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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

On algebraic characterization of SSC of the Jahangir’s graph 𝓙n,m

Zahid Raza / Agha Kashif / Imran Anwar
Published Online: 2018-03-20 | DOI: https://doi.org/10.1515/math-2018-0025

Abstract

In this paper, some algebraic and combinatorial characterizations of the spanning simplicial complex Δs(𝓙n,m) of the Jahangir’s graph 𝓙n,m are explored. We show that Δs(𝓙n,m) is pure, present the formula for f-vectors associated to it and hence deduce a recipe for computing the Hilbert series of the Face ring k[Δs(𝓙n,m)]. Finally, we show that the face ring of Δs(𝓙n,m) is Cohen-Macaulay and give some open scopes of the current work.

Keywords: Simplicial complexes; Spanning trees; Face ring; Hilbert series; f-vectors; Cohen Macaulay

MSC 2010: 13P10; 13H10; 13F20; 13C14

1 Introduction

The concept of spanning simplicial complex (SSC) associated with the edge set of a simple finite connected graph is introduced by Anwar, Raza and Kashif in [1]. They revealed some important algebraic properties of SSC of a unicyclic graph. Kashif, Raza and Anwar further established the theory and explored algebraic characterizations of some more general classes of n-cyclic graphs in [10, 11]. The problem of finding the SSC for a general simple finite connected graph is not an easy task to handle. Recently in [15] Zhu, Shi and Geng discussed the SSC of another class n−cyclic graphs with a common edge.

In this article, we discuss some algebraic and combinatorial properties of the spanning simplicial complex Δs(𝓙n,m) of a certain class of cyclic graphs, 𝓙n,m. For simplicity, we fixed n = 2 in our results. Here, 𝓙n,m is the class of Jahangir’s graph defined in [12] as follows:

The Jahangir’s graph Jn,m, for m ≥ 3, is a graph on nm + 1 vertices i.e., a graph consisting of a cycle Cnm with one additional vertex which is adjacent to m vertices of Cnm at distance n to each other on Cnm.

More explicitly, it consists of a cycle Cnm which is further divided into m consecutive cycles Ci of equal length such that all these cycles have one vertex common and every pair of consecutive cycles has exactly one edge common. For example the graph 𝓙2,3 is given in Figure 1. We fix the edge set of 𝓙2,m as follows:

E={e11,e12,e13,e21,e22,e23,,em1,em2,em3}.(1)

The graph 𝓙2,3
Figure 1

The graph 𝓙2,3

Here, {ek1, ek2, ek3, e(k+1)1} is the edge set of the cycle Ck for k ∈ {1, 2, …, m − 1} and {em1, em2, em3, e11} is the edge set of cycle Cm. Also ek1 always represents the common edge between Ck−1 and Ck for k ∈ {1, 2, …, m − 1} and e11 is the common edge between the cycle Cm and C1.

2 Preliminaries

In this section, we give some background and preliminaries of the topic and define some important notions to make this paper self-contained. However, for more details of the notions we refer the reader to [3, 4, 5, 6, 7, 13, 14].

Definition 2.1

A spanning tree of a simple connected finite graph G(V, E) is a subtree of G that contains every vertex of G.

We represent the collection of all edge-sets of the spanning trees of G by s(G), in other words;

s(G):={E(Ti)E,whereTiisaspanningtreeofG}.

Lemma 2.2

Let G = (V, E) be a simple finite connected graph containing m cycles. Then its spanning tree contains exactly |E| − m edges.

Proof

A spanning tree of a graph is its spanning subgraph containing no cycles and no disconnection. If G is a unicyclic graph then deletion of one edge from it results in a spanning tree. If more than one edge is removed from the cycle in G then a disconnection is obtained which is not a spanning tree. Therefore, spanning tree has exactly |E| − 1 edges.

If G has m disjoint cycles in it i.e. cycles sharing no common edges, then its spanning tree is obtained by removing exactly m edges from it, one from each of its cycle. Therefore, its spanning tree has |E| − m edges in it.

If any two cycles of G share one or more common edges and remaining are disjoint cycles, then one edge is needed to be removed from each cycle of G to obtain a spanning tree. However, if a common edge between two cycles is removed then exactly one edge from non common edges must be removed of the resulting big cycle. Therefore, its spanning tree has |E| − m edges in it. This can be extended to any number of cycles in G sharing common edges. This completes the proof. □

Applying Lemma 2.2, we can obtain the spanning tree of the Jahangir’s graph 𝓙2,m by removing exactly m edges from it keeping in view the following:

  • Not more than one edge can be removed from the non common edges of any cycle.

  • If a common edge between two or more consecutive cycles is removed then exactly one edge must be removed from the resulting big cycle.

  • Not all common edges can be removed simultaneously.

This method is referred as the cutting-down method. For example, by using the cutting-down method for the graph 𝓙2,3 given in Fig. 1 we obtain:

s(𝓙2,3) = {{e11, e21, e31, e12, e22, e32}, {e11, e21, e31, e12, e22, e33}, {e11, e21, e31, e12, e23, e32}, {e11, e21, e31, e12, e23, e33}, {e11, e21, e31, e13, e22, e32}, {e11, e21, e31, e13, e22, e33}, {e11, e21, e31, e13, e23, e32}, {e11, e21, e31, e13, e23, e33}, {e21, e31, e32, e33, e12, e22}, {e21, e31, e32, e33, e12, e23}, {e21, e31, e32, e33, e13, e22}, {e21, e31, e32, e33, e13, e23}, {e21, e31, e12, e13, e33, e22}, {e21, e31, e12, e13, e33, e23}, {e21, e31, e12, e13, e32, e22}, {e21, e31, e12, e13, e32, e23}, {e11, e31, e12, e13, e22, e32}, {e11, e31, e12, e13, e22, e33}, {e11, e31, e12, e13, e23, e32}, {e11, e31, e12, e13, e23, e33}, {e11, e31, e22, e23, e13, e32}, {e11, e31, e22, e23, e13, e33}, {e11, e31, e22, e23, e12, e32}, {e11, e31, e22, e23, e12, e33}, {e11, e21, e23, e22, e32, e12}, {e11, e21, e23, e22, e32, e13}, {e11, e21, e23, e22, e33, e12}, {e11, e21, e23, e22, e33, e13}, {e11, e21, e32, e33, e22, e12}, {e11, e21, e32, e33, e22, e13}, {e11, e21, e32, e33, e23, e12}, {e11, e21, e32, e33, e23, e13}, {e11, e13, e22, e23, e32, e33}, {e11, e12, e22, e23, e32, e33}, {e11, e12, e13, e23, e32, e33}, {e11, e12, e13, e22, e32, e33}, {e11, e12, e13, e22, e23, e33}, {e11, e12, e13, e22, e23, e32}, {e21, e13, e22, e23, e32, e33}, {e21, e12, e22, e23, e32, e33}, {e21, e12, e13, e23, e32, e33}, {e21, e12, e13, e22, e32, e33}, {e21, e12, e13, e22, e23, e33}, {e21, e12, e13, e22, e23, e32}, {e31, e13, e22, e23, e32, e33}, {e31, e12, e22, e23, e32, e33}, {e31, e12, e13, e23, e32, e33}, {e31, e12, e13, e22, e32, e33}, {e31, e12, e13, e22, e23, e33}, {e31, e12, e13, e22, e23, e32}}.

Definition 2.3

A simplicial complex Δ over a finite set [n] = {1, 2, …, n} is a collection of subsets of [n], with the property that {i} ∈ Δ for all i ∈ [n], and if FΔ then Δ will contain all the subsets of F (including the empty set). An element of Δ is called a face of Δ, and the dimension of a face F of Δ is defined as |F| − 1, where |F| is the number of vertices of F. The maximal faces of Δ under inclusion are called facets of Δ. The dimension of the simplicial complex Δ is:

dimΔ=max{dimF|FΔ}.

We denote the simplicial complex Δ with facets {F1, …, Fq} by

Δ=F1,,Fq

Definition 2.4

For a simplicial complex Δ having dimension d, its fvector is a d + 1-tuple, defined as:

f(Δ)=(f0,f1,,fd)

where fi denotes the number of idimensional faces of Δ.

Definition 2.5 (Spanning Simplicial Complex)

Let G(V, E) be a simple finite connected graph and s(G) = {E1, E2, …, Et} be the edge-sets of all possible spanning trees of G(V, E), then we defined (in [1]) a simplicial complex Δs(G) on E such that the facets of Δs(G) are precisely the elements of s(G), we call Δs(G) as the spanning simplicial complex of G(V, E). In other words;

Δs(G)=E1,E2,,Et.

For example, the spanning simplicial complex of the graph 𝓙2,3 given in Fig. 1 is:

Δs(𝓙2,3) = 〈 {e11, e21, e31, e12, e22, e32}, {e11, e21, e31, e12, e22, e33}, {e11, e21, e31, e12, e23, e32}, {e11, e21, e31, e12, e23, e33}, {e11, e21, e31, e13, e22, e32}, {e11, e21, e31, e13, e22, e33}, {e11, e21, e31, e13, e23, e32}, {e11, e21, e31, e13, e23, e33}, {e21, e31, e32, e33, e12, e22}, {e21, e31, e32, e33, e12, e23}, {e21, e31, e32, e33, e13, e22}, {e21, e31, e32, e33, e13, e23}, {e21, e31, e12, e13, e33, e22}, {e21, e31, e12, e13, e33, e23}, {e21, e31, e12, e13, e32, e22}, {e21, e31, e12, e13, e32, e23}, {e11, e31, e12, e13, e22, e32}, {e11, e31, e12, e13, e22, e33}, {e11, e31, e12, e13, e23, e32}, {e11, e31, e12, e13, e23, e33}, {e11, e31, e22, e23, e13, e32}, {e11, e31, e22, e23, e13, e33}, {e11, e31, e22, e23, e12, e32}, {e11, e31, e22, e23, e12, e33}, {e11, e21, e23, e22, e32, e12}, {e11, e21, e23, e22, e32, e13}, {e11, e21, e23, e22, e33, e12}, {e11, e21, e23, e22, e33, e13}, {e11, e21, e32, e33, e22, e12}, {e11, e21, e32, e33, e22, e13}, {e11, e21, e32, e33, e23, e12}, {e11, e21, e32, e33, e23, e13}, {e11, e13, e22, e23, e32, e33}, {e11, e12, e22, e23, e32, e33}, {e11, e12, e13, e23, e32, e33}, {e11, e12, e13, e22, e32, e33}, {e11, e12, e13, e22, e23, e33}, {e11, e12, e13, e22, e23, e32}, {e21, e13, e22, e23, e32, e33}, {e21, e12, e22, e23, e32, e33}, {e21, e12, e13, e23, e32, e33}, {e21, e12, e13, e22, e32, e33}, {e21, e12, e13, e22, e23, e33}, {e21, e12, e13, e22, e23, e32}, {e31, e13, e22, e23, e32, e33}, {e31, e12, e22, e23, e32, e33}, {e31, e12, e13, e23, e32, e33}, {e31, e12, e13, e22, e32, e33}, {e31, e12, e13, e22, e23, e33}, {e31, e12, e13, e22, e23, e32}〉.

3 Spanning trees of 𝓙2,m and Face ring Δs(𝓙2,m)

In this section, we give two lemmas which give an important characterization of the graph 𝓙2,m and its spanning simplicial complex s(𝓙2, m). We present a proposition which gives the f-vectors and the dimension of the 𝓙2,m. Finally, in Theorem 3.13 we give the formulation for the Hilbert series of the Face ring k[Δs(𝓙2,m)].

Definition 3.1

Let Ci1, Ci2, …, Cik be consecutive cycles in the Jahangirs graph 𝓙2,m. Then the cycle obtained by deleting the common edges between the consecutive cycles Ci1, Ci2, …, Cik is a new cycle of the Jahangris graph 𝓙2,m is denoted by Ci1,i2, …, ik. The cardinality count of the number of edges in the cycle Ci1,i2,…,ik is denoted by βi1,i2,…,ik = |Ci1,i2,…,ik|.

The following lemma computes the total number of cycles in the Jahangir’s graph 𝓙2,m and the cardinality count of the edges in these cycles.

Lemma 3.2 (Characterization of 𝓙2,m)

Let 𝓙2,m be the graph with the edges E as is defined in eq. (1) and C1, C2, ⋯, Cm be its m consecutive cycles of equal lengths, then the total number of cycles in the graph are

τ=m2

such that βi1,i2,…,ik = 2(k + 1).

Proof

The Jahangir’s graph 𝓙2,m contains more than just m consecutive cycles. The remaining cycles can be obtained by deleting the common edges between any number (included) of consecutive cycles and getting a cycle by their remaining edges. The cycle obtained in this way by adjoining consecutive cycles Ci1, Ci2, …, Cik is denoted by Ci1,i2, …, ik. Therefore, we get the following cycles

C1,2,C2,3,,Cm1,m,Cm,1,C1,2,3,,Cm2,m1,m,Cm1,m,1,Cm,1,2,,C1,2,3,,m,C2,3,4,,m,1,C3,4,5,,m,1,2,Cm,1,2,,m1.

Combining these with m cycles given we have total cycles in the graph 𝓙2,m,

Ci1,i2,,ikij{1,2,,m}and1km,

such that ij+1 = ij + 1 if ijm and ij+1 = 1 if ij = m.

Now for a fixed value of k, simple counting reveals that the total number of cycles Ci1,i2,…,ik is m for ik <m. Hence the total number of cycles in 𝓙2,m is τ. Also it is clear from the construction above that Ci1,i2,…,ik is obtained by deleting common edges between consecutive cycles Ci1,Ci2,…,Cik which are k − 1 in number. Therefore, the order of the cycle Ci1,i2,…,ik is obtained by adding orders of all Ci1,Ci2, …,Cik that is, 4k and subtracting 2(k − 1) from it, since the common edges are being counted twice in sum. This implies

βi1,i2,,ik=|Ci1,i2,,ik|=t=1k|Cit|2(k1)=2(k+1). □

In the following results, we fix Cu1,u2,…,up, Cv1,v2,…,vq to represent any two cycles from the cycles

Ci1,i2,,ikij{1,2,,m}and1km,

such that ij+1 = ij + 1 if ijm and ij+1 = 1 if ij = m, of the graph 𝓙2,m. Also we fix the notation "ab" if b immediately proceeds a i.e., the very next in order of preferences.

Proposition 3.3

Let 𝓙2,m be the graph with the edges E as defined in eq. (1) such that {u1, u2, …,up} ⊆ {v1,v2, …,vq} then we have

|Cu1,u2,,upCv1,v2,,vq|=βu1,u2,,up2,{u1,up}{v1,vq}βu1,u2,,up1,u1{v1,vq}&up{v1,vq}βu1,u2,,up1,up{v1,vq}&u1{v1,vq}βu1,u2,,up,u1=v1&up=vqoru1=vqup=vq

Proof

Since the cycles Cu1,u2,…,up and Cv1,v2,…,vq are obtained by deleting the common edges between cycles Cu1,Cu2, …,Cup and Cv1,Cv2, …,Cvq respectively, therefore, {u1, up} ⊈ {v1,vq} implies {u1,u2, …,up} ⊂ {v1, v2, …, vq}. Hence, the intersection Cu1,u2,…,upCv1,v2,…,vq will contain only the non common edges of the cycle Cu1,u2,…,up excluding its two edges common with the cycles on its each end. This gives the order of intersection in this case βu1,u2,…,up − 2. The remaining cases can be visualized in a similar manner. □

Proposition 3.4

Let 𝓙2,m be the graph with the edges E as defined in eq. (1) such that {u1, u2, …, uσ} ⊆ {v1, v2, …, vq} and ut ∈ {u1, u2, …, up} & ut−1ut with tσ < p then we have

|Cu1,u2,,upCv1,v2,,vq|=βu¯1,u¯2,,u¯σ1,u¯1=v1&vqu1βu¯1,u¯2,,u¯σ2,u¯1=v1&vqu1βu¯1,u¯2,,u¯σ1,u¯σ=vq&upv1βu¯1,u¯2,,u¯σ2,u¯σ=vq&upv1

Proof

Here, the cycles Cu1, Cu2, …, Cuσ are amongst σ consecutive adjoining cycles of the cycle Cu1,u2,…,up which are also overlapping with the σ consecutive adjoining cycles of the cycle Cv1,v2,…,vq. If the adjoining cycle Cu1 of the cycle Cu1,u2,…,up overlaps with the first adjoining cycle Cv1 of the cycle Cv1,v2,…,vq and the adjoining cycles Cvq and Cu1 are consecutive then by previous proposition the order of the intersection Cu1,u2,…,upCv1,v2,…,vq is indeed βu1,u2, …,uσ − 1. Similarly, if the adjoining cycles Cvq and Cu1 are not consecutive then they will have no common edge and the use of proposition 3.3 gives the order of the intersection Cu1,u2,…,upCv1,v2,…,vq as βu1,u2, …,uσ − 2. Similar can be done for the remaining cases. □

Remark 3.5

The case when there exists a t0 <σ < p such that ut0−1ut0 in above proposition i.e., when cycles Cu1, Cu2, …, Cut0−1, Cut0, …, Cuσ are not amongst σ consecutive adjoining cycles of the cycle Cu1,u2,…,up, the order of the intersection Cu1,u2,…,upCv1,v2,…,vq can be calculated by applying proposition 3.4 on the overlapping portions.

Proposition 3.6

Let 𝓙2,m be the graph with the edges E as defined in (1) such that {u1,u2, …,up} ⋂ {v1,v2, …,vq} = φ and pq. Then we have

|Cu1,u2,,upCv1,v2,,vq|=1,upv1&vqu11,upv1&vqu12,upv1&vqu10,otherwise.

Proof

In this case the adjoining cycles of Cu1,u2,…,up and Cv1,v2,…,vq have no common cycle. However, if the adjoining cycle on one of the extreme ends of the cycle Cu1,u2,…,up is consecutive with the adjoining cycles on one of the extreme ends of the other cycle Cv1,v2,…,vq then the intersection Cu1,u2,…,upCv1,v2,…,vq will have only one edge. The remaining cases are easy to see. □

In the following three propositions we give some characterizations of 𝓙2,m. We fix E(T(j1i1,j2i2,…,jmim)), where jα ∈ {1, 2, …, m} and iα ∈ {1, 2, 3}, as a subset of E. s(𝓙2,m).

Proposition 3.7

A subset E(T(j1i1,j2i2,…,jmim) of E with jα iαjα 1 for all α will belong to s(𝓙2,m) if and only if

E(T(j1i1,j2i2,,jmim))=E{e1i1,e2i2,,emim}

Proof

𝓙2,m is a graph with cycles C1, C2, …, Cm and e11, e21, …, em1 are the common edges between the consecutive cycles. The cutting down process explains we need to remove exactly m edges, keeping the graph connected and no cycles and no isolated edge left and no isolated vertices left in the graph. Therefore, in order to obtain a spanning tree of 𝓙2,m with none of common edges e11, e21, …, em1 to be removed, we need to remove exactly one edge from the non common edges from each cycle. This explains the proof of the proposition. □

Proposition 3.8

A subset E(T(j1i1,j2i2,…,jmim) of E with jα iα = jα 1 for any α will belong to s(𝓙2,m) if and only if

E(T(j1i1,j2i2,,jmim))=E{ej1i1,ej2i2,,ejmim}

where, {ej1i1, ej2i2, …, ejmim} will contain exactly one edge from C(jα−1)(jα) ∖ {e(jα−1)1, e(jα+1)1} other than ejα1.

Proof

For a spanning tree of 𝓙2,m such that exactly one common edge ejα1 is removed, we need to remove precisely m-1 edges from the remaining edges using the cutting down process. However, we cannot remove more than one edge from the non common edges of the cycle C(jα−1)(jα) (since this will result a disconnected graph. This explains the proof of the above case.

Proposition 3.9

A subset E(T(j1i1,j2i2,…,jmim)) ⊂ E, where jα iα = jα 1 for α ∈ {r1,r2, …, rρ} ⊂ {1, 2, …, m}, will belong to s(𝓙2,m) if and only if it satisfies any of the following:

  1. if ejr11, ejr21, …, ejrρ1 are common edges from consecutive cycles then

    E(T(j1i1,j2i2,,jmim))=E{ej1i1,ej2i2,,ejmim}

    such that {ej1i1, ej2i2, …, ejmim} will contain exactly exactly one edge from Cjr0jr1jrρ other than ejr11, ej(r2)1, …, ejrρ1, where jr0jr1.

  2. if none of ejr11, ejr21, …, ejrρ1 are common edges from consecutive cycles then

    E(T(j1i1,j2i2,,jmim))=E{ej1i1,ej2i2,,ejmim}

    such that for each edge ejrt1 proposition 3.6 holds.

  3. if some of ejr11,ejr21, …, ejrρ1 are common edges from consecutive cycles then

    E(T(j1i1,j2i2,,jmim))=E{ej1i1,ej2i2,,ejmim}

    such that proposition 3.9.1 is satisfied for the common edges of consecutive cycles and proposition 3.9.2 is satisfied for remaining common edges.

Proof

For the case 1, we need to obtain a spanning tree of 𝓙2,m such that |rρr1|m common edges must be removed from ρ consecutive cycles Cjr1, Cjr2, …, Cjrρ. The remaining m − |rρr1|m edges must be removed in such a way that exactly one edge is removed from the non common edges of the adjoining cycles Cjr0, Cjr1, …, Cjrρ and the remaining m − |rρr1|m cycles of the graph 𝓙2,m. This concludes the case.

The remaining cases of the proposition can be visualised in a similar manner using the propositions 3.7 and 3.8. This completes the proof. □

Remark 3.10

If we denote the disjoint classes of subsets of E discussed in propositions 3.7,3.8 and 3.9 by 𝓒𝓙1, 𝓒𝓙2, 𝓒𝓙3a, 𝓒𝓙3b, 𝓒𝓙3c respectively, then, we can write s(𝓙2,m) as follows:

s(J2,m)=CJ1CJ2CJ3aCJ3bCJ3c.

In our next result, we give an important characterization of the f-vectors of Δs(𝓙2,m).

Proposition 3.11

Let Δs(𝓙2,m) be a spanning simplicial complex of the graph 𝓙2,m, then the dim(Δs(𝓙2,m)) = 2m − 1 with fvector f(Δs(𝓙2,m)) = (f0, f1, ⋯, f2m−1) and

fi=3mi+1+t=1τ(1)t{i1,i2,,it}CIt3ms=1tβis+{iu,iv}{ip}p=1t|CiuCiv|i+1s=1tβis+{iu,iv}{ip}p=1t|CiuCiv|

where 0 ≤ i ≤ 2m − 1 I = {i1i2ik|ij ∈ {1, 2, …, m} and 1 ≤ km such that ij+1 = ij + 1 if ijm and ij+1 = 1 if ij = m} and CIt = {Subsets of I of cardinality t}.

Proof

Let E be the edge set of 𝓙2,m and 𝓒𝓙1, 𝓒𝓙2, 𝓒𝓙3a, 𝓒𝓙3b, 𝓒𝓙3c are disjoint classes of spanning trees of 𝓙2,m then from propositions 3.7, 3.8, 3.9 and the remark 3.10 we have

s(J2,m)=CJ1CJ2CJ3aCJ3bCJ3c.

Therefore, by definition 2.5 we can write Δs (𝓙2,m) = 〈𝓒𝓙1 ⋃𝓒𝓙2 ⋃𝓒𝓙3a ⋃𝓒𝓙3b ⋃𝓒𝓙3c〉. Since each facet Ê(j1i1,j2i2,…,jmim) = E(T(j1i1,j2i2,…,jmim) is obtained by deleting exactly m edges from the edge set of 𝓙2,m, keeping in view the propositions 3.7, 3.8 and 3.9, therefore dimension of each facet is the same i.e., 2m − 1 (since |Ê(j1i1,j2i2,…,jmim)| = 2m) and hence dimension of Δs(𝓙2,m) will be 2m − 1.

Also it is clear from the definition of Δs(𝓙2,m) that it contains all those subsets of E which do not contain the given sets of cycles {ek1, ek2, ek3, e(k+1)1} for k ∈ {1, 2, …, m − 1} and {em1, em2, em3, e11} in graph as well as any other cycle in the graph 𝓙2,m.

Now by Lemma 3.2 the total cycles in the graph 𝓙2,m are

Ci1,i2,,ikij{1,2,,m}and1km,

such that ij+1 = ij + 1 if ijm and ij+1 = 1 if ij = m, and their total number is τ. Let F be any subset of E of order i+1 such that it does not contain any Ci1,i2,…,ikij ∈ {1, 2, …, m} and 1 ≤ km, in it. The total number of such F is indeed fi for 0 ≤ i ≤ 2m − 1. We use inclusion exclusion principle to find this number. Therefore, fi = Total number of subsets of E of order i + 1 not containing Ci1,i2,…,ikij ∈ {1, 2, …, m} and 1 ≤ km such that ij+1 = ij + 1 if ijm and ij+1 = 1 if ij = m.

Therefore, using these notations and applying Inclusion Exclusion Principle we can write, fi = (Total number of subsets of E of order i + 1) − {i1}CI1 (subset of E of order i + 1 containing Cis for s = 1) + {i1,i2}CI2 (subset of E of order i + 1 containing both Cis for all 1 ≤ s ≤ 2 ) − ⋯ + (−1)τ {i1,i2,,iτ}CIτ (subset of E of order i + 1 simultaneously containing each Cis for all 1 ≤ sτ).

This implies

fi=3mi+1[{i1}CI13mβi1i+1βi1]+{i1,i2}CI23ms=12βis+{iu,iv}{ip}p=12|CiuCiv|i+1s=12βis+{iu,iv}{ip}p=12|CiuCiv|+(1)τ{i1,i2,,iτ}CIτ3ms=1τβis+{iu,iv}{ip}p=1τ|CiuCiv|i+1s=1τβis+{iu,iv}{ip}p=1τ|CiuCiv|.

This implies

fi=3mi+1+t=1τ(1)t{i1,i2,,it}CIt3ms=1tβis+{iu,iv}{ip}p=1t|CiuCiv|i+1s=1tβis+{iu,iv}{ip}p=1t|CiuCiv|.

Example 3.12

Let Δs (𝓙2,3) be a spanning simplicial complex of the Jahangirs graph 𝓙2,m given in Figure 1, then the dim(Δs (𝓙2,3)) = 5 and τ = 32 = 9. Therefore, fvectors f(Δs (𝓙2,3)) = (f0, f1, …, f5) and

fi=9i+1[{i1}CI19βi1i+1βi1]+{i1,i2}CI29s=12βis+{iu,iv}{ip}p=12|CiuCiv|i+1s=12βis+{iu,iv}{ip}p=12|CiuCiv|+(1)9{i1,i2,,i9}CI93ms=19βis+{iu,iv}{ip}p=19|CiuCiv|i+1s=19βis+{iu,iv}{ip}p=19|CiuCiv|,

where 0 ≤ i ≤ 5.

For a simplicial complex Δ over [n], one would associate to it the Stanley-Reisner ideal, that is, the monomial ideal I𝓝(Δ) in S = k[x1, x2,…, xn] generated by monomials corresponding to non-faces of this complex (here we are assigning one variable of the polynomial ring to each vertex of the complex). It is well known that the Face ring k[Δ] = S/I𝓝(Δ) is a standard graded algebra. We refer the readers to [7] and [14] for more details about graded algebra A, the Hilbert function H(A, t) and the Hilbert series Ht(A) of a graded algebra.

Our main result of this section is as follows;

Theorem 3.13

Let Δs(𝓙2,m) be the spanning simplicial complex of 𝓙2,m, then the Hilbert series of the Face ring k[Δs(𝓙2,m)] is given by,

H(k[Δs(J2,m)],t)=1+i=0dni+1ti+1(1t)i+1+i=0dk=1τ(1)k{i1,i2,,ik}CIk3ms=1kβis+{iu,iv}{ip}p=1k|CiuCiv|i+1s=1kβis+{iu,iv}{ip}p=1k|CiuCiv|ti+1(1t)i+1

Proof

From [14], we know that if Δ is a simplicial complex of dimension d and f(Δ) = (f0, f1, …,fd) its f-vector, then the Hilbert series of the face ring k[Δ] is given by

H(k[Δ],t)=1+i=0dfiti+1(1t)i+1.

By substituting the values of fi’s from Proposition 3.11 in this above expression, we get the desired result.

4 Cohen-Macaulayness of the face ring of Δs(𝓙2,m)

In this section, we present the Cohen-Macaulayness of the face ring of SSC Δs(𝓙2,m), using the notions and results from [2].

Definition 4.1

([2]). Let IS = k[x1,x2, …,xn] be a monomial ideal. We say that I has linear residuals, if there exists an ordered minimal monomial system of generators {m1, m2, …,mr} of I such that Res(Ii) is minimally generated by linear monomials for all 1 < ir, where Res(Ii) = {u1,u2, …, ui−1} such that uk =migcd(mk,mi) for all 1 ≤ ki − 1.

Theorem 4.2

([2]). Let Δ be a simplicial complex of dimension d over [n]. Then Δ will be a shellable if and only if I𝓕(Δ) has linear residuals.

Corollary 4.3

([2]). If the facet ideal I𝓕(Δ) of a pure simplicial complex Δ over [n] has linear residuals, then the face ring k[Δ] is Cohen Macaulay.

Here, we present the main result of this section.

Theorem 4.4

The face ring of Δs(𝓙2,m) is Cohen-Macaulay.

Proof

By corollary 4.3, it is sufficient to show that I𝓕(Δs(𝓙2,m)) has linear residuals in S = k[x11, x12, x13, x21, x22, x23, x31, …,xm1, xm2, xm3]. By propositions 3.7, 3.8, 3.9 and the remark 3.10, we have

s(J2,m)=CJ1CJ2CJ3aCJ3bCJ3c.

Therefore,

Δs(J2,m)=E^(j1i1,j2i2,,jmim)=E{ej1i1,ej2i2,,ejmim}E^(j1i1,j2i2,,jmim)s(J2,m)

and hence we can write,

IF(Δs(J2,m))=(xE^(j1i1,j2i2,,jmim)E^(j1i1,j2i2,,jmim)s(J2,m)).

Here, I𝓕(Δs(𝓙2,m)) is a pure monomial ideal of degree 2m − 1 with xÊ(j1i1,j2i2,…,jmim) as the product of all variables in S except xj1i1, xj2i2, …, xjmim. Now we will show that I𝓕(Δs(𝓙2,m)) has linear residuals with respect to the following orders in its monomials:

{xE^(j1i1,j2i2,,jmim)ir11;1r1m&ik=1;kr1},{xE^(j1i1,j2i2,,jmim)ir1,ir21;1r1,r2m&ik=1;kr1,r2},{xE^(j1i1,j2i2,,jmim)ir1,ir2,,irm1;1r1,r2rmm}.(2)

More explicitly, the monomials {xÊ(j1i1,j2i2,…,jmim)ir1 ≠ 1; 1 ≤ r1m & ik = 1; kr1} in the order 2, consists of monomials of the form xÊ(11,21,…,(m−1)1,jmim), xÊ(11,21,…,j(m−1)i(m−1),m1), …, xÊ(11,j2i2,31,…,(m−1)1,m1), xÊ(j1i1,21,…,(m−1)1,m1), where ik ∈ {2,3} and 1 ≤ jkm. Similarly, other monomials in order 2. Let us put

Res(xE^(j1i1,j2i2,,jmim))={xE^(j1i1,j2i2,,jmim)gcd(mk,xE^(j1i1,j2i2,,jmim))mkprceeds xE^(j1i1,j2i2,,jmim) wrt order 2}

For instance, for r1 = m in Res(xÊ(j1i1,j2i2,…,jmmim)) we have, Res(xÊ(11,21,…,(m−1)1,jmim)) ={xE^(11,21,,(m1)1,jmim)gcd(mk,xE^(11,21,,(m1)1,jmim))} where, mk in this case are all the monomials in S of the form xÊ(j1i1,j2i2,…,jmim) where imm and jm = 2, 3. Since all the monomials mk differ from xÊ(11,21,…,(m−1)1,jmim) at only one position, therefore, Res(xÊ(11,21,…,(m−1)1,jmim)) have all linear terms i.e., Res(xÊ(11,21,…,(m−1)1,jmim)) is minimally generated by linear monomials.

Continuing the same process the order 2 of the monomials of I𝓕(Δs(𝓙2,m)) guarantees that Res(xÊ(j1i1,j2i2,…,jmim)) is minimally generated by linear monomials for all xÊ(j1i1,j2i2,…,jmim)I𝓕(Δs(𝓙2,m)). Hence, I𝓕(Δs(𝓙2,m)) has linear residuals, and by Corollary 4.3 Δs(𝓙2,m) is Cohen-Macaulay. □

5 Conclusions and Scopes

We conclude this paper with some perspectives for further study as well as some constraints related to our work.

  • The results given in this paper can be naturally extended for any integer n ≥ 2.

  • The scope of SSC of a graph can be explored for some other classes of graphs like the wheel graph Wn etc. However, since finding spanning trees of a general graph is a NP-hard problem, therefore the results given here are not easily extendable for a general class of graph.

  • In view of the work done in [8, 9], we intend to find some perspectives for the SSC in studying sensor networks.

Acknowledgement

The authors are grateful to the reviewers and editor for their valuable suggestions to improve the manuscript. ZR is partially supported by the research Grant (1602144025 − P), from University of Sharjah, Sharjah, UAE.

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About the article

aghakashifkhan@hotmail.com


Received: 2016-07-11

Accepted: 2017-12-18

Published Online: 2018-03-20


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 250–259, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0025.

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© 2018 Raza et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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