Algebraic Properties of Clique Complexes of Line Graphs
AAlgebraic Properties of Clique Complexes ofLine Graphs
Ashkan Nikseresht
Department of Mathematics, College of Science, Shiraz University,71457-13565, Shiraz, IranE-mail: ashkan [email protected]
Abstract
Let H be a simple undirected graph and G = L( H ) be its linegraph. Assume that ∆( G ) denotes the clique complex of G . We showthat ∆( G ) is sequentially Cohen-Macaulay if and only if it is shellableif and only if it is vertex decomposable. Moreover if ∆( G ) is pure, weprove that these conditions are also equivalent to being strongly con-nected. Furthermore, we state a complete characterizations of those H for which ∆( G ) is Cohen-Macaulay, sequentially Cohen-Macaulayor Gorenstein. We use these characterizations to present linear timealgorithms which take a graph G , check whether G is a line graphand if yes, decide if ∆( G ) is Cohen-Macaulay or sequentially Cohen-Macaulay or Gorenstein. Keywords: Line graph; Cohen-Macaulay ring; Gorenstein ring; Simplicialcomplex; Edge ideal;Mathematics Subject Classification (2020): 13F55; 05E40; 05E45.
In this paper, K denotes a field and S = K [ x , . . . , x n ]. It is known that usingseveral transformations on a graded ideal I of S , such as taking generic initialideal, polarization, . . . , we can get a square-free monomial ideal J generatedin degree 2, such that S/I is Cohen-Macaulay (CM for short) if and only if
S/J is so (see [3] and also [7]). But then J is the edge ideal of a graph andthis shows why it is important to study algebraic properties of edge idealsof graphs. Let G be a simple graph on vertex set V( G ) = { v , . . . , v n } and1 a r X i v : . [ m a t h . A C ] J u l dge set E( G ). Then the edge ideal I ( G ) of G is the ideal of S generatedby { x i x j | v i v j ∈ E( G ) } . A graph G is called CM (resp. Gorenstein) when S/I ( G ) is CM (resp. Gorenstein) for every field K . Many researchers havetried to combinatorially characterize CM or Gorenstein graphs in specificclasses of graphs, see for example, [2–6, 10, 11, 13, 17, 18, 20]).The family of cliques of a graph G forms a simplicial complex which iscalled the clique complex of G and is denoted by ∆( G ). Algebraic propertiesof simplicial complexes in general also has got a wide attention recently,see for example [3, 7, 9, 12, 15] and the references therein. If we denotethe Stanley-Reisner ideal of ∆ by I ∆ , then we have I ∆( G ) = I ( G ), where G denotes the complement of the graph G . Thus studying clique complexes ofgraphs algebraically, is another way to study algebraic properties of graphs.Suppose that H is a simple undirected graph and G = L( H ) is the linegraph of H , that is, edges of H are vertices of G and two vertices of G are adjacent if they share a common endpoint in H . Line graphs are well-known in graph theory and have many applications (see for example [19,Section 7.1]). In particular, Theorems 7.1.16 to 7.1.18 of [19], state somecharacterizations of line graphs and methods that, given a line graph G , canfind a graph H for which G = L( H ). Indeed, in [8] a linear time algorithmis presented that, given a graph G , it checks if G is a line graph and if G isa line graph, it returns a graph H such that G = L( H ).Here we study some algebraic properties of ∆( G ). Because of the afore-mentioned results and algorithm, we do this in terms of the graph H forwhich G = L( H ). First in Section 2, we characterize combinatorially those H with a CM or a Gorenstein ∆(L( H )). As we will see, the class of suchgraphs is very limited.Then in Section 3, we characterize those H whose line graph has a se-quentially CM clique complex. Recall that a simplicial complex ∆ is calledsequentially CM, when its pure skeletons ∆ [ i ] = (cid:104) F ∈ ∆ (cid:12)(cid:12) | F | = i + 1 (cid:105) are CMfor all i (see also [15], for an equivalent algebraic definition). Our character-ization enables us to present a linear time algorithm which decides whether∆(L( H )), for a given graph H , is sequentially CM or not. Thus our algo-rithm enables us to efficiently decide whether a given graph is a line graphand if yes, whether its clique complex is sequentially CM.Here we say a simplicial complex ∆ is CM (resp. Gorenstein) over K ,when S/I ∆ is CM (resp. Gorenstein). If ∆ is CM (resp. Gorenstein) overevery field K , then we simply say that ∆ is CM (resp. Gorenstein). Fordefinitions and basic properties of simplicial complexes and graphs one cansee [3] and [19], respectively. In particular, all notations used in the sequelwithout stating the definitions are as in these two references.2 Line graphs with CM or Gorenstein cliquecomplexes
In what follows, H is a simple undirected graph with at least one edge and G = L( H ) is the line graph of H . We are going to study when ∆( G ) is CM.It is well-known that a CM complex is pure (see [3, Lemma 8.1.5]). Thefollowing lemma characterizes those H with pure ∆( G ). Here K n denotesthe complete graph on n vertices. Also we call a set of r edges of H adjacentto a common vertex v , an r -star (or simply, a star ) of H at v . Lemma 2.1. If H is connected, then the clique complex of G is pure, if andonly if one of the following holds. (i) H has no triangles and there is an integer r > such that every vertexof H has degree either one or r . (ii) The maximum degree of vertices of H is 3 and every vertex of H withdegree 2 is contained in a triangle. (iii) H is a path or a cycle.Proof. Note that since H is connected, G is also connected. Moreover, eachclique of G is either a set of edges sharing an endpoint v in H , that is, a starof H at v , or a triangle of H . Thus if ∆( G ) is pure and dim ∆( G ) >
2, theneach star of H should be contained in a star with size dim ∆( G ) + 1 and H cannot have any triangles. Thus case (i) occurs. If dim ∆( G ) = 2, then eachstar of H should be in a star of size 3 or a triangle, that is, H is as in case(ii). Finally, if dim ∆( G ) <
2, then vertices of H have degree at most 2 andcase (iii) happens. The converse is clear.If ∆ is pure and for any two facets F and G of ∆, there is a sequence F = F , . . . , F t = G of facets of ∆, such that | F i ∩ F i +1 | = | F i | − i , we say that ∆ is strongly connected (or connected in codimension 1).By [3, Lemma 9.1.12], every CM complex is strongly connected. Thus wenext investigate when ∆( G ) is strongly connected. Note that every stronglyconnected complex with dim > F and F of ∆( G ), adjacent when | F ∩ F | = | F | − | F | −
1, andby a strong path between facets F and F t , we mean a sequence F , . . . , F t offacets, with F i adjacent to F i +1 for all 1 ≤ i < t . Lemma 2.2.
Suppose that ∆( G ) is pure. Then ∆( G ) is strongly connectedif and only if H is either a star or a path or a cycle or one of the graphs inFigure 1. b)(a)(e) (f ) (g)(c) (d) Figure 1: Graphs with line graphs having strongly connected clique complexs
Proof.
It is easy to check that the line graph of each of the mentioned graphshas a strongly connected clique complex. Conversely, suppose that ∆( G ) isstrongly connected and d = dim ∆( G ). Note that two stars of H at differentvertices can have at most one edge in common. Also no two different trianglesare adjacent and a triangle can only be adjacent with a 3-star at a vertexof the triangle. So if d > H has a vertex v with degree d + 1, the( d + 1)-star at v cannot be adjacent to any other facet of ∆( G ) and henceshould be the only facet of ∆( G ). Therefore, by (2.1), all other vertices of H have degree 1 and H is a star. If d = 1, then connectedness and stronglyconnectedness for ∆( G ) are equivalent and by (2.1), H is either a path or acycle.Now assume that d = 2. If H has no triangle, then similar to the casethat d > H is a star. If H has just one triangle, then every 3-star of H must be adjacent to this triangle, which means, should be centered at avertex of the triangle. Thus H is either a triangle, that is, a 3-cycle or oneof the graphs (b), (c) or (d) of Figure 1. Now assume that H has exactly 2triangles T and T . Then there should be a 3-star F , such that both T and T are adjacent to F . Thus F is centered at a vertex of both T and T andmust share two edges with each of them. Since degree of each vertex is atmost three, it follows that T and T have a common edge. Again as every3-star of H is adjacent to either T or T , H should be one of the graphs (e),(f) or (g) in Figure 1.Note that if H has more than 2 triangles, again a 3-star should be adjacentto two triangles and hence H has a subgraph isomorphic to the graph ofFigure 1(e). Lets call this subgraph H , call the vertices of degree 2 of H , a and b and call the vertices of degree 3 of H , u and v . Suppose that H has a triangle not contained in H . Then there exists a strong path starting4ith a triangle in H and ending with a triangle not in H . Let T be thefirst triangle in this path which is not in H . Then T and a triangle of H are both adjacent to a 3-star and hence share an edge. But any edge in H in incident to either u or v , hence u or v is in T . Thus T is either uab or vab . This means that a and b are adjacent and since all vertices have degreeat most 3, H is the graph in Figure 1(a).Recall that for a face F of a simplicial complex ∆, we define link ∆ F = { G \ F | F ⊆ G ∈ ∆ } . Also for a vertex v of ∆, ∆ − v is the simplicial complexwith faces { F ∈ ∆ | v / ∈ F } . A vertex v of a nonempty simplicial complex ∆is called a shedding vertex , when no face of link ∆ ( v ) is a facet of ∆ − v . By [9,Lemma 3.1], if ∆ is pure, a vertex v is a shedding vertex if and only if ∆ − v is pure and dim(∆ − v ) = dim ∆. A nonempty simplicial complex ∆ is called vertex decomposable , when either it is a simplex or there is a shedding vertex v such that both link ∆ v and ∆ − v are vertex decomposable. The ( − {∅} is considered a simplex and hence vertexdecomposable. To characterize line graphs with pure vertex decomposableclique complexes, we use the following lemmas. Lemma 2.3.
Suppose that C is a graph without isolated vertices. Then ∆ = (cid:104) E( C ) (cid:105) is vertex decomposable if and only if C is connected.Proof. Note that every 0-dimensional simplicial complex is vertex decompos-able. Thus ∆ is vertex decomposable if and only if it can be transformed toa simplex by repeatedly deleting shedding vertices. Also a vertex v of ∆ is ashedding vertex, if and only if v has no neighbor with degree 1. Assume that C is connected. If C has a vertex v of degree 1, then v is a shedding vertexunless C is K , in which case ∆ is a simplex and vertex decomposable. Thuswe can delete v and a get a smaller connected graph, hence the result followsby induction. If C has no vertex of degree 1, then there is a (shedding)vertex v of C such that C − v is again connected. Again the result followsby induction.Conversely, deleting a shedding vertex from C does not decrease the num-ber of connected components of C . Hence if C is not connected, then afterdeleting any number of shedding vertices, the obtained graph C (cid:48) is still notconnected and hence (cid:104) E( C (cid:48) ) (cid:105) is not a simplex. Hence ∆ is not vertex decom-posable.It should be mentioned that [12, Lemma 3.1], states that for a graph C ,the simplicial complex (cid:104) E( C ) (cid:105) is “vertex decomposable” if and only if C isa tree, which is in contradiction with the above lemma. This is because, asmentioned in a corrigendum to [12], vertex decomposability as used in [12],5iffers slightly with that used in the literature and here. For more details,see the corrigendum at the end of the arXiv version of [12]. Lemma 2.4.
If each connected component of a graph C is a tree, then ∆( C ) is vertex decomposable.Proof. Immediate consequence of [20, Theorem 1].Another combinatorial property which is stronger than being CM is shella-bility. If there is an ordering F , . . . , F t of all facets of ∆ such that for each i we have (cid:104) F , . . . , F i (cid:105) ∩ (cid:104) F i +1 (cid:105) is a pure simplicial complex of dimension= dim F i +1 −
1, then ∆ is called shellable and such an order is called a shellingorder . It is well-known that a vertex decomposable complex is shellable (seefor example [20, Section 2]) and a pure shellable complex is CM ([3, Theorem8.2.6]).
Theorem 2.5.
Suppose that H is a graph with at least one edge and G =L( H ) . Then the following are equivalent for ∆ = ∆( G ) . (i) ∆ is pure vertex decomposable. (ii) ∆ is pure shellable. (iii) ∆ is CM (over some field). (iv) ∆ is pure and strongly connected. (v) H is either a star or a path or a cycle or one of the graphs in Figure1.Proof. The facts that (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) are well-known. Also (2.2)shows that (iv) ⇔ (v).(v) ⇒ (i): It is easy to check that in each case ∆ is pure. If H is a star,then G is a complete graph and ∆ is a simplex. If H is a path or a cycle,then G is also a path or a cycle and ∆( G ) = (cid:104) E( G ) (cid:105) (unless H = K , inwhich case ∆ is a simplex). Consequently, the result follows from (2.3). It isroutine to check that if H is any of the graphs in Figure 1, except the graphs(d) and (g), then each connected component of G is a tree and according to(2.4), ∆ = ∆( G ) is vertex decomposable.Now assume that H is the graph (d) in Figure 1. Then G is the graph(a) in Figure 2. If v is the vertex specified in Figure 2(a), then ∆( G ) − v =∆( G − v ) is pure with dimension 2. Hence v is a shedding vertex of ∆ andsince G − v is a tree, it follows that ∆ − v is vertex decomposable. Alsolink ∆ ( v ) is a simplex. Thus, ∆ is vertex decomposable. If H is the graph(g) in Figure 1, then G is the graph (b) in Figure 2 and a similar argument,with v as specified in Figure 2(b), shows that ∆ is vertex decomposable.6 v (a) (b) Figure 2: The graph G when H is the graph (a) in Figure (1d) (b) in Figure(1g)Now we can also answer the question “when ∆( G ) is Gorenstein?” Notethat this question is equivalent to asking when G is a Gorenstein graph.By [5, Lemma 3.1] or [6, Lemma 3.5], each Gorenstein graph C withoutisolated vertices is a W graph , that is, | V( C ) | ≥ C are contained in two disjoint maximum independentsets of C . Corollary 2.6.
Assume that H is a graph with at least one edge and G =L( H ) . Then ∆( G ) is Gorenstein if and only if H is either a star or a cycleor a path with length at most 3 or one of the graphs in Figure 3.Proof. Since every Gorenstein complex is CM, we must search for graphswith Gorenstein ∆( G ) between the graphs mentioned in (2.5)(v). Since asimplex is Gorenstein and also the complement of every cycle is Gorensteinby [13, Corollary 2.4], we deduce that if H is a star or a cycle, then ∆( G ) orequivalently G is Gorenstein. If H is a path of length n , then G is a path oflength n − n − ≥
3, so by [13, Corollary 2.4], G is not Gorenstein.If n − <
3, then each connected component of G is K or K and hence G is Gorenstein.If H is one of the graphs in Figure 1 and not in Figure 3, except for thecase that H is the graph (g) of Figure 1, then G has a vertex of degree 1 ina non-complete connected component. But according to Theorem 4 of [16]such a graph is not W and hence G is not Gorenstein. If H is the graph (g)in Figure 1, then the smallest cycle of G has length 5. But by [14, Theorem7], every W graph with girth at least five is either the 5-cycle or K . Thus inthis case also G is not W nor Gorenstein. If H is any of the graphs in 3, thenevery connected component of G is K or K and hence G is Gorenstein.At the end of this section, we should mention that we can use (2.5) and(2.6) to decide whether a given graph G is a line graph with a CM or aGorenstein clique complex or not. For Cohen-Macaulayness, we must check7igure 3: Some graphs with Gorenstein ∆( G ).whether G is a complete graph or a path or a cycle or the line graph of oneof the graphs depicted in Figure 1. And for being Gorenstein, G must beeither a complete graph, a cycle, a path length ≤ In this section, we consider the case that ∆( G ) is not pure and characterizethose H whose line graphs have sequentially CM clique complexes. Recallthat ∆ [ i ] = (cid:104) F | F ∈ ∆ , dim F = i (cid:105) is called the pure i -skeleton of ∆ andif each ∆ [ i ] is CM for i ≤ dim ∆, then ∆ is called sequentially CM . Notethat every 0-dimensional complex is CM and a pure 1-dimensional complexis CM if and only if it is connected (see for example [1, Exercise 5.1.26]). Thefollowing result considers ∆ [ i ] for i ≥
3. In this section, we always assumethat ∆ = ∆( G ). Proposition 3.1.
Suppose that H is connected. Then all nonempty ∆ [ i ] for i ≥ are CM if and only if H has at most one vertex v with degree ≥ .Proof. If H has no vertex with degree ≥
4, then ∆ [ i ] = ∅ for i ≥
3. If H hasexactly one vertex with degree r ≥
4, then for 3 ≤ i < r , ∆ [ i ] is the pure i -skeleton of the simplex with the r -star at v as the only facet. Hence ∆ [ i ] is CM. If v (cid:54) = v are two vertices of H with degree ≥ E and F are4-stars of H at v and v , respectively, then E and F are facets of ∆ [4] . Notethat every facet of ∆ [4] which is adjacent to E , is a 4-star at v , hence ∆ [4] is not strongly connected and hence not CM.Next we are going to introduce a graph H (cid:48) , such that ∆(L( H (cid:48) )) [2] = ∆ [2] and in H (cid:48) every vertex of degree 2 is in a triangle. For this we need thefollowing lemmas. It should be mentioned that in this paper, just one sideof the following lemmas are used, but we state and prove both sides forcompleteness. The first lemma is easy and its proof is left to the reader.8 emma 3.2. If Γ is a simplicial complex with an isolated vertex v , then Γ is vertex decomposable (resp. shellable, sequentially CM) if and only if Γ − v is so. Lemma 3.3.
Suppose that Γ is a connected simplicial complex and a is a ver-tex of Γ . Also assume that b is a new vertex. Then Γ is vertex decomposable(resp. shellable, sequentially CM) if and only if Γ + (cid:104) ab (cid:105) is so.Proof. ( ⇒ ): If Γ = (cid:104) a (cid:105) , then Γ (cid:48) = Γ + (cid:104) ab (cid:105) is a simplex and the result isclear. Assume that Γ (cid:54) = (cid:104) a (cid:105) , that is, { a } is not a facet of Γ. Then b is ashedding vertex of Γ (cid:48) with link Γ (cid:48) ( b ) = (cid:104) a (cid:105) and Γ (cid:48) − b = Γ. Hence if Γ is vertexdecomposable, then Γ (cid:48) is also vertex decomposable. If Γ is shellable, then itcan readily be checked that by adding { ab } at the end of a shelling order ofΓ, we get a shelling order of Γ (cid:48) . The case for sequentially CM follows fromthe facts that Γ [ i ] = Γ (cid:48) [ i ] for all i > i = 1, being CM is equivalentto being connected.( ⇐ ): For sequentially CM, the proof again follows the aforementionedfacts. Suppose that Γ (cid:48) is shellable. In any shelling order of Γ (cid:48) , either { ab } isthe first facet or there is a facet E containing a before { ab } . In the formercase, the second facet should be a facet E containing a . Thus in both casesfor each term F i after { ab } in the shelling order, F i ∩ { ab } ⊆ F i ∩ E andit follows that dropping { ab } from a shelling order of Γ (cid:48) , gives us a shellingorder for Γ.Now assume that Γ (cid:48) is vertex decomposable. If Γ (cid:48) is a simplex then wemust have Γ = (cid:104) a (cid:105) and is vertex decomposable. So suppose that v is a shed-ding vertex of Γ (cid:48) such that both link Γ (cid:48) ( v ) and Γ (cid:48) − v are vertex decomposable.If v = b , then Γ = Γ (cid:48) − v is vertex decomposable. Also v (cid:54) = a , because { b } is a facet of both link Γ (cid:48) ( a ) and Γ (cid:48) − a . Thus we assume v (cid:54) = a, b . Nowlink Γ ( v ) = link Γ (cid:48) ( v ) and Γ − v + (cid:104) ab (cid:105) = Γ (cid:48) − v . Thus by induction we candeduce that Γ − v , as well as link Γ ( v ), is vertex decomposable. Therefore,if v is a shedding vertex of Γ − v , we are done. Else, a facet F of link Γ ( v )is a facet of Γ − v but not a facet of Γ − v + (cid:104) ab (cid:105) . Thus F must be { a } and in Γ (cid:48) − v , a is only connected to b . Since Γ (cid:48) − v is vertex decomposableand hence sequentially CM, its pure 1-skeleton should be CM and connected.This means that Γ (cid:48) − v is just a set of isolated vertices along with (cid:104) ab (cid:105) orequivalently facets of Γ are either 0-dimensional or 1-dimensional containing v . Now the result follows from Lemmas 3.2 and 2.3.Recall that a free vertex of a simplicial complex is a vertex which iscontained in exactly one facet. Lemma 3.4.
Suppose that Γ is a simplicial complex and a, b are two freevertices of Γ contained in the facets E and F , respectively. Also assume hat F (cid:54) = E and | E | , | F | ≥ . If Γ is vertex decomposable (resp., shellable,sequentially CM), then Γ + (cid:104) ab (cid:105) is so. Moreover if | E | , | F | ≥ , then theconverse also holds.Proof. ( ⇒ ): Set Γ (cid:48) = Γ + (cid:104) ab (cid:105) . We use induction on the number of verticesof Γ. Suppose that Γ is vertex decomposable and v is a shedding vertexof Γ with both link Γ ( v ) and Γ − v vertex decomposable. If v = a , thenlink Γ (cid:48) ( v ) = link Γ ( v ) + (cid:104) b (cid:105) and b is isolated in this link. Hence by (3.2)link Γ (cid:48) ( v ) is vertex decomposable. Also Γ (cid:48) − v = Γ − v and as b is not a facetof Γ (cid:48) − v , v is a shedding vertex of Γ (cid:48) and Γ (cid:48) is vertex decomposable. Thecase that v = b is similar. Suppose v (cid:54) = a, b . If v ∈ E , then E \ { v } is a facetof link Γ ( v ) and hence there is a facet E (cid:48) of Γ − v , such that E \ { v } (cid:40) E (cid:48) . So E (cid:48) (cid:54) = E is a facet of Γ containing a , contradicting freeness of a . Consequently, v / ∈ E and similarly v / ∈ F and hence the size of the facet containing a or b does not differ in Γ and Γ − v . Therefore, Γ (cid:48) − v = (Γ − v ) + (cid:104) ab (cid:105) is vertexdecomposable by induction hypothesis and link Γ (cid:48) ( v ) = link Γ ( v ) is also vertexdecomposable. Clearly v is a shedding vertex of Γ (cid:48) and hence Γ (cid:48) is vertexdecomposable. The result for shellability and being sequentially CM followsan argument similar to the proof of the previous lemma.( ⇐ ): Now assume that | E | , | F | ≥
3. If Γ (cid:48) is shellable and in any of itsshelling orders one of E or F , say E , is after { ab } , then the intersection of (cid:104) E (cid:105) with the previous terms of the order has a facet { a } with 0 = dim { a } < dim E −
1, a contradiction. So { ab } appears after both E and F . Now if F i is a term after { ab } , then (cid:104) F i (cid:105) ∩ (cid:104) ab (cid:105) = (cid:104) F i (cid:105) ∩ (cid:104) E, F (cid:105) . Hence by deleting theterm { ab } from a shelling order of Γ (cid:48) , we get a shelling order for Γ. If Γ (cid:48) issequentially CM, then as Γ [ i ] = Γ (cid:48) [ i ] for i >
1, we just need to show that Γ [1] is CM or equivalently connected. Because Γ (cid:48) [1] = Γ [1] + (cid:104) ab (cid:105) is connected, itsuffices to show that a and b are connected in Γ. By our assumption, thereare a ∈ E ⊆ E and b ∈ F ⊆ F , with | E | = | F | = 3. Thus there is astrong path between E and F in Γ [2] = Γ (cid:48) [2] and it follows that a and b areconnected in Γ.Finally assume that Γ (cid:48) is vertex decomposable and v is a shedding vertexof Γ (cid:48) with both link Γ (cid:48) ( v ) and Γ (cid:48) − v vertex decomposable. If v = a , thenlink Γ (cid:48) ( v ) = link Γ ( v ) + (cid:104) b (cid:105) and b is isolated in this link. Hence according to(3.2), link Γ ( v ) is vertex decomposable. Also Γ (cid:48) − v = Γ − v and as b is nota facet of Γ − v , v is a shedding vertex of Γ and Γ is vertex decomposable.The case that v = b is similar. If v (cid:54) = a, b , then an argument similar tothe proof of ( ⇒ ), shows that v / ∈ E ∪ F . Thus link Γ ( v ) = link Γ (cid:48) ( v ) andΓ (cid:48) − v = Γ − v + (cid:104) ab (cid:105) is vertex decomposable by induction. If D is a facetof both link Γ ( v ) and Γ − v , then it is strictly contained in a facet of Γ (cid:48) − v ,since v is shedding in Γ (cid:48) . Thus we must have D (cid:40) { ab } so D = { a } or { b } .10ut then D is strictly contained in E \ { v } or F \ { v } which are facets oflink Γ ( v ), a contradiction. Consequently, v is a shedding face of Γ and Γ isvertex decomposable.We should mention that the conditions | F | , | G | ≥ (cid:104) av, bu (cid:105) is not sequentially CM butΓ + (cid:104) ab (cid:105) is vertex decomposable.Suppose that v is a vertex of H with degree 2 adjacent to vertices a and b . By splitting v , we get the graph H (cid:48) with vertex set (V( H ) \ { v } ) ∪ { v , v } ,where v and v are new vertices, and the same edge set as H , where weidentify the edges av and bv of H with av and bv in H (cid:48) . Note that v and v which we call the halves of v are both leaves (vertices of degree 1) in H (cid:48) .Here by a leaf edge we mean an edge incident to a leaf. Proposition 3.5.
Suppose that H is connected. Then the following areequivalent. (i) ∆( G ) is sequentially CM. (ii) If H (cid:48) is obtained by splitting all vertices of degree 2 of H which are notin a triangle, then every connected component of H (cid:48) is an edge exceptat most one component whose line graph has a sequentially CM cliquecomplex. (iii) H can be obtained by consecutively applying the following two operationson a graph H in which every vertex of degree two is in a triangle andwhose line graph has a sequentially CM clique complex: (a) attaching a new leaf to an old leaf of the graph; (b) unifying two leaves whose distance is at least 4.Moreover, if any the above statements holds, H is as in (iii) and ∆(L( H )) is vertex decomposable (resp. shellable), then ∆( G ) is vertex decomposable(resp. shellable).Proof. (i) ⇒ (ii): We assume that H and H (cid:48) have the same set of edges,according to the remarks before the proposition. If dim ∆ = 1, then H is acycle or a path and hence either every component of H (cid:48) is an edge or H (cid:48) isa triangle. Assume that dim ∆ >
1. Suppose that F and F are adjacentfacets of ∆ [2] . Then one of F and F , say F is a triangle and the other one, F , is a 3-star in H with two vertices (esp. the center of the star) on F . Notethat the vertices of the triangle are not split in H (cid:48) and hence F and F areon the same component of H (cid:48) . As ∆ [2] is CM and hence strongly connected,11t follows that all triangles and all 3-stars of H are in one component of H (cid:48) ,call it H . Noting that every vertex with degree r ≥ (cid:0) r (cid:1) ≥ H and also all triangles of H are on H . Hence ∆(L( H )) [ i ] = ∆ [ i ] for each i ≥ H )) issequentially CM. Since all vertices of H (cid:48) not on H have degree at most 2and H (cid:48) has no vertex of degree 2 not on a triangle, all other components of H (cid:48) are edges.(ii) ⇒ (iii): If every connected component of H (cid:48) is an edge, then H is apath or a cycle with length at least 4 and can be constructed by applying (a)and (b) starting with any of the connected components of H (cid:48) . Else let H bethe component of H (cid:48) which is not an edge. Then ∆(L( H )) is sequentially CMand H has no vertex of degree 2 not in a triangle (because all such verticeshave been split before). Thus we can construct H from H by “undoing” thevertex splits one by one. Note that when splitting vertices of degree 2 not ina triangle, the order of the vertices to split does not make any difference onthe final graph. So we can assume that one half of the last vertex v which issplit is on H . If both v and v , the halves of v , are on H , then unifying v and v which are leaves, makes the last split undone. Note that as v was notin a triangle when it got split, the distance of v and v is at least 4. If onlyone half of v , say v is on H , then the other half is on an edge componentof H (cid:48) . Thus undoing the last split in this case is indeed attaching a leaf to v . Hence “undoing” each vertex split is indeed applying (a) or (b).(iii) ⇒ (i) and the “moreover” statement: Let G = L( H ) and ∆ =∆( G ). Note that each leaf edge of H is a free vertex of ∆ . Let v , v beleaves of H , e , e be leaf edges containing them and E and E be facets of∆ containing e and e , respectively. If v and v have distance at least 4,then e and e neither are adjacent nor have a common neighbor in G andhence E (cid:54) = E . Also note that | E i | = 1 if and only if G is just one edge.Since the operations mentioned in (iii), just add a new facet { ee (cid:48) } to ∆ (for(a), e is a free vertex of ∆ and e (cid:48) a new vertex, and for (b), both of e and e (cid:48) are free vertices of ∆ ), the result follows from Lemmas 3.3 and 3.4.An instant consequence of (3.5) is the following characterization of graphswith no vertex of degree ≥ Corollary 3.6.
Suppose that H is a graph with at least one edge and G =L( H ) . If deg v ≤ for each vertex v of H , then the following are equivalent. (i) ∆( G ) is vertex decomposable. (ii) ∆( G ) is shellable. iii) ∆( G ) is sequentially CM (over some field). (iv) H can be constructed from a star or a path or a cycle or a graph inFigure 1, by consecutively applying (a) and (b) of (3.5)(iii).Proof. Let H be a subgraph of H in which every vertex of degree 2 is ina triangle. Then ∆( H ) [ i ] = ∅ for i >
2. Hence ∆( H ) is pure and beingsequentially CM is equivalent to being CM for ∆( H ). Consequently, theresult follows by (2.5) and (3.5).In the sequel, unless stated otherwise explicitly, we assume that H is aconnected graph with exactly one vertex v with degree r > H is in a triangle. We also let G = L( H )and ∆ = ∆( G ). According to (3.5) and its corollary, by characterizing those H for which ∆ is sequentially CM, we can derive a characterization of allgraphs whose line graphs have a sequentially CM clique complex. Notingthat for i >
2, ∆ [ i ]0 is either empty or the pure i -skeleton of a simplex andfor i <
2, ∆ [ i ]0 is CM since ∆ is connected, we just need to see when ∆ [2]0 isCM. First we study when ∆ [2]0 is strongly connected.Suppose that l = { v } and define L i = N H ( L i − ) \ ( ∪ i − j =0 L j ) to be the setof vertices of level i in H . Here N H ( A ) is the set of all vertices adjacent toa vertex in A inside the graph H . Thus indeed, the level of a vertex is itsdistance to v . Note that a vertex with level i can be adjacent only to verticeswith levels i − , i, i +1. Suppose that H [ L i ] is the induced subgraph of H onthe vertex set L i . Then if H (cid:48) = H [ L ], every u ∈ L has degree at most 2 in H (cid:48) , since it is also adjacent to v in H . Therefore each connected componentof H (cid:48) is either an isolated vertex or a cycle or a path of length ≥
1. We callthese isolated vertices, cycles and paths with positive lengths of H [ L ], the level 1 isolated vertices , level 1 cycles and level 1 paths , respectively. Lemma 3.7. If ∆ [2]0 is strongly connected, then every vertex x with level ≥ and deg( x ) = 3 , has level 2 and is adjacent to both endpoints of a level 1 pathwith length 1.Proof. Let F be the 3-star at x and F (cid:48) be a 3-star at v . Then as ∆ [2]0 isstrongly connected, there exist a strong path F = F , . . . , F k = F (cid:48) of 3-starsand triangles of H . We suppose that this strong path is the smallest possibleand in particular, there is no repetition in the path. Assume that t is thesmallest index such that v is incident to an edge in F t . Thus F t is a 3-star ata level 1 vertex a and F t = { av, ab, ac } . Since v is not on any edge of F t − , F t − must be the triangle { ab, ac, bc } and F t − is a 3-star at b or c . As each3-star at a level 1 vertex has an edge incident to v , the center of F t − has13 L L L Figure 4: An example of H satisfying conditions of (3.8)level 2. Also since F t +1 is a triangle and | F t +1 ∩ F t | = 2 is F t +1 is the edge setof either the triangle vab or the triangle vac . In particular, at least one of b, c has level 1. So we can suppose that b ∈ L and c ∈ L and F t − = { cb, ca, cd } is the 3-star at c . If F t − (cid:54) = F , then t ≥ F t − must be either { cb, cd, db } or { ca, cd, da } for a vertex d / ∈ { a, b, c, v } . But then a or b has degree > F t − = F and hence x = c has level 2. Moreover, x isadjacent to a, b which have level 1 and also are adjacent to each other. So ab is a level 1 path with length 1, as required. Proposition 3.8.
The complex ∆ [2]0 is strongly connected, if and only if H satisfies both of the following conditions (see an example in Figure 4). (i) Every level 3 vertex of H is a leaf. (ii) A level 2 vertex x of H satisfies one of the following: (a) x is a leaf adjacent to an endpoint of a level 1 path; (b) deg( x ) = 2 and x is adjacent to both endpoints of a level 1 pathwith length 1; (c) deg( x ) = 3 and x is adjacent to both endpoints of a level 1 pathwith length 1 and the other neighbor of x is either a level 3 vertexor a level 2 vertex with degree 3 or the endpoint of a level 1 path.Proof. ( ⇒ ): Assume that y is a level 1 isolated vertex. If deg( y ) = 2, then y is in a triangle yva and hence y is adjacent to the level 1 vertex a , acontradiction. If deg( y ) = 3 and y is adjacent to 2 level 2 vertices a, b , thenthere must exist a strong path F , . . . , F k starting with the 3-star at y andending with a 3-star at v . Then F must be a triangle containing two edgesof F , hence F = { ya, yb, ab } and F is the 3-star at one of a or b , say a .Then the level 2 vertex a has degree 3 but is not adjacent to the endpointsof a level 1 path, contradicting (3.7). Consequently, deg( y ) = 1. So the onlylevel 1 vertices that can be adjacent to a level 2 vertex are the endpoints of14evel 1 paths and since each such vertex has two neighbors in L ∪ L , eachcan have at most one neighbor in L .Now assume that x is a level 2 vertex, according to the above argument, x is adjacent to an endpoint a of a level 1 path. If deg( x ) = 1, then x is a leafand case (ii)(a) holds. If deg( x ) = 2, then x is in a triangle axb . If b is not in L , then a has two neighbors in L , contradicting the above remarks. Thus b has level 1 and is adjacent to a , that is a, b are the endpoints of a level 1path with length 1 as in (ii)(b). If deg( x ) = 3, then two of the neighbors of x are endpoints of a level 1 path with length 1 according to (3.7). The otherneighbor y has either level 1 and must be an endpoint of a level 1 path bythe above argument, or has level 3, or has level 2. In the last case, since y is adjacent to x and a level 1 vertex, deg( y ) ≥ L , deg( y ) (cid:54) = 2. Hence if thethird neighbor y of x has level 2, then deg( y ) = 3, as required.Finally suppose that x has level 3. According to (3.7), deg( x ) ≤
2. Ifdeg( x ) = 2, then x is in a triangle containing a level 2 vertex y . But then y has two neighbors in L ∪ L which contradicts part (ii) proved above. Thusdeg( x ) = 1.( ⇐ ): First note that these conditions ensure that the only level 1 verticeswith a neighbor in L are endpoints of level 1 paths and ∪ ∞ i =4 L i = ∅ . To provethat ∆ [2]0 is strongly connected, it suffices to show that from every 3-star ortriangle F of H , there is a strong path F = F , F , . . . , F t in ∆ [2]0 , where F t is a 3-star at v , because there exists strong paths between any two 3-stars at v . Clearly every triangle containing v is adjacent to a 3-star at v . Also alevel 1 vertex a with deg( a ) = 3, is adjacent to at least another level 1 vertex.Else a is a level 1 isolated vertex and is not adjacent to any level 2 vertex bycondition (ii), which means deg( a ) = 1. Therefore the 3-star at a is adjacentto a triangle containing v . Now assume that F is a triangle on which v doesnot lie. Since every level 3 vertex is a leaf and at most one of the neighborsof a level 2 vertex has level (cid:54) = 1, there must be at least one level 1 vertex a on F . Therefore, F is adjacent to the 3-star at a . Finally, any 3-star withcenter not on L ∪ L , must be centered at a level 2 vertex. Then by (ii)(c),this 3-star is adjacent to a triangle containing the two endpoints of a level1 path of length 1. This shows that starting from any 2-face of ∆ and bypassing from adjacent 2-faces, we can reach a 3-star at v .For simplicity, in the following definition we give a name to the graphs ofthe form H satisfying the conditions of the previous result. To consider thecases that the maximum degree of H is at most 3, we state the definition alittle bit more general. 15 efinition 3.9. Suppose that C is a graph, v is a vertex of C and r is apositive integer. We say that C is an r -graph rooted at v or simply an r -graph, if C is connected, deg( v ) = r , all other vertices of C have degree atmost min { r, } , all vertices of C with degree 2 are in some triangles and also C satisfies the conditions of (3.8), where the level of a vertex of C is definedby L = { v } and L i = N( L i − ) \ ( ∪ i − j =0 L j ).Thus (3.8) states that ∆ [2]0 is strongly connected if and only if H is an r -graph for some r > [2]0 is CM, we need an algebraic tool. Let Γ be asimplicial complex and denote by (cid:101) C d (Γ) = (cid:101) C d (Γ; K ) the free K -module whosebasis is the set of all d -dimensional faces of Γ. Consider the K -map ∂ d : (cid:101) C d (Γ) → (cid:101) C d − (Γ) defined by ∂ d ( { v , . . . , v d } ) = d (cid:88) i =0 ( − i { v , . . . , v i − , v i +1 , . . . , v d } , where v < · · · < v d is a linear order. Then ( (cid:101) C • , ∂ • ) is a complex of free K -modules and K -homomorphisms called the augmented oriented chain com-plex of Γ over K . We denote the i -th homology of this complex by (cid:101) H i (Γ; K ).The Reisner theorem states that Γ is CM over K , if and only if for allfaces F of Γ including the empty face and for all i < dim link Γ ( F ), one has (cid:101) H i (link Γ ( F ); K ) = 0 (see [3, Theorem 8.1.6]). In particular, applying thiswith F = ∅ we see that if Γ is CM, then for i = dim Γ −
1, we must have (cid:101) H i (Γ; K ) = 0. Theorem 3.10.
Suppose that H is a connected graph with at least 1 edge.Let ∆ = ∆(L( H )) . Then the following are equivalent. (i) ∆ is vertex decomposable. (ii) ∆ is shellable. (iii) ∆ is sequentially CM (over some field). (iv) For some positive integer r , there is an r -graph H in which every level2 vertex with degree 3 has a leaf neighbor and H can be constructed from H by consecutively applying the operations (a) and (b) of (3.5)(iii). (v) If H (cid:48) is the graph obtained by splitting all vertices of H with degree 2which are not in any triangle, then every connected component of H (cid:48) isan edge except at most one. The only non-edge connected componentof H (cid:48) , if exists, is an r -graph for a positive integer r , in which everylevel 2 vertex with degree 3 has a leaf neighbor. roof. Note that if r ≤
3, then there are only finitely many r -graphs (up toisomorphism) which are either a path with length 1 or a triangle or a 3-staror one of the graphs in Figure 1. Thus if the maximum degree of H is at most3, then the result follows from (3.6). So we assume that r and the maximumdegree of H are at least 4.(i) ⇒ (ii) ⇒ (iii) are well-known. Also the proof of (iv) ⇔ (v) followsthe fact that“undoing” each vertex splitting is indeed applying (a) or (b) of(3.5)(iii) which is proved in the proof of (ii) ⇒ (iii) in (3.5).(iii) ⇒ (iv): Suppose that ∆ is sequentially CM over a field K and r ≥ H . According to (3.1), there is exactly onevertex v of H with degree ≥
4. Further, by (3.5) and (3.8) and since everyCM complex is strongly connected, it follows that H can be constructed froman r -graph H rooted at v by consecutively applying the operations (a) and(b) of (3.5)(iii) and ∆ = ∆(L( H )) is sequentially CM. Thus we just have toshow that a level 2 vertex x of H with degree 3 has a leaf neighbor. Supposeit is not true. According to (3.8)(ii), one of the following cases may occur. Case 1: x has a level 2 neighbor y with degree 3. Assume that a, b are thelevel 1 neighbors of x and c is a level 1 neighbor of y . Let e = va , e = vc , e = ax , e = xy , e = yc and e = xb be vertices of ∆ with e i < e j ⇔ i < j (see Figure 5(a)). Consider σ = { e , e } + { e , e } − { e , e } − { e , e } + { e , e } ∈ (cid:101) C (∆ [2]0 ). Then ∂ ( σ ) = 0 and as (cid:101) H (∆ [2]0 ; K ) = 0, we must have σ ∈ Im ∂ , that is, there is a δ ∈ (cid:101) C (∆ [2]0 ) with ∂ ( δ ) = σ . Since { e , e } appears only in the 2-face F = { e , e , e } , the coefficient of F in δ must be1. But then in ∂ ( δ ) the coefficient of { e , e } is 1, because F is the only2-face containing { e , e } . This contradiction shows that case 1 cannot occur. Case 2:
The third neighbor y of x has level 1. Again assume that a, b are the other level 1 vertices adjacent to x and set e = va , e = vy , e = ax , e = xy and e = xb with e i < e j ⇔ i < j (see Figure 5(b)). By a similarargument as in case 1 with σ = { e , e }−{ e , e }−{ e , e } + { e , e } ∈ ker ∂ ,we again reach a contradiction.(iv) ⇒ (i): We show that ∆ = ∆(L( H )) is vertex decomposable. Thenthe result follows from the “moreover” statement of (3.5). For this, we useinduction on the number of edges of H . Assume that H is rooted at thevertex v . First we show that for every vertex e of ∆ not incident to v ,link ∆ ( e ) is vertex decomposable. Indeed, if e is in a triangle of H , whoseother two edges are f and f , then every edge of H adjacent to e in L( H ),is also adjacent to either f or f and since f and f are adjacent, the pure1-dimensional complex link ∆ ( e ) is connected and by (2.3), vertex decompos-able. If e is not in any triangle, then the conditions of part (iv), ensures that e is a leaf edge and so again link ∆ ( e ) is connected and vertex decomposable.17 e e e e e e e e e e vb a c yx v b ax y ( a ) ( b ) Figure 5: Illustrations for the proof of Theorem 3.10; (a) is case 1 and (b) iscase 2Assume that H has a level 3 vertex x . Then x is a leaf and is adjacentto a level 2 vertex y which is in a triangle. Let e = xy . Then the only facetof link ∆ ( e ) is strictly contained in this triangle and hence e is a sheddingvertex of ∆ . Also if H (cid:48) = H − e − x , then ∆ − e = ∆(L( H (cid:48) )) and H (cid:48) satisfies the conditions of (iv). So ∆ − e is vertex decomposable by inductionhypothesis and hence ∆ is also vertex decomposable. Thus we can assume H has no level 3 vertex. By a similar argument we can also assume that nolevel 2 vertex of H is a leaf.Since every level 2 vertex with degree 3 is adjacent to a level 3 vertex whichwe have assumed does not exist, if x is a level 2 vertex, then deg( x ) = 2.So by the assumptions of (iv), x is adjacent to the endpoints a, b of a level1 path with length 1. Let e = xa . Then the facets of link ∆ ( e ) are { va, ab } and { xb, ab } which are contained in the facets { va, ab, vb } and { xb, ab, vb } of∆ − e = ∆(L( H − e )), respectively. Because H − e satisfies the conditionsof (iv), ∆ − e is vertex decomposable by induction hypothesis and hence sois ∆ . Therefore, we can assume that H has no vertex with level 2.Let C be a level 1 cycle of H with two adjacent vertices x and y . Assumethat in C , a (cid:54) = y is adjacent to x and b (cid:54) = x is adjacent to y (we may have a = b ). Let e = xy . The facets of link ∆ ( e ) are { vx, xa } , { vy, yb } , { vx, vy } and if a = b , { xa, ay } , which are in the following facets of ∆ − e , respectively: { vx, xa, va } , { vy, yb, vb } , the r -star at v and the 3-star at a . Thus again itfollows that e is a shedding vertex of ∆ and ∆ − e is vertex decomposableby induction hypothesis and hence ∆ is also vertex decomposable.If P is a level 1 path of H starting with the vertices a and b , then similarto the above paragraphs one can see that e = ab is a shedding vertex with∆ − e vertex decomposable and it follows that ∆ is vertex decomposable.If H has no level 1 cycle or level 1 path, then it is just an star centered at v . In this case ∆ is a simplex and the result follows.18igure 6: A graph whose line graph has a sequentially CM clique complexFrom this theorem we see that if H is the graph in Figure 4 and ∆ =∆(L( H )), then ∆ is not sequentially CM, that is, ∆ [2] is not CM althoughit is strongly connected. Also as another example, Theorem 3.10 shows thatthe line graph of the graph in Figure 6 has a sequentially CM clique complex. An algorithm
At the end of this paper, we show that using Theorem 3.10, we can presenta linear time algorithm which takes as input a graph G and checks whether G is a line graph or not and if yes, says whether ∆( G ) is sequentially CM.Checking if G is a line graph and even returning an H such that G = L( H )has been previously done by Lehot in [8] in a linear time. Thus we canassume that H is given and we must find out if ∆(L( H )) is sequentiallyCM. We assume that H is given as lists of neighbors of vertices and n is thenumber of vertices of H . Here we state an algorithm, the correctness of whichis ensured by Theorem 3.10 and show that its worst case time complexityis Θ( n ). In this algorithm, we use breadth-first search (BFS) which can befound in for example [19]. Step 1:
Run through the vertices of H and compute the degree of eachvertex. If for a second time a vertex with degree more than three is visited,return false. Also for each vertex x with degree 2 and with neighbors a and b , check if a is a neighbor of b . If not, split the vertex x by removing theedge xb and adding a new vertex adjacent only to b .Note that if H has more than one vertex with degree >
3, then after atmost checking 3( n −
1) + r + 1 ≤ n edges where r is the degree of the firstvertex with degree > n/ n ) time complexity. Also note that the obtained graph has at most2 n vertices. 19 tep 2: Compute the connected components of the obtained graph (say,by BFS). If more than one connected component is not an edge return false.If all connected components are edges, return true. Else let H be the onlyconnected component which is not an edge. This step clearly needs Θ( n )time complexity. Step 3:
Find a vertex v with maximum degree in H . Run a BFS startingat v and mark each visited vertex with its level which is the distance of thevertex from v . When visiting a level 2 vertex y consider the following cases.deg( y ) = 1 : Let a be the neighbor of y (which has level 1). If a has no level 1neighbor (so that a is not the endpoint of a level 1 path), return false.deg( y ) = 2 : The neighbors of y should have level 1 and be adjacent. If not,return false.deg( y ) = 3 : Then its neighbors should be two level 1 adjacent vertices anda vertex not yet visited. If not, return false.Also when visiting a level 3 vertex x , if x has not degree 1, return false. Thisstep also consumes Θ( n ) running time. Step 4:
Return true.Note that steps 2 and 3 can be carried out simultaneously.We have to make two remarks regarding the correctness of the algorithm.First, if the maximum degree of H is 3 and ∆ is sequentially CM, then thegraph H is a 3-star or one of the graphs in Figure 1. All of these graphs are3-graphs rooted at v , where v can be any of the degree 3 vertices. Thus in step3 it does not differ which vertex with degree 3 we choose as v . Furthermorein step 3, we may visit a level 2 vertex with degree 3 which has a level 2neighbor z not still visited, without returning false at that moment. In thiscase, when visiting the vertex z , since z has a visited level 2 neighbor, thealgorithm returns false. References [1] W. Bruns and J¨urgen Herzog,
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