# The weak Lefschetz property of artinian algebras associated to paths and
cycles

Hop D. Nguyen  Institute of Mathematics, VAST, 18 Hoang Quoc Viet, Cau Giay,
10307 Hanoi, Vietnam<EMAIL_ADDRESS>and  Quang Hoa Tran  University of
Education, Hue University, 34 Le Loi St., Hue City, Vietnam.
<EMAIL_ADDRESS>Dedicated to Professor Ngo Viet Trung on the
occasion of his 70th birthday

###### Abstract.

Given a base field $\Bbbk$ of characteristic zero, for each graph $G$, we
associate the artinian algebra $A(G)$ defined by the edge ideal of $G$ and the
squares of the variables. We study the weak Lefschetz property of $A(G)$. We
classify some classes of graphs with relatively few edges, including paths and
cycles, such that its associated artinian ring has the weak Lefschetz
property.

###### Key words and phrases:

Artinian algebras; edge ideals; independence polynomials; weak Lefschetz
property

###### 2020 Mathematics Subject Classification:

13E10, 13F20, 13F55, 05C31

##  1\. Introduction

A graded artinian algebra $A=[A]_{0}\oplus[A]_{1}\oplus\cdots\oplus[A]_{D}$
over a field $\Bbbk$ has the weak Lefschetz property (WLP for short) if there
exists a linear form $\ell\in[A]_{1}$ such that the multiplication maps
$\cdot\ell:[A]_{i}\rightarrow[A]_{i+1}$ have maximal rank for all $i$, while
$A$ has the strong Lefschetz property (SLP for short) if there exists a linear
form $\ell$ such that multiplication map
$\cdot\ell^{j}:[A]_{i}\longrightarrow[A]_{i+j}$ has maximal rank for all $i$
and all $j$. The study of Lefschetz properties of graded algebras has
connections to several areas of mathematics. Many authors have studied the
problem from many different points of view, applying tools from representation
theory, algebraic topology, differential geometry, commutative algebra, among
others (see, for instance, [6, 16, 22, 25, 26, 27, 28, 29, 30, 31, 32, 33,
36]). Even the characteristic of $\Bbbk$ plays an interesting role in the
study of the Lefschetz properties; see, for example, [7, 8, 9, 20, 21, 27].

The case of artinian $\Bbbk$-algebras defined by monomial ideals, while being
rather accessible, is far from simple and the literature concerning their
Lefschetz properties is quite extensive; see, for instance, [2, 3, 11, 12, 23,
22, 27, 24] and the references therein. In this work, we focus on a special
class of artinian algebras defined by quadratic monomials, given as follows.
Let $G=(V,E)$ be a simple graph on the vertex set $V=\\{1,2,\ldots,n\\}$ and
$R=\Bbbk[x_{1},\ldots,x_{n}]$ be the standard graded polynomial ring over
$\Bbbk$. The _edge ideal_ of $G$ is given by

$I(G)=(x_{i}x_{j}\mid\\{i,j\\}\in E)\subset R.$

Then, we say that

$A(G)=\frac{R}{(x_{1}^{2},\ldots,x_{n}^{2})+I(G)}$

is the artinian algebra associated to $G$. We are interested the following
question.

######  Question 1.1.

For which graphs $G$ does $A(G)$ have the WLP or the SLP? If $A(G)$ does not
have the WLP or the SLP, in which degrees do the multiplication maps fail to
have maximal rank?

The algebra $A(G)$ has been studied in [37] where the second author classifies
the WLP/SLP for some special classes of graphs including the complete graphs,
the star graphs, the Barbell graphs, and wheel graphs. Note that artinian
algebras defined by quadratic monomial relations were considered in previous
work by Michałek–Miró-Roig [23], and Migliore–Nagel–Schenck [24]. These rings
can be regarded as special cases of a more general construction due to Dao and
Nair [11] where they associate to each simplicial complex $\Delta$ on $n$
vertices the ring

$A(\Delta)=\frac{R}{\left(x_{1}^{2},\ldots,x_{n}^{2}\right)+I_{\Delta}},$

in which $I_{\Delta}$ is the Stanley-Reisner ideal of $\Delta$. A recent work
due to Cooper et al. [10] also investigates the WLP of $A(G)$ where the focus
was on whiskered graphs.

Our main goal in this note is to classify some important classes of graphs $G$
where $A(G)$ has the weak Lefschetz property, such as paths, cycles and
certain tadpole graphs. More precisely, denote by
$P_{n},C_{n},{\operatorname{Pan}}_{n}$ the paths, cycles, pan graphs (namely a
cycle together with a pendant attached to one vertex), respectively. Our main
results are the following.

######  Theorem 1.2 (Theorems 4.2, 4.4 and 4.7).

Assume that $\operatorname{char}(\Bbbk)=0.$

  1. (1)

The ring $A(P_{n})$ has the WLP if and only if
$n\in\\{1,2,\ldots,7,9,10,13\\}.$

  2. (2)

The ring $A(C_{n})$ has the WLP if and only if
$n\in\\{3,4,\ldots,11,13,14,17\\}.$

  3. (3)

The ring $A({\operatorname{Pan}}_{n})$ has the WLP if and only if
$n\in\\{4,5,\ldots,10,12,13,16\\}.$

The proof combines Macaulay2 [13] computations with inductive arguments based
on the unimodality of the independence polynomials of the relevant graphs. We
hope that our main results will inspire further research on Question 1.1.

Our paper is structured as follows. In the next section we recall the relevant
terminology and results on artinian algebras, Lefschetz properties, and graph
theory. In Section 3, we investigate the unimodality and the mode of the
independence polynomials of familiar graphs, such as path, cycle and pan
graphs. These results are useful to study the WLP of artinian algebras
associated to these graphs. In Section 4, we prove the our main theorems (see
Theorems 4.2, 4.4 and 4.7) on the WLP of the artinian algebras associated to
paths, cycles and the pan graphs.

##  2\. Preliminaries

In this section we recall some standard terminology and notations from
commutative algebra and combinatorial commutative algebra, as well as some
results needed later on. For a general introduction to artinian rings and the
weak and strong Lefschetz properties we refer the readers to [17] and [29].

###  2.1.  The weak Lefschetz property

In this paper we consider artinian algebras defined by monomial ideals, and in
this case it suffices to chose the Lefschetz element to be the sum of the
variables.

######  Proposition 2.1.

[27, 34] Let $I\subset R=\Bbbk[x_{1},\ldots,x_{n}]$ be an artinian monomial
ideal. Then $A=R/I$ has the WLP if and only if $\ell=x_{1}+x_{2}+\cdots+x_{n}$
is a Lefschetz element for $A$.

A necessary condition for the WLP and SLP of an artinian algebra $A$ is the
unimodality of the Hilbert series of $A$.

######  Definition 2.2.

Let $A=\oplus_{j\geq 0}[A]_{j}$ be a standard graded $\Bbbk$-algebra. The
Hilbert series of $A$ is the power series $\sum\dim_{\Bbbk}[A]_{i}t^{i}$ and
is denoted by $HS(A,t)$. The Hilbert function of $A$ is the function
$h_{A}:\mathbb{N}\longrightarrow\mathbb{N}$ defined by
$h_{A}(j)=\dim_{\Bbbk}[A]_{j}$.
If $A$ is an artinian graded algebra, then $[A]_{i}=0$ for $i\gg 0.$ Denote

$D=\max\\{i\mid[A]_{i}\neq 0\\},$

the _socle degree_ of $A$.

######  Definition 2.3.

A polynomial $\sum_{k=0}^{n}a_{k}t^{k}\in\mathbb{R}[t]$ with non-negative
coefficients is called unimodal if there is some $m$, such that

$a_{0}\leq a_{1}\leq\cdots\leq a_{m-1}\leq a_{m}\geq a_{m+1}\geq\cdots\geq
a_{n}.$

Set $a_{-1}=0$. The mode of the unimodal polynomial $\sum_{k=0}^{n}a_{k}t^{k}$
is defined to be the unique integer $i$ between $0$ and $n$ such that

$a_{i-1}<a_{i}\geq a_{i+1}\geq\cdots\geq a_{n}.$

######  Proposition 2.4.

[17, Proposition 3.2] If $A$ has the WLP or SLP then the Hilbert series of $A$
is unimodal.

Finally, to study the failure of the WLP of tensor products of
$\Bbbk$-algebras, the following simple lemma turns out to be quite useful.

######  Lemma 2.5.

[5, Lemma 7.8] Let $A=A^{\prime}\otimes_{\Bbbk}A^{\prime\prime}$ be a tensor
product of two graded artinian $\Bbbk$-algebras $A^{\prime}$ and
$A^{\prime\prime}$. Let $\ell^{\prime}\in A^{\prime}$ and
$\ell^{\prime\prime}\in A^{\prime\prime}$ be linear elements, and set
$\ell=\ell^{\prime}+\ell^{\prime\prime}=\ell^{\prime}\otimes
1+1\otimes\ell^{\prime\prime}\in A$. Then

  1. (a)

If the multiplication maps
$\cdot\ell^{\prime}:[A^{\prime}]_{i}\longrightarrow[A^{\prime}]_{i+1}$ and
$\cdot\ell^{\prime\prime}:[A^{\prime\prime}]_{j}\longrightarrow[A^{\prime\prime}]_{j+1}$
are both not surjective, then neither is map

$\cdot\ell:[A]_{i+j+1}\longrightarrow[A]_{i+j+2}.$

  2. (b)

If the multiplication maps
$\cdot\ell^{\prime}:[A^{\prime}]_{i}\longrightarrow[A^{\prime}]_{i+1}$ and
$\cdot\ell^{\prime\prime}:[A^{\prime\prime}]_{j}\longrightarrow[A^{\prime\prime}]_{j+1}$
are both not injective, then neither is the map

$\cdot\ell:[A]_{i+j}\longrightarrow[A]_{i+j+1}.$

###  2.2. Graph theory

From now on, by a graph we mean a simple graph $G=(V,E)$ with the vertex set
$V=V(G)$ and the edge set $E=E(G)$. We start by recalling some basic
definitions.

######  Definition 2.6.

The disjoint union of the graphs $G_{1}$ and $G_{2}$ is a graph $G=G_{1}\cup
G_{2}$ having as vertex set the disjoint union of $V(G_{1})$ and $V(G_{2})$,
and as edge set the disjoint union of $E(G_{1})$ and $E(G_{2})$. In
particular, $\cup_{n}G$ denotes the disjoint union of $n>1$ copies of the
graph $G$.

######  Definition 2.7.

Let $G=(V,E)$ be a graph.

  1. (i)

A subset $X$ of $V$ is called an independent set of $G$ if for any $i,j\in X,\
\\{i,j\\}\notin E$, i.e., the vertices in $X$ are pairwise non-adjacent. If an
independent set $X$ has $k$ elements, then we say that $X$ is an independent
set of size $k$ or a $k$-independent set of $G$.

  2. (ii)

An independent set $X$ is called maximal if for every vertices $v\in
V\setminus X$, $X\cup\\{v\\}$ is not an independent set of $G$.

  3. (iii)

The independence number of a graph $G$ is the largest cardinality of an
independent set of $G$. We denote this value by $\alpha(G)$.

  4. (iv)

A graph $G$ is said to be well-covered if every maximal independent set of $G$
has the same size $\alpha(G).$

######  Definition 2.8.

The independence polynomial of a graph $G$ is a polynomial in one variable $t$
whose coefficient of $t^{k}$ given by the number of independent sets of size
$k$ of $G$. We denote this polynomial by $I(G;t)$, i.e.,

$I(G;t)=\sum_{k=0}^{\alpha(G)}s_{k}(G)t^{k},$

where $s_{k}(G)$ is the number of independent sets of size $k$ in $G$. Note
that $s_{0}(G)=1$ since $\emptyset$ is an independent set of any graph $G$.

The independence polynomial of a graph was defined by Gutman and Harary in
[14] as a generalization of the matching polynomial of a graph. For a vertex
$v\in V$, define $N(v)=\\{w\mid w\in V\;\text{and}\;\\{v,w\\}\in E\\}$ and
$N[v]=N(v)\cup\\{v\\}$. The following equalities are very useful for the
calculation of the independence polynomial for various families of graphs
(see, for instance, [14, 18]).

######  Proposition 2.9.

Let $G_{1},G_{2},G$ be the graphs. Assume that $G=(V,E),w\in V$ and
$e=\\{u,v\\}\in E$. Then the following equalities hold:

  1. (i)

$I(G;t)=I(G\setminus w;t)+t\cdot I(G\setminus N[w];t)$;

  2. (ii)

$I(G;t)=I(G\setminus e;t)-t^{2}\cdot I(G\setminus(N(u)\cup N(v));t)$;

  3. (iii)

$I(G_{1}\cup G_{2};t)=I(G_{1};t)I(G_{2};t)$.

###  2.3. Artinian algebras associated to graphs

Let $G=(V,E)$ be a graph, with the set of vertices $V=\\{1,2,\ldots,n\\}$. Let
$R=\Bbbk[x_{1},\ldots,x_{n}]$ be a standard polynomial ring over $\Bbbk$. The
edge ideal of $G$ is the ideal

$I(G)=(x_{i}x_{j}\mid\\{i,j\\}\in E)\subset R.$

Then, we say that

$A(G)=\frac{R}{(x_{1}^{2},\ldots,x_{n}^{2})+I(G)}$

is the artinian algebra associated to $G$. The algebra $A(G)$ contains
significant combinatorial information about $G$, as witnessed by

######  Proposition 2.10.

The Hilbert series of $A(G)$ is equal to the independence polynomial of $G$,
i.e.

$HS(A(G);t)=I(G;t)=\sum_{k=0}^{\alpha(G)}s_{k}(G)t^{k}.$

As a consequence, the Castelnuovo–Mumford regularity of $A(G)$ is
${\operatorname{reg}}(A(G))=\alpha(G)$ and $A(G)$ is level if and only $G$ is
well-covered.

Therefore, the WLP/SLP of $A(G)$ has strong consequences on the unimodality of
the independence polynomial of $G$. Indeed, if $I(G;t)$ is not unimodal, then
$A(G)$ fails the WLP by Proposition 2.4. Thus, to study the WLP/ SLP of
$A(G)$, it is enough to consider the graphs whose independence polynomials are
unimodal. Concerning the unimodality of the independence polynomial of graphs,
we have the following famous conjecture.

######  Conjecture 2.11.

[1] If $G$ is a tree or forest, then the independence polynomial of $G$ is
unimodal.

To our best knowledge, until now, the largest class of graphs for which the
independence polynomial is known to be unimodal is the class of claw-free
graphs [15]. Conjecture 2.11 remains widely open. The following example due to
Bhattacharyya and Kahn [4] shows that one cannot expect the statement of
Conjecture 2.11 to be true for bipartite graphs.

######  Example 2.12.

Given positive integers $m$ and $n>m$, let $G=(V,E)$ with $V=V_{1}\cup
V_{2}\cup V_{3}$, where $V_{1},V_{2},V_{3}$ are disjoint; $|V_{1}|=n-m$ and
$|V_{2}|=|V_{3}|=m;$ $E$ consists of the edges of the complete bipartite graph
with the bipartition $V_{1}\cup V_{2}$ and a perfect matching between $V_{2}$
and $V_{3}$. Then $G$ is a bipartite graph and for every $i\geq 0,\
s_{i}(G)=(2^{i}-1)\binom{m}{i}+\binom{n}{i}.$ Therefore, for $m\geq 95$ and
$n=\lfloor m\log_{2}(3)\rfloor$, $I(G;t)$ is not unimodal. As a consequence,
$A(G)$ fails the WLP.

It is known that the Lefschetz properties depend strongly on the
characteristic of the field.

######  Example 2.13.

An empty graph is simply a graph with no edges. We denote the empty graph on
$n$ vertices by $E_{n}$. Then

$A(E_{n})=R/(x_{1}^{2},\ldots,x_{n}^{2})\quad\text{and}\;I(E_{n};t)=(1+t)^{n}.$

A well-known result of Stanley on complete intersections says that $A(E_{n})$
has the SLP if $\operatorname{char}(\Bbbk)=0$ or
$\operatorname{char}(\Bbbk)>n$ [36, 35]. This result does not hold if
$\operatorname{char}(\Bbbk)\leq n$, even for the WLP. Indeed, in the case
where $\operatorname{char}(\Bbbk)=2$, it was known that $A(E_{n})$ has the WLP
if and only if $n=3$ [7, 20]. In [20] Kustin and Vraciu showed that if $n\geq
5$ and $\operatorname{char}(\Bbbk)=p$ is odd, then $A(E_{n})$ has the WLP if
and only if $p\geq\lfloor\frac{n+3}{2}\rfloor$.

The complete graph on $n$ vertices, denoted by $K_{n}$, is the graph where
each vertex is adjacent to every other. It follows that

$A(K_{n})=R/(x_{1},\ldots,x_{n})^{2}\quad\text{and}\;I(K_{n};t)=1+nt.$

It is easy to see that $A(K_{n})$ has the SLP for any field $\Bbbk$.
Concerning disjoint unions of complete graphs, we have the following.

######  Proposition 2.14.

[24, Theorem 4.8] Let $\operatorname{char}(\Bbbk)\neq 2$ and $A(G)$ be the
artinian algebra associated to $G=\cup_{i=1}^{r}K_{n_{i}}$. Assume $n_{1}\geq
n_{2}\geq\cdots\geq n_{r}\geq 1$. Then $A(G)$ has the WLP if and only if one
of the following holds:

  1. (1)

$n_{2}=\cdots=n_{r}=1$, i.e. $G$ is the disjoint union of a complete graph
$K_{n_{1}}$ and an empty graph on $r-1$ vertices.

  2. (2)

$n_{3}=\cdots=n_{r}=1$ and $r$ is odd.

In particular for every $n\geq 2$, if $G$ is the disjoint union of $n$
complete graphs, none of which is a singleton, then $A(G)$ does not have the
WLP.

##  3\. Independence polynomial of some graphs

In this section, we provide some results on the independence polynomial of
some familiar graphs, namely paths, cycles, and pan graphs. These results will
be useful to prove our main theorems in the next section.

###  3.1. Paths

Let $P_{n}$ be the path on $n$ vertices ($n\geq 1$).

123456 Figure 1. The path $P_{6}$

######  Proposition 3.1.

The independence polynomial of $P_{n}$ is

$I(P_{n};t)=\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}\binom{n+1-i}{i}t^{i}.$

Moreover, for every $n\geq 1$, $I(P_{n};t)$ is unimodal, with the mode

$\lambda_{n}=\left\lceil\dfrac{5n+2-\sqrt{5n^{2}+20n+24}}{10}\right\rceil.$

###### Proof.

Hopkins and Staton [19] showed that

$I(P_{n};t)=F_{n+1}(t),$

where $F_{n}(t),\ n\geq 0$, are the so-called Fibonacci polynomials, which are
defined recursively by

$F_{0}(t)=1;F_{1}(t)=1;F_{n}(t)=F_{n-1}(t)+tF_{n-2}(t).$

Based on this recurrence, one can deduce that

$I(P_{n};t)=\sum_{i=0}^{\lfloor\frac{n+1}{2}\rfloor}\binom{n+1-i}{i}t^{i}.$

The unimodality of the independence polynomial of $P_{n}$ is implied from the
fact that the independence polynomial of a claw-free graph is unimodal [15].
Now we determine the mode of $I(P_{n};t)$. Let $i$ be an integer such that
$0\leq i\leq\lfloor\frac{n+1}{2}\rfloor$ and

$\binom{n+1-i}{i}\geq\binom{n-i}{i+1}.$

This is clearly true if $i\geq n/2$. If $i<n/2$, we have

$\displaystyle\binom{n+1-i}{i}\geq\binom{n-i}{i+1}$
$\displaystyle\Leftrightarrow\frac{n+1-i}{(n-2i)(n-2i+1)}\geq\frac{1}{i+1}$
$\displaystyle\Leftrightarrow 5i^{2}-(5n+2)i+n^{2}-1\leq 0$
$\displaystyle\Leftrightarrow\frac{5n+2-\sqrt{5n^{2}+20n+24}}{10}\leq
i\leq\frac{5n+2+\sqrt{5n^{2}+20n+24}}{10}.$

As the inequality on the right holds for any
$i\leq\lfloor\frac{n+1}{2}\rfloor$, we have

$\binom{n+1-i}{i}\geq\binom{n-i}{i+1}\Leftrightarrow
i\geq\frac{5n+2-\sqrt{5n^{2}+20n+24}}{10}.$

This means that the mode of $I(P_{n};t)$ is equal to
$\lambda_{n}=\lceil\frac{5n+2-\sqrt{5n^{2}+20n+24}}{10}\rceil.$ ∎

We summarize the important properties of the mode of $I(P_{n};t)$.

######  Lemma 3.2.

For any $n\geq 1$, one has the following.

  1. (i)

$\lambda_{n+1}\geq\lambda_{n};$

  2. (ii)

$\lambda_{n+3}-1\leq\lambda_{n}\leq\lambda_{n+4}-1;$

  3. (iii)

$\lambda_{n+11}\geq\lambda_{n}+3.$

###### Proof.

Set $\alpha_{n}=\frac{5n+2-\sqrt{5n^{2}+20n+24}}{10}.$ A straightforward
computation shows that

$\displaystyle\alpha_{n+1}\geq\alpha_{n};\;\alpha_{n+3}-1\leq\alpha_{n}\leq\alpha_{n+4}-1\;\text{and}\;\alpha_{n+11}\geq\alpha_{n}+3.$

The lemma follows from basic properties of the ceiling function. ∎

Table 1 provides information about the initial values of the mode of the
independence polynomial $I(G;t)$ for the classes of graphs considered in this
paper. A dash indicates an undefined value.

Table 1. Graphs and modes of their independence polynomials $G$ |  mode of $I(G;t)$ $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$P_{n}$ | $\lambda_{n}$ | 0 | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 4 | 4
$C_{n}$ | $\rho_{n}$ | - | - | 1 | 1 | 1 | 2 | 2 | 2 | 3 | 3 | 3 | 3 | 4
${\operatorname{CE}}_{n}$ | $\chi_{n}$ | - | - | - | 1 | 1 | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 4
${\operatorname{Pan}}_{n}$ | $\zeta_{n}$ | - | - | 1 | 1 | 2 | 2 | 2 | 3 | 3 | 3 | 3 | 4 | 4

###  3.2. Cycles

Let $C_{n}$ be the cycle on $n$ vertices ($n\geq 3$).

216543 Figure 2. The cycle $C_{6}$

######  Proposition 3.3.

The independence polynomial of $C_{n}$ is

$\displaystyle I(C_{n};t)$ $\displaystyle=I(P_{n-1};t)+tI(P_{n-3};t)$
$\displaystyle=1+\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{i}\binom{n-i-1}{i-1}t^{i}.$

Moreover, $I(C_{n};t)$ is unimodal, with the mode
$\rho_{n}=\lceil\frac{5n-4-\sqrt{5n^{2}-4}}{10}\rceil$ for all $n\geq 3$.

###### Proof.

Hopkins and Staton [19] showed that

$I(C_{n};t)=1+\sum_{i=1}^{\lfloor\frac{n}{2}\rfloor}\frac{n}{i}\binom{n-i-1}{i-1}t^{i}.$

The unimodality of the independence polynomial of $C_{n}$ is implied from the
fact that the independence polynomial of a claw-free graph is unimodal [15].
Arguing as in the proof of Proposition 3.1, solving for $1\leq
i\leq\lceil\frac{n}{2}\rceil$ from

$\frac{n}{i}\binom{n-i-1}{i-1}\geq\frac{n}{i+1}\binom{n-i-2}{i}$

we get

$\displaystyle(i+1)(n-i-1)\geq(n-2i-1)(n-2i).$

Equivalently $5i^{2}-i(5n-4)+n^{2}-2n+1\leq 0$. Thus

$\frac{5n-4-\sqrt{5n^{2}-4}}{10}\leq i\leq\frac{5n-4+\sqrt{5n^{2}-4}}{10}.$

This implies that the mode of $I(C_{n};t)$ is equal to
$\rho_{n}=\left\lceil\frac{5n-4-\sqrt{5n^{2}-4}}{10}\right\rceil$, as desired.
∎

######  Lemma 3.4.

For all $n\geq 5$, there are inequalities
$\lambda_{n-1}\leq\rho_{n}\leq\lambda_{n-4}+1\leq\lambda_{n}$.

###### Proof.

By Lemma 3.2, $\lambda_{n-4}+1\leq\lambda_{n}$, hence it suffices to show that

$\lambda_{n-1}\leq\rho_{n}\leq\lambda_{n-4}+1.$

For the inequality on the left, we have to show that

$\displaystyle\dfrac{5(n-1)+2-\sqrt{5(n-1)^{2}+20(n-1)+24}}{10}\leq\dfrac{5n-4-\sqrt{5n^{2}-4}}{10}$
$\displaystyle\Leftrightarrow 5n-3-\sqrt{5n^{2}+10n+9}\leq
5n-4-\sqrt{5n^{2}-4}$
$\displaystyle\Leftrightarrow\sqrt{5n^{2}-4}+1\leq\sqrt{5n^{2}+10n+9}$
$\displaystyle\Leftrightarrow 5n^{2}-3+2\sqrt{5n^{2}-4}\leq 5n^{2}+10n+9$
$\displaystyle\Leftrightarrow\sqrt{5n^{2}-4}\leq
5n+6\Leftrightarrow(5n+6)^{2}-(5n^{2}-4)\geq 0$ $\displaystyle\Leftrightarrow
20n^{2}+60n+40\geq 0,$

which is clear.

For the inequality on the right, we have to show that

$\displaystyle\dfrac{5n-4-\sqrt{5n^{2}-4}}{10}\leq\dfrac{5(n-4)+2-\sqrt{5(n-4)^{2}+20(n-4)+24}}{10}+1$
$\displaystyle\Leftrightarrow 5n-4-\sqrt{5n^{2}-4}\leq
5n-8-\sqrt{5n^{2}-20n+24}$
$\displaystyle\Leftrightarrow\sqrt{5n^{2}-20n+24}+4\leq\sqrt{5n^{2}-4}$
$\displaystyle\Leftrightarrow 5n^{2}-20n+24+16+8\sqrt{5n^{2}-20n+24}\leq
5n^{2}-4\quad(\text{by squaring})$ $\displaystyle\Leftrightarrow
8\sqrt{5n^{2}-20n+24}\leq 20n-44$ $\displaystyle\Leftrightarrow
2\sqrt{5n^{2}-20n+24}\leq 5n-11$ $\displaystyle\Leftrightarrow
4(5n^{2}-20n+24)\leq(5n-11)^{2}$ $\displaystyle\Leftrightarrow
5n^{2}-30n+25\geq 0\Leftrightarrow 5(n-1)(n-5)\geq 0$

which is true for all $n\geq 5$. The proof is completed. ∎

###  3.3. Pans

The $n$-pan graph is the graph obtained by joining a cycle graph $C_{n}$ to a
singleton graph $K_{1}$ with a bridge. We denote this graph by
${\operatorname{Pan}}_{n}$.

Our goal is to show the independence polynomial of ${\operatorname{Pan}}_{n}$
is unimodal. For this, we consider a family of graphs formed by adding an edge
$\\{n-2,n\\}$ to the cycles $C_{n}$ ($n\geq 4$). We denote this graph by
${\operatorname{CE}}_{n}$.

2165437 | 216543
---|---
Figure 3. ${\operatorname{Pan}}_{6}$ and ${\operatorname{CE}}_{6}$

Note that ${\operatorname{CE}}_{n}$ is a claw-free graph, and hence its
independence polynomial of is unimodal [15].

######  Lemma 3.5.

The independence polynomial of ${\operatorname{CE}}_{n}$ is

$\displaystyle I({\operatorname{CE}}_{n};t)$
$\displaystyle=\sum_{i=0}^{\alpha({\operatorname{CE}}_{n})}s_{i}({\operatorname{CE}}_{n})t^{i}$
$\displaystyle=I(P_{n-1};t)+tI(P_{n-4};t)$
$\displaystyle=\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor}\bigg{[}\binom{n-i}{i}+\binom{n-i-2}{i-1}\bigg{]}t^{i}.$

Let $\chi_{n}$ be the mode of $I({\operatorname{CE}}_{n};t)$ and $\lambda_{n}$
be the mode of $I(P_{n};t)$ as in Proposition 3.1. For any $n\geq 5$, one has
$\lambda_{n-1}\leq\chi_{n}\leq\lambda_{n-4}+1.$

###### Proof.

The first assertion follows from applying Proposition 2.9(i) for the vertex
numbered $n$.

Let $1\leq i\leq\lambda_{n-1}$. We need to show that

$s_{i-1}({\operatorname{CE}}_{n})<s_{i}({\operatorname{CE}}_{n}),$

namely,

$\binom{n-i+1}{i-1}+\binom{n-i-1}{i-2}<\binom{n-i}{i}+\binom{n-i-2}{i-1}.$

This is clear for $i=1$, so we assume that $i\geq 2$.

Since $i\leq\lambda_{n-1}$, $\binom{n-i+1}{i-1}<\binom{n-i}{i}$. It suffices
to show that

$\displaystyle\binom{n-i-1}{i-2}\leq\binom{n-i-2}{i-1}$
$\displaystyle\Leftrightarrow(n-i-1)(i-1)\leq(n-2i)(n-2i+1)$
$\displaystyle\Leftrightarrow 5i^{2}-(5n+2)i+n^{2}+2n-1\geq 0$
$\displaystyle\Leftrightarrow i\leq\dfrac{5n+2-\sqrt{5n^{2}-20n+24}}{10}\
\text{or}\ i\geq\dfrac{5n+2+\sqrt{5n^{2}-20n+24}}{10}.$

As $i\leq\lambda_{n-1}$, it is enough to show that

$\displaystyle\dfrac{5(n-1)+2-\sqrt{5(n-1)^{2}+20(n-1)+24}}{10}\leq\dfrac{5n+2-\sqrt{5n^{2}-20n+24}}{10}-1$
$\displaystyle\Leftrightarrow 5n-3-\sqrt{5n^{2}+10n+9}\leq
5n-8-\sqrt{5n^{2}-20n+24}$ $\displaystyle\Leftrightarrow
5+\sqrt{5n^{2}-20n+24}\leq\sqrt{5n^{2}+10n+9}$
$\displaystyle\Leftrightarrow\sqrt{5n^{2}-20n+24}\leq 3n-4\quad(\text{after
squaring and simplifying})$ $\displaystyle\Leftrightarrow n^{2}-n-2\geq 0$
$\displaystyle\Leftrightarrow(n+1)(n-2)\geq 0,$

which is clear for any $n\geq 4$. It follows that $\lambda_{n-1}\leq\chi_{n}.$
It remains to show that if $\lfloor\frac{n}{2}\rfloor\geq
i\geq\lambda_{n-4}+1$ (note $\lfloor\frac{n}{2}\rfloor$ is the independence
number of ${\operatorname{CE}}_{n}$), then

$s_{i}({\operatorname{CE}}_{n})\geq
s_{i+1}({\operatorname{CE}}_{n})\Longleftrightarrow\binom{n-i}{i}+\binom{n-i-2}{i-1}\geq\binom{n-i-1}{i+1}+\binom{n-i-3}{i}.$

By Lemma 3.2, $i\geq\lambda_{n-4}+1\geq\lambda_{n-1}$, so
$\binom{n-i}{i}\geq\binom{n-i-1}{i+1}$ thanks to Proposition 3.1. We have to
show that

$\displaystyle\binom{n-i-2}{i-1}$ $\displaystyle\geq\binom{n-i-3}{i}$
$\displaystyle\Leftrightarrow i(n-i-2)\geq(n-2i-2)(n-2i-1)$
$\displaystyle\Leftrightarrow 5i^{2}-(5n-8)i+n^{2}-3n+2\leq 0$
$\displaystyle\Leftrightarrow\dfrac{5n-8-\sqrt{5n^{2}-20n+24}}{10}\leq
i\leq\dfrac{5n-8+\sqrt{5n^{2}-20n+24}}{10}$

Since $\lfloor\frac{n}{2}\rfloor\geq i\geq\lambda_{n-4}+1$, and by simple
computations,

$\left\lfloor\frac{n}{2}\right\rfloor\leq\dfrac{5n-8+\sqrt{5n^{2}-20n+24}}{10}\quad\text{for
all $n\geq 5$},$

the inequality on the right of the last chain is always true. Thus it is
enough to prove the inequality on the left, which would be true if

$\displaystyle\dfrac{5n-8-\sqrt{5n^{2}-20n+24}}{10}\leq\dfrac{5(n-4)+2-\sqrt{5(n-4)^{2}+20(n-4)+24}}{10}+1$
$\displaystyle\Leftrightarrow\dfrac{5n-8-\sqrt{5n^{2}-20n+24}}{10}\leq\dfrac{5n-18+\sqrt{5n^{2}-20n+24}}{10}+1,$

which is clear. Thus $\chi_{n}\leq\lambda_{n-4}+1.$ ∎

Note that ${\operatorname{Pan}}_{n}$ is not a claw-free graph. Hence, we need
to show the unimodality of its independence polynomial. We have the following.

######  Lemma 3.6.

The independence polynomial $I({\operatorname{Pan}}_{n};t)$ of $n$-pan graph
is unimodal. Let $\zeta_{n}$ be the mode of $I({\operatorname{Pan}}_{n};t)$.
Then for all $n\geq 5$, there are inequalities

(3.1)
$\chi_{n+1}\leq\zeta_{n}\leq\rho_{n}+1\leq\lambda_{n}+1\leq\chi_{n+1}+1,$

where $\lambda_{n},\rho_{n},\chi_{n}$ are as in Propositions 3.1, 3.3 and
Lemma 3.5.

###### Proof.

By using Proposition 2.9(i) for the vertex of $K_{1}$, Propositions 3.1 and
3.3, we have

$\displaystyle I({\operatorname{Pan}}_{n};t)$
$\displaystyle=\sum_{i=0}^{\alpha({\operatorname{Pan}}_{n})}s_{i}({\operatorname{Pan}}_{n})t^{i}$
$\displaystyle=I(C_{n};t)+tI(P_{n-1};t)$
$\displaystyle=I(P_{n-1};t)+t\big{(}I(P_{n-3};t)+I(P_{n-1};t)\big{)}$
$\displaystyle=\sum_{i=0}^{\lfloor\frac{n}{2}\rfloor+1}\bigg{[}\binom{n-i}{i}+\binom{n-i-1}{i-1}+\binom{n-i+1}{i-1}\bigg{]}t^{i}.$

Therefore, we have

$\displaystyle s_{i}({\operatorname{Pan}}_{n})$
$\displaystyle=\binom{n-i}{i}+\binom{n-i-1}{i-1}+\binom{n-i+1}{i-1}$
$\displaystyle=\binom{n-i+1}{i}+\binom{n-i-1}{i-1}+\binom{n-i}{i-2}$
$\displaystyle\qquad\left(\text{using
$\binom{n}{p}=\binom{n-1}{p}+\binom{n-1}{p-1}$}\right)$ (3.2)
$\displaystyle=s_{i}({\operatorname{CE}}_{n+1})+\binom{n-i}{i-2}\quad\text{(by
Lemma \ref{lem_compare_lambda_chi})}.$

We first have the following assertion.

Claim 1: $s_{i-1}({\operatorname{Pan}}_{n})<s_{i}({\operatorname{Pan}}_{n})$
for any $i\leq\chi_{n+1}.$

_Proof of Claim 1:_ For any $1\leq i\leq\chi_{n+1}$,
$s_{i-1}({\operatorname{CE}}_{n+1})<s_{i}({\operatorname{CE}}_{n+1}).$
Therefore by Equation (3.2), it suffices to show that

$\displaystyle\hskip 34.14322pt\binom{n-i+1}{i-3}\leq\binom{n-i}{i-2}$
$\displaystyle\Longleftrightarrow(n-i+1)(i-2)\leq(n-2i+3)(n-2i+4)$
$\displaystyle\Longleftrightarrow 5i^{2}-(5n+17)i+n^{2}+9n+14\geq 0$
$\displaystyle\Longleftrightarrow i\leq\frac{5n+17-\sqrt{5n^{2}-10n+9}}{10}\
\text{or}\ i\geq\frac{5n+17+\sqrt{5n^{2}-10n+9}}{10}.$

By Lemma 3.5,

$\displaystyle i\leq$ $\displaystyle\chi_{n+1}\leq\lambda_{n-3}+1$
$\displaystyle<\dfrac{5(n-3)+2-\sqrt{5(n-3)^{2}+20(n-3)+24}}{10}+2$
$\displaystyle=\dfrac{5n+7-\sqrt{5n^{2}-10n+9}}{10}.$

Consequently, it suffices to show

$\displaystyle\frac{5n+7-\sqrt{5n^{2}-10n+9}}{10}\leq\frac{5n+17-\sqrt{5n^{2}-10n+9}}{10},$

which is clear. Thus, we finish the proof of Claim 1.

Now, by again Proposition 2.9, we have

$\displaystyle I({\operatorname{Pan}}_{n};t)$
$\displaystyle=\sum_{i=0}^{\alpha({\operatorname{Pan}}_{n})}s_{i}({\operatorname{Pan}}_{n})t^{i}$
$\displaystyle=I(C_{n};t)+tI(P_{n-1};t),$

we get $s_{i}({\operatorname{Pan}}_{n})=s_{i}(C_{n})+s_{i-1}(P_{n-1}).$ Next
we have the following

Claim 2: $s_{i}({\operatorname{Pan}}_{n})\geq
s_{i+1}({\operatorname{Pan}}_{n})$ for any $i\geq\rho_{n}+1.$

_Proof of Claim 2:_ Since $i\geq\rho_{n}+1$ and $n\geq 5$,
$i-1\geq\rho_{n}\geq\lambda_{n-1}$ by Lemma 3.4. It follows that
$s_{i}(C_{n})\geq s_{i+1}(C_{n})$ and $s_{i-1}(P_{n-1})\geq s_{i}(P_{n-1}).$
Thus $s_{i}({\operatorname{Pan}}_{n})\geq s_{i+1}({\operatorname{Pan}}_{n})$,
as desired.

By Lemmas 3.4 and 3.5, $\rho_{n}\leq\lambda_{n}\leq\chi_{n+1},$ which yields
the last two inequalities in (3.1). Moreover, it follows from Claims 1 and 2
that $s_{i-1}({\operatorname{Pan}}_{n})<s_{i}({\operatorname{Pan}}_{n})$ for
any $i\leq\chi_{n+1}$ and $s_{i}({\operatorname{Pan}}_{n})\geq
s_{i+1}({\operatorname{Pan}}_{n})$ for any $i\geq\chi_{n+1}+1$. Thus, the
independence polynomial $I({\operatorname{Pan}}_{n};t)$ of $n$-pan graph is
unimodal. Moreover, $\chi_{n+1}\leq\zeta_{n}$ by Claim 1 and
$\zeta_{n}\leq\rho_{n}+1$ by Claim 2. This concludes the proof. ∎

##  4\. WLP for algebras associated to paths and cycles

In this section, we study the WLP for artinian algebras associated to paths
and cycles. From now on, we always assume $\operatorname{char}(\Bbbk)=0$ and
denote by $\ell$ the sum of variables in the polynomial ring we are working
with.

###  4.1. Paths

The artinian algebra associated to $P_{n}$ is

$A(P_{n})=R/K,$

where
$K=(x_{1}^{2},\ldots,x_{n}^{2})+(x_{1}x_{2},x_{2}x_{3},\ldots,x_{n-1}x_{n})\subset
R=\Bbbk[x_{1},\ldots,x_{n}]$. The following lemma is useful to an inductive
argument on the WLP of $A(P_{n})$.

######  Lemma 4.1.

For every integer $i$, there is a commutative diagram with exact rows

$\displaystyle\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n-2})]_{i-1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
106.3441pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
43.27203pt\raise-20.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\cdot\ell}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 54.13316pt\raise-29.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
106.3441pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n})]_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
177.43268pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
115.44948pt\raise-20.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\cdot\ell}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 126.31061pt\raise-29.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
177.43268pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n-1})]_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
249.61014pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
199.73251pt\raise-20.0pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern
0.0pt\raise-2.43056pt\hbox{$\scriptstyle{\cdot\ell}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 199.73251pt\raise-29.5pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
249.61014pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{0}$}}}}}}}{\hbox{\kern-5.5pt\raise-40.0pt\hbox{\hbox{\kern
0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
31.83333pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
31.83333pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n-2})]_{i}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
102.76633pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
102.76633pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n})]_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
173.85489pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
173.85489pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{[A(P_{n-1})]_{i+1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
249.61014pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
249.61014pt\raise-40.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

###### Proof.

Assume $A(P_{n})=R/K$ and set $I=K+(x_{n})$ and $J=(K\colon x_{n})$. Then
$A(P_{n-1})\cong R/I$ and $A(P_{n-2})\cong R/J$ and we have the following
exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R/J(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\quad\cdot
x_{n}}$$\textstyle{R/K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

The desired conclusion follows. ∎

We now prove our first main result.

######  Theorem 4.2.

The ring $A(P_{n})$ has the WLP if and only if
$n\in\\{1,2,\ldots,7,9,10,13\\}.$

###### Proof.

Using Macaulay2 to compute the Hilbert series of the rings $A(P_{n})$ and
$A(P_{n})/\ell A(P_{n})$ with $1\leq n\leq 17$, we see that $A(P_{n})$ has the
WLP for each $n\in\\{1,2,\ldots,7,9,10,13\\}.$ Furthermore, for each
$n\in\\{8,11,14,15,17\\},\ A(P_{n})$ only fails the surjectivity in the
multiplication map by $\ell$ from degree $\lambda_{n}$ to degree
$\lambda_{n}+1$. However, for $n\in\\{12,16\\},\ A(P_{n})$ only fails the
injectivity in the multiplication map by $\ell$ from degree $\lambda_{n}-1$ to
degree $\lambda_{n}$.

It remains to show the following

Claim: The multiplication map
$\cdot\ell:[A(P_{n})]_{\lambda_{n}}\longrightarrow[A(P_{n})]_{\lambda_{n}+1}$
is not surjective for all $n\geq 17$.

We will prove the above claim by induction on $n$, having just established the
case $n=17$. For $n\geq 18$, we consider the multiplication map
$\cdot\ell:[A(P_{n})]_{\lambda_{n}}\longrightarrow[A(P_{n})]_{\lambda_{n}+1}$.
To prove that this map is not surjective, we consider the following two cases.

Case 1: $\lambda_{n}=\lambda_{n-1}$. In the diagram of Lemma 4.1 where
$i=\lambda_{n-1}=\lambda_{n}$, the right vertical map is not surjective by the
induction hypothesis, so neither is the middle vertical map.

Case 2: $\lambda_{n}=\lambda_{n-1}+1$. By Lemma 3.2, one has
$\lambda_{n-1}=\lambda_{n-2}=\lambda_{n-3}$. In this case, we must have $n\geq
20$ since $\lambda_{16}=\lambda_{17}=\lambda_{18}=\lambda_{19}=5$.

Assume $A(P_{n})=R/K$ and set $I=K+(x_{n-2})$ and $J=K\colon x_{n-2}$. Then we
have the following exact sequence

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R/J(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
88.20705pt\raise 5.2pt\hbox{{}\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\hbox{\hbox{\kern 0.0pt\raise-0.91112pt\hbox{$\scriptstyle{\quad\cdot
x_{n-2}}$}}}\kern 3.0pt}}}}}}\ignorespaces{\hbox{\kern 131.45474pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
101.45474pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 131.45474pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{R/K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
187.77756pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
187.77756pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{R/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
240.0726pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
240.0726pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces,$

where $R/J\cong A(P_{n-4})\otimes_{\Bbbk}\Bbbk[x_{n}]/(x_{n}^{2})$ and
$R/I\cong A(P_{n-3})\otimes_{\Bbbk}A(P_{2})$, with
$A(P_{2})=\Bbbk[x_{n-1},x_{n}]/(x_{n-1},x_{n})^{2}.$ This exact sequence gives
rise to the following commutative diagram, with exact rows

To prove that the middle vertical map is not surjective, it suffices to show
that the right vertical map

$\cdot\ell:[R/I]_{\lambda_{n}}\longrightarrow[R/I]_{\lambda_{n}+1}$

is not surjective. By the inductive hypothesis, $A(P_{n-3})$ fails the
surjectivity from degree $\lambda_{n}-1$ to degree $\lambda_{n}$, as
$\lambda_{n-3}=\lambda_{n}-1$. Clearly, the Hilbert function of $A(P_{2})$ is
$(1,2)$, and hence $A(P_{2})$ fails the surjectivity from degree 0 to degree
1. Then by Lemma 2.5(a), $R/I\cong A(P_{n-3})\otimes_{\Bbbk}A(P_{2})$ fails
the surjectivity from degree $\lambda_{n}$ to degree $\lambda_{n}+1$, as
desired. ∎

The above theorem shows that $A(P_{n})$ fails the WLP since surjectivity fails
for any $n\geq 17$. The next result also prove that $A(P_{n})$ fails the
injectivity for some cases.

######  Proposition 4.3.

Recall the mode $\lambda_{n}$ of the independence polynomial of $I(P_{n};t)$.
If $n\geq 12$ is an integer such that $\lambda_{n}=\lambda_{n-1}+1,$ then
$A(P_{n})$ fails the injectivity from degree $\lambda_{n}-1$ to degree
$\lambda_{n}$.

###### Proof.

We prove the above proposition by induction on $n\geq 12$. A computation with
Macaulay2 shows that the proposition holds for $n\in\\{12,16,20\\}.$ This
covers all cases from $12$ to $20$ due to Lemma 3.2. Now consider an $n\geq
21$ such that $\lambda_{n}=\lambda_{n-1}+1.$ Set

$\displaystyle n_{1}$ $\displaystyle=\max\\{j\mid
j<n\;\text{and}\;\lambda_{j}=\lambda_{j-1}+1\\}$ $\displaystyle n_{2}$
$\displaystyle=\max\\{j\mid
j<n_{1}\;\text{and}\;\lambda_{j}=\lambda_{j-1}+1\\}$ $\displaystyle m$
$\displaystyle=\max\\{j\mid
j<n_{2}\;\text{and}\;\lambda_{j}=\lambda_{j-1}+1\\}.$

Then, by Lemma 3.2, $9\leq n-m\leq 11$. We have the following exact sequence

$0\to A(P_{m})\otimes_{\Bbbk}A(P_{n-m-3})(-1)\xrightarrow{\cdot
x_{m+2}}A(P_{n})\to A(P_{m+1})\otimes_{\Bbbk}A(P_{n-m-2})\to 0.$

By using this exact sequence, it suffices to show that

$\cdot\ell:[A(P_{m})\otimes_{\Bbbk}A(P_{n-m-3})]_{\lambda_{n}-2}\longrightarrow[A(P_{m})\otimes_{\Bbbk}A(P_{n-m-3})]_{\lambda_{n}-1}$

is not injective. By the inductive hypothesis, $A(P_{m})$ fails the
injectivity from degree $\lambda_{m}-1$ to $\lambda_{m}$. Observe that
$\lambda_{m}=\lambda_{n}-3$ and $6\leq n-m-3\leq 8$. Hence by Table 1,
$\lambda_{n-m-3}=2$ and consequently, $A(P_{n-m-3})$ fails the injectivity
from degree 2 to degree 3. By Lemma 2.5(b),
$A(P_{m})\otimes_{\Bbbk}A(P_{n-m-3})$ fails the injectivity from degree
$\lambda_{m}+1=\lambda_{n}-2$ to $\lambda_{n}-1$, as desired. ∎

###  4.2. Cycles

The artinian algebra associated to the cycle on $n$ vertices is

$A(C_{n})=R/K,$

where
$K=(x_{1}^{2},\ldots,x_{n}^{2})+(x_{1}x_{2},x_{2}x_{3},\ldots,x_{n-1}x_{n},x_{n}x_{1})\subset
R=\Bbbk[x_{1},\ldots,x_{n}]$. Our second main result is the following.

######  Theorem 4.4.

The algebra $A(C_{n})$ has the WLP if and only if $n\in\\{3,4,\ldots,11,13,\\\
14,17\\}.$

###### Proof.

Recall that $\rho_{n}$ is the mode of the independence polynomial of $C_{n}$.
Using Macaulay2 to compute the Hilbert series of $A(C_{n})$ and $A(C_{n})/\ell
A(C_{n})$ with $3\leq n\leq 20$, we can check that:

  * •

$A(C_{n})$ has the WLP for each $3\leq n\leq 17$ and $n\notin\\{12,15,16\\}$;

  * •

for $n\in\\{12,15,18,19\\}$, then $A(C_{n})$ fails the surjectivity from
degree $\rho_{n}$ to degree $\rho_{n}+1$;

  * •

for $n\in\\{16,20\\}$, then $A(C_{n})$ fails the injectivity from degree
$\rho_{n}-1$ to degree $\rho_{n}.$

Now assume that $n\geq 21$. By Lemmas 3.2 and 3.4,
$\lambda_{n-1}\leq\rho_{n}\leq\lambda_{n-4}+1\leq\lambda_{n-1}+1$. Consider
the following two cases.

Case 1: $\rho_{n}=\lambda_{n-1}$. In this case, we will show that $A(C_{n})$
fails the WLP by the failure of the surjectivity from degree $\rho_{n}$ to
degree $\rho_{n}+1.$ Indeed, write $A(C_{n})=R/K$, and let $I=K+(x_{n})$ and
$J=K\colon x_{n}$. Then $A(P_{n-1})\cong R/I$ and $A(P_{n-3})\cong R/J$ and we
have the following exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R/J(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\quad\cdot
x_{n}}$$\textstyle{R/K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R/I\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

This yields a commutative diagram

The proof of Theorem 4.2 shows that the multiplication map

$\cdot\ell:[A(P_{n-1})]_{\rho_{n}}\longrightarrow[A(P_{n-1})]_{\rho_{n}+1}$

is not surjective for any $n\geq 18.$ Hence the middle vertical map

$\cdot\ell:[A(C_{n})]_{\rho_{n}}\longrightarrow[A(C_{n})]_{\rho_{n}+1}$

is not surjective, as desired.

Case 2: $\rho_{n}=\lambda_{n-1}+1$. In this case, Lemmas 3.2 and 3.4 yield
$\lambda_{n-1}=\lambda_{n-4}$.

Denote $y_{1}=x_{n-1},y_{2}=x_{n-2}$. We have the following diagram

By the proof of Theorem 4.2 and the fact that $n-4\geq 17$, the map

$A(P_{n-4})\xrightarrow{\cdot\ell}A(P_{n-4})$

fails the surjectivity at degree $\lambda_{n-4}$. Since the map

$\Bbbk[y_{1},y_{2}]/(y_{1},y_{2})^{2}\xrightarrow{\cdot(y_{1}+y_{2})}\Bbbk[y_{1},y_{2}]/(y_{1},y_{2})^{2}$

fails the surjectivity at degree 0, Lemma 2.5(a) yields that the right
vertical map of the diagram fails the surjectivity at degree
$\lambda_{n-4}+1$.

By the surjectivity of the horizontal maps in the diagram, we conclude that
left vertical map in the diagram fails the surjectivity at degree
$\lambda_{n-4}+1=\rho_{n}$. Hence $A(C_{n})$ does not have the WLP. This
concludes the proof. ∎

Next, we consider a special case of the tadpole graphs. The _tadpole_ graph
$T_{3,n}$ is obtained by joining a cycle $C_{3}$ to a path $P_{n}$ with a
bridge.

$x_{1}$$x_{2}$$x_{3}$$y_{1}$$y_{2}$$y_{3}$$y_{4}$$y_{5}$ Figure 4.  Tadpole
$T_{3,5}$

Clearly, $T_{3,n}$ is a claw-free graph. Therefore, the independence
polynomial of $T_{3,n}$ is unimodal [15]. By Proposition 2.9 for either of the
vertices on the left of the three cycle, we have

$\displaystyle I(T_{3,n};t)=I(P_{n+2};t)+tI(P_{n};t)=I(C_{n+3};t).$

By Proposition 3.3, it follows that the mode of $I(T_{3,n};t)$ is equal to
that of $I(C_{n+3};t)$, which is

$\rho_{n+3}=\left\lceil\frac{5(n+3)-4-\sqrt{5(n+3)^{2}-4}}{10}\right\rceil.$

######  Corollary 4.5.

The algebra $A(T_{3,n})$ has the WLP if and only if
$n\in\\{1,2,\ldots,8,10,\\\ 11,14\\}.$

###### Proof.

The proof proceeds along the same lines as that of Theorem 4.4, replacing
$A(C_{n+3})$ by $A(T_{3,n})$. ∎

###  4.3. Pans

To study the WLP of rings associated to pans, we first examine the WLP of
rings associated to ${\operatorname{CE}}_{n}$. The latter is by definition

$A({\operatorname{CE}}_{n})=\frac{\Bbbk[x_{1},\ldots,x_{n}]}{(x_{1}^{2},\ldots,x_{n}^{2})+(x_{1}x_{2},x_{2}x_{3},\ldots,x_{n-1}x_{n},x_{n}x_{1})+(x_{n-2}x_{n})}.$

$x_{2}$$x_{1}$$x_{6}$$x_{5}$$x_{4}$$x_{3}$ Figure 5. ${\operatorname{CE}}_{6}$

######  Theorem 4.6.

The algebra $A({\operatorname{CE}}_{n})$ has the WLP if and only if
$n\in\\{4,5,\ldots,8,10,\\\ 11,14\\}.$

###### Proof.

By using Macaulay2 to compute the Hilbert series of
$A({\operatorname{CE}}_{n})/\ell A({\operatorname{CE}}_{n})$ and
$A({\operatorname{CE}}_{n})$ with $4\leq n\leq 20$, we can check that:

  * •

$A({\operatorname{CE}}_{n})$ has the WLP for each $4\leq n\leq 14$ and
$n\notin\\{9,12,13\\}$;

  * •

for $n\in\\{9,12,15,16,18,19\\}$, $A({\operatorname{CE}}_{n})$ fails the
surjectivity from degree $\chi_{n}$ to degree $\chi_{n}+1$;

  * •

for $n\in\\{9,13,17,20\\}$, $A({\operatorname{CE}}_{n})$ fails the injectivity
from degree $\chi_{n}-1$ to degree $\chi_{n}.$

Now assume that $n\geq 21$. We will prove that $A({\operatorname{CE}}_{n})$
fails the surjectivity from degree $\chi_{n}$ to degree $\chi_{n}+1$. The
proof is similar to that of Theorem 4.4. Recall that by Lemmas 3.2 and 3.5,

(4.1) $\lambda_{n-1}\leq\chi_{n}\leq\lambda_{n-4}+1\leq\lambda_{n-1}+1.$

We consider the following two cases.

Case 1: $\chi_{n}=\lambda_{n-1}$. In this case, we will show that
$A({\operatorname{CE}}_{n})$ fails the WLP by the failure of the surjectivity
from degree $\chi_{n}$ to degree $\chi_{n}+1.$ Consider the exact sequence

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A(P_{n-4})(-1)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
97.0798pt\raise 4.98611pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-1.125pt\hbox{$\scriptstyle{\quad\cdot x_{n}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 138.94815pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
108.94815pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 138.94815pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A({\operatorname{CE}}_{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
201.61508pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
201.61508pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A(P_{n-1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
262.72992pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
262.72992pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

The proof of Theorem 4.2 shows that the multiplication map

$\cdot\ell:[A(P_{n-1})]_{\chi_{n}}\longrightarrow[A(P_{n-1})]_{\chi_{n}+1}$

is not surjective for any $n\geq 18.$ Hence, the map

$\cdot\ell:[A({\operatorname{CE}}_{n})]_{\chi_{n}}\longrightarrow[A({\operatorname{CE}}_{n})]_{\chi_{n}+1}$

is also not surjective, as desired.

Case 2: $\chi_{n}=\lambda_{n-1}+1$. In this case, the chain (4.1) yields
$\lambda_{n-1}=\lambda_{n-4}$. As in the proof of Theorem 4.4, denoting
$y_{1}=x_{n-1},y_{2}=x_{n-2}$, we have the following diagram

Since the right vertical map of the diagram fails the surjectivity at degree
$\lambda_{n-4}+1$, we conclude that left vertical map in the diagram fails the
surjectivity at degree $\lambda_{n-4}+1=\chi_{n}$, as desired. ∎

Finally, we show the last main result.

######  Theorem 4.7.

The algebra $A({\operatorname{Pan}}_{n})$ associated to the pan graph
${\operatorname{Pan}}_{n}$ has the WLP if and only if
$n\in\\{4,5,\ldots,10,12,13,16\\}.$

###### Proof.

$x_{2}$$x_{1}$$x_{6}$$x_{5}$$x_{4}$$x_{3}$$x_{7}$ Figure 6.
${\operatorname{Pan}}_{6}$

The artinian algebra associated to ${\operatorname{Pan}}_{n}$ is

$A({\operatorname{Pan}}_{n})=\frac{\Bbbk[x_{1},\ldots,x_{n+1}]}{(x_{1}^{2},\ldots,x_{n+1}^{2})+(x_{1}x_{2},x_{2}x_{3},\ldots,x_{n-1}x_{n},x_{n}x_{n+1})+(x_{1}x_{n})}.$

By using Macaulay2 to compute the Hilbert series of
$A({\operatorname{Pan}}_{n})$ and $\dfrac{A({\operatorname{Pan}}_{n})}{\ell
A({\operatorname{Pan}}_{n})}$ with $4\leq n\leq 20$, we can check that:

  * •

$A({\operatorname{Pan}}_{n})$ has the WLP for each $4\leq n\leq 16$ and
$n\notin\\{11,14,15\\}$;

  * •

for $n\in\\{11,14,17,18,20\\}$, $A({\operatorname{Pan}}_{n})$ fails the
surjectivity from degree $\zeta_{n}$ to degree $\zeta_{n}+1$;

  * •

for $n\in\\{15,19\\}$, $A({\operatorname{Pan}}_{n})$ fails the injectivity
from degree $\zeta_{n}-1$ to degree $\zeta_{n}.$

Now assume that $n\geq 21$. Recall that by Lemma 3.6,

(4.2)
$\chi_{n+1}\leq\zeta_{n}\leq\rho_{n}+1\leq\lambda_{n}+1\leq\chi_{n+1}+1.$

We consider the following two cases.

Case 1: $\zeta_{n}=\chi_{n+1}$. In this case, $A({\operatorname{Pan}}_{n})$
fails the surjectivity from degree $\zeta_{n}$ to degree $\zeta_{n}+1$ by
using the exact sequence

$\lx@xy@svg{\hbox{\raise 0.0pt\hbox{\kern
5.5pt\hbox{\ignorespaces\ignorespaces\ignorespaces\hbox{\vtop{\kern
0.0pt\offinterlineskip\halign{\entry@#!@&&\entry<EMAIL_ADDRESS>0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
29.5pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A(P_{n-3})(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces\ignorespaces\ignorespaces{\hbox{\kern
92.2878pt\raise 5.36667pt\hbox{{}\hbox{\kern 0.0pt\raise
0.0pt\hbox{\hbox{\kern 3.0pt\hbox{\hbox{\kern
0.0pt\raise-0.74446pt\hbox{$\scriptstyle{\quad\cdot x_{1}x_{n+1}}$}}}\kern
3.0pt}}}}}}\ignorespaces{\hbox{\kern 138.94815pt\raise 0.0pt\hbox{\hbox{\kern
0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
108.94815pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{}$}}}}}}}{\hbox{\kern 138.94815pt\raise
0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern 3.0pt\raise
0.0pt\hbox{$\textstyle{A({\operatorname{Pan}}_{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
204.67065pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
204.67065pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise
0.0pt\hbox{$\textstyle{A({\operatorname{CE}}_{n+1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$}}}}}}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces{}{\hbox{\lx@xy@droprule}}\ignorespaces{\hbox{\kern
274.49313pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise
0.0pt\hbox{\lx@xy@tip{1}\lx@xy@tip{-1}}}}}}{\hbox{\lx@xy@droprule}}{\hbox{\lx@xy@droprule}}{\hbox{\kern
274.49313pt\raise 0.0pt\hbox{\hbox{\kern 0.0pt\raise 0.0pt\hbox{\hbox{\kern
3.0pt\raise 0.0pt\hbox{$\textstyle{0}$}}}}}}}\ignorespaces}}}}\ignorespaces.$

Indeed, the proof of Theorem 4.6 shows that the multiplication map

$\cdot\ell:[A({\operatorname{CE}}_{n+1})]_{\zeta_{n}}\longrightarrow[A({\operatorname{CE}}_{n+1})]_{\zeta_{n}+1}$

is not surjective for any $n\geq 21.$ Hence the map

$\cdot\ell:[A({\operatorname{Pan}}_{n})]_{\zeta_{n}}\longrightarrow[A({\operatorname{Pan}}_{n})]_{\zeta_{n}+1}$

is also not surjective, as desired.

Case 2: $\zeta_{n}=\chi_{n+1}+1$. In this case, the chain (4.2) yields
$\lambda_{n}=\rho_{n}=\zeta_{n}-1$. Since $\lambda_{n}-\lambda_{n-3}\leq 1$ by
Lemma 3.2, we have the following two subcases.

Subcase 2.1: $\lambda_{n}=\lambda_{n-3}$. As in the proof of Theorem 4.4,
denote $y_{1}=x_{n},y_{2}=x_{n+1}$, we have the following diagram

Since the right vertical map of the above diagram fails the surjectivity at
degree $\lambda_{n-3}+1=\zeta_{n}$, we conclude that the left vertical map
fails the surjectivity at the same degree.

Subcase 2.2: $\lambda_{n}=\lambda_{n-3}+1$. Set

$m=\max\\{j\mid j\leq n\;\text{and}\;\lambda_{j}=\lambda_{j-1}+1\\}.$

Then by Lemma 3.2, $n-2\leq m\leq n$ and $\lambda_{m}=\lambda_{n}$. First, we
consider the case where $m\neq n$. Set

$y=\begin{cases}x_{n-2}&\text{if}\;m=n-2\\\
x_{n+1}&\text{if}\;m=n-1.\end{cases}$

Then we have the following diagram

Since $\zeta_{n}-2=\lambda_{n}-1=\lambda_{m}-1$, we have the first vertical
map of the diagram fails the injectivity at degree $\lambda_{m}-1$ by
Proposition 4.3. It follows that the second vertical map of the diagram fails
the injectivity at degree $\zeta_{n}-1$.

To complete the proof of the theorem, we consider the case where $m=n$. In
this case, one has $\rho_{n}=\lambda_{n}=\lambda_{n-1}+1$. By Lemmas 3.2 and
3.4, $\lambda_{n-1}=\lambda_{n-4}=\lambda_{n-5}+1$. Hence
$\lambda_{n-4}=\zeta_{n}-2$. Now we consider the following diagram

By Proposition 4.3, the first vertical map of the diagram fails the
injectivity at degree $\lambda_{n-4}-1=\zeta_{n}-3$. It follows that the
second vertical map of the diagram fails the injectivity at degree
$\zeta_{n}-1$. Thus we complete the proof. ∎

## Acknowledgments

It is our pleasure to thank Prof. Le Tuan Hoa for his constant encouragement
during the preparation of this work. The first named author was partially
supported by the Vietnam Academy of Science and Technology (through the grant
NCXS02.01/22-23). The second named author was partially supported by the
NAFOSTED (Vietnam) under the grant number 101.04-2023.07. Parts of the paper
were written during our stay at the Vietnam Institute for Advanced Study in
Mathematics (VIASM). We would like to thank the VIASM for its generous support
and hospitality.

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