Boolean graphs are Cohen-Macaulay

Transcription

1 Boolean graphs are Cohen-Macaulay A-Ming Liu and Tongsuo Wu Department of Mathematics, Shanghai Jiaotong University Abstract. For each Boolean graph B n, it is proved that both B n and its complement B n are vertex decomposable. It is also proved that B n is an unmixed graph, thus it is also Cohen-Macaulay. Key Words: Boolean graphs, vertex decomposable graphs, unmixed graphs, Cohen-Macaulay graphs 2010 AMS Classification: Primary: 13H10; 05E45; Secondary: 13F55; 05E40. Throughout this paper, let [n] = {1,2,...,n} and 2 [n] the power set of [n]. Recall from [10] that a finite Boolean graph, denoted by B n, is a graph defined on the vertex set 2 [n] \{[n], }, in which two vertices M and N are adjacent if M N =. Clearly, B n is also the zero-divisor graph of the finite Boolean ring n i=1 Z 2. Note that the complement B n of B n is the intersection graph of proper subsets of [n]. Note also that a finite or an infinite Boolean graph has a unique corresponding commutative ring, as well as a unique corresponding zero-divisor commutative semigroup, see [10] and [9] for the related discussions and background materials. In the following, we draw the graphs B 3 and B 4 : This research was supported by the National Natural Science Foundation of China (Grant No ). aming8809@163.com Corresponding author. tswu@sjtu.edu.cn

2 u 2 u 1 v 5 v 4 v 3 v 2 v 6 v 1 u 3 u 6 v 7 v 14 v 8 v9 v 13 u 4 u 5 v 10 v 11 v 12 Figure 1. The graphs B 3 and B 4 Due to the importance of Cohen-Macaulay rings in commutative algebra, it will always be interesting and important to discover new families of (sequentially) Cohen-Macaulay simplicial complexes, as well as new families of Cohen-Macaulay graphs or clutters. The purpose of this paper is to show that both B n and B n are vertex decomposable, thus sequentially Cohen-Macaulay. Furthermore, it is shown that all Boolean graphs B n are Cohen-Macaulay. This paper is organized as follows. In section 1, we recall some basic concepts, facts and related backgrounds from combinatorial commutative algebra. In section 2, we first prove that B n is an unmixed graph, and then check that B n is vertex decomposable. This shows that B n is a Cohen-Macaulay graph for each n. In section 3, we show that the complement B n is vertex decomposable, and study the properties of the Alexander dual complex of the clique complex of B n (and B n, respectively). In section 4, we have a preliminary study on the unmixed property of a blow up of a Boolean graph. Throughout the remainder, whenever there is no ambiguity, we use 421 to denote the vertex {1,2,4} of V(B 4 ). We assume n > n 1 > > 2 > 1, and use the pure lexicographic order on the vertices of V(B n ), e.g., 5421 > 5321 in V(B 6 ). Throughout, an empty graph is a graph without edges. 1 Preliminaries In this part, we recall some definitions and results from combinatorial commutative algebra. For more concepts and more details without mention, one can refer to the recent monographs, [20, 7]. Recall that a simplicial complex over the set [n] is a subset of the power set 2 [n] of [n], such that is hereditary and, all singletons {v} (1 v n) are in. An element of is called a face, while a singleton {v} is called a vertex of the complex and it will

3 be denoted by v. A face F of a simplicial complex is called a facet if no face contains it as a proper subset. The set of facets of will be denoted by F( ). If F( ) = 1, then is called a simplex; if all the facets of have a same cardinality, then is said to be pure or unmixed. For some faces F 1,...,F r of, a subcomplex F 1,...,F r generated by F 1,...,F r is defined by F 1,...,F r = {F F F i for some 1 i r}. Recall that the important subcomplexes deletion and a link of a simplicial complex, are defined as follows: \x = {F x F}, lk (x) = {F x F, F {x} }. Recall the following inductive definition of vertex decomposable simplicial complex: Definition 1.1. Let be a simplicial complex over [n]. If one of the following inductive condition is satisfied, then is called vertex decomposable: (1) is a simplex, or (2) There is a vertex v such that the following requirements are fulfilled (α) Both \v and lk (v) are vertex decomposable. (β) No facet of lk (v) is a facet of \v, or equivalently, \v = {F v F F( )}. Such a vertex v satisfying conditions (α) and (β) is called a shedding vertex of. If v only satisfies the condition (β), then we call it a weak shedding vertex. Recall that a simplicial complex is called (nonpure) shellable, if there is a shelling order F 1,...,F r of all facets, such that for each 1 i r 1, the simplicial complex F 1,...,F i F i+1 is pure of dimension dimf i+1 1. Note that F 1,...,F r of all facets of is a shelling order if and only if for each pair (i,j) with 1 i < j r, there exists an integer k with 1 k < j, such that both F j \ F k = 1 and F j \ F k F j \ F i holds; if assume further F k \ F j F i, then F 1,...,F r is called a strong shelling order. By [5], a nonpure simplicial complex is called strongly shellable, provided that there exists a strong shelling order in F( ). Recall that a simplicial complex is called pure (strongly) shellable if it is pure, and (strongly) shellable in the afore mentioned sense. Now let us recall some classical and recent results related to vertex decomposable simplicial complexes. First, recall the following implications for a nonpure simplicial complex: vertex decomposable = shellable = strongly shellable. Recall the following implications for a simplicial complex: matroid = vertex decomposable and pure = pure shellable

4 = Cohen Macaulay = pure. Note that in[5], counterexamples are given to show that there is no implication between the concepts vertex decomposable and strongly shellable. Note also that by [5], if is strongly shellable, then both I and I( ) have linear quotients, where = {[n]\f F } and is called the Alexander dual complex of, I is the Stanley-Reisner ideal of, while I( ) = x F : F F( ) and is called the facet ideal of. The following result has interesting application in considering vertex decomposable graphs (for details, see Corollary 1.3): Proposition 1.2. Let 1 and 2 be complexes over [n] = [1,n] and [n + 1,n + m] respectively. Then the join simplicial complex 1 2 is vertex decomposable if and only if both simplicial complexes 1 and 2 are vertex decomposable. Note that a similar result holds for each of the following properties: shifted, strongly shellable, shellable, Cohen-Macaulay, sequentially Cohen-Macaulay. In [14], the authors define the concept of vertex splittable ideal and show that a simplicial complex is vertex decomposable if and only if I is vertex splittable ideal. Also, it is proved that the edge ideal of a graph is vertex splittable if and only if it has a linear resolution. For a graph G, recall that the edge ideal I(G) is identical with the Stanley-Reisner ideal I G of the clique complex G of the complement G. Recall that a graph G is called vertex decomposable (Cohen-Macaulay, or shellable, or unmixed, respectively) if the simplicial complex G has the corresponding property. In the remaining part of this section, we give a brief survey on some results related to vertex decomposable graphs. First, by Proposition 1.2, we have Corollary 1.3. ([21, Lemma 20]) A graph G is vertex decomposable if and only if all connected components of G are vertex decomposable. Recall that an independent vertex set S of a graph G is a subset of V(G) such that the subgraph induced on S contains no edges. For a vertex v in a graph G, let N G [v] = N G (v) {v}, the closed neighborhood of v in G. The following is a translation of vertex decomposable of a simplicial complex in the language of a graph: Definition 1.4. ([21,Lemma 4]) A graph G is vertex decomposable if either it has no edges, or else has some vertex x such that we have as follows: (1) Both G\N G [x] and G\x are vertex decomposable. (2) For every independent set S in G \ N G [x], there exists a vertex y N G (x) such that S {y} is independent in G\x.

5 The following result tells a way for enlarging the class of vertex decomposable graphs: Proposition 1.5. ([15,Proposition2.3]) Let G 1,...,G n be finite graphs, and assume V(G i ) 2, V(G i ) V(G j ) = for all i j. For a graph G with n vertices x i, let G(G i 1 i n) be a graph obtained by attaching x i with a vertex in G i. (x i is called a gluing vertex.) If the graphs G 1,...,G n are vertex decomposableand each gluing vertex x i is a shedding vertex of G i, then G(G i 1 i n) is vertex decomposable. Chordal graphs are an important class of vertex decomposable graphs. Recall that a graph is called chordal, if all cycles of four or more vertices have a chord, which is an edge that is not part of thecycle but connects two vertices of thecycle. Adam VanTuyl, Rafael H. Villarreal in [19] proved that all chordal graphs are (nonpure) shellable. Woodroofe in [21] proved that a chordal graph is further vertex decomposable and, studied chordal clutters in [22]. Recall the following theorem, which contains important results in the algebraic combinatorics of a chordal graph: Theorem 1.6. Let G be a graph and G the complement of G. Then the following three conditions are equivalent: (1) G is chordal. (2) (Fröberg [4]) The edge ideal I(G) of G has a linear minimal free resolution. (3) (Lyubeznik [11]) The cover ideal I c (G) is Cohen-Macaulay, where I c (G) is the edge ideal of the clutter consisting of all minimal vertex covers of G. Recall that a simplicial vertex of a graph is a vertex v such that the neighbourhood N(v) is a clique. Recall the following main theorem (by Dirac) characterizing chordal graphs: Theorem 1.7. A graph G is chordal if and only if every induced subgraph of G has a simplicial vertex. Recall the following remarkable result, which is related to Theorem 1.6 and holds for the edge ideal of a graph: Proposition 1.8. [8, Theorem 3.2] Let I be a monomial ideal generated by degree 2. Then the following conditions are equivalent: (1) I has a linear minimal free resolution. (2) I has linear quotients. (3) I s has a linear minimal free resolution, for each s 1.

6 2 Boolean graphs are Cohen-Macaulay Recall that a vertex cover C of a graph G is a subset of the vertex set V(G) such that C {i, j}, for all{i, j} E(G). A vertex cover is also called edge-dominated set of G, while the edge-dominated number of G is the least of cardinalities of all minimal vertex covers. Recall that a graph G is said to be unmixed, if all minimal vertex covers of G have the same cardinality. An unmixed graph is also often called well-covered. It is known that a graph G is unmixed if and only if the clique simplicial complex of G is pure, while a Cohen-Macaulay graph is always unmixed. Recall also that C is a minimal vertex cover if and only if V(G)\C is a maximal independent vertex set of G. Now we give the first main result of this paper: Theorem 2.1. The Boolean graph B n is unmixed for all n 1. Proof: Let G = B n. We give a proof by considering the maximal independent vertex set. Note that a vertex subset V 0 is a minimal vertex cover of G if and only if V0 c is a maximal independent vertex set of G, where V0 c = V(G)\V 0. A subset = {b 1,b 2,...,b t } of V(G) is an independent vertex set if and only if b i b j holds for any distinct b i,b j in. In the following, we proceed to prove that all maximal independent vertex set of V(G) have the same cardinality of 2 n 1 1 and for any b i V(G), only one of {b i,b c i} is in, where b c i = [n]\b i. As the cardinality of the vertex set V(G) is 2 n 2, observe that for any vertex b in V(G), the complement b c is also in V(G) and this is a one to one correspondence, we can decompose V(G) into two disjoint parts {b 1,b 2,...,b 2 n 1 1} and {b c 1,bc 2,...,bc 2 n 1 1 }. Thus, the cardinality of is not larger than 2 n 1 1. For any independent vertex set of G with < 2 n 1 1, we claim that more vertices can be added to and obtain a larger independent vertex set, until the cardinality reaches 2 n 1 1. Indeed, for any independent vertex set = {b 1,,b t } of G with = t < 2 n 1 1, clearly there exists b t+1 such that {b t+1,b c t+1 } =. If neither {b t+1} nor {b c t+1 } is independent vertex set, then there are distinct b i and b j in, such that both b i b t+1 = and b j b c t+1 = holds. Then b i b c t+1 and b j b t+1, contradict to b i b j. Then the above claim holds. This shows that the graph G is unmixed. By the proof, it is clear that the edge ideal of the graph B n has height 2 n 1 1, which is half of the number V(B n ). We remark that the authors of [2] considered an unmixed graph without isolated vertex with height height(i(g)) = V(G) /2, and they gave Cohen-Macaulay criteria for graphs with the property. Note that B n is not chordal when n 4, since the subgraph induced on the vertex set {1,2,32,41} is a cycle And Boolean graph B n is not matroidal for

7 any n 3. In fact, the clique complex of the complement B n is far from being a matroid in general, as the following example shows: Example 2.2. The clique complex of the complement B 3 is = {1,21,31},{2,21,32},{3,31,32},{21,31,32}. Note that the vertex set of is 2 [3] \ {[3], }, so if we take a subset of it as W = {1,2,21,31}, then W = {1,21,31},{2,21}. Since the induced subcomplex W is not pure, by [18,Proposition3.1], the complex is not a matroid. Next we want to prove that all Boolean graphs are vertex decomposable. In order to do so, recall that a vertex u in a graph G is said to have a whisker, if there is an end vertex adjacent to u ([20, Definition ]). We observe the following: Lemma 2.3. Any vertex in a graph G with whiskers is a weak shedding vertex. Proof: Let d be an end vertex adjacent to u. Clearly, u G\N G [u], d N G (u) and, any independent set of G\N G [u] can be extended to a larger independent vertex set D {d} in G\u. Thus u is a weak shedding vertex of G. Note that each of the vertices 1,...,n has a whisker in B n. If let G 1 = B n \1\2\ \n, then each ji has a whisker in the graph G 1 for all 1 i < j n; If let G 2 = G 1 \21\31\ \nn 1, then every kji (n k > j > i 1) has a whisker in the graph G 2 and continue in this way. Thus in order to show that B n is vertex decomposable, we will choose 1,..., n; 21,..., nn 1; 321,... as a sequence of weak shedding vertices. Note also that G\N G [v] G\v. In the following, we present a weak shedding vertex order to prove the second main result of this paper: Theorem 2.4. For any n 1, let G = B n be the Boolean graph. Then G is vertex decomposable, hence Cohen-Macaulay. Proof: Let G n+1 = B n,g n = G n+1 \n,g n 1 = G n \n 1,...,G 1 = G 2 \1. (1)

8 Note that G \ N G [n] = {A {n} A V(B n 1 )}, and that it is a empty graph, hence vertex decomposable by Corollary 1.3. Note also G n \N Gn [n 1] = G\N G [n 1]\n, G n 1 \N Gn 1 [n 2] = G\N G [n 2]\n\n 1,... G 2 \N G2 [1] = G\N G [1]\n\n 1\ \2, thus they are all empty graphs (i.e., graphs without an edge) and hence, vertex decomposable. Note that each of {i} is a weak shedding vertex of the graph G i+1, thus by Definition 1.4, the graph G is vertex decomposable if and only if the subgraph G 1 is vertex decomposable. In order to see that the graph G 1 is vertex decomposable, let G nn 1 = G 1 \nn 1, G nn 2 = G nn 1 \nn 2,..., G n1 = G n2 \n1 G n 1n 2 = G n1 \n 1n 2,..., G n 11 = G n 12 \n 11..., G 32 = G 41 \32, G 31 = G 32 \31, G 21 = G 31 \21 (2) Note that each vertex ij is a weak shedding vertex of the graph in front of it. Now consider the corresponding H \ N H [ij]. Let H = G 1 \ N G1 [nn 1]. We have H = (H 1 \nn 1) ( i 2 H 2i ), where H 1 = {A V(B n ) A 2,n A}, H 2i = {A V(B n ) A = i,n 1 A, n A}. Note that both H 1 and i 2 H 2i are empty graphs, and that each vertex of H 22 has a whisker in the graph H (surely, in H 1 ), thus any linear order of vertices of H 22 is a weak shedding order of H. Then we delete H 22, and consider H \H 22. Surely, each vertex of H 23 has a whisker in H\H 22 (again, in H 1 ), thus we delete H 23 fromh\h 22, and continue such a deletion, until we obtain a forest. This shows that H is vertex decomposable. In a similar way, we see that each of G n 1i \N Gn 1i [n 1i 1] is vertex decomposable. Thus the graph G 1 is vertex decomposable if and only if the graph G 21 is vertex decomposable. Now assume n 6. In order to see that G 21 is vertex decomposable, the next step is to consider the sequential deletions: G nn 1n 2 =: G 21 \nn 1n 2,..., G 321 =: G 421 \321 (3) and the related H \ N H [ijk]. In this process, we always take advantage of the vertices with whiskers. For the graph L = G 21, let H = L\N L [nn 1n 2]. Then V(H) = H 1 ( i 3 (H 2i H 3i )),

9 where H 1 = {A V(B n ) A 3,n A, A nn 1n 2}. H 2i = {A V(B n ) A = i,n 1 A, n A}. H 3i = {A V(B n ) A = i,n 2 A, A {n,n 1} = }. Note that the subgraphs induced on each H i is empty, and that each vertex of H 33 H 23 has a whisker in H, with an adjacent end vertex in H 1. Thus in order to see that H is vertex decomposable, we delete H 33 and H 23 from H, then going on to consider the vertices with whiskers. In this way, we show that the graph G 21 is vertex decomposable if and only if G 321 is vertex decomposable. We continue this process for both related H \ u and H \ N H [u], until it reaches a forest, which is chordal thus vertex decomposable. In this way, due to the fact that the related H\N H [u] always has enough weak shedding vertices (actually, vertices which have whiskers), at the end we are able to prove that B n is actually vertex decomposable. Finally, it is known that vertex decomposable implies shellability, while pure shellabilityimpliescohen-macaulayness. ThusbyTheorem2.1, thegraphb n iscohen-macaulay. We remark that very detailed check has been taken for 3 n < 6, showing that both B n and B n are vertex decomposable for each n < 6. This is further verified for n = 6,7 respectively by using a commonly used computer, but the present algorithm can not work for n = 8. In the next section, we will prove that the graph B n is also vertex decomposable. 3 The complement graph B n AlthoughB n hasmoreedgescomparedwithb n,itisnotchordalforn 4,forexamplethe subgraphinducedonthevertexsubset{21,32,43,41}isthe4cycle without any chord. Note that the graph B n is not matroidal for any n 3. In fact, the clique complex of B n is not pure for each n 3. Note that the complement B n is the intersection graph on the proper subsets of [n], thus it is relatively easy to operate with; also, the complement graph has some nice properties, as the following third main result of this paper reveals: Theorem 3.1. For any n 1, the complement B n is vertex decomposable. Proof: Let G = B n. We prove the claim by induction on the number of vertices of the graph G. If 1 n 3, then the result follows easily. In the following, we assume n 4.

10 Let A be any nonempty proper subset of [2,n] and consider the vertex v = {1} A of G. Note that {i} is a simplicial vertex of G for any i in [n]. Note also N G [v] = {w V(G) v w }, thus G\N G [v] is the complement of the Boolean graph on vertex set [n]\v. By induction hypothesis, G\N G [v] is vertex decomposable. Let v 1,...,v t be any order of neighbourhoods of {1} in the intersection graph G, where t = 2 n 1 1. Let G 0 = G and G i = G \ {v 1,...,v i } for 1 i t. Then it is straightforward to check that each v i is a weak shedding vertex of the graph G i 1, for all 1 i t. For each i, we have G i \ N Gi [v i+1 ] = G \ N G [v i+1 ], thus G i \ N Gi [v i+1 ] is vertex decomposable. This shows that G is vertex decomposable if and only if G t is vertex decomposable. Note that G t is the complement of the Boolean graph on the vertex set [2,n], thus G t is vertex decomposable by induction hypothesis. By mathematical induction hypothesis, this shows that the complement B n is indeed vertex decomposable. Recall that for 1 s dim 1, the s th skeleton complex (0,s) of consists of all faces F of with F s +1. Recall that pure s th skeleton (s,s) is generated by all faces of of dimension s. Recall that all skeletons and pure skeletons of a shellable complex are shellable. Note that each skeleton complex (0,s) of is vertex decomposable if is vertex decomposable by [22, Lemma 3.10], thus we have the following Corollary 3.2. Let G be either the Boolean graph B n or its complement B n, and let be the clique complex of the graph G. Then (1) Each skeleton complex (0,s) of is vertex decomposable. (2) Each pure skeleton complex (s,s) of is pure shellable, thus Cohen-Macaulay. Recall that a 2-flag complex is a complex such that each minimal nonface of has cardinality 2. Recall that a complex is a 2-flag complex if and only if is a clique complex of a graph ([7, Proposition 9.1.3]). Note that the Alexander dual of a 2-flag complex is pure of dimension V( ) 2. Let G = B 3 be the graph labeled in Figure 1, and let be the clique complex of G. Then the Stanley-Reisner ideal of is I = x 1 x 6,x 1 x 2,x 1 x 3,x 2 x 3,x 2 x 5,x 3 x 4. Using CoCoA, we obtain the minimal free resolution of I as follows 0 R( 4) 3 R( 3) 8 R( 2) 6 I 0. Note that I has 2-linear resolution, by Eagon-Reiner s theorem, I is Cohen-Macaulay. Further more, it follows from [7, Lemma 1.5.3] that the Stanley-Reisner ideal of is I = x 2 x 3 x 6,x 2 x 2 x 3,x 1 x 3 x 5,x 1 x 2 x 4, which has the following 3-linear resolution 0 R( 4) 3 R( 3) 4 I 0.

11 In fact, the monomial ideal I has linear quotients and this follows from the following computations x 2 x 3 x 6 : x 1 x 2 x 3 = x 6, x 2 x 3 x 6 : x 1 x 2 x 3 : x 1 x 3 x 5 = x 2, x 2 x 3 x 6 : x 1 x 2 x 3,x 1 x 3 x 5 : x 1 x 2 x 4 = x 3. Hence by [7, Proposition 8.2.5], the simplicial complex is pure shellable. A direct verification shows 541,654,642,653 is a shelling order of. For n 4, let G be the Boolean graph B n or its complement B n. By Theorem 1.6, the edge ideal I(G) does not have a linear minimal free resolution, and it implies that I(G) does not have linear quotients by Proposition 1.8. In the following, we provide another direct proof to this fact: Proposition 3.3. Let n 4. Let G be either the Boolean graph B n or its complement B n, and let be the clique complex of the graph G. Then the Alexander dual complex is not shellable. Proof: (1) Let be the clique complex of B n. Clearly, F( ) = {V \{a,b} a V(B n ),b V(B n ), a b = }, where V = 2 [n] \{[n], }. Since n 4, we can choose a,b V(B n ), say, a = {1,2},b = {2,3}, such that u v, for allu v c,u,v {a,b,a c,b c }. Let F i = V \{a,a c }, F k = V \{b,b c }. If assume that is shellable, we can assume F i < F k in the shelling of facets. Then by definition, there exist 1 j < k and x F k, such that F i F k F j F k = F k \{x}. If let F j = V \{c,d}, then we have the following two facts: (i) F i F k F j F k = F k \{x}, i.e., V \{a,a c,b,b c } V \{c,d,b,b c } = V \{x,b,b c }. It follows that {x,b,b c } = {c,d,b,b c } and {c,d,b,b c } {a,a c,b,b c } (ii) x / F i and x / F j, in which F i = V \{a,a c } and F j = V \{c,d}. By (ii), x / F i = V \ {a,a c }, thus x {a,a c }. Assume x = a, and assume further c = a by fact (i). Then d {b,b c } and this follows from {x,b,b c } = {c,d,b,b c }. But then c d by the choice of a and b, contradicting assumption that F j is a facet of. This contradiction shows that is not shellable, thus the edge ideal I(B n ) does not have linear quotients. (2) As for the clique complex of B n, clearly F( ) = {V \{a,b} a V(B n ),b V(B n ), a b }.

12 As n 4, we can take a,b,c,d V(B) with and consider a b =, a c = = a d, b c = = b d, F i = V \{a,b}, F k = V \{c,d}. If isshellable, wecanassumef i < F k intheshelling offacets. Thenasimilarargument leads to a contradiction. The details will be omitted. We finish this section by posing the following unsettled questions: Question 3.4. Let G be either the Boolean graph B n or its complement B n, and let be the clique complex of the graph G. (1) Are the pure skeleton complexes (s,s) of vertex decomposable? (2) Is strongly shellable? 4 Blow up of Boolean graphs and unmixed property In [13], an interesting concept expanding simplicial complex (of a simplicial complex) is introduced and studied, and the authors of [13, 17] show that expanding a simplicial complex keeps a lot of properties unchanged, e.g., chordal, vertex decomposable, shellable and Cohen-Macaulay. In particular, this induces the concept of expansion for a graph. Note that Theorem 1.7 also can be applied to verify that a graph expanding keeps the chordal property, a nice result first discovered by the authors in [13]. For a graph, there is another technique related to expanding, which is called a blow up. Recall that to get a finite blow up graph G T of a finite graph G, it is enough to replace every vertex v of G by a finite nonempty set T v to get a possibly new and larger graph G T, where T v 1. The induced subgraph of G T on T v is an empty graph, while for distinct vertices u,v of G, u is adjacent to v in G if and only if each vertex of T u is adjacent to all vertices of T v in G T, see [12, 16] for details. If we further let T v be a complete graph, then G T becomes an expanding graph G E of G. For a graph G, let G be its complement. Then the following observation holds: A graph H is a blow up of a graph G if and only if H is an expanding graph of the graph G. Consider the graph G with V(G) = {v 1,v 2,v 3 } and E(G) = {v 1 v 2,v 2 v 3 }. Let T v1 = {v 11,v 12 },T v2 = {v 21,v 22 } and T v3 = {v 3 }. We illustrate the observation by the following diagrams:

13 v 11 v 1 v 2 v 3 v 21 v 3 v 1 v 3 v 2 v 11 v 3 v 21 v 12 v 22 v 12 v 22 G G T G G T Figure 2. Note that in a non-empty Cohen-Macaulay bipartite graph, there exists an end vertex. Thus graph blow up does not keep Cohen-Macaulay property unchanged, since the graph K 1,1 is clearly Cohen-Macaulay while its blow up K 2,2 is not. In a similar way, it is easy to see that graph blow up does not keep anyone of the following properties of a graph: chordal, vertex decomposable, shellable. In general, a blow up of a Boolean graph is not unmixed. For example, the complete bipartite graph K m,n is a blow up of the Boolean graph B 2 and, it is unmixed if and only if m = n. Claim 4.1. Let G T be a finite blow up of the graph B n. For any vertex u V(B n ), let x u = T u. Then (1) For n = 2, G T is unmixed if and only if G T = K m,m for some m 1. (2) For n = 3, G T is unmixed if and only if x i = x jk for all distinct {i,j,k} = {1,2,3}. (3) For n = 4, G T is unmixed if and only if the following seven equalities hold true: x i = x jkl, x ij = x kl, for all distinct {i,j,k,l} = [4]. Proof: First, note the following observations: If a graph G contains a clique K of r vertices, then any minimal vertex cover of G contains at least r 1 vertices of K; also, G T has a minimal vertex cover which contains n i=1 T i. (i) For n = 2, the result is clear. (ii) For n = 3, consider the following four minimal vertex covers of G T : T 1 T 2 T 3, T i T j T {i,j} (1 i < j 3). Clearly, G T is unmixed if and only if the vector (x 1,x 2,x 3,x 11,x 22,x 33 ) is the positive solution in Z 6 of the following system of equations: x 1 +x 2 +x 12 = x 1 +x 3 +x 13 x 1 +x 2 +x 12 = x 2 +x 3 +x 23. (1) x 1 +x 2 +x 12 = x 1 +x 2 +x 3 Then the result follows. In particular, it shows that the Boolean graph B 3 is unmixed.

14 (iii) For n = 4, note that ( 4 i=1 T i) ( 3 i=1 T u i ) is a minimal vertex cover of G T, where u 1,u 2,u 3 are taken from distinct {ij,kl} with {i,j,k,l} = [4] respectively. There are totally eight such minimal vertex covers of G T. Also, there are four others, and one representative of them is ( 4 i=2t i ) T 234 T 23 T 24 T 34. Like the n = 3 case, it follows from the system of linear equations that x i = x jkl holds for all {i,j,k,l} = [4]. Then it follows easily x ij = x kl. The converse holds clearly. In particular, the results imply that Boolean graph B i (1 i 4) is unmixed. Remark 4.2. The facet ideal of B 3 is I(B 3 ) = x 1 x 2,x 1 x 3,x 2 x 3,x 1 x 6,x 2 x 5,x 3 x 4.We calculate the primary decomposition of I(B 3 ) by CoCoA as follows: I(B 3 ) = x 1,x 2,x 3 x 1,x 2,x 4 x 1,x 3,x 5 x 2,x 3,x 6. Remark 4.3. The facet ideal of B 4 is I(B 4 ) = x 1 x 2,x 1 x 3,x 1 x 4,x 1 x 8,x 1 x 9,x 1 x 10,x 1 x 14, x 2 x 3,x 2 x 4,x 2 x 6,x 2 x 7,x 2 x 10,x 2 x 13,x 3 x 4,x 3 x 5,x 3 x 7,x 3 x 9,x 3 x 12,x 4 x 5,x 4 x 6,x 4 x 8,x 4 x 11, x 5 x 10,x 6 x 9,x 7 x 8. We calculate the primary decomposition of I(B 4 ) by CoCoA as follows: I(B 4 ) = x 1,x 2,x 3,x 4,x 7,x 9,x 10 x 1,x 2,x 4,x 5,x 7,x 9,x 12 x 1,x 2,x 3,x 5,x 6,x 8,x 11 x 1,x 2,x 4,x 5,x 6,x 8,x 9 x 1,x 2,x 3,x 4,x 5,x 6,x 7 x 1,x 2,x 3,x 4,x 6,x 7,x 10 x 1,x 2,x 3, x 4,x 6,x 8,x 10 x 1,x 3,x 4,x 6,x 7,x 10,x 13 x 1,x 2,x 3,x 4,x 8,x 9,x 10 x 2,x 3,x 4,x 8,x 9, x 10,x 14. These remarks show another way for illustrating Theorem 2.1, as well as Claim 4.1. When n is large, things will become complicated. But a similar careful discussion shows thattheunmixedness oftheblowupg T ofthebooleangraphb n (n = 5,6,7,respectively) amounts to the solving of a system of linear equations with indeterminate labeled properly according to their position in the layers. The above claim shows that graph blow up is a goodconcept for studying the unmixedness property of graphs. We can even generalize it a little to obtain a finite generalized blow up G S of a finite graph G explained in what follows. For every vertex v of G, let S v be a disjoint union of S 1v with S 2v, in which v S 1v. Replace v by S v to get a possibly new and larger graph G S : For any u V(G), the induced subgraph of G S on each S u is a empty graph, while for distinct vertices u,v of G, u is adjacent to v in G if and only if each vertex of S 1u is adjacent to all vertices of S v and each vertex of S 1v is adjacent to all vertices of S u. Note that whenever none of S 2u,S 2v is empty, no vertices in S 2u is adjacent to a vertex in S 2v. By the definition, each blow up is a generalized blow up, of a graph; but the converse is clearly not true. Generalized blow up occur naturally when we consider deleting a vertex from the graph B n, as the following example shows.

15 Claim 4.4. B n \n\12... n 1 is a generalized blow up of B n 1. Proof: Clearly, the vertex 12...n 1 is isolated in the graph B n \n. Let G = B n \ n \ n 1. Then the vertex set of V(G) splits into two parts, {A,A {n}}, for all A V(B n 1 ). Thus if we add A {n} to the vertex A as the second part, then clearly, G is a generalized blow up of B n 1, where for each vertex v of B n 1, we have S 1v = S 2v = 1. We finish the paper with an easy observation on the unmixedness of a generalized blow up of the graph G = B 2. Claim 4.5. Let G S be a generalized blow up of the graph G = B 2. Then G S is unmixed if and only if either G S = K m,m or S 1{1} = S 2{2}, S 1{2} = S 2{1}. Proof: Assume that G S is a generalized blow up of the graph G = B 2, but not a blow up of B 2. Note that S 1{1} S 1{2}, S 1{2} S 2{2} and S 1{1} S 2{1} are minimal vertex covers of the graph G S. Thus if G S is unmixed, then we have S 1{1} + S 1{2} = S 1{2} + S 2{2} = S 1{1} + S 2{1} hence S 1{1} = S 2{2}, S 1{2} = S 2{1}. The converse holds clearly. ACKNOWLEDGMENTS. The authors express their sincere thanks to the reviewer for the helpful comments and suggestions on an earlier version of this paper. References [1] Bruns, W., Herzog, J. (1998).Cohen-Macaulay Rings. Revised version. Cambridge University Press. [2] Crupi, M., Rinaldo, G., and Terai, N. (2011). Cohen-Macaulay edge ideals whose height is half of the number of vertices. Nagoya Math. J. 201 : arxiv: v1. [3] Eisenbud, D. (2004).Commutative Algebra with a View Toward Algebraic Geometry. Springer Science, Business Media, Inc. [4] Fröberg, R. (1990). On Stanley-Reisner rings. Banach Center Publ. 26(2) : [5] Guo, J., Shen, Y.H., Wu, T.S. (2016). Strongly shellable of simplicial complexes. arxiv : (22 pages). [6] Guo, J. Wu, T.S., Ye, M. (2015). Complemented graphs and blow-ups of Boolean graphs, with applications to co-maximal ideal graphs. Filomat 29(4) :

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