Buchsbaumness of the second powers of edge ideals
BBUCHSBAUMNESS OF THE SECOND POWERS OF EDGEIDEALS
DO TRONG HOANG AND TRAN NAM TRUNG
Dedicated to Professor Le Tuan Hoa on the occasion of his sixtieth birthday
Abstract.
We graph-theoretically characterize the class of graphs G such that I ( G ) are Buchsbaum. Introduction
Throughout this paper let G = ( V ( G ) , E ( G )) be a finite simple graph withoutisolated vertices. An independent set in G is a set of vertices no two of which areadjacent to each other. The size of the largest independent set, denoted by α ( G ), iscalled the independence number of G . A graph is called well-covered if every maximalindependent set has the same size. A well-covered graph G is a member of the class W if the remove any vertex of G leaves a well-covered graph with the same independencenumber as G (see e.g. [14]).Let R = K [ x , . . . , x n ] be a polynomial ring of n variables over a given field K .Let G be a simple graph on the vertex set V ( G ) = { x , . . . , x n } . We associate to thegraph G a quadratic squarefree monomial ideal I ( G ) = ( x i x j | x i x j ∈ E ( G )) ⊆ R, which is called the edge ideal of G . We say that G is Cohen-Macaulay (resp.
Goren-stein ) if I ( G ) is a Cohen-Macaulay (resp. Gorenstein) ideal. It is known that G iswell-covered whenever it is Cohen-Macaulay (see e.g. [20, Proposition 6 . . G is in W whenever it is Gorenstein (see e.g. [10, Lemma 2.5]). It is a wide openproblem to characterize graph-theoretically the Cohen-Macaulay (resp. Gorenstein)graphs. This problem was considered for certain classes of graphs (see [5, 6, 9, 10]).Generally, we cannot read off the Cohen-Macaulay and Gorenstein properties of G just from its structure because these properties in fact depend on the characteristicof the base field K (see [20, Exercise 5 . .
31] and [10, Proposition 2.1]).If we move on to the higher powers of I ( G ), then we can graph-theoretically charac-terize G such that I ( G ) m is Cohen-Macaulay (or Buchsbaum, or generalized Cohen-Macaulay) for some m (cid:62) m (cid:62)
1) (see [4, 15, 19]). For the second power,we proved that I ( G ) is Cohen-Macaulay if and only if G is a triangle-free graph in Mathematics Subject Classification.
Key words and phrases. W graphs, Edge ideal, Cohen-Macaulay, Gorenstein, Buchsbaum. a r X i v : . [ m a t h . C O ] J un (see [10]). As a consequence one can easily answer the question when I ( G ) isgeneralized Cohen-Macaulay (see Theorem 1.1).The remaining problem is to characterize G such that I ( G ) is Buchsbaum. Fora vertex v of G , let G v be the induced subgraph G \ ( { v } ∪ N G ( v )) of G . In thispaper, we will call G a locally triangle-free graph if G v is triangle-free for any vertex v of G . It is worth mentioning that [16, Theorem 2 .
1] and [10, Theorem 4 .
4] suggestthat G may be a locally triangle-free Gorenstein graph if I ( G ) is Buchsbaum. So itis natural to characterize such graphs. Note that they are in W by [10, Proposition3 . W (see Theorem 3 . C cn be the complement ofthe cycle C n of length n . Then, Theorem 1 (Theorem 4.1) . Let G be a locally triangle-free graph. Then G isGorenstein if and only if G is either a triangle-free graph in W , or G is isomorphicto one of C cn ( n (cid:62) ), Q , Q , P or P (see Figure 1). a bca a b b c c a bca a b b c c a c b Q Q a bcyx c c a b b a bc dxy a b ztc c P P Figure 1.
For graphs Q , Q , P and P .Now let B n ( n (cid:62)
4) be the graph with the edge set { x i x j | (cid:54) i + 1 < j (cid:54) n } . Usingthis theorem we can characterize graphs G such that I ( G ) are Buchsbaum. heorem 2 (Theorem 4.3) . Let G be a graph. Then I ( G ) is Buchsbaum if andonly if G is either a triangle-free graph in W , or isomorphic to one of K n ( n (cid:62) ), C cn ( n (cid:62) ), B n ( n (cid:62) , Q , Q , P or P . The paper is organized as follows. In Section 1 we recall some basic notations, andterminologies from Graph theory. In Section 2 we investigate the local structure oflocally triangle-free graphs in W . Section 3 is devoted to classifying the class of locallytriangle-free graphs in W . In the last section we graph-theoretically characterizegraphs G for which I ( G ) are Buchbaum.1. Preliminaries
Let R := K [ x , . . . , x n ] be the polynomial ring over a field K and m := ( x , . . . , x n ) R the maximal homogeneous ideal of R . Let H i m ( R/I ) denote the i -th local cohomologymodule of R/I with respect to m . A residue class ring R/I is called a generalizedCohen-Macaulay (resp.
Buchsbaum ) ring if H i m ( R/I ) has finite length (resp. thecanonical map Ext iR ( R/ m , S/I ) → H i m ( R/I )is surjective) for all i < dim(
R/I ) (see [2, 18]).First we address the problem of characterizing graphs G such that I ( G ) are gen-eralized Cohen-Macaulay. Theorem 1.1.
Let G be a simple graph. Then, I ( G ) is generalized Cohen-Macaulayif and only if:(1) G is well-covered;(2) Every nontrivial component of G v is a triangle-free graph in W for any vertex v of G .Proof. Follow from [8, Corollaries 2 . .
10] and [10, Theorem 4 . (cid:3) Next we recall some terminologies from Graph theory. Let G be a simple graphon the vertex set V ( G ) and the edge set E ( G ). An edge e ∈ E ( G ) connecting twovertices u and v will also be written as uv (or vu ). In this case, we say that u and v are adjacent. For a subset S of V ( G ), the neighborhood of S in G is the set N G ( S ) := { v ∈ V ( G ) \ S | uv ∈ E ( G ) for some u ∈ S } , and the close neighborhood of S in G is N G [ S ] := S ∪ N G ( S ). We denote by G [ S ] theinduced subgraph of G on the vertex set S , denote G \ S by G [ V \ S ], and denote G S by G \ N G [ S ]. For an edge ab of G , we write G ab stands for G { a,b } . The numberdeg G ( v ) := | N G ( v ) | is called the degree of v in G . Lemma 1.2. ([3, Lemma 1]) If G is a well-covered graph and S is an independentset of G , then G S is well-covered. Moreover, α ( G S ) = α ( G ) − | S | . Lemma 1.3. ([7, Lemma 7])
Let G be a graph in W and S an independent set of G .If | S | < α ( G ) , then G S is in W . In particular, G S has no isolated vertices. graph G is called bipartite if its vertex set can be partitioned into subsets A and B so that every edge has one end in A and one end in B ; such a partition is called a bipartition of the graph G and denoted by ( A, B ). It is well known that G is bipartiteif and only if G has no odd cycles (see e.g. [1, Theorem 4.7]). Lemma 1.4. ([7, Lemma 12]) If G is a bipartite graph in W , then G consists ofdisjoint edges. An ( s − G is a sequence of its edges u u , u u , . . . , u s − u s and will bedenoted by u . . . u s . An s -cycle ( s (cid:62)
3) is a path u . . . u s u , where u , . . . , u s aredistinct vertices; it will be denoted by ( u . . . u s ). A 3-cycle is called a triangle . Agraph G is called triangle-free if it has no triangles; and G is a locally triangle-freegraph if G v is triangle-free for every vertex v . Lemma 1.5. ([7, Lemma 10])
Let G be a locally triangle-free graph in W and let ab be an edge of G . Then, G ab is either empty or well-covered with α ( G ab ) = α ( G ) − . Let G = ( V , E ) and G = ( V , E ) be two disjoint graphs, i.e. V ∩ V = ∅ . Then,the join of G and G , denoted by G ∗ G , is the graph with the vertex set V ∪ V andthe edge set E ∪ E ∪ { uv | u ∈ V , v ∈ V } . Note that G and G are two inducedsubgraphs of G ∗ G .A simplicial complex ∆ on the vertex set V is a collection of subsets of V closedunder taking subsets; that is, if σ ∈ ∆ and τ ⊆ σ then τ ∈ ∆. For a graph G , let∆( G ) be the set of independent sets of G . Then ∆( G ) is a simplicial complex whichis the so-called independence complex of G .The condition that G is a join of its two proper subgraphs can be represented viathe connectivity of ∆( G ). Lemma 1.6.
A graph G is a join of its two proper subgraphs if and only if ∆( G ) isdisconnected.Proof. Assume that G = G ∗ G , where G and G are two non-empty graphs. Then,∆( G ) = ∆( G ) ∪ ∆( G ) . Since ∆( G ) ∩ ∆( G ) = {∅} , we have ∆( G ) is disconnected.Conversely, if ∆( G ) is disconnected, then it can write as a union of two simplicialcomplexes ∆( G ) = ∆ ∪ ∆ such that ∆ ∩ ∆ = {∅} and V i (cid:54) = ∅ , where V i is the set of vertices of ∆ i , for i = 1 , G i := G [ V i ] for i = 1 ,
2. We will show that G = G ∗ G , or equivalently E ( G ) = E ( G ∗ G ). Indeed, it is obvious that E ( G ) ⊆ E ( G ∗ G ). We now provethe reverse inclusion. Let v v be an edge of G ∗ G . If v v is an edge of either G or G , then v v ∈ E ( G ). Hence, we may assume that v i ∈ V ( G i ) for i = 1 , v i is a vertex of ∆ i , we have { v , v } / ∈ ∆( G ) ∪ ∆( G ) = ∆( G ). In otherwords, v v ∈ E ( G ), so that E ( G ) = E ( G ∗ G ), and the lemma follows. (cid:3) Locally triangle-free graphs G in W with α ( G ) (cid:54) Proposition 1.7.
Let G be a locally triangle-free graph in W with n vertices. Then,
1) If α ( G ) = 1 , then G is K n with n (cid:62) ;(2) If α ( G ) = 2 , then G is C cn with n (cid:62) .Proof. If α ( G ) = 1, then G = K n is a complete graph. Since G is in W , n (cid:62) α ( G ) = 2, for each v ∈ V ( G ), G v is a triangle-free graph in W by Lemma 1.3 and α ( G v ) = 1 by Lemma 1 .
2. Thus, G v is just an edge. It follows that deg G ( v ) = n − v ∈ V ( G ). It yields deg G c ( v ) = 2 for all v ∈ V ( G c ), so G c is an n -cycle. Since α ( G ) = 2, we get n (cid:62)
4, as required. (cid:3) The structure of Neighborhoods
In this section we explore the local structure of locally triangle-free graphs G in W with α ( G ) (cid:62)
3. Namely, let ( abc ) be a triangle in G , and let A := N G ( a ) \ N G [ b ].Then,(1) Either A ∈ ∆( G ) or G [ A ] is one edge and α ( G ) − G is not a join of its two proper subgraphs, then G ab is not empty.The first result is the following. Lemma 2.1.
Let G be a locally triangle-free graph in W with α ( G ) (cid:62) . Let ( abc ) be a triangle in G such that G ab (cid:54) = ∅ . Let A := N G ( a ) \ N G [ b ] (see Figure 2). If A isnot an independent set of G , then(1) G [ A ] consists of one edge and α ( G ) − isolated vertices, and(2) G ab has α ( G ) − vertices.Proof. In order to prove the lemma we divide into the following claims: ab G ab A Figure 2.
The structure of graph G . Claim : If xy ∈ E ( G [ A ]) and v ∈ V ( G ab ) , then v is adjacent in G to exactly oneof the two vertices x and y . Indeed, if vx, vy / ∈ E ( G ), then G v contains a triangle ( axy ). If v is adjacent to both x and y , then ( vxy ) is the triangle in G b . In both cases G is not locally triangle-free,a contradiction. Claim : Assume that xy ∈ E ( G [ A ]) , uv ∈ E ( G ab ) and ux ∈ E ( G ) . Then, vy ∈ E ( G ) and uy, vx / ∈ E ( G ) . ndeed, by Claim 1, ux ∈ E ( G ) implies uy / ∈ E ( G ). If vx ∈ E ( G ), then G b has thetriangle ( xuv ), a contradiction. Hence vx / ∈ E ( G ), and vy ∈ E ( G ) by Claim 1. Claim : G ab is bipartite. Indeed, assume that G ab has an odd cycle of length 2 k + 1, say ( z . . . z k +1 ), forsome k (cid:62)
1. Let xy ∈ E ( G [ A ]). Since z z ∈ E ( G ab ), by Claims 1 and 2, we mayassume that z x ∈ E ( G ) and so z y ∈ E ( G ). Since z z ∈ E ( G ab ), by Claim 2 wehave z x ∈ E ( G ). Repeating this argument for z z , . . . , z k z k +1 , z k +1 z , we finallyobtain yz ∈ E ( G ). Then, G b has a triangle ( z xy ), a contradiction. Therefore, G ab must be bipartite. Claim : G [ A ] is bipartite. Indeed, assume that G [ A ] has an odd cycle, say ( z . . . z m +1 ), of length 2 m + 1for some m (cid:62)
1. Let v ∈ V ( G ab ). By Claim 2 we may assume that vz ∈ E ( G ) sothat vz / ∈ E ( G ). Repeating this argument for z z , . . . , z m z m +1 , z m +1 z , we finallyobtain vz / ∈ E ( G ), a contradiction. Hence, G [ A ] is bipartite, as claimed.Now let S be the set of isolated vertices of G [ A ] and let Γ , . . . , Γ t be the connectedcomponents of G [ A \ S ]. Note that t (cid:62) A is not an independent set of G .By Claim 4, each Γ i is bipartite and let ( A i , B i ) be its bipartition. Claim : S (cid:54) = ∅ and every vertex of G ab is adjacent to just one vertex in S . Indeed, since G ab (cid:54) = ∅ , let v ∈ V ( G ab ) and let H be a connected component of G ab ,which contains v . If H is just one point v , then set C = { v } , and D = ∅ . Otherwise,by Claim 3, H is a bipartite and we let ( C, D ) be a bipartition of H , where v ∈ C . Let x be an arbitrary element of C . By repeating Claim 1, we can see that for each Γ i , x is adjacent to all vertices in B i but not adjacent to any vertex in A i . If D (cid:54) = ∅ , thenthere is y ∈ D such that xy ∈ E ( G ). By Claim 2, y is adjacent to all vertices in A i butnot adjacent to any vertex in B i . Applying the same argument to all edges between C and D , we conclude that all vertices in C (resp. D ) have the above properties as x (resp. y ). Let X := ∪ ti =1 A i . Then X ∪ { b } ∈ ∆( G ) and we can see that G X ∪{ b } is abipartite graph with a bipartition ( S, V ( G ab \ N G ( X )). By Lemma 1 .
4, this graph is adisjoint union of edges. Note that v ∈ C ⊆ V ( G ab \ N G ( X )), thus | S | = | V ( G ab \ N G ( X )) | (cid:62) | C | (cid:62) , and thus v is is adjacent to just one vertex in S , as claimed.By Claim 5, G S ∪{ b } = Γ ∪ · · · ∪ Γ t , so G S ∪{ b } is a bipartite graph. Thus Γ i is anedge for i = 1 , . . . , t , by Lemma 1.4. Let S = { p , . . . , p m } for m (cid:62)
1; and let Γ i bethe edge a i b i for i = 1 , . . . , t . Observe that α ( G S ∪{ b } ) = α ( G [ A \ S ]) = t , and hence α ( G ) = t + m + 1 by Lemma 1.2.Since each vertex of G ab is adjacent to just one vertex in S , the set V ( G ab ) can bepartitioned into V ( G ab ) = V ∪ · · · ∪ V m , where V i is the set of all vertices of G ab whichare adjacent to v i . Moreover, V i (cid:54) = ∅ because every vertex in S is adjacent to somevertex of G ab . Now we show that V i ∈ ∆( G ) for all i . Indeed, if G [ V i ] has an edge, ay xy , for some i , then ( xyv i ) would be a triangle in G b , which is impossible as G b istriangle-free, and then V i ∈ ∆( G ). Claim : | V i | = t + 1 for i = 1 , . . . , m . Indeed, let v ∈ V i . we may assume that va , . . . , va s ∈ E ( G ). Let U := S \ { p i } ∪{ b, v } . Then, α ( G U ) = α ( G ) − | U | = ( t + m + 1) − ( m + 1) = t . On the other hand, G U is a bipartite graph with bipartition ( { b , . . . , b t } , V i \ { v } ). By Lemma 1 .
4, thisgraph is just disjoint edges, so | V i | = t + 1, as claimed.In summary, we have proved that G [ A ] consists of t disjoint edges and m isolatedvertices; | V ( G ab ) | = m ( t + 1) and α ( G ) = t + m + 1. Hence, it remains to prove t = 1.Assume on the contrary that t (cid:62)
2. Write V = { u , . . . , u t +1 } . By Claim 2 we mayassume that u t +1 a , . . . , u t +1 a t ∈ E ( G ) and u t +1 b , . . . , u t +1 b t / ∈ E ( G ). We also canassume that u i b i ∈ E ( G ) for all i = 1 , . . . , t ; and u i b j / ∈ E ( G ) for all 1 (cid:54) i (cid:54) = j (cid:54) t .These facts together with Claim 2 help us conclude u i a j ∈ E ( G ) for all i (cid:54) = j . Now v is an isolated vertex of G ( S \ v ) ∪{ b,a ,a } , but this fact contradicts Lemma 1.3. Hence, t = 1, and the proof of the lemma is complete. (cid:3) Lemma 2.2.
Let G be a triangle-free graph in W . Assume that V ( G ) can be parti-tioned into V ( G ) = S ∪ T ∪ U , such that(1) | S | + | T | (cid:62) ;(2) Every vertex in U is adjacent to all vertices in S ∪ T .Then, U = ∅ .Proof. Assume on the contrary that U (cid:54) = ∅ . As | S | + | T | (cid:62)
2, we may assume that S (cid:54) = ∅ . Since G is a triangle-free graph, U ∈ ∆( G ). Let u ∈ U . Then, G u = G [ U \{ u } ],so G u must be empty by Lemma 1 .
3. It follows that α ( G ) = 1, so G is just an edge,and | V ( G ) | = 2. On the other hand, | V ( G ) | = | S | + | T | + | U | (cid:62)
3, a contradiction. (cid:3)
Lemma 2.3.
Let G be a locally triangle-free graph in W with α ( G ) (cid:62) such that G is not a join of two proper subgraphs. Assume that v v is an edge of G such that G v v = ∅ . Then, N G ( v ) ∩ N G ( v ) = ∅ .Proof. Since α ( G v ) = α ( G ) − (cid:62) G v is in W by Lemma 1 .
3, there is an edge ab in G v . Let A := N G ( a ) \ N G [ b ] , B := N G ( b ) \ N G [ a ] , and C := N G ( a ) ∩ N G ( b ) . Note that v ∈ C and v ∈ V ( G ab ). Figure 3 depicts this situation. bv A BCG ab v Figure 3.
A configuration for graph G . Claim 1: Every vertex in C is adjacent to all vertices in V ( G ab ) . Indeed, assume on the contrary that cv / ∈ E ( G ) for some c ∈ C and v ∈ V ( G ab ),then ( abc ) would be a triangle in G v , a contradiction, as claimed. Claim 2: A and B are not empty sets. Indeed, if A = B = ∅ , by Claim 1 we obtain G = G [ C ] ∗ G [ { a, b } ∪ V ( G ab )], acontradiction. Hence, we are able to assume that A (cid:54) = ∅ .Assume that B = ∅ . Then, every vertex of G ab is adjacent to all vertices in A .Because if uv / ∈ E ( G ) for some u ∈ V ( G ab ) and v ∈ A , by Claim 1 we get b as anisolated vertex of G { u,v } . It, however, contradicts Lemma 1 .
3. Now if G ab has an edge,say xy , then ( xyv ) would be a triangle in G b for any v ∈ A , a contradiction. Hence, G ab is a totally disconnected graph. On the other hand, G ab = G a , so that G a is alsoa totally disconnected graph. But then it contradicts Lemma 1 .
3, thus B (cid:54) = ∅ , andthe claim follows.Let S be the set of isolated vertices of G [ A ] and let T be the set of isolated verticesof G [ B ]. By Lemma 2 .
1, we see that G [ A ] (resp. G [ B ]) is either totally disconnectedor one edge and α ( G ) − S (resp. T ) is not empty. Claim 3: If a vertex in C is adjacent to a vertex in S (resp. T ), it must be adjacentto all vertices in A (resp. B ). Indeed, assume that cv ∈ E ( G ) for some c ∈ C and v ∈ S . If G bc (cid:54) = ∅ , then G bc iswell-covered and α ( G bc ) = α ( G ) − .
5. Since cv ∈ E ( G ), V ( G bc ) ⊆ A \{ v } and G bc is well-covered, we have α ( G bc ) ≤ α ( G bc \ S ) + | S \ { v }| ≤ α ( G [ A ] \ S ) + | S | − α ( G [ A ]) − . It follows that α ( G ) = α ( G bc ) + 1 (cid:54) α ( G [ A ]). On the other hand, α ( G ) (cid:62) α ( G [ A ∪ { b } ]) = α ( G [ A ]) + 1 , a contradiction. Thus, G bc = ∅ , and thus c is adjacent to every vertex in A , as claimed. e now let C := { c ∈ C | c is adjacent to all vertices in A } ,C := { c ∈ C | c is not adjacent to any vertex in S } . By Claim 2, the set C has a partition C = C ∪ C . We next prove that C = ∅ . Claim 4: If C (cid:54) = ∅ , then every vertex in C is adjacent to all vertices in B . Indeed, assume on the contrary that cv / ∈ E ( G ) for some c ∈ C and v ∈ B . If B isan independent set, i.e. B = T , then c is not adjacent to any vertex in B by Claim 3.Combining with Lemma 1.5, we get G ac = G [ B ], and so | B | = α ( G ac ) = α ( G ) − ≥ B ∪ { c } is a maximal independent set of G , and thus C \ N G [ c ] ⊆ N G ( B ). ByClaim 3, every vertex in B is adjacent to all vertices in C \ N G [ c ]. Let z be a vertex in B \ { v } . Since V ( G { c,z } ) = B \{ z } , v is an isolated vertex of G { c,z } , which contradictsLemma 1 . B / ∈ ∆( G ), then by Lemma 2.1 we have G [ B ] is just one edge, say xy , and isolatedvertices, say q , . . . , q m , where m = α ( G ) −
2. Hence, T = { q , . . . , q m } and m (cid:62) cq i / ∈ E ( G ) for any i = 1 , . . . , m by Claim 3.If cx, cy ∈ E ( G ), then G q has a triangle ( cxy ), a contradiction.If cx, cy / ∈ E ( G ), then V ( G c ) = ( C \ N G [ c ]) ∪ B . Since G c has no isolated verticesby Lemma 1 .
3, one has q u ∈ E ( G ) for some u ∈ C \ N G [ c ]. Note that u is adjacentto all vertices in B by Claim 3. But then G c has a triangle ( uxy ), a contradiction.If c is adjacent to either x or y but not both, we may assume that cx ∈ E ( G ) and cy / ∈ E ( G ). Then, V ( G c ) = ( C \ N G [ c ]) ∪ { y, q , . . . , q m } . Hence, every edge of G c has an endpoint in C \ N G [ c ]. Together with Claim 3 it follows that any vertex of G c that is adjacent to q is adjacent to y as well, so N G c ( q ) ⊆ N G c ( y ). But then, q isan isolated vertex of G { c,y } , which contradicts Lemma 1 .
3, and then claim follows.
Claim 5: C (cid:54) = ∅ . Indeed, assume on the contrary that C = ∅ so that C = C (cid:54) = ∅ . By Claims 1 and4, every vertex in C is adjacent to all vertices in { a, b } ∪ A ∪ B ∪ V ( G ab ). It followsthat G = G [ C ] ∗ G [ { a, b } ∪ A ∪ B ∪ V ( G ab )] , a contradiction, so C (cid:54) = ∅ , as claimed. Claim 6: If A ∈ ∆( G ) , then | A | = α ( G ) − . Indeed, since C (cid:54) = ∅ by Claim 5, we can take c ∈ C . Then, c is not adjacent toany vertex in A by Claim 3, so that G bc = G [ A ]. Now by applying Lemma 1 . | A | = α ( G [ A ]) = α ( G bc ) = α ( G ) − , as claimed. Claim 7: C = ∅ and C ∈ ∆( G ) . y Lemma 2 . | S | (cid:62) α ( G ) −
2, so α ( G S ) = α ( G ) − | S | (cid:54) G S is a triangle-free graph in W , it must be an edge, or two disjoint edges ora pentagon by Proposition 1.7. Consequently, deg G S ( x ) (cid:54) x of G S .Since | C | (cid:54) deg G S ( b ), we obtain | C | (cid:54)
2. Together with Claim 5, this fact yields | C | = 1 or | C | = 2.If | C | = 1, then C = { c } for some vertex c . We can partition V ( G c ) into V ( G c ) = ( A \ N G ( c )) ∪ ( B \ N G ( c )) ∪ ( C \ N G ( c )) . Since | A \ N G ( c ) | ∪ | B \ N G ( c ) | ≥ | S | + | T | ≥
2, together with Lemma 2 . C \ N G ( c ) = ∅ . In other words, c is adjacent to all vertices in C . Thus,by Claims 1 and 3 we conclude that all vertices in C are adjacent to all vertices in { a, b, c } ∪ A ∪ B ∪ V ( G ab ). If C (cid:54) = ∅ , then G = G [ C ] ∗ G [ { a, b, c } ∪ A ∪ B ∪ V ( G ab )] , a contradiction. Hence, C = ∅ .If | C | = 2, in this case we have G S is a pentagon and | S | = α ( G ) −
2. By Claim6, A is not the independent set in G . Thus, by Lemma 2.1, G [ A ] is a disjoint unionof one edge, say xy , and isolated vertices in S . Let C = { c, c (cid:48) } for some c, c (cid:48) ∈ V ( G ).Since { b, c, c (cid:48) , x, y } ⊆ V ( G S ), we may assume that G S is the pentagon with the edgeset { bc, cx, xy, yc (cid:48) , c (cid:48) b } . In particular, C ∈ ∆( G ). Since we can partition V ( G C ) into V ( G C ) = S ∪ ( B \ N G ( C )) ∪ ( C \ N G ( C )) . By Lemma 2 . C \ N G ( C ) = ∅ . It follows that c (cid:48) is adjacentto all vertices in C \ N G ( c ), and c is adjacent to all vertices in C \ N G ( c (cid:48) ). Togetherwith Lemma 2 . V ( G c ), V ( G c ) = ( S ∪ { y } ∪ { c (cid:48) } ) ∪ ( B \ N G ( c )) ∪ ( C \ N G ( c )) , this fact yields C \ N G ( c ) = ∅ . Similarly, C \ N G ( c (cid:48) ) = ∅ , i.e. every vertex in C isadjacent to all vertices in C . Thus, if C (cid:54) = ∅ , then we have G = G [ C ] ∗ G [ { a, b, c, c (cid:48) } ∪ ( A ∪ B ∪ V ( G ab )] , a contradiction, and the claim follows.We now return to prove the lemma. Since C = C by Claim 7, we have v ∈ C .Consequently, v is not adjacent to any vertex in S ∪ T . Since G v v = ∅ , v is adjacentto all vertices in S ∪ T .Now assume on the contrary that N G ( v ) ∩ N G ( v ) (cid:54) = ∅ . Let v ∈ N G ( v ) ∩ N G ( v ).Note that C ∈ ∆( G ) and N G ( v ) ⊆ { a, b } ∪ V ( G ab ) ∪ ( A \ S ) ∪ ( B \ T ) , so either v ∈ V ( G ab ) or v ∈ ( A \ S ) ∪ ( B \ T ).Assume that v is a vertex of G ab . Then, v is not adjacent to any vertex in S ∪ T .Because assume on the contrary that v p ∈ E ( G ) for some p ∈ S (similarly, p ∈ T ).Then, ( pv v ) would be a triangle in G b , a contradiction. It follows that S ∪ { b, v } isan independent set in G , and so | S | ≤ α ( G ) −
2. Moreover, by Claim 6 and Lemma . | S | (cid:62) α ( G ) −
2, and so | S | = α ( G ) −
2; and G [ A \ S ] is just an edge, say xy .Since S ∪ { b, v } is an independent set of G , we imply that v is adjacent to both x and y , and so ( xyv ) is a triangle in G b , a contradiction.Assume that v ∈ ( A \ S ) ∪ ( B \ T ). We may assume that v ∈ A \ S . In this case A is not an independent set in G , so G [ A ] consists of one edge, say xy , and isolatedvertices in S with | S | = α ( G ) −
2. Then either v = x or v = y . If v is adjacentto both x and y , then G bv = G [ S ] (cid:54) = ∅ . Therefore, α ( G ) = α ( G bv ) + 1 = | S | + 1, acontradiction. We now may assume that v x / ∈ E ( G ), so that v = y . On the otherhand, since G v v = ∅ , v x ∈ E ( G ). Hence, ( v xy ) is a triangle in G b , a contradiction.Therefore, we must have N G ( v ) ∩ N G ( v ) = ∅ , and the proof of the lemma iscomplete. (cid:3) From Lemma 2.3 we obtain the following result:
Corollary 2.4.
Let G be a locally triangle-free graph in W with α ( G ) (cid:62) such that G is not a join of two proper subgraphs. Then, for any edge ab lying in a triangle in G we have α ( G ab ) = α ( G ) − . In particular, G ab (cid:54) = ∅ .Proof. Assume ab is in the triangle ( abc ) for some vertex c of G , so that c ∈ N G ( a ) ∩ N G ( b ). In particular, N G ( a ) ∩ N G ( b ) (cid:54) = ∅ , so G ab (cid:54) = ∅ by Lemma 2 .
3. The lemma nowfollows from Lemma 1 . (cid:3) Locally Triangle-free graphs in W In this section we characterize locally triangle-free graphs in W . First we dealwith such graphs that are not triangle-free. Thus, we assume that G is a locallytriangle-free graph in W it satisfies:(1) α ( G ) (cid:62) G is not a join of its two proper subgraphs;(3) G has a triangle ( abc ).Let A := N G ( a ) \ N G [ b ] , B := N G ( b ) \ N G [ a ] , and I := N G ( a ) ∩ N G ( b ) . Then, by Corollary 2 .
4, we have G ab , G bc and G ca are not empty, and α ( G ab ) = α ( G bc ) = α ( G ca ) = α ( G ) − . We will classify G via the structure of G [ A ] , G [ B ] and G ab . Lemma 3.1. (1) Every vertex in I is adjacent to all vertices of G ab .(2) Every isolated vertex of G [ A ] is not adjacent to any vertex in I .(3) If A is an independent set of G then I = { c } .Proof. (1) If there are x ∈ I and y ∈ V ( G ab ) such that x is not adjacent to y . Then, G y has a triangle ( abx ), a contradiction.(2) If there are an isolated vertex of G [ A ], say v , and a vertex in I , say u , suchthat uv ∈ E ( G ), then by statement (1) we would have G bu is an induced subgraph of G [ A \{ v } ]. In this case, α ( G bu ) (cid:54) α ( G [ A \{ v } ]) = α ( G [ A ]) −
1. Since ( abu ) is a triangle n G , by Corollary 2.4, we get α ( G bu ) = α ( G ) −
1. Thus, α ( G ) = α ( G bu )+1 (cid:54) α ( G [ A ]).On the other hand, α ( G [ A ]) = α ( G [ A ∪ { b } ]) − (cid:54) α ( G ) −
1, a contradiction.(3) By Statements (1) and (2) we get G bc = G [ A ]. In particular, | A | = α ( G ) − α ( G A ) = 1, so it is an edge. By Statement (2) we have G [ I ∪ { b } ] isan induced subgraph of G A , so G A is just the edge bc and so I = { c } . (cid:3) Lemma 3.2. If A / ∈ ∆( G ) , then:(1) α ( G ) = 3 ;(2) G [ A ] is just one edge and one isolated vertex;(3) G ab is just two isolated vertices.Proof. Let m = α ( G ) −
2. By Lemma 2 . G [ A ] consists of one edge and m isolated vertices, and G ab has 2 m vertices.If G ab is a totally disconnected graph, then α ( G ab ) = 2 m so that α ( G ) = 2 m + 1.Together with α ( G ) = m + 2, this equality yields m = 1; and the lemma follows.Assume that G ab is not a totally disconnected graph. We will prove that thisassumption leads to a contradiction. Let C := N G ( c ) \ N G [ a ]. Observe that c isadjacent to every vertex in G ab by Lemma 3 .
1. Thus, G ab is an induced subgraph of G [ C ], and thus C is not an independent set of G . By Lemma 2 . a and b by c and a respectively, G [ C ] consists of one edge and m isolated vertices.Consequently, G ab is just one edge and 2 m − α ( G ab ) = 2 m −
1. From α ( G ) = α ( G ab ) + 1 and α ( G ) = m + 2, we obtain m = 2.Hence, α ( G ) = 4 and G [ C ] = G ab .Recall that C / ∈ ∆( G ). If G ac is a totally-disconnected graph, then α ( G ) = 3 asabove. Therefore, G ac is not a totally disconnected graph. By Lemma 3 . G ac is also an induced subgraph of G [ B ].Since V ( G ac ) ⊆ B by Lemma 3 .
1, we have G ac is an induced subgraph of G [ B ], so B / ∈ ∆( G ). By the argument above we get G [ B ] = G ac consists of one edge and 2isolated vertices. By symmetry, we also have G [ A ] = G bc .Assume that G [ A ] is the edge xy and two isolated vertices a and a ; G [ B ] is theedge zt and two isolated vertices b and b ; and G [ C ] is one edge uv and two isolatedvertices c and c .Note that G b = G [ A ∪ C ]. We now explore the structure of this graph. Firstly, wehave all vertices in C are adjacent to exactly one of two vertices x and y . Indeed,if wx, wy ∈ E ( G ) for some w ∈ C , then G b has a triangle ( wxy ), a contradiction.If wx, wy / ∈ E ( G ), then ( axy ) is a triangle in G w , a contradiction. Thus, we mayassume that xv, yu, c x ∈ E ( G ) and xu, yv, c y / ∈ E ( G ). Since { b, v, c } ∈ ∆( G )and α ( G { b,v,c } ) = 1, G { b,v,c } is just an edge; and this edge must be yc , and so c x / ∈ E ( G ). Similarly, since c ∈ V ( G { b,v,c } ) (cid:40) { c , a , a } , we assume a c ∈ E ( G )and a c / ∈ E ( G ). Thus, V ( G { b,u,c } ) ⊆ { c , a } , and so c a ∈ E ( G ), a c / ∈ E ( G ).Furthermore, since a , c ∈ V ( G { b,x,a } ), we have ua ∈ E ( G ) and va / ∈ E ( G ). Next,since v, a ∈ V ( G { b,y,a } ), va ∈ E ( G ) and ua / ∈ E ( G ). It follows that E ( G b ) = { xy, uv, xv, xc , yu, yc , a u, a c , a v, a c } . n the same way we may assume E ( G a ) = { zt, uv, zv, zc , tu, tc , b u, b c , b v, b c } . By symmetry, G c = G [ A ∪ B ] has the same structure as G a and G b . It follows that z is adjacent to either a or a . If z is adjacent to a , then G c has the triangle( za c ). If z is adjacent to a , then G a has the triangle ( za v ). Thus, G is not locallytriangle-free in both cases, a contradiction, and the lemma follows. (cid:3) Lemma 3.3. If A, B, V ( G ab ) ∈ ∆( G ) , then G is isomorphic to either Q or Q .Proof. By Lemma 3 . G bc = G [ A ], G ac = G [ B ], and N G ( a ) ∩ N G ( b ) = { c } .In particular, | A | = | B | = α ( G ) − C := V ( G ab ) and s := α ( G ) −
1. Then N G ( c ) = { a, b } ∪ C , and | A | = | B | = | C | = s ≥
2. Assume that A = { a , . . . , a s } ; B = { b , . . . , b s } and C = { c , . . . , c s } .Note that G b is a bipartite graph with bipartition ( A, C ). So G b is disjoint edgesby Lemma 1 .
4. Hence, we may assume that E ( G b ) = { a c , . . . , a s c s } . Similarly, wemay assume that E ( G a ) = { b c , . . . , b s c s } . Together with this Lemma 1.4 again, E ( G c ) = { a σ (1) b , . . . , a σ ( s ) b s } , where σ is a permutation of the set { , . . . , s } .Therefore, G [ A ∪ B ∪ C ] consists of disjoint cycles, say C , . . . , C t . Moreover thelength of each C i is a multiple of 3, say 3 s i , for s i (cid:62)
1. If s i = 1 for some i , then C i is the form ( a j b j c j ) for some j , and so C i is a triangle of G c m for any m (cid:54) = j , acontradiction. Thus, s i (cid:62)
2. This yields α ( C i ) = (cid:98) s i / (cid:99) = s i + (cid:98) s i / (cid:99) (cid:62) s i + 1. Thus, α ( G [ A ∪ B ∪ C ]) = t (cid:88) i =1 (cid:98) s i / (cid:99) (cid:62) t (cid:88) i =1 ( s i + 1) = s + t. Combining with α ( G [ A ∪ B ∪ C ]) (cid:54) α ( G ) = s + 1, we obtain t = 1. This means that α ( G [ A ∪ B ∪ C ]) is a cycle of length 3 s . From the equality α ( G [ X ∪ Y ∪ Z ]) (cid:54) α ( G )we have (cid:98) s/ (cid:99) (cid:54) s + 1, or equivalently (cid:98) s/ (cid:99) (cid:54)
1. This forces either s = 2 or s = 3.If s = 2, G is isomorphic to Q . Otherwise, s = 3 and G is isomorphic to Q . (cid:3) Lemma 3.4. If A / ∈ ∆( G ) and B ∈ ∆( G ) , then G is isomorphic to P .Proof. By Lemma 3 .
2, we have α ( G ) = 3, G [ A ] consists of one edge and one isolatedvertex, and G ab is two isolated vertices. By Lemma 3 .
1, we imply that G [ B ] = G ac ,and N G ( a ) ∩ N G ( b ) = { c } . So | B | = α ( G ) − G [ B ] is just two isolatedvertices.Let G [ A ] be the edge xy and one isolated vertex a ; G [ B ] be two isolated vertices b and b ; G ab be two isolated vertices c , c . This yields G has a vertex set V ( G ) = { a, b, c, x, y, a , b , b , c , c } (see Figure 4). bca x y b b c c Figure 4.
The configuration for graph G .As G a is a bipartite graph with bipartition ( { b , b } , { c , c } ), it is just two disjointedges by Lemma 1 .
4, and so we may assume that E ( G a ) = { b c , b c } .Since V ( G b ) = { x, y, a , c , c } and G b is a triangle-free graph in W , it must be apentagon. Hence, we may assume E ( G b ) = { xy, xc , c a , a c , c y } .Note that V ( G c ) = { y, a, b, b , c } , so we have E ( G c ) = { ab, bb , b c , c y, ya } .Hence, b y / ∈ E ( G ). Since ( axy ) is not a triangle in G b , we must have x / ∈ V ( G b ),or equivalently b x ∈ E ( G ). Similarly, b x / ∈ E ( G ) and b y ∈ E ( G ).Observe that V ( G c ) ⊆ { x, y, a , b , b } . Since α ( G c ) = 2, we conclude that G c iseither two disjoint edges or a pentagon.Assume that G c is just two disjoint edges. Then, by Lemma 3 . a , b , b ∈ V ( G c ), so the remaining vertex is either x or y . By symmetry, we may assume itis x , i.e. cx ∈ E ( G ). It follows that N G ( x ) = { a, c, y, b , c } , so G x must be twodisjoint edges bb and a c . Thus, a b / ∈ E ( G ), and thus N G ( b ) = { b, y, c } , and | V ( G b ) | = 6. On the other hand, G b must be either two disjoint edges or a pentagon,so | V ( G b ) | (cid:54)
5, a contradiction.Therefore G c is a pentagon, and thus V ( G c ) = { x, y, a , b , b } . Recall that yb , xb / ∈ E ( G ), so E ( G c ) = { xy, xb , b a , a b , b y } . It follows that E ( G ) = { ab, ac, aa , ax, ay, bc, bb , bb , cc , cc , xy, xb , xc , yb , yc , a b , a b ,a c , a c , b c , b c } , so G is isomorphic to P . (cid:3) Lemma 3.5. If A, B / ∈ ∆( G ) , then G is isomorphic to P . Proof.
Let C := V ( G ab ). By Lemma 3 . α ( G ) = 3, both A and B consist ofone edge and one isolated vertex, and G [ C ] is two isolated vertices. We may assumethat G [ A ] is one edge xy and an isolated vertex a ; G [ B ] is one edge zt and an isolatedvertex b ; and G [ C ] is two isolated vertices c and c . Claim : I ∈ ∆( G ) . Indeed, assume on the contrary that uv ∈ E ( G ) for some u, v ∈ I . By Lemma 3.1, a is not adjacent to any vertex in I . Thus, ( buv ) is a triangle in G a , a contradiction.Hence, I ∈ ∆( G ), as claimed.Since G a is a triangle-free graph in W and V ( G a ) = { z, t, b , c , c } , G a must be apentagon. Because zt ∈ V ( G ), we may assume that E ( G a ) = { zt, tc , c b , b c , c z } . imilarly, by symmetry we may assume that E ( G b ) = { xy, yc , c a , a c , c x } (seeFigure 5). a bcdyx a b ztc c CA BI
Figure 5.
The configuration for graph G . Claim 2: yt, xz / ∈ E ( G ) and yz, xt ∈ E ( G ) . Indeed, as V ( G c ) = { a, b, t, c , y } , we have G c is a pentagon, and so E ( G c ) = { ab, bt, tc , c y, ya } . It follows that yt / ∈ E ( G ). Together with G y a triangle-free graph,we imply that yz ∈ E ( G ). Similarly, xz / ∈ E ( G ) and xt ∈ V ( G ). Claim 3: a b ∈ E ( G ) . Indeed, if a b / ∈ E ( G ), then I ∪{ a , b } is an independent set of G . Since α ( G ) = 3,we have | I | = 1, so that I = { c } . Observe that { a, a , c, z, t } ⊆ V ( G b ). Togetherwith the fact that G b is a triangle-free graph in W with α ( G b ) = 2, it implies that G b is a pentagon with V ( G b ) = { a, a , c, z, t } . Hence, b x, b y ∈ E ( G ). But then G a has the triangle ( b xy ), a contradiction, and the claim follows. Claim 4: a z, a t, b x, b y ∈ E ( G ) . Indeed, we only prove a z ∈ E ( G ) and others are proved similarly. Assume onthe contrary that a z / ∈ V ( G ). Then, G z has a triangle ( a b c ), a contradiction.Therefore, a z ∈ E ( G ), as claimed. Claim 5: | I | = 2 . Indeed, since α ( G { x,z } ) = 1, G { x,z } must be an edge. Since c ∈ V ( G { x,z } ), weimply that G { x,z } is the edge c d for some d ∈ I . Note that G d is a triangle-freegraph in W and α ( G d ) = 2, so it is either a pentagon or two disjoint edges. Since G d contains 3-path xb a z , G d must be a pentagon. On the other hand, E ( G d ) ⊆{ a , b , z, x } ∪ ( I \ { d } ). It follows that | I \ { d }| (cid:54) = 0, so | I | (cid:62)
2. By Claim 1 andLemma 3 .
1, we get I ∪ { a } is an independent set of G . Since α ( G ) = 2, | I | (cid:54)
2. Ityields | I | = 2, as claimed.So now we may assume that I = { d, c } . In particular, V ( G d ) = { c, a , b , z, x } , andhence, cx, cz ∈ E ( G ). Note also that dx, dz / ∈ E ( G ). Claim : cy, ct / ∈ E ( G ) . ndeed, if cy ∈ E ( G ), then G a has a triangle ( xyc ), a contradiction, and then cy / ∈ E ( G ). Similarly, ct / ∈ E ( G ).Now using Claim 6, from the graph G c we get dy, dt ∈ E ( G ). In summary, weobtain: E ( G ) = { ab, aa , ac, ad, ax, ay, bb , bc, bd, bt, bz, cx, cz, cc , cc , dy, dt, dc , dc , a b ,a z, a t, a c , a c , b x, b y, b c , b c , xy, xt, xc , yz, yc , zt, zc , tc } , so G is isomorphic to P . (cid:3) We are in position to prove the main result of this section.
Theorem 3.6.
Let G be a graph with α ( G ) ≥ which is not a join of its two propersubgraphs. Then, G is a locally triangle-free graph in W if and only if G is a triangle-free graph in W , or G is isomorphic to one of Q , Q , P , or P .Proof. If G is a triangle-free graph in W or one of Q , Q , P , or P , then we cancheck that G is also a locally triangle-free graph in W .Conversely, assume that G is a locally triangle-free graph in W . It suffices to provethat if G is not triangle-free, then G is one of Q , Q , P , or P . We now considertwo possible cases: Case 1: For every triangle ( uvw ) of G , we have N G ( u ) \ N G [ v ] ∈ ∆( G ) . Let ( abc ) be a triangle of G and let A := N G ( a ) \ N G [ b ] , B := N G ( b ) \ N G [ a ], and C := N G ( c ) \ N G [ a ] so that A, B, C ∈ ∆( G ). By Lemma 3 .
1, we have V ( G ab ) ⊆ N G ( c ),so G ab is an induced subgraph of G [ C ]. Thus, G ab is a totally disconnected graph. ByLemma 3 .
3, we have G is isomorphic to either Q or Q . Case 2: There is a triangle ( abc ) of G such that N G ( a ) \ N G [ b ] / ∈ ∆( G ) . Let A := N G ( a ) \ N G [ b ] and B := N G ( b ) \ N G [ a ] , so that A / ∈ ∆( G ). If B ∈ ∆( G ),then G is isomorphic to Q by Lemma 3 .
4. Otherwise,
B / ∈ ∆( G ), and so G isisomorphic to P by Lemma 3 .
5. The proof of the theorem is complete. (cid:3) Buchsbauness of second powers of edge ideals
Let R := K [ x , . . . , x n ] be the polynomial ring over a field K and let G be a graphwith vertex set { x , . . . , x n } . In this section we characterize graphs G such that I ( G ) are Buchsbaum. First we characterize locally triangle-free Grorenstein graphs. Theorem 4.1.
Let G be a locally triangle-free graph. Then G is Gorenstein if andonly if G is either a triangle-free graph in W , or G is isomorphic to one of C cn ( n (cid:62) ), Q , Q , P or P .Proof. We consider three cases:
Case 1: α ( G ) = 1 . Then G is a complete graph. It is well known that all Gorensteincomplete graphs are just K and K , so the theorem holds true in this case. ase 2: α ( G ) = 2 . Note that C cn is Gorenstein for any n (cid:62) C cn ) is isomorphic to the 1-dimensional sphere. Thus, the theoremfollows from [10, Lemma 2.5] and Proposition 1.7. Case 3: α ( G ) (cid:62) . If G is Gorenstein, by [17, Proposition 3.3 in Chapter 0 andCorollary 4.2 in Chapter II] the complex ∆( G ) is connected. Together with Lemma1.6, it follows that G is not a join of its two proper subgraphs. By Theorem 3 . G iseither a triangle-free graph in W or isomorphic to one of Q , Q , P or P .Conversely, if G a triangle-free graph in W , then G is Gorenstein by [10, Theorem3.4]. If G is isomorphic to Q (resp. P , and P ) as indicated in Figure 1, its edges arediagonals of a triaugmented triangular prism (resp. gyroelongated square bipyramid ,and icosahedron ) shown in Figure 6. In other words, all maximal independent sets of G are triangles of such polytopes, and the geometric realization of ∆( G ) is isomorphicto the 2-dimensional sphere S . Therefore, G is a Gorenstein graph. ac b b c b a a c aa c c b b b c yx cc xz c a ba b ytd Triaugmented Gyroelongated Icosahedrontriangular prism square bipyramid
Figure 6.
Three polytopes associated with graphs Q , P , P . Finally, if G is isomorphic to Q (see Figure 1), its maximal independent sets arelisted as follows: ab b b ab b c ab b c ab c c ab b c ab c c ab c c ac c c ba a a ba a c ba a c ba c c ba a c ba c c ba c c bc c c ca a a ca a b ca a b ca b b ca a b ca b b ca b b cb b b a a b c a a b c a b b c a b c c a a b c a b b c a b c c a b b c a b c c It implies that the geometric realization of ∆( G ) is a triangulation of 3-dimensionalsphere S with face vector (12 , , ,
33) (see [11]), so G is a Gorenstein graph. Theproof of the theorem is complete. (cid:3) Lemma 4.2.
Assume that G is a well-covered, locally triangle-free, connected graphsuch that(1) α ( G ) (cid:62) ; and(2) G is not a join of its two proper subgraphs; and
3) Each nontrivial connected component of G v is in W for every vertex v .Then, G is in W .Proof. Assume on the contrary that G is not in W . By [3, Lemma 2], there wouldbe an independent set S of G such that G S has an isolated vertex, say b . Let a bea vertex in S so that b is a vertex of G a . Let G (cid:48) be the connected component of G a such that b ∈ V ( G (cid:48) ). If G (cid:48) is nontrivial, then G (cid:48) is in W . Let S (cid:48) = S ∩ V ( G (cid:48) ). Then, S (cid:48) is an independent set of G (cid:48) and G (cid:48) S (cid:48) is an induced subgraph of G S . But then, b isan isolated vertex of G (cid:48) S (cid:48) which contradicts Lemma 1 .
3. Thus, G (cid:48) is a trivial graph.In other words, b is an isolated vertex of G a .Let A := N G ( a ) and B := N G ( b ). Then ab / ∈ E ( G ) and B ⊆ A . Note that A and B are not empty since the graph G is connected. Let H := G { a,b } . By Lemma 1.2, H is well-covered with α ( H ) = α ( G ) − (cid:62)
1. In particular, H (cid:54) = ∅ . Claim: Each vertex of H is adjacent to all vertices in B . Indeed, assume on the contrary that uv / ∈ E ( G ) for some u ∈ V ( H ) and v ∈ B .Since G v has a connected component, say Γ, which contains a 2-path avb , Γ ∈ W and b is an isolated vertex of Γ a , which contradicts Lemma 1 .
3, and the claim follows.Let Z := A \ B . Then, Z = N G b ( a ), and so Z ∈ ∆( G ) because G b is a triangle-freegraph. Now if either Z = ∅ or B ⊆ N G ( z ) for all z ∈ Z , then by Claim above wewould have G = G [ B ] ∗ G [ { a, b } ∪ Z ∪ V ( H )] , a contradiction. Hence, Z (cid:54) = ∅ and there is z ∈ Z such that B (cid:54)⊆ N G ( z ).Next we consider the graph G z . Let H (cid:48) := H \ N G ( z ) and B (cid:48) := B \ N G ( z ) (seeFigure 7). Then, B (cid:48) (cid:54) = ∅ . Let Z := { z ∈ Z \ { z } | z b (cid:48) ∈ E ( G ) for some b (cid:48) ∈ B (cid:48) } ,Z := { z ∈ Z \ { z } | z h (cid:48) ∈ E ( G ) for some h (cid:48) ∈ V ( H (cid:48) ) } , and Z := Z \ ( { z } ∪ Z ∪ Z ) . Z Z Z B H b Figure 7.
The graph G z . ote that all Z , Z , Z are independent sets of G . Inside the triangle-free graph G z we have B (cid:48) = N G z ( b ) and B (cid:48) ∈ ∆( G ). Furthermore, by Claim above we imply that Z ∩ Z = ∅ , N G ( Z ) ∩ B (cid:48) = ∅ , N G ( Z ) ∩ V ( H (cid:48) ) = ∅ , and H (cid:48) is totally disconnected.It follows that Z is the set of isolated vertices in G z and G z \ Z is a connectedbipartite graph with bipartition ( B (cid:48) ∪ Z , V ( H (cid:48) ) ∪ Z ∪ { b } ). Since this bipartite graphis a nontrivial component of G z , it is in W . By Lemma 1 .
4, it is just an edge. Thus, H (cid:48) = ∅ , Z = Z = ∅ and | B (cid:48) | = 1.Finally, since G b is a connected bipartite graph with bipartition ( V ( H ) ∪ { a } , Z ∪{ z } ), it is an edge by Lemma 1 .
4. It follows that V ( H ) = ∅ , i.e. H = ∅ , a contradic-tion. The proof of the lemma is complete. (cid:3) We are now in position to prove the main result of this paper.
Theorem 4.3.
Let G be a simple graph. Then, I ( G ) is Buchsbaum if and only if G is a triangle-free graph in W , or G is isomorphic to one of K n ( n ≥ ), C cn ( n ≥ ), B n ( n (cid:62) ), Q , Q , P or P .Proof. If α ( G ) = 1, then G is a complete graph, and so I ( G ) is always Buchsbaum.If α ( G ) = 2, by [12, Theorem 4.12], I ( G ) is Buchsbaum if and only if ∆( G ) is an n -cycle, or an ( n − n (cid:62) G is isomorphic to one of B n ( n ≥ C cn ( n ≥ α ( G ) (cid:62)
3. By [8, Theorem 3.12] we have I ( G ) is Buchsbaum if andonly if G is Cohen-Macaulay and I ( G v ) is Cohen-Macaulay for all v ∈ V ( G ).If I ( G ) is Buchsbaum, then G is Cohen-Macaulay, and then G is well-covered.Recall that I ( G v ) is Cohen-Macaulay for every vertex v . Since I ( G v ) is Cohen-Macaulay if and only if every nontrivial connected component of G v is triangle-free in W due to [10, Theorem 4 . G is a well-covered locally triangle-free graph.Since G is Cohen-Macaulay, ∆( G ) is connected by [17, Proposition 3.3 in Chapter 0and Corollary 4.2 in Chapter II]. Thus, G is not a join of its two proper subgraphsby Lemma 1 .
6. By Lemma 4.2, G is in W . Together with Theorem 3 .
6, we have G is either a triangle-free graph or isomorphic to one of Q , Q , P or P .Conversely, assume first that G is a triangle-free graph in W . Then, I ( G ) is Cohen-Macaulay (and so is Buchsbaum) by [10, Theorem 4 . G is isomorphic to one of Q , Q , P and P , then G is a locally triangle-freeGorenstein graph by Theorem 4.1. In particular, G is a Cohen-Macaulay graph andin W . Thus, for each vertex v we have G v is a triangle-free graph in W , so I ( G v ) is Cohen-Macaulay by [10, Theorem 4 . I ( G ) is Buchsbaum, and the proofof the theorem is complete. (cid:3) Acknowledgment.
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