Depth of Binomial Edge Ideals in terms of Diameter and Graph Connectivity]Depth of Binomial Edge Ideals in terms of Diameter and Graph Connectivity
A. V. Jayanthan]A. V. Jayanthan
Department of Mathematics, Indian Institute of Technology Madras, Chennai, INDIA - 600036
Rajib Sarkar]Rajib Sarkar
School of Mathematics, Tata Institute of Fundamental Research Bombay, Mumbai, INDIA - 400005

Let $G$ be a simple connected non-complete graph on $n$ vertices and $J_G$ be its binomial edge ideal. It is known that $d(G)+f(G)\leq \depth(S/J_G) \leq n+2-\kappa(G)$, where $d(G)$ denotes the diameter, $f(G)$ denotes the number of simplicial vertices and $\kappa(G)$ denotes the graph connectivity of $G$. Hibi and Saeedi Madani characterized $G$ such that $d(G)+f(G)=n+2-\kappa(G)$. In this article, we characterize graphs $G$ such that  $d(G)+f(G)+1=n+2-\kappa(G)$ and compute the depth of $S/J_G$ for all such $G$.

Dedicated to Jürgen Herzog on the occasion of his $80^{th}$ birthday

AMS Subject Classification (2010): 13D02, 13C13, 05E40

§ INTRODUCTION

Let $G$ be a simple graph on $[n] = \{1, \ldots, n\}$. The binomial edge ideal corresponding to $G$, which was introduced in [7] and [15] independently, is the ideal $J_G$ generated by $\{x_iy_j - x_jy_i ~:~ \{i, j\} \in E(G)\}$ in the polynomial ring $S=\mathbb{K}[x_1, \ldots, x_n, y_1, \ldots, y_n]$, where $\mathbb{K}$ is a field. The connection between combinatorial invariants associated to $G$ and algebraic invariants associated to $J_G$ has been of intense study in the past one decade (see [1, 3, 4, 6, 9, 10, 14, 18, 17, 19] for a partial list). In [1], Banerjee and Núñez Betancourt proved that for a connected non-complete graph $G$,

$$\depth(S/J_G) \leq n+2-\kappa(G),$$
where $\kappa(G) = \min\{|T| ~:~ T \subset [n] \text{ and } G_{[n]\setminus T} \text{ is disconnected}\}$ is the graph connectivity. For $i, j \in [n],$ let $\dist_G(i,j)$ denote the smallest length of paths connecting $i$ and $j$ in $G$. Let $d(G) = \max\{\dist_G(i,j) ~:~ i, j \in [n]\}$ denote the diameter of $G$. In [17], Rouzbahani Malayeri, Saeedi Madani and Kiani proved that

$$d(G) + f(G) \leq \depth (S/J_G),$$
where $f(G)$ denotes the number of simplicial vertices of $G$. Then the natural question comes that whether there exists a graph $G$ such that $d(G)+f(G)=n+2-\kappa(G)$. Recently, Hibi and Saeedi Madani showed the existence of such graphs and characterized connected non-complete graphs with such property, [9]. In this article, we consider the next case and characterize connected non-complete graphs $G$ with the property that

In Section 3, we first characterize graphs $G$ satisfying the property $d(G)+f(G)+1 = n+2-\kappa(G)$ in terms of these invariants, (<Ref>). We then completely classify these graphs in terms of the structural properties, (<Ref>). For such a graph $G$,  $d(G)+f(G)\leq \depth(S/J_G)\leq d(G)+f(G)+1$. In Section 4, we describe completely the graphs that attain the lower bound or upper bound.

In Section 5, we obtain sufficient conditions for certain chordal graphs not to attain the minimum depth, (<Ref>). For unicyclic and quasi-cycle graphs, we obtain necessary conditions for the depth to be the minimum, (<Ref>).

§ PRELIMINARIES

We recall some basic notation, definitions and some known results which will be used throughout this article.

Let $G$ be a simple graph with the vertex set $V(G)$ and the edge set $E(G)$. A subgraph $H$ of $G$ is said to be an induced subgraph if for $u, v \in V(H)$, $\{u,v\} \in E(H)$ if and only if $\{u,v\} \in E(G)$. If $A \subseteq V(G)$, then $G[A]$ denotes the induced subgraph on the vertex set $A$. For a vertex $v$ of $G$, $G \setminus v$ denotes the induced subgraph on the vertex set $V(G) \setminus \{v\}$. For $e \in E(G)$,  $G\setminus e$ denotes the graph with vertex set $V(G)$ and the edge set $E(G) \setminus \{e\}$.  The graph $G$ is called a chordal graph if every induced cycle of $G$ has length $3$. We denote by $K_m$, a complete graph on $m$ vertices. We recall the definition of the clique sum of two graphs.

Let $G_1$ and $G_2$ be two subgraphs of $G$ such that $G_1\cap G_2\cong K_m$, $V(G)=V(G_1)\cup V(G_2)$ and $E(G)=E(G_1)\cup E(G_2).$ Then $G$ is called the clique sum of $G_1$ and $G_2$ along the complete graph $K_m$ and it is denoted by $G_1\cup_{K_m}G_2$. If $m=1$, i.e., $K_1=\{v\}$, (respectively $m=2$ , i.e., $K_2 = e=\{v_1,v_2\}$), then we denote the clique sum by $G_1 \cup_v G_2$ (respectively $G_1 \cup_e G_2$).

Let $\Delta$ be a simplicial complex. A facet $F$ of $\Delta$ is called a leaf if either it is the only facet of $\Delta$ or there exists a facet  $G$, called a branch of $F$, such that for any facet $H(\neq F)$ of $\Delta$, $H \cap F \subseteq G \cap F$. A simplicial vertex or free vertex of $\Delta$ is a vertex that is contained in only one facet. The simplicial complex $\Delta$ is called a quasi-forest if its facets can be ordered $F_1, \ldots, F_r$, called a leaf order, such that $F_i$ is a leaf of the simplicial complex with facets $F_1, \ldots, F_{i-1}$.

If a graph $G$ is the clique sum of two subgraphs along a simplicial vertex, then $G$ is said to be a decomposable graph. A graph $G$ is called an indecomposable graph if it is not a decomposable graph.

For a graph $G$, a clique is a complete subgraph of $G$. The collection of cliques of $G$ forms a simplicial complex, called the clique complex of $G$, denoted by $\Delta(G)$. A vertex of $G$ is called a simplicial vertex if it is a simplicial vertex of $\Delta(G)$, i.e., a vertex $v$ is a simplicial vertex of $G$ if it is contained in only one maximal clique of $G$. A vertex of $G$ which is not a simplicial vertex is called an internal vertex of $G$ and the number of internal vertices is denoted by $\iv(G)$. For $T\subseteq V(G)$, $c_G(T)$ denotes the number of components of $G\setminus T$. A subset $T\subseteq V(G)$ is said to be a cutset if $c_G(T\setminus \{v\})<c_G(T)$ for every vertex $v\in T$. The set of all cutsets of $G$ is denoted by $\mathcal{C}(G)$. If $T=\{v\}\in \mathcal{C}(G)$, then $v$ is called a cut vertex. If $H$ is a  subgraph of $G$ and $v \in V(G)$, then $\dist(v,H):=\min\{\dist_G(u,v) : u \in V(H)\}$.

We call the collection of graphs $G$ satisfying the property $d(G)+f(G) = |V(G)|+2-\kappa(G)$, Hibi-Madani class. It may be noted that if $G$ belongs to the Hibi-Madani class, then by <cit.>, $\depth(S/J_G) = d(G)+f(G)$.

Let $A=\K[x_1,\dots,x_m]$, $A'=\K[y_1,\dots,y_n]$ and $B=\K[x_1,\dots,x_m,y_1,\dots,y_n]$ be
polynomial rings. Let $I\subseteq A$ and $J\subseteq A'$ be
homogeneous ideals. Then the minimal free resolution of $B/(I+J)B$ can
be obtained by the tensor product of the minimal free resolutions of
$A/I$ and $A'/J$. Therefore, for all $i,j$, we get

\begin{align}\label{Bettiproduct}
\beta_{i,i+j}\left(\frac{B}{(I+J)B}\right) =
\underset{{\substack{i_1+i_2=i \\
\end{align}

We recall the following lemma due to Ohtani.
Let $G$ be a  graph on $V(G)$ and $v\in V(G)$ such that $v$ is an internal vertex. Then $J_G$ can be written as

$$J_{G}=J_{G_v}\cap ((x_v, y_v) + J_{G\setminus v}).$$

Therefore, we have the following short exact sequence:

\begin{align}\label{ohtani-ses}
0\longrightarrow \frac{S}{J_{G}}\longrightarrow
\frac{S}{(x_v,y_v)+J_{G \setminus v}} \oplus \frac{S}{J_{{G_v}}}\longrightarrow \frac{S}{(x_v,y_v)+J_{G_v \setminus v}} \longrightarrow 0
\end{align}

and correspondingly the  long exact sequence of Tor modules:

\begin{align}\label{ohtani-tor}
\cdots & \longrightarrow \Tor_{i}^{S}\left( \frac{S}{J_G},\K\right)_{i+j}\longrightarrow \Tor_{i}^{S}\left( \frac{S}{(x_v,y_v)+J_{G\setminus v}},\K\right)_{i+j}
\oplus \Tor_{i}^{S}\left(\frac{S}{J_{G_v}},\K\right)_{i+j}\nonumber \\ & \longrightarrow \Tor_{i}^{S}\left(\frac{S}{(x_v,y_v)+J_{G_v\setminus v}},\K\right)_{i+j} \longrightarrow \Tor_{i-1}^{S}\left( \frac{S}{J_G},\K\right)_{i+j}\longrightarrow \cdots .
\end{align}

§ GRAPHS WITH $D(G)+F(G)+1=N+2-\KAPPA(G)$

Let $G$ be a simple graph on the vertex set $[n]$ with diameter $d(G)$, number of simplicial vertices $f(G)$ and graph connectivity $\kappa(G)$. The main aim of this article is to characterize graphs $G$ such that
and then to compute $\depth(S/J_G)$.

In <cit.>, the authors proved that for a graph $G$ on $[n]$, if $d(G)\geq 3$ and $\kappa(G)\geq 2$, then $d(G)+f(G) < n+2-\kappa(G)$. We begin with an analogues result.

Let $G$ be a graph on $[n]$. If $d(G) + f(G) + 1 = n+2-\kappa(G)$ and $\kappa(G) \geq 2$, then $d(G) \leq 3$.

Assume that $d(G)+f(G)+1=n+2-\kappa(G)$ and $\kappa(G) \geq 2$. For convenience, we set $d(G)=d, f(G) = f$ and $\kappa(G)=\kappa$. Then $\iv(G)=n-f=(d-2)+(\kappa+1).$ Let $u,v\in V(G)$ be such that $\dist_G(u,v)=d$. By <cit.>, $u$ and $v$ can be joined by $\kappa$ vertex disjoint paths, say

$$ P_i:u,u_1^{i},\dots,u_{j_i-1}^{i},v$$

for $i = 1, \ldots, \kappa$. Note that $u_j^i$'s are internal vertices for all $i,j$. Therefore, for each $1\leq i\leq \kappa$, $j_i\geq d$, counting the internal vertices of these paths, we get $\iv(G)\geq (d-1)\kappa$. Therefore, $(d-1)\kappa\leq (d-2)+(\kappa+1)$. This implies that $(d-2)(\kappa-1)\leq 1$. Hence $d \leq 3$.

Let $d\geq 5$ and $f\geq 3$ be integers. Let $P_{d+1}$ be the path on the vertex set $\{1,\dots,d+1\}$ and $K_{1,f-1}$ be the complete bipartite graph on the vertex set $\{d+2,\dots,d+f+1\}$. Let $\Lambda=P_{d+1}\cup_{3=d+f+1}K_{1,f-1}$ be the clique-sum of $K_{1,f-1}$ and $P_{d+1}$ along the vertex $3=d+f+1$. Then clearly, $\kappa(\Lambda)=1$, $d(\Lambda)=d$, $f(\Lambda)=f$, and $f(\Lambda)+d(\Lambda)+1=n+2-\kappa(\Lambda)$, where $|V(\Lambda)|=n$.

(-2.5,0.78)– (-1.14,0.78);
(-1.14,0.78)– (0.16,0.78);
(0.16,0.78)– (1.64,0.78);
(3.74,0.78)– (4.94,0.78);
(0.16,1.66)– (0.16,0.78);
(0.16,1.66)– (-1.14,1.66);
(0.16,1.66)– (-0.9,2.5);
(0.16,1.66)– (0.86,2.7);

(-2.5,0.78) circle (1.5pt);
(-2.34,1.04) node $1$;
(-1.14,0.78) circle (1.5pt);
(-0.98,1.04) node $2$;
(0.16,0.78) circle (1.5pt);
(0.85,1.04) node $3=d+f+1$;
(1.64,0.78) circle (1.5pt);
(1.8,1.04) node $4$;
(2.42,0.78) circle (1pt);
(2.86,0.78) circle (1pt);
(3.16,0.78) circle (1pt);
(3.74,0.8) circle (1.5pt);
(3.9,1.04) node $d$;
(4.94,0.78) circle (1.5pt);
(5.04,1.04) node $d+1$;
(0.16,1.66) circle (1.5pt);
(0.55,1.66) node $d+2$;
(-1.14,1.66) circle (1.5pt);
(-0.92,1.94) node $d+3$;
(-0.9,2.5) circle (1.5pt);
(-0.8,2.62) node $d+4$;
(-0.5,2.84) circle (1pt);
(-0.17,2.9) circle (1pt);
(0.4,2.9) circle (1pt);
(0.86,2.7) circle (1.5pt);
(1.3,2.6) node $d+f$;

The following example shows that $\kappa(G) \geq 2$ can not be removed from the previous lemma.

Let $f \geq 3$ and $d \geq 2$ be integers. Let $H=K_f\cup_e K_3$. Let $v\in e$ and $G=H\cup_v P_{d}$. Then it follows that $\kappa(G)=1$, $d(G)=d$, $f(G)=f$ and $d(G)+f(G)+1=n+2-\kappa(G)$.

This shows that if $\kappa(G)=1$, there the equality does not impose any condition on $d(G)$, implying that the condition $\kappa(G) \geq 2$ is necessary in <Ref>.

(-2.06,2.56)– (-0.92,2.56);
(-0.92,2.54)– (-0.92,1.86);
(-0.92,1.86)– (-2.06,1.86);
(-2.06,2.56)– (-2.06,1.86);
(-2.06,1.86)– (-0.92,2.54);
(-2.06,2.56)– (-0.96,1.86);
(-0.92,1.86)– (-1.45,1.14);
(-2.08,1.86)– (-1.45,1.14);
(-0.92,1.86)– (0.26,1.86);
(0.26,1.86)– (1.56,1.86);
(1.56,1.86)– (2.88,1.86);

(-2.06,2.56) circle (1.5pt);

(-0.92,2.56) circle (1.5pt);
(-0.76,1.7) node $v$;
(-1.45,1.7) node $e$;
(-2.06,1.86) circle (1.5pt);

(-0.92,1.86) circle (1.5pt);

(-1.45,1.14) circle (1.5pt);

(0.26,1.86) circle (1.5pt);

(1.56,1.86) circle (1.5pt);

(2.88,1.86) circle (1.5pt);
(0.5,1.2) node $G$;

Our first aim is to characterize graphs $G$ satisfying the property $d(G)+f(G)+1 = n+2-\kappa(G)$. It can be easily verified that if $n \leq 4$, then the above condition is never met. So, we assume that $n \geq 5$.

Given integers $n\geq 5$, $\kappa, f, d  \in \mathbb{N}$, satisfying
there exists a connected graph $G$ on $[n]$ with
$f(G)=f,d(G)=d$ and $\kappa(G)=\kappa $
if and only if these integers satisfy one of the following conditions:

* $\kappa =1, f\geq 2$ and $d \geq 2,$

* $\kappa =2, f \geq 2$ and $d \in \{2, 3\},$

* $\kappa \geq 3, f \geq 2$ and $d=2.$

First, observe that for a graph $G$, $d(G)=1$ if and only if $G$ is a complete graph for which the equality $d+f+1=n+2-\kappa$ does not hold. Therefore in our case $d\geq2$.

Assume that there exists a graph $G$ such that $d+f+1=n+2-\kappa$. Let $u, v \in V(G)$ such that $\dist_G(u,v) = d$ and let $P$ be a path of length $d$ joining $u$ and $v$.

Suppose $\kappa = 1$. If possible, let us assume that $f= 1$. Then $n=d+1$ and $\iv(G)=d$. Then $V(G) = V(P)$ and either $u$ or $v$ is an internal vertex. But this is not possible since $V(G)=V(P)$. Hence $f\neq 1$. Thus $\kappa =1, f\geq 2$ and $d \geq 2$. It may also be noted that the case $f = 2, d=2$ and $\kappa = 1$ is not possible since this would force $n = 4$.

Now suppose $\kappa = 2$. Then it follows from <Ref> that $d \leq 3$. Suppose $f = 1$. In this case, $n=d+2$ and $\iv(G)=d+1$. So there is only one vertex in $V(G)\setminus V(P).$ Also by <cit.>, there exist two vertex disjoint paths of length at least $d$ from $u$ to $v$. If $d = 3$, then $n \geq d+3$ which is not possible. Hence $d=2$. But then, $n=4$ which is a contradiction to our assumption that $n \geq 5$. Hence $f \geq 2$. Therefore, $\kappa =2, f \geq 2$ and $d \in \{2, 3\}$.

Assume that $\kappa \geq 3$. By <Ref>, $d \leq 3$. We need to show  that $f \geq 2$ and $d \neq 3$.

If possible, let $d = 3$. Then $\iv(G)=\kappa+2$. By <cit.>, there exist $\kappa$ vertex disjoint paths of length at least $3$ from $u$ to $v$. Each of these paths has at least two internal vertices of $G$. This implies that $2\kappa \leq \kappa+2$, and so $\kappa \leq 2$. This is a contradiction. Hence, $d = 2$. Now assume that $f = 1$. Then $d=2$, $\iv(G)=\kappa+1$ and $n=\kappa+2$. For $1 \leq i \leq \kappa$, let $P_i:u,u^i,v$ be a path from $u$ to $v$. Note that $u^1, u^2, \ldots, u^\kappa$ are distinct internal vertices. Therefore, precisely one among $u$ and $v$ is a simplicial vertex. For $u$ to be a simplicial vertex, $\{u, u^1, u^2, \ldots, u^\kappa\}$ must be a clique. This would imply that $\{v,u^1, u^2, \ldots, u^\kappa\}$ is a clique and consequently $v$ is also a simplicial vertex. This is a contradiction. Hence $f \neq 1$. Therefore $\kappa \geq 3, f\geq 2$ and $d =2$.

Conversely, suppose $(1), (2)$ or $(3)$ of the statement of the theorem is satisfied by $\kappa, f$ and $d$. Then the existence of a graph $G$ satisfying $d + f + 1  = n+2-\kappa$, follows from Examples <ref>, <ref>, <ref>, <ref> and <ref>.

Let $d\geq 3$ and $\Sigma=K_{3}\cup_{e}K_{3}\cup_{v} P_{d-1}$, where $v\notin e$. 	Then  $|V(\Sigma)|=d+2$, $\kappa(\Sigma)=1$, $d(\Sigma)=d$,$f(\Sigma)=2$  and $d(\Sigma)+f(\Sigma)+1=n+2-\kappa(\Sigma)$.

The graph $\Sigma$, given on the right, is $\Sigma=K_3\cup_{e} K_3\cup_{v} P_3$. Note that $\kappa(\Sigma)=1$, $f(\Sigma)=2$, $d(\Sigma)=4$, and $d(\Sigma)+f(\Sigma)+1=n+2-\kappa(\Sigma)$.

(-2.24,3.56)– (-1,3.56);
(-1,3.56)– (-1,2.7);
(-1,2.7)– (-2.24,2.7);
(-2.24,2.7)– (-2.24,3.56);
(-2.24,2.7)– (-1,3.54);
(-1,2.7)– (0.16,2.7);
(0.16,2.7)– (1.34,2.7);

(-2.24,3.56) circle (1.5pt);

(-1,3.56) circle (1.5pt);
(-1.7,3.2) node $e$;
(-2.24,2.7) circle (1.5pt);

(-1,2.7) circle (1.5pt);
(-0.84,2.9) node $v$;
(0.16,2.7) circle (1.5pt);

(1.34,2.7) circle (1.5pt);

(-0.7,2.5) node $\Sigma $;

Let $f\geq 2$ and $\Gamma=K_{3}\cup_{e}K_{f+1}\cup_{e'} K_{3}, \text{ where }  \mid e\cap e' \mid ~= 1.$ 		Then  $|V(\Gamma)|=f+3$, $\kappa(\Gamma)=2$, $d(\Gamma)=2$,$f(\Gamma)=f$  and $d(\Gamma)+f(\Gamma)+1=n+2-\kappa(\Gamma)$.

The graph $\Gamma$, given on the right, is $\Gamma=K_3\cup_{e} K_4\cup_{e'} K_3$. Note that $\kappa(\Gamma)=2$, $f(\Gamma)=3$, $d(\Gamma)=2$, and $d(\Gamma)+f(\Gamma)+1=n+2-\kappa(\Gamma)$.

(-1.5,3.72)– (-0.48,4.18);
(-0.48,4.18)– (-0.48,3.26);

(-1.5,3.72)– (-0.48,3.26);

(-0.48,4.18)– (0.86,4.18);
(0.86,4.18)– (0.86,3.26);
(0.86,3.26)– (-0.48,3.26);
(-0.48,3.26)– (0.86,4.18);
(-0.48,4.18)– (0.86,3.26);
(0.86,3.26)– (0.18,2.34);
(-0.48,3.26)– (0.18,2.34);

(-1.5,3.72) circle (1.5pt);

(-0.48,4.18) circle (1.5pt);

(-0.48,3.26) circle (1.5pt);
(-0.6,3.75) node $e$;
(0.86,4.18) circle (1.5pt);

(0.86,3.26) circle (1.5pt);
(0.25,3.1) node $e'$;
(0.18,2.34) circle (1.5pt);
(-0.5,2.6) node $\Gamma$;

Let $f\geq 2$ and $\Omega=K_{3}\cup_{e}K_{f+2}\cup_{e'} K_{3}, \text{ where } e\cap e'=\emptyset .$
Then $|V(G)|=f+4$, $\kappa(G)=2$, $d(\Omega)=3$, $f(\Omega)=f$, and $d(\Omega)+f(\Omega)+1=n+2-\kappa(\Omega)$.

The graph $\Omega$, given on the right, is $\Omega=K_3\cup_{e} K_4\cup_{e'} K_3$. Note that $\kappa(\Omega)=2$, $f(\Omega)=2$, $d(\Omega)=3$, and $d(\Omega)+f(\Omega)+1=n+2-\kappa(\Omega)$.

(-1.5,3.72)– (-0.48,4.18);
(-0.48,4.18)– (-0.48,3.26);

(-1.5,3.72)– (-0.48,3.26);

(-0.48,4.18)– (0.86,4.18);
(0.86,4.18)– (0.86,3.26);
(0.86,3.26)– (-0.48,3.26);
(-0.48,3.26)– (0.86,4.18);
(-0.48,4.18)– (0.86,3.26);

(0.86,4.18)– (1.8,3.72);

(0.86,3.26)– (1.8,3.72);

(-1.5,3.72) circle (1.5pt);

(-0.48,4.18) circle (1.5pt);

(-0.48,3.26) circle (1.5pt);
(-0.65,3.7) node $e$;
(0.86,4.18) circle (1.5pt);

(0.86,3.26) circle (1.5pt);
(1.05,3.7) node $e'$;

(0.25,3) node $\Omega$;
(1.8,3.72) circle (1.5pt);

Let $f\geq 2,\kappa \geq 3$ and
$\Delta =K_{\kappa +1} \cup_{K_\kappa }K_{f+\kappa -1}\cup_{K'_\kappa }K_{\kappa +1}, \text{ where } |V(K_\kappa )\cap V(K'_\kappa)|=\kappa-1.$
Note that $V(\Delta)=f+\kappa +1$, $d(\Delta)=2$, $f(\Delta)=f$, $\kappa(\Delta)=\kappa$ and $d(\Delta)+f(\Delta)+1=n+2-\kappa(\Delta).$

The graph, given on the right, is $\Delta=K_4\cup_{K_3} K_4\cup_{K'_3}  K_4$, where $V(K_3)=\{i,j,l\}$ and $V(K'_3)=\{j,k,l\}$. Note that $\kappa(\Delta)=3$, $f(\Delta)=2$, $d(\Delta)=2$, and $d(\Delta)+f(\Delta)+1=n+2-\kappa(\Delta)$.

(-1.5,3.72)– (-0.48,4.18);
(-0.48,4.18)– (-0.48,3.26);
(-1.5,3.72)– (-0.48,3.26);
(-0.48,4.18)– (0.86,4.18);
(0.86,4.18)– (0.86,3.26);
(0.86,3.26)– (-0.48,3.26);
(-0.48,3.26)– (0.86,4.18);
(-0.48,4.18)– (0.86,3.26);
(0.86,4.18)– (1.8,3.72);
(0.86,3.26)– (1.8,3.72);
(-1.5, 3.72)– (0.86,4.18);
(1.8, 3.72)– (-0.48,3.26);

(-1.5,3.72) circle (1.5pt);
(-0.48,4.18) circle (1.5pt);
(-0.32,4.42) node $i$;
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(0.86,4.18) circle (1.5pt);
(1.06,4.42) node $l$;
(0.86,3.26) circle (1.5pt);
(1.05,3) node $k$;
(0.2,3) node $\Delta$;
(1.8,3.72) circle (1.5pt);

Using <Ref>, we now give some necessary conditions for a graph to satisfy the property $d(G)+f(G)+1 = n+2-\kappa(G)$.

Let $G$ be a graph on $[n]$ such that $d(G)+f(G)+1=n+2-\kappa(G).$

* If $\kappa(G)=1$, then either $G$ is chordal or $G$ has precisely one induced $C_4$ and has no induced $C_l$ for $l \geq 5$.

* If $\kappa(G)\geq 2$ and $d(G)=2$, then $G$ is a chordal graph.

* If $\kappa(G)=2$ and $d(G)=3$, then either $G$ is  chordal or $G$ has precisely one induced $C_4$ and has no induced $C_l$ for $l \geq 5$.

First we treat the case when $\kappa(G)=1$. If $d(G) + f(G) + 1 = n+2-\kappa(G)$, then $\kappa(G) =  1$ if and only $d(G) + f(G) = n$.  Hence we study graphs $G$ with $d(G)+f(G)=n$. Set $d=d(G)$ and $\kappa = \kappa(G)$. Then, $\iv(G)=n-f(G)=d.$ Let $u,v\in V(G)$ be such that $\dist_G(u,v)=d$. Let $P:u=u_0,u_1,\dots,u_{d-1},u_d=v$ be an induced path from $u$ to $v$. Since $u_1, \ldots, u_{d-1}$ are internal vertices, there is precisely one more internal vertex in $[n]\setminus \{u_1,\dots,u_{d-1}\}.$ This can happen in two ways:

Case-I: Suppose either $u$ or $v$ is an internal vertex. Without loss of generality we assume that $u$ is an internal vertex.
Suppose $G$ contains an induced cycle of length $l\geq 4$, say $C_l$. Let $V(C_l)=\{v_1,\dots,v_l\}$. Since $v_i$'s are internal vertices, $\{v_1,\dots,v_l\}\subseteq \{u,u_1,\dots,u_{d-1}\}$. This implies that either there exists $3\leq i \leq d-1$ such that $\{u,u_i\}\in E(G)$ or there exist integers $j, k$, $k > j+2$ such that $\{u_j, u_k\} \in E(G)$. In either case,  this would force that $\dist_G(u,v) < d$ which is a contradiction. Therefore, $G$ can not contain an induced cycle of  length $\geq 4$. So, $G$ is a chordal graph.

Case-II: Suppose neither $u$ nor $v$ is an internal vertex. In this case we prove that either $G$ is chordal or $G$ can have at most one induced cycle of length 4 and can have no induced cycle of length bigger than 4. Assume that $G$ is not chordal. Suppose $G$ contains an induced cycle of length of $l\geq 5$, say $C_l$. Let $V(C_l)=\{v_1,\dots,v_l\}$. Since $v_i$'s are internal vertices, $u_1, \ldots, u_{d-1}$ is a path and outside $V(P)$ there is exactly one internal vertex, $|\{v_1,\dots,v_l\}\cap \{u_1,\dots,u_{d-1}\}|=l-1$. Without loss of generality, we may assume that $v_l\notin \{u_1,\dots,u_{d-1}\}$ and $u_i = v_i$ for $i=1, \ldots, l-1$.

Then $P':u,v_1,v_l,v_{l-1},u_l,\ldots,u_{d-1},v$ is a path of length $d-(l-2)+2= d-l+4 \leq d-1$ from $u$ to $v$ which contradicts the assumption that $\dist_G(u,v) = d$.

(-2.6,1.5)– (-0.5,1.5);
(-0.5,1.5)– (1.52,1.5);
[dotted] (1.52,1.5)– (3.46,1.5);
(5.24,1.5)– (3.46,1.5);
(7.4,1.5)– (9.12,1.5);
(-0.5,1.5)– (2.42,3.08);
(2.42,3.08)– (5.24,1.5);
[dotted] (5.3, 1.5) – (7.3,1.5);

(-2.6,1.5) circle (1.5pt);
(-2.44,1.7) node $u$;
(-0.5,1.5) circle (1.5pt);
(-0.7,1.7) node $u_1$;
(1.52,1.5) circle (1.5pt);
(1.68,1.7) node $u_2$;
(3.46,1.5) circle (1.5pt);
(3.62,1.7) node $u_{l-2}$;
(5.24,1.5) circle (1.5pt);
(5.4,1.7) node $u_{l-1}$;

(7.4,1.5) circle (1.5pt);
(7.54,1.7) node $u_{d-1}$;
(9.12,1.5) circle (1.5pt);
(9.28,1.7) node $v$;
(2.42,3.08) circle (1.5pt);
(2.58,3.34) node $v_l$;

Therefore, $G$ does not have an induced cycle of length $5$ or more.
Now if possible assume that $G$ has at least two induced cycles of length 4. Since $G$ has one internal vertex belonging to $[n]\setminus \{u_1,\dots,u_{d-1}\}$, all the vertices of one of these two cycles is a subset of $\{u_1,\dots,u_{d-1}\}$. This would again force that $\dist_G(u,v) \leq d-2$, which is a contradiction. Hence if $G$ is not chordal, then $G$ contains at most one induced $C_4$. This completes the proof of $(1)$.

Now we treat the case $\kappa(G)\geq 2$. Then it follows from <Ref> that $2\leq d\leq 3$, and $\iv(G)=n-f(G)=d-1+\kappa.$ Suppose $d = 2$. Then $\iv(G)=\kappa+1 $. Assume that $G$ has an induced cycle $C_l$ $(l\geq 4)$ with $V(C_l)=\{v_1,\dots,v_l\}$. By <cit.>, there exist $\kappa$ vertex-disjoint paths of length $\geq 2$ from $v_1$ to $v_3$. Each such path contains at least one internal vertex. Since $v_1$ and $v_3$ are vertices of an induced cycle of length $\geq 4$, these are internal vertices as well. Hence $\iv(G) \geq \kappa + 2$. This is a contradiction. Hence, $G$ is chordal. This proves $(2)$.

Assume that $d=3$. Then it follows from Theorem <ref> that $\kappa=2$, and so $\iv(G)=4$. Since all the vertices of cycles are internal vertices, $G$ has no induced cycle of length $\geq 5$ and $G$ can have at most one induced cycle of length 4. This completes the proof of the theorem.

§ DEPTH

In this section, we compute the depth of $S/J_G$ for graphs satisfying $d(G)+f(G)+1 = n+2-\kappa(G)$. In this computation, the graphs $G_v, G_v \setminus v$ and $G\setminus v$, for some internal vertex $v$, play important roles via the Ohtani Lemma, (<Ref>) and (<ref>). We begin this section with a couple of technical lemmas which are useful in proving the main results.

Let $G$ be a graph and $v$ be an internal vertex. Then,

* if $G \setminus v$ is connected, then $d(G) \leq d(G\setminus v)$;

* $f(G) \leq f(G\setminus v)$.

For any two vertices $z, z'$ in $G\setminus v$, a shortest induced path in $G\setminus v$ joining $z$ and $z'$ will be a path in $G$, but need not necessarily be an induced path. This implies that $\dist_{G}(z,z') \leq \dist_{G\setminus v}(z,z')$. Hence $d(G) \leq d(G\setminus v)$. This proves (1).

(2) follows directly from <cit.>.

We now prove a result connecting the diameter of a graph $G$ and the diameter of $G_v$ for some internal vertex $v$.

Let $G$ be any graph and $v$ be any internal vertex in $G$. Then for any two vertices $z,z'\in V(G)$,
$$\dist_G(z,z')-1\leq \dist_{G_v}(z,z') \leq \dist_{G}(z,z').$$

In particular, $d(G)-1 \leq d(G_v) \leq d(G).$

Note that an induced path in $G$ need not necessarily be an induced path in $G_v$. Thus, for any $z,z'\in V(G)$, $\dist_{G_v}(z,z')\leq \dist_{G}(z,z')$.

Let $P:z=z_1,\dots,z_{d'},z_{d'+1}=z'$ be a shortest induced path from $z$ to $z'$ in $G_v$.

If $E(P)\subseteq E(G)$, then $P$ remains an induced path in $G$. Hence, $\dist_G(z,z')\leq d'$. Observe that all the edges in $E(G_v)\setminus E(G)$ belong to a single clique in $G_v$. If $E(P)\nsubseteq E(G)$, then there exists only one edge $e\in E(P)\setminus E(G)$. Let $e=\{z_i,z_{i+1}\}$ for some $1\leq i\leq d'$ and $z_i,z_{i+1}\in N_G(v)$. Hence, in $G$, we have an induced path $z=z_1, \ldots, z_i, v, z_{i+1}, \ldots, z_{d'}=z'$ in $G$. Therefore $\dist_G(z,z') \leq d'+1$.

It is easy to observe that the structure of $G_v$ and $G_v\setminus v$ are similar and the inequalities in the above lemma remain the same if we replace $G_v$ by $G_v \setminus v$.

Let $G$ be a graph on $[n]$ such that $d(G)+f(G)=n+2-\kappa(G)$. Hibi and Saeedi Madani classified these graphs into two collection, $\mathcal{G}_d$ and $\mathcal{F}_q$ (see <cit.>). It follows from <cit.> and Auslander-Buchsbaum formula that if $G$ is either in $\mathcal{G}_d$ or in $\mathcal{F}_q$, then $\pd(S/J_G)=n-2+\kappa(G)$. To understand the depth of graphs satisfying $d(G)+f(G)+1=n+2-\kappa(G)$, we need to understand certain extremal Betti numbers of graphs in the Hibi-Madani class.

Let $G$ be a graph $[n]$ such that $d(G)+f(G)=n+2-\kappa(G)$. Then $\beta_{n-2+\kappa(G),n-2+\kappa(G)+d(G)}(S/J_G)$ is an extremal Betti number of $S/J_G$.

First we consider the case when $\kappa(G)=1$. Then $G\in \mathcal{G}_{d(G)}$. We proceed by induction on $d(G)$. If $d(G) = 2$, then $G$ is a block graph and the assertion follows from <cit.>, noting that $\iv(G)=d(G)-1$. Assume that $d(G)\geq 3$. Let $w$ be an internal vertex. Then it follows from <Ref> that $d(G_w) \geq d(G)-1$. From <cit.>, we get $f(G_w) \geq f(G)+1$. Therefore, $n+1=d(G)+f(G) \leq d(G_w)+f(G_w)\leq n+1$. The same way we can see that $d(G_w\setminus w) + f(G_w\setminus w) = n$. Hence $G_w,G_w\setminus w\in \mathcal{G}_{d-1}$. Also, $G\setminus w$ is a disconnected graph where each component of $G\setminus w$ is either a complete graph or belongs to $\mathcal{G}_i$ for some $i\leq d-2$. Therefore, $\depth(S/J_{G_w})=n+1,\depth(S/((x_w,y_w)+J_{G_w\setminus w}))=n$ and $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$. Hence by Auslander-Buchsbaum formula, we have $\pd(S/J_{G_w})=n-1,\pd(S/((x_w,y_w)+J_{G_w\setminus w}))=n$ and $\pd(S/((x_w,y_w)+J_{G\setminus w}))\leq n-1$. Also, note that $d(G_w)=d(G_w\setminus w)=d(G)-1$.  Therefore, by induction, $\beta_{n-1,n-1+d(G_w)}(S/J_{G_w})$, $\beta_{n,n+d(G_w\setminus w)}(S/((x_w,y_w)+J_{G_w\setminus w}))$ are extremal Betti numbers.

If $\pd(S/((x_w,y_w)+J_{G\setminus w})) < n-1$, then $\beta_{n-1,j}(S/((x_w,y_w)+J_{G\setminus w})) = 0$ for all $j$. If $\pd(S/((x_w,y_w)+J_{G\setminus w}))= n-1$, then $G\setminus w$ has precisely two components, say $G_1$ and $G_2$. Set $|V(G_1)|=n_1$ and $|V(G_2)|=n_2$. Since $d(G_1),d(G_2) < d$, by applying induction on $G_1$ and $G_2$, we get that $\beta_{n_i-1,n_i-1+d(G_i)}(S_{G_i}/J_{G_i})$ is an extremal Betti number for $i=1,2$. Observe that $d(G_1)+d(G_2)\leq d(G)$. Hence by (<ref>),  $\beta_{n-1,n-1+j}(S/((x_w,y_w)+J_{G\setminus w}))=0$ for $j\geq d(G)+1$. Therefore, it follows from (<ref>) for $i=n-1$ and $j=d(G)$ that

$$\beta_{n-1,n-1+d(G)}(S/J_G)\neq 0 \text{ and } \beta_{n-1,n-1+j}(S/J_G)=0 \text{ for } j\geq d(G)+1. $$

Since $\pd(S/J_G)=n-1$, $\beta_{n-1,n-1+d(G)}(S/J_G)$ is an extremal Betti number of $S/J_G$.

Now we consider $\kappa(G)\geq 2$. Then $G\in \mathcal{F}_{\kappa(G)}$ with $d(G)=2$. Take $w$ to be an internal vertex in $G$. Then it can be noted that $G_w, G_w\setminus w$ are complete graphs on $n$ and $n-1$ vertices respectively. So, $\beta_{n-1,n}(S/J_{G_w})$ and $\beta_{n,n+1}(S/((x_w,y_w)+J_{G_w\setminus w}))$ are extremal Betti numbers. We proceed by induction on $\kappa(G)$. Assume that $\kappa(G)=2$. Then $G\setminus w$ is a block graph on $n-1$ vertices, and hence by <cit.>, $\beta_{n,n+2}(S/((x_w,y_w)+J_{G\setminus w}))$ is an extremal Betti number as $\iv(G\setminus w)=1$. Therefore it follows from the long exact sequence (<ref>) that $\beta_{n,n+2}(S/J_G)$ is an extremal Betti number of $S/J_G$. Suppose now $\kappa(G)\geq 3$. In this case, note that $G\setminus w\in \mathcal{F}_{\kappa(G)-1}$. Therefore, by induction and (<ref>), $\beta_{n-1+\kappa(G\setminus w),n-1+\kappa(G\setminus w)+d(G\setminus w)}(S/((x_w,y_w)+J_{G\setminus w}))$ is an extremal Betti number. Since $\kappa(G\setminus w)=\kappa(G)-1$ and $d(G\setminus w)=d(G)$, the assertion follows from the long exact sequence (<ref>).

For the rest of the section, $G$ denotes a connected non-complete graph on $[n]$ with $d(G)+f(G)+1= n+2-\kappa(G)$. We now describe the structure of these graphs in more details and then compute the depth of $S/J_G$.

Assume that $\kappa(G)=1$. Then $d(G)+f(G)=n$. Let $u, v \in V(G)$ such that $\dist_G(u,v) = d(G) =d$. Let $P: u, u_1, \ldots, u_{d-1}, v$ be an induced path from $u$ to $v$. Since $\iv(G) = n-f(G) = d(G)$ and $u_1, \ldots, u_{d-1}$ are internal vertices, there  is precisely one internal vertex among $[n]\setminus \{u_1, \ldots, u_{d-1}\}$. This can occur in the following ways:

* either $u$ or $v$ is an internal vertex;

* there is an internal vertex $v' \in [n] \setminus V(P)$.

If either $u$ or $v$ is an internal vertex, then it follows from the proof of <Ref> that $G$ is a chordal graph having all its internal vertices contained in $V(P)$. If both $u$ and $v$ are simplicial vertices, then there is a unique internal vertex in $[n]\setminus V(P)$, say $v'$. Since $v'$ is the unique internal vertex in $[n] \setminus V(P)$, $\dist(v', P) = 1$. Let $j = \min\{\ell : \{u_\ell, v'\} \in E(G)\}$. It may also be noted that $\max\{k : u_k \in N_P(v')\} \leq j+2$, since otherwise it will contradict the assumption that $\dist_G(u,v) = d$. Hence $N_P(v') \subseteq \{u_j, u_{j+1},u_{j+2}\}$.
Consider the following graph $H$:

[dotted] (-6, 0) – (-5,0);
[dotted] (-3, 0) – (-2,0);
(-7,0)– (-6,0);
(-5,0)– (-4,0);
(-4,0)– (-3,0);
(-4,1)– (-5,0);
(-4,1)– (-4,0);
(-4,1)– (-3,0);
(-2,0)– (-1,0);

(-7,0) circle (1.5pt);
(-6.84,-0.3) node $u$;
(-6,0) circle (1.5pt);
(-5.84,-0.3) node $u_1$;
(-5,0) circle (1.5pt);
(-4.84,-0.3) node $u_j$;
(-4,0) circle (1.5pt);
(-3.84,-0.3) node $u_{j+1}$;
(-3,0) circle (1.5pt);
(-2.84,-0.3) node $u_{j+2}$;
(-2,0) circle (1.5pt);
(-1.84,-0.3) node $u_{d-1}$;
(-1,0) circle (1.5pt);
(-0.84,-0.3) node $v$;
(-4,1) circle (1.5pt);
(-4,1.3) node $v'$;

Note that if either $u$ or $v$ is an internal vertex of $G$, then $H$ cannot be an induced subgraph of $G$.

(-0.65,1)– (-1.3,0);
(0,0)– (-0.65,1);
(-1.3,0)– (-0.65,-0.8);
(-0.65,-0.8)– (0,0);
(-1.3,0)– (0,0);
(0,0)– (1.3,0);

(2.6,0)– (3.9,0);
(1.3,0)– (0.65,-0.8);
(1.3,0)– (1.95,-0.8);
(2.6,0)– (3.28,-0.64);
(3.9,0)– (3.28,-0.64);
(2.6,1)– (2.6,0);
(2.6,1)– (3.25,1);
(3.25,1)– (2.6,0);
(2.6,0)– (1.95,1);
(0,1)– (1.3,1);
(1.3,1)– (1.3,0);
(0,0)– (0,1);
(0,0)– (1.3,1);
(0,1)– (1.3,0);
[dotted] (1.3, 0) – (2.6,0);

(0,0) circle (1.5pt);
(0.16,-0.2) node $u_1$;
(-1.3,0) circle (1.5pt);
(-1.4,-0.2) node $u$;
(1.3,0) circle (1.5pt);
(1.7,-0.2) node $u_2$;
(2.6,0) circle (1.5pt);
(2.4,-0.2) node $u_{d-1}$;
(3.9,0) circle (1.5pt);
(4.06,0.26) node $v$;
(-0.65,1) circle (1.5pt);

(-0.65,-0.8) circle (1.5pt);

(0,1) circle (1.5pt);

(1.3,1) circle (1.5pt);

(0.65,-0.8) circle (1.5pt);

(1.95,-0.8) circle (1.5pt);

(1.95,1) circle (1.5pt);

(2.6,1) circle (1.5pt);

(3.25,1) circle (1.5pt);

(3.28,-0.64) circle (1.5pt);

$u$ is an internal vertex

If both $u$ and $v$ are simplicial vertices, then $H \setminus \left\{ \{v', u_{j+1}\}, \{v', u_{j+2}\}\right\}$ is necessarily an induced subgraph of $G$. It follows from <cit.> and <cit.> that $\depth (S_H/J_{H}) = |V(H)|.$
\[
\mathcal{D} := \big\{ G ~: G \text{ is chordal}, \kappa(G) = 1 \text{ and } d(G)+f(G) = n \big\}
\]

\[
\mathcal{D}_1 := \left\{ G \in \mathcal{D}:
\begin{array}{l}
H \text{ is an induced subgraph of } G \text{ and }
\{v'\}, \{u_{j+1}\} \notin \mathcal{C}(G)
\end{array}
\right\}.\]

For the benefit of the readers, we illustrate the graphs using some figures.

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,4.24);
(-2,3.16)– (-0.5,3.16);
(-2,4.24)– (-0.5,4.24);
(-0.5,4.24)– (-0.5,3.16);

(-2,3.16)– (-0.5,4.24);
(-0.5,4.24)– (1,3.16);

(-2.4,4.16) circle (1.5pt);
(-3,3.56) circle (1.5pt);
(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-1.7,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2,4.24) circle (1.5pt);
(-0.5,4.24) circle (1.5pt);
(-0.2,4.5) node $v'$;

$G_1\in \mathcal{D}_1$

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,4.24);
(-3.5,3.16)– (2.5,3.16);
(-2,3.16)– (-0.5,3.16);
(-2,4.24)– (-0.5,4.24);
(-0.5,4.24)– (-0.5,3.16);
(-2,4.24)– (-0.5,3.16);
(-2,3.16)– (-0.5,4.24);
(-0.5,4.24)– (1,3.16);
(-0.5,3.16)– (-0.5,2.32);
(-0.5,2.32)– (0.38,2.32);
(-0.5,3.16)– (0.38,2.32);
(-0.5,3.16)– (-1.18,2.52);

(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-1.7,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3.4) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2,4.24) circle (1.5pt);

(-0.5,4.24) circle (1.5pt);

(-0.5,2.32) circle (1.5pt);

(-1.18,2.52) circle (1.5pt);
(-0.2,4.5) node $v'$;

(0.38,2.32) circle (1.5pt);

$G_2\in \mathcal{D}\setminus \mathcal{D}_1$ with $H$ as an induced subgraph and $\{u_2\}\in \mathcal{C}(G)$

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,2.14);
(-2,2.14)– (-0.5,2.14);
(-0.5,3.16)– (-0.5,2.14);
(-3.5,3.16)– (-2.72,4.24);
(-2.72,4.24)– (-2,3.16);
(-2,3.16)– (-2,4.24);
(-2,3.16)– (-1.2,4.24);
(-1.2,4.24)– (-0.5,3.16);
(-0.5,3.16)– (0.3,4.24);
(0.3,4.24)– (1.18,4.24);
(0.3,4.24)– (1.1,5.08);
(1.1,5.08)– (0.32,5.08);
(0.32,5.08)– (0.3,4.24);
(0.3,4.24)– (-0.44,5.08);
(-2,3.16)– (-0.5,2.14);
(-2,2.14)– (-0.5,3.16);
(-3.5,3.16)– (2.5,3.16);
(-2,3.16)– (-0.5,3.16);

(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-2.2,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2.72,4.24) circle (1.5pt);

(-2,2.14) circle (1.5pt);

(-0.5,2.14) circle (1.5pt);

(-1.2,4.24) circle (1.5pt);

(-2,4.24) circle (1.5pt);

(0.3,4.24) circle (1.5pt);
(0.3,4) node $v'$;
(-0.44,5.08) circle (1.5pt);

(0.32,5.08) circle (1.5pt);

(1.1,5.08) circle (1.5pt);

(1.18,4.24) circle (1.5pt);

$G_3\in \mathcal{D}\setminus \mathcal{D}_1$ with $H$ is not an induced subgraph

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,2.14);
(-2,2.14)– (-0.5,2.14);
(-0.5,3.16)– (-0.5,2.14);
(-3.5,3.16)– (-2.72,4.24);
(-2.72,4.24)– (-2,3.16);
(-2,3.16)– (-2,4.24);
(-2,3.16)– (-1.2,4.24);
(-1.2,4.24)– (-0.5,3.16);
(-0.5,3.16)– (0.3,4.24);
(0.3,4.24)– (1.18,4.24);
(0.3,4.24)– (1.1,5.08);
(1.1,5.08)– (0.32,5.08);
(0.32,5.08)– (0.3,4.24);

(-2,3.16)– (-0.5,2.14);
(-2,2.14)– (-0.5,3.16);
(-3.5,3.16)– (2.5,3.16);
(-2,3.16)– (-0.5,3.16);
(1,3.16)– (0.3,4.24);

(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-2.2,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2.72,4.24) circle (1.5pt);

(-2,2.14) circle (1.5pt);

(-0.5,2.14) circle (1.5pt);

(-1.2,4.24) circle (1.5pt);

(-2,4.24) circle (1.5pt);

(0.3,4.24) circle (1.5pt);
(0.3,3.9) node $v'$;

(0.32,5.08) circle (1.5pt);

(1.1,5.08) circle (1.5pt);

(1.18,4.24) circle (1.5pt);

$G_4\in \mathcal{D}\setminus \mathcal{D}_1$ with $H$ is not an induced subgraph

We now proceed to compute the depth. Let $G$ be a connected non-complete graph on $[n]$ such that $d(G)+f(G)+1=n+2-\kappa(G)$. It follows from <cit.> and <cit.> that $d(G)+f(G)\leq \depth(S/J_G)\leq n+2-\kappa(G)=d(G)+f(G)+1.$

First we study the graphs $G$ with $\kappa(G)=1$. The main tool in our proofs is the use of Ohtani lemma (<Ref>). To help the reader imagine $G_w$, for an internal vertex $w$, we have given pictorial illustration of one such instance in each of the proofs. For example, given below are the graphs $(G_4)_{v'}$ and $G_4 \setminus v'$, where $G_4$ is the graph given above.

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,2.14);
(-2,2.14)– (-0.5,2.14);
(-0.5,3.16)– (-0.5,2.14);
(-3.5,3.16)– (-2.72,4.24);
(-2.72,4.24)– (-2,3.16);
(-2,3.16)– (-2,4.24);
(-2,3.16)– (-1.2,4.24);
(-1.2,4.24)– (-0.5,3.16);
(-0.5,3.16)– (0.3,4.24);
(0.3,4.24)– (1.18,4.24);
(0.3,4.24)– (1.1,5.08);
(1.1,5.08)– (0.32,5.08);
(0.32,5.08)– (0.3,4.24);

(-2,3.16)– (-0.5,2.14);
(-2,2.14)– (-0.5,3.16);
(-3.5,3.16)– (2.5,3.16);
(-2,3.16)– (-0.5,3.16);
(1,3.16)– (0.3,4.24);
[color=red] (0.32,5.08) – (1.18,4.24);
[color=red](0.32,5.08) – (-0.5,3.16);
[color=red](0.32,5.08) – (1,3.16);
[color=red](1.18,4.24) – (1.1,5.08);
[color=red](1.18,4.24) – (-0.5,3.16);
[color=red](1.18,4.24) – (1,3.16);
[color=red] (1.1,5.08)– (-0.5,3.16);
[color=red] (1.1,5.08)– (1,3.16);

(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-2.2,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2.72,4.24) circle (1.5pt);

(-2,2.14) circle (1.5pt);

(-0.5,2.14) circle (1.5pt);

(-1.2,4.24) circle (1.5pt);

(-2,4.24) circle (1.5pt);

(0.3,4.24) circle (1.5pt);
(0.3,3.9) node $v'$;

(0.32,5.08) circle (1.5pt);

(1.1,5.08) circle (1.5pt);

(1.18,4.24) circle (1.5pt);

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);
(-2,3.16)– (-2,2.14);
(-2,2.14)– (-0.5,2.14);
(-0.5,3.16)– (-0.5,2.14);
(-3.5,3.16)– (-2.72,4.24);
(-2.72,4.24)– (-2,3.16);
(-2,3.16)– (-2,4.24);
(-2,3.16)– (-1.2,4.24);
(-1.2,4.24)– (-0.5,3.16);

(1.1,5.08)– (0.32,5.08);

(-2,3.16)– (-0.5,2.14);
(-2,2.14)– (-0.5,3.16);
(-3.5,3.16)– (2.5,3.16);
(-2,3.16)– (-0.5,3.16);

(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-2.2,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_2$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;
(-2.72,4.24) circle (1.5pt);

(-2,2.14) circle (1.5pt);

(-0.5,2.14) circle (1.5pt);

(-1.2,4.24) circle (1.5pt);

(-2,4.24) circle (1.5pt);

(-0.44,5.08) circle (1.5pt);

(0.32,5.08) circle (1.5pt);

(1.1,5.08) circle (1.5pt);

(1.18,4.24) circle (1.5pt);

$G_4\setminus v'$

Let $G$ be a chordal graph with $\kappa(G) = 1$ and $d(G)+f(G) = n$. If $G \in \mathcal{D} \setminus \mathcal{D}_1$, then $\depth(S/J_G)=n+1.$

Note that $G \in \mathcal{D} \setminus \mathcal{D}_1$ if and only if one of the following conditions hold:

* either $u$ or  $v$ is an internal vertex;

* if both $u$ and $v$ are simplicial vertices, then $H$ is not  an induced subgraph of $G$, i.e., $\{v', u_{j+1}\}$ or $\{v', u_{j+2}\}$ is not an edge of $G$;

* if both $u$ and $v$ are simplicial vertices and $H$ is an induced subgraph of $G$, then $v'$ or $u_{j+1}$ is a cut-vertex of $G$, i.e., there are cliques attached along $v'$ or $u_{j+1}$.

First we identify an appropriate internal vertex $w$ so that Ohtani's lemma (<Ref>) can be applied to a short exact sequence of the type (<ref>). We show that, for a suitable choice of $w$, $\depth(S/J_{G_w}) \geq n+1$, $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) \geq n$ and $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$.
We choose $w$ as follows:

* If $u$ is an internal vertex in $G$, then set $w=u_1$ (if $v$ is an internal vertex, then set $w =u_{d-1}$).

* Suppose both $u,v$ are simplicial vertices.

* Assume that $H$ is not an induced subgraph of $G$, i.e.,  $N_P(v')\subseteq \{u_j,u_{j+1}\}$.

* If $N_P(v')\subseteq \{u_j,u_{j+1}\}$ and there are cliques attached to $v'$, then set $w=v'$.

* If $N_P(v')=\{u_j\}$ and there are no cliques attached to $v'$, then set $w = u_{j}$.

* If $N_P(v')= \{u_j,u_{j+1}\}$, and there are no cliques attached to $v'$, then set $w=u_{j+1}$ for $j\leq d-2$. If $j=d-1$ i.e., $u_{j+1}=v$, then choose $w=u_j$.

* Assume that $H$ is an induced subgraph of $G$. If cliques are attached to $u_{j+1}$, then set $w=u_{j+1}$ otherwise set $w=v'$.

(1) Since there are no cliques attached along the vertex $u$ and there are no internal vertices in $G$ other than those in $V(P)$, any clique containing $u$ will contain $u_1$ as well, i.e., $N_G[u] \subseteq N_G[u_1].$ Thus, in $G_w$, both $u$ and $u_1=w$ become simplicial vertices. Hence $f(G_w) \geq f(G)+2$. Therefore it follows from <Ref> that $n+1 = d(G)+f(G)+1 \leq d(G_w)+f(G_w) \leq n+1.$ This shows that $G_w$ belongs to the Hibi-Madani class. Since $G_w$ is in the Hibi-Madani class and $w$ is a simplicial vertex in $G_w$, $G_w\setminus w$ is also in the Hibi-Madani class. Hence $\depth(S/J_{G_w}) = n+1$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w}))=n$. Observe that $G \setminus w$ is a disconnected graph. The component containing $u$ is a clique sum of complete graphs along $u$. Hence this component belongs to the Hibi-Madani class. If there are cliques attached to $w$ in $G$, then there will be components in $G\setminus w$ which are cliques, not containing $u$. The component containing $v$ in $G\setminus w$ will be a repeated clique sum of complete graphs along the edges and vertices of the path on the vertices $u_2,\ldots,u_{d-1},v$. Hence this component also belongs to the Hibi-Madani class. Thus for each component $G_i$ with $V(G_i) =n_i$, we get $\depth(S_{G_i}/J_{G_i})=n_i+1$. Therefore, if $c$ denotes the number of components in $G\setminus w$, then $\depth(S/((x_w,y_w)+J_{G\setminus w})) = (n-1)+c \geq n+1.$

$n=d(G)+f(G)-1\leq d(G\setminus w) + f(G\setminus w) \leq n $
(2a(i)) Observe that $G\setminus w$ is a disconnected graph and has no internal vertices other than those in $V(P)$. Cliques attached along $w$ in $G$, will form disconnected components in $G\setminus w$ and the component containing the path $P$ is a clique sum of complete graphs along the vertices and edges of the path $P$. Hence each component of $G\setminus w$ belongs to the Hibi-Madani class. Therefore, $\depth(S/((x_w,y_w)+J_{G\setminus w})) \geq n+1$. Since $N_P(w) \subseteq \{u_j,u_{j+1}\}$, the path $P$ remains an induced path in $G_w$. Moreover, there are no internal vertices in $G_w$ other than those in $V(P)$, this is the only induced path in $G_w$, joining $u$ and $v$. Thus $\dist_{G_w}(u,v) = d$, and hence by <Ref>, we get $d(G_w) = d$. By <cit.>, we get $f(G) \leq f(G_w)-1$. Therefore, $n+1 = d(G)+f(G)+1 \leq d(G_w)+f(G_w) \leq n+1$. Hence $G_w$ belongs to the Hibi-Madani class. As we concluded in (1), we can see that $G_w \setminus w$ also belongs to the Hibi-Madani class. Therefore, $\depth(S/J_{G_w}) = n+1$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) = n$.

(2a(ii)) Since there are no cliques attached to $v'$ and $v'$ is an internal vertex, there are more than one clique along the edge $\{v',u_j\}$. Therefore, $N_G[v']\subseteq N_G[w]$ and hence, $v'$ becomes a simplicial vertex in $G_w$. Thus $f(G_w)\geq f(G)+2$ and $d(G)-1\leq d(G_w)$ yields $n+1=d(G)+f(G)+1\leq d(G_w)+f(G_w)=n+1$. So, $G_w$ belongs to the Hibi-Madani class and similarly, one can show that $G_w\setminus w$ also belongs to the Hibi-Madani class. Hence, $\depth(S/J_{G_w}) = n+1$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) = n$. Now $G\setminus w$ is a disconnected graph whose component containing $v'$ is a clique sum of complete graphs along $v'$, component containing $u$ is a repeated clique sum along the vertices and edges of the path $P_1:u,u_1,\ldots,u_{j-1}$, component containing $v$ is a repeated clique sum along the vertices and edges of the path $P_2:u_{j+1},\ldots,u_{d-1},v$ and if there are cliques attached along $u_j$, then we get those cliques  without the vertex $u_j$ as components. Thus each component of $G\setminus w$ belongs to the Hibi-Madani class and hence $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$.

(2a(iii)) If $N_G[v']\subseteq N_G[w]$ (for instance, if $j=d-1$), then $v'$ becomes a simplicial vertex in $G_w$ and one can show as in the above case that $G_w,G_w\setminus w$ and each component of $G\setminus w$ is a clique or belong to the Hibi-Madani class. Hence, $\depth(S/J_{G_w}) = n+1, \depth(S/((x_w,y_w)+J_{G_w\setminus w})) = n$ and $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$.

If $N_G[v']\nsubseteq N_G[w]$ i.e., $w=u_{j+1}$ for some $j\leq d-2$ and cliques are attached along the edge $\{v',u_j\}$, then $v'$ remains an internal vertex in $G_w$. Moreover, $u_i$ remains an internal vertex for $i \neq j+1$. Thus, $f(G_w)=f(G)+1$. Also since the internal vertices in $G_w$ are $\{u_1,\ldots,u_{j},u_{j+2},\ldots,u_{d-1},v'\}$ and $\{v',u_{j}\},\{v',u_{j+2}\}\in E(G_w)$, $d(G_w) < d$. Hence, by <Ref>, $d(G_w)=d(G)-1$. Therefore we can see that $d(G_w)+f(G_w)+1=n+1$. Also, $N_{P'}(v')=\{u_j,u_{j+2}\}$ and $N_{G_w}[v']\subseteq N_{G_w}[u_j],$ where $P':u,u_1,\ldots,u_j,u_{j+2},\ldots,u_{d-1},v$ in $G_w$. Therefore, $G_w$ is the graph discussed in the first paragraph of the proof of (2a(iii)) and hence  $\depth(S/J_{G_w})\geq n+1$ (follows from the depth conclusions in the first paragraph together with <cit.> applied to the short exact sequence (<ref>)). Similarly, one can get $\depth(S/((x_w,y_w)+J_{G_w\setminus w}))\geq n$. As we have seen earlier, $G\setminus w$ is a disconnected graph whose each component belongs to the Hibi-Madani class. So, $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$.

(2b) Assume that cliques are attached to $u_{j+1}$. If $N_G[v']\subseteq N_G[w]$, then $v'$ becomes simplicial vertex in $G_w$. Then arguments similar to those in the proof of (2a(ii)), we can show that $\depth(S/J_{G_w}) = n+1$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) = n$. Moreover, $G \setminus w$ is a disconnected graph whose each component is either a clique (coming from a clique attached to $u_{j+1}$) or a repeated clique sum of complete graphs along the path on the vertices $u,u_1,\ldots,u_j, v', u_{j+2},\ldots,u_{d-1},v$. Hence each such component is either a complete graph or belongs to the Hibi-Madani class. Therefore, $\depth(S/((x_w,y_w)+J_{G\setminus w})) \geq n+1$.

If $N_G[v']\nsubseteq N_G[w]$, i.e., cliques are attached along $v', \{v',u_j\}$ or $\{v',u_{j+2}\}$. In this case, $v'$ remains an internal vertex in $G_w$ and  $d(G_w)+f(G_w)+1=n+1$. By taking $P'':u,u_1,\ldots,u_j,u_{j+2},\ldots,u_{d-1},v$, we see that $N_{P''}(v')=\{u_j,u_{j+2}\}$. Therefore, it follows from the proof of (2a(i)) and (2a(iii)) that $\depth(S/J_{G_w})\geq n+1$. Similarly, one can show that $\depth(S/((x_w,y_w)+J_{G_w\setminus w}))\geq n$. In this case also, it is easy to see that $G\setminus w$ is a disconnected graph whose each component is a clique or belongs to the Hibi-Madani class. So, $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$.

If cliques are not attached to $u_{j+1}$, then by the definition of the class of graphs $\mathcal{D}\setminus \mathcal{D}_1$, cliques are necessarily attached to $v'$. Taking $w=v'$, we see that the graph is of the type discussed in the first two paragraph of the proof of (2b) with the roles of $u_{j+1}$ and $v'$ interchanged. Hence, arguments similar to those give us the required inequalities on the depth.

In each of the above cases, we have shown that $\depth(S/J_{G_w}) \geq n+1$, $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) \geq n$ and $\depth(S/((x_w,y_w)+J_{G\setminus w})) \geq n+1$. Therefore, by applying <cit.> to the short exact sequence (<ref>), we get $\depth(S/J_G)\geq n+1$.

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);

(-2,3.16)– (-2,4.24);
[color=red](-2,3.16)– (-0.5,3.16);

(-0.5,4.24)– (-0.5,3.16);

(-2,3.16)– (-0.5,4.24);

(-2.4,4.16) circle (1.5pt);
(-3,3.56) circle (1.5pt);
(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-1.7,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $u_3$;
(1,3.16) circle (1.5pt);
(1.3,3) node $v$;
(-1.25,4.78) circle (1.5pt);
(-1.25,5) node $v'$;
(-2,4.24) circle (1.5pt);
(-0.5,4.24) circle (1.5pt);
(-0.2,4.5) node $u_2$;

(-3.5,3.16)– (-2,3.16);
(-2,3.16)– (-0.5,3.16);
(-0.5,3.16)– (1,3.16);
(1,3.16)– (2.5,3.16);

(-2,3.16)– (-0.5,3.16);

(-2.4,4.16) circle (1.5pt);
(-3,3.56) circle (1.5pt);
(-3.5,3.16) circle (1.5pt);
(-3.34,3) node $u$;
(-2,3.16) circle (1.5pt);
(-1.7,3) node $u_1$;
(-0.5,3.16) circle (1.5pt);
(-0.2,3) node $v'$;
(1,3.16) circle (1.5pt);
(1.3,3) node $u_3$;
(2.5,3.16) circle (1.5pt);
(2.64,3) node $v$;

(-1.25,4.24) circle (1.5pt);

$G_1\setminus u_2$

If $G\in \mathcal{D}_1$, then $\depth(S/J_G)=n$.

Since $n=d(G)+f(G) \leq \depth (S/J_G)$, it is enough to prove that $\pd(S/J_G)\geq n.$ We show that $\beta_{n,n+d(G)}(S/J_G)\neq 0$.

Set $w=u_{j+1}$. Since $G$ is clique sum of complete graphs, $u_j$ and $u_{j+2}$ are adjacent to $v'$ and there are no cliques attached along $u_{j+1}$, $G\setminus w$ is a connected graph. By <Ref>, we get $d(G)\leq d(G\setminus w)$ and $f(G) \leq f(G\setminus w)$. Hence
\[n=d(G)+f(G) \leq d(G\setminus w) + f(G\setminus w) \leq (n-1)+2-\kappa(G\setminus w) = n.\]

Since both the above inequalities become equalities, we actually get $d(G) = d(G\setminus w)$, $f(G) = f(G\setminus w)$ and $d(G\setminus w)+f(G\setminus w)=|V(G\setminus w)|+1$. Therefore, by <Ref>, $\beta_{n,n+d}(S/((x_w,y_w)+J_{G\setminus w}))$ is an extremal Betti number.

Now we study the graphs $G_w$ and $G_w\setminus w$. If $N_G[v']\subseteq N_G[w]$, then both $w$ and $v'$ become simplicial vertices in $G_w$. Therefore,  $f(G_w)\geq f(G)+2$ and $f(G_w\setminus w)\geq f(G)+1$. If $d(G_w)=d$, then the induced path providing the diameter gives at least $d-1$ internal vertices in $G_w$ which contradicts the fact that $\iv(G_w)\leq d-2$. Thus, <Ref> yields that $d(G_w)=d(G)-1$. Similarly, we can see that $d(G_w\setminus w)=d(G)-1$. Now

\[ n+1=d(G)+f(G)+1\leq d(G_w)+f(G_w)\leq n+1.
\]

Hence, $G_w$ belongs to the Hibi-Madani class. Similarly, we can show that $G_w\setminus w$ also belongs to the Hibi-Madani class. Therefore it follows from <Ref> and the fact $d(G_w)=d(G_w\setminus w)=d(G)-1$ that

\[\beta_{n-1,n-1+d-1}(S/J_{G_w}) ~ ~ \text{ and } \beta_{n,n+d-1}(S/((x_w,y_w)+J_{G_w\setminus w})) \]

are extremal Betti numbers.

Suppose $N_G[v']\nsubseteq N_G[w]$, i.e., cliques are attached along $\{v',u_j\}$ or $\{v',u_{j+2}\}$.
In this case, $v'$ remains an internal vertex in $G_w$ and so $f(G_w)=f(G)+1$. The internal vertices of $G_w$ are $u_1,\ldots,u_j,u_{j+2},\ldots,u_{d-1},v'.$ Since any path of length $d$ has at least $d-1$ internal vertices and $\{v',u_j\},\{v',u_{j+2}\}\in E(G_w)$, there is no induced path of length $d$ in $G_w$. Thus $d(G_w)\leq d-1$ and hence by <Ref>, $d(G_w)=d-1$. Therefore, $d(G_w)+f(G_w)+1=n+1$. Taking $P': u,u_1,\ldots,u_j,u_{j+2},\ldots,u_{d-1},v$ in $G_w$ and observing that $v'$ is an internal vertex in $G_w$ with $N_{P'}(v')=\{u_j,u_{j+2}\}$, we can see that $G_w$ is a graph of the type considered in <Ref>(2a(iii)). It follows from <Ref> and Auslander-Buchsbaum formula that $\pd(S/J_{G_w})=n-1$.

Claim: $\beta_{n-1,n-1+d-1}(S/J_{G_w})$ is an extremal Betti number of $S/J_{G_w}$.

Proof of the Claim: Let $H=G_w$. Choose $w'=u_{j+2}$ if $0\leq j\leq d-3$, otherwise choose $w'=u_j$. If $N_H[v']\subseteq N_H[w']$ (for instance, if $j=d-2$), then it follows from the proof of <Ref>(2a(iii)) that $H_{w'},H_{w'}\setminus w'$ and each component of the disconnected graph $H\setminus w'$ is a clique or belong to the Hibi-Madani class with $d(H_{w'})=d(H_{w'}\setminus w')=d(H)-1=d-2$. Hence it follows from <Ref> and (<ref>) that
$$\beta_{n-1,n-1+d-2}(S/J_{H_{w'}}) \text{ and }\beta_{n,n+d-2}(S/((x_{w'},y_{w'})+J_{H_{w'}\setminus w'}))  $$
are extremal Betti numbers. Since $H\setminus w'$ is disconnected, by <cit.>, Auslander-Buchsbaum formula and (<ref>), $\pd(S/((x_{w'},y_{w'})+J_{H\setminus w'}))\leq n-1$  and as in the proof of <Ref>, we can show that $\beta_{n-1,n-1+j}(S/((x_{w'},y_{w'})+J_{H\setminus w'}))=0$ for $j\geq d$. Therefore, it follows from (<ref>) that for $j\geq d-2$

\begin{equation*}
\Tor_{n}^{S}\left( \frac{S}{(x_{w'},y_{w'})+J_{H_{w'}\setminus w'}},\K\right)_{n+j} \simeq \Tor_{n-1}^{S}\left( \frac{S}{J_{H}},\K\right)_{n+j}.
\end{equation*}

This implies that $\beta_{n-1,n+d-2}(S/J_H)$ is an extremal Betti number of $S/J_H$.

Suppose $N_H[v']\nsubseteq N_H[w']$, i.e., cliques are attached along $\{v',u_j\}$. Then it follows from the proof of <Ref>(2a(iii)) that $H_{w'}$ and $H_{w'}\setminus w'$ belong to the class of graphs discussed in the previous paragraph and hence $\beta_{n-1,n-1+d-2}(S/J_{H_{w'}})$ and $\beta_{n,n+d-2}(S/((x_{w'},y_{w'})+J_{H_{w'}\setminus w'}))$ are extremal Betti numbers. Also, $H\setminus w'$ is a disconnected graph whose each component is a clique or belongs to the Hibi-Madani class. Thus, it follows from <cit.>, Auslander-Buchsbaum formula and (<ref>) that $\pd(S/((x_{w'},y_{w'})+J_{H\setminus w'}))\leq n-1.$ Therefore, from long exact sequence (<ref>), we see that $\beta_{n-1,n-1+d-1}(S/J_H)$ is an extremal Betti number of $S/J_H$. This completes the proof of the claim.

Structurally, $G_w$ and $G_w\setminus w$ are very similar. Using arguments similar to the ones used above, we can show that $\beta_{n,n+d-1}(S/((x_w,y_w)+J_{G_w\setminus w}))$ is an extremal Betti number of $S/((x_w,y_w)+J_{G_w\setminus w})$.

Taking $i=n$ and $j=d$ in (<ref>), we get

\begin{equation*}
\Tor_{n}^{S}\left( \frac{S}{J_G},\K\right)_{n+d}\simeq \Tor_{n}^{S}\left( \frac{S}{(x_w,y_w)+J_{G\setminus w}},\K\right)_{n+d}.
\end{equation*}

This implies that $\beta_{n,n+d}(S/J_G)\neq 0$ so that $\pd(S/J_G) \geq n.$ Hence, $\depth(S/J_G)=n.$

Now assume that both $u$ and $v$ are simplicial vertices. There is precisely one internal vertex in $[n] \setminus V(P)$. Let $v' \notin V(P)$ denote the internal vertex.

First we assume that $G$ is chordal. Note that $\dist(v', P) = 1$, since otherwise there will be more internal vertices. Let $j = \min\{l : \{u_l, v'\} \in E(G)\}$. Observe that

* $N_P(v') \subseteq \{u_j, u_{j+1}, u_{j+2}\}$. Since otherwise, $\dist_G(u,v) \leq d-1$.

* If $\{v',u_{j+1}\} \notin E(G)$, then $\{v', u_{j+2}\} \notin E(G)$. Since otherwise, the chordality of $G$ will be contradicted.

Since $G$ is chordal, it is a repeated clique sum of complete graphs. The intersection of any such complete graphs will be contained in $\{u_1,\ldots, u_{d-1},v'\}$. If $\{v', u_{j+1}\}$ (respectively $\{v', u_{j+2}\}$) is an edge, then it is also possible to have a clique along the triangle $\{v', u_j, u_{j+1}\}$ (respectively $\{v', u_{j+1}, u_{j+2}\}$). There are unique cliques containing $u$ and $v$. Thus $G$ is a repeated clique sum of complete graphs along
the complete subgraphs of $G[u_1,\ldots,u_{d-1},v'].$
We categorize these graphs into the following two types:

* Type I: $N_P(v') \subseteq \{u_j, u_{j+1}\}$ or if $N_P(v') = \{u_j, u_{j+1}, u_{j+2}\}$, then there are cliques attached to $u_{j+1}$ or to $v'$.

* Type II: $N_P(v') = \{u_j, u_{j+1},u_{j+2}\}$ and there are no cliques attached to $v'$ and $u_{j+1}$.

There may be different longest induced paths in $G$. We first show that the property of a graph being Type I or Type II is independent of the longest induced path chosen. First we fix the path $P: u, u_1, \ldots, u_{d-1}, v$. In a path, all vertices, except the pendent vertices, are internal vertices. Moreover, in $G$, the internal vertices are $\{u_1, \ldots, u_{d-1}, v'\}$. Since $\dist(v',P)=1$, choose $j=\min \{l: \{u_l,v'\}\in E(G)\}.$

If $N_P(v')=\{u_j,u_{j+1},u_{j+2}\},$ then any other longest induced path of length $d$ must contain the induced path $P':u_1,\dots,u_j,v',u_{j+2},\dots,u_{d-1}$. We see that $N_{P'}(u_{j+1})=\{u_j,v',u_{j+2}\}$ and so $G$ is the same type with respect to both the induced paths in $G$. If $N_P(v')= \{u_j\}$, then there may be another induced path $P'=v'',v',u_j,\dots,v$ of length $d$. Then $j=2$ and $u_1$ is an internal vertex outside the path $P'$ with $N_{P'}(u_1)\subseteq \{v'\}$. Therefore, $G$ is Type I.

So we may assume that $N_P(v')= \{u_j,u_{j+1}\}.$ Therefore, it is enough to prove that if $G$ is Type I with respect to the path $P$, then $G$ is Type I with respect to any other longest induced paths in $G$.

First assume that $3\leq j\leq d-3$. If $N_P(v')=\{v_j,v_{j+1}\}$, then any induced path containing $v'$ has length $<d$, and so any longest induced path from a simplicial vertex to another simplicial vertex must contain $\{u_1,\dots,u_{d-1}\}$. Hence, $G$ is the same type with respect to any longest induced paths in $G$.

Suppose now $j=0$. Then $\{u_1,v'\}$ is necessarily an edge in $G$, and so $N_P(v')=\{u,u_1\}$. Note that there is no complete graphs attached to $v'$ otherwise, it gives a path of length $>d$. So, $N_G[v']\subseteq N_G[u_1]$. There is an induced path $P':v',u_1,\dots,v$ of length $d$ which case has been already considered when one of the vertices of $u$ and $v$ is internal in <ref>. Therefore, $1\leq j$. Assume that $j=1$. If $N_P(v')=\{u_1,u_2\}$, then any other longest induced path is of the form $P':v'',v',u_2,\dots,v$ where $v''\in N_G(v')\setminus \{u_1,u_2\}$. Now $u_1$ is an internal vertex from outside the path $P'$ with $N_{P'}(u_1)\subseteq \{v'',v',u_2\}$ and complete graph is attached at $u_1$. Therefore, $G$ is type I.

Suppose now $j=2$. Since $N_P(v')=\{u_2,u_3\}$, there is no induced path of length $d$ containing the vertex $v'$. Hence, $G$ is type I only.

If $j=d-2$ and $N_P(v')=\{u_{d-2},u_{d-1}\}$, then there is another path of length $d$ of the form $P''':u,u_1,\dots,u_{d-2},v',v'''$, where $v'''\in N_G(v')\setminus \{u_{d-2},u_{d-1}\}$. In this case, $u_{d-1}$ is the internal vertex not contained in $V(P''')$ with $N_{P'''}(u_{d-1})\subseteq \{u_{d-2},v',v'''\}$. Since $\{u_{d-1},u_d\}\in E(G)$, $G$ is type I only.

Suppose now $j=d-1$. Clearly, complete graphs are not attached at $v'$ otherwise it gives an induced path of length $>d$. Note that there is another induced path of length $d$ which ends with $v'$. Since the ending vertex is an internal vertex, which case has been already discussed in <ref>. Note that $j=d$ is not possible as if there are complete graphs along $v'$ or $\{v,v'\}$, then it contradicts the fact that $d(G)=d$.

It may be noted that if $j =0, 1, d-1$ or $d$, then there are no complete graphs attached to $\{v'\}$. If $j = 0$, then $\{u_1, v'\}$ is necessarily an edge in $G$. In both the above cases, assuming the contrary will imply that $G$ has an induced path of length at least $d+1$.

For the benefit of the readers, we illustrate the graphs listed above using some figures.

Now we consider non-chordal graphs, i.e., graphs having precisely one cycle, $C_4$, as an induced subgraph. Since $\kappa(G)=1$, there is only one internal vertex outside $V(P)$. Note also that all vertices of $C_4$ are internal. Hence $|V(C_4) \cap V(P)|=3$, say $V(C_4) \cap V(P) = \{u_j, u_{j+1}, u_{j+2}\}$ for some $j\in \{1, \ldots, d-3\}$. Consider the graph $H'$:

[dotted] (-6, 0) – (-5,0);
[dotted] (-3, 0) – (-2,0);
(-7,0)– (-6,0);
(-5,0)– (-4,0);
(-4,0)– (-3,0);
(-4,1)– (-5,0);

(-4,1)– (-3,0);
(-2,0)– (-1,0);

(-7,0) circle (1.5pt);
(-6.84,-0.3) node $u$;
(-6,0) circle (1.5pt);
(-5.84,-0.3) node $u_1$;
(-5,0) circle (1.5pt);
(-4.84,-0.3) node $u_j$;
(-4,0) circle (1.5pt);
(-3.84,-0.3) node $u_{j+1}$;
(-3,0) circle (1.5pt);
(-2.84,-0.3) node $u_{j+2}$;
(-2,0) circle (1.5pt);
(-1.84,-0.3) node $u_{d-1}$;
(-1,0) circle (1.5pt);
(-0.84,-0.3) node $v$;
(-4,1) circle (1.5pt);
(-4,1.3) node $v'$;

Note that $H'$ is an induced subgraph of $G$ in this situation.
Let $C: u_j, u_{j+1}, u_{j+2}, v', u_j$ be the induced $4$-cycle in $G$, where $v'  \notin V(P)$. Then $G \setminus v'$ is a chordal graph having internal vertices $\{u_1,\ldots,u_{d-1}\}$. Hence, $G\setminus v'$ is a clique sum of complete graphs along the complete subgraphs of $G[u_1,\ldots,u_{d-1}]$. It  may also be noted that there are no cliques of size at least 3 in the induced subgraph $G[u_1,\ldots,u_{d-1},v']$. Therefore, in $G$, intersection of any two cliques will either be a vertex or an edge of $G[u_1,\ldots,u_{d-1},v']$.

(-3,2.6)– (-1.5,2.6);
(-1.5,2.6)– (0,2.6);
(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);
(-1.5,2.6)– (0,3.84);
(0,3.84)– (1.5,2.6);
(-3,2.6)– (-2.24,3.84);
(-2.24,3.84)– (-1.5,2.6);
(1.5,3.84)– (0,3.84);
(0,4.72)– (0.98,4.72);
(0,3.84)– (0,4.72);
(0,3.84)– (0.98,4.72);
(0,2.6)– (0,1.8);
(0,2.6)– (1,1.8);
(0,1.8)– (1,1.8);

(-3,2.6) circle (1.5pt);
(-2.84,2.4) node $u$;
(-1.5,2.6) circle (1.5pt);
(-1.34,2.4) node $u_1$;
(0,2.6) circle (1.5pt);
(0.16,2.8) node $u_2$;
(0,1.8) circle (1.5pt);
(1,1.8) circle (1.5pt);
(1.5,2.6) circle (1.5pt);
(1.66,2.4) node $u_3$;
(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(0,3.84) circle (1.5pt);
(-0.25,4.1) node $v'$;
(-2.24,3.84) circle (1.5pt);

(1.5,3.84) circle (1.5pt);

(0,4.72) circle (1.5pt);

(0.98,4.72) circle (1.5pt);

(-3,2.6)– (-1.5,2.6);
(-1.5,2.6)– (0,2.6);
(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);
(-1.5,2.6)– (0,3.84);
(0,3.84)– (1.5,2.6);
(-3,2.6)– (-2.24,3.84);
(-2.24,3.84)– (-1.5,2.6);

(-1.5,3.84)– (-1.5,2.6);
(0,3.84)– (-1.5,3.84);

(-3,2.6) circle (1.5pt);
(-2.84,2.4) node $u$;
(-1.5,2.6) circle (1.5pt);
(-1.34,2.4) node $u_1$;
(-1.5,3.84) circle (1.5pt);
(0,2.6) circle (1.5pt);
(0.16,2.8) node $u_2$;
(1.66,2.4) node $u_3$;
(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(0,3.84) circle (1.5pt);
(-0.25,4.1) node $v'$;
(-2.24,3.84) circle (1.5pt);

(-1.5,2.6)– (0,2.6);
(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);
(-1.5,2.6)– (0,3.84);
(0,3.84)– (1.5,2.6);

(1.5,3.84)– (0,3.84);
(0,4.72)– (0.98,4.72);
(0,3.84)– (0,4.72);
(0,3.84)– (0.98,4.72);
(0,2.6)– (0,1.8);
(0,2.6)– (1,1.8);
(0,1.8)– (1,1.8);
(-2.5,3.84)– (-1.5,4.5);
(-2.5,3.84)– (-1.5,2.6);
[color=red](0,3.84)– (-1.5,4.5);
[color=red](0,3.84)– (0,2.6);
[color=red](-2.5,3.84)– (0,2.6);
[color=red](-1.5,4.5)– (0,2.6);
(-1.5,4.5)– (-1.5,2.6);
[color=red](0,3.84)– (-2.5,3.84);

(-2.5,3.84) circle (1.5pt);
(-2.84,3.7) node $u$;
(-1.5,2.6) circle (1.5pt);
(-1.34,2.4) node $u_1$;
(0,2.6) circle (1.5pt);
(0.25,2.8) node $u_2$;
(0,1.8) circle (1.5pt);
(1,1.8) circle (1.5pt);
(1.5,2.6) circle (1.5pt);
(1.66,2.4) node $u_3$;
(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(0,3.84) circle (1.5pt);
(-0.25,4.2) node $v'$;
(-1.5,4.5) circle (1.5pt);

(1.5,3.84) circle (1.5pt);

(0,4.72) circle (1.5pt);

(0.98,4.72) circle (1.5pt);

(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);

(0,3.84)– (0,2.6);
(0,3.84)– (-1.5,2.6);
(-2.5,2.6)– (-2.5,3.84);

(2.5,3.84)– (1.5,3.84);
(1.5,3.84)– (1.5,2.6);
(1.5,4.8)– (2.5,4.8);
(1.5,4.8) –(1.5,3.84);
(2.5,4.8) – (1.5,3.84);

(-2.5,3.84) circle (1.5pt);
(-2.5,2.6) circle (1.5pt);
(-2.4,2.4) node $u$;

(-1.5,2.6) circle (1.5pt);
(0,2.6) circle (1.5pt);
(0.16,2.4) node $u_2$;
(2.5,4.8) circle (1.5pt);
(1.5,4.8) circle (1.5pt);
(1.5,2.6) circle (1.5pt);
(1.66,2.4) node $u_3$;
(2.5,3.84) circle (1.5pt);
(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(1.5,3.84) circle (1.5pt);
(0,3.84) circle (1.5pt);
(1.3,4.1) node $v'$;

$G_5\setminus u_1$

We now compute the depth of $S/J_G$ for the above described class of graphs.

Let $G$ be a non-chordal graph with $\kappa(G)=1$ and $H'$ as an induced subgraph. If $\mathcal{C}(G) \cap \{\{u_{j+1}\},\{ v'\}\} \neq \emptyset$, then $\depth(S/J_G)=n+1.$

Set $w=u_{j},$ and in $G_w$ set $P':u,u_1,\ldots,u_{j-1},u_{j+1},\ldots,u_{d-1},v$. Then we see that $v'$ is an internal vertex in $G_w$ with $N_{P'}(v')=\{u_{j-1},u_{j+1},u_{j+2}\}$ and cliques are attached to $u_{j+1}$ or to $v'$. Since $d(G_w) < d$, by <Ref>, we get $d(G_w)=d-1$. Therefore, $d(G_w)+f(G_w)+1=d(G)+f(G)+1=n+1$ which implies that $G_w, G_w\setminus w\in \in \mathcal{D} \setminus \mathcal{D}_1$. Hence, by <Ref>, $\depth(S/J_{G_w})=n+1$ and $\depth(S/((x_w,y_w)+J_{G_w \setminus w}))=n$. Also, $G\setminus w$ is a disconnected graph whose each component is either a complete graph or belongs to the Hibi-Madani class (if either $u_{j+1}$ or $v'$ is a simplicial vertex in $G \setminus w$) or belongs to the class $ \mathcal{D} \setminus \mathcal{D}_1$ (if both $u_{j+1}$ and $v'$ are internal vertices in $G \setminus w$). In either case, $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq n+1$. Hence, by applying <cit.> to the short exact sequence $\eqref{ohtani-ses}$, we get $\depth(S/J_G)\geq n+1$. Thus $\depth(S/J_G)=n+1$.

(-1.5,2)– (0,2.6);
(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);
(-1.5,2)– (0,3.84);
(0,3.84)– (1.5,2.6);

[color=red](-2.5,3.84)– (-1.5,4.5);
(-2.5,3.84)– (-1.5,2);
(0,3.84)– (-1.5,4.5);
[color=red](0,3.84)– (0,2.6);
[color=red](-2.5,3.84)– (0,2.6);
[color=red](-1.5,4.5)– (0,2.6);
(-1.5,4.5)– (-1.5,2);
[color=red](0,3.84)– (-2.5,3.84);

(-2.5,2.6) circle (1.5pt);
(-2.84,2.7) node $u$;
(-1.5,2) circle (1.5pt);
(-1.34,1.8) node $w=u_1$;
(0,2.6) circle (1.5pt);
(0.25,2.8) node $u_2$;

(1.5,2.6) circle (1.5pt);
(1.66,2.4) node $u_3$;
(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(0,3.84) circle (1.5pt);
(0.2,4.1) node $v'$;
(-2.5,3.84) circle (1.5pt);
(-1.5,4.5) circle (1.5pt);

(0,2.6)– (1.5,2.6);
(1.5,2.6)– (3,2.6);

(-1,2.6)– (-1,3.84);

(1.5,3.84)– (1.5,2.6);

(1.5,4.8) –(1.5,3.84);

(-1,3.84) circle (1.5pt);
(-1,2.6) circle (1.5pt);
(-0.84,2.4) node $u$;

(0,2.6) circle (1.5pt);
(0.16,2.4) node $u_2$;

(1.5,4.8) circle (1.5pt);
(1.5,2.6) circle (1.5pt);
(1.66,2.4) node $u_3$;

(3,2.6) circle (1.5pt);
(3.16,2.4) node $u_4$;
(1.5,3.84) circle (1.5pt);

(1.3,4.1) node $v'$;

$G_6\setminus u_1$

Let $G$ be a non-chordal graph with $\kappa(G)=1$ and $H'$ as an induced subgraph. If $\mathcal{C}(G) \cap \{\{u_{j+1}\},\{ v'\}\} = \emptyset$, then $\depth(S/J_G)=n.$

As in the proof of <Ref>, we show that $\beta_{n,n+d(G)}(S/J_G)\neq 0.$ Taking $w=u_{j}$, we can see that $G_w,G_w\setminus w \in \mathcal{D}_1$. Therefore, it follows from the proof of <Ref> that $\beta_{n,n+d(G)-1}(S/J_{G_w})\neq 0$ and $\beta_{n+1,n+d(G)}(S/((x_w,y_w)+J_{G_w\setminus w}))\neq 0$ as $d(G)=d(G_w)+1=d(G_w\setminus w)+1$. Also, $G\setminus w$ is a disconnected graph such that each component is  either a complete graph or belongs to the Hibi-Madani class (if either $u_{j+1}$ or $v'$ is a simplicial vertex in $G \setminus w$) or belongs to the class $\mathcal{D} \setminus \mathcal{D}_1$ (if both $u_{j+1}$ and $v'$ are internal vertices in $G \setminus w$). Hence $\depth(S/((x_w,y_w)+J_{G\setminus w})) \geq n+1$ which implies that $\pd(S/((x_w,y_w)+J_{G\setminus w}))\leq n-1$. Therefore, it follows from  (<ref>) that

$\beta_{n,n+d(G)}(S/J_G)\neq 0$.

Now we assume that $\kappa(G) \geq 2$. By <Ref>, $2\leq d(G) \leq 3.$ First we assume that $d(G) = 2$. By <Ref>, $G$ is chordal. Moreover, any two vertices of $G$ are either adjacent or have a common neighbor. Note that $\iv(G) = \kappa+1$. Since $G$ is chordal with $\kappa(G) = \kappa$, it is a repeated clique sum of complete graphs along complete subgraphs of size $\kappa$ or ${\kappa+1}$.  Observe that any maximal clique in $G$ has at least $\kappa+1$ vertices, and $\kappa$ of those vertices must be internal vertices.

(0,1)– (1.3,1);
(1.3,1)– (1.3,0);
(0,0)– (0,1);
(0,0)– (1.3,0);
(1.3,0)– (0,1);
(0,0)– (1.3,1);
(1.3,1)– (2.6,1);
(2.6,0)– (1.3,0);
(2.6,1)– (2.6,0);
(1.3,1)– (2.6,0);

(0,0) circle (1.5pt);

(1.3,0) circle (1.5pt);

(2.6,0) circle (1.5pt);

(1.3,1) circle (1.5pt);
(1.46,1.26) node $w$;
(0,1) circle (1.5pt);

(2.6,1) circle (1.5pt);

(1.3,1)– (2.6,0);
(2.6,1)– (2.6,0);
(1.3,0)– (1.3,1);
(1.3,1)– (2.6,1);
(1.3,0)– (2.6,0);
(1.3,0)– (2.6,1);
(1.3,2)– (1.3,1);
(1.3,2)– (2.6,1);
(2.6,2)– (2.6,1);
(2.6,2)– (1.3,1);
(0,0.5)– (1.3,1);
(0,0.5)– (1.3,0);

(1.3,0) circle (1.5pt);
(1.46,-0.2) node $w$;
(2.6,0) circle (1.5pt);

(1.3,1) circle (1.5pt);

(2.6,1) circle (1.5pt);

(1.3,2) circle (1.5pt);

(2.6,2) circle (1.5pt);

(0,0.5) circle (1.5pt);

(0,1)– (1.3,1);
(1.3,1)– (1.3,0);
(0,0)– (0,1);
(0,0)– (1.3,0);
(1.3,0)– (0,1);
(0,0)– (1.3,1);
(1.3,1)– (2.6,0.7);
(2.6,0.3)– (1.3,0);
(2.6,0.7)– (2.6,0.3);
(1.3,1)– (2.6,0.3);
(2.6,0.3)– (2.6,0.7);
[color=red](0,1)– (2.6,0.7);
[color=red](2.6,0.7)– (0,0);
[color=red](1.3,0)– (2.6,0.7);
[color=red](2.6,0.3)– (0,1);
[color=red](2.6,0.3)– (0,0);

(0,0) circle (1.5pt);

(1.3,0) circle (1.5pt);

(2.6,0.3) circle (1.5pt);

(1.3,1) circle (1.5pt);
(1.46,1.26) node $w$;
(0,1) circle (1.5pt);

(2.6,0.7) circle (1.5pt);

(0,1)– (0,0);
(0,1)– (1.3,0);
(0,0)– (1.3,0);
(1.3,0)– (2.6,0);
(2.6,0)– (3.9,0);

(0,0) circle (1.5pt);

(1.3,0) circle (1.5pt);

(2.6,0) circle (1.5pt);

(0,1) circle (1.5pt);

(3.9,0) circle (1.5pt);

$G_7\setminus w$

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$G_8\setminus w$

Let $G$ be a graph with $d(G)=2$ and $\kappa(G) \geq 2$. Then

It is enough to prove that $\depth(S/J_G)\geq n+2-\kappa$. Let $w$ be an internal vertex.
If $w$ belongs to the intersection of all maximal cliques in $G$, then $J_{G_w}$ and $J_{G_w \setminus w}$ are complete graphs.  Hence  $\depth(S/J_{G_w})=n+1$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w}))=n$.
If there exists a maximal clique in $G$ not containing $w$, then $G_w$ is a clique sum of complete graphs along a fixed $K_\kappa$. Hence $G_w$ belongs to the Hibi-Madani class $\mathcal{F}_{\kappa}$ defined in [9]. Thus $G_w \setminus w$ is also in $\mathcal{F}_{\kappa}$.
Hence by <cit.>, $\depth(S/J_{G_w}) = n+2-\kappa$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w})) = n+1-\kappa.$

We now study $G\setminus w$.  Choose the vertex $w$ such that $G\setminus \{w=w_1,\ldots,w_\kappa\}$ is disconnected. We prove the assertion by induction on $\kappa$. Let $\kappa=2$. Hence $\iv(G) = 3$. Hence $G$ is obtained by taking clique sum of complete graphs along a fixed $K_3$ or its edges. Therefore, $G\setminus w$ is a clique sum of complete graphs along a fixed $K_2$ or vertices of the $K_2$. Since $\iv(G) = 3$ and $G\setminus w$ is not a complete graph, $1\leq \iv(G \setminus w) \leq  2$ and $d(G\setminus w) > 1$. We now observe that $d(G\setminus w) \leq 3$. Suppose $d(G\setminus w) \geq 4$, then there exists an induced path of length at least $4$. This implies that $\iv(G\setminus w) \geq 3$, which is a contradiction. Therefore, $2 \leq d(G\setminus w) \leq 3$. Suppose $\iv(G\setminus w)=2$ and $d(G\setminus w)=2$. Then $d(G\setminus w) + f(G \setminus w) + 1 = n = (n-1) + 2 - \kappa(G \setminus w)$. If $G\setminus w \in  \mathcal{D}_1$, then, following the notation set in the definition, $j=0$ with $u_{j+2} = v$. Moreover, $\kappa(G\setminus w) = 1$ and there are no cliques attached either to $v'$ or $u_{j+1}$, which are the only internal vertices. This contradicts the assumption that $\kappa(G  \setminus w) = 1$. Hence $G \setminus w \in \mathcal{D} \setminus \mathcal{D}_1$, and so $\depth (S/((x_w,y_w)+J_{G \setminus w})) = n$. If $\iv(G\setminus w)=1$ and $d(G\setminus w) =  2$, then $d(G\setminus w) + f(G\setminus w) = n = (n-1) + 2 - \kappa(G \setminus w)$. Suppose  $d(G\setminus w)=3$. This implies  that $\iv(G\setminus w)=2$ and hence $d(G\setminus w) + f(G\setminus w) = n $.
This implies that $G\setminus w$ belongs to the Hibi-Madani class.
Hence, by <cit.>, $\depth(S/((x_w,y_w)+J_{G\setminus w}))=n$. Thus <cit.> applied to the short exact sequence (<ref>) yields $\depth(S/J_G)\geq n$.

Assume that $\kappa\geq 3$. Then by the chosen $w$, we have $\kappa(G\setminus w) = \kappa -1$ and $\iv(G\setminus w) = \kappa$. Now we show that $d(G\setminus w) = 2$. Assume that $d(G\setminus w)=d'\geq 3$. Let $z,z'\in V(G\setminus w)$ such that $\dist_{G\setminus w}(z,z')=d'$. By <cit.>, there exist $\kappa -1$ vertex-disjoint paths of length $\geq d'$ from $z$ to $z'$ in $G\setminus w$. Each such path gives at least $(d'-1)$ internal vertices. This implies that $\iv(G \setminus w) \geq (d'-1)(\kappa-1) > \kappa$, a contradiction. Therefore, $d(G\setminus w) = 2$.  Since $\kappa(G\setminus w) = \kappa-1$ by induction, $\depth(S/((x_w,y_w)+J_{G\setminus w}))\geq (n-1)+2-\kappa(G\setminus w)=n+2-\kappa$. Hence, we obtain the required assertion by applying <cit.> to (<ref>).

Now we move on to the remaining part, i.e., $\kappa(G)=2$ and $d(G)=3$. Then $\iv(G)=4$. Since $\kappa(G)=2$, there exist two vertex disjoint paths joining $u$ and $v$. Let $P: u, u_1, u_2, v$ and $P':u,  v_1, v_2, v$ be disjoint paths in $G$ joining $u$ and $v$. Note that $u_1, u_2, v_1, v_2$ are the internal vertices of $G$. Since $u$ is a simplicial vertex, there exists a unique clique containing the edges $\{u,u_1\}$  and $\{u,v_1\}$. Similarly there exists a unique clique containing $\{u_2,v\}$ and $\{v_2,v\}$. Let $e_1 = \{u_1, v_1\}, e_2 = \{v_1, v_2\}, e_3=\{u_1, u_2\}$ and $e_4 = \{v_2, u_2\}$.
Consider the following graph $H''$:

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Now we assume that $G$ is chordal. Then the cycle on the vertices $\{u_1,u_2,v_2,v_1\}$ is not an induced cycle, i.e., $\{u_1,v_2\} \in E(G)$ or $\{u_2,v_1\} \in E(G)$. Then $G$ is a clique sum of complete graphs along cliques of size at least $2$ of the induced subgraph $G[u_1,u_2,v_1,v_2]$. Moreover, there are unique cliques containing $\{u,u_1,v_1\}$ and $\{v,u_2,v_2\}$.

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Let $G$ be a chordal graph $\kappa(G)=2$, $d(G)=3$ and $H''$ is an induced subgraph of $G$. If there are no cliques in $G$ along the edges $e_2$ and $e_3$ of $H''$, then $\depth(S/J_G)=n-1.$

First observe that $d(G)+f(G) = n-1 \leq \depth S/J_G$. For the reverse inequality,
we prove that $\beta_{n+1,n+4}(S/J_G)\neq 0$. Let $w=u_1$. Then, $G_w$ is a clique sum of complete graphs along the edge $e_4$. Therefore, $G_w$ and $G_w\setminus w$ belong to the Hibi-Madani class $\mathcal{F}_2$. Therefore, by  <Ref> and (<ref>), $\beta_{n,n+2}(S/J_{G_w})$ and $\beta_{n+1,n+3}(S/((x_w,y_w)+J_{G_w\setminus w}))$ are extremal Betti numbers.

We now describe the graph $G \setminus w$. The unique clique containing $\{u,v_1\}$ is attached to the rest of the graph along the vertex $v_1$. Since $\kappa(G) = 2$ and there are no cliques attached to $e_2$ and $e_3$ in $G$, it can be seen that there are no cliques attached to $v_2$ and $u_2$ in $G\setminus w$. Hence, in $G\setminus w$, complete graphs are attached along the complete subgraphs of $G[u_2,v_1,v_2]$, except along the vertices $u_2,v_2$ and the edge $e_2$. Moreover, $u_2,v_1,v_2$ are the only internal vertices in $G \setminus w$. Hence, $d(G\setminus w) = 3.$ Since $N_{G\setminus w}(u_2) = \{v_1,v_2,v\}$ and there are no cliques along the vertices $u_2$ and $v_2$,  $G\setminus w\in \mathcal{D}_1$. Therefore, by the proof of <Ref> and (<ref>), $\beta_{n+1,n+4}(S/((x_w,y_w)+J_{G\setminus w}))\neq 0$.
Now it follows from (<ref>) that $\beta_{n+1,n+4}(S/J_G)\neq 0$.

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Let $G$ be a chordal graph with $\kappa(G)=2$ and $d(G)=3$. Assume that either of the following statements hold:

* $G[u_1,u_2,v_1,v_2]$ is not a complete graph;

* $H''$ is an induced subgraph of $G$ and there are cliques in $G$ along the edges $e_2$ or $e_3$.

Then $\depth(S/J_G)=n.$

We show that $\depth(S/J_G)\geq n$.

First assume that $G[u_1,u_2,v_1,v_2]$ is not a complete graph. Assume that $\{u_1,v_2\} \in E(G)$ and $\{u_2,v_1\} \notin E(G)$. Take $w=u_1$. If there are cliques along $e_2$, then $G_w$ is a graph obtained by taking clique sum of the complete graph on $N_{G}[w]$ and complete graphs along $e_2$ and $e_4$. If $d(G_w) \geq 3$, then there exist two disjoint paths of length at least $d(G_w)$ connecting any two vertices $z$ and $z'$ such that $\dist_{G_w}(z, z') = d(G_w)$. This gives rise to at least $4$ internal vertices. This contradicts the fact that $\iv(G_w)\leq 3$. Therefore, $\kappa(G_w) = 2 = d(G_w)$. Since $f(G_w)\geq f(G)+1$, we can see that $n \leq d(G_w)+f(G_w) + 1 \leq n+2-\kappa(G_w)=n$. It may also be noted that $G_w \setminus w$ also satisfy the same property as $G_w$. Hence by <Ref>, $\depth (S/J_{G_w}) = n$ and $\depth (S/((x_w,y_w)+J_{G_w \setminus w}))=n-1$. If there are no cliques along $e_2$, then $G_w$ is a clique sum of the complete graph on $N_G[w]$ and complete graphs along $e_4$. Then $G_w$ and $G_w \setminus w$ belong to the Hibi-Madani class $\mathcal{F}_2$. Hence by <cit.>, we get $\depth (S/J_{G_w}) = n$ and $\depth (S/((x_w,y_w)+J_{G_w \setminus w})) = n-1.$
Now we study the graph $G \setminus w$. Suppose, in $G$, there are cliques along  $e_3$ or at least one clique other than the clique containing $\{u_2,v_2,v\}$ along $e_4$ or cliques along $G[u_1,u_2,v_2]$. Then $u_2$ remains an internal vertex in $G \setminus w$. By taking the path $P': u,v_1,v_2,v$, we can see that $N_{P'}(u_2) = \{v_2,v\}$. Therefore, $G\setminus w \in \mathcal{D}\setminus \mathcal{D}_1$, so that $\depth (S/((x_w,y_w)+J_{G\setminus w})) = n$, by <Ref>. If there are no cliques along $e_3$, no cliques other than the clique containing $\{u_2,v_2,v\}$ along $e_4$ and no cliques along $G[u_1,u_2,v_2]$, then $u_2$ becomes a simplicial vertex in $G\setminus w$. Therefore, $G\setminus w$ belongs to the Hibi-Madani class $\mathcal{G}_3$. Hence by <cit.>, $\depth (S/((x_w,y_w)+J_{G\setminus w})) = n$. Therefore, if $G[u_1,u_2,v_1,v_2]$ is not a complete graph, then it follows from <cit.> and (<ref>) that $\depth (S/J_G) \geq n$.

Now we assume that $G[u_1,u_2,v_1,v_2]$ is a complete graph and cliques are attached along $e_2$ or $e_3$. If there are cliques along the edge $e_3$, then take $w=u_1$, else take $w=v_1$. If there are no cliques along $e_2$, then $G_w$ is a clique sum of complete graphs along $e_4$. Hence $G_w$ and $G_w\setminus w$ belong to the Hibi-Madani class $\mathcal{F}_2$. If there are cliques along $e_2$, then $G_w$ is obtained by taking clique sum of the complete graph on $N_G[w]$ and complete graphs along $e_2$ and $e_4$. Hence $G_w$ and $G_w\setminus w$ have the property that $d(G_w) = d(G_w \setminus w) = \kappa(G_w) = \kappa(G_w \setminus w) = 2.$ Moreover, as seen in the earlier paragraph, we can see that $d(G_w) + f(G_w) +1 = n$. Therefore, by either <cit.> or <Ref>, we get $\depth (S/J_{G_w}) = n$ and $\depth (S/((x_w,y_w)+J_{G_w\setminus w}))=n-1.$ If $w = u_1,$ then $G\setminus w \in \mathcal{D}\setminus \mathcal{D}_1$ with the path $P':u,v_1,v_2,v$ and $u_2$ being an internal vertex. If $w = v_1$, then again $G \setminus w \in \mathcal{D}\setminus \mathcal{D}_1$ with the path $P'':u,u_1,u_2,v$ and $v_2$ being an internal vertex. Hence by <Ref>, $\depth (S/((x_w,y_w)+J_{G\setminus w})) = n$.
Hence by applying <cit.> to the short exact sequence (<ref>), we get $\depth (S/J_G) \geq n$.

Now assume that $G$ is not chordal. Then $G$ contains precisely one induced cycle of length $4$. Since $\iv(G) = 4$, the only possibility of a $C_4$ in $G$ is on the vertices $\{u_1, u_2, v_1, v_2\}$. Let $C$ denote the cycle on $u_1,u_2,v_1,v_2$. Then $G$ is a repeated clique sum of complete graphs along the edges of the cycle $C$.

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Let $G$ be a non-chordal graph with $\kappa(G)=2$ and $d(G)=3$. Then

Let $w=u_1$. We first describe the graph $G_w$. Since $G$ is the clique sum of complete graphs along the edges of the cycle $C$, $G_w$ is obtained by taking clique sum of complete graphs along the edges of the $K_3$ on the vertices $\{v_1,u_2,v_2\}$. Therefore, these are the only internal vertices of $G_w$ and hence $\iv(G_w) = 3$. If $d(G_w) \geq 3$, then there exist two vertex disjoint paths of length at least $d(G_w)$ which gives rise to at least $4$ internal vertices, a contradiction. Hence $d(G_w) = 2$. Since $G_w \setminus \{u_2,v_2\}$ is disconnected, $\kappa(G_w) = 2$. And, $d(G_w)+f(G_w)+1 = n$. Moreover, $G_w\setminus w$ also satisfies the same properties. Hence <Ref>, $\depth(S/J_{G_w})=n$ and $\depth(S/((x_w,y_w)+J_{G_w\setminus w}))=n-1$.

If there is a clique along $e_3$ or at least two cliques along $e_4$, then $G \setminus w \in \mathcal{D}\setminus \mathcal{D}_1$ with the path $P':u,v_1,v_2,v$ and $u_2$ being an internal vertex outside the path $P$. If there are no cliques along $e_3$ and there is a unique clique, namely the clique containing $v$, along the edge $e_4$, then $G\setminus w$ is a repeated clique sum of complete graphs along the vertices and edges of the path $P':u,v_1,v_2,v$. Hence, $G\setminus w$ belongs to the Hibi-Madani class $\mathcal{G}_3$. Therefore, either by using <Ref> or by using <cit.>, we conclude that $\depth(S/((x_w,y_w)+J_{G\setminus w}))=n$. Now the assertion follows from <cit.> applied to the short exact sequence (<ref>).

§ ON THE LOWER BOUND OF DEPTHS

It was proved by Rouzbahani Malayeri, Saeedi Madni and Kiani  that for any graph $G$, $d(G)+f(G) \leq \depth(S/J_G)$, [17].
There is a natural question arising from this result, namely, “What are the graphs $G$ satisfying the equality $\depth(S/J_G) = d(G)+f(G)$?”. While it is not clear, whether one can give a complete characterization of graphs with this minimal depth, in this section, we try to understand when a graph does not have this minimal depth.

A block of a graph is a non-trivial maximal connected subgraph which has no cut vertex. A graph $G$ is said to be a block graph if every block of $G$ is a complete subgraph. It is easy to observe that $G$ is a block graph if and only if G is a chordal graph
with the property that any two maximal cliques intersect in at most one vertex. A
connected chordal graph $G$ is called a generalized block graph if three maximal
cliques of $G$ intersect non-trivially, then the intersection of each pair of them is the

Let $G$ be a generalized block graph on $[n]$. The
maximum size of the maximal cliques of $G$ is called the clique number of a graph $G$, denoted by $\omega(G)$. For each $i=1,\dots,\omega(G)-1$, we set

\[ \mathcal{A}_i=\{ A\subseteq [n]: |A|=i \text{ and } A \text{ is a minimal cut set of } G\}.
\]

Also, we set $a_i=|\mathcal{A}_i|.$ Note that a generalized block graph $G$ is a block graph if and only if $a_i=0$ for $i\geq 2$. Let $m(G)$ denote the number of minimal cut sets of $G$, i.e., $m(G)=\sum_{i=1}^{\omega(G)-1}a_i(G).$

Let $G$ be a generalized block graph. Then $\depth(S/J_G)=d(G)+f(G)$ if and only if $m(G)+1=d(G).$

It follows from <cit.> that

\begin{equation*}
\depth(S/J_G)=n+1-\sum_{i=2}^{\omega(G)-1}(i-1)a_i(G) =n-\sum_{i=1}^{\omega(G)-1}ia_i(G)+1+m(G).
\end{equation*}

Note that $\iv(G)=\sum_{i=1}^{\omega(G)-1}ia_i(G)$. Hence, the assertion follows.

Let $G$ be a connected chordal graph and $\Delta(G)$ be its clique complex. Then $\Delta(G)$ is a quasi-tree i.e., there exists a leaf order of the facets in $\Delta(G)$. Let $G$ be such that $F_1,\dots,F_r$ be a leaf order in $\Delta(G)$, and $F_{i-1}$ is the unique branch of $F_{i}$ for $i=2,\dots,r$. Ene, Herzog and Hibi studied this class of graphs in [6] and called it a chain of cliques. Then we can write

\[ G=K_{r_1}\cup_{K_{q_1}}K_{r_2}\cup_{K_{q_2}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}, m\geq 2.
\]

We now characterize $G$ such that $\depth(S/J_G)=d(G)+f(G)$. If $F_{i-2}\cap F_i=\emptyset$ for all $i=3,\dots,r$, i.e., $V(K_{q_{j-1}})\cap V(K_{q_{j}})=\emptyset$ for all $j=2,\dots,r$, then $G$ is a generalized block graph with $m(G)+1=d(G)$, and hence by <Ref>, $\depth(S/J_G)=d(G)+f(G)$.

Let $G=K_{r_1}\cup_{K_{q_1}}K_{r_2}\cup_{K_{q_2}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}$ such that $q_i\geq 1$ for all $i=1,\dots,m-1$ and $V(K_{q_{j-1}})\cap V(K_{q_{j}})=\emptyset$ for all $j=2,\dots,r$. Then $S/J_G$ admits a unique extremal Betti number, namely $\beta_{2n-d(G)-f(G),2n-f(G)}(S/J_G)$. In particular, $\depth(S/J_G)=d(G)+f(G)$.

Note that $\iv(G)=\sum_{i=1}^{m-1}q_i$. We proceed by induction on $\iv(G)$. If $m=2$, then $G=K_{r_1}\cup_{K_{q_1}}K_{r_2}$ belong to the family of $\mathcal{F}_{q_1}$, defined in [9], and the assertion follows from <cit.>.
Let $m\geq 3$.
If $q_j = 1$ for some $j=i_1, \ldots, i_t$, then one can write $G = G_1 \cup_{v_1} \cdots \cup_{v_t}G_{t+1}$, where $K_{q_j} = \{v_j\}$ for $j=1, \ldots, t$. Suppose $\beta_{\ell_j, \ell_j+p_j}(S/J_{G_j})$ is an extremal Betti number of $G_j$ for $j  = 1, \ldots, t+1$, then
by <cit.>, we get that $\beta_{\sum\ell_j, \sum\ell_j+\sum+p_j}(S/J_G)$ is an extremal Betti number of $S/J_G$. Therefore, to compute the extremal Betti number of $S/J_G$, it is enough to compute the extremal Betti numbers of the indecomposable components $G_1, \ldots G_{t+1}$ of $G$.
Moreover, by <cit.>, $\depth(S/J_G) = [\sum_{j=1}^{t+1} \depth(S_{G_j}/J_{G_j})]-t-1.$ Now, $d(G) = \sum_{j=1}^{t+1} d(G_j)$ and $f(G) = [\sum_{j=1}^{t+1} f(G_i)]-t-1.$ This implies that if the assertions are true for indecomposable graphs, then the assertion is true for decomposable graphs as well.

Case 1: Suppose $q_i=1$ for some $1\leq i\leq m-1$.
Then $V(K_{q_i})$ is singleton set, say $V(K_{q_i})=\{v\}$. Then $G\setminus v$ is  a disconnected graph whose components are $G_1=K_{r_1}\cup_{K_{q_1}}\cdots \cup_{K_{q_{i-1}}}K_{r_i-1}$ and $G_2=K_{r_{i+1}-1}\cup_{K_{q_{i+1}}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}$. Then by induction, $\beta_{2n_1-d(G_1)-f(G_1),2n_1-f(G_1)}(S_{G_1}/J_{G_1})$ and $\beta_{2n_2-d(G_2)-f(G_2),2n_2-f(G_2)}(S_{G_2}/J_{G_2})$ are extremal Betti numbers where $|V(G_1)|=n_1$ and $|V(G_2)|=n_2$.
So, by (<ref>),
$$\beta_{2n-d(G_1)-d(G_2)-f(G_1)-f(G_2),2n-f(G_1)-f(G_2)}(S/((x_v,y_v)+J_{G\setminus v}))$$
is an extremal Betti number. Observe that $d(G)+f(G)\leq d(G_1+d(G_2)+f(G_1)+f(G_2)$. Therefore,

$$\beta_{2n-d(G)-f(G)+i,j}(S/((x_v,y_v)+J_{G\setminus v}))=0 \text{ for any } i,j\geq 1.$$

Also, note that
$$G_v=K_{r_1}\cup_{K_{q_1}}\cdots \cup_{K_{q_{i-1}}}K_{r_i+r_{i+1}-1}\cup_{K_{q_{i+1}}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}$$

$$G_v\setminus v=K_{r_1}\cup_{K_{q_1}}\cdots \cup_{K_{q_{i-1}}}K_{r_i+r_{i+1}-2}\cup_{K_{q_{i+1}}}\cdots \cup_{K_{q_{m-1}}}K_{r_m},$$

with $\iv(G_v)=\iv(G_v\setminus v)=\sum_{j=1,j\neq i}^{m}q_j$. So, by induction,

and by (<ref>),
$$\beta_{2n-d(G_v\setminus v)-f(G_v\setminus v),2n-f(G_v\setminus v)}(S/((x_v,y_v)+J_{G_v\setminus v}))$$
are extremal Betti numbers. Clearly, $f(G)=f(G_v\setminus v)=f(G_v)-1$ and $d(G)=d(G_v)+1=d(G_v\setminus v)+1$. Thus,
$$\beta_{2n-d(G)-f(G),2n-f(G)-1}(S/J_{G_v}) \text{ and }$$

$$\beta_{2n-d(G)-f(G)+1,2n-f(G)}(S/((x_v,y_v)+J_{G_v\setminus v}))$$
are extremal Betti numbers. From the long exact sequence corresponding to the short exact sequence (<ref>) we get

\begin{equation*}
\Tor_{2n-d(G)-f(G)+1}^{S}\left(\frac{S}{(x_v,y_v)+J_{G_v\setminus v}},\K\right)_{j}\simeq \Tor_{2n-d(G)-f(G)}^{S}\left( \frac{S}{J_G},\K\right)_{j} \text{ for }j\geq 2n-f(G).
\end{equation*}

This implies that $\beta_{2n-d(G)-f(G),2n-f(G)}(S/J_G)$ is an extremal Betti number of $S/J_G$.

We now prove the assertions when $q_i \geq 2$ for all $i$. Let $v\in V(K_{q_1})$ be an internal vertex in $G$. Then
$$G\setminus v=K_{r_1-1}\cup_{K_{q_1-1}}K_{r_2-1}\cup_{K_{q_2}}K_{r_3}\cup_{K_{q_3}}\cdots \cup_{K_{q_{m-1}}}K_{r_m},$$

$$G_v=K_{r_1+r_2-q_1}\cup_{K_{q_2}}K_{r_3}\cup_{K_{q_3}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}$$

$$G_v\setminus v=K_{r_1+r_2-q_1-1}\cup_{K_{q_2}}K_{r_3}\cup_{K_{q_3}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}.$$

Note that

\begin{align*}
\iv(G) & =\iv(G\setminus v) +1 = \iv(G_v)+q_1 = \iv(G_v\setminus v)+q_1, \\
d(G)& =d(G\setminus v)=d(G_v)+1=d(G_v\setminus v)+1, \\ f(G)&=f(G\setminus v)=f(G_v)-q_1=f(G_v\setminus v)-q_1+1.
\end{align*}

Therefore, by induction and (<ref>),

$$\beta_{2n-d(G)-f(G),2n-f(G)}(S/((x_v,y_v)+J_{G\setminus v})), \beta_{2n-d(G)-f(G)-(q_1-1),2n-f(G)-q_1}(S/J_{G_v})\text{ and }$$

$$\beta_{2n-d(G)-f(G)-(q_1-2),2n-f(G)-(q_1-1)}(S/((x_v,y_v)+J_{G_v\setminus v}))$$
are extremal Betti numbers of $S/((x_v,y_v)+J_{G\setminus v}), S/J_{G_v}, S/((x_v,y_v)+J_{G_v\setminus v})$ respectively.
As $q_1\geq 2$, it follows from (<ref>) that

\begin{equation*}
\Tor_{2n-d(G)-f(G)}^{S}\left( \frac{S}{J_G},\K\right)_{j}\simeq \Tor_{2n-d(G)-f(G)}^{S}\left(\frac{S}{(x_v,y_v)+J_{G\setminus v}},\K\right)_{j} \text{ for }j\geq 2n-f(G).
\end{equation*}

This implies that $\beta_{2n-d(G)-f(G),2n-f(G)}(S/J_G)$ is an extremal Betti number of $S/J_G$, and hence $S/J_G$ admits a unique extremal Betti number. Thus, $\pd(S/J_G) \geq 2n-d(G)-f(G)$ so that $\depth(S/J_G)\leq d(G)+f(G).$ Therefore, from <cit.>, we get $\depth(S/J_G) = d(G)+f(G)$.

We now prove that if $V(K_{q_{j-1}})\cap V(K_{q_{j}})\neq \emptyset$ for some $j=2,\dots,r$, then $d(G)+f(G)+1\leq \depth(S/J_G)$. First we observe that we need to prove the result only for indecomposable graphs, i.e., $q_i > 1$ for all $i = 1, \ldots, m-1$. Assume that the result is proved for indecomposable graphs. Let $G = G_1 \cup_{w} G_2$, where $G, G_1$ and $G_2$ are chain of cliques, $G_1, G_2$ indecomposable, and $w$ is a simplicial vertex in $G_1$ and in $G_2$. Write $G = G_1 \cup_{K_{q_i}} G_2$, where $G_1 = K_{r_1}\cup_{K_{q_1}} \cdots \cup_{K_{q_{i-1}}}K_{r_i}$ and $G_2 = K_{r_{i+1}}\cup_{K_{q_{i+1}}} \cdots \cup_{K_{q_{m-1}}}K_{r_m}.$ Then by <cit.>, $\depth (S/J_G)=\depth (S_{G_1}/J_{G_1}) + \depth (S_{G_2}/J_{G_2})-2$, where $S_{G_i}$ denotes the polynomial ring in $2|V(G_i)|$ variables. If $G$ satisfies the property that $V(K_{q_{j-1}}) \cap V(K_{q_j}) \neq \emptyset$, then it is clear that $G_1$ or $G_2$ satisfies this property. Without loss of generality, we may assume that $G_1$ satisfies this property.
If $K_{q_i} = \{w\}$, then $w$ is an internal vertex in $G$ while it is a simplicial vertex in $G_1$ and in $G_2$. Moreover, if $v \in V(G_i) \setminus \{w\}$, then $v$ is simplicial in $G_i$ if and only if $v$ is simplicial in $G$. Therefore, $f(G) = |V(G)|-\iv(G) = (|V(G_1)|+|V(G_2)|-1) - (\iv(G_1)+\iv(G_2)+1) = f(G_1)+f(G_2)-2$.  Since any longest induced path in $G$ must pass through $w$, $d(G) = d(G_1) + d(G_2)$. Therefore,

\begin{eqnarray*}
\depth (S/J_G) & = & \depth (S_{G_1}/J_{G_1}) + \depth (S_{G_2}/J_{G_2}) -2 \\
& \geq & [d(G_1) + f(G_1) + 1] + [d(G_2) +  f(G_2) ] -2 \\
& = & d(G) + f(G) + 1.
\end{eqnarray*}

Therefore, to prove our result, we only need to prove the same for indecomposable graphs. Hence, we assume that $q_i \geq 2$ for all $i =  1, \ldots, m-1$.

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Let $G=K_{r_1}\cup_{K_{q_1}}K_{r_2}\cup_{K_{q_2}}\cdots \cup_{K_{q_{m-1}}}K_{r_m}$ such that $q_i\geq 2$ for all $i=1,\dots,m-1$ and $V(K_{q_{j-1}})\cap V(K_{q_{j}})\neq \emptyset$ for some $j=2,\dots,r$. Then $\depth(S/J_G)\geq d(G)+f(G)+1$.

First, observe that $\iv(G)=|\cup_{i=1}^{m-1} V(K_{q_i})|$ and $\kappa(G)=\min\{q_i : i=1,\ldots,m-1\}$.
We prove the assertion by induction on $\iv(G)$. The base case is $\iv(G)=3$, i.e., $m=3, q_1=2$ and $q_2=2$. By construction, $d(G) = 2, f(G) = n-3$ and $\kappa(G)=2$. Hence $d(G)+f(G)+1 = n = n+2-\kappa(G)$. Therefore, by <Ref>, $\depth (S/J_G) = n=d(G)+f(G)+1$. Assume that $\iv(G)\geq 4$. Let $v\in V(K_{q_{j-1}})\cap V(K_{q_{j}})$.
\[G_v = K_{r_1} \cup_{K_{q_1}} \cdots \cup_{K_{q_{j-3}}} K_{r_{j-2}} \cup_{K_{q_{j-2}}} K_r \cup_{K_{q_{j+1}}}K_{r_{j+2}} \cup_{K_{q_{j+2}}} \cdots \cup_{K_{q_{m-1}}} K_{r_m},\]

where $K_r$ is the complete graph on $N_G[v]$.
If $V(K_{q_{k-1}})\cap V(K_{q_{k}}) = \emptyset$ for all $k\neq j$, then $G_v$ and $G_v \setminus v$ are generalized block graphs.
If $V(K_{q_{k-1}})\cap V(K_{q_{k}}) \neq \emptyset$ for some $k \neq j$, then
$G_v$ and $G_v\setminus v$ are graphs satisfying the hypothesis of the theorem with $\iv(G_v) < \iv(G)$ and $\iv(G_v\setminus v) < \iv(G)$.
Any longest induced path in $G$ intersects with $K_{r_t}$ for all $t = 1,\ldots,m$ and can have at most one edge in $K_{r_t}$ for each $t=1,\ldots,m$. In any of the longest induced path in $G$, there is a portion of length at least $2$ that passes through $K_{r_{j-1}} \cup K_{r_j} \cup K_{r_{j+1}}$. Since these three complete graphs have merged to become one complete graph in $G_v$, such a portion will be replaced with an edge in $G_v$. All the other edges in the path in $G$ will remain the same in $G_v$ as well. Hence,
$d(G_v) \leq d(G)-1$. Combining this inequality with <Ref>, we get $d(G_v) = d(G)-1$.

Note that the vertices of $V(K_{q_i})$ are internal vertices in $G$ for $i = 1, \ldots, m-1$. If $v  \in V(K_{q_i})$, then the vertices of $V(K_{q_i})$ become simplicial vertices in $G_v$. Hence $f(G_v) =f(G)+|\cup_{v \in V(K_{q_l})}V(K_{q_l})|.$
Since $|V(K_{q_i})| \geq 2$ for all $i$ and $v \in V(K_{q_{j-1}}) \cap V(K_{q_j})$, $|\cup_{v \in V(K_{q_l})}V(K_{q_l})|\geq 3$. Moreover, all these properties are satisfied by the graph $G_v \setminus v$ as well. Therefore,
$d(G_v)+f(G_v)\geq d(G)+f(G)+2$ and $d(G_v\setminus v)+f(G_v\setminus v)\geq d(G)+f(G)+1$. Therefore, by <cit.>, we have
$$\depth(S/J_{G_v})\geq d(G_v)+f(G_v)\geq d(G)+f(G)+2 \text{ and } $$

$$\depth(S/((x_v,y_v)+J_{G_v\setminus v}))\geq d(G_v\setminus v)+f(G_v\setminus v)\geq  d(G)+f(G)+1.$$

Now we describe $G \setminus v$. Suppose $V(K_{q_{j-1}}) \cap V(K_{q_j}) = \{v\}$. Since $G\setminus v$ is connected, $f(G\setminus v)=f(G)$. Note that any shortest induced path between two vertices, say $z$ and $z'$, at a distance $d(G)$ must pass through $v$. Thus $\dist_{G\setminus v}(z,z') \geq d(G)+1$. Therefore, $d(G\setminus v) \geq d(G)+1$. Hence

\[\depth (S/((x_v,y_v)+J_{G\setminus v})) \geq d(G\setminus v) + f(G\setminus v) \geq d(G)+f(G)+1,\]
where the first inequality follows from <cit.>. If $\{v\} \subsetneq V(K_{q_{j-1}})\cap V(K_{q_j}),$ then $G \setminus v$ satisfies the hypothesis of the theorem and $\iv(G\setminus v) = \iv(G)-1$. Hence by induction on the number of internal vertices, we get $\depth(S/((x_v,y_v)+J_{G\setminus v})) \geq d(G \setminus v) + f(G \setminus v) + 1.$ If $P'$ is a path in $G$ that does not pass through $v$, then it remains a path in $G\setminus v$. If $P':z=z_1,z_2,\ldots,z_{j-1},v,z_{j+1},\ldots,z_k,z'$ is a path in $G'$, then we can see that $P'':z=z_1, z_2,\ldots, z_{j-1},v',z_{j+1},\ldots,z_k,z'$ is a path in $G\setminus  v$, where $v' \in V(K_{q_{j-1}})\cap V(K_{q_j}) \setminus \{v\}$. Hence $d(G\setminus v)=d(G)$. Therefore, we get $\depth(S/((x_v,y_v)+J_{G\setminus v})) \geq d(G)+f(G)+1$.

Hence, the assertion follows by applying <cit.> to the short exact sequence (<ref>).

Now, for an arbitrary graph $G$, we give a sufficient condition for  $\depth(S/J_G) \geq d(G)+f(G)+1$.

Let $G$ be a graph such that there exists an internal non-cut vertex $v$ in $G$ with $d(G\setminus v)\geq d(G)+1$. Then $\depth(S/J_G)\geq d(G)+f(G)+1$.

Since $|N_G(v)|\geq 2$, $f(G_v)=f(G_v\setminus v)+1\geq f(G)+2$.
Therefore it follows from <Ref> that $d(G_v)+f(G_v)\geq d(G)+f(G)+1$ and $d(G_v\setminus v)+f(G_v\setminus v)\geq d(G)+f(G).$
Also, since $\iv(G\setminus v)<\iv(G)$, $f(G)\leq f(G\setminus v)$. This together with $d(G\setminus v)\geq d(G)-1$ implies that $d(G\setminus v)+f(G\setminus v)\geq d(G)+f(G)+1.$
Hence, the assertion follows by applying <cit.> to the short exact sequence (<ref>).

Let $G$ be a graph on $[n]$ such that $d(G)+f(G)=n$. Then either $G$ is  chordal or $G$ has precisely one induced $C_4$ and has no induced $C_l$ for $l \geq 5$.

Let $u,v\in V(G)$ such that $\dist_G(u,v)=d(G)$, and $P:u,u_1,\dots,u_{d-1},v$ be a path from $u$ to $v$ where $u_i$'s are internal vertices. Since, $d(G)+f(G)=n$, $\iv(G)=d(G)$. So, there is only one internal vertex in $[n]\setminus \{u_1,\dots,u_{d-1}\}.$ Suppose now that $G$ has an induced cycle $C_r$ for $r\geq 5$. Since $P:u,u_1,\dots,u_{d-1},v$ is a smallest length path from $u$ to $v$, at most $\lceil \frac{r}{2} \rceil$ vertices can be part of that path. Therefore, there are at least $\lfloor \frac{r}{2} \rfloor\geq 2$ internal vertices outside the path $P$, which is a contradiction. So, $G$ can not have an induced cycle of length $\geq 5$. Similarly, we can prove that if $G$ has more than $2$ cycles of length $4$, then $G$ has at least $2$ internal vertices outside the path $P$. Hence, either $G$ is chordal or $G$ has precisely one induced cycle of length $4$.

For a graph $G$, if $d(G)+f(G) = n+2-\kappa(G)$, then $G$ is chordal, [9]. And,  in that case, they proved that $\depth (S/J_G) = d(G)+f(G)$, (Theorem 2.4, [9]). If $d(G)+f(G)+1=n+2-\kappa(G)$, then it follows from <Ref> that $G$ is either chordal or has at most one induced cycle of length $4$ and no cycles of length $\geq 5$. Therefore, we raise the following question:

If $G$ is a graph containing an induced cycle of length at least $5$, then is $\depth (S/J_G) \geq d(G)+f(G)+1$?

While we believe that this question has an affirmative answer, we are unable to prove this in general. Below, we answer this question affirmatively when $G$ is a unicyclic graph or a quasi-cycle graph.

Let $G$ be a unicyclic graph or a quasi-cycle graph such that $\depth(S/J_G)=d(G)+f(G)$. Then either $G$ is  chordal or $G$ has an induced $C_4$.

Let $G$ be a unicyclic graph or a quasi-cycle graph. Then it follows from <cit.> and <cit.> that $n\leq \depth(S/J_G) \leq n+1$. If $G$ is a quasi-cycle, then $\kappa(G) \geq 2$. Hence $\depth (S/J_G) \leq n$ so that $\depth (S/J_G) = n = d(G) + f(G)$. Therefore, by <cit.>, $G$ is a chordal graph. Assume that $G$ is unicyclic but not a cycle. Then $\kappa(G)=1$. Now if $\depth (S/J_G) = n$, then $d(G)+f(G)+1=n+2-\kappa(G)$, and hence the assertion follows from <Ref>. If $\depth (S/J_G) = n+1$, then by <cit.> $G$ is chordal.

3mm Acknowledgement: We are thankful to the anonymous referee for going through an initial manuscript and making suggestions which improved the exposition extensively.

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