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

In this paper, some algebraic and combinatorial characterizations of the spanning simplicial complex ∆s(Jn,m) of the Jahangir’s graph Jn,m are explored. We show that ∆s(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[∆s(Jn,m)]. Finally, we show that the face ring of ∆s(Jn,m) is Cohen-Macaulay and give some open scopes of the current work.


Introduction
The concept of spanning simplicial complex (SSC) associated with the edge set of a simple nite 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 nding the SSC for a general simple nite 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 ∆ s (J n,m ) of a certain class of cyclic graphs, J n,m . For simplicity, we xed n = in our results. Here, J n,m is the class of Jahangir's graph de ned in [12] as follows: The Jahangir's graph J n,m , for m ≥ , is a graph on nm + vertices i.e., a graph consisting of a cycle C nm with one additional vertex which is adjacent to m vertices of C nm at distance n to each other on C nm .
More explicitly, it consists of a cycle C nm which is further divided into m consecutive cycles C i 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 J , is given in Figure 1. We x the edge set of J ,m as follows: Here, {e k , e k , e k , e (k+ ) } is the edge set of the cycle C k for k ∈ { , , ⋯, m − } and {e m , e m , e m , e } is the edge set of cycle C m . Also e k always represents the common edge between C k− and C k for k ∈

Preliminaries
In this section, we give some background and preliminaries of the topic and de ne some important notions to make this paper self-contained. However, for more details of the notions we refer the reader to [3-7, 13, 14]. 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 − 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 J ,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 J , given in Fig We denote the simplicial complex ∆ with facets {F , . . . , F q } by For example, the spanning simplicial complex of the graph J , given in Fig. 1 is:

Spanning trees of J ,m and Face ring ∆ s (J ,m )
In this section, we give two lemmas which give an important characterization of the graph J ,m and its spanning simplicial complex s(J ,m ). We present a proposition which gives the f -vectors and the dimension of the J ,m . Finally, in Theorem 3.13 we give the formulation for the Hilbert series of the Face ring k ∆ s (J ,m ) .
De nition 3.1. Let C i , C i , ⋯, C i k be consecutive cycles in the Jahangir's graph J ,m . Then the cycle obtained by deleting the common edges between the consecutive cycles C i , C i , ⋯, C i k is a new cycle of the Jahangri's graph The following lemma computes the total number of cycles in the Jahangir's graph J ,m and the cardinality count of the edges in these cycles.
Proof. The Jahangir's graph J ,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 C i , C i , ⋯, C i k is denoted by C i ,i ,⋯,i k . Therefore, we get the following cycles Combining these with m cycles given we have total cycles in the graph J ,m , Now for a xed value of k, simple counting reveals that the total number of cycles C i ,i ,⋯,i k is m for i k < m. Hence the total number of cycles in J ,m is τ . Also it is clear from the construction above that C i ,i ,⋯,i k is obtained by deleting common edges between consecutive cycles C i , C i , ⋯, C i k which are k − in number. Therefore, the order of the cycle C i ,i ,⋯,i k is obtained by adding orders of all C i , C i , ⋯, C i k that is, k and subtracting (k − ) from it, since the common edges are being counted twice in sum. This implies In the following results, we x C u ,u ,⋯,u p , C v ,v ,⋯,v q to represent any two cycles from the cycles such that i j+ = i j + if i j ≠ m and i j+ = if i j = m, of the graph J ,m . Also we x the notation "a → b" if b immediately proceeds a i.e., the very next in order of preferences.
Since the cycles C u ,u ,⋯,u p and C v ,v ,⋯,v q are obtained by deleting the common edges between cycles C u , C u , ⋯, C u p and C v , Hence, the intersection C u ,u ,⋯,u p ⋂ C v ,v ,⋯,v q will contain only the non common edges of the cycle C u ,u ,⋯,u p excluding its two edges common with the cycles on its each end. This gives the order of intersection in this case β u ,u ,⋯,u p − . The remaining cases can be visualized in a similar manner.

Proposition 3.4.
Let J ,m be the graph with the edges E as de ned in eq. (1) Proof. Here, the cycles C u , C u , ⋯, C u σ are amongst σ consecutive adjoining cycles of the cycle C u ,u ,⋯,u p which are also overlapping with the σ consecutive adjoining cycles of the cycle C v ,v ,⋯,v q . If the adjoining cycle C u of the cycle C u ,u ,⋯,u p overlaps with the rst adjoining cycle C v of the cycle C v ,v ,⋯,v q and the adjoining cycles C v q and C u are consecutive then by previous proposition the order of the intersection Similarly, if the adjoining cycles C v q and C u are not consecutive then they will have no common edge and the use of proposition 3.3 gives the order of the intersection C u ,u ,⋯,u p ⋂ C v ,v ,⋯,v q as β u ,u ,⋯,u σ − . Similar can be done for the remaining cases.
Remark 3.5. The case when there exists a t < σ < p such that u t − → u t in above proposition i.e., when cycles C u , C u , ⋯, C u t − , C u t , ⋯, C u σ are not amongst σ consecutive adjoining cycles of the cycle C u ,u ,⋯,u p , the order of the intersection C u ,u ,⋯,u p ⋂ C v ,v ,⋯,v q can be calculated by applying proposition 3.4 on the overlapping portions.
Proof. In this case the adjoining cycles of C u ,u ,⋯,u p and C v ,v ,⋯,v q have no common cycle. However, if the adjoining cycle on one of the extreme ends of the cycle C u ,u ,⋯,u p is consecutive with the adjoining cycles on one of the extreme ends of the other cycle C v ,v ,⋯,v q then the intersection C u ,u ,⋯,u p ⋂ C v ,v ,⋯,v q will have only one edge. The remaining cases are easy to see.
In the following three propositions we give some characterizations of J ,m . We   where, {e j i , e j i , ⋯, e j m i m } will contain exactly one edge from C (j α − )(j α ) ∖ {e (j α − ) , e (j α + ) } other than e j α .
Proof. For a spanning tree of J ,m such that exactly one common edge e j α is removed, we need to remove precisely m − 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 (j α − )(j α ) (since this will result a disconnected graph. This explains the proof of the above case. such that {e j i , e j i , ⋯, e j m i m } will contain exactly exactly one edge from C j r j r ⋯j rρ other than e j r , e j (r ) , ⋯, e j rρ , where j r → j r . 2. if none of e j r , e j r , ⋯, e j rρ are common edges from consecutive cycles then such that for each edge e j r t proposition 3.6 holds. 3. if some of e j r , e j r , ⋯, e j rρ are common edges from consecutive cycles then such that proposition 3.9.1 is satis ed for the common edges of consecutive cycles and proposition 3.9.2 is satis ed for remaining common edges.
Proof. For the case 1, we need to obtain a spanning tree of J ,m such that r ρ − r m common edges must be removed from ρ consecutive cycles C j r , C j r , ⋯, C j rρ . The remaining m − r ρ − r m edges must be removed in such a way that exactly one edge is removed from the non common edges of the adjoining cycles C j r , C j r , ⋯, C j rρ and the remaining m − r ρ − r m cycles of the graph J ,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.
In our next result, we give an important characterization of the f -vectors of ∆ s (J ,m ).
Therefore, by de nition 2.5 we can write ∆ s (J ,m ) = ⟨C J ⋃ C J ⋃ C J a ⋃ C J b ⋃ C J c ⟩. Since each facet . This implies Example 3.12. Let ∆ s (J , ) be a spanning simplicial complex of the Jahangir's graph J ,m given in Figure 1, then the dim(∆ s (J , )) = and τ = = . Therefore, f −vectors f (∆ s (J , )) = (f , f , ⋯, f ) and For a simplicial complex ∆ over [n], one would associate to it the Stanley-Reisner ideal, that is, the monomial ideal I N (∆) in S = k[x , x , . . . , x n ] 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[∆] = S I N (∆) 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 H t (A) of a graded algebra.
Our main result of this section is as follows;