THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS

SOOCHOW JOURNAL OF MATHEMATICS Volume 28, No. 4, pp. 347-355, October 2002 THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS BY HUA MAO 1,2 AND SANYANG LIU 2 Abstract. This paper first shows how the operations of direct sum, union and intersection may be defined for poset matroids. This paper then discusses the relation between the partition of a poset and a partition of the associated poset matroids. 1. Introduction M. Barnabei etc. develop in [1, 3] a more general theory of matroids by replacing the underlying set of a matroid by a poset. However, not all properties of matroids can be translated in a simple way into this new theory. Some significant examples of this fact are given in [1, 3]. In this paper, our definitions of rank, direct sum, union, intersection and partition in poset matroid theory allow the extension to this new setting of every notion of matroid theory. Our main results are the constructions about direct sum, union and intersection of poset matroids which devoted in Section 3 and Section 4; in addition, all results here are also the extension of the relevant results of matroid theory without losing some of the valuable properties of matroid theory. For the problem considered in this paper, at least to our knowledge, the concepts above and their properties have not been discussed previously. 2. Preliminaries The following definition is taken from [4]. Received February 20, 2001; revised June 27, 2002. AMS Subject Classification. 05B35. Key words. direct sum, union, intersection, partition. 347

348 HUA MAO AND SANYANG LIU Definition 1. (1) A poset is a set in which a binary relation x y is defined, which satisfies for all x, y, z the following conditions: (p1) For all x, x x; (p2) if x y and y x, then x = y; (p3) if x y and y z, then x z. (2) a and b are incomparable in a poset P, in notation, a//b. (3) Let P = (P, ) be a poset and T P. T := (T, T ) is called a subposet of P, if the binary relation x T y is defined as follows: x T y iff x y, for all x, y T. Lemma 1. Let P j = (P j, j ) (j = 1, 2) be two posets with P 1 P 2 =. Simply say that P 1 and P 2 are disjoint, in notation, P 1 P 2 =. Then P := (P 1 P 2, ) is a poset in which the binary relation is defined as follows: x y iff x 1 y or x 2 y; x//y iff none of x 1 y and x 2 y are true. Proof. We notice that if x y, then x, y P 1 or x, y P 2. Thus P satisfies (p1), (p2) and (p3) trivially. We denote the poset P in Lemma 1 by P 1 P 2. We next summarize the known facts of poset matroid theory that are needed in the present work. The Definition 2 and Lemma 2, 3, 4 and 5 come from [1, 3]. Definition 2. (1) A filter of a poset P is any subset A of P such that for every x, y P, if y x and y A, then x A. (2) A poset matroid on P is a family B of filters of P, called bases, satisfying the following axioms: (b.0) B ; (b.1) for every B 1, B 2 B : B 1 B 2 ; (b.2) for every B 1, B 2 B and for every pair of filters X, Y of P, such that X B 1, B 2 Y, X Y, there exists B B such that X B Y. (3) An independent set of a poset matroid on P is a filter I of P such that there exists a basis B B such that I B. (4) A spanning set of B on P to be a filter S of P, such that there exists a basis B such that B S. (5) Let I be the set of independent sets of B. The elements of the set of Inc(P)-I are called dependent sets. (6) A minimal dependent set of B is called a circuit. Lemma 2. Let B be a poset matroid on P. Then for every B 1, B 2 B, we have B 1 = B 2. Lemma 3. Let A, B Inc(P), where P is a poset and Inc(P) is the collection of all filters of P. Then A B, A B Inc(P).

THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS 349 Lemma 4. (1) Let B be a poset matroid on the poset P = (P, ). Then the family B := {P B; B B} is a poset matroid on the dual poset P = (P, ). (2) Spanning sets of B are independent sets of B, and conversely. Lemma 5. The family I of all independent sets of a poset matroid B on the poset P satisfies the following properties: (i.0) I ; (i.1) if X, Y are filters of P such that Y I and X Y, then X I; (i.2) for X, Y I with X < Y, there exists y Max(Y X) such that X y I. Conversely, let I be a family of filters of P satisfying properties (i.0), (i.1), (i.2); then the family B := {I I; I is a maximal element of I} is a poset matroid. Before proceeding, we give the concepts of rank associated with a poset matroid and the partition of a poset. Definition 3. Let B be a nonempty, incomparable family of filters of a poset P. The rank ρ associated with B is the nonnegative integer valued function on Inc(P) defined as follows: for every X Inc(P), ρ(x) := max{ Y ; Y Inc(P), Y X and Y B for some B B}. Definition 4. Let {P 1,..., P m } be a collection of disjoint, nonempty subposet of the poset P such that P = P 1 P m. Then {P 1,..., P m } is called a partition of P. The following Lemma concludes the section. Lemma 6. Let P j = (P j, j ) (j = 1, 2) be two disjoint posets. Then X Inc(P 1 P 2 ) iff there exist X j Inc(P j ) (j = 1, 2) such that X = X 1 X 2. Further, X j = X P j (j = 1, 2) for X Inc(P 1 P 2 ). Proof. Since X Inc(P 1 P 2 ), it follows that X P 1 P 2. According to P 1 P 2 =, there exist X j P j (j = 1, 2) satisfying X = X 1 X 2. Certainly, X 1 X 2 =. Suppose x, y P 1 P 2 and x 1 y, x X 1. Because of x 1 y, it follows that x y. Because X Inc(P 1 P 2 ), there is y X. If y X 2, then by Lemma 1, x//y which is a contradiction. Thus y X 1 ; and so X 1 Inc(P 1 ). Analogously, X 2 Inc(P 2 ).

350 HUA MAO AND SANYANG LIU Conversely, for every X j Inc(P j ) (j = 1, 2), we have X 1 X 2 P 1 P 2 = and X 1 X 2 P 1 P 2. Let x, y P 1 P 2, with x y and x X 1 X 2. If x X 1, then by Lemma 1, y X 1 X 1 X 2. Likewise, if x X 2, then y X 2. Hence, X 1 X 2 Inc(P 1 P 2 ). Actually, we notice that for X Inc(P 1 P 2 ), X P j = X j, (j = 1, 2). 3. Direct Sum We deal with the constructions and properties of direct sum of poset matroids in this section. Theorem 1. Let B 1, B 2 be two poset matroids on disjoint posets P 1, P 2 respectively and B = {B 1 B 2 ; B j B j (j = 1, 2)}. Then B is a poset matroid on P 1 P 2. Proof. By Lemma 6, we have B 1 B 2 Inc(P 1 P 2 ) for every B j B j (j = 1, 2); and so B Inc(P 1 P 2 ). For B, (b.0) is trivially satisfied. If there are B 1 B 2, B 3 B 4 B for which B 2k 1 B 1, B 2 B 2, (k = 1, 2), such that B 1 B 2 B 3 B 4, then B 1 B 2 < B 3 B 4. Further, by Lemma 2, B 3 + B 4 = B 1 B 2 < B 3 B 4 = B 3 + B 4, i.e. 0 < 0, a contradiction. Namely, (b.1) is true for B. To prove (b.2), let X, Y Inc(P 1 P 2 ), X B 1 B 2, B 3 B 4 Y, X Y, where B 2k 1 B 1, B 2k B 2, (k = 1, 2). X, Y Inc(P 1 P 2 ) with Lemma 6 implies that there exist X k, Y k Inc(P k ) (k = 1, 2) satisfying X = X 1 X 2, Y = Y 1 Y 2. Considering P 1 P 2 =, we get X 1 B 1, B 3 Y 1, X 1 Y 1 ; X 2 B 2, B 4 Y 2, X 2 Y 2. Thus there are B 5 B 1, B 6 B 2 such that X 1 B 5 Y 1 and X 2 B 6 Y 2 ; moreover, X = X 1 X 2 B 5 B 6 Y 1 Y 2 = Y and B 5 B 6 B. Hence (b.2) is true for B. Consequently, B is a poset matroid on the poset P 1 P 2. We call B in Theorem 1 the direct sum of B 1 and B 2, in notation, B 1 + B 2. Some properties of B 1 + B 2 will be discussed next. Property 1. Let ρ 1 and ρ 2 be the ranks associated with B 1, B 2 respectively, where B 1, B 2 are two poset matroids on disjoint posets P 1 and P 2 respectively.

THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS 351 Let ρ be the rank associated with B 1 + B 2. Then (1) ρ(a) = ρ 1 (A P 1 ) + ρ 2 (A P 2 ) for every A Inc(P 1 P 2 ). (2) I Inc(P 1 P 2 ) is an independent set of B 1 + B 2 iff I P 1 and I P 2 are independent sets of B 1 and B 2 respectively. (3) Let C, C 1, C 2 be the families of all circuits of B 1 + B 2, B 1, B 2 respectively. Then C C iff C C 1 or C C 2. Proof. (1) For every A Inc(P 1 P 2 ), by Definition 3 and Lemma 6, ρ(a) = max{ Y ; Y Inc(P 1 P 2 ), Y A, Y B 1 B 2 for some B 1 B 2 B 1 + B 2 } = max{ Y 1 Y 2 ; Y = Y 1 Y 2 Inc(P 1 P 2 ), Y 1 Y 2 A = A 1 A 2, Y 1 Y 2 B 1 B 2 for some B 1 B 1, B 2 B 2, where A j, Y j Inc(P j ) (j = 1, 2)} = max{ Y 1 + Y 2 ; Y j Inc(P j ), Y j A j, Y j B j for some B j B j, (j = 1, 2)} = 2 j=1 max{ Y j ; Y j Inc(P j ), Y j A j, Y j B j for some B j B j } = ρ 1 (A 1 ) + ρ 2 (A 2 ) = ρ 1 (A P 1 ) + ρ 2 (A P 2 ). (2) I Inc(P 1 P 2 ) is an independent set of B 1 +B 2 iff there exist B 1 B 2 B 1 +B 2 such that I B 1 B 2. On the other hand, I = (I P 1 ) (I P 2 ) B 1 B 2 iff I P j B j (j = 1, 2), i.e. I P j is an independent set of B j (j = 1, 2). (3) By (2), if C is a circuit of B 1 + B 2, then C P 1 and C P 2 are not independent sets of B 1 and B 2 respectively at the same time. There is no harm in supposing C P 1 to be a dependent set of B 1. Then there is a circuit C 1 of B 1 such that C 1 C P 1 ; further, C 1 (C P 2 ) is also a dependent set of B 1 + B 2 and C 1 C 1 (C P 2 ) C. On the other hand, C 1 = C 1 Inc(P 1 P 2 ) in virtue of Lemma 6. Therefore, by (2), C 1 is also a dependent set of B 1 + B 2. Thus C P 2 =. Namely, C = C 1 is a circuit of B 1 and only a circuit of B 1, i.e. C C 1. By Lemma 6 and C 1 C 2 P 1 P 2 = with (2), it is easy to obtain the converse. 4. Union and Intersection In this section, we examine the union and intersection of poset matroids. These lead to some new poset matroids. Another purpose is in the partition of a poset and its relation with union of poset matroids.

352 HUA MAO AND SANYANG LIU Theorem 2. Let B 1 and B 2 be two poset matroids on the poset P satisfying B 1 B 2 = for any B 1 B 1, B 2 B 2. Then B := {B 1 B 2 ; B j B j (j = 1, 2)} is a poset matroid on P. Proof. (b.0) is trivially satisfied. By Lemma 2 and Lemma 3, (b.1) is satisfied as well. To show B is a poset matroid, we have only to prove that B satisfies (b.2). For every B 1 B 2, B 3 B 4 B in which B 2k 1 B 1, B 2k B 2, (k = 1, 2) and for every pair of filters X, Y of P, such that X B 1 B 2, B 3 B 4 Y, X Y, from Lemma 3 and B 1 B 2 =, it follows that X B j Inc(P) (j = 1, 2). Since B 3, B 4 B 3 B 4 Y and B j (j = 1, 2) are poset matroids on P, there exist B 5 B 1, B 6 B 2 such that X B 1 B 5 Y, X B 2 B 6 Y ; and so X = (X B 1 ) (X B 2 ) B 5 B 6 Y ; in addition, B 5 B 6 B. Hence, (b.2) is satisfied; further, the theorem is proved. Next we give an example to show that if there exist B 1 B 1, B 2 B 2 such that B 1 B 2, then Theorem 2 will not be true. c a b d e Figure 1. Example. Let P be the poset shown in Figure 1 ([1, Figure 4.1]). Then B 1 := {{a, c}} is a poset matroid on P (This can be obtained with [1, Example 4.1]); B 2 := {{a, c, e}, {b, c, e}, {c, d, e}} is a poset matroid on P (This is proved in [1, Example 4.3]). Let B 1 = {a, c} and B 2 = {a, c, e}. We have B 1 B 2 = {a, c} and B := {{a, c} {a, c, e}, {a, c} {b, c, e}, {a, c} {c, d, e}} = {{a, c, e}, {a, b, c, e}, {a, c, d, e}}. Since {a, c, e} = 3, {a, c, d, e} = 4, by Lemma 2, B is not a poset matroid on P. We call B in Theorem 2 the union of B 1 and B 2, in notation, B 1 B 2. If B 1, B 2,..., B m are poset matroids on P such that B i B j = for every B i

THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS 353 B i, B j B j, (i j; i, j = 1, 2,..., m), then by induction on m with Theorem 2, B = {B 1 B 2 B m ; B j B j (j = 1, 2,..., m)} is a poset matroid on P. This poset matroid is called the union of B 1, B 2,..., B m, in notation, B 1 B 2 B m. The following property discusses the properties of B 1 B 2. The related result about B 1 B 2 B m can be gotten by induction. Property 2. Let B 1, B 2 be two poset matroids on the poset P, satisfying B 1 B 2 = for every B j B j (j = 1, 2). Then: (1) For every A Inc(P), ρ 12 (A) = max X A (ρ 1 (X)+ρ 2 (F X )), where ρ 1, ρ 2, ρ 12 are the ranks associated with B 1, B 2, B 1 B 2 respectively and F X Inc(P), F X (A\X) such that for every Z Inc(P), if Z (A\X), then Z F X (2) Let I 1, I 2, I 12 be the families of all independent sets of B 1, B 2, B 1 B 2 respectively. Then I 12 = I 1 I 2 in which I 1 I 2 = {X 1 X 2 ; X j I j (j = 1, 2)}. Proof. (1) By Lemma 3, F X exists in Inc(P) for any X Inc(P), X A. Since ρ 1 (X) = max{ X 1 ; X 1 Inc(P), X 1 X B 1 for some B 1 B 1 }, ρ 2 (F X ) = max{ X 2 ; X 2 Inc(P), X 2 F X B 2 for some B 2 B 2 }, ρ 12 (A) = max{ Y ; Y Inc(P), Y A (B 1 B 2 ) for some B 1 B 2 B 1 B 2 } and X 1 X 2 (X B 1 ) (F X B 2 ) A (B 1 B 2 ) for some B 1 B 1, B 2 B 2, we have ρ 1 (X)+ρ 2 (F X ) ρ 12 (A); moreover, max X A (ρ 1 (X)+ρ 2 (F X )) ρ 12 (A). On the other hand, Y A (B 1 B 2 ) means Y 1 = Y B 1 A, Y B 2 A. According to B 1 B 2 =, it follows that Y B 2 F Y1. Thus ρ 12 (A) max X A (ρ 1 (X) + ρ 2 (F X )). Hence, ρ 12 (A) = max X A (ρ 1 (X) + ρ 2 (F X )). (2) It is obvious that I 1 I 2 I 12. Let X I 12. Then, there exists B 1 B 2 B 1 B 2 in which B j B j (j = 1, 2) such that X B 1 B 2. X = (X B 1 ) (X B 2 ) together with B 1 B 2 = and X B 1 B 2 implies X B 1 I 1, X B 2 I 2, i.e. X I 1 I 2. Since a base is a minimal spanning set of a poset matroid, we can define the intersection of two poset matroids as follows. Theorem 3. Let B 1 and B 2 be two poset matroids on the poset P, whose spanning sets are S 1 and S 2 respectively and B 1 B 2 = for every B 1 B 1, B 2

354 HUA MAO AND SANYANG LIU B 2. Then S 1 S 2 = {X 1 X 2 ; X j S j (j = 1, 2)} is the family of spanning sets of a poset matroid on P. We call this poset matroid the intersection of B 1 and B 2, in notation, B 1 B 2. Proof. According to Definition 2, Lemma 4 and Lemma 5, we have only to prove that S 1 S 2 is the collection of all independent sets of some poset matroid on P. For any X j S j, by Lemma 4, X j is an independent set of Bj (j = 1, 2); moreover, by Property 2, X j = X j (j = 1, 2) are independent sets of B1 B 2 ; further, X 1 X 2 is an independent set of B 1 B 2. Thus, X 1 X 2 is a spanning set of (B 1 B 2 ). Conversely, for a spanning set S of (B 1 B 2 ), or say, an independent set of B1 B 2, there exist independent sets X 1, X 2 of B 1, B 2 respectively such that S = X1 X 2. On the other hand, by Lemma 4, X 1 and X 2 are spanning sets of B 1 and B 2 respectively; moreover, S is a spanning set of B 1 and B 2 respectively because of X j S (j = 1, 2), i.e. S S j (j = 1, 2), and so S = S S S 1 S 2. Hence, the theorem is proved. Corollary 1. Let B 1 and B 2 be two poset matroids on the poset P such that B 1 B 2 = for every B 1 B 1, B 2 B 2. Then B 1 B 2 = (B 1 B 2 ). We next discuss the partition. Theorem 4. Let B j (j = 1, 2) be poset matroids on P satisfying B 1 B 2 = for any B j B j (j = 1, 2) and ρ j (j = 1, 2) be the ranks associated with B j (j = 1, 2) respectively. Then P has a partition {P 1, P 2 } satisfying P 1 and P 2 are independent sets of B 1 and B 2 respectively iff ρ 1 (A) + ρ 2 (A) A for every A Inc(P). Proof. Suppose P has a partition {P 1, P 2 } and P j (j = 1, 2) are independent sets of B j (j = 1, 2). Since A = A P = (A P 1 ) (A P 2 ) for every A Inc(P), by Lemma 3 and Definition 2, it follows that A P 1 and A P 2 are independent sets of B 1 and B 2 respectively. Thus ρ j (A P j ) = A P j (j = 1, 2), and so A = A P 1 + A P 2 = ρ 1 (A P 1 ) + ρ 2 (A P 2 ). Because of A P j A (j = 1, 2) and Definition 3, there are ρ j (A P j ) ρ j (A) (j = 1, 2). Further, A ρ 1 (A) + ρ 2 (A).

THE DIRECT SUM, UNION AND INTERSECTION OF POSET MATROIDS 355 Conversely, since ρ 1 (A) + ρ 2 (A) A for every A Inc(P), especially, ρ 1 (P)+ρ 2 (P) P. From Definition 3, ρ j (P) = B j for every B j B j (j = 1, 2). Thus B 1 + B 2 P for every B j B j (j = 1, 2). It follows from B 1 B 2 P and B 1 B 2 = that there is B 1 + B 2 = B 1 B 2 P. Hence, B 1 B 2 = P ; moreover, P = B 1 B 2. Obviously, B 1 B 2 = and B j (j = 1, 2) are independent sets of B j (j = 1, 2). Namely, {B 1, B 2 } is a partition of P. Corollary 2. Let B j (j = 1, 2,..., k) be poset matroids on the poset P such that B 1 B 2 = (i j; i, j = 1, 2,..., k) and ρ j (j = 1, 2,..., k) be the ranks associated with B j (j = 1, 2,..., k) respectively. Then P has a partition {P 1, P 2,..., P k } satisfying P j is an independent set of B j (j = 1, 2,..., k) iff kj=1 ρ j (A) A for every A Inc(P). Proof. By induction on k and analogous to the proof of Theorem 4, the assertion is easily proved. Corollary 3. Let B j (j = 1, 2,..., k) be poset matroids on P such that B i B j = (i j; i, j = 1, 2,..., k) and ρ be the rank associated with B 1 B 2 B k. Then P has a partition {P 1, P 2,..., P k } satisfying P j (j = 1, 2,..., k) are independent sets of B j (j = 1, 2,..., k) respectively iff ρ(p) = P. Proof. By induction on k with Property 2 and Theorem 4, the assertion is easily proved. References 1 M. Barnabei, G. Nicoletti and L. Pezzoli, Matroids on partially ordered sets, Adv, in Appl. Math., 21(1998), 78-112. 2 D. J. A. Welsh, Matroid Theory, Academic Press, London, 1976. 3 M. Barnabei, G. Nicoletti and L. Pezzoli, The symmetric exchange property for poset matroids, Adv. in Math., 102(1993), 230-239. 4 G. Grätzer, General Lattice Theory, Birkhäuser, Basel, 1978. 1 Department of Mathematics, Zhengzhou University, Zhengzhou, 450052, P.R. China. E-mail: yushengmao@263.net 2 Department of Applied Mathematics, Xidian University, Xi an 710071, P.R. China.