Topological Properties of Graphical Arrangements

Honam Mathematical J. 36 (2014), No. 2, pp. 435–454 http://dx.doi.org/10.5831/HMJ.2014.36.2.435 TOPOLOGICAL PROPERTIES OF GRAPHICAL ARRANGEMENTS Thi A. Nguyen and Sangwook Kim∗ Abstract. We show that for any graph G, the proper part of the intersection poset of the corresponding graphical arrangement AG has the homotopy type of a wedge of spheres. Furthermore, we also indicate the number of spheres in the wedge, based on the number of spanning forests of G and other graphs that are obtained from G. 1. Introduction A hyperplane arrangement A is a finite set of affine hyperplanes in some vector space V ∼ = K n , where K is a field. Graphical arrangements are special hyperplane arrangements obtained from finite simple undirected graphs. We know that the graphical arrangement of a graph G on vertex set [n] := {1, 2, . . . , n} is a subarrangement of the braid arrangement Bn that is the graphical arrangement of the complete graph Kn . Graphical arrangement are studied in the connection with graph theory [3, 6]. Many of the invariants associated with a graphical arrangement can be computed directly from the corresponding graph such as the characteristic polynomial ([9]). One of the important invariants of a hyperplane arrangement A is the intersection poset L(A), that is, the collection of all nonempty intersections of hyperplanes in A ordered by reverse inclusion. The number of regions and bounded regions of a real hyperplane arrangement is determined by its intersection poset [10] and the Betti numbers of Received April 3, 2014. Accepted May 28, 2014. 2010 Mathematics Subject Classification. 05C88, 05C89. Key words and phrases. graphical arrangement, EL-shallablity. This research was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(NRF-2012R1A1A1013149). ∗ Corresponding author 436 Thi A. Nguyen and Sangwook Kim the complement of a complex hyperplane arrangement can be expressed using the intersection poset [4]. In this paper, we study the topological properties of the intersection poset of a graphical arrangement. Björner and Wach [2] show that if a bounded poset P is EL-shellable, then the order complex ∆(P ) has the homotopy type of a wedges of spheres, where P is the proper part of P and the order complex ∆(P ) is the abstract simplicial complex whose vertices are all elements of P and whose simplices are all finite chains of P , including the empty chain. They also show that the number of spheres in the wedge can be obtained from decreasing maximal chains. A well-known example of an EL-shellable poset is Πn , the proper part of the partition lattice Πn of the n-element set ordered by refinement. The poset Πn has the homotopy type of a wedge of spheres, and the number of spheres is computed by counting decreasing maximal chains (see [5, 11, 12]). The main purpose of this paper is to show that the intersection poset of a graphical arrangement is EL-shellable. We also find the number of spheres in the wedge by using the number of spanning forests of G and other graphs that are obtained from G. The rest of the paper is organized as follows. Section 2 provides the basic notions and definitions which will be used throughout the paper. In Section 3, an edge-labeling is defined and shown to be an EL-labeling. Section 4 describes the number of spheres in the wedge using the number of decreasing chains. The number of regions of a graphical arrangement and the enumerative results for special cases will be given in the last section. 2. Basic notions and definitions Let G be a simple undirected graph on a vertex set [n] = {1, 2, . . . , n} and E(G) = {e1 , e2 , . . . , em } denote the set of edges of G. We also use the notation ij for an edge of G with two vertices i and j. Let V = K n be a vector space over a field K with coordinates x1 , x2 , . . . , xn . Definition 2.1. The graphical arrangement AG in a vector space V is the hyperplane arrangement {xi − xj = 0|ij ∈ E(G)}. In the intersection poset of the graphical arrangement AG , we also label each hyperplane xi − xj = 0 by ij. There are two cases to label an intersection of two hyperplanes. If two hyperplanes are ij and kl, where Topological properties of Graphical arrangements 437 1234 3 1 2 1 e1 2 123 12−34 2 1 1 124 23−14 234 134 3 e4 e2 2 3 1 4 12 4 e3 3 2 1 4 23 1 34 2 3 4 2 14 3 4 1 0 G1 PG1 Figure 1. A graph and the intersection poset of its graphical arrangement with labeling values {i, j} ∩ {k, l} = ∅, then the common subspace is labeled by ij − kl. Otherwise, we may assume that l = j. Then the common subspace will be labeled by ijk. A general subspace in the intersection poset is denoted by A = A1 − A2 − · · · − Ar , where Ai , i = 1, 2, . . . , r, are the sets of indices of coordinates. We call Ai blocks of A, and those indices of coordinates indices of A. When we get the intersection of two subspaces H = H1 − H2 − · · · − Hp and K = K1 − K2 − · · · − Kq , where Hi and Ki are blocks of H and K respectively, we also have two cases to label their common subspace. It is similar to labeling common subspace of two hyperplanes. Furthermore, K < H if and only if each block Kj of K is contained in some block Hi of H. Example 2.2. Consider the graph G1 shown in Figure 1. The graphical arrangement AG1 of G1 consists of hyperplanes x1 − x2 = 0, x2 − x3 = 0, x3 − x4 = 0, and x1 − x4 = 0. In the intersection poset PG1 of AG1 , we label hyperplane x1 − x2 = 0 by 12, x2 − x3 = 0 by 23, x3 − x4 = 0 by 34, and x1 − x4 = 0 by 14. Figure 1 also shows the intersection poset PG1 . Definition 2.3. An edge-labeling of a poset P is a map λ : E(P ) → Λ, where E(P ) is the set of edges of the Hasse diagram of poset P and Λ is an ordered set. For an edge {s, t} ∈ E(P ), we denote λ(s, t) 438 Thi A. Nguyen and Sangwook Kim for λ({s, t}). A maximal chain m : x0 ⋖ x1 ⋖ x2 ⋖ · · · ⋖ xt of P is increasing if λ(x0 , x1 ) < λ(x1 , x2 ) < · · · < λ(xt−1 , xt ). Similarly, m is decreasing if λ(x0 , x1 ) > λ(x1 , x2 ) > · · · > λ(xt−1 , xt ). The maximal chain m′ := y0 ⋖ y1 ⋖ y2 ⋖ · · · ⋖ ys is less than m in lexicographic partial order if  λ(xi , xi+1 ) = λ(yi , yi+1 ) for i = 0, . . . , s − 1 and s < t, or  λ(xi , xi+1 ) ̸= λ(yi , yi+1 ) for some i and λ(xi , xi+1 ) < λ(yi , yi+1 ) for the least such i. Definition 2.4. An edge-labeling is called an EL-labeling (edge lexicographical labeling) if for every interval [x, y] in poset P , there is a unique increasing maximal chain, which lexicographically precedes all other maximal chains of [x, y]. Björner and Wachs [2] showed the following fundamental result about the EL-labeling. Theorem 2.5. If a bounded poset P admits an EL-labeling, then the order complex ∆(P ) has the homotopy type of a wedge of spheres. Furthermore, the number of i-spheres is the number of decreasing maximal (i + 2)-chains of P . 3. The edge-labeling In this section, we will define an edge-labeling of the intersection poset of the graphical arrangement for graph G, and show that it is an EL-labeling. Let PG be the intersection poset of the graphical arrangement AG of G and H and K be two elements in PG where H covers K. We label an edge {K, H} by λ(K, H) = min{i : K ∩ ei = H}. Theorem 3.1. The labeling λ is an EL-labeling. Proof. Let M and N be two elements in the poset PG such that N < M . We construct a maximal chain C = {N = N0 ⋖ N1 ⋖ N2 ⋖ · · · ⋖ Nl = M } where Ni+1 is an element covering Ni satisfying λ(Ni , Ni+1 ) = min{λ(Ni , N ′ ) : Ni ⋖ N ′ < M }. We claim that C is the only increasing maximal chain from N to M , which lexicographically precedes all other maximal chains of [N, M ]. Topological properties of Graphical arrangements 439 Let A, B and C be three elements in C such that A ⋖ B ⋖ C. If λ(A, B) = x and λ(B, C) = y, then A ∩ ex = B and B ∩ ey = C. Since the hyperplane ey doesn’t contain the subspace A, A ∩ ey is another covering of A. By the choice of B, x = λ(A, B) < λ(A, A ∩ ey ) ≤ y. Thus C is increasing. Let D = {N = M0 ⋖ M1 ⋖ M2 ⋖ · · · ⋖ Ml = M } be another maximal chain in [N, M ] and let k be the smallest index such that Nk ̸= Mk . Let λ(Nk−1 , Nk ) = z1 and λ(Mk−1 , Mk ) = z2 . If z1 = z2 , then Nk = Nk−1 ∩ ez1 = Nk−1 ∩ ez2 = Mk which is impossible. Thus z1 < z2 . Since ez1 < M and ez1 ≮ Mk in PG , we have Mk < Mt ∩ ez1 = Mt+1 < M for some t. But then λ(Mt , Mt+1 ) ≤ z1 < z2 = λ(Mk−1 , Mk ). Therefore the maximal chain D is not increasing. By the construction of the maximal chain C, it is obvious that C lexicographically precedes all other maximal chains of [N, M ]. Remark 3.2. Theorem 3.1 can be obtained from [7] since the intersection poset of a graphical arrangement is a geometric lattice. We provide an elementary proof for completeness. Example 3.3. Figure 1 shows a graph G1 and its intersection poset PG1 with the labeling values. Since PG is the intersection poset of a hyperplane arrangement, all maximal chains in PG have the same length n − c(G), where c(G) is the number of connected components in G. Thus the order complex ∆(PG ) has the homotopy type of a wedge of spheres of the same dimension n − c(G) − 2. Remark 3.4. In [6], Stanley gives a labeling rule for the graphical arrangement of a chordal graph, and the labeling values are from the set [l] where l is the rank of the intersection poset. In our paper, we also give a labeling rule for graphical arrangements, and the labeling values are from the set [n] where n is the number of hyperplanes in the arrangement. In general, the two labeling rules are different. For example, see the chordal graph G2 in Figure 2. By the labeling rule of this paper, the unique increasing maximal chain in [0̂, 1̂] of the intersection poset is 0̂ ⋖ 12 ⋖ 123 ⋖ 1234. If we choose that maximal chain for the M-chain in the labeling of Stanley, then we get 0̂ ⋖ 34 ⋖ 134 ⋖ 1234 for a decreasing maximal chain while it is not decreasing in the labeling of this paper. 440 Thi A. Nguyen and Sangwook Kim 1 4 e1 2 e4 e2 e3 3 Figure 2. A chordal graph G2 4. The number of decreasing maximal chains In this section, we will find the number of decreasing maximal chains in the poset PG . Let M := 0̂ ⋖ X1 ⋖ X2 ⋖ X3 ⋖ · · · ⋖ Xl ⋖ 1̂ be a maximal chain in [0̂, 1̂]. We associate M with the ordered string of labeling values: λ(M) = [λ(0̂, X1 ), λ(X1 , X2 ), . . . , λ(Xl , 1̂)]. Since PG is the intersection poset of a hyperplane arrangement, the labeling values λ(0̂, X1 ), λ(X1 , X2 ), . . . , λ(Xl , 1̂) are corresponding to hyperplanes or edges of G. For example, we can associate the maximal chain M := 0̂ ⋖ 12 ⋖ 123 ⋖ 1234 of Figure 1 with the string of edges [12, 23, 34]. Definition 4.1. A string X of l(PG ) edges in G is associated with a maximal chain if all the edges of X are used to find labeling values of a maximal chain. Furthermore, X is associated with a decreasing maximal chain if the labeling values corresponding to the edges are decreasing. Definition 4.2. A spanning tree of a connected graph G is a subgraph that connects all the vertices of G and has no cycle. For a general graph G, a subgraph of G formed by the union of spanning trees in all connected components is called a spanning forest of G. The following lemma describes the condition when a set corresponds to a decreasing maximal chain. Lemma 4.3. Let A be a set of edges in G. Then we can arrange egdes in A to get a string X that can be associated with a decreasing maximal chain if and only if  The edges in A form a spanning forest in G.  A doesn’t contain C\ej , the set of all edges of a cycle C except the edge ej with the smallest label in C. Topological properties of Graphical arrangements 441 Proof. If we can arrange edges in A to get a string X associated with a decreasing maximal chain M, then it is easy to see that A must contain edges that can make a spanning forest in G. Assume that A contains C\ej where C is a cycle of G and j is the smallest label of edges in C. Let S be an element of the poset PG , and S is the intersection of some hyperplanes (or edges) in A such that the set of edges making S contains all edges in cycle C except ej and another edge ek ∈ C. By the condition of j, we have j < k. Let v1 , v2 , . . . , vm be all vertices of C. If ej and ek share one vertex vt , then v1 v2 · · · vˆt · · · vm < S in PG , where vˆt means that we don’t have vt in that subspace. Let T be the smallest element in M such that T > S and v1 v2 · · · vm < T . If R is an element in M covered by T , then R ∩ ek = R ∩ ej = T . By the labeling rule, λ(R, T ) ≤ j < k. Therefore, ek is not used to find a labeling value. So X is not associated with a maximal chain. If ej and ek share no vertex, then A1 − A2 < S, where A1 ∪ A2 = C and ej , ek can connect these two blocks. Similar to the previous case, ek can not be used to find a labeling value. On the other side, if the edges in A form a spanning forest in G and A doesn’t contain C\ej , then we arrange edges of A in decreasing order of labels to get a string X. We claim that X can be associated with a decreasing maximal chain. Since edges in X are arranged in decreasing order of labels, we only need to prove that all edges in X are used to find labeling values. Let ew be an edge of X, and P = ew ∩ Q where Q is the subspace formed by getting the intersection of hyperlanes standing before ew in X. If ew doesn’t belong to any cycle of G, then it is obvious that λ(Q, P ) = w. If ew belongs to a cycle of G, then ew is not used to find a labeling value when there is another edge ez such that Q ∩ ez = P and z < w. That means ez , ew and some other edges appearing before ew in X form a cycle of G. Since X doesn’t contain a cycle, ez doesn’t belong to X. If there is an edge ez ′ appearing before ew in X and z ′ < z < w, we get a contradiction because all edges in X are in decreasing order of labels. So there doesn’t exist such edge ez ′ . Then X contains C\ez where C is the cycle above and z is the smallest label of edges in C. It is impossible. Therefore, ew is used to find a labeling value. Example 4.4. For A = {e4 = 14, e2 = 23, e3 = 34} in Figure 1, we cannot get a string that is associated with a decreasing maximal chain from A. 442 Thi A. Nguyen and Sangwook Kim By Lemma 4.3, the number of decreasing maximal chains of PG is the same as the the number of sets A satisfying conditions of Lemma 4.3. Let BG be the set of all spanning forests in G. For each cycle C of graph G. Let xC = min{j : ej is an edge of C}, and BxC be the set of spanning forests containing C\exC . Then the following theorem follows immediately. Theorem 4.5. The number of decreasing maximal chains of the poset PG is |BG | − | ∪C:cycle of G BxC |. Furthermore, the order complex ∆(PG ) has the homotopy type of a wedge of |BG | − | ∪C:cycle of G BxC | spheres of dimension n − c(G) − 2 where c(G) is the number of connected components of G. Example 4.6. Let G1 be the graph shown in Figure 1. There are three decreasing maximal chains 0̂⋖34⋖234⋖1234, 0̂⋖14⋖23−14⋖1234 and 0̂ ⋖ 14 ⋖ 134 ⋖ 1234. This coincides with the result of Theorem 4.5. Lemma 4.7. Let C1 , C2 , . . . , Ck be cycles of G. If there is no exCi that lies on Cj where i ̸= j, then |BxC1 ∩ BxC2 ∩ · · · ∩ BxCk | = |BG/C1 ,C2 ,...,Ck |, where G/C1 , C2 , . . . , Ck is the graph obtained from G by contracting all edges in C1 , C2 , . . . , Ck . Otherwise, |BxC1 ∩ BxC2 ∩ · · · ∩ BxCk | = 0. Proof. The first statement follows from the definition of BxC . For the second statement, assume that eCi lies on Cj where i ̸= j. If eCi ̸= eCj , then it is obvious that |BxCi ∩ BxCj | = 0. If eCi = eCj , then any subgraph of G that contains Ci \eCi and Cj \eCj always has a cycle (Ci ∪ Cj ) \ eCi and so |BxCi ∩ BxCj | = 0. Thus the result follows. Example 4.8. Let G3 be the graph shown in Figure 3(a). Let C be the cycle of G3 with vertices 1, 2, 3, 4. See Figure 3(c) for the spanning trees of G3 that doesn’t contain exC = e1 and contains all other edges of C. Those spanning trees are corresponding to spanning trees of graph G3 /C shown in Figure 3(d). Theorem 4.5 can be rephrased in the following form using Lemma 4.7 and Principle of inclusion and exclusion. Theorem 4.9. The number of decreasing maximal chains of the poset PG is c ∑ ∑ |BG | − |BG/Ci1 ,Ci2 ,...,Cij |, (−1)j−1 j=1 i1 ,i2 ,...,ij Topological properties of Graphical arrangements 1 e1 e2 4 e5 2 e4 e3 3 (a) 443 5 1 2 5 1 2 5 1 2 5 4 3 6 4 3 6 4 3 6 e6 e7 6 (c) 5 1,2,3,4 5 1,2,3,4 5 1,2,3,4 6 6 (b) 5 1,2,3,4 6 6 (d) Figure 3. G3 , G3 /C and their spanning trees where c is the number of cycles of G, and {i1 , i2 , . . . , ij } are the sets of labels of cycles such that there is no exCa that lies on Cb where a ̸= b and a, b ∈ {i1 , i2 , . . . , ij }. In the formula above, we need to find the number of spanning forests of a graph. The Matrix tree theorem will help us to find the number of spanning trees of a connected graph, and we can use it to find the number of spanning forests. Definition 4.10. Given an undirected graph G with the vertices set [n] = {1, 2, . . . , n}, the Laplacian matrix L(G) is defined to be n × n matrix with the following entries:    deg(i) if i = j, −m where m is the number of edges between i and j, li,j =  0 otherwise. When counting the degree of a vertex, all loops are excluded for multigraph. Theorem 4.11 (The Matrix Tree Theorem, [8]). For a given undirected connected graph G with n vertices, let λ1 , . . . , λn−1 be the nonzero eigenvalues of L(G). Then the number of distinct spanning trees of G is equal to n−1 1 ∏ λi . t(G) = n i=1 444 Thi A. Nguyen and Sangwook Kim If G is not connected with connected components G1 , G2 , . . . , Gd , then it is not difficult to see that d ∏ |BG | = t(Gi ). i=1 Example 4.12. We will use Theorem 4.9 to decreasing maximal chains in PG3 . We have  2 −1 −1 0 0 −1 2 0 −1 0  −1 0 3 −1 −1 L(G3 ) =   0 −1 −1 3 0  0 0 −1 0 2 0 0 0 −1 −1 find the number of  0 0  0  −1  −1 2 The 5 non-zero eigenvalues of L(G3 ) are 1, 2, 3, 3, 5. So the number of spanning trees of G3 is 1·2·3·3·5 = 15. Let C1 be the cycle with ver6 tices 1, 2, 3, 4, and C2 be the cycle with vertices 2, 3, 5, 6, and C3 be the cycle with vertices 1, 2, 5, 6, 3, 4. Since G3 /C1 and G3 /C3 are triangles, they both have 3 spanning trees. Since G3 /C3 is a loop, it has 1 spanning tree. Since e4 lies on C1 , |BG/C1 ,C2 | = 0. Similarly, |BG/C1 ,C3 | = |BG/C1 ,C2 ,C3 | = 0. It is obvious that |BG/C2 ,C3 | = 1. Theorem 4.9 shows that the number of decreasing maximal chains in PG3 is 15 − (3 + 3 + 1 − 1 − 0 − 0 + 0) = 9. Now we provide several enumerative results for some classes of graphs. Proposition 4.13. Let G be a graph such that cycles in G do not share edges. Then the number of decreasing maximal chains in PG is ∏ (k − 1)αk k where αk is the number of k-cycles, k ≥ 3. Proof. Let G′ be a spanning forest of G that satisfies the conditions of Lemma 4.3. Then G′ must contain exC for all cycles of G. Since G′ is a spanning forest, it doesn’t contain only one edge in C\exC for every C. If C is a cycle of length k, then we have k − 1 choices for that edge. Therefore, the number of spanning forests satisfying the conditions of Lemma 4.3 is ∏ (k − 1)αk . k Topological properties of Graphical arrangements 445 e7 e2 e1 e5 C1 e4 C2 e3 e6 C3 e8 e9 e10 Figure 4. An example of G Definition 4.14. A cycle C is called a minimal cycle if C doesn’t contain the whole vertices of another cycle. Proposition 4.15. Let G be a graph such that all minimal cycles in G can form a chain C1 , C2 , . . . , Cs where { ∅, a vertex, or an edge if j = i + 1, Ci ∩ Cj = ∅ if j ̸= i + 1. Then the number of decreasing maximal chains in PG is ∏ (k − 1)αk k where αk is the number of minimal cycles of length k. Proof. Let G be a graph satisfying the conditions above. Since G does not contain a subdivision of K5 or K3,3 , G is a planar graph. Thus the chain of minimal cycles can be labeled from left to right as Figure 4. We label the edges of G in increasing order from the left to the right. If we meet a cycle, we label all edges of that cycle first before we move to other edges. This rule guarantees that every common edge of consecutive minimal cycles is the smallest edge of latter cycle. Let G′ be a spanning forest that satisfies the condition of Lemma 4.3. If an edge e doesn’t belong to any cycle, then G′ must contain e. Let C be a minimal cycle of G of length k. Since G′ is a spanning forest, G′ can contain at most k − 1 edges of C. Since G′ doesn’t contain C\{exC }, there are k − 1 ways to choose (k − 1)-subsets of edges in C in which we can choose some edges to put into G′ . Those subsets contain the smallest edge of C. We construct a multiset M that will contain edges of G as follows. Firstly, we add all edges that are not in any cycle into M . In the next step, we choose a (k − 1)-subset for every minimal cycle C as described 446 Thi A. Nguyen and Sangwook Kim above to add into M . By the labeling rule of edges, every common edge of consecutive cycles must appear in M at least once. Furthermore, each common edge can appear in M at most twice because each edge can be a common edge of at most two cycles. Now we prove that if we remove every common edge one time from the multiset M , we will get a set of edges that can form a spanning forest of G satisfying the condition of Lemma 4.3. And also, whenever we change the subset of edges of a cycle, we will get another spanning forest. By the construction, M can not contain C\{exC } for all C. Consider a maximal chain of minimal cycles Cj , Cj+1 , . . . , Cj+t of G such that Cj+i and Cj+(i+1) have a common edge or a common vertex for all 0 ≤ i ≤ t − 1. By the rule of labeling edges, each common edge of the consecutive cycles are the smallest edges of the latter cycle and it will appear in M . If Cj+i has length kj+i , M contains a subset of kj+i − 1 edges of Cj+i . Let ∪i Cj+i be the union of Cj+i for 0 ≤ i ≤ t. If there are x common vertices ∑ and y common edges of minimal cycles in ∪i Cj+i , then ∪i Cj+i has ( i kj+i∑ )−x−2y vertices. The subsets of edges of Cj+i form a multiset N with ( i kj+i )−t elements including edges of ∪i Cj+i . When we remove the common edges of consecutive cycles in N one time, ∑ we get a ∑ set N ′ with ( i kj+i ) − t − y edges. Since t = x + y + 1, the set N ′ has ( i kj+i ) − ∑x − 2y − 1 edges. Those edges∑form a subgraph of ∪i Cj+i including ( i kj+i ) − x − 2y vertices and ( i kj+i ) − x − 2y − 1 edges with no isolated vertex. Thus N ′ form a spanning tree of ∪i Cj+i . Therefore we can conclude that when we remove every common edge one time in the multiset M , we will get a set of edges that can form a spanning forest of G satisfying the condition of Lemma 4.3. For each multiset M , the edges removed are the same, and they are all common edges in G. Since each edge except the common edges will appear in the subsets of edges of only one cycle, it will appear only once in M . Therefore, whenever we change the subset of edges of a minimal cycle, we will get another spanning forest. On the other hand, let T be a spanning tree of a connected component of G. We add common edges of cycles in that component into T one time to get a multiset N ′′ . Then we can partition N ′′ into a set of edges that are not in any cycle and (kj − 1)-subsets of cycles Cj in the component. Therefore, the number of spanning forests satisfying the condition of Lemma 4.3 is ∏ (k − 1)αk k Topological properties of Graphical arrangements 447 multisets M decreasing maximal chains {1, 10, 2, 3, 3, 5, 5, 7, 8} [10, 8, 7, 5, 3, 2, 1] {1, 10, 2, 3, 3, 5, 5, 7, 9} [10, 9, 7, 5, 3, 2, 1] {1, 10, 2, 3, 3, 5, 5, 8, 9} [10, 9, 8, 5, 3, 2, 1] {1, 10, 2, 3, 3, 6, 5, 7, 8} [10, 8, 7, 6, 3, 2, 1] {1, 10, 2, 3, 3, 6, 5, 7, 9} [10, 9, 7, 6, 3, 2, 1] [10, 9, 8, 6, 3, 2, 1] {1, 10, 2, 3, 3, 6, 5, 8, 9} {1, 10, 2, 4, 3, 5, 5, 7, 8} [10, 8, 7, 5, 4, 2, 1] [10, 9, 7, 5, 4, 2, 1] {1, 10, 2, 4, 3, 5, 5, 7, 9} [10, 9, 8, 5, 4, 2, 1] {1, 10, 2, 4, 3, 5, 5, 8, 9} [10, 8, 7, 6, 4, 2, 1] {1, 10, 2, 4, 3, 6, 5, 7, 8} {1, 10, 2, 4, 3, 6, 5, 7, 9} [10, 9, 7, 6, 4, 2, 1] [10, 9, 8, 6, 4, 2, 1] {1, 10, 2, 4, 3, 6, 5, 8, 9} Table 1. Multisets and decreasing maximal chains where αk is the number of minimal cycles of length k. Example 4.16. Let G be the graph with edge labeled shown in Figure 4. The subsets of edges of C1 are {2, 3} and {2, 4}. The subsets of edges of C2 are {3, 5} and {3, 6}. The subset of edges of C3 are {5, 8, 9}, {5, 7, 9} and {5, 7, 8}. The common edges are 3, 5. Table 1 shows multisets and corresponding decreasing maximal chains. Thus there are 12 different decreasing maximal chains. Proposition 4.17. Let G be a graph such that we can label minimal cycles in G to form a chain of cycles C1 , C2 , . . . , Cs of length k1 , k2 , . . . , ks respectively. If    ∅, a vertex, or an edge if i ̸= t and j = i + 1, Ci ∩ Cj =   α edges(α > 1) if i = t and j = t + 1, ∅ if j ̸= i + 1, then the number of decreasing maximal chains in PG is ∏ [(kt − 1)(kt+1 − 1) − α(α − 1)] (ki − 1). i̸=t,t+1 Proof. Since G is a planar graph, we can label cycles and edges in G as in the proof of Proposition 4.15. Similar to Proposition 4.15, we add each (ki − 1)-subset of each Ci into a multiset M , and remove common edges one time. If a connected component that doesn’t contain Ct and Ct+1 , the number of spanning trees is counted by Proposition 4.15. 448 Thi A. Nguyen and Sangwook Kim Consider the connected component that contains Ct and Ct+1 . Assume that this connected component of G contains cycles Ct−v , Ct−v+1 , . . . , Ct−1 , Ct , Ct+1 , Ct+2 , . . . , Ct+u . Consider the case when Ct−1 ∩Ct , Ct+1 ∩Ct+2 ∈ {∅, a vertex}. By the labeling of edges, the smallest edge of Ct+1 is a common edge of Ct and Ct+1 . Since edges are labeled from the left to the right, the smallest edge of Ct can not be a common edge of the two cycles. There are kt − α and kt+1 − α edges of Ct and Ct+1 respectively that are not common edges of Ct , Ct+1 . If (kt −1)-subset of Ct and (kt+1 −1)-subset of Ct+1 added into M contain all those edges, then after removing common edges we get a cycle. There are α ways to choose (kt − 1)-subsets of Ct that contain all non-common edges of Ct , Ct+1 . Since the smallest edge of Ct+1 is a common edge, there are α − 1 ways to choose (kt+1 − 1)-subsets of Ct+1 that contain all non-common edges of the two cycles. So we have α(α − 1) ways of combination of subsets of Ct and Ct+1 in which we have a cycle. Then the number of spanning trees of this component is [(kt − 1)(kt+1 − 1) − α(α − 1)]kt−v · · · kt−1 kt+2 · · · kt+u . The proof is similar to the proof of Proposition 4.15. When Ct−1 ∩ Ct = {an edge}, Ct+1 ∩ Ct+2 ∈ {∅, a vertex}, the common edge of Ct−1 and Ct must be the smallest edge of Ct . If M contains the (kt−1 − 1)-subset of Ct−1 that does not contain the common edge of Ct−1 and Ct , and (kt − 1)-subset, (kt+1 − 1)-subset of Ct and Ct+1 respectively such that they contain all non-common edges of Ct , Ct+1 , then we have a cycle. We also have α(α − 1) ways of combination of subsets of Ct and Ct+1 as the above situation. Therefore, the number of spanning trees is [(kt − 1)(kt+1 − 1) − α(α − 1)]kt−v · · · kt−1 kt+2 · · · kt+u . When Ct−1 ∩ Ct ∈ {∅, a vertex} and Ct+1 ∩ Ct+2 = {an edge}, that edge must be the smallest edge of Ct+2 . Every (kt+2 − 1)-subset of Ct+2 contains that common edge. Similarly, we also have [(kt − 1)(kt+1 − 1) − α(α − 1)]kt−v · · · kt−1 kt+2 · · · kt+u spanning trees. When Ct+1 ∩ Ct+2 = {an edge} and Ct−1 ∩ Ct = {an edge}, the common edge of Ct+1 and Ct+2 must be the smallest edge of Ct+2 . If M contains the (kt−1 − 1)-subset of Ct−1 that does not contain the common edge of Ct−1 and Ct , and (kt −1)-subset, (kt+1 −1)-subset of Ct and Ct+1 respectively such that they contain all non-common edges of Ct , Ct+1 , Topological properties of Graphical arrangements 449 then after removing the common edges we also have a cycle. So the number of spanning trees of this component is [(kt − 1)(kt+1 − 1) − α(α − 1)]kt−v · · · kt−1 kt+2 · · · kt+u . Consequently, the number of spanning forests that satisfy the condition of Lemma 4.3 is ∏ [(kt − 1)(kt+1 − 1) − α(α − 1)] (ki − 1). ki i̸=t,t+1 Proposition 4.18. Let G be a graph such that in each connected component all minimal cycles share only one common edge. Then the number of decreasing maximal chains of the intersection poset of graphical arrangement of G is equal to ∏ (k − 1)αk k where αk is the number of minimal cycles of length k. Proof. In each connected component, we label edges so that the common edge is the smallest edge of all minimal cycles in that component. We construct a multi set M by adding all edges that do not belong to any cycle and adding one (k − 1)-subset of each minimal cycle C of length k. If e is a common edge of x cycles, then e will appear x times in M . We remove e x − 1 times to get a set N of edges in G. We prove that edges in N form a spanning forest that satisfies the condition of Lemma 4.3. It is obvious that N doesn’t contain C\{exC } for all C. Let C1 , C2 , . . . , Ct be all minimal cycles of a connected component. Let ∪i Ci be the union of C1 , C2 , . . . , Ct . Assume that C1 , C2 , . . . , Ct have length k1 , k2 , . . . , kt respectively. Then the number of vertices in ∪i Ci is ∑ [ ti=1 ki ] − 2(t − 1). After removing edges, the number of edges of ∪Ci ∑ ∑ in N is ( ti=1 (ki − 1)) − (t − 1) = ( ti=1 ki ) − 2t + 1. Since the subgraph formed by edges in N doesn’t contain isolated vertices, the edges in N form a spanning tree of ∪i Ci . Whenever we change a (ki − 1)-subset of Ci in M , we get a new spanning tree of ∪i Ci . On the other hand, let T be a spanning tree of a connected component of G which has t cycles . We add the common edge of those cycles t − 1 times into T to get a multiset N ′ . Then we can partition N ′ into a set 450 Thi A. Nguyen and Sangwook Kim of edges that are not in any cycle and (ki − 1)-subsets of cycles Ci in the component. Therefore, the number of spanning forests of G satisfying condition of Lemma 4.3 is ∏ (k − 1)αk k where αk is the number of minimal cycles of length k. 5. The number of regions An application of finding the number of decreasing maximal chains is counting regions of an arrangement. Recall that a region of an arrangement A is a connected component of the complement X of the hyperplanes, X = Rn − ∪H∈A H. By Zaslavsky, the number of regions of an arrangement is found by the formula. Theorem 5.1 (see [10]). Let A be a hyperplane arrangement in Rn . Then ∑ #{regions} = |µ(0̂, x)|, x∈P where P is the intersection poset of arrangement A and µ is the Möbius function of P . The following theorem describes the number of regions of graphical arrangement. Theorem 5.2. The number of regions of the graphical arrangement of a graph G is equal to ∑ dH H where the sum is over all subgraphs H of G whose connected components are induced subgraphs with at least two vertices and dH is the number of decreasing maximal chains in the intersection poset PH . Proof. Let A = A1 − A2 − · · · − Ap be an element of the intersection poset PG where Ai , i = 1, 2, . . . , p, are blocks of A. Each Ai contains indices or vertices of G. Let Gi be the subgraph induced by Ai . Let GA be the union of subgraphs Gi , i = 1, 2, . . . , p. Then for each element A of PG , we get a subgraph GA of G. It is obvious that the intersection poset PGA is isomorphic to the interval [0̂, A] of PG (see [5, Lemma 2.11]). By Topological properties of Graphical arrangements 451 [9, Theorem 4.11], |µ(0̂, A)| is equal to the number dGA of decreasing maximal chains of PGA . By Theorem 5.1, the number of regions of AG is ∑ ∑ |µ(0̂, A)| = dGA . A∈PG A∈PG Since each block of an element A in PG can be associated with an induced subgraph of G with at least two vertices, the result follows. Now we provide some enumerative results. Proposition 5.3. The number of regions of graphical arrangement of a forest with m edges is 2m Proof. It is easy to see that the intersection poset of AG , where G is a forest, has only one decreasing maximal chain, and the intersection poset of arrangement of every its subgraph also has one decreasing maximal chain. To find the number of regions, we need to find the number of subgraphs of G satisfying the condition of Theorem 5.2. Since the union of any edges can give a subgraph satisfying the condition of Theorem 5.2, the number of regions is the number of subsets of edges, and equal to 2m . Remark 5.4. In general, a graphical arrangement is not in general position, but the graphical arrangement for a forest is in general position. It is known that if an arrangement of m hyperplanes in Rn is in general position, then the number of regions is found by the formula (see [9, Proposition 2.4]) ( ) ( ) m m 1+m+ + ··· + . 2 n So we can also use this formula to show Proposition 5.3. Proposition 5.5. Let G be a graph with m edges and a unique cycle C of length k. Then the number of regions of graphical arrangement of G is 2m − 2m−k+1 . Proof. We would like to count the number of subgraphs of G satisfying the conditions of Theorem 5.2. First, consider the case when the subgraph H does not contain C. If H contains all but one edge of C, the connected component of H containing those edges is not an induced subgraph. Since there are 2m−k subgraphs of G containing C and k2m−k subgraphs of G containing all but one edge of C, there are 2m − (k + 1)2m−k subgraphs of G with no cycle which satisfy the condition of Theorem 5.2. 452 Thi A. Nguyen and Sangwook Kim Since any subgraph of G containing C satisfies the condition of Theorem 5.2, there are 2m−k subgraphs of G containing C. By Proposition 4.13, the intersection poset of graphical arrangement of graph with no cycle has 1 decreasing maximal chain, and the number of decreasing maximal chains is k − 1 when the graph has one cycle of length k. Theorem 5.2 implies the number of regions is 2m − (k + 1)2m−k + (k − 1)2m−k = 2m − 2m−k+1 . Proposition 5.6. Let G be a graph with m edges and s cycles C1 , C2 , . . . , Cs of length k1 , k2 , . . . , ks respectively such that they do not share edges. Then the number of regions of graphical arrangement of G is ∑ ∏ (kj − 1)], [SY j∈Y Y ⊆[s] where ∑ SY = 2m− j∈Y kj ∑ − ∅̸=X⊆[s]\Y [( ∏ ∑ kj + 1)2m− j∈Y kj − ∑ j∈X kj ] j∈X for subsets X, Y of [s]. Proof. We would like to count the number of subgraphs of G satisfying the conditions of Theorem 5.2. Consider the case when the subgraph H does not contain any cycle. If H contains all but one edge of any cycle, the connected component of H containing those edges is not an induced subgraph. For r cycles Cx1 , Cx2 , . . . , Cxr , there are 2m−kx1 −···−kxr subgraphs containing Cx1 , Cx2 , . . . , Cxr and kx1 kx2 · · · kxr 2m−kx1 −···−kxr subgraphs containing all but one edge in each cycle Cx1 , Cx2 , . . . , Cxr . Thus the number of subgraphs of G with no cycle which satisfy the condition of Theorem 5.2 is s r ∑r ∑ ∑ ∏ S = 2m − [ [( kxq + 1)2m− q=1 kxq ] r=1 x1 ,x2 ,...,xr q=1 = 2m − ∑ [( ∏ ∑ kj + 1)2m− j∈X kj ] ∅̸=X⊆[s] j∈X When H contains cycles Cy1 , Cy2 , . . . , Cyu for y1 , y2 , . . . , yu ∈ Y ⊆ [s], we have ∑ ∑ ∑ ∑ ∏ SY = 2m− j∈Y kj − [( kj + 1)2m− j∈Y kj − j∈X kj ] ∅̸=X⊆[s]\Y j∈X Topological properties of Graphical arrangements 453 subgraphs of G containing Cy1 , Cy2 , . . . , Cyu which satisfy the condition of Theorem 5.2. By Proposition 4.15 and Theorem 5.2, the number of regions is ∑ ∏ S+ [SY (kj − 1)]. ∅̸=Y ⊆[s] j∈Y Since S = SY when Y = ∅, the result follows. References [1] Anders Björner, Shellable and Cohen-Macaulay partially ordered sets, Trans. Amer. Math. Soc., 260(1) (1980), 159-183. [2] Anders Björner and Michelle L. Wachs, Shellable nonpure complexes and posets. I, Trans. Amer. Math. Soc., 348(4) (1996), 1299-1327. [3] Curtis Greene and Thomas Zaslavsky, On the interpretation of Whitney numbers through arrangements of hyperplanes, zonotopes, non-Radon partitions, and orientations of graphs, Trans. Amer. Math. Soc., 280(1) (1983), 97-126. [4] Peter Orlik and Louis Solomon, Combinatorics and topology of complements of hyperplanes, Invent. Math., 56(2) (1980), 167-189. [5] Peter Orlik and Hiroaki Terao, Arrangements of hyperplanes, volume 300 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], Springer-Verlag, Berlin, 1992. [6] Richard P. Stanley, Supersolvable lattices, Algebra Universalis, 2 (1972), 197-217. [7] Richard P. Stanley, Finite lattices and Jordan-Hölder sets, Alg. Univ., 4 (1974), 361-371. [8] Richard P. Stanley, Enumerative Combinatorics, vol 2, Cambridge Studies in Advanced Mathematics, 62, Cambridge University Press, 1999. [9] Richard P. Stanley, An introduction to hyperplane arrangements, In Geometric combinatorics, volume 13 of IAS/Park City Math. Ser., pages 389-496. Amer. Math. Soc., Providence, RI, 2007. [10] Thomas Zaslavsky, Facing up to arrangements: face-count formulas for partitions of space by hyperplanes, Mem. Amer. Math. Soc., 1(issue 1, 154):vii+102, 1975. [11] Michelle L. Wachs, A basis for the homology of the d-divisible partition lattice, Adv. Math., 117(2) (1996), 294-318. [12] Michelle L. Wachs, Poset topology: tools and applications, In Geometric combinatorics, volume 13 of IAS/Park City Math. Ser., pages 497-615. Amer. Math. Soc., Providence, RI, 2007. Thi A. Nguyen Department of Mathematics, Chonnam National University, Gwangju 500-757, Republic of Korea. E-mail: [email protected] 454 Thi A. Nguyen and Sangwook Kim Sangwook Kim Department of Mathematics, Chonnam National University, Gwangju 500-757, Republic of Korea. E-mail: [email protected]