, which is called the local-global principle in [2], is valid also with respect to some other total orders [3]. In what concerns the SMP, the above approach can not be directly applied because of the necessity to maintain the level structure of a poset. Another difficulty in applying the results of [1] is that the function |∆(·)| is not decomposable in the sense above. 3
However, the applicability of compression and some general principles in the proof techniques for establishing the Macaulayness of posets and solving edge-isoperimetric problems (cf. [3]) gave us a starting point to look for a local-global principle with respect to the SMP. Our main result is a local-global principle with respect to the lexicographic order, which is formulated in Theorem 1 in the next section. It turns out that for the validity of this result it is important that the poset satisfies some additional conditions, which have no analogies for graphs yet (cf. Section 2 for precise definitions). In Theorem 2 we show that these conditions are essential for the lexicographic order to provide the Macaulayness of any cartesian power of a poset. The next section of our paper is devoted to the formulation of some poset properties and statements of our main results: Theorem 1 and Theorem 2. In Section 3 we present some auxiliary propositions which will be used throughout the text. Sections 4 and 5 are devoted to the proof of Theorems 1 and 2 respectively. An application of our technique to establishing the Macaulayness of a new class of posets is demonstrated in Section 6. Final remarks in Section 7 conclude the paper.
2
Strongly Macaulay posets
Let (P, ≤, ) be Macaulay. Consider the SMP for the cartesian powers P n . We represent the elements of P n as n-dimensional vectors (x1 , . . . , xn ) and denote them by bold letters. The lexicographic order n on the set P n is defined as follows: (x1 , . . . , xn ) n (y1 , . . . , yn ) iff there exists an i such that xi yi and xj = yj for all j < i. We often omit the superscript n in n if n is uniquely defined by the context. For a subset A ⊆ P n we refer to a minimal or a maximal element of A as the corresponding elements in the order n . We call a set F ⊆ Pi a segment if it consists of elements that are consecutive in Pi in the order . We introduce initial and final segments of Pi defined naturally. For a segment F ⊆ Pi let G be the set of elements in Pi that precede each element of F in order . The new shadow of F is defined as ∆new (F ) = ∆(F ) \ ∆(G). A Macaulay poset (P, ≤, ) is called additive if for any i ≥ 1 |∆new (F1 )| ≥ |∆new (F2 )| ≥ |∆new (F3 )|
(1)
for all segments F1 , F2 , F3 ⊆ Pi with F1 being initial, F3 being final and |F1 | = |F2 | = |F3 |. Proposition 2 (cf. [9]) A Macaulay poset is additive iff so is its dual. It is worth to mention another (equivalent) definition of additivity, which relates to submodularity. For a Macaulay poset (P, ≤, ) and i = 0, . . . , r(P ) define the shadow function sfi : 2Pi 7→ IN by setting, for F ⊆ Pi , sfi (F ) = |∆(F 0 )|, 4
with F 0 ⊆ Pi being an initial segment with |F 0 | = |F |. The shadow function is called little submodular if sfi (F1 ) + sfi (F2 ) ≥ sfi (F1 ∪ F2 ) + sfi (F1 ∩ F2 ) for all F1 , F2 ⊆ Pi
(2)
with F1 ∩ F2 = ∅ or F1 ∪ F2 = Pi .
(3)
The poset (P, ≤, ) is called little submodular if the shadow function sfi is little submodular for any i = 0, . . . , r(P ). It is known that a poset is little submodular iff it is additive (cf. [9]). If we do not claim the condition (3) then the shadow function will be submodular (cf. Section 1). It is, however, not the case for the powers of posets in Fig. 1 (cf. a counterexample on page 349 in [9]). It is worth to mention that the function |∆(·)| is submodular. Furthermore, a Macaulay poset (P, ≤, ) is called shadow increasing if for any i ≥ 1 |∆(F1 )| ≥ |∆(F2 )|
(4)
for any initial segments F1 ⊆ Pi and F2 ⊆ Pi−1 with |F1 | = |F2 |. Cartesian powers of the posets shown in Fig. 1 are additive and shadow increasing [9]. Similarly, the poset P is called final shadow increasing, if for any i ≥ 1 |∆new (F1 )| ≥ |∆new (F2 )|
(5)
for any final segments F1 ⊆ Pi and F2 ⊆ Pi−1 with |F1 | = |F2 |. This notion has been already appeared in [10], where extremal ideals in posets were studied. Obviously, if (P, ≤, ) is shadow increasing then (4) is valid for any initial segments F1 ⊆ Pi and F2 ⊆ Pj with |F1 | = |F2 | and i ≥ j. A similar remark concerns a final shadow increasing poset and (5). Denote by M the class of ranked posets having only one maximum and only one minimum element. The next proposition is a bit weaker form of Propositions 7 and 8 of [10]. Proposition 3 (cf. [10]). Let (P, ≤, ) ∈ M be Macaulay. a. If P is final shadow increasing then the dual P ∗ is shadow increasing; b. If P is additive and shadow increasing, and if the dual P ∗ is shadow increasing, then P is final shadow increasing. We call a Macaulay poset P strongly Macaulay if it is additive, shadow increasing and final shadow increasing. Proposition 4 A poset P ∈ M is strongly Macaulay iff so is its dual P ∗ .
5
Proof. Obviously, P ∗ ∈ M. Then Propositions 1, 2 and 3a respectively imply that P ∗ is Macaulay, additive and shadow increasing. Since (P ∗ )∗ = P , then Proposition 3b applied to P ∗ implies P ∗ is final shadow increasing. Thus, P ∗ is strongly Macaulay. The second part of the assertion follows from (P ∗ )∗ = P . 2 For F ⊆ P and x ∈ P we write F ≺ x if y ≺ x for any y ∈ F . A Macaulay poset (P, ≤, ) is called rank-greedy if the order is a linear extension of ≤, and if ∆(x) ≺ y and rP (x) > rP (y) imply x ≺ y.
(6)
Proposition 5 (cf. [9]). If P is a rank-greedy Macaulay poset, then so is its dual P ∗ . Our main result is the following theorem. Theorem 1 (the local-global principle for SMP). Let (P, ≤, ) ∈ M be strongly Macaulay and rank-greedy. Let the order 2 be Macaulay for P 2 . Then for any n ≥ 2 the order n is a Macaulay order for P n . The following result shows that the assumptions concerning the poset P in Theorem 1 are essential. Theorem 2 Let (P, ≤, ) be a Macaulay poset. Furthermore, let r(P ) ≥ 3 and assume the orders 2 and 3 are Macaulay for P 2 and P 3 , respectively. Then for any n ≥ 1 a. P n ∈ M; b. P n is rank-greedy; c. P n is additive; d. P n is shadow increasing; e. P n is final shadow increasing We prove Theorems 1 and 2 in Sections 4 and 5, respectively, after establishing some properties of the order n in the next section.
3
Some auxiliary results
The following property, which is called consistency, is straightforward. Lemma 1 Let x = (x1 , . . . , xn ) ∈ P n and y = (y1 , . . . , yn ) ∈ P n . Furthermore, let xi = yi for some i. Then x n y iff x0 n−1 y0 for x0 and y0 obtained from x and y, respectively, by omitting their ith entries. 2 6
Let A ⊆ P n . For a fixed i and z ∈ P denote P n (i, z) = {(ξ1 , . . . , ξn ) ∈ P n | ξi = z} A(i, z) = A ∩ P n (i, z). Denote by (P n (i, z), v) the subposet of P n with induced partial order (and the total order induced in it by n ). Obviously, this poset is isomorphic to P n−1 . Moreover, by the consistency of the order n , the natural isomorphism between the two posets sends initial segments to initial segments. Proposition 6 Let P be an additive and Macaulay poset. Then ∆(Pi ) = Pi−1
for i = 1, . . . , r(P ).
(7)
Proof. Indeed, assume it is not so for some i ≥ 1 and denote Z = Pi−1 \ ∆(Pi ). Consider the dual P ∗ . By Propositions 1 and 2, P ∗ is Macaulay and additive. Since Z is a final segment in P , it is an initial segment in P ∗ . Then, ∆(Z) = ∅ in P ∗ contradicts the additivity of P ∗ . 2 Lemma 2 Let (P, ≤, ) be additive, Macaulay and rank-greedy. Then P n satisfies the property N2 with respect to the order n for any n ≥ 1. Proof. We prove the lemma by induction on n. Since it is true for n = 1, we proceed with n ≥ 2 and t ≥ 1. Let A ⊆ Ptn be an initial segment and let x = (x1 , . . . , xn ) be the largest element of A in order n . Then A =
[
A(1, z) ∪ A(1, x1 ),
z≺x1
∆(A) =
[
∆new (A(1, z)) ∪ ∆new (A(1, x1 )),
(8)
z≺x1
where all the unions are disjoint. Note that for any z ≺ x1 one has Ptn ∩ P n (1, z) ⊆ A.
(9)
n Furthermore, note that (7) implies ∆(Pin ) = Pi−1 for any n ≥ 2 and any i = 1, . . . , r(P n ). n n This and (9) imply Pt−1 ∩ P (1, z) ⊆ ∆(A), i.e. n ∆new (A(1, z)) = Pt−1 ∩ P n (1, z).
(10)
Note that A(1, x1 ) is an initial segment in the subposet (P n (1, x1 ), v). Since this subposet is isomorphic to P n−1 and since the natural isomorphism sends initial segments to initial segments, then the induction hypothesis implies that ∆new (A(1, x1 )) is an initial segment in (P n (1, x1 ), v). This, (8) and (10) imply the lemma. 2 This lemma is essential for the proof of Lemma 3. Let A ⊆ Ptn and let i be fixed. We say that A is i-compressed if A(i, z) is an initial segment (with respect to the order n ) in the subposet (P n (i, z), v) for any z ∈ P . The set A is called compressed if it is i-compressed for i = 1, . . . , n. 7
Lemma 3 Let (P, ≤, ) be additive, Macaulay and rank-greedy. Suppose for some n ≥ 2 that P n−1 is Macaulay and that the order n−1 is a Macaulay order. Then for any A ⊆ Ptn and for a fixed i, 1 ≤ i ≤ n, there exists an i-compressed set B ⊆ Ptn such that |B| = |A| and |∆(B)| ≤ |∆(A)|. Proof. For z ∈ P with rP (z) < r(P ) denote ∇(z) = {x ∈ P | z ∈ ∆(x)}. Let A ⊆ Ptn . Then |∆(A)| = |
[ z∈P
(∆(A) ∩ P n (i, z))| ≥
X z∈P
max {|∆(A(i, z)) ∩ P n (i, z)|, |A(i, x)|}.
x∈∇(z)
(11)
Let B be the i-compression of A. Then B(i, z) and ∆(B(i, z)) ∩ P n (i, z) (cf. Lemma 2) are initial segments in the subposet (P n (i, z), v). Therefore, the lower bound (11) for the set B is tight. Since (P n (i, z), v) is isomorphic to the Macaulay poset P n−1 , then |∆(A(i, z)) ∩ P n (i, z)| ≥ |∆(B(i, z)) ∩ P n (i, z)|. This implies |∆(A)| ≥ |∆(B)|. 2 Applying the compression for i = 1, . . . , n sufficiently many times one gets a compressed set C ⊆ Ptn such that |C| = |A| and |∆(C)| ≤ |∆(A)|.
4
Proof of Theorem 1
Obviously, the total order provides a numbering of P n with numbers 1, . . . , |P n |. According to this, for z ∈ Ptn denote by N (z) the number of z and for A ⊆ Ptn let P N (A) = z∈A N (z). Note that with A and B as in Lemma 3 it holds N (B) ≤ N (A). We prove the theorem by induction on n. Since for n = 2 it is true, we proceed for n ≥ 3. Let A ⊆ Ptn be an optimal set. By Lemma 3 we can assume that A is compressed. Let x = (x1 , . . . , xn ) be the largest element of A, and let y = (y1 , . . . , yn ) be the smallest element of Pt \ A. If A is not an initial segment then y ≺ x. Without loss of generality we can assume that xi 6= yi for i = 1, . . . , n, since otherwise y ∈ A follows from Lemma 1 and the fact that A is compressed. In particular, y1 ≺ x1 . Lemma 4 If n ≥ 3 then rP (xn ) ≥ rP (yn ). Proof. Assume rP (xn ) < rP (yn ). We show that this assumption leads to a contradiction. For i = 0, . . . , r(P n ) denote by fin and lni the smallest and largest elements of Pin respectively. Since P ∈ M and P is rank-greedy, then fin = (0, . . . , 0, fg1 , 1, . . . , 1) |
{z h
}
lni = (1, . . . , 1, l1g , 0, . . . , 0), |
{z
}
h
1 where 0 = f01 , 1 = fr(P ) , and g and h are defined by i = r(P ) · h + g, 0 ≤ g < r(P ).
8
Denote t0 = t − rP (x1 ) − rP (xn ) and t00 = t − rP (y1 ) − rP (yn ), and let u = (y1 , ln−2 t00 , yn ),
v = (x1 , ftn−2 , xn ). 0
Case 1. Assume rP (x1 ) ≤ rP (y1 ). Now rP (xn ) < rP (yn ) implies t0 > t00 . Consider n 00 000 0 z = (y1 , ln−2 t000 , xn ) ∈ Pt with t < t ≤ t . Lemma 1 and the fact that P is rank-greedy imply y u ≺ z ≺ v x. (12) Since x and v agree in the first entry, and since v x, then v ∈ A because A is 1-compressed. By a similar argument {z, u, y} ⊆ A, and we have a contradiction. Case 2. Assume rP (y1 ) < rP (x1 ). If rP (x1 ) + rP (xn ) > rP (y1 ) + rP (yn ) then consider an element z1 ∈ P such that rP (z1 ) = rP (x1 )−(rP (yn )−rP (xn )). Obviously, rP (z1 ) < rP (x1 ). On the other hand, rP (z1 ) = (rP (x1 ) + rP (xn )) − rP (yn ) > rP (y1 ). Since P is rank-greedy, then z1 can be chosen so that y1 ≺ z1 ≺ x1 . Therefore, z = (z1 , ftn−2 , yn ) ∈ Ptn . Since n ≥ 3 then (12) holds and we have a contradiction as above. 0 Assume rP (x1 ) + rP (xn ) ≤ rP (y1 ) + rP (yn ). Now if rP (x1 ) > rP (y1 ) + 1 then consider , zn ) ∈ Ptn with some zn and some z1 , such that rP (z1 ) = rP (y1 ) + 1 w = (z1 , ftn−2 0 and y1 ≺ z1 ≺ x1 . Since P is rank-greedy, then such an element z1 does exist. If rP (x1 ) = rP (y1 ) + 1 then set w = v. Now for q = rP (yn ) − rP (zn ) one has 0 < q = (t − rP (y1 ) − t00 ) − rP (zn ) = (t − rP (z1 ) − rP (zn )) + 1 − t00 = t0 − t00 + 1. 00 Thus, if t00 ≥ 1 then q ≤ t0 . In this case for z = (z1 , ftn−2 0 −q , yn ) one has z ≺ w v. If t = 0 then take z = (y1 , fqn−2 , xn ). In both cases (12) holds and leads to a contradiction. 2
For z = (z1 , . . . , zn−1 ) ∈ P n−1 and A ⊆ P n denote P n (z) = {(ξ1 , . . . , ξn ) ∈ P n | ξi = zi , i = 1, . . . , n − 1} A(z) = A ∩ P n (z). Obviously, the subposet of (P n , ≤× ) with the element set P n (z) and the induced partial order is isomorphic to (P, ≤). Furthermore, denote x0 = (x1 , . . . , xn−1 ), y0 = (y1 , . . . , yn−1 ) and let k 0 = rP (xn ) and k 00 = rP (yn ). Then k 0 ≥ k 00 by Lemma 4. Moreover, the sets A(x0 ) and A(y0 ) are initial segments in Pkn0 (x0 ) and Pkn00 (y0 ) respectively (these sets are shown by bold lines in Fig. 2). Our goal is to take some elements from A(x0 ) and add them to A(y0 ). More precisely, denote δ = min{|Pkn00 (y0 ) \ A(y0 )|, |A(x0 )|}. Let Fx ⊆ Pkn0 (x0 ) be the segment consisting of the last δ elements of the segment A(x0 ) and let Fy ⊆ Pkn00 (y0 ) be the segment consisting of δ elements of Pkn00 (y0 ) \ A(y0 ) such that y ∈ Fy . Consider the set B = (A \ Fx ) ∪ Fy . Obviously, B ⊆ Ptn and |B| = |A|. We show |∆(A)| ≥ |∆(B)|. 9
(13)
'$
'$
P n (x0 )
P n (y0 )
s XX XXX XXX XX zy c 00
k0
x
k
'
$
n
s 0
P (n, 0)
s & %
(x1 , . . . , xn−1 , 0)
s & %
(y1 , . . . , yn−1 , 0)
&
%
Figure 2: Usage of the shadow increase property in the proof of Theorem 1 For this note that ∆(B \ A) \ P n (y0 ) ⊆ ∆(A). Indeed, if k 00 = r(P ) then k 0 = r(P ) as well. Since |Pr(P ) | = 1 by the assumption P ∈ M, then the nth entries of y and x agree. This implies y ∈ A, which contradicts the definition of y. Therefore k 00 < r(P ). Now any element z ∈ ∆(B \ A) \ P n (y0 ) is of the form z = (z1 , . . . , zn−1 , yn ) and the vectors (z1 , . . . , zn−1 ) and y0 differ just in one entry. Since z ∈ ∆(u) for u = (z1 , . . . , zn−1 , zn ) with yn ≤ zn , and since u ≺ y, then z ∈ ∆(A). Therefore, to prove (13) it is sufficient to show that |∆(A \ B) ∩ P n (x0 )| ≥ |∆(B \ A) ∩ P n (y0 )|. Equivalently, by the natural identification of P n (x0 ) and P n (y0 ) with P , it suffices to show that, in (P, ≤), |∆new (Fx )| ≥ |∆new (Fy )|. (14) Assume |Pkn00 (y0 ) \ A(y0 )| ≥ |A(x0 )|. Let F with |F | = δ be the initial segment of Pkn00 (y0 ). Since P is additive and shadow increasing, then (1) and (4) imply |∆new (Fx )| = |∆(Fx )| ≥ |∆(F )| = |∆new (F )| ≥ |∆new (Fy )|. Now assume |Pkn00 (y0 ) \ A(y0 )| < |A(x0 )|. Then Fy is a final segment in Pkn00 (y0 ). Denote by F , |F | = δ, the final segment of Pkn0 (x0 ). Since P is additive and final shadow increasing, then (1) and (5) imply |∆new (Fx )| ≥ |∆new (F )| ≥ |∆new (Fy )|. Therefore, (14) and, thus, (13) are established. Obviously, N (B) < N (A). Hence, by successive applications of the above arguments we can transform the set A into an initial segment without increasing the shadow. 2
10
1 s
s3
@ @
2@s
p+1 sa aa QQ Qaaa s Qs p a s s a · · ·Q 1 HH3 4 !s p + 2 !! @ ! HH@ ! H !! H @ s ! 2
a.
b.
Figure 3: Counterexamples to Theorem 2 for n = 2
5
Proof of Theorem 2
By Proposition 1 the inverse of n is a Macaulay order for the dual of P n , n = 1, 2, 3. It is easy to show that the inverse of n is the lexicographic order too. Thus, the assumptions of Theorem 2 are valid for (P ∗ )n for n = 2, 3. In the proof we often refer to this assertion. We use some ideas of [9] in the proof of propositions c) – e) of the theorem. a. It suffices to prove the assertion for n = 1 only. For this denote k = r(P ) and assume |Pk | ≥ 2. Let b be the first element of Pk . Then (b, b) is the first element of 2 P2k . Let c ∈ Pk be the successor of b, and let a ∈ ∆(b). If a ≺ b then (a, c) 6∈ ∆(b, b),
(a, c) ≺ (b, a) ∈ ∆(b, b).
This implies ∆(b, b) is not an initial segment, which contradicts N2 . 3 Assume a b. Similarly to above, (b, b, b) is the smallest element of P3k . One has
(b, c, a) 6∈ ∆(b, b, b),
(b, c, a) ≺ (a, b, b) ∈ ∆(b, b, b).
This implies ∆(b, b, b) is not an initial segment, which again contradicts N2 . Applying the above arguments to the dual P ∗ , one gets that it has an only maximum element. This implies |P0 | = 1. Note that |P0 | = |Pr(P ) | = 1 implies, in particular, that the Hasse diagram of P is connected. For n = 2 Theorem 2a is not true for poset shown in Fig. 3a together with a Macaulay order. b. It also suffices to consider the case n = 1 only. First assume the order is not a linear extension of ≤, i.e. there exist a, b ∈ P such that a ∈ ∆(b) and a b. We call such a pair {a, b} inverted pair and show that the existence of an inverted pair implies P is a chain. Denote by fi (resp. li ) the first (resp. last) element of Pi , i = 1, . . . , r(P ). Fact 1 ∆(z) = Pi−1 for any z ∈ Pi , i = 1, . . . , r(P ). Proof. Let us choose an inverted pair {a, b} satisfying if x ∈ ∆(y) and rP (y) > rP (b) then x ≺ y b is the smallest element of ∇(a). 11
(15) (16)
Assume there exists an element u ∈ Pi−1 \ ∆(fi ) for some i ≥ 1. Let t = rP (b) 2 and denote B = Ft+i (b, fi ) ⊆ Pt+i . Then (b, u) 6∈ ∆(B). Indeed, if (b, u) ∈ ∆(x, y) for some (x, y) ∈ B then either b ∈ ∆(x) or u ∈ ∆(y). If b ∈ ∆(x) then b ≺ x by (15) and, thus, (b, fi ) ≺ (x, y). If u ∈ ∆(y) then fi ≺ y and x = b. This implies (b, fi ) ≺ (x, y) again. Therefore, one has (b, u) 6∈ ∆(B) and (b, u) ≺ (a, fi ) ∈ ∆(B). Since B is an initial segment, then we have a contradiction with N2 . Therefore, the element u mentioned above does not exist, i.e. ∆(fi ) = Pi−1 for i = 1, . . . , r(P ). This and the Macaulayness of P imply the assertion for any z ∈ Pi . 2 Fact 2 Let {a, b} be an inverted pair, and let t = rP (b). Then a = lt−1 and b = ft . Proof. First we show that a = lt−1 . Since this is true if |Pt−1 | = 1, let us assume |Pt−1 | > 1. Let b be the smallest element in ∇(a) such that b ≺ a. Suppose a ≺ lt−1 . Then, by the choice of b, (a, lt−1 ) 6∈ ∆(F2t−1 (b, a)) and (a, lt−1 ) ≺ (lt−1 , a) ∈ ∆(b, a), thus contradicting N2 . Therefore, a = lt−1 . The second part of the statement follows by applying similar arguments to the dual poset (P ∗ )2 . 2 Fact 3 Let {a, b} be an inverted pair, and let t = rP (b). Then |Pt | = |Pt−1 | = 1. Proof. Assume the contrary, i.e. |Pt | > 1. By the part a) of the Theorem, t < r(P ). Fact 1 implies ∇(ft ) = ∇(lt ) = Pt+1 , and it follows from Fact 2 that b = ft ≺ Pt+1 . Therefore, (b, lt , a) 6∈ ∆(F3t (b, b, b)) and (b, lt , a) ≺ (a, b, b) ∈ ∆(b, b, b). This contradicts N2 . Similar arguments in the dual poset (P ∗ )3 imply |Pt−1 | = 1.
2
We are now ready to conclude the proof of the part b). Let {a, b} be an inverted pair 2 and let t = rP (b). Assume t < r(P ) and ft+1 b. Consider B = Ft+1 (b, f1 ) ⊆ Pt+1 . Now ∆new (b, f1 ) = {(b, f0 ), (a, f1 )}. Furthermore, (ft+1 , f0 ) 6∈ B and ∆(ft+1 , f0 ) = {(b, f0 )}. Therefore, |∆((B \ (b, f1 )) ∪ (ft+1 , f0 ))| = |∆(B)| − |∆new (b, f1 )| + 1 < |∆(B)|, which contradicts the optimality of B. Hence, ft+1 ≺ b and, thus, {b, ft+1 } is an inverted pair. Fact 3 implies |Pt+1 | = 1. By iterating the argument with respect to the inverted pair {ft+i , ft+i+1 } for 1 ≤ i ≤ r(P ) − t we get b = ft ft+1 · · · fr(P )
and |Pt | = |Pt+1 | = · · · |Pr(P ) | = 1. 12
Similar arguments in the dual (P ∗ )2 provide f0 f1 · · · ft−1 = a and |P0 | = |P1 | = · · · |Pt−1 | = 1. Therefore, P is a chain and f0 · · · fr(P ) . Denote k = r(P ) and consider 3 B = F2k−1 (fk−1 , fk , f0 ) ⊆ P2k−1 . Then |∆(B)| = 3k − 3, while the shadow of the 3 final segment in P2k−1 of the same size consists of 2k − 1 elements. This contradicts to the optimality of B if k ≥ 3. This proves that, if r(P ) ≥ 3 then the Macaulay order is a linear extension of the partial order ≤. Finally, assume (6) is not fulfilled for P . Thus, there exist a, b ∈ P such that ∆(b) ≺ a, rP (b) > rP (a) and b a. Denote q = rP (b) − rP (a) and consider the set B = FrP (a)+q (a, fq ) ⊆ Pr2P (a)+q . Since (b, f0 ) (a, fq ), then (b, f0 ) 6∈ B. Furthermore, ∆(b) ≺ a implies ∆(b, f0 ) ⊆ B. Since a 6= b and q ≥ 1, then ∆(b, f0 ) ∩ ∆(a, fq ) = ∅. Since the order is a linear extension of ≤, and since q ≥ 1, then (a, fq−1 ) ∈ ∆new (a, fq ). Thus, |∆new (a, fq )| ≥ 1. One has |∆((B \ (a, fq )) ∪ (b, f0 ))| = |∆(B)| − |∆new (a, fq )| < |∆(B)|, which contradicts the optimality of B. This concludes the proof. It is worth noting that Theorem 2b is not necessarily true for n = 2. The poset P shown in Fig. 3b has the Macaulay order presented in the figure which is not rank greedy, and P 2 admits the lexicographic order as Macaulay order for p = 1, 2. However, the lexicographic order is not Macaulay for P n for any n ≥ 3 and p ≥ 1. Moreover, the theorem is not true for r(P ) = 2, i.e. if P is a chain with 2 elements a and b with a ∈ ∆(b). In the case b ≺ a the order n is Macaulay for n = 2, 3, however, not for n ≥ 4. c. Taking into account that the assertion is true for the lattice of multisets [6, 9], let j be minimal index such that |Pj | ≥ 2. Furthermore, let y be the smallest element of Pj and let x ∈ Pj be the successor of y in Pj . Let n ≥ 1 and let F1 , F2 ⊆ Pin be segments with F1 being initial and |F2 | = |F1 |. Denote F10 = {(x, z) | z ∈ F1 } and F20 = {(y, z) | z ∈ F2 }.
(17)
n+1 Then F10 , F20 are segments in Pi+j . One has
∆new (F10 ) = {(x, w) | w ∈ ∆new (F1 )}, ∆new (F20 ) = {(y, w) | w ∈ ∆new (F2 )}. Hence, |∆new (F10 )| = |∆new (F1 )| and |∆new (F20 )| = |∆new (F2 )|.
(18)
n+1 Denote by G2 the set of elements in Pi+j that precede each element of F20 and let n+1 A = G2 ∪ F10 and B = G2 ∪ F20 . Note that B is an initial segment in Pi+j . Since n+1 P is Macaulay (Theorem 1), then
|∆(A)| = |∆(G2 )| + |∆new (F10 )| ≥ |∆(B)| = |∆(G2 )| + |∆new (F20 )|. 13
Thus, |∆new (F10 )| ≥ |∆new (F20 )| and the first inequality in (1) follows by taking (18) into account. To show the second inequality in (1), take segments F1 , F2 ∈ Pin with F2 being final n+1 and |F1 | = |F2 | as above. Construct the segments F10 , F20 ⊆ Pi+j according to (17). For these segments (18) is valid. Denote by G1 the set of elements of {(x, z) | z ∈ P n } that precede each element of F10 and consider the set G2 as above. Furthermore, denote A = G1 ∪ G2 ∪ F10 and n+1 B = G1 ∪ G2 ∪ F20 . Since B is an initial segment in Pi+j then |∆(A)| ≥ |∆(B)|, which is equivalent to |∆(G2 )| + |∆new (G1 )| + |∆new (F10 )| ≥ |∆(G2 )| + |∆new (G1 )| + |∆new (F20 )|, and the second inequality in (1) follows from (18). d. Let y be the only element of P0 and let x be the smallest element of P1 . Since y ≤ x n and P is rank-greedy by b) then y ≺ x. Let n ≥ 1 and let F1 ⊆ Pin and F2 ⊆ Pi−1 be initial segments such that |F1 | = |F2 |. Denote F10 = {(y, z) | z ∈ F1 } and F20 = {(x, z) | z ∈ F2 }.
(19)
Then F10 , F20 are segments in Pin+1 with F10 being initial. One has ∆new (F10 ) = {(y, w) | w ∈ ∆(F1 )}, ∆new (F20 ) = {(x, w) | w ∈ ∆(F2 )}. Hence, |∆new (F10 )| = |∆(F1 )| and |∆new (F20 )| = |∆(F2 )|.
(20)
Now since P n+1 is Macaulay by Theorem 1, then P n+1 is additive by c). Thus, the first inequality in (1) applied to P n+1 implies |∆new (F10 )| ≥ |∆new (F20 )|,
(21)
which, by taking (20) into account, implies (4). e. The proof of this proposition is similar to the last one. Let x be the only largest element of Pr(P ) and let y be the largest element of Pr(P )−1 . Then (7) implies y ≤ x. Moreover, y ≺ x, since P is rank-greedy by b). n Let n ≥ 1 and let F1 ⊆ Pin and F2 ⊆ Pi−1 be final segments such that |F1 | = |F2 |. n+1 0 0 We introduce F1 and F2 according to (19). Then F10 , F20 are segments in Pr(P )−1+i 0 0 0 with F2 being final. For the segments F1 and F2 equality (18) is valid.
Now since P n+1 is Macaulay, then P n+1 is additive by c). Thus, the second inequality in (1) applied to P n+1 implies (21), from where (5) follows. 2
14
a3 s
−T3
A A Asb2 a2 s @ @ a1 s @sb1 A A b0As
a3 s
6 s
−T2
A A Asb2 a2 s
A A As 5 3 s
−T1
a1 s
2 s
−T0
sb1
A A b0As
a.
b.
s4 A A 1 As
c.
Figure 4: Posets (T (3), ≤) (a.), 3 × 2 grid (b.), and torus C(3) (c.)
6
Applications
Firstly, let us mention that Theorem 1 being applied to chains implies the Kruskal-Katona theorem [12, 13] and a particular case of the Clements-Lindstr¨om theorem [8] if all the chains in the product are of the same length. Now consider the following poset (T (k), ≤) ∈ M of rank k. For 1 ≤ i ≤ k − 1 the ith level of T (k) consists of two elements ai and bi . Denote by b0 and ak the elements of T0 and Tk respectively. The partial order is defined as follows: x < y iff r(x) < r(y). The Hasse diagram of (T (3), ≤) is shown in Fig. 4a. We define the total order on T (k) by setting bi−1 ≺ ai for i = 1, . . . , k and ai ≺ bi for i = 1, . . . , k − 1. Obviously, the order is Macaulay on (T (k), ≤). Theorem 3 For any k ≥ 1 and any n ≥ 1 the poset (T n (k), ≤× , n ) is Macaulay. Proof. Denote T = T (k) for brevity. Since for k ≤ 2 the theorem is true, we proceed for k ≥ 3. Obviously, (T (k), ≤) is strongly Macaulay. Therefore, by Theorem 1 it suffices to prove Theorem 3 for n = 2 only. Note that if the properties N1 and N2 are satisfied with respect to a total order on a ∗ ranked poset P for some t > 0, then these properties are also valid for Pr(P )−t+1 and the inverse of (cf. the proof of Proposition 8.1.2. of [9]). Therefore, if P is isomorphic to P ∗ then P is Macaulay iff Pt satisfies the properties N1 and N2 for t ≤ dr(P )/2e. Applying this assertion to T 2 with r(T 2 ) = 2k, it suffices to consider the case t ≤ k only. 2 To avoid trivial cases, we assume that ∆(A) 6= Tt−1 . In particular, t > 1. By Lemma 2, 2 T satisfies the property N2 . Thus, we only have to check the property N1 . Let A ⊆ Tt2 be an optimal set and assume that A is not an initial segment. By Lemma 3 we may assume that A is compressed. In the sequel we construct a set B such that |B| = |A|, For S ⊆ Tt2 denote
|∆(B)| ≤ |∆(A)| and N (B) < N (A). ˜ ∆(S) = ∆(S) \ ∆(A \ S). 15
(22)
Note that if B = (A \ S) ∪ R for S ⊆ A and R ⊆ Tt2 \ A, then ˜ ˜ |∆(B)| = |∆(A)| + |∆(R)| − |∆(S) \ ∆(R)|.
(23)
Let y = (y1 , y2 ) be the smallest element in Tt2 \ A and let x = (x1 , x2 ) ∈ A with y ≺ x. ˜ Such an element exists because A is not an initial segment. We can assume |∆(y)| > 0, since otherwise B = (A \ {x}) ∪ {y} satisfies (22). Since A is compressed, then x1 > y 1 ,
x2 < y 2 .
(24)
˜ Note that if y2 = bj for some j ≥ 0, then |∆(y)| = 0, which contradicts our assumptions. Therefore, y2 = aj for some j ≥ 1. ˜ Note that, since t ≤ k, then |∆(y)| ≤ |∆new (y2 )| ≤ 2. We consider two cases. Case 1. Assume j < k. Since y 6∈ A and A is compressed, then (y1 , bj ) ∈ Tt2 \ A. Moreover, ˜ ˜ ∆({y, (y1 , bj )}) = ∆(y). (25) ˜ Case 1a. Assume |∆(y)| = 2. Then j ≥ 2. Let z ∈ A be the smallest element such that y ≺ z. Then z = (ap , aq ) for some q < j (cf. (24)) and p = t − q. We show j − q ≥ 2. Indeed, if j − q = 1 then p − rT (y1 ) = 1. Thus, (y1 , aq ) ∈ ∆(z) ∩ ∆new (y). This contradicts ˜ |∆(y)| = 2. ˜ Therefore, Z = {(ξ, aq ) | ξ ∈ ∆(ap )} ⊆ ∆(z). Since p ≥ rT (y1 ) + 2 ≥ 2, then, according to (24), |Z| = |∆(ap )| = 2. Consider (
B=
(A \ {z}) ∪ {y}, if (bp , aq ) 6∈ A (A \ {z, (bp , aq )}) ∪ {y, (y1 , bj )}, otherwise.
˜ ˜ In the first case |∆(z)| ≥ |Z| = 2, and in the second one |∆({z, (bp , aq )})| ≥ |Z| = 2. This, (23) and (25) imply the set B satisfies (22). ˜ Case 1b. Assume |∆(y)| = 1. Note that j ≥ 2. Indeed, if j = 1, then rT (y1 ) = t − 1 and, ˜ thus, rT (x1 ) = t according to (24). This contradicts |∆(y)| = 1. Denote i = rT (y1 ). Note that (ai+1 , aj−1 ) ∈ A and A ∩ {(ai+1 , bj−1 ), (bi+1 , bj−1 )} = ∅. Assume y = (ai , aj ). Then let (
B=
(A \ {(ai+1 , aj−1 )}) ∪ {y}, if (bi+1 , aj−1 ) 6∈ A (A \ {(ai+1 , aj−1 ), (bi+1 , aj−1 )}) ∪ {y, (y1 , bj )}, otherwise.
Assume y = (bi , aj ). Now if A ∩ {(ai+2 , bj−2 ), (bi+2 , bj−2 )} = ∅ then we construct B as above. Otherwise, if (ai+2 , bj−2 ) ∈ A and (bi+2 , bj−2 ) 6∈ A then let (
B=
(A \ {(ai+2 , bj−2 )}) ∪ {y}, if (bi+1 , aj−1 ) 6∈ A (A \ {(ai+2 , bj−2 ), (bi+1 , aj−1 )}) ∪ {y, (y1 , bj )}, otherwise. 16
Finally, if S = {(ai+2 , bj−2 ), (bi+2 , bj−2 )} ⊆ A then let (
B=
(A \ S) ∪ {y, (y1 , bj )}, if (bi+1 , aj−1 ) 6∈ A (A \ S \ {(bi+1 , aj−1 )}) ∪ {y, (y1 , bj ), (ai+1 , bj−1 )}, otherwise.
Case 2. Assume y2 = ak . Thus, t = k and y = (b0 , ak ). Let φ be the poset automorphism of T such that φ(ai ) = bi , 0 < i < k and φ(b0 ) = b0 , φ(ak ) = ak . It induces an automorphism ψ in T 2 defined as ψ(α, β) = (φ(β), φ(α)). Let I be the largest initial segment of Tt2 \ A and let F be the largest final segment of Tt2 \ A. Since |∆(A)| = |∆(ψ(A))|, we may assume that |F | ≥ |I| > 0. Denote Zi = {(α, β) ∈ Tk2 | rT (α) = i}, i = 0, . . . , k. Now |F | ≥ |I| > 0 implies (Z0 ∪ Zk ) ∩ A = ∅. Furthermore, if Z1 ∩ A = ∅ then Zk−1 ∩ A = ∅ too. In this case construct the set B by replacing each (αi , βj ) ∈ A with (αi+1 , βj−1 ) for α, β ∈ {a, b}. ˜ ˜ If A∩Z1 6= ∅, then |∆(y)| < 2. Thus, without loss of generality we can assume |∆(y)| = 1. This implies A ∩ {(a1 , bk−1 ), (b1 , bk−1 )} = ∅. Let x be the largest element in A. Now if ˜ x2 = bq then 1 ≤ q < k − 1. Thus, for S = {x, (x1 , aq )} one has: |∆(S)| ≥ 1 and ∆(S) ∩ ∆(y) = ∅. In this case set B = (A \ S) ∪ {y, (a1 , bk−1 )}. ˜ Finally, assume x2 = aj . If q = 1 then |∆(x)| = 1 and ∆(x) ∩ ∆(y) = ∅, because k ≥ 3. ˜ If q > 1 then |∆(x)| = 2 and |∆(x) ∩ ∆(y)| ≤ 1. We set B = (A \ {x}) ∪ {y}. It is easy to check that in all cases the set B satisfies (22). Therefore, by applying similar arguments (and the compression, if necessary) to the set B we transform it into an initial segment without increasing the shadow. 2 It is shown in [5] how to deduce a solution to an edge-isoperimetric problem on a graph G if a certain poset P is Macaulay. In fact, for any graph G one can construct the associated poset P . Without going into details (for which we refer readers to [3, 5]), let us mention that the poset (T (k), ≤) corresponds to the complete bipartite graph Kk,k . Moreover, it is shown in [3] that any cartesian power of the graph G corresponds in the sense above to the same cartesian power of P . Therefore, Theorem 3 implies a solution for an edgeisoperimetric problem on the graph (Kk,k )n . This extends a corresponding result in [2], where these graphs were studied with respect to the edge-isoperimetric problem.
7
Concluding remarks
Our first remarks concern the lexicographic order. Let us modify the partial order in T (k) a bit by making the elements ai and bi+1 incomparable for i = 1, . . . , k − 1. Then the resulting poset Q is isomorphic to the k × 2 grid (cf. Fig. 4b). Due to the ClementsLindstr¨om theorem [8] the cartesian product of any chains Z1 , . . . , Zn of lengths k1 , . . . , kn is Macaulay. Moreover, if k1 ≥ · · · ≥ kn (26)
17
then the lexicographic order over the set of n-dimensional vectors is a Macaulay order. However, there is no Macaulay order on Q such that the lexicographic order n is Macaulay for n ≥ 2. The reason for this is that regardless of the total order , the condition (26) is not satisfied. For n ≥ 3 this condition is essential for the lexicographic order to work for the lattice of multisets and, thus, for the poset Qn too. If we further modify the poset T (k) by making the elements bi and ai+1 incomparable for i = 1, . . . , k − 1, then this results in a cycle C(k) (cf. Fig. 4c). This poset is rank-greedy and the corresponding Macaulay order is essentially unique (it is shown in the figure). However, the lexicographic order does not work for C(k) × C(k) (the element (2, 6) is the smallest one in the fourth level, but its shadow is greater than one of (3, 3)). Nevertheless, any cartesian power of C(k) (which is isomorphic to a torus) is Macaulay. The Macaulay order can be derived from [11, 15], where a vertex-isoperimetric problem for tori was studied, and is not lexicographic. Although Theorem 2 provides some necessary conditions for posets, all whose cartesian powers are Macaulay, there are many infinite series of such posets. We illustrate this by providing a tool for constructing such series. Given a Macaulay poset (P, ≤, ), construct another poset Q = (P, v) as follows. Take an element a ∈ Pi for some i > 1 and consider Fi (a). Then ∆(Fi (a)) = Fi−1 (b) for some b ∈ Pi−1 . Let c ∈ Fi−1 (b) and assume c 6≤ a. Now we extend the partial order ≤ by setting c ≤ a (cf. Fig. 5). 9 s
9 s @ @
@ @
@sa 8@ @ @ @ @s 3 sc 5@ 7@s b @ @ @ @s 2@ 4@s @ 1@s
@ 8s a @ @ @ @ @s s s b 3 c 5@ 7@ @ @ @ @ s 2@ 4@s @ 1@s
a.
b.
6 s
6 s
Figure 5: Posets P (a) and Q (b) Denote ∆P (m, i) = min |∆(A)|, where the minimum runs over all A ⊆ Pi with |A| = m. Since the partial order ≤ is a suborder of v, then ∆P (A) ⊆ ∆Q (A) for any A ⊆ Pi , Thus, ∆P (m, i) ≤ ∆Q (m, i) for all m = 1, . . . , |Pi | and i = 1, . . . , r(P ).
(27)
However, ∆P (Fi (a)) = ∆Q (Fi (a)) for any a ∈ Pi . Therefore, since P is Macaulay, then the lower bound (27) for Q is tight. This implies (P, v, ) is Macaulay too. Moreover, if P is strongly Macaulay and rank-greedy then so is Q. Obviously, applying any number of the above extensions leads to a Macaulay poset.
18
Now consider P 2 . Since ∆P 2 (Fi ((x, y))) = {(x, ξ) | ξ ∈ ∆P (Fi−rP (x) (y))} ∪ {(ξ, y) | ξ ∈ ∆P (Fi−rP (y) (x))}, then ∆P 2 (Fi ((x, y)) = ∆Q2 (Fi ((x, y))). Therefore, if P satisfies the assumptions of Theorem 1, then so is for Q. On the other hand, since the lexicographic order is Macaulay for P 2 , then so is for P 4 , for example. Extending P 2 as shown above results in a new poset, for which Theorem 1 is applicable. In particular, the Macaulayness of the torus poset C n (k) implies a similar result for the powers of k × 2 grids (cf. Fig. 4b,c). It would be interesting to establish local-global principles with respect to some other extremal problems. In our forthcoming paper [7] we present a local-global principle for vertex-isoperimetric problems. Sperner theory [9] and the Erd¨os-Ko-Rado type results are closely related with Macaulay theory, and are further examples were local-global principles may exist.
Acknowledgements The authors would like to thank Konrad Engel (University of Rostock, Germany) for his helpful comments concerning the draft version of the manuscript.
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[9] K. Engel, Sperner theory, Cambridge University Press, 1997. [10] K. Engel, U. Leck, Optimal antichains and ideals in Macaulay posets, Preprint 96/21, University of Rostock, to appear in Graph Theory and Combinatorial Biology, Bolyai Soc. Math. Stud. 7, L. Lov´asz, A. Gyarfas, G.O.H. Katona, A. Recski, L. Sz´ekely eds., Budapest. [11] V.M. Karachanjan, A discrete isoperimetric problem on multidimensional torus, (in Russian), Doklady AN Arm. SSR, vol. LXXIV (1982), No. 2, 61–65. [12] G.O.H. Katona, A theorem of finite sets, in: Theory of graphs, Academia Kiado, Budapest, 1968, 187–207. [13] J.B. Kruskal, The optimal number of simplices in a complex, in: Math. Optimization Tech., University of California Press, Berkeley, California, 1963, 251–268. [14] U. Leck, Extremalprobleme f¨ ur den Schatten in Posets, Ph. D. Thesis, FU Berlin, 1995; Shaker-Verlag Aachen, 1995. [15] O. Riordan, An ordering on the discrete even torus, SIAM J. Discr. Math., 11 (1998), No. 1, 110–127.
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