Complete weight enumerators of a class of linear codes

Abstract

Let $\mathbb {F}_{q}$ be the finite field with $q=p^{m}$ elements, where p is an odd prime and m is a positive integer. For a positive integer t, let $D\subset \mathbb {F}^{t}_{q}$ and let ${\mathrm {Tr}}_{m}$ be the trace function from $\mathbb {F}_{q}$ onto $\mathbb {F}_{p}$. In this paper, let $D=\{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=0\},$ we define a p-ary linear code $\mathcal {C}_{D}$ by

$$\begin{aligned} \mathcal {C}_{D}=\{\mathbf {c}(a_{1},a_{2},\ldots ,a_{t}) : (a_{1},a_{2},\ldots ,a_{t})\in \mathbb {F}^{t}_{q}\}, \end{aligned}$$

where

$$\begin{aligned} \mathbf {c}(a_{1},a_{2},\ldots ,a_{t})=({\mathrm {Tr}}_{m}(a_{1}x^{2}_{1}+a_{2}x^{2}_{2}+\cdots +a_{t}x^{2}_{t}))_{(x_{1},x_{2},\ldots ,x_{t}) \in D}. \end{aligned}$$

We shall present the complete weight enumerators of the linear codes $\mathcal {C}_{D}$ and give several classes of linear codes with a few weights. This paper generalizes the results of Yang and Yao (Des Codes Cryptogr, 2016).

1 Introduction

Let $\mathbb {F}_{p}$ be the finite field with p elements where p is an odd prime. An [n, k, d] linear code $\mathcal {C}$ over $\mathbb {F}_{p}$ is k-dimensional subspace of $\mathbb {F}^{n}_{p}$ with minimum distance d. We recall the definition of the complete weight enumerator of linear code [14]. Suppose that the elements of $\mathbb {F}_{q}$ are $w_{0}=0,w_{1},\ldots ,w_{q-1}$, which are listed in some fixed order. The composition of a vector $\mathbf v =(v_{0},v_{1},\ldots ,v_{n-1})\in \mathbb {F}_{q}^{n}$ is defined to be comp(v)=$(t_{0},t_{1},\ldots ,t_{q-1}),$ where each $t_{i}=t_{i}(\mathbf v )$ is the number of components $v_{j}(0\leqslant j\leqslant n-1)$ of v that equal to $w_{i}$. Clearly, we have

$$\begin{aligned} \displaystyle \sum _{i=0}^{q-1}t_{i}=n. \end{aligned}$$

Definition 1.1

Let $\textit{C}$ be an [n, k] linear code over $\mathbb {F}_{q}$ and let $A(t_{0},t_{1}\ldots ,t_{q-1})$ be the number of codewords $\mathbf{c }\in \textit{C}$ with comp(c)=$(t_{0},t_{1},\ldots ,t_{q-1}).$ Then the complete weight enumerator of C is defined to be the polynomial

$$\begin{aligned} W_{\textit{C}}&=\sum _\mathbf{c \in \textit{C}}z_{0}^{t_{0}}z_{1}^{t_{1}}\cdots z_{q-1}^{t_{q-1}}\\&=\sum _{(t_{0},t_{1},\ldots ,t_{q-1})\in B_{n}}A(t_{0},t_{1},\ldots ,t_{q-1})z_{0}^{t_{0}}z_{1}^{t_{1}}\cdots z_{q-1}^{t_{q-1}}, \end{aligned}$$

$\mathrm{where}\;B_{n}=\lbrace (t_{0},t_{1},\ldots ,t_{q-1}):0\leqslant t_{i}\leqslant n, \displaystyle \sum _{i=0}^{q-1}t_{i}=n\rbrace .$

Recently, linear codes with a few weights have been investigated [1, 6–10, 13, 16, 19] by using exponential sums in some cases. They may have many applications in association schemes [3], strongly regular graphs [4], and secret sharing schemes [5, 18]. In addition, the complete weight enumerators of linear codes over finite fields can be applied to compute the deception probabilities of certain authentication codes constructed from linear codes [11, 12]. We begin to recall a class of two-weight and three-weight linear codes which were proposed by Ding and Ding [9]. Let $D=\{x \in \mathbb {F}^{*}_{q} : {\mathrm {Tr}}_{m}(x^{2})=0\},$ where ${\mathrm {Tr}}_{m}$ is the trace function from $\mathbb {F}_{q}$ onto $\mathbb {F}_{p}$. Then a linear code of length $n=|D|$ over $\mathbb {F}_{p}$ can be defined by

$$\begin{aligned} {\mathcal {C}}_{D}=\{\mathbf{c}(a)=({\mathrm {Tr}}_{m}(ax))_{\in D} : a\in \mathbb {F}_{q}\}. \end{aligned}$$

It was proved that $\mathcal {C}_{D}$ is a two-weight code if m is even and a three-weight code if m is odd. Motivated by the results given in [9], Bae et al. gave a generalization of Ding and Ding’s case [1]. Let $D = \{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}^{2}+x_{2}^{2}+\cdots +x_{t}^{2})=0\}.$ They define a p-ary linear code $\mathcal {C}_{D}$ by

$$\begin{aligned} {\mathcal {C}}_{D}=\{{\mathbf{c}}(a_{1},a_{2},\ldots ,a_{t}) : a_{1},a_{2},\ldots ,a_{t}\in \mathbb {F}_{p^{m}}\}, \end{aligned}$$

where

$$\begin{aligned} \mathbf c (a_{1},a_{2},\ldots ,a_{t})=({\mathrm {Tr}}_{m}(a_{1}x_{1}+a_{2}x_{2}+\cdots +a_{t}x_{t}))_{(x_{1},x_{2},\ldots ,x_{t})\in D.} \end{aligned}$$

It was also shown that $\mathcal {C}_{D}$ is two-weight if tm is even and three-weight if tm is odd. If $D=\{x \in \mathbb {F}^{*}_{q} : {\mathrm {Tr}}_{m}(x)=0\}$ and $\mathcal {C}_{D}=\{{\mathrm {Tr}}_{m}(ax^{2})_{x\in D} : a\in \mathbb {F}_{q}\},$ Yang and Yao [17] determined the complete weight enumerators of $\mathcal {C}_{D}$. In this paper, let $D = \{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=0\}.$ We define a p-ary linear code $\mathcal {C}_{D}$ by

$$\begin{aligned} \mathcal {C}_{D}=\{\mathbf {c}(a_{1},a_{2},\ldots ,a_{t}) : (a_{1},a_{2},\ldots ,a_{t})\in \mathbb {F}^{t}_{q}\}, \end{aligned}$$

(1)

where

$$\begin{aligned} \mathbf {c}(a_{1},a_{2},\ldots ,a_{t})=({\mathrm {Tr}}_{m}(a_{1}x^{2}_{1}+a_{2}x^{2}_{2}+\cdots +a_{t}x^{2}_{t}))_{(x_{1},x_{2},\ldots ,x_{t}) \in D}. \end{aligned}$$

(2)

We shall present the complete weight enumerators of this class of linear codes and get several linear codes with a few weights. In addition, this paper generalizes the results of Yang and Yao [17].

2 Preliminaries

Let p be an odd prime and $q=p^{m}$ for a positive integer m. For any $a\in \mathbb {F}_{q}$, we can define an additive character of the finite field $\mathbb {F}_{q}$ as follows:

$$\begin{aligned} \displaystyle \psi _{a}:\mathbb {F}_{q} \longrightarrow \mathbb {C}^{*}, \psi _{a}(x)=\zeta _{p}^{{\mathrm {Tr}}_{m}(ax)}, \end{aligned}$$

where $\zeta _p=e^{\frac{2\pi \sqrt{-1}}{p}}$ is a p-th primitive root of unity. It is clear that $\psi _0(x)=1$ for all $x \in \mathbb {F}_{q}$. Then $\psi _0$ is called the trivial additive character of $\mathbb {F}_{q}$. If $a=1$, we call $\psi :=\psi _1$ the canonical additive character of $\mathbb {F}_{q}$. It is easy to see that $\psi _a(x)=\psi (ax)$ for all $a, x \in \mathbb {F}_q$. The orthogonal property of additive characters which can be found in [14] is given by

$$\begin{aligned} \displaystyle \sum _{x\in \mathbb {F}_{q}}\psi _{a}(x) =&\left\{ \begin{array}{ll} q, &{} \text {if } a=0,\\ 0, &{} \text {if } a\in \mathbb {F}_{q}^{*}. \end{array} \right. \end{aligned}$$

Let $\lambda : \mathbb {F}_q^* \rightarrow \mathbb {C}^*$ be a multiplicative character of $\mathbb {F}_q^*$. Now we define the Gauss sum over $\mathbb {F}_{q}$ by

$$\begin{aligned} G(\lambda )=\sum _{x \in \mathbb {F}_{q}^{*}} \lambda (x) \psi (x). \end{aligned}$$

Let $q-1=sN$ for two positive integers $s>1$, $N>1$ and $\alpha$ be a fixed primitive element of $\mathbb {F}_{q}$. Let $\langle \alpha ^{N} \rangle$ denote the subgroup of $\mathbb {F}_{q}^{*}$ generated by $\alpha ^{N}$. The cyclotomic classes of order N in $\mathbb {F}_{q}$ are the cosets $C_{i}^{(N,q)}=\alpha ^{i}\langle \alpha ^{N}\rangle$ for $i=0,1,\ldots , N-1.$ We know that $|C_{i}^{(N,q)}|=\frac{q-1}{N}$. The Gaussian periods of order N are defined by

$$\begin{aligned} \displaystyle \eta _{i}^{(N,q)}=\sum _{x\in C_{i}^{(N,q)}}\psi (x). \end{aligned}$$

Lemma 2.1

[2, 14] Suppose that $q=p^m$ and $\eta$ is the quadratic character of $\mathbb {F}_{q}^{*}$ where p is an odd prime and $m \ge 1$. Then

$$\begin{aligned} \displaystyle G(\eta )=(-1)^{m-1} \sqrt{(p^*)^m}=\left\{ \begin{array}{ll} (-1)^{m-1} \sqrt{q}, &{} \text{ if } p \equiv 1 \pmod 4, \\ (-1)^{m-1} (\sqrt{-1})^m \sqrt{q}, &{} \text{ if } p \equiv 3 \pmod 4, \end{array}\right. \end{aligned}$$

where $p^*=\big (\frac{-1}{p}\big )p=(-1)^{\frac{p-1}{2}}p$.

Lemma 2.2

[14] If q is odd and $f(x)=a_{2}x^{2}+a_{1}x+a_{0}\in \mathbb {F}_{q}[x]$ with $a_{2}\ne 0,$ then

$$\begin{aligned} \displaystyle \sum _{x\in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(f(x))}=\zeta _p^{{\mathrm {Tr}}_{m}(a_{0}-a_{1}^2(4a_{2})^{-1})}\eta (a_{2})G(\eta ), \end{aligned}$$

where $\eta$ is the quadratic character of $\mathbb {F}_{q}^{*}$.

Lemma 2.3

[15] When $N=2$, the Gaussian periods are given by

$$\begin{aligned} \displaystyle \eta ^{(2,q)}_{0}=&\left\{ \begin{array}{ll} \frac{-1+(-1)^{m-1}}{2}\sqrt{q}, &{}\quad \text { if } p\equiv 1\pmod {4},\\ \frac{-1+(-1)^{m-1}(\sqrt{-1})^{m}\sqrt{q}}{2}, &{} \quad \text {if} p\equiv 3\pmod {4}, \end{array} \right. \end{aligned}$$

and $\eta ^{(2,q)}_{1}=-1-\eta ^{(2,q)}_{0}$.

3 Complete weight enumerators

In this section, we will investigate the complete weight enumerators of the linear codes $\mathcal {C}_{D}$ defined by (1) and (2), where

$$\begin{aligned} D=\{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=0\}. \end{aligned}$$

Let $\eta _{p}$ be the quadratic character of $\mathbb {F}^{*}_{p}$ and let $G(\eta _{p})$ denote the quadratic Gauss sum over $\mathbb {F}_{p}$. For $z\in \mathbb {F}^{*}_{p}$, it is easily checked that $\eta (z)=\eta _{p}(z)$ if m is odd and $\eta (z)=1$ if m is even, where $\eta$ is the quadratic character of $\mathbb {F}^{*}_{q}$ (see [9]). Since the trace function is balanced, we have the following lemma.

Lemma 3.1

Denote $n_{c}=|\{(x_{1}, x_{2}, \ldots , x_{t}) \in \mathbb {F}^{t}_{q} :{\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=c\}|$ for each $c \in \mathbb {F}_{p}$, then $n_{c}=p^{tm-1}.$

By Lemma 3.1 it is easy to see that the length of $\mathcal {C}_{D}$ is $n=n_{0}-1=p^{tm-1}-1.$ For a codeword $\mathbf {c}(a_{1},a_{2},\ldots ,a_{t})$ of $\mathcal {C}_{D}$ and $\rho \in \mathbb {F}_{p}^{*},$ let $N_{\rho }:=N_{\rho }(a_{1},a_{2},\ldots ,a_{t})$ be the number of components ${\mathrm {Tr}}_{m}(a_{1}x^{2}_{1}+\cdots +a_{t}x^{2}_{t})$ of $\mathbf {c}(a_{1},\ldots , a_{t})$ which are equal to $\rho$. Then

$$\begin{aligned} N_{\rho }&=\sum _{\begin{array}{c} x_{1}, x_{2}, \ldots , x_{t} \in \mathbb {F}_{q}\\ (x_{1}, x_{2}, \ldots , x_{t})\ne (0,0,\ldots ,0) \end{array}}\Big (\frac{1}{p}\sum _{y \in \mathbb {F}_{p}}\zeta _p^{y{\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})}\Big )\Big (\frac{1}{p}\sum _{z \in \mathbb {F}_{p}}\zeta _p^{z\mathrm {Tr_{m}}(a_{1}x^{2}_{1}+\cdots +a_{t}x^{2}_{t})-z\rho }\Big )\nonumber \\&=\frac{1}{p^{2}}\sum _{x_{1}, x_{2}, \ldots , x_{t} \in \mathbb {F}_{q}}\Big (1+\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _p^{y{\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})}\Big )\Big (1+\sum _{z \in \mathbb {F}^{*}_{p}}\zeta _p^{z{\mathrm {Tr}}_{m}(a_{1}x^{2}_{1}+\cdots +a_{t}x^{2}_{t})-z\rho }\Big )\nonumber \\&=p^{tm-2}+\frac{1}{p^{2}}(\Omega _{1}+\Omega _{2}+\Omega _{3}), \end{aligned}$$

(3)

where

$$\begin{aligned}&\Omega _{1}=\sum _{y \in \mathbb {F}_{p}^{*}}\sum _{x_{1} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(yx_{1})}\sum _{x_{2} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(yx_{2})}\cdots \sum _{x_{t} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(yx_{t})}=0,\\&\Omega _{2}=\sum _{z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\sum _{x_{1} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{1}x^{2}_{1})}\sum _{x_{2} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{2}x^{2}_{2})}\cdots \sum _{x_{t} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{t}x^{2}_{t})},\\&\mathrm {and}\\&\Omega _{3}=\sum _{y,z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\sum _{x_{1} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{1}x^{2}_{1}+yx_{1})}\sum _{x_{2} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{2}x^{2}_{2}+yx_{2})}\cdots \sum _{x_{t} \in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(za_{t}x^{2}_{t}+yx_{t})}. \end{aligned}$$

Lemma 3.2

Suppose that there are exactly k elements $a_{i_{1}},\ldots ,a_{i_{k}}\ne 0$ among $a_{1},\ldots ,a_{t}$ for $1\le k\le t$.

• (1) If m is even, then

  $$\begin{aligned} \displaystyle \Omega _{2}=-q^{t-k}\eta (a_{i_{1}}\cdots a_{i_{k}})G(\eta )^{k}. \end{aligned}$$

• (2) If m is odd, then

  $$\begin{aligned} \Omega _{2}=&\left\{ \begin{array}{ll} -q^{t-k}\eta (a_{i_{1}}\cdots a_{i_{k}})G(\eta )^{k}, &{}\quad k\;\mathrm{is\;even,}\\ q^{t-k}\eta (a_{i_{1}}\cdots a_{i_{k}})G(\eta )^{k}G(\eta _{p})\eta _{p}(-\rho ), &{}\quad k\;\mathrm{is\;odd.} \end{array} \right. \end{aligned}$$

Proof

If $a_{1}=a_{2}=\cdots =a_{t}=0$, then it is easy to see that $\Omega _{2}=-q^{t}$. Otherwise by Lemma 2.2, we get

$$\begin{aligned} \Omega _{2}&=q^{t-k}\sum _{z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta (za_{i_{1}})G(\eta )\eta (za_{i_{2}})G(\eta )\cdots \eta (za_{i_{k}})G(\eta )\\&=q^{t-k}\eta (a_{i_{1}}a_{i_{2}}\cdots a_{i_{k}})G(\eta )^{k}\sum _{z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta (z)^{k}. \end{aligned}$$

If m is even or if m is odd and k is even, then $\eta (z)^{k}=1$. Thus, we get the result.

If m is odd and k is odd, then $\eta _{p}(z)^{k}=\eta _{p}(z)$ and $\displaystyle \sum _{z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta _{p}(z)=G(\eta _{p})\eta _{p}(-\rho ).$ Thus, we get the result. $\square$

To simplify formulas, we denote $A=a_{1}\cdots a_{t}$ and $B=a^{-1}_{1}+\cdots +a^{-1}_{t}$ throughout this paper.

Lemma 3.3

If $a_{1}a_{2}\cdots a_{t}=0$, then $\Omega _{3}=0$. Assume that $a_{1}a_{2}\cdots a_{t}\ne 0$.

• (1) If tm is even, then

  $$\begin{aligned} \displaystyle \Omega _{3}=&\left\{ \begin{array}{ll} -(p-1)G(\eta )^{t}\eta (A), &{} \quad \mathrm{if}\;{\mathrm {Tr}}_{m}(B)=0,\\ G(\eta )^{t}\eta (A)\big (p\eta _{p}(-{\mathrm {Tr}}_{m}(B))\eta _{p}(\rho )+1\big ), &{}\quad \mathrm{if}\;{\mathrm {Tr}}_{m}(B)\ne 0. \end{array} \right. \end{aligned}$$

• (2) If tm is odd, then

  $$\begin{aligned} \displaystyle \Omega _{3}=&\left\{ \begin{array}{ll} (p-1)G(\eta )^{t}\eta (A)G(\eta _{p})\eta _{p}(-\rho ),\quad \;\;\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{if}\;{\mathrm {Tr}}_{m}(B)=0,\\ -G(\eta )^{t}\eta (A)G(\eta _{p})\big (\eta _{p}(-{\mathrm {Tr}}_{m}(B))+\eta (-\rho )\big ),\quad \;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{if}\;{\mathrm {Tr}}_{m}(B)\ne 0. \end{array} \right. \end{aligned}$$

Proof

By Lemma 2.2 we have

$$\begin{aligned} \Omega _{3}=&\sum _{y,z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\Big (\zeta _{p}^{{\mathrm {Tr}}_{m}(-y^{2}(4a_{1}z)^{-1})}\eta (za_{1})G(\eta )\Big )\cdots \Big (\zeta _{p}^{{\mathrm {Tr}}_{m}(-y^{2}(4a_{t}z)^{-1})}\eta (za_{t})G(\eta )\Big )\nonumber \\ =&\sum _{y,z \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\zeta _{p}^{{\mathrm {Tr}}_{m}(-y^{2}((4a_{1}z)^{-1})+\cdots +(4a_{t}z)^{-1}))}\eta (za_{1})\cdots \eta (za_{1})G(\eta )^{t}\nonumber \\ =&G(\eta )^{t}\eta (A)\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta (z)^{t}\sum _{y\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-y^{2}(4z)^{-1}{\mathrm {Tr}}_{m}(B)} \end{aligned}$$

(4)

$$\begin{aligned} =&G(\eta )^{t}\eta (A)\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta (z)^{t}\left( \sum _{y\in \mathbb {F}_{p}}\zeta _{p}^{-y^{2}(4z)^{-1}{\mathrm {Tr}}_{m}(B)}-1\right) . \end{aligned}$$

(5)

Now, we consider the case that tm is even. If ${\mathrm {Tr}}_{m}(B)=0$, then from (4) we have

$$\begin{aligned} \Omega _{3}=-(p-1)G(\eta )^{t}\eta (A). \end{aligned}$$

If ${\mathrm {Tr}}_{m}(B)\ne 0$, then from Lemma 2.2 and (5) we have

$$\begin{aligned} \Omega _{3}&=G(\eta )^{t}\eta (A)\eta _{p}({\mathrm {Tr}}_{m}(B))G(\eta _{p})\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta _{p}(-(4z)^{-1})+G(\eta )^{t}\eta (A)\\&= G(\eta )^{t}\eta (A)\eta _{p}({\mathrm {Tr}}_{m}(B))G(\eta _{p})\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta _{p}(-z)+G(\eta )^{t}\eta (A). \end{aligned}$$

Thus, we get the result.

Now, assume that tm is odd. If ${\mathrm {Tr}}_{m}(B)=0$, then it follows from (4) that

$$\begin{aligned} \Omega _{3}=(p-1)G(\eta )^{t}\eta (A)G(\eta _{p})\eta _{p}(-\rho ). \end{aligned}$$

Also, if ${\mathrm {Tr}}_{m}(B)\ne 0$, then it follows from Lemma 2.2 and (5) that

$$\begin{aligned} \Omega _{3}&=G(\eta )^{t}\eta (A)\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\eta _{p}(z)\Big (\eta _{p}(-(4z)^{-1})\eta _{p}({\mathrm {Tr}}_{m}(B))G(\eta _{p})-1\Big )\\&=G(\eta )^{t}\eta (A)\sum _{z\in \mathbb {F}^{*}_{p}}\zeta _{p}^{-z\rho }\Big (\eta _{p}(-1)\eta _{p}({\mathrm {Tr}}_{m}(B))G(\eta _{p})-\eta _{p}(z) \Big ). \end{aligned}$$

Thus, we get the results. $\square$

By the Lemmas 3.2 and 3.3, we obtain the values of $N_{\rho }$. To get the frequency of each composition, we need the following lemmas.

Lemma 3.4

For $c \in \mathbb {F}_{p}$, let

$$\begin{aligned} n^{'}_{c}=|\{(a_{1},\ldots ,a_{t})\in (\mathbb {F}^{*}_{q})^{t} : {\mathrm {Tr}}_{m}(B)=c\}|. \end{aligned}$$

Then we have

$$\begin{aligned} n^{'}_{c}=&\left\{ \begin{array}{ll} \frac{1}{p}\{(p^{m}-1)^{t}+(-1)^{t}(p-1)\}, &{}\quad \text {if } c=0,\\ \frac{1}{p}\{(p^{m}-1)^{t}-(-1)^{t}\}, &{} \quad \text {if } c\ne 0. \end{array} \right. \end{aligned}$$

Proof

By the orthogonal property of additive characters we get

$$\begin{aligned} n^{'}_{c}=&\sum _{a_{1},\ldots ,a_{t} \in \mathbb {F}_{q}^{*}}\frac{1}{p}\sum _{y \in \mathbb {F}_{p}}\zeta _{p}^{y({\mathrm {Tr}}_{m}(B)-c)}\\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}\left( \sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\sum _{a_{1},\ldots ,a_{t} \in \mathbb {F}_{q}^{*}}\zeta _{p}^{{\mathrm {Tr}}_{m}(y(B)-yc)}\right) \\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\left( \sum _{a_{1} \in \mathbb {F}_{q}^{*}}\zeta _{p}^{{\mathrm {Tr}}_{m}(ya^{-1}_{1})}\cdots \sum _{a_{t} \in \mathbb {F}_{q}^{*}}\zeta _{p}^{{\mathrm {Tr}}_{m}(ya^{-1}_{t})}\right) \\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}\left( \sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}(-1)^{t}\right) . \end{aligned}$$

Thus, we get the desired results. $\square$

Lemma 3.5

For $i\in \{-1,1\}$, let

$$\begin{aligned} n_{i}=|\{(a_{1},\ldots ,a_{t})\in (\mathbb {F}^{*}_{q})^{t} : \eta (A)=i\;\mathrm{{and}}\; {\mathrm {Tr}}_{m}(B)=0\}| \end{aligned}$$

• (1) If m is even, then

  $$\begin{aligned}&n_{1}=\frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}+G(\eta )^{t}\big )\big )\\&n_{-1}=\frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}-G(\eta )^{t}\big )\big ). \end{aligned}$$

• (2) If m is odd, then

  $$\begin{aligned} n_{1}=&\left\{ \begin{array}{ll} \frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}+G(\eta )^{t}\big )\big ), &{}\quad \mathrm{{if}}\;t\;\mathrm{{is\;even}},\\ \frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)(-1)^{t}\big ), &{}\quad \mathrm{{if}}\;t\;\mathrm{{is\;odd}}.\\ \end{array} \right. \\ n_{-1}=&\left\{ \begin{array}{ll} \frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}-G(\eta )^{t}\big )\big ), &{}\quad \mathrm{{if}}\;t\;\mathrm{{is\;even}},\\ \frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)(-1)^{t}\big ), &{}\quad \mathrm{{if}}\;t\;\mathrm{{is\;odd}}. \end{array} \right. \end{aligned}$$

Proof

It follows from Lemma 3.4 that $n_{-1}=n^{'}_{0}-n_{1}$. Thus, we only need to compute $n_{1}$. Let $\alpha$ be a primitive element of $\mathbb {F}_{q}$. Then $\mathbb {F}^{*}_{p}=\langle \alpha ^{\frac{q-1}{p-1}}\rangle$. Note that $\eta (A)=1$ if and only if $A\in C_{0}^{(2,q)}=\langle \alpha ^{2}\rangle$.

$$\begin{aligned} n_{1}=&\sum _{A \in C_{0}^{(2,q)} }\frac{1}{p}\sum _{y \in \mathbb {F}_{p}}\zeta _{p}^{y\mathrm {Tr_{m}}(B)}\nonumber \\ =&\sum _{A \in C_{0}^{(2,q)}}\frac{1}{p}\left( \sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{y\mathrm {Tr_{m}}(B)}+1\right) \nonumber \\ =&\frac{1}{p}\frac{(q-1)^{t}}{2}+\frac{1}{p}\sum _{A \in C_{0}^{(2,q)}}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{y\mathrm {Tr_{m}}(B)}\nonumber \\ =&\frac{1}{p}\frac{(q-1)^{t}}{2}+\frac{1}{p}\sum _{y \in \mathbb {F}^{*}_{p}}\left( \sum ^{[\frac{t}{2}]}_{j=0}\sum _{\begin{array}{c} a_{i_{1}},\ldots ,a_{i_{2j}} \in C_{1}^{(2,q)}\\ a_{1},\ldots ,a_{t}\setminus \{a_{i_{1}},\ldots ,a_{i_{2j}}\} \in C_{0}^{(2,q)} \end{array}}\zeta _{p}^{\mathrm {Tr_{m}}(ya^{-1}_{1})}\cdots \zeta _{p}^{\mathrm {Tr_{m}}(ya^{-1}_{t})}\right) \nonumber \\ =&\frac{1}{p}\frac{(q-1)^{t}}{2}+\frac{1}{p}\sum _{y \in \mathbb {F}^{*}_{p}}\left( \sum ^{[\frac{t}{2}]}_{j=0}\left( {\begin{array}{c}t\\ 2j\end{array}}\right) \sum _{\begin{array}{c} a_{i_{1}},\ldots ,a_{2j} \in C_{1}^{(2,q)}\\ a_{2j+1},\ldots ,a_{t} \in C_{0}^{(2,q)} \end{array}}\zeta _{p}^{\mathrm {Tr_{m}}(ya^{-1}_{1})}\cdots \zeta _{p}^{\mathrm {Tr_{m}}(ya^{-1}_{t})}\right) . \end{aligned}$$

(6)

Assume that m is even, then 2 divides $\frac{q-1}{p-1}$ and so $\mathbb {F}^{*}_{p}\subseteq C_{0}^{(2,q)}.$ By (6) we can get

$$\begin{aligned} n_{1}=\frac{1}{p}\frac{(q-1)^{t}}{2}+\frac{1}{p}(p-1)\left( \sum ^{[\frac{t}{2}]}_{j=0}\left( {\begin{array}{c}t\\ 2j\end{array}}\right) (\eta ^{(2,q)}_{0})^{t-2j}(\eta ^{(2,q)}_{1})^{2j}\right) \end{aligned}$$

Note that $\eta ^{(2,q)}_{1}+\eta ^{(2,q)}_{0}$=$-1$ and $\eta ^{(2,q)}_{0}-\eta ^{(2,q)}_{1}=G(\eta ).$ Thus, we get the result. Now suppose that m is odd, then $|\mathbb {F}^{*}_{p}\cap C^{(2,q)}_{0}|=|\mathbb {F}^{*}_{p}\cap C^{(2,q)}_{1}|=\frac{p-1}{2}.$ By (6) we have

$$\begin{aligned} n_{1}=&\frac{1}{p}\frac{(q-1)^{t}}{2}+\frac{1}{p} \left( \frac{p-1}{2}\sum ^{[\frac{t}{2}]}_{j=0}\left( {\begin{array}{c}t\\ 2j\end{array}}\right) (\eta ^{(2,q)}_{0})^{t-2j}(\eta ^{(2,q)}_{1})^{2j}\right. \\&\qquad \qquad \qquad \qquad \quad \left. +\frac{p-1}{2}\sum ^{[\frac{t}{2}]}_{j=0}\left( {\begin{array}{c}t\\ 2j\end{array}}\right) (\eta ^{(2,q)}_{1})^{t-2j}(\eta ^{(2,q)}_{0})^{2j}\right) \\ =&\frac{1}{p}\frac{(p^{m}-1)^{t}}{2}+\frac{p-1}{2p}\left( \frac{1}{2}\Big ((\eta ^{(2,q)}_{0}-\eta ^{(2,q)}_{1})^{t}+(\eta ^{(2,q)}_{1}-\eta ^{(2,q)}_{0})^{t}\Big )+(-1)^{t}\right) . \end{aligned}$$

Note that $n_{-1}=n^{'}-n_{1}$ and this completes the proof. $\square$

Recall that $\alpha$ is a fixed primitive element of $\mathbb {F}_{q}$.

Lemma 3.6

For $0\le k \le [t/2]$ and $c\in C_{0}^{(2,p)}$, let $n_{2k,c}=|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : {\mathrm {Tr}}_{m}(\alpha a_{1}^{2}+\cdots +\alpha a_{2k}^{2}+a_{2k+1}^{2}+\cdots +a_{t}^{2})=c\}|.$

• (1) If m is even, then

  $$\begin{aligned} n_{2k,c}=\frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )+1)^{2k}(G(\eta )-1)^{t-2k}. \end{aligned}$$

• (2) If m is odd and $k<t/4$, then

  $$\begin{aligned} n_{2k,c}=&\left\{ \begin{array}{ll} \displaystyle \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\ \left. \displaystyle +\,\eta _{p}(-c)\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right) ,\;\mathrm{{if}}\;t\;\mathrm{{is\;even}},\\ \displaystyle \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\left( \sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\ \left. \displaystyle +\,\eta _{p}(-c)\sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right) ,\;\mathrm{{if}}\;t\;\;\mathrm{{is\;odd}}. \end{array} \right. \end{aligned}$$

• (3) If m is odd and $k>t/4$, then

  $$\begin{aligned} n_{2k,c}=&\left\{ \begin{array}{ll} \displaystyle \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\left( -\sum ^{\frac{4k-t}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\ \left. \displaystyle +\,\eta _{p}(-c)\sum ^{\frac{4k-t}{2}-1}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right) ,\;\mathrm{{if}}\;t\;\mathrm{{is\;even}},\\ \displaystyle \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\left( -\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\ \left. \displaystyle +\,\eta _{p}(-c)\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right) ,\;\;\mathrm{{if}}\;t\;\mathrm{{is\;odd}}.\\ \end{array} \right. \end{aligned}$$

• (4) If m is odd, $t\equiv 0\pmod {4}$ and $k=t/4$, then

  $$\begin{aligned} n_{2k,c}=n_{\frac{t}{2},c}&=\frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{\frac{t}{2}}. \end{aligned}$$

Proof

By the orthogonal property of additive characters, we have

$$\begin{aligned} n_{2k,c}=&\sum _{a_{1},\ldots ,a_{t} \in \mathbb {F}_{q}^{*}}\frac{1}{p}\sum _{y \in \mathbb {F}_{p}}\zeta _{p}^{y({\mathrm {Tr}}_{m}(\alpha a_{1}^{2}+\alpha a_{2}^{2}+\cdots +\alpha a_{2k}^{2}+a_{2k+1}^{2}+\cdots +a_{t}^{2})-c)}\\ =&\sum _{a_{1},\ldots ,a_{t} \in \mathbb {F}_{q}^{*}}\frac{1}{p}\left( \sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{y({\mathrm {Tr}}_{m}(\alpha a_{1}^{2}+\alpha a_{2}^{2}+\cdots +\alpha a_{2k}^{2}+a_{2k+1}^{2}+\cdots +a_{t}^{t})-c)}+1\right) \\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\sum _{a_{1},\ldots ,a_{t} \in \mathbb {F}_{q}^{*}}\zeta _{p}^{{\mathrm {Tr}}_{m}(y\alpha a_{1}^{2})}\cdots \zeta _{p}^{{\mathrm {Tr}}_{m}(y\alpha a_{2k}^{2})}\zeta _{p}^{{\mathrm {Tr}}_{m}(ya_{2k+1}^{2})}\cdots \zeta _{p}^{{\mathrm {Tr}}_{m}(ya_{t}^{2})}\\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}\left( \sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\left( \sum _{a_{1}\in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(y\alpha a_{1}^{2})}-1\right) \cdots \left( \sum _{a_{2k}\in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(y\alpha a_{2k}^{2})}-1\right) \right. \\&\left. \times \,\left( \sum _{a_{2k+1}\in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(ya_{2k+1}^{2})}-1\right) \cdots \left( \sum _{a_{t}\in \mathbb {F}_{q}}\zeta _{p}^{{\mathrm {Tr}}_{m}(y a_{t}^{2})}-1\right) \right) . \end{aligned}$$

It follows from Lemma 2.2 that

$$\begin{aligned} n_{2k,c}=\frac{(q-1)^{t}}{p}+\frac{1}{p}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}(\eta (\alpha y)G(\eta )-1)^{2k}(\eta (y)G(\eta )-1)^{t-2k}. \end{aligned}$$

(7)

Now, if m is even, then we have

$$\begin{aligned} n_{2k,c}=\frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )+1)^{2k}(G(\eta )-1)^{t-2k}. \end{aligned}$$

Suppose that m is odd and $k<t/4$. Then by (7) we have

$$\begin{aligned} n_{2k,c}=\frac{(q-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\left( \sum ^{t-4k}_{i=0}\left( {\begin{array}{c}t-4k\\ i\end{array}}\right) \eta _{p}(y)^{i}G(\eta )^{i}(-1)^{t-4k-i}\right) . \end{aligned}$$

(8)

If t is even, then it follows from (8) that

$$\begin{aligned} n_{2k,c}=&\frac{(q-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\&\left. +\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) \eta _{p}(y)G(\eta )^{2i+1}(-1)\right) \\ =&\frac{(q-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\left( (-1)\sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right. \\&\left. +(-1)\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}\eta _{p}(y)\right) . \end{aligned}$$

Thus, we get the desired result.

It is similar to give the proof when t is odd or m is odd for $k>t/4$. Finally, if m is odd, $t\equiv 0\pmod {4}$ and $k=t/4$, it follows from (7) that

$$\begin{aligned} n_{2k,c}=n_{\frac{t}{2},c}&=\frac{(q-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{\frac{t}{2}}\sum _{y \in \mathbb {F}^{*}_{p}}\zeta _{p}^{-yc}=\frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{\frac{t}{2}}. \end{aligned}$$

$\square$

Lemma 3.7

For $i,j\in \{-1,1\},$ let

$$\begin{aligned} n_{i,j}=|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : \eta (A)=i\;\mathrm{and}\;\eta _{p}(-{\mathrm {Tr}}_{m}(B))=j\}| \end{aligned}$$

Then we have

$$\begin{aligned} n_{1,1}=\frac{1}{2^{t}}\frac{p-1}{2}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) n_{2i,-1}\;\mathrm{and}\; n_{1,-1}=\frac{1}{2^{t}}\frac{p-1}{2}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) n_{2i,-\beta }, \end{aligned}$$

where $\beta =\alpha ^{\frac{q-1}{p-1}}$. Moreover, if tm is even, then $n_{1,1}=n_{1,-1}.$

Proof

For $j\in \{-1,1\}$, we have

$$\begin{aligned} \displaystyle n_{1,j}=&|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : \eta (A)=1\;\mathrm{and}\;\eta _{p}(-{\mathrm {Tr}}_{m}(B))=j\}|\nonumber \\ =&|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : A\in C_{0}^{(2,q)},\;\eta _{p}(-{\mathrm {Tr}}_{m}(B))=j\}|\nonumber \\ =&\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) |\{a_{1},\ldots ,a_{2i}\in C_{1}^{(2,q)}, a_{2i+1},\ldots ,a_{t}\in C_{0}^{(2,q)} :\;\eta _{p}(-{\mathrm {Tr}}_{m}(B))=j\}|\nonumber \\ =&\frac{1}{2^{t}}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) |\{b_{1},\ldots ,b_{t}\in \mathbb {F}^{*}_{q} : \;\eta _{p}(-{\mathrm {Tr}}_{m}(\alpha b^{2}_{1}+\alpha b^{2}_{2}+\cdots +\alpha b^{2}_{2[\frac{t}{2}]}\nonumber \\&+b^{2}_{2[\frac{t}{2}]+1}+\cdots +b^{2}_{t})=j\}|. \end{aligned}$$

By Lemma 3.6 we have

$$\begin{aligned} \displaystyle n_{1,1}&=\frac{1}{2^{t}}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) \sum _{c \in C_{0}^{(2,p)}}n_{2i,-c}=\frac{1}{2^{t}}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) \frac{p-1}{2}n_{2i,-1}.\\ \mathrm{and}\\ n_{1,-1}&=\frac{1}{2^{t}}\sum ^{[\frac{t}{2}]}_{i=0}\left( {\begin{array}{c}t\\ 2i\end{array}}\right) \sum _{c \in C_{0}^{(2,p)}}n_{2i,-\beta } \end{aligned}$$

If m is even, from Lemma 3.6, then it is easy to see that $n_{2k,c}$ are independent of $c \in \mathbb {F}^{*}_{p}$. Thus $n_{2k,-1}=n_{2k,-\beta }$ and so $n_{1,1}=n_{1,-1}.$

Next assume that m is odd and $t\equiv 2\pmod 4$. By Lemma 3.6 we have

$$\begin{aligned} n_{1,1}&=\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) n^{'}_{2k,-1}+\sum ^{\frac{t}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) n^{'}_{2k,-1}\right) \\&=\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \right. \\&\quad \left. \times \left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p}\right) \sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1})\right\} \\&\quad +\,\sum ^{\frac{t}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\right. \\&\quad \left. \left. \times \,\left( \sum ^{\frac{4k-t}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}-G(\eta _{p})\sum ^{\frac{4k-t}{2}-1}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) . \end{aligned}$$

By changing k with $\frac{t}{2}-k$ in the second summation we have

$$\begin{aligned} n_{1,1}&=\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \right. \\&\quad \times \,\left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p})\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1})\right\} \\&\quad +\sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \\&\quad \left. \left. \times \left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}-G(\eta _{p})\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) \\&=\frac{p-1}{2^{t}} \left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k} \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right\} \right) . \end{aligned}$$

For $n_{1,-1}$, we similarly have

$$\begin{aligned} n_{1,-1}=&\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(q-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \right. \\&\left. \times \left( \sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}-G(\eta _{p})\sum ^{\frac{t-4k}{2}-1}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \\&+\sum ^{\frac{t}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(q-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\right. \\&\left. \left. \times \left( \sum ^{\frac{4k-t}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p})\sum ^{\frac{4k-t}{2}-1}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) . \end{aligned}$$

By changing k with $\frac{t}{2}-k$ in the second summation we have

$$\begin{aligned} n_{1,-1}=&\frac{p-1}{2^{t}}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{2k}\sum ^{\frac{t-4k}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}\right\} \right) . \end{aligned}$$

Thus, $n_{1,1}=n_{1,-1}.$ It is similar to get the desired results when m is odd and $t\equiv 0\pmod 4$. This completes the proof. $\square$

Lemma 3.8

For $i\in \{-1,1\}$, let

$$\begin{aligned} s_{i}=|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : \eta (A)\eta _{p}(-{\mathrm {Tr}}_{m}(B))=i\}| \end{aligned}$$

• (1) If tm is even, then we have

  $$\begin{aligned} s_{1}=s_{-1}=\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big ). \end{aligned}$$

• (2) If tm is odd, then we have

  $$\begin{aligned} s_{\pm 1}=&\pm \frac{p-1}{2^{t}}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{1}{p}(G(\eta )^{2}-1)^{2k}\sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right\} \right. \\&\left. +\sum ^{\frac{t-1}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}G(\eta _{p})\right\} \right) \\&+\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big ). \end{aligned}$$

Proof

It is easy to see that $s_{1}=n_{1,1}+n_{-1,-1}=n_{1,1}+T-n_{1,-1},$ where $T=|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} :\eta _{p}(-{\mathrm {Tr}}_{m}(B))=-1\}|.$ By Lemma 3.4 we have

$$\begin{aligned} T=&|\{a_{1},\ldots ,a_{t}\in \mathbb {F}^{*}_{q} : -{\mathrm {Tr}}_{m}(B) \in C^{(2,p)}_{1}\}|\\ =&\sum _{c \in C^{(2,p)}_{1}}n^{'}_{-c}=\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big ). \end{aligned}$$

Similarly, we have $s_{-1}=n_{1,-1}+T-n_{1,1}.$ If tm is even, then by Lemma 3.7, we get the desired result.

Next, if tm is odd, then by Lemmas 3.6 and 3.7 we have

$$\begin{aligned} n_{1,1}=&\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \right. \\&\left. \times \left( \sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p})\sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \\&+\sum ^{\frac{t-1}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\right. \\&\left. \left. \times \left( -\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p})\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) ,\\ n_{1,-1}=&\frac{1}{2^{t}}\frac{p-1}{2}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}+\frac{1}{p}(G(\eta )^{2}-1)^{2k}\right. \right. \\&\left. \left. \times \left( \sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i\end{array}}\right) G(\eta )^{2i}-G(\eta _{p})\sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) \\&+\sum ^{\frac{t-1}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{(p^{m}-1)^{t}}{p}-\frac{1}{p}(G(\eta )^{2}-1)^{t-2k}\right. \\&\left. \left. \times \left( \sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i\end{array}}\right) G(\eta )^{2i}+G(\eta _{p})\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) G(\eta )^{2i+1}\right) \right\} \right) . \end{aligned}$$

Now, it is easy to get $s_{1}$ and $s_{-1}$. This completes the proof. $\square$

Theorem 3.9

Let $\mathcal {C}_{D}$ be the linear code defined by (1) and (2), where $D=\{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=0\}$ and $\rho \in \mathbb {F}_{p}^{*}$. Then $\mathcal {C}_{D}$ is a $[p^{tm-1}-1,tm]$ linear code.

• (1) If m is even, then the complete weight enumerator of $\mathcal {C}_{D}$ is given as follows:

  $$\begin{aligned}&N_{\rho }=0\;\mathrm {occurs}\;1\;\mathrm {time,}\\&N_{\rho }=p^{tm-2}\pm p^{m(t-\frac{k}{2})-2}\;(0<k<t)\;\mathrm{{occurs}}\;\left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(p^{m}-1)^{k}}{2}\;\mathrm{{times,}}\\&N_{\rho }=p^{tm-2}\pm p^{\frac{tm-2}{2}}\;\mathrm{{occurs}}\;\frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}\mp p^{\frac{tm}{2}}\big )\big )\;\mathrm{{times,}}\\&N_{\rho }=p^{tm-2}\pm p^{\frac{tm-2}{2}}\eta _{p}(\rho )\;\mathrm{{occurs}}\;\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big )\;\mathrm{{times.}} \end{aligned}$$

• (2) If m is odd and t is even, then the complete weight enumerator of $\mathcal {C}_{D}$ is given as follows:

  $$\begin{aligned}&N_{\rho }=0\;\mathrm {occurs}\;1\;\mathrm {time,}\\&N_{\rho }=p^{tm-2}\pm p^{m(t-\frac{k}{2})-2}\;(0<k<t\;\mathrm{{and}}\;k\;\mathrm{{is\;even}})\;\mathrm{{occurs}}\;\left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(p^{m}-1)^{k}}{2}\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm p^{m(t-\frac{k}{2})-\frac{3}{2}}\eta _{p}(\rho )\;(0<k<t\;\mathrm{{and}}\;k\;\mathrm{{is\;odd}})\;\mathrm{{occurs}}\;\left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(p^{m}-1)^{k}}{2}\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm p^{\frac{tm-2}{2}}\;\mathrm{{occurs}}\;\frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)\big ((-1)^{t}\mp p^{\frac{tm}{2}}\big )\big )\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm p^{\frac{tm-2}{2}}\eta _{p}(\rho )\;\mathrm{{occurs}}\;\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big )\;\mathrm {times.} \end{aligned}$$

• (3) If m is odd and t is odd, then the complete weight enumerator of $\mathcal {C}_{D}$ is given as follows:

  $$\begin{aligned}&N_{\rho }=0\;\mathrm {occurs}\;1\;\mathrm {time,}\\&N_{\rho }=p^{tm-2}\pm p^{m(t-\frac{k}{2})-2}\;(0<k<t\;\mathrm{{and}}\;k\;\mathrm{{is\;even}})\;\mathrm{{occurs}}\;\left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(p^{m}-1)^{k}}{2}\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm p^{m(t-\frac{k}{2})-\frac{3}{2}}\eta _{p}(\rho )\;(0<k<t\;\mathrm{{and}}\;k\;\mathrm{{is\;odd}})\;\mathrm{{occurs}}\;\left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(p^{m}-1)^{k}}{2}\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm p^{\frac{tm-1}{2}}\eta _{p}(\rho )\;\mathrm{{occurs}}\;\frac{1}{2p}\big ((p^{m}-1)^{t}+(p-1)(-1)^{t}\big )\;\mathrm {times,}\\&N_{\rho }=p^{tm-2}\pm (-1)^{\frac{(tm+1)(p-1)}{4}}p^{\frac{tm-3}{2}}\;\mathrm {occurs}\\&\mp \frac{p-1}{2^{t}}\left( \sum ^{[\frac{t}{4}]}_{k=0}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{1}{p}\big ((-1)^{\frac{m(p-1)}{2}}p^{m}-1\big )^{2k}\sum ^{\frac{t-4k-1}{2}}_{i=0}\left( {\begin{array}{c}t-4k\\ 2i+1\end{array}}\right) (-1)^{\frac{((2i+1)m+1)(p-1)}{4}}p^{\frac{(2i+1)m+1}{2}} \right\} \right. \\&\left. \quad +\sum ^{\frac{t-1}{2}}_{k=[\frac{t}{4}]+1}\left( {\begin{array}{c}t\\ 2k\end{array}}\right) \left\{ \frac{1}{p}\big ((-1)^{\frac{m(p-1)}{2}}p^{m}-1\big )^{t-2k}\sum ^{\frac{4k-t-1}{2}}_{i=0}\left( {\begin{array}{c}4k-t\\ 2i+1\end{array}}\right) (-1)^{\frac{((2i+1)m+1)(p-1)}{4}}p^{\frac{(2i+1)m+1}{2}}\right\} \right) \\&\quad +\frac{p-1}{2p}\big ((p^{m}-1)^{t}-(-1)^{t}\big )\;\mathrm {times.} \end{aligned}$$

Proof

Recall that $N_{\rho }=p^{tm-2}+\frac{1}{p^{2}}(\Omega _{1}+\Omega _{2}+\Omega _{3})$. We employ Lemmas 3.2 and 3.3 to compute $N_{\rho }$. As computations for frequencies are done by Lemmas 3.4, 3.5, 3.6, 3.7 and 3.8 it is sufficient to give a proof for even m.

Suppose that there are exactly k elements $a_{i_{1}},\ldots ,a_{i_{k}}\ne 0$ among $a_{1},\ldots ,a_{t}$ for $1\le k\le t$.

If $1\le k \le t-1$, then we obtain

$$\begin{aligned} \displaystyle N_{\rho }=&\left\{ \begin{array}{ll} p^{tm-2}-\frac{1}{p^{2}}q^{t-k}G(\eta )^{k}, &{}\quad \mathrm{if}\;a_{i_{1}}\cdots a_{i_{k}}\in C_{0}^{(2,q)},\\ p^{tm-2}+\frac{1}{p^{2}}q^{t-k}G(\eta )^{k}, &{}\quad \mathrm{if}\;a_{i_{1}}\cdots a_{i_{k}}\in C_{1}^{(2,q)}. \end{array} \right. \end{aligned}$$

In this case, the frequencies are both $\displaystyle \left( {\begin{array}{c}t\\ k\end{array}}\right) \frac{(q-1)^{k}}{2}.$

If $k=t$ and ${\mathrm {Tr}}_{m}(B)=0$, then

$$\begin{aligned} \displaystyle N_{\rho }=&p^{tm-2}+\frac{1}{p^{2}}\big (-\eta (A)G(\eta )^{t}-(p-1)\eta (A)G(\eta )^{t}\big )\\ =&p^{tm-2}-\frac{1}{p}\eta (a_{1}\cdots a_{t})G(\eta )^{t}. \end{aligned}$$

Thus,

$$\begin{aligned} \displaystyle N_{\rho }=&\left\{ \begin{array}{ll} p^{tm-2}-\frac{1}{p}G(\eta )^{t}, &{} \quad \mathrm{if}\; \eta (A)=1\;\mathrm{{and}}\;{\mathrm {Tr}}_{m}(B)=0,\\ p^{tm-2}+\frac{1}{p}G(\eta )^{t}, &{} \quad \mathrm{if}\; \eta (A)=-1\;\mathrm{{and}}\;{\mathrm {Tr}}_{m}(B)=0. \end{array} \right. \end{aligned}$$

Now the frequencies follow from Lemma 3.5.

If $k=t$ and ${\mathrm {Tr}}_{m}(B)\ne 0$, then

$$\begin{aligned} \displaystyle N_{\rho }=&p^{tm-2}+\frac{1}{p^{2}}\big (-\eta (A)G(\eta )^{t}+G(\eta )^{t}\eta (A)\big (p\eta _{p}(-{\mathrm {Tr}}_{m}(B))\eta _{p}(\rho )+1\big )\big )\\ =&p^{tm-2}+\frac{1}{p}\eta (A)G(\eta )^{t}\eta _{p}(-{\mathrm {Tr}}_{m}(B))\eta _{p}(\rho ). \end{aligned}$$

Thus,

$$\begin{aligned} \displaystyle N_{\rho }=&\left\{ \begin{array}{ll} p^{tm-2}+\frac{1}{p}G(\eta )^{t}\eta _{p}(\rho ), &{} \mathrm{if}\; \eta (A)\eta _{p}(-{\mathrm {Tr}}_{m}(B))=1\\ &{} \mathrm{{and}}\;{\mathrm {Tr}}_{m}(B)\ne 0,\\ p^{tm-2}-\frac{1}{p}G(\eta )^{t}\eta _{p}(\rho ), &{} \mathrm{if}\; \eta (A)\eta _{p}(-{\mathrm {Tr}}_{m}(B))=-1\\ &{} \mathrm{{and}}\;{\mathrm {Tr}}_{m}(B)\ne 0. \end{array} \right. \end{aligned}$$

Now the frequencies follow from Lemma 3.7. $\square$

In fact, when $t=1$, the complete weight enumerators of $\mathcal {C}_{D}$ were given by [17]. Thus Theorem 3.9 can be viewed as a generalization of the results in [17]. From Theorem 3.9 we can also get the weight enumerators of $\mathcal {C}_{D}$ directly.

Corollary 3.10

Let $\mathcal {C}_{D}$ be a linear code defined by (1) and (2),  where $D = \{(x_{1},x_{2},\ldots ,x_{t}) \in \mathbb {F}_{q}^{t}\setminus \{(0,0,\ldots ,0)\} : {\mathrm {Tr}}_{m}(x_{1}+x_{2}+\cdots +x_{t})=0\}.$

• (1) If m is even, then the weight distribution of $\mathcal {C}_{D}$ is given by Table 1.

• (2) If m is odd and t is even, then the weight distribution of $\mathcal {C}_{D}$ is given by Table 2.

• (3) If m is odd and t is odd, then the weight distribution of $\mathcal {C}_{D}$ is given by Table 3.

Table 1 The weight distribution of $\textit{C}_{D}$ for even m

Table 2 The weight distribution of $\textit{C}_{D}$ for odd m and even t

Table 3 The weight distribution of $\textit{C}_{D}$ for odd m and odd t, where $s_{\pm 1}$ is given by Lemma 3.8(2)

Remark 3.11

By Corollary 3.10, we easily get several linear codes with a few weights. For example, we obtain 3-weight linear codes for $m=2$ and $t=2$, and 5-weight linear codes for even $m\ge 4$, $t=2$ and $m=2$, $t=3$. We also have 3-weight linear codes for odd m, $t=2$, and 5-weight linear codes for odd m, $t=3,4$.

Example 3.12

• (1) Let $p=3, m=2,$ and $t=3.$ Then $q=9$ and $n=242.$ By Theorem 3.9, the code $\textit{C}_{D}$ is a [242, 6, 108] linear code. Its complete weight enumerator is

  $$\begin{aligned}&z^{242}_{0}+12z_{0}^{134}(z_{1}z_{2})^{54}+190z_{0}^{98}(z_{1}z_{2})^{72}+171z_{0}^{80}z_{1}^{72}z_{2}^{90}+171z_{0}^{80}z_{1}^{90}z_{2}^{72}\\&\quad +172z_{0}^{62}(z_{1}z_{2})^{90}+12z^{26}_{0}(z_{1}z_{2})^{108}, \end{aligned}$$

  and its weight enumerator is

  $$\begin{aligned} 1+12x^{108}+190x^{144}+342x^{162}+172x^{180}+12x^{216}, \end{aligned}$$

  which are checked by Magma.

• (2) Let $p=3, m=3,$ and $t=3.$ Then $q=27$ and $n=6560.$ By Theorem 3.9, the code $\textit{C}_{D}$ is a [6560, 9, 4212] linear code. Its complete weight enumerator is

  $$\begin{aligned}&z^{6560}_{0}+1014z_{0}^{2348}(z_{1}z_{2})^{2106}+5940z^{2240}_{0}(z_{1}z_{2})^{2160}+39z^{2186}_{0}z_{1}^{1458}z_{2}^{2916}\\&\quad +39z_{0}^{2186}z_{1}^{2916}z_{2}^{1458}+2929z^{2186}_{0}z_{1}^{2106}z_{2}^{2268}+2929z^{2186}_{0}z_{1}^{2268}z_{2}^{2106}\\&\quad +5778z^{2132}_{0}(z_{1}z_{2})^{2214}+1014z_{0}^{2024}(z_{1}z_{2})^{2268}, \end{aligned}$$

  and its weight enumerator is

  $$\begin{aligned} 1+1014x^{4212}+5940x^{4320}+5936x^{4374}+5778x^{4428}+1014x^{4536}, \end{aligned}$$

  which are checked by Magma.