On jumped Wenger graphs

TL;DR

This paper introduces a new class of bipartite graphs called jumped Wenger graphs, providing bounds on their diameter and girth, and explicitly calculating diameters for specific parameter cases, advancing graph theory understanding.

Contribution

It defines jumped Wenger graphs, establishes their diameter and girth bounds, and computes exact diameters for particular parameter configurations, enriching the study of bipartite graphs.

Findings

Determined tight upper bounds for diameter and girth.

Calculated exact diameters for specific jumped Wenger graphs.

Established the relationship between jumped Wenger graphs and Wenger graphs.

Abstract

In this paper we introduce a new infinite class of bipartite graphs, called jumped Wenger graphs, which are closely related to Wenger graphs. An tight upper bound of the diameter and the exact girth of a jumped Wenger graph $J_m(q, i, j )$ for integers $i, j$, $1\leq i <j \leq m+2$, are determined. In particular, the exact diameter of the jumped Wenger graph $J_m(q, i, j)$ if $(i, j)=(m,m+2), (m+1,m+2)$ or $(m,m+1)$ is also obtained.

Equations

$$l_2+p_2$$

$$l_3+p_3$$

$$⋮$$

$$l_{m+1}+p_{m+1}$$

$$l_k+p_k$$

$$l_k^{\prime }+p_k$$

$$l_k-l_k^{\prime }=(l_1-l_1^{\prime })f_k(p_1),k=2,\dots ,m+1.$$

$$l_k+p_k$$

$$l_k+p_k^{\prime }$$

$$p_k-p_k^{\prime }=l_1(f_k(p_1)-f_k(p_1^{\prime })),k=2,\dots ,m+1.$$

$${\sigma }_k(x_1,\dots ,x_n)=x_n{\sigma }_{k-1}(x_1,\dots ,x_{n-1})+{\sigma }_k(x_1,\dots ,x_{n-1}).$$

$${\sigma }_{i,j}(x_1,\dots ,x_n)={\sigma }_i(x_1,\dots ,x_n){\sigma }_{j+1}(x_1,\dots ,x_n)-{\sigma }_j(x_1,\dots ,x_n){\sigma }_{i+1}(x_1,\dots ,x_n),$$

$$|M_{n,i,j}(x_1,\dots ,x_n)|=(-1{)}^{i+j-1}{\sigma }_{n-i,n-j}(x_1,\dots ,x_n)\prod _{1\le k<l\le n}(x_l-x_k)$$

$$|V_{n+2}(x_1,\dots ,x_n,x,y)|$$

$$|M_{n,i,n+1}(x_1,\dots ,x_n)|=(-1{)}^{i+n}{\sigma }_{n-i}(x_1,\dots ,x_n)\prod _{1\le k<l\le n}(x_l-x_k).$$

$${\sigma }_{n-i,n-j}(x_1,\dots ,x_n)$$

$${\sigma }_{n-i-1,n-j-1}(x_1,\dots ,x_{n-1})\ne 0$$

$${\sigma }_{n-i,n-j}(x_1,\dots ,x_n)$$

$$l_k^{(h+1)}-l_k^{(h)}$$

$$l_{m+1}^{(h+1)}-l_{m+1}^{(h)}$$

$$(L_{s+1}-L_1{)}^T=M_{m+1,i,j}(x_1,\dots ,x_s)(t_1{t}_2\dots t_s{)}^T.$$

$$(P_{s+1}-P_1{)}^T=M_{m+1,i,j}(x_1,\dots ,x_s)(t_1{t}_2\dots t_s{)}^T,$$

$$(L^{\prime }-L{)}^T=M_{s^{\prime },i,j}(x_1^{\prime },\dots ,x_{s^{\prime }}^{\prime })(t_1,\dots ,t_{s^{\prime }}{)}^T,$$

$$L_2-L_1$$

$$L_3-L_2$$

$$\dots \dots$$

$$L_s-L_{s-1}$$

$$L_1-L_s$$

$$\left(\begin{array}[]{cccc}1&1&\cdots&1\\ f_{2}(p^{(1)}_{1})&f_{2}(p^{(2)}_{1})&\cdots&f_{2}(p^{(s)}_{1})\\ f_{3}(p^{(1)}_{1})&f_{3}(p^{(2)}_{1})&\cdots&f_{3}(p^{(s)}_{1})\\ \lx@intercol\dotfill\hfil\hfil\lx@intercol\\ f_{m+1}(p^{(1)}_{1})&f_{m+1}(p^{(2)}_{1})&\cdots&f_{m+1}(p^{(s)}_{1})\end{array}\right)\left(\begin{array}[]{c}t_{1}\\ t_{2}\\ \vdots\\ t_{s}\end{array}\right)=0$$

$$M_{m+1,i,j}(p_1^{(1)},\dots ,p_1^{(s)})(t_1\dots t_s{)}^T=0.$$

$$P_1=(0,0,\dots ,0),$$

$$P_3=(0,1,\dots ,1),$$

$$L_1=(0,0,\dots ,0),$$

$$L_3=(0,1,\dots ,1),$$

$$a_1{x}_1^2+a_2{x}_2^2=b$$

$$x_1{x}_2+x_3{x}_4+\dots +x_{n-1}{x}_n+x_n^2=b$$

$$x_1{x}_2+x_3{x}_4+\dots +x_{n-1}{x}_n=b$$

Full text

1 Introduction

Recently, researchers focus on a class of bipartite graphs related to Wenger graphs because of their nice graphical theory properties [1, 2, 4, 5, 7, 8, 9, 10]. First we introduce these graphs.

Let $\mathbb{F}_{q}$ be a finite field of order $q=p^{e}$ where $p$ is a prime and $e$ is a positive integer. Let $m$ be a positive integer and let $\mathfrak{P}=\mathbb{F}_{q}^{m+1}$ and $\mathfrak{L}=\mathbb{F}_{q}^{m+1}$ be two copies of the $(m+1)$-dimensional vector space over $\mathbb{F}_{q}$, which are called the point set and the line set respectively. Let $\mathfrak{G}=(V,E)$ be the bipartite graph with the vertex set $V=\mathfrak{P}\cup\mathfrak{L}$ and the edge set $E$ is defined as follow: there is an edge from a point $P=(p_{1},p_{2},\cdots,p_{m+1})\in\mathfrak{P}$ to a line $L=[l_{1},l_{2},\cdots,l_{m+1}]\in\mathfrak{L}$, denoted by $PL$, if the following $m$ equalities hold:

[TABLE]

where $f_{i}(x)\in\mathbb{F}_{q}[x]$ for $2\leq i\leq m+1$.

If $(1,f_{2}(x),\ldots,f_{m+1}(x))=(1,x,x^{2},\ldots,x^{m})$, this class of graphs is called Wenger graphs [1]. If $(1,f_{2}(x),\ldots,f_{m+1}(x))=(1,x,x^{p},\ldots,x^{p^{m-1}})$, this class of graphs is called linearized Wenger graphs [2]. In this paper, we focus on the more general case when $(1,f_{2}(x),\ldots$, $f_{m+1}(x))=(1,x,\ldots$, $x^{i-1},x^{i+1},\ldots$, $x^{j-1},x^{j+1},\ldots,x^{m+2})$, with $1\leq i<j\leq m+2$, which is called the jumped Wenger graphs with jump points at $x^{i}$ and $x^{j}$, denoted by $J_{m}(q,i,j)$. In particular, if $j=m+2$, that is, $(1,f_{2}(x),\ldots$, $f_{m+1}(x))=(1,x,\ldots,x^{i-1},x^{i+1}$, $\ldots,x^{m},x^{m+1})$, the graphs has only one jump point at $x^{i}$. If $i=m+1$ and $j=m+2$, the graphs become Wenger graphs, denoted by $W_{m}(q)$.

It is easy to get some properties from [2].

Proposition 1**.**

The jumped Wenger graph $J_{m}(q,i,j)$ for $1\leq i<j\leq m+1$ is $q$-regular.

Proposition 2**.**

If $m+2<q$, the jumped Wenger graph $J_{m}(q,i,j)$ for $1\leq i<j\leq m+1$ is connect.

In Section 2 we give an upper bound of the diameter of a jumped Wenger graph $J_{m}(q,i,j)$ for any integers $i,j$, $1\leq i<j\leq m+2$ and the exact diameter for some special jumped Wenger graphs such as $(i,j)=(m,m+2),(m+1,m+2)$ or $(m,m+1)$. In Section 3, we determine the girth of a jumped Wenger graph $J_{m}(q,i,j)$ for $1\leq i<j\leq m+2$. Finally, we give the conclusions.

2 The diameter of jumped Wenger graphs

Recall that a sequence of distinct vertices $v_{1},\cdots,v_{s}$ in a simple connected graph $\mathfrak{G}=(V,E)$ defines a path of length $s-1$ if $(v_{i},v_{i+1})\in E$ for every $i,1\leq i\leq s-1$. The distance between $v_{i}$ and $v_{j}$ is the number of edges in a shortest path joining $v_{i}$ and $v_{j}$. The diameter of a connected graph $\mathfrak{G}$ is the maximum distance between any two vertices of $\mathfrak{G}$. In [9] it is shown that the diameter of the Wenger graph $W_{m}(q)$ is $2m+2$ if $1\leq m\leq q-1$. The result for the linearized Wenger graphs also hold if $m\leq e$ in [1]. By Proposition 2, for any $1\leq i<j\leq m+2$, the jumped Wenger graph $J_{m}(q,i,j)$ is connected if $m+2<q$. So we consider the diameter of these jumped Wenger graphs in this section.

Assume that $L=[l_{1},\ldots,l_{m+1}]$ and $L^{\prime}=[l^{\prime}_{1},\ldots,l^{\prime}_{m+1}]$ are two lines and they share a point $P=(p_{1},\ldots,p_{m+1})$. Then we have

[TABLE]

and so we get

[TABLE]

We obtain the following fact which will be used later.

Fact 1. $L$ and $L^{\prime}$ with a common point are the same line if $l_{1}=l^{\prime}_{1}$.

Assume $P=(p_{1},\ldots,p_{m+1})$ and $P^{\prime}=(p^{\prime}_{1},\ldots,p^{\prime}_{m+1})$ be two points, both of which share a line $L=[l_{1},\ldots,l_{m+1}]$. Then we have

[TABLE]

and so we get

[TABLE]

In this case we have $P-P^{\prime}=(p_{1}-p^{\prime}_{1},l_{1}(p_{1}-p^{\prime}_{1}),\ldots,l_{1}(p^{i-1}_{1}-(p^{\prime}_{1})^{i-1})$, $l_{1}(p^{i+1}_{1}-(p^{\prime}_{1})^{i+1}),\ldots,l_{1}(p^{j-1}_{1}-(p^{\prime}_{1})^{j-1}),\ldots,l_{1}(p^{m+2}_{1}-(p^{\prime}_{1})^{m+2}))$ and so we obtain the following fact.

Fact 2. The $P$ and $P^{\prime}$ sharing the same line are identical if $p_{1}=p^{\prime}_{1}$.

Next, we recall elementary symmetric polynomials. If $f(x)=\prod_{i=1}^{n}(x-x_{i})=x^{n}-a_{1}x^{n-1}+a_{2}x^{n-2}+\ldots+(-1)^{n}a_{n}$, then the $k$th elementary symmetric polynomials $\sigma_{k}(x_{1},\ldots,x_{n})=\sum_{1\leq i_{1}<\ldots<i_{k}\leq n}x_{i_{1}}\cdots x_{i_{k}}$, $k=1,\ldots,n$. In addition, we define $\sigma_{0}(x_{1},\ldots,x_{n})=1$ and $\sigma_{-1}(x_{1},\ldots,x_{n})=0$. So it is straightforward to have the following iterative identity

[TABLE]

For simplicity, we denote

[TABLE]

M_{l,i,j}(x_{1},\ldots,x_{n})=\left(\begin{array}[]{cccc}1&1&\ldots&1\\ x_{1}&x_{2}&\ldots&x_{n}\\ \lx@intercol\dotfill\hfil\hfil\lx@intercol\\ x^{i-1}_{1}&x^{i-1}_{2}&\ldots&x^{i-1}_{n}\\ x^{i+1}_{1}&x^{i+1}_{2}&\ldots&x^{i+1}_{n}\\ \lx@intercol\dotfill\hfil\hfil\lx@intercol\\ x^{j-1}_{1}&x^{j-1}_{2}&\ldots&x^{j-1}_{n}\\ x^{j+1}_{1}&x^{j+1}_{2}&\ldots&x^{j+1}_{n}\\ \lx@intercol\dotfill\hfil\hfil\lx@intercol\\ x^{l+1}_{1}&x^{l+1}_{2}&\ldots&x^{l+1}_{n}\end{array}\right) and

V_{l}(x_{1},\ldots,x_{n})=\left(\begin{array}[]{cccc}1&1&\ldots&1\\ x_{1}&x_{2}&\ldots&x_{n}\\ {x}^{2}_{1}&{x}^{2}_{2}&\ldots&{x}^{2}_{n}\\ \lx@intercol\dotfill\hfil\hfil\lx@intercol\\ x^{l-1}_{1}&x^{l-1}_{2}&\ldots&x^{l-1}_{n}\end{array}\right) for a vector $(x_{1},\ldots,x_{n})\in\mathbb{F}^{n}_{q}$ and $0\leq i<j\leq l+1$.

Lemma 1**.**

For a vector $(x_{1},\ldots,x_{n})\in\mathbb{F}^{n}_{q}$ and $1\leq i<j\leq n+1$, we have

[TABLE]

and so the determinant is not [math] if and only if the $x_{i}$ are distinct and $\sigma_{n-i,n-j}(x_{1}$, $\ldots$, $x_{n})\neq 0$.

Proof. Consider the following determinant

[TABLE]

So $(-1)^{i+j+1}|M_{n,i,j}(x_{1},\ldots,x_{n})|$ is the coefficient of $x^{i}y^{j}$ in $|V_{n+2}(x_{1},\ldots,x_{n}$, $x,y)|$, i.e., $(-1)^{i+j+1}\sigma_{n-i,n-j}(x_{1},\ldots,x_{n})\prod_{1\leq k<l\leq n}(x_{l}-x_{k})$. $\Box$

If $j=n+1$, it is reduced to the case of just deleting $x^{i}$ and so we get the following corollary.

Corollary 1**.**

For any $i$ with $1\leq i\leq n$, we have

[TABLE]

Lemma 2**.**

Let $q>n+1$. For any integer $i$, $0\leq i\leq n$, there are distinct $x_{1},\ldots,x_{n}\in\mathbb{F}_{q}$ such that $\sigma_{n-i}(x_{1},\ldots,x_{n})\neq 0$.

Proof. We proceed by induction on $n-i$. If $n-i=0$, that is, $i=n$, the result holds obviously since $\sigma_{0}(x_{1},\ldots,x_{n})=1$. Suppose that the result holds for $n-i-1$. Since $\sigma_{n-i}(x_{1},\ldots,x_{n})=x_{n}\sigma_{n-i-1}(x_{1},\ldots$, $x_{n-1})+\sigma_{n-i}(x_{1},\ldots,x_{n-1})$, by induction hypothesis, there are distinct $x_{1},\ldots,x_{n-1}$ such that $\sigma_{n-i-1}(x_{1},\ldots,x_{n-1})\neq 0$. Because of $q>n+1$, one can choose $x_{n}$ different from all other $x_{i}$, $1\leq i\leq n-1$, such that $\sigma_{n-i}(x_{1},\ldots,x_{n})\neq 0$. $\Box$

Lemma 3**.**

Let $q>n+1$. For any integers $i,j$, $0\leq i<j\leq n+1$, there are distinct $x_{1},\ldots,x_{n}\in\mathbb{F}_{q}$ such that $\sigma_{n-i,n-j}(x_{1},\ldots,x_{n})\neq 0$.

Proof. By Lemma 2, the result holds for only induction on $n-i$ or $n-j$. Then we proceed by induction on both $n-i$ and $n-j$. If $n-i=0$ and $n-j=-1$, we have $\sigma_{0,-1}(x_{1},\ldots,x_{n})=\sigma_{0}(x_{1},\ldots,x_{n})\sigma_{0}(x_{1},\ldots,x_{n})-\sigma_{1}(x_{1},\ldots,x_{n})\sigma_{-1}(x_{1},\ldots,x_{n})$ $=1$ and so the result holds . Suppose that the result holds for any $n-j-1$ and $n-i-1$. By Eq. (2), we have

[TABLE]

By induction hypothesis, there are distinct $x_{1},\ldots,x_{n-1}\in\mathbb{F}_{q}$ such that

[TABLE]

and so one can choose $x_{n}$ not only different from $x_{1}$, $\ldots,x_{n-1}$ but also different from any roots of the above quadratic Eq. (3) since $q>n+1$. $\Box$

Lemma 4**.**

Let $q>n+1$. For any fixed $x_{1}\in\mathbb{F}_{q}$, any integer $0\leq i<j\leq n+1$, there are distinct $x_{2},\ldots,x_{n}\in\mathbb{F}_{q}$, which are also different from $x_{1}$, such that $\sigma_{n-i,n-j}(x_{1},\ldots,x_{n})\neq 0$.

Proof. We also proceed by induction on both $n-i$ and $n-j$. If $n-i=0$ and $n-j=-1$, we have $\sigma_{0,-1}(x_{1},\ldots,x_{n})=1$ and so the result holds clearly. Suppose the result holds for $n-j-1$ and $n-i-1$ for fixed $x_{1}$. By Eq. (2), we have

[TABLE]

By induction hypothesis, there are distinct $x_{2},\ldots,x_{n-1}$, which is also different from $x_{1}$, such that $\sigma_{n-i-1,n-j-1}(x_{1},\ldots$, $x_{n-1})\neq 0$ and so one can choose $x_{n}$ not only different from $x_{1}$, $x_{2}$, $\ldots,x_{n-1}$ but also different from any roots of the above quadratic Eq. (4) since $q>n+1$. $\Box$

Theorem 1**.**

If $1\leq m<q-2$, the diameter of the jumped Wenger graph $J_{m}(q,i,j)$ for any integers $i,j$, $1\leq i<j\leq m+1$, is at most $2(m+1)$.

Proof. First, we consider the distance between any two vertices $L$ and $L^{\prime}$ in $\mathfrak{L}$ of the jumped Wenger graph $J_{m}(q,i,j)$. If $L_{1}P_{1}\ldots P_{s}L_{s+1}$ is a path in $J_{m}(q,i,j)$ between $L=L_{1}$ and $L^{\prime}=L_{s+1}$, where $L_{h}=[l_{1}^{(h)},\cdots,l_{m+1}^{(h)}]$ and $P_{h}=(p_{1}^{(h)},\cdots,p_{m+1}^{(h)})$, we have

[TABLE]

Therefore there are elements $t_{h}=l_{1}^{(h+1)}-l_{1}^{(h)}$, $x_{h}=p_{1}^{(h)}\in\mathbb{F}_{q}$, $1\leq h\leq s$, such that

[TABLE]

In addition, by Fact 1 and 2, $x_{1},\ldots,x_{s}$ should be distinct and $t_{i}\neq 0$ for all $i$, $1\leq i\leq s$.

For any integers $i,j$, $1\leq i<j\leq m+2$, take $s=m+1$ in Eq. (2). By Lemma 1 and 3, there are distinct $x_{1},\ldots,x_{m+1}\in\mathbb{F}_{q}$ such that $|M_{m+1,i,j}(x_{1},\ldots,x_{m+1})|\neq 0$, and thus Eq. (5) has a unique solution for $t_{1},t_{2},\ldots,t_{s}$. If some $t_{i}=0$, then $P_{i}$ and $L_{i+1}$ can be deleted from the path. Thus the distance of any two vertices in $\mathfrak{L}$ is at most $2(m+1)$.

Then, consider any vertices $P$ and $P^{\prime}$ in $\mathfrak{P}$. Let $P_{1}L_{1}\ldots L_{s}P_{s+1}$ be a path in $J_{m}(q,i,j)$ between $P=P_{1}$ and $P^{\prime}=P_{s+1}$, where $L_{h}=[l^{(h)}_{1},\ldots,l^{(h)}_{m+1}]$ and $P_{h}=(p^{(h)}_{1},\ldots,p^{(h)}_{m+1})$. Therefore, we have

[TABLE]

where $x_{h}=p_{1}^{(h)}$, $1\leq h\leq s$, $t_{h}=l_{1}^{(h)}-l_{1}^{(h+1)}$, $h=2,\ldots,s-1$, $t_{1}=-l^{(1)}_{1}$, $t_{s}=l^{(s)}_{1}$. Similarly, take $s=m+1$. By Lemma 1 and Lemma 3, there are distinct $x_{1},\ldots,x_{m+1}\in\mathbb{F}_{q}$ such that $|M_{m+1,i,j}(x_{1},\ldots,x_{m+1})|\neq 0$, and thus Eq. (6) has a unique solution for $t_{1},t_{2},\ldots,t_{s}$. Thus the distance of any two vertices in $\mathfrak{P}$ is at most $2(m+1)$.

Next, we consider the distance between a vertex $P=(p_{1},\ldots,p_{m+1})\in\mathfrak{P}$ and a vertex $L\in\mathfrak{L}$. First we choose a line $L_{1}\in\mathfrak{L}$ such that it is adjacent to $P$. From the earlier discussion, there exists a path from $L_{1}$ to $L=L_{s+1}$ with distance at most $2(m+1)$. We modify the earlier construction so that the path goes through the vertex $P$. Namely, In Eq. (5), we let $x_{1}=p_{1}$ and choose distinct $x_{2},\ldots,x_{n}$ so that $|M_{m+1,i,j}(x_{1},\ldots,x_{m+1})|\neq 0$ by Lemma 1 and 4 . Then there is a unique solution $t_{1},\ldots,t_{m+1}$ and so there is a path between $L_{1}$ and $L$ with length at most $2(m+1)$ passing through $P$ by Fact 1. Therefore the distance of $P$ and $L$ is less than or equal to $2(m+1)$.

Finally, we obtain the same result for the distance between a line $L$ and a point $P$. $\Box$

Theorem 2**.**

If $(i,j)=(m,m+2),(m+1,m+2),(m,m+1)$, then the diameter of $J_{m}(q,i,j)$ is $2(m+1)$.

Proof. By Theorem 1, it suffices to show that there are $L$ and $L^{\prime}$ in $\mathfrak{L}$ with $L-L^{\prime}=[0,\ldots,0,1]$ such that the distance between them is at least $2(m+1)$. Otherwise, there is a shortest path $LP_{1}\ldots P_{s}L^{\prime}$ between $L$ and $L^{\prime}$ with $s\leq m$, where $P_{h}=(p^{(h)}_{1},\ldots,p^{(h)}_{m+1})$ and $x_{h}=p^{(h)}_{1}$ for $1\leq h\leq s$. The Eq. (5) is easily reduced to

[TABLE]

where $s^{\prime}\leq s\leq m$ and $x^{\prime}_{1},\ldots,x^{\prime}_{s^{\prime}}$ are distinct. Since $(i,j)=(m,m+2),(m+1,m+2),(m,m+1)$, the first $s^{\prime}$ rows of $M_{s^{\prime},i,j}(x^{\prime}_{1},\ldots,x^{\prime}_{s^{\prime}})$ are Vandermond determinant and so $t_{i}=0$ for $1\leq i\leq s^{\prime}$, which contradicts with $t^{\prime}_{1}(x^{\prime}_{1})^{m+1}+\ldots+t_{s^{\prime}}(x_{s^{\prime}})^{m+1}=1$, $t^{\prime}_{1}(x^{\prime}_{1})^{m}+\ldots+t_{s^{\prime}}(x_{s^{\prime}})^{m}=1$, or $t^{\prime}_{1}(x^{\prime}_{1})^{m+2}+\ldots+t_{s^{\prime}}(x_{s^{\prime}})^{m+2}=1$, respectively. $\Box$

3 The girth of jumped Wenger graphs

In graph theory, the girth of a graph is the length of a shortest cycle contained in the graph. In [3, 7], it is proved that the Wenger graphs have girth 8 or $6$ up to the choices of $q$ and $m$. In [2] authors proved that the linearized Wenger graphs have girth $8$ or $6$ depending on the choices of $q$ and $m$. In this section we consider the girth of the jumped Wenger graphs.

It is no harm to assume that $L_{1}P_{1}L_{2}\ldots P_{s}L_{1}$ is a cycle of length $2s$, where $P_{h}=(p^{(h)}_{1},\ldots$, $p^{(h)}_{m+1})$ and $L_{h}=[l^{(h)}_{1},\ldots,l^{(h)}_{m+1}]$, $h=1,\ldots,s$.

For those $L_{i}$, we have

[TABLE]

where $t_{h}=l^{(h+1)}_{1}-l^{(h)}_{1}$, for $h=1,\ldots,s-1$, $t_{s}=l^{(1)}_{1}-l^{(s)}_{1}$.

Therefore we have

[TABLE]

Again according to Fact 1 and 2, if $L_{1}P_{1}\ldots P_{s}L_{1}$ is a cycle of length $2s$, then $t_{h}\neq 0$ for all $h=1,\ldots,s$, $p^{(h)}_{1}-p^{(h+1)}_{1}\neq 0$ for $h=1,\ldots,s-1$ and $p^{(s)}_{1}-p^{(1)}_{1}\neq 0$.

In this paper we consider $(1,f_{2}(x),\ldots,f_{m+1}(x))=(1,x,x^{2},\ldots,x^{i-1}$, $x^{i+1},\ldots$, $x^{j-1},x^{j+1},\ldots,x^{m+2})$ and (8) can be written as

[TABLE]

First we give an upper bound for the girth of the jumped Wenger graphs.

Theorem 3**.**

For any integers $i,j$, $1\leq i<j\leq m+2$, the girth of the jumped Wenger graph $J_{m}(q,i,j)$ is less than or equal to $8$.

Proof. It suffices to find a cycle of length $8$ in any graph $J_{m}(q,i,j)$ for $1\leq i<j\leq m+2$. Putting $s=4$, $p^{(1)}_{1}=p^{(3)}_{1}=0$, $p^{(2)}_{1}=p^{(4)}_{1}=1$ in (9), the rank of $M_{m+1,i,j}(0,1,0,1)$ is $2$ and $t_{1}+t_{3}=0$ and $t_{2}+t_{4}=0$. Let $l^{(1)}_{1}=0$, $l^{2)}_{1}=1$, $l^{(3)}_{1}=0$, $l^{(4)}_{1}=1$ and $P_{1}=(0,\ldots,0)$. A cycle $L_{1}P_{1}L_{2}P_{2}L_{3}P_{3}L_{4}P_{4}L_{1}$, that is,

[TABLE]

is required. $\Box$

Next, we determine the exact girth case by case.

Theorem 4**.**

If $m=1$, the girth of the jumped Wenger graph $J_{m}(q,i,j)$, $1\leq i<j\leq m+2$, is $4$ in the following cases :

(a) $(i,j)=(1,3)\mbox{ and }char(q)\neq 2$;

(b) $(i,j)=(1,2)\mbox{ and }3|(q-1)$.

For $m=1$, the girth of the jumped Wenger graph $J_{m}(q,i,j)$, $1\leq i<j\leq m+2$, is $6$ in the following cases:

(a) $(i,j)=(1,3)\mbox{ and }p=char(q)=2$;

(b) $(i,j)=(1,2)\mbox{ and }3\nmid(q-1)$;

(c) $(i,j)=(2,3)$ and $q\neq 2,3$.

The girth of $J_{2}(q,2,3)$ if $q=2$ or $3$ is $8$.

The proof is straightforward and so is omitted here.

Lemma 5** ([6], Lemma 6.24).**

For odd q, let $b\in\mathbb{F}_{q}$, $a_{1},a_{2}\in\mathbb{F}^{*}_{q}$, and $\eta$ be the quadratic character of $\mathbb{F}_{q}$. Then the number of solutions of the equation

[TABLE]

is $q+v(b)\eta(-a_{1}a_{2})$, where $v(b)=-1$ for $b\in\mathbb{F}^{*}_{q}$ and $v(0)=q-1$.

Lemma 6** ([6], Theorem 6.32 ).**

Let $\mathbb{F}_{q}$ be a finite field with q even and let $b\in\mathbb{F}_{q}$. Then for odd $n$, the number of solutions of the equation

[TABLE]

in $\mathbb{F}^{n}_{q}$ is $q^{n-1}$. For even $n$, the number of solutions of the equation

[TABLE]

in $\mathbb{F}^{n}_{q}$ is $q^{n-1}+v(b)q^{(n-2)/2}$. For even $n$ and $a\in\mathbb{F}_{q}$ with $tr_{\mathbb{F}_{q}/\mathbb{F}_{p}}(a)=1$, the number of solutions of the equation

[TABLE]

in $\mathbb{F}^{n}_{q}$ is $q^{n-1}-v(b)q^{(n-2)/2}$. The definition of $v(b)$ is the same as Lemma 5.

Lemma 7**.**

For distinct $a,b,c\in\mathbb{F}_{q}$, the system of equations

[TABLE]

has solution $(t_{1},t_{2},t_{3})$ and $t_{i}\neq 0$ for all $i$, $i=1,2,3$.

The proof is trivial and so omitted here.

Lemma 8**.**

There are distinct $a,b$ and $c\in\mathbb{F}_{q}$ such that $|M_{3,2,3}(a,b,c)|=0$ if and only if either $q=p^{l}$ with $p\neq 2$ and $q\neq 3$, or $q=2^{2l}$, where $l$ is a positive integer.

Proof. First, if $3|\,(q-1)$, i.e., $q=p^{2l}$ for a positive integer $l$, then let $a=1,b=g^{\frac{q-1}{3}},c=g^{\frac{2(q-1)}{3}}$, where $g$ is a primitive element in $\mathbb{F}_{q}$, and we are done.

Next, we consider the case $3\nmid\,(q-1)$. If $a=0$ then $(bc^{-1})^{3}=1$, which implies $b=c$. So we can assume that $a\neq 0$ and so we get

[TABLE]

If $|M_{3,2,3}(a,b,c)|=0$, we obtain

[TABLE]

for some $t\in\mathbb{F}^{*}_{q}$ and hence $a^{-1}b$ and $a^{-1}c$ are different roots of the equation $x^{3}+x^{2}+x+t+1=0$.

Suppose that $x_{1},x_{2}$ and $x_{3}$ are roots of $x^{3}+x^{2}+x+t+1=0$ and so we have

[TABLE]

Substitute $x_{3}$ in Eq. (10) by $x_{3}=-1-x_{1}-x_{2}$, and so we get

[TABLE]

Denote the set of the solutions $(x_{1},x_{2})$ of Eq. (11) by $X$ and let

[TABLE]

Let us check whether $S$ is null. Consider three cases:

Case 1: $p\neq 2,3$ and $q=p^{2l-1}$ where $l$ is a positive integer.

After multiplying $4$ on both sides of Eq. (11) and putting $y_{1}=2x_{1}+x_{2}$ and $y_{2}=x_{2}$, Eq. (11) becomes

[TABLE]

Multiplying $3$ on both sides of Eq. (13) and putting $z_{1}=y_{1}+1$ and $z_{2}=3y_{2}+1$, Eq. (13) becomes

[TABLE]

By Lemma 5, the number of roots of Eq. (14) is $q+v(-8)\eta(-3)$ and so $|X|\geq q-1$.

If $x_{1}=x_{2}$, then Eq. (11) becomes

[TABLE]

Thus Eq. (15) has at most two roots. Furthermore, it is required that both $x_{1}\neq 1$ and $x_{2}\neq 1$. If $x_{1}=1$, Eq. (11) becomes $(x_{2}+1)^{2}=-2$, which has at most two roots in $\mathbb{F}_{q}$. The same result is for $x_{2}=1$. However, $(x_{1},x_{2})=(1,1)\notin X$.

In addition, it is required that both $x_{1}$ and $x_{2}$ are not zero. If $x_{1}=0$, Eq. (11) becomes $x^{2}_{2}+x_{2}+1=0$ and so it has two roots at most. The result is the same if $x_{2}=0$. In addition, $(x_{1},x_{2})=(0,0)\notin X$.

Thus, $|S|\geq q-1-2-4-4=q-11$. So the result holds for any $q>11$. In addition, Putting $a=1$, $b=2$ and $c=3$ if $q=5$, $a=1$, $b=2$ and $c=-3$ if $q=7$ and $a=-1$, $b=2$ and $c=5$ if $q=11$, we get $|M_{3,2,3}(a,b,c)|=0$.

Case 2. $q=2^{2l-1}$, where $l$ is a positive integer.

Putting $y_{1}=x_{1}+1$ and $y_{2}=x_{2}+1$, Eq. (11) becomes

[TABLE]

According to Lemma 6, the number of solutions of Eq. (16) in $\mathbb{F}^{2}_{q}$ is $q-v(0)q^{0}=1$ and so Eq. (11) has only one solution, that is, $x_{1}=1$ and $x_{2}=1$. Therefore, there are no distinct $a$, $b$ and $c$ such that $|M_{3,2,3}(a,b,c)|=0$.

Case 3. $q=3^{2l-1}$, where $l$ is a positive integer.

Multiply $4$ on both sides of Eq. (11) and put $y_{1}=2x_{1}+x_{2}$ and $y_{2}=x_{2}$, and so Eq. (11) becomes

[TABLE]

The number of the solutions of Eq. (17) is $q$ and so $|X|=q$.

If $x_{1}=0$, Eq. (11) becomes $(2x_{2}+1)^{2}=0$ and so $x_{2}=1$. Similarly, we have $(1,0)\in X$, but $(0,0)\notin X$. If $x_{1}=1$, Eq. (11) becomes $x^{2}_{2}+2x_{2}=0$ and so $x_{2}=0$, or $x_{2}=1$. Thus, $(1,1)\in X$. Therefore, $|S|\geq q-2-1=q-3$ and so the result holds for $q>3$.

In addition, it is east to check that there are no distinct $a$, $b$ and $c$ in $\mathbb{F}_{q}$ such that $|M_{3,2,3}(a,b,c)|=0$ if $q=3$.

So the result is established in all cases. $\Box$

Lemma 9**.**

There are distinct $a,b$ and $c\in\mathbb{F}_{q}$ such that $|M_{3,1,2}(a,b,c)|=0$ if and only if either $q=p^{l}$ with $p\neq 2$ and $q\neq 3$, or $q=2^{2l}$, where $l$ is a positive integer.

Proof. The result holds for $3|(q-1)$ in the same way as the proof of Lemma 8. Similarly, assume that $a\neq 0$ and so we obtain

[TABLE]

We have

[TABLE]

So $a^{-1}b$ and $a^{-1}c$ are the solutions of the equation $x^{3}+(1-t)x^{2}+(1-t)x+(1-t)=0$. Thus $ab^{-1}$ and $ac^{-1}$ are roots of the equation $y^{3}+y^{2}+y+t^{\prime}=0$. Similar to the proof in Lemma 8, the result is acquired. $\Box$

Lemma 10**.**

There are distinct $a,b,c\in\mathbb{F}_{q}$ such that $M_{3,1,3}(a^{2},b^{2},c^{2})=0$ if and only if $p$ is odd.

The proof is obvious and so is omitted here.

By Lemma 7, 8, 9, 10 and Theorem 3, we obtain the following result.

Theorem 5**.**

If $m=2$, the girth of the Wenger graph $J_{m}(q,i,j)$, $1\leq i<j\leq m+2$, is $6$ under the following cases:

(a) $(i,j)=(1,2)$ , $q\neq 2^{2l-1}$ and $q\neq 3$ with a positive integer $l$;

(b) $(i,j)=(2,3)$, $q\neq 2^{2l-1}$ and $q\neq 3$ with a positive integer $l$;

(c) $(i,j)=(1,4)$ and $q\neq 2,3,5$;

(d) $(i,j)=(2,4)$ and $q\neq 2$.

If $m=2$, the girth of $J_{m}(1,3)$ with $char(q)\neq 2$ is 4.

If $m=2$, the girth of $J_{m}(q,i,j)$, $1\leq i<j\leq m+2$, is $8$ under the following cases:

(a) $(i,j)=(1,2)$ and $q=2^{2l-1}$ with a positive integer $l$ or $q=3$;

(b) $(i,j)=(2,3)$ and $q=2^{2l-1}$ with a positive integer $l$ or $q=3$;

(c) $(i,j)=(1,3)$ and $char(q)=2$.

(d) $(i,j)=(1,4)$ and $q=2,3,5$

(e) $(i,j)=(2,4)$ and $q=2$.

Lemma 11**.**

For $m=3,4,5$, there are distinct $a,b,c\in\mathbb{F}_{q}$ such that the rank of $M_{m+1,i,j}(a,b,c)$ is less than $3$ if and only if one of the following conditions holds:

(i) $(i,j)=(1,4)$ and $3|\,(q-1)$.

(ii) $(i,j)=(2,5)$ and $3|\,(q-1)$.

The proof is clear and so is omitted here.

Theorem 6**.**

For $m\geq 3$, the girth of the Wenger graph $J_{m}(q,i,j)$, $1\leq i<j\leq m+2$, is $8$ except the following cases:

(a) $m=3,4,5$, $3|\,(q-1)$ and $(i,j)=(1,4)$.

(b) $m=3,4,5$, $3|\,(q-1)$ and $(i,j)=(2,5)$.

(b) $m=6$, $3|(q-1)$ and $(i,j)=(2,5)$.

Furthermore, the girth of such graphs is $6$ in the above cases.

**Proof. ** If $3|(q-1)$, put $a=1$, $b=g^{\frac{q-1}{3}}$, and $c=g^{\frac{2(q-1)}{3}}$ where $g$ is a primitive element in $\mathbb{F}_{q}$. So the rank of $M_{m+1,i,j}(a,b,c)$ is less than $3$ in those cases by Lemma 11 and so the girth of those graphs is $6$ by Lemma 7. For other cases, the rank of the matrix $M_{m+1,i,j}(a,b,c)$ is $3$ and so the girth is $8$ by Theorem 3. $\Box$

4 Conclusions

In this paper we consider the diameter and girth of jumped Wenger graphs, whose functions jumps at two places. Using our similar method, we can get similar results for those graphs with more jumped places. In addition, determination of the exact diameter of any jumped Wenger graphs is our future task.