A Local-Global Principle for Macaulay PosetsSergei L. Bezrukov�Department of Mathematics and Computer ScienceUniversity of PaderbornF�urstenallee 11, 33102 Paderborn, GermanyXavier Portasy, Oriol SerraDepartament de Matem�atica Aplicada i Telem�aticaModul C3, Campus Nord, Universitat Polit�ecnica de CatalunyaJordi Girona 1, 08034 Barcelona, SpainAbstractWe consider the shadow minimization problem (SMP) for cartesian powers Pnof a Macaulay poset P . Our main result is a local-global principle with respectto the lexicographic order Ln. Namely, we show that under certain conditions theshadow of any initial segment of the order Ln for n � 3 is minimal i� it is sofor n = 2. These conditions include such poset properties as additivity , shadowincreasing , �nal shadow increasing and being rank-greedy . We also show that theseconditions are essentially necessary for the lexicographic order to provide nestednessin the SMP.1 IntroductionLet (P;�) be a �nite poset with a partial order �. The poset (P;�) is called ranked ifthere exists a function rP : P 7! IN such that minx2P rP (x) = 0 and rP (x) + 1 = rP (y)whenever x < y and there is no z 2 P yielding x < z < y. We call the numbers rP (x)and r(P ) = maxy2P r(y) the rank of x 2 P and of P , respectively. Usually we denote aposet and its element set by the same letter.The set Pt = fx 2 P j rP (x) = tg is called the tth level of P . For A � Pt and t � 0 de�nethe shadow of A as �(A) = fx 2 Pt�1 j x � y for some y 2 Ag:The shadow minimization problem (SMP for brevity) is one of the most important prob-lems in combinatorics and has a lot of applications: for a ranked poset P and givennatural numbers t > 0 and m, 1 � m � jPtj, �nd a set A � Pt such that jAj = m and�Partially supported by the Spanish Research Council under project TIC97-0963.yGranted by the Spanish Research Council, PN94 40433776.1

j�(A)j � j�(B)j for any B � Pt with jBj = m. We call such a set optimal . See [9] for anintroduction to the subject.Let us concentrate on posets decomposable under the cartesian product operation. Forposets (P 0;�P 0) and (P 00;�P 00) we de�ne their cartesian product as a poset with theelement set P 0�P 00 and with the partial order �� de�ned as follows: (x1; y1) �� (x2; y2)i� 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 de�ned. It can be easily shown that thecartesian product of ranked posets is a ranked poset too. We denote by P n the nthcartesian power of a poset P .ssa. ssss...b. s sc. ss��� QQQQQ � � � s s ss s ss s ss��� QQQQQ... ... ...� � �d.Figure 1: The basic posetsThe SMP for cartesian products of various posets was considered in the literature. Ex-amples 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], andthe 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 includesall the above mentioned posets as special cases. For all these posets, the solution to theSMP problem provides nested families of optimal subsets. An exact formulation of thisphenomenon leads to the notion of a Macaulay poset .Let � be a total order on P . For z 2 Pt denoteFt(z) = fx 2 Pt j x � zg:We call a subset A � Pi initial segment if A = Ft(z) for some z 2 Pt. A poset P is calledMacaulay, if there exists a total order � (called Macaulay order), such that for any t > 0the following properties hold:N1 (nestedness) : For any z 2 Pt the initial segment Ft(z) has minimal shadow amongall subsets of Pt of the same cardinality;N2 (continuity) : The shadow of an initial segment is an initial segment itself, i.e. forany z 2 Pt there exists z0 2 Pt�1 such that �(Ft(z)) = Ft�1(z0).2

Note that a Macaulay order � is not claimed to be a linear extension of �. We refer toa 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 withinverse partial order.Proposition 1 (cf. [9]) A poset P is Macaulay i� so is its dual P �. Moreover, if � isthe 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 resultsimply that the same holds for cartesian powers of these posets.There exists a powerful technique for establishing the Macaulayness of cartesian powers ofposets. This technique is based on compression (see below) and usually involves inductionon the number n of posets in the product. However, for this technique it is principal thatn � 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 hasbeen studied. For a �nite set S a function f : 2S 7! IR is called submodular if for anyA;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 separatinga set A � VG from VGnA is an example of a submodular function. Based on a submodularfunction f de�ned on 2VG, some special functions f (n) on the nth cartesian power of Gare considered in [1]. These functions f (n) are, in a sense, decomposable, i.e. they can berepresented as certain sums of functions f (n�1). Without going into details, consider, forexample, the function f (n) de�ned as the number of cut edges which separate a subsetof vertices of the n-cube from its complement. Then f (n) can be represented as the sumover 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 submodularityof the function f de�ned on 2VG provides a su�cient condition for applying the compres-sion for minimization of f (n) de�ned on the nth cartesian power of the graph G. Secondly,if the lexicographic order (see Section 2) provides nestedness (cf. N1) in minimization off (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 validalso with respect to some other total orders [3].In what concerns the SMP, the above approach can not be directly applied because of thenecessity to maintain the level structure of a poset. Another di�culty in applying theresults of [1] is that the function j�(�)j is not decomposable in the sense above.However, the applicability of compression and some general principles in the proof tech-niques for establishing theMacaulayness of posets and solving edge-isoperimetric problems(cf. [3]) gave us a starting point to look for a local-global principle with respect to theSMP. 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 ofthis result it is important that the poset satis�es some additional conditions, which haveno analogies for graphs yet (cf. Section 2 for precise de�nitions). In Theorem 2 we show3

that these conditions are essential for the lexicographic order to provide the Macaulaynessof any cartesian power of a poset.The next section of our paper is devoted to the formulation of some poset propertiesand statements of our main results: Theorem 1 and Theorem 2. In Section 3 we presentsome auxiliary propositions which will be used throughout the text. Sections 4 and 5 aredevoted to the proof of Theorems 1 and 2 respectively. An application of our techniqueto 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 posetsLet (P;�;�) be Macaulay. Consider the SMP for the cartesian powers P n. We representthe 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 de�ned as follows: (x1; : : : ; xn) �n (y1; : : : ; yn)i� there exists an i such that xi � yi and xj = yj for all j < i. We often omit thesuperscript n in �n if n is uniquely de�ned by the context. For a subset A � P n we referto 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 theorder �. We introduce initial and �nal segments of Pi de�ned naturally.For a segment F � Pi let G be the set of elements in Pi that precede each element of Fin order �. The new shadow of F is de�ned as�new(F ) = �(F ) n�(G):A Macaulay poset (P;�;�) is called additive if for any i � 1j�new(F1)j � j�new(F2)j � j�new(F3)j (1)for all segments F1; F2; F3 � Pi with F1 being initial, F3 being �nal and jF1j = jF2j = jF3j.Proposition 2 (cf. [9]) A Macaulay poset is additive i� so is its dual.It is worth to mention another (equivalent) de�nition of additivity, which relates to sub-modularity. For a Macaulay poset (P;�;�) and i = 0; : : : ; r(P ) de�ne the shadow func-tion sfi : 2Pi 7! IN by setting, for F � Pi,sfi(F ) = j�(F 0)j;with F 0 � Pi being an initial segment with jF 0j = jF j. The shadow function is called littlesubmodular ifsfi(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 sub-modular for any i = 0; : : : ; r(P ). It is known that a poset is little submodular i� it is4

additive (cf. [9]). If we do not claim the condition (3) then the shadow function will besubmodular (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 j�(�)jis submodular.Furthermore, a Macaulay poset (P;�;�) is called shadow increasing if for any i � 1j�(F1)j � j�(F2)j (4)for any initial segments F1 � Pi and F2 � Pi�1 with jF1j = jF2j. Cartesian powers of theposets shown in Fig. 1 are additive and shadow increasing [9]. Similarly, the poset P iscalled �nal shadow increasing, if for any i � 1j�new(F1)j � j�new(F2)j (5)for any �nal segments F1 � Pi and F2 � Pi�1 with jF1j = jF2j. This notion has beenalready appeared in [10], where extremal ideals in posets were studied.Obviously, if (P;�;�) is shadow increasing then (4) is valid for any initial segmentsF1 � Pi and F2 � Pj with jF1j = jF2j and i � j. A similar remark concerns a �nalshadow increasing poset and (5).Denote byM the class of ranked posets having only one maximumand only one minimumelement. The next proposition is a bit weaker form of Propositions 7 and 8 of [10].Proposition 3 (cf. [10]). Let (P;�;�) 2 M be Macaulay.a. If P is �nal 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 �nal shadow increasing.We call a Macaulay poset P strongly Macaulay if it is additive, shadow increasing and�nal shadow increasing.Proposition 4 A poset P 2 M is strongly Macaulay i� so is its dual P �.Proof. Obviously, P � 2 M. Then Propositions 1, 2 and 3a respectively imply that P � isMacaulay, additive and shadow increasing. Since (P �)� = P , then Proposition 3b appliedto P � implies P � is �nal shadow increasing. Thus, P � is strongly Macaulay. The secondpart of the assertion follows from (P �)� = P . 2For F � P and x 2 P we write F � x if y � x for any y 2 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. 5

Theorem 1 (the local-global principle for SMP).Let (P;�;�) 2 M be strongly Macaulay and rank-greedy. Let the order �2 be Macaulayfor 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 1are essential.Theorem 2 Let (P;�;�) be a Macaulay poset. Furthermore, let r(P ) � 3 and assumethe orders �2 and �3 are Macaulay for P 2 and P 3, respectively. Then for any n � 1a. P n 2 M;b. P n is rank-greedy;c. P n is additive;d. P n is shadow increasing;e. P n is �nal shadow increasingWe prove Theorems 1 and 2 in Sections 4 and 5, respectively, after establishing someproperties of the order �n in the next section.3 Some auxiliary resultsThe following property, which is called consistency, is straightforward.Lemma 1 Let x = (x1; : : : ; xn) 2 P n and y = (y1; : : : ; yn) 2 P n. Furthermore, letxi = yi for some i. Then x �n y i� x0 �n�1 y0 for x0 and y0 obtained from x and y,respectively, by omitting their ith entries. 2Let A � P n. For a �xed i and z 2 P denoteP n(i; z) = f(�1; : : : ; �n) 2 P n j �i = zgA(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 totalorder induced in it by �n). Obviously, this poset is isomorphic to P n�1. Moreover, bythe consistency of the order �n, the natural isomorphism between the two posets sendsinitial segments to initial segments.Proposition 6 Let P be an additive and Macaulay poset. Then�(Pi) = Pi�1 for i = 1; : : : ; r(P ): (7)6

Proof. Indeed, assume it is not so for some i � 1 and denote Z = Pi�1 n�(Pi). Considerthe dual P �. By Propositions 1 and 2, P � is Macaulay and additive. Since Z is a �nalsegment in P , it is an initial segment in P �. Then, �(Z) = ; in P � contradicts theadditivity of P �. 2Lemma 2 Let (P;�;�) be additive, Macaulay and rank-greedy. Then P n satis�es theproperty 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 proceedwith n � 2 and t � 1. Let A � P nt be an initial segment and let x = (x1; : : : ; xn) be thelargest element of A in order �n. ThenA = [z�x1 A(1; z) [A(1; x1);�(A) = [z�x1 �new(A(1; z)) [�new(A(1; x1)); (8)where all the unions are disjoint. Note that for any z � x1 one hasP nt \ P n(1; z) � A: (9)Furthermore, note that (7) implies �(P ni ) = P ni�1 for any n � 2 and any i = 1; : : : ; r(P n).This and (9) imply P nt�1 \ P n(1; z) � �(A), i.e.�new(A(1; z)) = P nt�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 subposetis isomorphic to P n�1 and since the natural isomorphism sends initial segments to initialsegments, then the induction hypothesis implies that �new(A(1; x1)) is an initial segmentin (P n(1; x1);v). This, (8) and (10) imply the lemma. 2This lemma is essential for the proof of Lemma 3.Let A � P nt and let i be �xed. We say that A is i-compressed if A(i; z) is an initialsegment (with respect to the order �n) in the subposet (P n(i; z);v) for any z 2 P . Theset A is called compressed if it is i-compressed for i = 1; : : : ; n.Lemma 3 Let (P;�;�) be additive, Macaulay and rank-greedy. Suppose for some n � 2that P n�1 is Macaulay and that the order �n�1 is a Macaulay order. Then for any A � P ntand for a �xed i, 1 � i � n, there exists an i-compressed set B � P nt such that jBj = jAjand j�(B)j � j�(A)j.Proof. For z 2 P with rP (z) < r(P ) denote r(z) = fx 2 P j z 2 �(x)g. Let A � P nt .Thenj�(A)j = j [z2P (�(A) \ P n(i; z))j � Xz2P maxx2r(z)fj�(A(i; z)) \ P n(i; z)j; jA(i; x)jg: (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) forthe set B is tight. Since (P n(i; z);v) is isomorphic to the Macaulay poset P n�1, thenj�(A(i; z))\ P n(i; z)j � j�(B(i; z))\ P n(i; z)j. This implies j�(A)j � j�(B)j. 2Applying the compression for i = 1; : : : ; n su�ciently many times one gets a compressedset C � P nt such that jCj = jAj and j�(C)j � j�(A)j.7

4 Proof of Theorem 1Obviously, the total order � provides a numbering of P n with numbers 1; : : : ; jP nj.According to this, for z 2 P nt denote by N(z) the number of z and for A � P nt letN(A) = Pz2AN(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 � P nt be an optimal set. By Lemma 3 we can assume that A is compressed. Letx = (x1; : : : ; xn) be the largest element of A, and let y = (y1; : : : ; yn) be the smallestelement of Pt n A. If A is not an initial segment then y � x. Without loss of generalitywe can assume that xi 6= yi for i = 1; : : : ; n, since otherwise y 2 A follows from Lemma 1and 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 contradic-tion. For i = 0; : : : ; r(P n) denote by fni and lni the smallest and largest elements of P nirespectively. Since P 2 M and P is rank-greedy, thenfni = (0; : : : ; 0; f1g ; 1; : : : ; 1| {z }h )lni = (1; : : : ; 1| {z }h ; l1g; 0; : : : ; 0);where 0 = f10 , 1 = f1r(P ), and g and h are de�ned by i = r(P ) � h+ g, 0 � g < r(P ).Denote t0 = t� rP (x1)� rP (xn) and t00 = t� rP (y1)� rP (yn), and letu = (y1; ln�2t00 ; yn); v = (x1; fn�2t0 ; xn):Case 1. Assume rP (x1) � rP (y1). Now rP (xn) < rP (yn) implies t0 > t00. Considerz = (y1; ln�2t000 ; xn) 2 P nt with t00 < t000 � t0. Lemma 1 and the fact that P is rank-greedyimply y � u � z � v � x: (12)Since x and v agree in the �rst entry, and since v � x, then v 2 A because A is1-compressed. By a similar argument fz;u;yg � A, and we have a contradiction.Case 2. Assume rP (y1) < rP (x1). If rP (x1) + rP (xn) > rP (y1) + rP (yn) then consider anelement z1 2 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; fn�2t0 ; yn) 2 P nt . Since n � 3 then (12) holds and we have a contradiction as above.Assume rP (x1) + rP (xn) � rP (y1) + rP (yn). Now if rP (x1) > rP (y1) + 1 then considerw = (z1; fn�2t0 ; zn) 2 P nt with some zn and some z1, such that rP (z1) = rP (y1) + 1and y1 � z1 � x1. Since P is rank-greedy, then such an element z1 does exist. IfrP (x1) = rP (y1) + 1 then set w = v. Now for q = rP (yn)� rP (zn) one has0 < q = (t� rP (y1)� t00)� rP (zn) = (t� rP (z1)� rP (zn)) + 1� t00 = t0 � t00 + 1:8

'& $%s s s'&$% '&$%s cx yP n(x0) P n(y0)P n(n; 0)0(x1; : : : ; xn�1; 0) (y1; : : : ; yn�1; 0)k0 k00XXXXXXXXXXzFigure 2: Usage of the shadow increase property in the proof of Theorem 1Thus, if t00 � 1 then q � t0. In this case for z = (z1; fn�2t0�q ; yn) one has z � w � v. If t00 = 0then take z = (y1; fn�2q ; xn). In both cases (12) holds and leads to a contradiction. 2For z = (z1; : : : ; zn�1) 2 P n�1 and A � P n denoteP n(z) = f(�1; : : : ; �n) 2 P n j �i = zi; i = 1; : : : ; n� 1gA(z) = A \ P n(z):Obviously, the subposet of (P n;��) with the element set P n(z) and the induced partialorder is isomorphic to (P;�). Furthermore, denote x0 = (x1; : : : ; xn�1), y0 = (y1; : : : ; yn�1)and let k0 = rP (xn) and k00 = rP (yn). Then k0 � k00 by Lemma 4. Moreover, the setsA(x0) and A(y0) are initial segments in P nk0(x0) and P nk00(y0) respectively (these sets areshown 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 � = minfjP nk00(y0) nA(y0)j; jA(x0)jg. Let Fx � P nk0(x0) be the segment consisting ofthe last � elements of the segment A(x0) and let Fy � P nk00(y0) be the segment consistingof � elements of P nk00(y0) nA(y0) such that y 2 Fy.Consider the set B = (A n Fx) [ Fy. Obviously, B � P nt and jBj = jAj. We showj�(A)j � j�(B)j: (13)For this note that �(B n A) n P n(y0) � �(A). Indeed, if k00 = r(P ) then k0 = r(P ) aswell. Since jPr(P )j = 1 by the assumption P 2 M, then the nth entries of y and x agree.This implies y 2 A, which contradicts the de�nition of y. Therefore k00 < r(P ). Nowany element z 2 �(B n A) n P n(y0) is of the form z = (z1; : : : ; zn�1; yn) and the vectors(z1; : : : ; zn�1) and y0 di�er just in one entry. Since z 2 �(u) for u = (z1; : : : ; zn�1; zn)with yn � zn, and since u � y, then z 2 �(A).Therefore, to prove (13) it is su�cient to show that j�(A n B) \ P n(x0)j � j�(B n A) \P n(y0)j. Equivalently, by the natural identi�cation of P n(x0) and P n(y0) with P , it su�cesto show that, in (P;�), j�new(Fx)j � j�new(Fy)j: (14)9

Assume jP nk00(y0) nA(y0)j � jA(x0)j. Let F with jF j = � be the initial segment of P nk00(y0).Since P is additive and shadow increasing, then (1) and (4) implyj�new(Fx)j = j�(Fx)j � j�(F )j = j�new(F )j � j�new(Fy)j:Now assume jP nk00(y0)nA(y0)j < jA(x0)j. Then Fy is a �nal segment in P nk00(y0). Denote byF , jF j = �, the �nal segment of P nk0(x0). Since P is additive and �nal shadow increasing,then (1) and (5) imply j�new(Fx)j � j�new(F )j � j�new(Fy)j:Therefore, (14) and, thus, (13) are established. Obviously, N(B) < N(A). Hence, bysuccessive applications of the above arguments we can transform the set A into an initialsegment without increasing the shadow. 25 Proof of Theorem 2By Proposition 1 the inverse of �n is a Macaulay order for the dual of P n, n = 1; 2; 3. It iseasy to show that the inverse of �n is the lexicographic order too. Thus, the assumptionsof 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 su�ces to prove the assertion for n = 1 only. For this denote k = r(P ) andassume jPkj � 2. Let b be the �rst element of Pk. Then (b; b) is the �rst element ofP 22k. Let c 2 Pk be the successor of b, and let a 2 �(b). If a � b then(a; c) 62 �(b; b); (a; c) � (b; a) 2 �(b; b):This implies �(b; b) is not an initial segment, which contradicts N2.Assume a � b. Similarly to above, (b; b; b) is the smallest element of P 33k. One has(b; c; a) 62 �(b; b; b); (b; c; a) � (a; b; b) 2 �(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 maximumelement. This implies jP0j = 1. Note that jP0j = jPr(P )j = 1 implies, in particular,that the Hasse diagram of P is connected. For n = 2 Theorem 2a is not true forposet shown in Fig. 3a together with a Macaulay order.b. It also su�ces to consider the case n = 1 only. First assume the order � is not alinear extension of �, i.e. there exist a; b 2 P such that a 2 �(b) and a � b. Wecall such a pair fa; bg inverted pair and show that the existence of an inverted pairimplies P is a chain.Denote by fi (resp. li) the �rst (resp. last) element of Pi, i = 1; : : : ; r(P ).Fact 1 �(z) = Pi�1 for any z 2 Pi, i = 1; : : : ; r(P ).10

s ss@@@ ���1 2 3a. s s s s sssHHHHHH @@@ �����!!!!!!!!aaaaaaaa������ ��� QQQQQ� � �1 23 4 pp+ 1 p+ 2b.Figure 3: Counterexamples to Theorem 2 for n = 2Proof. Let us choose an inverted pair fa; bg satisfyingif x 2 �(y) and rP (y) > rP (b) then x � y (15)b is the smallest element of r(a): (16)Assume there exists an element u 2 Pi�1 n �(fi) for some i � 1. Let t = rP (b)and denote B = Ft+i(b; fi) � P 2t+i. Then (b; u) 62 �(B). Indeed, if (b; u) 2 �(x; y)for some (x; y) 2 B then either b 2 �(x) or u 2 �(y). If b 2 �(x) then b � x by(15) and, thus, (b; fi) � (x; y). If u 2 �(y) then fi � y and x = b. This implies(b; fi) � (x; y) again. Therefore, one has(b; u) 62 �(B) and (b; u) � (a; fi) 2 �(B):Since B is an initial segment, then we have a contradiction with N2. Therefore, theelement 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 2 Pi. 2Fact 2 Let fa; bg 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 jPt�1j = 1, let us assumejPt�1j > 1. Let b be the smallest element in r(a) such that b � a. Suppose a � lt�1.Then, by the choice of b,(a; lt�1) 62 �(F2t�1(b; a)) and (a; lt�1) � (lt�1; a) 2 �(b; a);thus contradicting N2. Therefore, a = lt�1.The second part of the statement follows by applying similar arguments to the dualposet (P �)2. 2Fact 3 Let fa; bg be an inverted pair, and let t = rP (b). Then jPtj = jPt�1j = 1.Proof. Assume the contrary, i.e. jPtj > 1. By the part a) of the Theorem, t < r(P ).Fact 1 implies r(ft) = r(lt) = Pt+1, and it follows from Fact 2 that b = ft � Pt+1.Therefore, (b; lt; a) 62 �(F3t(b; b; b)) and (b; lt; a) � (a; b; b) 2 �(b; b; b):This contradicts N2.Similar arguments in the dual poset (P �)3 imply jPt�1j = 1. 211

We are now ready to conclude the proof of the part b). Let fa; bg be an inverted pairand let t = rP (b). Assume t < r(P ) and ft+1 � b. Consider B = Ft+1(b; f1) � P 2t+1.Now �new(b; f1) = f(b; f0); (a; f1)g. Furthermore, (ft+1; f0) 62 B and �(ft+1; f0) =f(b; f0)g. Therefore,j�((B n (b; f1)) [ (ft+1; f0))j = j�(B)j � j�new(b; f1)j+ 1 < j�(B)j;which contradicts the optimality of B.Hence, ft+1 � b and, thus, fb; ft+1g is an inverted pair. Fact 3 implies jPt+1j = 1.By iterating the argument with respect to the inverted pair fft+i; ft+i+1g for 1 �i � r(P )� t we getb = ft � ft+1 � � � � � fr(P ) and jPtj = jPt+1j = � � � jPr(P )j = 1:Similar arguments in the dual (P �)2 providef0 � f1 � � � � � ft�1 = a and jP0j = jP1j = � � � jPt�1j = 1:Therefore, P is a chain and f0 � � � � � fr(P ). Denote k = r(P ) and considerB = F2k�1(fk�1; fk; f0) � P 32k�1. Then j�(B)j = 3k � 3, while the shadow of the�nal segment in P 32k�1 of the same size consists of 2k� 1 elements. This contradictsto the optimality of B if k � 3.This proves that, if r(P ) � 3 then the Macaulay order � is a linear extension of thepartial order �.Finally, assume (6) is not ful�lled for P . Thus, there exist a; b 2 P such that�(b) � a, rP (b) > rP (a) and b � a. Denote q = rP (b) � rP (a) and considerthe set B = FrP (a)+q(a; fq) � P 2rP (a)+q. Since (b; f0) � (a; fq), then (b; f0) 62 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) 2�new(a; fq). Thus, j�new(a; fq)j � 1. One hasj�((B n (a; fq)) [ (b; f0))j = j�(B)j � j�new(a; fq)j < j�(B)j;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 posetP shown in Fig. 3b has the Macaulay order presented in the �gure which is notrank 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 elementsa and b with a 2 �(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], letj be minimal index such that jPj j � 2. Furthermore, let y be the smallest elementof Pj and let x 2 Pj be the successor of y in Pj . Let n � 1 and let F1; F2 � P ni besegments with F1 being initial and jF2j = jF1j. DenoteF 01 = f(x; z) j z 2 F1g and F 02 = f(y; z) j z 2 F2g: (17)12

Then F 01; F 02 are segments in P n+1i+j . One has�new(F 01) = f(x;w) j w 2 �new(F1)g;�new(F 02) = f(y;w) j w 2 �new(F2)g:Hence, j�new(F 01)j = j�new(F1)j and j�new(F 02)j = j�new(F2)j: (18)Denote by G2 the set of elements in P n+1i+j that precede each element of F 02 and letA = G2 [ F 01 and B = G2 [ F 02. Note that B is an initial segment in P n+1i+j . SinceP n+1 is Macaulay (Theorem 1), thenj�(A)j = j�(G2)j+ j�new(F 01)j � j�(B)j = j�(G2)j+ j�new(F 02)j:Thus, j�new(F 01)j � j�new(F 02)j and the �rst inequality in (1) follows by taking (18)into account.To show the second inequality in (1), take segments F1; F2 2 P ni with F2 being �naland jF1j = jF2j as above. Construct the segments F 01; F 02 � P n+1i+j according to (17).For these segments (18) is valid.Denote by G1 the set of elements of f(x; z) j z 2 P ng that precede each element ofF 01 and consider the set G2 as above. Furthermore, denote A = G1 [ G2 [ F 01 andB = G1 [ G2 [ F 02. Since B is an initial segment in P n+1i+j then j�(A)j � j�(B)j,which is equivalent toj�(G2)j+ j�new(G1)j+ j�new(F 01)j � j�(G2)j+ j�new(G1)j+ j�new(F 02)j;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 � xand P is rank-greedy by b) then y � x. Let n � 1 and let F1 � P ni and F2 � P ni�1be initial segments such that jF1j = jF2j. DenoteF 01 = f(y; z) j z 2 F1g and F 02 = f(x; z) j z 2 F2g: (19)Then F 01; F 02 are segments in P n+1i with F 01 being initial. One has�new(F 01) = f(y;w) j w 2 �(F1)g;�new(F 02) = f(x;w) j w 2 �(F2)g:Hence, j�new(F 01)j = j�(F1)j and j�new(F 02)j = j�(F2)j: (20)Now since P n+1 is Macaulay by Theorem 1, then P n+1 is additive by c). Thus, the�rst inequality in (1) applied to P n+1 impliesj�new(F 01)j � j�new(F 02)j; (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 largestelement 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).Let n � 1 and let F1 � P ni and F2 � P ni�1 be �nal segments such that jF1j = jF2j.We introduce F 01 and F 02 according to (19). Then F 01; F 02 are segments in P n+1r(P )�1+iwith F 02 being �nal. For the segments F 01 and F 02 equality (18) is valid.Now since P n+1 is Macaulay, then P n+1 is additive by c). Thus, the second inequalityin (1) applied to P n+1 implies (21), from where (5) follows. 213

sss sssAAA ������ AAA���@@@b0a1a2 b1b2a3 �T0�T1�T2�T3a. sss sssAAA ������ AAA���b0a1a2 b1b2a3b. sss sssAAA ������ AAA123 456 c.Figure 4: Posets (T (3);�) (a.), 3� 2 grid (b.), and torus C(3) (c.)6 ApplicationsFirstly, let us mention that Theorem 1 being applied to chains implies the Kruskal-Katonatheorem [12, 13] and a particular case of the Clements-Lindstr�om theorem [8] if all thechains in the product are of the same length.Now consider the following poset (T (k);�) 2 M of rank k. For 1 � i � k�1 the ith levelof T (k) consists of two elements ai and bi. Denote by b0 and ak the elements of T0 andTk respectively. The partial order is de�ned as follows: x < y i� r(x) < r(y). The Hassediagram of (T (3);�) is shown in Fig. 4a.We de�ne the total order � on T (k) by setting bi�1 � ai for i = 1; : : : ; k and ai � bi fori = 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 fork � 3. Obviously, (T (k);�) is strongly Macaulay. Therefore, by Theorem 1 it su�ces toprove Theorem 3 for n = 2 only.Note that if the properties N1 and N2 are satis�ed with respect to a total order � on aranked poset P for some t > 0, then these properties are also valid for P �r(P )�t+1 and theinverse of � (cf. the proof of Proposition 8.1.2. of [9]). Therefore, if P is isomorphic toP � then P is Macaulay i� Pt satis�es the properties N1 and N2 for t � dr(P )=2e.Applying this assertion to T 2 with r(T 2) = 2k, it su�ces to consider the case t � k only.To avoid trivial cases, we assume that �(A) 6= T 2t�1. In particular, t > 1. By Lemma 2,T 2 satis�es the property N2. Thus, we only have to check the property N1.Let A � T 2t be an optimal set and assume that A is not an initial segment. By Lemma 3we may assume that A is compressed. In the sequel we construct a set B such thatjBj = jAj; j�(B)j � j�(A)j and N(B) < N(A): (22)For S � T 2t denote ~�(S) = �(S) n�(A n S):Note that if B = (A n S) [R for S � A and R � T 2t nA, thenj�(B)j = j�(A)j+ j ~�(R)j � j ~�(S) n�(R)j: (23)14

Let y = (y1; y2) be the smallest element in T 2t n A and let x = (x1; x2) 2 A with y � x.Such an element exists because A is not an initial segment. We can assume j ~�(y)j > 0,since otherwise B = (A n fxg) [ fyg satis�es (22). Since A is compressed, thenx1 > y1; x2 < y2: (24)Note that if y2 = bj for some j � 0, then j ~�(y)j = 0, which contradicts our assumptions.Therefore, y2 = aj for some j � 1.Note that, since t � k, then j ~�(y)j � j�new(y2)j � 2. We consider two cases.Case 1. Assume j < k. Since y 62 A and A is compressed, then (y1; bj) 2 T 2t n A.Moreover, ~�(fy; (y1; bj)g) = ~�(y): (25)Case 1a. Assume j ~�(y)j = 2. Then j � 2. Let z 2 A be the smallest element such thaty � 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) 2 �(z)\�new(y). This contradictsj ~�(y)j = 2.Therefore, Z = f(�; aq) j � 2 �(ap)g � ~�(z). Since p � rT (y1) + 2 � 2, then, accordingto (24), jZj = j�(ap)j = 2. ConsiderB = ( (A n fzg) [ fyg; if (bp; aq) 62 A(A n fz; (bp; aq)g) [ fy; (y1; bj)g; otherwise:In the �rst case j ~�(z)j � jZj = 2, and in the second one j ~�(fz; (bp; aq)g)j � jZj = 2.This, (23) and (25) imply the set B satis�es (22).Case 1b. Assume j ~�(y)j = 1. Note that j � 2. Indeed, if j = 1, then rT (y1) = t� 1 and,thus, rT (x1) = t according to (24). This contradicts j ~�(y)j = 1.Denote i = rT (y1). Note that (ai+1; aj�1) 2 A and A \ f(ai+1; bj�1); (bi+1; bj�1)g = ;.Assume y = (ai; aj). Then letB = ( (A n f(ai+1; aj�1)g) [ fyg; if (bi+1; aj�1) 62 A(A n f(ai+1; aj�1); (bi+1; aj�1)g) [ fy; (y1; bj)g; otherwise:Assume y = (bi; aj). Now if A \ f(ai+2; bj�2); (bi+2; bj�2)g = ; then we construct B asabove. Otherwise, if (ai+2; bj�2) 2 A and (bi+2; bj�2) 62 A then letB = ( (A n f(ai+2; bj�2)g) [ fyg; if (bi+1; aj�1) 62 A(A n f(ai+2; bj�2); (bi+1; aj�1)g) [ fy; (y1; bj)g; otherwise:Finally, if S = f(ai+2; bj�2); (bi+2; bj�2)g � A then letB = ( (A n S) [ fy; (y1; bj)g; if (bi+1; aj�1) 62 A(A n S n f(bi+1; aj�1)g) [ fy; (y1; bj); (ai+1; bj�1)g; otherwise:Case 2. Assume y2 = ak. Thus, t = k and y = (b0; ak). Let � be the poset automorphismof T such that �(ai) = bi, 0 < i < k and �(b0) = b0, �(ak) = ak. It induces an15

automorphism in T 2 de�ned as (�; �) = (�(�); �(�)). Let I be the largest initialsegment of T 2t n A and let F be the largest �nal segment of T 2t n A. Since j�(A)j =j�( (A))j, we may assume that jF j � jIj > 0.Denote Zi = f(�; �) 2 T 2k j rT (�) = ig, i = 0; : : : ; k. Now jF j � jIj > 0 implies(Z0 [ Zk) \ A = ;. Furthermore, if Z1 \ A = ; then Zk�1 \ A = ; too. In this caseconstruct the set B by replacing each (�i; �j) 2 A with (�i+1; �j�1) for �; � 2 fa; bg.If A\Z1 6= ;, then j ~�(y)j < 2. Thus, without loss of generality we can assume j ~�(y)j = 1.This implies A \ f(a1; bk�1); (b1; bk�1)g = ;. Let x be the largest element in A. Now ifx2 = bq then 1 � q < k � 1. Thus, for S = fx; (x1; aq)g one has: j ~�(S)j � 1 and�(S) \�(y) = ;. In this case set B = (A n S) [ fy; (a1; bk�1)g.Finally, assume x2 = aj. If q = 1 then j ~�(x)j = 1 and �(x) \�(y) = ;, because k � 3.If q > 1 then j ~�(x)j = 2 and j�(x) \�(y)j � 1. We set B = (A n fxg) [ fyg.It is easy to check that in all cases the set B satis�es (22). Therefore, by applying similararguments (and the compression, if necessary) to the set B we transform it into an initialsegment without increasing the shadow. 2It is shown in [5] how to deduce a solution to an edge-isoperimetric problem on a graph Gif a certain poset P is Macaulay. In fact, for any graph G one can construct the associatedposet P . Without going into details (for which we refer readers to [3, 5]), let us mentionthat the poset (T (k);�) corresponds to the complete bipartite graph Kk;k. Moreover, itis shown in [3] that any cartesian power of the graph G corresponds in the sense aboveto the same cartesian power of P . Therefore, Theorem 3 implies a solution for an edge-isoperimetric 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 remarksOur �rst 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 theresulting poset Q is isomorphic to the k � 2 grid (cf. Fig. 4b). Due to the Clements-Lindstr�om theorem [8] the cartesian product of any chains Z1; : : : ; Zn of lengths k1; : : : ; knis Macaulay. Moreover, if k1 � � � � � kn (26)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 isMacaulay for n � 2. The reason for this is that regardless of the total order �, thecondition (26) is not satis�ed. For n � 3 this condition is essential for the lexicographicorder 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 fori = 1; : : : ; k � 1, then this results in a cycle C(k) (cf. Fig. 4c). This poset is rank-greedyand the corresponding Macaulay order is essentially unique (it is shown in the �gure).However, the lexicographic order does not work for C(k)�C(k) (the element (2; 6) is thesmallest 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 Macaulay16

order can be derived from [11, 15], where a vertex-isoperimetric problem for tori wasstudied, and is not lexicographic.Although Theorem 2 provides some necessary conditions for posets, all whose cartesianpowers are Macaulay, there are many in�nite series of such posets. We illustrate this byproviding a tool for constructing such series. Given a Macaulay poset (P;�;�), constructanother poset Q = (P;v) as follows. Take an element a 2 Pi for some i > 1 and considerFi(a). Then �(Fi(a)) = Fi�1(b) for some b 2 Pi�1. Let c 2 Fi�1(b) and assume c 6� a.Now we extend the partial order � by setting c � a (cf. Fig. 5).@@@@@@ sss @@@@@@ sss @@@@@@ sss������������������123 456 789 ac ba. @@@@@@ sss @@@@@@ sss @@@@@@ sss������������������123 456 789 ac b������������������b.Figure 5: Posets P (a) and Q (b)Denote �P (m; i) = min j�(A)j, where the minimum runs over all A � Pi with jAj = 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; : : : ; jPij and i = 1; : : : ; r(P ): (27)However, �P (Fi(a)) = �Q(Fi(a)) for any a 2 Pi. Therefore, since P is Macaulay, thenthe lower bound (27) for Q is tight. This implies (P;v;�) is Macaulay too. Moreover, ifP is strongly Macaulay and rank-greedy then so is Q. Obviously, applying any numberof the above extensions leads to a Macaulay poset.Now consider P 2. Since�P 2(Fi((x; y))) = f(x; �) j � 2 �P (Fi�rP (x)(y))g [ f(�; y) j � 2 �P (Fi�rP (y)(x))g;then �P 2(Fi((x; y)) = �Q2(Fi((x; y))). Therefore, if P satis�es the assumptions of Theo-rem 1, then so is for Q. On the other hand, since the lexicographic order is Macaulay forP 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 posetCn(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 otherextremal problems. In our forthcoming paper [7] we present a local-global principle forvertex-isoperimetric problems. Sperner theory [9] and the Erd�os-Ko-Rado type results areclosely related with Macaulay theory, and are further examples were local-global principlesmay exist. 17

AcknowledgmentsThe authors would like to thank Konrad Engel (University of Rostock, Germany) for hishelpful comments concerning the draft version of the manuscript.References[1] R. Ahlswede, N. Cai, General edge-isoperimetric inequalities, Part I: Information-theoretical method, Europ. J. Combin., 18 (1997), 355{372.[2] R. Ahlswede, N. Cai, General edge-isoperimetric inequalities, Part II: A local-globalprinciple for lexicographic solution, Europ. J. Combin., 18 (1997), 479{489.[3] S.L. Bezrukov, Edge isoperimetric problems on graphs, to appear in Graph Theoryand 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-isoperimetricproblems, preprint, 1998.[8] G.F. Clements, B. Lindstr�om, A generalization of a combinatorial theorem ofMacaulay, J. Comb. Th. 7 (1969), No. 2, 230{238.[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, BolyaiSoc. 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, (inRussian), Doklady AN Arm. SSR, vol. LXXIV (1982), No. 2, 61{65.[12] G.O.H. Katona, A theorem of �nite 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. OptimizationTech., 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. 18