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\title{\bf Non-classical hyperplanes of $DW(5,q)$}
\author{Bart De Bruyn\\
\small Department of Mathematics\\[-0.8ex]
\small Ghent University\\[-0.8ex] 
\small Belgium\\
\small\tt bdb@cage.ugent.be\\
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{June 8, 2012}{April ??, 2013}\\
\small Mathematics Subject Classifications: 51A45, 51A50}

\begin{document}

\maketitle

\begin{abstract}
The hyperplanes of the symplectic dual polar space $DW(5,q)$ arising from embedding, the so-called classical hyperplanes of $DW(5,q)$, have been determined earlier in the literature. In the present paper, we classify non-classical hyperplanes of $DW(5,q)$. If $q$ is even, then we prove that every such hyperplane is the extension of a non-classical ovoid of a quad of $DW(5,q)$. If $q$ is odd, then we prove that every non-classical ovoid of $DW(5,q)$ is either a semi-singular hyperplane or the extension of a non-classical ovoid of a quad of $DW(5,q)$. If $DW(5,q)$, $q$ odd, has a semi-singular hyperplane, then $q$ is not a prime number.

\bigskip\noindent \textbf{Keywords:}  symplectic dual polar space, hyperplane, projective embedding
\end{abstract}

\section{Introduction} \label{sec1}

The hyperplanes of the finite symplectic dual polar space $DW(5,q)$ that arise from some projective embedding, the so-called classical hyperplanes of $DW(5,q)$, have explicitly been determined earlier in the literature, see Cooperstein \& De Bruyn \cite{Co-DB}, De Bruyn \cite{bdb:qeven} and Pralle \cite{Pr:3}. In the present paper, we give a rather complete classification for the non-classical hyperplanes of $DW(5,q)$. There are two standard constructions for such hyperplanes.

\medskip (1) Suppose $x$ is a point of $DW(5,q)$ and $O$ is a set of points of $DW(5,q)$ at distance 3 from $x$ such that every line at distance 2 from $x$ has a unique point in common with $O$. Then $x^\perp \cup O$ is a non-classical hyperplane of $DW(5,q)$, the so-called semi-singular hyperplane with deepest point $x$. 
 
\medskip (2) Suppose $Q$ is a quad of $DW(5,q)$. Then the points and lines contained in $Q$ define a generalized quadrangle $\widetilde{Q}$ isomorphic to $Q(4,q)$. If $O$ is a non-classical ovoid of $\widetilde{Q}$, then the set of points of $DW(5,q)$ at distance at most 1 from $O$ is a non-classical hyperplane of $DW(5,q)$, the so-called {\em extension of} $O$. Several classes of non-classical ovoids of $Q(4,q)$ are known, see Section \ref{sec2.2} for a discussion.

\bigskip \noindent The following is our main result.

\begin{theorem} \label{theo1.1}
$(1)$ If $q$ is even, then every non-classical hyperplane of $DW(5,q)$ is the extension of a non-classical ovoid of a quad of $DW(5,q)$.

$(2)$ If $q$ is odd, then every non-classical hyperplane of $DW(5,q)$ is either a semi-singular hyperplane or the extension of a non-classical ovoid of a quad of $DW(5,q)$.
\end{theorem}

\bigskip \noindent Up to present, no semi-singular hyperplane of $DW(5,q)$ is known to exist. If a semi-singular hyperplane of $DW(5,q)$ exists, then $q$ must be odd (Theorem \ref{theo3.11}) and not a prime (Corollary \ref{co3.10}).

\medskip \noindent The lines and quads through a given point $x$ of $DW(5,q)$ define a projective plane isomorphic to $\PG(2,q)$ which we denote by $Res(x)$. If $H$ is a hyperplane of $DW(5,q)$ and $x$ is a point of $H$, then $\Lambda_H(x)$ denotes the set of lines through $x$ contained in $H$. We regard $\Lambda_H(x)$ as a set of points of $Res(x)$. If $\Lambda_H(x)$ is the whole set of points of $Res(x)$, then $x$ is called {\em deep with respect to} $H$.

The dual polar space $DW(5,q)$ has a nice full projective embedding $e$ in the projective space $\PG(13,q)$, which is called the {\em Grassmann embedding} of $DW(5,q)$, see e.g. Cooperstein \cite[Proposition 5.1]{Co}. A hyperplane of $DW(5,q)$ whose image under $e$ is contained in a hyperplane of of $\PG(13,q)$ is said to arise from $e$. For a proof of the following proposition, we refer to Pasini \cite[Theorem 9.3]{Pas} or Cardinali \& De Bruyn \cite[Corollary 1.5]{Ca-DB}.

\begin{proposition} \label{prop1.2}
If $H$ is a hyperplane of $DW(5,q)$ arising from the Grassmann embedding of $DW(5,q)$, then for every point $x$ of $H$, $\Lambda_H(x)$ is one of the following sets of points of $Res(x)$: $(1)$ a point; $(2)$ a line; $(3)$ the union of two distinct lines; $(4)$ a nonsingular conic; $(5)$ the whole set of points of $Res(x)$. 
\end{proposition}

\bigskip \noindent If $q \not= 2$, then the Grassmann embedding of $DW(5,q)$ is the so-called absolutely universal embedding of $DW(5,q)$ (Cooperstein \cite[Theorem B]{Co}, Kasikova \& Shult \cite[Section 4.6]{Ka-Sh}, Ronan \cite{Ro}), implying that the classical hyperplanes of $DW(5,q)$ are precisely those hyperplanes arising from the Grassmann embedding. Combining Theorem \ref{theo1.1} with Proposition \ref{prop1.2}, we easily find:

\begin{corollary} \label{co1.3}
If $H$ is a hyperplane of $DW(5,q)$, $q \not= 2$, then for every point $x$ of $H$, $\Lambda_H(x)$ is one of the following sets of points of $Res(x)$: $(1)$ the empty set; $(2)$ a point; $(3)$ a line; $(4)$ the union of two distinct lines; $(5)$ a nonsingular conic; $(6)$ the whole set of points of $Res(x)$. If $\Lambda_H(x)$ is the empty set, then $H$ is a semi-singular hyperplane whose deepest point lies at distance $3$ from $x$. If $H$ is not a semi-singular hyperplane, then case $(1)$ cannot occur.
\end{corollary}

\bigskip \noindent The conclusion of Corollary \ref{co1.3} is false for the dual polar space $DW(5,2)$. If $x$ is a point of $DW(5,2)$, then for every set $Y$ of points of $Res(x) \cong \PG(2,2)$, there exists a hyperplane $H$ through $x$ such that $\Lambda_H(x)=Y$, see Pralle \cite[Table 1]{Pr:3}.

\bigskip \noindent If $n \geq 4$, then the symplectic dual polar space $DW(2n-1,q)$ has many full subgeometries isomorphic to $DW(5,q)$. So, Corollary \ref{co1.3} reveals information on the local structure of any hyperplane of any symplectic dual polar space $DW(2n-1,q)$, where $q \not= 2$ and $n \geq 4$.

\bigskip \noindent Theorem \ref{theo1.1} will be proved in Section \ref{sec3}. In Section \ref{sec2}, we give the basic definitions (including some of the notions already mentioned above) and basic properties which will play a role in the proof of Theorem \ref{theo1.1}.

\section{Preliminaries} \label{sec2}

\subsection{The dual polar space $DW(5,q)$} \label{sec2.1}

Let $\mathcal{S} = (\mathcal{P},\mathcal{L},\mathrm{I})$ be a point-line geometry with nonempty point-set $\mathcal{P}$, line set $\mathcal{L}$ and incidence relation $\mathrm{I} \subseteq \mathcal{P} \times \mathcal{L}$. A set $H \subsetneq \mathcal{P}$ is called a {\em hyperplane} of $\mathcal{S}$ if every line of $\mathcal{S}$ has either one or all of its points in $H$. A {\em full projective embedding} of $\mathcal{S}$ is an injective mapping $e$ from $\mathcal{P}$ to the point-set of a projective space $\Sigma$ satisfying (i) $\langle e(\mathcal{P}) \rangle_\Sigma = \Sigma$; (ii) $\{ e(x) \, | \, (x,L) \ \in \ \mathrm{I} \}$ is a line of $\Sigma$ for every line $L$ of $\mathcal{S}$. If $e:\mathcal{S} \to \Sigma$ is a projective embedding of $\mathcal{S}$ and $\Pi$ is a hyperplane of $\Sigma$, then $e^{-1}(e(\mathcal{P}) \cap \Pi)$ is a hyperplane of $\mathcal{S}$. A hyperplane of $\mathcal{S}$ is said to be {\em classical} if it is of the form $e^{-1}(e(\mathcal{P}) \cap \Pi)$, where $e$ is some full projective embedding of $\mathcal{S}$ into a projective space $\Sigma$ and $\Pi$ is some hyperplane of $\Sigma$.

Distances $\d(\cdot,\cdot)$ in $\mathcal{S}$ will be measured in its collinearity graph. If $x$ is a point of $\mathcal{S}$ and $i \in \N$, then $\Gamma_i(x)$ denotes the set of points of $\mathcal{S}$ at distance $i$ from $x$. Similarly, if $X$ is a nonempty set of points and $i \in \N$, then $\Gamma_i(X)$ denotes the set of all points at distance $i$ from $X$, i.e. the set of all points $y$ for which $\min\{ \d(y,x) \, | \, x \in X \} = i$.

Let $W(5,q)$ be the polar space whose subspaces are the subspaces of $\PG(5,q)$ that are totally isotropic with respect to a given symplectic polarity of $\PG(5,q)$, and let $DW(5,q)$ denote the associated dual polar space. The points and lines of $DW(5,q)$ are the totally isotropic planes and lines of $\PG(5,q)$, with incidence being reverse containment. The dual polar space $DW(5,q)$ belongs to the class of {\em near polygons} introduced by Shult and Yanushka in \cite{Sh-Ya}. This means that for every point $x$ and every line $L$, there exists a unique point on $L$ nearest to $x$. The maximal distance between two points of $DW(5,q)$ is equal to 3. The dual polar space $DW(5,q)$ has $(q+1)(q^2+1)(q^3+1)$ points, $q+1$ points on each line and $q^2+q+1$ lines through each point.

If $x$ and $y$ are two points of $DW(5,q)$ at distance 2 from each other, then the smallest convex subspace $\langle x,y \rangle$ of $DW(5,q)$ containing $x$ and $y$ is called a {\em quad}. A quad $Q$ of $DW(5,q)$ consists of all totally isotropic planes of $W(5,q)$ that contain a given point $x_Q$ of $W(5,q)$. Any two lines $L$ and $M$ of $DW(5,q)$ that meet in a unique point are contained in a unique quad. We denote this quad by $\langle L,M \rangle$. Obviously, we have $\langle L,M \rangle = \langle x,y \rangle$ where $x$ and $y$ are arbitrary points of $L \setminus M$ and $M \setminus L$, respectively. The points and lines of $DW(5,q)$ that are contained in a given quad $Q$ define a point-line geometry $\widetilde{Q}$ isomorphic to the generalized quadrangle $Q(4,q)$ of the points and lines of a nonsingular parabolic quadric of $\PG(4,q)$. If $Q$ is a quad of $DW(5,q)$ and $x$ is a point not contained in $Q$, then $Q$ contains a unique point $\pi_Q(x)$ collinear with $x$ and $\d(x,y) = 1 + \d(\pi_Q(x),y)$ for every point $y$ of $Q$. If $Q_1$ and $Q_2$ are two distinct quads of $DW(5,q)$, then $Q_1 \cap Q_2$ is either empty or a line of $DW(5,q)$. If $Q_1 \cap Q_2 = \emptyset$, then the map $Q_1 \to Q_2; x \mapsto \pi_{Q_2}(x)$ is an isomorphism between $\widetilde{Q_1}$ and $\widetilde{Q_2}$.

\subsection{Hyperplanes of $Q(4,q)$} \label{sec2.2}

By Payne and Thas \cite[2.3.1]{Pa-Th}, every hyperplane of the generalized quadrangle $Q(4,q)$ is either the perp $x^\perp$ of a point $x$, a $(q+1) \times (q+1)$-subgrid or an ovoid. An ovoid of $Q(4,q)$ is {\em classical} if it is an elliptic quadric $Q^-(3,q) \subseteq Q(4,q)$. For many values of $q$, non-classical ovoids of $Q(4,q)$ do exist: (i) $q=p^h$ with $p$ an odd prime and $h \geq 2$ \cite{Ka}; (ii) $q=2^{2h+1}$ with $h \geq 1$ \cite{Ti}; (iii) $q=3^{2h+1}$ with $h \geq 1$ \cite{Ka}; (iv) $q=3^h$ with $h \geq 3$ \cite{Th-Pa}; (v) $q=3^5$ \cite{Pe-Wi}. For several prime powers $q$, it is known that all ovoids of $Q(4,q)$ are classical:

\begin{proposition} \label{prop2.1} $\bullet$ {\rm \textbf{(\cite{Ba,Pa})}} Every ovoid of $Q(4,4)$ is classical.

$\bullet$ {\rm \textbf{(\cite{OK-Pe:1,OK-Pe:2})}} Every ovoid of $Q(4,16)$ is classical.

$\bullet$ {\rm \textbf{(\cite{Ba-Go-St})}} Every ovoid of $Q(4,q)$, $q$ prime, is classical.
\end{proposition}

A set $\mathcal{G}$ of hyperplanes of $Q(4,q)$ is called a {\em pencil of hyperplanes} if every point of $Q(4,q)$ is contained in either 1 or all elements of $\mathcal{G}$. The following lemma is precisely Lemma 3.2 and Corollary 3.3 of De Bruyn \cite{bdb:sym}.

\begin{lemma} \label{lem2.2} If $G_1$ and $G_2$ are two distinct classical hyperplanes of $Q(4,q)$, then through every point $x$ of $Q(4,q)$ not contained in $G_1 \cup G_2$, there exists a unique classical hyperplane $G_x$ satisfying $G_x \cap G_1 = G_1 \cap G_2 = G_2 \cap G_x$. As a consequence, any two distinct classical hyperplanes of $Q(4,q)$ are contained in a unique pencil of classical hyperplanes of $Q(4,q)$.
\end{lemma}

\subsection{Hyperplanes of $DW(5,q)$} \label{sec2.3} 

Since $DW(5,q)$ is a near polygon, the set of points of $DW(5,q)$ at distance at most 2 from a given point $x$ is a hyperplane of $DW(5,q)$, the so-called {\em singular hyperplane with deepest point} $x$. If $O$ is a set of points of $DW(5,q)$ at distance 3 from a given point $x$ such that every line at distance 2 from $x$ has a unique point in common with $O$, then $x^\perp \cup O$ is a hyperplane of $DW(5,q)$, a so-called {\em semi-singular hyperplane of $DW(5,q)$ with deepest point} $x$. If $Q$ is a quad of $DW(5,q)$ and $G$ is a hyperplane of $\widetilde{Q} \cong Q(4,q)$, then $Q \cup \{ x \in \Gamma_1(Q) \, | \, \pi_Q(x) \in G \}$ is a hyperplane of $DW(5,q)$, the so-called {\em extension} of $G$.

If $H$ is a hyperplane of $DW(5,q)$ and $Q$ is a quad, then either $Q \subseteq H$ or $Q \cap H$ is a hyperplane of $Q \cong Q(4,q)$. If $Q \subseteq H$, then $Q$ is called a {\em deep quad}. If $Q \cap H = x^\perp \cap Q$ for some point $x \in Q$, then $Q$ is called {\em singular} with respect to $H$ and $x$ is called the {\em deep point} of $Q$. The quad $Q$ is called {\em ovoidal} (respectively, {\em subquadrangular}) with respect to $H$ if and only if $Q \cap H$ is an ovoid (respectively, a $(q+1) \times (q+1)$-subgrid) of $Q$. A hyperplane $H$ of $DW(5,q)$ is called {\em locally singular} ({\em locally subquadrangular}, respectively {\em locally ovoidal}) if every non-deep quad of $DW(5,q)$ is singular (subquadrangular, respectively ovoidal) with respect to $H$. A hyperplane that is locally singular, locally ovoidal or locally subquadrangular is also called a {\em uniform hyperplane}. In the following proposition, we collect a number of known results which we will need to invoke later in the proof of the Main Theorem.

\begin{proposition} \label{prop2.3}
$(1)$ The dual polar space $DW(5,q)$, $q \not=2$, has no locally subquadrangular hyperplanes.

$(2)$ The dual polar space $DW(5,q)$ has no locally ovoidal hyperplanes.

$(3)$ Every nonuniform hyperplane of $DW(5,q)$ admits a singular quad.
\end{proposition}

\medskip \noindent Proposition \ref{prop2.3}(1) is due to Pasini \& Shpectorov \cite{Pa-Sh}. Locally ovoidal hyperplanes of $DW(5,q)$ are just ovoids and cannot exist by Thomas \cite[Theorem 3.2]{Th}, see also Cooperstein and Pasini \cite{Co-Pa}. Proposition \ref{prop2.3}(3) is due to Pralle \cite{Pr:1}.

\medskip \noindent The classical hyperplanes of the dual polar space $DW(5,q)$ have already been classified in the literature. The dual polar space $DW(5,q)$, $q \not= 2$, has six isomorphism classes of classical hyperplanes by Cooperstein \& De Bruyn \cite{Co-DB} and De Bruyn \cite{bdb:qeven}. This fact is not true if $q=2$. The dual polar space $DW(5,2)$ has twelve isomorphism classes of hyperplanes by Pralle \cite{Pr:3}, see also De Bruyn \cite[Section 9]{bdb:qeven}. Observe that all these hyperplanes are classical by Ronan \cite[Corollary 2]{Ro}. By De Bruyn \cite{bdb:sym}, the classical hyperplanes of $DW(5,q)$ can be characterized as follows.

\begin{proposition} \label{prop2.4}
The classical hyperplanes of $DW(5,q)$ are precisely those hyperplanes $H$ of $DW(5,q)$ that satisfy the following property: if $Q$ is an ovoidal quad, then $Q \cap H$ is a classical ovoid of $Q$. 
\end{proposition}

\subsection{Hyperbolic sets of quads of $DW(5,q)$} \label{sec2.4}

As in Section \ref{sec2.1}, let $W(5,q)$ be the polar space associated with a symplectic polarity of $\PG(5,q)$. If $L$ is a hyperbolic line of $\PG(5,q)$ (i.e. a line of $\PG(5,q)$ that is not a line of $W(5,q)$), then the set of the $q+1$ (mutually disjoint) quads of $DW(5,q)$ corresponding to the points of $L$ satisfy the property that every line that meets at least two of its members meets each of its members in a unique point. Any set of $q+1$ quads that is obtained in this way will be called a {\em hyperbolic set of quads} of $DW(5,q)$. Every two disjoint quads $Q_1$ and $Q_2$ of $DW(5,q)$ are contained in a unique hyperbolic set of quads of $DW(5,q)$. We will denote this hyperbolic set of quads by $\mathcal{H}(Q_1,Q_2)$. Considering all the lines meeting $Q_1$ and $Q_2$, we easily see that the following holds.

\begin{lemma} \label{lem2.5} 
Let $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$ be a hyperbolic set of quads of $DW(5,q)$ and let $H$ be a hyperplane of $DW(5,q)$ such that $H \cap Q_1$ and $\pi_{Q_1}(H \cap Q_2)$ are distinct hyperplanes of $\widetilde{Q_1}$. Then $\{ \pi_{Q_1}(H \cap Q_i) \, | \, 1 \leq i \leq q+1 \}$ is a pencil of hyperplanes of $\widetilde{Q_1}$.
\end{lemma}

\section{Proof of Theorem \ref{theo1.1}} \label{sec3}

Throughout this section, we suppose that $H$ is an arbitrary hyperplane of $DW(5,q)$. In De Bruyn \cite{bdb:deep}, we classified for every field $\K$ of size at least three the hyperplanes of $DW(5,\K)$ containing a quad. The main theorem of \cite{bdb:deep} implies the following:

\begin{proposition} \label{prop3.1}
Every non-classical hyperplane of $DW(5,q)$, $q \not= 2$, containing a quad is the extension of a non-classical ovoid of a quad.
\end{proposition}

\bigskip \noindent We have already mentioned above that every hyperplane of $DW(5,2)$ is classical by Ronan \cite[Corollary 2]{Ro}. Since we are interested in the classification of all non-classical hyperplanes of $DW(5,q)$, we may by the above assume that the following holds:
\begin{quote}
\textbf{Assumption:} We have $q \geq 3$ and the hyperplane $H$ does not contain quads.
\end{quote}
We denote by $v$ the total number of points of $H$ and by $l$ the total number of lines of $DW(5,q)$ contained in $H$. In Section \ref{sec3.1}, we prove that there are only three possible values for $v$, namely $q^5+q^3+q^2+q+1$, $q^5+q^4+q^3+q^2+2q+1$ or $q^5+q^4+q^3+q^2+q+1$. In Section \ref{sec3.2}, we prove that if $v=q^5+q^3+q^2+q+1$, then $H$ is a semi-singular hyperplane. We also prove there that semi-singular hyperplanes cannot exist if $q$ is even. In \cite{DB-Va} (see also Corollary \ref{co3.10}), the nonexistence of semi-singular hyperplanes was already shown for prime values of $q$. In Section \ref{sec3.3}, we prove that the case $v=q^5+q^4+q^3+q^2+2q+1$ cannot occur and in Section \ref{sec3.4}, we prove that $H$ must be classical if $v=q^5+q^4+q^3+q^2+q+1$. All these results together imply that Theorem \ref{theo1.1} must hold.

\subsection{The possible values of $v$} \label{sec3.1}

The following lemma is an immediate consequence of Proposition \ref{prop2.3}.

\begin{lemma} \label{lem3.2}
The hyperplane admits singular quads.
\end{lemma}

\begin{lemma} \label{lem3.3}
We have $l = \frac{v \cdot (q^2+q+1) - (q^2+1)(q^3+1)(q^2+q+1)}{q}$.
\end{lemma}
\pr
We count the number of lines not contained in $H$. There are $(q+1)(q^2+1)(q^3+1)-v$ points outside $H$ and each of these points is contained in $q^2+q+1$ lines which contain a unique point of $H$. Hence, the total number of lines not contained in $H$ is equal to $\frac{((q+1)(q^2+1)(q^3+1)-v)(q^2+q+1)}{q}$. Since the total number of lines of $DW(5,q)$ equals $(q^2+1)(q^3+1)(q^2+q+1)$, we have $l=(q^2+1)(q^3+1)(q^2+q+1)- \frac{((q+1)(q^2+1)(q^3+1)-v)(q^2+q+1)}{q} = \frac{v \cdot (q^2+q+1) - (q^2+1)(q^3+1)(q^2+q+1)}{q}$.
\eop

\begin{lemma} \label{lem3.4}
If $Q$ is a singular quad with deep point $x$, then one of the following cases occurs:

$(1)$ $x^\perp \cap H = x^\perp \cap Q$;

$(2)$ there exists a line $L$ through $x$ not contained in $Q$ such that $x^\perp \cap H = (x^\perp \cap Q) \cup L$;

$(3)$ there exists a quad $R$ through $x$ distinct from $Q$ such that $x^\perp \cap H = (x^\perp \cap Q) \cup (x^\perp \cap R)$;

$(4)$ $x^\perp \subseteq H$.
\end{lemma}
\pr
Since $x^\perp \cap Q \subseteq x^\perp \cap H$, $|\Lambda_H(x)| \geq q+1$. If $|\Lambda_H(x)| \in \{ q+1,q+2 \}$, then either case (1) or (2) of the lemma occurs. Suppose therefore that $|\Lambda_H(x)| \geq q+3$ and let $L_1$ and $L_2$ be two distinct lines through $x$ that are contained in $H$, but not in $Q$. Put $R := \langle L_1,L_2 \rangle$. Since $L_1 \subseteq R \cap H$, $L_2 \subseteq R \cap H$ and $R \cap Q \subseteq R \cap H$, $R$ is singular with deep point $x$ and hence every line of $R$ through $x$ is contained in $H$. So, $|\Lambda_H(x)| \geq 2q+1$. 

If $|\Lambda_H(x)| = 2q+1$, then obviously case (3) of the lemma occurs. Suppose therefore that $|\Lambda_H(x)| \geq 2q+2$. Then there exists a line $L_3 \subseteq H$ through $x$ not contained in $Q \cup R$. If $Q'$ is a quad through $L_3$ distinct from $\langle L_3,Q \cap R \rangle$, then since $Q' \cap Q \subseteq H$, $Q' \cap R \subseteq H$ and $L_3 \subseteq H$, $Q'$ is singular with deep point $x$ and hence every line of $Q'$ through $x$ is contained in $H$. It follows that all lines of $DW(5,q)$ through $x$ are contained in $H$, except maybe for the $q-1$ lines through $x$ contained in $\langle L_3,Q \cap R \rangle$ and distinct from $L_3$ and $Q \cap R$. Let $L'$ be one of these $q-1$ lines and let $Q''$ be a quad through $L'$ distinct from $\langle L_3,Q \cap R \rangle$. Since $q \geq 3$ lines of $Q''$ through $x$ are contained in $H$, $Q''$ is singular with deep point $x$ and hence also $L'$ is contained in $H$. So, $x^\perp \subseteq H$ and case (4) of the lemma occurs. \eop

\begin{lemma} \label{lem3.5}
If $Q$ is a singular quad with deep point $x$, then $|\Gamma_3(x) \cap H| = q^5$.
\end{lemma}
\pr
Every point of $\Gamma_3(x) \cap H$ is collinear with a unique point of $\Gamma_2(x) \cap Q$. Conversely, every point $u$ of $\Gamma_2(x) \cap Q$ is collinear with precisely $q^2$ points of $\Gamma_3(x) \cap H$. (One on each line through $u$ not contained in $Q$.) Hence, $|\Gamma_3(x) \cap H| = |\Gamma_2(x) \cap Q| \cdot q^2 = q^5$. 
\eop

\begin{lemma} \label{lem3.6} 
Suppose $Q$ is a singular quad with deep point $x$. 
\begin{itemize}
\item If case $(1)$ of Lemma \ref{lem3.4} occurs, then $v=q^5+q^4+q^3+q^2+q+1$ and $l=q^5+q^4+q^3+q^2+q+1$.
\item If case $(2)$ of Lemma \ref{lem3.4} occurs, then $v=q^5+q^4+q^3+q^2+2q+1$ and $l=(q^2+q+1)(q^3+2)$.
\item If case $(3)$ of Lemma \ref{lem3.4} occurs, then $v=q^5+q^4+q^3+q^2+q+1$ and $l=q^5+q^4+q^3+q^2+q+1$.
\item If case $(4)$ of Lemma \ref{lem3.4} occurs, then $v=q^5+q^3+q^2+q+1$ and $l=q^2+q+1$.
\end{itemize}
\end{lemma}
\pr
Suppose case (1) of Lemma \ref{lem3.4} occurs. Then $x$ is contained in 1 singular quad that has $x$ as deep point (namely $Q$) and $q^2+q$ singular quads that do not have $x$ as deep point. In this case, $|\Gamma_0(x) \cap H|=1$, $|\Gamma_1(x) \cap H|=q^2+q$, $|\Gamma_2(x) \cap H| = 1 \cdot 0 + (q^2+q) \cdot q^2$ and $|\Gamma_3(x) \cap H| = q^5$. Hence, $v= 1 + (q^2+q) + (q^2+q) \cdot q^2 + q^5 = q^5+q^4+q^3+q^2+q+1$.

Suppose case (2) of Lemma \ref{lem3.4} occurs. Then $x$ is contained in 1 singular quad with deep point equal to $x$, $q+1$ subquadrangular quads and $q^2-1$ singular quads with deep point different from $x$. In this case, $|\Gamma_0(x) \cap H|=1$, $|\Gamma_1(x) \cap H|=(q+2)q = q^2+2q$, $|\Gamma_2(x) \cap H| = 1 \cdot 0 + (q+1) \cdot q^2 + (q^2-1) \cdot q^2 = q^4+q^3$ and $|\Gamma_3(x) \cap H|=q^5$. Hence, $v=1+(q^2+2q)+(q^4+q^3)+q^5 = q^5+q^4+q^3+q^2+2q+1$. 

Suppose case (3) of Lemma \ref{lem3.4} occurs. Then $x$ is contained in 2 singular quads with deep point $x$, $q-1$ singular quads with deep point different from $x$ and $q^2$ subquadrangular quads. In this case, $|\Gamma_0(x) \cap H|=1$, $|\Gamma_1(x) \cap H| = (2q+1)q = 2q^2+q$, $|\Gamma_2(x) \cap H| = 2 \cdot 0 + (q-1) \cdot q^2 + q^2 \cdot q^2 = q^4+q^3-q^2$ and $|\Gamma_3(x) \cap H|=q^5$. Hence, $v=1+(2q^2+q)+(q^4+q^3-q^2)+q^5 = q^5+q^4+q^3+q^2+q+1$.

Suppose case (4) of Lemma \ref{lem3.4} occurs. Then $x$ is contained in $q^2+q+1$ singular quads that have $x$ as deep point. Hence, $v=|\Gamma_0(x) \cap H| + |\Gamma_1(x) \cap H| + |\Gamma_2(x) \cap H| + |\Gamma_3(x) \cap H| = 1 + q(q^2+q+1) + 0 + q^5 = q^5+q^3+q^2+q+1$.

In each of the four cases, the value of $l$ can be derived from Lemma \ref{lem3.3}. \eop

\bigskip \noindent By Lemmas \ref{lem3.2}, \ref{lem3.4} and \ref{lem3.6}, we have:

\begin{corollary} \label{co3.7}
$v \in \{ q^5+q^3+q^2+q+1, q^5+q^4+q^3+q^2+q+1, q^5+q^4+q^3+q^2+2q+1 \}$.
\end{corollary}

\bigskip \noindent We see that if case (2) of Lemma \ref{lem3.4} occurs for one singular quad $Q$, then case (2) occurs for all singular quads $Q$. A similar remark holds applies to case (4) of Lemma \ref{lem3.4}. 


\subsection{The case $v=q^5+q^3+q^2+q+1$} \label{sec3.2}

Let $Q^\ast$ denote a singular quad and $x^\ast$ its deep point. 

\begin{lemma} \label{lem3.8}
If $v=q^5+q^3+q^2+q+1$, then $H$ is a semi-singular hyperplane of $DW(5,q)$ with deepest point $x^\ast$.
\end{lemma}
\pr
If $v=q^5+q^3+q^2+q+1$, then case (4) of Lemma \ref{lem3.4} occurs for the pair $(Q^\ast,x^\ast)$. So, we have that ${x^\ast}^\perp \subseteq H$ and $\Gamma_2(x^\ast) \cap H = \emptyset$ (no deep quad through $x^\ast$). Since $\Gamma_2(x^\ast) \cap H = \emptyset$, every line at distance 2 from $x^\ast$ contains a unique point of $\Gamma_3(x^\ast) \cap H$. It follows that $H$ is a semi-singular hyperplane of $DW(5,q)$ with deepest point $x^\ast$.
\eop

\medskip \noindent The following proposition was proved in De Bruyn and Vandecasteele \cite[Corollary 6.3]{DB-Va}. 

\begin{proposition} \label{prop3.9}
If $q$ is a prime power such that every ovoid of $Q(4,q)$ is classical, then $DW(5,q)$ does not have semi-singular hyperplanes.
\end{proposition}

\bigskip \noindent By Propositions \ref{prop2.1} and \ref{prop3.9}, we have

\begin{corollary} \label{co3.10}
If $q$ is prime, then $DW(5,q)$ has no semi-singular hyperplanes.
\end{corollary}

\bigskip \noindent We will now use hyperbolic sets of quads of $DW(5,q)$ to prove the nonexistence of semi-singular hyperplanes of $DW(5,q)$, $q$ even.

\begin{theorem} \label{theo3.11}
The dual polar space $DW(5,q)$, $q$ even, has no semi-sin\-gu\-lar hyperplanes.
\end{theorem}
\pr
Suppose $H$ is a semi-singular hyperplane of $DW(5,q)$, $q$ even, and as before let $x^\ast$ denote the deepest point of $H$. Let $Q$ be a quad through $x^\ast$, let $G$ be a $(q+1) \times (q+1)$-subgrid of $\widetilde{Q}$ not containing $x^\ast$, let $L_1$ and $L_2$ be two disjoint lines of $G$ and let $Q_i$, $i \in \{ 1,2 \}$, be a quad through $L_i$ distinct from $Q$. Then $Q_1$ and $Q_2$ are disjoint. Put $\mathcal{H} = \mathcal{H}(Q_1,Q_2)$. Every $Q_3 \in \mathcal{H}$ intersects $Q$ in a line of $G$ and hence $x^\ast \not\in Q_3$. It follows that every $Q_3 \in \mathcal{H}$ is ovoidal with respect to $H$. Suppose $Q_3 \in \mathcal{H} \setminus \{ Q_1 \}$ and $x_3 \in Q_3 \cap H$ such that $x_1 = \pi_{Q_1}(x_3) \in Q_1 \cap H$. Then the line $x_1 x_3$ is contained in $H$ and hence $x^\ast \in x_1 x_3$. But this is impossible, since no quad of $\mathcal{H}$ contains $x^\ast$. Hence, $\pi_{Q_1}(Q_3 \cap H)$ is disjoint from $Q_1 \cap H$. By Lemma \ref{lem2.5}, the set $\{ \pi_{Q_1}(Q_3 \cap H) \, | \, Q_3 \in \mathcal{H} \}$ is a partition of $Q_1$ into ovoids. This is however impossible since the generalized quadrangle $Q(4,q)$, $q$ even, has no partition in ovoids by Payne and Thas \cite[Theorem 1.8.5]{Pa-Th}. \eop

\subsection{The case $v=q^5+q^4+q^3+q^2+2q+1$} \label{sec3.3}

We suppose that $v=q^5+q^4+q^3+q^2+2q+1$ and $l=(q^2+q+1)(q^3+2)$. Recall that if $Q$ is a singular quad and $x$ is the deep point of $Q$, then case (2) of Lemma \ref{lem3.4} occurs for the pair $(Q,x)$. 

\begin{lemma} \label{lem3.12}
Let $Q$ be a singular quad, let $x$ be the deep point of $Q$, let $L$ be the line through $x$ not contained in $Q$ such that $x^\perp \cap H = (x^\perp \cap Q) \cup L$ and let $y$ be a point of $L \setminus \{ x \}$. Then there are $q+1$ lines $L_1,L_2,\ldots,L_{q+1}$ through $y$ different from $L$ that are contained in $H$. The $q+2$ lines $L,L_1,L_2,\ldots,L_{q+1}$ form a hyperoval of the projective plane $Res(y) \cong \PG(2,q)$. (Hence, $q$ must be even.) 
\end{lemma}
\pr
The $q+1$ quads $R_1,\ldots,R_{q+1}$ through $L$ determine a partition of the set of lines through $y$ different from $L$. Each of these quads is subquadrangular. Hence, $R_i$, $i \in \{ 1,2,\ldots,q+1 \}$, contains a unique line $L_i \not= L$ through $y$ that is contained in $H$.

For all $i,j \in \{ 1,2,\ldots,q+1 \}$ with $i \not= j$, the lines $L$, $L_i$ and $L_j$ are not contained in a quad since the quad $\langle L,L_i \rangle$ is subquadrangular. Suppose there exist mutually distinct $i,j,k \in \{ 1,2,\ldots,q+1 \}$ such that $L_i$, $L_j$ and $L_k$ are contained in a quad $Q'$. Then $L$ is not contained in $Q'$ and hence $Q \cap Q' = \emptyset$. Since $L_i$, $L_j$ and $L_k$ are contained in $H$, $Q'$ is singular with deep point $y$. Let $z' \in Q' \setminus y^\perp$ and $z := \pi_Q(z')$. Since $z$ and $z'$ are not contained in $H$, the line $zz'$ contains a unique point $z'' \in H$. Let $Q''$ denote the unique quad through $z''$ intersecting $L$ in a point $u$. Then $Q'' \in \mathcal{H}(Q,Q')$. So, every point of $u^\perp \cap Q''$ is contained in a line joining a point of $y^\perp \cap Q'$ with a point of $x^\perp \cap Q$ and hence is contained in $H$. Since also $z'' \in H$, $Q'' \subseteq H$, contradicting the fact that there are no deep quads.
\eop

\begin{lemma} \label{lem3.13}
There are four possible types of points in $H$:

$(A)$ points $x$ for which $\Lambda_H(x)$ is the union of a line of $Res(x)$ and a point of $Res(x)$ not belonging to that line;

$(B)$ points $x$ for which $\Lambda_H(x)$ is a hyperoval of $Res(x)$;

$(C)$ points $x$ for which $|\Lambda_H(x)|=2$;

$(D)$ points $x$ for which $\Lambda_H(x)$ is empty.

\noindent Moreover, we have:

$(i)$ Every point of Type $(A)$ has distance $1$ from precisely $q^2-1$ points of Type $(A)$, $q$ points of Type $(B)$ and $q+1$ points of Type $(C)$.

$(ii)$ Every point of Type $(B)$ has distance $1$ from precisely $q+2$ points of Type $(A)$, $(q+2)(q-1)$ points of Type $(B)$ and $0$ points of Type $(C)$.

$(iii)$ Every point of Type $(C)$ has distance $1$ from precisely $2q$ points of Type $(A)$, $0$ points of Type $(B)$ and $0$ points of Type $(C)$.
\end{lemma}
\pr
Suppose $Q^\ast$ is a singular quad and $x^\ast$ is its deep point. Consider the collinearity graph $\Gamma$ of $DW(5,q)$ and let $\Gamma_H$ denote the subgraph of $\Gamma$ induced on the vertex set $H$. Suppose $x$ is a point of $H$ such that $x$ and $x^\ast$ belong to different connected components of $\Gamma_H$. We prove that $\Lambda_H(x)$ is empty. Suppose to the contrary that there exists a line $L$ through $x$ contained in $H$. If $L$ meets $Q^\ast$, then $L \cap Q^\ast$ must be contained in ${x^\ast}^\perp$, contradicting the fact that $x^\ast$ and $x$ belong to different connected components of $\Gamma_H$. So, $L$ is disjoint from $Q^\ast$. Then $\pi_{Q^\ast}(L)$ meets ${x^\ast}^\perp$ and hence $x^\ast$ and $x$ are connected by a path of $\Gamma_H$, again a contradiction.

Notice that by Lemma \ref{lem3.6} and the fact that $v=q^5+q^4+q^3+q^2+2q+1$, $x^\ast$ is a point of Type (A). So, in order to prove the first part of the lemma, it suffices to verify that every vertex $x$ of Type ($X$), $X \in \{ A,B,C \}$, of $\Gamma_H$ is adjacent with only vertices of Type (A), (B) or (C). As a by-product of our verification, also the conclusions of the second part of the lemma will be obtained.

First, suppose that $x$ is a point of Type (A). Without loss of generality, we may suppose that $x=x^\ast$. Let $L^\ast$ denote the unique line through $x^\ast$ such that ${x^\ast}^\perp \cap H = ({x^\ast}^\perp \cap Q^\ast) \cup L^\ast$. By Lemma \ref{lem3.12}, every point of $L^\ast \setminus \{ x^\ast \}$ has Type (B). Now, let $L$ be a line through $x^\ast$ contained in $Q^\ast$. Then $\langle L,L^\ast \rangle$ is a subquadrangular quad. Any quad through $L$ different from $\langle L,L^\ast \rangle$ and $Q^\ast$ is singular with deep point contained in $L \setminus \{ x^\ast \}$. By Lemmas \ref{lem3.4} and \ref{lem3.6} and the fact that $v=q^5+q^4+q^3+q^2+2q+1$, every point of $L \setminus \{ x^\ast \}$ is the deep point of at most 1 such singular quad. Hence, $q-1$ points of $L \setminus \{ x^\ast \}$ have Type (A) and the remaining point of $L \setminus \{ x^\ast \}$ has type (C).

Suppose $x$ is a point of Type (C). Let $L_1$ and $L_2$ denote the two lines through $x$ that are contained in $H$. Then $\langle L_1,L_2 \rangle$ is a subquadrangular quad. If $Q$ is a quad through $L_1$ distinct from $\langle L_1,L_2 \rangle$, then $Q$ is singular with deep point on $L_1 \setminus \{ x \}$. By Lemmas \ref{lem3.4} and \ref{lem3.6} and the fact that $v=q^5+q^4+q^3+q^2+2q+1$, every point of $L_1 \setminus \{ x \}$ is the deep point of at most 1 such singular quad. It follows that every point of $L_1 \setminus \{ x \}$ has Type (A). In a similar way, one shows that every point of $L_2 \setminus \{ x \}$ has Type (A).

Suppose $x$ is a point of Type (B). Let $L$ be an arbitrary line through $x$ contained in $H$. Every quad through $L$ is subquadrangular. It follows that through every point $u \in L$ there are precisely $q+2$ lines that are contained in $H$. If at least three of these lines are contained in a certain quad $R$, then $R$ is singular with deep point $u$ and hence $u$ is of type (A). Otherwise, $u$ is of type (B). By Lemma \ref{lem3.12}, there are two possibilities.
\begin{enumerate}
\item[(1)] $L$ contains a unique point of Type (A) and $q$ points of Type (B).
\item[(2)] $L$ contains $q+1$ points of Type (B).
\end{enumerate}
We show that case (2) cannot occur. Suppose it does occur. Then $|\Gamma_0(L) \cap H|=q+1$ and $|\Gamma_1(L) \cap H| = (q+1)^2 q$. Each quad intersecting $L$ in a unique point is either ovoidal or subquadrangular and contributes $q^2$ to the value of $|\Gamma_2(L) \cap H|$. Since every point of $\Gamma_2(L)$ is contained in a unique quad that intersects $L$ in a unique point, $|\Gamma_2(L) \cap H| = (q+1) q^2 \cdot q^2$. It follows that $|H| = |\Gamma_0(L) \cap H| + |\Gamma_1(L) \cap H| + |\Gamma_2(L) \cap H| = (q+1) + (q+1)^2 q + (q+1)q^4 = q^5+q^4+q^3+2q^2+2q+1$, contradicting the fact that $|H|=q^5+q^4+q^3+q^2+2q+1$. \eop

\bigskip \noindent Now, let $n_A$, $n_B$, $n_C$ respectively $n_D$, denote the total number of points of $H$ of Type (A), (B), (C), respectively (D). Then by Lemma \ref{lem3.13}, we have $n_A \cdot q = n_B \cdot (q+2)$ and $n_A \cdot (q+1) = n_C \cdot 2q$. Hence,
\begin{eqnarray}
n_B & = & \frac{n_A \cdot q}{q+2}, \label{eq1} \\
n_C & = & \frac{n_A \cdot (q+1)}{2q}. \label{eq2}
\end{eqnarray}
Now, counting in two different ways the number of pairs $(x,L)$, with $x \in H$ and $L$ a line through $x$ contained in $H$, we obtain
\begin{equation}
n_A \cdot (q+2) + n_B \cdot (q+2) + n_C \cdot 2 = l \cdot (q+1) = (q^2+q+1)(q+1)(q^3+2). \label{eq3}
\end{equation}
From equations (\ref{eq1}), (\ref{eq2}) and (\ref{eq3}), we find $n_A = \frac{(q^2+q+1)(q^3+2)q}{2q+1}$, $n_B = \frac{(q^2+q+1)(q^3+2)q^2}{(q+2)(2q+1)}$ and $n_C = \frac{(q^2+q+1)(q^3+2)(q+1)}{2(2q+1)}$. If $q = 3$, then $n_A \not\in \N$. If $q \geq 4$, then 
\begin{eqnarray}
n_A + n_B + n_C & = & (q^2+q+1)(q^3+2) \cdot \frac{5q^2+7q+2}{2(q+2)(2q+1)} \nonumber \\
                & > & (q^5+q^4+q^3+q^2+2q+1) \cdot 1 \nonumber \\ 
                & = & v, \nonumber
\end{eqnarray}
a contradiction. Hence, the case $v=q^5+q^4+q^3+q^2+2q+1$ cannot occur.

\subsection{The case $v=q^5+q^4+q^3+q^2+q+1$} \label{sec3.4}

Suppose $v=q^5+q^4+q^3+q^2+q+1$. 

\begin{lemma} \label{lem3.14}
There are five possible types of points in $H$:

$(A)$ points $x$ for which $|\Lambda_H(x)|=1$;

$(B)$ points $x$ for which $\Lambda_H(x)$ is a line of $Res(x)$;

$(C)$ points $x$ for which $\Lambda_H(x)$ is the union of two distinct lines of $Res(x)$;

$(D)$ points $x$ for which $\Lambda_H(x)$ is an oval of $Res(x)$;

$(E)$ points $x$ for which $\Lambda_H(x)$ is empty.
\end{lemma}
\pr
Suppose $Q^\ast$ is a singular quad and $x^\ast$ is its deep point. Consider the collinearity graph $\Gamma$ of $DW(5,q)$ and let $\Gamma_H$ denote the subgraph of $\Gamma$ induced on the vertex set $H$. Suppose $x$ is a point of $H$ such that $x$ and $x^\ast$ belong to different connected components of $\Gamma_H$. Then we prove that $\Lambda_H(x)$ is empty. Suppose to the contrary that there exists a line $L$ through $x$ contained in $H$. If $L$ meets $Q^\ast$, then $L \cap Q^\ast$ must be contained in ${x^\ast}^\perp$, contradicting the fact that $x^\ast$ and $x$ belong to different connected components of $\Gamma_H$. So, $L$ is disjoint from $Q^\ast$. Then $\pi_{Q^\ast}(L)$ meets ${x^\ast}^\perp$ and hence $x^\ast$ and $x$ are connected by a path of $\Gamma_H$, again a contradiction.

By Lemmas \ref{lem3.4} and \ref{lem3.6} applied to the pair $(Q^\ast,x^\ast)$, $x^\ast$ is a point of Type (B) or (C). So, in order to prove the lemma, it suffices to prove that if $x$ is a point of Type $(X) \in \{ (A),(B),(C),(D) \}$ and $y$ is a point of $H \setminus \{ x \}$ collinear with $x$, then $y$ is of Type (A), (B), (C) or (D). Put $L := xy$. Since $x$ is of Type (A), (B), (C) or (D), one of the following two possibilities occurs:  
\begin{enumerate}
\item[(1)] $L$ is contained in $q+1$ singular quads with deep point on $L$.
\item[(2)] $L$ is contained in a unique singular quad with deep point on $L$ and $q$ subquadrangular quads.
\end{enumerate}
Observe that case (1) can only occur if $x$ has Type (A), (B) or (C), while case (2) can only occur if $x$ has Type (C) or (D).

Suppose case (1) occurs. Then $\Lambda_H(y)$ is the union of a number of lines of $Res(y)$ through a given point of $Res(y)$, union this point. Since every quad through $y$ is singular, subquadrangular or ovoidal, every line of $Res(y)$ intersects $\Lambda_H(y)$ in either 0, 1, 2 or $q+1$ points. Notice also that the point $y$ cannot be deep with respect to $H$, since otherwise Lemmas \ref{lem3.4} and \ref{lem3.6} applied to any singular quad through $y$ would yield that $v=q^5+q^3+q^2+q+1$, which is impossible. It follows that $y$ is of Type (A), (B) or (C).

If case (2) occurs, then there are two possibilities:
\begin{enumerate}
\item[(2a)] $\Lambda_H(y)$ is a line of $Res(y)$ + $q$ extra points. By Lemma \ref{lem3.4}, $y$ necessarily is a point of Type (C).
\item[(2b)] $|\Lambda_H(y)|=q+1$. If at least three of the points of $\Lambda_H(y)$ are collinear, then $\Lambda_H(y)$ is necessarily a line of $Res(y)$. But this is impossible since $y$ is not the deep point of a singular quad through $L$. So, no three points of $\Lambda_H(y)$ are collinear. This implies that $\Lambda_H(y)$ is an oval of $Res(y)$, i.e. $y$ is a point of Type (D). \eop
\end{enumerate}

\bigskip \noindent \textbf{Definition.} As we have already noticed in the proof of Lemma \ref{lem3.14}, every line $L \subseteq H$ must be contained in either $q+1$ singular quads or one singular quad and $q$ subquadrangular quads. If all quads on $L$ are singular, then $L$ is said to be {\em special}.

\begin{lemma} \label{lem3.15}
If $L$ is a special line, then $L$ contains only points of Type $(A)$, $(B)$ and $(C)$. Moreover, the number of points of Type $(A)$ on $L$ equals the number of points of Type $(C)$ on $L$. 
\end{lemma}
\pr
Since every quad through $L$ is singular, there are $(q+1)q$ lines contained in $H$ that meet $L$ in a unique point. Moreover, for every $y \in L$, $\Lambda_H(y)$ is the union of a number of lines of $Res(y)$, union the point of $Res(y)$ corresponding to $L$. It follows that every point of $L$ is of Type (A), (B) or (C). Let $n_1$, $n_2$, respectively $n_3$, denote the number of points of Type (A), (B), respectively (C), contained in $L$. Then $n_1 + n_2 + n_3 = q+1$ and $n_1 \cdot 0 + n_2 \cdot q + n_3 \cdot 2q = q(q+1)$. It follows that $n_1=n_3$.
\eop

\bigskip \noindent The proof of the following lemma is straightforward.

\begin{lemma} \label{lem3.16}
Every point of Type $(A)$ is contained in a unique special line. Every point of Type $(C)$ is contained in a unique special line.
\end{lemma}

\bigskip Let $n_A$, $n_B$, $n_C$, $n_D$, respectively $n_E$, denote the total number of points of $H$ of Type (A), (B), (C), (D), respectively (E). The following is an immediate corollary of Lemmas \ref{lem3.15} and \ref{lem3.16}.

\begin{corollary} \label{co3.17}
We have $n_C=n_A$.
\end{corollary}

\begin{lemma} \label{lem3.18}
We have $n_E=0$.
\end{lemma}
\pr
We count in two different ways the number of pairs $(x,L)$ with $x \in H$ and $L$ a line of $H$ through $x$. We find
\[ n_A \cdot 1 + n_B \cdot (q+1) + n_C \cdot (2q+1) + n_D \cdot (q+1) + n_E \cdot 0 = l(q+1). \]
Using the facts that $n_A=n_C$ and $l=(q^2+q+1)(q^3+1)=v$, we find $n_A + n_B + n_C + n_D = v$. Hence, $n_E=0$. \eop

\begin{lemma} \label{lem3.19}
We have $n_D = \frac{2q^2}{q+1} n_A$.
\end{lemma}
\pr
We count in two different ways the number of pairs $(x,Q)$ where $Q$ is a singular quad and $x$ is its deep point. We find
\begin{equation} \label{eq4}
Si = n_B + 2 \cdot n_C,
\end{equation}
where $Si$ denotes the total number of singular quads. We count in two different ways the number of pairs $(x,Q)$ where $Q$ is a singular quad and $x$ is a point of $Q \cap H$ distinct from the deep point of $Q$. We find
\begin{equation} \label{eq5}
(q+1)q \cdot Si = (q+1) n_A + q(q+1) n_B + (q-1) n_C + (q+1) n_D.
\end{equation}
From (\ref{eq4}) and (\ref{eq5}) and the fact that $n_A = n_C$, it readily follows that $n_D = \frac{2q^2}{q+1} n_A$. \eop

\bigskip \noindent Now, put $\delta := n_A$. Then we have $n_A = n_C = \delta$, $n_D = \frac{2q^2}{q+1} \cdot \delta$ and $n_B = (q^2+q+1)(q^3+1) - \frac{2(q^2+q+1)}{q+1} \cdot \delta$.

\begin{lemma} \label{lem3.20}
We have $0 \leq \delta \leq \lfloor \frac{1}{2} (q+1)(q^3+1) \rfloor$.
\end{lemma}
\pr
This follows from the fact that $n_B \geq 0$.
\eop

\bigskip \noindent \textbf{Remark.} If $q \geq 4$ is even, then by De Bruyn \cite{bdb:qeven}, the dual polar space $DW(5,q)$ has up to isomorphism two hyperplanes not containing quads. The values of $\delta$ corresponding to these two hyperplanes are respectively equal to 0 and $\frac{q^3(q+1)}{2}$. If $q$ is odd, then by Cooperstein and De Bruyn \cite{Co-DB}, the dual polar space $DW(5,q)$ has up to isomorphism two hyperplanes not containing quads. The values of $\delta$ corresponding to these two hyperplanes are respectively equal to $\frac{1}{2}(q+1)(q^3-1)$ and $\frac{1}{2}(q+1)(q^3+1)$. So, the lower and upper bounds in Lemma \ref{lem3.20} can be tight.

\bigskip \noindent \textbf{Definition}. Recall that if $Q$ is a quad of $DW(5,q)$ then the points and lines of $DW(5,q)$ contained in $Q$ bijectively correspond to the points and lines of $\PG(4,q)$ that are contained in a given nonsingular parabolic quadric $Q(4,q)$ of $\PG(4,q)$. A {\em conic} of $Q$ is a set of $q+1$ points of $Q$ that corresponds to a nonsingular conic of $Q(4,q)$, i.e. with a set of $q+1$ points of $Q(4,q)$ contained in a plane $\pi$ of $\PG(4,q)$ intersecting $Q(4,q)$ in a nonsingular conic of $\pi$.

\begin{lemma} \label{lem3.21}
Let $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$ be a hyperbolic set of quads of $DW(5,q)$ such that $Q_1$ is ovoidal with respect to $H$ and $|\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)| \geq 2$. Then:
\begin{enumerate}
\item[$(1)$] $\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)$ is a conic of $Q_1$.
\item[$(2)$] The number of ovoidal quads of $\{ Q_1,\ldots,Q_{q+1} \}$ is bounded above by $\frac{q+1}{2}$. If the number of these ovoidal quads is precisely $\frac{q+1}{2}$, then the remaining $\frac{q+1}{2}$ quads of $\{ Q_1,\ldots,Q_{q+1} \}$ are subquadrangular with respect to $H$.
\end{enumerate}
\end{lemma}
\pr
We first prove that $\pi_{Q_1}(Q_2 \cap H) \not= Q_1 \cap H$. Suppose to the contrary that $\pi_{Q_1}(Q_2 \cap H) = Q_1 \cap H$. Let $u$ be a point of $Q_1 \setminus H$, let $L$ be the unique line through $u$ meeting each quad of $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$, let $v$ denote the unique point of $L$ contained in $H$, and let $i$ be the unique element of $\{ 3,\ldots,q+1 \}$ such that $v \in Q_i$. Now, since $Q_i \cap H$ contains $\pi_{Q_i}(Q_2 \cap H)$ and the point $v \in Q_i \setminus \pi_{Q_i}(Q_2 \cap H)$, we must have $Q_i \subseteq H$. This is however impossible since no quad is contained in $H$.

So, $\pi_{Q_1}(Q_2 \cap H) \not= Q_1 \cap H$. By Lemma \ref{lem2.5}, $\{ \pi_{Q_1}(Q_i \cap H) \, | \, 1 \leq i \leq q+1 \}$ is a pencil of hyperplanes of $\widetilde{Q_1}$. Let $\alpha_1$, $\alpha_2$, respectively $\alpha_3$, denote the number of quads of $\{ Q_1,\ldots,Q_{q+1} \}$ that are ovoidal, singular, respectively subquadrangular, with respect to $H$. Put $\beta : = |\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)| \geq 2$. We prove that $\beta = q+1$.

If $\alpha_1 = q+1$ and $\alpha_2=\alpha_3=0$, then $(q+1)(q^2+1) = |Q_1| = \beta + (q+1)(q^2+1 - \beta) = (q+1)(q^2+1) - q \beta < (q+1)(q^2+1)$, a contradiction. So, without loss of generality, we may suppose that $Q_2$ is not ovoidal with respect to $H$. If $Q_2$ is subquadrangular with respect to $H$, then $\beta = |\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)| = q+1$. If $Q_2$ is singular with respect to $H$ with deep point $u$ such that $\pi_{Q_1}(u) \not\in Q_1 \cap H$, then also $\beta = |\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)| = q+1$. If $Q_1$ were singular with respect to $H$ with deep point $u$ such that $\pi_{Q_1}(u) \in Q_1 \cap H$, then $\beta = |\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)| = 1$, a contradiction. Hence, $\beta=q+1$ as claimed.

Now, we have $\alpha_1 + \alpha_2 + \alpha_3 = q+1$ and $(q+1)(q^2+1) = |Q_1| = (q+1) + \alpha_1(q^2-q) + \alpha_2 q^2 + \alpha_3(q^2+q) = (q+1)+(q+1)q^2 + q(\alpha_3 - \alpha_1)$, i.e. $\alpha_1 + \alpha_2 + \alpha_3 = q+1$ and $\alpha_1 = \alpha_3$. Hence, $\alpha_1 = \alpha_3 \leq \frac{q+1}{2}$. Moreover, if $\alpha_1 = \alpha_3 = \frac{q+1}{2}$, then $\alpha_2 = 0$. This proves claim (2).
 
Now, $\alpha_2 + \alpha_3 \geq \frac{q+1}{2}$. So, $\alpha_2 + \alpha_3 \geq 2$. Without loss of generality, we may suppose that the quads $Q_2$ and $Q_3$ are singular or subquadrangular with respect to $H$.

The points and lines contained in $Q_1$ can be identified (in a natural way) with the points and lines lying on a given nonsingular parabolic quadric $Q(4,q)$ of $\PG(4,q)$. Now, each of $\pi_{Q_1}(Q_2 \cap H)$ and $\pi_{Q_1}(Q_3 \cap H)$ is either a singular hyperplane or a subgrid of $\widetilde{Q_1}$ and hence arises by intersecting $Q(4,q)$ with a hyperplane of $\PG(4,q)$. Since $\pi_{Q_1}(Q_2 \cap H) \cap \pi_{Q_1}(Q_3 \cap H) = \pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)$ is a set of $q+1$ mutually noncollinear points, $\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)$ must be a conic of $Q_1$.
\eop

\begin{lemma} \label{lem3.22}
If $Q_1$ is an ovoidal quad, then through every two points of $Q_1 \cap H$, there is a conic of $Q_1$ that is completely contained in $Q_1 \cap H$.
\end{lemma}
\pr
Let $x_1$ and $x_2$ be two distinct points of $Q_1 \cap H$. By Lemmas \ref{lem3.14} and \ref{lem3.18}, there exists a line $L_i$, $i \in \{ 1,2 \}$ through $x_i$ that is contained in $H$. Let $Q_2$ be a quad distinct from $Q_1$ that meets $L_1$ and $L_2$, and let $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$ be the unique hyperbolic set of quads of $DW(5,q)$ containing $Q_1$ and $Q_2$. Since $\{ x_1,x_2 \} \subseteq \pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)$, Lemma \ref{lem3.21} applies. We conclude that $\pi_{Q_1}(Q_2 \cap H) \cap (Q_1 \cap H)$ is a conic containing $x_1$ and $x_2$. 
\eop

\begin{lemma} \label{lem3.23}
For every quad $Q_1$ that is ovoidal with respect to $H$, there is a quad $Q_2$ disjoint from $Q_1$ that is singular with respect to $H$ such that $\pi_{Q_1}(u) \not\in Q_1 \cap H$ where $u$ is the deepest point of the singular hyperplane $Q_2 \cap H$ of $\widetilde{Q_2}$.
\end{lemma}
\pr
The number of points $x \in \Gamma_1(Q_1) \cap H$ for which $\pi_{Q_1}(x) \not\in Q_1 \cap H$ is equal to $(|Q_1| - |Q_1 \cap H|) \cdot q^2 = q^3(q^2+1)$. Now, since $n_D = \frac{2q^2}{q+1} \delta \leq \frac{2q^2}{q+1} \cdot \frac{1}{2}(q+1)(q^3+1)=q^2(q^3+1) < q^3(q^2+1)$, there exists a point $y \in \Gamma_1(Q_1) \cap H$ not of type (D) for which $\pi_{Q_1}(y) \not\in Q_1 \cap H$. Let $L \subseteq H$ be a special line through $y$ and let $z$ denote the unique point of $L$ for which $\pi_{Q_1}(z) \in Q_1 \cap H$. By Lemma \ref{lem3.14}, there are at most two quads $R$ through $L$ for which $z$ is the deep point of the singular hyperplane $R \cap H$ of $\widetilde{R}$. Hence, there exists a quad $Q_2$ through $L$ for which the deep point $u$ of the singular hyperplane $Q_2 \cap H$ of $\widetilde{Q_2}$ is distinct from $z$. Since $u$ is not collinear with a point of $Q_1 \cap H$, $Q_1$ and $Q_2$ are disjoint.
\eop

\begin{lemma} \label{lem3.24}
If $Q_1$ is ovoidal with respect to $H$, then $Q_1 \cap H$ is a classical ovoid of $\widetilde{Q_1}$.
\end{lemma}
\pr
By Lemma \ref{lem3.23}, there exists a quad $Q_{q+1}$ disjoint from $Q_1$ that is singular with respect to $H$ such that $\pi_{Q_1}(u) \not\in Q_1 \cap H$ where $u$ is the deep point of the singular hyperplane $Q_{q+1} \cap H$ of $\widetilde{Q_{q+1}}$. Let $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$ denote the unique hyperbolic set of quads of $DW(5,q)$ containing $Q_1$ and $Q_{q+1}$. By Lemma \ref{lem3.21}, we then have:

(1) $X := \pi_{Q_1}(Q_{q+1} \cap H) \cap (Q_1 \cap H)$ is a conic of $Q_1$;

(2) the number $k$ of ovoidal quads of the set $\{ Q_1,Q_2,\ldots,Q_{q+1} \}$ is at most $\frac{q}{2}$.

\noindent Without loss of generality, we may suppose that $Q_1,\ldots,Q_k$ are the quads of $\{ Q_1,Q_2,\ldots,$ $Q_{q+1} \}$ that are ovoidal with respect to $H$. Since $(q+1)-\frac{q}{2} \geq 2$, $Q_q$ and $Q_{q+1}$ are not ovoidal with respect to $H$. By Lemmas \ref{lem2.2} and \ref{lem2.5}, $\pi_{Q_1}(Q_q \cap H)$ and $\pi_{Q_1}(Q_{q+1} \cap H)$ are contained in a unique pencil of classical hyperplanes of $\widetilde{Q_1}$. Moreover, this pencil contains the hyperplanes $\pi_{Q_1}(Q_i \cap H)$, $i \in \{ k+1,\ldots,q+1 \}$. Let $A_1,\ldots,A_k$ denote the remaining elements of this pencil. Then $X \subseteq A_1 \cap \cdots \cap A_k$ and $A_1 \cup \cdots \cup A_k = \pi_{Q_1}(Q_1 \cap H) \cup \cdots \cup \pi_{Q_1}(Q_k \cap H)$. Now, $|A_1 \cup \cdots \cup A_k| \geq |X| + k (q^2+1-|X|) = (q+1) + k (q^2-q)$ and equality holds if and only if every $A_j$, $j \in \{ 1,\ldots,k \}$, is a classical ovoid of $\widetilde{Q_1}$. Now, since $|\pi_{Q_1}(Q_1 \cap H) \cup \cdots \cup \pi_{Q_1}(Q_k \cap H)| = |X| + k (q^2+1 - |X|) = (q+1) + k (q^2-q)$, we can conclude that every $A_j$, $j \in \{ 1,\ldots,k \}$, is a classical ovoid of $\widetilde{Q_1}$.

Now, let $i \in \{ 1,\ldots,k \}$ and suppose there exists no $j \in \{ 1,\ldots,k \}$ such that $\pi_{Q_1}(Q_i \cap H) = A_j$. Then $X \subseteq \pi_{Q_1}(Q_i \cap H) \subseteq A_1 \cup \cdots \cup A_k$ and there exist two distinct $j_1,j_2 \in \{ 1,\ldots,k \}$ such that $\pi_{Q_1}(Q_i \cap H) \cap (A_{j_1} \setminus X) \not= \emptyset$ and $\pi_{Q_1}(Q_i \cap H) \cap (A_{j_2} \setminus X) \not= \emptyset$. Let $y_1$ be an arbitrary point of $\pi_{Q_1}(Q_i \cap H) \cap (A_{j_1} \setminus X)$ and let $y_2$ be an arbitrary point of $\pi_{Q_1}(Q_i \cap H) \cap (A_{j_2} \setminus X)$. By Lemma \ref{lem3.22}, there exists a conic $C$ through $y_1$ and $y_2$ that is completely contained in $\pi_{Q_1}(Q_i \cap H)$ and hence also in $A_1 \cup \cdots \cup A_k$. Since $|C| = q+1$ and $k \leq \frac{q}{2}$, there exists a $j_3 \in \{ 1,\ldots,k \}$ such that $|C \cap A_{j_3}| \geq 3$. Since $A_{j_3}$ is a classical ovoid of $\widetilde{Q_1}$, this necessarily implies that $C \subseteq A_{j_3}$, contradicting the fact that $y_1 \in A_{j_1} \setminus X$, $y_2 \in A_{j_2} \setminus X$ and $j_1 \not= j_2$. Hence, there exists a $j \in \{ 1,\ldots,k \}$ such that $\pi_{Q_1}(Q_i \cap H) = A_j$. This implies that the ovoid $Q_i \cap H$ of $\widetilde{Q_i}$ is classical. \eop

\begin{corollary} \label{co3.25}
The hyperplane $H$ is classical.
\end{corollary}
\pr
This is an immediate corollary of Proposition \ref{prop2.4} and Lemma \ref{lem3.24}.
\eop

\bigskip \noindent \textbf{Remark.} With the terminology of Cooperstein \& De Bruyn \cite{Co-DB} and De Bruyn \cite{bdb:qeven}, the hyperplane $H$ is either a hyperplane of Type V or a hyperplane of Type VI.

\subsection*{Acknowledgment}

At the time of the writing of this paper, the author was a Postdoctoral Fellow of the Research Foundation - Flanders (Belgium).

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