# Generalized quadrangles of order $(s,s^{2})$. IV
Translations, Moufang and Fong-Seitz

J. A. THAS
Ghent University  Ghent University, Department of Mathematics, Krijgslaan 281,
S25, B-9000 Ghent, Belgium<EMAIL_ADDRESS>
###### Abstract.

In the period 1994-1999 Thas wrote a series of three papers on generalized
quadrangles of order $(s,s^{2})$. In this Part IV we classify all finite
translation generalized quadrangles of order $(s,s^{2})$ having a kernel of
size at least 3, containing a regular line not incident with the translation
point. There are several applications on generalized quadrangles of order
$(s,s^{2})$ having at least 2 translation points, on Moufang quadrangles, and
concerning the theorem of Fong and Seitz classifying all groups with a
$BN$-pair of rank 2.

Keywords: generalized quadrangle, projective space, translation, Moufang,
generalized ovoid

##  1\. Introduction

In 1976 Tits [19] shows that each finite Moufang generalized quadrangle
$\mathcal{S}$ is classical or dual classical. This follows from the
classification of all groups with a $BN$-pair of rank 2 by Fong and Seitz [4,
5]. In their monograph ”Finite Generalized Quadrangles”, the first edition of
which appeared in 1984, Payne and Thas [15, 16] did an almost successful
attempt to prove the Moufang theorem in a pure geometric way, using just some
elementary group theory. As a corollary there would follow a pure geometric
proof of the theorem of Fong and Seitz, namely the case where the Weyl group
is $D_{8}$.

The case that was missing in the work of Payne and Thas is when the
generalized quadrangle has order $(s,s^{2}),s\not=1$, and each point is a
translation point. In 1991 Kantor [10] succeeded in handling this and compared
to Fong and Seitz the amount of group theory employed was very small:
information was required concerning a relatively restricted type of
2-transitive permutation group, together with a fact about very small degree
representations of such a group. In [28] K. Thas gave an alternative proof
relying on the classification of finite groups with a split $BN$-pair of rank
1, a classification independently obtained in 1972 by Hering, Kantor and Seitz
[6] and Shult [17].

In this paper we will give a pure geometric proof, without any group theory.
In the even case we exclude the case where the ”kernel” has just 2 elements;
we hope to be able to solve also that case. In fact we obtain much more
general results, characterizing classes of generalized quadrangles which are
not necessarily classical nor dual classical, and yielding as corollary the
”missing part” of our monograph (up to the mentioned exception). In particular
we will classify the translation generalized quadrangles of order $(s,s^{2})$
containing a regular line which is not incident with the translation point. As
we start from a local situation the use of group theory is excluded. Hence,
together with the results in the monograph of Payne and Thas [15, 16], this
will provide a geometric proof, relying just on a few elementary results on
groups, of a large part of the deep group theoretical result of Fong and
Seitz.

Also we mention that the classification of all Moufang generalized polygons,
including the infinite ones, was obtained by Tits and Weiss, and this huge
achievement is the theme of their monograph ”Moufang Polygons” [20]. In this
context we also mention the excellent monograph on Moufang quadrangles by De
Medts [3].

Finally we mention that in the period 1994-1999 Thas wrote a series of three
papers [23, 24, 25] on generalized quadrangles of order $(s,s^{2})$. This
paper can be considered as Part IV of that series.

##  2\. Finite generalized quadrangles

###  2.1. Definitions

A (finite) generalized quadrangle (GQ) is an incidence structure
$\mathcal{S}=(P,B,\mathbf{I})$ in which $P$ and $B$ are disjoint (nonempty)
sets of objects called points and lines (respectively) and for which
$\mathbf{I}$ is a symmetric point-line incidence relation satisfying the
following axioms:

  * (i)

Each point is incident with $1+t$ lines $(t\geq 1)$ and two distinct points
are incident with at most one line.

  * (ii)

Each line is incident with $1+s$ points $(s\geq 1)$ and two distinct lines are
incident with at most one point.

  * (iii)

If $x$ is a point and $L$ is a line not incident with $x$, then there is a
unique pair $(y,M)\in P\times B$ for which
$x\mathbf{I}M\mathbf{I}y\mathbf{I}L$.

The integers $s$ and $t$ are the parameters of the GQ and $\mathcal{S}$ is
said to have order $(s,t)$; if $s=t$, then $\mathcal{S}$ is said to have order
$s$. There is a point-line duality for GQ for which in any definition or
theorem the words ”point” and ”line” are interchanged and the parameters $s$
and $t$ are interchanged. Normally, we assume without further notice that the
dual of a given theorem or definition has also been given.

Given two (not necessarily distinct) points $x,y$ of the GQ $\mathcal{S}$, we
write $x\sim y$ and say that $x$ and $y$ are collinear, provided that there is
some line $L$ for which $x\mathbf{I}L\mathbf{I}y$. And $x\not\sim y$ means
that $x$ and $y$ are not collinear. Dually, for $L,M\in B$, we write $L\sim M$
or $L\not\sim M$ according as $L$ and $M$ are concurrent or nonconcurrent,
respectively. The line (respectively point) which is incident with distinct
collinear points $x,y$ (respectively distinct concurrent lines $L,M$) is
denoted by $xy$ (respectively $LM$ or $L\cap M$).

For $x\in P$, put $x^{\perp}=\\{y\in P\ |\ y\sim x\\}$, and note that $x\in
x^{\perp}$. If $A\subseteq P$, then A ”perp” is defined by
$A^{\perp}=\cap\\{x^{\perp}\ |\ x\in A\\}$. So for $x,y\in P$, with $x\not=y$,
we have $\\{x,y\\}^{\perp}=x^{\perp}\cap y^{\perp}$ and
$\\{x,y\\}^{\perp\perp}=\\{u\in P\ |\ u\in z^{\perp}\ \forall z\in
x^{\perp}\cap y^{\perp}\\}$. So $|\\{x,y\\}^{\perp}|=s+1$ or $t+1$ according
as $x\sim y$ or $x\not\sim y$, and $|\\{x,y\\}^{\perp\perp}|=s+1$ or
$|\\{x,y\\}^{\perp\perp}|\leq t+1$ according as $x\sim y$ or $x\not\sim y$.

A triad (of points) is a triple of pairwise noncollinear points. Given a triad
$T=\\{x,y,z\\}$, a center of $T$ is just a point of $T^{\perp}$.

For terminology, notations, results, etc., concerning finite generalized
quadrangles and not explicitly given here, see the monograph of Payne and Thas
[15, 16].

###  2.2. Restrictions on parameters

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a GQ of order $(s,t)$. If $|P|=v$ and
$|B|=b$, then $v=(s+1)(st+1)$ and $b=(t+1)(st+1)$; see 1.2.1 of [15, 16].
Also, $s+t$ divides $st(s+1)(t+1)$; see 1.2.2 of [15, 16].

If $s>1$ and $t>1$, then $t\leq s^{2}$, and dually $s\leq t^{2}$; see 1.2.3 of
[15, 16]. Also, for $s>1$ and $t>1$, $s^{2}=t$ if and only if each triad (of
points) has a constant number of centers, in which case this constant number
of centers is $s+1$; see 1.2.4 of [15, 16].

###  2.3. Regularity

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a GQ of order $(s,t)$. If $x\sim
y,x\not=y$, or if $x\not\sim y$ and $|\\{x,y\\}^{\perp\perp}|=t+1$, we say
that the pair $\\{x,y\\}$ is regular. The point $x$ is regular provided
$\\{x,y\\}$ is regular for all $y\in P,y\not=x$. A point $x$ is coregular
provided each line incident with $x$ is regular.

If $\mathcal{S}$ contains a regulair pair $\\{x,y\\}$, then either $s=1$ or
$s\geq t$; see 1.3.6 of [15, 16]. A coregular point in a GQ of order $s,s>1,$
is regular if and only $s$ is even; see 1.5.2 of [15, 16].

##  3\. Translation generalized quadrangles

###  3.1. Elations, translations, symmetries

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a GQ of order $(s,t)$, with
$s\not=1,t\not=1$. A collineation $\theta$ of $\mathcal{S}$ is an elation
about the point $p$ if $\theta$ = id or if $\theta$ fixes all lines incident
with $p$ and fixes no point of $P\setminus p^{\perp}$. If there is a group $H$
of elations about $p$ acting regularly on $P\setminus p^{\perp}$, we say
$\mathcal{S}$ is an elation generalized quadrangle (EGQ) with elation group
$H$ and base point $p$. Briefly we say that $(\mathcal{S}^{(p)},H)$ or
$\mathcal{S}^{(p)}$ is an EGQ. If the group $H$ is abelian, then we say that
the EGQ $(\mathcal{S}^{(p)},H)$ is a translation generalized quadrangle (TGQ)
with base point or translation point $p$ and translation group $H$. In such a
case $H$ is the set of all elations about $p$; see 8.6.4 of [15, 09]. For any
TGQ $\mathcal{S}^{(p)}$ the point $p$ is coregular, hence the parameters $s$
and $t$ of a TGQ satisfy $s\leq t$; see Section 2.3. Also, by 8.5.2 of [15,
16], for any TGQ with $s\not=t$ we have $s=q^{a}$ and $t=q^{a+1}$ with $q$ a
prime power and $a$ an odd integer; if $s$ (or $t$) is even, then by 8.6.1(iv)
of [15, 16] either $s=t$ or $s^{2}=t$.

If a collineation $\theta$ of the GQ $\mathcal{S}=(P,B,\mathbf{I})$ of order
$(s,t),s\not=1\not=t$, fixes each line of $L^{\perp}$, with $L\in B$, then
$\theta$ is a symmetry about $L$. By 8.1 of [15, 09] any symmetry about $L$ is
an elation about $L$. The line $L$ is called an axis of symmetry provided its
group of symmetries has order $s$. If $\mathcal{S}$ is a TGQ with translation
point $p$, then the lines $L_{0},L_{1},\ldots,L_{t}$ incident with $p$ are
axes of symmetry and the translation group $G$ contains the $t+1$ full groups
of symmetries about these lines; in fact $G$ is generated by these symmetries;
see 8.3.1 of [15, 09]. Conversely, if $p$ is a point of the GQ $\mathcal{S}$
for which $t+1$ lines incident with $p$ are axes of symmetry, then
$(\mathcal{S}^{(P},G)$ is a TGQ, where $G$ is generated by the symmetries
about the lines through $p$; see 8.3.1 of [15, 16].

For a detailed study of TGQ we refer to the monograph ”Translation Generalized
Quadrangles” by Thas, Thas and Van Maldeghem [27].

###  3.2. The projective model

In $\mathbf{PG}(2n+m-1,q)$ consider a set $O=O(n,m,q)$ of $q^{m}+1$
$(n-1)$-dimensional subspaces
$\mathbf{PG}^{(0)}(n-1,q),\mathbf{PG}^{(1)}(n-1,q),\ldots,\mathbf{PG}^{(q^{m})}(n-1,q)$,
every three of which generate a $\mathbf{PG}(3n-1,q)$ and such that each
element $\mathbf{PG}^{(i)}(n-1,q)$ of $O(n,m,q)$ is contained in a
$\mathbf{PG}^{(i)}(n+m-1,q)$ having no point in common with any
$\mathbf{PG}^{(j)}(n-1,q)$, for $j\not=i$. It is easy to check that
$\mathbf{PG}^{(i)}(n+m-1,q)$ is uniquely determined, $i=0,1,\ldots,q^{m}$. The
space $\mathbf{PG}^{(i)}(n+m-1,q)$ is called the tangent space of $O(n,m,q)$
at $\mathbf{PG}^{(i)}(n-1,q)$. For $n=m$ such a set $O(n,n,q)$ is called a
pseudo-oval or a generalized oval or an $[n-1]$-oval of $\mathbf{PG}(3n-1,q)$;
a generalized oval of $\mathbf{PG}(2,q)$ is just an oval of
$\mathbf{PG}(2,q)$. For $n\not=m$ such a set $O(n,m,q)$ is called a pseudo-
ovoid or a generalized ovoid or an $[n-1]$-ovoid or an egg of
$\mathbf{PG}(2n+m-1,q)$; a $[0]$-ovoid of $\mathbf{PG}(3,q)$ is just an ovoid
of $\mathbf{PG}(3,q)$.

Now embed $\mathbf{PG}(2n+m-1,q)$ in a $\mathbf{PG}(2n+m,q)$ and construct a
point-line geometry $T(n,m,q)=T(O)$ as follows.
Points are of three types:

  * (i)

the points of $\mathbf{PG}(2n+m,q)\setminus\mathbf{PG}(2n+m-1,q)$;

  * (ii)

the $(n+m)$-dimensional subspaces of $\mathbf{PG}(2n+m,q)$ wich intersect
$\mathbf{PG}(2n+m-1,q)$ in one of the spaces $\mathbf{PG}^{(i)}(n+m-1,q)$;

  * (iii)

the symbol $(\infty)$.

Lines are of two types:

  * (a)

the $n$-dimensional subspaces of $\mathbf{PG}(2n+m,q)$ which intersect
$\mathbf{PG}(2n+m-1,q)$ in one of the spaces $\mathbf{PG}^{(i)}(n-1,q)$;

  * (b)

the elements of $O(n,m,q)$.

Incidence in $T(n,m,q)$ is defined as follows. A point of type (i) is incident
only with lines of type (a); here the incidence is that of
$\mathbf{PG}(2n+m,q)$. A point of type (ii) is incident with all lines of type
(a) contained in it and with the unique element of $O(n,m,q)$ contained in it.
The point $(\infty)$ is incident with no line of type (a) and with all lines
of type (b).

######  Theorem 3.1 (8.7.1 of Payne and Thas [15, 16]).

The point-line geometry $T(O)=T(n,m,q)$ is a TGQ of order $(q^{n},q^{m})$ with
base point $(\infty)$. Conversely, every TGQ is isomorphic to a $T(n,m,q)$. It
follows that the theory of TGQ is equivalent to the theory of the sets
$O(n,m,q)$.

######  Corollary 3.2.

The following hold for every $O(n,m,q)$:

  * (i)

$n=m$ or $n(a+1)=ma$ with $a$ odd;

  * (ii)

if $q$ is even, then $n=m$ or $m=2n$.

Each TGQ $\mathcal{S}$ of order $(s,t)$ with base point $(\infty)$, with
$s\not=1\not=t$, has a kernel $\mathbb{K}$, where $\mathbb{K}$ is a field the
multiplicative group of which is isomorphic to the group of all collineations
of $\mathcal{S}$ fixing linewise the point $(\infty)$ and any given point not
collinear with $(\infty)$; see 8.6.5 of Payne and Thas [15, 16].

We have $|\mathbb{K}|\leq s$; see Payne and Thas [15, 16]. The field GF$(q)$
is a subfield of $\mathbb{K}$ if and only if $\mathcal{S}$ is isomorphic to a
$T(n,m,q)$; see 8.7.1 of Payne and Thas [15, 16].

Consider any GQ $T(n,m,q)=T(O)$, with $n=m$ and $q$ odd or with $n\not=m$. By
8.7.2 of [15, 16] or 3.9 and 3.10 of [27] the $q^{m}+1$ tangent spaces of
$O=O(n,m,q)$ form an $O^{\star}=O^{\star}(n,m,q)$ in the dual space of
$\mathbf{PG}(2n+m-1,q)$. So in addition to $T(O)$ there arises a TGQ
$T(O^{\star})$, also denoted $T^{\star}(O)$, with the same parameters. The TGQ
$T^{\star}(O)$ is called the translation dual of the TGQ $T(O)$. Examples are
known for which $T(O)\cong T^{\star}(O)$ and examples are known for which
$T(O)\not\cong T^{\star}(O)$; see Thas [23] and Thas, Thas and Van Maldeghem
[27].

Let $O(n,2n,q)$ be an egg of $\mathbf{PG}(4n-1,q)$. We say that $O(n,2n,q)$ is
good at the element $\mathbf{PG}^{(i)}(n-1,q)$ of $O(n,2n,q)$ if any
$\mathbf{PG}(3n-1,q)$ containing $\mathbf{PG}^{(i)}(n-1,q)$ and at least two
other elements of $O(n,2n,q)$, contains exactly $q^{n}+1$ elements of
$O(n,2n,q)$.

##  4\. Flocks and generalized quadrangles

Let $F$ be a flock of the quadratic cone $K$ of $\mathbf{PG}(3,q)$ with vertex
$x$, that is, a partition of $K\setminus\\{x\\}$ into $q$ disjoint irreducible
conics. Independently, Walker [29] and Thas [21] show that with each flock
there corresponds a translation plane $\mathcal{P}$ of order $q^{2}$. If
$\mathcal{P}$ is a semifield plane, then $F$ is called a semifield flock; see
e.g. [27].

Relying on the work of Payne [12, 13] and Kantor [9], Thas [22] proves that
with $F$ there corresponds a GQ $\mathcal{S}(F)$ of order $(q^{2},q)$. Such a
GQ $\mathcal{S}(F)$ will be called a flock GQ.

For the following theorem one has to rely on work of Casse, Thas and Wild [2],
Lunardon [11] and Payne [14]; see also 4.7 and 4.8 of Thas, Thas and Van
Maldeghem [27].

######  Theorem 4.1.

The flock GQ $\mathcal{S}(F)$ is a TGQ if and only if $F$ is a semifield
flock.

######  Theorem 4.2 (Thas [23, 25]).

If the egg $O$ is good, then the point-line dual of the translation dual
$T(O^{\star})$ of $T(O)$ is a flock GQ $\mathcal{S}(F)$.

######  Corollary 4.3.

If the egg $O$ is good, then the point-line dual of the translation dual
$T(O^{\star})$ of $T(O)$ is a semifield flock GQ $\mathcal{S}(F)$.

##  5\. Moufang quadrangles

###  5.1. Moufang conditions

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a GQ of order $(s,t)$. For a fixed point
$p$ define the following condition.

$(M)_{p}$: For any two distinct lines $A$ and $B$ of $\mathcal{S}$ incident
with $p$, the group of collineations of $\mathcal{S}$ fixing $A$ and $B$
pointwise and $p$ linewise is transitive on the set of lines distinct from $A$
and incident with a given point $x$ on $A$, with $x\not=p$.

The GQ $\mathcal{S}$ is said to satisfy condition $(M)$ provided it satsifies
$(M)_{p}$ for all points $p\in P$. For a fixed line $L\in B$ let
$(\widehat{M})_{L}$ be the condition that is the dual of $(M)_{p}$, and let
$(\widehat{M})$ be the dual of $(M)$. If $s\not=1\not=t$ and $\mathcal{S}$
satisfies both $(M)$ and $(\widehat{M})$ it is said to be a Moufang GQ; if
$s\not=1\not=t$ and $\mathcal{S}$ satisfies either $(M)$ or $(\widehat{M})$ it
is said to be a half Moufang GQ.

A celebrated result of Tits [19] is that all finite Moufang GQ are classical
or dual classical. In fact, Tits observes that his result easily follows from
the classification of all groups with a $BN$-pair of rank 2 by Fong and Seitz
[4, 5]. A huge achievement of Tits and Weiss [20] is the classification of all
(finite and infinite) Moufang GQ. In [26] Thas, Payne and Van Maldeghem prove
that half Moufang implies Moufang for finite GQ; in [18] Tent proves the
analogue for infinite GQ.

In their monograph Payne and Thas [15, 16] did an almost successful attempt to
prove the Moufang theorem (in the finite case) in a pure geometric way. To
complete a geometric proof of the theorem of Tits (in the finite case) it
would be sufficient to show in a geometric way that if, for each point $p$ of
$\mathcal{S}$, the GQ $\mathcal{S}^{(p)}$ is a TGQ of order
$(s,s^{2}),s\not=1$, then $\mathcal{S}$ is isomorphic to the classical GQ
$Q(5,s)$ arising from a non-singular elliptic quadric in $\mathbf{PG}(5,s)$.
As a corollary there would also follow a pure geometric proof of a large part
of the theorem of Fong and Seitz, namely in the case where the Weyl group is
$D_{8}$.

That ”missing part” will be a corollary of the much stronger results that
follow.

##  6\. Translation generalized quadrangles of order $(s,s^{2})$ with a
regular line not containing the base point

###  6.1. Introduction

In this section let $\mathcal{S}=(P,B,\mathbf{I})$ be a TGQ of order
$(s,s^{2}),s\not=1$, with base point $(\infty)$, and having a regular line $M$
not incident with $(\infty)$. Assume that $\mathcal{S}=T(n,2n,q)=T(O)$, with
$O=\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{2n}}\\}$, and where $M$ is concurrent
with the line $\pi_{0}$ of $\mathcal{S}$. By transitivity it is clear that all
lines concurrent with $\pi_{0}$ are regular. The projective space containing
$O$ will be denoted by $\mathbf{PG}(4n-1,q)$ and the projective space
containing $T(O)$ by $\mathbf{PG}(4n,q)$. The tangent space of $O$ at
$\pi_{i}$ will be denoted by $\tau_{i}$, with $i=0,1,\ldots,q^{2n}$. Further,
let $\widetilde{O}=\pi_{0}\cup\pi_{1}\cup\ldots\cup\pi_{q^{2n}}$.

###  6.2. Conics on $\widetilde{O}$

Consider lines $L_{0},L_{1}$ of $\mathcal{S}$, with
$L_{0}\\!\not\\!\mathbf{I}(\infty)\\!\not\\!\mathbf{I}L_{1},L_{0}\sim\pi_{0},L_{1}\sim\pi_{1},L_{0}\not\sim
L_{1}$. Let
$\\{L_{0},L_{1}\\}^{\perp}=\\{M_{0},M_{1},\ldots,M_{q^{n}}\\},\\{L_{0},L_{1}\\}^{\perp\perp}=\\{L_{0},L_{1},\ldots,L_{q^{n}}\\}$
and assume that notations are chosen in such a way that $L_{i}\sim\pi_{i}\sim
M_{i}$, with $i=0,1,\ldots q^{n}$. Then $\pi_{i},L_{i},M_{i}$ contain a common
point $r_{i}$ of $\mathcal{S}$, with $r_{i}$ a $3n$-dimensional space
containing $\tau_{i},i=0,1,\ldots,q^{n}$. So the $(n+1)$-dimensional space
$\langle L_{i},M_{i}\rangle$ generated by $L_{i}$ and $M_{i}$ intersects
$\mathbf{PG}(4n-1,q)$ in an $n$-dimensional space $\gamma_{i}$ containing
$\pi_{i}$ and contained in $\tau_{i}$.

As $\langle L_{i},M_{i}\rangle\cap\langle L_{j},M_{j}\rangle$, with $i\not=j$
and $i,j\in\\{0,1,\ldots,q^{n}\\}$, is a line $U_{ij}=U_{ji}$ we have $\langle
L_{i},M_{i}\rangle\cap\langle
L_{j},M_{j}\rangle\cap\mathbf{PG}(4n-1,q)=\gamma_{i}\cap\gamma_{j}=U_{ij}\cap\mathbf{PG}(4n-1,q)=\\{u_{ij}\\}$.
Hence $\gamma_{i}$ and $\gamma_{j}$ have exactly one point $u_{ij}=u_{ji}$ in
common. This point $u_{ij}$ belongs to $\tau_{i}\cap\tau_{j}$.

The grid, that is, the subquadrangle of $\mathcal{S}$ of order $(s,1)$,
containing $L_{1},L_{0}$ will be denoted by $\mathcal{G}$. Let us consider the
lines $L_{0},L_{1},L_{2},M_{0},M_{1},M_{2}$ of $\mathcal{G}$, and put
$L_{i}\cap M_{j}=n_{ij}$, with $i\not=j$ and $i,j\in\\{0,1,2\\}$. The space
$\langle L_{0},L_{1}\rangle\cap\mathbf{PG}(4n-1,q)=\langle
M_{0},M_{1}\rangle\cap\mathbf{PG}(4n-1,q)=\rho$ is the $2n$-dimensional space
$\langle\gamma_{0},\gamma_{1}\rangle$ containing
$\langle\pi_{0},\pi_{1}\rangle$. The line $n_{20}n_{21}$ of
$\mathbf{PG}(4n,q)$ is contained in $\langle L_{0},L_{1}\rangle$, so
intersects $\rho$ is a point $s_{2}$. Notice that $\langle
L_{0},L_{1}\rangle\cap L_{2}=n_{20}n_{21}$. Also $\langle
L_{0},L_{1},L_{2},M_{0},M_{1},M_{2}\rangle$ is $3n$-dimensional. The point
$s_{2}$ is the unique point of $\rho$ in $\pi_{2}$. Hence also the line
$n_{02}n_{12}$ contains $s_{2}$, and so $n_{20}n_{21}\cap
n_{02}n_{12}=\\{s_{2}\\}$. Analogously, let $n_{01}n_{02}\cap
n_{10}n_{20}=\\{s_{0}\\}$ and $n_{10}n_{12}\cap n_{01}n_{21}=\\{s_{1}\\}$.
Consequently the lines
$n_{01}n_{02},n_{10}n_{12},n_{20}n_{21},n_{10}n_{20},n_{01}n_{21},n_{02}n_{12}$
belong to a common hyperbolic quadric $\mathcal{H}$ of some
$\mathbf{PG}(3,q)$.

Let $n\in n_{20}n_{21}\setminus\\{n_{20},n_{21},s_{2}\\}$. Through $n$ there
is a line $M$ of the GQ intersecting $L_{0}$ and $L_{1}$. Let $m_{i}$ be the
common point of $M$ and $L_{i}$, $i=0,1$. Assume, by way of contradiction,
that $n,m_{0},m_{1}$ are not collinear in $\mathbf{PG}(4n,q)$. Then
$M\cap\rho$ contains the line $\langle
n,m_{0},m_{1}\rangle\cap\mathbf{PG}(4n-1,q)$ (as $n,m_{0},m_{1}$ belong to
$\langle L_{0},L_{1}\rangle=\langle M_{0},M_{1}\rangle$), so $M$ contains a
point of $\langle\pi_{0},\pi_{1}\rangle$, so
$M\cap\mathbf{PG}(4n-1,q)=\pi_{i},i\not=0,1$, contains a point of
$\langle\pi_{0},\pi_{1}\rangle$, clearly a contradiction. Hence the points
$n,m_{0},m_{1}$ are collinear in $\mathbf{PG}(4n,q)$. Let
$n_{3},n_{4},\ldots,n_{q}$ be the points of $n_{20}n_{21}$ distinct from
$n_{20},n_{21},s_{2}$, let $m_{0}^{i}$ be the point of $L_{0}$ collinear in
$\mathcal{S}$ with $n_{i}$, let $m_{1}^{i}$ be the point of $L_{1}$ collinear
in $\mathcal{S}$ with $n_{i}$, and let
$m_{0}^{i}m_{1}^{i}\cap\rho=\\{s_{i}\\}$, with $i=3,4,\ldots,q$. Notations are
chosen in such a way that $s_{i}\in\pi_{i}$, with $i=3,4,\ldots,q$. From the
foregoing it follows that the $q+1$ lines of $\mathcal{H}$ which intersect
$L_{0},L_{1},L_{2}$ are contained in the $n$-dimensional spaces
$M_{0},M_{1},\ldots,M_{q}$ which respectively contain
$\pi_{0},\pi_{1},\ldots,\pi_{q}$. Similarly the $q+1$ lines of $\mathcal{H}$
intersecting $M_{0},M_{1},M_{2}$ are contained in respectively
$L_{0},L_{1},\ldots,L_{q}$. Let $N_{i}$ be the line of $\mathcal{H}$ contained
in $L_{i}$, and let $R_{i}$ be the line of $\mathcal{H}$ contained in
$M_{i},i=0,1,\ldots,q$. Then $N_{i}\cap R_{i}=\\{s_{i}\\},i=0,1,\ldots,q$. So
$\\{s_{0},s_{1},\dots,s_{q}\\}$ is a non-singular conic $\mathcal{C}$
contained in $\widetilde{O}$.

The tangent line $T_{i}$ of $\mathcal{C}$ at $s_{i}$ is contained in $\langle
L_{i},M_{i}\rangle\cap\mathbf{PG}(4n-1,q)$, it is the line $\langle
R_{i},N_{i}\rangle\cap\mathbf{PG}(4n-1,q)$. If $\sigma$ is the plane
containing $\mathcal{C}$, then
$T_{i}=\gamma_{i}\cap\sigma=\tau_{i}\cap\sigma$, with $i=0,1,\ldots,q$.

In the foregoing the lines $L_{0}$ and $L_{1}$ may be replaced by any two
distinct lines of $\\{L_{0},L_{1}\\}^{\perp\perp}$.

######  Lemma 6.1.

The conic $\mathcal{C}$ is uniquely defined by $\pi_{0},\pi_{1}$ and $s_{2}$.

Proof We have $T_{i}=\gamma_{i}\cap\sigma=\tau_{i}\cap\sigma$, with
$\gamma_{i}=\langle L_{i},M_{i}\rangle\cap\mathbf{PG}(4n-1,q),i=0,1,\ldots,q$.
Also, $\rho=\langle\gamma_{0},\gamma_{1}\rangle,\rho=\langle
L_{0},L_{1}\rangle\cap\mathbf{PG}(4n-1,q)=\langle\pi_{0},\pi_{1},s_{2}\rangle,\gamma_{i}=\rho\cap\tau_{i},i=0,1$.
So $\gamma_{0},\gamma_{1}$ and also $T_{0}\cap
T_{1}=\gamma_{0}\cap\gamma_{1}=\\{u_{01}\\}$ are uniquely defined by
$\pi_{0},\pi_{1}$ and $s_{2}$.

Now we project from $u_{01}$ onto a $(2n-1)$-dimensional space $\bar{\rho}$ of
$\rho$, not containing $u_{01}$. Let the image of $\pi_{i}$ be
$\bar{\pi}_{i}$, with $i=0,1$, and let the image of $s_{2}$ be $\bar{s}_{2}$.
There is a unique line $\bar{U}$ in $\bar{\rho}$ containing $\bar{s}_{2}$ and
intersecting $\bar{\pi}_{0}$ and $\bar{\pi}_{1}$. Then $\langle
u_{01},\bar{U}\rangle$ is the unique plane in $\rho$ containing $u_{01},s_{2}$
and intersecting $\pi_{0}$ and $\pi_{1}$. Hence $\langle
u_{01},\bar{U}\rangle=\sigma$ and $\\{s_{i}\\}=\sigma\cap\pi_{i},i=0,1$.
Consequently $s_{0}$ and $s_{1}$ are uniquely defined by $\pi_{0},\pi_{1}$ and
$s_{2}$. So $\mathcal{C}$ is the unique conic containing $s_{0},s_{1},s_{2}$
and having $s_{0}u_{01}$ and $s_{1}u_{01}$ as tangent lines. $\blacksquare$

######  Remark 6.2.

In 6.1 the subscripts $0,1,2$ may be replaced by any three distinct subscripts
$i,j,k\in\\{0,1,\ldots,q^{n}\\}$.

Definition. A conic $\mathcal{C}$ of the type described in 6.1 or in 6.2,
where we start from any three distinct lines of some
$\\{L_{0},L_{1}\\}^{\perp\perp}$, will be called a $\widetilde{\pi}_{0}$-conic
of $\widetilde{O}$. If $\mathcal{C}$ intersects $\pi_{0}$ it will also be
called a $\pi_{0}$-conic of $\widetilde{O}$.

######  Lemma 6.3.

Let $\pi_{i},\pi_{j},\pi_{k}$ be distinct elements of $O$, with
$\pi_{0}\in\\{\pi_{i},\pi_{j},\pi_{k}\\}$, and let $s$ be a point of
$\pi_{k}$. Then $s$ is contained in exactly one $\pi_{0}$-conic intersecting
$\pi_{i}$ and $\pi_{j}$.

Proof Let $L_{i}$ and $L_{j}$ be nonconcurrent lines of $\mathcal{S}$ which
are concurrent with respectively $\pi_{i}$ and $\pi_{j}$, but not incident
with the point $(\infty)$. Assume also that $L_{i}$ and $L_{j}$ are chosen in
such a way that $\langle
L_{i},L_{j}\rangle\cap\mathbf{PG}(4n-1,q)=\langle\pi_{i},\pi_{j},s\rangle$. If
$\pi_{0}\in\\{\pi_{i},\pi_{j}\\}$, then the lines $L_{i}$ and $L_{j}$ define a
$\pi_{0}$-conic containing $s$ and intersecting $\pi_{i}$ and $\pi_{j}$. By
6.1 this $\pi_{0}$-conic is unique. Now assume that $\pi_{0}=\pi_{k}$. As in
$\mathbf{PG}(4n,q)$ there is a line containing $s$ and intersecting $L_{i}$
and $L_{j}$, there is also a line $M$ in $\mathcal{S}$ concurrent with
$\pi_{k}$ and intersecting $L_{i}$ and $L_{j}$. As the line $M$ is regular in
$\mathcal{S}$, also the pair $\\{L_{i},L_{j}\\}$ is regular in $\mathcal{S}$.
Interchanging roles of $L_{0}$ and $L_{i}$, see also 6.2, the result follows.
$\blacksquare$

###  6.3. Segre varieties defined by $\\{O,\pi_{0}\\}$

For Segre varieties we refer to the monograph ”General Galois Geometries” by
Hirschfeld and Thas [7, 8].

######  Lemma 6.4.

If $\pi_{i}$ and $\pi_{j}$ are distinct elements of $O\setminus\\{\pi_{0}\\}$,
then all $\pi_{0}$-conics intersecting $\pi_{0},\pi_{i},\pi_{j}$ belong to a
common Segre variety $\mathcal{S}_{2;n-1}$.

Proof Let $\langle\pi_{0},\pi_{i},\pi_{j}\rangle=\mathbf{PG}(3n-1,q)$ and
$\tau_{k}\cap\mathbf{PG}(3n-1,q)=\bar{\tau}_{k}$, with $k\in\\{0,i,j\\}$. Then
$\bar{\tau}_{0},\bar{\tau}_{i},\bar{\tau}_{j}$ are $(2n-1)$-dimensional.
Further, let $\bar{\tau}_{i}\cap\bar{\tau}_{j}=\eta$. So $\eta$ is
$(n-1)$-dimensional. Let $s$ be any point of $\pi_{0}$, and let $\mathcal{C}$
be the $\pi_{0}$-conic containing $s$ and intersecting $\pi_{i}$ and
$\pi_{j}$. Let $\mathcal{C}\cap\pi_{i}=\\{s_{i}\\}$ and
$\mathcal{C}\cap\pi_{j}=\\{s_{j}\\}$. The tangent lines of $\mathcal{C}$ at
respectively $s_{i}$ and $s_{j}$ have a point $u_{ij}$ in common, which is the
common point of $\langle s,\pi_{i},\pi_{j}\rangle\cap\tau_{i}=\gamma_{i}$ and
$\langle s,\pi_{i},\pi_{j}\rangle\cap\tau_{j}=\gamma_{j}$; so $u_{ij}$ is
contained in $\bar{\tau}_{i}\cap\bar{\tau}_{j}=\eta$. Hence the plane $\sigma$
containing $\mathcal{C}$ intersects the spaces $\pi_{0},\pi_{i},\pi_{j},\eta$.

It follows that the $(q^{n}-1)/(q-1)$ planes $\sigma$ are exactly the
$(q^{n}-1)/(q-1)$ planes of $\mathbf{PG}(3n-1,q)$ which intersect
$\pi_{0},\pi_{i},\pi_{j},\eta$. It follows that the union of these planes is a
Segre variety $\mathcal{S}_{2;n-1}$ of a plane and an $(n-1)$-dimensional
space; see [7, 8].

We conclude that the $(q^{n}-1)/(q-1)$ $\pi_{0}$-conics intersecting
$\pi_{0},\pi_{i}$ and $\pi_{j}$ belong to $\mathcal{S}_{2;n-1}$.
$\blacksquare$

######  Lemma 6.5.

Let $\pi_{i}$ and $\pi_{j}$ be distinct elements of $O\setminus\\{\pi_{0}\\}$.
Further, let $\langle\pi_{0},\pi_{i},\pi_{j}\rangle=\mathbf{PG}(3n-1,q)$, and
$\tau_{l}\cap\mathbf{PG}(3n-1,q)=\bar{\tau}_{l}$, with $l\in\\{0,i,j\\}$.
Finally, let
$\bar{\tau}_{0}\cap\bar{\tau}_{j}=\eta_{i},\bar{\tau}_{0}\cap\bar{\tau}_{i}=\eta_{j},\bar{\tau}_{i}\cap\bar{\tau}_{j}=\eta_{0}$.
If $q$ is even, then $\eta_{0}=\eta_{i}=\eta_{j}\subset\mathbf{PG}(3n-1,q)$;
if $q$ is odd, then
$\langle\pi_{0},\eta_{0}\rangle,\langle\pi_{i},\eta_{i}\rangle,\langle\pi_{j},\eta_{j}\rangle$
have an $(n-1)$-dimensional space in common.

Proof Let $q$ be even. By the proof of 6.4 the $(q^{n}-1)/(q-1)$ nuclei of the
corresponding $\pi_{0}$-conics intersecting $\pi_{0},\pi_{i},\pi_{j}$ belong
to $\eta_{0}$, respectively $\eta_{i},\eta_{j}$. Hence
$\eta_{0}=\eta_{i}=\eta_{j}$. Clearly that space belongs to
$\mathbf{PG}(3n-1,q)$.

Next, let $q$ be odd. Let $\mathcal{C}$ be a $\pi_{0}$-conic intersecting
$\pi_{0},\pi_{i},\pi_{j}$, let $\mathcal{C}\cap\pi_{l}=\\{s_{l}\\}$ and let
$T_{l}$ be the tangent line of $\mathcal{C}$ at $s_{l}$, with $l=0,i,j$.
Further, let $T_{0}\cap T_{i}=\\{u_{0i}\\},T_{0}\cap
T_{j}=\\{u_{0j}\\},T_{i}\cap T_{j}=\\{u_{ij}\\}$. Then, by an elementary
property of conics, the lines $s_{i}u_{0j},s_{j}u_{oi},s_{0}u_{ij}$ have a
point $z$ in common. As the $(q^{n}-1)/(q-1)$ points $s_{0}$, respectively
$s_{i},s_{j},u_{0i},u_{0j},u_{ij}$ belong to a maximal $(n-1)$-space of
$\mathcal{S}_{2;n-1}$, also the $(q^{n}-1)/(q-1)$ points $z$ will belong to a
maximal $(n-1)$-space of $\mathcal{S}_{2;n-1}$. It follows that the
$(2n-1)$-dimensional spaces
$\langle\pi_{0},\eta_{0}\rangle,\langle\pi_{i},\eta_{i}\rangle,\langle\pi_{j},\eta_{j}\rangle$
have an $(n-1)$-dimensional space in common. $\blacksquare$

Consider two distinct elements of $O\setminus\\{\pi_{0}\\}$, say $\pi_{1}$ and
$\pi_{2}$. Let $\langle\pi_{0},\pi_{1},\pi_{2}\rangle=\mathbf{PG}(3n-1,q)$.
The $\pi_{0}$-conics intersecting $\pi_{0},\pi_{1},\pi_{2}$ contain
$(q^{n}-1)(q+1)/(q-1)$ points of $\widetilde{O}$. By 6.4 these points belong
to a common Segre variety $\mathcal{S}_{2;n-1}$. Let $\mathcal{C}$ be some
$\pi_{0}$-conic intersecting $\pi_{0},\pi_{1},\pi_{2}$ and let $\mathcal{V}$
be the set of all points on the $(q^{n}-1)/(q-1)$ $\pi_{0}$-conics
intersecting $\pi_{0},\pi_{1},\pi_{2}$. Further, let $s\in\mathcal{C}$.

######  Lemma 6.6.

The maximal $(n-1)$-space of $\mathcal{S}_{2;n-1}$ (distinct from the plane of
$\mathcal{C}$ if $n=3$) containing $s$ is contained in $\mathcal{V}$.

Proof Let $\sigma$ be the plane of $\mathcal{C}$ and let $\sigma^{\prime}$ be
the plane of a second $\pi_{0}$-conic $\mathcal{C}^{\prime}$ intersecting
$\pi_{0},\pi_{1},\pi_{2}$. The maximal $(n-1)$-spaces of $\mathcal{S}_{2;n-1}$
(of the family containing $\pi_{0},\pi_{1},\pi_{2}$ if $n=3$) define a linear
projectivity $\theta$ from $\sigma$ onto $\sigma^{\prime}$. If
$\mathcal{C}\cap\pi_{i}=\\{s_{i}\\},\mathcal{C}^{\prime}\cap\pi_{i}=\\{s^{\prime}_{i}\\}$,
then $s_{i}^{\theta}=s^{\prime}_{i},i=0,1,2$. Let
$\bar{\tau}_{i}=\tau_{i}\cap\langle\pi_{0},\pi_{1},\pi_{2}\rangle$, with
$i=0,1,2$. If $\bar{\tau}_{i}\cap\bar{\tau}_{j}=\eta_{k}$ with
$\\{i,j,k\\}=\\{0,1,2\\}$, and
$\sigma\cap\eta_{k}=\\{l_{k}\\},\sigma^{\prime}\cap\eta_{k}=\\{l^{\prime}_{k}\\},k=0,1,2$,
then $l_{k}$ is the intersection of the tangent lines of $\mathcal{C}$ at
$s_{i}$ and $s_{j}$, and $l^{\prime}_{k}$ is the intersection of the tangent
lines of $\mathcal{C}^{\prime}$ at $s^{\prime}_{i}$ and $s^{\prime}_{j}$, with
$\\{i,j,k\\}=\\{0,1,2\\}$. Also $\l^{\theta}_{k}=l^{\prime}_{k}$, with
$k=0,1,2$. It follows that $\mathcal{C}^{\theta}=\mathcal{C}^{\prime}$. Hence
for any point $s\in\mathcal{C}$ the maximal $(n-1)$-space of
$\mathcal{S}_{2;n-1}$ (distinct from the plane of $\mathcal{C}$ if $n=3$)
containing $s$ is contained in $\mathcal{V}$. $\blacksquare$

######  Lemma 6.7.

The set $\mathcal{V}$ is the union of $q+1$ elements of $O$.

Proof From 6.6 follows that $\mathcal{V}$ is the union of $q+1$
$(n-1)$-dimensional spaces. Let $\pi$ be such an $(n-1)$-dimensional space. As
$\pi\subset\widetilde{O}$, any line of $\pi$ belongs to some element of $O$.
Hence $\pi\in O$, so $\mathcal{V}$ is the union of $q+1$ elements of $O$.
$\blacksquare$

Definition. The set $C$ consisting of the $q+1$ elements of $O$ which
intersect some $\pi_{0}$-conic $\mathcal{C}$ of $\widetilde{O}$ is called a
$\Pi_{0}$-conic of $O$. Also, $\langle C\rangle\cap\tau_{i}=\bar{\tau}_{i}$,
with $\pi_{i}\in C$, is called the tangent space of $C$ at $\pi_{i}$. Clearly
any three distinct elements of $O$ are contained in exactly one
$\Pi_{0}$-conic.

Let $\delta$ be the projection from $\pi_{0}$ onto a $(3n-1)$-dimensional
subspace $\Phi$ of $\mathbf{PG}(4n-1,q)$ which is skew to $\pi_{0}$. Let
$\tau_{0}\cap\Phi=\tau^{\star}_{0}$ and let
$\langle\pi_{0},\pi_{j}\rangle\cap\Phi=\pi^{\star}_{j}$, with
$j=1,2,\ldots,q^{2n}$. Then
$\tau^{\star}_{0}\cup\pi^{\star}_{1}\cup\pi^{\star}_{2}\cup\ldots\cup\pi^{\star}_{q^{2n}}=\Phi$.
Further, let
$\langle\pi^{\star}_{i},\pi^{\star}_{j}\rangle\cap\tau^{\star}_{0}=\pi^{\star}_{ij}=\pi^{\star}_{ji}$,
with $0\not=i\not=j\not=0$. The space $\tau^{\star}_{0}$ is
$(2n-1)$-dimensional, the spaces $\pi^{\star}_{j}$ are $(n-1)$-dimensional,
and the spaces $\pi^{\star}_{ij}$ are $(n-1)$-dimensional.

######  Lemma 6.8.

For any $i,j$, with $0\not=i\not=j\not=0$, the Segre variety
$\mathcal{S}_{1;n-1}$ defined by
$\pi^{\star}_{i},\pi^{\star}_{j},\pi^{\star}_{ij}$, contains exactly $q$
elements $\pi^{\star}_{k}$.

Proof Consider a $\pi_{0}$-conic $\mathcal{C}$ intersecting
$\pi_{0},\pi_{i},\pi_{j}$. The $q+1$ elements of the $\Pi_{0}$-conic $C$
defined by $\mathcal{C}$ are maximal $(n-1)$-spaces of a Segre variety
$\mathcal{S}_{2;n-1}$; see 6.6 and 6.7. Let
$\langle\pi_{0},\pi_{i},\pi_{j}\rangle\cap\tau_{0}=\bar{\tau}_{0}$ and let
$\pi_{0},\pi_{i},\pi_{j},\pi_{k_{1}},\ldots,\pi_{k_{q-2}}$ be the elements of
the $\Pi_{0}$-conic $C$. Then
$\bar{\tau}_{0}\cap\Phi=\pi^{\star}_{ij},\pi^{\star}_{i},\pi^{\star}_{j},\pi^{\star}_{k_{1}},\ldots,\pi^{\star}_{k_{q-2}}$
are the elements of a system of maximal $(n-1)$-dimensional spaces of a
$\mathcal{S}_{1;n-1}$. So the Segre variety $\mathcal{S}_{1;n-1}$ defined by
$\pi^{\star}_{i},\pi^{\star}_{j},\pi^{\star}_{ij}$ contains exactly $q$
elements $\pi^{\star}_{k},k\in\\{1,2,\ldots,q^{2n}\\}$. $\blacksquare$

######  Remark 6.9.

For $n\not=2$ the elements of the family of maximal $(n-1)$-dimensional spaces
of a $\mathcal{S}_{1;n-1}$, and for $n=2$ the elements of any of the two
families of lines of $\mathcal{S}_{1;1}$, form an $(n-1)$-regulus of
$\langle\mathcal{S}_{1;n-1}\rangle$; see Chapter 4 of [8].

###  6.4. $\Pi_{0}$-sets $(\Pi,\Gamma)$

Consider again
$\\{L_{0},L_{1}\\}^{\perp\perp}=\\{L_{0},L_{1},\ldots,L_{q^{n}}\\},L_{0}\\!\not\\!\mathbf{I}(\infty)\\!\not\mathbf{I}L_{1}$,
with $L_{i}\sim\pi_{i},i=0,1,\ldots,q^{n}$. Also, let
$\\{L_{0},L_{1}\\}^{\perp}=\\{M_{0},M_{1},\ldots,M_{q^{n}}\\}$, with
$\pi_{i}\sim M_{i}\sim L_{i}\sim\pi_{i},i=0,1,\ldots,q^{n}$. Further, $\langle
L_{i},M_{i}\rangle\cap\tau_{i}=\gamma_{i},i=0,1,\ldots,q^{n}$. Then
$\pi_{i}\subset\gamma_{i}$.

Definition. With $\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{n}}\\}=\Pi$ and
$\\{\gamma_{0},\gamma_{1},\ldots,\gamma_{q^{n}}\\}=\Gamma$, the pair
$(\Pi,\Gamma)$ is called a $\Pi_{0}$-set.

######  Lemma 6.10.

  * (i)

$\gamma_{i}\cap\gamma_{j}=\\{u_{ij}\\}$, with $i,j$ distinct elements of
$\\{0,1,\ldots,q^{n}\\}$.

  * (ii)

Any three distinct elements $\pi_{i},\pi_{j},\pi_{k}$, with
$0\in\\{i,j,k\\}\subset\\{0,1,\ldots,q^{n}\\}$, are contained in exactly one
$\Pi_{0}$-conic of $\Pi$.

  * (iii)

If $\pi_{i},\pi_{j},\pi_{k}$, with $0\in\\{i,j,k\\}$, are distinct elements of
$\Pi$, then $|\gamma_{i}\cap\langle\pi_{j},\pi_{k}\rangle|=1$.

  * (iv)

If we know $\Pi$ and some $\gamma_{i}\in\Gamma$, then all $\gamma_{j}$ of
$\Gamma$ are determined.

Proof

  * (i)

Was proved in Section 6.2.

  * (ii)

From 6.3, 6.4, 6.6, 6.7 follows (ii).

  * (iii)

The spaces $\pi_{i},\pi_{j},\pi_{k}$, with $0\in\\{i,j,k\\}$, are contained in
exactly one $\Pi_{0}$-conic $C$. The tangent space $\bar{\tau}_{i}$ of $C$ at
$\pi_{i}$ contains the space $\gamma_{i}$. Hence the $n$-space $\gamma_{i}$
has exactly one point in common with the $(2n-1)$-space
$\langle\pi_{j},\pi_{k}\rangle$ of the $(3n-1)$-space $\langle C\rangle$.

  * (iv)

Assume we know $\Pi$ and some $\gamma_{i}\in\Gamma$. Let
$\pi_{j}\in\Pi,i\not=j$. Then $\tau_{j}$ has exactly one point $x$ in common
with $\gamma_{i}$. So $\gamma_{j}=\langle\pi_{j},x\rangle.$

$\blacksquare$

######  Lemma 6.11.

The $\Pi_{0}$-set $(\Pi,\Gamma)$ is uniquely defined by
$\\{\gamma_{0},\pi_{j}\\}$ or $\\{\gamma_{j},\pi_{0}\\}$, with
$j\in\\{1,2,\ldots,q^{n}\\}$. The elements of
$\Pi\setminus\\{\pi_{0},\pi_{j}\\}$ are the $q^{n}-1$ elements of
$O\setminus\\{\pi_{0},\pi_{j}\\}$ containing the common points of
$\widetilde{O}\setminus(\pi_{0}\cup\pi_{j})$ and
$\langle\gamma_{0},\pi_{j}\rangle$.

Proof By the proof of 6.10(iv) it is sufficient to assume that
$\\{\gamma_{0},\pi_{j}\\}$ is given. Each $n$-dimensional space containing
$\pi_{0}$, distinct from $\gamma_{0}$, and contained in
$\langle\gamma_{0},\pi_{j}\rangle$ has exactly one point in common with
$\widetilde{O}\setminus\pi_{0}$. In this way there arise $q^{n}-1$ points of
$\widetilde{O}\setminus(\pi_{0}\cup\pi_{j})$. Consider
$\pi_{k}\in\Pi,\pi_{k}\not\in\\{\pi_{0},\pi_{j}\\}$. Then by 6.10
$\gamma_{0}\subset\langle\pi_{0},\pi_{j},\pi_{k}\rangle$, so
$\langle\gamma_{0},\pi_{j}\rangle$ has exactly one point in common with
$\pi_{k}$. Hence the $q^{n}-1$ common points of
$\widetilde{O}\setminus(\pi_{0}\cup\pi_{j})$ and
$\langle\gamma_{0},\pi_{j}\rangle$ determine the $q^{n}-1$ elements of
$\Pi\setminus\\{\pi_{0},\pi_{j}\\}$. $\blacksquare$

######  Lemma 6.12.

Let $\gamma_{0}$ be any $n$-dimensional space containing $\pi_{0}$ and let
$\pi_{k}\in O\setminus\\{\pi_{0}\\}$. Then $\\{\gamma_{0},\pi_{k}\\}$
determines exactly one $\Pi_{0}$-set $(\Pi,\Gamma)$.

Proof Let $L_{0},M_{0}$ be two $n$-dimensional spaces of $\mathbf{PG}(4n,q)$
for which $\langle L_{0},M_{0}\rangle\cap\mathbf{PG}(4n-1,q)=\gamma_{0}$. Let
$L_{k}$ be an $n$-dimensional space containing $\pi_{k}$ and a point of
$M_{0}\setminus\pi_{0}$. Then the $\Pi_{0}$-set $(\Pi,\Gamma)$ corresponding
to $\\{L_{0},L_{k}\\}^{\perp\perp}$ is uniquely determined by the pair
$\\{\gamma_{0},\pi_{k}\\}$. $\blacksquare$

######  Lemma 6.13.

Let $\gamma_{0}$ be any $n$-dimensional space containing $\pi_{0}$. Then the
$\Pi_{0}$-sets $(\Pi,\Gamma)$ with $\gamma_{0}\in\Gamma$ determine a partition
of $O\setminus\\{\pi_{0}\\}$.

Proof By 6.11 and 6.12 all $\Pi_{0}$-sets $(\Pi,\Gamma)$ with
$\gamma_{0}\in\Gamma$ give rise to sets $\Pi\setminus\\{\pi_{0}\\}$ defining a
partition of $O\setminus\\{\pi_{0}\\}$. $\blacksquare$

######  Remark 6.14.

Consider any $\Pi_{0}$-set $(\Pi,\Gamma)$. From Section 6.3 follows that for
$q$ even all elements of $\Gamma$ have a point $n$ in common. This point $n$
is called the kernel of $(\Pi,\Gamma)$. The tangent spaces of $O$ containing
the elements of $\Pi$ are exactly the $q^{n}+1$ tangent spaces of $O$
containing the point $n$. Now let $q$ be odd. Consider
$\pi_{0},\pi_{i},\pi_{j}\in\Pi$, with $0\not=i\not=j\not=0$. Let
$\gamma_{0}\cap\gamma_{i}=\\{u_{0i}\\},\gamma_{0}\cap\gamma_{j}=\\{u_{0j}\\},\gamma_{i}\cap\gamma_{j}=\\{u_{ij}\\}$.
Then the plane $\sigma=\langle u_{0i},u_{0j},u_{ij}\rangle$ contains a
$\pi_{0}$-conic $\mathcal{C}$ and let
$\mathcal{C}\cap\pi_{0}=\\{s_{0}\\},\mathcal{C}\cap\pi_{i}=\\{s_{i}\\},\mathcal{C}\cap\pi_{j}=\\{s_{j}\\}$.
Then the lines $u_{0i}s_{j},u_{0j}s_{i}$ and $u_{ij}s_{0}$ have a point $x$ in
common.

###  6.5. Projections and projectivities

In Section 6.3 we introduced the projection $\delta$ from $\pi_{0}$ onto a
$(3n-1)$-dimensional subspace $\Phi$ of $\mathbf{PG}(4n-1,q)$ which is skew to
$\pi_{0}$. Let
$\tau_{0}\cap\Phi=\tau^{\star}_{0}=(\tau_{0}\setminus\pi_{0})^{\delta}$ and
let $\langle\pi_{0},\pi_{j}\rangle\cap\Phi=\pi^{\star}_{j}=\pi^{\delta}_{j}$,
with $j=1,2,\ldots,q^{2n}$. Then
$\tau^{\star}_{0}\cup\pi^{\star}_{1}\cup\pi^{\star}_{2}\cup\ldots\cup\pi^{\star}_{q^{2n}}=\Phi$.
Further, let
$\langle\pi^{\star}_{i},\pi^{\star}_{j}\rangle\cap\tau^{\star}_{0}=\pi^{\star}_{ij}=\pi^{\star}_{ji}$,
with $0\not=i\not=j\not=0$. The space $\tau^{\star}_{0}$ is
$(2n-1)$-dimensional, the spaces $\pi^{\star}_{j}$ are $(n-1)$-dimensional,
and the spaces $\pi^{\star}_{ij}$ are $(n-1)$-dimensional.

Let $C$ be a $\Pi_{0}$-conic. Then for $\pi_{i}\in C,\langle
C\rangle\cap\tau_{i}=\bar{\tau}_{i}(C)$, or $\bar{\tau}_{i}$ if no confusion
is possible, is the tangent space of $C$ at $\pi_{i}$. The projection of
$C\setminus\\{\pi_{0}\\}$ together with $\bar{\tau}_{0}\cap\tau^{\star}_{0}$
is an $(n-1)$-regulus $\mathcal{R}$. It is a family of maximal spaces of a
Segre variety $\mathcal{S}_{1;n-1}$. The space
$\bar{\tau}^{\star}_{0}=\bar{\tau}_{0}\cap\tau^{\star}_{0}$ is the space at
infinity of $\mathcal{R}$ (or $\mathcal{S}_{1;n-1}$). If
$\pi^{\star}_{i},\pi^{\star}_{j}\in\mathcal{R}\setminus\\{\bar{\tau}^{\star}_{0}\\},i\not=j$,
then
$\pi^{\star}_{ij}=\langle\pi^{\star}_{i},\pi^{\star}_{j}\rangle\cap\tau^{\star}_{0}=\bar{\tau}^{\star}_{0}$.

Let $\mathcal{R}_{0},\mathcal{R}_{1},\ldots$, respectively
$\mathcal{S}^{(o)}_{1;n-1},\mathcal{S}^{(1)}_{1;n-1},\ldots$, be the
$(n-1)$-reguli, respectively Segre varieties, defined by the pairs of elements
of $\\{\pi^{\star}_{1},\pi^{\star}_{2},\ldots,\pi^{\star}_{q^{2n}}\\}$.
Further, let $\xi_{1},\xi_{2},\dots$ be the $(q^{n}-1)/(q-1)$ $2n$-dimensional
subspaces of $\Phi$ which contain $\tau^{\star}_{0}$.

Then $\xi_{i}\cap\pi^{\star}_{k},k\in\\{1,2,\ldots,q^{2n}\\}$, is a point and
$\xi_{i}\cap\mathcal{S}^{(j)}_{1;n-1}$ consists of the space at infinity
$\langle\mathcal{S}^{(j)}_{1;n-1}\rangle\cap\tau^{\star}_{0}$ of
$\mathcal{S}^{(j)}_{1;n-1}$ and a maximal line of $\mathcal{S}^{(j)}_{1;n-1}$
(distinct from the $\mathcal{R}_{j}$’s if $n=2$), that is, a line intersecting
the $q+1$ elements of $\mathcal{R}_{j}$.

Let the affine space $\xi_{i}\setminus\tau^{\star}_{0}$ be denoted by
$\mathbf{AG}^{(i)}(2n,q),i=1,2,\dots,(q^{n}-1)/(q-1)$. Let
$\alpha_{i}:x\mapsto\pi^{\star}$, with $x$ any point of
$\mathbf{AG}^{(i)}(2n,q)$ and $\pi^{\star}$ the element of
$\\{\pi^{\star}_{1},\pi^{\star}_{2},\ldots,\pi^{\star}_{q^{2n}}\\}$ which
contains $x$. Then on
$\\{\pi^{\star}_{1},\pi^{\star}_{2},\ldots,\pi^{\star}_{q^{2n}}\\}$ a
$2n$-dimensional affine space $\bar{\mathbf{AG}}^{(i)}(2n,q)$ is induced
having as lines the $(n-1)$-reguli $\mathcal{R}_{k}$ minus their space at
infinity. For $q\not=2$ we will prove that $\bar{\mathbf{AG}}^{(i)}(2n,q)$ is
independent from $i$, which will appear not to be obvious for $q=2$.

The $3n$-dimensional space $\langle\pi_{0},\xi_{i}\rangle$, which contains
$\tau_{0}$, will be denoted by $\bar{\xi}_{i}$, with $i=1,2,\ldots$. Each
space $\pi_{j}\in O\setminus\\{\pi_{0}\\}$, intersects $\bar{\xi}_{i}$ in
exactly one point $z_{j}$. Let $Z_{i}$ be the set of all points $z_{j}$; then
$|Z_{i}|=q^{2n}$.

######  Lemma 6.15.

Any two points of $Z_{i}$ are contained in exactly one $\pi_{0}$-conic
$\mathcal{C}$. Also, $\mathcal{C}\setminus\pi_{0}\subset Z_{i}$ and
$(\mathcal{C}\setminus\pi_{0})^{\delta}$ is an affine line of
$\mathbf{AG}^{(i)}(2n,q)$. In such a way all affine lines of
$\mathbf{AG}^{(i)}(2n,q)$ are obtained.

Proof Let $x,y$ be two distinct points of $Z_{i}$. Let $\mathcal{C}$ be the
unique $\pi_{0}$-conic which contains $x$ and intersects the space
$\pi_{j}\in\mathbf{O}\setminus\\{\pi_{0}\\}$ containing $y$. Let
$\mathcal{C}\cap\tau_{0}=\\{u\\}$. The tangent line of $\mathcal{C}$ at $u$ is
contained in $\tau_{0}\subset Z_{i}$, so $\langle\mathcal{C}\rangle$ is
contained in $Z_{i}$, so $\mathcal{C}\cap\pi_{j}\in Z_{i}$, so
$y\in\mathcal{C}$. Hence $x$ and $y$ are contained in a unique $\pi_{0}$-conic
$\mathcal{C}$, and $\mathcal{C}\setminus\pi_{0}$ belongs to $Z_{i}$. Clearly
$(\mathcal{C}\setminus\pi_{0})^{\delta}$ is an affine line of
$\mathbf{AG}^{(i)}(2n,q)$ and each affine line of $\mathbf{AG}^{(i)}(2n,q)$ is
obtained in such a way. $\blacksquare$

Consider distinct affine spaces $\mathbf{AG}^{(i)}(2n,q)$ and
$\mathbf{AG}^{(j)}(2n,q)$. Let $\alpha_{ij}$ be the bijection from
$\mathbf{AG}^{(i)}(2n,q)$ onto $\mathbf{AG}^{(j)}(2n,q)$ which maps the point
$\xi_{i}\cap\pi^{\star}_{k}$ onto the point $\xi_{j}\cap\pi^{\star}_{k}$, with
$k=1,2,\ldots,q^{2n}$.

######  Lemma 6.16.

For $q\not=2$ the bijection $\alpha_{ij}$ is a linear projectivity from
$\mathbf{AG}^{(i)}(2n,q)$ onto $\mathbf{AG}^{(j)}(2n,q)$.

Proof If $L$ is a line of $\mathbf{AG}^{(i)}(2n,q)$, then $L$ and
$L^{\alpha_{ij}}$ are maximal lines of one of the Segre varieties
$\mathcal{S}_{1;n-1}$. As $q\not=2$ the mapping $\alpha_{ij}$ is a
projectivity from $\mathbf{AG}^{(i)}(2n,q)$ onto $\mathbf{AG}^{(j)}(2n,q)$.

Let $\alpha_{i}:x\mapsto\pi^{\star}$, with $x\in\mathbf{AG}^{(i)}(2n,q)$ and
$\pi^{\star}$ the element of
$\\{\pi^{\star}_{1},\pi^{\star}_{2},\ldots,\pi^{\star}_{q^{2n}}\\}$ which
contains $x$. Then on
$\\{\pi^{\star}_{1},\pi^{\star}_{2},\ldots,\pi^{\star}_{q^{2n}}\\}$ a
$2n$-dimensional space $\bar{\mathbf{AG}}^{(i)}(2n,q)$ is induced having as
lines the $(n-1)$-reguli $\mathcal{R}_{k}$ minus their space at infinity. From
the preceding paragraph also follows that
$\bar{\mathbf{AG}}^{(i)}(2n,q)=\bar{\mathbf{AG}}^{(j)}(2n,q)$. As
$\alpha_{i}\alpha^{-1}_{j}=\alpha_{ij}$ it follows that $\alpha_{ij}$
preserves the cross-ratio of any four points on any affine line of
$\mathbf{AG}^{(i)}(2n,q)$. Hence $\alpha_{ij}$ is linear. $\blacksquare$

######  Corollary 6.17.

$\bar{\mathbf{AG}}^{(i)}(2n,q)=\bar{\mathbf{AG}}^{(j)}(2n,q)=\bar{\mathbf{AG}}(2n,q)$
and all $(n-1)$-reguli $\mathcal{R}_{k}$ which define a set of parallel lines
of $\bar{\mathbf{AG}}(2n,q)$ have a common space at infinity (in any
$\mathbf{AG}^{(l)}(2n,q)$ the maximal lines of the corresponding Segre
varieties $\mathcal{S}^{(k)}_{1;n-1}$ intersect $\tau^{\star}_{0}$ in a common
point). Hence the corresponding $\Pi_{0}$-conics on $O$ have a common tangent
space at $\pi_{0}$.

Let us consider a $\Pi_{0}$-set $(\Pi,\Gamma)$, with
$\Pi=\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{n}}\\}$ and
$\Gamma=\\{\gamma_{0},\gamma_{1},\ldots,\gamma_{q^{n}}\\}$. The set of all
points $\xi_{i}\cap\pi^{\star}_{k}$, with $k=1,2,\ldots,q^{n}$, will be
denoted by $\xi^{\star}_{i}$.

######  Lemma 6.18.

For $q\not=2$ the set $\xi^{\star}_{i}$ is an $n$-dimensional affine subspace
$\mathbf{AG}^{(i)}(n,q)$ of $\mathbf{AG}^{(i)}(2n,q)$, with
$i=1,2,\ldots,(q^{n}-1)/(q-1)$.

Proof Let $z^{\star}_{u}\in\pi^{\star}_{u},z^{\star}_{v}\in\pi^{\star}_{v}$,
with $u,v\in\\{1,2,\ldots,q^{n}\\}$ and $u\not=v$. With the line
$z^{\star}_{u}z^{\star}_{v}$ corresponds a $\pi_{0}$-conic $\mathcal{C}$,
which is contained in $\bar{\xi}_{i}$ and which intersects the $q+1$ elements
of the $\Pi_{0}$-conic determined by $\pi_{0},\pi_{u},\pi_{v}$; see 6.10 and
6.15. Hence the affine line
$z^{\star}_{u}z^{\star}_{v}\setminus\tau^{\star}_{0}$ belongs to
$\xi^{\star}_{i}$. As $q\not=2$, It follows that $\xi^{\star}_{i}$ is an
affine $n$-dimensional subspace $\mathbf{AG}^{(i)}(n,q)$ of
$\mathbf{AG}^{(i)}(2n,q)$, with $i=1,2,\ldots,(q^{n}-1)/(q-1)$. $\blacksquare$

######  Remark 6.19.

Consider again the $\Pi_{0}$-set $(\Pi,\Gamma)$, with
$\Pi=\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{n}}\\}$ and
$\Gamma=\\{\gamma_{0},\gamma_{1},\ldots,\gamma_{q^{n}}\\}$. Assume that
$\gamma_{0}\cap\tau^{\star}_{0}=\\{x^{\star}\\}$. Then the (projective
completions of the ) lines of $\xi^{\star}_{i}$ through $x^{\star}$ correspond
to the $\pi_{0}$-conics $\mathcal{C}$ in
$\bar{\xi}_{i}=\langle\pi_{0},\xi_{i}\rangle$ intersecting $q$ elements of
$\langle\pi_{1},\pi_{2},\ldots.\pi_{q^{n}}\\}$ and having their tangent line
at $\mathcal{C}\cap\pi_{0}$ in the space $\gamma_{0}$.

######  Corollary 6.20.

By 6.18 the tangents of the $\pi_{0}$-conics in $\bar{\xi}_{i}$ of a
$\Pi_{0}$-set all belong to a $(2n-1)$-dimensional space $\beta_{i}$
containing $\pi_{0}$.

###  6.6. MAIN THEOREMS

######  Theorem 6.21.

For $q\not=2$ the pseudo-ovoid $O$ is good at $\pi_{0}$.

Proof We use the notations of Section 6.5. Let $q\not=2$.

We introduced the projectivity $\alpha_{ij}$ from
$\mathbf{AG}^{(i)}(2n,q)=\xi_{i}\setminus\tau^{\star}_{0}$ onto
$\mathbf{AG}^{(j)}(2n,q)=\xi_{j}\setminus\tau^{\star}_{0}$. With each
$(n-1)$-regulus $\mathcal{R}_{k}$ (or Segre variety
$\mathcal{S}^{(k)}_{1;n-1}$) there corresponds a line $L$ of
$\mathbf{AG}^{(i)}(2n,q)$ and a line $L^{\prime}$ of
$\mathbf{AG}^{(j)}(2n,q)$; also $L^{\alpha_{ij}}=L^{\prime}$.

With a set of parallel lines of $\mathbf{AG}^{(i)}(2n,q)$ (and
$\mathbf{AG}^{(j)}(2n,q)$) there corresponds a set of $(n-1)$-reguli
$\mathcal{R}_{k}$ with a common space at infinity in $\tau^{\star}_{0}$, and
also a set of $\Pi_{0}$-conics with a common tangent space at $\pi_{0}$.

With any $\Pi_{0}$-set $(\Pi,\Gamma)$ there corresponds an affine
$n$-dimensional subspace $\xi^{\star}_{i}=\mathbf{AG}^{(i)}(n,q)$ of
$\mathbf{AG}^{(i)}(2n,q)$. Clearly
$\mathbf{AG}^{(i)}(n,q)^{\alpha_{ij}}=\mathbf{AG}^{(j)}(n,q)$. Now let
$\Pi=\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{n}}\\}$,
$\Gamma=\\{\gamma_{0},\gamma_{1},\ldots,\gamma_{q^{n}}\\}$, and
$\gamma_{0}\cap\tau^{\star}_{0}=\\{x^{\star}_{0}\\}$. The $\pi_{0}$-conics
$\mathcal{C}$ in $\bar{\xi}^{\star}_{i}=\langle\pi_{0},\xi_{0}\rangle$
intersecting $q$ elements of $\Pi\setminus\\{\pi_{0}\\}$ and having their
tangent line at $\mathcal{C}\cap\pi_{0}$ in $\gamma_{0}$, correspond to the
lines of $\mathbf{AG}^{(i)}(n,q)$ having $x^{\star}_{0}$ as point at infinity.
The $\Pi_{0}$-conics $C$ containing the conics $\mathcal{C}$ have a common
$(2n-1)$-space $\eta_{i}$ as tangent space at $\pi_{0}$. Let
$\eta_{i}\cap\tau^{\star}_{0}=\eta^{\star}_{i}$. These $\Pi_{0}$-conics are
denoted by $C^{i}_{1},C^{i}_{2},\ldots,C^{i}_{q^{n-1}}$.

Consider the $n$-dimensional space $\widetilde{\gamma}_{0}$ through $\pi_{0}$
in $\eta_{i}$ and consider the $\Pi_{0}$-set
$(\widetilde{\Pi},\widetilde{\Gamma})$ defined by the pair
$\\{\widetilde{\gamma}_{0},\pi_{1}\\}$, with $\pi_{1}\in C^{i}_{1}$. Let
$\widetilde{\gamma}_{0}\cap\tau^{\star}_{0}=\\{\widetilde{x}^{\star}_{0}\\}$,
with $\widetilde{x}^{\star}_{0}\in\eta^{\star}_{i}$. As
$\widetilde{\gamma}_{0}\subset\eta_{i}$ and $\eta_{i}$ is the tangent space of
$C^{i}_{1}$ at $\pi_{0}$, all elements of $C^{i}_{1}$ are in
$\widetilde{\Pi}$. The $\pi_{0}$-conic $\mathcal{C}_{1}$ in $C^{i}_{1}$ with
tangent line at $\mathcal{C}_{1}\cap\pi_{0}$ in $\widetilde{\gamma}_{0}$, is
projected onto a line containing $\widetilde{x}^{\star}_{0}$. This line
belongs to some space $\xi_{j}$, and its affine part is contained in the
$n$-dimensional affine space $\widetilde{\xi}^{\star}_{j}$ defined by
$(\widetilde{\Pi},\widetilde{\Gamma})$. Notice that
$\widetilde{x}^{\star}_{0}=(x^{\star}_{0})^{\alpha_{ij}}$ and that all lines
of $\xi_{j}\setminus\tau^{\star}_{0}$ on $\widetilde{x}^{\star}_{0}$ are
images for $\alpha_{ij}$ of all lines in $\xi_{i}\setminus\tau^{\star}_{0}$ on
$x^{\star}_{0}$.

With $(\widetilde{\Pi},\widetilde{\Gamma})$ correspond affine $n$-dimensional
spaces $\widetilde{\xi}^{\star}_{i}$ (in $\xi_{i}$) and
$\widetilde{\xi}^{\star}_{j}$ (in $\xi_{j}$). The mapping $\alpha_{ji}$ sends
all lines of $\widetilde{\xi}^{\star}_{j}$ on $\widetilde{x}^{\star}_{0}$ to
all lines of $\widetilde{\xi}^{\star}_{i}$ on $x^{\star}_{0}$. The
$(n-1)$-reguli defined by these lines, that is
$\widetilde{\mathcal{R}}_{1}=\mathcal{R}_{1}$ (the $(n-1)$-regulus defined by
$\mathcal{C}_{1}$ = the $(n-1)$-regulus defined by the $\pi_{0}$-conic of
$C_{1}$ in $\bar{\xi}_{i}$),
$\widetilde{\mathcal{R}}_{2},\ldots,\widetilde{\mathcal{R}}_{q^{n-1}}$, have
at infinity the common space $\eta^{\star}_{i}$.

By 6.13 the space $\gamma_{0}$ defines a partition $\mathcal{P}$ of
$O\setminus\\{\pi_{0}\\}$ into sets of size $q^{n}$.

Let $(\Pi_{k},\Gamma_{k})$, with $k=1,2,\ldots,q^{n}$, be the $q^{n}$
$\Pi_{0}$-sets defined by $\mathcal{P}$ (one of these $\Pi_{0}$-sets is
$(\Pi,\Gamma))$. For each $(\Pi_{k},\Gamma_{k})$ there are $q^{n-1}$
$\pi_{0}$-conics $\mathcal{C}$ in $\bar{\xi}_{i}$ intersecting $q$ elements of
$\Pi_{k}\setminus\\{\pi_{0}\\}$ and having their tangent line at
$\mathcal{C}\cap\pi_{0}$ in $\gamma_{0}$. The corresponding $\Pi_{0}$-conics
have a common tangent space $\eta_{i}$ at $\pi_{0}$ (the corresponding lines
of $\xi^{\star}_{i}$ have $x^{\star}_{0}$ as point at infinity). Hence each
$n$-dimensional space $\gamma_{0}$ through $\pi_{0}$ determines for each $i$ a
unique tangent space $\eta_{i}$. Any $n$-dimensional space
$\widetilde{\gamma}_{0}$ of $\eta_{i}$ through $\pi_{0}$ also determines the
tangent space $\eta_{i}$.

Assume that $\eta_{i}$ and $\eta^{\prime}_{i}$ are two distinct such tangent
spaces at $\pi_{0}$ and that $\eta_{i}\cap\eta^{\prime}_{i}$ contains an
$n$-dimensional subspace $\zeta$ containing $\pi_{0}$. By the foregoing
paragraphs $\zeta$ defines (for each $i$) a unique such tangent space, and so
$\eta_{i}=\eta^{\prime}_{i}$. Hence for each $i$ all such tangent spaces
corresponding to $\Pi_{0}$-sets have two by two just $\pi_{0}$ in common.

So for each $i$ there arise $(q^{n}-1)/(q-1)$ $(2n-1)$-dimensional subspaces
$\eta_{i}$ containing $\pi_{0}$ having two by two just $\pi_{0}$ in common,
and $(q^{n}-1)/(q-1)$ $(n-1)$-dimensional subspaces $\eta^{\star}_{i}$ which
form a partition of $\tau^{\star}_{0}$. The set of all $\eta_{i}$’s will be
denoted by $\Sigma_{i}$, the partition containing the $\eta^{\star}_{i}$’s by
$\Sigma^{\star}_{i}$.

Let $x^{\star}_{0}\in\tau^{\star}_{0}$ and let
$\eta^{\star}_{i}\in\Sigma^{\star}_{i}$ contain $x^{\star}_{0}$. Then
$(x^{\star}_{0})^{\alpha_{ij}}\in\eta^{\star}_{i}$. Let
$\eta^{\star}_{j}\in\Sigma^{\star}_{j}$ contain
$(x^{\star}_{0})^{\alpha_{ij}}$. Then
$(x^{\star}_{0})^{\alpha_{ij}\alpha_{ji}}=x^{\star}_{0}\in\eta^{\star}_{j}$.
The spaces $\eta^{\star}_{i}$ and $\eta^{\star}_{j}$ are the spaces at
infinity of a common $(n-1)$-regulus. Hence
$\eta^{\star}_{i}=\eta^{\star}_{j}$. So
$\Sigma^{\star}_{i}=\Sigma^{\star}_{j}=\Sigma^{\star}$.

Consider again $(\Pi,\Gamma)$, with
$\Pi=\\{\pi_{0},\pi_{1},\ldots,\pi_{q^{n}}\\},\Gamma=\\{\gamma_{0},\gamma_{1},\ldots,\gamma_{q^{n}}\\}$,
and let $(\Pi,\Gamma)$ be determined by $\\{\gamma_{0},\pi_{1}\\}$. The
corresponding $\Pi_{0}$-conics, defined by the $\pi_{0}$-conics in
$\bar{\xi}_{i}$, are $C^{i}_{1},C^{i}_{2},\ldots,C^{i}_{q^{n-1}}$, with
$C^{i}_{1}=\\{\pi_{0},\pi_{1},\ldots,\pi_{q}\\}$. Let $C^{\prime}$ be the
$\Pi_{0}$-conic in $\Pi$ containing $\pi_{0},\pi_{1},\pi_{k}$, with
$k\in\\{q+1,q+2,\ldots,q^{n}\\}$. We have:
$\gamma_{0}\subset\tau_{0},\gamma_{0}$ intersects
$\langle\pi_{1},\pi_{k}\rangle$ by 6.10 , so
$\gamma_{0}\subset\langle\pi_{0},\pi_{1},\pi_{k}\rangle$, and hence the
tangent space $\eta^{\prime\prime}$ of $C^{\prime}$ at $\pi_{0}$, that is
$\langle\pi_{0},\pi_{1},\pi_{k}\rangle\cap\tau_{0}$, contains $\gamma_{0}$.

There is a $\pi_{0}$-conic $\mathcal{C}$ intersecting
$\pi_{0},\pi_{1},\pi_{k}$ and having its tangent line at
$\mathcal{C}\cap\pi_{0}$ in $\gamma_{0}$. Projecting $\mathcal{C}$ there
arises a line $L$ of $\Phi$. Assume that $L$ is in $\xi_{l}$, so
$\eta^{\prime\prime}\in\Sigma^{\star}_{l}=\Sigma^{\star}$. Hence
$\eta^{\prime\prime}$ is tangent space at $\pi_{0}$ of each of the
$\Pi_{0}$-conics defined by $\\{\gamma_{0},\pi_{1}\\}$. So
$\eta^{\prime\prime}$ is the common tangent space at $\pi_{0}$ of the
$\Pi_{0}$-conics defined by all triples $\pi_{0},\pi_{1},\pi_{k}$, and hence
clearly of all $\Pi_{0}$-conics contained in $\Pi$. It follows that all
$\Pi_{0}$-conics in $\Pi$ are contained in the $(3n-1)$-dimensional space
$\langle\eta^{\prime\prime},\pi_{0}\rangle$.

Each $\Pi_{0}$-conic $C^{\prime\prime}$ is contained in a $(\Pi,\Gamma)$-set.
Hence the $(3n-1)$-dimensional space $\langle C^{\prime\prime}\rangle$
contains $q^{n}+1$ elements of $O$.

We conclude that $O$ is good at $\pi_{0}$. $\blacksquare$

######  Theorem 6.22.

Let $\mathcal{S}=T(n,2n,q)$, with $q\not=2$, be a translation generalized
quadrangle with base point $(\infty)$, and having a regular line $L$ not
incident wit $(\infty)$.

  * (i)

If $q$ is odd, then $\mathcal{S}$ is the point-line dual of the translation
dual of a semifield flock translation generalized quadrangle.

  * (ii)

If $q$ is even, then $\mathcal{S}$ is classical.

Proof Let $\mathcal{S}=T(O)$, with
$O=\\{\pi_{0},\pi_{1},\ldots,\pi_{q{2n}}\\}$ and where the regular line $L$ is
concurrent with $\pi_{0}$. By 6.21 the pseudo-ovoid $O$ is good at $\pi_{0}$.

If $q$ is odd, then by 4.3 the GQ $\mathcal{S}$ is the point-line dual of the
translation dual of a semifield flock TGQ.

Let $q$ be even, let $T(O)$ be contained in $\mathbf{PG}(4n,q)$ and let
$\langle O\rangle=\mathbf{PG}(4n-1,q)$. Further, let
$\langle\pi_{0},\pi_{1},\pi_{2}\rangle=\mathbf{PG}(3n-1,q)$ and let $\langle
L,\mathbf{PG}(3n-1,q)\rangle=\mathbf{PG}(3n,q)$. In $\mathbf{PG}(3n,q)$ a
subquadrangle $\mathcal{S}^{\prime}=T(O^{\prime})$ of order $q^{n}$ is
induced, with $O^{\prime}$ the pseudo-oval consisting of the $q^{n}+1$
elements of $O$ in $\mathbf{PG}(3n-1,q)$.The line $L$ of
$\mathcal{S}^{\prime}$ is regular, so by the action of the translation group
of the TGQ $T(O^{\prime})$ (with base point $(\infty)$) on $L$ we see that all
lines of $\mathcal{S}^{\prime}$ concurrent with $\pi_{0}$ are regular. Now by
1.3.6(iv) of [16], all lines of $\mathcal{S}^{\prime}$ are regular. Hence, by
the well-known theorem of Benson, see 5.2.1 of [16], the GQ
$\mathcal{S}^{\prime}$ is isomorphic to the classical GQ arising from a
nonsingular quadric of $\mathbf{PG}(4,q^{n})$. Now by Brown and Lavrauw [1] it
follows that $\mathcal{S}$ is isomorphic to the classical GQ arising from a
nonsingular elliptic quadric of $\mathbf{PG}(5,q^{n})$; see also 7.8 of [27]
$\blacksquare$

##  7\. Applications

######  Theorem 7.1.

Let $\mathcal{S}$ be a GQ of order $(s,s^{2}),s\not=1$, having two distinct
translation points. Assume that for at least one of the corresponding TGQ’s
the kernel $\mathbb{K}$ has size at least 3.

  * (i)

If $s$ is odd, then $\mathcal{S}$ is the point-line dual of the translation
dual of a semifield flock TGQ.

  * (ii)

If $s$ is even, then $\mathcal{S}$ is classical.

Proof First assume that the two translation points $(\infty)$ and
$(\infty)^{\prime}$ are collinear. Then all lines incident with
$(\infty)^{\prime}$ are regular, so we can apply 6.22 to the TGQ
$\mathcal{S}^{(\infty)}$.

Next let $(\infty)$ and $(\infty)^{\prime}$ be not collinear. There is a
translation about $(\infty)$ which maps $(\infty)^{\prime}$ onto a point
$(\infty)^{\prime\prime}\not=(\infty)^{\prime}$ collinear with
$(\infty)^{\prime}$. So $\mathcal{S}$ has distinct collinear translation
points, and the first part of the proof can be applied. $\blacksquare$

######  Remark 7.2.

Result 7.1, without the restriction $|\mathbb{K}|>2$, was first proved by K.
Thas [28], relying on the classification of finite groups with a split
$BN$-pair of rank 1 due independently to Hering, Kantor and Seitz [6] and
Shult [17].

######  Theorem 7.3.

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a GQ of order $(s,s^{2})$, with
$s\not=1$, for which each point is a translation point. If no one of the TGQ’s
$\mathcal{S}^{(p)}$, with $p\in P$, has a kernel with 2 elements, then
$\mathcal{S}$ is classical.

Proof It is clear that all lines of $\mathcal{S}$ are regular.

Let $\mathcal{S}=T(O)$. By 6.21 the pseudo-ovoid $O$ is good at each of its
elements. By 8.7.4(iii) of [16], the GQ $\mathcal{S}$ is isomorphic to a
$T_{3}(\bar{O})$ of Tits, with $\bar{O}$ some ovoid of $\mathbf{PG}(3,s)$. As
all lines of $\mathcal{S}$ are regular, by 3.3.3(iii) of [16], the GQ is
classical. $\blacksquare$

######  Remark 7.4.

So for $|\mathbb{K}|\not=2$ this is the ”missing part” in the monograph of
Payne and Thas [15, 16], so that now, for $|\mathbb{K}|\not=2$, there is a
pure geometrical proof of one of the main cases (the most complicated one) in
the theorem of Fong and Seitz [4, 5] on $BN$-pairs of rank 2.

######  Theorem 7.5.

Let $\mathcal{S}=(P,B,\mathbf{I})$ be a TGQ of order $(s,t)$, with
$s\not=1\not=t$, and $s\not=t$, having $(\infty)$ as base point. If
$\mathcal{S}$ has a regular line $L$ not incident with $(\infty)$, then
$t=s^{2}$. Hence if the kernel has more than two elements, then 6.22 can be
applied.

Proof By 3.2 we know that for $s$ or $t$ even we necessarily have $t=s^{2}$.
So from now on assume that $s$ and $t$ are odd. Let $L$ be concurrent with the
line $\pi_{0}\in O$ of $\mathcal{S}=T(O)=T(n,m,q)$. Similar to the case
$t=s^{2}$ we prove that any $(3n-1)$-dimensional space containing $\pi_{0}$
and two distinct elements $\pi_{i},\pi_{j}\in O\setminus\\{\pi_{0}\\}$,
contains exactly $q^{n}+1$ elements of $O$. Notice that
$\langle\pi_{0},\pi_{i},\pi_{j}\rangle$ intersects no one of the remaining
$q^{m}-q^{n}$ elements of $O$. So the spaces
$\langle\pi_{0},\pi_{i},\pi_{j}\rangle$, with $\pi_{j}\in
O\setminus\\{\pi_{0},\pi_{i}\\}$ and $\pi_{i}$ fixed, determine a partition of
$O\setminus\\{\pi_{0},\pi_{i}\\}$ into sets of size $q^{n}-1$. Hence $q^{n}-1$
divides $q^{m}-1$. It follows that $n$ divides $m$, and as $n\not=m$ we have
$m\geq 2n$. As $m\leq 2n$ by Section 2.2, there follows that $m=2n$, and so
$t=s^{2}$. $\blacksquare$

##  8\. the case $q=2$

All results in this paper also hold for $q=2$ if 6.16 can be proved for this
case. In the proof of 6.18 we also rely on $q\not=2$, but for $q=2$ this lemma
can be proved relying on 6.16.

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