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\title{Intersections of the  Hermitian surface with irreducible
  quadrics in even characteristic}
\author{A.~Aguglia\footnote{Dipartimento di Meccanica, Matematica e Management, Politecnico di Bari,
  Via Orabona 4, I-70126 Bari, Italy}
,\,  \,L.~Giuzzi\footnote{DICATAM, Section of Mathematics, Universit\`a di
Brescia, Via Branze 43, I-25123 Brescia, Italy}}
 %\thanks{Research supported by  the Italian
 %   Ministry MIUR, Strutture geometriche, combinatoria e loro
 %   applicazioni.}}
\date{}
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\begin{document}

\maketitle


\begin{abstract}
We determine the possible intersection sizes of a Hermitian surface $\cH$
with an irreducible quadric of $\PG(3,q^2)$
sharing at least a tangent plane at a common non-singular point
when $q$ is even.
\end{abstract}
{\bf Keywords}: Hermitian surface; quadrics; functional codes.
\par\noindent
{\bf MSC(2010)}: 05B25; 51D20; 51E20.

\section{Introduction}
The study of intersections of geometric objects is a classical problem
in geometry; see e.g. \cite{EH,fulton}.
In the case of combinatorial geometry, it has
several possible applications either
to characterize configurations
or to construct new codes.
\par
Let $\cC$ be a projective $[n,k]$-linear code over $\GF(q)$.
It is always possible to consider the set of points
$\Omega$ in $\PG(k-1,q)$ whose coordinates correspond to the
columns of any generating matrix for $\cC$. Under this setup
the problem of determining the minimum weight of $\cC$ can be
reinterpreted, in a purely geometric setting, as finding
the largest hyperplane sections of $\Omega$.
More in detail, any codeword $c\in\cC$ corresponds to a linear functional
evaluated on the points of $\Omega$; see \cite{L,tvn}.
For examples of applications of these techniques see \cite{IG1,IG2,IG3}.

Clearly, it is not necessary to restrict the study to hyperplanes.
The higher weights of $\cC$ correspond to sections of $\cC$
with subspaces of codimension larger than $1$; see \cite{HTV} and
also \cite{tv} for  Hermitian varieties.

A different generalization consists in studying codes arising
from the evaluation
on $\Omega$ of functionals of degree $t>1$; see \cite{L}.
These constructions  yield, once more, linear codes, whose
weight distributions depend on the intersection patterns of $\Omega$ with
all possible algebraic hypersurfaces of $\PG(k-1,q)$ of degree $t$.

The case of quadratic functional codes on Hermitian varieties has been
extensively investigated in recent years; see \cite{BBFS,E1,EHRS,E2,E3,HS}.
It is however still an open problem to classify all possible intersection
numbers and patterns
between a quadric surface $\cQ$ in $\PG(3,q^2)$ and
a Hermitian surface $\cH=\cH(3,q^2)$.

In \cite{AG}, we determined the possible intersection
numbers between $\cQ$ and  $\cH$ in $\PG(3,q^2)$ under
the assumption that $q$ is an odd prime power and $\cQ$ and $\cH$
share at least one tangent plane.
The same problem has been studied independently also in \cite{CP} for
$\cQ$ an elliptic quadric; this latter work contains
also  some results for $q$ even.

In this paper we fully extend the arguments of \cite{AG} to the case of
$q$ even. It turns out that the geometric properties being considered, as
well as the algebraic conditions to impose, are
different and more involved than in the odd $q$ case.
Our main result is contained in the following theorem.
\begin{theorem}\label{main1}
In $\PG(3,q^2)$, with $q$ even, let $\cH$ and $\cQ$
respectively be a Hermitian surface and
an irreducible quadric sharing at least a tangent plane at  one common
non-singular point $P$.
Then, the possible sizes of the intersection $\cH\cap\cQ$ are
as follows.
\begin{itemize}
\item For $\cQ$ elliptic:
\[q^3-q^2+1,q^3-q^2+q+1, q^3-q+1, q^3+1, q^3+q+1, q^3+q^2-q+1, q^3 + q^2+1. \]
\item For $\cQ$ a quadratic cone:
\[ q^3-q^2+q+1,  q^3-q+1, q^3+q+1, q^3+q^2-q+1,   q^3 + 2q^2-q+1.\]
\item For $\cQ$ hyperbolic:
\[ q^2+1,q^3-q^2+1,  q^3-q^2+q+1, q^3-q+1, q^3+1,  q^3+q+1,
q^3+q^2-q+1, q^3+ q^2+1, \]
\[   q^3+2q^2-q+1,  q^3+2q^2+1,  q^3+3q^2-q+1,   2q^3+q^2+1.  \]
\end{itemize}
\end{theorem}
We remark that, as we are dealing with irreducible quadrics in $PG(3,q^2)$, by quadratic cone
(or, in short, cones) we shall always mean in dimension $3$ the quadric projecting an irreducible
conic contained in a plane $\pi$ from a point (vertex) $V\not\in\pi$.

%In dimension $4$, we shall consider several possibilities as either
%\begin{itemize}
 % \item $V$ is a line, $\cC$ is a conic in a plane $\pi$ disjoint from $V$ and
 %   the cone is the set of all lines projecting $\cC$ from $V$, or
  %\item $V$ is a point, $\cC$ is a non-singular quadric in a $3$-dimensional subspace $\Sigma$
   % and the cone is the set of all lines projecting $\cC$ from $V$. Clearly, here there
    %are $2$ different cases according as $\cC$ is elliptic or hyperbolic.
%\end{itemize}
Our methods are mostly algebraic in nature, based upon
the $\GF(q)$--linear representation of vector spaces over $\GF(q^2)$,
but in order to  rule out some cases some geometric and combinatorial
arguments are needed, as well as some considerations on the action of the
unitary groups.

For generalities on Hermitian varieties in projective spaces the reader is referred to
\cite{BK, PGOFF, HT, S}.
Basic notions on quadrics over finite fields are found in \cite{HT,PGOFF}.

\section{Invariants of quadrics}
\label{s2}

In this section we recall some basic invariants of quadrics in
even characteristic; our main reference for these results is \cite[\S 1.1, 1.2]{HT},
whose notation and approach we closely follow.

Recall that a quadric $\cQ$ in $\PG(n,q)$ is just the set of points
$(x_0,\ldots,x_n)\in\PG(n,q)$ such that $F(x_0,\ldots,x_n)=0$ for
some non-null quadratic form
\[ F(x_0,\ldots,x_n)=\sum_{i=0}^n a_ix_i^2+\sum_{i<j} a_{ij}x_ix_j. \]
If there is no change of coordinates reducing $F$ to a form in
fewer variables, then $\cQ$ is called \emph{non-degenerate} or \emph{non-singular}; otherwise
$\cQ$ is said to be \emph{degenerate} or \emph{singular}. If the polynomial $F$ splits into
linear factors in the algebraic closure of $\GF(q)$,
then $\cQ$ is \emph{reducible}; otherwise
$\cQ$ is \emph{irreducible}. It is well known that if $\cQ$ is reducible,
then $F(x_1,\ldots,x_n)=L_1(x_1,\ldots,x_n)L_2(x_1,\ldots,x_n)$ with
$L_1$ and $L_2$ linear polynomials defined over $\GF(q^2)$.

The minimum number of indeterminates
which may appear in an equation for $\cQ$ is the \emph{rank} of
the quadric, denoted by $\rank(\cQ)$; see \cite[\S 15.3]{FPS3D}.


Suppose $q$ even and
consider the  quadric $\cQ$ in $\PG(3,q)$  of equation $\sum_{i=0}^3 a_ix_i^2+\sum_{i<j} a_{ij}x_ix_j=0$;
define
\[ A:=\begin{pmatrix}
   2a_0 & a_{01} & a_{02} & a_{03} \\
   a_{01} & 2a_{1} & a_{12} & a_{13} \\
   a_{02} & a_{12} & 2a_{2} & a_{23} \\
   a_{03} & a_{13} & a_{23} & 2a_{3}
   \end{pmatrix},\quad
   B:=\begin{pmatrix}
   0  & a_{01} & a_{02} & a_{03} \\
   -a_{01} & 0 & a_{12} & a_{13} \\
   -a_{02} & -a_{12} & 0 & a_{23} \\
   -a_{03} & -a_{13} & -a_{23} & 0
   \end{pmatrix} \]
and, for $\det B\neq0$,
\begin{equation}\label{eqAlpha} \alpha:=
  \frac{\det A-\det B}{4\det B}.
\end{equation}
%We regard $\alpha$ as a rational function in $a_{0},\ldots,a_{23},a_3$; this expression
%makes sense also when the formal entries $a_0,\ldots,a_{23},a_3$ are replaced
%with elements of $\GF(q)$ with $q$ even and it provides an invariant for the quadric $\cQ$.
%  Indeed in this case
% \[ \alpha=\frac{(a_{01}^2a_2+a_0a_{12}^2+a_{01}a_{02}a_{12}+a_{02}^2a_1)a_3+a_0a_1a_{23}^2+
%     (a_0a_{12}a_{13}+a_{01}a_{03}a_{12}+a_{02}a_{03}a_1)a_{23}+
%     (a_0a_{13}^2+a_{01}a_{03}a_{13}+a_{03}^2a_1)a_2}{(a_{01}a_{23}+a_{02}a_{13}+a_{03}a_{12})^2}. \]

The values
 $\det A$, $\det B$ and $\alpha$ should be interpreted as follows. In $A$ and $B$ we replace the terms
 $a_i$ and  $a_{ij}$ by indeterminates  $Z_i$ and $Z_{ij}$ and we evaluate  $\det A$, $\det B$ and $\alpha$
 as rational functions over the integer ring $\mathbb{Z}$. Then we specialize $Z_i$ and  $Z_{i}$ to $a_i$ and $a_{i,j}$.
Furthermore, as $q$ is even, the quadric $\cQ$ induces a  symplectic polarity which is non-degenerate if,
and only if, $\det B\neq 0$ (this is actually equivalent to $\det A\neq 0$ in odd projective dimension), so the formula~\eqref{eqAlpha} giving the invariant
$\alpha$ is well defined for non-singular quadrics.

By \cite[Theorem 1.2]{HT}, a non-singular quadric
$\cQ$ of $PG(3,2^h)$ is hyperbolic or elliptic
according as
\[ \Tr_{q}(\alpha)=0 \mbox{ or } \Tr_{q}(\alpha)=1, \]
respectively, where
$\Tr_{q}$ denotes the absolute trace $\GF(q)\to \GF(2)$ which maps $x\in \GF(q)$
to $x+x^2+x^{2^2}+\ldots+x^{2^{h-1}}$.




\section{Some technical tools}
In this section we are going to prove a series of  lemmas that shall
be useful to prove our main result, namely  Theorem \ref{main1}.


Henceforth, we shall always assume $q$ to be even; $x,y,z$ will denote affine coordinates
in $\AG(3,q^2)$ and the corresponding homogeneous coordinates will be
$J,X,Y,Z$. The hyperplane at infinity
of $\AG(3,q^2)$, denoted as $\Sigma_{\infty}$,  is taken with equation
$J=0$.

Since all non-degenerate Hermitian surfaces of $\PG(3,q^2)$ are projectively
equivalent,
we can assume, without loss of generality,
$\cH$ to have affine equation
\begin{equation}\label{eqH}
z^q+z=x^{q+1}+y^{q+1}.
\end{equation}
Since $\PGU(4,q)$ is transitive on $\cH$, see \cite[\S 35]{S}, we can also suppose that a point
where $\cH$ and $\cQ$ have a common tangent plane is
$P=P_{\infty}(0,0,0,1) \in\cH$; so, the tangent plane at $P_{\infty}$ to $\cH$ and $\cQ$
is $\Sigma_{\infty}$.
Under the aforementioned assumptions,   $\cQ$
has affine equation of the form
\begin{equation}\label{eqQ}
z=ax^{2}+by^{2}+cxy+dx+ey+f
\end{equation}
with $a,b,c,d,e,f\in\GF(q^2)$.
%where $c\neq 0$.
A straightforward computation proves that $\cQ$ is non-singular if and only if $c\neq0$; furthermore
$\cQ$ is hyperbolic or elliptic according as the value of
$$\Tr_{q^2}(ab/c^2)$$ is $0$ or $1$ respectively.
When $c=0$ and $(a,b)\neq (0,0)$, the quadric $\cQ$ is a cone with vertex
a single point $V$.
Write now \begin{equation} \label{infy}
\cC_{\infty}:=\cQ\cap\cH\cap\Sigma_{\infty}.\end{equation}
If $\cQ$ is elliptic,
the point $P_{\infty}$ is, clearly, the only point at infinity of
$\cQ\cap\cH$, that is
$\cC_{\infty}=\{P_{\infty}\}$.
The nature of
$\cC_{\infty}$ when $\cQ$ is either hyperbolic or a  cone, is
detailed by the following lemma.
\begin{lemma}
\label{l0}
 If $\cQ$ is a  cone, then $\cC_{\infty}$
  consists of either $1$ point or $q^2+1$ points on a line.
  When $\cQ$ is a hyperbolic quadric, then $\cC_{\infty}$ consists of
  either $1$ point, or $q^2+1$ points on a line or $2q^2+1$ points on two lines. All cases may
  actually occur.
\end{lemma}
\begin{proof}
As both $\cH\cap\Sigma_{\infty}$ and $\cQ\cap\Sigma_{\infty}$ split in
lines through $P_{\infty}$, it is straightforward to
see that the only possibilities for $\cC_{\infty}$ are
those outlined above; in particular, when $\cQ$ is hyperbolic,
 $\cC_{\infty}$  consists of either $1$ point or $1$ or $2$ lines.
It is straightforward to see that all cases may actually occur, as given
any two lines $\ell,m$ in $\PG(3,q^2)$ there always exist
at least one hyperbolic quadric containing both $m$ and $\ell$.
Likewise, given a line $\ell\in\Sigma_{\infty}$ with $P\in\ell$
there always is at least one cone with vertex $V\in\ell$ and $V\neq P$
meeting $\Sigma_{\infty}$ just in $\ell$.
\end{proof}
%We now provide some explicit examples yielding the various configurations
%of Lemma \ref{l0}.
%\begin{enumerate}[a)]
%\item Suppose  $\cQ$ to be a hyperbolic quadric,
%that is $c\neq 0$, and $\Tr_{q^2}(ab/c^2)=0$.
%\begin{enumerate}[1.]
%\item
%If $a=b=0$, then  $\cC_{\infty}=\{P_{\infty}\}$.
%\item
%If $b=0$ and $a^{q+1}=c^{q+1}$, then
%$\cC_{\infty}$ consists of $q^2+1$ points on the line
%$J=0=aX+cY$.
%\item
 % Finally, if $a\neq 0$, $b=\beta a$, $c=(\beta+1)a$ where $\beta\neq 1$
  %and $\beta^{q+1}=1$,
  %then   $\cC_{\infty}$ consists of the $2q^2+1$ points on the union of the lines
%$J=0=X+Y$ and
%$J=0=X+\beta Y$.
%\end{enumerate}
%\item
%Suppose now that  $\cQ$ is a cone; thus $c=0$ and $(a,b)\neq (0,0)$.
%\begin{enumerate}
%\item
%If $a^{q+1}\neq b^{q+1}$, then $\cC_{\infty}=\{P_{\infty}\}$.
%\item
%When  $a=b$ the conic $\cC_{\infty}$ consists of $q^2+1$ points on the line
%$J=0=X+Y.$
%\end{enumerate}
%\end{enumerate}


Now we are going to use  the same group theoretical
arguments as in \cite[Lemma 2.3]{AG}
in order to be able to fix
the values of some of the parameters in~\eqref{eqQ} without
losing in generality.
\begin{lemma}\label{main4}
  If $\cQ$ is a hyperbolic quadric, we can assume without
  loss of generality:
  \begin{enumerate}
  \item $b=0$, and $a^{q+1}\neq c^{q+1}$
    when $\cC_{\infty}$ is just the point $P_\infty$;
  \item $b=0$, $a=c$ when $\cC_{\infty}$ is a line;
  \item $b=\beta a$, $c=(\beta+1)a$,
    $a\neq 0$ and $\beta^{q+1}=1$, with $\beta\neq 1$ when $\cC_{\infty}$ is the union of
    two lines.
  \end{enumerate}
  If $\cQ$ is a cone, we can assume without loss of generality:
  \begin{enumerate}
    \item $b=0$ when $\cC_{\infty}$ is a point;
    \item $a=b$ when $\cC_{\infty}$ is a line.
    \end{enumerate}
\end{lemma}
\begin{proof}
  Let $\Lambda$ be the set of all lines of $\Sigma_{\infty}$ through
  $P_{\infty}$.
  The action of the stabilizer $G$ of $P_{\infty}$ in $\PGU(4,q)$
  on $\Lambda$
  is the same as the action
  of $\PGU(2,q)$ on the points of $\PG(1,q^2)$.
  This can be easily seen by considering the action on $\PGU(2,q)$
  on the line $\ell$ spanned by $(0,1,0,0)$ and $(0,0,1,0)$ fixing the
  equation $X^{q+1}+Y^{q+1}=0$. Indeed, if $M$ is a $2\times 2$
  matrix representing any $\sigma\in\PGU(2,q)$, then
  $M':=\begin{pmatrix}
    1 & 0 & 0 \\
    0 & M & 0 \\
    0 & 0 & 1
  \end{pmatrix}$
  represents an element of $\PGU(4,q)$ fixing $P_{\infty}=(0,0,0,1)$.
    The action of $\PGU(2,q)$ on $\ell$ is analyzed in detail in \cite[\S 42]{S}.
  So, we see that the group $G$ has
  two orbits on $\Lambda$, say $\Lambda_1$ and
  $\Lambda_2$ where $\Lambda_1$ consists of the
  totally isotropic lines of $\cH$ through $P_{\infty}$
  while $\Lambda_2$ contains
  the remaining $q^2-q$ lines of $\Sigma_{\infty}$ through $P_{\infty}$.
  Furthermore, $G$ is doubly transitive on
  $\Lambda_1$ and the stabilizer of any $m\in\Lambda_1$ is
  transitive on $\Lambda_2$.

  Let now $\cQ_{\infty}=\cQ\cap\Sigma_{\infty}$.
  If $\cQ$ is hyperbolic and $\cC_{\infty}=\{P_{\infty}\}$ we can assume
  $\cQ_{\infty}$ to be the union of the line $\ell:J=X=0$ and
  another line, say $u:J=aX+cY=0$ with $a^{q+1}\neq c^{q+1}$. Thus, $b=0$.


  Otherwise,
  up to the choice of a suitable element $\sigma\in G$,
  we can always take
  $\cQ_{\infty}$ as the union of
  any two lines in $\{\ell,s,t \}$
  where
  \[ \ell\!:\, J=X=0, \qquad s\!:\, J=X+Y=0,\qquad
  t\!:\,J= X+\beta Y=0
  \]
  with $\beta^{q+1}=1$ and $\beta\neq 1$.


  Actually,  when $\cC_{\infty}$ contains just
  one line we take $\cQ_{\infty}: X(X+Y)=0$,
  while if $\cC_{\infty}$ is the union of two lines we have
  $\cQ_{\infty}: (X+Y)(X+\beta Y)=0$.
  When $\cQ$ is a cone, we get either $\cQ_{\infty}: X^2=0$  or $\cQ_{\infty}: (X+Y)^2=0$.
  The lemma follows.
  \end{proof}

\section{Proof of Theorem~\ref{main1}}

%We use the same setup as in the previous section.
%Lemma~\ref{main2}  gives the possible values of $N$ that is, the intersection sizes of the quadric
%$\cQ$ and the non-singular Hermitian variety $\cH$  in $\AG(3,q^2)$. Recall that  we have computed $N$ as the difference
%$|\Xi|-|\Xi_{\infty}|$ where $\Xi$ is the quadric of $\PG(4,q)$ of  equation~\eqref{eqeven1},  whereas $\Xi_{\infty}$ is its section at infinity.
We use the same setup as in the previous section. Thus, the Hermitian surface $\cH$ has equation~\eqref{eqH} whereas  the quadric $\cQ$ has  equation~\eqref{eqQ}.
We  first  determine the number of affine points that $\cQ$ and $\cH$ have in common, that is the size of $(\cQ \cap \cH )\setminus \cC_{\infty}$, where $\cC_{\infty}$ is defined in~\eqref{infy}. Hence we  study the following system of equations
\begin{dmath}
  \label{sis1}
  \begin{cases}
    z^q+z=x^{q+1}+y^{q+1}     \\
       z=ax^{2}+by^{2}+cxy+dx+ey+f.
    \end{cases}
  \end{dmath}
In order to
solve~\eqref{sis1}, recover the value of $z$ from the second
equation and substitute it in the first. This gives
\begin{dmath}
\label{eqc}
a^qx^{2q}+b^qy^{2q}+c^qx^qy^q+d^qx^q+e^qy^q+f^q
+ax^{2}+by^{2}\\+cxy+dx+ey+f
=x^{q+1}+y^{q+1}.
\end{dmath}
Consider $\GF(q^2)$ as a vector space over $\GF(q)$ and fix a
basis $\{1,\varepsilon\}$ with $\varepsilon\in\GF(q^2)\setminus
\GF(q)$. Write any element in $\GF(q^2)$ as a linear
combination with respect to this basis, that is, for any $x\in\GF(q^2)$
let $x=x_0+x_1\varepsilon$, where
$x_0,x_1\in\GF(q)$.
Analogously write also $a=a_0+\varepsilon a_1$,
$b=b_0+\varepsilon b_1$ and so on.
Thus,~\eqref{eqc} can be studied as a quadratic equation over
$\GF(q)$ in the indeterminates $x_0,x_1,y_0,y_1$.

As $q$ is even, it is always possible to choose
$\varepsilon \in \GF(q^2) \setminus\GF(q)$
such that $\varepsilon^2+\varepsilon+\nu=0$, for some
$\nu \in \GF(q)\setminus \{1\}$ and $\Tr_{q}(\nu)=1$.
Then, also, $\varepsilon^{2q}+\varepsilon^q+\nu=0$. Therefore,
$(\varepsilon^q+\varepsilon)^2+(\varepsilon^q+\varepsilon)=0$,
whence $\varepsilon^q+\varepsilon+1=0$. With
this choice of $\varepsilon$,~\eqref{eqc} reads as
\begin{dmath}
\label{eqeven1}
(a_1+1)x_0^2+x_0x_1+[a_0+(1+\nu)a_1+\nu]x_1^2+(b_1+1)y_0^2+y_0y_1
+[b_0+(1+\nu)b_1+\nu]y_1^2+c_1x_0y_0+(c_0+c_1)x_0y_1+(c_0+c_1)x_1y_0
+[c_0+(1+\nu)c_1]x_1y_1
+d_1x_0+(d_0+d_1)x_1+e_1y_0+(e_0+e_1)y_1+f_1=0.
\end{dmath}
As~\eqref{eqeven1} is a non-homogeneous quadratic equation in $(x_0,x_1,y_0,y_1)$,
its solutions
correspond to the affine points of a (possibly degenerate)
 quadratic hypersurface $\Xi$ of $\PG(4,q)$.
Recall that
the number $N$ of affine points of $\Xi$ equals
the number of points of $\cH\cap\cQ$ which lie in
$\AG(3,q^2)$; we shall use the formulas of
\cite[\S 1.5]{HT} in order to actually count the number of these points.

To this purpose, we  first  determine the number of  points
at infinity of $\Xi$.
These points are those of the quadric $\Xi_\infty$ of
$\PG(3,q)$ with equation
\begin{dmath}\label{eq00}
f(x_0,x_1,y_0,y_1)=(a_1+1)x_0^2+x_0x_1+[a_0+(1+\nu)a_1+\nu]x_1^2+(b_1+1)y_0^2+y_0y_1
+[b_0+(1+\nu)b_1+\nu]y_1^2+c_1x_0y_0+(c_0+c_1)x_0y_1+(c_0+c_1)x_1y_0
+[c_0+(1+\nu)c_1]x_1y_1=0.
\end{dmath}

Following the approach outlined in Section \ref{s2}, we  write the matrix associated to $\cQ_{\infty}$
\begin{equation}\label{matr}
 A_{\infty}=\begin{pmatrix}
    2(a_1+1) &1&c_1& c_0+c_1 \\
1 & 2[a_0+(1+\nu)a_1+\nu]  & c_0+c_1&c_0+(1+\nu)c_1  \\
   c_1 & c_0+c_1& 2(b_1+1) &1 & \\
    c_0+c_1&c_0+(1+\nu)c_1&1 & 2[b_0+(1+\nu)b_1+\nu]
   \end{pmatrix}
   \end{equation}
As $q$ is even, a direct computation gives  \[\det A_{\infty}=1+c^{2(q+1)};\] so, the quadric  $\Xi_{\infty}$ is  non-singular  if and only if
$\det A_{\infty}\neq 0$, that is $c^{q+1}\neq1$.





\begin{lemma}\label{hyp}
If $\cQ$ is a cone, then
 $\rank(\Xi_{\infty})=4$.
When $\cQ$ is non-singular then $\rank(\Xi_\infty)\geq 2$ and
if $\rank(\Xi_{\infty})=2$, then
 the quadric  $\cQ$  is hyperbolic.
\end{lemma}
\begin{proof}
Let $\cQ$ be a cone,  namely $c=0$. It turns out that $\det A_{\infty}\neq 0$ and hence
 $\rank(\Xi_{\infty})=4$.

Now assume  that  $\cQ$ is non-singular.
If the equation  of $\Xi_{\infty}$ were to be of the form
$f(x_0,x_1,y_0,y_1)=(lx_0+mx_1+ny_0+ry_1)^2$ with $l,m,n,r$ over some extension of $\GF(q)$,
then $c=0$; this is a contradiction. So $\rank(\Xi_\infty)\geq 2$.
Finally, suppose
 $\rank(\Xi_{\infty})=2$, that is $\Xi_{\infty}$ splits into two planes. We need to prove
 $\Tr_{q^2}(ab/c^2)=0$.
 First observe that $c^{q+1}=1$ since the quadric $\Xi_{\infty}$ is degenerate.

Consider now the following $4$
intersections  $\cC_0:\Xi_{\infty}\cap [x_0=0] $, $\cC_1:\Xi_{\infty}\cap [x_1=0] $,
$\cC_2:\Xi_{\infty}\cap [y_0=0]$, $\cC_3:\Xi_{\infty}\cap [y_1=0]$.
Clearly, as $\Xi_{\infty}$ is, by assumption, reducible in the union of two planes,
all of these conics are  degenerate; thus
we get the following four formal equations
\[ \frac{1}{2}\det\begin{pmatrix}
 2[a_0+(1+\nu)a_1+\nu]  & c_0+c_1& c_0+(1+\nu)c_1  \\
    c_0+c_1& 2(b_1+1) &1 & \\
    c_0+(1+\nu)c_1&1 & 2[b_0+(1+\nu)b_1+\nu]
   \end{pmatrix}=0, \]
\[ \frac{1}{2}\det\begin{pmatrix}
    2(a_1+1) &c_1& c_0+c_1 \\
   c_1 & 2(b_1+1) &1 & \\
    c_0+c_1&1 & 2[b_0+(1+\nu)b_1+\nu]
   \end{pmatrix}=0, \]
\[ \frac{1}{2}\det\begin{pmatrix}
    2(a_1+1) &1& c_0+c_1 \\
1 & 2[a_0+(1+\nu)a_1+\nu]  &c_0+(1+\nu)c_1  \\
       c_0+c_1&[c_0+(1+\nu)c_1]& 2[b_0+(1+\nu)b_1+\nu]
   \end{pmatrix}=0,\]
\[\frac{1}{2}\det\begin{pmatrix}
    2(a_1+1) &1&c_1\\
1 & 2[a_0+(1+\nu)a_1+\nu]  & c_0+c_1  \\
   c_1 & c_0+c_1& 2(b_1+1)
   \end{pmatrix}=0. \]
Using the condition $c^{q+1}=1$, these give
\begin{equation}\label{sis2}
\begin{cases}
 a_0+(1+\nu)a_1+(c_0^2+c_1^2)b_0+\nu (c_0^2+c_1^2+c_1^2\nu)b_1=0\\
  a_1+c_1^2b_0+(c_0^2+\nu c_1^2)b_1=0\\
  (c_0^2+c_1^2)a_0+\nu(c_0^2+c_1^2+\nu c_1^2)a_1+b_0+(1+\nu)b_1=0\\
  c_1^2a_0+(c_0^2+\nu c_1^2)a_1+b_1=0.
\end{cases}
\end{equation}
Since $c^{q+1}=1$,  if $c_1=0$, then $c_0=1$.
Solving~\eqref{sis2}, we obtain
 \begin{equation}\label{eq6}
 \begin{cases}
   a_1=b_1 \\
   a_0+a_1+b_0=0.
  \end{cases}
 \end{equation}
Therefore   $\Tr_{q^2}(ab/c^2)=\Tr_{q^2}((a_0+\varepsilon a_1)(a_0+(\varepsilon+1) a_1))=\Tr_{q^2}(a^{q+1})=0$ as $a^{q+1}\in\GF(q)$.

Suppose now  $c_1\neq 0$.
Then $c^{q+1}=(c_0^2+c_0c_1+\nu c_1^2)=1$ and, after some elementary algebraic manipulations,
 System~\eqref{sis2} becomes
\begin{equation}
\label{eq7}
\begin{cases}
a_0=\left(\frac{c_0^2}{c_1^2}+\nu\right)a_1+\frac{b_1}{c_1^2}\\
b_0=\frac{a_1}{c_1^2}+\left(\frac{c_0^2}{c_1^2}+\nu\right)b_1;\\
\end{cases}
\end{equation}
 hence,
\[\frac{ab}{c^2}=\frac{(a_1^2+b_1^2)(c_1^2\nu+c_1^2\varepsilon+c_0^2)+a_1b_1(c_1^2\nu+c_1^2\varepsilon+c_0^2+1)^2}{c_1^4(c_0+\varepsilon c_1)^2}.\]
Since $ \varepsilon^2=\varepsilon +\nu $ and $c_1^2\nu+c_0^2+1=c_0c_1$, we get $$\frac{ab}{c^2}=\frac{a_1^2+b_1^2}{c_1^4}+\frac{a_1b_1}{c_1^2}\in\GF(q),$$
which gives $\Tr_{q^2}(ab/c^2)=0$ once more.
 Hence if $\rank(\Xi_{\infty})=2$, then $\cQ$ is hyperbolic.
\end{proof}


\begin{lemma}
\label{2:5}
Suppose $\cQ$ to be a hyperbolic quadric with $\cC_{\infty}$ being  the union
  of two lines. If
$\rank(\Xi_{\infty})=2$, then $\Xi_{\infty}=\Pi_1\cup\Pi_2$ is a plane pair over $\GF(q)$.
\end{lemma}
\begin{proof}
By Lemma \ref{main4} we can assume that $b=\beta a$, $c=(\beta+1)a$, $a\neq 0$ and $\beta^{q+1}=1$ with $\beta\neq 1$.

Furthermore, since $\rank(\Xi_{\infty})=2$, we have
$\Xi=\Pi_1\cup\Pi_2$ where the planes
$\Pi_1$ and $\Pi_2$ have respectively equations
$lx_0+mx_1+ny_0+ry_1=0$ and
$l'x_0+m'x_1+n'y_0+r'y_1=0$,
for some values of $l,m,n,r$ and $l',m',n',r'$ in $\GF(q^2)$.
Clearly, in this case,
\begin{equation}
\label{eq8}
f(x_0,x_1,y_0,y_1)=(lx_0+mx_1+ny_0+ry_1)(l'x_0+m'x_1+n'y_0+r'y_1).
\end{equation}

Then, up to a scalar multiple, the following must be satisfied:
\begin{equation}
\label{eqsys}
\begin{cases}
  ll'=a_1+1\\
  lm'+l'm=1 \\
  l'n+ln'=c_1 \\
  l'r+lr'=c_0+c_1 \\
  mm'=a_0+(1+\nu)a_1+\nu\\
  mn'+nm'=c_0+c_1\\
  mr'+rm'=c_0+(1+\nu)c_1\\
  nr'+rn'=1\\
  nn'=b_1+1\\
  rr'=b_0+(1+\nu)b_1+\nu.
\end{cases}
\end{equation}

If $c_1=0$, then $c_0=1$ as $c^{q+1}=1$; in particular, as $c=b+a$, we have
$a_0+b_0=c_0=1$ and, consequently, as~\eqref{eq6} holds  we get $a_1=1=b_1$.
So System~\eqref{eqsys} becomes
\[
\begin{cases}
  ll'=0\\
  lm'+l'm=1 \\
  l'n+ln'=0 \\
  l'r+lr'=1 \\
  mm'=a_0+1\\
  mn'+nm'=1\\
  mr'+rm'=1\\
  nr'+rn'=1\\
  nn'=0\\
  rr'=b_0+1.
\end{cases}
\]
We can assume without loss of generality
either $l=0$ or $l'=0$. Suppose the former; then $l'\neq0$ and
$m={l'\,}^{-1}$. We also have $n=0$ and $r={l'\,}^{-1}$.
It follows that
$\Pi_1$ is the plane of equation $x_1+y_1=0$. In particular, $\Pi_1$ is
defined over $\GF(q)$ and, consequently, also  $\Pi_2$ is. If $l'=0$, an
analogous argument leads to $\Pi_2:x_1+y_1=0$ and, once more, $\Xi_{\infty}$
splits into two planes defined over $\GF(q)$.

Now suppose
$c_1\neq 0$. From~\eqref{eqsys} we get
\begin{equation}\label{eqprime}\begin{cases}
    ll'=a_1+1 \\
    ln'+l'n=c_1 \\
    nn'=b_1+1 \\
    mm'=a_0+(1+\nu)a_1+\nu \\
    mr'+rm'=c_0+(1+\nu)c_1 \\
    rr'=b_0+(1+\nu)b_1+\nu \\
    nr'+rn'=1 \\
    lm'+l'm=1.
  \end{cases}\end{equation}
We obtain $ll'+nn'=a_1+b_1=c_1=ln'+l'n$ and
$mm'+rr'=c_0+(1+\nu)c_1=mr'+rm'$.
Hence,
\[ (l'+n')(l+n)=0,\qquad (m+r)(m'+r')=0. \]
There are the following two cases to consider:
\begin{enumerate}
\item  $l=n$ and $m=r$ or, equivalently,
  $l'=n'$ and $m'=r'$
\item  $l=n$ and $m'=r'$ or, equivalently,
  $l'=n'$ and $m=r$.
\end{enumerate}
Suppose first $l=n$ and $m=r$; then
$n(n'+l')=c_1\neq 0$; consequently, $n=l\neq 0$ and also $n'\neq l'$.
If $m=0$, then $\Pi_1$ has equation $x_0+x_1=0$ and is defined over
$\GF(q)$; then also $\Pi_2$ is defined over $\GF(q)$ and we are done.


Suppose now $m\neq 0$ (and hence $r\neq 0$). We claim $l/m\in\GF(q)$. This would give that
$\Pi_1$ is defined over $\GF(q)$, whence the thesis.
From~\eqref{eqprime} we have
\[ \begin{cases}
    n'=\frac{b_1+1}{n} \\
    r'=\frac{b_0+(1+\nu)b_1+\nu}{r} \\
    nr'+rn'=1.
    \end{cases} \]
Replacing the values of $n'$ and $r'$ in the last equation we obtain
\begin{equation}
\label{eqA}
 n^2\left(b_0+(1+\nu)b_1+\nu\right)+r^2\left({b_1+1}\right)+nr=0;
 \end{equation}
if we consider
\[ \begin{cases}
    ll'=\frac{a_1+1}{n} \\
    lm'+lm'=1\\
    mm'=a_0+(1+\nu)a_1+\nu
    \end{cases} \]
a similar argument on $l,l',m,m'$ gives
\begin{equation}
l^2\left(a_0+(1+\nu)a_1+\nu\right)+m^2(a_1+1)+lm=0.
\end{equation}
Since, by assumption, $l=n$ and $m=r$ we get
\[ l^2\left(a_0+b_0+(1+\nu)(a_1+b_1)\right)+m^2(a_1+b_1)=0, \]
whence $l^2/m^2\in\GF(q)$. As $q$ is even, this gives $l/m\in\GF(q)$.
The case
$l'=n'$ and $m'=r'$ is clearly analogous and can be obtained by switching the roles of $\Pi_1$ and
$\Pi_2$.

Suppose now $l=n$ and $m'=r'$. Since $c_1\neq 0$, we also have $l\neq 0$; furthermore,
\[ n'=\frac{b_1+1}{l}. \]
If $m'=r'=0$, then $\Pi_2$ has equation $(a_1+1)x_0+(b_1+1)y_0=0$ and, consequently, is
defined over $\GF(q)$.
Suppose then $m'=r'\neq 0$. There are several subcases to consider:
\begin{itemize}
\item
if $m=0$, then $m'=r'=l^{-1}$ and $\Pi_2$ has equation
$(a_1+1)x_0+x_1+(b_1+1)y_0+y_1=0$, which is defined over $\GF(q)$;
\item
if $r=0$, then $m'=r'=n^{-1}=l^{-1}$ and we deduce, as above, that $\Pi_2$
is defined over $\GF(q)$;
\item
finally, suppose $m\neq 0\neq r$; then $b_0+(1+\nu)b_1+\nu\neq 0$ and from~\eqref{eqsys} we get
\[ m'=\frac{a_0+(1+\nu)a_1+\nu}{m},\qquad
   r'=\frac{b_0+(1+\nu)b_1+\nu}{r}. \]
Since $m'=r'$ we deduce
\begin{equation}\label{m/r} \frac{m}{r}=\frac{a_0+(1+\nu)a_1+\nu}{b_0+(1+\nu)b_1+\nu}\in\GF(q).
\end{equation}
Observe that $l'=(a_1+1)l^{-1}$ and also $r'=(b_0+(1+\nu)b_1+\nu)r^{-1}$; thus from~\eqref{eqsys} we obtain
\begin{equation}\label{fi}
l^2(b_0+(1+\nu)b_1+\nu)+r^2(a_1+1)+(c_0+c_1)lr=0.
\end{equation}
On the other hand, since $lm'+l'm=1$,
\[ \frac{l}{m}(a_0+(1+\nu)a_1+\nu)+\frac{m}{l}(a_1+1)=1; \]
using~\eqref{m/r} we obtain
\[ \frac{l}{r}(b_0+(1+\nu)b_1+\nu)+\frac{r}{l}(a_1+1)\left(\frac{a_0+(1+\nu)a_1+1}{b_0+(1+\nu)b_1+1}\right)=1, \]
whence
\begin{equation}
\label{e16}
l^2(b_0+(1+\nu)b_1+\nu)+lr+r^2(a_1+1)\left(\frac{a_0+(1+\nu)a_1+1}{b_0+(1+\nu)b_1+1}\right)=0;
\end{equation}
thus, adding~\eqref{fi} to~\eqref{e16}, we get
\[ \frac{l}{r}=\frac{a_1+1}{c_0+c_1+1}\left(\frac{a_0+(1+\nu)a_1+1}{b_0+(1+\nu)b_1+1}\right)\in\GF(q) \]
and the plane $\Pi_1$ is defined over $\GF(q)$.
The case $l'=n'$ and $m=r$ is analogous.
\end{itemize}
\end{proof}



\begin{lemma}
\label{lc}
Suppose that $\cQ$ is a hyperbolic quadric $\cC_{\infty}=\{P_{\infty}\}$. If
$\rank(\Xi_{\infty})=2$, then $\Xi_{\infty}$ is a line.
\end{lemma}
\begin{proof}
By Lemma~\ref{main4} we can assume $b=0$. Since $\rank(\Xi_{\infty})=2$, we have  $\det A_{\infty}=0$, that is $c^{q+1}=1$ and~\eqref{sis2} holds.
If $c_1=0$, then solving~\eqref{sis2} we obtain ~\eqref{eq6} and, since $b=0$,
we get $a=0$.
In the case in which $c_1\neq 0$ from~\eqref{eq7} we again obtain $a=0$.
We have now to show that $\Xi_{\infty}$ is the union of two conjugate planes.
In order to obtain this result, it suffices to prove  that   the coefficients
$l, m,n, r$ in~\eqref{eq8}  belong to some extension of $\GF(q)$ but
are not in $\GF(q)$.
Since ~\eqref{eqsys} holds we have
\[
\begin{cases}
  ll'=1\\
  lm'+l'm=1 \\
  mm'=\nu;
 \end{cases}
\]
thus, $\frac{l\nu}{m}+\frac{m}{l}=1$; hence $\nu l^2+l m +m^2=0$.
Since $\Tr_{q}(\nu)=1$ this implies that $\frac{l}{m} \notin\GF(q)$.
\end{proof}









Now, set $N=|(\cH\cap\cQ)\cap \AG(3,q^2)|$. First we observe that $N=|\Xi|-|\Xi_{\infty}|$.
By Lemma \ref{hyp} we see that  $ \rank(\Xi_{\infty})\geq 2$.
Thus, the following possibilities for $N$  may occur according as:
\begin{enumerate}[(C1)]
\item\label{C1}
  $ \rank(\Xi)=5$ and  $ \rank(\Xi_{\infty})=4$;
    \begin{enumerate}[\mbox{(C\ref{C1}.}1)]
\renewcommand{\theenumi}{\relax}
  \item\label{C1.1}  $\Xi$ is a parabolic quadric and  $\Xi_{\infty}$ is a hyperbolic quadric.  Then,
\[N= (q+1)(q^2+1)-(q+1)^2=q^3-q. \]

 \item\label{C1.2}  $\Xi$ is a parabolic quadric and the quadric $\Xi_{\infty}$ is  elliptic.  Then,
   \[N=  (q+1)(q^2+1)-(q^2+1)=q^3+q. \]
 \end{enumerate}

\item\label{C2} $\rank(\Xi)=5$ and $\rank(\Xi_{\infty})=3$;

  $\Xi$ is a parabolic quadric and the hyperplane at infinity is tangent to $\Xi$ whereas $\Xi_{\infty}$ is a cone
comprising the join of a point to a conic. Then,
 \[ N= (q+1)(q^2+1)-(q^2+q+1)=q^3. \]

\item\label{C3}
  $\rank(\Xi)=4$ and $ \rank(\Xi_{\infty})=4$;
\begin{enumerate}[\mbox{(C\ref{C3}.}1)]
\renewcommand{\theenumi}{\relax}
  \item\label{C3.1} $\Xi$ is a cone  projecting a hyperbolic quadric of $\PG(3,q)$ and the quadric $\Xi_{\infty}$ is  hyperbolic. Then,
 \[ N= q(q+1)^2+1-(q+1)^2=q^3+q^2-q. \]
\item\label{C3.2}  $\Xi$ is a cone  projecting an elliptic quadric of $\PG(3,q)$ and the quadric $\Xi_{\infty}$ is  elliptic. Then,
\[ N= q(q^2+1)+1-(q^2+1)=q^3-q^2+q. \]
 \end{enumerate}

\item\label{C4} $\rank(\Xi)=4$, $\rank(\Xi_{\infty})=3$;
\begin{enumerate}[\mbox{(C\ref{C4}.}1)]
\renewcommand{\theenumi}{\relax}
  \item\label{C4.1}$\Xi$ is a cone  projecting
a hyperbolic quadric and
 $\Xi_{\infty}$ is a cone comprising the join of a point to a conic. Then,
\[N=q(q+1)^2+1-[q(q+1)+1]=q^3+q^2.\]

\item \label{C4.2}
$\Xi$ is a cone  projecting
 an elliptic  quadric and $\Xi_{\infty}$ is a cone comprising the join of a point to a conic.
 Then,
\[N=q(q^2+1)+1-[q(q+1)+1]=q^3-q^2.\]
\end{enumerate}

\item\label{C5} $\rank(\Xi)=4$ and $\rank(\Xi_{\infty})=2$;
\begin{enumerate}[\mbox{(C\ref{C5}.}1)]
\renewcommand{\theenumi}{\relax}
\item\label{C5.1}$\Xi$ is a cone projecting a hyperbolic quadric and
$\Xi_{\infty}$ is the union of two planes defined over $\GF(q)$. Then,
 \[ N=q(q+1)^2+1-(2q^2+q+1)=q^3.\]
\item\label{C5.2} $\Xi$ is a cone projecting an elliptic quadric and and
$\Xi_{\infty}$  is a line (i.e. the union of two planes defined over the extension $\GF(q^2)$
but not over $\GF(q)$).
 Then,
\[N=q(q^2+1)+1-(q+1)=q^3. \]
\end{enumerate}

\item\label{C6} $\rank(\Xi)=\rank(\Xi_{\infty})=3$;

 $\Xi$  is the join of a line to a conic and  $\Xi_{\infty}$ is a cone comprising the join of a point to a conic. Then,
 \[N= q^3+q^2+q+1-(q^2+q+1)=q^3.\]

\item\label{C7} $\rank(\Xi)=3$, $\rank(\Xi_{\infty})=2$;
\begin{enumerate}[\mbox{(C\ref{C7}.}1)]
\renewcommand{\theenumi}{\relax}
\item\label{C7.1} $\Xi$ is the join of a line to a conic whereas $\Xi_{\infty}$ is a pair of planes over $\GF(q)$. Then,
\[N=q^3+q^2+q+1-(2q^2+q+1)=q^3-q^2.\]
\item\label{C7.2} $\Xi$ is the join of a line to a conic whereas $\Xi_{\infty}$ is a line. Then,
\[N=q^3+q^2+q+1-q-1=q^3+q^2.\]
\end{enumerate}

\item\label{C8} $\rank(\Xi)=\rank(\Xi_{\infty})=2$;
\begin{enumerate}[\mbox{(C\ref{C8}.}1)]
\renewcommand{\theenumi}{\relax}
\item\label{C8.1}
  $\Xi$ is a pair of solids and $\Xi_{\infty}$ is a pair of planes over $\GF(q)$. Then,
  \[N=2q^3+q^2+q+1-(2q^2+q+1)=2q^3-q^2.\]
\item\label{C8.2} $\Xi$ is a plane and
  $\Xi_{\infty}$ is a line. Then,
  \[N=q^2+q+1-(q+1)=q^2.\]
\end{enumerate}
\end{enumerate}
We are going to determine which cases (C\ref{C1})--(C\ref{C8}) may occur according as $\cQ$ is either elliptic or a cone or hyperbolic. In order to do this we need to
 establish the nature of $\Xi_{\infty}$, when $\Xi_{\infty}$ is non-singular. Hence we shall to
compute
 the trace of $\alpha$ as given by~\eqref{eqAlpha}
% $$\alpha=\frac{\det A_{\infty}-\det B}{4\det B}$$
 where  the matrix  $A$ is defined as $A_{\infty}$ in~\eqref{matr} and
\[B=\begin{pmatrix}
    0 &1&c_1& c_0+c_1 \\
-1 & 0  & c_0+c_1&c_0+(1+\nu)c_1  \\
   -c_1 & -(c_0+c_1)& 0 &1 & \\
    -(c_0+c_1)&-c_0-(1+\nu)c_1&-1 & 0
   \end{pmatrix} .\]
Write  $\gamma=(1+c^{q+1})=(1+c_0^2+\nu c_1^2+c_0c_1)$. A straightforward
computation shows
\begin{multline}
\label{alfa}
\alpha=\frac{a_0+a_1+b_0+b_1}{\gamma}+
   \frac{1}{\gamma^2}\big[(1+\nu)(a_1^2+b_1^2)+\\ (c_0^2+c_1^2\nu)(a_0b_1+a_1b_0)+
a_0a_1+b_0b_1+a_0b_0c_1^2+a_1b_1c_0^2\big].
\end{multline}

\subsection{The elliptic case}
Let $\cQ$ be an elliptic quadric. By Lemma~\ref{hyp}, we have
$\rank(\Xi_{\infty})\geq 3$  and hence cases (C\ref{C1}), (C\ref{C2}), (C\ref{C3}), (C\ref{C4}) and (C\ref{C6}) may occur.   Whence
\[N \in \{q^3-q^2, q^3-q^2+q, q^3-q, q^3, q^3+q, q^3+q^2-q, q^3+q^2 \}.\]
In this case $\cC_{\infty}=\{P_{\infty}\}$, hence
\[|\cQ\cap\cH|=N+1 \in \{\!\!\begin{array}[t]{l}q^3-q^2+1, q^3-q^2+q+1, q^3-q+1, q^3+1, q^3+q+1,\\ q^3+q^2-q+1, q^3+q^2+1 \}. \end{array} \]






\subsection{The degenerate case}
Let $\cQ$ be a cone. By Lemma~\ref{hyp},  $N$ falls in one of cases (C\ref{C1}) or (C\ref{C3}) and hence
\[N \in \{q^3-q^2+q, q^3-q, q^3+ q, q^3+q^2-q\}.\]
Here, by Lemma ~\ref{l0} $\cC_{\infty}$ is
either a point or $q^2+1$ points on a line. We distinguish these two cases.
\begin{itemize}
\item \framebox{$\cC_{\infty}=\{P_{\infty}\}$.} By Lemma~\ref{main4}, we can assume $b=0$  in~\eqref{eqQ}. 
Thus~\eqref{alfa} becomes $\alpha=a_0+a_1+(1+\nu)a_1^2+a_0a_1$ and $\Tr_{q}(\alpha)$ may be either  zero or one.

Hence
 cases (C\ref{C1}) and (C\ref{C3})  may happen; so
\[|\cQ\cap\cH|=N+1 \in \{q^3-q^2+q+1, q^3-q+1, q^3+q+1, q^3+q^2-q+1 \}. \]
\item \framebox{$\cC_{\infty}$ is a line.} By Lemma~\ref{main4},
we can assume $a=b$  in~\eqref{eqQ}. Furthermore as $\cQ$ is a cone $c=0$ in~\eqref{eqQ}; thus,
in this case,~\eqref{alfa} gives $\alpha=0$ that is, $\Tr_{q}(\alpha)=0$; this means that  only subcases  (C\ref{C1}.\ref{C1.1}) of (C\ref{C1}) and (C\ref{C3}.\ref{C3.1}) of (C\ref{C3})  may occur.
In particular,
 \[|\cQ\cap\cH|=N+q^2+1 \in \{q^3+q^2-q+1, q^3+2q^2-q+1 \}. \]

\end{itemize}



\subsection{The hyperbolic case}


Let $\cQ$ be a hyperbolic quadric. Then, by Lemma~\ref{hyp}, $\rank(\Xi_{\infty})\geq 2$ and  all  cases (C\ref{C1})--(C\ref{C8})
might occur.
\[ N\in\{\!\!\begin{array}[t]{l} q^2, q^3-q^2, q^3-q^2+q, q^3-q,  q^3, q^3+q,   q^3+q^2-q, q^3+q^2, 2q^3-q^2 \}.\end{array}\]
We have three possibilities
for $\cC_{\infty}$ from Lemma~\ref{l0},  that is $\cC_{\infty}$ is either a point, or $q^2+1$ points on a line or $2q^2+1$ points in the union of two lines. We now analyze these cases.
\begin{itemize}
\item
\framebox{$\cC_{\infty}=\{P_{\infty}\}$.}  We are going to show that some subcases of (C\ref{C1})--(C\ref{C8}) can be excluded.
Indeed, when $\rank(\Xi_{\infty})=2$, from Lemma
\ref{lc} we have that  subcases (C\ref{C7}.\ref{C7.1})   and (C\ref{C8}.\ref{C8.1}) cannot occur.
So,
\[|\cQ\cap\cH|=N+1 \in\{\!\!\begin{array}[t]{l} q^2+1, q^3-q^2+1, q^3-q^2+q+1, q^3-q+1,  q^3+1, \\  q^3+q+1,   q^3+q^2-q+1, q^3+q^2+1 \}.\end{array}\]

\item\framebox{$\cC_{\infty}$ is one line.}
  By Lemma~\ref{main4}
we can assume $b=0$ and $a=c$  in~\eqref{eqQ}.
\begin{enumerate}
\item
 When  $\Xi_{\infty}$ is non-degenerate,  only cases  (C\ref{C1}) and (C\ref{C3})  may occur.  Observe that~\eqref{alfa} becomes
\[ \alpha=\frac{c_0+c_1}{\gamma}+
   \frac{1}{\gamma^2}\big[(1+\nu)(c_1^2)+c_0c_1 \big]. \]
 Since $\gamma=c_0^2+\nu c_1^2+c_0c_1+1$ we have  $c_1^2+\nu c_1^2+c_1c_0=\gamma+c_0^2+c_1^2+1$. Therefore
\[ \alpha=\frac{(c_0+c_1)}{\gamma}+\frac{c_0^2+c_1^2}{\gamma^2}+\frac{1}{\gamma}+\frac{1}{\gamma^2} \]
and
$\Tr_{q}(\alpha)=0$.
 Hence,  subcases  (C\ref{C1}.\ref{C1.2}) and (C\ref{C3}.\ref{C3.2})  cannot  happen;
so,
\[ |\cQ\cap\cH|=N+q^2+1 \in\{ q^3+q^2-q+1, q^3+2q^2-q+1 \}. \]

\item Assume now that $\Xi_{\infty}$ is degenerate, that is
$2\leq\rank(\Xi_{\infty})\leq 3$.  Cases (C\ref{C2}) and (C\ref{C4})--(C\ref{C8}) occur.
 We are going to show that
$\rank(\Xi_{\infty})=3$.
Suppose, on the contrary, $\rank(\Xi_{\infty})=2$; then we end up
with considering a system identical to~\eqref{sis2}, as it appears
in the proof of Lemma~\ref{hyp}, and its consequences
 \eqref{eq6} and
\eqref{eq7};
so we shall not repeat explicitly these equations here.
First observe that~\eqref{sis2} holds.
 If it were $c_1=0$, from~\eqref{eq6} we would
 have $a_1=a_0=0$, that is $a=0$, which is impossible. So, $c_1\neq 0$;
since we are assuming  $b=b_0+\varepsilon b_1=0$, that is, $b_1=b_0=0$,
we would now have from~\eqref{eq7} $a_1=a_0=0$---again  a contradiction.

Thus, only cases (C\ref{C2}) and (C\ref{C4})
  might happen;  in particular,
\[ |\cQ\cap\cH|=N+q^2+1\in\{q^3+1, q^3+q^2+1, q^3+2q^2+1 \}. \]
\end{enumerate}

\item \framebox{$\cC_{\infty}$ consists of two lines.} By Lemma~\ref{main4}  we can assume
$b=\beta a$, $c=(\beta+1)a$ where $a\neq 0$ and $\beta^{q+1}=1$  in~\eqref{eqQ}.

\begin{enumerate}
\item
Suppose now $\Xi_{\infty}$ to be non-degenerate. Cases (C\ref{C1}) and (C\ref{C3})  occur.

From $c=a+b$ we get $c^{q+1}=a^{q+1}+a^qb+b^qa+b^{q+1}$. Since $b^{q+1}=a^{q+1}$, we have
$c^{q+1}=a^qb+b^qa$, that is
  \[a_0b_1+a_1b_0=c_0^2+\nu c_1^2+c_0c_1,\]

 On the other hand,  from $c_0=a_0+b_0$ and $c_1=a_1+b_1$ we obtain
  $c_0c_1=a_0a_1+a_0b_1+a_1b_0+b_0b_1$
 that is \[  a_0a_1+b_0b_1=c_0^2+\nu c_1^2. \]
 Now $(c_0^2+\nu c_1^2+c_0c_1)(c_0^2+\nu c_1^2)=
(a_0b_1+a_1b_0)( a_0a_1+b_0b_1)= a_0b_0(a_1^2+b_1^2)+a_1b_1(a_0^2+b_0^2)=a_0b_0c_1^2+a_1b_1c_0^2$ and
thus~\eqref{alfa} becomes
 \[\alpha = \frac{c_0+c_1}{1+c_0^2+\nu c_1^2+c_0c_1}+ \frac{(c_0+c_1)^2}{(1+c_0^2+\nu c_1^2+c_0c_1)^2};\]
so, $\alpha$ has trace $0$.

Hence just  subcases (C\ref{C1}.\ref{C1.1}) and (C\ref{C3}.\ref{C3.1})
 may occur and
\[ |\cQ\cap\cH|=N+2q^2+1\in\{q^3+2q^2-q+1,q^3+3q^2-q+1\}. \]
\item
If $\Xi_{\infty}$ is degenerate, then cases (C\ref{C2}) and  (C\ref{C4})--(C\ref{C8})  may happen.

When $\rank(\Xi_{\infty})=2$,
it follows from Lemma~\ref{2:5} that  only subcases
(C\ref{C7}.\ref{C7.1}) in (C\ref{C7})  and (C\ref{C8}.\ref{C8.1})  in (C\ref{C8})  may occur.

Now, we need a preliminary lemma.
Recall that a quadric $\cQ$
meeting a Hermitian surface $\cH$ in at least $3$ lines
of a regulus is permutable with $\cH$; see
\cite[\S 19.3, pag. 124]{FPS3D}.
We have the following statement.
\begin{lemma} \label{hard}
Suppose $\cQ$ to be hyperbolic and $\cC_{\infty}$ to be the union of two lines;
then  $|\cH\cap\cQ|=q^3+3q^2+1$ cannot happen.
\end{lemma}
\begin{proof} The case $|\cH\cap\cQ|=q^3+3q^2+1$ may happen just for  case (C\ref{C4}.\ref{C4.1}).
% of  Lemma~\ref{main2}.
Let $\cR$ be a regulus of $\cQ$ and denote by $r_1,r_2,r_3$ respectively
the numbers of $1$--tangents, $(q+1)$--secants and $(q^2+1)$-secants to $\cH$ in $\cR$.
A direct counting gives
\begin{equation}\label{r3} \begin{cases}
 r_1+r_2+r_3=q^2+1 \\
 r_1+(q+1)r_2+r_3(q^2+1)=q^3+3q^2+1.
\end{cases}\end{equation}
By straightforward algebraic manipulations we obtain
\[ qr_1+(q-1)r_2=q(q^2-q-1). \]
In particular, $r_2=qt$ with $t\leq q$.
If it were $t=q$, then $r_1=-1$ --- a contradiction;
so $r_2\leq q(q-1)$.
Solving~\eqref{r3} in $r_1$ and $r_3$, we obtain
\[ r_3=\frac{q^2+2q-r_2}{q}\geq\frac{q^2+2q-q^2+q}{q}=3. \]
In particular,
there are at least $3$ lines of $\cR$ contained in $\cH$.
This means that $\cQ$ is  permutable with $\cH$,
see \cite[\S 19.3, pag. 124]{FPS3D} or \cite[\S 86, pag. 154]{S} and
$|\cQ\cap\cH|=2q^3+q^2+1$, a contradiction.
\end{proof}

So, by Lemma ~\ref{hard}  only subcase
(C\ref{C4}.\ref{C4.2}) in (C\ref{C4}) is possible.

Thus we get
\[ |\cQ\cap\cH|=N+2q^2+1 \in\{ q^3+q^2+1, q^3+2q^2+1, 2q^3+q^2+1 \} \]
 and the proof is completed.
\end{enumerate}
\end{itemize}

It is straightforward to see, by means of a computer aided computation for small values of $q$, that
all the cardinalities enumerated above may occur.

% Finally we remark that a
% computer aided computation shows for small values of $q$ that all different cardinalities occur.
% ;

\section{Extremal configurations}
\label{EC}
As in the case of odd characteristic, it is possible to provide
a geometric description of the intersection configuration when
the size is either $q^2+1$ or $2q^3+q^2+1$.
These values are
respectively the minimum and the maximum yielded by Theorem~\ref{main1}.
and they can happen only when $\cQ$ is an hyperbolic quadric.
Throughout this section we assume that the hypotheses of
Theorem~\ref{main1} hold, namely that $\cH$ and $\cQ$ share a tangent
plane at some point $P$.

\begin{theorem}
\label{ppt1}
Suppose $|\cH\cap\cQ|=q^2+1$. Then, $\cQ$ is a hyperbolic quadric
and $\Omega=\cH\cap\cQ$ is an ovoid of $\cQ$.
\end{theorem}
\begin{proof}
  By Theorem~\ref{main1}, $\cQ$ is hyperbolic.
  Fix a regulus $\cR$ on $\cQ$. The  $q^2+1$ generators of $\cQ$ in $\cR$ are pairwise disjoint
  and each has non-empty intersection with $\cH$; so there can be at most
  one point of $\cH$ on each of them. It follows that $\cH\cap\cQ$ is an
  ovoid.
  % We first show that $\Omega$ must be an ovoid of $\cQ$.
  % Indeed, suppose there is a generator $r$ of $\cQ$ meeting $\cH$ in
  % more than $1$ point. Then, $|r\cap\cH|\geq q+1$.
  % On the other hand, any generator $\ell\neq r$ of $\cQ$ belonging to the same
  % regulus $\cR$ as $r$
  % necessarily meets $\cH$ in at least one point. As there
  % are $q^2$ such generators, we get $|\Omega|\geq q^2+q+1$ --- a
  % contradiction.
   In particular,
   by the above argument, any generator of $\cQ$ through a point of
   $\Omega$ must be tangent to $\cH$. Thus, at all points of $\Omega$
   the tangent planes   to $\cH$ and to $\cQ$  are the same.
\end{proof}

\begin{theorem}
\label{ppt2}
Suppose $|\cH\cap\cQ|=2q^3+q^2+1$. Then, $\cQ$ is a hyperbolic quadric
permutable with $\cH$.
\end{theorem}
Theorem~\ref{ppt2} can be obtained as a consequence of
the analysis contained in \cite[\S 5.2.1]{E1},
in light of \cite[Lemma 19.3.1]{FPS3D}.


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\end{document}


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%  LocalWords:  Hermitian quadric quadrics combinatorial projective
%  LocalWords:  cardinality hyperplane codeword functionals affine dx
%  LocalWords:  hypersurfaces nonsingular indeterminates symplectic
%  LocalWords:  projectively tangency cxy ey qx 2q qy 0x 0y 1x 1y C1
%  LocalWords:  hypersurface bilinear regulus permutable Aguglia Appl
%  LocalWords:  Giuzzi Bartoli De Boeck Fanali Storme Des Cryptogr de
%  LocalWords:  Cardinali Grassmannians Pasini Grassmann preprint al
%  LocalWords:  Cossidente Pavese Edoukou rensen's Hallez Rodier Thas
%  LocalWords:  codewords Combin Springer Verlag Hirschfeld Lachaud
%  LocalWords:  Luminy Gruyter Forme geometrie Hermitiane particolare
%  LocalWords:  riguardo caso finito Pura Tsfasman Vl du Nogin MSC 2q


%%% CHECKS

A:=matrix [[2*(a1+1),1,c1,c0+c1],[1,2*(a0+(1+v)*a1+v),c0+c1,c0+(1+v)*c1],[c1,c0+c1,2*(b1+1),1],[c0+c1,c0+(1+v)*c1,1,2*(b0+(1+v)*b1+v)]]
B:=matrix[[0,1,c1,c0+c1],[-1,0,c0+c1,c0+(1+v)*c1],[-c1,-(c0+c1),0,1],[-(c0+c1),-c0-(1+v)*c1,-1,0]]
alpha0:=(determinant A-determinant B)/(4*determinant B)
alpha:=alpha::Fraction(Polynomial(FF(2,4)))



d1:=((determinant A([2,3,4],[2,3,4]))/2)::Polynomial(FiniteField(2,4))
d2:=((determinant A([1,3,4],[1,3,4]))/2)::Polynomial(FiniteField(2,4))
d3:=((determinant A([1,2,4],[1,2,4]))/2)::Polynomial(FiniteField(2,4))
d4:=((determinant A([1,2,3],[1,2,3]))/2)::Polynomial(FiniteField(2,4))

Dlist:=[d1=0,d2=0,d3=0,d4=0]
eval(Dlist,[c1=0,c0=1])

a0:=(c0^2/c1^2+v)*a1+b1/c1^2
b0:=(c0^2/c1^2+v)*b1+a1/c1^2

ss:=numer ((a0+e*a1)*(b0+e*b1))::Polynomial(FiniteField(2,4))
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