# On the minimum size of linear sets

Sam Adriaensen
Vrije Universiteit Brussel Paolo Santonastaso
Università degli Studi della Campania
“Luigi Vanvitelli”

###### Abstract

Recently, a lower bound was established on the size of linear sets in
projective spaces, that intersect a hyperplane in a canonical subgeometry.
There are several constructions showing that this bound is tight. In this
paper, we generalize this bound to linear sets meeting some subspace $\pi$ in
a canonical subgeometry. We obtain a tight lower bound on the size of any
${\mathbb{F}}_{q}$-linear set spanning $\mathrm{PG}(d,q^{n})$ in case that
$n\leq q$ and $n$ is prime. We also give constructions of linear sets
attaining equality in the former bound, both in the case that $\pi$ is a
hyperplane, and in the case that $\pi$ is a lower dimensional subspace.

Keywords: Projective geometry, Linear set, Subgeometry.
MSC2020: 51E20, 05B25.

##  1 Introduction

Linear sets are certain point sets in projective spaces, generalizing the
notion of a subgeometry. They have proven themselves to be very useful in
constructing interesting objects in projective spaces, such as blocking sets
[26] and KM-arcs [9], and have been used to construct Hamming and rank metric
codes [20, 18, 1, 24, 25, 23, 27]. For a survey on linear sets, we refer the
reader to [13, 19].

Given the usefulness of linear sets, their recent spurt in popularity within
the field of finite geometry is far from surprising. One of the most natural
questions arising in the study of linear sets is establishing lower and upper
bounds on their size. There is a quite trivial upper bound on the size of
linear sets, and the study of linear sets attaining equality in this bound can
be traced back to a paper by Blokhuis and Lavrauw [5]. However, finding good
lower bounds on the size of linear sets seems to be a harder problem. Yet it
is an interesting endeavor, e.g. due to its connection with the weight
distribution of linear rank metric codes [21].

As a consequence of the celebrated result on the number of directions
determined by a function over a finite field [4, 2], Bonoli and Polverino
established a lower bound on the size of certain linear sets on a projective
line. More specifically, they proved the following result (for the
definitions, we refer to Section 2).

######  Result 1.1 ([6, Lemma 2.2]).

If $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of rank $n$ on
$\mathrm{PG}(1,q^{n})$, and $L_{U}$ contains at least one point of weight 1,
then $|L_{U}|\geq q^{n-1}+1$.

De Beule and Van de Voorde managed to remove the condition on the rank from
this bound. We note that linear sets of rank greater than $n$ on
$\mathrm{PG}(1,q^{n})$ are not interesting to study, since they necessarily
contain all the points of the projective line. Hence it is natural to limit
the study to linear sets whose rank is at most $n$.

######  Result 1.2 ([8, Theorem 1.2]).

If $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of rank $k$, with $1<k\leq n$
on $\mathrm{PG}(1,q^{n})$, and $L_{U}$ contains at least one point of weight
1, then $|L_{U}|\geq q^{k-1}+1$.

Using an inductive argument, they obtained a bound on the size of a linear set
in a higher dimensional projective space. Using Lemma 3.2, which we prove
later in this paper, this is equivalent to the following result.

######  Result 1.3 ([8, Theorem 4.4]).

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set of rank $k>d$ in
$\mathrm{PG}(d,q^{n})$. If $L_{U}$ meets some hyperplane $\Omega$ in a
canonical ${\mathbb{F}}_{q}$-subgeometry of $\Omega$, then

$|L_{U}|\geq q^{k-1}+q^{k-2}+\ldots+q^{k-d}+1.$

De Beule and Van de Voorde note directly after their statement of the above
result that they would like to find lower bounds on linear sets satisfying
less restrictive conditions. Furthermore, Jena and Van de Voorde [11, §2.5
(B)] state that they believe the above lower bound to hold for all
${\mathbb{F}}_{q}$-linear sets of rank $k$ that span $\mathrm{PG}(d,q^{n})$,
if $n$ is prime and $k\leq d+n$.

In this article we will generalize the above result by dropping the condition
that $\Omega$ is a hyperplane.

######  Theorem 1.4.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set of rank $k$ in
$\mathrm{PG}(d,q^{n})$. Suppose that there exists some $(r-1)$-space $\Omega$,
with $r<k$, such that $L_{U}$ meets $\Omega$ in a canonical
${\mathbb{F}}_{q}$-subgeometry of $\Omega$. Then

$|L_{U}|\geq q^{k-1}+\ldots+q^{k-r}+I_{\Omega},$

where $I_{\Omega}$ denotes the number of $r$-spaces through $\Omega$,
containing a point of $L_{U}\setminus\Omega$.

The theorem leads us to wonder, given a linear set, how we can assure the
existence of a large subspace intersecting it in a canonical subgeometry. This
question turns out to be closely related to studying which linear sets must
certainly have a point of weight 1. Csajbók, Marino, and Pepe [7] recently
proved the following seminal result.

######  Result 1.5 ([7, Theorem 2]).

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set of $\mathrm{PG}(d,q^{n})$ of
rank $k\leq dn$, such that the following assumptions are satisfied:

  1. 1.

$n\leq q$;

  2. 2.

every point of $L_{U}$ has weight at least $w\geq 2$.

Then there exists an integer $t$ with $w\leq t\mid n$ such that
$L_{U}=L_{U^{\prime}}$, with $U^{\prime}=\langle
U\rangle_{{\mathbb{F}}_{q^{t}}}$.

Especially when $n$ is prime, this is a powerful result. In that case, a
linear sets without points of weight 1 must coincide with a subspace as point
sets. This allows us to prove the following theorem.

######  Theorem 1.6.

Suppose that $n$ is a prime number with $n\geq q$. Let $L_{U}$ be an
${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d,q^{n})$ spanning the whole
space. Define $r=d-\left\lfloor\frac{k-(d+2)}{n-1}\right\rfloor$. Then $L_{U}$
meets some $(r-1)$-space in a canonical subgeometry and

$|L_{U}|\geq q^{k-1}+\ldots+q^{k-r}+\frac{q^{n(d-r+1)}-1}{q^{n}-1}.$

Moreover, this lower bound is tight.

Note in particular that this confirms the previously mentioned belief of Jena
and Van de Voorde [11, §2.5 (B)] – that all ${\mathbb{F}}_{q}$-linear sets of
rank $k\leq d+n$ spanning $\mathrm{PG}(d,q^{n})$, $n$ prime, satisfy the lower
bound of 1.3 – in case that $n\leq q$.

Also in the case where $n$ is not prime, Result 1.5 is interesting from the
point of view of lower bounding the size of a linear set. It ensures that we
can take $r=1$ in Theorem 1.4, i.e. $\Omega$ is a point, in case that
${\mathbb{F}}_{q}$ is the maximum geometric field of linearity of $L_{U}$.

In $\mathrm{PG}(1,q^{n})$, the bound of De Beule and Van de Voorde is tight.
For every rank $k\leq n$, there exist so-called $(k-1)$-clubs of rank $k$.
These linear sets contain (an abundance of) points of weight 1, and their size
matches the bound in Result 1.2. Lunardon and Polverino [15] provided the
first less trivial family of linear sets of rank $n$ reaching equality in
Result 1.1. Their example was extended by Jena and Van de Voorde [11] to a
very large family of linear sets of general rank, attaining equality in Result
1.2. More recently, there have been other constructions of such linear sets,
and partial classification results, see Napolitano et al. [16]. Moreover, Jena
and Van de Voorde generalized their constructions to higher dimensions, to
obtain linear sets attaining equality in the bound of Result 1.3, some of
which also satisfy the conditions of Result 1.3 [11, §2.5 (B)].

In this article, we study the construction by Jena and Van de Voorde in
general dimension, and we provide a sufficient condition for these linear sets
to satisfy the hypothesis of Result 1.3. We also generalize the construction
of Napolitano et al. to higher dimensions. Furthermore, we construct linear
sets in $\mathrm{PG}(d,q^{n})$ satisfying the conditions of Theorem 1.4, and
attaining equality in the corresponding bound, where $n$ is not prime. The
size of these linear sets is smaller than the bound from Result 1.3, hence
this illustrates the necessity of the conditions imposed in Result 1.3 in case
$n$ is not prime.

Structure of the paper. Section 2 contains preliminary results on linear sets.
Section 3 contains the proof of Theorems 1.4 and 1.6. In addition, we deduce
from Theorem 1.4 that the rank of a linear set is determined by its size and
the minimum weight of its points, and that it is spanned by its points of
minimum weight. In Section 4 we discuss linear sets attaining equality in
Result 1.3. More specifically, we show a sufficient condition for the minimum
size linear sets of [11] to satisfy the hypothesis of Result 1.3, and we
generalize the construction from [16] to higher dimension. Section 5 contains
constructions of linear sets attaining equality in Theorem 1.4.

##  2 Preliminaries

Throughout this article, $q$ will always denote a prime power, and
${\mathbb{F}}_{q}$ will denote the finite field of order $q$. The
$d$-dimensional projective space over ${\mathbb{F}}_{q}$ will be denoted by
$\mathrm{PG}(d,q)$. If the projective space is constructed from a
$(d+1)$-dimensional ${\mathbb{F}}_{q}$-vector space $V$, and we want to
emphasize the underlying vector space, we might also denote the projective
space as $\mathrm{PG}(V,{\mathbb{F}}_{q})$. We note that the number of points
in $\mathrm{PG}(d,q)$ equals $\frac{q^{d+1}-1}{q-1}=q^{d}+q^{d-1}+\ldots+q+1$.

######  Notation 2.1.

Throughout the article, when working in
$\mathrm{PG}(d,q)=\mathrm{PG}({\mathbb{F}}_{q}^{d+1},{\mathbb{F}}_{q})$, we
denote the vectors of ${\mathbb{F}}_{q}^{d+1}$ as $(x_{0},\dots,x_{d})$, i.e.
we label the coordinate positions from 0 to $d$. The $i$th standard basis
vector will be denoted as

$\mathbf{e}_{i}=(0,\dots,0,\underbrace{1}_{i^{\text{th}}\text{
position}},0\dots,0),$

and the corresponding point in $\mathrm{PG}(d,q)$ will be denoted as $E_{i}$.

###  2.1 Linear sets

Let $V$ be a $(d+1)$-dimensional vector space over $\mathbb{F}_{q^{n}}$. Then
$V$ is also a $(d+1)n$-dimensional vector space over ${\mathbb{F}}_{q}$. Let
$U$ denote an ${\mathbb{F}}_{q}$-subspace of $V$. Then

$L_{U}=\\{\langle u\rangle_{\mathbb{F}_{q^{n}}}\colon u\in
U\setminus\\{\mathbf{0}\\}\\}$

is a set of points in $\mathrm{PG}(d,q^{n})$. Sets of this type are called
_${\mathbb{F}}_{q}$ -linear sets_, and the ${\mathbb{F}}_{q}$-dimension of $U$
is called the _rank_ of $L_{U}$.

We note that if $U_{1}$ and $U_{2}$ are ${\mathbb{F}}_{q}$-subspaces, and
$L_{U_{1}}$ and $L_{U_{2}}$ are equal as point set in $\mathrm{PG}(d,q^{n})$,
this need not imply that
$\dim_{{\mathbb{F}}_{q}}U_{1}=\dim_{{\mathbb{F}}_{q}}U_{2}$. Hence, the rank
of a linear set $L_{U}$ is generally defined ambiguously by $L_{U}$ as point
set in $\mathrm{PG}(d,q^{n})$, without taking into account the underlying
subspace $U$.

Given an $\mathbb{F}_{q^{n}}$-subspace $W\leq V$, we define the _weight_ of
$\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$ to be

$\operatorname{w}_{L_{U}}(\Omega)=\dim_{{\mathbb{F}}_{q}}(U\cap W).$

Note that $\operatorname{w}_{L_{U}}(\Omega)$ equals the rank of the linear set
$L_{U\cap W}=L_{U}\cap\Omega$.

For each $i\in\\{1,\dots,n\\}$, let $N_{i}(L_{U})$ denote the number of points
in $\mathrm{PG}(d,q^{n})$ of weight $i$. We will simply denote this as $N_{i}$
if $L_{U}$ is clear from context. The numbers $N_{1},\dots,N_{n}$ are called
the _weight distribution_ of $L_{U}$. In addition, the _weight spectrum_ of
$L_{U}$ is the ordered tuple $(i_{1},\dots,i_{t})$ with $i_{1}<\ldots<i_{t}$
and

$\\{i_{1},\dots,i_{t}\\}=\\{\operatorname{w}_{L_{U}}(P):P\in
L_{U}\\}=\\{i\in\\{1,\dots,n\\}:N_{i}>0\\}.$

Let $k>0$ denote the rank of $L_{U}$. Then the weight distribution satisfies
the following properties.

$\displaystyle|L_{U}|$ $\displaystyle=N_{1}+\ldots+N_{n},$ (1)
$\displaystyle\sum_{i=1}^{n}N_{i}\frac{q^{i}-1}{q-1}$
$\displaystyle=\frac{q^{k}-1}{q-1},$ (2) $\displaystyle|L_{U}|$
$\displaystyle\leq\frac{q^{k}-1}{q-1},$ (3) $\displaystyle|L_{U}|$
$\displaystyle\equiv 1\ (\hbox{{\rm mod}\,}q).$ (4)

Let $T$ be an ${\mathbb{F}}_{q}$-subspace of $V$ with
$\dim_{{\mathbb{F}}_{q}}(T)=r\leq d+1$. If
$\dim_{{\mathbb{F}}_{q^{n}}}(\langle T\rangle_{{\mathbb{F}}_{q^{n}}})=r$, we
will say that $L_{T}\cong\mathrm{PG}(T,{\mathbb{F}}_{q})=\mathrm{PG}(r-1,q)$
is an ${\mathbb{F}}_{q}$-_subgeometry_ of
$\mathrm{PG}(V,{\mathbb{F}}_{q^{n}})$ and $r$ is the _rank_ of the subgeometry
$L_{T}$. When $r=d+1$, we say that $L_{T}$ is a _canonical subgeometry_ of
$\mathrm{PG}(V,\mathbb{F}_{q^{n}})=\mathrm{PG}(d,q^{n})$. Note that each point
of a subgeometry $L_{T}$ has weight 1 and hence $\lvert
L_{T}\rvert=\frac{q^{r}-1}{q-1}$, if $L_{T}$ has rank $r$.

Regarding the linearity of a linear set, we recall the following definitions
explored in [12].

######  Definition 2.2 ([12, Definitions 1.1, 1.2]).

An ${\mathbb{F}}_{q}$-linear set $L_{U}$ is an _${\mathbb{F}}_{q^{s}}$ -linear
set_ if $U$ is also an ${\mathbb{F}}_{q^{s}}$-vector space. We say that
${\mathbb{F}}_{q^{s}}$ is the _maximum field of linearity_ of $L_{U}$ if $s$
is the largest exponent such that $L_{U}$ is ${\mathbb{F}}_{q^{s}}$-linear.

######  Definition 2.3 ([12, Definitions 1.3, 1.4]).

An ${\mathbb{F}}_{q}$-linear set $L_{U}$ has _geometric field of linearity_
${\mathbb{F}}_{q^{s}}$ if there exists an ${\mathbb{F}}_{q^{s}}$-linear set
$L_{U^{\prime}}$ such that $L_{U}=L_{U^{\prime}}$. An
${\mathbb{F}}_{q}$-linear set $L_{U}$ has _maximum geometric field of
linearity_ ${\mathbb{F}}_{q^{s}}$ if $s$ is the largest integer such that
$L_{U}$ has geometric field of linearity ${\mathbb{F}}_{q^{s}}$.

The maximum field of linearity and the maximum geometric field of linearity do
not always coincide. Clearly if $L_{U}$ is an ${\mathbb{F}}_{q^{s}}$-linear
set, it has geometric field of linearity ${\mathbb{F}}_{q^{s}}$, but the
converse need not hold, see e.g. [12, Example 1.5].

######  Remark 2.4.

Note that if there exists a line $\ell$ that is $(q+1)$-secant to a linear set
$L_{U}$, then by (4) $L_{U}$ has maximum geometric field of linearity
${\mathbb{F}}_{q}$, see also [12].

We refer to [19] and [13] for comprehensive references on linear sets.

###  2.2 Subspaces of complementary weights

Recently, there has been an interest in linear sets admitting subspaces of
complementary weights (see below for the definition), due to their application
in coding theory, see e.g. [20, 18, 28]. Linear sets on the projective line
admitting two points of complementary weights have been studied in [17] (see
also [12, 16]). The higher dimensional analogue has been studied in [28]. For
the sake of completeness, we state the definition and prove the structural
description of such linear sets here in full generality.

Call subspaces $W_{1},\dots,W_{m}\leq_{q^{n}}\mathbb{F}_{q^{n}}^{d+1}$
_independent_ if each subspace $W_{i}$ intersects $\langle W_{j}:j\neq
i\rangle_{\mathbb{F}_{q^{n}}}$ trivially, or equivalently if
$\dim_{\mathbb{F}_{q^{n}}}\langle
W_{i}:i=1,\dots,m\rangle=\dim_{\mathbb{F}_{q^{n}}}W_{1}+\ldots+\dim_{\mathbb{F}_{q^{n}}}W_{m}$.

######  Lemma 2.5.

Let $W_{1},\dots,W_{m}$ be independent subspaces in
$\mathbb{F}_{q^{n}}^{d+1}$, and let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear
set in $\mathrm{PG}(d,q^{n})$ of rank $k$, that spans the entire space. Then

$\operatorname{w}_{L_{U}}(\mathrm{PG}(W_{1},\mathbb{F}_{q^{n}}))+\ldots+\operatorname{w}_{L_{U}}(\mathrm{PG}(W_{m},\mathbb{F}_{q^{n}}))\leq
k.$

If equality holds, then $\mathbb{F}_{q^{n}}^{d+1}=W_{1}\oplus\ldots\oplus
W_{m}$.

###### Proof.

Since no $W_{i}$ intersects the span of the others, it is permitted to
consider the direct sum $W_{1}\oplus\ldots\oplus W_{m}$. Then

$\displaystyle k$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}U\geq\dim_{{\mathbb{F}}_{q}}(U\cap(W_{1}\oplus\ldots\oplus
W_{m}))\geq\dim_{{\mathbb{F}}_{q}}(U\cap
W_{1})+\ldots+\dim_{{\mathbb{F}}_{q}}(U\cap W_{m})$
$\displaystyle=\operatorname{w}_{L_{U}}(\mathrm{PG}(W_{1},\mathbb{F}_{q^{n}}))+\ldots+\operatorname{w}_{L_{U}}(\mathrm{PG}(W_{m},\mathbb{F}_{q^{n}}))$

If equality holds, then

$U\cap(W_{1}\oplus\ldots\oplus W_{m})=U.$

Since $W_{1}\oplus\ldots\oplus W_{m}$ is an $\mathbb{F}_{q^{n}}$-subspace, and
$\langle U\rangle_{\mathbb{F}_{q^{n}}}=\mathbb{F}_{q^{n}}^{d+1}$, we get that

$W_{1}\oplus\ldots\oplus W_{m}=\mathbb{F}_{q^{n}}^{d+1}.\qed$

######  Definition 2.6.

If the subspaces $W_{1},\dots,W_{m}$ attain equality in Lemma 2.5, we say that
$\mathrm{PG}(W_{1},\mathbb{F}_{q^{n}}),\dots,\mathrm{PG}(W_{m},\mathbb{F}_{q^{n}})$
are _subspaces of complementary weight_ (w.r.t. $L_{U}$).

######  Lemma 2.7.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set spanning
$\mathrm{PG}(d,q^{n})$. Then there exist subspaces
$\Omega_{1},\dots,\Omega_{m}$ in $\mathrm{PG}(d,q^{n})$ of complementary
weight, with $\dim\Omega_{i}=d_{i}$ and
$\operatorname{w}_{L_{U}}(\Omega_{i})=k_{i}$, if and only if $U$ is
$\operatorname{GL}(d+1,q^{n})$-equivalent to an ${\mathbb{F}}_{q}$-subspace
$U_{1}\times\ldots\times U_{m}$, with each $U_{i}$ a $k_{i}$-dimensional
${\mathbb{F}}_{q}$-subspace of $\mathbb{F}_{q^{n}}^{d_{i}+1}$ satisfying
$\langle U_{i}\rangle_{\mathbb{F}_{q^{n}}}=\mathbb{F}_{q^{n}}^{d_{i}+1}$.

###### Proof.

First suppose that such subspaces
$\Omega_{i}=\mathrm{PG}(W_{i},\mathbb{F}_{q^{n}})$ exist. Then there exists a
map $\varphi\in\operatorname{GL}(d+1,q^{n})$ such that
$\varphi(W_{1})=\langle\mathbf{e}_{0},\dots,\mathbf{e}_{d_{1}}\rangle_{\mathbb{F}_{q^{n}}}$,
$\varphi(W_{2})=\langle\mathbf{e}_{d_{1}+1},\dots,\mathbf{e}_{d_{1}+d_{2}+1}\rangle_{\mathbb{F}_{q^{n}}}$,
and so on. As can be seen in the proof of the previous lemma,

$\displaystyle\varphi(U)=\varphi(U\cap W_{1})\oplus\ldots\oplus\varphi(U\cap
W_{m}),$

which equals $U_{1}\times\ldots\times U_{m}$, with

$U_{i}=\\{u\in\mathbb{F}_{q^{n}}^{d_{i}+1}:(0,\ldots,0,u,0,\dots,0)\in\varphi(U)\cap\varphi(W_{i})\\}.$

Clearly,

$\dim_{{\mathbb{F}}_{q}}U_{i}=\dim_{{\mathbb{F}}_{q}}\varphi(U\cap
W_{i})=\dim_{{\mathbb{F}}_{q}}U\cap
W_{i}=\operatorname{w}_{L_{U}}(\Omega_{i})=k_{i}.$

Vice versa, suppose that $\varphi(U)=U_{1}\times\ldots\times U_{m}$, with each
$U_{i}$ a $k_{i}$-dimensional ${\mathbb{F}}_{q}$-subspace of
$\mathbb{F}_{q^{n}}^{d_{i}+1}$, for some
$\varphi\in\operatorname{GL}(d+1,q^{n})$. Then define

$\displaystyle
W_{1}=\langle\mathbf{e}_{0},\ldots,\mathbf{e}_{d_{1}}\rangle_{\mathbb{F}_{q^{n}}},$
$\displaystyle
W_{2}=\langle\mathbf{e}_{d_{1}+1},\ldots,\mathbf{e}_{d_{1}+d_{2}+1}\rangle_{\mathbb{F}_{q^{n}}},$
$\displaystyle\dots$

Then clearly
$\mathrm{PG}(W_{1},\mathbb{F}_{q^{n}}),\dots,\mathrm{PG}(W_{m},\mathbb{F}_{q^{n}})$
are subspaces of complementary weights w.r.t. $L_{U}$. Having subspaces of
complementary weights is $\operatorname{GL}(d+1,q^{n})$-invariant, which
finishes the proof. ∎

##  3 General bounds

This section is devoted to proving Theorems 1.4 and 1.6. From Theorem 1.4 we
derive that if a linear set $L_{U}$ contains a point of weight 1, its rank
equals $\lceil\log_{q}(|L_{U}|)\rceil$, and $\langle L_{U}\rangle$ is spanned
by the points of $L_{U}$ of weight 1.

###  3.1 Proof of Theorem 1.4

De Beule and Van de Voorde proved the following bound.

######  Result 3.1 ([8, Theorem 4.4]).

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set spanning
$\mathrm{PG}(d,q^{n})$ of rank $k$. Suppose that $L_{U}$ meets some hyperplane
$\Omega$ in exactly $\frac{q^{d}-1}{q-1}$ points, spanning $\Omega$. Then

$|L_{U}|\geq q^{k-1}+q^{k-2}+\ldots+q^{k-d}+1.$

Note that if $d=1$, this result is exactly Result 1.2. We now prove that this
result is equivalent to Result 1.3. This follows directly from the following
lemma.

######  Lemma 3.2.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d-1,q^{n})$,
with $d\geq 2$. Then $L_{U}$ spans $\mathrm{PG}(d-1,q^{n})$ and satisfies
$|L_{U}|=\frac{q^{d}-1}{q-1}$ if and only if $L_{U}$ is a canonical
${\mathbb{F}}_{q}$-subgeometry.

###### Proof.

If $L_{U}$ is a canonical subgeometry, then it immediately follows that
$L_{U}$ spans the entire space, and $|L_{U}|=\frac{q^{d}-1}{q-1}$. So suppose
that $L_{U}$ spans the space, and that $|L_{U}|=\frac{q^{d}-1}{q-1}$. We need
to prove that all points of $L_{U}$ have weight 1. Indeed in that case, by
equations (1) and (2), $L_{U}$ must then have rank $d$, which proves that
$L_{U}$ is a canonical subgeometry. So suppose by way of contradiction that
$L_{U}$ has points of weight greater than 1. Note that by Equations (1) and
(2), the rank of $L_{U}$ is some number $k>d$. Let

$\sigma=\langle P\in L_{U}\colon\operatorname{w}_{L_{U}}(P)>1\rangle$

denote the subspace of $\mathrm{PG}(d-1,q^{n})$ spanned by points of weight
greater than 1.

Suppose that $\sigma$ is not $\mathrm{PG}(d-1,q^{n})$. Then
$L_{U}\not\subseteq\sigma$, and every point in $L_{U}\setminus\sigma$ is a
point of weight 1. Hence, there are at least $q^{k-1}$ points in
$L_{U}\setminus\sigma$ corresponding to (necessarily distinct) points of
weight 1 of $L_{U}$. Thus, $|L_{U}|>q^{k-1}>\frac{q^{d}-1}{q-1}$ since $k>d$,
a contradiction.

Hence $\sigma$ equals $\mathrm{PG}(d-1,q^{n})$. Let

$m=\max_{P\in L_{U}}\operatorname{w}_{L_{U}}(P)$

denote the maximum weight of the points of $L_{U}$. Then we can choose $d$
independent points $P_{1},\dots,P_{d}$ in $L_{U}$ such that
$\operatorname{w}_{L_{U}}(P_{1})=m$, and $\operatorname{w}_{L_{U}}(P_{i})\geq
2$ for each $i$. By Lemma 2.5,

$k\geq\sum_{i=1}^{d}\operatorname{w}_{L_{U}}(P_{i})\geq m+2(d-1).$

Let $N_{1},\ldots,N_{m}$ denote the weight distribution of $L_{U}$. Then by
Equations (1) and (2),

$\frac{q^{m}-1}{q-1}|L_{U}|=\sum_{i=1}^{m}N_{i}\frac{q^{m}-1}{q-1}\geq\sum_{i=1}^{m}N_{i}\frac{q^{i}-1}{q-1}=\frac{q^{k}-1}{q-1}\geq\frac{q^{m+2(d-1)}-1}{q-1}.$

This implies that

$\frac{q^{d}-1}{q-1}=|L_{U}|\geq\frac{q^{m+2(d-1)}-1}{q^{m}-1},$

which yields a contradiction if $d\geq 2$. ∎

We will now prove Theorem 1.4.

Proof of Theorem 1.4. Consider the $r$-spaces $\Pi_{1},\Pi_{2},\ldots$ of
$\mathrm{PG}(d,q^{n})$ through $\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$,
with $\Pi_{i}=\mathrm{PG}(W_{i},{\mathbb{F}}_{q^{n}})$, for each $i$. We can
order the $r$-spaces in such a way that $\Pi_{i}$ contains a point of
$L_{U}\setminus\Omega$ if and only if $i\leq I_{\Omega}$. Let

$k_{i}=\dim_{{\mathbb{F}}_{q}}(U\cap W_{i})$

denote the rank of the ${\mathbb{F}}_{q}$-linear set $L_{U\cap W_{i}}$. Then
the sets $W_{i}\cap U\setminus W$ partition the vectors in $U\setminus W$.
Since $L_{U}$ intersects $\Omega$ in a canonical subgeometry,
$\dim_{{\mathbb{F}}_{q}}U\cap W=r$. This yields

$\displaystyle q^{k}-q^{r}$
$\displaystyle=\sum_{i=1}^{I_{\Omega}}(q^{k_{i}}-q^{r})$
$\displaystyle\implies$ $\displaystyle
q^{k-r}=1+\sum_{i=1}^{I_{\Omega}}(q^{k_{i}-r}-1).$ (5)

Analogously, the points of $\Pi_{i}\setminus\Omega$ partition the points of
$L_{U}\setminus\Omega$. Note that for $i\leq I_{\Omega}$, we have that
$L_{U}\cap\Pi_{i}=L_{U\cap W_{i}}$ is an ${\mathbb{F}}_{q}$-linear set in
$\Pi_{i}$ of rank $k_{i}$, satisfying the hypothesis of Result 1.3. Hence,

$\displaystyle|L_{U}|$
$\displaystyle=|L_{U}\cap\Omega|+\sum_{i=1}^{I_{\Omega}}\left(|L_{U}\cap\Pi_{i}|-|L_{U}\cap\Omega|\right)$
$\displaystyle\geq\frac{q^{r}-1}{q-1}+\sum_{i=1}^{I_{\Omega}}\left((q^{k_{i}-1}+\ldots+q^{k_{i}-r}+1)-\frac{q^{r}-1}{q-1}\right)$
$\displaystyle=\frac{q^{r}-1}{q-1}+\sum_{i=1}^{I_{\Omega}}\left(q^{k_{i}-r}\frac{q^{r}-1}{q-1}-\frac{q^{r}-1}{q-1}+1\right)$
$\displaystyle=\frac{q^{r}-1}{q-1}\left(1+\sum_{i=1}^{I_{\Omega}}(q^{k_{i}-r}-1)\right)+I_{\Omega}.$

Using Equation (5) this implies that

$\lvert
L_{U}\rvert\geq\frac{q^{r}-1}{q-1}q^{k-r}+I_{\Omega}=q^{k-1}+q^{k-2}+\ldots+q^{k-r}+I_{\Omega}.$

######  Remark 3.3.

If one wants to apply Theorem 1.4 to a particular linear set $L_{U}$,
different choices of the $(r-1)$-space $\Omega$ can yield different bounds. In
other words, $I_{\Omega}$ need not be the same for all $(r-1)$-spaces meeting
$L_{U}$ in a canonical ${\mathbb{F}}_{q}$-subgeometry. This is illustrated in
the example below.

######  Example 3.4.

Consider the $(n+1)$-dimensional ${\mathbb{F}}_{q}$-subspace

$U=\\{(x,x^{q}):x\in\mathbb{F}_{q^{n}}\\}\times{\mathbb{F}}_{q}$

of $\mathbb{F}_{q^{n}}^{3}$. Consider the corresponding
${\mathbb{F}}_{q}$-linear set $L_{U}$ of rank $n+1$ in $\mathrm{PG}(2,q^{n})$.
Every point of $L_{U}$ has weight 1, so we can apply Theorem 1.4 with $\Omega$
any point of $L_{U}$. However, for a point $P\in L_{U}$, $I_{P}=q^{n-1}+1$ if
$P$ lies on the line $X_{2}=0$, and $I_{P}=\frac{q^{n}-1}{q-1}$ if $P$ does
not lie on $X_{2}=0$. These numbers are distinct if $n>2$.

We also remark that in Theorem 1.4 the number $I_{\Omega}$ of $r$-spaces
through $\Omega$ containing a point of $L_{U}\setminus\Omega$ equals the size
of a certain linear set.

######  Definition 3.5.

Consider an ${\mathbb{F}}_{q}$-linear set $L_{U}$ in
$\mathrm{PG}(V,\mathbb{F}_{q^{n}})$ and a subspace
$\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$. Let $\overline{U}$ denote the
subspace $(U+W)/W$ of the quotient space $V/W$. Then the _projection_ of
$L_{U}$ from $\Omega$ is the ${\mathbb{F}}_{q}$-linear set $L_{\overline{U}}$
of $\mathrm{PG}(V/W,\mathbb{F}_{q^{n}})$.

######  Lemma 3.6.

Suppose that $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of rank $k$ in
$\mathrm{PG}(V,\mathbb{F}_{q^{n}})$ and let $L_{\overline{U}}$ be the
projection of $L_{U}$ from an $(r-1)$-space
$\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$. Then for each
$\mathbb{F}_{q^{n}}$-subspace $W^{\prime}\leq V$ through $W$,

$\operatorname{w}_{L_{\overline{U}}}(\mathrm{PG}((W^{\prime}+W)/W,\mathbb{F}_{q^{n}}))=\operatorname{w}_{L_{U}}(\mathrm{PG}(W^{\prime},\mathbb{F}_{q^{n}}))-\operatorname{w}_{L_{U}}(\Omega).$

In particular, $L_{\overline{U}}$ has rank
$k-\operatorname{w}_{L_{U}}(\Omega)$, and $|L_{\overline{U}}|$ equals the
number of $r$-spaces in $\mathrm{PG}(V,\mathbb{F}_{q^{n}})$ through $\Omega$
that contain a point of $L_{U}\setminus\Omega$. Furthermore, if $L_{U}$ spans
$\mathrm{PG}(V,\mathbb{F}_{q^{n}})$, then $L_{\overline{U}}$ spans
$\mathrm{PG}(V/W,\mathbb{F}_{q^{n}})$.

###### Proof.

We can find ${\mathbb{F}}_{q}$-subspaces $U_{1},U_{2},U_{3}$ of $U$ such that

  * $U_{1}=W\cap U$,

  * $U_{1}\oplus U_{2}=W^{\prime}\cap U$,

  * $U_{1}\oplus U_{2}\oplus U_{3}=U$.

Then

$\displaystyle\operatorname{w}_{L_{\overline{U}}}(\mathrm{PG}((W^{\prime}+W)/W,\mathbb{F}_{q^{n}}))$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}(\overline{U}\cap((W^{\prime}+W)/W))=\dim_{{\mathbb{F}}_{q}}(\langle
U,W\rangle_{{\mathbb{F}}_{q}}\cap W^{\prime})-\dim_{{\mathbb{F}}_{q}}W$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}(W\oplus
U_{2})-\dim_{{\mathbb{F}}_{q}}(W)=\dim_{{\mathbb{F}}_{q}}(U_{2})$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}(U_{1}\oplus
U_{2})-\dim_{{\mathbb{F}}_{q}}(U_{1})=\operatorname{w}_{L_{U}}(\mathrm{PG}(W^{\prime},\mathbb{F}_{q^{n}}))-\operatorname{w}_{L_{U}}(\Omega).$

If we put $W^{\prime}=V$, we see that $L_{\overline{U}}$ has rank
$k-\operatorname{w}_{L_{U}}(\Omega)$. It also follows that the points of
$L_{\overline{U}}$ are in 1-1 correspondence with the $(r+1)$-spaces
$W^{\prime}$ of $V$ with
$\operatorname{w}_{L_{U}}(\mathrm{PG}(W,\mathbb{F}_{q^{n}}))>\operatorname{w}_{L_{U}}(\Omega)$,
which are exactly the $r$-spaces through $\Omega$ in
$\mathrm{PG}(V,\mathbb{F}_{q^{n}})$ containing a point of
$L_{U}\setminus\Omega$. ∎

This tells us the following about the quantity $I_{\Omega}$ in Theorem 1.4.

######  Proposition 3.7.

In the hypothesis of Theorem 1.4, let $L_{\overline{U}}$ be the projection of
$L_{U}$ from $\Omega$. Then

$\lvert L_{U}\rvert\geq q^{k-1}+q^{k-2}+\ldots+q^{k-r}+\lvert
L_{\overline{U}}\rvert.$

Moreover, $L_{\overline{U}}$ has rank $k-r$.

In the next sections, we will investigate linear sets attaining equality in
the bound of Theorem 1.4. To this end, we introduce some relevant terminology.

######  Definition 3.8.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set of rank $k$ in
$\mathrm{PG}(d,q^{n})$. If

$|L_{U}|=q^{k-1}+\ldots+q^{k-d}+1,$

we say that $L_{U}$ is of _$d$ -minimum size_. If there is some $(r-1)$-space
$\Omega$ such that $L_{U}$ and $\Omega$ satisfy the hypothesis of Theorem 1.4,
and

$|L_{U}|=q^{k-1}+\ldots+q^{k-r}+I_{\Omega}\leq q^{k-1}+\ldots+q^{k-d}+1,$

then we say that $L_{U}$ is of _$(r,d,\Omega)$ -minimum size_, or simply of
_$(r,d)$ -minimum size_. A linear set of $(d,d)$-minimum size, will also be
called of _proper $d$-minimum size_.

######  Remark 3.9.

By Remark 2.4, an $(r,d)$-minimum size linear set has maximum geometric field
of linearity ${\mathbb{F}}_{q}$ whenever $r\geq 2$.

In the next proposition, we also prove that if a linear set is of
$(r,d)$-minimum size, it is of $(r^{\prime},d)$-minimum size for every
$r^{\prime}\leq r$.

######  Proposition 3.10.

Let $L_{U}$ be an $(r,d)$-minimum size ${\mathbb{F}}_{q}$-linear set of rank
$k$ in $\mathrm{PG}(d,q^{n})$. Then $L_{U}$ is of $(r^{\prime},d)$-minimum
size as well, for every $0<r^{\prime}\leq r$.

###### Proof.

It is enough to prove the statement for $r^{\prime}=r-1$. By hypothesis, we
know that there is some $(r-1)$-space
$\Omega=\mathrm{PG}(W,{\mathbb{F}}_{q^{n}})$ of $\mathrm{PG}(d,q^{n})$ meeting
$L_{U}$ in a canonical subgeometry, such that

$\lvert L_{U}\rvert=q^{k-1}+q^{k-2}+\ldots+q^{k-r}+\lvert
L_{\overline{U}}\rvert,$ (6)

where $L_{\overline{U}}$ is the ${\mathbb{F}}_{q}$-linear set in
$\mathrm{PG}(V/W,{\mathbb{F}}_{q^{n}})=\mathrm{PG}(d-r,q^{n})$ defined by
$\overline{U}=U+W\subseteq V/W$. Let
$\Omega^{\prime}=\mathrm{PG}(W^{\prime},{\mathbb{F}}_{q^{n}})$ be an
$(r-2)$-space of $\Omega$ that meets $L_{U}$ in a canonical subgeometry. So,
by Proposition 3.7, we have that

$\lvert L_{U}\rvert\geq q^{k-1}+q^{k-2}+\ldots+q^{k-r+1}+\lvert
L_{\overline{U^{\prime}}}\rvert,$ (7)

where $L_{\overline{U^{\prime}}}$ is the ${\mathbb{F}}_{q}$-linear set of rank
$k-r+1$ in
$\mathrm{PG}(V/W^{\prime},{\mathbb{F}}_{q^{n}})=\mathrm{PG}(d-r+1,q^{n})$
defined by $\overline{U^{\prime}}=U+W^{\prime}\subseteq V/W^{\prime}$.
Therefore, by (6), it follows $q^{k-r}+\lvert L_{\overline{U}}\rvert\geq\lvert
L_{\overline{U^{\prime}}}\rvert$. On the other hand, since
$w_{L_{U}}(\Omega)=r$, we get
$w_{L_{\overline{U^{\prime}}}}(\mathrm{PG}(W/W^{\prime},{\mathbb{F}}_{q^{n}}))=1$
and so $L_{\overline{U^{\prime}}}$ has a point of weight 1. Now, by
Proposition 3.7, we get that

$\lvert L_{\overline{U^{\prime}}}\rvert\geq q^{k-r}+\left\lvert
L_{\overline{\overline{U^{\prime}}}}\right\rvert$

with
$\overline{{\overline{U^{\prime}}}}=U/W^{\prime}+W/W^{\prime}\leq(V/W^{\prime})/W$,
which is equal to $\overline{U}=U+W\leq V/W$. Hence,

$|L_{\overline{U^{\prime}}}|\geq q^{k-r}+\lvert L_{\overline{U}}\rvert.$

Then, by (6), equality holds in (7) and so $L_{U}$ is of
$(r-1,d,\Omega^{\prime})$-minimum size. ∎

We conclude this subsection by giving a sufficient condition to apply Result
1.3.

######  Theorem 3.11.

Let $k,d$ and $r$ be non negative integers with $r<k,d$. Let $L_{U}$ be an
${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d,q^{n})$ of rank $k+d-r$
spanning $\mathrm{PG}(d,q^{n})$. Suppose that there is an $r$-space $\Omega$
of $\mathrm{PG}(d,q^{n})$ such that $\operatorname{w}_{L_{U}}(\Omega)=k$, and
$\Omega$ contains an $(r-1)$-space $\Omega^{\prime}$ that meets $L_{U}$ in a
canonical ${\mathbb{F}}_{q}$-subgeometry. Then some hyperplane $\Pi$ of
$\mathrm{PG}(d,q^{n})$ meets $L_{U}$ in a canonical
${\mathbb{F}}_{q}$-subgeometry, implying $\lvert L_{U}\rvert\geq
q^{k+d-r-1}+q^{k+d-r-2}+\ldots+q^{k-r}+1$.

###### Proof.

Suppose that $\mathrm{PG}(d,q^{n})=\mathrm{PG}(V,\mathbb{F}_{q^{n}})$,
$\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$, and
$\Omega^{\prime}=\mathrm{PG}(W^{\prime},\mathbb{F}_{q^{n}})$. Consider the
projection of $L_{U}$ from $\Omega^{\prime}$, which equals the linear set
$L_{\overline{U}}$, with $\overline{U}=U+W^{\prime}\subseteq V/W^{\prime}$.
Write $P_{0}=W/W^{\prime}$, and choose any point $P_{1}\in
L_{\overline{U}}\setminus\\{P_{0}\\}$. Since $L_{U}$ spans
$\mathrm{PG}(d,q^{n})$, we can extend $P_{0},P_{1}$ to a subset
$P_{0},P_{1},\dots,P_{d-r}$ of $L_{\overline{U}}$ that spans
$\mathrm{PG}(V/W^{\prime},\mathbb{F}_{q^{n}})$. Also,
$\operatorname{w}_{L_{\overline{U}}}(P_{0})=k-r$, and the rank of
$L_{\overline{U}}$ equals $k+d-2r=(k-r)+(d-r)$. Hence, by Lemma 2.5,

$(k-r)+(d-r)\geq\sum_{i=0}^{d-r}\operatorname{w}_{L_{\overline{U}}}(P_{i})=(k-r)+\sum_{i=1}^{d-r}\operatorname{w}_{L_{\overline{U}}}(P_{i}),$

which implies that $\operatorname{w}_{L_{\overline{U}}}(P_{i})=1$ for all
$i\geq 1$, and $P_{0},\ldots,P_{d-r}$ are points of complementary weights.
Therefore, $L_{\overline{U}}$ meets $\langle P_{1},\ldots,P_{d-r}\rangle$ in a
canonical subgeometry. There is a unique $\mathbb{F}_{q^{n}}$-space
$W^{\prime\prime}$ through $W^{\prime}$ such that $\langle
P_{1},\ldots,P_{d-r}\rangle=\mathrm{PG}(W^{\prime\prime}+W^{\prime},\mathbb{F}_{q^{n}})$.
It follows that $\mathrm{PG}(W^{\prime\prime},\mathbb{F}_{q^{n}})$ meets
$L_{U}$ in a canonical subgeometry. ∎

###  3.2 Proof of Theorem 1.6

Our next goal it to prove Theorem 1.6. Let us start by recalling two well-
known results concerning linear sets.

######  Lemma 3.12.

Suppose that $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of rank $k$ of
$\mathrm{PG}(n,q)$. Let $m=\min\\{\operatorname{w}_{L_{U}}(P)\,:\,P\in
L_{U}\\}$ denote the minimum weight of the points of $L_{U}$. Then there
exists an ${\mathbb{F}}_{q}$-linear set $L_{U^{\prime}}$ of rank $k-m+1$ in
$\mathrm{PG}(n,q)$ containing points of weight 1 such that $L_{U}$ and
$L_{U^{\prime}}$ coincide as point sets.

###### Proof.

Take a vector $u\in U$ such that $P=\langle u\rangle_{\mathbb{F}_{q^{n}}}$ has
weight $m$ in $L_{U}$. Then there exists a $(k-m+1)$-dimensional
${\mathbb{F}}_{q}$-subspace $U^{\prime}$ of $U$ that intersects $\langle
u\rangle_{{\mathbb{F}}_{q^{n}}}$ in a 1-dimensional subspace. Then
$\operatorname{w}_{L_{U^{\prime}}}(P)=1$. It remains to show that $L_{U}$ and
$L_{U^{\prime}}$ coincide as points sets. The inclusion
$L_{U^{\prime}}\subseteq L_{U}$ is evident. On the other hand, take a non-zero
$v\in U$. Then, by Grassmann’s identity

$\displaystyle\operatorname{w}_{L_{U^{\prime}}}(\langle
v\rangle_{\mathbb{F}_{q^{n}}})$ $\displaystyle=\dim_{{\mathbb{F}}_{q}}(\langle
v\rangle_{\mathbb{F}_{q^{n}}}\cap U^{\prime})=\dim_{{\mathbb{F}}_{q}}((\langle
v\rangle_{\mathbb{F}_{q^{n}}}\cap U)\cap U^{\prime})$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}(\langle
v\rangle_{\mathbb{F}_{q^{n}}}\cap
U)+\dim_{{\mathbb{F}}_{q}}(U^{\prime})-\dim_{{\mathbb{F}}_{q}}(\langle\langle
v\rangle_{{\mathbb{F}}_{q^{n}}}\cap U,U^{\prime}\rangle_{{\mathbb{F}}_{q}})$
$\displaystyle\geq m+(k-m+1)-\dim_{{\mathbb{F}}_{q}}(U)=1.$

This shows that $L_{U}\subseteq L_{U^{\prime}}$. Thus, $L_{U}$ and
$L_{U^{\prime}}$ coincide as point sets. ∎

######  Lemma 3.13.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d,q^{n})$ of
rank $k$. Then $L_{U}$ contains all points of $\mathrm{PG}(d,q^{n})$ if and
only if $k>dn$. In that case, $L_{U}$ contains a point of weight 1 if and only
if $k=dn+1$.

###### Proof.

If $k\leq dn$, then $|L_{U}|\leq\frac{q^{dn}-1}{q-1}$ by (3), so it can’t
contain all points of $\mathrm{PG}(d,q^{n})$. If $k>dn$, then $U$ intersects
every ${\mathbb{F}}_{q}$-subspace of $\mathbb{F}_{q^{n}}^{d+1}$ of dimension
$n$ non-trivially by Grassmann’s identity. In particular, $U$ intersects all
one-dimensional $\mathbb{F}_{q^{n}}$-subspaces of $\mathbb{F}_{q^{n}}^{d+1}$
non-trivially. Therefore, $L_{U}$ contains all points of
$\mathrm{PG}(d,q^{n})$.

If $k=dn+1$, then $L_{U}$ contains points of weight 1 by Lemma 3.12, since we
have proven that $L_{U}$ can’t coincide with a linear set of lower rank. If
$k>dn+1$, then we can use Grassmann’s identity again to prove that $L_{U}$ has
no points of weight 1. ∎

Combining this with Result 1.5, we obtain the following lemma.

######  Lemma 3.14.

Suppose that $n$ is a prime number with $n\leq q$. Let $L_{U}$ be an
${\mathbb{F}}_{q}$-linear set of rank $k$ spanning $\mathrm{PG}(d,q^{n})$.
Then $L_{U}$ does not contain any points of weight 1 if and only if $dn+2\leq
k\leq dn+n$.

###### Proof.

By Lemma 3.13, it suffices to prove that $L_{U}$ contains a point of weight 1
if $k\leq dn$. Suppose the contrary, i.e. $k\leq dn$ and $L_{U}$ contains no
points of weight 1. Then by Result 1.5 and $n$ being prime, $L_{U}$ as point
set must be a subspace of $\mathrm{PG}(d,q^{n})$. Since $L_{U}$ spans the
entire space, this means that $L_{U}$ contains all points of
$\mathrm{PG}(d,q^{n})$. Using Lemma 3.13, we see that this contradicts $k\leq
dn$. ∎

Before proving the lower bound of Theorem 1.6, let us provide examples
attaining equality in the bound.

######  Construction 3.15.

Choose integers $0\leq r\leq d$. Let $U_{1}$ be a $k_{1}$-dimensional
${\mathbb{F}}_{q}$-subspace of $\mathbb{F}_{q^{n}}^{d-r+1}$ with $(d-r)n+2\leq
k_{1}\leq(d-r+1)n$. Define
$U=U_{1}\times{\mathbb{F}}_{q}^{r}\leq_{{\mathbb{F}}_{q}}\mathbb{F}_{q^{n}}^{d+1}$.
Then $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set spanning
$\mathrm{PG}(d,q^{n})$ of rank $k=k_{1}+r$ of size

$q^{k-1}+\ldots+q^{k-r}+\frac{q^{(d-r+1)n}-1}{q^{n}-1}.$

Moreover, $r=d-\left\lfloor\frac{k-(d+2)}{n-1}\right\rfloor$.

###### Proof.

It follows immediately that $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of
rank $k$ spanning $\mathrm{PG}(d,q^{n})$. Moreover, since $k=k_{1}+r$, one can
check that

$(d-r)n+2\leq k_{1}\leq(d-r+1)n\implies d-r\leq\frac{k-(d+2)}{n-1}\leq
d-r+1-\frac{1}{n-1},$

which yields $r=d-\left\lfloor\frac{k-(d+2)}{n-1}\right\rfloor$.

It remains to determine the size of $L_{U}$. Let $u=(u_{1},u_{2})\in
U_{1}\times{\mathbb{F}}_{q}^{r}$ be a non-zero vector of $U$. If
$u_{2}\neq\mathbf{0}$, then clearly $\alpha(u_{1},u_{2})\in U$ if and only if
$\alpha\in{\mathbb{F}}_{q}$. Hence, the point $\langle(u_{1},u_{2})\rangle$
has weight 1 in $L_{U}$. To make such a vector, we have $q^{k_{1}}$ choices
for $u_{1}$ and $q^{r}-1$ choices for $u_{2}$. Since this always gives points
of weight 1, this gives us

$q^{k_{1}}\frac{q^{r}-1}{q-1}=q^{k-1}+\ldots+q^{k-r}$

points of $L_{U}$.

The number of points of $L_{U}$ represented by a vector of the form
$(u_{1},\mathbf{0})$ equals the size of $L_{U_{1}}$, which equals the number
of points of $\mathrm{PG}(d-r,q^{n})$ by Lemma 3.13. Putting this together, we
find that the size of $L_{U}$ is as claimed in the lemma. ∎

Now we are ready to finish the proof of Theorem 1.6.

###### Proof of Theorem 1.6.

Let $s$ denote the largest integer such that $L_{U}$ meets an $(s-1)$-space
$\sigma$ in a canonical subgeometry. Then the projection $L_{\overline{U}}$ of
$L_{U}$ from $\sigma$ is an ${\mathbb{F}}_{q}$-linear set of rank $k-s$
spanning $\mathrm{PG}(d-s,q^{n})$. Moreover, Lemma 3.6 implies that
$L_{\overline{U}}$ does not have any points of weight 1, since no $s$-space
through $\sigma$ can intersect $L_{U}$ in a canonical subgeometry. By Lemma
3.14, this means that

$(d-s)n+2\leq k-s\leq(d-s)n+n,$

which is equivalent to

$s=d-\left\lfloor\frac{k-(d+2)}{n-1}\right\rfloor=r.$

Hence, there exists an $(r-1)$-space $\rho$ intersecting $L_{U}$ in a
canonical subgeometry. Moreover, the projection $L_{\overline{U}}$ of $L_{U}$
from $\rho$ is an $\mathbb{F}_{q^{n}}$-linear set in $\mathrm{PG}(d-r,q^{n})$
whose rank exceeds $(d-r)n$. By Lemma 3.13, $L_{\overline{U}}$ contains all
points of $\mathrm{PG}(d-r,q^{n})$ and

$|L_{U}|\geq q^{k-1}+\ldots+q^{k-r}+\frac{q^{n(d-s+1)}-1}{q^{n}-1}$

by Theorem 1.4 and Proposition 3.7. The bound is tight by 3.15. ∎

###  3.3 Consequences of Theorem 1.4

When $r=1$, Theorem 1.4 looks as follows.

######  Corollary 3.16.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set of rank $k\geq 2$ in
$\mathrm{PG}(d,q^{n})$, admitting at least one point of weight 1. Let $I$ be
the number of secant lines through some point of weight 1. Then $\lvert
L_{U}\rvert\geq q^{k-1}+I$.

In particular, this result implies that the rank of a linear set is determined
by its size and the minimum weight of its points.

######  Proposition 3.17.

Let $L_{U}$ be an ${\mathbb{F}}_{q}$-linear set spanning
$\mathrm{PG}(d,q^{n})$, containing more than one point. Denote $m=\min_{P\in
L_{U}}\operatorname{w}_{L_{U}}(P)$. Then the rank of $L_{U}$ is the unique
integer $k$ satisfying

$q^{k-m}+1\leq|L_{U}|\leq\frac{q^{k}-1}{q^{m}-1},$

i.e. $k=\lceil\log_{q}(|L_{U}|)\rceil+m-1=\lfloor\log_{q}(|L_{U}|)\rfloor+m$.

###### Proof.

By Lemma 3.12, $L_{U}$ coincides as point set with an
${\mathbb{F}}_{q}$-linear set $L_{U^{\prime}}$ of rank $k-m+1$, containing
points of weight 1. By Theorem 1.4, $|L_{U^{\prime}}|\geq q^{k-m}+1$. The
lower bound follows from Lemma 3.12 and Corollary 3.16. By Equation 2,

$(q^{m}-1)\lvert
L_{U}\rvert=(q^{m}-1)\sum_{i=m}^{n}N_{i}\leq\sum_{i=m}^{n}N_{i}(q^{i}-1)=q^{k}-1.\qed$

Another consequence of Corollary 3.16 is that any ${\mathbb{F}}_{q}$-linear
set is spanned by its points of minimum weight, (cf. [6, Lemma 2.2] for linear
sets on $\mathrm{PG}(1,q^{n})$).

######  Proposition 3.18.

If an ${\mathbb{F}}_{q}$-linear set $L_{U}$ spans $\mathrm{PG}(d,q^{n})$, then
its points of minimum weight also span $\mathrm{PG}(d,q^{n})$.

###### Proof.

Suppose that $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of rank $k$, spanning
$\mathrm{PG}(d,q^{n})$, and denote $m=\min_{P\in
L_{U}}{\operatorname{w}_{L_{U}}(P)}$. By Corollary 3.16, $|L_{U}|>q^{k-m}$.
Now assume that the points of weight $m$ of $L_{U}$ lie in a hyperplane
$\pi=\mathrm{PG}(W,{\mathbb{F}}_{q^{n}})$ of $\mathrm{PG}(d,q^{n})$. Suppose
that $U_{1}=U\cap W$. Then there exits a subspace $U_{2}$ of $U$ such that
$U=U_{1}\oplus U_{2}$. Now let $U_{1}^{\prime}$ be an
${\mathbb{F}}_{q}$-subspace of $U_{1}$ of codimension $m-1$, and let
$U_{2}^{\prime}$ be an ${\mathbb{F}}_{q}$-subspace of $U_{2}$ of codimension
1. Let $U^{\prime}=U_{1}^{\prime}\oplus U_{2}^{\prime}$, then $L_{U^{\prime}}$
and $L_{U}$ coincide as point sets. Indeed, if $P\in L_{U}\cap\pi$, then
$\operatorname{w}_{L_{U}}(P)=\operatorname{w}_{L_{U_{1}}}(P)\geq m$. As in the
proof of Lemma 3.12, this implies that
$\operatorname{w}_{L_{U^{\prime}}}(P)=\operatorname{w}_{L_{U_{1}^{\prime}}}(P)\geq
1$. If $P\in L_{U}\setminus\pi$, then $\operatorname{w}_{L_{U}}(P)\geq m+1$,
and as in the proof of Lemma 3.12, $\operatorname{w}_{L_{U^{\prime}}}(P)\geq
1$. But

$\dim_{q}U^{\prime}=(\dim_{q}U_{1}-(m-1))+(\dim_{q}U_{2}-1)=k-m.$

Hence, by Equation 3,
$|L_{U}|=|L_{U^{\prime}}|\leq\frac{q^{k-m}-1}{q-1}<q^{k-m}$, a contradiction.
∎

##  4 Constructions of $d$-minimum size linear sets

###  4.1 Exploring the Jena-Van de Voorde construction

Recently, Jena and Van de Voorde constructed $d$-minimum size linear sets
admitting points of complementary weights, and they completely determined
their weight spectrum and weight distribution. Recall that if
$\lambda\in\mathbb{F}_{q^{n}}$, then the _degree_ of $\lambda$ over
${\mathbb{F}}_{q}$ equals the degree of the minimal polynomial of $\lambda$
over ${\mathbb{F}}_{q}$, or equivalently the smallest integer $t$ such that
$\lambda\in{\mathbb{F}}_{q^{t}}$.

######  Construction 4.1 ([11, Theorem 2.17]).

Suppose that $\lambda\in\mathbb{F}_{q^{n}}$ has degree $t>1$ over
${\mathbb{F}}_{q}$. Choose positive integers $k_{0}\geq\ldots\geq k_{d}$ such
that $k_{0}+k_{1}\leq t+1$. Define

$\displaystyle JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})$ $\displaystyle=\langle
1,\lambda,\ldots,\lambda^{k_{0}-1}\rangle_{{\mathbb{F}}_{q}}\times\ldots\times\langle
1,\lambda,\ldots,\lambda^{k_{d}-1}\rangle_{{\mathbb{F}}_{q}}$
$\displaystyle=\\{(f_{0}(\lambda),\dots,f_{d}(\lambda))\colon
f_{i}\in{\mathbb{F}}_{q}[X],\,\deg(f_{i})<k_{i}\\}.$

Then $L_{JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})}$ is a $d$-minimum size
${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d,q^{n})$ of rank
$k_{0}+\ldots+k_{d}$.

Note that since $JV_{q,n}(\lambda,t;k_{0},\ldots,k_{d})$ is a Cartesian
product of ${\mathbb{F}}_{q}$-subspaces of ${\mathbb{F}}_{q^{n}}$, it indeed
admits points of complementary weights.

Before proceeding, we make some conventions regarding polynomials.

######  Definition 4.2.

Given two polynomials $f,g\in{\mathbb{F}}_{q}[X]$, let $\gcd(f,g)$ denote the
unique monic polynomial of maximal degree that divides $f$ and $g$. We call
$f$ and $g$ _coprime_ if $\gcd(f,g)=1$. Furthermore, we will use the
convention that the degree of the zero polynomial is $-\infty$, so that the
equality $\deg(f\cdot g)=\deg f+\deg g$ still holds if $f$ or $g$ is the zero
polynomial.

######  Remark 4.3 ([11, Remark 2.19]).

Jena and Van de Voorde also determined the weight spectrum of the above linear
set. It is $(1,\dots,k_{0})$ if $k_{1}=k_{0}$, and $(1,\dots,k_{1},k_{0})$ if
$k_{1}<k_{0}$, in which case $E_{0}$ is the unique point of weight $k_{0}$.
They also described the weight distribution, but since it is rather involved,
we omit it here. It follows from their arguments that if
$\gcd(f_{0},\dots,f_{d})=1$, then

$\operatorname{w}_{L_{U}}\left(\langle
f_{0}(\lambda),\dots,f_{d}(\lambda)\rangle_{\mathbb{F}_{q^{n}}}\right)=\min_{0\leq
i\leq d}\\{k_{i}-\deg(f_{i})\\}.$ (8)

This makes it relatively easy to determine $N_{i}$ for some large values of
$i$. For instance, let

$U=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})\subseteq{\mathbb{F}}_{q^{n}}^{d+1},$

and assume that $k_{1}<k_{0}$. As stated above, $E_{0}$ is the unique point of
weight $k_{0}$, and the second largest weight of $L_{U}$ is $k_{1}$. We can
determine $N_{k_{1}}(L_{U})$. Let $m$ denote the number of indices $j$ with
$k_{j}=k_{1}$, i.e. $k_{1}=\ldots=k_{m}>k_{m+1}$. Let $P=\langle
f_{0}(\lambda),\dots,f_{d}(\lambda)\rangle_{\mathbb{F}_{q^{n}}}\in L_{U}$,
with $\gcd(f_{0},\dots,f_{d})=1$. Then, by (8), $P$ has weight $k_{1}$ if and
only if $\deg(f_{0})\leq k_{0}-k_{1}$, $\deg(f_{i})\leq 0$, for $i=1,\ldots,m$
and $f_{i}=0$, for $i>m$ and there exists some $j\in\\{1,\ldots,m\\}$ such
that $\deg(f_{j})>0$. Then,

$N_{k_{1}}(L_{U})=q^{k_{0}-k_{1}+1}\frac{q^{m}-1}{q-1}.$

The above construction has the following consequence on the existence of
$d$-minimum size linear sets in $\mathrm{PG}(d,q^{n})$.

######  Corollary 4.4 ([11, Corollary 2.18]).

There exists a $d$-minimum size ${\mathbb{F}}_{q}$-linear set of rank $k$ in
$\mathrm{PG}(d,q^{n})$ whenever

$d<k\leq\begin{cases}(d+1)\frac{n+1}{2}&\text{if $n$ is odd,}\\\
(d+1)\frac{n}{2}+1&\text{if $n$ is even.}\end{cases}$

We now present a sufficient condition for the linear set of 4.1 to be of
proper $d$-minimum size.

######  Theorem 4.5.

Consider $U=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})$ as in 4.1. Suppose that
there exist pairwise coprime polynomials
$g_{0},\dots,g_{d}\in{\mathbb{F}}_{q}[X]$ such that for each $i$, $\deg
g_{i}=k_{i}-1$. If $k_{0}+\ldots+k_{d}\leq t+d$, then $L_{U}$ is of proper
$d$-minimum size.

###### Proof.

By 4.1, we know that $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set in
$\mathrm{PG}(d,q^{n})$ of rank $k=k_{0}+\ldots+k_{d}$ of $d$-minimum size. So
it remains to prove that there exists a hyperplane of $\mathrm{PG}(d,q^{n})$
meeting $L_{U}$ in a canonical subgeometry. Consider the points
$P_{i}=\langle\mathbf{e}_{0}+g_{i}(\lambda)\mathbf{e}_{i}\rangle_{{\mathbb{F}}_{q^{n}}}$
for $i=1,\dots,d$. Clearly, $P_{1},\dots,P_{d}$ are independent, hence they
span a hyperplane. Define the polynomial

$G(X)=\prod_{i=1}^{d}g_{i}(X).$

Note that for each $i\geq 1$, the polynomial

$\left(\frac{G}{g_{i}}\right)(X)=\prod_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{d}g_{j}(X)$

is well-defined. Then the equation of the hyperplane $\Pi=\langle
P_{1},\dots,P_{d}\rangle_{\mathbb{F}_{q^{n}}}$ of $\mathrm{PG}(d,q^{n})$ is

$G(\lambda)X_{0}=\sum_{i=1}^{d}\left(\frac{G}{g_{i}}\right)(\lambda)X_{i}.$
(9)

Let $k=k_{0}+\ldots+k_{d}$ denote the rank of $L_{U}$. Then

$\displaystyle\deg G=\sum_{i=1}^{d}(k_{i}-1)=k-k_{0}-d<t,$

hence $G(\lambda)\neq 0$, and Equation (9) does indeed define a hyperplane.
Now take a non-zero vector $v=(f_{0}(\lambda),\dots,f_{d}(\lambda))\in U$, and
suppose that $\langle v\rangle_{\mathbb{F}_{q^{n}}}\in\Pi$. Then

$G(\lambda)f_{0}(\lambda)=\sum_{i=1}^{d}\left(\frac{G}{g_{i}}\right)(\lambda)f_{i}(\lambda).$
(10)

Every term in Equation 10 is a polynomial in $\lambda$, and

$\displaystyle\deg(Gf_{0})$ $\displaystyle=\deg G+\deg f_{0}=(k-k_{0}-d)+\deg
f_{0}<t,$ $\displaystyle\deg((G/g_{i})f_{i})$ $\displaystyle=\deg G+\deg
f_{i}-\deg(g_{i})\leq\deg(G)<t.$

Since $1,\lambda,\dots,\lambda^{t-1}$ are ${\mathbb{F}}_{q}$-linearly
independent, Equation 10 implies that

$G(X)f_{0}(X)=\sum_{i=1}^{d}\left(\frac{G}{g_{i}}\right)(X)f_{i}(X).$

On the one hand, this implies that $f_{0}$ is a constant polynomial.
Otherwise, the left-hand side has degree greater than $\deg(G)$, but the
degree of the right-hand side is at most $\deg(G)$, a contradiction. On the
other hand, for each $i$,

$g_{i}(X)\mid\left(G(X)f_{0}(X)-\sum_{1\leq j\neq
i}\left(\frac{G}{g_{j}}\right)(X)f_{j}(X)\right)=\left(\frac{G}{g_{i}}\right)(X)f_{i}(X).$

Since $G/g_{i}$ is coprime with $g_{i}$, and $\deg(f_{i})\leq\deg(g_{i})$ this
is only possible if $f_{i}$ is an ${\mathbb{F}}_{q}^{*}$-multiple of $g_{i}$.
Hence,

$v=(\alpha_{0},\alpha_{1}g_{1}(\lambda),\dots,\alpha_{d}g_{d}(\lambda)),$

for some scalars $\alpha_{0},\dots,\alpha_{d}\in{\mathbb{F}}_{q}$. Moreover,
since $\langle v\rangle_{\mathbb{F}_{q^{n}}}\in\Pi$,
$\alpha_{0}=\alpha_{1}+\ldots+\alpha_{d}$. Hence, $L_{U}$ intersects $\Pi$ in
the linear set $L_{W}$, with

$W=\left\\{\sum_{i=1}^{d}\alpha_{i}(\mathbf{e}_{0}+g_{i}(\lambda)\mathbf{e}_{i})\colon\alpha_{i}\in{\mathbb{F}}_{q}\right\\}.$

Therefore, $L_{U}$ intersects $\Pi$ in a canonical subgeometry. ∎

A sufficient condition to ensure the existence of pairwise coprime polynomials
$g_{0},\ldots,g_{d}\in{\mathbb{F}}_{q}[X]$ such that $\deg(g_{i})=k_{i}-1$, is
to choose the size of the ground field large enough.

######  Proposition 4.6.

Consider $U=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})$ as in 4.1, with
$k_{0}+\ldots+k_{d}\leq t+d$. Assume that

$\sum_{i=0}^{d}k_{i}-d-1\leq q.$

Then $L_{U}$ is of proper $d$-minimum size.

###### Proof.

By the hypothesis, we can consider $d+1$ subsets $S_{0},\ldots,S_{d}$ of
${\mathbb{F}}_{q}$ that are pairwise disjoint and such that $\lvert
S_{i}\rvert=k_{i}-1$, for each $i\in\\{0,\ldots,d\\}$. Then, we can define
$g_{i}(x)=\prod_{\alpha\in S_{i}}(x-\alpha_{i})$. So the assertion follows by
Theorem 4.5. ∎

Another sufficient condition to ensure the existence of pairwise coprime
polynomials $g_{0},\ldots,g_{d}\in{\mathbb{F}}_{q}[X]$ such that
$\deg(g_{i})=k_{i}-1$, is that the $g_{i}$’s are different monic irreducible
polynomials over ${\mathbb{F}}_{q}$. It is well known, see e.g. [14, Theorem
3.25], that the number of monic irreducible polynomials of degree $s$ over the
finite field ${\mathbb{F}}_{q}$ is given by Gauss’s formula

$\frac{1}{s}\sum_{h\mid s}\mu(s/h)q^{h},$

where $h$ runs over the set of all positive divisors of $s$ and $\mu$ denotes
the Möbius function.

######  Remark 4.7.

We note the following lower bound on the number of monic irreducible
polynomials of degree $s$ over ${\mathbb{F}}_{q}$, see e.g. [3]:

$\frac{1}{s}\sum_{h\mid
s}\mu\left(s/h\right)q^{h}\geq\frac{q^{s}-2q^{s/2}}{s}.$

So we get the following corollary.

######  Corollary 4.8.

Consider $U=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})$ as in 4.1, with
$k_{0}+\ldots+k_{d}\leq t+d$. For each $s=1,\dots,t$, suppose that

$|\\{i\colon k_{i}-1=s\\}|\leq\frac{q^{s}-2q^{s/2}}{s}.$

Then $L_{U}$ is of proper $d$-minimum size.

Clearly, if the rank of a linear set $L_{U}$ obtained from 4.1 is greater than
$n+d$, then every hyperplane has weight at least $d+1$ in $L_{U}$, so $L_{U}$
cannot be of proper $d$-minimum size. In case the rank exceeds $n+d$, we can
prove that $L_{U}$ is of $(1,d)$-minimum size under some constraints on the
rank.

######  Proposition 4.9.

Let $U=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d})$ be as in 4.1. If
$k_{0}+k_{d-1}+k_{d}\leq t+2$, then $L_{U}$ is a $(1,d)$-minimum size
${\mathbb{F}}_{q}$-linear set.

###### Proof.

Let

$U^{\prime}=\\{(f_{0}(\lambda),\ldots,f_{d-1}(\lambda)+\lambda^{k_{d-1}-1}f_{d}(\lambda),f_{d}(\lambda))\colon
f_{i}\in{\mathbb{F}}_{q}[X],\,\deg(f_{i})<k_{i}\\}.$

Then $U^{\prime}$ is $\mathrm{GL}(d+1,q^{n})$-equivalent to $U$ via the
$\mathbb{F}_{q^{n}}$-linear map

$\varphi:v=(v_{0},\dots,v_{d})\mapsto
v+v_{d}\lambda^{k_{d-1}-1}\mathbf{e}_{d-1}.$

The point
$\langle(0,\ldots,0,-\lambda^{k_{d-1}-1},1)\rangle_{{\mathbb{F}}_{q^{n}}}$ has
weight $1$ in $L_{U}$ and it is mapped to point $E_{d}$ by $\varphi$. So
$E_{d}$ has weight $1$ in $L_{U^{\prime}}$. We prove that $\lvert
L_{U^{\prime}}\rvert=q^{k-1}+\lvert L_{\overline{U}}\rvert$, where
$\overline{U}=U^{\prime}+E_{d}\leq_{q}{\mathbb{F}}_{q^{n}}^{d+1}/E_{d}$. Note
that ${\mathbb{F}}_{q^{n}}^{d+1}/E_{d}$ can be identified with
${\mathbb{F}}_{q^{n}}^{d}$ and
$\overline{U}=U^{\prime}+E_{d}\leq_{q}{\mathbb{F}}_{q^{n}}^{d+1}/E_{d}$ with

$\overline{U}=JV_{q,n}(\lambda,t;k_{0},\dots,k_{d-2},k_{d-1}+k_{d}-1).$

By hypothesis $k_{0}+k_{d-1}+k_{d}-1\leq t+1$, and so
$k_{0},\dots,k_{d-2},k_{d-1}+k_{d}-1$ indeed satisfy the hypothesis of 4.1
when rearranged in descending order. Therefore, $\lvert
L_{\overline{U}}\rvert=q^{k-2}+\ldots+q^{k-d}+1$. Since $\lvert
L_{U}\rvert=\lvert
L_{U^{\prime}}\rvert=q^{k-1}+\ldots+q^{k-d}+1=q^{k-1}+|L_{\overline{U}}|$, we
have the assertion. ∎

The above proposition together with Corollary 4.4, allows to construct
$(1,d)$-minimum size linear sets whose ranks exceed $n+d$.

######  Corollary 4.10.

There exist $(1,d)$-minimum size ${\mathbb{F}}_{q}$-linear sets in
$\mathrm{PG}(d,q^{n})$, $d\geq 2$, of rank $k$, whenever

$d<k\leq\begin{cases}d\frac{n+1}{2}+1&\text{if $n$ is odd},\\\
d\frac{n}{2}+2&\text{if $n$ is even.}\end{cases}$

###  4.2 Generalizing the Caserta construction

In [16, Theorem 4.1], a construction is given of linear sets on the projective
line, based on the more general framework exploited in [10] and [22]. In this
subsection, we generalize this to higher dimensions. The construction starts
from an ${\mathbb{F}}_{q}$-linear set $L_{U^{\prime}}$ in
$\mathrm{PG}(d,q^{t})$, and yields an ${\mathbb{F}}_{q}$-linear set in
$\mathrm{PG}(d,q^{st})$. Moreover, the weight distribution of $L_{U}$ is
completely determined by the weight distribution of $L_{U^{\prime}}$.

######  Construction 4.11.

Suppose that $n=st$ with $s,t>1$. Let $U^{\prime}$ be an
${\mathbb{F}}_{q}$-subspace of
${\mathbb{F}}_{q^{t}}^{d+1}\subseteq{\mathbb{F}}_{q^{n}}^{d+1}$ with
$\dim_{{\mathbb{F}}_{q}}(U^{\prime})=k^{\prime}>0$. Let $Z$ be an
${\mathbb{F}}_{q^{t}}$-subspace of ${\mathbb{F}}_{q^{n}}$ of dimension $r>0$,
such that $1\notin Z$. Define

$C_{q,s,t}(Z,U^{\prime}):=\\{(z+u_{0},u_{1},\ldots,u_{d})\colon z\in
Z,(u_{0},\ldots,u_{d})\in U^{\prime}\\}\subseteq{\mathbb{F}}_{q^{n}}^{d+1},$

which we will simply denote by $U$. Then

  1. (1)

the ${\mathbb{F}}_{q}$-linear set $L_{U}\subseteq\mathrm{PG}(d,q^{n})$ has
rank $rt+k^{\prime}$,

  2. (2)

$|L_{U}|=q^{rt}|L_{U^{\prime}}\setminus\\{E_{0}\\}|+1$,

  3. (3)

$\operatorname{w}_{L_{U}}(E_{0})=rt+\operatorname{w}_{L_{U^{\prime}}}(E_{0})$,

  4. (4)

$N_{i}(L_{U})=q^{rt}(N_{i}(L_{U^{\prime}})-\delta_{i,\operatorname{w}_{L_{U^{\prime}}}(E_{0})})+\delta_{i,\operatorname{w}_{L_{U}}(E_{0})}$,

where $\delta_{i,j}$ denotes the Kronecker symbol.

###### Proof.

(1) Since $Z$ is an ${\mathbb{F}}_{q^{t}}$-subspace of $\mathbb{F}_{q^{n}}$,
and $1\notin Z$, $Z\cap{\mathbb{F}}_{q^{t}}=\\{0\\}$. Furthermore, since
$U^{\prime}$ is an ${\mathbb{F}}_{q}$-subspace of
${\mathbb{F}}_{q^{t}}^{d+1}$, $Z\cap\\{u_{0}:(u_{0},\dots,u_{d})\in
U^{\prime}\\}=\\{0\\}$. Hence,

$U=(Z\times\\{0\\}^{d})\oplus_{{\mathbb{F}}_{q}}U^{\prime}.$

Therefore,

$\dim_{{\mathbb{F}}_{q}}U=\dim_{{\mathbb{F}}_{q}}Z+\dim_{{\mathbb{F}}_{q}}U^{\prime}=rt+k^{\prime}.$

(3) Similarly,

$\displaystyle\operatorname{w}_{L_{U}}(E_{0})$
$\displaystyle=\dim_{{\mathbb{F}}_{q}}\left(Z\oplus_{{\mathbb{F}}_{q}}\\{u_{0}:u_{0}\mathbf{e}_{0}\in
U^{\prime}\\}\right)=\dim_{{\mathbb{F}}_{q}}Z+\dim_{{\mathbb{F}}_{q}}(\\{u_{0}:u_{0}\mathbf{e}_{0}\in
U^{\prime}\\})$ $\displaystyle=rt+\operatorname{w}_{L_{U^{\prime}}}(E_{0}).$

(2,4) Suppose that

$\langle z\mathbf{e}_{0}+u\rangle_{\mathbb{F}_{q^{n}}}=\langle
z^{\prime}\mathbf{e}_{0}+v\rangle_{\mathbb{F}_{q^{n}}},$

with $z,z^{\prime}\in Z$, and $u,v\in
U^{\prime}\setminus\langle\mathbf{e}_{0}\rangle_{\mathbb{F}_{q^{n}}}$. Then
$z\mathbf{e}_{0}+u=\alpha(z^{\prime}\mathbf{e}_{0}+v)$ for some
$\alpha\in\mathbb{F}_{q^{n}}$. Since $u,v$ are not multiples of
$\mathbf{e}_{0}$, there must exist some position $j>0$ such that
$u_{j},v_{j}\neq 0$. This implies that
$\alpha=v_{j}/u_{j}\in{\mathbb{F}}_{q^{t}}$. We also have that
$z+u_{0}=\alpha(z^{\prime}+v_{0})$, hence

$z-\alpha z^{\prime}=\alpha v_{0}-u_{0}$

Recall that $Z$ is an ${\mathbb{F}}_{q^{t}}$-subspace, and that
$u_{0},v_{0},\alpha\in{\mathbb{F}}_{q^{t}}$. Therefore, the left-hand side of
the above equality is in $Z$, and the right-hand side is in
${\mathbb{F}}_{q^{t}}$. Since $Z\cap{\mathbb{F}}_{q^{t}}=\\{0\\}$, this
implies that $z=\alpha z^{\prime}$ and therefore $u=\alpha v$.

Vice versa, if $z\in Z$, $u\in
U^{\prime}\setminus\langle\mathbf{e}_{0}\rangle_{\mathbb{F}_{q^{n}}}$ and
$\alpha u\in U^{\prime}$ for some $\alpha\in{\mathbb{F}}_{q^{t}}$, then
$\langle z\mathbf{e}_{0}+u\rangle_{\mathbb{F}_{q^{n}}}=\langle\alpha
z\mathbf{e}_{0}+\alpha u\rangle_{\mathbb{F}_{q^{n}}}$. This proves that

$\operatorname{w}_{L_{U}}(\langle
z+u\rangle_{\mathbb{F}_{q^{n}}})=\dim_{{\mathbb{F}}_{q}}\\{\alpha\in{\mathbb{F}}_{q^{t}}:\alpha
u\in U^{\prime}\\}=\operatorname{w}_{L_{U^{\prime}}}(\langle
u\rangle_{\mathbb{F}_{q^{n}}}).$

Hence, varying $z$, we see that every point of
$L_{U^{\prime}}\setminus\\{E_{0}\\}$ gives rise to $|Z|$ points of
$L_{U}\setminus\\{E_{0}\\}$ of the same weight, and this accounts for all
points of $L_{U}\setminus\\{E_{0}\\}$. Points (2) and (4) follow directly from
this observation and the fact that $E_{0}\in L_{U}$. ∎

######  Remark 4.12.

We remark that $L_{U^{\prime}}$ is contained in $L_{U}$ and the weight
distribution and rank of $L_{U}$ in the above construction only depends on the
weight distribution of $L_{U^{\prime}}$ and
$\operatorname{w}_{L_{U^{\prime}}}(E_{0})$, but not on the specific structure
of $U^{\prime}$. In particular, if
$\varphi\in\operatorname{\Gamma\mathrm{L}}(d+1,q^{t})$, and $\varphi$ fixes
$\langle\mathbf{e}_{0}\rangle$, then $C_{q,s,t}(Z,U^{\prime})$ and
$C_{q,s,t}(Z,\varphi(U^{\prime}))$ have the same rank and weight distribution.

Given some minor conditions, the above construction preserves the property of
being $(r,d)$-minimum size.

######  Proposition 4.13.

Let $U=C_{q,s,t}(Z,U^{\prime})$ be as in 4.11. If $L_{U^{\prime}}$ is an
${\mathbb{F}}_{q}$-linear set of $(r,d,\Omega)$-minimum size, and $E_{0}\in
L_{U^{\prime}}\setminus\Omega$, then $L_{U}$ is also of $(r,d,\Omega)$-minimum
size.

###### Proof.

Suppose that $\Omega=\mathrm{PG}(W,\mathbb{F}_{q^{n}})$, and that the rank of
$L_{U^{\prime}}$ is $k^{\prime}$. Since $L_{U^{\prime}}$ is of
$(r,d,\Omega)$-minimum size, $L_{U^{\prime}}$ meets $\Omega$ in a canonical
subgeometry of $\Omega$, and

$\lvert L_{U^{\prime}}\rvert=q^{k^{\prime}-1}+\ldots+q^{k^{\prime}-r}+\lvert
L_{\overline{U^{\prime}}}\rvert,$

where
$\overline{U^{\prime}}:=U^{\prime}+W\leq_{q}{\mathbb{F}}_{q^{n}}^{d+1}/W$.
Since $E_{0}\notin\Omega$, up to $\mathrm{GL}(d+1,q^{n})$-equivalence, we can
suppose that $\Omega$ is defined by the equations $X_{0}=\ldots=X_{d-r}=0$.
Hence ${\mathbb{F}}_{q^{n}}^{d+1}/W$ can be identified with
${\mathbb{F}}_{q^{n}}^{d-r+1}$ in an obvious way. Now, an element $z+u\in U$
belongs to $W$ if and only if $z+u_{0}=u_{1}=\ldots=u_{d-r}=0$ if and only if
$z=u_{0}=\ldots=u_{d-r}=0$. Therefore, $U\cap W=U^{\prime}\cap W$ and so
$\Omega$ also meets $L_{U}$ in a canonical subgeometry. Moreover, by 4.11,

$\overline{U}=U+W=\\{z+\overline{u}\colon\overline{u}\in\overline{U^{\prime}}\\}\subseteq{\mathbb{F}}_{q^{n}}^{d+1}/W,$

has size $q^{rt}(\lvert L_{\overline{U^{\prime}}}\rvert-1)+1$. Therefore we
have

$\displaystyle|L_{U}|$ $\displaystyle=q^{rt}(\lvert
L_{U^{\prime}}\rvert-1)+1=q^{rt}(q^{k^{\prime}-1}+\ldots+q^{k^{\prime}-r}+\lvert
L_{\overline{U^{\prime}}}\rvert-1)+1$
$\displaystyle=q^{k-1}+\ldots+q^{k-r}+\lvert L_{\overline{U}}\rvert,$

with $k=rt+k^{\prime}$ the rank of $L_{U}$. ∎

We can apply 4.11 with $U^{\prime}$ as in 4.1, obtaining the following
families of $d$-minimum size linear sets.

######  Theorem 4.14.

Consider $U^{\prime}=JV_{q,t}(\lambda,t^{\prime};k_{0},\dots,k_{d})$ where
$t^{\prime}\mid t$ as in 4.1, and choose
$\varphi\in\operatorname{GL}(d+1,q^{t})$ such that $E_{0}\in
L_{\varphi(U^{\prime})}$. Now define $U=C_{q,s,t}(Z,\varphi(U^{\prime}))$ as
in 4.11, with $Z$ an ${\mathbb{F}}_{q^{t}}$-subspace of rank $r>0$, not
containing 1. Then $L_{U}$ is a $d$-minimum size ${\mathbb{F}}_{q}$-linear set
of rank $k=rt+k_{0}+\ldots+k_{d}$. Moreover, the weight spectrum of $L_{U}$ is

$\begin{cases}\left(1,\ldots,k_{1},k_{0},rt+\operatorname{w}_{L_{\varphi(U^{\prime})}}(E_{0})\right)&\mbox{if
}\operatorname{w}_{L_{\varphi(U^{\prime})}}(E_{0})<k_{0}\text{ and
}k_{1}<k_{0},\\\
\left(1,\ldots,k_{1},rt+\operatorname{w}_{L_{\varphi(U^{\prime})}}(E_{0})\right)&\mbox{otherwise}.\end{cases}$

###### Proof.

The ${\mathbb{F}}_{q}$-linear set $L_{\varphi(U^{\prime})}$ has the same
weight spectrum, weight distribution, and size as $L_{U^{\prime}}$. So the
assertions follow by applying 4.11 and Remark 4.3. ∎

######  Remark 4.15.

Using [11, Remark 2.19] and 4.11 (3,4), one could in fact also determine the
weight distribution of the linear set in the above theorem.

The above construction gives new examples of proper $d$-minimum size linear
sets.

######  Corollary 4.16.

In the hypothesis of Theorem 4.14, suppose that $L_{U^{\prime}}$ is a
$(d,d,\Pi)$-minimum size ${\mathbb{F}}_{q}$-linear set, with
$\Pi=\mathrm{PG}(W,{\mathbb{F}}_{q^{n}})$. Suppose that $E_{0}\in
L_{\varphi(U^{\prime})}\setminus\tilde{\Pi}$, with
$\tilde{\Pi}=\mathrm{PG}(\varphi(W),{\mathbb{F}}_{q^{n}})$. Then $L_{U}$ is a
proper $d$-minimum size ${\mathbb{F}}_{q}$-linear set in
$\mathrm{PG}(d,q^{n})$.

###### Proof.

The linear set $L_{U^{\prime}}$ is of $(d,d,\Pi)$-minimum size, so the
hyperplane $\Pi=\mathrm{PG}(W,{\mathbb{F}}_{q^{n}})$ of $\mathrm{PG}(d,q^{n})$
meets $L_{U^{\prime}}$ in a canonical subgeometry of $\Pi$ and

$\lvert L_{U^{\prime}}\rvert=q^{m-1}+\ldots+q^{m-d}+1.$

It follows that $\tilde{\Pi}$ also meets $L_{\varphi(U^{\prime})}$ in a
canonical subgeometry of $\tilde{\Pi}$, that is $L_{\varphi(U^{\prime})}$ is
of proper $d$-minimum size as well. The assertion follows by Proposition 4.13.
∎

4.1 provides constructions of $d$-minimum size linear sets admitting points of
complementary weights. Using Theorem 4.14, it is possible to construct proper
$d$-minimum size linear sets that do not have this property, as we will see in
the next example. This proves that in general a $d$-minimum size linear set
need not contain independent points whose weights sum to the rank of the
linear set. So in general, as already observed in [16] for the projective
line, being minimum size does not determine the weight spectrum and
distribution of a linear set.

######  Example 4.17.

Consider

$U^{\prime}=JV_{q,6}(\lambda,6;2,2,2)$

as in 4.1. Then $L_{U^{\prime}}$ is an ${\mathbb{F}}_{q}$-linear set of rank
$6$ in $\mathrm{PG}(2,q^{6})$ having size $q^{5}+q^{4}+1$ and points of weight
at most $2$. Moreover,
$\operatorname{w}_{L_{U^{\prime}}}(E_{0})+\operatorname{w}_{L_{U^{\prime}}}(E_{1})+\operatorname{w}_{L_{U^{\prime}}}(E_{2})=2+2+2=6$
is equal to the rank of $L_{U^{\prime}}$. Define

$\varphi\in\operatorname{GL}(3,q^{6}):(x,y,z)\mapsto(x,y-\lambda x,z).$

Then the ${\mathbb{F}}_{q}$-linear set $L_{U^{\prime\prime}}$ in
$\mathrm{PG}(2,q^{6})$, with

$U^{\prime\prime}=\varphi(U^{\prime})=\\{(\alpha_{0}+\alpha_{1}\lambda,\beta_{0}+\beta_{1}\lambda-\alpha_{1}\lambda^{2},\gamma_{0}+\gamma_{1}\lambda)\colon\alpha_{i},\beta_{i},\gamma_{i}\in{\mathbb{F}}_{q}\\}\subseteq{\mathbb{F}}_{q^{6}}^{3}$

has the same rank, weight spectrum and weight distribution as
$L_{U^{\prime}}$. Note that
$\operatorname{w}_{L_{U^{\prime\prime}}}(E_{0})=1$. Choose a 1-dimensional
${\mathbb{F}}_{q^{6}}$-subspace $Z\neq{\mathbb{F}}_{q^{6}}$ of
${\mathbb{F}}_{q^{12}}$. By Theorem 4.14, the ${\mathbb{F}}_{q}$-linear set
$L_{U}$ of $\mathrm{PG}(2,q^{12})$, with

$U=C_{q,2,6}(Z,U^{\prime\prime})$

has rank $12$ and size $q^{11}+q^{10}+1$. So, it is a $2$-minimum size linear
set. Note that the weight spectrum of $L_{U}$ is $(1,2,7)$, and so there do
not exist three points of complementary weights. In particular, $L_{U}$ cannot
be obtained from 4.1.

In some cases, Theorem 4.14 gives us linear sets admitting points of
complementary weights, but with a different weight distribution than those of
4.1, as stated in the following theorem.

######  Theorem 4.18.

Consider

$U^{\prime}=JV_{q,t}(\lambda,t;k_{0},\dots,k_{d})$

and let $\varphi_{i}$ be the linear map swapping coordinates 0 and
$i\in\\{0,\dots,d\\}$ (with $\varphi_{0}=\mathrm{id}$). Consider

$U=C_{q,s,t}(Z,\varphi_{i}(U^{\prime})),$

  1. (1)

If $k_{i}=k_{0}$, then there exists an ${\mathbb{F}}_{q}$-linear set obtained
from 4.1 with the same weight distribution as $L_{U}$.

  2. (2)

If $k_{i}<k_{0}-1$, then there does not exist an ${\mathbb{F}}_{q}$-linear set
obtained from 4.1 with the same weight distribution as $L_{U}$.

###### Proof.

Write $n=st$.

(1) If $k_{i}=k_{0}$, choose a primitive element $\mu$ of
$\mathbb{F}_{q^{n}}$. Consider
$U_{2}=JV_{q,n}(\mu,n;k_{0}+rt,k_{1},\dots,k_{d})$. Then $L_{U}$ and
$L_{U_{2}}$ have the same weight distribution by Remark 4.3 (see also [11,
Remark 2.19]) and 4.11 (3,4).

(2) Now assume that $k_{i}<k_{0}-1$, and suppose that there exists some
${\mathbb{F}}_{q}$-subspace

$U_{3}=JV_{q,n}(\mu,t^{\prime};k_{0}^{\prime},\dots,k_{d}^{\prime})\subseteq{\mathbb{F}}_{q^{n}}^{d+1}$

such that $L_{U}$ and $L_{U_{3}}$ have the same weight distribution. Since
$L_{U}$ has a unique point of weight $rt+k_{i}$ by 4.11 (3,4), we see that by
Remark 4.3, $k_{0}^{\prime}=rt+k_{i}$. Furthermore, the second largest weight
of $L_{U}$ and $L_{U_{3}}$ is respectively $k_{0}$ and $k_{1}^{\prime}$, hence
$k_{0}=k_{1}^{\prime}$. Let $m^{\prime}$ denote the number of indices $j$ with
$k_{j}^{\prime}=k_{0}$, i.e.
$k_{1}^{\prime}=\ldots=k_{m^{\prime}}^{\prime}>k_{m^{\prime}+1}^{\prime}$.
Then, using Remark 4.3 (see also [11, Remark 2.19]),

$N_{k_{0}}(L_{U_{3}})=q^{k_{0}^{\prime}-k_{1}^{\prime}+1}\frac{q^{m^{\prime}}-1}{q-1}=q^{rt+k_{i}-k_{0}+1}\frac{q^{m^{\prime}}-1}{q-1}.$

On the other hand, by 4.11 (4) and Remark 4.3 (see also [11, Remark 2.19]),

$N_{k_{0}}(L_{U})=q^{rt}N_{k_{0}}(L_{U^{\prime}})=q^{rt}\frac{q^{m}-1}{q-1},$

with $m$ the number indices $j$ with $k_{j}=k_{0}$. Therefore,

$q^{rt+k_{i}-k_{0}+1}\frac{q^{m^{\prime}}-1}{q-1}=q^{rt}\frac{q^{m}-1}{q-1}.$

Since $\frac{q^{m}-1}{q-1}$ and $\frac{q^{m^{\prime}}-1}{q-1}$ are coprime
with $q$, this implies that $k_{i}=k_{0}-1$. ∎

The case $k_{i}=k_{0}-1$ is a bit more complicated, and we will not discuss it
here.

###  4.3 Regarding equivalence

We show that the two different types of ${\mathbb{F}}_{q}$-subspaces that
define the $d$-minimum size linear sets defined in 4.1 and Theorem 4.14 are
$\Gamma\mathrm{L}(d+1,q^{n})$-inequivalent, even if the associated linear sets
have the same weight spectrum and distribution (see Theorem 4.18 (1)), when
the dimension of $Z$ is the maximum possible. The trace function
$\mathrm{Tr}_{q^{n}/q}$ of $\mathbb{F}_{q^{n}}$ over ${\mathbb{F}}_{q}$,
defines a non-degenerate symmetric bilinear form as follows:

$(a,b)\in{\mathbb{F}}_{q^{n}}\times{\mathbb{F}}_{q^{n}}\mapsto\mathrm{Tr}_{q^{n}/q}(ab)\in{\mathbb{F}}_{q}.$

Hence, for any subset $S$ of $\mathbb{F}_{q^{n}}$ we can define the orthogonal
complement as

$S^{\perp}=\\{a\in\mathbb{F}_{q^{n}}\colon\mathrm{Tr}_{q^{n}/q}(ab)=0,\,\,\,\forall
b\in S\\}.$

Note that if $S$ is an ${\mathbb{F}}_{q^{t}}$-subspace of
${\mathbb{F}}_{q^{n}}$, then $S^{\perp}$ is an ${\mathbb{F}}_{q^{t}}$-subspace
as well.

Given an ordered ${\mathbb{F}}_{q}$-basis
$\mathcal{B}=(\xi_{0},\ldots,\xi_{n-1})$ of ${\mathbb{F}}_{q^{n}}$, there
exists a unique ordered ${\mathbb{F}}_{q}$-basis
$\mathcal{B}^{*}=(\xi_{0}^{*},\ldots,\xi_{n-1}^{*})$ of ${\mathbb{F}}_{q^{n}}$
such that $\mathrm{Tr}_{q^{n}/q}(\xi_{i}\xi_{j}^{*})=\delta_{ij}$, for
$i,j\in\\{0,\ldots,n-1\\}$, called the _dual basis_ of $\mathcal{B}$, see e.g.
[14, Definition 2.30].

######  Lemma 4.19 ([17, Corollary 2.7]).

Let $\lambda\in\mathbb{F}_{q^{n}}$ and suppose that
$\mathcal{B}=(1,\lambda,\ldots,\lambda^{n-1})$ is an ordered
${\mathbb{F}}_{q}$-basis of $\mathbb{F}_{q^{n}}$. Let
$f(x)=a_{0}+a_{1}x+\ldots+a_{n-1}x^{n-1}+x^{n}$ be the minimal polynomial of
$\lambda$ over ${\mathbb{F}}_{q}$. Then the dual basis $\mathcal{B}^{*}$ of
$\mathcal{B}$ is

$\mathcal{B}^{*}=(\delta^{-1}\gamma_{0},\ldots,\delta^{-1}\gamma_{n-1}),$

where $\delta=f^{\prime}(\lambda)$ and
$\gamma_{i}=\sum_{j=1}^{n-i}\lambda^{j-1}a_{i+j}$, for every
$i\in\\{0,\ldots,n-1\\}$.

######  Theorem 4.20.

Suppose that $n=(s+1)t$, with $s,t>1$. Consider
$U^{\prime}=JV_{q,t}(\mu,t;k_{0},\dots,k_{d})$ as in 4.1, with $k_{0}<t-1$.
Let $\varphi_{i}$ be the linear map swapping coordinates 0 and
$i\in\\{0,\dots,d\\}$ (with $\varphi_{0}=\mathrm{id}$) and define

$U_{1}=C_{q,s,t}(Z,\varphi_{i}(U^{\prime})),$

as in 4.11, with $Z$ an ${\mathbb{F}}_{q^{t}}$-subspace of dimension $s$, not
containing 1. Consider $U_{2}=JV_{q,n}(\lambda,n;h_{0},k_{1},\dots,k_{d})$ as
in 4.1, with $h_{0}=st+k_{i}$. Then the ${\mathbb{F}}_{q}$-subspaces $U_{1}$
are $U_{2}$ are $\Gamma\mathrm{L}(d+1,q^{n})$-inequivalent.

###### Proof.

Suppose that $k_{i}<k_{0}-1$. Then, by Theorem 4.18, $L_{U_{1}}$ and
$L_{U_{2}}$ have a distinct weight distribution, hence $U_{1}$ and $U_{2}$
cannot be $\Gamma\mathrm{L}(d+1,q^{n})$-equivalent. So suppose that
$k_{i}\in\\{k_{0}-1,k_{0}\\}$ and suppose by contradiction that $U_{1}$ and
$U_{2}$ are $\Gamma\mathrm{L}(d+1,q^{n})$-equivalent via an element $\varphi$.
Since $h_{0}>k_{i}$, for every $i\in\\{1,\ldots,d\\}$, the point $E_{0}$ is
the only point in $L_{U_{1}}$ and in $L_{U_{2}}$ of weight $h_{0}$. So, we
have that $\varphi(U_{1}\cap E_{0})=U_{2}\cap E_{0}$, that is

$aS_{1}^{\rho}=S_{2},$

some $a\in\mathbb{F}_{q^{n}}^{*}$ and
$\rho\in\mathrm{Aut}(\mathbb{F}_{q^{n}})$, with $S_{1}=Z\oplus\langle
1,\mu,\ldots,\mu^{k_{i}-1}\rangle_{{\mathbb{F}}_{q}}$ and $S_{2}=\langle
1,\lambda,\ldots,\lambda^{h_{0}-1}\rangle_{{\mathbb{F}}_{q}}$. In particular,
we have that $aZ^{\rho}\subseteq S_{2}$ and so $(aZ^{\rho})^{\perp}\supseteq
S_{2}^{\perp}$. Note that
$\dim_{{\mathbb{F}}_{q^{t}}}(aZ^{\rho})=\dim_{{\mathbb{F}}_{q^{t}}}(Z)=s$.
This implies that $\dim_{{\mathbb{F}}_{q}}((aZ^{\rho})^{\perp})=n-st=t$ and
hence $(aZ^{\rho})^{\perp}$ is an ${\mathbb{F}}_{q^{t}}$-subspace of
$\mathbb{F}_{q^{n}}$ of dimension one. Consider the ordered ${\mathbb{F}}_{q}$
basis $\mathcal{B}=(1,\lambda,\ldots,\lambda^{n-1})$ of ${\mathbb{F}}_{q^{n}}$
and its dual basis
$\mathcal{B}^{*}=(\lambda_{0}^{*},\ldots,\lambda_{n-1}^{*})$. So we have that
$S_{2}^{\perp}=\langle\lambda_{h_{0}}^{*},\ldots,\lambda_{n-1}^{*}\rangle_{{\mathbb{F}}_{q}}$
and since $k_{0}<t-1$, we have that $h_{0}<n-1$. By Lemma 4.19 it follows that

$\lambda_{n-2}^{*}=\delta^{-1}(a_{n-1}+\lambda),$

and

$\lambda_{n-1}^{*}=\delta^{-1},$

where $f(x)=a_{0}+a_{1}x+\ldots+a_{n-1}x^{n-1}+x^{n}$ is the minimal
polynomial of $\lambda$ over ${\mathbb{F}}_{q}$ and
$\delta=f^{\prime}(\lambda)$. Now, since
$\lambda_{n-2}^{*},\lambda_{n-1}^{*}\in(aZ^{\rho})^{\perp}$ and since
$(aZ^{\rho})^{\perp}$ has dimension one over ${\mathbb{F}}_{q^{t}}$, it
follows

$\frac{\lambda_{n-2}^{*}}{\lambda_{n-1}^{*}}=a_{n-1}+\lambda\in{\mathbb{F}}_{q^{t}},$

that is $\lambda\in{\mathbb{F}}_{q^{t}}$, a contradiction. ∎

##  5 Below the De Beule-Van de Voorde bound

In this section, we will provide constructions of linear sets $L_{U}$ in
$\mathrm{PG}(d,q^{n})$, with $d>2$, that are of $(r,d)$-minimum size but not
of $d$-minimum size. They have maximum geometric field of linearity
${\mathbb{F}}_{q}$, and admit two subspaces of complementary weights.

For our aims, we will suppose that one of these subspaces intersects $L_{U}$
in a linear set with greater field of linearity. This gives us the following
constructions.

######  Theorem 5.1.

Let $n=st$, with $s,t>1$, and suppose that

  * $U_{1}$ is a $k_{1}$-dimensional ${\mathbb{F}}_{q^{t}}$-subspace of $\mathbb{F}_{q^{n}}^{d_{1}+1}$,

  * $U_{2}$ is a $k_{2}$-dimensional ${\mathbb{F}}_{q}$-subspace of ${\mathbb{F}}_{q^{t}}^{d_{2}+1}\subseteq{\mathbb{F}}_{q^{n}}^{d_{2}+1}$.

Define $U=U_{1}\times U_{2}$, and $d=d_{1}+d_{2}+1$. Then $L_{U}$ is an
${\mathbb{F}}_{q}$-linear set of $\mathrm{PG}(d,q^{n})$ of rank
$k_{1}t+k_{2}$, with

$|L_{U}|=|L_{U_{1}}|+q^{k_{1}t}|L_{U_{2}}|.$

Moreover, its weight distribution satisfies

$N_{i}(L_{U})=N_{i}(L_{U_{1}})+q^{k_{1}t}N_{i}(L_{U_{2}}).$

###### Proof.

Take a vector $u\in U_{1}$ and $v\in U_{2}$ with $(u,v)\neq\mathbf{0}$. Then

$\operatorname{w}_{L_{U}}(\langle(u,v)\rangle_{\mathbb{F}_{q^{n}}})=\dim_{{\mathbb{F}}_{q}}\\{\alpha\in\mathbb{F}_{q^{n}}:\alpha(u,v)\in
U\\}.$

Evidently, $\alpha(u,v)\in U$ if and only if $\alpha u\in U_{1}$ and $\alpha
v\in U_{2}$. If $v\neq\mathbf{0}$, then $\alpha v\in U_{2}$ implies that
$\alpha\in{\mathbb{F}}_{q^{t}}$, and since $U_{1}$ is an
${\mathbb{F}}_{q^{t}}$-subspace, $\alpha u$ is automatically in $U_{1}$.
Therefore, every point $\langle v\rangle_{\mathbb{F}_{q^{n}}}$ of $L_{U_{2}}$
gives rise to the $q^{k_{1}t}$ points
$\\{\langle(u,v)\rangle_{\mathbb{F}_{q^{n}}}:u\in U_{1}\\}$ of $L_{U}$ with
the same weight. If $v=\mathbf{0}$, then we just need that $\alpha u\in
U_{1}$, hence in this way, every point of $L_{U_{1}}$ gives rise to one point
of $L_{U}$ of the same weight. Since this accounts for all points of $L_{U}$,
the statement of the theorem follows. ∎

Using the above theorem, we are able to obtain constructions of linear sets in
$\mathrm{PG}(d,q^{n})$, with $d\geq 3$, having maximum geometric field of
linearity ${\mathbb{F}}_{q}$ that are $(r,d)$-minimum size with $2\leq r<d$
and that are not $d$-minimum size.

######  Theorem 5.2.

Let $n=st$, with $s,t>1$, and suppose that

  * $U_{1}$ is a $k_{1}$-dimensional ${\mathbb{F}}_{q^{t}}$-subspace of $\mathbb{F}_{q^{n}}^{d_{1}+1}$, with $k_{1}\leq d_{1}s$,

  * $U_{2}$ is a $k_{2}$-dimensional ${\mathbb{F}}_{q}$-subspace of ${\mathbb{F}}_{q^{t}}^{d_{2}+1}$, such that $L_{U_{2}}$ is a proper $d_{2}$-minimum size ${\mathbb{F}}_{q}$-linear set.

Define $U=U_{1}\times U_{2}$, $d=d_{1}+d_{2}+1$, and $k=k_{1}t+k_{2}$. Then
$L_{U}$ is a $(d_{2},d)$-minimum size ${\mathbb{F}}_{q}$-linear set of size

$|L_{U}|=q^{k-1}+q^{k-2}+\ldots+q^{k-d_{2}}+q^{k_{1}t}+|L_{U_{1}}|.$

Hence, $L_{U}$ is not $d$-minimum size if $k_{2}\geq d_{2}+2$. Furthermore, if
$d_{2}\geq 2$, then ${\mathbb{F}}_{q}$ is the maximum geometric field of
linearity of $L_{U}$.

###### Proof.

The ${\mathbb{F}}_{q}$-linear set $L_{U_{2}}\subseteq\mathrm{PG}(d_{2},q^{n})$
is of proper $d_{2}$-minimum size, and so its size is

$|L_{U_{2}}|=q^{k_{2}-1}+\ldots+q^{k_{2}-d_{2}}+1.$

By Theorem 5.1, $L_{U}$ has rank $k=k_{1}t+k_{2}$ and size

$\lvert
L_{U_{1}}\rvert+q^{k_{1}t}(q^{k_{2}-1}+\ldots+q^{k_{2}-d_{2}}+1)=\lvert
L_{U_{1}}\rvert+q^{k-1}+q^{k-2}+\ldots+q^{k-d_{2}}+q^{k_{1}t}.$

Moreover there exists a $(d_{2}-1)$-space
$\Gamma=\mathrm{PG}(W,{\mathbb{F}}_{q^{n}})$ of $\mathrm{PG}(d_{2},q^{n})$,
with $W\subseteq{\mathbb{F}}_{q^{n}}^{d_{2}+1}$, meeting $L_{U_{2}}$ in a
canonical subgeometry. Now, let $W^{\prime}=\\{0\\}^{d_{1}+1}\times W$. Then
$W^{\prime}$ defines a $d_{2}$-space of $\mathrm{PG}(d,q^{n})$ meeting $L_{U}$
in a canonical subgeometry. Identifying ${\mathbb{F}}_{q^{n}}^{d+1}/W$ with
${\mathbb{F}}_{q^{n}}^{d_{1}+1}$ we have that
$\overline{U}=U_{1}+W\leq_{q}{\mathbb{F}}_{q^{n}}^{d+1}/W$ with

$\overline{U}=U_{1}\times U^{\prime},$

where $U^{\prime}$ is an ${\mathbb{F}}_{q}$-subspace of ${\mathbb{F}}_{q^{n}}$
of dimension $k_{2}-d_{2}$. So again, by Theorem 5.1, we have $\lvert
L_{\overline{U}}\rvert=q^{k_{1}t}+\lvert L_{U_{1}}\rvert$. Moreover, by (3),
$|L_{U_{1}}|\leq q^{(k_{1}-1)t}+\ldots+q^{t}+1$, and since $k_{2}>d_{2}+1$ it
follows that

$|L_{U_{1}}|<q^{k-d_{2}-1}+\ldots+q^{k-d}+1-q^{k_{1}t}.$

This implies that $L_{U}$ is not of $d$-minimum size. Finally, the assertion
on the geometric field of linearity follows from Remark 3.9. ∎

By the above corollary and Proposition 4.5, we get the following construction.

######  Corollary 5.3.

Let $n=st$, with $s,t>1$, and suppose that

  * $U_{1}=JV_{q^{t},n}(\lambda,n;l_{0},\dots,l_{d_{1}})$, and denote $k_{1}=l_{0}+\ldots+l_{d_{1}}$,

  * $U_{2}=JV_{q,t}(\mu,t;m_{0},\dots,m_{d_{2}})$, and denote $k_{2}=m_{0}+\ldots+m_{d_{2}}$,

with $L_{U_{2}}$ satisfying the condition of Theorem 4.5. Define
$U=U_{1}\times U_{2}$. Then $L_{U}$ is a $(d_{2},d)$-minimum size
${\mathbb{F}}_{q}$-linear set, but not of $d$-minimum size. Moreover, if
$d_{2}\geq 2$, then ${\mathbb{F}}_{q}$ is the maximum geometric field of
linearity of $L_{U}$.

###### Proof.

Note that

$\lvert
L_{U_{1}}\rvert=q^{(k_{1}-1)t}+\ldots+q^{(k_{1}-d_{1})t}+1<q^{k-d_{2}-1}+\ldots+q^{k-d}+1-q^{k_{1}t},$

and then the assertion follows by Theorem 5.2. ∎

######  Remark 5.4.

Other examples of $(d_{2},d)$-minimum size linear set can be obtained by using
the minimum size linear sets constructed in Corollary 4.14 as $L_{U_{1}}$ or
$L_{U_{2}}$ in Theorem 5.2 .

######  Remark 5.5.

It is natural to consider $\mathrm{PG}(d,q^{n})$, $n$ not prime, and wonder
what the maximal value of $d_{2}$ is such that the above corollary implies the
existence of an ${\mathbb{F}}_{q}$-linear set in $\mathrm{PG}(d,q^{n})$ that
is of $(d_{2},d)$-minimum size, but not of $d$-minimum size, and has maximum
geometric field of linearity ${\mathbb{F}}_{q}$. So let $t$ be the largest
proper divisor of $n$. Note that $t\geq\sqrt{n}$. We want to construct a set
$U_{2}=JV_{q,t}(\mu,t;m_{0},\dots,m_{d_{2}})$ with $d_{2}$ maximal, such that
it satisfies the conditions of Theorem 4.5. Hence, there must exist pairwise
coprime polynomials $g_{i}$ of degree $m_{i}-1$ such that
$(m_{0}-1)+\ldots+(m_{d_{2}}-1)\leq t-1$. Let $\delta(x)$ denote the maximum
number of distinct monic irreducible polynomials over ${\mathbb{F}}_{q}$ such
that the sum of their degrees is smaller than $x$. Then for any $m\geq 1$,
$\delta(q^{m})\geq\frac{q^{m}-1}{m}$. Indeed, consider the minimal polynomials
of the elements of ${\mathbb{F}}_{q^{m}}^{*}$. Since every element of
${\mathbb{F}}_{q^{m}}^{*}$ is the root of a unique such polynomial, their
degrees sum to $q^{m}-1$. Furthermore, the maximum degree equals $m$, so there
are at least $\frac{q^{m}-1}{m}$ such polynomials. Hence, to answer the
original question, asymptotically,
$d_{2}=\Omega(t/\log_{q}(t))=\Omega(\sqrt{n}/\log_{q}(n))$.

We conclude this subsection with examples of linear sets of $(1,2)$-minimum
size that are not of $(2,2)$-minimum size and have maximum geometric field of
linearity ${\mathbb{F}}_{q}$.

######  Proposition 5.6.

Let $n=st$, with $s>1$, and $t>2$ prime. Suppose that the smallest prime that
divides $s$ is at least $t$. Let

  * $U_{1}$ be a $k_{1}$-dimensional ${\mathbb{F}}_{q^{t}}$-subspace of ${\mathbb{F}}_{q^{n}}$,

  * $U_{2}=JV_{q,t}(\mu,t;m_{0},m_{1})$, with $t=m_{0}+m_{1}$.

Define $U=U_{1}\times U_{2}$. Then $L_{U}$ is a $(1,2)$-minimum size
${\mathbb{F}}_{q}$-linear set, but not of $2$-minimum size. Moreover,
${\mathbb{F}}_{q}$ is the maximum geometric field of linearity of $L_{U}$.

###### Proof.

By Theorem 5.1, $L_{U}$ is an ${\mathbb{F}}_{q}$-linear set of
$\mathrm{PG}(2,q^{n})$ of rank $(k_{1}+1)t$ having size

$\lvert L_{U}\rvert=q^{(k_{1}+1)t-1}+q^{k_{1}t}+1,$

that is not of $2$-minimum size. Since $L_{U_{2}}$ has a point of weight $1$,
there exists $\varphi\in\mathrm{GL}(2,q^{t})$, such that the
${\mathbb{F}}_{q}$-linear set $L_{U^{\prime}}$, with
$U^{\prime}=U_{1}\times\varphi(U_{2})$ has $E_{2}$ as a point of weight $1$.
Hence ${\mathbb{F}}_{q^{n}}^{3}/E_{2}$ can be identified with
${\mathbb{F}}_{q^{n}}^{2}$ in an obvious way. Clearly, $L_{U}$ and
$L_{U^{\prime}}$ are $\mathrm{GL}(3,q^{n})$-equivalent. In this way,
$U^{\prime}/E_{2}$ can be identified as an ${\mathbb{F}}_{q}$-subspace
$\overline{U}=U_{1}\times U_{2}^{\prime}$, where $U_{2}^{\prime}$ is an
$(t-1)$-dimensional ${\mathbb{F}}_{q}$-subspace of ${\mathbb{F}}_{q^{t}}$.
Again, by Theorem 5.1, we have that $\lvert
L_{\overline{U}}\rvert=q^{k_{1}t}+1$ and hence $L_{U}$ is a $(1,2)$-minimum
size ${\mathbb{F}}_{q}$-linear set. Suppose now, that $L_{U}=L_{W}$ for some
${\mathbb{F}}_{q^{r}}$-linear set $L_{W}$. If $r<t$, then by our hypothesis,
$r$ is coprime with $s$ and $t$, hence $r$ is coprime with $n=st$, and
${\mathbb{F}}_{q^{r}}$ is not a subfield of $\mathbb{F}_{q^{n}}$. Therefore,
$r\geq t$. Let $\ell$ be the line of $\mathrm{PG}(2,q^{n})$ having equation
$X_{0}=0$. Then

$q^{t-1}+1=\lvert L_{U_{2}}\rvert=\lvert\ell\cap L_{U}\rvert=\lvert\ell\cap
L_{W}\rvert.$

Since $\ell\cap L_{W}$ is an ${\mathbb{F}}_{q^{r}}$-linear set we have that
$\lvert\ell\cap L_{W}\rvert\geq q^{r}+1$. So $t-1\geq r$, a contradiction. ∎

######  Remark 5.7.

Choosing $s,t$ be prime numbers with $s>t>2$, Proposition 5.6 gives examples
of ${\mathbb{F}}_{q}$-linear sets in $\mathrm{PG}(2,q^{st})$ having rank $st$
and with size $q^{st}+q^{st-t}+1$. Moreover, for such a linear set, the
maximum geometric field of linearity if ${\mathbb{F}}_{q}$ and there cannot
exist a line meeting it in a subline, since its size is less than
$q^{st}+q^{st-1}+1$.

## Acknowledgment

We would like to thank Jan De Beule, Olga Polverino and Ferdinando Zullo for
fruitful discussions. Paolo Santonastaso is very grateful for the hospitality
of the Department of Mathematics and Data Science, Vrije Universiteit Brussel,
Brussels, Belgium, where he was a visiting PhD student for 2 months during the
preparation of this paper. Paolo Santonastaso was supported by the project
“VALERE: VAnviteLli pEr la RicErca” of the University of Campania “Luigi
Vanvitelli” and by the Italian National Group for Algebraic and Geometric
Structures and their Applications (GNSAGA - INdAM).

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Sam Adriaensen
Vrije Universiteit Brussel
Department of Mathematics and Data Science
Pleinlaan 2, 1050 Elsene, Belgium
<EMAIL_ADDRESS>

Paolo Santonastaso
Università degli Studi della Campania “Luigi Vanvitelli”
Dipartimento di Matematica e Fisica
Viale Lincoln, 5, I– 81100 Caserta, Italy
<EMAIL_ADDRESS>