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\def \PG{\mathrm{PG}}
\def\AG{\mathrm{AG}}
\def\V{\mathrm{V}}
\def\Co{\mathrm{C}}
\def\B{\mathcal{B}}
\def\D{\mathcal{D}}
\def\N{\mathcal{N}}
\def\F{\mathbb{F}}
\def\R{\mathcal{R}}
\def\P{\mathcal{P}}
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\def\K{\mathbb{K}}
\def\Se{\mathbb{S}}
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\def\PGammaL{\mathrm{P}\Gamma\mathrm{L}}
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\title{\bf Scattered linear sets and pseudoreguli}
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\author{M. Lavrauw\\
\small Dipartimento di Tecnica e Gestione dei Sistemi Industriali \\[-0.8ex]
\small Universit\'a degli Studi di Padova\\[-0.8ex]
\small Vicenza, Italy.\\
\small\tt michel.lavrauw@unipd.it\\
\and
Geertrui Van de Voorde\\
\small Departement Wiskunde\\[-0.8ex]
\small Vrije Universiteit Brussel\\[-0.8ex]
\small Brussel, Belgium\\
\small\tt gvdevoor@vub.ac.be
}
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\date{\dateline{Jan 30, 2012}{November 2, 2012}\\
\small Mathematics Subject Classifications: 51E20}
\begin{document}
\maketitle
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\begin{abstract}In this paper, we show that one can associate a pseudoregulus with every scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$. We construct a scattered linear set having a given pseudoregulus as associated pseudoregulus and prove that there are $q-1$ different scattered linear sets that have the same associated pseudoregulus. Finally, we give a characterisation of reguli and pseudoreguli in $\PG(3,q^3)$.
\end{abstract}
\section{Motivation and preliminaries}
\subsection{Motivation}
Linear sets in projective spaces have gained attention in recent years because of their connection with other geometrical structures (e.g. blocking sets, translation ovoids, $\ldots$). For an overview of the use of linear sets in these topics, we refer to \cite{Polverino2010}. The motivation for the study of the particular linear sets studied in this paper arose from the relation between linear sets and finite semifields.
In \cite{Lavrauw2011} it was shown that to any semifield $\mathbb{S}$ of order $q^{nt}$, with left nucleus containing $\F_{q^t}$ and center containing $\F_q$, there corresponds an $\F_q$-linear set of rank $nt$ in the projective space
$\PG(n^2-1,q)$, disjoint from the $(n-2)$-nd secant variety of a Segre variety, and conversely.
This result was previously proved for $n=2$ by Lunardon \cite{Lunardon2003}, and is crucial in the
classification of semifields with $n=2$, $t=2$ obtained in \cite{CaPoTr2006}. It was applied again in \cite{MaPoTr2007}, where the case $n=2,t=3$ is considered, and the authors prove that there exist eight non-isotopic families of such semifields, according to the different configurations of the associated linear sets of $\PG(3,q^3)$. Also, they prove that to any scattered semifield, there is associated an $\F_q$-pseudoregulus of $\PG(3,q^3)$ and they characterise the known examples of scattered semifields in terms of the associated $\F_q$-pseudoregulus. In this paper, we show that one can associate an $\F_q$-pseudoregulus to any scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$.
In the case that $n=3$, this provides a tool to study {\em symplectic} scattered semifields of order $q^9$, with left nucleus containing $\F_{q^3}$ and center containing $\F_q$. (See \cite{LaMaPoTr2011}, for a study of such semifields
when $n=2$.) For more applications of the connection between linear sets and semifields we refer to \cite{LaPo2011} and the references contained therein.
\subsection{Preliminaries}
If $V$ is a vector space, then we denote by $\PG(V)$ the corresponding projective space. If $V$ has dimension $n$ over the finite field $\F_q$ with $q$ elements, then we also write $\PG(n-1,q)$.
Let $V$ be an $r$-dimensional vector space over a finite field ${\mathbb{F}}$.
A set $\L$ of points of $\PG(V)$ is called a {\em linear set} ({\it of rank $t$}) if there exists a subset $U$ of $V$ that forms a ($t$-dimensional) $\F_q$-vector space for some $\F_q \subset {\mathbb{F}}$, such that $\L=\B(U)$, where
$$\B(U):=\{\langle u \rangle_{\mathbb{F}}~:~u \in U\setminus \{0\}\}.$$
If we want to specify the subfield we call $\L$ an {\it $\F_q$-linear set}.
In other words, if ${\mathbb{F}}={\mathbb{F}}_{q^n}$, we have the following diagram
\begin{displaymath}
\begin{array}{ccccccc}
& & {\mathbb{F}}_{q^n}^r & \longleftrightarrow & \F_q^{rn} & \supseteq & U\\
\\
& & \updownarrow & & \updownarrow & & \updownarrow \\
\\
\B(U) & \subseteq &\PG(r-1,q^n) & \longleftrightarrow & \PG(rn-1,q) & \supseteq & \PG(U)\\
\end{array}
\end{displaymath}
We also use the notation $\B(\pi)$ for the set of points of $\PG(r-1,q^n)$ induced by $\pi=\PG(U)$. Since the points of $\PG(r-1,q^n)$ correspond to $1$-dimensional subspaces of ${\mathbb{F}}_{q^n}^r$, and by field reduction to $n$-dimensional subspaces of $ \F_q^{rn}$, they correspond to a set $\D$ of $(n-1)$-dimensional subspaces of $\PG(rn-1,q)$, which partitions the point set of $\PG(rn-1,q)$. The set $\D$ is called a {\it Desarguesian spread}, and we have a one-to-one correspondence between the points of $\PG(r-1,q^n)$ and the elements of $\D$. This gives us a more geometric perspective on the notion of a linear set; namely, an $\F_q$-linear set is a set $\L$ of points of $\PG(r-1,q^n)$ for which there exists a subspace $\pi$ in
$\PG(rn-1,q)$ such that the points of $\L$ correspond to the elements of $\D$ that have a non-empty intersection with $\pi$. Also in what follows, we will often identify the elements of $\D$ with the points of $\PG(r-1,q^n)$, which allows us to view $\B(\pi)$ as a subset of $\D$. To avoid confusion, we denote subspaces of $\PG(r-1,q^n)$ by capital letter and subspaces of $\PG(rn-1,q)$ by lowercase letters. For more on this approach to linear sets, we refer to \cite{Lavrauw2001} and \cite{LaVa2010}.
If the subspace $\pi$ intersects each spread element in at most a point, then $\pi$ is called {\it scattered} with respect to $\D$ (see \cite{Lavrauw2001}, \cite{BlLa2000}). In this case we also call the associated linear set $\B(\pi)$ {\it scattered}.
Note that if $\pi$ is $(t-1)$-dimensional and scattered, then
the associated $\F_q$-linear set $\B(\pi)$ has rank $t$ and has exactly $\frac{q^t-1}{q-1}$ points, and conversely.
In this paper, we will make use of the following bound on the rank of a scattered linear set, which follows from \cite[Theorem 4.3]{BlLa2000}.
\begin{theorem}\label{max}
A scattered $\F_q$-linear set in $\PG(r-1,q^t)$ has rank $\leq rt/2$.
\end{theorem}
\begin{proof}
Immediate from the definition and \cite[Theorem 4.3]{BlLa2000}.
\end{proof}
In this paper, we focus on scattered $\F_q$-linear sets of rank $3n$ in $\PG(2n-1,q^3)$. By Theorem \ref{max}, these scattered linear sets are {\em maximum scattered}.
\section{Projectively equivalent scattered linear sets}
In this section, we show that all scattered $\F_q$-linear sets of rank $3n$ in $\PG(2n-1,q^3)$ are projectively equivalent.
Desarguesian spreads, introduced in the previous section, are well-known and frequently used in finite geometry. We recall another classic construction of a Desarguesian spread based
on the following lemma (see e.g. \cite[Lemma 1]{Lunardon1999}).
\begin{lemma} \label{wie}
A subspace of $\PG(hn-1,q^h)$ of dimension $d$ is fixed by the mapping $x\mapsto x^q$ if and only if it intersects the subgeometry $\PG(hn-1,q)$ in a subspace of dimension $d$.
\end{lemma}
Now, let $\Pi$ be an $(n-1)$-space, disjoint from the subgeometry $\rho=\PG(hn-1,q)$ of $\PG(hn-1,q^h)$, such that $\dim\langle \Pi,\Pi^q,\ldots,\Pi^{q^{h-1}}\rangle$ is maximal, i.e. spans $\PG(hn-1,q^h)$. Let $P$ be a point of $\Pi$ and let $\tau(P)$ denote the $(h-1)$-dimensional subspace generated by the conjugates of $P$, i.e., $\tau(P) = \langle P,P^q,...,P^{q^{h-1}}\rangle$. Then $\tau(P)$ is fixed by $x\mapsto x^q$ and so, by Lemma \ref{wie}, it intersects $\PG(hn-1,q)$ in an $(h-1)$-dimensional subspace over $\F_q$. If we do this for every point of $\Pi$ we obtain a Desarguesian $(h-1)$-spread of $\PG(hn-1,q)$ (see Segre \cite{Segre1964}). For future reference, we denote this spread by $\D(\Pi)$. Moreover, every Desarguesian spread can be constructed this way (\cite{Segre1964}), and all Desarguesian $(h-1)$-spreads in $\PG(hn-1,q)$ are projectively equivalent (see e.g. \cite{BaCo1974}).
In order to prove that the Desarguesian spread $\D(\Pi)$ determines the subspace $\Pi$ up to conjugacy, we need to
introduce the following terminology.
A set $\R$ of $q+1$ mutually disjoint $(n-1)$-dimensional subspaces of $\PG(2n-1,q)$, such that a line meeting $3$ elements of $\R$, meets all elements of $\R$, is called a {\em regulus} (or {\it $(n-1)$-regulus}). A line meeting each element of a regulus $\R$ is called a {\it transversal of $\R$}.
The following theorem is considered as folklore, but, by lack of a reference, we include a proof.
\begin{theorem} \label{th1} If $\D(\Pi_1)=\D(\Pi_2)$, then $\Pi_1$ and $\Pi_2$ are conjugate.
\end{theorem}
\begin{proof} Let $\Pi_1$ and $\Pi_2$ be two different $(n-1)$-spaces determining the spread $\D$, and suppose $\Pi_2$ is not conjugated to $\Pi_1$. Then there exist lines $L$ in $\Pi_1$ and $M$ in $\Pi_2$ such that $L$ and $M$ are not conjugated and they determine the same $(h-1)$-subspread $\D_1\subset \D$ in a $(2h-1)$-space $\tau$.
Let $\bar{X}$ denote the extension of $X \in \D_1$ to a subspace over $\F_{q^h}$. Let $m$ be minimal such that
$M\subset \langle L,L^q,\ldots, L^{q^{m-1}}\rangle$, and choose an $X \in \D_1$ such that
$\{x_i~:~ i=0,\ldots,m\}$ is a frame where
$$
x_i:=\bar{X}\cap L^{q^i},~ i=0,\ldots, m-1~\mbox{and} ~x_m:=\bar{X} \cap M.
$$
Observe that $x_i=x_0^{q^i}$, for $i>0$, and $U:=\langle x_0,\ldots, x_m\rangle$ is the unique $(m-1)$-space through $x_m$ which intersects all lines
$L, L^q,\ldots, L^{q^{m-1}}$. Now choose a line $\ell$ in $\tau$ disjoint from $\bar{X}$, and let ${\mathcal{R}}$ denote the associated regulus induced by the elements $R_0, R_1, \ldots, R_{q}$ of $\D_1$ that intersect $\ell$. Since $x\mapsto x^q$ preserves the regulus $\mathcal R$, it follows that for each $R \in {\mathcal{R}}$ we have $R^q=R$ (when $R\cap \ell \neq \emptyset$) or $R^q\cap R=\emptyset$ (when $R \cap \ell = \emptyset$). Also, the lines $L,L^q, \ldots, L^{q^{m-1}}, M$ are transversals to the regulus, since each such line intersects the elements $R_0,\ldots,R_q$. The uniqueness of $U$ implies that $U \subset R$ for some $R \in {\mathcal{R}}$. But then $x_1=x_0^q \in R\cap R^q$ and $R=R^q$. This implies that $R\cap \ell \neq \emptyset$, and hence $R=R_j$ for some $j \in \{0,\ldots,q\}$. Since
$U\subset R_j\cap \bar{X}$, this implies that $\bar{R_j}=\bar{X}$, contradicting $\ell \cap \bar{X}=\emptyset$.
\end{proof}
The next theorem generalises Proposition 2.7 from \cite{MaPoTr2007}, where the theorem is proved for $n=2$.
\begin{theorem} \label{equivalent} All scattered $\F_q$-linear sets of rank $3n$ in $\PG(2n-1,q^3)$, spanning the whole space, are $\PGammaL$-equivalent.
\end{theorem}
\begin{proof} Let $\L_1$, $\L_2$ be two scattered $\F_q$-linear sets of rank $3n$ in $\PG(2n-1,q^3)$, spanning the whole space. By \cite[Theorem 2]{LaVa2010}, for $i=1,2$, there exist a subgeometry $\rho_i\cong \PG(3n-1,q)$ of $\PG(3n-1,q^3)$, and an $(n-1)$-space $\Pi_i$ in $\PG(3n-1,q^3)$, with $\Pi_i\cap \rho_i= \emptyset$, such that
$$\alpha_i(\L_i)=\{\langle x,\Pi_i\rangle / \Pi_i~:~x \in \rho_i\},$$
for some collineation $\alpha_i$ from $\PG(2n-1,q^3)$ to $\PG(3n-1,q^3)/\Pi_i$.
Suppose $\langle \Pi_i,\Pi_i^q,\Pi_i^{q^2}\rangle$ is a space of dimension $d$. Then projecting the $d$-dimensional subspace $\rho_i\cap \langle \Pi_i,\Pi_i^q,\Pi_i^{q^2}\rangle$ from $\Pi_i$ gives rise to a scattered linear set of rank $d+1$ contained in a projective space $\cong \PG(d-n,q^3)$, and hence $d\geq 3n-1$ by Theorem \ref{max}.
Since all $(3n-1)$-dimensional $\F_q$-subgeometries of $\PG(3n-1,q^3)$ are $\PGL$-equivalent to the canonical subgeometry $\rho=\{\langle (x_0,x_1,\ldots,x_{3n-1})\rangle\vert x_j\in \F_q\}$, there is, for $i=1,2$ an element $\phi_i$ of $\PGL(3n,q^3)$ such that $\phi_i(\rho_i)=\rho$. The set $$\{\langle P,P^q,P^{q^{2}}\rangle \cap \rho\vert P \in \phi_i(\Pi_i)\}, \ i=1,2,$$ is a Desarguesian $2$-spread $\D_i$ of $\rho$. Since all Desarguesian $2$-spreads of $\PG(3n-1,q)$ are projectively equivalent, and, by Theorem \ref{th1}, the spaces $\Pi_i,\Pi_i^q,\Pi_i^{q^2}$ determining $\D_i$ are uniquely determined up to conjugacy, this implies that there is an element $\psi$ of $\PGammaL(3n,q^3)$ such that $\psi(\D_1)=\D_2$ and $\psi(\phi_1(\Pi_1))=\phi_2(\Pi_2)$. Now $\xi=\phi_2^{-1}\circ \psi\circ \phi_1$ is an element of $\PGammaL(3n,q^3)$, and
\begin{eqnarray*} \xi(\rho_1)&=&(\phi_2^{-1}\circ \psi\circ \phi_1)(\rho_1)\\
&=&(\phi_2^{-1}\circ \psi)(\rho)\\
&=&\phi_2^{-1}(\rho)=\rho_2;
\end{eqnarray*}
\begin{eqnarray*} \xi(\Pi_1)&=&(\phi_2^{-1}\circ \psi\circ \phi_1)(\Pi_1)\\
&=&\phi_2^{-1}(\phi_2(\Pi_1))=\Pi_2.
\end{eqnarray*}
Now $\xi$ induces a collineation $\tau$ from $\PG(3n-1,q^3)/\Pi_1$ to $\PG(3n-1,q^3)/\Pi_2$ defined by
$$\tau~:~\langle x,\Pi_1\rangle /\Pi_1 \mapsto \langle \xi(x),\xi(\Pi_1)\rangle /\xi(\Pi_1)
= \langle \xi(x),\Pi_2\rangle /\Pi_2,$$
and
$$\tau(\alpha_1(\L_1))=\{\langle \xi(x),\Pi_2\rangle / \Pi_2~:~x \in \rho_1\}=\{\langle y,\Pi_2\rangle / \Pi_2~:~y \in \rho_2\}=\alpha_2(\L_2).$$
This shows that $\L_1$ and $\L_2$ are $\PGammaL$-equivalent.\end{proof}
\section{Scattered linear sets of rank $3n$ in $\PG(2n-1,q^3)$ and the associated pseudoregulus}
In this section, we show that we can associate a pseudoregulus to a scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$ and that there exist exactly two transversal spaces to this pseudoregulus.
\subsection{The $(q^2+q+1)$-secants to a scattered linear set}
\begin{lemma} \label{le}
Let $\L$ be a scattered $\F_q$-linear set of rank $3n$ in $\PG(2n-1,q^3)$, i.e. $\L=\B(\mu)$, with $\mu$ a $(3n-1)$-space of $\PG(6n-1,q)$.
\begin{itemize}
\item[(i)] A line of $\PG(2n-1,q^3)$ meets $\L$ in $0,1,q+1$ or $q^2+q+1$ points.
\item[(ii)] Every point of $\L$ lies on exactly one $(q^2+q+1)$-secant to $\L$ and two different $(q^2+q+1)$-secants to $\L$ are disjoint.
\item[(iii)] If $\vert L\cap \L\vert=q^2+q+1$ for some line $L$, then $L=\B(\pi)$, for a unique plane $\pi$ contained in $\mu$.
\end{itemize}
\end{lemma}
\begin{proof}
(i) Immediate, since by Theorem \ref{max} every line of $\PG(2n-1,q^3)$ meets a scattered $\F_q$-linear set in a scattered $\F_q$-linear set of rank at most 3.
(ii) By Theorem \ref{max}, $\mu$ is a maximum scattered space. This implies that if $\nu$ is a $3n$-space of $\PG(6n-1,q)$ through $\mu$, then there is at least one line, say $\ell_1$, contained in $\nu$ such that $\ell_1\subset \B(p_1)$, for some $p_1\in \mu$. Now if there is a second line, say $\ell_2$, contained in $\nu$ and $\B(p_2)$ with $p_2\in \mu$, then the $3$-space $\langle \ell_1,\ell_2\rangle$ is contained in $\nu$ and meets $\mu$ in a plane $\pi$. Hence, by part $(i)$, $\langle \B(\ell_1),\B(\ell_2)\rangle$ meets $\B(\mu)$ in exactly $q^2+q+1$ points, the set $\B(\pi)$. If we count the number of pairs $(\ell,\nu)$, where $\ell$ is a line contained in an element of $\B(\mu)$ and $\nu$ is a $3n$-space through $\mu$ containing $\ell$, we get that, on average, such a $3n$-space $\nu$ contains $q+1$ such lines $\ell$.
Now suppose that there is a $3n$-space $\nu$ containing a set $\S$ of more than $q+1$ such lines, say $\S=\{\ell_1,\ell_2,\ldots,\ell_s\}$. If the lines of $\S$ span a subspace of dimension at least 5, then this subspace meets $\mu$ in a scattered space of dimension at least $4$ with respect to a plane-spread in $\PG(8,q)$. By Theorem \ref{max}, this is a contradiction. If the lines of $\S$ span a $4$-dimensional space, then each line of $\S$ intersects $\langle l_1,l_2\rangle$, and hence
$\langle \B(\ell_1),\B(\ell_2), \ldots,\B(\ell_s)\rangle$ corresponds to a line over $\F_{q^3}$ with $q^3+q^2+q+1$ points of $\L$, a contradiction. Hence, all the lines of $\S$ are contained in the $3$-space $\langle \ell_1,\ell_2\rangle$.
But then by \cite[Lemma 10]{LaVa2010}, there are $q^2+1$ lines contained in $\langle \ell_1,\ell_2\rangle$ inducing an $\F_{q^2}$-subline of $\langle \B(\ell_1),\B(\ell_2)\rangle$, and
we get that $2\vert 3$, again a contradiction. This implies that every $3n$-space through $\mu$ contains exactly $q+1$ lines $\ell_i$ with $\ell_i\in \B(p_i)$ for some $p_i\in \mu$, $i=1\ldots q+1$.
Now let $P=\B(r)$ be a point of $\L=\B(\mu)$, where $r\in \mu$. Let $\ell_1$ be a line through $r$ in $\B(r)$, then the $3n$-space $\langle \mu,\ell_1\rangle$ contains $q+1$ lines $\ell_i$ with $\ell_i\in \B(p_i)$, $p_i$ in $\mu$. As seen before, this implies that there is a plane through $r$, contained in $\langle \B(\ell_1),\B(\ell_2)\rangle\cap \mu$, hence $\langle \B(\ell_1),\B(\ell_2)\rangle$ is a $(q^2+q+1)$-secant to $\L$ through $P$. This shows that every point of $\L$ lies on at least one $(q^2+q+1)$-secant.
Suppose that two $(q^2+q+1)$-secants, $M$ and $N$, intersect. Then the plane $\langle M,N\rangle$ meets $\L$ in a scattered linear set of rank at least $5$, contradicting Theorem \ref{max}. This concludes the proof of part (ii).
(iii) This follows from the proof of part (ii) where we have shown that every point of $\mu$ lies on a unique plane $\pi\subset \mu$ such that $\B(\pi)=L\cap \L$, where $L$ is a $(q^2+q+1)$-secant.
\end{proof}
\begin{definition} Let $\L$ be a scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$. In the spirit of the pseudoregulus defined by Freeman in \cite{Freeman1980}, and extending the definition in \cite{MaPoTr2007}, we define the {\em pseudoregulus} $\P$ associated with $\L$ as the set $\P$ of $\frac{q^{3n}-1}{q^3-1}$ lines meeting $\L$ in $q^2+q+1$ points. The set of points lying on the lines of $\P$ is denoted by $\tilde{\P}$. \end{definition}
\subsection{The transversal spaces to a pseudoregulus}
Let $\P$ denote the pseudoregulus associated to a scattered linear set $\L=\B(\mu)$ of rank $3n$ in
$\PG(2n-1,q^3)$.
A subspace whose point set is contained in $\tilde{\P}$ and which intersects all lines of $\P$ in at most a point, is called a {\em transversal space} to the pseudoregulus $\P$. In this section (Theorem \ref{unique}) we prove that there exist exactly two $(n-1)$-dimensional transversal spaces to $\P$.
\begin{lemma} \label{le2}
There exist two disjoint transversal $(n-1)$-spaces to $\P$.
\end{lemma}
\begin{proof} Since $\L$ is a scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$, it can be obtained in the quotient
geometry over an $(n-1)$-space $\Pi$ of $\PG(3n-1,q^3)$
by considering an appropriate subgeometry $\Sigma=\PG(3n-1,q)$ disjoint from $\Pi$ (see \cite[Theorem 2]{LaVa2010}). Since $\L$ is scattered, the space $\langle\Pi,\Pi^q,\Pi^{q^2}\rangle$ is $(3n-1)$-dimensional, as seen in the proof of Theorem \ref{equivalent}. For every $P\in \Pi$, the plane $\langle P,P^q,P^{q^2}\rangle$ meets $\Sigma$ in a subplane $\cong\PG(2,q)$. This implies that the lines $\langle P,P^q,P^{q^2},\Pi\rangle/\Pi$ are exactly the $(q^2+q+1)$-secants to $\L$. Moreover, $\Pi_1:=\langle \Pi^q,\Pi \rangle/\Pi$ and $\Pi_2:=\langle \Pi^{q^2},\Pi\rangle /\Pi$ are two disjoint $(n-1)$-spaces intersecting each of these $(q^2+q+1)$-secants to $\L$, whose point sets are contained in $\tilde{\P}$.
\end{proof}
In what follows, $\Pi_1$ and $\Pi_2$ denote the transversal spaces constructed in Lemma \ref{le2}.
\begin{lemma} \label{drie}
If $P_1,P_2,P_3$ are three collinear points in $\Pi_1$, then the intersection points $Q_i$ of the lines of $\P$ through $P_i$, $i=1,2,3$, with $\Pi_2$ are collinear. Moreover, the only points of $\tilde{\P}$, contained in $\langle P_1,P_2,Q_1,Q_2\rangle$, are the $(q^3+1)^2$ points on the lines of $\P$ in $\langle P_1,P_2,Q_1,Q_2\rangle$.
\end{lemma}
\begin{proof} Let $S_i$ denote the line of $\P$ through $P_i$, $i=1,2$, and put $T_1:=\langle P_1,P_2\rangle$ and $T_2:=\langle Q_1,Q_2\rangle$. By Lemma \ref{le}(iii), the point $P_i$ corresponds to a spread element lying in $\langle \B(\pi_i)\rangle$, with $\pi_i$ a plane of $\mu$, where $\L=\B(\mu)$. Since the subspace $\langle S_1,S_2,S_3\rangle$ has dimension at most 4 and intersects $\L$ in a scattered linear set, it follows from the upper bound on the dimension of the subspace $\langle \pi_1,\pi_2,\pi_3 \rangle$ (Theorem \ref{max}), that there exists a line $\ell$ in $\mu$, meeting $\pi_1,\pi_2$ and $\pi_3$. Hence the line $L:=\langle\B(\ell)\rangle$ meets $S_1,S_2$, and $S_3$, and these lines are contained in the $3$-space $\langle T_1,L\rangle$. Since $\Pi_1$ and $\Pi_2$ are disjoint, $\langle T_1,L\rangle$ meets $\Pi_2$ in the line $T_2$, and hence $Q_1,Q_2$, and $Q_3$ are collinear.
Now, suppose that there is a point $R$ of $\P$, lying in the 3-space $\langle T_1,T_2\rangle$, but not on a line of $\P$ in $\langle T_1,T_2\rangle$, then $R$ lies on a line of $\P$ meeting $\Pi_1$, resp. $\Pi_2$ in a point $R_1$, resp $R_2$, not lying on $T_1$ or $T_2$. But then the planes $\langle T_1,R_1\rangle$, and $\langle T_2, R_2\rangle$ must intersect since both are contained in the 4-space $\langle T_1,T_2,R_1, R_2\rangle$. This contradicts $\Pi_1\cap \Pi_2=\emptyset$.
\end{proof}
\begin{theorem} \label{rechte} All transversal lines to $\P$ lie in one of the transversal spaces $\Pi_1$ or $\Pi_2$.
\end{theorem}
\begin{proof} Suppose that there exists a transversal line $L=R_1R_2$ to $\P$, not in $\Pi_1$ or $\Pi_2$. Let $S_i$ be the line of $\P$ through $R_i$ and let $P_i$, resp. $Q_i$, be the intersection of $S_i$ with $\Pi_1$, resp. $\Pi_2$. It follows from Lemma \ref{drie} that $R_1R_2$ meets the $q^3+1$ lines of $\P$ that are contained in the $3$-space $\rho=\langle P_1,P_2,Q_1,Q_2\rangle$. If $R_1,R_2$ meets $\Pi_1$ or $\Pi_2$, the lines of $\P$ in $\rho$ would intersect, a contradiction. Hence, $P_1P_2,R_1R_2,Q_1Q_2$ are three disjoint lines in $\rho$, defining a regulus $\R$.
By Lemma \ref{le}(iii) the $q^3+1$ lines of $\P$ contained in the 3-dimensional space $\rho$ correspond to $q^3+1$ two by two disjoint planes contained in a $5$-dimensional subspace $\zeta$ of $\mu$, i.e. they form a plane spread of $\zeta$.
Let $P=\B(r)$ be a point of $\L$ on the line $P_1Q_1$ with $r \in \zeta$, then connecting $r$ with the $q^2+q+1$ points of the plane $\pi_2\subset \zeta$ corresponding to the $(q^2+q+1)$-secant $S_2$ shows that there are at least $q^2+q+1$ lines through $P$ meeting at least $q+1$ lines of the regulus $\R$, a contradiction unless $\B(\zeta)$ is a line, which contradicts Theorem \ref{max}.
\end{proof}
\begin{theorem}\label{unique}
There are exactly two $(n-1)$-dimensional transversal spaces to $\P$.
\end{theorem}
\begin{proof} This follows immediately from Lemma \ref{le2} and Theorem \ref{rechte}.
\end{proof}
\subsection{The stabiliser of a pseudoregulus}
\begin{lemma}\label{stab} The stabiliser in $\PGL(2n,q^3)$ of the pseudoregulus $\P$ in $\PG(2n-1,q^3)$ acts transitively on the points of a line of $\P$ that do not lie on one of the transversal $(n-1)$-spaces to $\P$.
\end{lemma}
\begin{proof} Let $\Pi_1$ and $\Pi_2$ be the transversal spaces to the pseudoregulus $\P$ and let $P$ be a point on one of the lines $L$ of $\P$ but not contained in $\Pi_i$, $i=1,2$. Let $P_1,\ldots,P_{2n+1}$ be the points of a standard frame of $\PG(2n-1,q^3)$, chosen in such a way that $P_1,\ldots, P_n$ lie in $\Pi_1$, $P_{n+1}\ldots,P_{2n}$ lie in $\Pi_2$ and $P=P_{2n+1}$. It follows that the intersection point $Q_1$ of the line $L$ with $\Pi_1$ is $\langle e_1+\ldots+e_n\rangle$ and the intersection point $Q_2$ of the line $L$ with $\Pi_2$ is $\langle e_{n+1}+\ldots+e_{2n}\rangle$. If $Q$ is a point on $L$, different from $Q_1,Q_2$, then $Q$ has coordinates $\langle e_1+\ldots+e_n+s(e_n+\ldots+e_{2n})\rangle$. It is easy to check that the element $\phi$ of $\PGL(2n,q^3)$ corresponding to the matrix $A=(a_{ij})$, with $a_{ij}=0$ if $i\neq j$, $a_{ii}=1$ if $1\leq i\leq n$ and $a_{ii}=s$ if $n+1\leq i\leq 2n$, stabilises $\P$ and maps $P$ onto $Q$.
\end{proof}
\section{The reconstruction of a linear set having a fixed pseudoregulus}
If $\L$ is a scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$, then we have seen in the previous section that there exists a unique associated pseudoregulus $\P$. The aim of this section is to construct a scattered linear set of rank $3n$ having a given pseudoregulus $\P$ as associated pseudoregulus, and show that there are $q-1$ different scattered linear sets of rank $3n$ giving rise to the same pseudoregulus $\P$.
\begin{theorem} \label{th3}Let $\L$ be a scattered linear set of rank $3n$ in $\PG(2n-1,q^3)$.
\begin{itemize}
\item[(i)] A plane meets $\L$ in $0$, $1$, $q+1$, $q^2+q+1$ or $q^3+q^2+q+1$ points.
\item[(ii)] A plane $\Gamma$ meeting $\L$ in $q^3+q^2+q+1$ points contains exactly one line with $q^2+q+1$ points of $\L$.
\end{itemize}
\end{theorem}
\begin{proof}
(i) Immediate, since a plane meets the scattered linear set $\L$ in a scattered linear set of rank at most 4, by Theorem \ref{max}.
(ii) In this case, the plane $\Gamma$ meets $\L$ in a set $\B(\rho)$, where $\rho$ has dimension $3$. Since a line of $\Gamma$ corresponds to a $5$-space in $\PG(8,q)$ and a $3$-space and $5$-space always meet in $\PG(8,q)$, all lines of $\Gamma$ meet $\L$ in at least one point. If we denote the number of lines in $\Gamma$ meeting $\L$ in $i$ points by $\ell_i$, we get that $\sum_i \ell_i=q^6+q^3+1$, $\sum_i i\ell_i=(q^3+q^2+q+1)(q^3+1)$ and $\sum_i i(i-1)\ell_i=(q^3+q^2+q+1)(q^3+q^2+q)$.
If we suppose that all lines meet in $1$ or $q+1$ points, then we obtain that $\sum_i(i-1)(i-(q+1))\ell_i=0$, a contradiction if we use the previously found values for $\sum_i \ell_i$, $\sum_i i\ell_i$ and $\sum_i i(i-1)\ell_i$. Hence, there is a line meeting $\L$ in more than $q+1$ points, which then, by Lemma \ref{le}(i), meets $\L$ in $q^2+q+1$ points. Suppose that $L_1$ and $L_2$ are two different lines in $\Pi$ meeting $\L$ in $q^2+q+1$ points, then there would be two intersecting $(q^2+q+1)$-secants to $\L$, a contradiction by Lemma \ref{le}(ii).
\end{proof}
\begin{remark} In the case that $n=2$, every plane meets $\L$ in $q^2+q+1$ points or $q^3+q^2+q+1$ points. This follows also from \cite[Theorem 2.4]{BlLa2000}. \end{remark}
Let us fix some more notation. Let $\P$ denote a pseudoregulus in $\PG(2n-1,q^3)$ corresponding to the scattered linear set $\L$ of rank $3n$.
Let $\mu$ be a $(3n-1)$-space such that $\B(\mu)=\L$. A $(q^2+q+1)$-secant to $\L$ defines a $5$-space in $\PG(6n-1,q)$ meeting $\mu$ in a plane. Since every point of $\L$ lies on a unique $(q^2+q+1)$-secant by Lemma \ref{le}(ii), the $(q^{3n}-1)/(q^3-1)$ planes defined in this way determine a spread of $\mu$. Let us denote this spread by $\Sigma$.
\begin{lemma} \label{SigmaDes}The spread $\Sigma$ is Desarguesian.
\end{lemma}
\begin{proof}
As in the proof of Lemma \ref{le2}, we see that $\L$ is the projection of a subgeometry $\rho=\PG(3n-1,q)$ of $\PG(3n-1,q^3)$ from an $(n-1)$-space $\Pi$ onto $\PG(2n-1,q^3)$, and the planes $\langle P,P^q,P^{q^2}\rangle$, with $P$ a point from $\Pi$ form a Desarguesian spread $\D$ in $\rho$. If we now return to the spread representation, we get that $\mu$ is the projection of $\rho$ from the $(3n-1)$-space $\langle\B(\Pi)\rangle$. Every plane $\langle P,P^q,P^{q^2}\rangle$, with $P$ on $\Pi$ corresponds to an $8$-dimensional space, meeting $\rho$ in a plane of $\D$. The projection of this $8$-space from $\langle \B(\Pi)\rangle$ is a $5$-space $\lambda$, meeting $\mu$ in a plane. Since $\lambda$ corresponds to a $(q^2+q+1)$-secant, this plane is an element of the spread $\Sigma$. This shows that $\Sigma$ is the projection of the Desarguesian spread $\D$, from which the statement follows (see e.g. \cite[Theorem 1.5.4]{Lavrauw2001}).\end{proof}
\begin{lemma} \label{three} If $\pi_1,\pi_2,\pi_3$ are planes of $\Sigma$ defining a regulus with elements $\pi_1,\ldots,\pi_{q+1}$, then the $5$-spaces $\langle \B(\pi_1)\rangle,\langle \B(\pi_2)\rangle,\langle \B(\pi_3)\rangle$ determine the regulus with elements $\langle \B(\pi_i)\rangle$, $i=1,\ldots,q+1$.
\end{lemma}
\begin{proof}
Each plane $\pi_i,$ $i=1,\ldots, q+1$, is contained in some element of the regulus
defined by $\langle \B(\pi_1)\rangle,\langle \B(\pi_2)\rangle,\langle \B(\pi_3)\rangle$, since a line $\ell$ through $\pi_1,\pi_2$ and $\pi_3$ meets the elements of the regulus defined by $\pi_1,\pi_2,\pi_3$, say $\ell\cap \pi_i=\{p_i\}$.
Now $\B(p_1),\B(p_2)$ and $\B(p_3)$ form a regulus of the Desarguesian spread $\D$, and the other spread elements in this regulus are $\B(p_i)$. Since a line meeting $\B(p_i)$, $i=1,2,3$ meets $\B(p_i)$ for all $i=1,\ldots,q+1$, $\B(p_i)$ is contained in some element of the regulus defined by $\langle \B(\pi_1)\rangle,\langle \B(\pi_2)\rangle,\langle \B(\pi_3)\rangle$. Since $\pi_i$ and $\B(p_i)$ meet in a point, $\langle \pi_i,\B(p_i)\rangle$ is contained in an element of this regulus. The same reasoning holds for a different transversal line $\ell'$, meeting $\pi_i$ in a point $p_i'$, and hence $\langle \pi_i,\B(p_i')\rangle$ is contained in an element of this regulus. This implies that $\langle \B(\pi_i)\rangle$ is an element of the regulus defined by $\langle \B(\pi_1)\rangle,\langle \B(\pi_2)\rangle,\langle \B(\pi_3)\rangle$.
\end{proof}
\begin{lemma} \label{full}Let $q>2$. A set of points $\S$ in $\PG(1,q^3)$, $\vert \S \vert \geq 3$, such that the subline through any $3$ of them is contained in $\S$ is either a subline or a full line.
\end{lemma}
\begin{proof} Let $\D$ be the Desarguesian $2$-spread of $\PG(5,q)$ obtained from $\PG(1,q^3)$. Suppose $\S$ has at least $q+2$ points, and let $\rho_1,\ldots,\rho_{q+1}$ be the regulus corresponding to a $(q+1)$-secant to $\S$ and let $\rho_{q+2}$ be a spread element, not in this regulus, corresponding to a point of $\S$. Let $\ell_1$ be the transversal line through the point $p_1$ of $\rho_1$ to the regulus $\rho_1,\rho_2,\ldots,\rho_{q+1}$. Let $\ell_2$ be the transversal line through $p_1$ of the regulus through $\rho_1,\rho_2$ and $\rho_{q+2}$, then $\B(\ell_2)\subset \S$ by the hypothesis. We will now show that $\B(\langle \ell_1,\ell_2\rangle)\subset \S$. The plane $\langle\ell_1,\ell_2\rangle$ meets $\rho_2$ in a line $m$. Now every line $n$ in $\langle \ell_1,\ell_2\rangle$, not through any of the three points $\ell_1\cap m$, $\ell_2\cap m$, $\ell_1\cap \ell_2$, meets $\ell_1,\ell_2$ and $m$ in a point, and hence, $\B(n)$ contains $3$ elements of $\S$. This implies that $\B(n)\subset \S$ for all such lines $n$. Since $q>2$, all lines through one of the intersection points of $m,\ell_1$ and $\ell_2$ now contain at least $3$ points of $\S$, hence, this argument shows that $\B(\langle \ell_1,\ell_2\rangle)\subset \S$. If $\S=\B(\langle \ell_1,\ell_2\rangle)$, then this linear set is a linear set of size $q^2+1$ in $\PG(1,q^3)$, which is not isomorphic to $\PG(1,q^2)$. By Corollary 13 of \cite{LaVa2010}, through two points of such a linear set, there is exactly one subline that is completely contained in this linear set, a contradiction by our assumption on $\S$. Hence, there is an element $\rho_{q+3}$ of $\S$, not in $\B(\langle \ell_1,\ell_2\rangle)$. Repeating the same argument with a transversal line $\ell_3$ through $\rho_1,\rho_2$ and $\rho_{q+3}$ and a line of $\langle \ell_1,\ell_2\rangle$ shows that $\B(\langle \ell_1,\ell_2,\ell_3\rangle)\subset \S$, hence, $\S$ is a full line.
\end{proof}
\begin{theorem} \label{interrechte} Let $q>2$. A line $L$ in $\PG(2n-1,q^3)$ meets the point set $\tilde{\P}$ of a pseudoregulus $\P$ in $0,1,2,q+1$ or $q^3+1$ points. If $\vert L\cap \tilde{\P}\vert=q+1$, then $L$ meets $\tilde{P}$ in a subline.
\end{theorem}
\begin{proof} Let $L$ be a line meeting $3$ points of $\tilde{\P}$, say $P_1,P_2,P_3$, and suppose that the points $P_1,P_2,P_3$ are not contained in the same line of $\P$. Let $\rho_1,\rho_2,\rho_3$ be the corresponding spread elements, then they determine $3$ elements of $\Sigma$, say $\pi_1$, $\pi_2$, $\pi_3$, and $\rho_i\in \langle \B(\pi_i)\rangle$. A line through $\rho_1,\rho_2,\rho_3$ meets $\langle \B(\pi_1)\rangle,\langle \B(\pi_2)\rangle$ and $\langle \B(\pi_3)\rangle$, and by Lemma \ref{three}, also $\langle \B(\pi_i)\rangle$, $i=4,\ldots,q+1$.
From this, it follows that the line $L$ meets $\tilde{\P}$ in a set of points $\mathcal{K}$ such that the subline through any 3 of them is contained in $\mathcal{K}$. Such a set is either a subline, or a full line by Lemma \ref{full}.
\end{proof}
%
\begin{lemma} \label{lemdir} Let $q>2$. Let $\tilde{\S}$ be the point set of a set $\S$ of $q^3+1$ mutually disjoint lines in $\PG(3,q^3)$ with the property that the subline through $3$ collinear points of $\tilde{\S}$ is contained in $\tilde{\S}$. Then a plane $\Pi$ through a line $L$ of $\S$ contains $q^3$ points of $\tilde{\S}$, not on $L$ and this set of $q^3$ points determines a set $D$ of either 1 or $q^2+q+1$ directions on $L$. Moreover, $I\cup D=\B(\nu)$, where $\nu$ is a $3$-space of $\PG(11,q)$.
\end{lemma}
\begin{proof} Since the lines of $\S$ are mutually disjoint, the plane $\Pi$ meets the $q^3$ lines of $\S$, different from $L$ in a point. Let $I=\{P_1,\ldots,P_{q^3}\}$ this set of $q^3$ points. Let $D=\{D_1,\ldots,D_d\}$ be the set of directions determined by the set $I$. We claim that $d=1$ or $d=q^2+q+1$ and that the set $I\cup D$ is an $\F_q$-linear set of rank $4$.
Let $\rho_i$ be the spread element corresponding to $P_i$. If the $q^3$ points in $I$ are collinear, say they lie on the line $M$, then we are in the first case and $\B(\nu)=M=I\cup D$ for all $3$-spaces contained in $\langle \rho_1,\rho_2\rangle$. Otherwise, every line in $\Pi$, different from the line $L$ meets $\tilde{\S}$ in $0,1,2$ or $q+1$ points by Lemma \ref{full}. The line through $P_i$ and $P_j$, $j\neq i$, meets $L$, and hence, contains a third point of $\tilde{\S}$, say $R_{ij}$. It follows that $P_iP_j$ meets $\tilde{\S}$ in $q+1$ points, forming a subline. Let $\ell_i$ be the transversal line through a point $p_1$ of $\rho_1$ to the regulus defined by $\rho_1$, $\rho_i$ and the spread element corresponding to $R_{1i}$. We claim that $\B(\langle \ell_2,\ell_3\rangle)\subset \tilde{\S}$. Each line $m$ in $\langle \ell_2,\ell_3\rangle$, for which the points $\B(\ell_2\cap m)$, $\B(\ell_3\cap m)$ and $\langle \B(m)\rangle \cap L$ are different points of $\tilde{\S}$, induces the subline $\B(m)$ contained in $\tilde{\S}$ and since $q>2$, repeating this argument for the other lines in $\langle \ell_2,\ell_3\rangle$ and $m$ implies that $\B(\langle \ell_2,\ell_3\rangle)\subset \tilde{S}$. Similarly, we get that $\B(\langle \ell_i,\ell_j\rangle)\subset \tilde{\S}$ for all $i\neq j>1$, hence $\nu:=\langle \ell_2,\ell_3,\ell_4,\ldots\rangle\subset \tilde{S}$, and $\nu$ is a $3$-dimensional space, since $\vert I \vert=q^3$. If a spread element $\rho$ would intersect $ \nu $ in more than a point, every line in $\Pi$ through the point corresponding to $\rho$ and a point of $\tilde{S}$, would contain more than $q+1$ points of $\tilde{\S}$, a contradiction. From this, it follows that $\B(\nu)$ is scattered, hence, there are $q^2+q+1$ determined directions.
\end{proof}
%
\begin{lemma} \label{directions} Let $q>2$. A plane through a line $L$ of a pseudoregulus $\P$ and a point of $\tilde{\P}$, outside $L$ meets $q^3$ other lines of $\P$ in a point, and this set of $q^3$ points determines either 1 or $q^2+q+1$ directions on $L$.
\end{lemma}
\begin{proof} Let $\Pi$ be a plane through one of the lines $L$ of $\P$, and the point $R$ of $\tilde{\P}$, not on $L$. Let $M$ be the line of $\P$ through $R$. From Lemma \ref{drie}, we get that there are exactly $q^3+1$ lines of $\P$ in $\langle L,M\rangle$, and $\langle L,M\rangle$
does not contain other points of $\tilde{\P}$. Hence, $\Pi$ meets exactly $q^3$ of the lines of $\P$ in a point. The statement now follows from Lemma \ref{lemdir}.
\end{proof}
\begin{lemma}\label{counting} Let $q>2$. If $P$ is a point of $\tilde{\P}$, not on the transversal spaces $\Pi_1$ and $\Pi_2$, then the number of $(q+1)$-secants to $\tilde{\P}$ through $P$ is $q^2(q^{3n-3}-1)/(q-1)$.
Moreover, if $\L\ni P$ is a linear set with $\P$ as associated pseudoregulus, then each $(q+1)$-secant of $\P$ through $P$ is a $(q+1)$-secant to $\L$.
\end{lemma}
\begin{proof} By Lemma \ref{stab}, we may assume that the point $P$ is contained in the linear set $\L$ defining $\tilde{\P}$. Now $\vert \L\vert=(q^{3n}-1)/(q-1)$ and $P$ lies on a unique $(q^2+q+1)$-secant to $\L$, namely the line $S_1$ of $\P$ through $P$, hence, there are $q^2(q^{3n-3}-1)/(q-1)$ $(q+1)$-secants through $P$ to $\L$, which are necessarily also $(q+1)$-secants to $\tilde{\P}$ by Theorem \ref{rechte} and Theorem \ref{interrechte}. Suppose now that there is a $(q+1)$-secant $M$ through $P$ to $\tilde{\P}$ which is not a $(q+1)$-secant to $\L$.
Then a plane $\langle P,S_2\rangle$, with $S_2$ a line of $\P$ through a point of $M$ different from $P$, contains $q^3$ points of $\L\cap \tilde{\P}$, not on $S_2$, and $q$ points of $M$, the plane $\langle P,S_2\rangle$ contains more than $q^3+q^2+q+1$ points of $\tilde{\P}$, a contradiction by Lemma \ref{directions}.
\end{proof}
\begin{lemma}\label{subplane} Let $q>2$. Let $L_1$ and $L_2$ be two $(q+1)$-secants to $\tilde{\P}$ through a point $P$ of $\tilde{\P}$. Then the subplane, defined by the intersection of $L_1$ and $L_2$ with $\tilde{\P}$ is contained in $\tilde{\P}$.
\end{lemma}
\begin{proof} By Lemma \ref{stab}, we may assume that the point $P$ is contained in the linear set $\L$ defining $\P$, and from Lemma \ref{counting}, we get that the $(q+1)$-secants to $\L$ through $P$ are the $(q+1)$-secants to $\tilde{\P}$. Hence, the subplane, defined by the intersection of $L_1$ and $L_2$ with $\tilde{\P}$, is the subplane defined by the intersection of $L_1$ and $L_2$ with the linear set $\L$. This subplane is entirely contained in $\L$, hence, in $\tilde{\P}$.
\end{proof}
In the following theorem, we show, given a pseudoregulus, how to construct a linear set defining this pseudoregulus.
\begin{theorem} Let $q>2$. Let $\P$ be a pseudoregulus in $\PG(2n-1,q^3)$, let $P$ be a point of $\tilde{\P}$, on the line $L$ of $\P$, not lying on one of the transversal spaces to $\P$. Let $T=\{L_1,L_2,\ldots \}$ be the set of $(q+1)$-secants through $P$ to $\tilde{\P}$, let $P(T)$ be the set of points on the lines of $T$ in $\tilde{\P}$. Let $\Pi_i$ be the plane $\langle L,L_i\rangle$, and let $D_i$ be the set of directions on $L$, determined by the intersection of $\Pi_i$ with $\tilde{\P}$. Then $D_i=D_1$, for all $i$, and $P(T)$, together with the points of $D_1$ form a linear set $\L$ of rank $3n$ determining the pseudoregulus $\P$.
\end{theorem}
\begin{proof} By Lemma \ref{counting}, there are $q^2(q^{3n-3}-1)/(q-1)$ lines in $T$, each defining a subline through $P$, that is contained in $\tilde{P}$. In the spread representation, this implies that there are $q^2(q^{3n-3}-1)/(q-1)$ lines $\ell_i$ through a point $x$ of the spread element corresponding to $P$, such that $\B(\ell_i) \subset \tilde{\P}$. By Lemma \ref{subplane}, $\B(\langle \ell_i,\ell_j\rangle)\subset \tilde{\P}$, and since the number of $(q+1)$-secants through $P$ is exactly $q^2(q^{3n-3}-1)/(q-1)$, this implies that $\nu:=\langle \ell_1,\ell_2,\ldots\rangle$ is a subspace of dimension $3n-1$. Then $P(T)\subset \B(\nu)$, by construction.
Each plane $\langle L,L_i\rangle$ contains $q^3$ points of $\tilde{\P}$ and $q^2$ $(q+1)$-secants $\langle \B(\ell_{i_1})\rangle$, $\langle \B(\ell_{i_2})\rangle, \ldots,$ $\langle \B(\ell_{i_{q^2}})\rangle$
through $P$, and determines a set $D_i$ of directions on $L$. The lines $\ell_{i_1}, \ldots, \ell_{i_{q^2}}$
span a subspace $\nu_i$ of $\nu$ and each direction of $D_i$ is of the form $\B(y)$, for some $y\in \nu_i$, and hence each set of directions $D_i$ on $L$ determined by the points of $P(T)$ is contained in $\B(\nu)\cap L$.
Since $\B(\nu)$ intersects $L$ in a linear set, and each $D_i$ contains at least $q^2+q+1$ points, by Lemma \ref{directions}, $\B(\nu)\cap L$ is a linear set of rank at least $3$. On the other hand, since $\B(\nu)$ contains
the $(q^{3n}-q^3)/(q-1)$ points of $P(T)$ and $\nu$ has dimension $3n-1$, it follows that $\B(\nu)$ is
a scattered linear set $\L$ of rank $3n$ and $\L\cap L=D_i$.
The scattered linear set $\L$ of rank $3n$ defines a pseudoregulus $\P(\L)$, so we need to show that $\P=\P(\L)$. The $(q^{3n}-1)/(q-1)$ points of $\L$ all lie on one of the lines of $\P$, hence, a line of $\P$ has on average $q^2+q+1$ points of $\L$, and by Lemma \ref{le}(i), it is not possible that one of the lines of $\P$ contains more than $q^2+q+1$ points of $\L$. This implies that $\P=\P(\L)$.
\end{proof}
\begin{corollary} Let $q>2$. If $\P$ is a pseudoregulus, then there are $q-1$ scattered linear sets having $\P$ as associated pseudoregulus.
\end{corollary}
\begin{proof} Counting the number of couples $(P,\L)$, where $P$ is a point of the pseudoregulus, not on one of the transversal spaces and $\L$ is a scattered linear set through $P$ having $\P$ as pseudoregulus yields that the number of scattered linear sets having $\P$ as pseudoregulus is equal to $\frac{q^{3n}-1}{q^3-1}(q^3-1)\frac{q-1}{q^{3n}-1}$.
\end{proof}
\section{A characterisation of reguli and pseudoreguli in $\PG(3,q^3)$}
\begin{theorem} Let $q>2$. Let $\tilde{\S}$ be the point set of a set $\mathcal{S}$ of $q^3+1$ mutually disjoint lines in $\PG(3,q^3)$ such that the subline defined by three collinear points of $\tilde{\S}$ is contained in $\tilde{\S}$, then $\S$ is a regulus or pseudoregulus.
\end{theorem}
\begin{proof} By Lemma \ref{full}, a line meets $\tilde{\S}$ in $0,1,2,q+1$ or $q^3+1$ points.
{\bf Case 1: Suppose first that every line meets $\tilde{\S}$ in $0,1,2$ or $q^3+1$ points}.
Let $L$ be a line of $\S$ and let $\Pi$ be a plane through $L$. Since $\Pi$ meets all lines of $\S$ and all lines of $\S$ are disjoint, there are exactly $q^3$ points of $\tilde{\S}$ in $\Pi$, not on $L$. Let $P$ and $Q$ be two points of $\tilde{\S}\setminus L$ in $\Pi$. Since the line $PQ$ has to contain $q^3$ points of $\tilde{\S}\setminus L$, the $q^3$ points of $\tilde{\S}$ in $\Pi$ are collinear. In this way, we find a line $\notin \S$ contained in $\tilde{\S}$, in every of the $q^3+1$ planes through $L$. If two of those lines meet, then the lines of $\S$ would not be disjoint, a contradiction. Hence, we find a set of $q^3+1$ mutually disjoint lines $\S'$, meeting the lines of $\S$. This shows that $\S$ is the opposite regulus to $\S'$ and vice versa.
{\bf Case 2: There is a line $M$ meeting $\tilde{\S}$ in exactly $q+1$ points}. Let $P$ be a point of $M$, let $L_0$ be the line of $\S$ through $P$ and let $L_1,\ldots,L_{q^3}$ be the other lines of $\S$. A plane $\langle L_i,P\rangle$, $i=1,\ldots,q^3,$ meets $q^3$ points of $\tilde{\S}$ that do not lie on $L_i$. Suppose that in one of the planes, these $q^3$ points are collinear, say on $N$, then the plane $\langle M,N\rangle$ meets $q$ lines of $\S$ in $2$ different points, a contradiction since the lines of $\S$ are mutually disjoint. By Lemma \ref{lemdir}, this implies that in every plane $\langle P,L_i\rangle$, there are exactly $q^2+q+1$ $(q+1)$-secants through $P$. Let $p$ be a point of the spread element corresponding to $P$. By Lemma \ref{lemdir}, there is a $3$-space $\nu_i$ such that $\B(\nu_i)\subset \langle P,L_i\rangle \cap \tilde{S}$; w.l.o.g. we may choose $\nu_i$ through $p$. Let $\mu_i$ be the plane $\nu_i\cap \langle \B(L_i)\rangle$. The $3$-space $\nu_i$ is the unique $3$-space through $p$ such that $\B(\nu_i)\subset \langle P,L_i\rangle \cap \tilde{S}$ since $pr_j$, with $r_j\in \mu_i$, is the unique transversal line to the regulus $\langle P,\B(r_j)\rangle \cap \tilde{S}$.
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The $q^3$ planes $\mu_1,\ldots,\mu_{q^3}$ are mutually disjoint and satisfy the condition that the line $\langle p,x\rangle$, where $x$ is a point on one of the planes $\mu_i$, corresponds to a subline contained in $\tilde{\S}$. We get that the $3$-space $\langle p,\mu_i\rangle$ intersects the plane $\mu_j$ for all $j$ non-trivially, and hence, since the planes $\mu_i$ are mutually disjoint, $\langle p,\mu_i\rangle$ and $\mu_j$ meet in a point if $i\neq j$. This implies that $\langle p,\mu_1,\mu_2\rangle$ is $5$-dimensional.
We will prove that $\langle p,\mu_1,\ldots,\mu_{q^3}\rangle$ is $5$-dimensional. W.l.o.g. suppose that $\mu_3$ does not go through the line $\langle p,\mu_1\rangle \cap \langle p,\mu_2\rangle$.
It is clear that the space $\rho:=\langle p,\mu_1,\mu_2,\mu_3\rangle$ is at most $6$-dimensional, so assume that $\rho$ is $6$-dimensional. Since every plane $\mu_i$ has to meet the spaces $\langle p,\mu_1\rangle$, $\langle p,\mu_2\rangle$, and $\langle p,\mu_3\rangle$, it is clear that if $\mu_i$ is not going through one of the $3$ lines $\ell_1:=\langle p,\mu_1\rangle\cap \langle p,\mu_2\rangle$, $\ell_2:=\langle p,\mu_1\rangle\cap\langle p,\mu_3\rangle$ or $\ell_3:=\langle p,\mu_2\rangle\cap\langle p,\mu_3\rangle$, then $\mu_i$ is contained in $\langle p,\mu_1,\mu_2,\mu_3\rangle$. This means that at least $q^3-3q$ planes $\mu_i$ are in $\rho$. Let $\mu_i$ be a plane, through one of the lines $\ell_j$, $j=1,2,3$. Repeating the same argument with $3$ planes in $\rho$ such that $\mu_i$ is not on the intersection lines of the cones defined by $p$ and these $3$ planes shows that all planes $\mu_i$ are contained in $\rho$.
Now let $m$ be a line through $p$, such that $\langle \B(m)\rangle$ is not the line $L_0$. Suppose that $m$ does not meet any of the planes $\mu_i$. There are $q^4+q^2+q+1$ planes through $m$ in $\rho$ and there are $q^3(q^2+q+1)$ points in $\rho$ contained in one of the planes $\mu_i$. This implies that there is a plane $\nu$ through $m$ containing at least $3$ points lying on one of the planes $\mu_i$. Since $m$ does not meet any of the planes $\mu_i$, the $3$ points belong to different planes, say $\mu_1$, $\mu_2$ and $\mu_3$. Hence, in the plane $\nu$, there are $3$ lines $n_1,n_2,n_3$ through $p$ such that $\B(n_i)$ is contained in $\tilde{S}$. Let $n_4$ be a line meeting $n_1,n_2,n_3$ in different points. As $\B(n_4)$ is a subline containing $3$ points of $\tilde{S}$, $\B(\pi_4)$ is contained in $\tilde{S}$. This implies that the intersection point $p':=n_4\cap m$ has necessarily $\B(p')$ contained in a line, say $L_1$ of $\S$. Since we have assumed that $p'$ is not on one of the planes $\mu_i$, $p'$ does not lie on $\mu_1$ and the $3$-space $\langle p',\mu\rangle$ is contained in $\langle \B(L_1)\rangle \cap \rho$, which means that $L_1$ is entirely contained in $\B(\rho)$. Repeating the same argument for a line meeting $n_1,n_2,n_3$ in three distinct points and meeting $n_4$ in a point $p''$, different from $p'$ shows that, if $p''$ is not on $\mu_i$, there is a second line of $\S$, say $L_2$ contained in $\B(\rho)$. But then $L_1\cap \B(\rho)=\sigma_1$ and $L_2\cap \B(\rho)=\sigma_2$ with $\sigma_1$ and $\sigma_2$ three-spaces in the $6$-space $\rho$. Since $\sigma_1$ and $\sigma_2$ necessarily meet in a point, the lines $L_1$ and $L_2$ meet in a point, a contradiction. This implies that every line through $p$ in $\rho$ such that $\langle \B(m)\rangle$ is not the line $L_0$, meets one of the planes $\mu_i$. There are at least $q^5+q^4+q^3$ such lines, but as there are only $q^3$ planes and every line through a point of $p$ and a point of a plane $\mu_i$ contains $q$ points, lying on a plane $\mu_i$, the number of these lines is exactly $q^2(q^2+q+1)$, a contradiction. Hence, $\rho$ is $5$-dimensional.
Let $r$ be a point of the $5$-space $\rho$, not on one of the $q^3$ planes $\mu_i$, then there is a line through $r$ meeting at least $3$ different planes of $\{ \mu_i\vert i=1,\ldots,q^3\}$. This gives rise to a subline meeting $3$ points of $\tilde{S}$, hence, contained in $\tilde{S}$, which implies that $\B(r)$ is on the line $L_0$. We conclude that $\rho$ meets the space $\langle \B(L_0)\rangle$ in a plane.
Now $\rho$ is scattered: suppose that there is a spread element $\B(\pi)$ meeting $\rho$ in a subspace $\pi$ of dimension at least one, then every line through $\B(\pi)$ would contain $q^2+1$ points of $\tilde{S}$, a contradiction. As seen in Lemma \ref{le}, the scattered linear set $\rho$ of rank $6$ defines a pseudoregulus in $\PG(3,q^3)$ and the lines of $\S$ are the $(q^2+q+1)$-secants to $\B(\rho)$, hence, $\S$ is the associated pseudoregulus.
\end{proof}
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