Small clique number graphs with three trivial critical ideals
aa r X i v : . [ m a t h . C O ] N ov SMALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICALIDEALS
CARLOS A. ALFARO AND CARLOS E. VALENCIA
Abstract.
The critical ideals of a graph are the determinantal ideals of the generalized Laplacianmatrix associated to a graph. In this article we provide a set of minimal forbidden graphs for the setof graphs with at most three trivial critical ideals. Then we use these forbidden graphs to characterizethe graphs with at most three trivial critical ideals and clique number equal to 2 and 3. Introduction
Given a connected graph G = ( V, E ) and a set of indeterminates X G = { x u : u ∈ V ( G ) } , the generalized Laplacian matrix L ( G, X G ) of G is the matrix with rows and columns indexed by thevertices of G given by L ( G, X G ) uv = ( x u if u = v, − m uv otherwise , where m uv is the number of the edges between vertices u and v of G . Definition 1.1.
For all ≤ i ≤ | V | , the i -th critical ideal of G is the determinantal ideal given by I i ( G, X G ) = h{ det( m ) : m is an i × i submatrix of L ( G, X G ) }i ⊆ Z [ X G ] . We say that a critical ideal is trivial when it is equal to h i . Critical ideals were firstly definedin [8] and have been studied in a more general framework in [2]. The algebraic co-rank γ ( G ) of G is the number of trivial critical ideals of G . The algebraic co-rank allows to separate the set of allsimple connected graphs in the following graph families:Γ ≤ i = { G : G is a simple connected graph with γ ( G ) ≤ i } . In [8] it was proven that if H is an induced subgraph of G , then I i ( H, X H ) ⊆ I i ( G, X G ) for all i ≤ | V ( H ) | . Thus γ ( H ) ≤ γ ( G ). This implies that the set Γ ≤ i is closed under induced subgraphs.Therefore we can say that a graph G is forbidden for Γ ≤ k when γ ( G ) ≥ k + 1. Moreover, we definethe set of minimal forbidden as follows: Definition 1.2.
Let
Forb (Γ ≤ k ) be the set of minimal (under induced subgraphs property) forbiddengraphs for Γ ≤ k . A graph G is called γ - critical if γ ( G − v ) < γ ( G ) for all v ∈ V ( G ). Then Forb (Γ ≤ k ) is equal tothe set of γ -critical graphs with γ ( G ) ≥ k + 1 and γ ( G − v ) ≤ k for all v ∈ V ( G ). Given a family F of graphs, a graph G is called F -free if no induced subgraph of G is isomorphic to a member of F .Thus G ∈ Γ ≤ k if and only if G is Forb (Γ ≤ k )-free. Therefore, characterizing Forb (Γ ≤ k ) leads to acharacterization of Γ ≤ k . Mathematics Subject Classification.
Primary 05C25; Secondary 05C50, 05E99.
Key words and phrases.
Critical ideal, Critical group, Laplacian matrix, Forbidden induced subgraph.Both authors were partially supported by CONACyT grant 166059, the first author was partially supported byCONACyT and the second author was partially supported by SNI.
In [1], these ideas were used to obtain a characterization of Γ ≤ and Γ ≤ . More precisely, it wasfound that Forb (Γ ≤ ) = { P } and thus Γ ≤ consists of the complete graphs. On the other hand, Forb (Γ ≤ ) consists of 5 graphs: P , K \ P , K \ M , cricket and dart . And Γ ≤ consists of thegraphs isomorphic to an induced subgraph of one of the following graphs: tripartite complete graph K n ,n ,n or T n ∨ ( K n + K n ), where G + H denote the disjoint union of the graphs of G and H ,and G ∨ H denote the join of G and H .The main goal of this paper is to provide a set of minimal forbidden graphs for Γ ≤ and a partialdescription of Γ ≤ . Specifically, we will characterize the graphs in Γ ≤ with clique number equalto 2 and 3. Therefore, we prove that if a graph G ∈ Γ ≤ has clique number at most 3, then G isan induced subgraph of one graph in the family F (see Figure 1). The converse is stronger, eachinduced subgraph of a graph in F belongs to Γ ≤ , but not all graphs in F have clique number lessthan 4. n n n n n n n n (i) graph G (ii) family of graphs F (iii) family of graphs F Figure 1.
The family of graphs F . A black vertex represents a clique of cardinality n v , a white vertex represents a stable set of cardinality n v and a gray vertex representsa single vertex.This article is divided as follows. In Section 1.1, we will show how the characterization of Γ ≤ i leads to a characterization of the graphs having critical group with i invariant factors equal to 1.In Section 2, we will construct a graph G d by replacing its vertices by cliques and stable sets. Andwe will show a novel method to verify whether the k - th critical ideal of G d is trivial or not. Thismethod will be applied to prove that each induced subgraph of a graph in the family F belongs toΓ ≤ . In Section 3, we will give a family of minimal forbidden graphs for Γ ≤ , and it will be used toprove that a graph G ∈ Γ ≤ has clique number 2 if and only if G is an induced subgraph of a graphin F (see Figure 3). Section 4 is devoted to prove that if a graph G ∈ Γ ≤ and has clique number3, then G is an induced subgraph of one graph in the family F .1.1. Applications to the critical group.
The
Laplacian matrix L ( G ) of G is the evaluation of L ( G, X G ) at X G = D G , where D G is the degree vector of G . By considering L ( G ) as a linear map L ( G ) : Z V → Z V , the cokernel of L ( G ) is the quotient module Z V / Im L ( G ). The torsion part ofthis module is the critical group K ( G ) of G . The critical group has been studied intensively onseveral contexts over the last 30 years: the group of components [14, 15], the Picard group [3, 4],the
Jacobian group [3, 4], the sandpile group [7], chip-firing game [4, 17], or
Laplacian unimodularequivalence [10, 18].It is known [13, Theorem 1.4] that the critical group of a connected graph G with n vertices canbe described as follows: K ( G ) ∼ = Z d ⊕ Z d ⊕ · · · ⊕ Z d n − , where d , d , ..., d n − are positive integers with d i | d j for all i ≤ j . These integers are called invariantfactors of the Laplacian matrix of G . Besides, if ∆ i ( G ) is the greatest common divisor of the i -minors MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 3 of the Laplacian matrix L ( G ) of G , then the i - th invariant factor d i is equal to ∆ i ( G ) / ∆ i − ( G ), where∆ ( G ) = 1 (for details see [12, Theorem 3.9]). Definition 1.3.
Given an integer number k , let f k ( G ) be the number of invariant factors of thecritical group of G equal to k . Also, let G i = { G : G is a simple connected graph with f ( G ) = i } . The study and characterization of G i is of great interest. In particular, some results and conjectureson the graphs with cyclic critical group can be found in [15, Section 4] and [20, Conjectures 4.3 and4.4]. On the other hand, it is easy to see [7, 14, 18] that G consists only of the complete graphs.Besides, several people (see [17]) posed interest on the characterization of G and G . In this sense,few attempts have been done. In [19] it was characterized the graphs in G whose third invariantfactor is equal to n , n − n −
2, or n −
3. Later in [6] the characterizations of the graphs in G with a cut vertex, and the graphs in G with number of independent cycles equal to n − G was given in [1]. However, nothing is known about G .A crucial result linking critical groups and critical ideals is [8, Theorem 3.6], which states if D G is the degree vector of G , and d | · · · | d n − are the invariant factors of K ( G ), then I i ( G, D G ) = * i Y j =1 d j + = h ∆ i ( G ) i for all 1 ≤ i ≤ n − . Thus if the critical ideal I i ( G, X G ) is trivial, then ∆ i ( G ) and d i are equal to 1. Equivalently, if∆ i ( G ) and d i are not equal to 1, then the critical ideal I i ( G, X G ) is not trivial.The critical ideals behave better than critical group under induced subgraph property. It is notdifficult to see that unlike of Γ ≤ k , G k is not closed under induced subgraphs. For instance, the coneof the claw graph c ( K , ) belongs to G , but the claw graph K , belongs to G . Also, K \ { P } belongs to G , meanwhile K \ { P } belongs to G . Moreover, if H is an induced subgraph of G ,then it is not always true that K ( H ) E K ( G ). For example, K ( K ) ∼ = Z K ( K ) ∼ = Z .On the other hand, a consequence of [8, Theorem 3.6] is that G i ⊆ Γ ≤ i for all i ≥
0. Therefore,after an analysis of the i - th invariant factor of the Laplacian matrix of the graphs in Γ ≤ i , thecharacterization of G i can be obtained. See for instance [1] for the characterizations of G and G .2. Cliques, stable sets and critical ideals
Let G = ( V, E ) a simple graph. Suppose that V ′ is a subset of V , the induced subgraph G [ V ′ ]of G is the subgraph of G whose vertex set is V ′ and whose edge set is the set of those edges of G that have both ends in V ′ . If E ′ is a subset of E , the edge-induced subgraph G [ E ′ ] is the subgraphof G whose edge set is E ′ and whose vertex set consists of all ends of edges of E ′ . Let P, Q be twosubsets of V , we denote by E ( P, Q ) the set of edges of G with one end in P and the other end in Q . A clique of G is a subset S of V of mutually adjacent vertices, and the maximum size of a cliqueof G is the clique number ω ( G ) of G . A subset S of V is called an independent set , or stable set , of G if no two vertices of S are adjacent in G . The graph with n vertices whose vertex set induces astable set is called the trivial graph, and denoted by T n . The cardinality of a maximum stable setin G is called the stability number of G and is denoted by α ( G ).Given a simple graph G = ( V, E ) and a vector d ∈ Z V , the graph G d is constructed as follows.For each vertex u ∈ V is associated a new vertex set V u , where V u is a clique of cardinality − d u if d u is negative, and V u is a stable set of cardinality d u if d u is positive. And each vertex in V u isadjacent with each vertex in V v if and only if u and v are adjacent in G . Then the graph G is calledthe underlying graph of G d . CARLOS A. ALFARO AND CARLOS E. VALENCIA
A convenient way to visualize G d is by means of a drawing of G , where the vertex u is colored inblack if d u is negative, and colored in white if d u is positive. We indicate the cardinality of V u bywriting it inside the drawing of vertex u . When | d u | = 1, we may color u in gray and avoid writingthe cardinality (see Figure 1). On the other hand, it will be useful to avoid writing the cardinalitywhen | d u | = 2 (see Figure 2).In general, the computation of the Gr¨obner bases of the critical ideals is more than complicated.However, in the rest of this section we will show a novel method, developed in [2], to decide for i ≤ | V ( G ) | whether the i - th critical ideal of G d is trivial or not.For V ′ ⊆ V ( G ) and d ∈ Z V ′ , we define φ ( d ) as follows: φ ( d ) v = ( d v > , − d v < . Theorem 2.1. [2, Theorem 3.7]
Let n ≥ and G be a graph with n vertices. For V ′ ⊆ V ( G ) , ≤ j ≤ n and d ∈ Z V ′ , the critical ideal I j ( G d , X G d ) is trivial if and only if the evaluation of I j ( G, X G ) at X G = φ ( d ) is trivial. Thus the procedure of verify whether a family of graphs belongs to Γ ≤ i becomes in an evaluationof the i - th critical ideal of the underlying graph of the family.Let G be the underlying graph of the family of graphs F (see Figure 1.ii) with vertex set V = { v , v , v , v , v , v , v } . Let d ∈ Z V such that d , d , d are positive integers and d , d , d , d are negative integers. Thus φ ( d ) = (0 , , , − , − , − , − I ( G , X G ) = h , x , x , x , x + 1 , x + 1 , x + 1 , x + 1 i . Since the evaluation I ( G , X G ) at X G = φ ( d ) is equal to h i , then by Theorem 2.1 the criticalideal I ( G d , X G d ) is non-trivial. Therefore, each graph in this family of graphs has algebraic co-rankat most 3. Now let G be the underlying graph of the family of graphs F (see Figure 1.iii) withvertex set V = { v , v , v , v , v , v } . Let d be positive integer. By using a computer algebra systemwe can check that I ( G , X G ) = h x + 5 x + 5 , x + x + 3 , x + x + 3 , x + x + 3 , x + x + 3 , x + x + 3 i . Since the evaluation of I ( G , X G ) at x = 0 is non-trivial, then by Theorem 2.1 the critical ideal I ( G d , X G d ) is non-trivial. Therefore, each graph in F has algebraic co-rank at most 3.On the other hand, it can be verified that γ ( G ), γ ( G ) and γ ( G ) are equal to 3. Then thegraphs in F have algebraic co-rank 3, and so each induced subgraph of a graph in F has algebraicco-rank at most 3. Proposition 2.2.
Each graph in F belongs to Γ ≤ . A description of Γ ≤ It is possible to compute the algebraic co-rank of all connected graphs with at most 9 verticesusing the software Macaulay2 [9] and Nauty [16]. The computation of the algebraic co-rank of theconnected graphs with at most 8 vertices required at most 3 hours on a MacBookPro with a 2.8 GHzIntel i7 quad core processor and 16 GB RAM. Besides, the computation of the algebraic co-rank ofthe connected graphs with 9 vertices required 4 weeks of computation on the same computer.Let F be the family of graphs shown in Figure 2. This family represents the graphs in Forb (Γ ≤ )with at most 8 vertices. Since there exists no minimal forbidden graph with 9 vertices for Γ ≤ , thenit is likely that F = Forb (Γ ≤ ). MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 5 P G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , G , Figure 2.
The family of graphs F . A black vertex represents a clique of cardinality2, a white vertex represents a stable set of cardinality 2 and a gray vertex representa single vertex. Proposition 3.1.
Each graph in F belongs to Forb (Γ ≤ ) .Proof. It can easily checked, using a computer algebra system, that each graph in F is γ -criticaland has algebraic co-rank equal to 4. (cid:3) One of the main results of this article is the following:
Theorem 3.2.
If a graph G ∈ Γ ≤ has clique number at most 3, then G is an induced subgraph ofa graph in F . We divide the proof in two characterizations: the graphs in Γ ≤ with clique number equal to 2 and3. The converse of Theorem 3.2 is stronger, by Proposition 2.2 we have that each induced subgraphof a graph in F belongs to Γ ≤ . However, it is not difficult to recognize the graphs in F withclique number greater or equal than 4. Theorem 3.3.
Let G be a simple connected graph with ω ( G ) = 2 . Then, G is F -free if and only if G is isomorphic to an induced subgraph of a graph in F (see Figure 3). n n n n n n n n (i) K n ,n (ii) F (iii) F (iv) F Figure 3.
The family of graphs F . A white vertex represents a stable set ofcardinality n v and a gray vertex represents a single vertex. CARLOS A. ALFARO AND CARLOS E. VALENCIA
Since each graph in F is isomorphic to an induced subgraph of a graph in F , then Proposition2.2 implies that each graph in F belongs to Γ ≤ . Note that the graphs in Γ ≤ and Γ ≤ are inducedsubgraph of a graph in F (see Figure 1.ii). Theorem 3.4.
Let G be a simple connected graph with ω ( G ) = 3 . Then, G is F -free if and only if G is isomorphic to an induced subgraph of a graph in F with clique number 3. Section 4 is devoted to the proof of the Theorem 3.4. Now we give the proof of Theorem 3.3.
Proof of Theorem 3.3.
Since each graph in F belongs to Γ ≤ , then each graph in F is F -free.Thus, we get one implication.Suppose G is F -free. Let a, b ∈ V ( G ) such that ab ∈ E ( G ). Since ω ( G ) = 2, then there is novertex adjacent with a and b at the same time. For a vertex v ∈ V ( G ), the neighbor set N G ( v ) of v in G is the set of all vertices adjacent with v . Let A = N G ( a ) − b and B = N G ( b ) − a . Clearly, eachof the sets A and B induces a trivial graph. Let us define A = { u ∈ A : ∃ v ∈ B such that uv ∈ E } , A ′ = { u ∈ A : ∄ v ∈ B such that uv ∈ E } B = { u ∈ B : ∃ v ∈ A such that uv ∈ E } , and B ′ = { u ∈ B : ∄ v ∈ A such that uv ∈ E } . Thus we have two possible cases: when A and B are not empty and when A and B are empty. First we consider when A and B are not empty. In this case we have the following statements:
Claim 3.5.
One of the sets A ′ or B ′ is empty, and the other has cardinality at most one.Proof. Suppose A ′ and B ′ are not empty. Let u ∈ A , v ∈ B , s ∈ A ′ and t ∈ B ′ . Then the vertexset { a, b, u, v, s, t } induces a graph isomorphic to G , , which is impossible. Now suppose A ′ hascardinality more than 1. Take w , w ∈ A ′ . Then { x, y, u, v, w , w } induces a graph isomorphic to G , ; a contradiction. Thus A ′ has cardinality at most one. (cid:3) Claim 3.6.
The edge set E ( A, B ) induces either a complete bipartite graph or a complete bipartitegraph minus an edge.Proof. First note that each vertex in A is incident with each vertex in B , except for at most onevertex. It is because if u ∈ A and v , v , v ∈ B such that uv ∈ E ( G ) and uv , uv / ∈ E ( G ), thenthe induced subgraph G [ { x, y, u, v , v , v } ] is isomorphic to G , ; which is impossible. In a similarway, each vertex in B is incident with each vertex in A , except for at most one vertex. Thus theedge set E ( A, B ) must be equal to the edge set { uv : u ∈ A, v ∈ B } minus a matching. In fact,the cardinality of this matching must be at most one. Otherwise, if u, v ∈ A and s, t ∈ B such that us, vt / ∈ E ( G ) and ut, vs ∈ E ( G ), then the induced subgraph G [ { t, u, a, v, s } ] is isomorphic to P ;which is a contradiction. (cid:3) Claim 3.7.
It is not possible that, at the same time, the edge set E ( A, B ) induces a complete bipartitegraph minus an edge and A ′ ∪ B ′ = ∅ .Proof. Suppose both situations occur at the same time. Let u , u ∈ A and v , v ∈ B such that u v , u v , u v ∈ E ( G ) and u v / ∈ E ( G ). And let w ∈ A ′ . Then the vertex set { u , u , v , v , y, w } induces a graph isomorphic to G , ; which is a contradiction. (cid:3) Thus there are three possible cases: (a) E ( A, B ) = { uv : u ∈ A, v ∈ B } and A ′ ∪ B ′ = ∅ , (b) E ( A, B ) = { uv : u ∈ A, v ∈ B } and A ′ = T , B ′ = ∅ or (c) E ( A, B ) induces a bipartite complete graph minus an edge and A ′ ∪ B ′ = ∅ . MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 7
Let V ∅ denote the set of vertices that are not adjacent with a nor b . In what follows we willdescribe the vertex set V ∅ . Case (a).
Let w , w ∈ V ∅ , u , u ∈ A , and v ∈ B . Note that it is not possible that a vertex in V ∅ is adjacent with a vertex in A and a vertex in B at the same time, because then we get ω ( G ) ≥ Claim 3.8.
There exist no two vertices in V ∅ such that one is adjacent with a vertex in A and theother one is adjacent with a vertex in B .Proof. Suppose w u , w v ∈ E ( G ). There are two cases: either w w ∈ E ( G ) or w w / ∈ E ( G ).If w w ∈ E ( G ), then G [ { w , w , v , b, a } ] ≃ P ; which is impossible. And if w w / ∈ E ( G ), then G [ { w , w , v , u , b, a } ] ≃ G , ; which is a contradiction. (cid:3) Without loss of generality, suppose w ∈ V ∅ is adjacent with u ∈ A . Thus Claim 3.9.
The vertex set V ∅ has cardinality at most 1.Proof. There are two possible cases: either each vertex in V ∅ is adjacent with a common vertexin A or not. Suppose there exists w ∈ V ∅ such that u is adjacent with both w and w . Then w w / ∈ E ( G ), because otherwise w , w and u induce a K . Thus the vertex set { w , w , u , v , a, b } induces a graph isomorphic to G , ; which is forbidden. Then this case is not possible. Now suppose w and w are not adjacent with a common vertex in A . We have the following possible cases:(1) w u , w u ∈ E ( G ),(2) w u , w u , w w ∈ E ( G ), or(3) w w ∈ E ( G ).This yields a contradiction since in case (1) the vertex set { w , u , v , u , w } induces a graph iso-morphic to P , in case (2) the vertex set { w , w , u , v , b } induces a graph isomorphic to P , and incase (3) the vertex set { w , w , u , v , b } induces a graph isomorphic to P . (cid:3) If | A | = 2, there are two possibilities: either w is adjacent with each vertex in A or w is adjacentwith only one vertex of A . If | A | ≥
3, then w is adjacent with either all vertex in A or only onevertex in A . It is because if w is adjacent with both u , u ∈ A and w is not adjacent with u ∈ A ,then the vertex set { w , u , u , u , a, b } induces a graph isomorphic to G , . Note that when w isadjacent with each vertex in A , then the graph is isomorphic to an induced subgraph of a graph in F . Meanwhile, when w is adjacent only with one vertex of A , then the graph is isomorphic to aninduced subgraph of a graph in F . Finally, when V ∅ = ∅ , the graph is a complete bipartite graph. Case (b).
Suppose without loss of generality that A ′ = ∅ . By Claims 3.8 and 3.9, the vertex set V ∅ has cardinality at most one. Let w ∈ V ∅ , u ∈ A , u ∈ A ′ and v ∈ B . We have two cases: w is adjacent with either a vertex in A or a vertex in B . Let us consider when w is adjacent with avertex in A . Here we have two possibilities: either wu ∈ E ( G ) or wu / ∈ E ( G ). However, none ofthe two cases is allowed, since in the first case we get that the vertex set { w, u , a, b, v } induces agraph isomorphic to P , and in the second case the vertex set { w, u , a, b, u , v } induces a graphisomorphic to G , . Thus the remaining case is that w is adjacent with a vertex in B . In this case w must be adjacent with u and each vertex in B , because otherwise the graph P appears as inducedsubgraph. Note that this graph is isomorphic to an induced subgraph of a graph in F . Case(c).
Let w ∈ V ∅ , u , u ∈ A , and v , v ∈ B such that u v / ∈ E ( G ) and u v , u v , u v ∈ E ( G ).Note that if w is adjacent with u or v , then w is adjacent with both u and v , because since u v / ∈ E ( G ), we would obtain P as induced subgraph; which is not possible. In this case, when w is adjacent with u and v , the graph is isomorphic to an induced subgraph of F . Now we consider CARLOS A. ALFARO AND CARLOS E. VALENCIA when w is adjacent with a vertex in ( A − u ) ∪ ( B − v ). Without loss of generality, we can suppose w is adjacent with u . The vertex w is not adjacent with v or v , because otherwise a clique ofcardinality 3 is obtained. On the other hand, w is not adjacent with u , because otherwise w wouldbe adjacent with both vertices u and v . Thus w is adjacent only with u , but the vertex set { w, u , u , v , a, b, } induces a graph isomorphic to G , ; a contradiction. Thus V ∅ is empty, and thegraph is isomorphic to an induced subgraph of F . Now we consider the case when A and B are empty. One of the vertex sets A ′ or B ′ hascardinality at most one, because otherwise A ′ ∪ B ′ ∪ { a, b } would contain G , as induced subgraph.Thus, let us assume that A ′ = { u } and | B ′ | >
1. Let V ∅ denote the vertex set whose vertices are notadjacent with both a and b . Claim 3.10. If A ′ and B ′ are not empty, and w ∈ V ∅ is adjacent with a vertex in A ′ ∪ B ′ , then w is adjacent with each vertex in A ′ ∪ B ′ .Proof. Let v ∈ B ′ . Suppose one of the edges uw or vw does not exist. Then { w, u, a, b, v } induces agraph isomorphic to P ; which is a contradiction. (cid:3) Note that the vertex set V ∅ induces a stable set, because otherwise ω ( G ) >
2. Thus when A ′ and B ′ are not empty, the graph G is isomorphic to an induced subgraph of a graph in F .Now consider the case when B = ∅ and | A | > Claim 3.11.
Each vertex w ∈ V ∅ is adjacent with either a unique vertex in A or each vertex in A .Proof. The result is easy to check when | A | ≤
2. So suppose A has cardinality greater or equal than3. Let u , u , u ∈ A such that w is adjacent with both u , u , and w is not adjacent with u . Sincethe vertex set { w, u , u , u , a, b } induces a graph isomorphic to G , , then we get a contradictionand the result follows. (cid:3) Claim 3.12.
Let w ∈ V ∅ such that it is adjacent with u ∈ A . If V ∅ has cardinality greater than 1,then each vertex in V ∅ is adjacent either with u or with each vertex in A .Proof. Suppose there exists w ′ ∈ V ∅ such that w ′ is not adjacent with u . Let u ′ ∈ A such that w ′ isadjacent with w ′ . There are four possible cases: • wu ′ , ww ′ ∈ E ( G ), • ww ′ ∈ E ( G ) and wu ′ / ∈ E ( G ), • wu ′ ∈ E ( G ) and ww ′ / ∈ E ( G ), • wu ′ , ww ′ / ∈ E ( G ).In the cases when ww ′ ∈ E ( G ), the vertex set { w, u, a, u ′ , w ′ } induces a graph isomorphic to P ;those cases do not occur. On the other hand if wu ′ ∈ E ( G ) and ww ′ / ∈ E ( G ), then { w, w ′ , u, u ′ , a, b } induces a graph isomorphic to G , , and this case can not occur. Finally, if wu ′ , ww ′ / ∈ E ( G ), thenthe vertex set { w, u, a, u ′ , w ′ } induces a graph isomorphic to P . Which is a contradiction. Thuseach pair of vertices in V ∅ are adjacent with the same vertices in A . And the result follows. (cid:3) Thus there are two cases: when each vertex in V ∅ is adjacent with a unique vertex u in A , andwhen each vertex in V ∅ is adjacent with each vertex in A . Note that V ∅ induces a stable set, becauseotherwise ω ( G ) >
2. In the first case we have that either | V ∅ | ≥ A − u is empty. Itis because otherwise G would have G , as induced subgraph. Therefore, in each case G is isomorphicto a graph in F . (cid:3) MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 9 Proof of Theorem 3.4
One implication is easy because Proposition 2.2 implies that each graph in F is F -free. Theother implication is much more complex.Suppose G is F -free. Let W = { a, b, c } be a clique of cardinality 3. For each X ⊆ { a, b, c } , let V X = { u ∈ V ( G ) : N G ( u ) ∩ { a, b, c } = X } . Note that V ∅ denote the vertex set whose vertices arenot adjacent with a vertex in { a, b, c } .Since ω ( G ) = 3, then the vertex set V a,b,c is empty, and for each pair { x, y } ⊂ { a, b, c } , the vertexset V x,y induces a stable set. Furthermore Claim 4.1.
For x ∈ { a, b, c } , the induced subgraph G [ V x ] is isomorphic to either K m,n , K , K + K ,or T n .Proof. First consider K , P , K + T and P + T as induced subgraphs of G [ V x ]. Since G [ K ∪ x ] ≃ K , G [ P ∪ { x, y } ] ≃ G , , G [ K + T ∪ { a, b, c } ] ≃ G , , G [ P + T ∪ { x, y } ] ≃ G , and all of themare forbidden for G , then the graphs K , P , K + T and P + T are forbidden in G [ V x ]. Thus ω ( G [ V x ]) ≤ ω ( G [ V x ]) = 2, then there exist u, v ∈ V x such that uv ∈ E ( G [ V x ]). Clearly, N G [ V x ] ( u ) ∩ N G [ V x ] ( v ) = ∅ , and each vertex set N G [ V x ] ( u ) and N G [ V x ] ( u ) induces a trivial subgraph. Since G [ V x ]is P -free, then st ∈ E ( V x ) for all s ∈ N V x ( u ) \ { v } and t ∈ N V x ( v ) \ { u } . Therefore, each componentin G [ V x ] is a complete bipartite subgraph.If a component in G [ V x ] has cardinality at least three, then G [ V x ] does not have another com-ponent, because the existence of another component makes that P + T appears as an inducedsubgraph in G [ V x ]; which is impossible. If there is a component in G [ V x ] of cardinality at least two,then there is at most another component in G [ V x ] since K + T is forbidden in G [ V x ]. And thus theresult turns out. (cid:3) Claim 4.2.
If there is x ∈ { a, b, c } such that the induced subgraph G [ V x ] has at least 2 componentsor is isomorphic to a complete bipartite with at least three vertices, then the vertex set V y is emptyfor each y ∈ { a, b, c } − x .Proof. Let G [ V x ] be as above. Then there are two vertices u, v ∈ V x that are not adjacent. Suppose V y is not empty, that is, there is w ∈ V y . There are three possibilities: • uw, vw ∈ E ( G ), • uw ∈ E ( G ) and vw / ∈ E ( G ), and • uw, vw / ∈ E ( G ).But in each case the vertex set { a, b, c, u, v, w } induces a graph isomorphic to G , , G , , and G , , respectively. Since these graphs are forbidden, then we obtain a contradiction and then V y isempty. (cid:3) Claim 4.3.
If there is x ∈ { a, b, c } such that G [ V x ] has at least 2 components or is isomorphicto a complete bipartite with at least three vertices, then the vertex set V x,y is empty for each y ∈{ a, b, c } − x .Proof. Let G [ V x ] be as above. Then there are two vertices u , u ∈ V x that are not adjacent. Suppose V x,y = ∅ . Let v ∈ V x,y . There are three possibilities: • u v ∈ E ( G ) and u v ∈ E ( G ), • u v ∈ E ( G ) and u v / ∈ E ( G ), and • u v / ∈ E ( G ) and u v / ∈ E ( G ). Then in each case, G [ { a, b, c, u , u , v } ] is isomorphic to G , , G , and G , , respectively. Sincethese graphs are forbidden, we have that V x,y is empty. (cid:3) Remark 4.4.
Claims 4.2 and 4.3 imply that when V x has connected 2 components or is a completebipartite graph of at least 3 vertices, then the only non-empty vertex sets are V ∅ , V x and V y,z , where x, y and z are different elements of { a, b, c } Moreover, by Claim 4.1, the vertex set V x is one of thefollowing vertex sets: • T n , where n ≥ , • complete bipartite graph with cardinality at least 3, • K + K or K . Next result describes the induced subgraph G [ V a ∪ V b ∪ V c ] when each set V x is connected ofcardinality at most 2. Claim 4.5.
Suppose for each x ∈ { a, b, c } the vertex set V x is connected of cardinality at most 2. Iffor all x ∈ { a, b, c } the set V x is not empty, then G [ V a ∪ V b ∪ V c ] is isomorphic (where x, y and z aredifferent elements of { a, b, c } ) to one of the following sets: • V x ∨ ( V y + V z ) where V x = K , V y = K m , V z = K n and m, n ∈ { , } , • V x ∨ ( V y + V z ) where V x = K , V y = K , V z = K , or • V x ∨ ( V y ∨ V z ) where V x = K , V y = K , V z = K .If V z = ∅ , then G [ V x ∪ V y ] is isomorphic to one of the following sets: • V x + V y , where V x = K m , V y = K n and m, n ∈ { , } , or • V x ∨ V y , where V x = K , V y = K m and m ∈ { , } .If V y = V z = ∅ , then V x is isomorphic to K or K .Proof. It is not difficult to prove that either E ( V x , V y ) is empty or E ( V x , V y ) induces a completebipartite graph. The result follows by checking the possibilities with a computer algebra system. (cid:3) In the rest of the proof, for each case obtained in Remark 4.4 and Claim 4.5, we will analyze theremaining edges sets and the vertex set V ∅ . As before, we may refer to x, y or z as different elementsof { a, b, c } . Each case will be consider in a subsection.4.1. When V a ∪ V b ∪ V c = ∅ . Now we describe the induced subgraph G [ V a,b ∪ V a,c ∪ V b,c ]. Claim 4.6.
If the vertex sets V x,y and V y,z are not empty, and the edge set E ( V x,y , V y,z ) is empty,then | V x,y | = | V y,z | = 1 .Proof. Suppose | V x,y | ≥ V y,z = ∅ . Take u, u ′ ∈ V x,y and v ∈ V y,z . The result follows since thevertex set { a, b, c, u, u ′ , v } induces a graph isomorphic to the forbidden graph G , . (cid:3) Claim 4.7. If E ( V x,y , V y,z ) = ∅ , E ( V x,y , V x,z ) = ∅ and E ( V y,z , V x,z ) = ∅ , then V x,z is empty.Proof. Let u ∈ V x,y and v ∈ V y,z such that uv ∈ E ( G ). Suppose there exists w ∈ V x,z . Then uw, vw / ∈ E ( G ). Since the induced subgraph G [ { a, b, c, u, w, v } ] is isomorphic to G , , then we geta contradiction. Then V x,z is empty. (cid:3) Claim 4.8. If E ( V x,y , V y,z ) = ∅ , then E ( V x,y , V y,z ) induces either a complete bipartite graph or acomplete bipartite graph minus an edge. MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 11
Proof.
Let u ∈ V x,y . Suppose there exist v, v ′ ∈ V y,z such that uv, uv ′ / ∈ E ( G ). Since the inducedsubgraph G [ { a, b, c, u, v, v ′ } ] is isomorphic to G , , which is forbidden, then the vertex u is adjacentwith at least all but one vertices in V y,z . In a similar way, we have that each vertex in V y,z is adjacentwith at at least all but one vertices in V x,y . Therefore, the edge set E ( V x,y , V y,z ) induces a completebipartite graph minus a matching. Now suppose this matching has cardinality greater or equal to 2.Then there exist u, u ′ ∈ V x,y and v, v ′ ∈ V y,z such that uv ′ , u ′ v / ∈ E ( G ) and uv, u ′ v ′ ∈ E ( G ). Since G [ { v, u, x, u ′ , v ′ } ] is isomorphic to P , then we get a contradiction and the matching has cardinalityat most 1. Therefore, E ( V xy , V y,z ) induces a complete bipartite graph or a complete bipartite graphminus an edge. (cid:3) Claim 4.9. If E ( V x,y , V y,z ) = ∅ , E ( V y,z , V x,z ) = ∅ and E ( V x,y , V x,z ) = ∅ , then V x,y = { v } , V x,z = { v } and one of the following two statements holds: • E ( v , V y,z ) = { v v : v ∈ V y,z } and E ( v , V y,z ) = { v v : v ∈ V y,z } , or • there exists v ∈ V y,z such that E ( v , V y,z ) = { v v : v ∈ V y,z − v } and E ( v , V y,z ) = { v v : v ∈ V y,z − v } .Proof. Since there is no edge joining a vertex in V x,y and a vertex in V x,z , then Claim 4.6 impliesthat | V x,y | = | V x,z | = 1. Let V x,y = { v } and V x,z = { v } . By Claim 4.8, each edge set E ( V x,y , V y,z )and E ( V y,z , V x,z ) induces either a complete bipartite graph or a complete bipartite graph minus anedge. If both sets E ( V x,y , V y,z ) and E ( V y,z , V x,z ) induce either a complete bipartite graph, then weare done. So it remains to check two cases: • when one edge set induces a complete bipartite graph and the other one induces a completebipartite graph minus an edge, and • when both edge sets induce a complete bipartite graph minus an edge.First consider the former case. Suppose there exists v ∈ V y,z such that E ( v , V y,z ) = { v v : v ∈ V y,z − v } and E ( v , V y,z ) = { v v : v ∈ V y,z } . Since G [ { a, b, c, v , v , v } ] is isomorphic to G , , then thiscase is not possible. Now consider second case. Let v , v ∈ V y,z . There are two possible cases: either v v , v v / ∈ E ( G ), or v v , v v / ∈ E ( G ). Suppose v v , v v / ∈ E ( G ). Since G [ { v , v , z, v , v } ] isisomorphic to P , then this case is impossible. And therefore v v , v v / ∈ E ( G ). (cid:3) Claim 4.10.
If each edge set E ( V x,y , V y,z ) is not empty, then the induced subgraph G [ V a,b ∪ V a,c ∪ V b,c ] is isomorphic to one of the following graphs: • a complete tripartite graph, • a complete tripartite graph minus an edge, or • a complete tripartite graph minus the edges v v , v v , v v , where v ∈ V x,y , v ∈ V y,z , v ∈ V x,z .Proof. By Claim 4.8, the induced subgraph G [ V a,b ∪ V a,c ∪ V b,c ] is a complete tripartite graph minusat most three edges. We will analyze the cases where 2 or 3 edges have been removed.Suppose G [ V a,b ∪ V a,c ∪ V b,c ] induces a complete tripartite graph minus 2 edges. Since E ( V x,y , V y,z )induces complete bipartite graph or complete bipartite graph minus an edge, then the 2 edges cannotbe removed from a unique edge set E ( V x,y , V y,z ). Then there are two possibilities: • there exist u, u ′ ∈ V x,y , v ∈ V y,z and w ∈ V x,z such that uv, u ′ w / ∈ E ( G ), or • there exist u ∈ V xy , v ∈ V y,z and w ∈ V x,z such that uv, uw / ∈ E ( G ).In the first case the graph induced by the set { x, v, u, u ′ , w, z } is isomorphic to the forbidden graph G , ; which is impossible. And in the second case, the the induced subgraph G [ { a, b, c, u, v, w } ] is isomorphic to G , , which is impossible. Thus the case where 2 edges are removed from G [ V a,b ∪ V a,c ∪ V b,c ] is not possible.Suppose G [ V a,b ∪ V a,c ∪ V b,c ] is a complete tripartite graph minus 3 edges. Since E ( V x,y , V y,z )induces a complete bipartite graph or a complete bipartite graph minus an edge, then the 3 edgescannot be removed from a unique edge set E ( V x,y , V y,z ). Thus there are four possible cases:(a) v v , v v , v v / ∈ E ( G ), where v , v ∈ V x,y , v , v ∈ V y,z and v , v ∈ V x,z (b) v v , v v , v v / ∈ E ( G ), where v , v ∈ V x,y , v , v ∈ V y,z and v ∈ V x,z (c) v v , v v , v v / ∈ E ( G ), where v , v ∈ V x,y , v , v ∈ V y,z and v ∈ V x,z and(d) v v , v v , v v / ∈ E ( G ), where v , v ∈ V x,y , v , v ∈ V y,z and v ∈ V x,z .Cases ( a ), ( b ) and ( d ) are impossible, the argument is the following. Case ( a ) is impossible becausethe induced subgraph G [ { v , v , v , v , x, y } ] is isomorphic to the forbidden graph G , . In case ( b ),the induced subgraph G [ { v , v , v , a, b, c } ] is isomorphic to the forbidden graph G , . And in case( d ), the induced subgraph G [ { v , v , v , a, b, c } ] is isomorphic to the forbidden graph G , . Thuswhen 3 edges are removed the only possible case is ( c ). (cid:3) By previous Claims we have the following cases:(1) when the set V x,y is the only not empty set,(2) when V x,z = ∅ , V x,y = { v xy } , V y,z = { v yz } and v xy v yz / ∈ E ( G ),(3) when V x,z = ∅ and E ( V x,y , V y,z ) induces a bipartite complete graph,(4) when V x,z = ∅ and E ( V x,y , V y,z ) induces a bipartite complete graph minus an edge,(5) when V a,b = { v ab } , V a,c = { v ac } , V b,c = { v bc } and v ab v ac , v ab v bc , v ac v bc / ∈ E ( G ),(6) when V x,z = { v xz } , V y,z = { v yz } , v xz v yz / ∈ E ( G ), and the edge sets E ( v xz , V x,y ) and E ( v yz , V x,y ) induce a complete bipartite graph,(7) when V x,z = { v xz } , V y,z = { v yz } , v xz v yz / ∈ E ( G ), and there exists v xy ∈ V x,y such that E ( v xz , V x,y ) = { v xz v : v ∈ V x,y − v xy } and E ( v yz , V x,y ) = { v yz v : v ∈ V x,y − v xy } .(8) when G [ V a,b ∪ V a,c ∪ V b,c ] is isomorphic to a complete tripartite graph,(9) when G [ V a,b ∪ V a,c ∪ V b,c ] is isomorphic to a complete tripartite graph minus an edge,(10) when G [ V a,b ∪ V a,c ∪ V b,c ] is isomorphic to a complete tripartite graph, where v ∈ V x,y , v ∈ V y,z , v ∈ V x,z and v v , v v , v v / ∈ E ( G ).Now we describe the vertex set V ∅ . Remark 4.11.
Let w, w ′ ∈ V ∅ . Suppose w is adjacent with a vertex in V x,y and with w ′ . Then thevertex w ′ is adjacent with a vertex in V x,y , because otherwise the shortest path from w ′ to z wouldcontains a graph isomorphic to P . Thus each vertex in V ∅ is adjacent with a vertex in V x,y for some { x, y } ⊂ { a, b, c } . Claim 4.12. If w ∈ V ∅ is adjacent with v ∈ V x,y , then either w is adjacent only with v and withno other vertex in V x,y , or w is adjacent with each vertex in V x,y . Moreover, if each vertex in V ∅ is adjacent with a vertex in V x,y , then either exists a vertex v ∈ V v,x such that each vertex in V ∅ isadjacent with v , or each vertex in V ∅ is adjacent with each vertex in V x,y .Proof. Since the first statement is easy when V x,y has cardinality at most 2, then we assume that V x,y has cardinality at least 3. Let v ′ , v ′′ ∈ V x,y . Suppose w is adjacent with v and v ′ but not adjacentwith v ′′ . Since G [ { x, z, v, v ′ , v ′′ , w } ] is isomorphic to G , , then we get a contradiction. And then w is adjacent only with v or with v, v ′ and v ′′ .Let w, w ′ ∈ V ∅ . Suppose there is v ∈ V x,y such that wv ∈ E ( G ) and w ′ v / ∈ E ( G ). The vertex w isnot adjacent with w ′ , because otherwise by Remark 4.11 we get a contradiction. Let v ′ ∈ V x,y suchthat w ′ is adjacent with w ′ . Thus there are two possible cases: MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 13 • ww ′ , wv ′ / ∈ E ( G ), and • wv ′ ∈ E ( G ) and ww ′ / ∈ E ( G ).Since in the first case G [ { w, v, x, v ′ , w ′ } ] is isomorphic to P and in the second case G [ { x, z, v, v ′ , w, w ′ } ]is isomorphic to G , , then we get a contradiction and thus there is no vertex in V x,y adjacent witha vertex in V ∅ and not adjacent with other vertex in V ∅ . (cid:3) Claim 4.13.
Let v ∈ V x,y . If each vertex in V ∅ is adjacent with v , then V ∅ induces either K or itis a trivial graph. If furthermore there exists v ′ ∈ V x,y such that no vertex in V ∅ is adjacent with v ′ ,then V ∅ is a clique of cardinality at most 2.Proof. First note that P is forbidden as induced subgraph in G [ V ∅ ]. It is because if the vertices w , w , w ∈ V ∅ induce a graph isomorphic to P , then G [ { x, y, v, w , w , w } ] ≃ G , . Now we willsee that each component in G [ V ∅ ] is a clique. Let C be a component in G [ V ∅ ]. Suppose C is nota clique, then there are two vertices not adjacent in C , say w and w ′ . Let P be the smallest pathcontained in C between w and w ′ . The length of P is greater or equal to 3. So P is an inducedsubgraph of P , and hence of C . Which is a contradiction, and therefore, C is a clique. On the otherhand, the graph K + K is forbidden as induced subgraph in G [ V ∅ ]. It is because if w , w , w ∈ V ∅ such that G [ { w , w , w } ] ≃ K + K , then G [ { x, y, v, w , w , w } ] ≃ G , . Therefore, if G [ V ∅ ] hasmore than one component, then each component has cardinality one.Let v, v ′ ∈ V x,y such that each vertex in V ∅ is adjacent with v , and no vertex in V ∅ is adjacentwith v ′ . Suppose V ∅ induces a stable set of cardinality at least 2. Take w, w ′ ∈ V ∅ . Then we get acontradiction since the induced graph G [ { w, w ′ , v, v ′ , x, z } ] is isomorphic to G , . Thus V ∅ is a cliqueof cardinality at most 2. (cid:3) Thus by Claims 4.12 and 4.13, in case (1) we have the following possible cases: • V ∅ is a clique of cardinality at most 2, and each vertex in V ∅ is adjacent with only one vertexin V x,y , and • V ∅ is a trivial graph and E ( V ∅ , V x,y ) induces a complete bipartite graph.Note that these graphs are isomorphic to an induced subgraph of a graph in F . Claim 4.14. If E ( V x,y , V y,z ) = ∅ and E ( V ∅ , V x,y ∪ V y,z ) = ∅ , then V ∅ is a clique of cardinality atmost 2, and each vertex in V ∅ is adjacent each vertex in V x,y ∪ V y,z Proof.
Let v xy ∈ V x,y , v yz ∈ V y,z such that v xy v yz / ∈ E ( G ). It is easy to see that if w is adjacentwith v xy or v yz , then w is adjacent with both vertices, because otherwise G has an induced subgraphisomorphic to P . Thus each vertex in V ∅ is adjacent with each vertex in V x,y ∪ V y,z . Now suppose w, w ′ ∈ V ∅ such that w and w ′ are not adjacent. Since the induced subgraph G [ { w, w ′ , x, y, v xy , v yz } ]is isomorphic to G , , then we get a contradiction and V ∅ induces a clique of cardinality at most2. (cid:3) By previous Claim we get that in case (2), V ∅ is a clique of cardinality at most 2 and each vertexin V ∅ is adjacent each vertex in V x,y ∪ V y,z . This graph is isomorphic to an induced subgraph of agraph in F . Claim 4.15.
If the edge set E ( V x,y , V y,z ) induces a complete bipartite graph and E ( V ∅ , V x,y ∪ V y,z ) = ∅ ,then V ∅ is a clique of cardinality at most 2, and each vertex in V ∅ is adjacent only to one vertex in V x,y ∪ V y,z . Proof.
First we prove that there each vertex in V ∅ is adjacent with a vertex in only one of the sets V x,y or V y,z . Suppose there exists a vertex w ∈ V ∅ adjacent with v ∈ V x,y and u ∈ V y,z . Since theinduced subgraph G [ { w, a, b, c, v, u } ] is isomorphic to G , , then we get a contradiction and eachvertex in V ∅ is adjacent only with vertices of one of the vertex sets V x,y or V y,z . Suppose there aretwo vertices w, w ′ ∈ V ∅ such that w is adjacent with v ∈ V x,y , and w ′ is adjacent with u ∈ V y,z .Since G [ { w, w ′ , u, v, x, z } ] is isomorphic to G , , then this is impossible and the vertices of V ∅ areadjacent only to vertices in one of the vertex sets either V x,y or V y,z . Now suppose that w ∈ V ∅ isadjacent with two vertices in V x,y , say v and v ′ . Take u ∈ V y,z . So u is adjacent with both v and v ′ . Since the induced subgraph G [ { w, v, v ′ , u, a, b, c } ] is isomorphic to G , , then this cannot occur.Thus each vertex in V ∅ is adjacent only with one vertex in V x,y ∪ V y,z . Finally suppose V ∅ induces atrivial graph of cardinality at least 2. Let w, w ′ ∈ V ∅ adjacent with v ∈ V x,y . Take u ∈ V y,z , so u isadjacent with both v . Since the induced subgraph G [ { w, w, v, u, x, z } ] is isomorphic to G , , we geta contradiction and the result follows. (cid:3) By previous Claim we have that in case (3) the vertex set V ∅ induces a clique of cardinality atmost 2 and each vertex in V ∅ is adjacent only to one vertex in V x,y ∪ V y,z . In this case, the graph isisomorphic to an induced subgraph of a graph in F . Claim 4.16.
Let u ∈ V x,y and v, v ′ ∈ V y,z such that u is adjacent with v but not with v ′ . If w ∈ V ∅ ,then w is not adjacent with v .Proof. Suppose w is adjacent with v . Note that if w is adjacent with u or v ′ , then w is adjacent withboth u and v ′ . Thus there are two cases: either w is adjacent only with v , or w is adjacent with v, v ′ and u . Both cases are impossible because in the former case G [ { x, z, u, v, v ′ , w } ] is isomorphic to G , ,meanwhile in the second case G [ { y, z, u, v, v ′ , w } ] is isomorphic to G , ; which is a contradiction. (cid:3) Consider case (4). Let u ∈ V x,y and v ∈ V y,z such that u is not adjacent with v . By Claim 4.16,each vertex in V ∅ is adjacent with u or with v . It is not difficult to see that in fact each vertex in V ∅ is adjacent with both u and v , because otherwise G would contain P as induced subgraph. Byapplying Claim 4.14 to the induced subgraph G [ { u, v } ∪ V ∅ ], we get that V ∅ is a clique of cardinalityat most 2. And this graph is isomorphic to an induced subgraph of a graph in F .Now consider case (5), by Claim 4.14, we get that V ∅ is a clique of cardinality at most 2, and eachvertex in V ∅ is adjacent each vertex in V a,b ∪ V b,c ∪ V a,c . And this graph is isomorphic to an inducedsubgraph of a graph in F . Claim 4.17.
Let u ∈ V x,y , u ∈ V y,z and u ∈ V x,z such that u is adjacent with both u and u ,and u u / ∈ E ( G ) . If w ∈ V ∅ , then w is not adjacent with u .Proof. Suppose w is adjacent with u . Note that if w is adjacent with u or u , then w is adjacentwith both u and u . Thus there are two cases: either w is adjacent only with u , or w is adjacentwith u , u and u . Both cases are impossible because in the former case G [ { x, y, u , u , u , w } ]is isomorphic to G , , meanwhile in the second case G [ { x, z, u , u , u , w } ] is isomorphic to G , ;which is a contradiction. (cid:3) Consider the Cases (6) and (9). Let v xz ∈ V x,z and v yz ∈ V y,z such that v xz is not adjacent with v yz . By Claims 4.14 and 4.17, each vertex in V ∅ is adjacent only with both v xz and v yz . And byClaim 4.14, we get that V ∅ is a clique of cardinality at most 2. And these graphs are isomorphic toan induced subgraph of a graph in F .Now consider the Case (7) and (10). Let v xy ∈ V x,y , v xz ∈ V x,z and v yz ∈ V y,z such that v xy is notadjacent with v xz and v yz , and v xz is not adjacent with v yz . By Claims 4.17 and 4.14, each vertex in MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 15 V ∅ is adjacent only with v xy , v xz and v yz . And by Claim 4.14, we get that V ∅ is a clique of cardinalityat most 2. And these graphs are isomorphic to an induced subgraph of a graph in F .Finally in the case (8), by Claim 4.15 we have that the vertex set V ∅ is a clique of cardinalityat most 2 and each vertex in V ∅ is adjacent with only one vertex in V a,b ∪ V b,c ∪ V a,c . This casecorresponds to a graph that is isomorphic to an induced subgraph of a graph in F .4.2. Case V x = T n , where n ≥ . First we will obtain that E ( V x , V y,z ) satisfies one of the followingstatements: • it induces a complete bipartite graph, • there exists a vertex s ∈ V x , we called the apex, such that E ( V x , V y,z ) = { uv : u ∈ V x − s and v ∈ V y,z } , or • there exists a vertex s ∈ V y,z , we called the apex, such that E ( V x , V y,z ) = { uv : u ∈ V x and v ∈ V y,z − s } .To do it, we will prove the following claims. Claim 4.18.
Let u , u ∈ V x and v , v ∈ V y,z such that u v , u v / ∈ E ( G ) . Then either u = u or v = v .Proof. Suppose u = u and v = v . There are three possible cases: • u v , u v / ∈ E ( G ), • u v / ∈ E ( G ) and u v ∈ E ( G ), or • u v , u v ∈ E ( G ).The first two cases are not possible since the induced subgraph G [ { x, y, u , u , v , v } ] would beisomorphic to G , and G , , respectively. In the last case, the induced subgraph G [ { x, u , u , v , v } ]is isomorphic to P ; which is impossible. Thus the result follows. (cid:3) The last claim implies that all non-edges in E ( V x , V y,z ) are incident to a one vertex: the apex s .Suppose the vertex s is in V x , and there are vertices v , v ∈ V y,z such that sv ∈ E ( G ), and sv / ∈ E ( G ). By Claim 4.18, each vertex in V x − s is adjacent with v and v . Then the inducedsubgraph G [ { a, b, c, u , u , v , v } ] ≃ G , , that is impossible. This implies that if the vertex s ∈ V x isnot adjacent with a vertex in V y,z , then s is not adjacent with all vertices in V y,z . A similar argumentyields that if the apex vertex s is in V y,z and s is not adjacent with a vertex in V x , then s is notadjacent with all vertices in V x .Thus, we have three cases:(a) E ( V x , V y,z ) is complete bipartite minus the edges between a vertex s (the apex) in V x and allvertices of V y,z ,(b) E ( V x , V y,z ) is complete bipartite minus the edges between a vertex s (the apex) in V y,z and allvertices in V x , and(c) E ( V x , V y,z ) is complete bipartite. Claim 4.19. If | V x | ≥ and v ∈ V y,z , then E ( v, V x ) either it induces a complete bipartite graph orit is empty.Proof. Let u , u , u ∈ V x . Suppose one of the two following possibilities happen: vu ∈ E ( G ) and vu , vu / ∈ E ( G ) , or vu , vu ∈ E ( G ) and vu / ∈ E ( G ). In the first case the induced subgraph G [ { u , u , u , v, x, y } ] is isomorphic to G , , meanwhile in the second case the induced subgraph G [ { u , u , u , v, x, y, z } ] is isomorphic to G , . Then both cases cannot occur, and we get the result. (cid:3) Thus case (a) occur only when | V x | = 2.In what follows we describe the vertex set V ∅ , that is, the vertex set whose vertices have no edgein common with the vertex set { a, b, c } . Remark 4.20.
Let w ∈ V ∅ . The vertex w is adjacent with a vertex in V x ∪ V y,z , because otherwise theshortest path from w to { a, b, c } would contains the graph P as induced subgraph. Let u , u ∈ V x .If w is adjacent with u or u , then w is adjacent with both vertices, because otherwise the inducedsubgraph G [ { a, b, c, u , u , w } ] would be isomorphic to G , , which is forbidden. In case (a), we will see that each vertex in V ∅ is adjacent with each vertex in V x ∪ V y,z . Let w in V ∅ . Supppose s ∈ V x is the vertex that is not adjacent with any vertex in V y,z . If w ∈ V ∅ is adjacentwith one of the vertices in { s } ∪ V y,z , then w must to be adjacent with s and each vertex in V y,z ,because otherwise let v ∈ V y,z , then the induced subgraph G [ { x, y, s, w, v } ] would be isomorphic to P . Then by Remark 4.20, w is adjacent with each vertex in V x ∪ V y,z . Claim 4.21.
The vertex set V ∅ induces a stable set.Proof. Suppose w , w ∈ V ∅ are adjacent. Since both w and w are adjacent with u ∈ V x − s and v ∈ V y,z , then the induced subgraph G [ { u, v, w , w } ] is isomorphic to K ; which is forbidden. Thus w and w are not adjacent, and therefore V ∅ is a stable set. (cid:3) Thus this case corresponds to a graph that is isomorphic to an induced subgraph of a graph in F .Now consider case (b). Let u , u ∈ V x and s ∈ V y,z such that s is not adjacent with u and u . Claim 4.22. If w ∈ V ∅ is adjacent with a vertex in V y,z \ { s } , then each vertex in V ∅ is adjacentwith each vertex in V x ∪ V y,z .Proof. Let w ∈ V ∅ . Suppose w is adjacent with v ∈ V y,z \ { s } . If w is not adjacent with both vertices u , u ∈ V x , then we get that the induced subgraph G [ { a, b, c, w, u , u , v } ] is isomorphic to G , ,which cannot be. Thus w is adjacent with at least one vertex in V x . Then by Remark 4.20, w isadjacent with both vertices u and u . If there exist v ′ ∈ V y,z \ { s } such that w is not adjacentwith v ′ , then the vertex set { w, u , x, y, v, v ′ } would induce a graph isomorphic to G , . On theother hand, if w is not adjacent with s , then the vertex set { w, u , x, y, s } induces the subgraph P .Therefore, w is adjacent with each vertex in V x ∪ V y,z .Suppose there is another vertex w ′ ∈ V ∅ . By the above argument if w ′ is adjacent with a vertexin V y,z \ { s } , then it must be adjacent with each vertex in V x ∪ V y,z . Also if w ′ is adjacent witha vertex in { s } ∪ V x , then w ′ must be adjacent with each vertex in { s } ∪ V x . So suppose w ′ isadjacent with each vertex in s ∪ V x and w ′ is not adjacent with each vertex in V y,z \ { s } . Thenthere are two possibilities: either ww ′ / ∈ E ( G ) or ww ′ ∈ E ( G ). Let v ∈ V y,z \ { s } . In the first case G [ { x, y, s, v, w, w ′ } ] ≃ G , and in the second case G [ { x, y, v, w, w ′ } ] ≃ P . Since both graphs areforbidden, then we get a contradiction and thus w ′ is adjacent with v . And therefore w ′ is adjacentwith each vertex in V x ∪ V y,z . (cid:3) Claim 4.23.
Either each vertex in V ∅ is adjacent with each vertex in V x ∪ { s } , or each vertex in V ∅ is adjacent with each vertex in V x ∪ V y,z .Proof. Let w ∈ V ∅ . Clearly, if w is adjacent with a vertex in V x ∪ { v } , then w is adjacent with eachvertex in V x ∪ { v } , because otherwise P would be an induced subgraph. By Claim 4.22, we havethat if there is a vertex w ∈ V ∅ adjacent with a vertex in V y,z , different to the apex s , then eachvertex in V ∅ is adjacent with each vertex in V x and V y,z \ { s } . Thus we get the result. (cid:3) MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 17
Claim 4.24.
The vertex set V ∅ induces either a clique of cardinality at most 2 or a trivial graph.Proof. First note that P is forbidden as induced subgraph in V ∅ . It is because if w , w , w ∈ V ∅ induce P , then G [ { x, y, u , w , w , w } ] ≃ G , . Now we will get that each component in G [ V ∅ ] isa clique. Let C be a component in G [ V ∅ ]. Suppose C is not a clique, then it has two vertices notadjacent, say u and v . Let P be the smallest path in C between u and v . Thus the length of P isgreater or equal to 3. So P is an induced subgraph of P , and hence of C . Therefore, C is a completegraph.On the other hand, the graph K + K is forbidden for G [ V ∅ ]. It is because if w , w , w ∈ V ∅ such that G [ { w , w , w } ] ≃ K + K , then G [ { x, y, u , w , w , w } ] ≃ G , ; which cannot happen.Therefore, if G [ V ∅ ] has more than one component, then each component has cardinality one. (cid:3) In the first case of Claim 4.23, if | V y,z | ≥
2, then V ∅ is either K or K , because if u ∈ V x , v ∈ V y,z \ { s } , and w , w ∈ V ∅ are adjacent, then G [ { w , w , u , v, x, y } ] ≃ G , . Otherwise if V y,z = { s } , then both possibilities in Claim 4.24 are allowed. In the second case of Claim 4.23, if | V y,z | ≥
3, we have that V ∅ = ∅ , because if w ∈ V ∅ is adjacent with u , u ∈ V x and with two vertices v , v ∈ V y,z \ { s } , we get G [ { x, y, u , u , v , v , w } ] ≃ G , as forbidden subgraph. If | V y,z | = 2, then V ∅ is trivial since ω ( G ) = 3. And if V y,z = { s } , then both possibilities in Claim 4.24 are allowed.With no much effort the reader can see that each of these cases corresponds to a graph isomorphicto an induced subgraph of a graph in F .Case (c). By Claim 4.23, there are two possible cases: • either each vertex in V ∅ is adjacent with each vertex in V x ∪ V y,z , or • each vertex in V ∅ is adjacent with each vertex in V x , and no vertex in V ∅ is adjacent withany vertex in V y,z .In first case when | V y,z | ≥
2, the vertex set V ∅ is empty. Because if w ∈ V ∅ is adjacent with thevertices v , v ∈ V y,z , and u , u ∈ V x , then the induced subgraph G [ { x, y, u , u , v , v , w } ] ≃ G , which is forbidden. Then the vertex set V ∅ is empty. Otherwise when | V y,z | = 1, then the vertex set V ∅ must be a stable set, because if there exist two adjacent vertices in V ∅ , then by taking a vertexin V x and a vertex in V y,z we get K that is forbidden. Finally, in the case when each vertex in V ∅ is adjacent with each vertex in V x , and no vertex in V ∅ is adjacent with a vertex in V y,z , we get that V ∅ is a clique of cardinality at most 2. It is because if w , w ∈ V ∅ are such that w w / ∈ E ( G ), thenwe get G [ { w , w , u , v, x, y } ] ≃ G , that is forbidden. And each of these graphs are isomorphic toan induced subgraph of a graph in F .4.3. Case V x is a complete bipartite graph of cardinality at least 3. Assume V x is a completebipartite graph of cardinality at lest three with ( A, B ) the bipartition of V x . Claim 4.25. If v ∈ V y,z , then E ( v, V x ) = ∅ .Proof. Suppose E ( v, V x ) = ∅ . There exist u , u , u ∈ V x such that G [ { u , u , u } ] ≃ P . And thenwe get contradiction since the induced subgraph G [ { u , u , u , x, y, v } ] ≃ G , ; which is forbidden. (cid:3) Claim 4.26.
Let u ∈ V x and v ∈ V y,z . If u and v are adjacent, then v is adjacent with each vertexin the part ( A or B ) containing u .Proof. Suppose u ∈ A and v is not adjacent with any vertex in A \ { u } . If | A | = 1, then theresult follows, so we may assume | A | ≥
2. Since G [ V x ] is connected and has cardinality at least3, then there exists u ′ ∈ A and u ′′ ∈ B such that G [ { u, u ′′ , u ′ } ] ≃ P . We have u ′ and v arenot adjacent. There are two possibilities: either u ′′ v ∈ E ( G ) or u ′′ v / ∈ E ( G ). In the first case the induced subgraph G [ { u ′ , u ′′ .u, v, y } ] is isomorphic to P , and in the second case the inducedsubgraph G [ { x, y, u, u ′ , u ′′ , v } ] is isomorphic to G , . Since both cases are forbidden, we get acontradiction. (cid:3) Previous Claims suggest to divide the vertex set V y,z in three subsets: • V Ay,z , the vertices in V y,z that are adjacent with each vertex in A , • V By,z , the vertices in V y,z that are adjacent with each vertex in B , and • V ABy,z , the vertices in V y,z that are adjacent with each vertex in A ∪ B .In what follows, we assume | A | ≥ | B | . Claim 4.27.
The cardinality of the sets V Ay,z and V By,z is no more than 1.Proof.
Suppose there exist v, v ′ ∈ V Ay,z . Let u ∈ A , and u ′ ∈ B . Since G [ { u, u ′ , v, v ′ , x, y } ] isisomorphic to G , , which is forbidden, then we get a contradiction. The case V By,z is similar. (cid:3)
Claim 4.28. If | B | ≥ and V y,z = ∅ , then one of the sets V By,z or V ABy,z is empty.Proof.
Suppose v ∈ V By,z and v ′ ∈ V ABy,z . Let u, u ′ ∈ B , and u ′′ ∈ A . Thus G [ { v, v ′ , u, u ′ , u ′′ , x, y } ] ≃ G , . Which is impossible. (cid:3) Since | A | ≥
2, then by applying previous Claim to A , one of the sets V Ay,z or V ABy,z is empty.Thus the possible cases we have are the following:(a) V y,z = ∅ ,(b) V By,z ∪ V ABy,z = ∅ and | V Ay,z | = 1,(c) V Ay,z ∪ V ABy,z = ∅ and | V By,z | = 1,(d) V ABy,z = ∅ and | V Ay,z | = | V By,z | = 1,(e) V Ay,z = ∅ , | B | = 1, | V By,z | = 1 and | V ABy,z | ≥
1, and(f) V Ay,z ∪ V By,z = ∅ and | V ABy,z | ≥ V ∅ , that is, the set of vertices not adjacent with any vertex in { a, b, c } . Let w ∈ V ∅ . The vertex w is adjacent with a vertex in V x ∪ V y,z , because otherwise the shortest pathfrom w to { x, y } would contains the graph P as induced subgraph. Claim 4.29.
Let w ∈ V ∅ . If w is adjacent with a vertex in V x , then w is adjacent with each vertexin the parts of the partition ( A, B ) with cardinality greater or equal to 2.Proof. Let v ∈ V x be a vertex adjacent with w . Suppose v ∈ B . we will prove two things: (1) if | B | ≥
2, then w is adjacent with each vertex in B , and (2) w is adjacent with each vertex in A .Let us consider case when | B | ≥
2. We will see that w is adjacent with each vertex in B . Supposethere is a vertex v ′ ∈ B not adjacent with w . Take u ∈ A . Thus there are two possibilities: either u and w are adjacent or not. The case uw ∈ E ( G ) is impossible because G [ { u, v, v ′ , w, x, y } ] ≃ G , ,which is forbidden. Meanwhile, the case uw / ∈ E ( G ) is impossible because G [ { u, v, v ′ , w, x, y } ] ≃ G , , which is forbidden. Thus w is adjacent with v ′ , and therefore w is adjacent with each vertexin B .Now we see that w is adjacent with each vertex in A . Note that in this case | B | may be equalto 1. Suppose w is not adjacent with any vertex in A . Let u, u ′ ∈ A . Since the induced subgraph G [ { w, v, u, u ′ , x, y } ] is isomorphic to the forbidden graph G , , then we get a contradiction. Thus w is adjacent with a vertex in A . Now applying the previous case (1) to A , we get that w is adjacentwith each vertex in A . (cid:3) MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 19
Next Claim show us what happens in the case when | B | = 1. Claim 4.30. If | B | = 1 and E ( V ∅ , V x ) = ∅ , then only one of the edges sets E ( V ∅ , V x ) or E ( V ∅ , A ) induces a complete bipartite graph.Proof. Let w, w ′ ∈ V ∅ , u ∈ B , and u ′ ∈ A . By Claim 4.29, vertices w and w ′ are adjacent with eachvertex in A . Suppose w is adjacent with u , and w ′ is not adjacent with u . There are two possibilities:either w and w ′ are adjacent or not. If ww ′ ∈ E ( G ), then the induced subgraph G [ w, w ′ , u, x, y ] isisomorphic to P ; which is impossible. On the other hand, if ww ′ / ∈ E ( G ), then G [ { w, w ′ , u, u ′ , x, y } ]is isomorphic to G , , which is forbidden. So we get a contradiction, and the result follows. (cid:3) Claim 4.31.
Let w ∈ V ∅ . If w is adjacent with v ∈ V y,z , then w is adjacent with each vertex inthe parts of partition ( A, B ) with cardinality greater or equal to 2. Moreover, if v ∈ V Ay,z ∪ V ABy,z and | B | = 1 , then w is adjacent with the unique vertex in B .Proof. Let w ∈ V ∅ and v ∈ V y,z such that w and v are adjacent. By Claim 4.25, E ( v, V x ) = ∅ , andtherefore there are three cases: v ∈ V Ay,z , v ∈ V ABy,z , or v ∈ V By,z .Suppose v ∈ V Ay,z . If E ( w, V x ) = ∅ , then by taking u, u ∈ A , the forbidden induced subgraph G [ W ∪ { w, v, u, u ′ } ] ≃ G , would appear and we get a contradiction. Thus E ( w, V x ) = ∅ . By Claim4.29, w is adjacent with each vertex in the parts of V x with cardinality greater or equal to 2. When | B | = 1, take u ′′ ∈ B . If wu ′′ / ∈ E ( G ), then G [ { w, v, y, x, u ′′ } ] would be isomorphic to the forbiddengraph P . Therefore, wu ′′ ∈ E ( G ).Suppose v ∈ V ABy,z . In a similar way than in previous case we get that E ( w, V x ) = ∅ , and v isadjacent with each vertex in the parts of the partition ( A, B ) of cardinality greater or equal to 2. Inthe case when | B | = 1, suppose u ∈ B with wu / ∈ E ( G ). Take u ′ ∈ A , we know that u ′ w ∈ E ( G ).Since G [ { w, u, u ′ , x, y, v } ] ≃ G , , then we get a contradiction. Thus wu ∈ E ( G ).Now suppose v ∈ V By,z . It is easy to see that v is adjacent with each vertex in A , because otherwise P would appear. Therefore E ( w, V x ) = ∅ , and by Claim 4.29, we have that w is adjacent with eachvertex in the parts of the partition ( A, B ) of cardinality greater or equal to 2. (cid:3)
Claim 4.32. If E ( V ∅ , B ) = ∅ , then E ( V ∅ , V ABy,z ) = ∅ . Moreover, if E ( V ∅ , B ) = ∅ and E ( V ∅ , V By,z ) = ∅ ,then V ABy,z = ∅ .Proof. Let w ∈ V ∅ and v ∈ V ABy,z . Suppose w and v are adjacent. Take u ∈ A and u ∈ B . Since v is in V ABy,z and u is adjacent with u , then G [ { v, u , u } ] is isomorphic to K . On the other hand,by Claims 4.29 and 4.30, w is adjacent with each vertex in A ∪ B . Thus the induced subgraph G [ { w, v, u , u } ] is isomorphic to K that is forbidden, and therefore w cannot be adjacent with v .Let w is adjacent with u ∈ B and v ′ ∈ V By,z , and suppose there exists a vertex v ∈ V ABy,z . Since w is not adjacent with v , then G [ { w, v, v ′ , u , x, y } ] is isomorphic to G , ; which it is not possible. (cid:3) Claim 4.33.
Let w ∈ V ∅ and v ∈ V y,z . If w and v are adjacent, then each vertex in V ∅ is adjacentwith v .Proof. Suppose w ′ ∈ V ∅ such that w ′ is not adjacent with v . By Claims 4.29 and 4.30, both vertices w and w ′ are adjacent with each vertex in A . Let u ∈ A . There are four cases obtained by thecombinations of the following possible cases: either ww ′ ∈ E ( G ) or ww ′ / ∈ E ( G ), and either v ∈ V Ay,z or v ∈ V By,z . When ww ′ ∈ E ( G ) and v ∈ V By,z , the induced subgraph G [ { w ′ , w, v, y, x } ] is isomorphic to P ; which is not possible. When ww ′ / ∈ E ( G ) and v ∈ V By,z , the induced subgraph G [ { w ′ , u, w, v, y } ]is isomorphic to P ; which is not possible. When ww ′ ∈ E ( G ) and v ∈ V Ay,z , the induced subgraph G [ { w ′ , w, v, y, x } ] is isomorphic to P ; which is not possible. Finally, when ww ′ / ∈ E ( G ) and v ∈ V Ay,z ,the induced subgraph G [ { w ′ , w, x, y, v, u } ] is isomorphic to G , ; which is not possible. Thus w ′ isadjacent with v . (cid:3) Claim 4.34.
Let v ∈ V Ay,z and v ′ ∈ V By,z . If w ∈ V ∅ is adjacent with v or v ′ , then w is adjacent withboth v and v ′ .Proof. First suppose w is adjacent with v and not with v ′ . Let u ∈ A . By Claim 4.31, w is adjacentwith a vertex u ∈ A . Thus G [ { w, u, x, y, v ′ } ] is isomorphic to P that is a contradiction. Therefore w and v ′ are adjacent. Now suppose w is adjacent with v ′ and not with v . Let u ∈ B . If u is adjacentwith w , then a P is obtained in a similar way than previous case. Thus assume u is not adjacentwith w . Then G [ { w, x, y, u, v, v ′ } ] is isomorphic to G , ; which is impossible. Therefore, w and v areadjacent. (cid:3) Claim 4.35. If V ABy,z = ∅ , then E ( V ∅ , B ) = ∅ .Proof. Let u , u ∈ A , u ∈ B , w ∈ V ∅ and v ∈ V ABy,z . By Claim 4.29, w is adjacent with u and u . Suppose w is adjacent with u . By Claim 4.32, the vertices w and v are not adjacent. Thus G [ { x, y, u , u , u , v, w } ] is isomorphic to G , , which is a contradiction. Therefore, E ( V ∅ , B ) = ∅ . (cid:3) Thus applying previous Claims to cases (a) - (f), we obtain the following possibilities:(1) V y,z = ∅ , | B | = 1, and E ( V ∅ , A ) induces a complete bipartite graph,(2) V y,z = ∅ , | B | ≥
1, and E ( V ∅ , V x ) induces a complete bipartite graph,(3) V By,z ∪ V ABy,z = ∅ , | V Ay,z | = 1, | B | ≥ E ( V ∅ , V x ∪ V y,z ) induces a complete bipartite graph,(4) V Ay,z ∪ V ABy,z = ∅ , | V By,z | = 1, | B | = 1 and E ( V ∅ , A ∪ V y,z ) induces a complete bipartite graph,(5) V Ay,z ∪ V ABy,z = ∅ , | V By,z | = 1, | B | ≥ E ( V ∅ , V x ∪ V y,z ) induces a complete bipartite graph,(6) V ABy,z = ∅ , | V Ay,z | = | V By,z | = 1, | B | ≥ E ( V ∅ , V x ∪ V y,z ) induces a complete bipartite graph,(7) V Ay,z ∪ V By,z = ∅ , | V ABy,z | ≥ | B | = 1 and E ( V ∅ , A ) induces a complete bipartite graph,(8) V Ay,z ∪ V By,z = ∅ , | V ABy,z | ≥ | V By,z | = 1, | B | = 1 and E ( V ∅ , A ∪ V By,z ) induces a complete bipartitegraph.With a similar argument as in Claim 4.24, we obtain that V ∅ is either trivial or K . In cases(1), (4), (7) and (8), V ∅ cannot be T n with n ≥
2, since taking w, w ′ ∈ V ∅ , u ∈ A , u ′ ∈ B , then G [ { w, w ′ , u, u ′ , x, y } ] ≃ G , . On the other hand, since ω ( G ) = 3, then V ∅ is not isomorphic to K in the cases (2), (3), (5) and (6). It is not difficult to see that in each case G is isomorphic to aninduced subgraph of a graph in F .4.4. Case when V x induces K + K or K . Through this case we assume that V x = { u , u , u , u } such that u u , u u ∈ E ( G ). That is, G [ { u , u , u , u } ] ≃ K . Let A = { u , u } and B = { u , u } .The following discussion also applies when one of the vertex set A or B has cardinality 1. Claim 4.36. If v ∈ V y,z , then E ( v, V x ) = ∅ . Moreover, if v ∈ V y,z , then v is adjacent with eachvertex in one of the following sets A , B or V x .Proof. Suppose there is no edge joining v and a vertex in V x . Since G [ { u , u , u , x, y, v } ] ≃ G , is aforbidden induced subgraph, then a contradiction is obtained. Then v is adjacent with some vertexin V x .Suppose v is adjacent with one of the vertices in A . We will prove that v cannot be adjacent onlywith one vertex of A , say u . Thus suppose v is adjacent with u , and v is not adjacent with u , u and u . Since G [ { u , u , u , v, x, y } ] ≃ G , , then v is adjacent with u or a vertex in B , say u . If v is adjacent with u , then we are done. So we assume v is adjacent with u , but v is not adjacentwith u . This is not possible because G [ { u , u , u , v, y, z } ] ≃ G , . Therefore, either v is adjacentwith both u and u , or v is not adjacent with neither of u and u (cid:3) Claim 4.37. If | V y,z | ≥ , then each vertex in V y,z is adjacent with each vertex of only one of thefollowing sets A , B or V x .Proof. Let v, v ′ ∈ V y,z . Consider the following cases:(a) v is adjacent with each vertex in A and v ′ is adjacent with each vertex in B ,(b) v is adjacent with each vertex in V x and v ′ is adjacent with each vertex in B , and(c) v is adjacent with each vertex in V x and v ′ is adjacent with each vertex in A .Case (a) is impossible because G [ { u , v, y, v ′ , u } ] ≃ P , which is forbidden. On the other hand,cases (b) and (c) are not allowed because G [ { a, b, c, v, v ′ , u , u } ] ≃ G , ; which is forbidden. Thus,the result follows. (cid:3) Thus E ( V x , V y,z ) satisfies only one of the following three cases:(1) E ( V y,z , A ) induces a complete bipartite graph,(2) E ( V y,z , B ) induces a complete bipartite graph, or(3) E ( V y,z , V x ) induces a complete bipartite graph.Now we describe V ∅ , the set of vertices not adjacent with any vertex in { a, b, c } . Let w ∈ V ∅ . Thevertex w is adjacent with a vertex in V x ∪ V y,z , because otherwise the shortest path from w to { x, y } would contains the graph P as induced subgraph. Claim 4.38. If w ∈ V ∅ is adjacent with a vertex in V x , then w is adjacent with each vertex in V x .Proof. Suppose w is adjacent with u and not with u . Since G [ W ∪ { w, u , u } ] ≃ G , is forbidden, w also must be adjacent with u . In a similar way we get the opposite case and it turns out theresult. (cid:3) Consider Cases (1) and (2). The following arguments works on both cases. First note that if thereexists w ∈ V ∅ adjacent with v ∈ V y,z , then w is adjacent with each vertex in V x . The reason is thefollowing. Suppose w is not adjacent with any vertex in V x . Since v is not adjacent with a vertexin V x , say u , and the vertices w and u are not adjacent, then we have that G [ { w, v, y, x, u } ] ≃ P ;which is a contradiction. Then w and u are adjacent. And by Claim 4.38, the edge set E ( w, V x )induces a complete bipartite graph. On the other hand, we can prove in a similar way that if thereexists w ∈ V ∅ adjacent with each vertex in V x , then w is adjacent with each vertex in V y,z . Therefore,each vertex in V ∅ is adjacent with each vertex in V x ∪ V y,z . Furthermore, the set V ∅ is a stable set.It is because if w, w ′ ∈ V ∅ were adjacent, then by taking u ∈ V x and v ∈ V y,z such that u and v are adjacent, the induced subgraph G [ { w, w ′ , u, v, x, y } ] would be isomorphic to G , ; which isimpossible. These cases correspond to a graph isomorphic to an induced subgraph of a graph in F .Now let us consider case (3). Claim 4.39. If w ∈ V ∅ is adjacent with a vertex in V y,z , then each vertex in V ∅ is adjacent witheach vertex in V x ∪ V y,z .Proof. Let w ∈ V ∅ and v ∈ V y,z such that wv ∈ E ( G ). Suppose w is not adjacent with any vertexin V x . Take u , u ∈ V x such that u u ∈ E ( G ). Since the induced subgraph G [ { a, b, c, w, u , u , v } ] is isomorphic to G , , then we get a contradiction; and w is adjacent with a vertex in V x . Thus byClaim 4.38, w is adjacent with each vertex in V x .Suppose that there exists v ′ ∈ V y,z such that w is not adjacent with v ′ . Since the vertex set { w, u , x, y, v, v ′ } would induce G , , this does not occur. Therefore, w is adjacent with each vertexin V x ∪ V y,z .Suppose there is another vertex w ′ ∈ V ∅ . By the above argument, if w ′ is adjacent with a vertexin V y,z , then it must be adjacent with each vertex in V x ∪ V y,z . And we are done. On the otherhand, by Claim 4.38, if w ′ is adjacent with a vertex in V x , then w ′ must be adjacent with eachvertex in V x . So suppose w ′ is adjacent with each vertex in V x , but not adjacent with each vertexin V y,z . Then there are two possibilities: either ww ′ / ∈ E ( G ) or ww ′ ∈ E ( G ). In the first case G [ { x, y, u , v, w, w ′ } ] ≃ G , and in the second case G [ { x, y, v, w, w ′ } ] ≃ P . Since both graphs areforbidden, then w ′ is adjacent with v . And therefore w ′ is adjacent with each vertex in V x ∪ V y,z . (cid:3) By Claims 4.38 and 4.39, we obtain that there are two possible cases: either each vertex in V ∅ is adjacent with each vertex in V x ∪ V y,z , or each vertex in V ∅ is adjacent only with each vertex in V x . Consider first case. If there exists a vertex w ∈ V ∅ , then w adjacent with the vertices v ∈ V y,z ,and u , u ∈ V x . Thus G [ { w, v, u , u } ] is isomorphic to K that is not allowed, then V ∅ is empty.Now consider second case. Let w ∈ V ∅ and v ∈ V y,z . Thus w is adjacent with u , u and u . Since G [ { w, u , u , u , v, x, y } ] is isomorphic to G , , we get a contradiction. Then V ∅ = ∅ . The graph inthis case is isomorphic to an induced subgraph of a graph in F .4.5. Cases when G [ V a ∪ V b ∪ V c ] = V x ∨ ( V y + V z ) is one of the following graphs: K ∨ K , K ∨ ( K + K ) , K ∨ K , or K ∨ K . For the sake of clarity, we suppose E ( V a , V b ) and E ( V a , V c )induce a complete bipartite graph, and E ( V b , V c ) is empty. Now we are going to obtain some claimsthat describe the edge sets joining V x and V y,z . Claim 4.40.
Let x, y ∈ { a, b, c } . If E ( V x , V y ) is not empty, then E ( V x , V xy ) and E ( V y , V xy ) areempty.Proof. Let v x ∈ V x , v y ∈ V y and v xy ∈ V x,y . Suppose v xy is adjacent with both v x and v y . Then G [ { a, b, c, v xy , v x , v y } ] ≃ G , ; which is a contradiction. Now suppose v xy is adjacent with v x andnot with v y . In this case G [ { a, b, c, v xy , v x , v y } ] ≃ G , ; which is impossible. And therefore resultturns out. (cid:3) Claim 4.40 implies that E ( V a,b , V a ∪ V b ) = ∅ and E ( V a,c , V a ∪ V c ) = ∅ . Claim 4.41.
Let x, y ∈ { a, b, c } . If V x,y = ∅ and E ( V x , V y ) = ∅ , then each edge set E ( V x , V x,y ) and E ( V y , V x,y ) induces a complete bipartite graph.Proof. Let v x ∈ V x , v y ∈ V y and v xy ∈ V x,y . Suppose v xy is not adjacent with both v x and v y . Then G [ { a, b, c, v xy , v x , v y } ] ≃ G , ; which is a contradiction. Finally, suppose v xy is adjacent with v x andnot with v y . Since G [ { a, b, c, v x,y , v x , v y } ] ≃ G , is forbidden, then we get a contradiction. And theresult turns out. (cid:3) Claim 4.41 implies that each vertex in V b,c is adjacent with each vertex in V b ∪ V c . Claim 4.42. If V b,c = ∅ , then E ( V a , V b,c ) is empty.Proof. Let v a ∈ V a , v b ∈ V b and v bc ∈ V b,c . Suppose v bc is adjacent with v a . Then G [ { a, b, c, v a , v b , v bc } ]is isomorphic to G , that is forbidden. Therefore, there is no edge joining a vertex in V a with avertex in V b,c . (cid:3) MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 23
Claim 4.43.
Let x ∈ { b, c } . If V a,x = ∅ , then E ( V a,x , V x ) induces a complete bipartite graph.Proof. Let v b ∈ V b , v c ∈ V c and v ax ∈ V a,x . Suppose v ax is not adjacent with v x . Since G [ { a, b, c, v ax , v b , v c } ]is isomorphic to G , that is forbidden, then each vertex in V x is adjacent with each vertex in V a,x . (cid:3) Claim 4.44.
The edge set E ( V a,b , V a,c ) induces a complete bipartite graph.Proof. Let v a ∈ V a , v b ∈ V b , v ab ∈ V a,b , and v ac ∈ V a,c . We know that v a is not adjacent with both v ab and v ac , and v b is adjacent with v ac , but v b is not adjacent with v ab . Suppose v ab and v ac arenot adjacent. Then G [ { b, c, v a , v b , v ab , v ac , } ] ≃ G , ; which is impossible. And therefore, each vertexin V a,b is adjacent with each vertex in V a,c . (cid:3) Claim 4.45.
Let x ∈ { b, c } The edge set E ( V a,x , V b,c ) induces a complete bipartite graph.Proof. Let v b ∈ V b , v c ∈ V c , v ax ∈ V a,x , and v bc ∈ V b,c . We know that v b v bc , v bc v c , v c v ab , v b v ac ∈ E ( G )and v x v ax , v b v c / ∈ E ( G ). Suppose v ax and v bc are not adjacent. Then G [ { b, c, v b , v c , v ax , v bc , } ] ≃ G , ;which is impossible. Therefore each vertex in V a,x is adjacent with each vertex in V b,c . (cid:3) Now we analyze V ∅ . Claim 4.46. If w ∈ V ∅ , then w cannot be adjacent with any vertex in V a ∪ V b ∪ V c .Proof. Let x ∈ { b, c } , y ∈ { b, c } − x , v a ∈ V a and v x ∈ V x . Suppose w ∈ V ∅ such that E ( w, V a ∪ V b ∪ V c ) = ∅ . Consider the following cases:(a) w is adjacent only with v a ,(b) w is adjacent only with v x , or(c) w is adjacent with both v a and v x .Cases (a) and (b) are impossible, because G would have P as induced subgraph obtained by G [ { w, v a , v x , x, y } ] and G [ { w, v x , v a , a, y } ], respectively. Finally in case (c), the induced subgraph G [ { a, b, c, w, v a , v x } ] is isomorphic to G , ; which is impossible. (cid:3) Claim 4.47.
There exists no vertex in V ∅ adjacent with a vertex in V b,c .Proof. Let w ∈ V ∅ , v a ∈ V a , v b ∈ V b , v c ∈ V c , and v bc ∈ V b,c . Suppose w is adjacent with v bc . Acontradiction is obtained since G [ { w, v a , v b , v c , v bc , w } ] ≃ G , . Thus w is not adjacent with anyvertex in V b,c (cid:3) Claim 4.48.
There exists no vertex in V ∅ adjacent with a vertex in V a,b ∪ V a,c .Proof. Let x ∈ { b, c } , y ∈ { b, c } − x , v a ∈ V a , v b ∈ V b and v ax ∈ V a,x . Suppose w is adjacent with v ax .Since the induced subgraph G [ { w, v ax , v a , v b , v c } ] is isomorphic to G , , then we get a contradiction.And then the result follows. (cid:3) Claims 4.46, 4.47 and 4.48 imply that no vertex in V ∅ is adjacent with a vertex in G \ V ∅ , whichimplies that V ∅ = ∅ . Thus the graph is isomorphic to an induced subgraph of a graph in F .4.6. Case G [ V a ∪ V b ∪ V c ] = K , , , where each vertex set V x = K . Let V a = { v a } , V b = { v b } ,and V c = { v c } . By Claim 4.40, the edge sets E ( V x,y , V x ) and E ( V x,y , V y ) are empty for x, y ∈ { a, b, c } .By Claim 4.42, the edge set E ( V x,y , V z ) is empty for x, y, z ∈ { a, b, c } . Now let v xy ∈ V x,y . Since G [ { v y , v z , z, x, v x,y } ] is isomorphic to P which is a forbidden, then V xy is empty for each pair x, y ∈{ a, b, c } . On the other hand, by Claim 4.46 the edge set V ∅ is empty. This graph is isomorphic to G , see Figure 1.i. Case V z = ∅ and G [ V x ∪ V y ] = V x + V y , where V x = K m , V y = K n and m, n ∈ { , } . Without loss of generality, suppose V c = ∅ , and E ( V a , V b ) is empty. By Claim 4.41, each vertex in V a,b is adjacent with each vertex in V a ∪ V b . Claim 4.49.
Let x ∈ { a, b } . If V x,c = ∅ , then E ( V x , V x,c ) = ∅ .Proof. Let y ∈ { a, b } − x , v xc ∈ V x,c , v a ∈ V a and v b ∈ V b . Suppose v x and v xc are adja-cent. There are two possible cases: either v y and v xc are adjacent or not. Since in the first case G [ { a, b, c, v a , v b , v xc } ] ≃ G , and in the second case G [ { v a , v b , y, v xc , c } ] is isomorphic to P , then weget a contradiction. Thus v x and v xc are not adjacent. (cid:3) By Claim 4.43, each vertex in V a,c is adjacent with each vertex in V b , and each vertex in V b,c isadjacent with each vertex in V a . Claim 4.50.
Each vertex in V a,b is adjacent with each vertex in V a,c ∪ V b,c .Proof. Let v ab ∈ V a,b , v a ∈ V a and v b ∈ V b . Suppose there exists v xc ∈ V xc with x ∈ { a, b } such that v ab v xc / ∈ E ( G ). Since G [ { v a , v b , v ab , v xc , c } ] is isomorphic to P , then we get a contradiction; and thevertices v ab and v xc are adjacent. And the result turns out. (cid:3) Claim 4.51.
Each vertex in V a,c is adjacent with each vertex in V b,c .Proof. Suppose there are v ac ∈ V a,c and v bc ∈ V b,c such that v bc v ac / ∈ E ( G ). Let v a ∈ V a and v b ∈ V b Since G [ { v b , v ac , c, v bc , v a } ] ≃ P , then we get a contradiction. (cid:3) Now we describe the vertex set V ∅ , that is, the set of vertices that are not adjacent with anyvertex in { a, b, c } . Claim 4.52. If w ∈ V ∅ , then w cannot be adjacent with any vertex in V a ∪ V b .Proof. Let v a ∈ V a and v b ∈ V b . Suppose w ∈ V ∅ such that E ( w, V a ∪ V b ) = ∅ . Consider the followingcases:(a) w is adjacent only with v a , or(b) w is adjacent with both v a and v b .Cases (a) and (b) are impossible, because G would have P as induced subgraph obtained by G [ { w, v b , b, a, v a } ] and G [ { v a , w, v b , b, c } ], respectively. Thus w is not adjacent with any vertex in V a ∪ V b . (cid:3) Claim 4.53.
There is no vertex w ∈ V ∅ adjacent with a vertex in V a,b .Proof. Suppose w ∈ V ∅ is adjacent with v ab ∈ V a,b . Let v a ∈ V a and v b ∈ V b . Since G [ { v a , v b , v a,b , w, a, c } ]is isomorphic to G , , then we get a contradiction and w and v ab are not adjacent. (cid:3) Claim 4.54.
There is no vertex w ∈ V ∅ adjacent with a vertex in V a,c ∪ V b,c .Proof. Let x ∈ { a, b } , v xc ∈ V x,c , y ∈ { a, b } − x and v y ∈ V y . Suppose w ∈ V ∅ is adjacent with v xc .Since G [ { v y , y, c, v xc , w } ] is isomorphic to P , then we get a contradiction and w is not adjacent with v xc . (cid:3) Thus there is no edge between W and G \ W , and therefore W is empty. Therefore, the graph isisomorphic to an induced subgraph of a graph in F . MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 25
Case V z = ∅ and G [ V x ∪ V y ] = V x ∨ V y , where V x = K , V y = K m and m ∈ { , } . Withoutloss of generality, suppose V c = ∅ , and E ( V a , V b ) is complete. By Claim 4.40, E ( V a,b , V x ) = ∅ for x ∈ { a, b } . Claim 4.55.
Let x ∈ { a, b } . Either E ( V x,c , V a ) induces a complete bipartite graph and E ( V x,c , V b ) = ∅ , or E ( V x,c , V a ) = ∅ and E ( V x,c , V b ) induces a complete bipartite graph.Proof. Let v a ∈ V a , v b ∈ V b , v xc ∈ V x,c and y ∈ { a, b } − x . First note that v xc cannot be not adjacentwith v a and v b at the same time, because otherwise G [ { v a , v b , y, c, v xc } ] would be isomorphic to P .Also v xc cannot be adjacent with v a and v b at the same time, because otherwise G [ { a, b, c, v a , v b , v xc } ]would be isomorphic to G , . Thus v xc is adjacent only with one vertex either v a or v b .Suppose v xc is adjacent with v y . Let v ′ y ∈ V y and v ′ xc ∈ V x,c . If v xc is not adjacent with v ′ y , then G [ { v y , v ′ y , x, c, v x , v xc } ] is isomorphic to G , . Thus E ( V y , v xc ) induces a complete bipartite graphs.On the other hand, if v y and v ′ xc are not adjacent, then G [ { v a , v b , v xc , c, v ′ xc } ] is isomorphic to P .Then E ( V x,c , V y ) induces a complete bipartite graph.Suppose v xc is adjacent with v x . Let v ′ x ∈ V x and v ′ xc ∈ V x,c . If v ′ x and v xc are not adjacent, then G [ { v a , v b , v ′ x , x, c, v xc } ] is isomorphic to G , . Thus E ( V x , v xc ) induces a complete bipartite graph.On the other hand, if v ′ xc and v x are not adjacent, then G [ { v y , v x , v xc , c, v ′ xc } ] is isomorphic to P .Thus E ( V x,c , V x ) induces a complete bipartite graph. (cid:3) Claim 4.56. If V a,c and V b,c are not empty, then each vertex in V a,c ∪ V b,c is adjacent with eachvertex in either V a or V b .Proof. Let v ac ∈ V a,c , v bc ∈ V b,c , v a ∈ V a and v b ∈ V b .Suppose v ac is adjacent with v a , and v bc is adjacent with v b . Then there are two cases: either v ac and v bc are adjacent or not. In the first case G [ { a, b, c, v b , v ac , v bc } ] is isomorphic to G , ; then thiscase is impossible. And in the second case G [ { a, b, v a , v b , v ac , v bc } ] is isomorphic to G , ; then thiscase is not possible.Suppose v ac is adjacent with v b and v bc is adjacent with v a . Then there are two cases: either v ac and v bc are adjacent or not. In the first case G [ { a, b, c, v a , v b , v ac , v bc } ] is isomorphic to G , ; thenthis case is impossible. And in the second case G [ { a, b, c, v b , v ac , v bc } ] is isomorphic to G , ; thenthis case is not possible.Thus v ac and v bc are adjacent with the same vertex: either v a or v b . And the result follows. (cid:3) Claim 4.57. If V a,c and V b,c are not empty, then the set E ( V ac , V bc ) induces a complete bipartitegraph.Proof. Let x ∈ { a, b } and y ∈ { a, b } − x , v ac ∈ V a,c , v bc ∈ V b,c , v a ∈ V a and v b ∈ V b . Suppose v xc and V yc are not adjacent, and v xc and v yc are adjacent with v x . Since G [ { x, y, v x , v y , v x,c , v y,c } ] ≃ G , ,we get a contradiction. And then E ( V ac , V bc ) induces a complete bipartite graph. (cid:3) Claim 4.58.
Let x ∈ { a, b } . If V x,c = ∅ , then E ( V x,c , V a,b ) induces a complete bipartite graph.Proof. Let y ∈ { a, b } − x , v xc ∈ V x,c , v ab ∈ V a,b , v a ∈ V a and v b ∈ V b . Suppose v xc and v ab are notadjacent. There are two cases: either v xc is adjacent with v x or v xc is adjacent with v y . If v xc isadjacent with v x , then the induced subgraph G [ { v ab , y, v y , v x , v xc } ] is isomorphic to P ; which is acontradiction. Then this case is impossible. On the other hand, if v xc is adjacent with v y , then theinduced subgraph G [ { v ac , v a , v b , a, b, v xc } ] is isomorphic to G , ; which is a contradiction. Thus thiscase is also impossible, and therefore E ( V x,c , V a,b induces a complete bipartite graph. (cid:3) Now let us describe V ∅ . By Claim 4.46, there exists no vertex in V ∅ adjacent with a vertex in V a ∪ V b . Claim 4.59.
There is no vertex in V ∅ adjacent with a vertex in V a,b .Proof. Let v ab ∈ V a,b and w ∈ V ∅ . Suppose v ab and w are adjacent. Since G [ { w, v a,b , a, v a , v b } ] isisomorphic to P , then we get a contradiction. And therefore v ab and w are not adjacent. (cid:3) Claim 4.60.
Let x ∈ { a, b } . There is no vertex in V ∅ adjacent with a vertex in V x,c .Proof. Let y ∈ { a, b } − x , v xc ∈ V x,c , v ab ∈ V a,b , v a ∈ V a and v b ∈ V b . Suppose the vertex w ∈ V ∅ isadjacent with v xc . There are two cases: either v xc is adjacent with x or v xc is adjacent with y . Firstcase is impossible since G [ { w, v xc , v x , v y , y } ] is isomorphic to P ; which is forbidden. And secondcase cannot occur because G [ { w, y, v xc , c, v a , v b } ] ≃ G , . Then it follows that w is adjacent with novertex in V x,c . (cid:3) Thus by previous Claims, the vertex set V ∅ is empty, because there is no vertex in V ∅ adjacentwith a vertex in G \ V ∅ . Then G is isomorphic to an induced subgraph of a graph in F .4.9. Case V y ∪ V z = ∅ and V x is K or K . Without loss of generality, suppose V b = V c = ∅ and V a = { u , u } . Claim 4.61.
Let x ∈ { b, c } . If V a,x = ∅ , then either E ( V a , V a,x ) induces a complete bipartite graphor E ( V a , V a,x ) is empty.Proof. Let u ∈ V a . Suppose there exist v , v ∈ V a,x such that uv ∈ E ( G ) and uv / ∈ E ( G ). Sincethe induced subgraph G [ { a, b, c, u, v , v } ] is isomorphic to G , , then we get a contradiction. Thusthis case is not possible and therefore u is adjacent with either each vertex in V a,x or no vertex in V a,x . Now we are going to discard the possibility that E ( u , V a,x ) induces a complete bipartite graphand E ( u , V a,x ) = ∅ . Suppose there is a vertex v ∈ V a,x such that u v ∈ E ( G ) and u v / ∈ E ( G ).Since the induced subgraph G [ { c, b, v, u , u } ] is isomorphic to P , then we get a contradiction. Thuseither E ( V a , V a,x ) induces a complete bipartite graph or E ( V a , V a,x ) is empty. (cid:3) Claim 4.62.
Let u ∈ V a . If V b,c = ∅ , then E ( u, V b,c ) satisfies only one of the following: • it induces a complete bipartite graph, • it is an empty edge set, or • it induces a complete bipartite graph minus an edge.Proof. Suppose there exist v , v , v ∈ V b,c such that uv ∈ E ( G ) and uv , uv / ∈ E ( G ). Since theinduced subgraph G [ { a, b, u, v , v , v } ] is isomorphic to G , , then we get a contradiction. Thus thiscase is not possible and the result follows. (cid:3) Claim 4.63. If V a = { u , u } and V b,c = ∅ , then E ( V a , V b,c ) satisfies one of the following: • it induces a complete bipartite graph, • it is an empty edge set, • it induces a complete bipartite graph minus an edge, • it induces a perfect matching and | V a | = | V b,c | = 2 , or • it induces a complete bipartite graph minus two edges u v and u v , where v ∈ V b,c .Proof. Since cases where | V b,c | ≤ | V b,c | ≥
3. By Claim 4.62, we only have
MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 27 to check the possibilities of the edge sets E ( u , V b,c ) and E ( u , V b,c ). The possible cases we have todiscard are the following: • E ( u , V b,c ) = ∅ and E ( u , V b,c ) induces a complete bipartite graph, • E ( u , V b,c ) = ∅ and E ( u , V b,c ) induces a complete bipartite graph minus an edge, and • each edge set E ( u , V b,c ) and E ( u , V b,c ) induces a complete bipartite graph minus an edgeand the two removed edges don’t share a common vertex.Let v , v , v ∈ V b,c . Suppose we are in the first case. Thus u is adjacent with no vertex in V b,c ,and u is adjacent with each vertex in V b,c . Since the induced subgraph G [ { a, b, u , u , v , v } ] isisomorphic to G , , then we get a contradiction and this case is not possible. Suppose we are in thesecond case. Thus u is adjacent with no vertex in V b,c , and u is adjacent only with each vertexin V b,c − v . Since the induced subgraph G [ { v , b, v , u , u } ] is isomorphic to P , then we get acontradiction and this case is not possible. Finally, suppose we are in the third case. Thus u isadjacent with each vertex in V b,c − v , and u is adjacent with each vertex in V b,c − v . Since theinduced subgraph G [ { a, u , u , v , v , v } ] is isomorphic to G , , then we get a contradiction and thiscase is not possible. And the result turns out. (cid:3) Claim 4.64. If E ( V a , V a,b ) and E ( V a , V a,c ) are empty, then E ( V a,b , V a,c ) induces a complete bipartitegraph.Proof. Let v ab ∈ V a,b and v ac ∈ V a,c . Suppose v ab and v ac are not adjacent. Since the inducedsubgraph G [ { a, b, c, u , v ab , v ac } ] is isomorphic to G , , then we get a contradiction. And thus eachvertex in V a,b is adjacent with each vertex in V a,c . (cid:3) Claim 4.65.
If each of the edge sets E ( V a , V a,b ) and E ( V a , V a,c ) induces a complete bipartite graph,then | V a | = | V a,b | = | V a,c | = 1 and E ( V a,b , V a,c ) = ∅ .Proof. First we will prove that E ( V a,b , V a,c ) = ∅ . Let v ab ∈ V a,b and v ac ∈ V a,c . Suppose v ab and v ac are adjacent. Since the induced subgraph G [ { a, b, c, v ab , v ac , v a } ] is isomorphic to G , , then we geta contradiction. And thus E ( V a,b , V a,c ) = ∅ .Now let x ∈ { b, c } . Suppose V a,x has cardinality at least 2. Let y ∈ { b, c } − x , v ax , v ′ ax ∈ V a,x and v ∈ V a,y . Then v ax v, v ax v / ∈ E ( G ). Since the induced subgraph G [ { a, y, v ax , v ′ ax , v a , v } ] is isomorphicto G , , then we get a contradiction. And then | V a,b | = | V a,c | = 1Finally suppose that | V a | = 2. Let v ab ∈ V a,b and v ac ∈ V a,c . Since G [ { v ab , v ac , u , u , a, x } ] isisomorphic to G , , then we get a contradiction. Thus V a has cardinality at most 1. (cid:3) Claim 4.66.
Let x ∈ { b, c } and y ∈ { b, c } − x . If E ( V a , V a,x ) = ∅ and E ( V a , V a,y ) induces a completebipartite graph, then E ( V a,b , V a,c ) induces a complete bipartite graph.Proof. Let v ab ∈ V a,b and v ac ∈ V a,c . Suppose v ab and v ac are not adjacent. Since the induced sub-graph G [ { v a , v ay , y, x, v ax } ] is isomorphic to P , then we get a contradiction. And thus E ( V a,b , V a,c )induces a complete bipartite graph. (cid:3) Claim 4.67.
Let x ∈ { b, c } . If V a,x = ∅ , V b,c = ∅ and E ( V a , V a,x ) = ∅ , then only one of the followingstatements is true: • E ( V a , V b,c ) = ∅ and E ( V a,x , V b,c ) induces a complete graph, or • each edge set E ( V a , V b,c ) and E ( V a,x , V b,c ) induce a complete bipartite graph.Proof. We will analyze the following four cases: • when E ( V a , V b,c ) = ∅ , • when E ( V a , V b,c ) induces a complete bipartite graph, • when there exist u ∈ V a and v , v ∈ V b,c such that uv ∈ E ( G ) and uv / ∈ E ( G ), and • when there exists v ∈ V b,c such that u v ∈ E ( G ) and u v / ∈ E ( G ), where u , u ∈ V a .Consider the first case. Let v ax ∈ V a,x and v bc ∈ V b,c . Suppose v ax and v bc are not adjacent. Sincethe induced subgraph G [ { a, b, c, u , v ax , v bc } ] is isomorphic to G , , then we get a contradiction. Andthus E ( V a,x , V b,c ) induces a complete bipartite graph.Now consider the second case. Let v ax ∈ V a,x and v bc ∈ V b,c . Suppose v ax and v bc are notadjacent. Since the induced subgraph G [ { a, b, c, u , v ax , v bc } ] is isomorphic to G , , then we get acontradiction. And thus E ( V a,x , V b,c ) induces a complete bipartite graph.In what follows we will prove that the last two cases are impossible which implies that the restpossibilities of Claim 4.63 are impossible.Now consider the third case. Let v ax ∈ V a,x . Suppose there exist u ∈ V a and v , v ∈ V b,c suchthat uv ∈ E ( G ) and uv / ∈ E ( G ). There are four possible cases: • v ax v , v ax v ∈ E ( G ), • v ax v , v ax v / ∈ E ( G ), • v ax v ∈ E ( G ) and v ax v / ∈ E ( G ), or • v ax v / ∈ E ( G ) and v ax v ∈ E ( G ).Since in first case G [ { a, b, c, u, v ax , v , v } ] ≃ G , , in second case G [ { b, c, u, v ax , v , v } ] ≃ G , , inthird case G [ { b, c, u, v ax , v , v } ] ≃ G , and in fourth case G [ { u, v , v y , v , v ax } ] ≃ P , then we get acontradiction.In the last case, we get a contradiction because otherwise by first case: v a,x v / ∈ E ( G ), but bysecond case: v a,x v ∈ E ( G ). Thus this case is impossible. (cid:3) Claim 4.68.
Let x ∈ { b, c } . If V a,x = ∅ , V b,c = ∅ , and E ( V a , V a,x ) induces a complete bipartitegraph, then the edge sets E ( V a , V b,c ) and E ( V a,x , V b,c ) induce complete bipartite graphs.Proof. Let v ax ∈ V a,x , v ∈ V b,c and u ∈ V a . Suppose uv / ∈ E ( G ). There are two cases: either vv ax ∈ E ( G ) or vv ax / ∈ E ( G ). Since the induced subgraph G [ { a, b, c, u, v, v ax } ] is isomorphic to G , and G , , in the first case and in the second case, respectively, then we get a contradiction. Thus uv ∈ E ( G ) and therefore the edge set E ( V a , V b,c ) induces a complete bipartite graph.Now suppose vv ax / ∈ E ( G ). By previous result, uv ∈ E ( G ). Since the induced subgraph G [ { a, b, c, u, v, v ax } ] is isomorphic to G , , then we get a contradiction and thus vv ax ∈ E ( G ).Therefore E ( V a,x , V b,c ) induces a complete bipartite graph. (cid:3) By previous Claims we have the following cases:(1) E ( V a , V a,b ∪ V ac ∪ V b,c ) = ∅ , and each edge set E ( V a,b , V a,c ) and E ( V a,b ∪ V a,c , V b,c ) induces acomplete bipartite graph,(2) E ( V a , V a,b ∪ V ac ) = ∅ , and each edge set E ( V a,b , V a,c ) and E ( V a ∪ V a,b ∪ V a,c , V b,c ) induces acomplete bipartite graph,(3) E ( V a , V a,b ) = ∅ , and each edge set E ( V a ∪ V a,b , V a,c ) and E ( V a ∪ V a,b ∪ V a,c , V b,c ) induces acomplete bipartite graph,(4) E ( V a,b , V a,c ) = ∅ , and each edge set E ( V a , V a,b ∪ V a,c ∪ V b,c ) and E ( V a,b ∪ V a,c , V b,c ) induces acomplete bipartite graph and | V a | = | V a,b | = | V a,c | = 1,(5) V b,c = ∅ , E ( V a , V a,b ∪ V ac ) = ∅ , and E ( V a,b , V a,c ) induces a complete bipartite graph,(6) V b,c = ∅ , E ( V a , V a,x ) = ∅ , and E ( V a ∪ V a,x , V a,y ) induces a complete bipartite graph, where x ∈ { a, b } and y ∈ { a, b } − x , MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 29 (7) V b,c = ∅ , E ( V a , V a,b ∪ V a,c ) induce a complete bipartite graph, E ( V a,b , V a,c ) = ∅ and | V a | = | V a,b | = | V a,c | = 1,(8) V a,y = ∅ , E ( V a , V a,x ∪ V b,c ) = ∅ , and E ( V a,x , V b,c ) induces a complete bipartite graph, where x ∈ { a, b } and y ∈ { a, b } − x ,(9) V a,y = ∅ , E ( V a , V a,x ) = ∅ , and E ( V a ∪ V a,x , V b,c ) induces a complete bipartite graph, where x ∈ { a, b } and y ∈ { a, b } − x ,(10) V a,y = ∅ , E ( V a , V a,x ∪ V b,c ) and E ( V a,x , V b,c ) induce a complete bipartite graph, where x ∈{ a, b } and y ∈ { a, b } − x ,(11) V a,b ∪ V a,c = ∅ , and E ( V a , V b,c ) induces a complete bipartite graph,(12) V a,b ∪ V a,c = ∅ and E ( V a , V b,c ) = ∅ ,(13) V a,b ∪ V a,c = ∅ , and E ( V a , V b,c ) induces a complete bipartite graph minus an edge,(14) V a,b ∪ V a,c = ∅ , | V b,c | ≥ E ( V a , V b,c ) induces a complete bipartite graph minus two edges u v and u v , where v ∈ V b,c ,(15) V a,b ∪ V a,c = ∅ , | V a | = | V b,c | = 2 and E ( V a , V b,c ) induces a perfect matching,(16) V a,y ∪ V b,c = ∅ and E ( V a , V a,x ) induces a complete bipartite graph, where x ∈ { a, b } and y ∈ { a, b } − x , and(17) V a,y ∪ V b,c = ∅ and E ( V a , V a,x ) = ∅ , where x ∈ { a, b } and y ∈ { a, b } − x .Now we describe V ∅ , the set of vertices that are not adjacent with any vertex in { a, b, c } . Let w ∈ V ∅ . The vertex w is adjacent with a vertex in V a ∪ V a,b ∪ V a,c ∪ V b,c , because otherwise theshortest path from w to { a, b, c } would contains the graph P as induced subgraph. Claim 4.69. If V a,x = ∅ for some x ∈ { b, c } , then E ( V ∅ , V a ∪ V a,x ) = ∅ .Proof. Let w ∈ V ∅ , v ax ∈ V a,x and u ∈ V a . Suppose w is adjacent with u or v ax . Then there arethree possible cases:(a) w is adjacent only with u ,(b) w is adjacent only with v ax , or(c) w is adjacent with both vertices u and v ax .First consider case (a). This case has two subcases: either u is adjacent with v ax or not. If they areadjacent, then G [ { u, w, v ax , b, c } ] is isomorphic to P . Since it is forbidden, then v a is not adjacentwith v ax . On the other hand, if they are not adjacent, then G [ { a, b, c, w, u, v ax } ] is isomorphic to G , . Therefore, case (a) is impossible. Now consider case (b). There are two possible cases: either u is adjacent with v ax or not. If they are adjacent, then G [ { a, b, c, w, u, v ax } ] is isomorphic to G , ;which is forbidden. Thus u is not adjacent with v ax . But if they are not adjacent, then we have that G [ { a, b, c, w, u, v ax } ] is isomorphic to G , . Thus case (b) cannot occur. Finally, consider case (c).The subcases are: either u is adjacent to v ax or not. If they are adjacent, then G [ { a, b, c, w, u, v ax } ] isisomorphic to G , ; which is forbidden. Then u is not adjacent with v ax But if they are not adjacent,then the induced subgraph G [ { u, w, v ax , b, c } ] is isomorphic to P ; which is forbidden. Thus, we getthat case (c) is impossible. And therefore, w is not adjacent with u or v ax . From which the resultfollows. (cid:3) Claim 4.70.
Let x ∈ { b, c } . If V a,x = ∅ , then E ( V ∅ , V b,c ) = ∅ . Moreover, if V a,x = ∅ , then V ∅ = ∅ .Proof. Let w ∈ V ∅ , v ax ∈ V a,x , v a ∈ V a and v bc ∈ V b,c . Suppose w is adjacent with v bc . By Claim4.69, we have that w is not adjacent with a vertex in V a ∪ V a,x . By Claims 4.67 and 4.68, the vertex v ax is adjacent with v bc , and one of the following three cases occur:(a) v a is adjacent with v ax and v bc ,(b) v a is adjacent only with v bc , and (c) v a is not adjacent with both v ax and v bc .Case (a) is not possible because the induced subgraph G [ { v a , v ax , b, c, v bc , w } ] would be isomorphicto G , ; which is impossible. In case (b), we have that G [ { v a , v ax , b, c, v bc , w } ] is isomorphic to G , . Since it is forbidden, then this case is not possible. Finally, case (c) is not possible since G [ { v a , a, x, v bc , w } ] is isomorphic to P that is forbidden. Thus w is not adjacent with v bc . And thereis no vertex in V ∅ adjacent with a vertex in G \ V ∅ . Therefore V ∅ is empty. (cid:3) Thus by previous Claim, in cases (1) to (10), (16) and (17) the vertex set V ∅ is empty. The caseswhen V a,b ∪ V a,c = ∅ (cases (4) and (7)) correspond to a graph isomorphic to an induced subgraphof F . The rest of these cases correspond to a graph isomorphic to an induced subgraph of F . Claim 4.71.
Let w a ∈ V ∅ such that w a is adjacent with u ∈ V a , and w a is not adjacent with a vertexin V b,c . Let w bc ∈ V ∅ such that w bc is adjacent with v ∈ V b,c and w bc is not adjacent with a vertex in V a . Let w ∈ V ∅ such that w is adjacent with u ′ ∈ V a and v ′ ∈ V b,c . If E ( V a , V b,c ) induces a completebipartite graph, then no two vertices of { w a , w bc , w } exist at the same time.Proof. Suppose w a and w bc exist at the same time. There there are two possible cases: either w a and w bc are adjacent or not. If w a and w bc are adjacent, then G [ { b, a, u, w a , w bc } ] is isomorphic to P ; which is impossible. If w a and w bc are not adjacent, then G [ { b, c, u, v, w a , w bc } ] is isomorphic to G , ; which is impossible. Then w a and w bc do not exist at the same time.Suppose w a and w exist at the same time. There there are two possible cases: either u = u ′ or u = u ′ . Suppose u = u ′ , then we have two possible cases: either w a and w are adjacent or not. If w a and w are adjacent, then G [ { w a , w, v ′ , b, a } ] is isomorphic to P ; which is impossible. But if w a and w are not adjacent, then G [ { b, c, u, v ′ , w a , w } ] is isomorphic to G , ; which is impossible. Then u = u ′ . Thus suppose u = u ′ . Note that in the induced subgraph G [ { a, b, c, u, v ′ , w a , w } ], the vertex w is only adjacent with v ′ , and w a is only adjacent with u . Then applying previous ( w a , w bc ) casein this induced subgraph, we get that w and w a do not exist at the same time.Suppose w bc and w exist at the same time. There are two possible cases: either v = v ′ or v = v ′ .Suppose v = v ′ , then we have two possible cases: either w bc and w are adjacent or not. If w a and w bc are adjacent, then G [ { w bc , w, u, a, b } ] is isomorphic to P ; which is impossible. But if w a and w bc are not adjacent, then G [ { b, c, u, v, w bc , w } ] is isomorphic to G , ; which is impossible. Then v = v ′ . Suppose v = v ′ . Note that in the induced subgraph G [ { a, b, c, u, v, w a , w } ], the vertex w isonly adjacent with u , and w bc is only adjacent with v . Thus by applying first case in this inducedsubgraph, we get that w and w bc do not exist at the same time. (cid:3) Claim 4.72.
Let w ∈ V ∅ , u ∈ V a and v ∈ V b,c . If E ( V a , V b,c ) induces a complete bipartite graph and w is adjacent with u and v , then E ( w, V a ∪ V b,c ) induces a complete graph.Proof. First we see that if v ′ ∈ V b,c − v , then w is adjacent with v ′ . Suppose w and v ′ are notadjacent. Since the induced subgraph G [ { a, b, u, v, v ′ , w } ] is isomorphic to the forbidden graph G , ,then we get a contradiction. Thus w is adjacent with each vertex in V b,c . Now we see that if u ′ ∈ V a − u , then w is adjacent with u ′ . Suppose w and u ′ are not adjacent. Since the inducedsubgraph G [ { a, b, u, u, u ′ , w } ] is isomorphic to G , , then we get a contradiction. Therefore, w isadjacent with each vertex in V a ∪ V b,c . (cid:3) Claim 4.73. If | V a | = 2 and E ( V ∅ , V a ) = ∅ , then either each vertex in V ∅ is adjacent only with onevertex in V a and V ∅ is a clique, or each vertex in V ∅ is adjacent with u and u , and V ∅ is trivial.Proof. Let w, w ′ ∈ V ∅ and i, j ∈ { , } with i = j . By proving that the following cases are not possi-ble, it follows that the only possible cases are that either E ( { u , u } , { w, w ′ } ) is equal to { wu i , w ′ u j } MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 31 and ww ′ ∈ E ( G ), or E ( { u , u } , { w, w ′ } ) is equal to { wu i , wu j , w ′ u i , w ′ u j } and ww ′ ∈ E ( G ). Whichimplies the result.(a) ww ′ , wu j , w ′ u j / ∈ E ( G ) and wu i , w ′ u i ∈ E ( G ),(b) ww ′ , wu j , w ′ u i / ∈ E ( G ) and wu i , w ′ u j ∈ E ( G ),(c) wu j , w ′ u i / ∈ E ( G ) and ww ′ , wu i , w ′ u j ∈ E ( G ),(d) ww ′ , w ′ u j / ∈ E ( G ) and wu i , wu j , w ′ u i ∈ E ( G ),(e) w ′ u j / ∈ E ( G ) and ww ′ , wu i , wu j , w ′ u i ∈ E ( G ), and(f) w ′ u j , ww ′ , wu i , wu j , w ′ u i ∈ E ( G ).Since in cases (a), (b) and (d) the induced subgraph G [ { a, b, u , u , w, w ′ } ] is isomorphic to G , , G , , G , , respectively, then these cases are impossible. On the other hand, in case (c) the induced graph G [ { w, w ′ , u j , a, b } ] is isomorphic to P : which is impossible. Also in case (e) the induced subgraph G [ { w ′ , w, u j , a, b } ] is isomorphic to P ; which is not possible. Finally in case (f) the induced subgraph G [ { a, b, c, u , u , w, w ′ } ] is isomorphic to G , ; which is not possible. (cid:3) Claim 4.74. If E ( V a , V b,c ) induces a complete bipartite graph, and each vertex in V ∅ is adjacent onlywith one vertex in V a , then there is no vertex in V ∅ adjacent with a vertex in V b,c .Proof. Let w ∈ V ∅ , v ∈ V b,c . Suppose w is adjacent with u and v , but w is not adjacent with u .Since the induced subgraph G [ { a, b, w, u , u , v } ] is isomorphic to G , , then we get a contradictionand the result follows. (cid:3) Claim 4.75.
Let w ∈ V ∅ and u ∈ V a . If E ( V a , V b,c ) induces a complete bipartite graph, E ( w, V b,c ) = ∅ and E ( w, V a ) = ∅ , then each vertex in V ∅ is adjacent with only one vertex in V b,c , and V ∅ is a cliqueof cardinality at most 2.Proof. Let v, v ′ ∈ V b,c . Suppose w is adjacent with v and v ′ . Since G [ { a, b, c, u, v, v ′ , w } ] is isomorphicto G , , then we get a contradiction. And we have that w is adjacent only with one vertex in V b,c Now let w, w ′ ∈ V ∅ and v, v ′ ∈ V b,c such that wv ∈ E ( G ) and w ′ v ′ / ∈ E ( G ). Suppose v = v ′ . Thereare two cases: either w and w ′ are adjacent or not. If w and w ′ are adjacent, then the inducedsubgraph G [ { b, c, v, u, w, w ′ } ] is isomorphic to G , , then we get a contradiction and ww ′ / ∈ E ( G ).Now if w and w ′ are not adjacent, then the induced subgraph G [ { w, v, u, v ′ , w ′ } ] is isomorphic to P and we get a contradiction. Thus v = v ′ .Finally, let w, w ′ ∈ V ∅ adjacent with v ∈ V b,c . Suppose w and w ′ are not adjacent. Sincethe induced subgraph G [ { a, b, v, u, w, w ′ } ] is isomorphic to G , , then we get a contradiction and ww ′ ∈ E ( G ). (cid:3) In case (11), by Claims 4.71, 4.72, 4.73, 4.74 and 4.75, we have the following possible cases: • each vertex in V ∅ is adjacent with each vertex in V a ∪ V b,c , and V ∅ is a trivial graph, • each vertex in V ∅ is adjacent only with a vertex in V a , and V ∅ is a clique of cardinality atmost 2, or • each vertex in V ∅ is adjacent only with a vertex in V b,c , and V ∅ is a clique of cardinality atmost 2.Each of these cases corresponds to a graph isomorphic to an induced subgraph of F . Remark 4.76.
Let w ∈ V ∅ , u ∈ V a and v ∈ V b,c . If uv / ∈ E ( G ) , and w adjacent with u or v , then w is adjacent with both vertices u and v . In case (12), by Claim 4.73 and Remark 4.76, we have that each vertex in V ∅ is adjacent witheach vertex in V a ∪ V b,c , and • V ∅ is a clique or a trivial graph when | V a | = 1, or • V ∅ is a clique when | V a | = 2.It is not difficult to see that each of these graphs are isomorphic to an induced subgraph of F Claim 4.77.
Let w ∈ V ∅ , u ∈ V a and v ∈ V b,c . Suppose u is adjacent with each vertex in V b,c − v .If E ( w, V b,c ∪ { u } ) = ∅ , then one of the following statements holds: • each vertex in V ∅ is adjacent with u and v ,or • E ( w, V b,c ∪ { u } ) induces a complete bipartite graph and | V b,c − v | = 1 .Proof. First case follows by Remark 4.76. Now we check when w is adjacent with a vertex in V b,c − v .Let v ′ ∈ V b,c . Then u is adjacent with v ′ . Suppose w is adjacent only with v ′ . Since the inducedsubgraph G [ { a, b, u, v, v ′ , w } ] is isomorphic to G , , then we have a contradiction and w is adjacentalso with u or v . Thus by Remark 4.76, w is adjacent with u , v and v ′ . Finally by applying Claim4.72 to the induced subgraph G [ { w, u } ∪ ( V b,c − v )], we get that w also is adjacent with each vertexin V b,c − v . Thus w is adjacent with each vertex in { u } ∪ V b,c .Suppose the cardinality of V b,c − v is at least 2. Take v, v ′ ∈ V b,c − v . Thus w is adjacent with v, v ′ and v ′′ . Since the induced subgraph G [ { u, v, v ′ , v ′′ , a, w } ] is isomorphic to G , , then we get acontradiction. Thus the cardinality of V b,c − v is at most 1. (cid:3) Claim 4.78.
Let w ∈ V ∅ and v ∈ V b,c . Suppose u is adjacent with each vertex in V b,c , and u isadjacent with each vertex in V b,c − v . If E ( w, V b,c ) = ∅ , then either w is adjacent only with u and v , or | V b,c | = 1 and E ( w, V a ∪ V b,c ) induces a complete bipartite graph.Proof. First case follows by Remark 4.76. Now we prove that if w is adjacent with u , then w is adjacent with u and v . Suppose w is adjacent only with u . Since the induced subgraph G [ { a, b, u , u , v, w } ] is isomorphic to G , , then we have a contradiction and w is adjacent also with u or v . Thus by Remark 4.76, w is adjacent with u , u and v . Finally, suppose | V b,c | ≥
2. Let v ′ ∈ V b,c . By applying Claim 4.72 to the induced subgraph G [ { w, u } ∪ V b,c ], we get that w alsois adjacent with each vertex in V b,c . Thus w is adjacent with each vertex in V a ∪ V b,c . But since G [ { a, u , u , v, v ′ , w } ] is isomorphic to G , , then we get a contradiction, and | V b,c | = 1. (cid:3) Let u ∈ V a and v ∈ V b,c such that uv / ∈ E ( G ). Therefore, in case (13), by Claims 4.73, 4.77 and4.78, we have that one of the following cases holds: • V ∅ is a clique of cardinality at most 2, and each vertex in V ∅ is adjacent with u and v , • V a = { u } , | V b,c \ { v }| ≤ V ∅ is trivial and each vertex in V ∅ is adjacent with each vertex in { u } ∪ V b,c , or • | V a | = 2, V ∅ is trivial and V b,c = { v } and each vertex in V ∅ is adjacent with u , u and v .It can be checked that each of these graphs is isomorphic to an induced subgraph or a graph in F . Claim 4.79.
Let w ∈ V ∅ and v ∈ V b,c such that v is not adjacent with u and u . Suppose | V b,c | ≥ and E ( V a , V b,c − v ) induces a complete bipartite graph. If E ( w, V b,c ) = ∅ , then V ∅ = { w } , and w isadjacent with u , u and v .Proof. By Remark 4.76, we have that if w is adjacent with one vertex of { v, u , u } , then w isadjacent with v, u and u . Now we see that w is not adjacent to any other vertex. Suppose w isadjacent with v ′ ∈ V b,c − v . First consider the case when w is not adjacent with v, u and u . Since G [ { b, v, v ′ , u , u , w } ] is isomorphic to G , , then we get a contradiction and thus w is adjacent with v, v ′ , u , and u . But if w is adjacent with v, v ′ , u , and u , then G [ { b, v, v ′ , u , u , w } ] is isomorphicto G , ; which is impossible. Then w is not adjacent with any vertex in V b,c − v . MALL CLIQUE NUMBER GRAPHS WITH THREE TRIVIAL CRITICAL IDEALS 33
Suppose there exist another vertex w ′ ∈ V ∅ . By previous discussion w ′ is adjacent with v, u and u . Since G [ { a, b, w, w ′ , v, v ′ } ] is isomorphic to G , , then we get a contradiction. Thus V ∅ hascardinality at most 1. (cid:3) Let v ∈ V b,c such that u v, u v / ∈ E ( G ). Therefore, in case (14), by Claims 4.73 and 4.79, we havethat V ∅ = { w } and w is adjacent with u , u and v . Then G is isomorphic to an induced subgraphof F . Claim 4.80.
Let V b,c = { v, v ′ } such that vu , v ′ u ∈ E ( G ) and vu , v ′ u / ∈ E ( G ) . If w ∈ V ∅ and E ( w, V a ∪ V b,c ) = ∅ , then w is adjacent with u , u , v and v ′ .Proof. Since w is adjacent with a vertex in { u , u , v, v ′ } , then w is adjacent with u and v ′ , or w isadjacent with u and v . Suppose w is adjacent with u and v ′ , but w is not adjacent with u and v . Since G [ { a, u , u , v, v ′ , w } ] is isomorphic to G , , then we get a contradiction, and therefore w is also adjacent with u and v . (cid:3) Let V b,c = { v, v ′ } such that vu , v ′ u ∈ E ( G ) and vu , v ′ u / ∈ E ( G ). Thus in case (15), by Claims4.73 and 4.80, we have that V ∅ is trivial and each vertex in V ∅ is adjacent only with u , u , v and v ′ .Therefore, G is isomorphic to an induced subgraph of F . References [1] C.A. Alfaro and C.E. Valencia, Graphs with two trivial critical ideals, to appear in Discrete Appl. Math.[2] C.A. Alfaro, H.H. Corrales and C.E. Valencia, Critical ideals of signed graphs with twin vertices, preprint.[3] R. Bacher, P. de la Harpe and T. Nagnibeda, The lattice of integral flows and the lattice of integral cuts on afinite graph, Bull. Soc. Math. France 125 (1997) 167-198.[4] N. Biggs, Chip-firing and the critical group of a graph, J. Alg. Combin. 9 (1999) 25-46.[5] J.A. Bondy and U.S.R. Murty, Graph Theory, Grad. Texts in Math., vol. 244, Springer, 2008.[6] W.H. Chan, Y. Hou, W.C. Shiu, Graphs whose critical groups have larger rank, Acta Math. Sinica 27 (2011)1663–1670.[7] R. Cori and D. Rossin, On the sandpile group of dual graphs, European J. Combin. 21 (4) (2000) 447–459.[8] H.H. Corrales and C.E. Valencia, On the critical ideals, to appear.[9] D.R. Grayson and M.E. Stillman,
Macaulay2, a software system for research in algebraic geometry nauty Users Guide (Version 2.4) , available at http://cs.anu.edu.au/ ∼ bdm/nauty/.[17] C. Merino, The chip-firing game, Discrete Math. 302 (2005) 188–210.[18] R. Merris, Unimodular Equivalence of Graphs, Linear Algebra Appl. 173 (1992),181-189[19] Y. Pan and J. Wang, A note on the third invariant factor of the Laplacian matrix of a graph, preprint,math.CO/0912.3608v1[20] D.G. Wagner, The critical group of a directed graph, preprint, arXiv:math/0010241v1 [math.CO] Departamento de Matem´aticas, Centro de Investigaci´on y de Estudios Avanzados del IPN, ApartadoPostal 14–740, 07000 Mexico City, D.F.
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