# Powers of edge ideals of edge-weighted trees

Jiaxin Li, Guangjun Zhu∗ and Shiya Duan  School of Mathematical Sciences,
Soochow University, Suzhou, Jiangsu, 215006, P. R. China
lijiaxinworking@163.com(Jiaxin Li), zhuguangjun@suda.edu.cn(Corresponding
author:Guangjun Zhu),3136566920@qq.com(Shiya Duan).

###### Abstract.

This paper gives exact formulas for the regularity of edge ideals of edge-
weighted integrally closed trees. In addition, we provide some linear upper
bounds on the regularity of powers of such ideals.

∗ Corresponding author

2020 Mathematics Subject Classification. Primary 13A15,13D02; Secondary 05E40

Keywords: Regularity, integrally closed, powers of the edge ideal, edge-
weighted tree

##  1\. Introduction

Let $G$ be a graph with vertex set $V(G)=\\{x_{1},\ldots,x_{n}\\}$ and edge
set $E(G)$. We write $xy$ for $\\{x,y\\}$ if $\\{x,y\\}\in E(G)$ is an edge of
$G$ with $x$ and $y$ as endpoints. Suppose $w:E(G)\rightarrow\mathbb{Z}_{>0}$
is an edge weight function on $G$. We write $G_{\omega}$ for the pair
$(G,\omega)$ and call it an edge-weighted graph with the underlying graph $G$.
For a weighted graph $G_{\omega}$, its edge-weighted ideal (or simply edge
ideal), was introduced in [21], is the ideal of the polynomial ring
$S={\mathbb{K}}[x_{1},\dots,x_{n}]$ in $n$ variables over a field
${\mathbb{K}}$ given by

$I(G_{\omega})=(x^{\omega(xy)}y^{\omega(xy)}\mid xy\in E(G_{\omega})).$

If $w$ is the constant function defined by $w(e)=1$ for all $e\in E(G)$, then
$I(G_{\omega})$ is the classical edge ideal of the underlying graph $G$ of
$G_{\omega}$, which has been studied extensively in the literature [1, 2, 16,
18, 24, 25].

Recently, there has been some interest in characterizing weights for which the
edge ideals of edge-weighted graphs are Cohen-Macaulay. For example, Paulsen
and Sather-Wagstaff in [21] classified Cohen-Macaulay edge-weighted graphs
$G_{\omega}$ where the underlying graph $G$ is a cycle, a tree, or a complete
graph. Seyed Fakhari et al. in [22] continued this study by classifying Cohen-
Macaulay edge-weighted graphs $G_{\omega}$ if $G$ is a very well-covered
graph. Recently, Diem et al. in [5] gave a complete characterization of
sequentially Cohen-Macaulay edge-weighted graphs. In [23], Wei classified all
Cohen-Macaulay weighted chordal graphs from a purely graph-theoretic point of
view. Hien in [14] classified Cohen-Macaulay edge-weighted graphs $G_{\omega}$
if $G$ has girth at least $5$.

Integral closure of monomial ideals is also an interesting topic. In [6], we
gave a complete characterization of integrally closed edge-weighted graphs
$G_{\omega}$ and showed that if their underlying graph $G$ is a star graph, a
path, or a cycle, then $G_{\omega}$ is normal. Later, in [26], we gave some
exact formulas for the regularity of powers of edge ideals of edge-weighted
star graphs and integrally closed paths.

The study of edge ideals of edge-weighted graphs is much more recent and
consequently there are fewer results in this direction. In this paper, we
decide to focus on the regularity of powers of the edge ideals of integrally
closed edge-weighted trees. Recall that the regularity is an important
invariant associated to a homogeneous ideal $I$. It is well-known that
$\operatorname{reg}(I^{t})$, as a function in $t$, is asymptotically linear
for $t\gg 0$ (cf. [4, 17]). In general, it is very difficult to decide when
this function starts to be linear. To find the exact form of the linear
function is also not easy (cf. [1, 7, 12, 27]).

This paper is organized as follows. In the next section, we recall several
definitions and terminology that we will need later. In Section 3, by using
the Betti splitting and polarization approaches, we give precise formulas for
the regularity of the edge ideals of edge-weighted integrally closed trees. In
Section 4, by classifying based on the distance from a vertex of the tree to
its longest path containing all non-trivial edges, we give some linear upper
bounds on the regularity of powers of edge ideals of edge-weighted integrally
closed trees.

##  2\. Preliminaries

In this section, we provide the definitions and basic facts that will be used
throughout this paper. For detailed information we refer to [13].

###  2.1. Notions of simple graphs

Throughout the paper, all graphs will be finite and simple, i.e., undirected
graphs with no loops nor multiple edges. Given an edge-weighted graph
$G_{\omega}$, we denote its vertex and edge sets by $V(G_{\omega})$ and
$E(G_{\omega})$, respectively. Any concept valid for graphs automatically
applies to edge-weighted graphs. For example, the _neighborhood_ of a vertex
$v$ in an edge-weighted graph $G_{\omega}$ with the underlying graph $G$ is
defined as $N_{G}(v)\\!:=\\{u\in V(G)\mid uv\in E(G)\\}$. Given a subset $W$
of $V(G_{\omega})$, its _neighborhood_ is defined as
$N_{G}(W)\\!:=\bigcup\limits_{v\in W}N_{G}(v)$. The _induced subgraph_ by $W$
in $G_{\omega}$ is the graph $G_{\omega}[W]$ with vertex set $W$, and for any
$u,v\in V(G_{\omega}[W])$, $uv$ is an edge in $G_{\omega}[W]$ if and only if
$uv$ is an edge in $G_{\omega}$, and the weight function $\omega^{\prime}$
satisfies $\omega^{\prime}(uv)=\omega(uv)$. At the same time, the induced
subgraph of $G_{\omega}$ on the set $V(G_{\omega})\setminus W$ is denoted by
$G_{\omega}\setminus W$ or $G\setminus W$ for simplicity. In particular, if
$W=\\{v\\}$, we write $G\setminus v$ instead of $G\setminus\\{v\\}$ for
simplicity. For any subset $A$ of $E(G_{\omega})$, $G_{\omega}\setminus A$ is
a subgraph of $G_{\omega}$ obtained by removing all edges in $A$. In
particular, if $A=\\{e\\}$ then we also write $G_{\omega}\setminus e$ or
$G\setminus e$ instead of $G_{\omega}\setminus\\{e\\}$.

A _walk_ $W$ of length $n$ in a graph $G$ is a sequence of vertices
$(w_{1},\ldots,w_{n},w_{n+1})$, such that $w_{i}w_{i+1}\in E(G)$ for $1\leq
i\leq n$. The vertices $w_{1}$ and $w_{n+1}$ are connected by $W$ and are
called its ends, the vertices $w_{2},\ldots,w_{n}$ are the inner vertices of
$W$. The walk $W$ is _closed_ if $w_{1}=w_{n+1}$. Furthermore, the walk $W$ is
called a _cycle_ if it is closed and the points $w_{1},\ldots,w_{n}$ are
distinct. At the same time, a _path_ denoted by $P$ is a walk where all points
are distinct. A _tree_ is a connected simple graph without cycles. For a tree
$T$, let $L(T)=\\{v\in V(T)|\deg_{T}(v)=1\\}$ be the set of all leaves of $T$.
The tree $T$ is called to be a _caterpillar_ if $T\backslash L(T)$ is either
empty or is a simple path. A longest path in a caterpillar is called the
_spine_ of the caterpillar. Note that given any spine, every edge of a
caterpillar is incident on it. With respect to a fixed spine $P$, the the
pendant edges incident on $P$ are called _whiskers_.

A matching in a graph is a subset of edges no two of which share of vertex. A
matching is induced if no two vertices belonging to different edges of the
matching are adjacent. In other words, an induced matching in a graph $G$ is
formed by the edges of a 1-regular induced subgraph of $G$. If $G_{\omega}$ is
an edge-weighted graph, its induced matching refers to the induced matching of
its underlying graph $G$. The _induced matching number_ of $G_{\omega}$,
denoted by $\nu(G_{\omega})$ or $\nu(G)$, is the maximum size of an induced
matching in $G$.

An edge-weighted graph is said to be _non-trivial_ if there is at least one
edge with a weight greater than $1$, otherwise, it is said to be trivial. An
edge $e\in E(G_{\omega})$ is said to be non-trivial if its weight $w(e)\geq
2$. Otherwise, it is said to be trivial.

###  2.2. Notions from commutative algebra

Let $S={\mathbb{K}}[x_{1},\ldots,x_{n}]$ be a polynomial ring over a field
${\mathbb{K}}$. Let $M$ be a graded $S$-module with minimal free resolution

$0\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{p,j}}\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{p-1,j}}\rightarrow\cdots\rightarrow\bigoplus\limits_{j}S(-j)^{\beta_{0,j}}\rightarrow
M\rightarrow 0,$

where the maps are exact, $p\leq n$, and $S(-j)$ is the free module obtained
by shifting the degrees in $S$ by $j$. The numbers $\beta_{i,j}$’s are
positive integers and are called the $(i,j)$-th graded Betti number of $M$. An
important homological invariant related to these numbers are the Castelnuovo-
Mumford regularity (or simply regularity), denoted by $\operatorname{reg}(M)$,

$\displaystyle\operatorname{reg}(M)$ $\displaystyle=\mbox{max}\,\\{j-i\ |\
\operatorname{Tor}_{i}(M,{\mathbb{K}})_{j}\neq 0\\}$

The following lemmas are often used to compute the regularity of a module or
ideal.

######  Lemma 2.1.

([15, Lemma 3.1]) Let $0\longrightarrow M\longrightarrow N\longrightarrow
P\longrightarrow 0$ be a short exact sequence of finitely generated graded
$S$-modules. Then

$\operatorname{reg}(N)\leq\max\\{\operatorname{reg}(M),\operatorname{reg}(P)\\}.$

The equality holds if $\operatorname{reg}(P)\neq\operatorname{reg}(M)-1$.

######  Lemma 2.2.

([24, Lemma 3.1]) and [15, Lemma 3.1])  Let
$S_{1}=k\left[x_{1},\ldots,x_{m}\right]$ and
$S_{2}=k\left[x_{m+1},\ldots,x_{n}\right]$ be two polynomial rings,
$I\subseteq S_{1}$ and $J\subseteq S_{2}$ be two nonzero homogeneous ideals.
Then

  * (1)

$\operatorname{reg}(IJ)=\operatorname{reg}(I)+\operatorname{reg}(J)$,

  * (2)

$\operatorname{reg}((I+J)^{t})=\max_{\begin{subarray}{c}i\in[t-1]\\\
j\in[t]\end{subarray}}\\{\operatorname{reg}(I^{t-i})+\operatorname{reg}(J^{i}),\operatorname{reg}(I^{t-j+1})+\operatorname{reg}(J^{j})-1\\}$
for any $t\geq 1$.

Calculating or even estimating the regularity for a general ideal is a
challenging problem. Formulas for $\operatorname{reg}(I)$ in special cases
will be provided using methods developed in [9] and [8]. For a monomial ideal
$I$, let $\mathcal{G}(I)$ denote its unique minimal set of monomial
generators.

######  Definition 2.3.

([9, Definition 1.1]) Let $I$ be a monomial ideal. If there exist monomial
ideals $J$ and $K$ such that $\mathcal{G}(I)=\mathcal{G}(J)\cup\mathcal{G}(K)$
and $\mathcal{G}(J)\cap\mathcal{G}(K)=\emptyset$. Then $I=J+K$ is a Betti
splitting if

$\beta_{i,j}(I)=\beta_{i,j}(J)+\beta_{i,j}(K)+\beta_{i-1,j}(J\cap K)\text{ for
all }i,j\geq 0,$

where $\beta_{i-1,j}(J\cap K)=0$ for $i=0$.

######  Lemma 2.4.

([9, Corollary 2.7]) Suppose $I=J+K$, where $\mathcal{G}(J)$ contains all the
generators of $I$ that are divisible by some variable $x_{i}$, and
$\mathcal{G}(K)$ is a nonempty set containing the remaining generators of $I$.
If $J$ has a linear resolution, then $I=J+K$ is a Betti splitting.

Definition 2.3 states that
$\operatorname{reg}(I)=\max\\{\operatorname{reg}(J),\operatorname{reg}(K),\operatorname{reg}(J\cap
K)-1\\}$, as a result of Betti splitting.

######  Definition 2.5.

([8, Definition 2.1]) Let $I\subset S$ be a monomial ideal with
$\mathcal{G}(I)=\\{u_{1},\ldots,u_{m}\\}$ where
$u_{i}=\prod\limits_{j=1}^{n}x_{j}^{a_{ij}}$ for $i=1,\ldots,m$. The
polarization of $I$, denoted by $I^{\mathcal{P}}$, is a squarefree monomial
ideal in the polynomial ring $S^{\mathcal{P}}$

$I^{\mathcal{P}}=(\mathcal{P}(u_{1}),\ldots,\mathcal{P}(u_{m}))$

where
$\mathcal{P}(u_{i})=\prod\limits_{j=1}^{n}\prod\limits_{k=1}^{a_{ij}}x_{jk}$
is a squarefree monomial in
$S^{\mathcal{P}}={\mathbb{K}}[x_{j1},\ldots,x_{ja_{j}}\mid j=1,\ldots,n]$ and
$a_{j}=\max\\{a_{ij}|i=1,\ldots,m\\}$ for $1\leq j\leq n$.

A monomial ideal and its polarization share many homological and algebraic
properties. The following is a useful property of the polarization.

######  Lemma 2.6.

([13, Corollary 1.6.3]) Let $I\subset S$ be a monomial ideal and
$I^{\mathcal{P}}\subset S^{\mathcal{P}}$ be its polarization. Then

$\beta_{ij}(I)=\beta_{ij}(I^{\mathcal{P}})$

for all $i$ and $j$. In particular,
$\operatorname{reg}(I)=\operatorname{reg}(I^{\mathcal{P}})$.

##  3\. Regularity of the edge ideal of an edge-weighted integrally closed
tree

In this section, we will give precise formulas for the regularity of the edge
ideals of edge-weighted integrally closed trees. We first recall the
definition of the integral closure of an ideal.

######  Definition 3.1.

([13, Definition 1.4.1]) Let $I$ be an ideal in a ring $R$. An element $f\in
R$ is said to be integral over $I$ if there exists an equation

$f^{k}+c_{1}f^{k-1}+\dots+c_{k-1}f+c_{k}=0\text{\ \ with\ \ }c_{i}\in I^{i}.$

The set $\overline{I}$ of elements in $R$ which are integral over $I$ is the
_integral closure_ of $I$. If $I=\overline{I}$, then $I$ is said to be
integrally closed. An edge-weighted graph $G_{\omega}$ is said to be
integrally closed if its edge ideal $I(G_{\omega})$ is integrally closed.

According to [13, Theorem 1.4.6], every edge-weighted graph $G_{\omega}$ with
trivial weights is integrally closed. The following lemma gives a complete
characterization of a non-trivial edge-weighted graph that is integrally
closed.

######  Lemma 3.2.

([6, Theorem 3.6]) If $G_{\omega}$ is a non-trivial edge-weighted graph, then
$I(G_{\omega})$ is integrally closed if and only if $G_{\omega}$ does not
contain any of the following three graphs as induced subgraphs.

  1. (1)

A path $P_{\omega}$ of length $2$ where all edges have non-trivial weights.

  2. (2)

The disjoint union $P_{\omega}\sqcup P_{\omega}$ of two paths $P_{\omega}$ of
length $1$ where all edges have non-trivial weights.

  3. (3)

A $3$-cycle $C_{\omega}$ where all edges have non-trivial weights.

From the lemma above, we can derive

######  Corollary 3.3.

Let $P_{\omega}$ be a non-trivial integrally closed path with $n$ vertices,
then it can have at most two edges with non-trivial weights.

For a trivially weighted tree, we have

######  Lemma 3.4.

([2, Theorem 4.7]) If $G_{\omega}$ is a trivial weighted tree, then

$\operatorname{reg}(S/I(G_{\omega})^{t})=2t+\nu(G)-1$

for all $t\geq 1$, where $\nu(G)$ is the induced matching number of the
underlying graph $G$ of $G_{\omega}$.

Thus, we will now consider a non-trivial edge-weighted integrally closed tree
which satisfies the following conditions.

######  Remark 3.5.

Let $G_{\omega}$ be a non-trivial edge-weighted integrally closed tree, and
let $P_{\omega}$ be the longest path containing all non-trivial edges in
$G_{\omega}$. Then $P_{\omega}$ is an induced subgraph of $G_{\omega}$.
Suppose $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$,
$\omega_{i}=\max\\{\omega_{t}\mid\omega_{t}=\omega(e_{t})\text{\ and
}e_{t}=x_{t}x_{t+1}\text{\ for each }1\leq t\leq k-1\\}$ with $\omega_{i}\geq
2$ and $\omega_{i}\geq\omega_{i+2}$ if $e_{i+2}\in E({P_{\omega}})$ for
simplicity. In this case, $G_{\omega}\backslash e_{i}$ is the disjoint union
of the two trees $G_{\omega}^{1}$ and $G_{\omega}^{2}$ with $x_{i}\in
V({G_{\omega}^{1}})$ and $x_{i+1}\in V({G_{\omega}^{2}})$.

Furthermore, let $s_{i}(G_{\omega})=\max\\{|M|\\!:M\text{\ is an induced
matching of\ }G_{\omega}\text{\ containing}$ $\text{the edge\ }e_{i}\\}$,
$s_{i+2}(G_{\omega})=\max\\{|M|\\!:M\text{\ is an induced matching of\
}G_{\omega}\text{\ containing}$ $\text{the edge\ }e_{i+2}\\}$ if $e_{i+2}\in
E({G_{\omega}})$ and $s_{i+2}(G_{\omega}^{2})=\max\\{|M|\\!:M\text{\ is an
induced matching}$ $\text{of\ }G_{\omega}^{2}\text{\ containing the edge\
}e_{i+2}\\}$ if $e_{i+2}\in E({G_{\omega}^{2}})$.

######  Lemma 3.6.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5. Let
$A=\\{x_{i},x_{i+1}\\}$ and $G_{\omega}^{3}=G_{\omega}\backslash N_{G}(A)$,
then $\nu(G_{\omega}^{3})=s_{i}(G_{\omega})-1$.

###### Proof.

Let $M$ be an induced matching of $G_{\omega}^{3}$ with cardinality
$\nu(G_{\omega}^{3})$, then $M\sqcup\\{e_{i}\\}$ is an induced matching of
$G_{\omega}$. Hence $s_{i}(G_{\omega})\geq\nu(G_{\omega}^{3})+1$. On the other
hand, if $M^{\prime}$ is an induced matching of $G_{\omega}$ containing
$e_{i}$ with cardinality $s_{i}(G_{\omega})$, then
$M^{\prime}\backslash\\{e_{i}\\}$ is an induced matching of $G_{\omega}^{3}$.
Thus, $\nu(G_{\omega}^{3})\geq s_{i}(G_{\omega})-1$.

######  Lemma 3.7.

If $G$ is a tree and $e$ is its edge, then

$\nu(G\backslash e)-1\leq\nu(G)\leq\nu(G\backslash e)+1.$

Further, if $\nu(G)=\nu(G\backslash e)-1$ or $\nu(G)=\nu(G\backslash e)+1$,
then $s(G)=\nu(G)$, where $s(G)=\max\\{|M|\\!:M\text{\ is an induced matching
of $G$ containing\ }e\\}$.

###### Proof.

Let $\mathcal{A}$ be the collection of induced matchings of $G\backslash e$
with cardinality $\nu(G\backslash e)$ and $M\in\mathcal{A}$. If $M$ is an
induced matching of $G$, then $\nu(G)\geq\nu(G\backslash e)$. Otherwise, there
are two edges, say $e_{1},e_{2}$, in $M$, which incident on two endpoints of
$e$. Thus every $M\backslash\\{e_{i}\\}$ is an induced matching of $G$, which
enforces $\nu(G)\geq\nu(G\backslash e)-1$. In particular, if this equality
holds, then, by the definition of $\nu(G)$, for any $M\in\mathcal{A}$, there
are two edges $e_{1},e_{2}\in M$ which are incident on two endpoints of $e$.
Thus $(M\backslash\\{e_{1},e_{2}\\})\sqcup\\{e\\}$ is an induced matching of
$G$. Hence $s(G)\geq\nu(G)$. Thus $s(G)=\nu(G)$, since $s(G)\leq\nu(G)$ always
holds.

Next we prove that $\nu(G)\leq\nu(G\backslash e)+1$. Conversely, if
$\nu(G)\geq\nu(G\backslash e)+2$, then for any $M\in\mathcal{B}$, where
$\mathcal{B}$ is the collection of induced matchings of $G$ with cardinality
$\nu(G)$, we have $e\in M$. Indeed, if $e\notin M$, then $M$ is an induced
matching of $G\backslash e$, which implies $\nu(G\backslash
e)\geq\nu(G)\geq\nu(G\backslash e)+2$, a contradiction. So
$M\backslash\\{e\\}$ is an induced matching of $G\backslash e$, forcing
$\nu(G\backslash e)\geq\nu(G)-1\geq\nu(G\backslash e)+1$, a contradiction. If
$\nu(G)=\nu(G\backslash e)+1$, then, for any induced matching $M$ of $G$ with
cardinality $\nu(G)$, we have $e\in M$. It follows that $s(G)\geq\nu(G)$. So
$s(G)=\nu(G)$, since $s(G)\leq\nu(G)$ always holds.

######  Lemma 3.8.

Let $G_{\omega}$ be a non-trivial weighted integrally closed tree as in Remark
3.5. If $e_{i+2}\in E({G_{\omega}})$, then
$s_{i+2}(G_{\omega})=\nu(G_{\omega}^{1})+s_{i+2}(G_{\omega}^{2})$.

###### Proof.

Let $M$ be an induced matching of $G_{\omega}$ containing $e_{i+2}$ with
$|M|=s_{i+2}(G_{\omega})$, then $e_{i}\notin M$. Thus $M=(M\cap
E(G_{\omega}^{1}))\sqcup(M\cap E(G_{\omega}^{2}))$, since
$G_{\omega}\backslash e_{i}=G_{\omega}^{1}\sqcup G_{\omega}^{2}$. It follows
that $|M|=|M\cap E(G_{\omega}^{1})|+|M\cap E(G_{\omega}^{2})|$. On the one
hand, $M\cap E(G_{\omega}^{1})$ and $M\cap E(G_{\omega}^{2})$ are an induced
matching of $G_{\omega}^{1}$ and $G_{\omega}^{2}$, respectively, and
$e_{i+2}\in M\cap E(G_{\omega}^{2})$. Thus $\nu(G_{\omega}^{1})\geq|M\cap
E(G_{\omega}^{1})|$ and $s_{i+2}(G_{\omega}^{2})\geq|M\cap
E(G_{\omega}^{2})|$, which forces
$\nu(G_{\omega}^{1})+s_{i+2}(G_{\omega}^{2})\geq|M\cap
E(G_{\omega}^{1})|+|M\cap E(G_{\omega}^{2})|=|M|=s_{i+2}(G_{\omega})$. On the
other hand, if $M_{1}$ and $M_{2}$ are an induced matching of $G_{\omega}^{1}$
and $G_{\omega}^{2}$ with cardinalities $\nu(G_{\omega}^{1})$ and
$s_{i+2}(G_{\omega}^{2})$, respectively, and $e_{i+2}\in M_{2}$, then
$M_{1}\cup M_{2}$ is an induced matching of $G_{\omega}$ containing the edge
$e_{i+2}$. Therefore, $s_{i+2}(G_{\omega})\geq|M_{1}\cup
M_{2}|=\nu(G_{\omega}^{1})+s_{i+2}(G_{\omega}^{2})$.

To fully complete the proof of Theorem 3.10, we have a small result to show.

######  Theorem 3.9.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $k=4$, $V(P_{\omega})=\\{x_{1},\ldots,x_{4}\\}$,
$\omega_{2}=\max\\{\omega_{t}\mid\omega_{t}=\omega(e_{t})\text{\ and
}e_{t}=x_{t}x_{t+1}\text{\ for any }1\leq t\leq 3\\}$, then
$\operatorname{reg}(I(G_{\omega}))=2\omega_{2}$.

###### Proof.

Let $I=I(G_{\omega})$ and $I^{\mathcal{P}}$ be its polarization, then
$I^{\mathcal{P}}=J^{\mathcal{P}}+K^{\mathcal{P}}$ is a Betti splitting by
Lemma 2.4, where $\mathcal{G}(J)=\\{x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}}\\}$
and $\mathcal{G}(K)=\mathcal{G}(I)\backslash\mathcal{G}(J)$. It follows from
Lemma 2.4 and Lemma 2.6 that

$\displaystyle\operatorname{reg}(I)$
$\displaystyle=\operatorname{reg}(I^{\mathcal{P}})=\max\\{\operatorname{reg}(J^{\mathcal{P}}),\operatorname{reg}(K^{\mathcal{P}}),\operatorname{reg}(J^{\mathcal{P}}\cap
K^{\mathcal{P}})-1\\}$ (1)
$\displaystyle=\max\\{\operatorname{reg}(J),\operatorname{reg}(K),\operatorname{reg}(J\cap
K)-1\\},$

where $K=I(G_{\omega}^{1})+I(G_{\omega}^{2})$, $J\cap K=JL$, where $L$ is a
prime ideal with minimal generators set
$N_{G}(\\{x_{i},x_{i+1}\\})\backslash\\{x_{i},x_{i+1}\\}$. Note that
$\operatorname{reg}(J)=2\omega_{2}$, $\operatorname{reg}(J\cap
K)-1=\operatorname{reg}(JL)-1=2\omega_{2}$, and $G_{\omega}^{1}\cup
G_{\omega}^{2}$ is a disjoint union of two trivial star, implying that
$\operatorname{reg}(K)=3$. Therefore, $\operatorname{reg}(I)=2\omega_{2}$ from
formula (1).

######  Theorem 3.10.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges, with $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$,
$\omega_{i}=\max\\{\omega_{t}\mid\omega_{t}=\omega(e_{t})\text{\ and
}e_{t}=x_{t}x_{t+1}\text{\ for each }1\leq t\leq k-1\\}$ where $\omega_{i}\geq
2$ and $\omega_{i}\geq\omega_{i+2}$ if $e_{i+2}\in E({P_{\omega}})$. Then,

  * (1)

if $e_{i+2}\notin E(G_{\omega})$, then
$\operatorname{reg}(I(G_{\omega}))=2\omega_{i}$,

  * (2)

if $e_{i+2}\in E(G_{\omega})$ and $\omega_{i+2}=1$, then
$\operatorname{reg}(I(G_{\omega}))=\max\\{\nu(G_{\omega})+1,2\omega_{i}+(s_{i}(G_{\omega})-1)\\}$,

  * (3)

if $e_{i+2}\in E(G_{\omega})$ and $\omega_{i+2}\geq 2$, then
$\operatorname{reg}(I(G_{\omega}))=\max\\{\nu(G_{\omega})+1,2\omega_{i}+(s_{i}(G_{\omega})-1),2\omega_{i+2}+(s_{i+2}(G_{\omega})-1)\\}$.

###### Proof.

If $k\leq 3$, or $k=4$ and $i=2$, then the desired results are obtained from
[26, Theorem 3.1] and Theorem 3.9. In the following, we assume that $k=4$ and
$i\neq 2$, or $k\geq 5$. Since $\omega_{i}\geq 2$ and $G_{\omega}$ is
integrally closed, $G_{\omega}$ has at most two edges with non-trivial
weights. If $G_{\omega}$ has only one edge with non-trivial weights, then by
rearranging the subscripts of elements in $V(P_{\omega})$, we can say
$i\leq\lfloor\frac{k}{2}\rfloor$. Thus we always have $e_{i+2}\in
E(G_{\omega})$. Let $I=I(G_{\omega})$ and $I^{\mathcal{P}}$ be its
polarization, then $I^{\mathcal{P}}=J^{\mathcal{P}}+K^{\mathcal{P}}$ is a
Betti splitting by Lemma 2.4, where
$\mathcal{G}(J)=\\{x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}}\\}$,
$\mathcal{G}(K)=\mathcal{G}(I)\backslash\mathcal{G}(J)=\mathcal{G}(I(G_{\omega}^{1}))+\mathcal{G}(I(G_{\omega}^{2}))$.
Meanwhile, $J\cap K=JL$, where $L$ is an ideal with minimal generators set
$(N_{G}(A)\backslash A)\sqcup\mathcal{G}(I(G_{\omega}^{3}))$ with
$A=\\{x_{i},x_{i+1}\\}$ and $G_{\omega}^{3}=G_{\omega}\backslash N_{G}(A)$,
and $\operatorname{reg}(J)=2\omega_{i}$. We consider the following two cases:

(1) If $\omega_{i+2}=1$, then
$\operatorname{reg}(K)=\operatorname{reg}(I(G_{\omega}^{1}))+\operatorname{reg}(I(G_{\omega}^{2}))-1=\nu(G_{\omega}^{1})+\nu(G_{\omega}^{2})+1=\nu(G_{\omega}\backslash
e_{i})+1$ and $\operatorname{reg}(J\cap
K)=\operatorname{reg}(JL)=2\omega_{i}+\nu(G_{\omega}^{3})+1=2\omega_{i}+s_{i}(G_{\omega})$
by Lemmas 2.2, 3.4 and 3.6. It follows from formula (1) and Lemma 3.7 that

$\displaystyle\operatorname{reg}(I)$
$\displaystyle=\max\\{\operatorname{reg}(J),\operatorname{reg}(K),\operatorname{reg}(J\cap
K)-1\\}$ $\displaystyle=\max\\{2\omega_{i},\nu(G_{\omega}\backslash
e_{i})+1,2\omega_{i}+s_{i}(G_{\omega})-1\\}$
$\displaystyle=\max\\{\nu(G_{\omega}\backslash
e_{i})+1,2\omega_{i}+s_{i}(G_{\omega})-1\\}$
$\displaystyle=\begin{cases}\max\\{\nu(G_{\omega})+1,2\omega_{i}+(s_{i}(G_{\omega})-1)\\},&\text{if
$\nu(G_{\omega}\backslash e_{i})=\nu(G_{\omega})$}\\\
2\omega_{i}+s_{i}(G_{\omega})-1,&\text{otherwise}\end{cases}$
$\displaystyle=\max\\{\nu(G_{\omega})+1,2\omega_{i}+(s_{i}(G_{\omega})-1)\\}.$

(2) If $\omega_{i+2}\geq 2$, then

$\displaystyle\operatorname{reg}(K)$
$\displaystyle=\operatorname{reg}(I(G_{\omega}^{1}))+\operatorname{reg}(I(G_{\omega}^{2}))-1=\nu(G_{\omega}^{1})+\operatorname{reg}(I(G_{\omega}^{2}))$
$\displaystyle=\nu(G_{\omega}^{1})+\max\\{\nu(G_{\omega}^{2})+1,2\omega_{i+2}+(s_{i+2}(G_{\omega}^{2})-1)\\}$
(2) $\displaystyle=\max\\{\nu(G_{\omega}\backslash
e_{i})+1,2\omega_{i+2}+(s_{i+2}(G_{\omega})-1)\\}$

where the third and last equalities follow from the proof of the above case
(1), $G_{\omega}\backslash e_{i}=G_{\omega}^{1}\sqcup G_{\omega}^{2}$ and
Lemma 3.8, respectively. Additionally, $\operatorname{reg}(J\cap
K)=\operatorname{reg}(JL)=\operatorname{reg}(J)+\operatorname{reg}(L)=2\omega_{i}+(\nu(G_{\omega}^{3})+1)=2\omega_{i}+s_{i}(G_{\omega})$
by Lemma 3.6. Therefore, applying formula (2) and Lemma 3.7, we can determine
that

$\displaystyle\operatorname{reg}(I)$
$\displaystyle=\max\\{\operatorname{reg}(J),\operatorname{reg}(K),\operatorname{reg}(J\cap
K)-1\\}$ $\displaystyle=\max\\{2\omega_{i},\nu(G_{\omega}\backslash
e_{i})+1,2\omega_{i+2}+s_{i+2}(G_{\omega})-1,2\omega_{i}+s_{i}(G_{\omega})-1\\}$
$\displaystyle=\max\\{\nu(G_{\omega}\backslash
e_{i})+1,2\omega_{i+2}+s_{i+2}(G_{\omega})-1,2\omega_{i}+s_{i}(G_{\omega})-1\\}$
$\displaystyle=\max\\{\nu(G_{\omega})+1,2\omega_{i+2}+s_{i+2}(G_{\omega})-1,2\omega_{i}+s_{i}(G_{\omega})-1\\}.$

Another immediate consequence of Theorem 3.10 is [6, Theorems 4.8 and 4.9].

######  Corollary 3.11.

Let $P_{\omega}$ be a non-trivial integrally closed path with $n$ vertices,
and $\omega=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$. If $n\leq 4$, then
$\operatorname{reg}(I(P_{\omega}))=2\omega$. If $n\geq 5$, then, by symmetry
and Corollary 3.3, we can assume that $\omega_{i}\geq\omega_{i+2}$ and
$\omega_{i}\geq 2$ for some $i\in[n-3]$, where $e_{i}=x_{i}x_{i+1}$ and
$\omega_{i}=\omega(e_{i})$ for all $i\in[n-1]$. Then

$\operatorname{reg}(I(P_{\omega}))=\max\\{2\omega_{i}+\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor,2\omega_{i+2}+\lfloor\frac{i-2}{3}\rfloor+\lfloor\frac{n-i}{3}\rfloor\\}.$

###### Proof.

If $k\leq 3$, or $k=4$ and $i=2$, then
$\operatorname{reg}(I(P_{\omega}))=2\omega$ according to [26, Theorem 3.1] and
Theorem 3.9. Now we assume that $k=4$ and $i\neq 2$, or $k\geq 5$.. In this
case, we have $\nu(P_{\omega})=\lfloor\frac{n+1}{3}\rfloor$,
$s_{i}(P_{\omega})=\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor+1$
and
$s_{i+2}(P_{\omega})=\lfloor\frac{i+1}{3}\rfloor+\lfloor\frac{n-(i+3)}{3}\rfloor+1$
by Lemmas 3.6 and 3.8. We consider two subcases: (i) If $\omega_{i+2}=1$,
then, by Theorem 3.10(2), we get
$\operatorname{reg}(I(P_{\omega}))=\max\\{\nu(P_{\omega})+1,2\omega_{i}+(s_{i}(P_{\omega})-1)\\}=2\omega_{i}+\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor$,
since
$2\omega_{i}+\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor>\lfloor\frac{n+1}{3}\rfloor+1$;
(ii) If $\omega_{i+2}\geq 2$, then, by Theorem 3.10(3), we can conclude that
$\operatorname{reg}(I(P_{\omega}))=\max\\{\nu(P_{\omega})+1,2\omega_{i}+(s_{i}(P_{\omega})-1),2\omega_{i+2}+(s_{i+2}(P_{\omega})-1)\\}=\max\\{2\omega_{i}+\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor,2\omega_{i+2}+\lfloor\frac{i-2}{3}\rfloor+\lfloor\frac{n-i}{3}\rfloor\\}$,
since
$2\omega_{i}+\lfloor\frac{i-1}{3}\rfloor+\lfloor\frac{n-(i+1)}{3}\rfloor>\lfloor\frac{n+1}{3}\rfloor+1$.

##  4\. Regularity of powers of the edge ideal of an edge-weighted integrally
closed tree

This section presents some linear upper bounds on the regularity of powers of
the edge ideal of an edge-weighted integrally closed tree. To support our
argument, we use the following Lemma, which is similar to [26, Lemma 4.10].
The proof is omitted.

######  Lemma 4.1.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $x$ be its leaf with $N_{G}(x)=\\{y\\}$ and $\omega(xy)=1$. Then

  * (1)

$(I(G_{\omega})^{t}:xy)=I(G_{\omega})^{t-1}$;

  * (2)

$(I(G_{\omega})^{t},x)=(I(G_{\omega}\backslash x)^{t},x)$;

  * (3)

$((I(G_{\omega})^{t}:x),y)=(I(G_{\omega}\backslash y)^{t},y)$;

  * (4)

$(I(G_{\omega})^{t},xy)=(I(G_{\omega}\backslash x)^{t},xy)$;

  * (5)

$(I(G_{\omega})^{t},y)=(I(G_{\omega}\backslash y)^{t},y)$;

  * (6)

if $x^{\prime}$ is also a leaf of $G_{\omega}$ such that
$N_{G}(x^{\prime})=\\{y\\}$ and $\omega(x^{\prime}y)=1$, then
$((I(G_{\omega})^{t},xy):x^{\prime}y)=(I(G_{\omega}\backslash x)^{t-1},x)$.

Let $G_{\omega}$ be an edge-weighted graph on the vertex set $\mathbf{X}$. For
any subset $\mathbf{Y}\subset\mathbf{X}$, let $H_{\omega}$ be its induced
subgraph on $\mathbf{Y}$. Let
${\mathbb{K}}[\mathbf{X}]={\mathbb{K}}[x|x\in\mathbf{X}]$ be the polynomial
ring over a field ${\mathbb{K}}$, then edge ideals $I(G_{\omega})$ and
$I(H_{\omega})$ are monomial ideals in the polynomial ring
${\mathbb{K}}[\mathbf{X}]$ and ${\mathbb{K}}[\mathbf{Y}]$, respectively.

######  Lemma 4.2.

Let $G_{\omega}$ be an edge-weighted graph and $H_{\omega}$ its induced
subgraph, then

$\beta_{ij}({\mathbb{K}}[\mathbf{Y}]/I(H_{\omega})^{t})\leq\beta_{ij}({\mathbb{K}}[\mathbf{X}]/I(G_{\omega})^{t})$

for all $i,j$ and $t\geq 1$. In particular,
$\operatorname{reg}(I(H_{\omega})^{t})\leq\operatorname{reg}(I(G_{\omega})^{t})$
for all $t\geq 1$.

###### Proof.

First, we show that
$I(H_{\omega})^{t}=I(G_{\omega})^{t}\cap{\mathbb{K}}[\mathbf{Y}]$ for all
$t\geq 1$. Since the natural generators of $I(H_{\omega})^{t}$ are
automatically contained in $I(G_{\omega})^{t}$, we have
$I(H_{\omega})^{t}\subseteq I(G_{\omega})^{t}\cap{\mathbb{K}}[\mathbf{Y}]$.
For the converse inclusion, let $g\in
I(G_{\omega})^{t}\cap{\mathbb{K}}[\mathbf{Y}]$. We can write $g$ as a finite
sum

$g=\sum_{\begin{subarray}{c}u_{\ell i}\in I(G_{\omega}),\\\ 1\leq i\leq
t\end{subarray}}h_{\ell 1,\ldots,\ell t}u_{\ell 1}\cdots u_{\ell t}$

where each $h_{\ell 1,\ldots,\ell t}\in{\mathbb{K}}[\mathbf{X}]$. Now,
consider the ${\mathbb{K}}$-algebra homomorphism
$\pi:{\mathbb{K}}[\mathbf{X}]\rightarrow{\mathbb{K}}[\mathbf{Y}]$ by setting

$\pi(x)=\begin{cases}x,&\text{if $x$ is a variable in $\mathbf{Y}$,}\\\
0,&\text{otherwise}.\end{cases}$

Thus,

$\pi(u_{\ell i})=\begin{cases}u_{\ell i},&\text{if $u_{\ell i}\in
I(H_{\omega})$,}\\\ 0,&\text{otherwise.}\end{cases}$

Since $g\in{\mathbb{K}}[\mathbf{Y}]$, we have $\pi(g)=g$. Therefore, we get

$g=\sum_{\begin{subarray}{c}u_{\ell i}\in I(G_{\omega}),\\\ 1\leq i\leq
t\end{subarray}}\pi(h_{\ell 1,\ldots,\ell t})\pi(u_{\ell 1})\cdots\pi(u_{\ell
t})=\sum_{\begin{subarray}{c}u_{\ell i}\in I(G_{\omega}),\\\ 1\leq i\leq
t\end{subarray}}\pi(h_{\ell 1,\ldots,\ell t})u_{\ell 1}\cdots u_{\ell t}.$

Thus, $g\in I(H_{\omega})^{t}$. This completes our proof for
$I(H_{\omega})^{t}=I(G_{\omega})^{t}\cap{\mathbb{K}}[\mathbf{Y}]$.

Consequently, ${\mathbb{K}}[\mathbf{Y}]/I(H_{\omega})^{t}$ is a
${\mathbb{K}}$-subalgebra of ${\mathbb{K}}[\mathbf{X}]/I(G_{\omega})^{t}$. Let
$\bar{\pi}:{\mathbb{K}}[\mathbf{X}]/I(G_{\omega})^{t}\\\
\rightarrow{\mathbb{K}}[\mathbf{Y}]/I(H_{\omega})^{t}$ be the homomorphism
induced by $\pi$. Since $\pi(I(G_{\omega})^{t})\subseteq I(H_{\omega})^{t}$,
the map $\bar{\pi}$ is well-defined. Notice that the restriction of
$\bar{\pi}$ to ${\mathbb{K}}[\mathbf{Y}]/I(H_{\omega})^{t}$ is the identity
map. Thus, $\bar{\pi}$ is surjective, and
${\mathbb{K}}[\mathbf{Y}]/I(H_{\omega})^{t}$ is an algebra retract of
${\mathbb{K}}[\mathbf{X}]/I(G_{\omega})^{t}$. Now, the expected inequalities
follow from [20, Corollary 2.5].

For any two vertices $x,y\in V(G)$, their _distance_ $d_{G}(x,y)$ is the
length of the shortest path in $G$ with $x$ and $y$ as endpoints; if no such
path exists, we set $d_{G}(x,y)=\infty$.

######  Notation 4.3.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. Suppose $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$. Given an
$x\in V(G_{\omega})$, we set $d(x)=\min\\{d_{G}(x,x_{j})|1\leq j\leq k\text{\
and\ }x\in V(G_{\omega})\\}$ and $d=\max\\{d(x)|\text{\ for any\ }x\in
V(G_{\omega})\\}$.

We first determine the regularity of powers of the edge ideal of a non-trivial
integrally closed caterpillar graph under the condition that $k=4$, $d=1$ and
$\omega_{2}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$.

######  Theorem 4.4.

Let $G_{\omega}$ be a non-trivial integrally closed caterpillar as in Remark
3.5, and let $P_{\omega}$ be its path of length $(k-1)$ containing all non-
trivial edges. If $k=4$, $d=1$ and $\omega_{2}=\max\\{\omega(e)\mid e\in
E(P_{\omega})\\}$, then $\operatorname{reg}(I(G_{\omega})^{t})=2\omega_{2}t$
for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$. We will prove the statements by induction on $t$. The
case where $t=1$ follows from Theorem 3.9. So we can assume that $t\geq 2$.
Let $V(P_{\omega})=\\{x_{1},x_{2},x_{3},x_{4}\\}$,
$N_{G}(x_{2})=\\{x_{1},x_{3}\\}\cup A$ and $N_{G}(x_{3})=\\{x_{2},x_{4}\\}\cup
B$, where $A=\\{y_{1},\ldots,y_{\ell}\\}$ and $B=\\{u_{1},\ldots,u_{m}\\}$,
see the graph displayed in Figure $1$. By convention, $A=\emptyset$ if
$\ell=0$, and $B=\emptyset$ if $m=0$. Since $d=1$, we have $\ell+m\geq 1$.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$y_{1}$$y_{\ell}$$u_{1}$$u_{m}$

Figure 1. $Caterpillar\ graph\ with\ k=4,d=1,\omega_{2}\geq 2$

Suppose $\gamma\\!:=\ell+m$. We prove the statements by induction on $\gamma$.
Since $G_{\omega}$ is a non-trivially weighted graph and
$\omega_{2}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$, we obtain that
$\omega_{2}\geq 2$ and $\omega_{1}=\omega_{3}=1$ by Lemma 3.2. Consider the
exact sequences:

$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{I^{t}:u_{m}}(-1)&\stackrel{{\scriptstyle\cdot
u_{m}}}{{\longrightarrow}}&\frac{S}{I^{t}}&\rightarrow&\frac{S}{(I^{t},u_{m})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{(I^{t}:u_{m}):x_{3}}(-1)&\stackrel{{\scriptstyle\cdot
x_{3}}}{{\longrightarrow}}&\frac{S}{I^{t}:u_{m}}&\rightarrow&\frac{S}{((I^{t}:u_{m}),x_{3})}&\rightarrow&0.\end{matrix}$

Note that $(I^{t}:u_{m}):x_{3}=I^{t-1}$,
$((I^{t}:u_{m}),x_{3})=(I(G_{\omega}\backslash x_{3})^{t},x_{3})$ and
$(I^{t},u_{m})=(I(G_{\omega}\backslash u_{m})^{t},u_{m})$ by Lemma 4.1, it
follows that
$\operatorname{reg}((I^{t}:u_{m}):x_{3})=\operatorname{reg}(I^{t-1})=2\omega_{2}(t-1)$
and
$\operatorname{reg}(((I^{t}:u_{m}),x_{3}))=\operatorname{reg}((I(G_{\omega}\backslash
x_{3})^{t},x_{3}))=\operatorname{reg}(I(G_{\omega}\backslash x_{3})^{t})=2t$
by the inductive hypothesis on $t$ and [26, Theorem 3.3].

If $\gamma=1$, then by symmetry we can assume that $m=1,\ell=0$. In this case,
we have
$\operatorname{reg}((I^{t},u_{1}))=\operatorname{reg}((I(G_{\omega}\backslash
u_{1})^{t},u_{1}))=\operatorname{reg}(I(G_{\omega}\backslash
u_{1})^{t})=2\omega_{2}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash
u_{1}))=2\omega_{2}(t-1)+\operatorname{reg}(I(P_{\omega}))=2\omega_{2}t$ by
[26, Theorem 4.16] and Corollary 3.11. The desired results follow from Lemma
2.1 and the above exact sequences with $m=1$.

If $\gamma\geq 2$, then
$\operatorname{reg}((I^{t},u_{m}))=\operatorname{reg}((I(G_{\omega}\backslash
u_{m})^{t},u_{m}))=\operatorname{reg}(I(G_{\omega}\backslash
u_{m})^{t})=2\omega_{2}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash
u_{m}))=2\omega_{2}t$ by the inductive hypothesis on $\gamma$ and Theorem
3.10. Therefore, the expected results follow from Lemma 2.1 and the above
exact sequences.

######  Lemma 4.5.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $k\geq 5$ or $d\geq 2$, then there exists a trivial induced
star subgraph $H_{\omega}$. Let $V(H_{\omega})=\\{x,y\\}\cup C$, where
$C=\\{z_{1}\ldots,z_{p}\\}$ with $p\geq 1$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for each $i\in[p]$. If $N_{G}(x)=\\{y\\}\cup C$ and
$N_{G}(y)=\\{x\\}\cup D$, where $D=\\{w_{1},\ldots,w_{q}\\}$ with $q\geq 1$.
Let $J=I(G_{\omega}\backslash\\{x,y\\})$, then
$\operatorname{reg}(J)\leq\operatorname{reg}(I(G_{\omega}))-1$.

###### Proof.

Let $\mathcal{A}$ be the collection of induced matchings of $G_{\omega}$ with
cardinality $\nu(G_{\omega})$. For any $M\in\mathcal{A}$, there exists an edge
$e$ which is taken from the set $\\{xy,xz_{1},\ldots,xz_{p},\\\
yw_{1},\ldots,yw_{q}\\}$. Thus $M\backslash\\{e\\}$ is an induced matching of
the graph $G_{\omega}\backslash\\{x,y\\}$. It follows that
$\nu(G_{\omega}\backslash\\{x,y\\})\geq\nu(G_{\omega})-1$. Furthermore, we can
see that $\nu(G_{\omega}\backslash\\{x,y\\})=\nu(G_{\omega})-1$. Indeed, if
$\nu(G_{\omega}\backslash\\{x,y\\})>\nu(G_{\omega})-1$, then there exists an
induced matching $N$ of $G_{\omega}\backslash\\{x,y\\}$ such that
$|N|>\nu(G_{\omega})-1$. It is easy to check that $N\cup\\{xz_{1}\\}$ is an
induced matching of $G_{\omega}$. So
$\nu(G_{\omega})\geq|N|+1>\nu(G_{\omega})$, a contradiction.

Similarly, we can show that
$s_{i}(G_{\omega}\backslash\\{x,y\\})=s_{i}(G_{\omega})-1$ if $e_{i}\in
E(G_{\omega}\backslash\\{x,y\\})$ and
$s_{i+2}(G_{\omega}\backslash\\{x,y\\})=s_{i+2}(G_{\omega})-1$ if $e_{i+2}\in
E(G_{\omega}\backslash\\{x,y\\})$. Therefore it follows from Theorem 3.10 that
$\operatorname{reg}(J)=\operatorname{reg}(I(G_{\omega}))-1$ if
$e_{i},e_{i+2}\in E(G_{\omega}\backslash\\{x,y\\})$. On the other hand, if
$e_{i}\notin E(G_{\omega}\backslash\\{x,y\\})$ or $e_{i+2}\notin
E(G_{\omega}\backslash\\{x,y\\})$, then we can deduce that
$\operatorname{reg}(J)\leq\operatorname{reg}(I(G_{\omega}))-1$ by adopting the
same technique.

The following remark is often used in the following sections.

######  Remark 4.6.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $k\geq 4$, then there exists an induced trivial star
subgraph $H_{\omega}$. Let $V(H_{\omega})=\\{x,y\\}\cup C$, where
$C=\\{z_{1}\ldots,z_{p}\\}$ with $p\geq 1$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for each $i\in[p]$. If $N_{G}(x)=\\{y\\}\cup C$ and
$N_{G}(y)=\\{x\\}\cup D$, where $D=\\{w_{1},\ldots,w_{q}\\}$ with $q\geq 1$.
Let
$L_{1}=(I^{t}:xz_{1}),T_{1}=(I^{t},xz_{1}),L_{j}=(T_{j-1}:xz_{j}),T_{j}=(T_{j-1},xz_{j})$
with $2\leq j\leq p$, then, by repeatedly applying Lemma 4.1, we can conclude
that $L_{1}=I^{t-1}$, $T_{1}=(I(G_{\omega}\backslash z_{1})^{t},xz_{1})$,
$L_{2}=(T_{1}:xz_{2})=(I(G_{\omega}\backslash z_{1})^{t-1},z_{1})$,
$T_{2}=(T_{1},xz_{2})=(I(G_{\omega}\backslash\\{z_{1},z_{2}\\})^{t},xz_{1},xz_{2})$,
and
$L_{j}=(T_{j-1}:xz_{j})=((I(G_{\omega}\backslash\\{z_{1},\ldots,z_{j-1}\\})^{t},xz_{1},\ldots,xz_{j-1}):xz_{j})=(I(G_{\omega}\backslash\\{z_{1},\ldots,z_{j-1}\\})^{t-1},z_{1},\\\
\ldots,z_{j-1})$,
$T_{j}=(T_{j-1},xz_{j})=(I^{t},xz_{1},\ldots,xz_{j})=(I(G_{\omega}\backslash\\{z_{1},\ldots,z_{j}\\})^{t},xz_{1},\ldots,xz_{j})$
for any $3\leq j\leq p$. At the same time, we also have
$(T_{p}:x)=((I(G_{\omega}\backslash C)^{t}:x),z_{1},\ldots,z_{p})$,
$(T_{p},x)=(I(G_{\omega}\backslash x)^{t},x)$,
$((T_{p}:x):y)=(I(G_{\omega}\backslash C)^{t-1},z_{1},\ldots,z_{p})$ and
$((T_{p}:x),y)=((I(G_{\omega}\backslash
C)^{t}:x),z_{1},\ldots,z_{p},y)=(I(G_{\omega}\backslash\\{x,y\\})^{t},z_{1},\ldots,z_{p},y)$.
It follows from Lemma 2.2 that
$\operatorname{reg}(L_{1})=\operatorname{reg}(I^{t-1})$,
$\operatorname{reg}(L_{j})=\operatorname{reg}(I(G_{\omega}\backslash\\{z_{1},\ldots,z_{j-1}\\})^{t-1})$
for any $2\leq j\leq p$,
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})$,
$\operatorname{reg}(((T_{p}:x):y))=\operatorname{reg}(I(G_{\omega}\backslash
C)^{t-1})$ and
$\operatorname{reg}(((T_{p}:x),y))=\begin{cases}\operatorname{reg}((z_{1},\ldots,z_{p},y)),&\text{if
$k=4,d=1$,}\\\
\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t}),&\text{otherwise}.\\\
\end{cases}$

Next, we compute the regularity of powers of the edge ideal of a non-trivial
integrally closed caterpillar graph under the condition that $k=4$, $d=1$ and
$\omega_{1}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$.

######  Theorem 4.7.

Let $G_{\omega}$ be a non-trivial integrally closed caterpillar as in Remark
3.5, and let $P_{\omega}$ be its spine of length $(k-1)$ containing all non-
trivial edges. If $k=4$, $d=1$ and $\omega_{1}=\max\\{\omega(e)\mid e\in
E(P_{\omega})\\}$, then $\operatorname{reg}(I(G_{\omega})^{t})=2\omega_{1}t$
for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$. We prove the statements by induction on $t$. The case
where $t=1$ follows from Theorem 3.10. So we can assume that $t\geq 2$. Let
$V(P_{\omega})=\\{x_{1},x_{2},x_{3},x_{4}\\}$,
$N_{G}(x_{2})=\\{x_{1},x_{3}\\}\cup A$ and $N_{G}(x_{3})=\\{x_{2},x_{4}\\}\cup
B$, where $A=\\{y_{1},\ldots,y_{\ell}\\}$, $B=\\{u_{1},\ldots,u_{m}\\}$, and
$A=\emptyset$ if $\ell=0$, and $B=\emptyset$ if $m=0$ by convention, see
Figure $2$ for the case $\omega_{3}\geq 2$. Since $d=1$, we have $\ell+m\geq
1$. Set $\gamma\\!:=\ell+m$. Since $G_{\omega}$ is a non-trivially weighted
graph and $\omega_{1}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$, we have
$\omega_{1}\geq 2$ and $\omega_{2}=1$ by Lemma 3.2. We distinguish between the
following two cases:

(1) If $\omega_{3}=1$, then there exists an induced trivial star subgraph
$H_{\omega}$ with $V(H_{\omega})=\\{x_{3},x_{2}\\}\cup C$, where
$N_{G}(x_{3})=\\{x_{2}\\}\cup C$ and $C=\\{x_{4}\\}\cup B$. We rearrange the
elements in $C$ and assume that $C=\\{z_{1},\ldots,z_{p}\\}$ with
$z_{1}=x_{4}$ and $p=m+1$, see Figure $2$ for the case $\omega_{3}=1$.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$y_{1}$$y_{\ell}$$u_{1}$$u_{m}$ (A) The case
$\omega_{3}\geq 2$

$x_{1}$$x_{2}=y$$x_{3}=x$$x_{4}=z_{1}$$y_{1}$$y_{\ell}$$u_{1}=z_{2}$$u_{m}=z_{p}$
(B) The case $\omega_{3}=1$

Figure 2. $Caterpillar\ graph\ with\ k=4,d=1,\omega_{1}\geq 2$

Adopting the notations in Remark 4.6 and replacing $x$ and $y$ by $x_{3}$ and
$x_{2}$, respectively, we can obtain from Theorem 3.10(2), [26, Theorems 3.1
and 3.3] and the inductive hypothesis on $t$ that
$\operatorname{reg}(L_{1})=\operatorname{reg}(I^{t-1})=2\omega_{1}(t-1)$,
$\operatorname{reg}(L_{j})=\operatorname{reg}(I(G_{\omega}\backslash\\{z_{1},\ldots,z_{j-1}\\})^{t-1})=2\omega_{1}(t-1)$
for any $2\leq j\leq p$,
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})=2\omega_{1}t$,
$\operatorname{reg}(((T_{p}:x):y))=\operatorname{reg}(I(G_{\omega}\backslash
C)^{t-1})=2\omega_{1}(t-1)$ and $\operatorname{reg}(((T_{p}:x),y))=1$, since
$k=4$ and $d=1$. Thus, the expected results follow from Lemma 2.1 and the
following exact sequences

(3)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{L_{1}}(-2)&\stackrel{{\scriptstyle\cdot
xz_{1}}}{{\longrightarrow}}&\frac{S}{I^{t}}&\rightarrow&\frac{S}{T_{1}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{L_{2}}(-2)&\stackrel{{\scriptstyle\cdot
xz_{2}}}{{\longrightarrow}}&\frac{S}{T_{1}}&\rightarrow&\frac{S}{T_{2}}&\rightarrow&0,\\\
&&\vdots&&\vdots&&\vdots&&\\\
0&\rightarrow&\frac{S}{L_{p}}(-2)&\stackrel{{\scriptstyle\cdot
xz_{p}}}{{\longrightarrow}}&\frac{S}{T_{p-1}}&\rightarrow&\frac{S}{T_{p}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{T_{p}:x}(-1)&\stackrel{{\scriptstyle\cdot
x}}{{\longrightarrow}}&\frac{S}{T_{p}}&\rightarrow&\frac{S}{(T_{p},x)}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{(T_{p}:x):y}(-1)&\stackrel{{\scriptstyle\cdot
y}}{{\longrightarrow}}&\frac{S}{T_{p}:x}&\rightarrow&\frac{S}{((T_{p}:x),y)}&\rightarrow&0.\\\
\end{matrix}$

(2) Suppose $\omega_{3}\geq 2$. Let
$J=(I^{t},x_{2}y_{1},\ldots,x_{2}y_{\ell},x_{3}u_{1},\ldots,x_{3}u_{m})$, then
$J=(I(G_{\omega}\backslash(A\cup
B))^{t},x_{2}y_{1},\ldots,x_{2}y_{\ell},x_{3}u_{1},\ldots,x_{3}u_{m})$, where
$G_{\omega}\backslash(A\cup B)$ is the path $P_{\omega}$ of length $3$. It
follows from [26, Lemma 4.11] that

$\displaystyle J:(x_{2}x_{3})^{t-1}$
$\displaystyle=(I(G_{\omega}\backslash(A\cup
B)),y_{1},\ldots,y_{\ell},u_{1},\ldots,u_{m}),$ $\displaystyle(J,x_{2}x_{3})$
$\displaystyle=((x_{1}x_{2})^{t\omega_{1}},x_{2}x_{3},(x_{3}x_{4})^{t\omega_{3}},x_{2}y_{1},\ldots,x_{2}y_{\ell},x_{3}u_{1},\ldots,x_{3}u_{m}),$
$\displaystyle((J:(x_{2}x_{3})^{i}),x_{2}x_{3})$
$\displaystyle=((x_{1}x_{2})^{(t-i)\omega_{1}},x_{2}x_{3},(x_{3}x_{4})^{(t-i)\omega_{3}},y_{1},\ldots,y_{\ell},u_{1},\ldots,u_{m})$

for any $i\in[t-2]$. Therefore, by Lemma 2.2, Lemma 3.10 and [26, Theorem
4.8], we can deduce that
$\operatorname{reg}(J:(x_{2}x_{3})^{t-1})=\operatorname{reg}(I(G_{\omega}\backslash(A\cup
B)))=2\omega_{1}$ and
$\operatorname{reg}(((J:(x_{2}x_{3})^{i}),x_{2}x_{3}))=2\omega_{1}(t-i)$ for
all $i=0,1,\ldots,t-2$. Therefore, $\operatorname{reg}(J)=2\omega_{1}t$ by
Lemma 2.1 and the following exact sequences

(4)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{J:x_{2}x_{3}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}x_{3}}}{{\longrightarrow}}&\frac{S}{J}&\rightarrow&\frac{S}{(J,x_{2}x_{3})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{J:(x_{2}x_{3})^{2}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}x_{3}}}{{\longrightarrow}}&\frac{S}{J:x_{2}x_{3}}&\rightarrow&\frac{S}{((J:x_{2}x_{3}),x_{2}x_{3})}&\rightarrow&0,\\\
&&\vdots&&\vdots&&\vdots&&\\\
0&\rightarrow&\frac{S}{J:(x_{2}x_{3})^{t-2}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}x_{3}}}{{\longrightarrow}}&\frac{S}{J:(x_{2}x_{3})^{t-3}}&\rightarrow&\frac{S}{((J:(x_{2}x_{3})^{t-3}),x_{2}x_{3})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{J:(x_{2}x_{3})^{t-1}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}x_{3}}}{{\longrightarrow}}&\frac{S}{J:(x_{2}x_{3})^{t-2}}&\rightarrow&\frac{S}{((J:(x_{2}x_{3})^{t-2}),x_{2}x_{3})}&\rightarrow&0.\\\
\end{matrix}$

In order to compute $\operatorname{reg}(I^{t})$, we distinguish into the
following three cases:

(i) If $\ell,m\geq 1$, then let
$F_{1}=(I^{t}:x_{3}u_{1}),K_{1}=(I^{t},x_{3}u_{1}),F_{j}=(K_{j-1}:x_{3}u_{j}),K_{j}=(K_{j-1},x_{3}u_{j})$
for all $2\leq j\leq m$,
$H_{1}=(K_{m}:x_{2}y_{1}),L_{1}=(K_{m},x_{2}y_{1}),H_{s}=(L_{s-1},x_{2}y_{s}),L_{s}=(L_{s-1},x_{2}y_{s})$
for all $2\leq s\leq\ell$, then by Lemma 4.1, we have

$\displaystyle F_{1}$ $\displaystyle=I^{t-1},\ \ \
K_{1}=(I^{t},x_{3}u_{1})=(I(G_{\omega}\backslash u_{1})^{t},x_{3}u_{1}),$
$\displaystyle F_{j}$
$\displaystyle=(K_{j-1}:x_{3}u_{j})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{j-1}\\})^{t-1},u_{1},\ldots,u_{j-1}),$
$\displaystyle K_{j}$
$\displaystyle=(K_{j-1},x_{3}u_{j})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{j}\\})^{t},x_{3}u_{1},\ldots,x_{3}u_{j}),$
$\displaystyle H_{1}$
$\displaystyle=(K_{m}:x_{2}y_{1})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m}\\})^{t-1},x_{3}u_{1},\ldots,x_{3}u_{m}),$
$\displaystyle L_{1}$
$\displaystyle=(F_{m},x_{2}y_{1})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m},y_{1}\\})^{t},x_{3}u_{1},\ldots,x_{3}u_{m},x_{2}y_{1}),$
$\displaystyle H_{s}$
$\displaystyle=(L_{s-1}:x_{2}y_{s})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m},y_{1},\ldots,y_{s-1}\\})^{t-1},x_{3}u_{1},\ldots,x_{3}u_{m},y_{1},\ldots,y_{s-1}),$
$\displaystyle L_{s}$
$\displaystyle=(L_{s-1},x_{2}y_{s})=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m},y_{1},\ldots,y_{s}\\})^{t},x_{3}u_{1},\ldots,x_{3}u_{m},x_{2}y_{1},\ldots,x_{2}y_{s}).$

Thus we have $\operatorname{reg}(F_{j})=2\omega_{1}(t-1)$ for all $1\leq j\leq
m$, and
$\operatorname{reg}(L_{\ell})=\operatorname{reg}((I^{t},x_{3}u_{1},\ldots,\\\
x_{3}u_{m},x_{2}y_{1},\ldots,x_{2}y_{\ell}))=2\omega_{1}t$ by similar
arguments as in the proof of the regularity of $J$ above.

Now, we compute $\operatorname{reg}(H_{1})$. For any $1\leq j\leq m-1$, by
Lemma 4.1, we have

$\displaystyle H_{1}:u_{1}$ $\displaystyle=(I(G_{\omega}\backslash
x_{3})^{t-1},x_{3}),$ $\displaystyle(H_{1},u_{1},\ldots,u_{j})$
$\displaystyle=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m}\\})^{t-1},u_{1},\ldots,u_{j},x_{3}u_{j+1},\ldots,x_{3}u_{m}),$
$\displaystyle(H_{1},u_{1},\ldots,u_{j}):u_{j+1}$
$\displaystyle=(I(G_{\omega}\backslash
x_{3})^{t-1},u_{1},\ldots,u_{j},x_{3}),$
$\displaystyle(H_{1},u_{1},\ldots,u_{m})$
$\displaystyle=(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m}\\})^{t-1},u_{1},\ldots,u_{m}).$

It follows from Theorem 3.10 and the inductive hypothesis on $t$ that
$\operatorname{reg}(H_{1}:u_{1})=\operatorname{reg}(I(G_{\omega}\backslash
x_{3})^{t-1})=2\omega_{1}(t-1)$,
$\operatorname{reg}((H_{1},u_{1},\ldots,u_{m}))=\operatorname{reg}(I(G_{\omega}\backslash\\{u_{1},\ldots,u_{m}\\})^{t-1})=2\omega_{1}(t-1)$,
$\operatorname{reg}((H_{1},u_{1},\ldots,u_{j}):u_{j+1})=\operatorname{reg}(I(G_{\omega}\backslash
x_{3})^{t-1})=2\omega_{1}(t-1)$. Thus, by Lemma 2.1 and the following exact
sequences

(5)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{H_{1}:u_{1}}(-1)&\stackrel{{\scriptstyle\cdot
u_{1}}}{{\longrightarrow}}&\frac{S}{H_{1}}&\rightarrow&\frac{S}{(H_{1},u_{1})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{(H_{1},u_{1}):u_{2}}(-1)&\stackrel{{\scriptstyle\cdot
u_{2}}}{{\longrightarrow}}&\frac{S}{(H_{1},u_{1})}&\rightarrow&\frac{S}{(H_{1},u_{1},u_{2})}&\rightarrow&0,\\\
&&\vdots&&\vdots&&\vdots&&\\\
0&\rightarrow&\frac{S}{(H_{1},u_{1},\ldots,u_{m-2}):u_{m-1}}(-1)&\stackrel{{\scriptstyle\cdot
u_{m-1}}}{{\longrightarrow}}&\frac{S}{(H_{1},u_{1},\ldots,u_{m-2})}&\rightarrow&\frac{S}{(H_{1},u_{1},\ldots,u_{m-1})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{(H_{1},u_{1},\ldots,u_{m-1}):u_{m}}(-1)&\stackrel{{\scriptstyle\cdot
u_{m}}}{{\longrightarrow}}&\frac{S}{(H_{1},u_{1},\ldots,u_{m-1})}&\rightarrow&\frac{S}{(H_{1},u_{1},\ldots,u_{m})}&\rightarrow&0,\\\
\end{matrix}$

we obtain $\operatorname{reg}(H_{1})\leq 2\omega_{1}(t-1)+1$. Similarly, we
also can deduce that $\operatorname{reg}(H_{s})\leq 2\omega_{1}(t-1)+1$ for
any $1\leq s\leq\ell$. Hence, by Lemma 2.1 and the following exact sequences

(6)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{F_{1}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}u_{1}}}{{\longrightarrow}}&\frac{S}{I^{t}}&\rightarrow&\frac{S}{K_{1}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{F_{2}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}u_{2}}}{{\longrightarrow}}&\frac{S}{K_{1}}&\rightarrow&\frac{S}{K_{2}}&\rightarrow&0,\\\
&&\vdots&&\vdots&&\vdots&&\\\
0&\rightarrow&\frac{S}{F_{m-1}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}u_{m-1}}}{{\longrightarrow}}&\frac{S}{K_{m-2}}&\rightarrow&\frac{S}{K_{m-1}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{F_{m}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}u_{m}}}{{\longrightarrow}}&\frac{S}{K_{m-1}}&\rightarrow&\frac{S}{K_{m}}&\rightarrow&0,\end{matrix}$

and

(7)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{H_{1}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}y_{1}}}{{\longrightarrow}}&\frac{S}{K_{m}}&\rightarrow&\frac{S}{L_{1}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{H_{2}}(-2)&\stackrel{{\scriptstyle\cdot
x_{2}y_{2}}}{{\longrightarrow}}&\frac{S}{L_{1}}&\rightarrow&\frac{S}{L_{2}}&\rightarrow&0,\\\
&&\vdots&&\vdots&&\vdots&&\\\
0&\rightarrow&\frac{S}{H_{\ell-1}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}y_{\ell-1}}}{{\longrightarrow}}&\frac{S}{L_{\ell-2}}&\rightarrow&\frac{S}{L_{\ell-1}}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{H_{\ell}}(-2)&\stackrel{{\scriptstyle\cdot
x_{3}y_{\ell}}}{{\longrightarrow}}&\frac{S}{L_{\ell-1}}&\rightarrow&\frac{S}{L_{\ell}}&\rightarrow&0,\\\
\end{matrix}$

we can obtain $\operatorname{reg}(I^{t})=2\omega_{1}t$.

(ii) If $\ell=0$, then
$\operatorname{reg}(K_{m})=\operatorname{reg}((I^{t},x_{3}u_{1},\ldots,x_{3}u_{m}))=2\omega_{1}t$
and $\operatorname{reg}(F_{j})=2\omega_{1}(t-1)$ for all $1\leq j\leq m$ as
shown in the proof above. It follows from by Lemma 2.1 and the exact sequences
(6) above that $\operatorname{reg}(I^{t})=2\omega_{1}t$.

(iii) If $m=0$, then
$\operatorname{reg}(L_{\ell})=\operatorname{reg}((I^{t},x_{2}y_{1},\ldots,x_{2}y_{\ell}))=2\omega_{1}t$
and $\operatorname{reg}(H_{j})=2\omega_{1}(t-1)$ for all $1\leq j\leq\ell$ as
shown in the proof above. It follows from by Lemma 2.1 and the exact sequences
(7)above that $\operatorname{reg}(I^{t})=2\omega_{1}t$.

In the following, we compute the regularity of powers of the edge ideal of a
non-trivial integrally closed caterpillar graph under the condition that
$k=5$, or $k=6$ and $i=2$, and $d=1$.

######  Theorem 4.8.

Assuming $d=1$, let $G_{\omega}$ be a non-trivial integrally closed
caterpillar, as described in Remark 3.5, and let $P_{\omega}$ be the longest
path of length $(k-1)$ containing all non-trivial edges. Also, assume that
$\omega_{i}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$ with $\omega_{i}\geq
2$ and $\omega_{i}\geq\omega_{i+2}$ if $e_{i+2}\in E({P_{\omega}})$. If $k=5$,
or $k=6$ and $i=2$, then $\operatorname{reg}(I(G_{\omega})^{t}))\leq
2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}))$ for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$ and
$V(G_{\omega})=V(P_{\omega})\cup\\{y_{1},\ldots,y_{q}\\}$, where
$V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$, then it follows from Theorem 3.10
that

$\operatorname{reg}(I)=\begin{cases}2\omega_{1}+1,&\text{if $k=5,i=1$,}\\\
2\omega_{2},&\text{if $k=5,i=2$ and $\omega_{2}>\omega_{4}$,}\\\
2\omega_{2}+1,&\text{if $k=5,i=2$ and $\omega_{2}=\omega_{4}$, or
$k=6,i=2$.}\end{cases}$

We prove the statements by induction on $t$. The case where $t=1$ is trivial.
Thus, we may assume that $t\geq 2$. If $k=5$, then $i=1$ or $i=2$ due to
symmetry. We consider the following three cases:

(1) If $k=5$ and $i=1$, then, by Lemmas 2.2 and 4.1, [26, Theorems 3.1 and
3.3] and the inductive hypothesis on $t$, we obtain that
$\operatorname{reg}(((I^{t}:x_{5}):x_{4}))=\operatorname{reg}(I^{t-1})\leq
2\omega_{1}(t-1)+1$ and
$\operatorname{reg}(((I^{t}:x_{5}),x_{4}))=\operatorname{reg}((I(G_{\omega}\backslash
x_{4})^{t},x_{4}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{4})^{t})=2\omega_{1}t$.

The following two subcases are being considered. (i) If $q=1$, then, by some
simple calculations, we obtain

$\operatorname{reg}((I^{t},x_{5}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{5})^{t})=\begin{cases}2\omega_{1}t+1,&\text{if
$N_{G}(y_{1})=\\{x_{4}\\}$,}\\\ 2\omega_{1}t,&\text{ otherwise.}\\\
\end{cases}$

Replacing $x_{5}$ and $x_{4}$ for $x_{u}$ and $x_{v}$ respectively in the
following short exact sequences,

(8)
$\displaystyle\begin{matrix}0&\rightarrow&\frac{S}{I^{t}:x_{u}}(-1)&\stackrel{{\scriptstyle\cdot
x_{u}}}{{\longrightarrow}}&\frac{S}{I^{t}}&\rightarrow&\frac{S}{(I^{t},x_{u})}&\rightarrow&0,\\\
0&\rightarrow&\frac{S}{(I^{t}:x_{u}):x_{v}}(-1)&\stackrel{{\scriptstyle\cdot
x_{v}}}{{\longrightarrow}}&\frac{S}{I^{t}:x_{u}}&\rightarrow&\frac{S}{((I^{t}:x_{u}),x_{v})}&\rightarrow&0,\\\
\end{matrix}$

we obtain from Lemma 2.1 that $\operatorname{reg}(I^{t})\leq 2\omega_{1}t+1$.

(ii) If $q\geq 2$, then, by Lemmas 2.2, 4.1, 4.2 and 4.7 and the inductive
hypothesis on $q$, we get
$\operatorname{reg}((I^{t},x_{5}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{5})^{t})\leq 2\omega_{1}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash
x_{5}))\leq 2\omega_{1}t+1$. Again using the exact sequences (8), replacing
$x_{5}$ and $x_{4}$ for $x_{u}$ and $x_{v}$ respectively, we have
$\operatorname{reg}(I^{t})\leq 2\omega_{1}t+1$.

(2) If $k=5$ and $i=2$, it is sufficient to assume $\omega_{2}>\omega_{4}$,
since the case where $\omega_{2}=\omega_{4}$ follows from case (1) above. In
this case, by Lemmas 2.2 and 4.1, [26, Theorems 3.1 and 3.3] and the inductive
hypothesis on $t$, we can deduce that
$\operatorname{reg}(((I^{t}:x_{1}):x_{2}))=\operatorname{reg}(I^{t-1})\leq
2\omega_{2}(t-1)$,
$\operatorname{reg}(((I^{t}:x_{1}),x_{2}))=\operatorname{reg}((I(G_{\omega}\backslash
x_{2})^{t},x_{2}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{2})^{t})=2\omega_{4}t$.

If $q=1$, then
$\operatorname{reg}((I^{t},x_{1}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{1})^{t})=2\omega_{2}t$ by Lemmas 2.2, 4.1 and 4.7, and [26, Theorem 4.16].
Therefore, replacing $x_{5}$ and $x_{4}$ for $x_{u}$ and $x_{v}$,
respectively, in the exact sequences (8), it can be concluded that
$\operatorname{reg}(I^{t})\leq 2\omega_{2}t$. If $q\geq 2$, then,
$\operatorname{reg}((I^{t},x_{1}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{1})^{t})\leq 2\omega_{2}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash
x_{1}))=2\omega_{2}t$ by Lemmas 2.2 and 4.1, Theorem 4.7, and the inductive
hypothesis on $q$. Substituting $x_{u}$ and $x_{v}$ for $x_{1}$ and $x_{2}$ in
the exact sequences (8), we can deduce $\operatorname{reg}(I^{t})\leq
2\omega_{2}t$.

(3) If $k=6$ and $i=2$, it follows from Lemmas 2.2 and 4.1, Theorem 4.4, case
(2) above, and the inductive hypothesis on $t$ that
$\operatorname{reg}(((I^{t}:x_{6}):x_{5}))=\operatorname{reg}(I^{t-1})\leq
2\omega_{2}(t-1)+1$ and
$\operatorname{reg}(((I^{t}:x_{6}),x_{5}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{5})^{t})\leq 2\omega_{2}t$.

If $q=1$, then
$\operatorname{reg}((I^{t},x_{6}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{6})^{t})\leq 2\omega_{2}t+1$ by Lemmas 2.2, 4.1 and 4.2, and [26, Theorem
4.16] and cases (1) and (2) above. Therefore, by replacing $x_{u}$ and $x_{v}$
for $x_{6}$ and $x_{5}$ in the exact sequences (8), we can obtain
$\operatorname{reg}(I^{t})\leq 2\omega_{2}t+1$. If $q\geq 2$, it follows from
Lemmas 2.2, 4.1 and 4.2, cases (1) and (2) above, and the inductive hypothesis
on $q$ that
$\operatorname{reg}((I^{t},x_{6}))=\operatorname{reg}(I(G_{\omega}\backslash
x_{6})^{t})\leq 2\omega_{2}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash
x_{6}))\leq 2\omega_{2}t+1$. Again, substituting $x_{u}$ and $x_{v}$ for
$x_{6}$ and $x_{5}$ in the exact sequences (8), we can conclude that
$\operatorname{reg}(I^{t})\leq 2\omega_{2}t+1$. This completes the proof.

Next, we prove a main result of this section.

######  Theorem 4.9.

Assuming $d=1$, let $G_{\omega}$ be a non-trivial integrally closed
caterpillar, as described in Remark 3.5, and let $P_{\omega}$ be the longest
spine of length $(k-1)$ containing all non-trivial edges. Also, assume that
$\omega_{i}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$ with $\omega_{i}\geq
2$ and $\omega_{i}\geq\omega_{i+2}$ if $e_{i+2}\in E({P_{\omega}})$. Then
$\operatorname{reg}(I(G_{\omega})^{t})\leq
2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}))$ for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$ and $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$. We prove
the statements by induction on $t$. The case where $t=1$ is trivial. So we can
assume that $t\geq 2$. By Remark 4.6, there exists an induced trivial star
subgraph $H_{\omega}$ with $V(H_{\omega})=\\{x,y,z_{1}\ldots,z_{p}\\}$, where
$p\geq 1$ and $N_{G}(x)=\\{y,z_{1}\ldots,z_{p}\\}$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for all $i\in[p]$. We distinguish between the following
two cases:

(1) If $i=1$. We prove the statements by induction on $k$, and the case where
$k\leq 5$ is verified separately by Theorem 4.7, Theorem 4.8 and [26, Theorems
3.1 and 3.3]. Now we can assume that $k\geq 6$. Adopting the notations in
Remark 4.6, we choose $x=x_{k-1}$, $y=x_{k-2}$, $L_{1}=(I^{t}:xz_{1})$,
$T_{1}=(I^{t},xz_{1})$, $L_{j}=(T_{j-1}:xz_{j})$, $T_{j}=(T_{j-1},xz_{j})$ for
all $2\leq j\leq p$, then, by Lemma 4.2, Remark 4.6 and the inductive
hypothesis on $t$, we can conclude that $\operatorname{reg}(L_{1})\leq
2\omega_{1}(t-2)+\operatorname{reg}(I)$,
$\operatorname{reg}(((T_{p}:x):y))\leq
2\omega_{1}(t-2)+\operatorname{reg}(I)$, $\operatorname{reg}(L_{j})\leq
2\omega_{1}(t-2)+\operatorname{reg}(I)$ for all $2\leq j\leq p$. Meanwhile, by
Lemma 4.5 and the inductive hypothesis on $k$, we can get
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq
2\omega_{1}(t-1)+\operatorname{reg}(I),\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})\leq
2\omega_{1}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega_{1}(t-1)+\operatorname{reg}(I)-1$. Therefore, we obtain from that
Lemma 2.1 and the short exact sequences (3) that
$\operatorname{reg}(I^{t})\leq 2\omega_{1}(t-1)+\operatorname{reg}(I)$.

(2) If $i=2$ and $k\leq 6$, then the desired results follow from [26, Theorems
3.1 and 3.3], Theorems 4.4 and 4.8. If $i=2$ and $k\geq 7$, we choose
$x=x_{k-1},y=x_{k-2}$, see Figure $3$ for the case $\omega_{1}\geq 2$.
Similarly, if $i\geq 3$, we choose $x=x_{2},y=x_{3}$, see Figure $3$ for the
case $\omega_{3}\geq 2$. In these two cases, the desired results can be shown
by similar arguments as in case (1) above, so we omit its proof.

$x_{1}$$x_{2}$$x_{3}$$x_{4}$$x_{5}=y$$x_{6}=x$$x_{7}=z_{1}$$y_{1}$$y_{q}=z_{p}$
(A) The case $\omega_{1}\geq 2$

$x_{1}=z_{1}$$x_{2}=x$$x_{3}=y$$x_{4}$$x_{5}$$x_{6}$$x_{7}$$y_{1}=z_{p}$$y_{q}$
(B) The case $\omega_{3}\geq 2$

Figure 3.  $Caterpillar\ graph\ with\ k=7,d=1$

Now, we provide some linear upper bounds on the regularity of powers of the
edge ideal of a non-trivial integrally closed tree under the condition that
$k=4$, $d=2$ and $\omega_{1}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$.

######  Theorem 4.10.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $k=4$, $d=2$ and $\omega_{1}=\max\\{\omega(e)\mid e\in
E(P_{\omega})\\}$, then $\operatorname{reg}(I(G_{\omega})^{t})\leq
2\omega_{1}(t-1)+\operatorname{reg}(I(G_{\omega}))$ for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$, $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$,
$d(x)=\min\\{d_{G}(x,x_{j})|1\leq j\leq k\text{\ and\ }x\in V(G_{\omega})\\}$
and $\gamma=|\\{x\in V(G_{\omega})\mid d(x)=2\\}|$. We prove the statements by
induction on $t$. The case where $t=1$ is trivial. So we can assume that
$t\geq 2$. Since $d=2$, there exists an induced trivial star subgraph
$H_{\omega}$ with $V(H_{\omega})=\\{x,y\\}\cup C$, where $d(x)=1$,
$N_{G}(x)=\\{y\\}\cup C$, $C=\\{z_{1}\ldots,z_{p}\\}$ with $p\geq 1$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for all $i\in[p]$, see Figure $4$ for the case $k=4$ and
$d=2$. From the definition of $P_{\omega}$, we know that if $\omega_{3}=1$,
then $y=x_{2}$, and if $\omega_{3}\geq 2$, then $y=x_{2}$ or $y=x_{3}$. For
the sake of consistency, we will always set $y=x_{2}$, since the case where
$y=x_{3}$ can be shown by similar arguments. Adapting the notations in Remark
4.6, let
$L_{1}=(I^{t}:xz_{1}),T_{1}=(I^{t},xz_{1}),L_{j}=(T_{j-1}:xz_{j}),T_{j}=(T_{j-1},xz_{j})$
for any $2\leq j\leq p$, then, by Lemma 4.2, the inductive hypothesis on $t$,
we have $\operatorname{reg}(L_{1})\leq
2\omega_{1}(t-2)+\operatorname{reg}(I)$,
$\operatorname{reg}(((T_{p}:x):y))\leq 2\omega_{1}(t-2)+\operatorname{reg}(I)$
and $\operatorname{reg}(L_{j})\leq 2\omega_{1}(t-2)+\operatorname{reg}(I)$ for
all $2\leq j\leq p$.

To compute $\operatorname{reg}(I^{t})$ from the exact sequences (3), we have
to compute $\operatorname{reg}((T_{p},x))$ and
$\operatorname{reg}(((T_{p}:x),y))$. From Remark 4.6, Theorem 4.7 and [26,
Theorem 4.16], we can deduce that if $\gamma=1$, then
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})=2\omega_{1}t$. If $\gamma\geq 2$, then
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq 2\omega_{1}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash x))\leq
2\omega_{1}(t-1)+\operatorname{reg}(I)$ again by Lemma 4.2 and the inductive
hypothesis on $\gamma$. From Remark 4.6, Lemmas 2.2, 3.4 and 4.5 and [26,
Theorems 3.1, 3.3 and 4.16], we can deduce that
$\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})=2\omega_{3}t$
when $\gamma=1$, and
$\operatorname{reg}(((T_{p}:x),y))=2(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega_{1}(t-1)+\operatorname{reg}(I)-1$ if $\gamma\geq 2$ and
$\omega_{3}=1$, and $\operatorname{reg}(((T_{p}:x),y))\leq
2\omega_{3}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega_{1}(t-1)+\operatorname{reg}(I)-1$ if $\omega_{3}\geq 2$ and
$\gamma\geq 2$, again by the case $\omega_{3}=1$ above and Theorem 4.7.

In any case, we can always obtain $\operatorname{reg}(I^{t})\leq
2\omega_{1}(t-1)+\operatorname{reg}(I)$ by Lemma 2.1 and the exact sequences
(3).

$x_{1}$$x_{2}=y$$x_{3}$$x_{4}$$x$$z_{1}$$z_{p}$ (A) The case $k=4,d=2$

$x_{1}$$x_{2}=y$$x_{3}$$x_{4}$$x_{5}$$y$$x$$z_{1}$$z_{p}$ (B) The case
$k=5,d=3$

Figure 4. $Caterpillar\ graph$

Next, we prove another main result of this section.

######  Theorem 4.11.

Assuming $d=2$, let $G_{\omega}$ be a non-trivial integrally closed tree, as
described in Remark 3.5, and let $P_{\omega}$ be the longest path of length
$(k-1)$ containing all non-trivial edges. Also, assume that
$\omega_{i}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$ with $\omega_{i}\geq
2$ and $\omega_{i}\geq\omega_{i+2}$ if $e_{i+2}\in E({P_{\omega}})$. Then
$\operatorname{reg}(I(G_{\omega})^{t})\leq
2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}))$ for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$, $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$,
$d(x)=\min\\{d_{G}(x,x_{j})|1\leq j\leq k\text{\ and\ }x\in V(G_{\omega})\\}$,
and $\gamma=|\\{x\in V(G_{\omega})\mid d(x)=2\\}|$. We prove the statements by
induction on $t$ and $k$. The case where $t=1$ is trivial and where $t\geq 2$
and $k=4$ is verified by Theorem 4.10. So we can assume that $t\geq 2$ and
$k\geq 5$. As in the proof of Theorem 4.10, there exists an induced trivial
star subgraph $H_{\omega}$ with $V(H_{\omega})=\\{x,y,z_{1}\ldots,z_{p}\\}$,
where $d(x)=1$, $N_{G}(x)=\\{y,z_{1}\ldots,z_{p}\\}$ and $p\geq 1$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for all $i\in[p]$. By the definition of $P_{\omega}$,
$y=x_{j}$ with $2\leq j\leq k-1$. For consistency, we choose $y=x_{3}$, since
other $y$ can be shown by similar arguments. Adapting the notations in Remark
4.6, let
$L_{1}=(I^{t}:xz_{1}),T_{1}=(I^{t},xz_{1}),L_{j}=(T_{j-1}:xz_{j}),T_{j}=(T_{j-1},xz_{j})$
for all $2\leq j\leq p$, then, by Lemma 4.2, the inductive hypothesis on $t$,
we can obtain that $\operatorname{reg}(L_{1})\leq
2\omega_{i}(t-2)+\operatorname{reg}(I)$,
$\operatorname{reg}(((T_{p}:x):y))\leq 2\omega_{i}(t-2)+\operatorname{reg}(I)$
and $\operatorname{reg}(L_{j})\leq 2\omega_{i}(t-2)+\operatorname{reg}(I)$ for
all $2\leq j\leq p$.

If $\gamma=1$, then by Lemmas 2.2, 3.4, 4.2 and 4.5, Theorem 4.9, and [26,
Theorems 3.1, 3.3 and 4.16] and Lemma 4.2, we can deduce that
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq 2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash x))\leq
2\omega_{i}(t-1)+\operatorname{reg}(I)$ and
$\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})\\\
\leq 2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega_{i}(t-1)+\operatorname{reg}(I)-1$.

If $\gamma\geq 2$, then, by Lemmas 2.2, 3.4, 4.2 and 4.5, [26, Theorems 3.1,
3.3 and 4.16], and the inductive hypothesis on $\gamma$, we have
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq 2\omega_{i}(t-1)+\operatorname{reg}(I(G_{\omega}\backslash x))\leq
2\omega_{i}(t-1)+\operatorname{reg}(I)$ and
$\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})\leq
2\omega_{i}(t-1)+\operatorname{reg}(I)-1$.

Therefore, in every case, we can always obtain $\operatorname{reg}(I^{t})\leq
2\omega_{i}(t-1)+\operatorname{reg}(I)$ by Lemma 2.1 and the exact sequences
(3).

It is time for the important result of this section.

######  Theorem 4.12.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $\omega=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$, then
$\operatorname{reg}(I(G_{\omega})^{t})\leq
2\omega(t-1)+\operatorname{reg}(I(G_{\omega}))$ for all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$, $V(P_{\omega})=\\{x_{1},\ldots,x_{k}\\}$,
$d(x)=\min\\{d_{G}(x,x_{j})|1\leq j\leq k\text{\ and\ }x\in V(G_{\omega})\\}$,
and $\gamma=|\\{x\in V(G_{\omega})\mid d(x)=d\\}|$. We prove the statements by
induction on $t$ and $d$. The case where $t=1$ is trivial and the case where
$t\geq 2$ and $d\leq 2$ is proved separately in [26, Theorem 4.16], Theorems
4.9 and 4.11. Thus, in the following, we can assume that $t\geq 2$ and $d\geq
3$. As in the proof of Theorem 4.10, there exists an induced trivial star
subgraph $H_{\omega}$ with $V(H_{\omega})=\\{x,y,z_{1}\ldots,z_{p}\\}$, where
$d(x)=d-1$, $N_{G}(x)=\\{y,z_{1}\ldots,z_{p}\\}$ and $p\geq 1$, and
$E(H_{\omega})=\\{xy,xz_{1},\ldots,xz_{p}\\}$ with $\deg_{G}(y)\geq 2$ and
$\deg_{G}(z_{i})=1$ for all $i\in[p]$ (, see Figure $4$ for the case $k=5$ and
$d=3$). Adapting the notations in Remark 4.6, let
$L_{1}=(I^{t}:xz_{1}),T_{1}=(I^{t},xz_{1}),L_{j}=(T_{j-1}:xz_{j}),T_{j}=(T_{j-1},xz_{j})$
for all $2\leq j\leq p$, then, by Lemma 4.2, the inductive hypothesis on $t$,
we have $\operatorname{reg}(L_{1})\leq 2\omega(t-2)+\operatorname{reg}(I)$,
$\operatorname{reg}(((T_{p}:x):y))\leq 2\omega(t-2)+\operatorname{reg}(I)$,
$\operatorname{reg}(L_{j})\leq 2\omega(t-2)+\operatorname{reg}(I)$ for all
$2\leq j\leq p$.

If $\gamma=1$, then, by Lemmas 4.2 and 4.5, and the inductive hypothesis on
$d$, we obtain
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq 2\omega(t-1)+\operatorname{reg}(I(G_{\omega}\backslash x))\leq
2\omega(t-1)+\operatorname{reg}(I)$ and
$\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})\leq
2\omega(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega(t-1)+\operatorname{reg}(I)-1$.

If $\gamma\geq 2$, then, by Lemmas 4.2 and 4.5, and the inductive hypothesis
on $\gamma$ and $d$, we can deduce
$\operatorname{reg}((T_{p},x))=\operatorname{reg}(I(G_{\omega}\backslash
x)^{t})\leq 2\omega(t-1)+\operatorname{reg}(I(G_{\omega}\backslash x))\leq
2\omega(t-1)+\operatorname{reg}(I)$ and
$\operatorname{reg}(((T_{p}:x),y))=\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\})^{t})\leq
2\omega(t-1)+\operatorname{reg}(I(G_{\omega}\backslash\\{x,y\\}))\leq
2\omega(t-1)+\operatorname{reg}(I)-1$.

Therefore, in each case, we can always obtain $\operatorname{reg}(I^{t})\leq
2\omega(t-1)+\operatorname{reg}(I)$ by Lemma 2.1 and the exact sequences (3).

In special cases, the upper bound in Theorem 4.12 can reach.

######  Theorem 4.13.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, and
let $P_{\omega}$ be its longest path of length $(k-1)$ containing all non-
trivial edges. If $\omega_{i}=\max\\{\omega(e)\mid e\in E(P_{\omega})\\}$ and
$\operatorname{reg}(I(G_{\omega}))=2\omega_{i}+(s_{i}(G_{\omega})-1)$, then
$\operatorname{reg}(I(G_{\omega})^{t})=2\omega_{i}t+(s_{i}(G_{\omega})-1)$ for
all $t\geq 1$.

###### Proof.

Let $I=I(G_{\omega})$ and $I^{t}=J^{t}+K$, where
$J=(x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}})$ and
$K=\mathcal{G}(I^{t})\backslash\mathcal{G}(J^{t})$, and let
$(J^{t})^{\mathcal{P}}$, $K^{\mathcal{P}}$ and $(I^{t})^{\mathcal{P}}$ be the
polarizations of $J^{t}$, $K$ and $I^{t}$, respectively. Then
$(I^{t})^{\mathcal{P}}=J^{\mathcal{P}}+K^{\mathcal{P}}$ and
$(J^{t})^{\mathcal{P}}\cap K^{\mathcal{P}}=(J^{t}\cap K)^{\mathcal{P}}$ by [8,
Proposition 2.3]. By Lemmas 2.4 and 2.6, we know that
$(I^{t})^{\mathcal{P}}=(J^{t})^{\mathcal{P}}+K^{\mathcal{P}}$ is a Betti
splitting and that

$\displaystyle\operatorname{reg}(I^{t})$
$\displaystyle=\operatorname{reg}((I^{t})^{\mathcal{P}})=\max\\{\operatorname{reg}((J^{t})^{\mathcal{P}}),\operatorname{reg}(K^{\mathcal{P}}),\operatorname{reg}((J^{t})^{\mathcal{P}}\cap
K^{\mathcal{P}})-1\\}$ (9)
$\displaystyle=\max\\{\operatorname{reg}(J^{t}),\operatorname{reg}(K),\operatorname{reg}(J^{t}\cap
K)-1\\}.$

Since $\omega_{i}=\max\\{\omega(e)\mid e\in E(G_{\omega})\\}$, we have

$\displaystyle J^{t}\cap K$
$\displaystyle=(x_{i}^{t\omega_{i}}x_{i+1}^{t\omega_{i}})\cap(\mathcal{G}(I^{t})\backslash\mathcal{G}(J^{t}))$
$\displaystyle=(x_{i}^{t\omega_{i}}x_{i+1}^{t\omega_{i}})\cap[(x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}})^{t-1}(\mathcal{G}(I)\backslash\mathcal{G}(J))+(x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}})^{t-2}(\mathcal{G}(I)\backslash\mathcal{G}(J))^{2}+\cdots$
$\displaystyle+(x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}})(\mathcal{G}(I)\backslash\mathcal{G}(J))^{t-1}+(\mathcal{G}(I)\backslash\mathcal{G}(J))^{t}]$
$\displaystyle=(x_{i}^{t\omega_{i}}x_{i+1}^{t\omega_{i}})\cap((x_{i}^{\omega_{i}}x_{i+1}^{\omega_{i}})^{t-1}(\mathcal{G}(I)\backslash\mathcal{G}(J)))$
$\displaystyle=J^{t}L$

where $L$ is an ideal, its minimal generator set is $(N_{G}(A)\backslash
A)\sqcup\mathcal{G}(I(G_{\omega}^{3}))$ with $A=\\{x_{i},x_{i+1}\\}$ and
$G_{\omega}^{3}=G_{\omega}\backslash N_{G}(A)$. So
$\operatorname{reg}(J^{t}\cap
K)=\operatorname{reg}(J^{t}L)=\operatorname{reg}(J^{t})+\operatorname{reg}(L)=2\omega_{i}t+\nu(G_{\omega}^{3})+1=2\omega_{i}t+s_{i}(G_{\omega})$
by Lemmas 2.2 and 3.6.

On the other hand, let $H$ and $H^{\prime}$ be hypergraphs associated with
$\mathcal{G}((I^{t})^{\mathcal{P}})$ and $\mathcal{G}(K^{\mathcal{P}})$,
respectively, then $H^{\prime}$ is an induced subhypergraph of $H$. Hence,
$\operatorname{reg}(K)=\operatorname{reg}(K^{\mathcal{P}})\leq\operatorname{reg}((I^{t})^{\mathcal{P}})\leq
2\omega_{i}t+(s_{i}(G_{\omega})-1)$ by Lemma 2.6, Theorem 4.12 and [11, Lemma
3.1]. Therefore,
$\operatorname{reg}(I(G_{\omega})^{t})=2\omega_{i}t+(s_{i}(G_{\omega})-1)$
from formula (9).

The following examples show that the upper bound in Theorem 4.12 can be
strict.

######  Example 4.14.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, its
edge ideal is
$I(G_{\omega})=(x_{1}x_{2},x_{2}^{2}x_{3}^{2},x_{3}x_{4},x_{4}x_{5},x_{3}x_{6},x_{6}x_{7},x_{3}x_{8},x_{8}x_{9})$.
Let $P_{\omega}$ be its longest path containing all non-trivial edges with
$V(P_{\omega})=\\{x_{1},x_{2},x_{3},x_{4},x_{5}\\}$, then
$\omega_{2}=\max\\{\omega_{t}\mid\omega_{t}=\omega(e_{t})\text{ and
}e_{t}=x_{t}x_{t+1}\text{ for any }1\leq t\leq 4\\}=2$, $s_{2}(G_{\omega})=1$
and $\nu(G_{\omega})=4$. Thus
$\operatorname{reg}(I(G_{\omega}))=\nu(G_{\omega})+1$ by Theorem 3.10. By
using CoCoA, we obtain
$\operatorname{reg}(I(G_{\omega})^{2})=8<2\omega_{2}+\operatorname{reg}(I(G_{\omega}))$.

######  Example 4.15.

Let $G_{\omega}$ be a non-trivial integrally closed tree as in Remark 3.5, its
edge ideal is
$I(G_{\omega})=(x_{1}^{3}x_{2}^{3},x_{2}x_{3},x_{3}^{2}x_{4}^{2},x_{2}x_{5},x_{5}x_{6},x_{2}x_{7},x_{7}x_{8},x_{2}x_{9},x_{9}x_{10})$.
Let $P_{\omega}$ be its longest path containing all non-trivial edges with
$V(P_{\omega})=\\{x_{1},x_{2},x_{3},x_{4}\\}$, then
$\omega_{1}=\max\\{\omega_{t}\mid\omega_{t}=\omega(e_{t})\text{ and
}e_{t}=x_{t}x_{t+1}\text{ for any }1\leq t\leq 4\\}=3$, $\omega_{3}=2$,
$s_{1}(G_{\omega})=1$, $s_{3}(G_{\omega})=3$ and $\nu(G_{\omega})=4$. Thus
$\operatorname{reg}(I(G_{\omega}))=2\omega_{3}+(s_{3}(G_{\omega})-1)=7$ by
Theorem 3.10. By using CoCoA, we obtain
$\operatorname{reg}(I(G_{\omega})^{2})=12<2\omega_{1}+\operatorname{reg}(I(G_{\omega}))$.

Acknowledgments

This research is supported by the Natural Science Foundation of Jiangsu
Province (No. BK20221353). The authors are grateful to the software systems
[3] and [10] for providing us with a large number of examples.

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