# Self-dual twisted generalized Reed-Solomon codes

Canze Zhu Qunying Liao  Corresponding author.
E-mail<EMAIL_ADDRESS>(Q. Liao<EMAIL_ADDRESS>(C. Zhu).
Supported by National Natural Science Foundation of China (Grant No.
12071321).

Abstract. In this paper, by using some properties for linear algebra methods,
the parity-check matrixs for twisted generalized Reed-Solomon codes with any
given hook $h$ and twist $t$ are presented, and then a sufficient and
necessary condition for that a twisted generalized Reed-Solomon code with
dimension $h+t$ $(h\geq t)$ to be self-dual is given. Furthermore, several
classes of self-dual codes with small singleton defect are constructed based
on twisted generalized Reed-Solomon codes.

Keywords. Twisted generalized Reed-Solomon code; self-dual code; $m$-MDS code;
small singleton defect.

##  1 Introduction

In recent years, study of self-dual maximum distance separable (in short, MDS)
codes have attracted a lot of attention [1, 4, 9, 10, 11, 12, 13, 18, 19, 28,
29, 31]. On the one hand, due to their nice structures, self-dual codes have
been attracted attention by coding theorists, cryptographers and
mathematicians, or rather, self-dual codes have been found various
applications in cryptography (in particular secret sharing) [5, 7, 25] and
combinatorics [16, 24]. On the other hand, the MDS code is with optimal
parameters, it can correct maximal number of errors for a given code rate
[16]. In addition, MDS codes are closely connected to combinatorial designs
and finite geometry [16, 24]. Thus, it is natural to consider the intersection
of these two classes of codes, namely, self-dual MDS codes. Especially,
generalized Reed-Solomon (in short, GRS) codes are a class of MDS codes and a
lot of self-dual MDS codes from GRS codes are constructed in [8, 17, 21, 32,
33].

For a non-negative integer $m$, the $m$-MDS code is introduced independently
in [23, 34]. A linear code $\mathcal{C}$ is called an $m$-MDS code if both
$\mathcal{C}$ and its dual code have $m$ singleton defect from being an MDS
code. Especially, for $m=0$ or $1$, $\mathcal{C}$ is MDS or Near MDS (in
short, NMDS) [6], respectively. It is easy to see that a self-dual code with
$m$ singleton defect is also an $m$-MDS code. For small $m$, $m$-MDS codes
approximate maximal minimum Hamming distance for a given code rate.
Futhermore, $m$-MDS codes also have been applicated in secret sharing scheme
[30], the index coding problem [35] and the informed source coding problem
[36]. From both theoretical and practical points of view, to study self-dual
$m$-MDS codes with small $m$ is interesting.

In 2017, inspired by twisted Gabidulin codes [27], twisted Reed-Solomon (in
short, TRS) codes were firstly introduced by Beelen, et al. [2], as a
generalization for Reed-Solomon codes, they also showed that TRS codes could
be well decoded. Different from GRS codes, the twisted generalized Reed-
Solomon (in short, TGRS) code is not necessarily MDS [2]. The authors also
showed that most of TGRS codes are not GRS codes [3, 22]. Furthermore, by TGRS
codes, Lavauzelle et, al. presented an efficient key-recovery attack used in
the McEliece cryptosystem [20]. TGRS codes are also used to construct linear
complementary dual (in short, LCD) MDS codes by their applications in
cryptography [14, 22]. Recently, Liu et, al. give the parity-check matrix for
a TGRS code with the hook $k-1$ and twist $1$, and then constructed several
classes of self-dual MDS or NMDS codes from these TGRS codes [15].

In this paper, we extend the main results in [15]. In section 2, some basic
notations and results about TGRS codes are given. In section 3, the parity-
check matrix for a TGRS code with any given hook $h$ and twist $t$ is
presented. In section 4, a sufficient and necessary condition for a TGRS code
with dimension $h+t$ $(h\geq t)$ to be self-dual is given, and then several
classes of self-dual $m$-MDS codes with small $m$ are constructed. In section
5, we conclude the whole paper.

##  2 Preliminaries

In this section, we give some notations and lemmas about TGRS codes and self-
dual $m$-MDS codes.

###  2.1 TGRS codes

Let $\mathbb{F}_{q}$ be the finite field with $q$ elements and
$\mathbb{F}_{q}^{*}=\mathbb{F}_{q}\backslash\\{0\\}$, where $q$ is a prime
power. Let $\mathbb{F}_{q}[x]$ be the polynomial ring over $\mathbb{F}_{q}$.
Now some related definitions are given in the following.

######  Definition 2.1 ([2, 3, 20])

Let $t$, $h$ and $k$ be positive integers with $0\leq h<k\leq q$ and
$\eta\in\mathbb{F}_{q}^{*}$. Define the set of $(k,t,h,\eta)$-twisted
polynomials as

$\displaystyle\mathcal{V}_{k,t,h,\eta}=\Big{\\{}f(x)=\sum_{i=0}^{k-1}a_{i}x^{i}+\eta
a_{h}x^{k-1+t}\Big{\\}},$

which is a $k$-dimensional $\mathbb{F}_{q}$-linear subspace. We call $h$ the
hook and $t$ the twist.

######  Definition 2.2 ([2, 3, 20])

Let $t$, $h$, $k$ and $n$ be positive integers, $\eta\in\mathbb{F}_{q}^{*}$,
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{F}_{q}^{n}$ with
$\alpha_{i}\neq\alpha_{j}$ $(i\neq j)$ and
$\bm{v}=(v_{1},\ldots,v_{n})\in(\mathbb{F}_{q}^{*})^{n}$. The TGRS code with
length $n$ and dimension $k$ is defined as

$\displaystyle\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)=\\{(v_{1}f({\alpha_{1}}),\ldots,v_{n}f(\alpha_{n}))~{}|~{}f(x)\in\mathcal{V}_{k,t,h,\eta}\\}.$

Especially, for $\bm{1}=(1,\ldots,1)$,
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{1},t,h,\eta)$ is called a TRS code.

By Definition 2.1, it is easy to see that the generator matrix of
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is

$\displaystyle G_{k}=\left(\begin{matrix}&v_{1}~{}&v_{2}~{}&\ldots~{}&v_{n}\\\
&v_{1}\alpha_{1}~{}&v_{2}\alpha_{2}~{}&\ldots~{}&v_{n}\alpha_{n}\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&v_{1}\alpha_{1}^{h}+\eta\alpha_{1}^{k-1+t}~{}&v_{2}\alpha_{2}^{h}+\eta\alpha_{2}^{k-1+t}~{}&\ldots~{}&v_{n}\alpha_{n}^{h}+\eta\alpha_{h}^{k-1+t}\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&v_{1}\alpha_{1}^{k-1}~{}&v_{2}\alpha_{2}^{k-1}~{}&\cdots~{}&v_{n}\alpha_{n}^{k-1}\end{matrix}\right).$

The following lemma gives a sufficient condition for a TGRS code to be MDS.

######  Lemma 2.1 (Theorem 17, in [2])

Let $\mathbb{F}_{s}$ be a proper subfield of $\mathbb{F}_{q}$ and
$\bm{\alpha}\in\mathbb{F}_{s}^{n}$. If
$\eta\in\mathbb{F}_{q}\backslash\mathbb{F}_{s}$, then
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is MDS.

An $[n,k,d]$ linear code $\mathcal{C}$ over $\mathbb{F}_{q}$ is a
$k$-dimensional subspace of $\mathbb{F}_{q}^{n}$ with minimum (Hamming)
distance $d$ and length $n$. For $\mathbf{a}=(a_{1},\ldots,a_{n})$,
$\mathbf{b}=(b_{1},\ldots,b_{n})$ $\in\mathbb{F}_{q}^{n}$, the inner product
is defined as $\langle\mathbf{a},\mathbf{b}\rangle=\sum_{i=1}^{n}a_{i}b_{i}$.
And then the dual code of $\mathcal{C}$ is defined as

$\mathcal{C}^{\perp}=\\{\mathbf{c}^{{}^{\prime}}\in\mathbb{F}_{q}^{n}~{}|~{}\langle\mathbf{c}^{{}^{\prime}},\mathbf{c}\rangle=0,\text{for
any}~{}\mathbf{c}\in\mathcal{C}\\}.$

We say that $\mathcal{C}$ is self-dual if $\mathcal{C}=\mathcal{C}^{\perp}$.

The following lemma gives the parameters for
$\mathcal{C}_{k,n}^{\perp}(\bm{\alpha},\bm{v},t,h,\eta)$ and a bound for the
minimum distance.

###### Lemma 2.2

$\mathcal{C}_{k,n}^{\perp}(\bm{\alpha},\bm{v},t,h,\eta)=[n,n-k,d]$ with
$\max\\{h+1,k-h\\}\leq d\leq k+1$.

Proof. If $d\leq h$, then there are $1\leq i_{1}<\cdots<i_{d}\leq n$, such
that

$\displaystyle G_{d}=\left(\begin{matrix}&1~{}&1~{}&\ldots~{}&1\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&\alpha_{i_{1}}^{h}+\eta\alpha_{i_{1}}^{k-1+t}~{}&\alpha_{i_{2}}^{h}+\eta\alpha_{i_{2}}^{k-1+t}~{}&\ldots~{}&\alpha_{i_{d}}^{h}+\eta\alpha_{i_{d}}^{k-1+t}\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&\alpha_{i_{1}}^{k-1}~{}&\alpha_{i_{2}}^{k-1}~{}&\cdots~{}&\alpha_{i_{d}}^{k-1}\end{matrix}\right),$

with $d$ linearly dependent columns, and then $Rank(G_{d})<d$, which leads
that $\mathbf{1}$, $\bm{\alpha}_{d}^{1}$,$\ldots$,$\bm{\alpha}_{d}^{d-1}$ are
linearly dependent, where

$\displaystyle\bm{\alpha}_{d}^{j}=(\alpha_{i_{1}}^{j},\alpha_{i_{2}}^{j},\ldots,\alpha_{i_{d}}^{j})\quad(j=1,\ldots,d-1).$

Thus there exist not all zero elements $b_{0},b_{1},\ldots,b_{d-1}$ in
$\mathbb{F}_{q}$, such that $\alpha_{i_{1}},\ldots,\alpha_{i_{d}}$ are
pairwise different roots of the polynomial
$f(x)=\sum\limits_{i=0}^{d-1}b_{i}x^{i}$, it contradicts with the assumption
$\deg f\leq d-1$. Thus $d\geq h+1$. Similarly, we can obtain that $d\geq k-h$.
$\hfill\Box$

###  2.2 Self-dual $m$-MDS codes

The $m$-MDS code is defined in the following

######  Definition 2.3 ([23])

For a linear code $\mathcal{C}$ with parameters $[n,k,d]$, the Singleton
defect is defined as $S(\mathcal{C})=n+1-k-d$. $\mathcal{C}$ is called an
$m$-MDS code if $S(\mathcal{C})=S(\mathcal{C^{\perp}})=m$. Especially,
$\mathcal{C}$ is called an MDS or NMDS code if $m=0$ or $1$, respectively.

It is easy to see that a self-dual code is always an $m$-MDS code, the
following lemma gives the tight upper bound for the Singleton defect of the
self-dual TGRS code.

###### Lemma 2.3

If $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is self-dual, then it is
an $m$-MDS code with

$m\leq\min\\{t,k-h,h+1\\}.$

Proof. By Lemma 2.3, we have

$S(\mathcal{C}_{k,n}^{\perp}(\bm{\alpha},\bm{v},t,h,\eta))\leq
k+1-\max\\{h+1,k-h\\}.$

Furthermore, by the definition for the TGRS code, one has

$S(\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta))\leq t.$

Thus if $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is a self-dual code,
then

$\displaystyle\qquad\qquad
S(\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta))=S(\mathcal{C}_{k,n}^{\perp}(\bm{\alpha},\bm{v},t,h,\eta))\leq\min\\{t,k-h,h+1\\}.\qquad\qquad\hfill\Box$

The following lemma gives a sufficient and necessary condition for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},1,k-1,\eta)$ to be MDS or NMDS,
respectively.

######  Lemma 2.4 (Lemma 2.6 in [15])

For positive integers $k$ and $n$ with $k<n$ and
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{F}_{q}^{n}$ with
$\alpha_{i}\neq\alpha_{j}$ $(i\neq j)$, denote

$S_{k}(1,\bm{\alpha})=\Big{\\{}\sum_{i\in I}\alpha_{i}~{}\bigg{|}~{}\forall
I\subsetneq\\{1,\ldots,n\\}\text{~{}with~{}}|I|=k\Big{\\}},$

then we have

$(1)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is an MDS code if and
only if $\eta^{-1}\notin S_{k}(\bm{\alpha})$;

$(2)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is an NMDS code if
and only if $\eta^{-1}\in S_{k}(\bm{\alpha})$.

In the following lemma, a sufficient and necessary condition for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ to be MDS, NMDS or $2$-MDS
is given, respectively.

###### Lemma 2.5

For positive integers $k$ and $n$ with $k<n$ and
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{n})\in\mathbb{F}_{q}^{n}$ with
$\alpha_{i}\neq\alpha_{j}$ $(i\neq j)$, denote

$\displaystyle S_{k}(2,\bm{\alpha})=\left\\{\sum_{i\in
I}\alpha_{i}\sum_{\begin{subarray}{c}{i,j\in I}\\\ i\neq
j\end{subarray}}\alpha_{i}\alpha_{j}-\\!\\!\\!\\!\sum_{\begin{subarray}{c}{i,j,l\in
I}\\\ {(i-j)(i-l)(j-l)\neq
0}\end{subarray}}\\!\\!\\!\\!\alpha_{i}\alpha_{j}\alpha_{l}~{}\bigg{|}~{}\forall
I\subsetneq\\{1,\ldots,n\\}\text{~{}with~{}}|I|=k\right\\}$

and

$\displaystyle\tilde{S}_{k}(2,\bm{\alpha})=\left\\{-\\!\\!\\!\\!\sum_{\begin{subarray}{c}{i,j,l\in\tilde{I}}\\\
{(i-j)(i-l)(j-l)\neq
0}\end{subarray}}\\!\\!\\!\\!\alpha_{i}\alpha_{j}\alpha_{l}~{}\bigg{|}~{}\forall\tilde{I}\subseteq\\{1,\ldots,n\\}\text{~{}with~{}}|\tilde{I}|=k+1~{}\text{and}~{}\sum\limits_{i\in\tilde{I}}\alpha_{i}=0\right\\},$

then we have

$(1)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ is an MDS code if and
only if $\eta^{-1}\notin S_{k}(2,\bm{\alpha})$;

$(2)$ $S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=1$ if
and only if $\eta^{-1}\in
S_{k}(2,\bm{\alpha})\backslash\tilde{S}_{k}(2,\bm{\alpha})$;

$(3)$ $S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=2$ if
and only if $\eta^{-1}\in
S_{k}(2,\bm{\alpha})\cap\tilde{S}_{k}(2,\bm{\alpha})$.

###### Remark 2.1

Taking $k=3$, $n=5$, $q=5$ and $\bm{\alpha}=(0,1,2,3,4)$ in Lemma 2.4, by
ccomputing directly, we have

$\displaystyle
S_{k}(2,\bm{\alpha})=\\{0,1,2,3\\}~{}~{}\text{and}~{}~{}\tilde{S}_{k}(2,\bm{\alpha})=\\{1,2,3\\},$

which means that $S_{k}(2,\bm{\alpha})\backslash
S_{k}(2,\bm{\alpha})=\\{0\\},$ and so it is non-empty.

The proof for Lemma 2.5. By the definition for the TGRS code, it is easy to
see that $S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}\leq
2$. In the following, we first give the proof for $(1)$, and then show that
$(3)$ holds, finally, by $(1)$ and $(3)$, we can get $(2)$ directly.

For the convience, we denote $w_{H}(\mathbf{c})$ to be the Hamming weight for
a vector $\mathbf{c}\in\mathbb{F}_{q}^{n}$.

The proof for $(1)$. It is enought to prove that the converse-negative
proposition for $(1)$ holds. In fact, if
$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}\neq 0$, then
there exist some $I\subsetneq\\{1,\ldots,n\\}$ with $|I|=k$, and
$f(x)=a_{0}+a_{1}+\ldots+a_{k-1}x^{k-1}+\eta a_{k-2}\eta
x^{k+1}\in\mathbb{F}_{q}[x]$, such that $f(a_{i})=0~{}(\forall i\in I)$, which
implies $a_{k-2}\neq 0$, and then there exists some $b\in\mathbb{F}_{q}$ such
that

$\displaystyle f(x)=\eta a_{k-2}(x-b)\prod_{i\in I}(x-\alpha_{i}).$

Thus one has

$\displaystyle-b-\sum_{i\in
I}\alpha_{i}=0\quad\text{and}\quad-\eta\left(b\sum_{\begin{subarray}{c}{i,j\in
I}\\\ i\neq
j\end{subarray}}\alpha_{i}\alpha_{j}+\sum_{\begin{subarray}{c}{i,j,l\in I}\\\
{(i-j)(i-l)(j-l)\neq
0}\end{subarray}}\alpha_{i}\alpha_{j}\alpha_{l}\right)=1,$

and so $\eta^{-1}\in S_{k}(\bm{\alpha})$.

Conversely, for $\eta^{-1}\in S_{k}(\bm{\alpha})$, there exists some
$I\subsetneq\\{1,\ldots,n\\}$ with $|I|=k$, such that

$\displaystyle\eta=\left(\sum_{i\in
I}\alpha_{i}\sum_{\begin{subarray}{c}{i,j\in I}\\\ i\neq
j\end{subarray}}\alpha_{i}\alpha_{j}-\sum_{\begin{subarray}{c}{i,j,l\in I}\\\
{(i-j)(i-l)(j-l)\neq
0}\end{subarray}}\alpha_{i}\alpha_{j}\alpha_{l}\right)^{-1},$

thus $f(x)=\eta(x+\sum_{i\in I}\alpha_{i})\prod_{i\in
I}(x-\alpha_{i})\in\mathcal{V}_{k,t,h,\eta}$ and
$w_{H}(ev_{\bm{\alpha},\bm{v}}(f))\leq n-k$, and then

$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}\neq 0.$

By the above discussions,
$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=0$ if and
only if $\eta^{-1}\notin S_{k}(2,\bm{\alpha})$.

The proof for $(3)$. If
$S(\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta))=2$, then there exist some
$\tilde{I}\subseteq\\{1,\ldots,n\\}$ with $|\tilde{I}|=k+1$, and
$f(x)=a_{0}+a_{1}+\ldots+a_{k-1}x^{k-1}+\eta a_{k-2}\eta
x^{k+1}\in\mathbb{F}_{q}[x]$, such that $f(a_{i})=0~{}(\forall
i\in\tilde{I})$, which implies $a_{k-2}\neq 0$ and then

$\displaystyle f(x)=\eta a_{k-2}\prod_{i\in\tilde{I}}(x-\alpha_{i}).$

Thus one has

$\displaystyle\eta^{-1}=-\sum_{\begin{subarray}{c}{i,j,l\in\tilde{I}}\\\
{(i-j)(i-l)(j-l)\neq 0}\end{subarray}}\alpha_{i}\alpha_{j}\alpha_{l}.$

Conversely, for $\eta^{-1}\in\tilde{S}_{k}(2,\bm{\alpha})$, there exists some
$\tilde{I}\subseteq\\{1,\ldots,n\\}$ with $|\tilde{I}|=k+1$, such that

$\displaystyle\eta=\left(-\sum_{\begin{subarray}{c}{i,j,l\in\tilde{I}}\\\
{(i-j)(i-l)(j-l)\neq
0}\end{subarray}}\\!\\!\\!\\!\alpha_{i}\alpha_{j}\alpha_{l}\right)^{-1},$

thus $f(x)=\eta\prod_{i\in\tilde{I}}(x-\alpha_{i})\in\mathcal{V}_{k,t,h,\eta}$
and $w_{H}(ev_{\bm{\alpha},\bm{v}}(f))\leq n-k-1$, and then

$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=2.$

So far, we know that
$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=2$ if and
only if

$\eta^{-1}\in S_{k}(2,\bm{\alpha})\cap\tilde{S}_{k}(2,\bm{\alpha}).$

By $(1)$ and $(3)$,
$S\big{(}\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)\big{)}=1$ if and
only if $\eta^{-1}\in
S_{k}(2,\bm{\alpha})\backslash\tilde{S}_{k}(2,\bm{\alpha})$. $\hfill\Box$

By Lemma 2.5, the following corollary is immediate.

###### Corollary 2.1

If $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ is self-dual, then

$(1)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ is MDS if and only if
$\eta^{-1}\notin S_{k}(2,\bm{\alpha})$;

$(2)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ is NMDS if and only
if $\eta^{-1}\in S_{k}(2,\bm{\alpha})\backslash\tilde{S}_{k}(2,\bm{\alpha})$;

$(3)$ $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},2,k-2,\eta)$ is $2$-MDS if and
only if $\eta^{-1}\in\tilde{S}_{k}(2,\bm{\alpha})$.

##  3 The parity-check matrix for TGRS codes

By the theory of Vandermonde matrices [26], one has the following lemma, which
is necessary to calculate the parity-check matrix for the code
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$.

###### Lemma 3.1

Denote

$\displaystyle G=\left(\begin{matrix}&1~{}&1~{}&\ldots~{}&1\\\
&\alpha_{1}~{}&\alpha_{2}~{}&\ldots~{}&\alpha_{n}\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&\alpha_{1}^{n-1}~{}&\alpha_{2}^{n-1}~{}&\cdots~{}&\alpha_{n}^{n-1}\end{matrix}\right),$

then for the system of equations over $\mathbb{F}_{q}$

$\displaystyle G(u_{1},u_{2},\ldots,u_{n})^{T}=(0,0,\ldots,1)^{T},$ (3.1)

there is an unique solution $(u_{1},u_{2},\ldots,u_{n})^{T}$, where
$u_{i}=\prod\limits_{j=1,j\neq i}^{n}(\alpha_{i}-\alpha_{j})^{-1}$ $(1\leq
i\leq n)$.

The parity-check matrix for $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$
is given in Theorem 3.1. The following nonations are needed to prove Theorem
3.1.

For $r\in\mathbb{F}_{q}$, $1\leq i_{1},i_{2}\leq n$ and $n\times n$ matrix
$M$, let $P\big{(}i_{1},i_{2}(r)\big{)}$, $Q\big{(}i_{1},i_{2}(r)\big{)}$ and
$T\big{(}i_{1},i_{2}\big{)}$ be $n\times n$ elementary matrixs satisfing the
following conditions, respectively.

$\bullet$ the $i_{1}$th row of $P\big{(}i_{1},i_{2}(r)\big{)}M$ is replaced by
the sum of $r$ times $i_{2}$th row and $i_{1}$th row of $M$;

$\bullet$ the $i_{1}$th column of $P\big{(}i_{1},i_{2}(r)\big{)}M$ is replaced
by the sum of $r$ times $i_{2}$th column and $i_{1}$th column of $M$;

$\bullet$ $MT(i_{1},i_{2})$ is the matrix by exchanging $k$th column and
$(k+1)$th column of $M$.

###### Theorem 3.1

For $1\leq i\leq n$ and any integer $m$, denote $u_{i}=\prod_{j=1,j\neq
i}^{n}(\alpha_{i}-\alpha_{j})^{-1}$ and
$L_{m}=\sum\limits_{l=1}^{n}u_{l}\alpha_{l}^{n-1+m}$, then the TGRS code
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ has a parity-check matrix in
the form

$\tilde{H}_{n-k}(\bm{v})=\left(\begin{matrix}\tilde{\bm{\beta}}_{1},\tilde{\bm{\beta}}_{2},\ldots,\tilde{\bm{\beta}}_{n}\end{matrix}\right),$

where

$\displaystyle\tilde{\bm{\beta}}_{j}=\left(\begin{matrix}&\frac{u_{j}}{v_{j}}\\\
&\vdots\\\ &\frac{u_{j}}{v_{j}}\alpha_{j}^{n-(k+t+1)}\\\
&\frac{u_{j}}{v_{j}}\big{(}\tilde{L}\alpha_{j}^{n-(k+t)}+\alpha_{j}^{n-(h+1)}-\sum\limits_{m=1}^{k-h-1}L_{m}\alpha_{j}^{n-(h+1)-m}\big{)}\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+t-1)}-L_{1}\alpha_{j}^{n-(k+t)}\big{)}\\\
&\vdots\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+1)}-L_{t-1}\alpha_{j}^{n-(k+t)}\big{)}\\\
\end{matrix}\right)_{(n-k)\times n}$

and

$\displaystyle\tilde{L}=\sum\limits_{m=1}^{k-h-1}L_{m}L_{k+t-h-1-m}-\eta^{-1}(1+\eta
L_{k+t-h-1}).$

Proof. To calculate the parity-check matrix $\tilde{H}_{n-k}(\bm{v})$ for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$, we investigate the parity-
check matrix $\tilde{H}_{n-k}(\mathbf{1})$ for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{1},t,h,\eta)$. It is easy to see that

$\displaystyle\tilde{H}_{n-k}(\bm{v})=\left(\begin{matrix}&v_{1}^{-1}~{}&\ldots~{}&v_{1}^{-1}~{}&\ldots~{}&v_{1}^{-1}\\\
&v_{2}^{-1}~{}&\ldots~{}&v_{2}^{-1}~{}&\ldots~{}&v_{2}^{-1}\\\
&\vdots~{}&~{}&\vdots~{}&~{}&\vdots\\\
&v_{n}^{-1}~{}&\ldots~{}&v_{n}^{-1}~{}&\ldots~{}&v_{n}^{-1}\\\
\end{matrix}\right)\tilde{H}_{n-k}(\mathbf{1}).$ (3.2)

Now, denote

$\displaystyle G=\left(\begin{matrix}&1~{}&1~{}&\ldots~{}&1\\\
&\alpha_{1}~{}&\alpha_{2}~{}&\ldots~{}&\alpha_{n}\\\
&\vdots~{}&\vdots&~{}&\vdots\\\
&\alpha_{1}^{n-1}~{}&\alpha_{2}^{n-1}~{}&\cdots~{}&\alpha_{n}^{n-1}\end{matrix}\right)$

and

$\displaystyle
H=\left(\begin{matrix}&u_{1}\alpha_{1}^{n-1}~{}&\ldots~{}&u_{1}\alpha_{1}^{n-k-1}~{}&\ldots~{}&u_{1}\\\
&u_{2}\alpha_{2}^{n-1}~{}&\ldots~{}&u_{2}\alpha_{2}^{n-k-1}~{}&\ldots~{}&u_{2}\\\
&\vdots~{}&~{}&\vdots~{}&~{}&\vdots\\\
&u_{n}\alpha_{n}^{n-1}~{}&\ldots~{}&u_{n}\alpha_{n}^{n-k-1}~{}&\ldots~{}&u_{n}\\\
\end{matrix}\right),$

then one has

$\displaystyle GH=(l_{i,j})_{1\leq i,j\leq n}$

with $l_{i,j}=L_{i-j}=\sum\limits_{l=1}^{n}u_{l}\alpha_{l}^{n-1+i-j}$. Thus by
Lemma 3.1, we can get

$\displaystyle l_{i,j}=\begin{cases}0,\quad&i<j;\\\ 1,\quad&i=j.\\\
\end{cases}$

For the matrix $GH$, by replacing the $(h+1)$th row by the sum of $\eta$ times
$(k+t)$th row and $(h+1)$th row, one has

$\displaystyle P\big{(}h+1,(k+t)(\eta)\big{)}GH$
$\displaystyle\quad\begin{array}[]{ccccccccccc}1&&~{}\quad h&~{}\quad
h+1&&~{}\quad k+1&&~{}~{}~{}k+t&&&\end{array}$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{ccccccccccc}1&\ldots&0&0&\cdots&0&\ldots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&~{}\vdots\\\ &\cdots&*&1+\eta
l_{k+t,h+1}&\cdots&\eta l_{k+t,k+1}&\ldots&\eta&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&0&\cdots&0&0&\cdots&~{}0\\\
&\cdots&*&*&\cdots&1&\cdots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&1&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&*&*&\cdots&~{}1\\\
\end{array}\right),\begin{array}[]{rrrrrrrrrr}\\\ \\\ h+1\\\ \\\ k{}{}\\\
k+1\\\ \\\ k+t\\\ \\\ \\\ \end{array}$

Noting that $h+1<k+1$ and $\eta\neq 0$, then we have the following column
transform for the matrix $P\big{(}h+1,(k+t)(\eta)\big{)}GH$,

$\displaystyle P\big{(}h+1,(k+t)(\eta)\big{)}GHT_{1}$
$\displaystyle\quad\begin{array}[]{cccccccccccc}1&{}\hfil&h&&&~{}k+1&&k+t&&&\end{array}$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccccccccc}1&\ldots&0&0&\cdots&0&\ldots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&~{}\vdots\\\
&\cdots&*&0&\cdots&0&\ldots&\eta&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&0&\cdots&0&0&\cdots&~{}0\\\
&\cdots&*&*&\cdots&1&\cdots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&1&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&*&*&\cdots&~{}1\\\
\end{array}\right),\begin{array}[]{rrrrrrrrrr}\\\ \\\ h+1\\\ \\\ k{}{}\\\
k+1\\\ \\\ k+t\\\ \\\ \\\ \end{array}$

where

$\displaystyle T_{1}=Q\big{(}h+1,(k+t)(-\eta^{-1}(1+\eta
l_{k+t,h+1}))\big{)}\prod_{i=h+2}^{k+t-1}Q\big{(}i,(k+t)(-l_{k+t,i})\big{)}.$

Hence, one has

$\displaystyle P_{h+1,k+t}(\eta)GHT_{1}T_{2}$
$\displaystyle\quad\begin{array}[]{cccccccccccc}1&{}\hfil&h&&&~{}k+1&&k+t&&&\end{array}$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccccccccc}1&\ldots&0&0&\cdots&0&\ldots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&~{}\vdots\\\
&\cdots&*&0&\cdots&0&\ldots&\eta&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&0&\cdots&0&\cdots&0&0&\cdots&~{}0\\\
&\cdots&*&*&\cdots&1&\cdots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&1&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&*&*&\cdots&~{}1\\\
\end{array}\right),\begin{array}[]{rrrrrrrrrr}\\\ \\\ h+1\\\ \\\ k{}{}\\\
k+1\\\ \\\ k+t\\\ \\\ \\\ \end{array}$

where

$\displaystyle
T_{2}=\prod_{i=h+1}^{k-1}Q\big{(}i,k(-l_{k,i})\big{)}\cdots\prod_{i=h+1}^{h+2}Q\big{(}i,(h+3)(-l_{h+3,i})\big{)}\cdot
Q\big{(}h+1,(h+2)(-l_{h+2,h+1})\big{)}.$

By exchanging the $(h+1)$th column and the $(k+t)$th column of the matrix
$P\big{(}(h+1),(k+t)(\eta)\big{)}GHT_{1}T_{2}$, one has

$\displaystyle P_{h+1,k+t}(\eta)GHT_{1}T_{2}T(h+1,k+t)$
$\displaystyle\quad\begin{array}[]{cccccccccccc}1&{}\hfil&h&&&~{}k+1&&k+t&&&\end{array}$
$\displaystyle=$
$\displaystyle\left(\begin{array}[]{cccccccccccc}1&\ldots&0&0&\cdots&0&\ldots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&~{}\vdots\\\
&\cdots&*&\eta&\cdots&0&\ldots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&0&\cdots&0&\cdots&0&0&\cdots&~{}0\\\
&\cdots&*&*&\cdots&1&\cdots&0&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&1&0&\cdots&~{}0\\\
\vdots&&\vdots&\vdots&&\vdots&&\vdots&\vdots&&\vdots\\\
&\cdots&*&*&\cdots&*&\cdots&*&*&\cdots&~{}1\\\
\end{array}\right).\begin{array}[]{rrrrrrrrrr}\\\ \\\ h+1\\\ \\\ k{}{}\\\
k+1\\\ \\\ k+t\\\ \\\ \\\ \end{array}$

Now by calculating the $j$th $(j=k+1,\ldots,n)$ column of
$HT_{1}T_{2}T(h+1,k+t)$, we can get $\tilde{H}_{n-k}(\mathbf{1})$ directly.
And then by $(\ref{HH1})$, the parity-check matrix for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is given by

$\displaystyle\tilde{H}_{n-k}(\bm{v})=\left(\begin{matrix}\tilde{\bm{\beta}}_{1},\tilde{\bm{\beta}}_{2},\ldots,\tilde{\bm{\beta}}_{n}\end{matrix}\right),$

where

$\displaystyle\tilde{\bm{\beta}}_{j}=\left(\begin{matrix}&\frac{u_{j}}{v_{j}}\\\
&\vdots\\\ &\frac{u_{j}}{v_{j}}\alpha_{j}^{n-(k+t+1)}\\\
&\frac{u_{j}}{v_{j}}\big{(}\tilde{L}\alpha_{j}^{n-(k+t)}+\alpha_{j}^{n-(h+1)}-\sum\limits_{m=1}^{k-h-1}L_{m}\alpha_{j}^{n-(h+1)-m}\big{)}\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+t-1)}-L_{1}\alpha_{j}^{n-(k+t)}\big{)}\\\
&\vdots\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+1)}-L_{t-1}\alpha_{j}^{n-(k+t)}\big{)}\\\
\end{matrix}\right)_{(n-k)\times n}$

and

$\displaystyle\tilde{L}=\sum\limits_{m=1}^{k-h-1}L_{m}L_{k+t-h-1-m}-\eta^{-1}(1+\eta
L_{k+t-h-1}).$

$\hfill\Box$

###### Remark 3.1

Let
${H}_{n-k}(\bm{v})=\prod\limits_{i=1}^{t-1}P\big{(}n-(k+t)+1,(n-(k+t)+1+i)(L_{k-h-1-i})\big{)}\tilde{H}_{n-k}(\bm{v})$,
then the TGRS code $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ has a
parity-check matrix in the form

$H_{n-k}(\bm{v})=\left(\begin{matrix}\bm{\beta}_{1},\bm{\beta}_{2},\ldots,\bm{\beta}_{n}\end{matrix}\right)$

with

$\displaystyle\bm{\beta}_{j}=\left(\begin{matrix}&\frac{u_{j}}{v_{j}}\\\
&\vdots\\\ &\frac{u_{j}}{v_{j}}\alpha_{j}^{n-(k+t+1)}\\\
&\frac{u_{j}}{v_{j}}\bigg{(}\alpha_{j}^{n-(k+t)}+\eta\Big{(}-\alpha_{j}^{n-(h+1)}+\sum\limits_{m=1}^{k+t-(h+1)}L_{m}\alpha_{j}^{n-(h+1)-m}\Big{)}\bigg{)}\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+t-1)}-L_{1}\alpha_{j}^{n-(k+t)}\big{)}\\\
&\vdots\\\
&\frac{u_{j}}{v_{j}}\big{(}\alpha_{j}^{n-(k+1)}-L_{t-1}\alpha_{j}^{n-(k+t)}\big{)}\\\
\end{matrix}\right)_{(n-k)\times n}.$

##  4 Self-dual TGRS codes

###  4.1 A sufficient and necessary condition for a TGRS code to be self-dual

Before giving a sufficient and necessary condition for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$$(h+t=k,~{}h\geq t)$ to be
self-dual, we need the following lemma.

###### Lemma 4.1

For any integer $m\geq 1$, denote

$u_{l}=\prod\limits_{j=1,j\neq l}^{n}(\alpha_{l}-\alpha_{j})^{-1}~{}(1\leq
l\leq n),~{}~{}~{}~{}L_{m}=\sum\limits_{l=1}^{n}u_{l}\alpha_{l}^{n-1+m}$

and

$a_{j}=(-1)^{n-j}\sum\limits_{1\leq l_{1}<\cdots<l_{n-j}\leq
n}\alpha_{l_{1}}\cdots\alpha_{l_{n-j}}~{}(1\leq j\leq n-1),$

then $L_{1}=\cdots=L_{s-1}=0$ if and only if $a_{n-1}=\cdots=a_{n-(s-1)}=0$.
Furthermore, if $L_{1}=\cdots=L_{s-1}=0$, then $L_{2s-1}=-a_{n-(2s-1)}$.

Proof. Let
$P(x)=\prod\limits_{l=1}^{n}(x-\alpha_{l})=x^{n}+\sum\limits_{j=0}^{n-1}a_{j}x^{j}$,
by $P(\alpha_{l})=0~{}(l=1,\ldots,n)$, we have
$\alpha_{l}^{n}=-\sum\limits_{j=0}^{n-1}a_{j}\alpha_{l}^{j}$, and then by
Lemma 3.1, for any $1\leq m\leq n$, it holds that

$\displaystyle\begin{aligned}
L_{m}&=-\sum_{j=1}^{n-1}a_{j}\sum_{l=1}^{n}u_{l}\alpha_{l}^{j+m-1}=-\sum_{j=n-m}^{n-1}a_{j}\sum_{l=1}^{n}u_{l}\alpha_{l}^{j+m-1}=-a_{n-m}-\sum_{j=1}^{m-1}a_{n-m+j}L_{j}.\end{aligned}$
(4.1)

If $L_{1}=\cdots=L_{s-1}=0$, then $(\ref{lm})$ yields that

$\displaystyle L_{i}=-a_{n-i}-\sum_{j=1}^{i-1}a_{n-i+j}L_{j}=-a_{n-i}~{}(1\leq
i\leq s-1),$

thus $a_{n-1}=\cdots=a_{n-(s-1)}=0$.

Conversely, if $a_{n-1}=\cdots=a_{n-(s-1)}=0$, then $(\ref{lm})$ yields
$L_{1}=\cdots=L_{s-1}=0$ directly.

Furthermore, if $L_{1}=\cdots=L_{s-1}=0$, then

$\displaystyle L_{2s-1}$
$\displaystyle=-a_{n-(2s-1)}-\sum_{j=1}^{2s-2}a_{n-(2s-1)+j}L_{j}$
$\displaystyle=-a_{n-(2s-1)}-\sum_{j=1}^{s-1}a_{n-(2s-1)+j}L_{j}-\sum_{j=1}^{s-1}a_{n-s+j}L_{s-1+j}$
$\displaystyle=-a_{n-(2s-1)}.$

$\hfill\Box$

###### Theorem 4.1

Let $n$, $k$, $t$ and $h$ be non-negative integers with $n=2k$, $k\geq 4$,
$t\geq 1$, $k-1+t\leq n$ and $h+t=k$. For $\eta\in\mathbb{F}_{q}^{*}$,
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is self-dual if the following
two conditions hold,

$(1)$ there exists some $\lambda\in\mathbb{F}_{q}^{*}$ such that
$\frac{v_{i}^{2}}{u_{i}}=\lambda$ $(1\leq i\leq n)$;

$(2)$ $a_{n-m}=0~{}(m=1,\ldots,t-1)$ and $2-\eta a_{n-(2t-1)}=0$.

Conversely, if $h\geq t$ and $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$
is self-dual, then $(1)$-$(2)$ hold.

Proof. In the case $h=k-1$, the proof is the same as that for Theorem $2.8$
[15]. Now, we assume $h\leq k-2$. It is easy to check that
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is self-dual if conditions
$(1)$-$(2)$ hold.

Suppose $h\geq t$ and $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is
self-dual, denote

$\displaystyle G_{k}=\left(\begin{matrix}\mathbf{g}_{0}~{}\\\
\mathbf{g}_{1}~{}\\\ \vdots~{}\\\ \mathbf{g}_{k-1}~{}\\\
\end{matrix}\right),\qquad H_{n-k}=\left(\begin{matrix}\mathbf{h}_{0}~{}\\\
\mathbf{h}_{1}~{}\\\ \vdots~{}\\\ \mathbf{h}_{n-k-1}~{}\\\
\end{matrix}\right).$

Since $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$ is self-dual, then for
any $0\leq i\leq k-1$, we have
$\mathbf{g}_{i}\in\mathrm{Span}\\{\mathbf{h}_{0},\ldots,\mathbf{h}_{k-1}\\}$,
which means there is a vector
$(a_{i,0},a_{i,1},\ldots,a_{i,k-1})\in\mathbb{F}_{q}^{k}\backslash\\{(0,\ldots,0)\\}$,
such that $\mathbf{g}_{i}=(a_{i,0},a_{i,1},\ldots,a_{i,k-1})H_{n-k}$. Now for
any $0\leq i\leq k-1$, denote

$\displaystyle f_{i}(x)=$ $\displaystyle
a_{i,0}+a_{i,1}x+\ldots+a_{i,h-1}x^{h-1}+a_{i,h}\Big{(}-\eta
x^{2k-(h+1)}+\eta\sum\limits_{m=1}^{k+t-(h+1)}L_{m}x^{2k-(h+1)-m}+x^{k-t}\Big{)}$
$\displaystyle+a_{i,h+1}\big{(}x^{h+1}-L_{1}x^{k-t}\big{)}+\ldots+a_{i,k-1}\big{(}x^{k-1}-L_{t-1}x^{k-t}\big{)},$

then for any $1\leq j\leq n$, we have

$\displaystyle\frac{v_{j}^{2}}{u_{j}}=f_{0}(\alpha_{j}),~{}~{}\frac{v_{j}^{2}}{u_{j}}\big{(}\alpha_{j}^{h}+\eta\alpha_{j}^{k-1+t}\big{)}=f_{h}(\alpha_{j})~{}~{}\text{and}~{}~{}\frac{v_{j}^{2}}{u_{j}}\alpha_{j}^{i}=f_{i}(\alpha_{j})(i\neq
h),$

which leads

$\displaystyle
f_{0}(\alpha_{j})\big{(}\alpha_{j}^{h}+\eta\alpha_{j}^{k-1+t}\big{)}-f_{h}(\alpha_{j})=0$
(4.2)

and

$\displaystyle f_{0}(\alpha_{j})\alpha_{j}^{i}-f_{i}(\alpha_{j})=0\quad(i\neq
h).$ (4.3)

Thus by the assumption $h+t=k$, we have
$\mathrm{\text{deg}}\big{(}f_{0}(x)x^{h-1}-f_{h-1}(x)\big{)}\leq n-2$. Noting
that $\alpha_{1},\ldots,\alpha_{n}$ are pairwise distinct roots of
$f_{0}(x)x^{h-1}-f_{h-1}(x)$, so $f_{0}(x)x^{h-1}-f_{h-1}(x)=0$, which implies
that $a_{0,h}=0$ and

$\displaystyle\begin{aligned}
f_{0}(x)=&a_{0,0}+a_{0,1}x+\ldots+a_{0,h-1}x^{h-1}\\\
&+a_{0,h+1}\big{(}x^{h+1}-L_{1}x^{h}\big{)}+\ldots+a_{0,k-1}\big{(}x^{k-1}-L_{t-1}x^{h}\big{)},\end{aligned}$
(4.4)

Namely, $\mathrm{\text{deg}}\big{(}f_{0}(x)x^{k-1}-f_{k-1}(x)\big{)}\leq n-2$.
While $\alpha_{1},\ldots,\alpha_{n}$ are pairwise distinct roots of
$f_{0}(x)x^{k-1}-f_{k-1}(x)$, so $f_{0}(x)x^{k-1}-f_{k-1}(x)=0$, which implies
that

$\displaystyle a_{0,j}=0\quad(k-h+1\leq j\leq k-1),$ (4.5)

thus we have $\deg f_{0}(x)\leq k-h$.

Noting that $h\geq t$, one has $(k-1+t)+(k-h)=n+t-h-1<n$, so

$\displaystyle\mathrm{deg}\Big{(}f_{0}(x)\big{(}x^{h}+\eta
x^{k-1+t}\big{)}-f_{h}(x)\Big{)}\leq n-1.$

Since $\alpha_{1},\ldots,\alpha_{n}$ are pairwise distinct roots of
$f_{0}(x)\big{(}x^{h}+\eta x^{k-1+t}\big{)}-f_{h}(x)$, one has

$\displaystyle f_{0}(x)\big{(}x^{h}+\eta x^{k-1+t}\big{)}-f_{h}(x)=0,$ (4.6)

it means that

$\displaystyle f_{0}(x)=a_{0,0}\neq 0,$

thus $(1)$ holds. And then

$\displaystyle a_{0,0}\big{(}x^{h}+\eta x^{k-1+t}\big{)}-f_{h}(x)=0,$

it implies that

$\displaystyle a_{0,0}\big{(}x^{h}+\eta x^{k+t-1}\big{)}-$ $\displaystyle
a_{h,h}\Big{(}-\eta
x^{k+t-1}+\eta\sum\limits_{m=1}^{2t-1}L_{m}x^{k+t-1-m}+x^{h}\Big{)}$
$\displaystyle+$ $\displaystyle
a_{h,h+1}\big{(}x^{h+1}-L_{1}x^{h}\big{)}+\ldots+a_{h,k-1}\big{(}x^{k-1}-L_{t-1}x^{h}\big{)}=0,$

and so

$\displaystyle\begin{cases}L_{m}=0~{}(i=1,\ldots,t-1);\\\
a_{0,0}+a_{h,h}=0;\\\ a_{0,0}-a_{h,h}(1+\eta L_{2t-1})=0,\end{cases}$ (4.7)

thus we have

$\displaystyle\begin{cases}L_{m}=0~{}(i=1,\ldots,t-1);\\\ 2+\eta
L_{2t-1}=0.\end{cases}$ (4.8)

Now by $(\ref{E2})$ and Lemma 4.1, $(2)$ holds. $\hfill\Box$

###### Remark 4.1

If $2\mid q$, then Theorem 4.1 $(2)$ can be replaced by the equation

$a_{n-2t+1}=a_{n-m}=0~{}(m=1,\ldots,t-1).$

###  4.2 The existence for a self-dual TGRS code over $\mathbb{F}_{q}$ $(2\mid
q)$

###### Theorem 4.2

let $t$, $k$, $s$ and $m$ be positive integers with $s\mid m$ and
$2^{s-1}-t>0$, $\mathbb{F}_{2^{s}}=\\{\alpha_{1},\ldots,\alpha_{2^{s}}\\}$ be
the subfield of $\mathbb{F}_{2^{m}}$ and $\eta\in\mathbb{F}_{2^{m}}^{*}$. If
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{2^{s}})$ and
$\bm{v}=(v_{1},\ldots,v_{2^{s}})$ with $v_{i}=\sum_{j=1,j\neq
i}^{2^{s}}(\alpha_{i}-\alpha_{j})^{-2^{m-1}}$ $(i=1,\ldots,2^{s})$, then
$\mathcal{C}_{2^{s-1},2^{s}}(\bm{\alpha},\bm{v},t,2^{s-1}-t,\eta)$ is a self-
dual $m$-MDS code with $m\leq\min\\{t,2^{s-1}-t+1\\}$.

Proof. It is easy to see that $v_{i}^{2}=\sum_{j=1,j\neq
i}^{2^{s}}(\alpha_{i}-\alpha_{j})^{-1}$ $(j=1,\ldots,2^{s})$. By
$\mathbb{F}_{2^{s}}=\\{\alpha_{1},\ldots,\alpha_{2^{s}}\\}$ we have
$\prod\limits_{i=1}^{2^{s}}(x-\alpha_{i})=x^{2^{s}}-x$, and so

$a_{r}=(-1)^{2^{s}-r}\sum\limits_{1\leq l_{1}<\cdots<l_{2^{s}-r}\leq
n}\alpha_{l_{1}}\cdots\alpha_{l_{2^{s}-r}}=0~{}~{}(1\leq r\leq 2^{s}-1).$

Now by Theorem 4.1 and Lemma 2.3, we complete the proof. $\hfill\Box$

By Lemma 2.1 and Theorem 4.2, a class of self-dual MDS codes can be get from
the following

###### Corollary 4.1

By taking $s\mid m$ with $s<m$ and
$\eta\in\mathbb{F}_{2^{m}}^{*}\backslash\mathbb{F}_{2^{s}}$ in Theorem 4.2,
then $\mathcal{C}_{2^{s-1},2^{s}}(\bm{\alpha},\bm{v},t,2^{s-1}-t,\eta)$ is a
self-dual MDS code.

###### Theorem 4.3

For positive integers $m$ and $l$ with $l\leq m$, let $w_{1},\ldots,w_{m}$ be
a basis of $\mathbb{F}_{2^{m}}$ over $\mathbb{F}_{2}$, denote

$\\{\alpha_{1},\ldots,\alpha_{2^{l}}\\}=\\{b_{1}w_{1}+\cdots+b_{l}w_{l}|b_{i}\in\mathbb{F}_{2}~{}(i=1,\ldots,l)\\}.$

Then for any positive integer $k\geq 2^{l-1}$,
$\eta\in\mathbb{F}_{2^{m}}^{*}$, $\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{2k})$
and $\bm{v}=(v_{1},\ldots,v_{2k})$ with $v_{i}=\sum\limits_{j=1,j\neq
i}^{2k}(\alpha_{i}-\alpha_{j})^{-2^{m-1}}$ $(i=1,\ldots,2k)$,
$\mathcal{C}_{k,2k}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is self-dual if one of the
following conditions holds.

$(1)$ $k=2^{l-1}$;

$(2)$ $l_{1}$ is odd with $3\leq l_{1}\leq m-l$ and
$k=\frac{2^{l}+l_{1}+1}{2}$, $\alpha_{2^{l}+j}=w_{l+j}(j=1,\ldots,l_{1})$ and
$\alpha_{2^{l}+l_{1}+1}=\sum\limits_{j=1}^{l_{1}}w_{l+j}$;

$(3)$ $l_{1}$ is even with $4\leq l_{1}\leq m-l$ and
$k=\frac{2^{l}+l_{1}}{2}$,
$\alpha_{2^{l}+j}=w_{l+j}+w_{l+j+1}(j=1,\ldots,l_{1}-1)$ and
$\alpha_{2^{l}+l_{1}}=w_{l+1}+w_{l+l_{1}}$.

Proof. If one of the conditions $(1)$-$(3)$ holds, it is easy to verify that
$\sum_{i=1}^{2k}\alpha_{i}=0$ and $v_{i}^{2}=\sum_{j=1,j\neq
i}^{2k}(\alpha_{i}-\alpha_{j})^{-1}$ $(j=1,\ldots,2k)$. Then by Theorem 4.1,
$\mathcal{C}_{k,2k}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is self-dual.$\hfill\Box$

By Lemma 2.4 and Theorem 4.3, we know that the self-dual code
$\mathcal{C}_{k,2k}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is MDS or NMDS, namely,

###### Corollary 4.2

In Theorem 4.3, if $\eta\notin S_{l}(1,\bm{\alpha})$, then
$\mathcal{C}_{k,2k}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is a self-dual MDS.
Otherwise, $\mathcal{C}_{k,2k}(\bm{\alpha},\bm{v},1,k-1,\eta)$ is a self-dual
NMDS code.

###### Theorem 4.4

Let $s$, $\lambda$, $l$, $t$ be positive integers with $t<l$, $q=2^{s}$,
$q_{1}=2^{\lambda}$ and $\eta\in\mathbb{F}_{q}^{*}$. Assume that
$\mathbb{F}_{q}$ is the splitting field of
$A(x)=x^{2l}+bx+c~{}(b,c\in\mathbb{F}_{q_{1}}^{*})$ over $\mathbb{F}_{q_{1}}$
and $\alpha_{1},\ldots,\alpha_{2l}$ are all roots of $A(x)$ over
$\mathbb{F}_{q}$, then $C_{l,2l}(\bm{\alpha},\mathbf{1},t,l-t,\eta)$ is a
self-dual $m$-MDS code with $m\leq\min\\{t,l-t+1\\}$, where
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{2l})$.

Proof. On the one hand, by $A(x)=x^{2l}+bx+c$ we have
$A^{{}^{\prime}}(x)=b\neq 0$, and then $\gcd(A(x),A^{{}^{\prime}}(x))=1$,
namely, $\alpha_{i}\neq\alpha_{j}$ $(i\neq j)$. On the other hand, by the
assumptions, we have $A(x)=\prod_{i=1}^{2l}(x-\alpha_{i})$, so
$a_{2l-(2t-1)}=a_{2l-i}=0$ $(i=1,\ldots,{t-1})$ and
$A^{{}^{\prime}}(x)=\sum\limits_{i=1}^{2l}\prod\limits_{j\neq
i,j=1}^{2l}(x-\alpha_{j})$, which means

$\displaystyle u_{i}=m^{{}^{\prime}}(\alpha_{i})^{-1}=-b^{-1}\neq 0.$

Now by Theorem 4.1 and Lemma 2.3, we complete the proof. $\hfill\Box$

By Theorem 4.4 and Lemmas 2.4-2.5, we have the following two corollaries.

###### Corollary 4.3

By taking $t=1$ in Theorem 4.4, if $\eta\notin S_{l}(1,\bm{\alpha})$, then
$\mathcal{C}_{l,2l}(\bm{\alpha},\bm{1},1,l-1,\eta)$ is a self-dual MDS code.
Otherwise, $\mathcal{C}_{l,2l}(\bm{\alpha},\bm{1},1,l-1,\eta)$ is a self-dual
NMDS code.

###### Corollary 4.4

By taking $t=2$ in Theorem 4.4, the following assertions hold.

$(1)$ If $\eta\notin S_{l}(2,\bm{\alpha})$, then
$\mathcal{C}_{l,2l}(\bm{\alpha},\bm{1},2,l-2,\eta)$ is a self-dual MDS code;

$(2)$ If $\eta^{-1}\in
S_{l}(2,\bm{\alpha})\backslash\tilde{S}_{l}(2,\bm{\alpha})$, then
$\mathcal{C}_{l,2l}(\bm{\alpha},\bm{1},2,l-2,\eta)$ is a self-dual NMDS code;

$(3)$ If $\eta^{-1}\in\tilde{S}_{l}(2,\bm{\alpha})\cap S_{l}(2,\bm{\alpha})$,
then $\mathcal{C}_{l,2l}(\bm{\alpha},\bm{1},2,l-2,\eta)$ is a self-dual
$2$-MDS code.

###  4.3 The existence for a self-dual TGRS code over $\mathbb{F}_{q}$
$(2\nmid q)$

###### Theorem 4.5

Let $s$, $\lambda$, $l$, $t$ be positive integers, $p$ an odd prime with
$lp-t>0$, $p\nmid(2t-1)$, $q=p^{s}$ and $q_{1}=p^{\lambda}$. Assume that
$\mathbb{F}_{q}$ is the splitting field of
$m(x)=x^{2lp}+bx^{2lp-(2t-1)}+c~{}(b,c\in\mathbb{F}_{q_{1}}^{*})$ over
$\mathbb{F}_{q_{1}}$, $\alpha_{1},\ldots,\alpha_{2lp}$ are all roots of $m(x)$
and $v_{i}=(\alpha_{i}^{lp-t})^{-1}$ $(i=1,\ldots,2lp)$. Then for
$\eta=2{b^{-1}}$, $C_{lp,2lp}(\bm{\alpha},\bm{v},t,lp-t,\eta)$ is a self-dual
$m$-MDS code with $m\leq\min\\{t,lp-t+1\\}$, where
$\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{2lp})$ and
$\bm{v}=(v_{1},\dots,v_{2lp})$.

Proof. On the one hand, since
$m(x)=x^{2lp}+bx^{2lp-(2t-1)}+c~{}(b,c\in\mathbb{F}_{q_{1}}^{*})$, we have
$m^{{}^{\prime}}(x)=-(2t-1)bx^{2lp-2t}$, and then
$\gcd(m(x),m^{{}^{\prime}}(x))=1$, namely, $\alpha_{i}\neq\alpha_{j}$ $(i\neq
j)$. On the other hand, by the assumptions, we have
$m(x)=\prod\limits_{i=1}^{2pl}(x-\alpha_{i})$, so $a_{2pl-i}=0$
$(i=1,\ldots,{t-1})$, $2\\!-\\!\eta a_{2pl-(2t-1)}=0$ and
$m^{{}^{\prime}}(x)=\sum\limits_{i=1}^{2pl}\prod\limits_{j\neq
i,j=1}^{2pl}(x-\alpha_{j})$, which means

$\displaystyle
u_{i}=m^{{}^{\prime}}(\alpha_{i})^{-1}=-((2t-1)b)^{-1}\alpha_{i}^{2t-2lp}=-((2t-1)b)^{-1}v_{i}^{2}\neq
0.$

By Theorem 4.1 and Lemma 2.3, we complete the proof. $\hfill\Box$

By Theorem 4.5 and Lemma 2.5, we have the following corollary.

###### Corollary 4.5

By taking $t=2$ in Theorem 4.5, the following asserctions hold.

$(1)$ If $\eta\notin S_{lp}(2,\bm{\alpha})$, then
$\mathcal{C}_{lp,2lp}(\bm{\alpha},\bm{v},2,lp-2,\eta)$ is a self-dual MDS
code.

$(2)$ If $\eta\in
S_{lp}(2,\bm{\alpha})\backslash\tilde{S}_{lp}(2,\bm{\alpha})$, then
$\mathcal{C}_{lp,2lp}(\bm{\alpha},\bm{v},2,lp-2,\eta)$ is a self-dual NMDS
code.

$(3)$ If $\eta\in S_{lp}(2,\bm{\alpha})\cap\tilde{S}_{lp}(2,\bm{\alpha})$,
then $\mathcal{C}_{lp,2lp}(\bm{\alpha},\bm{v},2,lp-2,\eta)$ is a self-dual
$2$-MDS code.

###### Theorem 4.6

Let $p$ be an odd prime, $t$, $s$ and $m$ be positive integers with
$s\mid\frac{m}{2}$,
$\beta\in\mathbb{F}_{p^{m}}^{*}\backslash\mathbb{F}_{p^{s}}$ and
$\eta=-2\beta^{-1}$. Denote
$\\{\alpha_{1},\ldots,\alpha_{p^{s}-1}\\}=\\{\beta+a|a\in\mathbb{F}_{p^{s}}^{*}\\}$,
then
$\mathcal{C}_{\frac{p^{s}-1}{2},p^{s}-1}(\bm{\alpha},\bm{v},1,\frac{p^{s}-1}{2}-1,\eta)$
is self-dual, where $\bm{\alpha}=(\alpha_{1},\ldots,\alpha_{p^{s}-1})$ and
$\bm{v}=(v_{1},\ldots,v_{p^{s}-1})$ with $v_{i}^{2}=-\alpha_{i}$
$(i=1,\ldots,2^{s})$.

Proof. By calculating directly, one has

$\sum_{j=1}^{p^{s}-1}\alpha_{i}=\sum_{a\in\mathbb{F}_{p^{s}}^{*}}(\beta+a)=-\beta,$

and then by $\eta=-2\beta^{-1}$, we have
$2-\eta(\sum_{j=1}^{p^{s}-1}\alpha_{i})=0$. Furthermore,

$\displaystyle\sum_{j=1,j\neq
i}^{p^{s}-1}(\alpha_{i}-\alpha_{j})^{-1}=-\alpha_{i}~{}(i=1,\ldots,p^{s}-1),$

By $s\mid\frac{m}{2}$, we know that $-\alpha_{i}$ $(i=1,\ldots,p^{s}-1)$ is a
square element in $\mathbb{F}_{p^{m}}$, and then,

$v_{i}^{2}=\sum_{j=1,j\neq i}^{p^{s}-1}(\alpha_{i}-\alpha_{j})^{-1}.$

By Theorem 4.1,
$\mathcal{C}_{\frac{p^{s}-1}{2},p^{s}-1}(\bm{\alpha},\bm{v},1,\frac{p^{s}-1}{2}-1,\eta)$
is self-dual. $\hfill\Box$

By Theorem 4.5 and Lemma 2.4, we know that the self-dual code
$\mathcal{C}_{\frac{p^{s}-1}{2},p^{s}-1}(\bm{\alpha},\bm{v},1,\frac{p^{s}-1}{2}-1,\eta)$
is MDS or NMDS, namely,

###### Corollary 4.6

In Theorem 4.6, if $\eta\notin S_{l}(1,\bm{\alpha})$, then
$\mathcal{C}_{\frac{p^{s}-1}{2},p^{s}-1}(\bm{\alpha},\bm{v},1,\frac{p^{s}-1}{2}-1,\eta)$
is a self-dual MDS code. Otherwise,
$\mathcal{C}_{\frac{p^{s}-1}{2},p^{s}-1}(\bm{\alpha},\bm{v},1,\frac{p^{s}-1}{2}-1,\eta)$
is a self-dual NMDS code.

##  5 Conclusions

In [15], the authors gave the parity-check matrix for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},1,k-1,\eta)$, and then obtained some
self-dual MDS or NMDS codes. In our works, we give the following main results
for $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$, which extends the main
results in [15],

$(1)$ the parity-check matrix for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$;

$(2)$ a sufficient and necessary condition for
$\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)\\!$ $(h+t=k,~{}h\geq t)\\!$ to
be self-dual;

$(3)$ construction for several classes of self-dual $m$-MDS codes with small
$m$ from $\mathcal{C}_{k,n}(\bm{\alpha},\bm{v},t,h,\eta)$.

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$\displaystyle+\frac{H^{2}q_{3}}{3}\left[-27\partial^{2}\gamma_{ij}\frac{1}{\partial^{4}}(\gamma_{kl}\partial^{2}\gamma_{kl})-12\gamma_{ij}\frac{1}{\partial^{4}}(\gamma_{kl}\partial^{4}\gamma_{kl})\right.$
$\displaystyle\hskip
44.0pt\left.+6\partial^{2}\gamma_{ij}\frac{1}{\partial^{4}}({\gamma^{\prime}_{kl}}^{2})+2\gamma^{\prime}_{ij}\frac{1}{\partial^{4}}(\gamma_{kl}\partial^{2}\gamma^{\prime}_{kl})-4\gamma_{ij}\frac{1}{\partial^{4}}(\gamma^{\prime}_{kl}\partial^{2}\gamma^{\prime}_{kl})\right.$
(3.57) $\displaystyle\left.\hskip
44.0pt-10\gamma^{\prime}_{ij}\frac{1}{\partial^{4}}(\gamma^{\prime}_{kl}\partial^{2}\gamma_{kl})-30\partial^{2}\gamma_{ij}\frac{1}{\partial^{4}}(\partial_{m}\gamma_{kl}\partial_{m}\gamma_{kl})-60\partial_{m}\gamma_{ij}\frac{1}{\partial^{4}}(\gamma_{kl}\partial^{2}\partial_{m}\gamma_{kl})\right].$

We could also construct similar non-local field redefinitions that generate
energy-dependent coefficients of the
$(\epsilon_{1}\cdot\epsilon_{3})(\epsilon_{1}\cdot\epsilon_{4})(\epsilon_{2}\cdot\epsilon_{3})(\epsilon_{2}\cdot\epsilon_{4})$
tensor structure involving inverse powers of $s$. This would simply require
different contractions between the fields in the above field redefinitions. We
did not include terms of this form in our ansatz when we imposed the MLT
condition since for this tensor structure no inverse powers of $s$ arose after
steps 1 and 2. They can however be dealt with similarly and the resulting free
parameters can be removed by field redefinitions. If we were to make a
particular choice of field variables, e.g., by enforcing a particular
representation of the Ward identity or soft theorem, then we could use this to
fix some of these free parameters, as in the quartic scalar calculation.

##  4 Conclusions

In this work, we have computed the graviton four-point correlator and quartic
wavefunction coefficient at tree level in de Sitter space for pure Einstein
gravity. We derived our result in several different ways and found agreement
among them. A few natural directions to pursue present themselves:

  * •

A consistency check of our result is to ensure that it respects the
appropriate soft-graviton theorem, e.g., from [88]. We have found it
challenging to prove this, for a few reasons. First, the relevant soft theorem
in [88] for a correlator with soft and hard tensors is written in components.
To compare to our result one needs to make a specific choice of polarizations,
such as the helicity basis, and take derivatives of polarization tensors with
respect to their spatial momentum argument, which is subtle because of the
arbitrariness in defining this dependence and the necessary discontinuities.
This problem appears first for the trispectrum and it is not present in the
soft theorem for the bispectrum, where all contractions of polarization
tensors can be written in terms of the external energies (see, e.g., [13]).
Finally, a naive matching of the coefficients of the polarization tensor
structures does not seem to satisfy the trispectrum soft theorem and we could
not exclude that different structures cancel each other due to dimension-
dependent identities. We postpone further investigations to the future.

  * •

The graviton four-point function should display appropriate invariance under
the isometries of de Sitter spacetime. For correlators at the future conformal
boundary, this invariance should reduce to the set of conformal (and current
conservation) Ward identities. It would be nice to check explicitly that our
result indeed satisfies these constraints.

  * •

In general we expect that in simple and highly constrained theories—such as
pure gravity—higher-order observables should be fixed in terms of the lowest-
order interaction. For example, in flat space all $n$-particle graviton
amplitudes can in principle be derived from the three-graviton amplitude via
BCFW recursion relations [89, 90]. Because of this, one might hope to find
variables in which the graviton four-point function we compute is not much
more complicated than the graviton three-point function. Our final result
(LABEL:final) does not yet realize this hope. It would be nice to find a more
natural set of variables, analogous to spinor-helicity variables for maximally
helicity-violating amplitudes [91]. Cosmological spinor-helicity variables
[14], as used in [16], do not seem to achieve this simplification. One
possibility is that simplifications can be achieved by using a Mellin space
representation [34, 35, 45]; this would also facilitate a comparison to the
anti-de Sitter space result [17], as reviewed in [54]. It would be interesting
to further pursue these directions.

We close with a small historical parable. In a truly heroic computation, Parke
and Taylor computed the six-gluon scattering amplitude in 1985, and presented
it in [92], ending their discussion with the phrase “…we hope to obtain a
simple analytic form for the answer, making our result not only an
experimentalist’s, but also a theorist’s delight.” Less than a year later,
they found a spectacular such formula [91], which catalyzed the modern
amplitudes revolution. We share a similar hope to find a drastic
simplification of the formulas presented in this paper. We should be so lucky.

##### Acknowledgements

We thank Soner Albayrak, Daniel Baumann, Giovanni Cabass, Zongzhe Du, Carlos
Duaso Pueyo, Aaron Hillman, Kurt Hinterbichler, Sadra Jazayeri, Savan Kharel,
Hayden Lee, Guilherme Pimentel, and Akhil Premkumar for useful discussions. HG
is supported jointly by the Science and Technology Facilities Council through
a postgraduate studentship and the Cambridge Trust Vice Chancellor’s Award.
The work of AJ is supported in part by DOE (HEP) Award DE-SC0009924. EP has
been supported in part by the research program VIDI with Project No.
680-47-535, which is (partly) financed by the Netherlands Organisation for
Scientific Research (NWO). DS is supported by a UKRI Stephen Hawking
Fellowship [grant number EP/W005441/1] and a Nottingham Research Fellowship
from the University of Nottingham. For the purpose of open access, the authors
have applied a CC BY public copyright licence to any Author Accepted
Manuscript version arising.

##  Appendix A Polarisation sums

In this appendix, we introduce some of the formulas and formalism required to
deal with exchanges of spinning particles in de Sitter space. See [79, 32, 43,
46] for further details.

The transverse-traceless spin-2 projector is

$\Pi_{ijkl}^{{\bf k}}=\frac{1}{2}\sum_{s}e^{s}_{ij}({\bf k})e^{s}_{kl}(-{\bf
k})=\frac{1}{2}\pi_{ik}^{{\bf k}}\pi_{jl}^{{\bf k}}+\frac{1}{2}\pi_{il}^{{\bf
k}}\pi_{jk}^{{\bf k}}-\frac{1}{2}\pi_{ij}^{{\bf k}}\pi_{kl}^{{\bf k}},$ (A.1)

where $\pi_{ij}^{{\bf k}}=\delta_{ij}-\frac{1}{k^{2}}k_{i}k_{j}$. It is useful
to define the following contraction of this:

$\displaystyle\Pi_{2,2}^{(s)}\equiv\,$
$\displaystyle\frac{24}{s^{4}}\Pi_{ijkl}^{\bf
s}k^{1}_{i}k^{2}_{j}k^{3}_{k}k^{4}_{l}$ (A.2) $\displaystyle=\,$
$\displaystyle-\frac{3}{4}+3\frac{\sum_{a=1}^{4}k_{a}^{2}}{2s^{2}}-3\frac{4(k_{1}^{2}+k_{2}^{2})(k_{3}^{2}+k_{4}^{2})-2k_{1}^{2}k_{2}^{2}-2k_{3}^{2}k_{4}^{2}+\sum_{a=1}^{4}k_{a}^{4}-2(t^{2}-u^{2})^{2}}{4s^{4}}$
$\displaystyle+3\frac{2(t^{2}-u^{2})(k^{2}_{1}-k^{2}_{2})(k^{2}_{3}-k^{2}_{4})+(k_{3}^{2}+k_{4}^{2})(k_{1}^{2}-k_{2}^{2})^{2}+(k_{1}^{2}+k_{2}^{2})(k_{3}^{2}-k_{4}^{2})^{2}}{2s^{6}}$
$\displaystyle+\frac{3(k_{1}-k_{2})^{2}(k_{3}-k_{4})^{2}k_{12}^{2}k_{34}^{2}}{4s^{8}}.$
(A.3)

We also define

$\displaystyle\Pi_{2,1}^{(s)}$
$\displaystyle\equiv\frac{3(k_{1}-k_{2})(k_{3}-k_{4})}{s^{4}}\pi^{\bf
s}_{ij}(k_{1}^{i}-k_{2}^{i})(k_{3}^{j}-k_{4}^{j})$ (A.4)
$\displaystyle=\frac{3(k_{1}-k_{2})(k_{3}-k_{4})}{s^{6}}\left[s^{2}(t^{2}-u^{2})+k_{12}k_{34}(k_{1}-k_{2})(k_{3}-k_{4})\right],$
(A.5) $\displaystyle\Pi_{2,0}^{(s)}$
$\displaystyle\equiv\frac{1}{4}\left(1-\frac{3(k_{1}-k_{2})^{2}}{s^{2}}\right)\left(1-\frac{3(k_{3}-k_{4})^{2}}{s^{2}}\right),$
(A.6)

which can be written in terms of contractions of the rank-2 projectors onto
lower helicities. Lastly, we also define

$\displaystyle\Pi_{1,1}^{(s)}$
$\displaystyle\equiv\frac{k_{12}k_{34}(k_{1}-k_{2})(k_{3}-k_{4})+s^{2}(t^{2}-u^{2})}{s^{4}},$
(A.7) $\displaystyle\Pi_{1,0}^{(s)}$
$\displaystyle\equiv\frac{(k_{1}-k_{2})(k_{3}-k_{4})}{s^{2}}.$ (A.8)

When checking that we recover the correct amplitude limit on the leading
total-energy pole, it is useful to have expressions for these sums in that
limit. Individually the limits are not particularly illuminating but the
following combinations have neat limits:

$\displaystyle
s^{4}\Pi_{2,2}^{(s)}-E_{L}E_{R}s^{2}\Pi_{2,1}^{(s)}+E_{L}^{2}E_{R}^{2}\Pi^{(s)}_{2,0}$
$\displaystyle\xrightarrow{k_{T}\rightarrow
0}S^{2}P_{2}\left(1+\frac{2T}{S}\right),$ (A.9) $\displaystyle
s^{2}\Pi_{1,1}^{(s)}-E_{L}E_{R}\Pi_{1,0}^{(s)}$
$\displaystyle\xrightarrow{k_{T}\rightarrow
0}-SP_{1}\left(1+\frac{2T}{S}\right).$ (A.10)

Here we recognise the Legendre polynomials associated with the exchange of
spinning particles.

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