Vertex Decomposability of the Stanley–Reisner Complex of a Path Ideal

Abstract

The t-path ideal $I_t(G)$ of a graph G is the square-free monomial ideal generated by the monomials which correspond to the paths of length t in G. In this paper, we prove that the Stanley–Reisner complex of the 2-path ideal $I_2(G)$ of an (undirected) tree G is vertex decomposable. As a consequence, we show that the Alexander dual $I_2(G)^{\vee }$ of $I_2(G)$ has linear quotients. For each $t \ge 3$, we provide a counterexample of a tree for which the Stanley–Reisner complex of $I_t(G)$ is not vertex decomposable.

1 Introduction

Let $R = \mathbb {K}[x_1,\dots ,x_n]$ be a polynomial ring, where $\mathbb {K}$ is a field and n is a positive integer. In literature, there are several ways to associate a square-free monomial ideal in R to algebraic objects. In [11], Villarreal introduced the concept of edge ideal of a graph. Let G be a simple graph on the vertex set [n]. Then the edge ideal of G is the square-free monomial ideal given by

$$\begin{aligned} I(G) = \langle x_ux_w: \{u,w\} \in E(G) \rangle . \end{aligned}$$

The concept of edge ideal is generalized by Conca and De Negri [4]. They introduced the concept of t-path ideal of a graph G. A path of length t in G is a sequence $w_1,\dots ,w_{t+1}$ of vertices in G such that $\{w_1,w_2\},\dots ,\{w_t,w_{t+1}\}$ are distinct edges of G. The t-path ideal of G is the square-free monomial ideal given by

$$\begin{aligned} I_t(G) = \langle x_{w_1}\cdots x_{w_{t+1}}: w_1,\dots ,w_{t+1} ~\text{ is } \text{ a } \text{ path } \text{ in }~ G \rangle . \end{aligned}$$

Observe that $I_1(G) = I(G).$

Billera and Provan [10] introduced the concept of vertex decomposability for pure simplicial complexes. Björner and Wachs [3] extended this concept for non-pure simplicial complexes (see Definition 2.1). Numerous researchers have investigated the vertex decomposability of the Stanley–Reisner complex $\Delta _I$ for different classes of square-free monomial ideals. Ajdani and Jahan [1] proved that $\Delta _{I_t(C_n)}$ is vertex decomposable if and only if $t = n-1,t = n$, or n is odd and $t = (n-1) / 2.$ Authors in [2] proved that the Stanley–Reisner complex of the facet ideal $I(\Delta )$ of a simplicial tree $\Delta$ is vertex decomposable (see [5] for the definition of facet ideal). In [8], He and Van Tuyl showed that if G is a rooted tree, then there exists a simplicial tree $\Delta$ such that $I_t(G) = I(\Delta ).$ Combining these results, we obtain that if G is a rooted tree, then the complex $\Delta _{I_t(G)}$ is vertex decomposable. These results motivate us to explore the vertex decomposability of the complex $\Delta _{I_t(G)}$ for a broader class of graphs G. This paper explores the vertex decomposability of the Stanley–Reisner complex $\Delta _{I_t(G)}$ when G is an undirected tree. We prove that the complex $\Delta _{I_t(G)}$ is vertex decomposable when $t = 2$ (see Theorem 3.5). For each $t \ge 3$, we give counter-example of a tree G for which the complex $\Delta _{I_t(G)}$ is not vertex decomposable (see Example 3.8).

Let $I \subset R$ be a square-free monomial ideal. If $\Delta _I$ is a vertex decomposable simplicial complex, then the Alexander dual $I^{\vee }$ of I has linear quotients, and hence it is componentwise linear (see Definitions 2.5, 2.6, Lemma 2.7). Consequently, we obtain that the Alexander dual $I_2(G)^{\vee }$ of the path ideal $I_2(G)$ has linear quotients, and hence it is componentwise linear.

We now give a brief overview of this paper. In the upcoming section, we introduce basic notions of graph theory and commutative algebra. In Sect. 3, we prove the main result of this paper (Theorem 3.5, Corollary 3.6).

2 Preliminaries

In this section, we discuss some fundamental notation and terminology used in the paper. The vertex set and the edge set of a finite simple graph G, are denoted throughout by V(G) and E(G), respectively. For $A \subset V(G)$, the induced subgraph of G on A, denoted by G[A], is the graph with vertex set A and edge set $\{\{u,w\} \in E(G):u,w \in A\}.$ For simplicity, we use notation $G {\setminus } A$ for the induced subgraph of G on $V(G) {\setminus } A.$ For a vertex $v \in V(G)$, the set

$$N_G(v) = \{u \in V(G):\{u,v\} \in E(G)\}$$

is called the neighborhood of v in G. The set $N_G[v] = N_G(v) \cup \{v\}$ is called the closed neighborhood of v in G. The degree of vertex v in G is defined by $\deg _G(v) = |N_G(v)|.$

By a walk in G, we mean a sequence $w_1,\dots ,w_{t+1}$ of vertices in G such that $\{w_i,w_{i+1}\} \in E(G)$ for all $1 \le i \le t.$ If a walk $w_1,w_2,\dots ,w_{t+1}$ has all distinct vertices, except possibly $w_1 = w_{t+1}$, then it is called a path of length t. A path $w_1,w_2,\dots ,w_{t+1}$ with $w_1 = w_{t+1}$ is called a cycle. A connected graph without cycles is called a tree. Let G be a tree on the vertex set [n]. Then G is called a star graph if there exists $v \in [n]$ such that $\{v,w\} \in E(G)$ for all $w \in [n] {\setminus } \{v\}.$

Notation: Let G be a simple graph. Then $G^{\circ }$ denotes the graph obtained by removing isolated vertices of G.

Let $\Delta$ be a simplicial complex on the vertex set [n]. A subset $\sigma \subset [n]$ is called a face of $\Delta$ if $\sigma \in \Delta .$ Otherwise, $\sigma$ is called a nonface of $\Delta .$ A maximal face of $\Delta$ with respect to inclusion is called a facet of $\Delta .$ The set of all minimal (with respect to inclusion) nonfaces of $\Delta$ is denoted by $\mathcal {N}(\Delta ).$

Let $\sigma$ be a face of a simplicial complex $\Delta .$ The simplicial complex defined by the formula

$$\begin{aligned} {{\,\textrm{del}\,}}_{\Delta }(\sigma ) = \{\tau \in \Delta : \tau \cap \sigma = \emptyset \} \end{aligned}$$

is called the deletion of $\sigma$, while the simplicial complex defined by the formula

$$\begin{aligned} {{\,\textrm{link}\,}}_{\Delta }(\sigma ) = \{\tau \in \Delta : \tau \cap \sigma = \emptyset , \tau \cup \sigma \in \Delta \} \end{aligned}$$

is called the link of $\sigma .$ Let $v \in [n]$ be a vertex of $\Delta .$ Then we write ${{\,\textrm{link}\,}}_{\Delta }(v)$ and ${{\,\textrm{del}\,}}_{\Delta }(v)$ for ${{\,\textrm{link}\,}}_{\Delta }(\{v\})$ and ${{\,\textrm{del}\,}}_{\Delta }(\{v\})$, respectively.

Definition 2.1

(see [3]) Let $\Delta$ be a simplicial complex. A vertex v of $\Delta$ is called a shedding vertex of $\Delta$ if every facet of ${{\,\textrm{del}\,}}_{\Delta }(v)$ is a facet of $\Delta .$ Further, $\Delta$ is said to be vertex decomposable if either it has only one facet, or it has a shedding vertex v such that both ${{\,\textrm{del}\,}}_{\Delta }(v)$ and ${{\,\textrm{link}\,}}_{\Delta }(v)$ are vertex decomposable.

Observe that v is a shedding vertex of $\Delta$ if and only if no face of ${{\,\textrm{link}\,}}_{\Delta }(v)$ is a facet of ${{\,\textrm{del}\,}}_{\Delta }(v).$ By using inductive argument for the vertex decomposability, we make the following simple observation.

Lemma 2.2

Let $\Delta$ be a simplicial complex on the vertex set [n] and $\{w_1,\dots ,w_r\} \subset [n].$ Further, let $\Omega _0 = \Delta$, $\Omega _i = {{\,\textrm{del}\,}}_{\Omega _{i-1}}(w_i)$ and $\Phi _i = {{\,\textrm{link}\,}}_{\Omega _{i-1}}(w_i)$ for all $1 \le i \le r.$ If

• (i) $\Omega _r$ is a vertex decomposable simplicial complex,

• (ii) $\Phi _i$ is a vertex decomposable simplicial complex, for all $1 \le i \le r$, and

• (iii) $w_i$ is a shedding vertex of $\Omega _{i-1}$, for all $1 \le i \le r$,

then $\Delta$ is a vertex decomposable simplicial complex.

Let $\mathbb {K}$ be a field. Throughout, we write R for the polynomial ring $\mathbb {K}[x_1,\dots ,x_n].$ For a subset $\sigma$ of [n], $x^{\sigma }$ denotes the monomial $\prod _{i \in \sigma } x_i$ in R. Let $\Delta$ be a simplicial complex on the vertex set [n]. Then the Stanley–Reisner ideal of $\Delta$, denoted by $I_{\Delta }$, is the square-free monomial ideal of R given by

$$I_{\Delta } = \langle x^{\sigma }: \sigma \in \mathcal {N}(\Delta ) \rangle .$$

Further, let I be a square-free monomial ideal of R. Then the Stanley–Reisner complex of I is the simplicial complex given by $\Delta _I = \{\sigma \subset [n]: x^{\sigma } \notin I\}.$

Remark 2.3

(see [7]) Let $\Delta$ be a simplicial complex on the vertex set [n] and v be a vertex in $\Delta .$ Then, we have $I_{{{\,\textrm{del}\,}}_{\Delta }(v)} = I_\Delta + \langle x_v \rangle$ and $I_{{{\,\textrm{link}\,}}_{\Delta }(v)} = (I_{\Delta }: \langle x_v \rangle ) + \langle x_v \rangle .$

The t-path ideal of a graph G is the main object of study in this paper.

Definition 2.4

Let G be a graph on the vertex set [n] and $t \ge 1$ be an integer. Then the t -path ideal of G, denoted by $I_t(G)$, is the square-free monomial ideal of R defined by

$$\begin{aligned} I_t(G) = \langle x_{w_1} \cdots x_{w_{t+1}}: w_1,\dots ,w_{t+1} ~\text{ is } \text{ a } \text{ path } \text{ in }~ G \rangle . \end{aligned}$$

Specifically, the ideal $I_1(G)$, which is denoted by I(G), is known as the edge ideal of G.

Let P be a finitely generated $\mathbb {Z}$-graded R-module and $\beta _{i,j}^R(P)$ denote $(i,j)^{th}$ graded Betti number of P. We say that P has a linear resolution if there exists a integer d such that $\beta _{i,i+b}^R(P) = 0$ for all i and for all $b \ne d.$

Definition 2.5

Let I be a homogeneous ideal of R. For $j \in \mathbb {N}$, let $I_{<j>}$ denote the ideal generated by all homogeneous elements of I of degree j. We say that I is componentwise linear if $I_{<j>}$ has a linear resolution for all j.

Definition 2.6

Let $I \subset R$ be a monomial ideal. We say that I has linear quotients if there exists an ordering $M_1,\dots ,M_r$ of minimal generators of I such that the ideal $\langle M_1,\dots ,M_{i-1} \rangle : \langle M_i \rangle$ is generated by a subset of $\{x_1,\dots ,x_n\}$ for all $2 \le i \le r.$

The following lemma is an important tool to determine componentwise linearity of a square-free monomial ideal.

Lemma 2.7

Let $I \subset R$ be a square-free monomial ideal. If $\Delta _I$ is vertex decomposable, then the Alexander dual $I^{\vee }$ of I has linear quotients, and hence it is componentwise linear.

Proof

The conclusion follows directly from [9, Theorem 8.2.5].

3 The Stanley–Reisner Complex of t-Path Ideals

The main aim of this section is to prove that the Stanley–Reisner complex $\Delta _{I_2(G)}$ is vertex decomposable when G is a tree. We start by making the following observation.

Lemma 3.1

Let $w \in [n]$ and $I \subset R$ be a square-free monomial ideal. If there exists $u \in [n]$ with $u \ne w$ such that $x_ux_w \in I$ and $x_w | M$ for all monomials M in I with $x_u | M$, then w is a shedding vertex of $\Delta _I.$

Proof

Let $\sigma$ be a facet of ${{\,\textrm{del}\,}}_{\Delta _I}(w).$ Then there exists a facet $\sigma '$ of $\Delta _I$ such that $\sigma \subset \sigma '.$ It is sufficient to prove that $\sigma = \sigma '.$ For a contradiction, suppose $\sigma \ne \sigma '$ and $z \in \sigma ' {\setminus } \sigma .$ If $z \ne w$, then $\sigma \cup \{z\} \in {{\,\textrm{del}\,}}_{\Delta _I}(w)$ which contradicts the fact that $\sigma$ is a facet of ${{\,\textrm{del}\,}}_{\Delta _I}(w).$ Thus we must have $z = w$ or, equivalently, $\sigma ' = \sigma \cup \{w\}.$ Since $x_ux_w \in I$, it follows that $\{u,w\} \notin \Delta _I$, and hence $u \notin \sigma .$ This implies that $\sigma \cup \{u\} \notin {{\,\textrm{del}\,}}_{\Delta _I}(w).$ Now the fact $u \ne w$ implies that $\sigma \cup \{u\} \notin \Delta _I$ which means $x^{\sigma \cup \{u\}} \in I.$ By given hypothesis, $x_w | x^{\sigma \cup \{u\}}$, i.e. $w \in \sigma$, a contradiction. Therefore, $\sigma = \sigma '.$

Now we introduce some notation that will be used repeatedly in the remaining part of this article. Let G be a graph on the vertex set [n]. For a vertex $v \in [n]$, we set

$$\begin{aligned} \mathcal {O}_G(v) = \{w \in N_G(v): \deg _G(w) = 1\} \end{aligned}$$

and

$$\begin{aligned} \mathcal {T}_G(v) = \{w \in N_G(v): \deg _G(w) = 2, \deg _G(u) = 1 ~\text{ for } \text{ all }~ u \in N_G(w) {\setminus } \{v\}\}. \end{aligned}$$

Further, we set

$$\begin{aligned} \mathcal {U}_G(v) = \{u \in V(G): u \in N_G(w) {\setminus } \{v\} ~\text{ for } \text{ some }~ w \in \mathcal {T}_G(v)\}. \end{aligned}$$

It is important to note that $\mathcal {U}_G(v) = \left( \bigcup _{w \in \mathcal {T}_G(v)} N_G(w) \right) {\setminus } \{v\}.$ Thus, we have $|\mathcal {T}_G(v)| = |\mathcal {U}_G(v)|.$

The following lemma works as a tool for the proof of Theorem 3.5.

Lemma 3.2

Let G be a tree on the vertex set [n] with $n \ge 4.$ Then there exists a vertex $v \in V(G)$ satisfying at least one of the following conditions:

• (i) $\deg _G(v) \ge 3$, $|\mathcal {O}_G(v)| \ge 2$ and $|N_G(v) {\setminus } \mathcal {O}_G(v)| \le 1.$

• (ii) $\deg _G(v) = 2$ and $\mathcal {T}_G(v) \ne \emptyset .$

• (iii) $\deg _G(v) \ge 3$, $\mathcal {T}_G(v) \ne \emptyset$ and $|N_G(v) {\setminus } (\mathcal {O}_G(v) \cup \mathcal {T}_G(v))| \le 1.$

Proof

We prove the result by using induction on $n = |V(G)|.$ If $n = 4$, then either G is a path graph, or G is a star graph. Hence the result holds.

Now suppose that $n > 4.$ Let $\xi \in V(G)$ be such that $\deg _G(\xi ) = 1$ and $G' = G {\setminus } \{\xi \}.$ Further, let $N_G(\xi ) = \{\zeta \}.$ By induction hypothesis, there exists a vertex $v \in V(G')$ which satisfies at least one of the conditions (i), (ii), or (iii) as stated in the lemma. We consider the following three cases.

Case 1. When $\deg _{G'}(v) \ge 3$, $|\mathcal {O}_{G'}(v)| \ge 2$ and $|N_{G'}(v) {\setminus } \mathcal {O}_{G'}(v)| \le 1.$ In this case, $\deg _G(v) \ge 3.$ If $\zeta \notin \mathcal {O}_{G'}(v)$, then $\mathcal {O}_{G'}(v) \subset \mathcal {O}_G(v)$ and $N_G(v) {\setminus } \mathcal {O}_G(v) = N_{G'}(v) {\setminus } \mathcal {O}_{G'}(v).$ This implies that v satisfies the condition (i). On the other hand, if $\zeta \in \mathcal {O}_{G'}(v)$, then $\{\zeta \} \subset \mathcal {T}_G(v)$ and $N_G(v) \setminus (\mathcal {O}_G(v) \cup \mathcal {T}_G(v)) \subset N_{G'}(v) \setminus \mathcal {O}_{G'}(v).$ Thus, v satisfies the condition (iii).

Case 2. When $\deg _{G'}(v) = 2$ and $\mathcal {T}_{G'}(v) \ne \emptyset .$ Then $|\mathcal {U}_{G'}(v)| = 1.$ Let $\mathcal {T}_{G'}(v) = \{w\}$ and $\mathcal {U}_{G'}(v) = \{u\}.$ If $\zeta \notin \{u,v,w\}$, then $\deg _G(v) = 2$ and $\mathcal {T}_G(v) \ne \emptyset .$ Thus, v satisfies the condition (ii). Now suppose that $\zeta \in \{u,v,w\}.$ If

• $u = \zeta$, then $\deg _G(w) = 2$ and $\{u\} \subset \mathcal {T}_G(w)$ which means w satisfies the condition (ii).

• $w = \zeta$, then $\mathcal {O}_G(w) = \{u,\xi \}$ and $N_G(w) {\setminus } \mathcal {O}_G(w) =\{v\}$ which means w satisfies the condition (i).

• $v = \zeta$, then $\deg _G(v) = 3$, $\mathcal {T}_G(v) = \{w\}$ and $\mathcal {O}_G(v) = \{\xi \}$, means v satisfies the condition (iii).

Case 3. When $\deg _{G'}(v) \ge 3$, $\mathcal {T}_{G'}(v) \ne \emptyset$ and $|N_{G'}(v) {\setminus } (\mathcal {O}_{G'}(v) \cup \mathcal {T}_{G'}(v))| \le 1.$ If $\zeta \notin \mathcal {T}_{G'}(v) \cup \mathcal {U}_{G'}(v)$, then $\deg _G(v) \ge \deg _{G'}(v)$, $\mathcal {T}_{G'}(v) \subset \mathcal {T}_G(v)$ and

$$\begin{aligned} N_G(v) \setminus (\mathcal {O}_G(v) \cup \mathcal {T}_G(v)) \subset N_{G'}(v) \setminus (\mathcal {O}_{G'}(v) \cup \mathcal {T}_{G'}(v)). \end{aligned}$$

This implies that v satisfies the condition (iii). Now suppose that $\zeta \in \mathcal {T}_{G'}(v) \cup \mathcal {U}_{G'}(v).$ If

• $\zeta \in \mathcal {T}_{G'}(v)$, then $N_{G'}(\zeta ) {\setminus } \{v\} = \{u\}$ for some $u \in V(G')$ with $\deg _G(u) = 1.$ Thus $\deg _G(\zeta ) = 3$, $\mathcal {O}_G(\zeta ) = \{\xi ,u\}$ which means $\zeta$ satisfies the condition (i).

• $\zeta \in \mathcal {U}_{G'}(v)$, then there exists $w \in \mathcal {T}_{G'}(v)$ such that $N_{G'}(w) = \{\zeta ,v\}.$ Note that $\deg _G(w) = 2$ and $\mathcal {T}_G(w) = \{\zeta \}$ which means w satisfies the condition (ii).

This completes the proof of the lemma.

Let G be a tree on the vertex set [n] with $n \ge 3$ and $I_2(G)$ be the 2-path ideal of G. For a monomial $M = x_{w_1}x_{w_2}x_{w_3} \in I_2(G)$ with $\deg _G(w_1) = 1$, we set $M^{*} = x_{w_1}x_{w_2}$ and $M_{*} = x_{w_2}x_{w_3}.$ We use the notation $\Lambda [G]$ for the set

$$\begin{aligned} \{x_{w_1}x_{w_2}x_{w_3}: w_1,w_2,w_3 ~\text{ is } \text{ a }~ 2 \text{-path } \text{ in }~ G, \deg _G(w_1) = 1\} \end{aligned}$$

of monomial in $I_2(G).$ If $\chi ^{*},\chi _{*} \subset \Lambda [G]$ with $\chi ^{*} \cap \chi _{*} = \emptyset$, then we set

$$\begin{aligned} I_G[\chi ^{*},\chi _{*}] = I_2(G) + \langle M^{*}: M \in \chi ^{*} \rangle + \langle M_{*}: M \in \chi _{*} \rangle . \end{aligned}$$

Note that $I_G[\emptyset ,\emptyset ] = I_2(G).$ Furthermore, if M is a square-free monomial in R, then we set $\sigma _M = \{w \in [n]: x_w | M\}.$

Remark 3.3

With the above notation, we consider the sets $\mathcal {N}_1 = \{\sigma _{M^{*}}: M \in \chi ^{*}\}$, $\mathcal {N}_2 = \{\sigma _{M_{*}}: M \in \chi _{*}\}$ and

$$\begin{aligned} \mathcal {N}_3= & {} \{\{w_1,w_2,w_3\}: w_1,w_2,w_3 ~\text{ is } \text{ a }~ 2 \text{-path } \text{ in }~ G, \sigma \not \subset \{w_1,w_2,w_3\} \\{} & {} \quad ~\text{ for } \text{ all }~ \sigma \in \mathcal {N}_1 \cup \mathcal {N}_2\}. \end{aligned}$$

• (i) We have $\mathcal {N}(\Delta _{I_G[\chi ^{*},\chi _{*}]}) = \mathcal {N}_1 \cup \mathcal {N}_2 \cup \mathcal {N}_3.$

• (ii) If $\{w_1,w_2\} \in \mathcal {N}_1 \cup \mathcal {N}_2$, then $\{w_1,w_2\} \in E(G).$

• (iii) Let $w_1,w_2 \in [n]$ with $\deg _G(w_1) = 1$ and $w_2 \in N_G(w_1).$ Then $w_2 \in \sigma$ for all $\sigma \in \mathcal {N}(\Delta _{I_G[\chi ^{*},\chi _{*}]})$ with $w_1 \in \sigma .$

We need the following lemma for the proof of Theorem 3.5.

Lemma 3.4

Let G be a tree on the vertex set [n] with $n \ge 3$ and $\chi ^{*},\chi _{*} \subset \Lambda [G].$ Further, let v be a vertex of G.

• (i) If $|\mathcal {O}_G(v)| \ge 2$, then v is a shedding vertex of $\Delta _{I_G[\chi ^{*},\chi _{*}]}.$

• (ii) Let $w \in \mathcal {T}_G(v)$ and $N_G(w) = \{u,v\}.$ If $\{u,w\} \in \mathcal {N}(\Delta _{I_G[\chi ^{*},\chi _{*}]})$, then w is a shedding vertex of $\Delta _{I_G[\chi ^{*},\chi _{*}]}.$ Otherwise, v is a shedding vertex of $\Delta _{I_G[\chi ^{*},\chi _{*}]}.$

Proof

For convenience, we write $\mathcal {D}= \Delta _{I_G[\chi ^{*},\chi _{*}]}.$

(i) Let $w,w' \in \mathcal {O}_G(v)$ with $w \ne w'.$ Then $x_wx_vx_{w'} \in I_G[\chi ^{*},\chi _{*}]$, or equivalently $\{w,v,w'\}$ is a nonface of $\mathcal {D}.$ It follows from Remark 3.3(i) and (ii) that $\{w,v\} \in \mathcal {N}(\mathcal {D})$, $\{w',v\} \in \mathcal {N}(\mathcal {D})$, or $\{u,v,w'\} \in \mathcal {N}(\mathcal {D}).$ If $\{w,v\} \in \mathcal {N}(\mathcal {D})$, or $\{w',v\} \in \mathcal {N}(\mathcal {D})$, then it follows from Lemma 3.1 and Remark 3.3(iii) that v is a shedding vertex of $\mathcal {D}.$ Now assume that $\{w,v,w'\} \in \mathcal {N}(\mathcal {D}).$ Let $\sigma \in {{\,\textrm{link}\,}}_{\mathcal {D}}(v).$ Then $\{w,v,w'\} \not \subset \sigma \cup \{v\}$, i.e. $\{w,w'\} \not \subset \sigma .$ Let $z \in \{w,w'\} {\setminus } \sigma .$ We claim that $\sigma \cup \{z\} \in \mathcal {D}.$ For a contradiction, suppose $\sigma \cup \{z\} \notin \mathcal {D}.$ Then there exists $\sigma ' \in \mathcal {N}(\mathcal {D})$ such that $\sigma ' \subset \sigma \cup \{z\}.$ Since $\sigma \in \mathcal {D}$, we obtain $z \in \sigma '.$ Observe that $\deg _G(z) = 1$ and $v \in N_G(z).$ Thus, by Remark 3.3(iii), we obtain that $v \in \sigma '.$ This implies that $v \in \sigma$, a contradiction. Thus we must have $\sigma \cup \{z\} \in \mathcal {D}.$ Since $v \notin \sigma \cup \{z\}$, we obtain that $\sigma \cup \{z\} \in {{\,\textrm{del}\,}}_{\mathcal {D}}(v)$, and hence v is a shedding vertex of $\mathcal {D}.$

(ii) If $\{u,w\} \in \mathcal {N}(\mathcal {D})$, then it follows from Lemma 3.1 and Remark 3.3(iii) that w is a shedding vertex of $\mathcal {D}.$ Now assume that $\{u,w\} \notin \mathcal {N}(\mathcal {D}).$ Since $\{u,w,v\}$ is a nonface of $\mathcal {D}$, it follows from Remark 3.3(i) and (ii) that $\{w,v\} \in \mathcal {N}(\mathcal {D})$ or $\{u,w,v\} \in \mathcal {N}(\mathcal {D}).$ Let $\sigma \in {{\,\textrm{link}\,}}_{\mathcal {D}}(v).$ Then $\{u,w,v\} \not \subset \sigma \cup \{v\}.$ Let $z \in \{u,w\}$ be such that $z \notin \sigma .$ We claim that $\sigma \cup \{z\} \in \mathcal {D}.$ For a contradiction, suppose $\sigma \cup \{z\} \notin \mathcal {D}.$ Then there exists $\sigma ' \in \mathcal {N}(\mathcal {D})$ such that $\sigma ' \subset \sigma \cup \{z\}.$ Since $\sigma \in \mathcal {D}$, we obtain $z \in \sigma '.$ Further, since $\sigma ' \in \mathcal {N}(\mathcal {D})$, by Remark 3.3(i), we obtain that $\sigma ' \in \mathcal {N}_1 \cup \mathcal {N}_2$, or $\sigma ' \in \mathcal {N}_3.$ In the former case, it follows from Remark 3.3(ii) that $\sigma ' = \{z,z'\}$ for some $z' \in N_G(w) = \{u,v\}.$ Since $\{u,w\} \notin \mathcal {N}(\mathcal {D})$, we get $z' = v$, i.e $v \in \sigma '.$ In the later case, $\sigma '$ contains precisely vertices of a 2-path in G. Since very 2-path in G passing through $z \in \{u,w\}$ contains v, we obtain that $v \in \sigma '.$ Thus in both cases, we get $v \in \sigma '$ which implies that $v \in \sigma$, a contradiction. Thus $\sigma \cup \{z\} \in \mathcal {D}.$ Now the fact $v \notin \sigma \cup \{z\}$ implies that $\sigma \cup \{z\} \in {{\,\textrm{del}\,}}_{\mathcal {D}}(v)$, and hence v is a shedding vertex of $\mathcal {D}.$

Theorem 3.5

Let G be a tree on the vertex set [n] with $n \ge 3.$ Then $\mathcal {D}= \Delta _{I_G[\chi ^{*},\chi _{*}]}$ is vertex decomposable for all $\chi ^{*},\chi _{*} \subset \Lambda [G].$ In particular, $\Delta _{I_2(G)}$ is vertex decomposable.

Proof

We prove the result by using induction on $n = |V(G)|.$ For $n = 3$, the result follows from [2, Lemma 3.9]. Assume that $n > 3$ and that the result is true for all $n' < n.$ Let v be a vertex of G established in Lemma 3.2. Now, we have the following two cases.

Case 1. When $|\mathcal {O}_G(v)| \ge 2$ and $|N_G(v) {\setminus } \mathcal {O}_G(v)| \le 1.$ In this case, it follows from Lemma 3.4 that v is a shedding vertex of $\mathcal {D}.$

Let $\Upsilon ^{*} = \{M \in \chi ^{*}: \sigma _M \subset V((G {\setminus } \{v\})^\circ )\}$ and $\Upsilon _{*} = \{M \in \chi _{*}: \sigma _M \subset V((G {\setminus } \{v\})^\circ )\}.$ Then we have $\Upsilon ^{*},\Upsilon _{*} \subset \Lambda [(G {\setminus } \{v\})^\circ ]$ with $\Upsilon ^{*} \cap \Upsilon _{*} = \emptyset .$ Using Remark 2.3, we obtain ${{\,\textrm{del}\,}}_{\mathcal {D}}(v) = \Delta _I$, where

$$\begin{aligned} I = \langle x_v \rangle + I_{(G {\setminus } \{v\})^\circ }[\Upsilon ^{*},\Upsilon _{*}]. \end{aligned}$$

Thus, by induction hypothesis, ${{\,\textrm{del}\,}}_{\mathcal {D}}(v)$ is vertex decomposable.

Now assume that $N_G(v) = \{w_1,w_2,\dots ,w_r\}$ with $\deg _G(w_i) = 1$ for all $1 \le i \le s$, where $s \in \{r-1,r\}.$ Let

$$\begin{aligned} \Theta ^{*} = \{M \in \chi ^{*}: \sigma _M \subset V((G {\setminus } N_G(v))^\circ )\} \end{aligned}$$

and

$$\begin{aligned} \Theta _{*} = \{M \in \chi _{*}: \sigma _M \subset V((G {\setminus } N_G(v))^\circ )\}. \end{aligned}$$

Then $\Theta ^{*},\Theta _{*} \subset \Lambda [(G {\setminus } N_G(v))^\circ ]$ with $\Theta ^{*} \cap \Theta _{*} = \emptyset .$ Further, let $A = \{w \in N_G(v): x_wx_v \in I_G[\chi ^{*},\chi _{*}]\}$ and K be the complete graph on the vertex set $N_G(v) {\setminus } A.$ If either $s = r$, or $s = r-1$ and $w_r \in A$, then by Remark 2.3, we get ${{\,\textrm{link}\,}}_{\mathcal {D}}(v) = \Delta _J$, where

$$\begin{aligned} J = \langle x_w: w \in A \cup \{v\} \rangle + I(K) + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

Thus, by [12, Theorem 3.6] and induction hypothesis, ${{\,\textrm{link}\,}}_{\mathcal {D}}(v)$ is vertex decomposable. On the other hand, suppose $s = r-1$ and $w_r \notin A.$ If $N_G(v) {\setminus } A = \{w_{i_1},\dots ,w_{i_t} = w_r\}$, then we set $\Omega _1 = {{\,\textrm{link}\,}}_{\mathcal {D}}(v)$, $\Omega _m = {{\,\textrm{del}\,}}_{\Omega _{m-1}}(w_{i_m})$ and $\Phi _m = {{\,\textrm{link}\,}}_{\Omega _{m-1}}(w_{i_m})$ for all $2 \le m \le t.$ Let $2 \le m \le t$ and $K_{m-1}$ be the complete graph on the vertex set $\{w_{i_1},w_{i_m},\dots ,w_{i_t}\}.$ Then by Remark 2.3, we get $\Omega _{m-1} = \Delta _{J_{m-1}}$, where

$$\begin{aligned} J_{m-1} =&~\langle x_w: w \in A \cup \{v\} \rangle + \langle x_{w_{i_2}},\dots ,x_{w_{i_{m-1}}} \rangle + I(K_{m-1})\\&+ \langle x_{w_r}x_u: u \in N_G(w_r) {\setminus } \{v\} \rangle + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

Since $x_{w_{i_1}}x_{w_{i_m}} \in J_{m-1}$ and $(x_{w_{i_m}}M) / x_{w_{i_1}} \in J_{m-1}$ for all monomials M in $J_{m-1}$ with $x_{w_{i_1}} | M$, one can easily prove that $w_{i_m}$ is a shedding vertex of $\Omega _{m-1}.$ Again, it follows from Remark 2.3 that $\Phi _m = \Delta _{J'_m}$ and $\Omega _t = \Delta _{J_t}$, where

$$\begin{aligned} J'_m = \langle x_w: w \in A \cup \{v\} \rangle + \langle x_{w_{i_1}},\dots ,x_{w_{i_t}} \rangle + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}] \end{aligned}$$

and

$$\begin{aligned} J_t = \langle x_w: w \in A \cup \{v\} \rangle + \langle x_{w_{i_2}},\dots ,x_{w_{i_t}} \rangle + I_{(G \setminus N_G(v))^\circ }[\Theta ^{*}, \Theta _{*}]. \end{aligned}$$

By induction hypothesis, both $\Omega _t$ and $\Phi _m$ are vertex decomposable for all $2 \le m \le t.$ Using Lemma 2.2, we obtain that ${{\,\textrm{link}\,}}_{\mathcal {D}}(v)$ is vertex decomposable.

Case 2. When $\deg _G(v) \ge 2$, $\mathcal {T}_G(v) \ne \emptyset$ and $|N_G(v) {\setminus } (\mathcal {O}_G(v) \cup \mathcal {T}_G(v))| \le 1.$ If $|\mathcal {O}_G(v)| \ge 2$, then the result holds in view of Case 1. Therefore, we assume that $|\mathcal {O}_G(v)| \le 1.$ Consider the sets of vertices $A = \{w \in N_G(v): x_wx_v \in I_G[\chi ^{*},\chi _{*}]\}$ and

$$\begin{aligned} U = \{w \in \mathcal {T}_G(v): x_ux_w \in I_G[\chi ^{*},\chi _{*}] ~\text{ for } \text{ some }~ u \in N_G(w) {\setminus } \{v\}\}. \end{aligned}$$

We split the proof into the following two subcases.

Subcase 2(a). When $U = \emptyset .$ In this case, it follows from Lemma 3.4 that v is a shedding of $\mathcal {D}.$ Let $\Upsilon ^{*} = \{M \in \chi ^{*}: \sigma _M \subset V((G {\setminus } \{v\})^\circ )\}$ and $\Upsilon _{*} = \{M \in \chi _{*}: \sigma _M \subset V((G {\setminus } \{v\})^\circ )\}.$ Then $\Upsilon ^{*},\Upsilon _{*} \subset \Lambda [(G {\setminus } \{v\})^\circ ]$ and $\Upsilon ^{*} \cap \Upsilon _{*} = \emptyset .$ By Remark 2.3, we get ${{\,\textrm{del}\,}}_{\mathcal {D}}(v) = \Delta _I$, where

$$\begin{aligned} I = \langle x_v \rangle + I_{(G {\setminus } \{v\})^\circ }[\Upsilon ^{*},\Upsilon _{*}]. \end{aligned}$$

Thus, by induction hypothesis, ${{\,\textrm{del}\,}}_{\mathcal {D}}(v)$ is vertex decomposable.

Let $\Theta ^{*} = \{M \in \chi ^{*}: \sigma _M \subset V((G {\setminus } N_G(v))^\circ )\}$ and $\Theta _{*} = \{M \in \chi _{*}: \sigma _M \subset V((G {\setminus } N_G(v))^\circ )\}.$ Then $\Theta ^{*},\Theta _{*} \subset \Lambda [(G {\setminus } N_G(v))^\circ ]$ and $\Theta ^{*} \cap \Theta _{*} = \emptyset .$ We set $B = \mathcal {T}_G(v) \cup \mathcal {O}_G(v) \cup A.$ If K is the complete graph on the vertex set $N_G(v) {\setminus } A$, then by Remark 2.3, ${{\,\textrm{link}\,}}_{\mathcal {D}}(v) = \Delta _J$, where

$$\begin{aligned} J = ~&\langle x_w: w \in A \cup \{v\} \rangle + \langle x_ux_z: z \in \mathcal {T}_G(v) {\setminus } A, u \in N_G(z) {\setminus } \{v\} \rangle \\&+ \langle x_{\xi }x_{\phi }: \xi \in N_G(v) {\setminus } B, \phi \in N_G(\xi ) {\setminus } \{v\} \rangle + I(K) + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

Note that $|N_G(v) {\setminus } B| \le 1.$ First suppose that $\mathcal {O}_G(v) \not \subset A$ and $\mathcal {O}_G(v) {\setminus } A = \{\zeta \}.$ If $N_G(v) {\setminus } A = \{\zeta ,w_1,\dots ,w_r\}$ (with assumption $w_r = \xi$ when $N_G(v) {\setminus } B = \{\xi \}$), then we set $\Omega _0 = {{\,\textrm{link}\,}}_{\mathcal {D}}(v)$, $\Omega _k = {{\,\textrm{del}\,}}_{\Omega _{k-1}}(w_k)$ and $\Phi _k = {{\,\textrm{link}\,}}_{\Omega _{k-1}}(w_k)$ for all $1 \le k \le r.$ By proceeding as in Case 1, we can show that $\Omega _r$ and $\Phi _k$ are vertex decomposable for all $1 \le k \le r$, and $w_k$ is a shedding vertex of $\Omega _{k-1}$ for all $1 \le k \le r.$ Using Lemma 2.2, we obtain that ${{\,\textrm{link}\,}}_{\mathcal {D}}(v)$ is vertex decomposable.

Now suppose that $\mathcal {O}_G(v) \subset A.$ Let $\mathcal {T}_G(v) {\setminus } A = \{w_1,\dots ,w_r\}.$ Then we set $\Omega _0 = {{\,\textrm{link}\,}}_{\mathcal {D}}(v)$, $\Omega _m = {{\,\textrm{del}\,}}_{\Omega _{m-1}}(w_m)$ and $\Phi _m = {{\,\textrm{link}\,}}_{\Omega _{m-1}}(w_m)$ for all $1 \le m \le r.$ Further, let $1 \le m \le r$ and $K_{m-1}$ be the complete graph on the vertex set $N_G(v) {\setminus } (A \cup \{w_1,\dots ,w_{m-1}\})$, where $\{w_1,\dots ,w_{m-1}\} = \emptyset$ for $m = 1.$ Then, by Remark 2.3, we get $\Omega _{m-1} = \Delta _{J_{m-1}}$, where

$$\begin{aligned} J_{m-1} = ~&\langle x_w: w \in A \cup \{v,w_1,\dots ,w_{m-1}\} \rangle \\&+ \langle x_ux_z: z \in \mathcal {T}_G(v) {\setminus } (A \cup \{v,w_1,\dots ,w_{m-1}\}), u \in N_G(z) {\setminus } \{v\} \rangle \\&+ \langle x_{\xi }x_{\phi }: \xi \in N_G(v) {\setminus } B, \phi \in N_G(\xi ) {\setminus } \{v\} \rangle + I(K_{m-1}) + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

Suppose $u \in N_G(w_m) {\setminus } \{v\}.$ Then $x_ux_{w_m} \in J_{m-1}$ and $x_{w_m} | M$ for all monomials M in $J_{m-1}$ with $x_u | M.$ Thus, it follows from Lemma 3.1 that $w_m$ is a shedding vertex of $\Omega _{m-1}.$ In view of Remark 2.3, we also conclude that $\Phi _m = \Delta _{J'_m}$, where

$$\begin{aligned} J'_m = \langle x_w: w \in N_G(v) \cup \{u\} \rangle + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

By induction hypothesis, $\Phi _m$ is vertex decomposable. Finally, using Remark 2.3, we see that $\Omega _r = \Delta _{J_r}$, where

$$\begin{aligned} J_r = \langle x_w: w \in B \cup \{v\} \rangle + \langle x_{\xi }x_{\phi }: \xi \in N_G(v) {\setminus } B, \phi \in N_G(\xi ) {\setminus } \{v\} \rangle + I_{(G {\setminus } B)^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

To verify that $\Omega _r$ is vertex decomposable, let $v' \notin V(G)$ be a vertex and $G'$ be the tree on the vertex set $V(G') = V((G {\setminus } B)^\circ ) \cup \{v'\}$ and the edge set $E(G') = E((G {\setminus } B)^\circ ) \cup \{\{v',\xi \}\}$, where $\xi \in N_G(v) {\setminus } B.$ Then we can write

$$\begin{aligned} J_r = \langle x_w: w \in B \cup \{v\} \rangle + I_{G'}[\Theta ^{*},\Theta _{*} \cup \{x_{v'}x_{\xi }x_{\phi }: \phi \in N_G(\xi ) {\setminus } \{v\}\}]. \end{aligned}$$

Thus, by induction hypothesis, $\Omega _r$ is vertex decomposable. By Lemma 2.2, we deduce that ${{\,\textrm{link}\,}}_{\mathcal {D}}(v)$ is vertex decomposable.

Subcase 2(b). When $U \ne \emptyset .$ Let $w' \in U$ and $u' \in N_G(w') {\setminus } \{v\}.$ Then $x_{u'}x_{w'} \in I_G[\chi ^{*},\chi _{*}].$ It follows from Lemma 3.4 that $w'$ is a shedding vertex of $\mathcal {D}.$ By using similar arguments as in Subcase 2(a), we can prove that ${{\,\textrm{del}\,}}_{\mathcal {D}}(w')$ is vertex decomposable.

Let $\Theta ^{*} = \{M \in \chi ^{*}: \sigma _M \subset V((G {\setminus } W)^\circ )\}$ and $\Theta _{*} = \{M \in \chi _{*}: \sigma _M \subset V((G {\setminus } W)^\circ )\}$, where $W = \mathcal {O}_G(v) \cup \mathcal {T}_G(v).$ Then $\Theta ^{*},\Theta _{*} \subset \Lambda [(G {\setminus } W)^\circ ]$ and $\Theta ^{*} \cap \Theta _{*} = \emptyset .$ Using Remark 2.3, we get ${{\,\textrm{link}\,}}_{\mathcal {D}}(w') = \Delta _J$, where

$$\begin{aligned} J = ~&\langle x_{u'},x_{w'} \rangle + \langle x_ux_w: w \in U {\setminus } \{w'\}, u \in N_G(w) {\setminus } \{v\} \rangle \\&+ \langle x_vx_{\phi }: \phi \in N_G(v) {\setminus } \{w'\} \rangle + I_{(G {\setminus } W)^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

First suppose that $U = \{w'\}.$ Then, we have

$$\begin{aligned} J = ~&\langle x_{u'},x_{w'} \rangle + \langle x_vx_{\phi }: \phi \in N_G(v) {\setminus } \{w'\} \rangle + I_{(G {\setminus } W)^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

We further distinguish between two possibilities:

• If $N_G(v) {\setminus } \{w'\} = \{\xi \}$ for some $\xi \in [n]$, then $J = \langle x_{u'},x_{w'} \rangle + \langle x_vx_{\xi } \rangle + I_{(G {\setminus } W)^\circ }[\Theta ^{*},\Theta _{*}].$ Choose $\zeta \in N_G(\xi )$ such that $x_vx_{\xi }x_{\zeta } \notin \chi ^{*}.$ Then we can rewrite

  $$\begin{aligned} J = \langle x_u,x_w \rangle + I_{(G {\setminus } \{w\})^\circ }[\Theta ^{*} \cup \{x_vx_{\xi }x_{\zeta }\},\Theta _{*}]. \end{aligned}$$

  Thus, by induction hypothesis, ${{\,\textrm{link}\,}}_{\mathcal {D}}(w')$ is vertex decomposable.

• If $\deg _G(v) > 2$, then it follows from Lemma 3.1 that v is a shedding vertex of ${{\,\textrm{link}\,}}_{\mathcal {D}}(w').$ Write $\Omega = {{\,\textrm{link}\,}}_{\mathcal {D}}(w').$ Using Remark 2.3, we obtain ${{\,\textrm{del}\,}}_{\Omega }(v) = \Delta _{L_1}$ and ${{\,\textrm{link}\,}}_{\Omega }(v) = \Delta _{L_2}$, where

  $$\begin{aligned} L_1 = \langle x_{u'},x_{w'},x_v \rangle + I_{(G {\setminus } W)^\circ }[\Theta ^{*},\Theta _{*}] \end{aligned}$$

  and

  $$\begin{aligned} L_2 = \langle x_{u'},x_{w'},x_v \rangle + \langle x_{\phi }: \phi \in N_G(v) {\setminus } \{w'\} \rangle + I_{(G {\setminus } N_G(v))^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

  By induction hypothesis, both ${{\,\textrm{del}\,}}_{\Omega }(v)$ and ${{\,\textrm{link}\,}}_{\Omega }(v)$ are vertex decomposable, and hence $\Omega$ is vertex decomposable.

Now, suppose that $\{w'\} \subsetneq U$ and $(U \setminus \{w'\}) \cup \{v\} = \{w_1, \dots ,w_r\}$ with $w_r = v.$ For each $1 \le k \le r$, let $N_G(w_k) = \{u_k,v\}.$ We set $\Omega _0 = {{\,\textrm{link}\,}}_{\mathcal {D}}(w')$, $\Omega _k = {{\,\textrm{del}\,}}_{\Omega _{k-1}}(w_k)$ and $\Phi _k = {{\,\textrm{link}\,}}_{\Omega _{k-1}}(w_k)$ for all $1 \le k \le r.$ Let $1 \le k \le r.$ Then, using Remark 2.3, we obtain $\Phi _k = \Delta _{J_k}$, where

$$\begin{aligned} J_k = ~&\langle x_{u'},x_{w'},x_{w_1},\dots ,x_{w_k},x_{u_k},x_v \rangle + \langle x_{u_t}x_{w_t}: k < t \le r \rangle \\&+ I_{(G \setminus W)^\circ }[\Theta ^{*}, \Theta _{*}]. \end{aligned}$$

Note that $U \cap V((G {\setminus } W)^\circ ) = \emptyset .$ Thus, by induction hypothesis and [2, Lemma 3.9], $\Phi _k$ is vertex decomposable. Further, in view of Remark 2.3, we observe that $\Omega _r = \Delta _L$, where

$$\begin{aligned} L = ~&\langle x_{u'},x_{w'},x_{w_1},\dots ,x_{w_r} \rangle + \langle x_vx_{\phi }: \phi \in N_G(v) {\setminus } \{w',w_1,\dots ,w_r\} \rangle + I_{(G {\setminus } W)^\circ }[\Theta ^{*},\Theta _{*}]. \end{aligned}$$

By following the same method as above, we can prove that $\Omega _r$ is vertex decomposable. Also, it follows from Lemma 3.1 that $w_k$ is a shedding vertex of $\Omega _{k-1}$ for all $1 \le k \le r.$ Thus, it follows from Lemma 2.2 that ${{\,\textrm{link}\,}}_{\mathcal {D}}(w')$ is vertex decomposable.

The following corollary is a direct consequence of Lemma 2.7 and Theorem 3.5.

Corollary 3.6

Let G be a tree on the vertex set [n] with $n \ge 3.$ Then the Alexander dual $I_2(G)^{\vee }$ of $I_2(G)$ has linear quotients, and hence it is componentwise linear.

We illustrate Theorem 3.5 with the help of following example.

Example 3.7

Let G be the tree as shown in Fig. 1. Then

$$\begin{aligned} I_2(G) = \left\langle \begin{array}{ll} &{} x_1x_2x_3,x_1x_3x_4,x_1x_3x_5,x_2x_3x_4,x_2x_3x_5,\\ &{} x_3x_4x_5,x_3x_5x_6,x_3x_5x_{11},x_5x_6x_{11},x_5x_6x_7,\\ &{} x_6x_7x_8,x_7x_8x_9,x_7x_8x_{10},x_8x_9x_{10}, \\ &{} x_5x_{11}x_{12},x_5x_{11}x_{13},x_{11}x_{12}x_{13},x_{11}x_{13}x_{14} \end{array}\right\rangle . \end{aligned}$$

Let $\mathcal {D}= \Delta _{I_G[\chi ^{*},\chi _{*}]}$, where $\chi ^{*} = \{x_1x_3x_5\}$ and $\chi _{*} = \{x_{11}x_{13}x_{14}\}.$ In view of Lemma 3.4, 3 is a shedding vertex of $\mathcal {D}.$ In order to show that $\mathcal {D}$ is vertex decomposable, we have to show that both ${{\,\textrm{del}\,}}_{\mathcal {D}}(3)$ and ${{\,\textrm{link}\,}}_{\mathcal {D}}(3)$ are vertex decomposable.

Fig. 1

$\Delta _{I_2(G)}$ is vertex decomposable

By Remark 2.3, we obtain ${{\,\textrm{del}\,}}_{\mathcal {D}}(3) = \Delta _I$ and ${{\,\textrm{link}\,}}_{\mathcal {D}}(3) = \Delta _J$, where

$$\begin{aligned} I = \left\langle x_3,x_{11}x_{13},x_5x_6x_{11},x_5x_6x_7,x_6x_7x_8,x_7x_8x_9,x_7x_8x_{10},x_8x_9x_{10},x_5x_{11}x_{12}\right\rangle \end{aligned}$$

and

$$\begin{aligned} J = \left\langle x_1,x_3,x_{11}x_{13},x_2x_4,x_2x_5,x_4x_5,x_5x_6,x_5x_{11},x_6x_7x_8,x_7x_8x_9,x_7x_8x_{10},x_8x_9x_{10} \right\rangle . \end{aligned}$$

One can use inductive argument to check that ${{\,\textrm{del}\,}}_{\mathcal {D}}(3)$ is vertex decomposable. We can rewrite $I = \langle x_3 \rangle + I_{(G {\setminus } \{3\})^\circ }[\emptyset ,\chi _{*}]$ and

$$\begin{aligned} J = \langle x_1,x_3 \rangle + I(K) + \langle x_5x_6,x_5x_{11} \rangle + I_{(G {\setminus } N_G(3))^\circ }[\emptyset ,\chi _{*}], \end{aligned}$$

where K is the complete graph on vertex set $\{2,4,5\}.$ Now, set $\Omega _0 = {{\,\textrm{link}\,}}_{\mathcal {D}}(3)$, $\Omega _k = {{\,\textrm{del}\,}}_{\Omega _{k-1}}(w_k)$ and $\Phi _k = {{\,\textrm{link}\,}}_{\Omega _{k-1}}(w_k)$, for $k = 1,2$, where $w_1 = 4$ and $w_2 = 5.$ One can check that $w_k$ is a shedding vertex of $\Omega _{k-1}.$ Now, by Remark 2.3, we obtain that $\Phi _k = \Delta _{J_k}$, where

$$\begin{aligned} J_1 = \langle x_1,x_2,x_3,x_4,x_5 \rangle + I_{(G {\setminus } N_G(3))^\circ }[\emptyset ,\chi _{*}] \end{aligned}$$

and

$$\begin{aligned} J_2 = \langle x_1,x_2,x_3,x_4,x_5,x_6,x_{11} \rangle + I_{(G {\setminus } N_G(3))^\circ }[\emptyset ,\chi _{*}]. \end{aligned}$$

Also, we have $\Omega _2 = \Delta _L$, where $L = \langle x_1,x_3,x_4,x_5 \rangle + I_{(G {\setminus } N_G(3))^\circ }[\emptyset ,\chi _{*}].$ Again, by using inductive argument, we see that $\Omega _2$, $\Phi _1$ and $\Phi _2$ are vertex decomposable. Using Lemma 2.2, we obtain that ${{\,\textrm{link}\,}}_{\mathcal {D}}(3)$ is vertex decomposable.

The following example shows that $\Delta _{I_t(G)}$ does not need to be vertex decomposable for $t \ge 3.$

Example 3.8

Let $t \ge 3.$ We write $G_t$ for the tree on the vertex set $[t+3]$ with edge set

$$\begin{aligned} E(G) = \{\{1,3\},\{2,3\},\{3,4\},\dots ,\{t,t+1\},\{t+1,t+2\},\{t+1,t+3\}\}. \end{aligned}$$

We show that $\Delta _{I_t(G_t)}$ is not vertex decomposable. We use induction on t. One can check that $\Delta _{I_3(G_3)}$ is not vertex decomposable. Let $t > 3$ and $\mathcal {D}= \Delta _{I_t(G_t)}.$ First we show that $w \in [t+3]$ is a shedding vertex of $\mathcal {D}$ if and only if $w \notin \{1,2,t+2,t+3\}.$ Suppose that $w \notin \{1,2,t+2,t+3\}.$ Let $\sigma \in {{\,\textrm{link}\,}}_{\mathcal {D}}(w).$ Since $x_1x_3x_4\cdots x_{t+1}x_{t+2} \in I_t(G_t)$, there exists $z \in \{1,3,4,\dots ,t+1,t+2\}$ such that $z \notin \sigma .$ The fact $x_w | M$ for all monomials $M \in I_t(G)$ implies that $\sigma \cup \{z\} \in \mathcal {D}.$ Thus, w is a shedding vertex of $\mathcal {D}.$ Conversely, suppose that $w \in \{1,2,t+2,t+3\}.$ By symmetry, we may assume that $w = 1.$ Note that $\{2,\dots ,t+1\} \in {{\,\textrm{link}\,}}_{\mathcal {D}}(w)$ and $\{2,\dots ,t+1\}$ is a facet in ${{\,\textrm{del}\,}}_{\mathcal {D}}(w).$ This implies that w is not a shedding vertex of $\mathcal {D}.$

Now, let $w \in [t+3] {\setminus } \{1,2,t+2,t+3\}.$ Then, by Remark 2.3, we obtain that ${{\,\textrm{link}\,}}_{\mathcal {D}}(w) \simeq \Delta _J$, where $J = I_{t-1}(G_{t-1}) + \langle x_{t+3} \rangle$ (By considering $I_{t-1}(G_{t-1})$ as an ideal in $\mathbb {K}[x_1,\dots ,x_{t+3}]$). By induction hypothesis, ${{\,\textrm{link}\,}}_{\mathcal {D}}(w)$ is not vertex decomposable, and hence $\mathcal {D}$ is not vertex decomposable.