# Equivalence and automorphism groups of two families of maximum scattered
linear sets

Wei Tang 1 and  Yue Zhou 1 1Department of Mathematics, National University of
Defense Technology, 410073 Changsha, China<EMAIL_ADDRESS>
###### Abstract.

Linear set in projective spaces over finite fields plays central roles in the
study of blocking sets, semifields, rank-metric codes and etc. A linear set
with the largest possible cardinality and the maximum rank is called maximum
scattered. Despite two decades of study, there are only a few number of known
maximum scattered linear sets in projective lines, including two families
constructed in [5], and [6, 14], respectively. This paper aims to solve the
equivalence problem of the linear sets in each of these families and to
determine their automorphism groups.

###### Key words and phrases:

linear set; rank-metric code; finite geometry; linearized polynomial

This is an English translation of the original Chinese version published in
Scientia Sinica Mathematica. DOI: 10.1360/SSM-2022-0073

##  1\. Introduction

The concept of _linear set_ was introduced by Lunardon [12], which is natural
generalization of subgeometrie. In the pasting two decades after this work,
linear sets have been intensively investigated and applied to construct and
characterize various objects in finite geometry and coding theory, including
blocking sets, two-intersection sets, translation spreads of the Cayley
generalized Hexagon, translation ovoids of polar spaces, semifields and rank-
metric codes. We refer to [2, 8, 16, 17, 18] and the references therein.

In this paper, we only concern with the linear sets in a projective line. Let
$\Lambda=\mathrm{PG}(V)=\mathrm{PG}(1,q^{n})$, where $V$ is a $2$-dimensional
vector space over $\mathbb{F}_{q^{n}}$. Then $\Lambda$ is a projective line. A
point set $L$ of $\Lambda$ is said to be an _$\mathbb{F}_{q}$ -linear set_ of
rank $k$ if it is defined by the non-zero vectors of a $k$-dimensional
$\mathbb{F}_{q}$-vector subspace $U$ of $V$, i.e.

$L=L_{U}:=\\{\langle{\bf u}\rangle_{\mathbb{F}_{q^{n}}}:{\bf u}\in
U\setminus\\{{\bf 0}\\}\\}.$

For any _$\mathbb{F}_{q}$ -linear set_ $L_{U}$ with rank $k$, it is not
difficult to see that

$|L|\leq\frac{q^{k}-1}{q-1}.$

If the equality holds, $L_{U}$ is called a scattered linear set. In [3],
Blokhuis and Lavrauw show that a scattered linear set has rank at most $n$,
and they call a scattered linear set of rank $k=n$ in $\Lambda$ _maximum_.

Under the action of $\mathrm{P\Gamma L}(2,q^{n})$, we can always assume that
the linear set $L_{U}$ of rank $n$ in $\Lambda$ does not contain the point
$\langle(0,1)\rangle_{\mathbb{F}_{q^{n}}}$. Consequently, $U$ and $L_{U}$ can
be written as

$U_{f}=\\{(x,f(x))\colon x\in\mathbb{F}_{q^{n}}\\},$

and

$L_{f}=\left\\{\langle(x,f(x))\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\right\\},$

where $f$ is a $q$-polynomial over $\mathbb{F}_{q^{n}}$, that is
$f=\sum_{i=0}^{n-1}a_{i}X^{q^{i}}\in\mathbb{F}_{q^{n}}[X]$. It is not
difficult to show that $L_{f}$ is scattered if and only if for any
$z,y\in\mathbb{F}_{q^{n}}^{*}$ the condition

$\frac{f(z)}{z}=\frac{f(y)}{y}$

implies that $z$ and $y$ are $\mathbb{F}_{q}$-linearly dependent. Hence,
$q$-polynomials satisfying this condition are called _scattered_ by Sheekey in
[18].

Up to now, there are only three families of maximum scattered linear sets in
$\mathrm{PG}(1,q^{n})$ for infinitely many $n$. We list the corresponding
scattered polynomials over $\mathbb{F}_{q^{n}}$ below.

  1. (i)

$f=X^{q^{s}}$ with $\gcd(s,n)=1$, and $L_{f}$ is called the linear set of
pseudoregulus type; see [3].

  2. (ii)

$f=X^{q^{s}}+\delta X^{q^{n-s}}$ with $n\geq 4$,
$N_{q^{n}/q}(\delta)\notin\\{0,1\\}$ and $\gcd(s,n)=1$ which was found by
Lunardon and Polverino [13] and later generalized by Sheekey [18].

  3. (iii)

$f=X^{q^{s}}+X^{q^{s(m-1)}}+h^{1+q^{s}}X^{q^{s(m+1)}}+h^{1-q^{2m-s}}X^{q^{2m-s}}$
with $q$ odd, $n=2m\geq 6$, $h^{q^{m}+1}=-1$ and $\gcd(s,n)=1$. This family
was first introduced by Longobardi and Zanella in [11] and later generalized
to a big family in [10] and [15].

In this paper, we are interested in several families of maximum scattered
linear sets in $\mathrm{PG}(1,q^{6})$ and $\mathrm{PG}(1,q^{8})$. In the
following we list their corresponding scattered polynomials.

  1. (a)

$x^{q}+x^{q^{3}}+\delta x^{q^{5}}$ over $\mathbb{F}_{q^{6}}$ with
$\delta^{2}+\delta=1$ and $q$ is odd; see [6, 14].

  2. (b)

$x^{q^{s}}+\delta x^{q^{s+m}}$ over $\mathbb{F}_{q^{2m}}$ with
$m\in\\{3,4\\}$, $\gcd(s,m)=1$ and some conditions on $\delta$; see [5].

For convenience, we use $\mathscr{F}_{6}$ and $\mathscr{F}_{8}$ to denote the
families of linear sets given in (b) for $m=3$ and $4$, respectively. Let
$f_{\delta,m}$ stand for a member with parameter $(\delta,m)$ in
$\mathscr{F}_{6}$ or $\mathscr{F}_{8}$.

For $m=3$, it has been shown in [5] that for $q>4$ and
$N_{{q^{6}}/{q^{3}}}(\delta)\neq 1$ we can always find
$\delta\in\mathbb{F}_{q^{2}}^{*}$ such that $f_{\delta,3}$ is scattered. The
exact number of such $\delta$ has been determined by Bartoli, Csajbók and
Montanucci in [1].

For $m=4$, it also has been proved in [5] that for odd $q$,
$N_{{q^{8}}/{q^{4}}}(\delta)\neq 1$ and $\delta^{2}=-1$, $f_{\delta,4}$ is
scattered. Moreover, Timpanella and Zini [20] show that $L_{\delta,4}$ is
maximum scattered if and only if $\delta^{1+q^{4}}=-1$, under the extra
assumption that the odd prime power $q\geq 1039891$.

For two linear sets $L_{U}$ and $L_{W}$ of $\mathrm{PG}(1,q^{n})$, if there is
an element $\varphi$ in $\mathrm{P\Gamma L}(2,q^{n})$ such that
$L_{U}^{\varphi}=L_{W}$, then we say $L_{U}$ and $L_{W}$ are _$\mathrm{P\Gamma
L}$ -equivalent_. For given $q$ and $n$, it is obvious that there is only one
maximum scattered linear set of pseudoregulus type, because

$\\{\langle(1,x^{q^{s}-1})\rangle:x\in\mathbb{F}_{q^{n}}^{*}\\}=\\{\langle(1,x^{q-1})\rangle:x\in\mathbb{F}_{q^{n}}^{*}\\}$

for any integer $s$ satisfying $\gcd(s,n)=1$. The equivalence of different
members in the Lunardon-Polverino family has been recently solved by the
authors and Zullo in [19]. The equivalence for the scattered quadrinomials
from the third infinite family has been partially determined in [11].

In this paper, we continue to consider the equivalence among the members in
the families (a) and (b) defined on $\mathrm{PG}(1,q^{6})$ and
$\mathrm{PG}(1,q^{8})$, respectively. Firstly, in Section 3, we prove that (a)
contains only one element up to $\mathrm{P\Gamma L}$-equivalence, and we
completely determine its automorphism group. Then, in Section 4, the necessary
and sufficient conditions for the equivalence of the elements in
$\mathscr{F}_{6}$ and $\mathscr{F}_{8}$ are given, and their automorphism
groups are further determined. The main ideas for proving the above results
are as follows: First, we transform the $\mathrm{P\Gamma L}$-equivalence
problem into the equivalence of image sets of the corresponding polynomial
maps. By further analyzing and verifying some necessary conditions on the
coefficients of these polynomials, and considering some properties of their
adjoint polynomials, we can determine the equivalence between the linear sets
and their automorphism groups.

The rest part of this paper is organized as follows. In Section 2, we
introduce some preliminaries and auxiliary results for the equivalence of
maximum scattered linear sets. In Section 3, we focus on the equivalence
problem for the linear sets derived from the scattered polynomials given in
(a) and determine their automorphism groups. In Section 4, we turn to the
equivalence problem of linear sets in $\mathscr{F}_{6}$ and $\mathscr{F}_{8}$.
In section 5, we briefly summarized the contents of this paper.

##  2\. Preliminaries

A $q$-_polynomial_ (or a _linearized polynomial_) over $\mathbb{F}_{q^{n}}$ is
a polynomial of the following form

$f=\sum_{i=0}^{k}a_{i}X^{q^{i}},$

where $a_{i}\in\mathbb{F}_{q^{n}}$ and $k$ is a positive integer.

We will denote by $\mathcal{L}_{n,q}$ the set of all $q$-polynomials over
$\mathbb{F}_{q^{n}}$ with degree less than $q^{n}$. There is a bijection
between $\mathcal{L}_{n,q}$ and the $\mathbb{F}_{q}$-linear map defined over
$\mathbb{F}_{q^{n}}$. For more details on linearized polynomials we refer to
[9, Chapter 3, Section 4].

Consider the non-degenerate symmetric bilinear form of $\mathbb{F}_{q^{n}}$
over $\mathbb{F}_{q}$ defined by

$\langle x,y\rangle=\mathrm{Tr}_{q^{n}/q}(xy),$

for every $x,y\in\mathbb{F}_{q^{n}}$ and where
$\mathrm{Tr}_{q^{n}/q}(x)=x+\cdots+x^{q^{n-1}}$. The _adjoint polynomial_
$\hat{f}$ of the $q$-polynomial $\displaystyle
f=\sum_{i=0}^{n-1}a_{i}X^{q^{i}}$ with respect to the bilinear form
$\langle\cdot,\cdot\rangle$, i.e. the unique function over
$\mathbb{F}_{q^{n}}$ satisfying

$\mathrm{Tr}_{q^{n}/q}(yf(z))=\mathrm{Tr}_{q^{n}/q}(z\hat{f}(y))$

for every $y,z\in\mathbb{F}_{q^{n}}$, is given by

$\hat{f}=\sum_{i=0}^{n-1}a_{i}^{q^{n-i}}X^{q^{n-i}}.$

The following result concerning the linear set defined by the adjoint of a
$q$-polynomial was proved in [2, Lemma 2.6] and [19, Lemma 2.1] in different
ways.

######  Lemma 2.1.

Let $f$ be a $\sigma$-polynomial in $\mathcal{L}_{n,\sigma}$.Then for any
$b\in\mathbb{F}_{q^{n}}$, there is

$\\#\left\\{x\in\mathbb{F}_{q^{n}}^{*}:\frac{f(x)}{x}=b\right\\}=\\#\left\\{y\in\mathbb{F}_{q^{n}}^{*}:\frac{\hat{f}(y)}{y}=b\right\\}.$

In particular, $L_{f}=L_{\hat{f}}$.

Two linear sets $L_{U}$ and $L_{U^{\prime}}$ of $\mathrm{PG}(1,q^{n})$ are
said to be $\mathrm{P\Gamma L}$-equivalent (or projectively equivalent) if
there exists $\varphi\in\mathrm{P\Gamma L}(2,q^{n})$ such that
$L_{U}^{\varphi}=L_{U^{\prime}}$.

Let $L_{U}$ be an $\mathbb{F}_{q}$-linear set of rank $n$ in $\Lambda$. Under
the action of ${\rm P\Gamma L}(1,q^{n})$ on $\Lambda$, we may assume that
$L_{U}$ does not contain the point $\langle(0,1)\rangle_{\mathbb{F}_{q^{n}}}$,
so that $L_{U}$ can be written as

$L_{f}=\\{\langle(x,f(x))\rangle_{\mathbb{F}_{q^{n}}}\colon
x\in\mathbb{F}_{q^{n}}^{*}\\},$

where $f$ is a $q$-polynomial over $\mathbb{F}_{q^{n}}$, that is
$f=\sum_{i=0}^{n-1}a_{i}X^{q^{i}}\in\mathbb{F}_{q^{n}}[X]$. Sheekey in [18]
called $f$ _scattered_ if $L_{f}$ turns out to be a scattered
$\mathbb{F}_{q}$-linear set.

For two $q$-polynomials $f$ and $g$ over $\mathbb{F}_{q^{n}}$, we say
$U_{f}=\\{(x,f(x)):x\in\mathbb{F}_{q^{n}}^{*}\\}$ and
$U_{g}=\\{(x,g(x)):x\in\mathbb{F}_{q^{n}}^{*}\\}$ are $\mathrm{\Gamma
L}$-equivalent if there exists $\varphi\in\mathrm{\Gamma L}(2,q^{n})$ such
that $U_{f}^{\varphi}=U_{g}$. From the definitions, we get the following
straightforward result.

######  Lemma 2.2.

For $q$-polynomials $f$ and $g$ over $\mathbb{F}_{q^{n}}$, if $U_{f}$ and
$U_{g}$ are $\mathrm{\Gamma L}$-equivalent, then $L_{f}$ and $L_{g}$ are
$\mathrm{P\Gamma L}$-equivalent.

However, the converse statement is not always true. For instance, if
$f(x)=x^{q}$ and $g(x)=x^{q^{s}}$ with $s\neq 1$ and $\gcd(s,n)=1$, then
$U_{f}$ and $U_{g}$ are not $\mathrm{\Gamma L}(2,q^{n})$-equivalent, but
obviously
$L_{f}=\left\\{\langle(1,x^{q-1})\rangle_{\mathbb{F}_{q^{n}}}:x\in\mathbb{F}_{q^{n}}^{*}\right\\}=L_{g}$.
For more results on the equivalence problems, we refer to [5, 7].

In general, the equivalence problem between $U_{f}$ and $U_{g}$ is easier than
the equivalence between $L_{f}$ and $L_{g}$. For instance, the equivalence
between $U_{f}$ and $U_{g}$ with $f$, $g$ from family (b) has been completely
determined; see the following lemma. However, the equivalence between $L_{f}$
and $L_{g}$ is still open and will be solved in Section 4.

######  Lemma 2.3.

[5, Proposition 5.1] Define two $\mathbb{F}_{q}$-subspaces
$U_{\delta,s_{1}}=\\{(x,x^{q^{s_{1}}}+\delta
x^{q^{s_{1}+m}}):x\in\mathbb{F}_{q^{2m}}\\}$ and
$U_{\theta,s_{2}}=\\{(x,x^{q^{s_{2}}}+\theta
x^{s_{2}+m}):x\in\mathbb{F}_{q^{2m}}\\}$ of $\mathbb{F}_{q^{2m}}^{2}$ with
$\delta$, $\theta\in\mathbb{F}_{q^{2m}}^{*}$. Assume that
$N_{q^{2m}/{q^{m}}}(\delta)\neq 1$, $N_{q^{2m}/{q^{m}}}(\theta)\neq 1$, $1\leq
s_{1},s_{2}<m$ and $\gcd(m,s_{1})=\gcd(m,s_{2})=1$. The
$\mathbb{F}_{q}$-subspaces $U_{\delta,s_{1}},U_{\theta,s_{2}}$ are
$\mathrm{\Gamma L}(2,q^{2m})$-equivalent if and only if either

$s_{1}=s_{2}\text{ and
}N_{q^{2m}/{q^{m}}}(\delta)=N_{q^{2m}/{q^{m}}}(\theta)^{\sigma},$

or

$s_{1}+s_{2}=m\text{ and
}N_{q^{2m}/{q^{m}}}(\delta)N_{q^{2m}/{q^{m}}}(\theta)^{\sigma}=1,$

for some automorphism $\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{m}})$.

Given two (multi-)subsets $S$ and $T\subseteq\mathbb{F}_{q^{n}}$, $S=T$
implies that

$\sum_{x\in S}x^{d}=\sum_{x\in T}x^{d},$

for any non-negative integer $d$. As a direct consequence, we can get the
following necessary result for two identical linear sets.

######  Lemma 2.4.

[4, Lemma 3.4] Let $f$ and $g$ be two linearized polynomials. If
$L_{f}=L_{g}$, then for each non-negative integer $d$ the following holds

$\sum_{x\in\mathbb{F}_{q^{n}}^{*}}\left(\frac{g(x)}{x}\right)^{d}=\sum_{x\in\mathbb{F}_{q^{n}}^{*}}\left(\frac{f(x)}{x}\right)^{d}.$

The following result is well-known.

######  Lemma 2.5.

For any prime power $q$ and integer $d$ we have

$\sum_{x\in\mathbb{F}_{q^{n}}^{*}}x^{d}=\begin{cases}-1,&q-1\mid d;\\\
0,&\text{otherwise}.\end{cases}$

By applying Lemmas 2.4 and 2.5, one can prove the following criterion on the
identity of two linear sets.

######  Lemma 2.6.

[4, Lemma 3.6] Let $f(x)=\sum_{i=1}^{n}a_{i}x^{q^{i}}$ and
$g(x)=\sum_{i=1}^{n}b_{i}x^{q^{i}}$ be two $q$-polynomials over
$\mathbb{F}_{q^{n}}$, such that $L_{f}=L_{g}$. Then

(1) $a_{0}=b_{0};$

for $k=1,2,\cdots,n-1$,

(2) $a_{k}a_{n-k}^{q^{k}}=b_{k}b_{n-k}^{q^{k}};$

for $k=2,3,\cdots,n-1$,

(3)
$a_{1}a_{k-1}^{q}a_{n-k}^{q^{k}}+a_{k}a_{n-1}^{q}a_{n-k+1}^{q^{k}}=b_{1}b_{k-1}^{q}b_{n-k}^{q^{k}}+b_{k}b_{n-1}^{q}b_{n-k+1}^{q^{k}}.$

In general, the restrictions on the coefficients of $f$ and $g$ in Lemma 2.6
are not enough to determine $g$ from $f$. In Sections 3 and 4, we will provide
some extra restrictions for $f$ and $g$ in (a) or (b).

##  3\. Csajbók-Marino-Montanucci-Zullo family

In this section, we consider the equivalence problem for the family of maximum
scattered linear sets which is found for odd prime power $q\equiv\pm
1,0\pmod{5}$ by Csajbók, Marino and Zullo in [6] and later completely proved
for any odd prime power $q$ by Marino, Montanucci and Zullo in [14]. Recall
that the linear sets in this family are of the shape

$\left\\{\langle(1,,x^{q-1}+x^{q^{3}-1}+\theta
x^{q^{5}-1})\rangle_{\mathbb{F}_{q^{6}}}:x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

where $\theta^{2}+\theta=1$. By definition, there are at most two elements in
this family for given $q$ which is odd. The following simple lemma tells us
whether $\theta$ is in the prime field $\mathbb{F}_{p}$ of $\mathbb{F}_{q}$ or
in $\mathbb{F}_{p^{2}}$.

######  Lemma 3.1.

For positive integer $n$ and odd prime power $q$, suppose that
$x_{1}^{2}+x_{1}=x_{2}^{2}+x_{2}=1$, where $x_{1},x_{2}\in\mathbb{F}_{q^{n}}$.
Then

  1. (i)

$x_{1},x_{2}\in\mathbb{F}_{q}$, if $q\equiv 0,\pm 1\pmod{5}$;

  2. (ii)

$x_{1},x_{2}\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$, if $q\equiv\pm
2\pmod{5}$.

Moreover, $x_{1}^{2q-1}=x_{2}^{2q-1}$ if and only if $x_{1}=x_{2}$. In
particular, when $q\equiv 0\pmod{5}$, $x_{1}=x_{2}=2$.

###### Proof.

Let $q=p^{m}$ such that $p$ is prime. By the law of quadratic reciprocity, $5$
is a square in $\mathbb{F}_{p}$ if and only if $p=0,1$ or $4\pmod{5}$.

Suppose that $p=5$. The two roots $x_{1}$ and $x_{2}$ of the polynomial
$X^{2}+X-1$ are identical and
$x_{1}=x_{2}=2\in\mathbb{F}_{p}\subseteq\mathbb{F}_{q}$.

Suppose that $q\equiv\pm 1\pmod{5}$. When $p=1$ or $4\pmod{5}$, $5$ is a
square in $\mathbb{F}_{p}$ whence $5$ is a square in $\mathbb{F}_{q}$. When
$p=2$ or $3\pmod{5}$ which means $q=p^{2k}$ for some $k\in\mathbb{Z}$, $5$ is
also a square in $\mathbb{F}_{q}$.

Suppose that $q\equiv\pm 2\pmod{5}$. Then $p=2$ or $3\pmod{5}$ and
$q=p^{2k+1}$ for some $k\in\mathbb{Z}$. Hence $5$ is not a square in
$\mathbb{F}_{q}$.

Next we show that the necessary and sufficient condition of
$x_{1}^{2q-1}=x_{2}^{2q-1}$ is $x_{1}=x_{2}$. The sufficiency is obvious.

Assume that $x_{1}^{2q-1}=x_{2}^{2q-1}$. If $q\equiv 0,\pm 1\pmod{5}$, by (a)
we get $x_{1},x_{2}\in\mathbb{F}_{q}$. From
$x_{1}=x_{1}^{2q-1}=x_{2}^{2q-1}=x_{2}$, we derive $x_{1}=x_{2}$.

If $q\equiv\pm 2\pmod{5}$, by (b) we get
$x_{1},x_{2}\in\mathbb{F}_{q^{2}}\setminus\mathbb{F}_{q}$. As $x_{1},x_{2}$
are two (not necessarily distinct) roots of $x^{2}+x-1$, $x_{1}=x_{2}$ or
$x_{1}=x_{2}^{q}$. Assume that $x_{1}=x_{2}^{q}$, by
$x_{1}^{2q-1}=x_{2}^{2q-1}$ and $x_{2}^{q+1}=-1$ we get $x_{2}^{6}=-1$. As
$x_{2}^{2}+x_{2}=1$,

$(1-x_{2})^{3}=-x_{2}^{3}+3x_{2}^{2}-3x_{2}+1=x_{2}(4x_{2}-4)+1=5-8x_{2}=-1,$

which implies

$x_{2}=3/4.$

However, this contradicts $x_{2}^{2}+x_{2}=1$ under the assumption that $p\neq
2,5$. Therefore $x_{1}$ must be identical to $x_{2}$. ∎

Before looking at maximum scattered linear sets in the Csajbók-Marino-
Montanucci-Zullo family, we first consider the equivalence between the
$6$-dimensional $\mathbb{F}_{q}$-subspaces $U_{f}$ associated with different
$L_{f}$ from this family. Depending on the value of $q$ mod $5$, our result
are divided into two lemmas.

######  Lemma 3.2.

For odd prime power $q$ with $q\equiv\pm 2\pmod{5}$, define functions
$f(x)=x^{q}+x^{q^{3}}+\theta x^{q^{5}}$, $g(x)=x^{q}+x^{q^{3}}+\delta
x^{q^{5}}$ on $\mathbb{F}_{q^{6}}$ with $\theta$,
$\delta\in\mathbb{F}_{q^{6}}$ satisfying $\theta^{2}+\theta=1$ and
$\delta^{2}+\delta=1$. Then $\mathbb{F}_{q}$-subspaces $U_{f}$ and $U_{g}$ are
$\mathrm{\Gamma L}$-equivalent.

###### Proof.

As $\theta,\delta$ are the roots of $x^{2}+x-1$ which is irreducible over
$\mathbb{F}_{q}$, $\theta=\delta$ or $\theta=\delta^{q}$.

If $\theta=\delta$, then $U_{f}=U_{g}$; if $\theta=\delta^{q}$, then

$\left\\{(x^{q},(x^{q}+x^{q^{3}}+\delta
x^{q^{5}})^{q}):x\in\mathbb{F}_{q^{6}}\right\\}=\left\\{(x,x^{q}+x^{q^{3}}+\delta^{q}x^{q^{5}}):x\in\mathbb{F}_{q^{6}}\right\\}=U_{f}.$

Therefore, $U_{f}$ and $U_{g}$ are $\mathrm{\Gamma L}$-equivalent. ∎

######  Lemma 3.3.

For odd prime power $q$ with $q\equiv 0,\pm 1\pmod{5}$, define functions
$f(x)=x^{q}+x^{q^{3}}+\theta x^{q^{5}}$, $g(x)=x^{q}+x^{q^{3}}+\delta
x^{q^{5}}$ on $\mathbb{F}_{q^{6}}$ with $\theta$,
$\delta\in\mathbb{F}_{q^{6}}$ satisfying $\theta^{2}+\theta=1$ and
$\delta^{2}+\delta=1$. If $\mathbb{F}_{q}$-subspace $U_{f}$ and $U_{g}$ are
$\mathrm{\Gamma L}$-equivalent, then $\theta=\delta.$

###### Proof.

Assume that $\mathbb{F}_{q}$-subspace $U_{f}$ and $U_{g}$ are $\mathrm{\Gamma
L}$-equivalent. By Lemma 3.1, $\theta,\delta\in\mathbb{F}_{q}$. We only have
to consider the existence of invertible matrix $M=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}$ over $\mathbb{F}_{q^{6}}$ such that for each
$x\in\mathbb{F}_{q^{6}}$ there exists $y\in\mathbb{F}_{q^{6}}$ satisfying

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\begin{pmatrix}x\\\
f(x)\end{pmatrix}=\begin{pmatrix}y\\\ g(y)\end{pmatrix}.$

This is equivalent to

(4) $cX+df(X)\equiv g\left(aX+bf(X)\right)\pmod{X^{q^{6}}-X}.$

Then the right-hand-side of (4) is

$\displaystyle\left(b^{q}\theta+b^{q^{3}}+b^{q^{5}}\delta\right)X+a^{q}X^{q}+\left(b^{q}+b^{q^{3}}\theta+b^{q^{5}}\delta\right)X^{q^{2}}$
$\displaystyle+a^{q^{3}}X^{q^{3}}+\left(b^{q}+b^{q^{3}}+b^{q^{5}}\theta\delta\right)X^{q^{4}}+a^{q^{5}}\delta
X^{q^{5}}.$

By the coefficients of $X^{q},X^{q^{3}}$ and $X^{q^{5}}$ in (4), we get

(5) $\begin{cases}d=a^{q}=a^{q^{3}},\\\ d\theta=a^{q^{5}}\delta.\end{cases}$

Depending on the value of $b$, we consider two cases.

Case 1: $b=0$. By the coefficients of $X$ in (4), we get $c=0$ and $ad\neq 0$.
Then (5) is equivalent to $a,d\in\mathbb{F}_{q^{2}}$ and

$\begin{cases}d=a^{q},\\\ \theta=\delta.\end{cases}$

Case 2: $b\neq 0$. By the coefficients of $X^{q^{2}}$ and $X^{q^{4}}$ in (4),
we get

(6) $\begin{cases}b^{q}+b^{q^{3}}\theta+b^{q^{5}}\delta=0,\\\
b^{q}+b^{q^{3}}+b^{q^{5}}\theta\delta=0.\end{cases}$

By (6), we obtain

$\left(b^{q^{3}}-b^{q^{5}}\delta\right)\left(1-\theta\right)=0,$

which means at least one of $b^{q^{3}}-b^{q^{5}}\delta$ and $1-\theta$ equals
$0$. Clearly $1-\theta=0$ contradicts the condition $\theta^{2}+\theta=1$. If
$b^{q^{3}}-b^{q^{5}}\delta=0$, then $\delta=b^{q^{3}-q^{5}},$ which implies

$\delta^{q^{4}+q^{2}+1}=N_{{q^{6}}/{q^{2}}}\left(b^{q^{3}(1-q^{2})}\right)=1.$

However, by $\delta^{2}+\delta=1$ and $\delta\in\mathbb{F}_{q^{2}}$, we get

$\delta^{q^{4}+q^{2}+1}=\delta^{3}=(1-\delta)\delta=2\delta-1.$

Thus the above two equations on $\delta^{q^{4}+q^{2}+1}$ tells us $\delta=1$
which contradicts $\delta^{2}+\delta=1$. Therefore, there is no
$b\in\mathbb{F}_{q^{6}}^{*}$ such that $\delta=b^{q^{3}-q^{5}}$.

To summarize, we have shown that $\theta$ must be identical to $\delta$ and
the matrix $M=\begin{pmatrix}a&0\\\ 0&a^{q}\end{pmatrix}$. ∎

The next result completely solve the equivalence of the members in the
Csajbók-Marino-Montanucci-Zullo family. In particular, compared with Lemma
3.3, we can see that the equivalence of $U_{f}$ and $U_{g}$ are not the same
as the equivalence of $L_{f}$ and $L_{g}$ in this family.

######  Theorem 3.4.

Let $q$ be an odd prime power. All elements of the Csajbók-Marino-Montanucci-
Zullo family are $\mathrm{P\Gamma L}$-equivalent.

###### Proof.

By Lemma 3.1, when $q\equiv 0\pmod{5}$ there are only one element in this
family and when $q\equiv\pm 1,\pm 2\pmod{5}$ there are only two elements.

For $q\not\equiv 0\pmod{5}$, let
$f_{i}(x)=x^{q}+x^{q^{3}}+\theta_{i}x^{q^{5}}$ with $i=1,2$ stand for the two
members in the family.

When $q\equiv\pm 2\pmod{5}$, Lemma 3.2 tells us that $U_{f_{1}}$ and
$U_{f_{2}}$ are $\mathrm{\Gamma L}$-equivalent, which means that $L_{f_{1}}$
and $L_{f_{2}}$ are equivalent by Lemma 2.2.

When $q\equiv\pm 1\pmod{5}$, by Lemma 3.1 we get
$\theta_{1},\theta_{2}\in\mathbb{F}_{q}$. Let
$h(x)=(-\theta_{1}-1)x^{q}+x^{q^{3}}+x^{q^{5}}$. Then it is straightforward to
verify $h(f_{1}(x))=x$, i.e. $h(x)$ is the inverse map of $f_{1}(x)$. Hence

$\left\\{\frac{x}{f_{1}(x)}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{(-\theta_{1}-1)x^{q}+x^{q^{3}}+x^{q^{5}}}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{h(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

which means $L_{f_{1}}$ is $\mathrm{P\Gamma L}$-equivalent to $L_{h}$.

By a simple calculation,

$\hat{h}(x)=x^{q}+x^{q^{3}}+\left(-\theta_{1}-1\right)^{q^{5}}x^{q^{5}}=x^{q}+x^{q^{3}}+\theta_{2}x^{q^{5}}=f_{2}(x).$

By Lemma 2.1, $L_{h}=L_{\hat{h}}=L_{f_{2}}$. Therefore $L_{f_{1}}$ is
$\mathrm{P\Gamma L}$-equivalent to $L_{f_{2}}$. ∎

By Theorem 3.4, the Csajbók-Marino-Montanucci-Zullo family contains only one
member up to equivalence. Next we determine the automorphism group of this
element.

######  Theorem 3.5.

Let $q$ be an odd prime power. Let $f(x)=x^{q}+x^{q^{3}}+\theta x^{q^{5}}$
over $\mathbb{F}_{q^{6}}$ with $\theta^{2}+\theta=1$. The automorphism group
of $L_{f}$ is

$\mathrm{Aut}(L_{f})=\begin{cases}\mathcal{D},&q\equiv\pm 1\pmod{5};\\\
\mathcal{D}\cup\mathcal{C},&q\equiv 0,\pm 2\pmod{5},\end{cases}$

where

$\mathcal{D}:=\left\\{M\tau:M=\begin{pmatrix}1&0\\\
0&d\end{pmatrix}\in\mathrm{PGL}(2,q^{6}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}}),\theta^{\tau-1}=1,d^{q+1}=1\right\\},$

and

$\mathcal{C}:=\left\\{M\tau:M=\begin{pmatrix}0&1\\\
c&0\end{pmatrix}\in\mathrm{PGL}(2,q^{6}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}}),\theta^{\tau+1}=-1,c^{q+1}=1\right\\}.$

###### Proof.

Let $M=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ be an element in
$\mathrm{PGL}(2,q^{6})$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}}/\mathbb{F}_{q})$. Assume that
$M\tau\in\mathrm{P\Gamma L}(2,q^{6})$ is an automorphism of $L_{f}$, that is

(7)
$\left\\{\frac{f(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{cx+d\bar{f}(x)}{ax+b\bar{f}(x)}:x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

where $\bar{f}(x)=x^{q}+x^{q^{3}}+\theta^{\tau}x^{q^{5}}$.

Next we determine $M$ and $\tau$. Depending on the value of $b$, we separate
the rest part into two cases.

Case 1: $b=0$. As $M$ is viewed as an element in $\mathrm{PGL}(2,q^{6})$, we
can assume that $a=1$ and (7) becomes

(8) $\left\\{x^{q-1}+x^{q^{3}-1}+\theta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{c+d\frac{\bar{f}(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By (1), (2) for $k=1$ and $3$ in Lemma 2.6, respectively, we derive $c=0$,

(9) $d^{q+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q},$

and

(10) $d^{q^{3}+1}=1.$

By (9) and (10), we can get

(11) $\theta^{\tau\left(q^{3}-q^{2}+q\right)}=\theta^{q^{3}-q^{2}+q}.$

By Lemma 3.1, (11) is equivalent to

(12) $\theta^{\tau}=\theta.$

By (9), (10) and (12),

(13) $d^{q+1}=1.$

Then (12) and (13) are the necessary conditions. Next, we show that (12) and
(13) are also sufficient. By (13), it is easy to see that there exists
$\alpha\in\mathbb{F}_{q^{2}}^{*}$ such that $d=\alpha^{q-1}$. Replacing $x$ by
$\alpha x$ in (8), we obtain

(14) $\left\\{(\alpha x)^{q-1}+(\alpha x)^{q^{3}-1}+\theta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+x^{q^{3}-1}+\theta
x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By (12), (14) is equivalent to

$\left\\{x^{q-1}+x^{q^{3}-1}+\theta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{d\left(x^{q-1}+x^{q^{3}-1}+\theta^{\tau}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

Therefore, $L_{f}$ is mapped to itself under $M\tau$ where
$M=\begin{pmatrix}1&0\\\ 0&d\end{pmatrix}$ and $\tau$ satisfy conditions (12)
and (13).

Case 2: $b\neq 0$. Without loss of generality, we assume that $b=1$. First we
concentrate on the proof of the following claim:

Claim: $a=0$.

Assume, by way of contradiction, that $a\neq 0$. Now

$\frac{cx+d\bar{f}(x)}{ax+b\bar{f}(x)}=d+\frac{(c-ad)x}{ax+\bar{f}(x)},$

where $\bar{f}(x)=x^{q}+x^{q^{3}}+\theta^{\tau}x^{q^{5}}.$

As $M$ is nonsingular, $c-ad\neq 0$. Furthermore, by definition of maximum
scattered linear sets in $\mathrm{PG}(1,q^{6})$, the map $x\mapsto
ax+\bar{f}(x)$ must be invertible; otherwise the point
$\langle(0,1)\rangle_{\mathbb{F}_{q^{6}}}$ would be in $L_{f}$ which is
impossible.

Let $h(x)=\sum_{i=0}^{5}r_{i}x^{q^{i}}$ denote the inverse map of
$x\mapsto\frac{ax+\bar{f}(x)}{c-ad}$. Setting $y=\frac{ax+\bar{f}(x)}{c-ad}$,
we get

$d+\frac{(c-ad)x}{ax+\bar{f}(x)}=d+\frac{h(y)}{y}.$

By comparing the coefficients of
$ah(Y)+\bar{f}(h(Y))\equiv(c-ad)Y\pmod{Y^{q^{6}}-Y}$, we obtain

(15) $ar_{i}+r_{i-1}^{q}+r_{i+3}^{q^{3}}+\theta^{\tau}r_{i+1}^{q^{5}}=0,$

for $i=1,2,3,4,5$.

Now (7) becomes

$\left\\{x^{q-1}+x^{q^{3}-1}+\theta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{d+\sum_{i=0}^{5}r_{i}y^{q^{i}-1}:y\in\mathbb{F}_{q^{5}}^{*}\right\\}.$

By (2) for $k=1$,

$r_{1}r_{5}^{q}=\theta^{q},$

which means $r_{1},r_{5}\neq 0$ . By (2) for $k=2$,

$r_{2}r_{4}^{q^{2}}=0,$

which means at least one of $r_{2}$ and $r_{4}$ equals $0$. Without loss of
generality, we assume that $r_{2}=0$. By (3) for $k=2$,

$r_{1}^{q+1}r_{4}^{q^{2}}+r_{2}r_{5}^{q^{2}+q}=0$

which means $r_{1}^{q+1}r_{4}^{q^{2}}=0$, then $r_{4}=0$.

By (15) for $i=1,3,5$ and the assumption that $a\neq 0$, we get

(16) $\begin{cases}r_{1}=-\frac{r_{0}^{q}}{a},\\\
r_{3}=-\frac{r_{0}^{q^{3}}}{a},\\\
r_{5}=-\frac{\theta^{\tau}r_{0}^{q^{5}}}{a}.\end{cases}$

By (15) for $i=2$, we get

$r_{1}^{q}+r_{5}^{q^{3}}+\theta^{\tau}r_{3}^{q^{5}}=0,$

and since $\theta^{\tau}\in\mathbb{F}_{q^{2}}$, which is equivalent to

(17) $r_{1}^{q^{5}}+r_{5}^{q}+\theta^{\tau}r_{3}^{q^{3}}=0,$

Similarly by (15) for $i=4$, we get

(18) $r_{1}^{q^{5}}+r_{3}^{q^{3}}+\theta^{\tau}r_{5}^{q}=0.$

By (17) and (18), we get
$\left(r_{3}^{q^{3}}-r_{5}^{q}\right)(\theta^{\tau}-1)=0$, which implies

$r_{3}^{q^{3}}=r_{5}^{q}.$

Plugging (16) into the above equation, we have $\theta^{\tau}=a^{q-q^{3}}$
which implies

(19)
$\theta^{3\tau}=N_{{q^{6}}/{q^{2}}}(\theta^{\tau})=N_{{q^{6}}/{q^{2}}}\left(a^{q(1-q^{2})}\right)=1.$

By $\theta^{2}+\theta=1$ and $\theta\in\mathbb{F}_{q^{2}}$, we get

$\theta^{3\tau}=(1-\theta^{\tau})\theta^{\tau}=\theta^{\tau}+\theta^{\tau}-1=2\theta^{\tau}-1,$

together with (19), we get $\theta=1$, which is a contradiction to the
assumption that $\theta^{2}+\theta=1$.

Therefore, we have proved the claim that $a$ must be $0$.

As

$\begin{pmatrix}0&1\\\ c&d\end{pmatrix}^{-1}=\begin{pmatrix}-d/c&1/c\\\
1&0\end{pmatrix},$

by applying Claim on the inverse map of $M$ with $a=0$, we get $d=0$. Thus

$M=\begin{pmatrix}0&1\\\ c&0\end{pmatrix}.$

Let $h(x)=-(\theta^{\tau q}+1)x^{q}+x^{q^{3}}+x^{q^{5}}.$ It can be readily
verified that $h(\bar{f}(x))=x$, i.e. $h(x)$ is the inverse map of
$\bar{f}(x)$. Consequently, (7) equals

$\left\\{\frac{ch(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{f(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

By (2) for $k=1$ in Lemma 2.6, we get

(20) $c^{q+1}=\frac{-\theta^{q}}{\left(\theta^{\tau q}+1\right)},$

and by (2) for $k=3$ we obtain

(21) $c^{q^{3}+1}=1.$

By (20) and (21), we get

(22) $\theta^{2q-1}=\left(-\theta^{\tau}-1\right)^{2q-1}.$

As $\theta$ and $-\theta^{\tau}-1$ are both roots of $X^{2}+X-1$, by Lemma
3.1, (22) is equivalent to $-\theta=\theta^{\tau}+1$. It follows that

(23) $\theta^{\tau+1}=-1.$

By (20) and $\theta^{\tau}+1=-\theta$,

(24) $c^{q+1}=1.$

Then (23) and (24) are the necessary conditions.

Next we show that (23) and (24) are sufficient. Suppose (23) and (24) hold.
Let $h(x)=-(\theta^{\tau q}+1)x^{q}+x^{q^{3}}+x^{q^{5}}.$ Then
$h(\bar{f}(x))=x$ which means

(25)
$\left\\{\frac{cx}{\bar{f}(x)}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{ch(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By a simple calculation, the adjoint $\hat{h}(x)$ of $h(x)$ is

$\hat{h}(x)=x^{q}+x^{q^{3}}+\left(-\theta^{\tau
q}-1\right)^{q^{5}}x^{q^{5}}=x^{q}+x^{q^{3}}+\theta x^{q^{5}}=f(x).$

By Lemma 2.1, $L_{h}=L_{\hat{h}}$, which means

(26)
$\left\\{\frac{ch(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{c\hat{h}(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{cf(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

The Case 1 of the proof tells us that if $c$ satisfies (24), then

(27)
$\left\\{\frac{cf(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{f(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By (25), (26) and (27), we get that (23) and (24) are the sufficient
conditions.

Now the automorphism group of $L_{f}$ is

$\mathrm{Aut}(L_{f})=\mathcal{D}\cup\mathcal{C},$

where

$\mathcal{D}:=\left\\{M\tau:M=\begin{pmatrix}1&0\\\
0&d\end{pmatrix},\theta^{\tau}=\theta,d^{q+1}=1\right\\},$

and

$\mathcal{C}:=\left\\{M\tau:M=\begin{pmatrix}0&1\\\
c&0\end{pmatrix},\theta^{\tau+1}=-1,c^{q+1}=1\right\\}.$

By Lemma 3.1 and $\theta^{2}+\theta=1$, we get that the equation
$\theta^{\tau+1}=-1$ holds if and only if one of the following collections of
conditions is satisfied:

  * •

$q\equiv 0\pmod{5}$, $\theta^{\tau}=\theta$

  * •

$q\equiv\pm 2\pmod{5}$, $\theta^{\tau}=\theta^{q}$.

In particular, when $q\equiv\pm 1\pmod{5}$, $\theta^{\tau+1}=-1$ can never be
satisfied and $\mathcal{C}=\varnothing$. ∎

##  4\. Csjabók-Marino-Polverino-Zanella Families

In this section, we consider the equivalence problem for the two families
$\mathscr{F}_{6}$ and $\mathscr{F}_{8}$ of maximum scattered linear sets
constructed by Csjabók, Marino, Polverino and Zanella in [5].

Recall that $\mathscr{F}_{2m}$ contains maximum linear sets in
$\mathrm{PG}(1,q^{2m})$ of the following shape

$\left\\{\langle(1,x^{q^{s}-1}+\delta
x^{q^{s+m}-1})\rangle_{\mathbb{F}_{q^{2m}}}:x\in\mathbb{F}_{q^{2m}}^{*}\right\\}$

for $m=3,4$, where $\gcd(s,m)=1$ and $\delta$, $q$ satisfy certain conditions.

By Lemma 2.2, the following result tells us that we only have to consider the
members in $\mathscr{F}_{2m}$ with $s=1$.

######  Lemma 4.1.

[5] Every subspace
$U_{b^{\prime},s}=\\{(x,x^{q^{s}}+b^{\prime}x^{q^{s+m}}):x\in\mathbb{F}_{q^{2m}}\\}$
with $\gcd(s,m)=1$, is $\mathrm{GL}(2,q^{2m})$-equivalent to a subspace

$U_{b}=\\{(x,x^{q}+bx^{q^{1+m}}):x\in\mathbb{F}_{q^{2m}}\\},$

for some $b\in\mathbb{F}_{q^{2m}}$.

To handle the equivalence problem for $\mathscr{F}_{6}$ and $\mathscr{F}_{8}$,
we need to use Lemma 2.6. However, it is not enough to provide us a complete
answer, and we need the following several extra lemmas.

######  Lemma 4.2.

Let $f(x)=a_{1}x^{q}+a_{4}x^{q^{4}}$ and $g(x)=b_{1}x^{q}+b_{4}x^{q^{4}}$ be
two q-polynomials over $\mathbb{F}_{q^{6}}$. If $L_{f}=L_{g}$, then

(28) $a_{4}^{q^{4}+q^{2}+1}=b_{4}^{q^{4}+q^{2}+1}.$

###### Proof.

Let $D=(q^{2}-q+1)(q^{2}+q+1)=q^{4}+q^{2}+1$. Then

$\sum_{x\in\mathbb{F}_{q^{6}}^{*}}\left(\frac{f(x)}{x}\right)^{D}=\sum_{u_{1},u_{2},u_{3}\in\\{1,4\\}}a_{u_{1}}a_{u_{2}}^{q^{2}}a_{u_{3}}^{q^{4}}\sum_{x\in\mathbb{F}_{q^{6}}^{*}}x^{U},$

with $U={q^{u_{1}}+q^{u_{2}+2}+q^{u_{3}+4}-(1+q^{2}+q^{4})}$. By Lemma 2.4, we
obtain $\sum_{x\in\mathbb{F}_{q^{6}}^{*}}x^{k}=-1$ if and only if $q^{6}-1|k$.
When $\sum_{x\in\mathbb{F}_{q^{6}}^{*}}x^{U}=-1$ ,

$q^{u_{1}}+q^{u_{2}+2}+q^{u_{3}+4}\equiv 1+q^{2}+q^{4}\pmod{q^{6}-1},$

with $u_{1},u_{2},u_{3}\in\\{1,4\\}.$ A simple calculation shows that
$u_{1}=u_{2}=u_{3}=4$.

By $L_{f}=L_{g}$ and Lemma 2.5, we get

$\sum_{x\in\mathbb{F}_{q^{6}}^{*}}\left(\frac{f(x)}{x}\right)^{D}=\sum_{x\in\mathbb{F}_{q^{6}}^{*}}\left(\frac{g(x)}{x}\right)^{D}.$

which means,

$a_{4}a_{4}^{q^{2}}a_{4}^{q^{4}}=b_{4}b_{4}^{q^{2}}b_{4}^{q^{4}}.\qed$

######  Lemma 4.3.

Let $f(x)=a_{1}x^{q}+a_{5}x^{q^{5}}$ and $g(x)=b_{1}x^{q}+b_{5}x^{q^{5}}$ be
two q-polynomials over $\mathbb{F}_{q^{8}}$. If $L_{f}=L_{g}$, then we have

(29) $a_{1}^{q^{2}+q+1}a_{5}^{q^{3}}=b_{1}^{q^{2}+q+1}b_{5}^{q^{3}},$

and

(30) $a_{1}a_{5}^{q^{6}+q^{3}+q}=b_{1}b_{5}^{q^{6}+q^{3}+q}.$

###### Proof.

Let $D_{1}=q^{3}+q^{2}+q+1$, we get

$\sum_{x\in\mathbb{F}_{q^{8}}^{*}}\left(\frac{f(x)}{x}\right)^{D_{1}}=\sum_{u_{1},u_{2},u_{3},u_{4}\in\\{1,5\\}}a_{u_{1}}a_{u_{2}}^{q}a_{u_{3}}^{q^{2}}a_{u_{4}}^{q^{3}}\sum_{x\in\mathbb{F}_{q^{8}}^{*}}x^{U},$

with $U=q^{u_{1}}+q^{u_{2}+1}+q^{u_{3}+2}+q^{u_{4}+3}-(1+q+q^{2}+q^{3}).$ By
Lemma 2.4, we obtain $\sum_{x\in\mathbb{F}_{q^{8}}^{*}}x^{k}=-1$ if and only
if $(q^{8}-1)|k$. When $\sum_{x\in\mathbb{F}_{q^{8}}^{*}}x^{U}=-1$,

$q^{u_{1}}+q^{u_{2}+1}+q^{u_{3}+2}+q^{u_{4}+3}\equiv
1+q+q^{2}+q^{3}\pmod{q^{8}-1},$

with $u_{1},u_{2},u_{3},u_{4}\in\\{1,5\\}$. A direct computation shows that
$u_{1}=u_{2}=u_{3}=1,u_{4}=5$.

Let $D_{2}=(q^{3}-q^{2}+1)(q^{3}+q^{2}+q+1)=q^{6}+q^{3}+q+1$, we get

$\sum_{x\in\mathbb{F}_{q^{8}}^{*}}\left(\frac{f(x)}{x}\right)^{D_{2}}=\sum_{i_{1},i_{2},i_{3},i_{4}\in\\{1,5\\}}a_{i_{1}}a_{i_{2}}^{q}a_{i_{3}}^{q^{3}}a_{i_{4}}^{q^{6}}\sum_{x\in\mathbb{F}_{q^{8}}^{*}}x^{I},$

with $I=q^{i_{1}}+q^{i_{2}+1}+q^{i_{3}+3}+q^{i_{4}+6}-(1+q+q^{3}+q^{6})$. When
$\sum_{x\in\mathbb{F}_{q^{8}}^{*}}x^{I}=-1$,

$q^{i_{1}}+q^{i_{2}+1}+q^{i_{3}+3}+q^{i_{4}+6}\equiv
1+q+q^{3}+q^{6}\pmod{q^{8}-1},$

with $i_{1},i_{2},i_{3},i_{4}\in\\{1,5\\}$. Hence we must have
$i_{1}=1,i_{2}=i_{3}=i_{4}=5$.

By $L_{f}=L_{g}$ and Lemma 2.5, we get

$\sum_{x\in\mathbb{F}_{q^{6}}^{*}}\left(\frac{f(x)}{x}\right)^{D_{k}}=\sum_{x\in\mathbb{F}_{q^{6}}^{*}}\left(\frac{g(x)}{x}\right)^{D_{k}},$

with $k=1,2$, which means

$a_{1}a_{1}^{q}a_{1}^{q^{2}}a_{5}^{q^{3}}=b_{1}b_{1}^{q}b_{1}^{q^{2}}b_{5}^{q^{3}},$

and

$a_{1}a_{5}^{q}a_{5}^{q^{3}}a_{5}^{q^{6}}=b_{1}b_{5}^{q}b_{5}^{q^{3}}b_{5}^{q^{6}}.\qed$

######  Lemma 4.4.

Let $f(x)=x^{q^{s_{1}}}+\delta x^{q^{s_{1}+m}}$, $g(x)=x^{q^{s_{2}}}+\theta
x^{q^{s_{2}+m}}$ with $\delta,\theta\in\mathbb{F}_{q^{2m}}$, $1\leq
s_{1},s_{2}<m$ and $\gcd(m,s_{1})=\gcd(m,s_{2})=1$. Let
$M=\begin{pmatrix}a&1\\\ c&d\end{pmatrix}$ be an invertible matrix over
$\mathbb{F}_{q^{2m}}$. If

(31) $\left\\{x^{q^{s_{1}}-1}+\delta
x^{q^{s_{1}+m}-1}:x\in\mathbb{F}_{q^{2m}}^{*}\right\\}=\left\\{\frac{cx+dg(x)}{ax+g(x)}:x\in\mathbb{F}_{q^{2m}}^{*}\right\\},$

then $a=d=0$.

###### Proof.

Now

$\frac{cx+dg(x)}{ax+bg(x)}=d+\frac{(c-ad)x}{ax+g(x)}.$

As $M$ is nonsingular, $c-ad\neq 0$. Furthermore, the map $x\mapsto ax+g(x)$
must be invertible; otherwise the point
$\langle(0,1)\rangle_{\mathbb{F}_{q^{2m}}}$ would be in $L_{f}$ which is
impossible.

Let $h(x)=\sum_{i=0}^{2m-1}r_{i}x^{q^{i}}$ denote the inverse map of
$x\mapsto\frac{ax+g(x)}{c-ad}$. Setting $y=\frac{ax+g(x)}{c-ad}$, we get

$d+\frac{(c-ad)x}{ax+g(x)}=d+\frac{h(y)}{y}.$

By comparing the coefficients of
$ah(Y)+g(h(Y))\equiv(c-ad)Y\pmod{Y^{q^{2m}}-Y}$, we obtain

(32) $ar_{i}+r_{i-s_{2}}^{q^{s_{2}}}+\theta r_{i-s_{2}+m}^{q^{s_{2}+m}}=0,$

for $i=1,2,\cdots,2m-1$. Now (31) becomes

$\left\\{x^{q^{s_{1}}-1}+\delta
x^{q^{s_{1}+m}-1}:x\in\mathbb{F}_{q^{2m}}^{*}\right\\}=\left\\{d+\sum_{i=0}^{2m-1}r_{i}y^{q^{i}-1}:y\in\mathbb{F}_{q^{2m}}^{*}\right\\}.$

We apply Lemma 2.6 on the above equation. By (1),

(33) $r_{0}=-d.$

Moreover, by (2),

(34) $r_{k}r_{2m-k}^{q^{k}}=0,$

for $k=1,2,\cdots,2m-1$. By (34) with $k=s_{2}$,

$r_{s_{2}}r_{2m-s_{2}}^{q^{s_{2}}}=0,$

which means at least one of $r_{s_{2}}$ and $r_{2m-s_{2}}$ equals $0$. Without
loss of generality, we assume that $r_{s_{2}}=0$. By (34) with $k=m$,
$r_{m}^{q^{m}+1}=0$, which means

(35) $r_{m}=0.$

By (32) with $i=s_{2}$, that is

(36) $r_{0}^{q^{s_{2}}}+\theta r_{m}^{q^{s_{2}+m}}=0,$

By (33), (35) and (36), we get $d=-r_{0}=0$.

As

$\begin{pmatrix}a&1\\\ c&0\end{pmatrix}^{-1}=\begin{pmatrix}0&1/c\\\
1&-a/c\end{pmatrix},$

by considering the inverse of $M$ acting on $L_{g}$, we get $a=0$. ∎

Now we are ready to consider the equivalence problem for $\mathscr{F}_{6}$.
The following theorem completely determine the equivalence between different
members in $\mathscr{F}_{6}$.

######  Theorem 4.5.

Let $f(x)=x^{q}+\delta x^{q^{4}}$ and $g(x)=x^{q}+\theta x^{q^{4}}$. Then
$L_{f}$ and $L_{g}$ are $\mathrm{P\Gamma L}$-equivalent if and only if

(37) $N_{q^{6}/{q^{3}}}(\delta)=N_{q^{6}/{q^{3}}}(\theta)^{\sigma},$

for some automorphism $\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{3}})$.

###### Proof.

If $\theta=\delta=0$, the above conclusion clearly holds. Hence, we can assume
that at least one of $\theta$ and $\delta$ is not equal to $0$.

First, we prove the necessity of (37).

Assume that $L_{f}$ and $L_{g}$ are equivalent which implies the existence of
invertible matrix $M=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ over
$\mathbb{F}_{q^{6}}$ and $\tau\in\mathrm{Aut}(\mathbb{F}_{q^{3}})$ such that

(38) $\left\\{x^{q-1}+\delta
x^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{cx+d\bar{g}(x)}{ax+b\bar{g}(x)}:x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

where $\bar{g}(x)=x^{q}+\theta^{\tau}x^{q^{4}}$.

Depending on the value of $b$, we separate the rest part into two cases.

Case 1: $b=0$. As $M$ is viewed as an element in $\mathrm{PGL}(2,q^{6})$, we
can assume that $a=1$ and (38) becomes

(39) $\left\\{x^{q-1}+\delta
x^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{c+d\frac{\bar{g}(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By (1) in Lemma 2.6, we derive $c=0$. Then (39) becomes

$\left\\{x^{q-1}+\delta
x^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By (3) with $k=2$ in Lemma 2.6, we get

(40) $d^{q^{2}+q+1}\theta^{\tau q^{2}}=\delta^{q^{2}}.$

If exactly one of $\theta$ and $\delta$ equals $0$, then it has to be $\delta$
and $d$ must be $0$. However, this implies $M$ is singular which leads to a
contradiction. Hence $\theta$ and $\delta\neq 0$. By (28) in Lemma 4.2, we get

(41)
$d^{(q^{2}-q+1)(q^{2}+q+1)}=\frac{\delta^{q^{4}+q^{2}+1}}{\theta^{\tau\left(q^{4}+q^{2}+1\right)}}.$

Then by (40) and (41), we get

$\delta^{q^{3}+1}=\left(\theta^{q^{3}+1}\right)^{\tau},$

let $\sigma=\tau|_{\mathbb{F}_{q^{3}}}$, then

$N_{q^{6}/{q^{3}}}(\delta)=N_{q^{6}/{q^{3}}}(\theta)^{\sigma}.$

Case 2: $b\neq 0$. Without loss of generality, we assume that $b=1$. By Lemma
4.4, we get $a=d=0$, which means

$M=\begin{pmatrix}0&1\\\ c&0\end{pmatrix}.$

Let $h_{0}(x)=-\theta^{\tau q^{5}}x^{q^{2}}+x^{q^{5}}$, then
$h_{0}(\bar{g}(x))=\left(1-\theta^{\tau(q^{2}+q^{5})}\right)x$.

If $N_{q^{6}/{q^{3}}}(\theta)=1$, which is equivalent to
$1-\theta^{\tau(q^{2}+q^{5})}=0$, then the map $x\mapsto g(x)$ must be not
invertible. By (38), the point $\langle(0,1)\rangle_{\mathbb{F}_{q^{6}}}$
would be in $L_{f}$ which is impossible. Hence $N_{q^{6}/{q^{3}}}(\theta)\neq
1$.

Let

$h(x)=\frac{h_{0}(x)}{1-\theta^{\tau(q^{2}+q^{5})}}=-\frac{\theta^{\tau
q^{5}}x^{q^{2}}}{1-\theta^{\tau(q^{2}+q^{5})}}+\frac{x^{q^{5}}}{1-\theta^{\tau(q^{2}+q^{5})}}.$

Then $h(\bar{g}(x))=x$, i.e. $h(x)$ is the inverse map of $\bar{g}(x)$. Thus
(38) equals

(42) $\left\\{x^{q-1}+\delta
x^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\frac{ch(x)}{x}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

By a simple calculation, the adjoint $\hat{h}_{1}(x)$ of $h_{1}(x)$ is

$\hat{h}_{1}(x)=x^{q}+(-\theta^{q^{5}\tau})^{q^{4}}x^{q^{4}},$

where $h_{1}(x)=h(x)(1-\theta^{\tau(q^{2}+q^{5})})$. By Lemma 2.1,
$L_{h_{1}}=L_{\hat{h}_{1}}$, then (42) equals

(43) $\left\\{x^{q-1}+\delta
x^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\hat{c}\left(x^{q-1}+(-\theta^{q^{5}\tau})^{q^{4}}x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\},$

where $\hat{c}=\frac{c}{1-\theta^{\tau(q^{2}+q^{5})}}$.

By (3) in Lemma 2.6, we get

(44) $\hat{c}^{q^{2}+q+1}(-\theta^{q^{5}\tau})=\delta^{q^{2}}.$

If exactly one of $\theta$ and $\delta$ equals $0$, then it has to be $\delta$
and $c$ must be $0$ which means $M$ is singular. However, this contradicts our
assumption on $M$. Hence $\theta$ and $\delta\neq 0$. By (28) in Lemma 4.2, we
get

(45)
$\hat{c}^{(q^{2}-q+1)(q^{2}+q+1)}=\frac{\delta^{q^{4}+q^{2}+1}}{(-\theta^{q^{5}\tau})^{q^{4}\left(q^{4}+q^{2}+1\right)}}.$

Then by (44) and (45), we get

$\delta^{q^{3}+1}=\left(-\theta^{q^{5}\tau}\right)^{q^{4}(q^{3}+1)},$

let $\sigma=\tau|_{\mathbb{F}_{q^{3}}}$, then

$N_{q^{6}/{q^{3}}}(\delta)=N_{q^{6}/{q^{3}}}(\theta)^{\sigma}.$

Therefore, we have finished the proof of the necessity of (37).

Next, we prove the sufficiency of (37). Assume that
$N_{q^{6}/{q^{3}}}(\delta)=N_{q^{6}/{q^{3}}}(\theta)^{\tau}$ and
$\theta,\delta\neq 0$.

We only have to prove that there exists an invertible matrix $M$ such that
(38) holds. Let $M=\begin{pmatrix}1&0\\\ 0&d\end{pmatrix}$ where $d$ satisfies
(40). Hence

$d^{(q^{3}+1)(q^{2}+q+1)}=\left(\frac{\delta}{\theta^{\tau}}\right)^{q^{2}(q^{3}+1)}=1.$

and there exists $\alpha\in\mathbb{F}_{q^{6}}^{*}$ such that $d=\alpha^{q-1}$.
Replacing $x$ by $\alpha x$, we obtain

(46) $L_{f}=\left\\{(\alpha x)^{q-1}+\delta(\alpha
x)^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{4}-q}\delta
x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

As
$\alpha^{q^{4}-q}\delta=d^{q(q^{2}+q+1)}\delta=\frac{\delta^{q^{3}+1}}{\theta^{q^{3}\tau}}=\theta^{\tau}$
and $d=\alpha^{q-1}$, (46) is equivalent to

$L_{f}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}=L_{d\bar{g}}.\qed$

######  Remark 4.6.

In fact, if $N_{q^{6}/{q^{3}}}(\delta)=N_{q^{6}/{q^{3}}}(\theta)^{\sigma}\neq
1$, we can use Lemma 2.3 directly to prove the sufficient condition in Theorem
4.5.

######  Theorem 4.7.

Let $f(x)=x^{q}+\theta x^{q^{4}}$ over $\mathbb{F}_{q^{6}}$ with $\theta\neq
0$. The automorphism group of $L_{f}$ is

$\displaystyle\mathrm{Aut}(L_{f})=\begin{cases}\mathcal{D},&N_{q^{6}/{q^{3}}}(\theta)=1;\\\
\mathcal{D}\cup\mathcal{C},&N_{q^{6}/{q^{3}}}(\theta)\neq 1,\end{cases}$

where

$\mathcal{D}=\left\\{M\tau:N_{q^{6}/{q^{3}}}(\theta)^{\tau-1}=1,M=\begin{pmatrix}1&0\\\
0&d\end{pmatrix}\in\mathrm{PGL}(2,q^{6}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}}),d^{q^{2}+q+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{2}}\right\\},$

and

$\mathcal{C}=\left\\{M\tau:N_{q^{6}/{q^{3}}}(\theta)^{\tau-1}=1,M=\begin{pmatrix}0&1\\\
c&0\end{pmatrix}\in\mathrm{PGL}(2,q^{6}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}}),\left(\frac{c}{1-\theta^{\tau(q^{2}+q^{5})}}\right)^{q^{2}+q+1}=\frac{\theta^{q^{2}}}{-\theta^{\tau
q^{5}}}\right\\}.$

###### Proof.

We only have to follow the proof of Theorem 4.5 under the assumption that
$\theta=\delta$. Depending on the value of $b$, we consider two cases.

Case 1: $b=0$. When $b=0$ in (38), we always have $a=1$ and $c=0$. Hence, we
just need to determine $d\in\mathbb{F}_{q^{6}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}})$ such that $L_{d\bar{f}}=L_{f}$ where
$\bar{f}(x)=x^{q}+\theta^{\tau}x^{q^{4}}$. By (40) and (41), if
$L_{d\bar{f}}=L_{f}$, then $d,\tau$ satisfy

(47) $d^{q^{2}+q+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{2}},$

and

(48) $N_{q^{6}/{q^{3}}}(\theta)=N_{q^{6}/{q^{3}}}(\theta)^{\tau}.$

Thus (47) and (48) are the necessary condition.

Next, we show that (47) and (48) are also sufficient. By (47) and (48), we get

$d^{(q^{3}+1)(q^{2}+q+1)}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{2}(q^{3}+1)}=N_{q^{6}/{q^{3}}}\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{2}}=1.$

Hence there exists $\alpha\in\mathbb{F}_{q^{6}}^{*}$ such that
$d=\alpha^{q-1}$. Replacing $x$ by $\alpha x$, we obtain

(49) $\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{4}-q}\theta
x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

As
$\alpha^{q^{4}-q}\theta=d^{q(q^{2}+q+1)}\theta=\frac{\theta^{q^{3}+1}}{\theta^{q^{3}\tau}}=\theta^{\tau}$,
(49) is equivalent to

$\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

Case 2: $b\neq 0$. When $b\neq 0$ in (38), we have
$N_{q^{6}/{q^{3}}}(\theta)\neq 1$, $b=1$, and $a=d=0$. Hence, we just need to
determine $c\in\mathbb{F}_{q^{6}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{6}})$ such that $L_{ch}=L_{f}$ where

$h(x)=-\frac{\theta^{\tau
q^{5}}x^{q^{2}}}{1-\theta^{\tau(q^{2}+q^{5})}}+\frac{x^{q^{5}}}{1-\theta^{\tau(q^{2}+q^{5})}}.$

By (44) and (45), if $L_{ch}=L_{f}$, then $c,\tau$ satisfy

(50) $\hat{c}^{q^{2}+q+1}=\frac{\theta^{q^{2}}}{-\theta^{q^{5}\tau}},$

where $\hat{c}=\frac{c}{1-\theta^{\tau(q^{2}+q^{5})}}$, and

(51) $N_{q^{6}/{q^{3}}}(\theta)=N_{q^{6}/{q^{3}}}(\theta)^{\tau}.$

Hence (50) and (51) are the necessary condition.

Next, we show that (50) and (51) are also sufficient. By (42), (43) and
$\delta=\theta$, we only need to proof that the sufficiency for (43) with
$\delta=\theta$. By (50) and (51), we get

$\hat{c}^{(q^{3}+1)(q^{2}+q+1)}=\left(\frac{\theta^{q^{2}}}{-\theta^{q^{5}\tau}}\right)^{q^{3}+1}=\frac{N_{q^{6}/{q^{3}}}(\theta)^{q^{2}}}{N_{q^{6}/{q^{3}}}(\theta)^{q^{2}\tau}}=1.$

Hence there exists $\alpha\in\mathbb{F}_{q^{6}}^{*}$ such that
$\hat{c}=\alpha^{q-1}$. Replacing $x$ by $\alpha x$, we obtain

(52) $\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{4}-q}\theta
x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.$

As
$\alpha^{q^{4}-q}\theta=\hat{c}^{q(q^{2}+q+1)}\theta=\frac{\theta^{q^{3}+1}}{(-\theta^{q^{5}\tau})^{q}}=(-\theta^{q^{5}\tau})^{q^{4}}$,
(52) is equivalent to

$\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{4}-1}:x\in\mathbb{F}_{q^{6}}^{*}\right\\}=\left\\{\hat{c}\left(x^{q-1}+(-\theta^{q^{5}\tau})^{q^{4}}x^{q^{4}-1}\right):x\in\mathbb{F}_{q^{6}}^{*}\right\\}.\qed$

######  Remark 4.8.

It is worth pointing out that $L_{f}$ and $L_{g}$ are not required to be
maximum scattered in Theorem 4.5 and Theorem 4.7. In another word, we have
determined the equivalence and the automorphism groups of the members in a
larger family of linear sets in $\mathrm{PG}(1,q^{6})$ which contains the
family $\mathscr{F}_{6}$ of maximum scattered ones discovered by Csjabók,
Marino, Polverino and Zanella in [5].

Next we turn to the investigation of equivalence problem for the second family
$\mathscr{F}_{8}$ constructed in [5].

######  Theorem 4.9.

Let $f(x)=x^{q}+\delta x^{q^{5}}$ and $g(x)=x^{q}+\theta x^{q^{5}}$. The
linear sets $L_{f}$ and $L_{g}$ are $\mathrm{P\Gamma L}-$equivalent is if and
only if

(53) $N_{q^{8}/{q^{4}}}(\delta)=N_{q^{8}/{q^{4}}}(\theta)^{\sigma},$

for some automorphism $\sigma\in\mathrm{Aut}(\mathbb{F}_{q^{4}})$.

###### Proof.

If $\theta=\delta=0$, the above conclusion is clearly correct. Hence, we can
assume that at least one of $\theta$ and $\delta$ is not equal to $0$.

First, we prove the necessity of (53).

Assume that $L_{f}$ and $L_{g}$ are equivalent which implies the existence of
invertible matrix $M=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}$ over
$\mathbb{F}_{q^{8}}$ and $\tau\in\mathrm{Aut}(\mathbb{F}_{q^{4}})$ such that

(54) $\left\\{x^{q-1}+\delta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\frac{cx+d\bar{g}(x)}{ax+b\bar{g}(x)}:x\in\mathbb{F}_{q^{2m}}^{*}\right\\},$

where $\bar{g}(x)=x^{q}+\theta^{\tau}x^{q^{5}}$.

Next, we investigate (54) in two different cases.

Case 1: $b=0$. As $M$ is viewed as an element in $\mathrm{PGL}(2,q^{8})$, we
can assume that $a=1$. By (1) in Lemma 2.6, we derive $c=0$. Then (54) becomes

$\left\\{x^{q-1}+\delta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

By (29) in Lemma 4.3, we get

(55) $d^{1+q+q^{2}+q^{3}}\theta^{\tau q^{3}}=\delta^{q^{3}},$

If exactly one of $\theta$ and $\delta$ equals $0$, then it has to be $\delta$
and $d$ must be $0$. However, this implies $M$ is singular which leads to a
contradiction. Hence $\theta$ and $\delta\neq 0$.

By (30) in Lemma 4.3, we get

(56)
$d^{(q^{3}-q^{2}+1)(1+q+q^{2}+q^{3})}=\frac{\delta^{q^{6}+q^{3}+q}}{\theta^{\tau(q^{6}+q^{3}+q)}}.$

Then by (55) and (56), we get

$\delta^{q^{5}+q}=\theta^{\tau(q^{5}+q)},$

let $\sigma=\tau|_{\mathbb{F}_{q^{4}}}$, then

$N_{q^{8}/{q^{4}}}(\delta)=N_{q^{8}/{q^{4}}}(\theta)^{\tau}.$

Case 2: $b\neq 0$. Without loss of generality, we assume that $b=1$. By Lemma
4.4, we get $a=d=0$ which implies

$M=\begin{pmatrix}0&1\\\ c&0\end{pmatrix}.$

Let $h_{0}(x)=-\theta^{\tau q^{7}}x^{q^{3}}+x^{q^{7}}$, then
$h_{0}(\bar{g}(x))=\left(1-\theta^{\tau(q^{3}+q^{7})}\right)x$.

If $N_{q^{8}/{q^{4}}}(\theta)=1$ which is equivalent to
$1-\theta^{\tau(q^{3}+q^{7})}=0$, then the map $x\mapsto g(x)$ must be not
invertible. By (54), the point $\langle(0,1)\rangle_{\mathbb{F}_{q^{8}}}$
would be in $L_{f}$ which is impossible. Hence $N_{q^{8}/{q^{4}}}(\theta)\neq
1$.

Let

$h(x)=-\frac{\theta^{\tau
q^{7}}x^{q^{3}}}{1-\theta^{\tau(q^{3}+q^{7})}}+\frac{x^{q^{7}}}{1-\theta^{\tau(q^{3}+q^{7})}}.$

Then $h(\bar{g}(x))=x$, i.e. $h(x)$ is the inverse map of $\bar{g}(x)$ with
$\bar{g}(x)=x^{q}+\theta^{\tau}x^{q^{5}}$. Thus (54) equals

(57) $\left\\{x^{q-1}+\delta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\frac{ch(x)}{x}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

By a simple calculation, the adjoint $\hat{h}_{1}(x)$ of $h_{1}(x)$ is

$\hat{h}_{1}(x)=x^{q}+(-\theta^{q^{7}\tau})^{q^{5}}x^{q^{5}},$

where $h_{1}(x)=(1-\theta^{\tau(q^{3}+q^{7})})h(x)$.

By Lemma 2.1, $L_{h_{1}}=L_{\hat{h}_{1}}$, (57) equals

(58) $\left\\{x^{q-1}+\delta
x^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\hat{c}\left(x^{q-1}+(-\theta^{q^{7}\tau})^{q^{5}}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\},$

where $\hat{c}=\frac{c}{1-\theta^{\tau(q^{3}+q^{7})}}$.

By (29) in Lemma 4.3, we get

(59) $\hat{c}^{1+q+q^{2}+q^{3}}(-\theta^{q^{7}\tau})=\delta^{q^{3}},$

If one of $\theta$ and $\delta$ equals $0$, then it has to be $\delta$ from
which it follows $\hat{c}=0$. It contradicts the assumption that $M$ is
invertible. Hence $\theta$ and $\delta$ must be nonzero.

Furthermore, by (30) in Lemma 4.3, we get

(60)
$\hat{c}^{(q^{3}-q^{2}+1)(1+q+q^{2}+q^{3})}=\frac{\delta^{q^{6}+q^{3}+q}}{(-\theta^{q^{7}\tau})^{q^{5}(q^{6}+q^{3}+q)}}.$

Then by (59) and (60), we get

$\delta^{q^{5}+q}=(-\theta^{q^{7}\tau})^{q^{5}(q^{5}+q)},$

let $\sigma=\tau|_{\mathbb{F}_{q^{4}}}$, then

$N_{q^{8}/{q^{4}}}(\delta)=N_{q^{8}/{q^{4}}}(\theta)^{\tau}.$

Therefore, we have finished the proof of the necessity of (53).

Now, we prove the sufficiency of (53). Assume that
$N_{q^{8}/{q^{4}}}(\delta)=N_{q^{8}/{q^{4}}}(\theta)^{\tau}$ with
$\theta,\delta\neq 0$.

We only have to prove that there exists an invertible matrix $M$ such that
(54) holds. Let $M=\begin{pmatrix}1&0\\\ 0&d\end{pmatrix}$ where $d$ satisfies
(55). Hence

$d^{(q^{4}+1)(q^{3}+q^{2}+q+1)}=\left(\frac{\delta}{\theta^{\tau}}\right)^{q^{3}(q^{4}+1)}=1.$

and there exists $\alpha\in\mathbb{F}_{q^{8}}^{*}$ such that $d=\alpha^{q-1}$.
Replacing $x$ by $\alpha x$, we obtain

(61) $L_{f}=\left\\{(\alpha x)^{q-1}+\delta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{5}-q}\delta
x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

As
$\alpha^{q^{5}-q}\delta=d^{q(q^{3}+q^{2}+q+1)}\delta=\frac{\delta^{q^{3}+1}}{\theta^{q^{3}\tau}}=\theta^{\tau}$
and $d=\alpha^{q-1}$, (61) is equivalent to

$L_{f}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}=L_{d\bar{g}}.\qed$

######  Theorem 4.10.

Let $f(x)=x^{q}+\theta x^{q^{5}}$ over $\mathbb{F}_{q^{8}}$ with $\theta\neq
0$. The automorphism group of $L_{f}$ is

$\displaystyle\mathrm{Aut}(L_{f})=\begin{cases}\mathcal{D},&N_{q^{8}/{q^{4}}}(\theta)=1;\\\
\mathcal{D}\cup\mathcal{C},&N_{q^{8}/{q^{4}}}(\theta)\neq 1,\end{cases}$

where

$\mathcal{D}=\left\\{M\tau:N_{q^{8}/{q^{4}}}(\theta)^{\tau-1}=1,M=\begin{pmatrix}1&0\\\
0&d\end{pmatrix}\in\mathrm{PGL}(2,q^{8}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{8}}),d^{q^{3}+q^{2}+q+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{3}}\right\\},$

and

$\mathcal{C}=\left\\{M\tau:N_{q^{8}/{q^{4}}}(\theta)^{\tau-1}=1,M=\begin{pmatrix}0&1\\\
c&0\end{pmatrix}\in\mathrm{PGL}(2,q^{8}),\tau\in\mathrm{Aut}(\mathbb{F}_{q^{8}}),\left(\frac{c}{1-\theta^{\tau(q^{3}+q^{7})}}\right)^{q^{3}+q^{2}+q+1}=\frac{\theta^{q^{3}}}{-\theta^{\tau
q^{7}}}\right\\}.$

###### Proof.

We only have to follow the proof of Theorem 4.9 under the assumption that
$\theta=\delta$. Depending on the value of $b$, we separate the rest part into
two cases.

Case 1: $b=0$. When $b=0$ in (54), we always have $a=1$ and $c=0$. Hence, we
just need to determine $d\in\mathbb{F}_{q^{8}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{8}})$ such that $L_{d\bar{f}}=L_{f}$ where
$\bar{f}(x)=x^{q}+\theta^{\tau}x^{q^{5}}$. By (55) and (56), if
$L_{d\bar{f}}=L_{f}$, then $d,\tau$ satisfy

(62) $d^{q^{3}+q^{2}+q+1}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{3}},$

and

(63) $N_{q^{8}/{q^{4}}}(\theta)=N_{q^{8}/{q^{4}}}(\theta)^{\tau}.$

Then (62) and (63) are the necessary condition.

Next, we show that (62) and (63) are also sufficient. By (62) and (63), we get

$d^{(q^{4}+1)(q^{3}+q^{2}+q+1)}=\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{3}(q^{4}+1)}=N_{q^{8}/{q^{4}}}\left(\frac{\theta}{\theta^{\tau}}\right)^{q^{3}}=1.$

Hence there exists $\alpha\in\mathbb{F}_{q^{8}}^{*}$ such that
$d=\alpha^{q-1}$. Replacing $x$ by $\alpha x$, we obtain

(64) $\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{5}-q}\theta
x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

As
$\alpha^{q^{5}-q}\theta=d^{q(q^{3}+q^{2}+q+1)}\theta=\frac{\theta^{q^{4}+1}}{\theta^{q^{4}\tau}}=\theta^{\tau}$,
(64) is equivalent to

$\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{d\left(x^{q-1}+\theta^{\tau}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

Case 2: $b\neq 0$. When $b\neq 0$ in (54), we have
$N_{q^{8}/{q^{4}}}(\theta)\neq 1$, $b=1$, and $a=d=0$. Hence, we just need to
determine $c\in\mathbb{F}_{q^{8}}$ and
$\tau\in\mathrm{Aut}(\mathbb{F}_{q^{8}})$ such that $L_{ch}=L_{f}$ where
$h(x)=-\frac{\theta^{\tau
q^{7}}x^{q^{3}}}{1-\theta^{\tau(q^{3}+q^{7})}}+\frac{x^{q^{7}}}{1-\theta^{\tau(q^{3}+q^{7})}}$.
By (59) and (60), if $L_{ch}=L_{f}$, then $c,\tau$ satisfy

(65) $\hat{c}^{q^{3}+q^{2}+q+1}=\frac{\theta^{q^{3}}}{-\theta^{q^{7}\tau}},$

where $\hat{c}=\frac{c}{1-\theta^{\tau(q^{3}+q^{7})}}$, and

(66) $N_{q^{8}/{q^{4}}}(\theta)=N_{q^{8}/{q^{4}}}(\theta)^{\tau}.$

Thus (65) and (66) are the necessary condition.

Next, we show that (65) and (66) are also sufficient. By (57), (58) and
$\delta=\theta$, we only need to proof that the sufficiency for (58) with
$\delta=\theta$. By (65) and (66), we get

$\hat{c}^{(q^{4}+1)(q^{3}+q^{2}+q+1)}=\left(\frac{\theta^{q^{3}}}{-\theta^{q^{7}\tau}}\right)^{q^{4}+1}=\frac{N_{q^{8}/{q^{4}}}(\theta)^{q^{3}}}{N_{q^{8}/{q^{4}}}(\theta)^{q^{3}\tau}}=1.$

Hence there exists $\alpha\in\mathbb{F}_{q^{8}}^{*}$ such that
$\hat{c}=\alpha^{q-1}$. Replacing $x$ by $\alpha x$, we obtain

(67) $\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\alpha^{q-1}\left(x^{q-1}+\alpha^{q^{5}-q}\theta
x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

As
$\alpha^{q^{5}-q}\theta=\hat{c}^{q(q^{3}+q^{2}+q+1)}\theta=\frac{\theta^{q^{4}+1}}{(-\theta^{q^{7}\tau})^{q}}=(-\theta^{q^{7}\tau})^{q^{5}}$,
(67) is equivalent to

$\left\\{(\alpha x)^{q-1}+\theta(\alpha
x)^{q^{5}-1}:x\in\mathbb{F}_{q^{8}}^{*}\right\\}=\left\\{\hat{c}\left(x^{q-1}+(-\theta^{q^{7}\tau})^{q^{5}}x^{q^{5}-1}\right):x\in\mathbb{F}_{q^{8}}^{*}\right\\}.$

∎

######  Remark 4.11.

As pointed our in Remark 4.8 on the equivalence problem of family
$\mathscr{F}_{6}$, we also have determined the equivalence and the
automorphism groups of the members in a larger family of linear sets in
$\mathrm{PG}(1,q^{8})$ which contains $\mathscr{F}_{8}$, because $f$ and $g$
in Theorem 4.9 and Theorem 4.10 are not required to be scattered.

##  5\. Concluding remarks

In this paper, we consider the equivalence of the members in the family of
maximum scattering linear sets constructed by Csjabók, Marino, Montanucci and
Zullo in [6, 14] and the same problem for two families $\mathscr{F}_{6}$ and
$\mathscr{F}_{8}$ constructed by Csjabók, Marino, Polverino and Zanella in
[5]. Their automorphism groups are determined, respectively.

For the Csjabók-Marino-Montanucci-Zullo family, it is proved that there is
only one element in it up to $\mathrm{P\Gamma L}$-equivalence, and its
automorphism group is determined. For the Csjabók-Marino-Polverino-Zanella
family, according to the result in [5], we only need to consider the linear
set equivalence problem for $s=1$ in $\mathscr{F}_{2m}$ for which the
necessary and sufficient conditions for their $\mathrm{P\Gamma L}$-equivalence
are given. The approach adopted in this paper for the maximum scattered linear
set can be used to solve the equivalence problems of other linear sets which
will be considered in our future work.

## Acknowledgment

This work is partially supported by the Natural Science Foundation of Hunan
Province (No. 2019RS2031) and the Training Program for Excellent Young
Innovators of Changsha (No. kq2106006).

## References

  * [1] D. Bartoli, B. Csajbók, M, Montanucci.  On a conjecture about maximum scattered subspaces of $\mathbb{F}_{q^{6}}\times\mathbb{F}_{q^{6}}$[J]. arXiv preprint arXiv:2004.13101, 2020.
  * [2] D. Bartoli, M. Giulietti, G. Marino and O. Polverino.  Maximum scattered linear sets and complete caps in Galois spaces,  _Combinatorica_ 38(2) (2018), 255–278.
  * [3] A. Blokhuis and M. Lavrauw.  Scattered spaces with respect to a spread in $\mathrm{PG}(n,q)$.  _Geom. Dedicata_ 81 (2000), 231–243.
  * [4] B. Csajbók, G. Marino, and O. Polverino.  Classes and equivalence of linear sets in $\mathrm{PG}(1,q^{n})$.  Journal of Combinatorial Theory, Series A, 157:402 – 426, 2018\.
  * [5] B. Csajbók, G. Marino, O. Polverino and C. Zanella.  A new family of MRD-codes,  Linear Algebra Appl. 548 (2018):203 – 220.
  * [6] B. Csajbók, G. Marino, and F. Zullo.  New maximum scattered linear sets of the projective line.  Finite Fields Appl. 54 (2018): 133 – 150.
  * [7] B. Csajbók and C. Zanella.  On the equivalence of linear sets.  Designs, Codes and Cryptography, 81(2):269–281, 2016.
  * [8] M. Lavrauw and G. Van de Voorde.  Field reduction and linear sets in finite geometry.  In G. Kyureghyan, G. Mullen, and A. Pott, editors, Contemporary Mathematics, volume 632, pages 271–293. American Mathematical Society, 2015\.
  * [9] R. Lidl and H. Niederreiter.  Finite fields, volume 20 of Encyclopedia of Mathematics and its Applications,  Cambridge University Press, Cambridge, second edition, 1997.
  * [10] G. Longobardi, G. Marino, R. Trombetti and Y. Zhou.  A large family of maximum scattered linear sets of PG$(1,q^{n})$ and their associated MRD codes,  arXiv:2102.08287, preprint, 2021.
  * [11] G. Longobardi and C. Zanella.  Linear sets and MRD-codes arising from a class of scattered linearized polynomials.  J Algebr Comb., 2021.
  * [12] G. Lunardon.  Normal Spreads.  Geometriae Dedicata, 75(3):245–261, 1999.
  * [13] G. Lunardon and O. Polverino.  Blocking sets and derivable partial spreads.  Journal of Algebraic Combinatorics, 14(1):49–56, 2001.
  * [14] G. Marino, M. Montanucci, and F. Zullo.  MRD-codes arising from the trinomial $x^{q}+x^{q^{3}}+cx^{q^{5}}\in\mathbb{F}_{q^{6}}$.  Linear Algebra Appl. 591 (2020): 99 – 114.
  * [15] A. Neri, P. Santonastaso, and F. Zullo.  Extending two families of maximum rank distance codes. arXiv: 2104.07602, preprint 2021.
  * [16] O. Polverino.  Linear sets in finite projective spaces.  Discrete Mathematics, (22):3096–3107, 2010.
  * [17] O. Polverino and F. Zullo.  Connections between scattered linear sets and MRD-codes.  Bulletin of the ICA, 89, 2020.
  * [18] J. Sheekey.  A new family of linear maximum rank distance codes.  _Adv. Math. Commun._ 10(3) (2016), 475 – 488.
  * [19] Tang W, Zhou Y, Zullo F.  On the automorphism groups of Lunardon-Polverino scattered linear sets. arXiv:2201.12777, preprint, 2022.
  * [20] Timpanella M, Zini G.  On a family of linear MRD codes with parameters $[8\times 8,16,7]_{q}$. arXiv:2108.13082, preprint, 2021.


# Optimal Distributed Control for Leader-Follower Networks: A Scalable Design

Jalal Arabneydi, Mohammad M. Baharloo, and Amir G. Aghdam  This work has been
supported in part by the Natural Sciences and Engineering Research Council of
Canada (NSERC) under Grant RGPIN-262127-17, and in part by Concordia
University under Horizon Postdoctoral Fellowship.Jalal Arabneydi, Mohammad M.
Baharloo, and Amir G. Aghdam are with the Department of Electrical and
Computer Engineering, Concordia University, 1455 de Maisonneuve Blvd. West,
Montreal, QC, Canada, Postal Code: H3G 1M8.
<EMAIL_ADDRESS><EMAIL_ADDRESS>and
<EMAIL_ADDRESS>
###### Abstract

The focus of this paper is directed towards optimal control of multi-agent
systems consisting of one leader and a number of followers in the presence of
noise. The dynamics of every agent is assumed to be linear, and the
performance index is a quadratic function of the states and actions of the
leader and followers. The leader and followers are coupled in both dynamics
and cost. The state of the leader and the average of the states of all
followers (called mean-field) are common information and known to all agents;
however, the local state of the followers are private information and unknown
to other agents. It is shown that the optimal distributed control strategy is
linear time-varying, and its computational complexity is independent of the
number of followers. This strategy can be computed in a distributed manner,
where the leader needs to solve one Riccati equation to determine its optimal
strategy while each follower needs to solve two Riccati equations to obtain
its optimal strategy. This result is subsequently extended to the case of the
infinite horizon discounted and undiscounted cost functions, where the optimal
distributed strategy is shown to be stationary. A numerical example with $100$
followers is provided to demonstrate the efficacy of the results.

Proceedings of IEEE Canadian Conf. on Elec. and Comp. Eng., 2018.

##  I Introduction

There has been a growing interest in networked control systems in recent
years, due to their applications in emerging areas such as control of a
platoon of autonomous vehicles, environmental monitoring using sensor
networks, and surveillance using a team of UAVs [1, 2, 3]. In particular, in
the control of multi-agent systems, every agent exchanges some information
with a subset of agents in order to properly coordinate its position and
movement such that a global objective is achieved. To this end, each agent
requires some computational effort in order to compute its control action
based on the information available to it.

The leader-follower structure is particularly very common in the coordination
control of multi-agent networks. In this type of system, each agent is either
a leader or a follower, where the movement of the followers is dependent on
the trajectory of the leader(s). A global objective that is of special
interest is consensus, where the states of the agents are desired to converge
to a common value [4, 5, 6]. If the communication graph of the network is
connected, then consensus can be reached using a linear strategy. However,
such strategy suffers from the curse of dimensionality, in general. In
addition, the required amount of communication between agents under this type
of strategy is typically high. This can lead to some practical problems,
specially given that the battery consumption of each node is closely related
to the amount of its required communication. An optimal control strategy can
therefore be very important for the efficient use of resources in the network.
Inspired by this objective, the authors in [7, 8] study distributed linear
quadratic control. Although the above techniques are effective in many
cooperative control applications, they are limited by the computational cost,
making them unsuitable for large-scale networks. A large-scale network of
homogeneous agents with decoupled dynamics is investigated in [9], for which
the infinite-horizon optimal control strategy is obtained by solving two
scalable coupled algebraic Riccati equations.

The present work aims to address the above shortcomings for a leader-follower
multi-agent network with a large number of followers. It is assumed that the
graph representing the network is such that the average of the states of the
followers and the local state of the leader are available to every agent. In
contrast to [9], the network considered in this paper has a leader, the
dynamics of agents are coupled, and the cost function can be either finite-
horizon or infinite-horizon. The mean-field team approach [10] is used in this
paper to obtain the optimal control strategy by solving two scalable decoupled
Riccati equations.

This paper is organized as follows. In Section II, a leader-follower multi-
agent network with the mean-field information structure is formulated. The
main result of the paper is developed in Section III for the finite-horizon
cost function, and is then extended to the infinite-horizon case in Section
IV. Simulation results are provided in Section V, and the paper is concluded
in Section VI.

##  II Problem Formulation

In this paper, $\mathbb{N}$ and $\mathbb{R}$ represent natural and real
numbers, respectively, and given any $k\in\mathbb{N}$, the finite set of
integers $\\{1,2,\ldots,k\\}$ is denoted by $\mathbb{N}_{k}$.

Consider a multi-agent network with one leader and $n\in\mathbb{N}$ followers,
operating over a finite control horizon $T\in\mathbb{N}$. Let
$x^{i}_{t}\in\mathbb{R}^{d_{x}}$ and $u^{i}_{t}\in\mathbb{R}^{d_{u}},$
$d_{x},d_{u}\in\mathbb{N}$, denote the state and action of follower
$i\in\mathbb{N}_{n}$ at time $t\in\mathbb{N}_{T}$, respectively. Denote the
average of the states of the followers at time $t$ by

$\bar{x}_{t}=\frac{1}{n}\sum_{i=1}^{n}x^{i}_{t}.$

Following the terminology of mean-field teams [10], we refer to the average of
the states of the followers as _mean-field_ in the sequel. Let also
$x^{0}_{t}\in\mathbb{R}^{d_{x}}$ and $u^{0}_{t}\in\mathbb{R}^{d_{u}}$ denote
the state and action of the leader at time $t\in\mathbb{N}_{T}$, respectively.
The dynamics of the leader at time $\mathbb{N}_{T}$ is given by

$x^{0}_{t+1}=A^{0}_{t}x^{0}_{t}+B^{0}_{t}u^{0}_{t}+D^{0}_{t}\bar{x}_{t}+w^{0}_{t},$
(1)

where $w^{0}_{t}\in\mathbb{R}^{d_{x}}$ is the state noise of the leader.
Similarly, the dynamics of follower $i\in\mathbb{N}_{n}$ is described by

$x^{i}_{t+1}=A_{t}x^{i}_{t}+B_{t}u^{i}_{t}+D_{t}\bar{x}_{t}+E_{t}x^{0}_{t}+w^{i}_{t},$
(2)

where $w^{i}_{t}\in\mathbb{R}^{d_{x}}$ is the state noise of the $i$-th
follower. In general, the leader’s dynamics may depend on the states of the
followers. Similarly, the followers’ dynamics may depend on the state of the
leader as well as the states of followers.

At each time $t\in\mathbb{N}_{T}$, the leader observes its local state and the
mean-field, i.e.,

$u^{0}_{t}=g^{0}_{t}(x^{0}_{t},\bar{x}_{t}),$ (3)

where $g^{0}_{t}:(\mathbb{R}^{d_{x}})^{2}\rightarrow\mathbb{R}^{d_{u}}$.
Furthermore, each follower $i\in\mathbb{N}_{n}$ observes its local state, the
state of the leader, and the mean-field, i.e.,

$u^{i}_{t}=g^{i}_{t}(x^{i}_{t},\bar{x}_{t},x^{0}_{t}),$ (4)

where $g^{i}_{t}:(\mathbb{R}^{d_{x}})^{3}\rightarrow\mathbb{R}^{d_{u}}$. Under
this information structure, the privacy of each follower is preserved, i.e.,
the local state of each follower is only known to itself. Note that there are
different ways to share the mean-field $\bar{x}_{t}$ among the agents,
depending on the structure of the communication graph. For example, all agents
can send their states to the leader and then the leader computes the mean and
sends it back to every follower (in which case, a link is required between the
leader and every follower). Alternatively, each agent can run a consensus
algorithm to compute the mean-field within the control time interval.111In
practical applications, the control operation has a much longer time-scale
compared to the communication operation.

It is desired that the leader and followers minimize a prescribed quadratic
cost function, while achieving a global objective (such as consensus) as a
group. This cost function can, for instance, reflect the coordination error of
the agents as well as the energy consumption of the actuators, collectively.
To this end, consider the following optimization problem.

######  Problem 1.

Given the dynamics (1) and (2) as well as the information structures (3) and
(4), find the optimal strategy that minimizes the following performance index:

$J_{T}=\mathbb{E}\Big{[}\sum_{t=1}^{T}(x^{0}_{t})^{\intercal}Q^{0}_{t}x^{0}_{t}+(u^{0}_{t})^{\intercal}R^{0}_{t}u^{0}_{t}\\\
+\frac{1}{n}\sum_{i=1}^{n}(x^{i}_{t})^{\intercal}Q_{t}x^{i}_{t}+(x^{i}_{t}-x^{0}_{t})^{\intercal}P_{t}(x^{i}_{t}-x^{0}_{t})+(u^{i}_{t})^{\intercal}R_{t}u^{i}_{t}\\\
+\frac{1}{2n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}(x^{i}_{t}-x^{j}_{t})^{\intercal}H_{t}(x^{i}_{t}-x^{j}_{t})\Big{]},$
(5)

where matrices $Q^{0}_{t},R^{0}_{t},Q_{t},P_{t},R_{t},$ and $H_{t}$ are
symmetric.

Note that the rate of convergence of the followers to the leader is directly
dependent on matrix $P_{t}$ in (5). Similarly, the movement of the followers
as a group depends on matrix $H_{t}$.

######  Remark 1.

_In the special case when matrices $B^{0}_{t}$, $D^{0}_{t}$, $Q^{0}_{t},$ and
$R^{0}_{t}$ are zero, Problem 1 becomes the optimal control of a leaderless
multi-agent system, where it is desired that the followers track the reference
signal $x^{0}_{t}$._

In general, Problem 1 is difficult to solve due to its complex information
structure. Since neither the information structure is partially nested [11]
nor the problem is quadratic invariant [12], and in addition the noise
processes are not necessarily Gaussian, one can not assume that the optimal
strategy is linear.222When the information structure is non-classical, the
optimal strategy may not be linear [13]. Moreover, the dimension of the
augmented matrices, which are fully dense, increases with the number of
followers, i.e., solving Problem 1 using existing techniques can be
computationally expensive for a large number of followers.

##  III Main Result

In this section, the main result of this paper is presented. It is assumed
that the primitive random variables satisfy the following standard assumption.

######  Assumption 1.

The initial states and noise processes are mutually independent in time.

At any time $t\in\mathbb{N}_{T}$, define the following matrices:

$\bar{A}_{t}:=\left[\begin{array}[]{cc}A^{0}_{t}&D^{0}_{t}\\\
E_{t}&A_{t}+D_{t}\end{array}\right],\bar{B}_{t}:=\left[\begin{array}[]{cc}B^{0}_{t}&0_{d_{x}\times
d_{u}}\\\ 0_{d_{x}\times d_{u}}&B_{t}\end{array}\right],$ (6)
$\bar{Q}_{t}:=\left[\begin{array}[]{cc}Q^{0}_{t}+P_{t}&-P_{t}\\\
-P_{t}&Q_{t}+P_{t}\end{array}\right],\bar{R}_{t}:=\left[\begin{array}[]{cc}R^{0}_{t}&0_{d_{u}\times
d_{u}}\\\ 0_{d_{u}\times d_{u}}&R_{t}\end{array}\right].$ (7)

######  Assumption 2.

Matrices $Q_{t}+P_{t}+H_{t}$ and $\bar{Q}_{t}$ are positive semi-definite and
matrices $R^{0}_{t}$ and $R_{t}$ are positive definite.

For any $t\in\mathbb{N}_{T}$, define the following Riccati equation:

$\breve{M}_{t}=-A_{t}^{\intercal}\breve{M}_{t+1}B_{t}\left(B_{t}^{\intercal}\breve{M}_{t+1}B_{t}+R_{t}\right)^{-1}B_{t}^{\intercal}\breve{M}_{t+1}A_{t}\\\
+A_{t}^{\intercal}\breve{M}_{t+1}A_{t}+Q_{t}+P_{t}+H_{t},$ (8)

where $\breve{M}_{T+1}=0_{d_{x}\times d_{x}}$. Define also the following
Riccati equation:

$\bar{M}_{t}=-\bar{A}_{t}^{\intercal}\bar{M}_{t+1}\bar{B}_{t}\left(\bar{B}_{t}^{\intercal}\bar{M}_{t+1}\bar{B}_{t}+\bar{R}_{t}\right)^{-1}\bar{B}_{t}^{\intercal}\bar{M}_{t+1}\bar{A}_{t}\\\
+\bar{A}_{t}^{\intercal}\bar{M}_{t+1}\bar{A}_{t}+\bar{Q}_{t},$ (9)

for any $t\in\mathbb{N}_{T}$, with
$\bar{M}_{T+1}=\left[\begin{array}[]{cc}0_{d_{x}\times d_{x}}&0_{d_{x}\times
d_{x}}\\\ 0_{d_{x}\times d_{x}}&0_{d_{x}\times d_{x}}\end{array}\right]$.

######  Theorem 1.

Let Assumptions 1 and 2 hold. The optimal strategy for Problem 1 is linear and
unique, i.e.,

$\displaystyle{u^{0}_{t}}$
$\displaystyle=\bar{L}^{1,1}_{t}x^{0}_{t}+\bar{L}^{1,2}_{t}\bar{x}_{t},$
$\displaystyle{u^{i}_{t}}$
$\displaystyle=\breve{L}_{t}x^{i}_{t}+\bar{L}_{t}^{2,1}x^{0}_{t}+(\bar{L}_{t}^{2,2}-\breve{L}_{t})\bar{x}_{t},$
(10)

where the gains $\\{\breve{L}_{t},\bar{L}_{t}\\}_{t=1}^{T-1}$ are obtained by
solving Riccati equations (8) and (9) as follows:

$\displaystyle\breve{L}_{t}$
$\displaystyle=-\left(B_{t}^{\intercal}\breve{M}_{t+1}B_{t}+R_{t}\right)^{-1}B_{t}^{\intercal}\breve{M}_{t+1}A_{t},$
$\displaystyle\bar{L}_{t}$
$\displaystyle=\left[\begin{array}[]{cc}\bar{L}^{1,1}_{t}&\bar{L}^{1,2}_{t}\\\
\bar{L}^{2,1}_{t}&\bar{L}^{2,2}_{t}\end{array}\right]=-\left(\bar{B}_{t}^{\intercal}\bar{M}_{t+1}\bar{B}_{t}+\bar{R}_{t}\right)^{-1}\bar{B}_{t}^{\intercal}\bar{M}_{t+1}\bar{A}_{t}.$
(13)

###### Proof.

Suppose every agent knows the centralized information
$\\{x^{0}_{t},x^{1}_{t},\ldots,x^{n}_{t}\\}$. Let
$\bar{u}_{t}=\frac{1}{n}\sum_{i=1}^{n}u^{i}_{t}$ and
$\bar{w}_{t}=\frac{1}{n}\sum_{i=1}^{n}w^{i}_{t}$. We first transform the
problem by using an isomorphic transformation and solve the transformed
problem. Then, we use the inverse transformation to transform the obtained
solution to the solution of the original problem, and show that the resultant
(centralized) solution is implementable under the decentralized information
structure. Define $\breve{x}^{i}_{t}:=x^{i}_{t}-\bar{x}_{t}$,
$\breve{u}^{i}_{t}:=u^{i}_{t}-\bar{u}_{t}$, and
$\breve{w}^{i}_{t}:=w^{i}_{t}-\bar{w}_{t}$. From (1) and (2),

$\displaystyle\breve{x}^{i}_{t+1}$
$\displaystyle=A_{t}\breve{x}^{i}_{t}+B_{t}\breve{u}^{i}_{t}+\breve{w}^{i}_{t},$
$\displaystyle\left[\begin{array}[]{c}x^{0}_{t+1}\\\
\bar{x}_{t+1}\end{array}\right]$
$\displaystyle=\bar{A}_{t}\left[\begin{array}[]{c}x^{0}_{t}\\\
\bar{x}_{t}\end{array}\right]+\bar{B}_{t}\left[\begin{array}[]{c}u^{0}_{t}\\\
\bar{u}_{t}\end{array}\right]+\left[\begin{array}[]{c}w^{0}_{t}\\\
\bar{w}_{t}\end{array}\right].$ (22)

Rewrite the cost function $J_{T}$, given by (5), in terms of the new variables
as follows:

$J_{T}=\mathbb{E}\Big{[}\sum_{t=1}^{T}(x^{0}_{t})^{\intercal}Q^{0}_{t}x^{0}_{t}+(u^{0}_{t})^{\intercal}R^{0}_{t}u^{0}_{t}+\frac{1}{n}\sum_{i=1}^{n}(\breve{x}^{i}_{t}+\bar{x}_{t})^{\intercal}Q_{t}(\breve{x}^{i}_{t}+\bar{x}_{t})\\\
+\frac{1}{n}\sum_{i=1}^{n}(\breve{x}^{i}_{t}+\bar{x}_{t}-x^{0}_{t})^{\intercal}P_{t}(\breve{x}^{i}_{t}+\bar{x}_{t}-x^{0}_{t})+(\breve{u}^{i}_{t}+\bar{u}_{t})^{\intercal}R_{t}(\breve{u}^{i}_{t}+\bar{u}_{t})\\\
+\frac{1}{2n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}(\breve{x}^{i}_{t}-\breve{x}^{j}_{t})^{\intercal}H_{t}(\breve{x}^{i}_{t}-\breve{x}^{j}_{t})\Big{]}.$
(23)

The above relation can be simplified by exploiting the fact that
$\frac{1}{n}\sum_{i=1}^{n}\breve{x}^{i}_{t}=0_{d_{x}\times 1}$ and
$\frac{1}{n}\sum_{i=1}^{n}\breve{u}^{i}_{t}=0_{d_{u}\times 1}$. This leads to
the following simplified equation:

$\displaystyle
J_{T}=\mathbb{E}\Big{[}\sum_{t=1}^{T}\left[\begin{array}[]{c}x^{0}_{t}\\\
\bar{x}_{t}\end{array}\right]^{\intercal}\bar{Q}_{t}\left[\begin{array}[]{c}x^{0}_{t}\\\
\bar{x}_{t}\end{array}\right]+\left[\begin{array}[]{c}u^{0}_{t}\\\
\bar{u}_{t}\end{array}\right]^{\intercal}\bar{R}_{t}\left[\begin{array}[]{c}u^{0}_{t}\\\
\bar{u}_{t}\end{array}\right]$ (32)
$\displaystyle+\frac{1}{n}\sum_{i=1}^{n}(\breve{x}^{i}_{t})^{\intercal}(Q_{t}+P_{t}+H_{t})(\breve{x}^{i}_{t})+(\breve{u}^{i}_{t})^{\intercal}R_{t}(\breve{u}^{i}_{t})\Big{]}.$
(33)

The cost function $J_{T}$ in (32) is the sum of the cost functions of $n$
systems with state and action $(\breve{x}^{i}_{t},\breve{u}^{i}_{t})$,
$i\in\mathbb{N}_{n},$ and one system with the state and action
$(\\{x^{0}_{t},\bar{x}_{t}\\},\\{u^{0}_{t},\bar{u}_{t}\\})$. These $n+1$
systems are decoupled due to the certainty equivalence theorem [14]. As a
result, $J_{T}$ is minimized when the cost functions of the $n+1$ systems with
decoupled dynamics and cost are minimized, i.e.,

$\left[\begin{array}[]{c}u^{0}_{t}\\\
\bar{u}_{t}\end{array}\right]=\bar{L}_{t}\left[\begin{array}[]{c}x^{0}_{t}\\\
\bar{x}_{t}\end{array}\right]\quad\text{and}\quad\breve{u}^{i}_{t}=\breve{L}_{t}\breve{x}^{i}_{t},\quad
i\in\mathbb{N}_{n}.$ (34)

We now transform the above solution to the solution of the original problem:

$u^{i}_{t}=\breve{u}^{i}_{t}+\bar{u}_{t}=\breve{L}_{t}(x^{i}_{t}-\bar{x}_{t})+[\bar{L}^{2,1}_{t}\quad\bar{L}^{2,2}_{t}]\left[\begin{array}[]{c}x^{0}_{t}\\\
\bar{x}_{t}\end{array}\right].$ (35)

The transformed solution is optimal for the original decentralized problem
because it is implementable under the information structure in Section II. ∎

According to Theorem 1, the leader must solve the Riccati equation (9) to
determine its optimal strategy ($\bar{L}^{1,1}_{t}$, $\bar{L}^{1,2}_{t}$)
whereas each follower must solve the Riccati equations (8) and (9) to find its
optimal strategy ($\breve{L}_{t},\bar{L}^{2,1}_{t},$ $\bar{L}^{2,2}_{t}$): one
for the local adjustment with their average (i.e., mean-field) and one for the
global adjustment with the leader.

######  Corollary 1.

Let matrices $\breve{L}_{t}$ and $\bar{L}^{1,1}_{t}$ in Theorem 1 be
invertible. The optimal strategy can be rewritten in the form of the solution
of a standard consensus problem as follows:

$\displaystyle{u^{0}_{t}}$
$\displaystyle=\sum_{i=1}^{n}\alpha_{t}(x^{0}_{t}-\beta_{t}x^{i}_{t}),$
$\displaystyle{u^{i}_{t}}$
$\displaystyle=\sum_{j=1}^{n}\gamma_{t}(x^{i}_{t}-\mu_{t}x^{j}_{t})+\sum_{i=1}^{n}\lambda_{t}(x^{0}_{t}-x^{i}_{t}),$
(36)

where $\alpha_{t}=\frac{1}{n}\bar{L}^{1,1}_{t}$,
$\beta_{t}=-(\bar{L}^{1,1}_{t})^{-1}\bar{L}^{1,2}_{t},\gamma_{t}=\frac{1}{n}\breve{L}_{t},\mu_{t}=-(\breve{L}_{t})^{-1}(\bar{L}^{2,2}_{t}+\bar{L}^{2,1}_{t}-\breve{L}_{t}),$
and $\lambda_{t}=\frac{1}{n}\bar{L}^{2,1}_{t}$.

######  Remark 2.

_Corollary 1 provides the optimal information flow topology between the leader
and followers for the case when such a topology is not pre-specified._

##  IV Infinite Horizon

In this section, the result of Theorem 1 is extended to the case of infinite
horizon. To this end, it is assumed that the dynamics of the agents as well as
the cost function are time-homogeneous; therefore, the subscript $t$ is
omitted to simplify the notation. Given $\beta\in(0,1]$, define

$J_{\infty}=\mathbb{E}\Bigg{[}\sum_{t=1}^{\infty}\beta^{t-1}\Big{(}(x^{0}_{t})^{\intercal}Q^{0}x^{0}_{t}+(u^{0}_{t})^{\intercal}R^{0}u^{0}_{t}\\\
+\frac{1}{n}\sum_{i=1}^{n}(x^{i}_{t})^{\intercal}Qx^{i}_{t}+(x^{i}_{t}-x^{0}_{t})^{\intercal}P(x^{i}_{t}-x^{0}_{t})+(u^{i}_{t})^{\intercal}Ru^{i}_{t}\\\
+\frac{1}{2n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}(x^{i}_{t}-x^{j}_{t})^{\intercal}H(x^{i}_{t}-x^{j}_{t})\Big{)}\Bigg{]}.$
(37)

When $\beta<1$, $J_{\infty}$ is called infinite-horizon discounted cost and
when $\beta=1$, it is called infinite horizon undiscounted cost. The following
standard assumption is imposed on the model.

######  Assumption 3.

Let $(\sqrt{\beta}A,\sqrt{\beta}B)$ and
$(\sqrt{\beta}\bar{A},\sqrt{\beta}\bar{B})$ be stabilizable and
$(\sqrt{\beta}A,Q^{1/2})$ and $(\sqrt{\beta}\bar{A},\bar{Q}^{1/2})$ be
detectable.

Define the following two algebraic Riccati equations:

$\breve{M}=-\beta
A^{\intercal}\breve{M}B\left(B^{\intercal}\breve{M}B+\beta^{-1}R\right)^{-1}B^{\intercal}\breve{M}A\\\
+\beta A^{\intercal}\breve{M}A+Q+P+H,$ (38)
$\bar{M}=-\beta\bar{A}^{\intercal}\bar{M}\bar{B}\left(\bar{B}^{\intercal}\bar{M}\bar{B}+\beta^{-1}\bar{R}\right)^{-1}\bar{B}^{\intercal}\bar{M}\bar{A}\\\
+\beta\bar{A}^{\intercal}\bar{M}\bar{A}+\bar{Q}.$ (39)

######  Theorem 2.

Let Assumptions 1, 2 and 3 hold. Then, the optimal control strategy is given
by:

$\displaystyle{u^{0}_{t}}$
$\displaystyle=\bar{L}^{1,1}x^{0}_{t}+\bar{L}^{1,2}\bar{x}_{t},$ (40)
$\displaystyle{u^{i}_{t}}$
$\displaystyle=\breve{L}x^{i}_{t}+\bar{L}^{2,1}x^{0}_{t}+(\bar{L}^{2,2}-\breve{L})\bar{x}_{t},$
(41)

where the gains $\\{\breve{L},\bar{L}\\}$ are obtained by solving algebraic
Riccati equations (38) and (39) as follows:

$\displaystyle\breve{L}$
$\displaystyle=-\left(B^{\intercal}\breve{M}B+\beta^{-1}R\right)^{-1}B^{\intercal}\breve{M}A,$
$\displaystyle\bar{L}$
$\displaystyle=\left[\begin{array}[]{cc}\bar{L}^{1,1}&\bar{L}^{1,2}\\\
\bar{L}^{2,1}&\bar{L}^{2,2}\end{array}\right]=-\left(\bar{B}^{\intercal}\bar{M}\bar{B}+\beta^{-1}\bar{R}\right)^{-1}\bar{B}^{\intercal}\bar{M}\bar{A}.$
(44)

###### Proof.

By a simple change of variables, an infinite-horizon discounted cost problem
with the 4-tuple $(A,B,Q,R)$ and discount factor $\beta$ can be transformed to
an infinite-horizon undiscounted cost problem with 4-tuple
$(\sqrt{\beta}A,\sqrt{\beta}B,Q,R)$. By applying the same isomorphic
transformation as in the proof of Theorem 1 on the resultant undiscounted
formulation, and using a similar argument as in that proof, the $n+1$
decoupled systems are obtained. ∎

##  V Numerical Example

Example 1. In this section, we present an example of a multi-agent system with
a leader and $100$ followers to verify our theoretical results. Let the
initial state of the leader be $30$, i.e., $x^{0}_{1}=30$, and the initial
states of the followers be uniformly distributed random variables in the
interval $[0,20]$. Let also the dynamics of the agents be driven by (1) and
(2) with the following scalar parameters:
$A^{0}_{t}=1,B^{0}_{t}=0.3,A_{t}=1,B_{t}=0.2,D^{0}_{t}=0.05,D_{t}=0.01,E_{t}=0.01,$
and noises:

$w^{0}_{t}\sim\mathcal{N}(0,0.1),\quad
w^{i}_{t}\sim\mathcal{N}(0,0.2),\quad\forall i\in\mathbb{N}_{n}.$ (45)

Consider the cost function (5) with the following parameters:

$Q^{0}_{t}=1,R^{0}_{t}=100,Q_{t}=0.1,P_{t}=50,R_{t}=50,H_{t}=1.$ (46)

Assume first that $T=80$ (the finite-horizon case). Using Theorem 1, the
optimal trajectories of the leader and followers shown in Figure 1 are
obtained. Figure 1 shows that the states of the followers (thin colored
curves) converge to a small neighborhood of the state of the leader (thick
black curve). The size of this neighborhood depends, in fact, on the noise
variance. In the special case when there is no noise, all followers’ states
approach the state of the leader.

Assume now that $T=\infty$ (the infinite-horizon case with undiscounted cost).
Using Theorem 2 in this case, the results demonstrated in Figure 2 are
obtained, analogously to Figure 1. Figure 2 shows good convergence results for
the followers in the presence of noise.

Figure 1: The trajectories of the leader and followers in Example 1 for the
finite-horizon case.

Figure 2: The trajectories of the leader and followers in Example 1 for the
infinite-horizon case.

##  VI Conclusions

In this paper, optimal distributed control of a multi-agent system with a
leader-follower structure is investigated. It is assumed that the average of
the states of the followers and the local state of the leader are available to
every agent. The optimal solution is obtained by solving two decoupled Riccati
equations whose computational complexities are independent of the number of
followers. In the infinite-horizon case, the Riccati difference equations
become algebraic Riccati equations. As an interesting future work, one can
consider the case where the number of followers is sufficiently large. In this
case, the average of the states of followers can be efficiently approximated
by the law of large numbers, which means that the only information to be
shared is the local state of the leader. This implies that the communication
topology of the network is described by a directed graph with paths from the
leader to followers.

## References

  * [1] J. A. Fax and R. M. Murray, “Information flow and cooperative control of vehicle formations,” _IEEE Transactions on Automatic Control_ , vol. 49, no. 9, pp. 1465–1476, 2004.
  * [2] P. Rawat, K. D. Singh, H. Chaouchi, and J. M. Bonnin, “Wireless sensor networks: a survey on recent developments and potential synergies,” _The Journal of supercomputing_ , vol. 68, no. 1, pp. 1–48, 2014.
  * [3] L. Gupta, R. Jain, and G. Vaszkun, “Survey of important issues in uav communication networks,” _IEEE Communications Surveys Tutorials_ , vol. 18, no. 2, pp. 1123–1152, 2016.
  * [4] A. Jadbabaie, J. Lin, and A. S. Morse, “Coordination of groups of mobile autonomous agents using nearest neighbor rules,” _IEEE Transactions on Automatic Control_ , vol. 48, no. 6, pp. 988–1001, 2003.
  * [5] R. Olfati-Saber, J. A. Fax, and R. M. Murray, “Consensus and cooperation in networked multi-agent systems,” _Proceedings of the IEEE_ , vol. 95, no. 1, pp. 215–233, 2007.
  * [6] W. Ren, R. W. Beard, and E. M. Atkins, “Information consensus in multivehicle cooperative control,” _IEEE Control Systems_ , vol. 27, no. 2, pp. 71–82, 2007.
  * [7] K. H. Movric and F. L. Lewis, “Cooperative optimal control for multi-agent systems on directed graph topologies,” _IEEE Transactions on Automatic Control_ , vol. 59, no. 3, pp. 769–774, 2014.
  * [8] Y. Cao and W. Ren, “Optimal linear-consensus algorithms: An LQR perspective,” _IEEE Transactions on Systems, Man, and Cybernetics, Part B (Cybernetics)_ , vol. 40, no. 3, pp. 819–830, 2010.
  * [9] F. Borrelli and T. Keviczky, “Distributed LQR design for identical dynamically decoupled systems,” _IEEE Transactions on Automatic Control_ , vol. 53, no. 8, pp. 1901–1912, 2008.
  * [10] J. Arabneydi, “New concepts in team theory: Mean field teams and reinforcement learning,” Ph.D. dissertation, McGill University, 2016.
  * [11] Y. C. Ho and K. h. Chu, “Team decision theory and information structures in optimal control problems–part I,” _IEEE Transactions on Automatic Control_ , vol. 17, no. 1, pp. 15–22, 1972.
  * [12] M. Rotkowitz and S. Lall, “A characterization of convex problems in decentralized control,” _IEEE Transactions on Automatic Control_ , vol. 51, no. 2, pp. 274–286, 2006.
  * [13] H. Witsenhausen, “A counterexample in stochastic optimum control,” _SIAM Journal Of Control And Optimization_ , vol. 6, pp. 131–147, 1968.
  * [14] P. E. Caines, _Linear stochastic systems_. John Wiley and Sons, Inc., 1987.