A New Approach to Macaulay Posets - Semantic Scholar
1
Introduction
Let (P, ≤P ) be a poset with a partial order ≤P . We say P is ranked if there exists a function rP : P 7→ IN such that rP (z) = 0 for some minimal element z ∈ P , and rP (x) = rP (y) − 1 whenever x
Z 1 b. Let a ∈ 4(b). Case 1. Assume a Z 1 1 and d ∈ P1 . For this assume the contrary and let C = {x ∈ P1 | 1 Z 1 ft−1 = a and |P0 | = |P1 | = · · · |Pt−1 | = 1. Therefore, P is a chain and f0 >Z 1 · · · >Z 1 fp for p ≥ 2. If p ≥ 3 then P22 = {(p − 2, p), (p, p − 2), (p − 1, p − 1)}, where the elements are listed in increasing order Z 2 . Consider B = F22 ((p − 2, p)). One has |B| = 1 and 4(B) = {(p − 1, p)} is not an initial segment, since (p, p − 1) 6∈ 4(B). This contradicts the continuity property. For p = 2 one has P22 = {(1, 1), (0, 2), (2, 0)}. Denoting A = {(1, 1)} and B = {(0, 2)} we have | 4 (A)| = 2 > 1 = | 4 (B)|, which is a contradiction. The above arguments show that the Macaulay order Z 1 is a linear extension of the partial order ≤. To complete the proof we have to show that the second claim in the definition of rank-greediness is fulfilled for P . Suppose this is not true. Then there exist a, b ∈ P such that z r(a) and b >Z 1 a. Since 4(b) 6= ∅ then a 6= 0. So, (b, a) 6= (1, 0). Furthermore, b 6= p since otherwise b is the top element of P because Z 1 is a linear extension of ≤. Hence, (b, a) 6= (p, p − 1). 2 2 Let q = r(b) − r(a) and consider the set B = Fr(a)+q ((a, fq )) ⊆ Pr(a)+q . Since (b, f0 ) >Z 2 (a, fq ), then (b, f0 ) 6∈ B. Furthermore, z rP n (y) then rP (x1 ) > rP (y1 ). Since x1 >Z 1 y1 and the order Z 1 is rank-greedy, there exists z ∈ 4(x1 ) such that z ≥Z 1 y1 . Let z be the maximum element satisfying these conditions. Since z ∈ 4(x1 ) then z 6= p. Assume y1 >Z 1 0. Now, if z >Z 1 y1 then (z, x2 , . . . , xn ) >Z n y, which contradicts (5). If z = y1 then rP n (x) = rP (x1 ) = rP (y1 ) + 1 and rP n (x) > rP n (y) imply rP n (x) = rP n (y) + 1 and r(y2 ) = · · · = r(yn ) = 0, so (y2 , . . . , yn ) = (0, . . . , 0). Hence, y ∈ 4(x) which contradicts (5). Assume y1 = 0. If z >Z 1 1 then we have a similar contradiction as above. Now, if z = 1 then 4(x1 ) = {z}. If rP n (y) = 0 then 4(x) = {(1, 0, . . . , 0)} contradicts (5). Therefore, rP n (y) = 1 which implies rP n−1 ((y2 , . . . , yn )) = 1. Now, if n ≥ 3 then x and y have a common zero entry and the lemma is true by induction and the consistency of Z n . If n = 2 then the first element of P22 is z = (1, 1) and | 4 (z)| = 2 > 1 = | 4 (x)| contradicting the Macaulayness of Z 2 . 12
Case 2. Assume x1 = y1 + 1, x1 6= p, and y1 6= 0. By similar arguments to the above one has r(xi ) = 0 for 2 ≤ i ≤ n, r(x1 ) > r(y1 ), and z >Z 1 y1 for some z ∈ 4(x1 ). Therefore, y1 Z n−1 (y2 , . . . , yn ) and rP n−1 (x2 , . . . , xn ) > rP n−1 (y2 , . . . , yn ). By induction there exists an element (z2 , . . . , zn ) ∈ 4((x2 , . . . , xn )) such that (z2 , . . . , zn ) ≥Z n−1 (y2 , . . . , yn ). But then for z = (x1 , z2 , . . . , zn ) ∈ 4(x) one has z ≥Z n y contradicting (5). 2 Later in this section we will obtain necessary conditions for a poset P in order for the zigzag order Z n to be Macaulay for P n for any n ≥ 2. If Z n is Macaulay for P n then by Lemma 3 it is an MRI-order. The MRI problem is easier to analyze than the SMP. As we show there exists a limited number of posets that satisfy this necessary condition. The proof that these posets are Macaulay will be done in Sections 4 and 5. Note that the proof of Lemma 10 and Theorem 1 can be slightly simplified by adding an additional condition concerning Z 3 and using Lemma 7. However, we did not do this in order to present a method that can be applied to a more general situation. Throughout this section we assume the MRI problem on a connected poset P has nested solutions provided by some total order Z 1 and that order Z 2 is an MRI-order for P 2 . For any integer n ≥ 1 and m ≥ 0 denote by F n (m) ⊆ P n the initial segment of length m of the order Z n on P n . Let δ(P ) = (δ0 , . . . , δp ). Lemma 10 Let P with |P | ≥ 4 be a connected poset and Z 1 and Z 2 be MRI-orders for P and P 2 , respectively. Then the following three conditions hold for δ(P ): a. δi+1 ≤ δi + 1 for 0 ≤ i < p; b. δ1 = 1 and 0 ≤ δ2 ≤ 1; c. If δ2 = 0 then P is the N-poset. Proof. To show the first assertion assume the contrary, i.e. δi+1 ≥ δi + 2 for some i ≥ 0. Let I1 = F 1 (i), I2 = F 1 (i + 1) and I3 = F 1 (i + 2). One has I1 ⊂ I2 ⊂ I3 . Let x = I2 \ I1 and y = I3 \ I2 . Since δi+1 − δi ≥ 2 then r(y) − r(x) ≥ 2. This implies that either x and y are incomparable, or there exists z ∈ P with x
0}. If |P | > 4 consider the ideal A = ({0, 1, . . . , i − 2} × P ) ∪ {(i − 1, 0), (i − 1, 1), (i, 0), (i, 1)}. One has R(A) > R(F 2 (|A|)), which is a contradiction, and so |P | ≥ 4 implies that |P | = 4. By part (a) one has 0 ≤ δ3 ≤ 1. Therefore, either δ(P ) = (0, 1, 0, 1) or δ(P ) = (0, 1, 0, 0). In the first case P is the N-poset, while in the second case P is not connected. 2 Denote by H n the Boolean lattice (or the Hypercube) of dimension n. We need the last auxiliary lemma before proving Theorem 1. Lemma 11 If Z 2 is an MRI-order on P 2 then the lexicographic order is an MRI-order on P × H k for any k ≥ 1. Proof. The lemma is true for k = 1 since P × H 1 with the lexicographic order is isomorphic to an initial segment of Z 2 on P 2 . Assume k ≥ 2, and let A be a compressed set which is a solution to the MRI problem. Let a = (a, α1 , . . . , αk ) be the last element (in lexicographic order) of A and let b = (b, β1 , . . . , βk ) be the the first element of (P × H k ) \ A. Without loss of generality we can assume that a and b have no equal entry. If a Z 1 bi . Suppose there exists an i ≥ 2, such that r(ai ) = r(bi ) + 2 in which case ai = 4 and bi = 0. If ai = 4 then ak = 4 for i < k ≤ n. Indeed, if ak i then choose a0k such that r(a0k ) = r(ak ) + 1. One has a = (2, a2 , . . . , ai = 4, . . . , ak , . . . , an ) >Z n (2, a2 , . . . , 2, . . . , a0k , . . . , an ) >Z n (0, b2 , . . . , 2, bi+1 , . . . , bn ) >Z n (1, b2 , . . . , bi = 0, . . . , bn ) = b, where the first inequality follows from (4, ak ) >Z 2 (2, a0k ) and the last one from (0, 2) >Z 2 (1, 0). We may now assume that a = (2, . . . , aj , . . . , 4, . . . , 4) and b = (1, . . . , bj , . . . , 0, . . . , 0). Moreover, since rP n (a) = rP n (b), there exists a j ≥ 2 such that r(aj ) < r(bj ). If aj = 0 then, since bj >Z 1 0, a = (2, . . . , aj = 0, . . . , 4, 4, . . . , 4) >Z n (2, . . . , aj = 1, . . . , 1, 4, . . . , 4) >Z n (0, . . . , bj , . . . , 1, 0, . . . , 0) >Z n (1, . . . , bj , . . . , 0, 0, . . . , 0) = b, where the first inequality follows from (0, 4) >Z 2 (1, 1). If aj >Z 1 0, then, since bj = 4, a = (2, . . . , aj , . . . , 4, 4, . . . , 4) >Z n (0, . . . , 3, . . . , 4, 4, . . . , 4) >Z n (1, . . . , 3, . . . , 1, 4, . . . , 4) >Z n (1, . . . , bj = 4, . . . , 0, 0, . . . , 0) = b, where the last inequality follows from (3, 1) >Z 2 (4, 0). Suppose that r(ai ) = r(bi ) for all i ≥ 2, which then implies ai ∈ {1, 2} and bi ∈ {2, 3}. If ai = 2 for all i ≥ 2 then a = (2, 2, . . . , 2) >Z n (2, 1, 3, 2, . . . , 2) >Z n (1, b2 , b3 , . . . , bn ) = b. Assume ai = 1 for some i ≥ 2. In this case (a1 , . . . , ai−1 , 0, ai+1 , . . . , an ) ∈ 4new (a), so | 4new (a)| ≥ 1. On the other hand, 4new (b) = {(0, b2 , . . . , bn )}, thus | 4new (a)| ≥ | 4new (b)|, and so | 4 ((A \ {a}) ∪ {b})| ≤ | 4 (A)|, and we can swap a and b. Case 3. Assume a = (0, a2 , . . . , an ) and b = (1, b2 , . . . , bn ). Then b0 = (0, b2 , . . . , bn ) ∈ 4(A) by Lemma 12. Hence, there exists a vector d = (0, b2 , . . . , di , . . . , bn ) ∈ A, where d and b0 differ in only the i-th component, with di >Z 1 bi . But now d = (0, b2 , . . . , di , . . . , bn ) > (1, b2 , . . . , bn ) = b, since (0, di ) >Z 2 (1, bi ), so b ∈ A.
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Case 4. Assume b = (0, b2 , . . . , bn ). In this case the assumptions of Lemma 12 are satisfied for the set Ftn (b). This implies 4(A ∪ {b}) = 4(A), hence, | 4 ((A \ {a}) ∪ {b})| ≤ | 4 (A)|, and we can swap a and b. Therefore, in all cases either we have b ∈ A, a contradiction, or we can swap a and b without increasing the shadow of A. After a finite number of such operations A will be transformed into an initial segment. 2
5
A new construction for Macaulay posets
Let P be a Macaulay poset with associated Macaulay order O. The poset P is called additive if for any t > 0 and any m and m0 the following condition is satisfied | 4 (Ft (m0 ))| + | 4 (Ft (m00 ))| ≥ | 4 (Ft (m0 + κ))| + | 4 (Ft (m00 − κ))|,
(9)
where κ = κ(m0 , m00 ) = min{m00 , |Pt | − m0 }. In other words, if we take two copies P 0 and P 00 of P , and initial segments Ft (m0 ) ⊆ Pt0 and Ft (m00 ) ⊆ Pt00 , we should be able to move some vertices from one initial segment to the other one without increasing the sum of their shadows. This transformation is schematically shown in Fig. 3, where the initial segments are depicted in bold. '$ '$'$ '$'$ '$
P0
P 00
m0
m00
P0 ⇒
P 00
m0 + m00
or
P0
P 00
|Pt |
m00 − κ
&% &%&% &%&% &%
a.
b.
c.
Figure 3: The additivity of P . (a) The original configuration. The resulting sets: (b) if κ = m00 and (c) if κ = |Pt | − m0 .
Examples of additive Macaulay posets include the Boolean lattice and the lattice of multisets, the star poset and colored complexes (see [11] for more details). Let P be an additive Macaulay poset. First we construct a poset P (k) whose Hasse diagram consists of k disjoint copies of P , assuming that the t-th level of P (k) is the union of the t-th levels of the corresponding copies. This poset can be viewed as cartesian product of P with a trivial poset T (k) with k elements q1 , . . . , qk of rank 0, i.e., P (k) = T (k) × P . Note that T (k) is a Macaulay poset with a Macaulay order T given by q1
