The Weak Lefschetz Property of Artinian Algebras Associated to Paths and Cycles

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.

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 each 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 each 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, 24,25,26,27,28, 30,31,32,33, 37]). Even the characteristic of $\Bbbk$ plays an interesting role in the study of the Lefschetz properties; see, for example, [7,8,9, 20, 21, 26].

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, 22, 23, 26, 29] 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

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

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 in 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 [38] 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 the 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 [29]. 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 $\varDelta$ on n vertices the ring

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

in which $I_{\varDelta }$ is the Stanley-Reisner ideal of $\varDelta$. 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 $P_n, C_n, {{\text {Pan}}}_n$ 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 $\textrm{char}(\Bbbk )=0$.

• (1) For an integer $n\ge 1$, the ring $A(P_n)$ has the WLP if and only if $n\in \{1,2,\ldots ,7,9,10,13\}.$

• (2) For an integer $n\ge 3$, the ring $A(C_n)$ has the WLP if and only if $n\in \{3,4,\ldots ,11,13,14,$ $17\}.$

• (3) For an integer $n\ge 3$, the ring $A({{\text {Pan}}}_n)$ has the WLP if and only if $n\in \{3,4,\ldots ,10,12,13,$ $16\}.$

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 relevant terminologies 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 paths, cycles and pan graphs. These results are useful to study the WLP of artinian algebras associated to these graphs. In Section 4, we prove 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 terminologies 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 [28].

2.1 The Weak Lefschetz Property

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

Proposition 2.1

[26, 35] 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 $A=\oplus _{j\ge 0} [A]_j$ 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

$$ D=\max \{i\mid [A]_i\ne 0\}, $$

the socle degree of A. In this case, the Hilbert series of A is a polynomial

$$ HS(A,t)=1+h_1t+\cdots + h_Dt^D, $$

where $h_i=\dim _\Bbbk [A]_i>0$. By definition, the degree of the Hilbert series for an artinian graded algebra A is equal to its socle degree D. Since A is artinian and non-zero, this number also agrees with the Castelnuovo-Mumford regularity of A, i.e.,

$$ \text {reg}\, (A)= D =\text {deg} (HS(A,t)). $$

The algebra A is called level if its socle is concentrated in one degree.

Definition 2.3

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

$$a_0\le a_1\le \cdots \le a_{m-1}\le a_m \ge a_{m+1}\ge \cdots \ge a_n.$$

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

$$ a_{i-1}< a_i \ge a_{i+1}\ge \cdots \ge a_n. $$

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

• (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 the map

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

• (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.

• (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.

• (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.

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

• (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 $t^k$ is 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:

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

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

• (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 graded polynomial ring over $\Bbbk$. The edge ideal of G is the ideal

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

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 ${{\text {reg}}}(A(G))=\alpha (G)$ 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]. Recall that a graph is said to be claw-free if it does not admit the complete bipartite graph $K_{1,3}$ as an induced subgraph. 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 $i\ge 0,\ s_i(G) =(2^i-1)\left( {\begin{array}{c}m\\ i\end{array}}\right) +\left( {\begin{array}{c}n\\ i\end{array}}\right) .$ Therefore, for $m\ge 95$ 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 $\textrm{char}(\Bbbk )=0$ or $\textrm{char}(\Bbbk )> n$ [36, 37]. This result does not hold if $\textrm{char}(\Bbbk )\le n$, even for the WLP. Indeed, in the case where $\textrm{char}(\Bbbk )=2$, 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 $n\ge 5$ and $\textrm{char}(\Bbbk )=p$ is odd, then $A(E_n)$ has the WLP if and only if $p\ge \lfloor \frac{n+3}{2}\rfloor$.

Fig. 1

The path $P_6$

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

[29, Theorem 4.8] Let $\textrm{char}(\Bbbk )\ne 2$ and A(G) be the artinian algebra associated to $G=\cup _{i=1}^r K_{n_i}$. Assume $n_1\ge n_2\ge \cdots \ge n_r\ge 1$. Then A(G) has the WLP if and only if one of the following holds:

• (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) $n_3=\cdots =n_r=1$ and r is odd.

In particular for every $n\ge 2$, 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 ($n\ge 1$) (Fig. 1).

Proposition 3.1

The independence polynomial of $P_n$ is

$$ I(P_n;t)=\sum _{i=0}^{\lfloor \frac{n+1}{2}\rfloor } \left( {\begin{array}{c}n+1-i\\ i\end{array}}\right) t^i. $$

Moreover, for every $n\ge 1$, $I(P_n;t)$ 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 $F_n(t),\ n\ge 0$, 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

$$ I(P_n;t)=\sum _{i=0}^{\lfloor \frac{n+1}{2}\rfloor } \left( {\begin{array}{c}n+1-i\\ i\end{array}}\right) t^i. $$

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 $0\le i\le \lfloor \frac{n+1}{2}\rfloor$ and

$$ \left( {\begin{array}{c}n+1-i\\ i\end{array}}\right) \ge \left( {\begin{array}{c}n-i\\ i+1\end{array}}\right) . $$

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

$$\begin{aligned}&\left( {\begin{array}{c}n+1-i\\ i\end{array}}\right) \ge \left( {\begin{array}{c}n-i\\ i+1\end{array}}\right) \\ \Leftrightarrow&\frac{n+1-i}{(n-2i)(n-2i+1)}\ge \frac{1}{i+1}\\ \Leftrightarrow&5i^2-(5n+2)i+n^2-1\le 0\\ \Leftrightarrow&\frac{5n+2-\sqrt{5n^2+20n+24}}{10}\le i\le \frac{5n+2+\sqrt{5n^2+20n+24}}{10}. \end{aligned}$$

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

$$ \left( {\begin{array}{c}n+1-i\\ i\end{array}}\right) \ge \left( {\begin{array}{c}n-i\\ i+1\end{array}}\right) \Leftrightarrow i\ge \frac{5n+2-\sqrt{5n^2+20n+24}}{10}. $$

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

We summarize the important properties of the mode of $I(P_n;t)$.

Lemma 3.2

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

• (i) $\lambda _{n+1}\ge \lambda _n;$

• (ii) $\lambda _{n+3}-1\le \lambda _n\le \lambda _{n+4}-1;$

• (iii) $\lambda _{n+11}\ge \lambda _n+3.$

Proof

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

$$\begin{aligned} \alpha _{n+1}\ge \alpha _n;\; \alpha _{n+3}-1\le \alpha _n\le \alpha _{n+4}-1 \; \text {and}\;\alpha _{n+11}\ge \alpha _n+3. \end{aligned}$$

The lemma follows from basic properties of the ceiling function.$\square$

Fig. 2

The cycle $C_6$

Table 1 Graphs and modes of their independence polynomials

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, by using Macaulay2 [34]. A dash indicates an undefined value.

3.2 Cycles

Let $C_n$ be the cycle on n vertices ($n\ge 3$) (Fig. 2).

Proposition 3.3

The independence polynomial of $C_n$ is

$$\begin{aligned} I(C_n;t)&=I(P_{n-1};t)+tI(P_{n-3};t)\\&=1+\sum _{i=1}^{\lfloor \frac{n}{2}\rfloor } \frac{n}{i}\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) t^i. \end{aligned}$$

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

Proof

Hopkins and Staton [19] showed that

$$ I(C_n; t) =1+\sum _{i=1}^{\lfloor \frac{n}{2}\rfloor } \frac{n}{i}\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) t^i. $$

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 $1\le i \le \lceil \frac{n}{2} \rceil -1$ from

$$ \frac{n}{i}\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) \ge \frac{n}{i+1}\left( {\begin{array}{c}n-i-2\\ i\end{array}}\right) $$

we get

$$\begin{aligned} (i+1)(n-i-1)\ge (n-2i-1)(n-2i). \end{aligned}$$

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

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

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.$\square$

Lemma 3.4

For all $n\ge 5$, there are inequalities $\lambda _{n-1}\le \rho _n\le \lambda _{n-4}+1 \le \lambda _n$.

Proof

By Lemma 3.2, $\lambda _{n-4}+1 \le \lambda _n$, hence it suffices to show that

$$ \lambda _{n-1}\le \rho _n\le \lambda _{n-4}+1. $$

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

$$\begin{aligned}&\dfrac{5(n-1)+2-\sqrt{5(n-1)^2+20(n-1)+24}}{10} \le \dfrac{5n-4-\sqrt{5n^2-4}}{10}\\ \Leftrightarrow&5n-3-\sqrt{5n^2+10n+9} \le 5n-4-\sqrt{5n^2-4}\\ \Leftrightarrow&\sqrt{5n^2-4}+1 \le \sqrt{5n^2+10n+9} \\ \Leftrightarrow&5n^2-3+2\sqrt{5n^2-4} \le 5n^2+10n+9\\ \Leftrightarrow&\sqrt{5n^2-4} \le 5n+6 \Leftrightarrow (5n+6)^2-(5n^2-4) \ge 0\\ \Leftrightarrow&20n^2+60n+40 \ge 0, \end{aligned}$$

which is clear.

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

$$\begin{aligned}&\dfrac{5n-4-\sqrt{5n^2-4}}{10} \le \dfrac{5(n-4)+2-\sqrt{5(n-4)^2+20(n-4)+24}}{10}+1 \\ \Leftrightarrow&5n-4-\sqrt{5n^2-4} \le 5n-8-\sqrt{5n^2-20n+24}\\ \Leftrightarrow&\sqrt{5n^2-20n+24}+4 \le \sqrt{5n^2-4} \\ \Leftrightarrow&5n^2-20n+24+16+8\sqrt{5n^2-20n+24} \le 5n^2-4 \quad (\text {by squaring})\\ \Leftrightarrow&8\sqrt{5n^2-20n+24} \le 20n-44 \\ \Leftrightarrow&2\sqrt{5n^2-20n+24} \le 5n-11\\ \Leftrightarrow&4(5n^2-20n+24) \le (5n-11)^2 \\ \Leftrightarrow&5n^2-30n+25 \ge 0 \Leftrightarrow 5(n-1)(n-5)\ge 0 \end{aligned}$$

which is true for all $n\ge 5$. The proof is complete.$\square$

Fig. 3

Pan$_{6}$ and CE$_{6}$

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 ${Pan }_n$.

Our goal is to show the independence polynomial of ${Pan }_n$ is unimodal. For this, we consider a family of graphs formed by adding an edge $\{n-2,n\}$ to the cycles $C_n$ ($n\ge 4$). We denote this graph by ${CE }_n$ (Fig. 3).

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

Lemma 3.5

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

$$\begin{aligned} I({CE }_n;t)&=\sum _{i=0}^{\alpha ({CE }_n)} s_i({CE }_n)t^i\\&=I(P_{n-1};t)+tI(P_{n-4};t)\\&=\sum _{i=0}^{\lfloor \frac{n}{2} \rfloor }\bigg [\left( {\begin{array}{c}n-i\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-2\\ i-1\end{array}}\right) \bigg ]t^i. \end{aligned}$$

Let $\chi _n$ be the mode of $I(CE _n;t)$ and $\lambda _n$ be the mode of $I(P_n;t)$ as in Proposition 3.1. For any $n\ge 5$, one has $\lambda _{n-1}\le \chi _n\le \lambda _{n-4}+1.$

Proof

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

Let $1\le i\le \lambda _{n-1}$. We need to show that

$$ s_{i-1}({CE }_n)<s_i({CE }_n), $$

namely,

$$ \left( {\begin{array}{c}n-i+1\\ i-1\end{array}}\right) +\left( {\begin{array}{c}n-i-1\\ i-2\end{array}}\right) < \left( {\begin{array}{c}n-i\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-2\\ i-1\end{array}}\right) . $$

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

Since $i\le \lambda _{n-1}$, $\left( {\begin{array}{c}n-i+1\\ i-1\end{array}}\right) \le \left( {\begin{array}{c}n-i\\ i\end{array}}\right)$. It suffices to show that

$$\begin{aligned}&\left( {\begin{array}{c}n-i-1\\ i-2\end{array}}\right)< \left( {\begin{array}{c}n-i-2\\ i-1\end{array}}\right) \\ \Leftrightarrow&(n-i-1)(i-1)< (n-2i)(n-2i+1)\\ \Leftrightarrow&5i^2-(5n+2)i+n^2+2n-1> 0\\ \Leftrightarrow&i< \dfrac{5n+2-\sqrt{5n^2-20n+24}}{10}\ \text {or}\ i> \dfrac{5n+2+\sqrt{5n^2-20n+24}}{10}. \end{aligned}$$

As $i\le \lambda _{n-1}$, it is enough to show that

$$\begin{aligned}&\dfrac{5(n-1)+2-\sqrt{5(n-1)^2+20(n-1)+24}}{10}< \dfrac{5n+2-\sqrt{5n^2-20n+24}}{10} -1\\ \Leftrightarrow&5n-3-\sqrt{5n^2+10n+9}< 5n-8-\sqrt{5n^2-20n+24}\\ \Leftrightarrow&5+\sqrt{5n^2-20n+24}< \sqrt{5n^2+10n+9}\\ \Leftrightarrow&\sqrt{5n^2-20n+24}< 3n-4 \quad (\text {after squaring and simplifying})\\ \Leftrightarrow&n^2-n-2> 0\\ \Leftrightarrow&(n+1)(n-2)> 0, \end{aligned}$$

which is clear for any $n\ge 4$. It follows that $\lambda _{n-1}\le \chi _n.$ It remains to show that if $\lfloor \frac{n}{2} \rfloor \ge i\ge \lambda _{n-4}+1$ (note that $\lfloor \frac{n}{2} \rfloor$ is the independence number of ${CE }_n$), then

$$ s_{i}({CE }_n)\ge s_{i+1}({CE }_n)\Longleftrightarrow \left( {\begin{array}{c}n-i\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-2\\ i-1\end{array}}\right) \ge \left( {\begin{array}{c}n-i-1\\ i+1\end{array}}\right) +\left( {\begin{array}{c}n-i-3\\ i\end{array}}\right) . $$

By Lemma 3.2, $i\ge \lambda _{n-4}+1\ge \lambda _{n-1}$, so $\left( {\begin{array}{c}n-i\\ i\end{array}}\right) \ge \left( {\begin{array}{c}n-i-1\\ i+1\end{array}}\right)$ thanks to Proposition 3.1. We have to show that

$$\begin{aligned} \left( {\begin{array}{c}n-i-2\\ i-1\end{array}}\right)&\ge \left( {\begin{array}{c}n-i-3\\ i\end{array}}\right) \\&\Leftrightarrow i(n-i-2)\ge (n-2i-2)(n-2i-1)\\&\Leftrightarrow 5i^2-(5n-8)i+n^2-3n+2\le 0\\&\Leftrightarrow \dfrac{5n-8-\sqrt{5n^2-20n+24}}{10}\le i\le \dfrac{5n-8+\sqrt{5n^2-20n+24}}{10}. \end{aligned}$$

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

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

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

$$\begin{aligned}&\dfrac{5n-8-\sqrt{5n^2-20n+24}}{10}\le \dfrac{5(n-4)+2-\sqrt{5(n-4)^2+20(n-4)+24}}{10}+1\\ \Leftrightarrow&\dfrac{5n-8-\sqrt{5n^2-20n+24}}{10}\le \dfrac{5n-18 +\sqrt{5n^2-20n+24}}{10}+1, \end{aligned}$$

which is clear. Thus $\chi _n \le \lambda _{n-4}+1.$ The proof is complete. $\square$

Note that ${Pan }_n$ 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 $I({Pan }_n;t)$ of the n-pan graph is unimodal. Let $\zeta _n$ be the mode of $I({Pan }_n;t)$. Then for all $n\ge 5$, there are inequalities

$$\begin{aligned} \chi _{n+1}\le \zeta _n\le \rho _n+1\le \lambda _n+1\le \chi _{n+1}+1, \end{aligned}$$

(3.1)

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

$$\begin{aligned} I({Pan }_n;t)&=\sum _{i=0}^{\alpha ({Pan }_n)} s_i({Pan }_n)t^i\\&=I(C_{n};t)+tI(P_{n-1};t)\\&=I(P_{n-1};t)+t\big (I(P_{n-3};t)+I(P_{n-1};t)\big )\\&=\sum _{i=0}^{\lfloor \frac{n}{2}\rfloor +1}\bigg [\left( {\begin{array}{c}n-i\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) +\left( {\begin{array}{c}n-i+1\\ i-1\end{array}}\right) \bigg ]t^i. \end{aligned}$$

Therefore, we have

$$\begin{aligned} s_i({Pan }_n)&=\left( {\begin{array}{c}n-i\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) +\left( {\begin{array}{c}n-i+1\\ i-1\end{array}}\right) \nonumber \\&=\left( {\begin{array}{c}n-i+1\\ i\end{array}}\right) +\left( {\begin{array}{c}n-i-1\\ i-1\end{array}}\right) +\left( {\begin{array}{c}n-i\\ i-2\end{array}}\right) \nonumber \\&\qquad \left( \text {using} \left( {\begin{array}{c}n\\ p\end{array}}\right) =\left( {\begin{array}{c}n-1\\ p\end{array}}\right) +\left( {\begin{array}{c}n-1\\ p-1\end{array}}\right) \right) \nonumber \\&=s_i({CE }_{n+1})+\left( {\begin{array}{c}n-i\\ i-2\end{array}}\right) \quad \text {(by Lemma 3.5)}. \end{aligned}$$

(3.2)

We first have the following assertion.

Claim 1: $s_{i-1}({Pan }_n)<s_i({Pan }_n)$ for any $i\le \chi _{n+1}.$

Proof of Claim 1: For any $1\le i\le \chi _{n+1}$, $s_{i-1}({CE }_{n+1})\le s_i({CE }_{n+1}).$ Therefore by (3.2), it suffices to show that

$$\begin{aligned}&\left( {\begin{array}{c}n-i+1\\ i-3\end{array}}\right)< \left( {\begin{array}{c}n-i\\ i-2\end{array}}\right) \\ \Longleftrightarrow&(n-i+1)(i-2)< (n-2i+3)(n-2i+4) \\ \Longleftrightarrow&5i^2-(5n+17)i+n^2+9n+14> 0\\ \Longleftrightarrow&i< \frac{5n+17-\sqrt{5n^2-10n+9}}{10}\ \text {or}\ i> \frac{5n+17+\sqrt{5n^2-10n+9}}{10}. \end{aligned}$$

By Lemma 3.5,

$$\begin{aligned} i&\le \chi _{n+1}\le \lambda _{n-3}+1 \\&< \dfrac{5(n-3)+2-\sqrt{5(n-3)^2+20(n-3)+24}}{10} +2 \\&= \dfrac{5n+7-\sqrt{5n^2-10n+9}}{10}. \end{aligned}$$

Consequently, it suffices to show that

$$\begin{aligned} \frac{5n+7-\sqrt{5n^2-10n+9}}{10}\le \frac{5n+17-\sqrt{5n^2-10n+9}}{10}, \end{aligned}$$

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

Now, by again Proposition 2.9, we have

$$\begin{aligned} I({\text {Pan}}_n;t)&=\sum _{i=0}^{\alpha ({\text {Pan}}_n)} s_i({\text {Pan}}_n)t^i\\&=I(C_{n};t)+tI(P_{n-1};t), \end{aligned}$$

we get $s_i({\text {Pan}}_n)=s_i(C_n)+s_{i-1}(P_{n-1}).$ Next we have the following.

Claim 2: $s_{i}({\text {Pan}}_n)\ge s_{i+1}({\text {Pan}}_n)$ for any $i\ge \rho _{n}+1.$

Proof of Claim 2: Since $i\ge \rho _{n}+1$ and $n\ge 5$, $i-1\ge \rho _n\ge \lambda _{n-1}$ by Lemma 3.4. It follows that $s_i(C_n)\ge s_{i+1}(C_n)$ and $s_{i-1}(P_{n-1})\ge s_i(P_{n-1}).$ Thus $s_i({\text {Pan}}_n)\ge s_{i+1}({\text {Pan}}_n)$, as desired.

By Lemmas 3.4 and 3.5, $\rho _n\le \lambda _n\le \chi _{n+1},$ which yields the last two inequalities in (3.1). Moreover, it follows from Claims 1 and 2 that $s_{i-1}({{\text {Pan}}}_n)< s_{i}({{\text {Pan}}}_n)$ for any $i\le \chi _{n+1}$ and $s_i({{\text {Pan}}}_n)\ge s_{i+1}({{\text {Pan}}}_n)$ for any $i\ge \chi _{n+1}+1$. Thus, the independence polynomial $I({{\text {Pan}}}_n;t)$ of the n-pan graph is unimodal. Moreover, $\chi _{n+1}\le \zeta _n$ by Claim 1 and $\zeta _n\le \rho _{n}+1$ by Claim 2. This concludes the proof.$\square$

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 $\textrm{char}(\Bbbk )=0$ 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 $K=(x_1^2,\ldots ,x_n^2)+(x_1x_2,x_2x_3,\ldots ,x_{n-1}x_n)\subset R=\Bbbk [x_1,\ldots ,x_n]$. 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

Proof

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

The desired conclusion follows.$\square$

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 [34] to compute the Hilbert series of the rings $A(P_n)$ and $A(P_n)/\ell A(P_n)$ with $1\le n\le 17$, 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 $n\ge 17$.

We will prove the above claim by induction on n, having just established the case $n=17$. For $n\ge 18$, we consider the multiplication map

$$ \cdot \ell : [A(P_n)]_{\lambda _n}\longrightarrow [A(P_n)]_{\lambda _n+1}. $$

By Lemma 3.2, one has $\lambda _{n-1}\le \lambda _n\le \lambda _{n-1}+1$, hence we consider the following two cases.

$\underline{{\textbf {Case 1:}}~ \lambda _{n}=\lambda _{n-1}.}$ 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.

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

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

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.

$\square$

The above theorem shows that $A(P_n)$ fails the WLP since surjectivity fails for any $n\ge 17$. 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 $n\ge 12$ 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 $n\ge 12$. A computation with Macaulay2 [34] 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 $n\ge 21$ such that $\lambda _n=\lambda _{n-1}+1.$ Set

$$\begin{aligned} n_1&=\max \{j \mid j<n \text { and } \lambda _j=\lambda _{j-1}+1\},\\ n_2&=\max \{j \mid j<n_1 \text { and } \lambda _j=\lambda _{j-1}+1\},\\ m&=\max \{j \mid j<n_2 \text { and } \lambda _j=\lambda _{j-1}+1\}. \end{aligned}$$

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

$$ 0\rightarrow A(P_m)\otimes _{\Bbbk } A(P_{n-m-3}) (-1)\xrightarrow {\cdot x_{m+2}} A(P_n)\rightarrow A(P_{m+1})\otimes _{\Bbbk } A(P_{n-m-2})\rightarrow 0. $$

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 $6\le n-m-3\le 8$. 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.$\square$

4.2 Cycles

The artinian algebra associated to the cycle on n vertices is

$$ (C_n)=R/K, $$

where $K=(x_1^2,\ldots ,x_n^2)+(x_1x_2,x_2x_3,\ldots ,x_{n-1}x_n,x_nx_1)\subset R=\Bbbk [x_1,\ldots ,x_n]$. 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 [34] to compute the Hilbert series of $A(C_n)$ and $A(C_n)/\ell A(C_n)$ with $3\le n\le 20$, we can check that:

• $A(C_n)$ has the WLP for each $3\le n\le 17$ 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 $n\ge 21$. By Lemmas 3.2 and 3.4, $\lambda _{n-1}\le \rho _n\le \lambda _{n-4}+1 \le \lambda _{n-1}+1$. Consider the following two cases.

$\underline{{\textbf {Case 1:}}~ \rho _n=\lambda _{n-1}.}$ In this case, we will show that $A(C_n)$ fails the WLP due to 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 $J=K:x_n$. Then $A(P_{n-1})\cong R/I$ and $A(P_{n-3})\cong R/J$ and we have the following exact sequence

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 $n\ge 18.$ Hence the middle vertical map

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

is not surjective, as desired.

$\underline{{\textbf {Case 2:}}~ \rho _n=\lambda _{n-1}+1.}$ 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

Fig. 4

Tadpole $T_{3,5}$

By the proof of Theorem 4.2 and the fact that $n-4\ge 17$, 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.$\square$

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 (Fig. 4).

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

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

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 $n\in \{1,3,4,7\}.$

Proof

The artinian algebra associated to $T_{3,n}$ is

$$ A(T_{3,n})=\frac{\Bbbk [x_1,x_2,x_3,y_1,\ldots ,y_{n}]}{(x_1^2,x_2^2,x_3^2,y_1^2,\ldots ,y_{n}^2)+(x_1x_2,x_2x_3, x_3x_1, x_3y_1, y_1y_2,\ldots ,y_{n-1}y_n)}. $$

Using Macaulay2 [34] to compute the Hilbert series of $A(T_{3,n})$ and $A(T_{3,n})/\ell A(T_{3,n})$ with $1\le n\le 17$, we can check that:

• $A(T_{3,n})$ has the WLP for each $n\in \{1,3,4, 7\}$;

• for $n \in \{2,5,8,9,11, 12, 14, 15, 16, 17\}$, then $A(T_{3,n})$ fails the surjectivity from degree $\rho _{n+3}$ to degree $\rho _{n+3}+1$;

• for $n \in \{2, 6, 10, 13, 14, 17\}$, then $A(T_{3,n})$ fails the injectivity from degree $\rho _{n+3} -1$ to degree $\rho _{n+3}.$

Now assume that $n\ge 18$. By Lemmas 3.2 and 3.4, $\lambda _{n+2}\le \rho _{n+3}\le \lambda _{n+2}+1$. We consider the following commutative diagram

The proof proceeds along the same lines as that of Theorem 4.4, replacing $A(C_{n+3})$ by $A(T_{3,n})$ and noting that $n+3\ge 21$.$\square$

4.3 Pans

To study the WLP of rings associated to pans, we first examine the WLP of rings associated to ${{\text {CE}}}_n$ (Fig. 5). The latter is by definition

$$ A({{\text {CE}}}_n)=\frac{\Bbbk [x_1,\ldots ,x_n]}{(x_1^2,\ldots ,x_n^2)+(x_1x_2,x_2x_3,\ldots ,x_{n-1}x_n,x_nx_1)+(x_{n-2}x_{n})}. $$

Theorem 4.6

For an integer $n\ge 4$, the algebra $A({{\text {CE}}}_n)$ has the WLP if and only if $n\in \{4,5,\ldots ,8,10,11,14\}.$

Proof

By using Macaulay2 [34] to compute the Hilbert series of $A({{\text {CE}}}_n)/\ell A({{\text {CE}}}_n)$ and $A({{\text {CE}}}_n)$ with $4\le n\le 20$, we can check that:

• $A({{\text {CE}}}_n)$ has the WLP for each $4\le n\le 14$ and $n\notin \{9,12,13\}$;

• for $n \in \{9,12,15,16,18,19\}$, $A({{\text {CE}}}_n)$ fails the surjectivity from degree $\chi _n$ to degree $\chi _n+1$;

• for $n \in \{13,17,20\}$, $A({{\text {CE}}}_n)$ fails the injectivity from degree $\chi _n -1$ to degree $\chi _n.$

Now assume that $n\ge 21$. We will prove that $A({{\text {CE}}}_n)$ 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,

$$\begin{aligned} \lambda _{n-1}\le \chi _n\le \lambda _{n-4}+1 \le \lambda _{n-1}+1. \end{aligned}$$

(4.1)

We consider the following two cases.

$\underline{{\textbf {Case 1:}}~ \chi _n=\lambda _{n-1}.}$ Consider the exact sequence

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 $n\ge 18.$ Hence, the map

$$ \cdot \ell : [A({{\text {CE}}}_{n})]_{\chi _n}\longrightarrow [A({{\text {CE}}}_{n})]_{\chi _n+1} $$

is also not surjective, as desired.

$\underline{{\textbf {Case 2:}}~ \chi _n = \lambda _{n-1}+1.}$ 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 the left vertical map in the diagram fails the surjectivity at degree $\lambda _{n-4}+1=\chi _n$, as desired.$\square$

Fig. 5

The graph ${{\text {CE}}}_6$

Fig. 6

${{\text {Pan}}}_6$

Finally, we show the last main result.

Theorem 4.7

The algebra $A({{\text {Pan}}}_n)$ associated to the pan graph ${{\text {Pan}}}_n$ has the WLP if and only if $n\in \{3,4,\ldots ,10,12,13,16\}.$

Proof

The artinian algebra associated to ${{\text {Pan}}}_n$ (see Fig. 6) is

$$ A({{\text {Pan}}}_n)=\frac{\Bbbk [x_1,\ldots ,x_{n+1}]}{(x_1^2,\ldots ,x_{n+1}^2)+(x_1x_2,x_2x_3,\ldots ,x_{n-1}x_n,x_nx_{n+1})+(x_1x_n)}. $$

By using Macaulay2 [34] to compute the Hilbert series of $A({{\text {Pan}}}_n)$ and $\dfrac{A({{\text {Pan}}}_n)}{\ell A({{\text {Pan}}}_n)}$ with $3\le n\le 20$, we can check that:

• $A({{\text {Pan}}}_n)$ has the WLP for each $3\le n\le 16$ and $n\notin \{11,14,15\}$;

• for $n \in \{11,14,17,18,20\}$, $A({{\text {Pan}}}_n)$ fails the surjectivity from degree $\zeta _n$ to degree $\zeta _n+1$;

• for $n \in \{15,19\}$, $A({{\text {Pan}}}_n)$ fails the injectivity from degree $\zeta _n -1$ to degree $\zeta _n.$

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

$$\begin{aligned} \chi _{n+1}\le \zeta _n\le \rho _{n}+1 \le \lambda _{n}+1\le \chi _{n+1}+1. \end{aligned}$$

(4.2)

We consider the following two cases.

$\underline{{\textbf {Case 1:}}~ \zeta _n=\chi _{n+1}.}$ In this case, $A({{\text {Pan}}}_n)$ fails the surjectivity from degree $\zeta _n$ to degree $\zeta _n+1$ by using the exact sequence

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

$$ \cdot \ell : [A({{\text {CE}}}_{n+1})]_{\zeta _n}\longrightarrow [A({{\text {CE}}}_{n+1})]_{\zeta _n+1} $$

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

$$ \cdot \ell : [A({{\text {Pan}}}_n)]_{\zeta _n}\longrightarrow [A({{\text {Pan}}}_n)]_{\zeta _n+1} $$

is also not surjective, as desired.

$\underline{{\textbf {Case 2:}}~ \zeta _n=\chi _{n+1}+1.}$ In this case, the chain (4.2) yields $\lambda _{n}=\rho _{n}=\zeta _n-1$. Since $\lambda _n-\lambda _{n-3}\le 1$ by Lemma 3.2, we have the following two subcases.

$\underline{{\textbf {Subcase 2.1:}}~ \lambda _n=\lambda _{n-3}.}$ 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.

$\underline{{\textbf {Subcase 2.2:}}~ \lambda _n=\lambda _{n-3}+1.}$ Set

$$ m=\max \{j \mid j\le n\; \text {and}\; \lambda _j=\lambda _{j-1}+1\}. $$

Then by Lemma 3.2, $n-2\le m\le n$ and $\lambda _m=\lambda _n$. First, we consider the case where $m\ne n$. Set

$$ y={\left\{ \begin{array}{ll} x_{n-2}& \text {if } m=n-2,\\ x_{n+1}& \text {if } m=n-1. \end{array}\right. } $$

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.$\square$