# Regularity in polar spaces of infinite rank

Antonio Pasini

###### Abstract

In this paper we propose a definition of regularity suited for polar spaces of
infinite rank and we investigate to which extent properties of regular polar
spaces of finite rank can be generalized to polar spaces of infinite rank.

##  1 Introduction

This paper is a continuation of [9]. In [9] I mainly stressed on differences
between polar spaces of infinite rank and those of finite rank, focusing on
properties which hold for all polar spaces of finite rank but fail to hold in
many polar spaces of infinite rank, thus unwillingly suggesting that the
infinite rank case might be too wild for nice theories can be composed for it.
In this paper I support the opposite view. Focusing on regularity and related
properties, I shall set down pieces of a theory suited for all polar spaces,
including those of infinite rank. As expected, the picture we can get in the
infinite rank case is not so neat and simple as for polar spaces of finite
rank. Nevertheless, it looks nicer than I dared to hope and more interesting
than in the finite rank case.

###  1.1 Premise

We are not going to recall all basics on polar spaces; we presume the reader
knows them. We only fix here some notation and terminology to be used in this
introductory section. More information will be offered in Section 2.

We use the symbol $\perp$ to denote collinearity, with the convention that
every point is collinear with itself. Given a point $x$ of a polar space
${\cal S}$ we denote by $x^{\perp}$ the set of points of ${\cal S}$ collinear
with $x$ and, for a set of points $X$ of ${\cal S}$, we set
$X^{\perp}:=\cap_{x\in X}x^{\perp}$. Two singular subspaces $X$ and $Y$ are
said to be opposite when $X^{\perp}\cap Y=Y^{\perp}\cap X=\emptyset$. In
particular, two points are opposite if and only if they are non-collinear and
two maximal singular subspaces, henceforth called generators, are opposite if
and only if they are disjoint.

The symbol $\langle.\rangle$ stands for spans, but we use it for spans in
polar spaces as well as in projective and vector spaces. Thus, if $X$ is a set
of points of a polar space ${\cal S}$ then $\langle X\rangle$ is the subspace
of ${\cal S}$ generated by $X$ and, if ${\cal S}$ admits a projective
embedding $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$, then
$\langle\varepsilon(X)\rangle$ is the projective subspace of $\mathrm{PG}(V)$
spanned by $\varepsilon(X)$.

In this paper we will often use an informal simplified notation, writing
$\\{X,Y\\}^{\perp}$ for $(X\cup Y)^{\perp}$, $\langle X,Y\rangle$ for $\langle
X\cup Y\rangle$, $\langle X\cap Y,Z^{\perp}\rangle$ for $\langle(X\cap Y)\cup
Z^{\perp}\rangle$ and so on. We trust this notation will not confuse the
reader.

We recall that the rank of a non-degenerate polar space ${\cal S}$, henceforth
called $\mathrm{rank}({\cal S})$, is the least upper bound of the set of the
ranks of the generators of ${\cal S}$, the rank of a projective space being
its dimension augmented by 1. Following a well established custom, we write
$\mathrm{rank}({\cal S})=\infty$ and $\mathrm{rank}({\cal S})<\infty$ as
shortenings of the sentences “$\mathrm{rank}({\cal S})$ is an infinite
cardinal number” and “$\mathrm{rank}({\cal S})$ is finite” respectively. We
warn that, while when $\mathrm{rank}({\cal S})<\infty$ all generators of
${\cal S}$ have the same dimension, when $\mathrm{rank}({\cal S})=\infty$ all
generators of ${\cal S}$ are infinite dimensional but in general not all of
them have the same dimension.

All polar spaces to be considered in the sequel are non-degenerate and thick-
lined of rank at least 2, by assumption. All embeddings are full projective
embeddings.

###  1.2 A definition of regularity

Different but equivalent ways exists to define regularity for pairs of
opposite points of a generalized quadrangle. For instance, the following is a
rephrasing of the special case $m=2$ of a definition stated in [18, 6.4.1] for
generalized $m$-gons: a pair $\\{a,b\\}$ of opposite points of a generalized
quadrangle is regular if, for any point $c$ opposite both $a$ and $b$, if
$|\\{a,b,c\\}^{\perp}|>1$ then $c\in\\{a,b\\}^{\perp\perp}$; equivalently, if
$x,y\in\\{a,b\\}^{\perp}$ and $x\neq y$ then
$\\{x,y\\}^{\perp}=\\{a,b\\}^{\perp\perp}$. The following is also equivalent
to this definition: for every line $\ell$, if
$\ell\cap\\{a,b\\}^{\perp}\neq\emptyset$ then
$\ell\cap\\{a,b\\}^{\perp\perp}\neq\emptyset$. In particular, in the finite
case the latter amounts to say that $|\\{a,b\\}^{\perp\perp}|$ is equal to the
number of lines through a point (compare Payne and Thas [13, 1.3]).

A generalization of these definitions to polar spaces of arbitrary but finite
rank is proposed in [12] and [3]: two opposite points of a polar space ${\cal
S}$ of finite rank are said to form a regular pair if

  * $({\mathrm{R}}1)$

$(X\cup Y)^{\perp}=\\{a,b\\}^{\perp\perp}$ for any two opposite generators $X$
and $Y$ of $\\{a,b\\}^{\perp}$; equivalently, if $c$ is a point opposite both
$a$ and $b$ and $\\{a,b,c\\}^{\perp}$ contains two opposite generators of
$\\{a,b\\}^{\perp}$, then $c\in\\{a,b\\}^{\perp\perp}$.

It is proved in [12, Lemma 5.5] (also [3, Proposition 5.1]) that
$({\mathrm{R}}1)$ is equivalent to the following:

  * $({\mathrm{R}}2)$

if a generator $M$ of ${\cal S}$ contains a generator of the polar space
$\\{a,b\\}^{\perp}$, then $M\cap\\{a,b\\}^{\perp\perp}\neq\emptyset$.

Property $({\mathrm{R}}2)$ still makes sense when $\mathrm{rank}({\cal
S})=\infty$. In fact $({\mathrm{R}}2)$ is involved in a characterization of
symplectic polar spaces of arbitrary (possibly infinite) rank (see [3]).
Property $({\mathrm{R}}1)$ also makes sense when $\mathrm{rank}({\cal
S})=\infty$, however it might possibly be vacuous in certain cases. Indeed it
is still an open problem whether polar spaces of infinite rank exist which
admit no pair of opposite generators. If $\\{a,b\\}^{\perp}$ admits no pair of
opposite generators then $({\mathrm{R}}1)$ is (trivially true but) vacuous for
$\\{a,b\\}$. This is not the unique problem we face with $({\mathrm{R}}1)$
when $rk({\cal S})=\infty$; the following is another one. It is easy to see
that $({\mathrm{R}}2)$ implies $({\mathrm{R}}1)$, even if $\mathrm{rank}({\cal
S})=\infty$. On the other hand, if $({\mathrm{R}}1)$ holds for a pair
$\\{a,b\\}$ of opposite points and every generator of $\\{a,b\\}^{\perp}$
admits an opposite in $\\{a,b\\}^{\perp}$, then $({\mathrm{R}}2)$ holds for
$\\{a,b\\}$ if and only if $X^{\perp}\cap M\neq\emptyset$ for any generator
$X$ of $\\{a,b\\}^{\perp}$ and every generator $M$ of ${\cal S}$ containing a
generator of $\\{a,b\\}^{\perp}$ (see the proofs of [12, Lemma 5.5] and [3,
Proposition 5.1]). So, if every generator of $\\{a,b\\}^{\perp}$ admits an
opposite in $\\{a,b\\}^{\perp}$ and ${\cal S}$ satisfies the following
property $(\mathrm{GS})$, then we can prove that if $({\mathrm{R}}1)$ holds
for $\\{a,b\\}$ then $({\mathrm{R}}2)$ also holds, otherwise we get stuck.

  * $(\mathrm{GS})$

$X^{\perp}\cap M\neq\emptyset$ for every generator $M$ and every non-maximal
singular subspace $X$ of ${\cal S}$.

It is well known that $(\mathrm{GS})$ holds true when $\mathrm{rank}({\cal
S})<\infty$ but in general it fails when $\mathrm{rank}(S)=\infty$. This
considered, in [3] we looked at $({\mathrm{R}}2)$ as a possible definition of
regularity when $\mathrm{rank}({\cal S})=\infty$, discarding
$({\mathrm{R}}1)$. In support of this proposal, consider the following
sharpening of $({\mathrm{R}}1)$:

  * $({\mathrm{R}}3)$

we have $(X\cup Y)^{\perp}=\langle X\cap Y,\\{a,b\\}^{\perp\perp}\rangle$ for
any two generators $X$ and $Y$ of $\\{a,b\\}^{\perp}$.

Obviously, $({\mathrm{R}}3)$ implies $({\mathrm{R}}1)$. In Section 3.1
(Theorem 3.3), without assuming that $\mathrm{rank}({\cal S})<\infty$, we
shall prove the following:

###### Theorem 1.1

Properties $({\mathrm{R}}2)$ and $({\mathrm{R}}3)$ are equivalent.

As previously recalled, properties $({\mathrm{R}}1)$ and $({\mathrm{R}}2)$ are
equivalent when $\mathrm{rank}({\cal S})<\infty$. In this case
$({\mathrm{R}}3)$ and $({\mathrm{R}}1)$ are equivalent. We believe that when
$\mathrm{rank}({\cal S})=\infty$ property $({\mathrm{R}}3)$ is stronger than
$({\mathrm{R}}1)$, although we have no example at hand which shows that this
is indeed the case.

We are now ready to state our definition of regularity for polar spaces of
arbitrary, possibly infinite rank.

###### Definition 1.2

We say that a pair of opposite points of ${\cal S}$ is regular if it satisfies
property $({\mathrm{R}}3)$ (equivalently, $({\mathrm{R}}2)$). If all pairs of
opposite points of ${\cal S}$ are regular then ${\cal S}$ is said to be
regular.

###  1.3 Tight embeddings

We say that an embedding $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ of a
polar space ${\cal S}$ is tight if $\langle\varepsilon(M\cup
M^{\prime})\rangle=\mathrm{PG}(V)$ for at least one pair of (necessarily
opposite) generators $M$ and $M^{\prime}$ of ${\cal S}$. When
$\mathrm{rank}({\cal S})=n<\infty$ an embedding $\varepsilon:{\cal
S}\rightarrow\mathrm{PG}(V)$ is tight if and only if $\dim(V)=2n$. If this is
the case then $\langle\varepsilon(M\cup M^{\prime})\rangle=\mathrm{PG}(V)$ for
every pair pair of opposite generators $M$ and $M^{\prime}$ of ${\cal S}$. The
following is proved in [10, Theorem 1.2]:

###### Proposition 1.3

An embeddable polar space of finite rank is regular if and only if it admits a
tight embedding.

In particular, when ${\cal S}$ (has finite rank and) admits a unique
embedding, as it is the case when ${\cal S}$ is defined over a division ring
of characteristic different from $2$, then ${\cal S}$ is regular if and only
if it can be spanned by the union of (any) two opposite generators [10,
Corollary 1.3].

As we shall prove in Section 3.3 (Theorem 3.8), Proposition 1.3 admits the
following generalization:

###### Theorem 1.4

An embeddable polar space ${\cal S}$ of possibly infinite rank admits a tight
embedding if and only if the following holds for at least one (equivalently,
every) pair $\\{a,b\\}$ of opposite points of ${\cal S}$:

  * $({\mathrm{R}}4)$

the subspace $\\{a,b\\}^{\perp}$ contains two singular subspaces $X$ and $Y$
such that $X$ and $Y$ are opposite generators of $\\{a,b\\}^{\perp}$, we have

$(X\cup Y)^{\perp}~{}=~{}\\{a,b\\}^{\perp\perp}$ (1)

and, if $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ is an embedding of
${\cal S}$, then

$\langle\varepsilon(X\cup Y\cup(X\cup
Y)^{\perp})\rangle~{}=~{}\mathrm{PG}(V).$ (2)

When ${\cal S}$ is embeddable the group $\mathrm{Aut}({\cal S})$ acts
transitively on the set of ordered pairs of opposite points. Accordingly, if
$({\mathrm{R}}4)$ holds for at least one pair $\\{a,b\\}$ of opposite points
of ${\cal S}$ then it holds for any such pair. If furthermore
$\mathrm{rank}({\cal S})<\infty$ then $\mathrm{Aut}({\cal S})$ is also
transitive on the set of pairs of opposite singular subspaces of any given
rank. Hence, if condition (1) holds for at least one choice of opposite points
$a$ and $b$ of ${\cal S}$ and opposite generators $X$ and $Y$ of
$\\{a,b\\}^{\perp}$, then it holds for any such choice. In this case
$({\mathrm{R}}4)$ is equivalent to ${\cal S}$ being regular.

In contrast, if $\mathrm{rank}({\cal S})$ is infinite then in general
$\mathrm{Aut}({\cal S})$ is intransitive on the set of pairs of opposite
generators and, similarly, the stabilizer of $\\{a,b\\}$ in
$\mathrm{Aut}({\cal S})$ acts intransitively on the set of pairs of opposite
generators of $\\{a,b\\}^{\perp}$. So, it can happen that (1) holds for a
particular choice of opposite generators $X$ and $Y$ of $\\{a,b\\}^{\perp}$
but not for all of them. If this is the case then ${\cal S}$ is not regular.

Moreover, when $\mathrm{rank}({\cal S})<\infty$ then (2) holds for any choice
of opposite singular subspaces $X$ and $Y$. So, when $\mathrm{rank}({\cal
S})<\infty$ condition (2) is redundant. On the other hand, when
$\mathrm{rank}({\cal S})=\infty$ then in general $\\{a,b\\}^{\perp}$ admits
opposite generators which do not satisfy (2) (but might possibly satisfy (1)).
Notably, this also can happen when ${\cal S}$ is regular.

Suppose that ${\cal S}$ admits a tight embedding $\varepsilon:{\cal
S}\rightarrow\mathrm{PG}(V)$ and $\mathrm{rank}({\cal S})$ is infinite. Then
in general opposite generators $M$ and $N$ also exist such that
$\langle\varepsilon(M\cup N)\rangle\subset\mathrm{PG}(V)$. This fact also
explains why, when $\mathrm{rank}({\cal S})=\infty$, conditions (1) and (2) of
$({\mathrm{R}}4)$ might hold for certain pairs of opposite generators of
$\\{a,b\\}^{\perp}$ but not for all of them. Indeed, as one can prove, if $X$
and $Y$ are opposite generators of $\\{a,b\\}^{\perp}$ then $\varepsilon(X\cup
Y)$ spans $\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$ if and only if $X$
and $Y$ satisfy both properties (1) and (2).

Summarizing the above discussion, when $\mathrm{rank}({\cal S})$ is infinite
$({\mathrm{R}}4)$ is not equivalent to ${\cal S}$ being regular. In fact, as
we shall see in Section 5.4, polar spaces of infinite rank exist which admit a
tight embedding (hence they satisfy $({\mathrm{R}}4)$) but are not regular.

###### Conjecture 1.5

Every regular polar space admits a tight embedding.

###  1.4 The three-generators property

Another characterization of regularity has been obtained in [12] by means of
the so-called three-generators property. Let $\mathrm{rank}({\cal S})<\infty$.
When $\mathrm{rank}({\cal S})$ is odd we assume that ${\cal S}$ is thick. So,
${\cal S}$ admits triples of mutually opposite generators and every pair of
opposite generators belongs to at least one such triple. Let $M,M_{1},M_{2}$
be such a triple. For every subspace $X$ of $M$ we put

$\pi^{M}_{1,2}(X)~{}:=~{}((X^{\perp}\cap M_{1})^{\perp}\cap M_{2})^{\perp}\cap
M.$ (3)

Thus we obtain a duality $\pi^{M}_{1,2}$ of $M$. The duality $\pi_{2,1}$ is
defined in the same way but switching $M_{1}$ and $M_{2}$. Clearly,
$\pi_{2,1}^{M}$ is the inverse of $\pi_{1,2}^{M}$. Hence $\pi_{1,2}^{M}$ is a
polarity if and only if $\pi_{1,2}^{M}=\pi_{2,1}^{M}$.

###### Definition 1.6

We say that a pair $\\{M_{1},M_{2}\\}$ of opposite generators of ${\cal S}$
satisfies the three-generators property if $\pi_{1,2}^{M}$ is a polarity for
every generator $M$ of ${\cal S}$ opposite to both $M_{1}$ and $M_{2}$.

The following has been proved in [12].

###### Proposition 1.7

A polar space ${\cal S}$ of finite rank is regular if and only if every pair
of opposite generators of ${\cal S}$ satisfies the three-generators property.

###### Remark 1

Definition 1.6 also makes sense when ${\cal S}$ is non-tick of odd rank, but
it is vacuous in this case. Indeed in this case no triples of mutually
opposite generators exist; consequently, every pair of opposite generators
trivially satisfies the 3-generators condition. However ${\cal S}$ is regular
in this case (by Proposition 1.3 when ${\cal S}$ is embeddable and [12,
Proposition 5.11] when it isn’t). Hence, when ${\cal S}$ is non-thick of odd
rank Proposition 1.7 trivially holds true.

The first obstacle we meet when looking for a generalization of Proposition
1.7 suited for polar spaces of infinite rank is the fact that, when
$\mathrm{rank}({\cal S})=\infty$, the mapping $\pi_{1,2}^{M}$ defined as in
(3) cannot be a duality. Indeed in this case $\dim(M)$ is infinite and
infinite dimensional projective spaces admit no dualities. Moreover, as
property $(\mathrm{GS})$ might fail to hold in ${\cal S}$, it can happen that
for some point $x\in M$ we have $\pi_{1,2}^{M}(x)=M$. We must change our
setting, replacing dualities with something weaker. We do as follows.

Given a projective space $\mathbb{P}$, let $\mathbb{P}^{*}$ be its dual. Let
$P$ and $P^{*}$ be the point-set of $\mathbb{P}$ and the set of hyperplanes of
$\mathbb{P}$ respectively.

###### Definition 1.8

A partial duality of $\mathbb{P}$ is a mapping $\pi:P\rightarrow
P^{*}\cup\\{P\\}$ such that, for every choice of distinct points $x$ and $y$
of $\mathbb{P}$, if both $\pi(x)$ and $\pi(y)$ belong to $P^{*}$ then
$\pi(x)\neq\pi(y)$ and $\pi$ induces a bijection from the line of $\mathbb{P}$
through $x$ and $y$ to the line of $\mathbb{P}^{*}$ through $\pi(x)$ and
$\pi(y)$.

Let $\pi$ be a partial duality of $\mathbb{P}$. The set
$\pi^{-1}(P^{*}):=\\{x\in P~{}|~{}\pi(x)\neq P\\}$ is a (possibly empty)
subspace of $\mathbb{P}$. If $\pi^{-1}(P^{*})=P$ then we say that $\pi$ is
non-degenerate. Obviously, $\pi$ is non-degenerate if and only if it is
injective. In contrast, when $\pi^{-1}(P^{*})=\emptyset$ we say that $\pi$ is
trivial. If for every two distinct points $x,y\in P$ we have $x\in\pi(y)$ if
and only if $y\in\pi(x)$, then we say that $\pi$ is reflexive. If $\pi$ is
reflexive then $P\setminus\pi^{-1}(P^{*})\subseteq\pi(x)$ for every
$x\in\pi^{-1}(P^{*})$. However no hyperplane of $\mathbb{P}$ contains the
complement of a proper subspace of $\mathbb{P}$. Therefore $\pi$ is reflexive
only if it is either non-degenerate or trivial. A non-trivial reflexive
partial duality is called a polarity.

When $\dim(\mathbb{P})<\infty$ the dualities of $\mathbb{P}$ are precisely the
(mappings from the poset of subspaces of $\mathbb{P}$ to the poset of
subspaces of $\mathbb{P}^{*}$ defined by) non-degenerate partial dualities; a
duality is involutory if and olny if it is reflexive.

Turning back to our polar space ${\cal S}$ but now allowing
$\mathrm{rank}({\cal S})$ to be infinite, let $M,M_{1},M_{2}$ be three mutualy
opposite generators of ${\cal S}$. Equation (3), whith $X$ ranging in the set
of points of $M$ instead of the set of subspaces of $M$, defines a partial
duality $\pi_{1,2}^{M}$ of $M$ (Section 4.4, Lemma 4.6). As in the finite rank
case, $\pi_{1,2}^{M}$ is a polarity if and only if
$\pi_{1,2}^{M}=\pi_{2,1}^{M}$ (Section 4.4, Lemma 4.7). Definition 1.6 still
makes sense in the present setting. We are not going to repeat it here.
However we need one more definition.

###### Definition 1.9

A generator $M$ of ${\cal S}$ is hyperbolic if every hyperplane of $M$ is
contained in at most one generator other than $M$.

As we shall prove in Section 4.5 (Theorem 4.11), the following holds:

###### Theorem 1.10

Let ${\cal S}$ be a polar space of infinite rank, defined over a division ring
of characteristic different from $2$. Let $M_{1}$ and $M_{2}$ be opposite
generators of ${\cal S}$ such that at least one generator of ${\cal S}$ exists
which is opposite to both $M_{1}$ and $M_{2}$. Suppose moreover that neither
$M_{1}$ nor $M_{2}$ are hyperbolic. Under these hypoyheses,
$\\{M_{1},M_{2}\\}$ satisfies the three-generators property if and only if
$\langle M_{1}\cup M_{2}\rangle={\cal S}$.

###### Conjecture 1.11

The hypothesis that the underlying division ring of ${\cal S}$ has
characteristic different from $2$ can be dropped from Theorem 1.10, provided
that the conclusion of that theorem is rephrased as follows: the pair
$\\{M_{1},M_{2}\\}$ satisfies the three-generators property if and only if
${\cal S}$ admits a (necessarily tight) embedding $\varepsilon:{\cal
S}\rightarrow\mathrm{PG}(V)$ such that $\langle\varepsilon(M_{1}\cup
M_{2})\rangle=\mathrm{PG}(V)$.

Organization of the paper. In Section 2 we state some terminology and recall
some basics on projective embeddings of polar spaces. Section 3 contains the
proofs of Theorems 1.1 and 1.4 and more results on regularity and tight
embeddings. Section 4 contains the proof of Theorem 1.10 and one more result
in the same vein as Theorem 1.10. A number of examples of polar spaces of
infinite rank are discussed in Section 5. Most of them are regular.

##  2 Preliminaries

As in the Introduction, throughout this section ${\cal S}$ is a non-degenerate
thick-lined polar space of rank at least 2, possibly $\mathrm{rank}({\cal
S})=\infty$.

###  2.1 A survey of elementary properties

In this subsection we recall a few well known properties of subspaces of
${\cal S}$, to be be freely used in the sequel of this paper.

We have $\langle X\rangle\subseteq X^{\perp\perp}$ for every set $X$ of points
of ${\cal S}$. Also, if $X\subseteq Y$ then $X^{\perp}\supseteq Y^{\perp}$.
Consequently, $X^{\perp}=X^{\perp\perp\perp}$ for every set $X$ of points of
${\cal S}$. If moreover $X$ is a singular subspace of ${\cal S}$ then
$X=X^{\perp\perp}$ if and only if $X$ is the intersection of a family of
generators (as it is always the case when $\mathrm{rank}({\cal S})<\infty$).
Note that, in general, when $\mathrm{rank}({\cal S})=\infty$ non-maximal
singular subspaces exist which are contained in a unique generator. If $X$ is
one of them then $X^{\perp\perp}$ is the unique generator containing $X$.

Let $X$ be a singular subspace of ${\cal S}$ and $M$ a generator containing
$X$. Let $\mathrm{cod}_{M}(X)$ be the codimension of $X$ in $M$. If
$\mathrm{cod}_{M}(X)<\infty$ then $\mathrm{cod}_{N}(X)=\mathrm{cod}_{M}(X)$
for every generator $N$ containing $X$. Indeed the star of $X$ is a (possibly
degenerate) polar space of finite rank equal to $\mathrm{cod}_{M}(X)$. If
$\mathrm{cod}_{M}(X)=1$ for some (hence every) generator $M$ containing $X$
then we say that $X$ is a sub-generator of ${\cal S}$.

Let $a$ and $b$ be two opposite points of ${\cal S}$. Then
$\\{a,b\\}^{\perp}\cong\mathrm{St}(c)$ for every $c\in\\{a,b\\}^{\perp\perp}$,
where $\mathrm{St}(c)$ (the star of $c$) is the polar space with the lines and
the planes of ${\cal S}$ through $c$ as points and lines respectively. All
generators of $\\{a,b\\}^{\perp}$ are sub-generators of ${\cal S}$. So, if $N$
and $N^{\prime}$ are generators of $\\{a,b\\}^{\perp}$ then $M=\langle
N,a\rangle$ and $M^{\prime}=\langle N^{\prime},b\rangle$ are generators of
${\cal S}$. Note that $M\cap M^{\prime}=N\cap N^{\prime}$. Indeed, if $c\in
M\cap M^{\prime}$ then $c\in\\{a,b\\}^{\perp}\cap N^{\perp}\cap
N^{\prime\perp}=N\cap N^{\prime}$. Hence $M$ and $M^{\prime}$ are opposite if
and only if $N$ and $N^{\prime}$ are opposite. Conversely, if $M$ and
$M^{\prime}$ are opposite generators of ${\cal S}$ and $a\in M$, $b\in
M^{\prime}$ are opposite, then $N:=M\cap b^{\perp}$ and
$N^{\prime}:=M^{\prime}\cap a^{\perp}$ are opposite generators of
$\\{a,b\\}^{\perp}$.

Sets as $\\{a,b\\}^{\perp\perp}$ for $a$ and $b$ opposite points are called
hyperbolic lines. Note that the points of a hyperbolic line
$\\{a,b\\}^{\perp\perp}$ are mutually opposite (in particular,
$\\{a,b\\}^{\perp}\cap\\{a,b\\}^{\perp\perp}=\emptyset$) and if $c$ and $d$
are distinct points of $\\{a,b\\}^{\perp\perp}$ then
$\\{c,d\\}^{\perp\perp}=\\{a,b\\}^{\perp\perp}$. Note also that
$\langle\\{a,b\\}^{\perp},c\rangle=c^{\perp}$ for every point
$c\in\\{a,b\\}^{\perp\perp}$. Accordingly,
$\langle\\{a,b\\}^{\perp}\cup\\{a,b\\}\rangle$ contains both $a^{\perp}$ and
$b^{\perp}$. However $a^{\perp}$ and $b^{\perp}$ are distinct maximal
subspaces of ${\cal S}$. Hence
$\langle\\{a,b\\}^{\perp}\cup\\{a,b\\}\rangle={\cal S}$.

###  2.2 Projective embeddings of point-line geometries

In the present subsection we recall some generalities and fix some terminology
on projective embeddings of point-line geometries. This will be helpful in the
next subsection, where projective embeddings of polar spaces will be
discussed.

As in Shult [15], a projective embedding of a point-line geometry ${\cal
G}=(P,{\cal L})$, henceforth called just embedding of $\cal G$ for short, is
an injective mapping $\varepsilon:P\rightarrow\mathbb{P}$ from the point-set
$P$ of $\cal G$ to the point-set of a projective geometry $\mathbb{P}$ such
that $\varepsilon(P)$ spans $\mathbb{P}$ and
$\varepsilon(\ell):=\\{\varepsilon(p)\\}_{p\in\ell}$ is a line of $\mathbb{P}$
for every line $\ell\in{\cal L}$ of $\cal G$. We take the dimension of
$\mathbb{P}$ as the dimension $\dim(\varepsilon)$ of $\varepsilon$. Note that
if $\cal G$ admits skew lines then necessarily $\dim(\mathbb{P})>2$, hence
$\mathbb{P}$ is desarguesian. If $\mathbb{P}$ is desarguesian and $\mathbb{K}$
is the underlying division ring of $\mathbb{P}$ then we say that $\varepsilon$
is defined over $\mathbb{K}$. If all embeddings of $\cal G$ are defined over
the same division ring $\mathbb{K}$ we say that $\cal G$ is defined over
$\mathbb{K}$.

Given two embeddings $\varepsilon:{\cal G}\rightarrow\mathbb{P}$ and
$\varepsilon^{\prime}:{\cal G}\rightarrow\mathbb{P}^{\prime}$ of $\cal G$
defined over the same division ring $\mathbb{K}$, a morphism from
$\varepsilon$ to $\varepsilon^{\prime}$ is a morphism of projective geometries
$\varphi:\mathbb{P}\rightarrow\mathbb{P}^{\prime}$ such that
$\varepsilon^{\prime}=\varphi\circ\varepsilon$. (See Faure and Frölicher [6,
Chaper 6]) for morphisms of projective geometries.) Note that the condition
$\varepsilon^{\prime}=\varphi\circ\varepsilon$ forces the morphism
$\varphi:\mathbb{P}\rightarrow\mathbb{P}^{\prime}$ to be surjective (and, when
$\cal G$ is connected, it uniquely determines $\varphi$). If $\varphi$ is also
injective then we say that $\varphi$ is an isomorphism from $\varepsilon$ to
$\varepsilon^{\prime}$ and we write $\varepsilon\cong\varepsilon^{\prime}$. In
general, if a morphism exists from $\varepsilon$ to $\varepsilon^{\prime}$ we
say that $\varepsilon$ covers $\varepsilon^{\prime}$ and
$\varepsilon^{\prime}$ is a quotient of $\varepsilon$. A motivation for this
terminology is the following: given a morphism
$\varphi:\varepsilon\rightarrow\varepsilon^{\prime}$, let
$K:=\mathrm{Ker}(\varphi)$ be the kernel of $\varphi$ (notation and
terminology as in [6]), let $p_{K}$ be the projection of $\mathbb{P}$ onto the
star $\mathbb{P}/K$ of $K$ in $\mathbb{P}$ and put
$\varepsilon/K:=p_{K}\circ\varepsilon$. Then
$\mathbb{P}^{\prime}\cong\mathbb{P}/K$ and
$\varepsilon/K\cong\varepsilon^{\prime}$.

Following Shult [15], we say that an embedding is relatively universal if it
admits no proper cover. As proved by Ronan [14], every embedding $\varepsilon$
of a geometry $\cal G$ is covered by a relatively universal embedding
$\tilde{\varepsilon}$ of $\cal G$, uniquely determined by $\varepsilon$ up to
isomorphisms and characterized by the following property:
$\tilde{\varepsilon}$ covers all embeddings which cover $\varepsilon$. We call
$\tilde{\varepsilon}$ the hull of $\varepsilon$. Thus, an embedding is
relatively universal if and only if it is its own hull.

An embedding of $\cal G$ is said to be absolutely universal if it covers all
embeddings of $\cal G$. So, $\cal G$ admits the absolutely universal embedding
if and only if all embeddings of $\cal G$ admit the same hull; equivalently,
up to isomorphism, $\cal G$ admits a unique relatively universal embedding.
Clearly, if $\cal G$ admits the absolutely universal embedding and at least
one of its embeddings is defined over a given division ring $\mathbb{K}$, then
$\cal G$ itself is defined over $\mathbb{K}$.

Finally, we say that an embedding is minimal if it admts no proper quotients.
For instance, tight embeddings of polar spaces, as defined in Section 1.3, are
minimal.

###  2.3 Projective embeddings of polar spaces

As proved by Tits [17, chapters 8 and 9] (see also Buekenhout and Cohen [1,
chapters 7-11] and Cuypers et al. [4]) all polar spaces of rank at least 3 are
embeddable but for two families of polar spaces of rank 3. The non-embeddable
ones are the line-grassmannians of 3-dimensional projective spaces defined
over non-commutative division rings and certain polar spaces with Moufang but
non-desarguesian planes, described in [17, Chapter 9].

Let $\cal S$ be an embeddable polar space and, when $\mathrm{rank}({\cal
S})=2$, assume that ${\cal S}$ is neither a grid of order at least $4$ nor a
generalized quadrangle as in Tits [17, 8.6(II)(a)]. Then ${\cal S}$ admits the
absolutely universal embedding as well as a unique minimal embedding (Tits
[17, Chapter 8]; see also Johnson [7] and [8] and Cuypers et al. [5]). In the
two excluded cases, all embeddings are 3-dimensional, hence they are both
relatively universal and minimal.

In any case, all embeddings of ${\cal S}$ have dimension at least
$2\cdot\mathrm{rank}({\cal S})-1$ ($\geq 3$ as $\mathrm{rank}({\cal S})\geq 2$
by assumption); hence they embed ${\cal S}$ in desarguesian projective spaces.
When ${\cal S}$ is not an infinite grid, these projective spaces are defined
over the same division ring (so ${\cal S}$ is defined over that ring).

Let $\varepsilon:{\cal S}\rightarrow\mathbb{P}=\mathrm{PG}(V)$ be an embedding
of a polar space ${\cal S}$. The $\varepsilon$-image $\varepsilon({\cal S})$
of ${\cal S}$ is a full subgeometry of $\mathrm{PG}(V)$ and there exists a
unique quasi-polarity $\pi_{\varepsilon}$ of $\mathrm{PG}(V)$ (see Buekenhout
and Cohen [1, Definition 7.1.9] for the definition of quasi-polarities) such
that all points of $\varepsilon({\cal S})$ are absolute for
$\pi_{\varepsilon}$ and for any two points $x,y\in{\cal S}$ we have $x\perp y$
if and only if $\varepsilon(x)\perp_{\varepsilon}\varepsilon(y)$, where
$\perp_{\varepsilon}$ is the orthogonality relation associated to
$\pi_{\varepsilon}$ (Buekenhout and Cohen [1, Chapter 9]).

More explictly, let ${\cal S}_{\pi_{\varepsilon}}$ be the (possibly
degenerate) polar space defined by $\pi_{\varepsilon}$ on $\mathrm{PG}(V)$.
Then $\varepsilon({\cal S})$ is a subspace of ${\cal S}_{\pi_{\varepsilon}}$,
possibly $\varepsilon({\cal S})={\cal S}_{\pi_{\varepsilon}}$. Note that, as
${\cal S}$ is non-degenerate by assumption, if $\varepsilon({\cal S})={\cal
S}_{\pi_{\varepsilon}}$ then $\pi_{\varepsilon}$ is a polarity, namely its
radical is trivial. (Recall that the radical of a quasi-polarity $\pi$ of a
projective geometry $\mathbb{P}$ is the subspace of $\mathbb{P}$ formed by the
points $p\in\mathbb{P}$ such that $\pi(p)=\mathbb{P}$.)

Suppose that $\varepsilon$ is relatively universal and let $\mathbb{K}$ be its
underlying division ring. Then one of the following occurs (Tits [17, Chaper
8]).

  * (1)

$\mathrm{char}(\mathbb{K})\neq 2$, the polarity $\pi_{\varepsilon}$ is defined
by a non-degenerate alternating form and $\varepsilon({\cal S})={\cal
S}_{\pi_{\varepsilon}}$.

  * (2)

The quasi-polarity $\pi_{\varepsilon}$ is defined by the sesquilinearized of a
non-degenerate $\sigma$-quadratic form
$q:V\rightarrow\mathbb{K}/\mathbb{K}_{\sigma,1}$ as defined by Tits [17,
Chapter 8] for an involutory anti-automorphism $\sigma$ of $\mathbb{K}$ and
$\varepsilon({\cal S})$ is the polar space ${\cal S}_{q}$ associated to $q$.
In this case, if either $\mathrm{char}(\mathbb{K})\neq 2$ or $\sigma$ acts
non-trivially on the center $Z(\mathbb{K})$ of $\mathbb{K}$, then
$\pi_{\varepsilon}$ is a polarity and $\varepsilon({\cal S})={\cal
S}_{q}={\cal S}_{\pi_{\varepsilon}}$.

Suppose moreover that $\varepsilon$ is absolutely universal (as when
$\mathrm{rank}({\cal S})>2$ or case (1) occurs). In case (1) and in case (2)
with $\pi_{\varepsilon}$ a polarity $\varepsilon$ is the unique embedding of
${\cal S}$. When $\pi_{\varepsilon}$ is not a polarity we can factorize
$\varepsilon$ over any subspace $K$ of the radical $R_{\varepsilon}$ of
$\pi_{\varepsilon}$, thus obtaining quotients $\varepsilon/K$ of
$\varepsilon$. In particular, $\varepsilon/R_{\varepsilon}$ is the minimum
embedding of ${\cal S}$.

####  2.3.1 Relations between $\perp$ and $\perp_{\varepsilon}$

Let $X$ be a set of points of ${\cal S}$ and $\varepsilon:{\cal
S}\rightarrow\mathrm{PG}(V)$ an embedding of ${\cal S}$. Clearly
$\langle\varepsilon(X^{\perp})\rangle\subseteq\varepsilon(X)^{\perp_{\varepsilon}}$.

###### Proposition 2.1

If $X^{\perp}\not\subseteq X^{\perp\perp}$ then
$\varepsilon(X)^{\perp_{\varepsilon}}=\langle\varepsilon(X^{\perp})\rangle$.

Proof. Let $\cal X$ be a projective subspace of $\mathrm{PG}(V)$ and suppose
that ${\cal X}\cap\varepsilon({\cal S})\not\subseteq{\cal
X}^{\perp_{\varepsilon}}$. Then $\cal X$ is spanned by ${\cal
X}\cap\varepsilon({\cal S})$. When $\varepsilon$ is relatively universal this
claim follows from Tits [17, §§8.1.6, 8.2.7]. Otherwise, $\varepsilon$ is a
quotient of the absolutely universal embedding $\tilde{\varepsilon}:{\cal
S}\rightarrow\mathrm{PG}(\widetilde{V})$ of ${\cal S}$, say
$\varepsilon=\tilde{\varepsilon}/K$ for a subspace $K$ of the radical
$R_{\tilde{\varepsilon}}$ of $\pi_{\tilde{\varepsilon}}$. Let $p_{K}$ be the
projection of $\mathrm{PG}(\widetilde{V})$ onto
$\mathrm{PG}(V)=\mathrm{PG}(\widetilde{V})/K$ and put $\widetilde{\cal
X}:=p_{K}^{-1}({\cal X})$. Then ${\cal X}\cap\varepsilon({\cal
S})=p_{K}(\widetilde{\cal X}\cap\tilde{\varepsilon}({\cal S}))$. However
${\cal X}\cap\varepsilon({\cal S})\not\subseteq{\cal X}^{\perp_{\varepsilon}}$
by assumption. Hence $\widetilde{\cal X}\cap\tilde{\varepsilon}({\cal
S})\not\subseteq{\cal X}^{\perp_{\tilde{\varepsilon}}}$. By Tits [17, §§8.1.6,
8.2.7], the set $\widetilde{\cal X}\cap\tilde{\varepsilon}({\cal S})$ spans
$\widetilde{\cal X}$. Hence ${\cal X}\cap\varepsilon({\cal S})$ spans $\cal
X$.

Let now $X^{\perp}\not\subseteq X^{\perp\perp}$ as in the hypotheses of the
lemma and put ${\cal X}=\varepsilon(X)^{\perp_{\varepsilon}}$. Then ${\cal
X}\cap\varepsilon({\cal S})=\varepsilon(X^{\perp})$, which is not contained in
${\cal X}^{\perp_{\varepsilon}}$ because $X^{\perp}\not\subseteq
X^{\perp\perp}$. By the previous paragraph, ${\cal X}\cap\varepsilon({\cal
S})$ spans $\cal X$, namely $\varepsilon(X^{\perp})$ spans
$\varepsilon(X)^{\perp_{\varepsilon}}$. $\Box$

###### Corollary 2.2

If $X^{\perp}\not\subseteq X^{\perp\perp}\not\subseteq X^{\perp}$ then
$\langle\varepsilon(X^{\perp\perp})\rangle=\varepsilon(X)^{\perp_{\varepsilon}\perp_{\varepsilon}}$.

Proof. Let $X^{\perp}\not\subseteq X^{\perp\perp}$. Then
$\langle\varepsilon(X^{\perp})\rangle=\varepsilon(X)^{\perp_{\varepsilon}}$ by
Proposition 2.1. Suppose moreover that $X^{\perp\perp}\not\subseteq
X^{\perp}$. Then, since $X^{\perp}=X^{\perp\perp\perp}$, we have that
$\langle\varepsilon(X^{\perp\perp})\rangle=\varepsilon(X^{\perp})^{\perp_{\varepsilon}}$
by Lemma 2.1 with $X^{\perp}$ in place of $X$. However
$\varepsilon(X^{\perp})^{\perp_{\varepsilon}}=\langle\varepsilon(X^{\perp})\rangle^{\perp_{\varepsilon}}=\varepsilon(X)^{\perp_{\varepsilon}\perp_{\varepsilon}}$,
since
$\langle\varepsilon(X^{\perp})\rangle=\varepsilon(X)^{\perp_{\varepsilon}}$.
Finally
$\langle\varepsilon(X^{\perp\perp})\rangle=\varepsilon(X)^{\perp_{\varepsilon}\perp_{\varepsilon}}$,
as claimed. $\Box$

###### Corollary 2.3

We have
$\langle\varepsilon(\\{a,b\\}^{\perp})\rangle=\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}}$
and
$\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}\perp_{\varepsilon}}$
for any two opposite points $a$ and $b$ of ${\cal S}$.

Proof. We have
$\\{a,b\\}^{\perp}\cap\\{a,b\\}^{\perp\perp}=\\{a,b\\}^{\perp\perp\perp}\cap\\{a,b\\}^{\perp\perp}=\emptyset$.
The conclusion follows from Proposition 2.1 and Corollary 2.2. $\Box$

####  2.3.2 Subspaces of an embeddable polar space

Given an embedding $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$, a subspace
$X$ of ${\cal S}$ arises from $\varepsilon$ if $X=\varepsilon^{-1}({\cal X})$
for a projective subspace $\cal X$ of $\mathrm{PG}(V)$; equivalently,
$\langle\varepsilon(X)\rangle\cap\varepsilon({\cal S})=\varepsilon(X)$.

###### Definition 2.4

Given a singular subspace $K$ of ${\cal S}$ (possibly $K=\emptyset$) let
$\\{X_{i}\\}_{i\in I}$ be a family of singular subspaces such that each of
them contains $K$ as a hyperplane, $X_{i}^{\perp}\cap X_{j}=K$ for any choice
of $i,j\in I$ with $i\neq j$ and $|I|>1$. Then $X:=\cup_{i\in I}X_{i}$ is a
subspace of ${\cal S}$, henceforth called a rosette. Clearly, $K$ is the
radical of $X$. In particular, if $K=\emptyset$ then $X$ is a set of mutually
opposite points.

###### Proposition 2.5

Let ${\cal S}$ be embeddable and let $X$ be a subspace of ${\cal S}$. Then $X$
arises from an embedding of ${\cal S}$, except possibly when $X$ a rosette.

Proof. When $\mathrm{rank}({\cal S})<\infty$ the above is just the main result
of [2]. Suppose that ${\cal S}$ has infinite rank. Then ${\cal S}$ admits the
universal embedding, say $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$. We
shall prove that, if $X$ is not a rosette, then $X$ arises from $\varepsilon$.

When $X$ is a singular subspace there is nothing to prove. So, suppose that
$X$ is neither a singular subspace nor a rosette. By contradiction, suppose
that $X\subset\varepsilon^{-1}(\langle\varepsilon(X)\rangle)=:X^{\prime}$.
Choose a point $x_{0}\in X^{\prime}\setminus(X\cup X^{\perp})$. As
$\varepsilon(x_{0})\in\langle\varepsilon(X)\rangle$, there exists a finite
subset $A\subseteq X$ such that
$\varepsilon(x_{0})\in\langle\varepsilon(A)\rangle$. As $X$ is not a rosette,
we can always choose $A$ in such a way that $X_{0}:=\langle A\rangle$ contains
a pair of mutually opposite lines. Hence $X_{0}$ is not a rosette.

Put $X^{\prime}_{0}:=\varepsilon^{-1}(\langle\varepsilon(X_{0})\rangle)$.
Clearly, $X^{\prime}_{0}$ has finite rank, since
$\langle\varepsilon(X_{0})\rangle$ is finite dimensional. If $X^{\prime}_{0}$
is non-degenerate, then put $X_{1}=X^{\prime}_{0}$. Otherwise, let
$R:=X^{\prime}_{0}\cap X_{0}^{\prime\perp}$ be the radical of
$X^{\prime}_{0}$. Then $R$ has finite rank, since
$\mathrm{rank}(X^{\prime}_{0})<\infty$. However ${\cal S}$ is non-degenerate
of infinite rank. An easy argument exploiting induction on $\mathrm{rank}(R)$
shows that we can find a finite subset $B$ of ${\cal S}$ such that
$\varepsilon^{-1}(\langle\varepsilon(X_{0}\cup B)\rangle)$ is non-degenerate.
With $B$ chosen in this way, put
$X_{1}:=\varepsilon^{-1}(\langle\varepsilon(X_{0}\cup B)\rangle)$. Again,
$X_{1}$ has finite rank, since $\langle\varepsilon(X_{0}\cup
B)\rangle=\langle\varepsilon(A\cup B)\rangle$ has finite dimension (less than
$|A\cup B|$). Moreover $X_{1}$ is non-degenerate, thanks to the addition of
$B$ to $A$. By construction, $X_{1}$ arises from $\varepsilon$. Therefore
$\varepsilon(X_{1})$ is the polar space defined on
$\langle\varepsilon(X_{0}\cup B)\rangle=\langle\varepsilon(X_{1})\rangle$ by
the pseudoquadratic (or alternating) form induced on
$\langle\varepsilon(X_{1})\rangle$ by the pseudoquadratic (respectively,
alternating) form which describes $\varepsilon({\cal S})$. This induced form
is non-degenerate, since $X_{1}$ is non-degenerate. Moreover the embedding of
$X_{1}$ in $\langle\varepsilon(X_{1})\rangle$, say $\varepsilon_{1}$, is
universal by the above and since $\varepsilon$ is universal by assumption
(which means that in the characteristic 2 case $\varepsilon({\cal S})$ cannot
be described by an alternating form). Thus we can apply Theorem 1 of [2] to
$X_{0}$ as a subspace of $X_{1}$. By that theorem, $X_{0}$ arises from
$\varepsilon_{1}$, namely
$X_{0}=\varepsilon_{1}^{-1}(\langle\varepsilon_{1}(X_{0})\rangle)$. However
$\varepsilon_{1}^{-1}(\langle\varepsilon_{1}(X_{0})\rangle)=\varepsilon^{-1}(\langle\varepsilon(X_{0}\rangle)$
and $x_{0}\in\varepsilon^{-1}(\langle\varepsilon(X_{0})\rangle$ by definition
of $X_{0}$. In the end, $x_{0}\in X_{0}$. However $X_{0}\subseteq X$.
Therefore $x_{0}\in X$. We have reached a final contradiction. $\Box$

####  2.3.3 Optimally embeddable subspaces

Suppose that ${\cal S}$ is embeddable and let $X$ be a subspace of ${\cal S}$.
Then
$\langle\varepsilon(X),\varepsilon(X)^{\perp_{\varepsilon}}\rangle\subseteq\varepsilon(X\cap
X^{\perp})^{\perp_{\varepsilon}}$ for every embedding $\varepsilon$ of ${\cal
S}$. Since both $\varepsilon(X)^{\perp_{\varepsilon}}$ and $\varepsilon(X\cap
X^{\perp})^{\perp_{\varepsilon}}$ contain the radical $R_{\varepsilon}$ of
$\pi_{\varepsilon}$ and all quotients of $\varepsilon$ arise by factorizing
$\varepsilon$ over subspaces of $R_{\varepsilon}$, we have

$\langle\varepsilon(X),\varepsilon(X)^{\perp_{\varepsilon}}\rangle~{}=~{}\varepsilon(X\cap
X^{\perp})^{\perp_{\varepsilon}}$ (4)

if and only if the same holds for a quotient of $\varepsilon$. Consequently,
if (4) holds for $\varepsilon$ then it also holds for all covers and all
quotients of $\varepsilon$. If $\varepsilon({\cal S})={\cal
S}_{\pi_{\varepsilon}}$ then (4) holds for every subspace $X$ such that
$\dim(\varepsilon(X))<\infty$. (This follows from the fact that the reflexive
sesquilinear form $f$ associated to $\pi_{\varepsilon}$ is trace-valued and if
$\pi_{\varepsilon}$ is a polarity then $f$ is non-degenerate.) In particular,
if ${\cal S}$ is one of the exceptional embeddable generalized quadrangles
which do not admit the absolutely universal embedding then (4) holds for every
subspace $X$ and every embedding $\varepsilon$ of ${\cal S}$. So, if (4) holds
for a subspace $X$ of ${\cal S}$ and at last one embedding of ${\cal S}$ then
it holds for $X$ and all embeddings of ${\cal S}$. We are now ready to state
the following definition.

###### Definition 2.6

If $X$ satisfies property (4) for some (equivalently, every) embedding of
${\cal S}$ then we say that $X$ is optimally embeddable.

Clearly, al singular subspaces of ${\cal S}$ are optimally embeddable. As
previously recalled, if ${\cal S}$ admits an embedding $\varepsilon$ such that
$\varepsilon({\cal S})={\cal S}_{\pi_{\varepsilon}}$ then every finitely
generated subspace of ${\cal S}$ is optimally embeddable. Note also that a
non-degenerate subspace $X$ of ${\cal S}$ is optimally embeddabe if and only
if $\varepsilon(X)\cup\varepsilon(X)^{\perp_{\varepsilon}}$ spans
$\mathrm{PG}(V)$ for every embedding $\varepsilon:{\cal
S}\rightarrow\mathrm{PG}(V)$ of ${\cal S}$.

###### Proposition 2.7

Let $Y$ and $Z$ be finite-dimensional opposite singular subspaces of ${\cal
S}$ and put $X:=\langle Y,Z\rangle$. Then $X\cap X^{\perp}=\emptyset$ and
$\langle\varepsilon(X),\varepsilon(X)^{\perp_{\varepsilon}}\rangle=\mathrm{PG}(V)$
for every embedding $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ of ${\cal
S}$.

Proof. Let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be an embedding of
${\cal S}$. The projective subspace $\langle\varepsilon(X)\rangle$ is the
union of the lines of $\mathrm{PG}(V)$ joining a point of $\varepsilon(Y)$
with a point of $\varepsilon(Z)$. Let $\ell$ be one of these lines, say
$\ell=\langle\varepsilon(y),\varepsilon(z)\rangle$ for $y\in Y$ and $z\in Z$.
Suppose that $\ell$ meets $\varepsilon(X)^{\perp_{\varepsilon}}$ non-
trivially. Then at least one of $y$ and $z$ belongs to $X^{\perp}$. This
contradicts the assumption that $Y$ and $Z$ are opposite. Therefore
$\langle\varepsilon(X)\rangle\cap\varepsilon(X)^{\perp_{\varepsilon}}=\emptyset$.
Accordingly, $X\cap X^{\perp}=\emptyset$.

By assumption, $\dim\langle\varepsilon(X)\rangle<\infty$. We have
$\langle{\cal X},{\cal X}^{\perp_{\varepsilon}}\rangle=({\cal X}\cap{\cal
X}^{\perp_{\varepsilon}})^{\perp_{\varepsilon}}$ for every finite-dimensional
subspace $\cal X$ of $\mathrm{PG}(V)$. Therefore

$\langle\varepsilon(X),\varepsilon(X)^{\perp_{\varepsilon}}\rangle~{}=~{}(\langle\varepsilon(X)\rangle\cap\varepsilon(X)^{\perp_{\varepsilon}})^{\perp_{\varepsilon}}.$

However
$\langle\varepsilon(X)\rangle\cap\varepsilon(X)^{\perp_{\varepsilon}}=\emptyset$,
as shown in the previous paragraph. Therefore
$\langle\varepsilon(X),\varepsilon(X)^{\perp_{\varepsilon}}\rangle=\mathrm{PG}(V)$,
as claimed. $\Box$

###### Lemma 2.8

Let $a$ and $b$ be opposite points of ${\cal S}$. If $\varepsilon$ is minimal
then
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}\cap\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}=\emptyset$.

Proof. Let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be minimal, namely
$\pi_{\varepsilon}$ is a polarity. Hence
$\varepsilon(\\{a,b\\}^{\perp_{\varepsilon}}$ has codimention 2 in
$\mathrm{PG}(V)$ and, consequently,
$L:=\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$ is a line
of $\mathrm{PG}(V)$. By way of contradiction, suppose that
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}\cap L\neq\emptyset$. Then
$L\subseteq L^{\perp_{\varepsilon}}$, since $L$ contains
$\varepsilon(a),\varepsilon(b)$ and $p$, which are mutually distinct, each of
these points is absolute for $\pi_{\varepsilon}$ and $\pi_{\varepsilon}(p)$
contains $\varepsilon(a)$ and $\varepsilon(b)$. Consequently
$\varepsilon(a)\perp_{\varepsilon}\varepsilon(b)$, namely $a\perp b$, while
$a\not\perp b$ by assumption. $\Box$

###### Proposition 2.9

Let $a$ and $b$ be opposite points of ${\cal S}$. Then both
$\\{a,b\\}^{\perp}$ and $\\{a,b\\}^{\perp\perp}$ are optimally embeddable.

Proof. As
$\\{a,b\\}^{\perp}\cap\\{a,b\\}^{\perp\perp}=\\{a,b\\}^{\perp\perp\perp}\cap\\{a,b\\}^{\perp\perp}=\emptyset$,
we need to prove that the following holds for at least one embedding
$\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ of ${\cal S}$.

$\begin{array}[]{rcl}\langle\varepsilon(\\{a,b\\}^{\perp}),\varepsilon(\\{a,b\\}^{\perp})^{\perp_{\varepsilon}}\rangle&=&\mathrm{PG}(V),\\\
\langle\varepsilon(\\{a,b\\}^{\perp\perp}),\varepsilon(\\{a,b\\}^{\perp\perp})^{\perp_{\varepsilon}}\rangle&=&\mathrm{PG}(V).\end{array}$
(5)

We have
$\langle\varepsilon(\\{a,b\\}^{\perp})\rangle=\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}}$
and
$\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}\perp_{\varepsilon}}$
by Corollary 2.3. Moreover
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}\perp_{\varepsilon}}=\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}$.
Hence both equations of (5) are equivalent to the following:

$\langle\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}},\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}\rangle~{}=~{}\mathrm{PG}(V).$
(6)

We can assume that $\varepsilon$ is minimal. Hence
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$ is a line,
as noticed in the proof of Lemma 2.8. Therefore

$\langle\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}},\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}\perp_{\varepsilon}}\rangle=(\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}\cap\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}\perp_{\varepsilon}})^{\perp_{\varepsilon}}.$

However
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}\perp_{\varepsilon}}=\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}$.
Hence the above equality amounts to the following:

$\langle\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}},\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}\rangle~{}=~{}(\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}\cap\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}})^{\perp_{\varepsilon}}.$
(7)

We know that
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}\cap\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}=\emptyset$
by Lemma 2.8 and, obviously, $\emptyset^{\perp_{\varepsilon}}=\mathrm{PG}(V)$.
So, (7) is the same as (6). $\Box$

##  3 Regularity

Throughout this section ${\cal S}$ is a (possibly non-embeddable) non-
degenerate thick-lined polar space of rank at least $2$.

###  3.1 Definitions and a proof of Theorem 1.1

Lat $a$ and $b$ be two opposite points of ${\cal S}$ and $N,N^{\prime}$
generators of $\\{a,b\\}^{\perp}$. Clearly,
$\\{N,N^{\prime}\\}^{\perp}\supseteq\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$.

###### Definition 3.1

We say that two generators $N,N^{\prime}$ of $\\{a,b\\}^{\perp}$ form a
$\perp$-minimal pair if $\\{N,N^{\prime}\\}^{\perp}=\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$. A generator $N$ of
$\\{a,b\\}^{\perp}$ will be said to be $\perp$-minimal if $\\{N,N^{\prime}\\}$
is $\perp$-minimal for every generator $N^{\prime}$ of $\\{a,b\\}^{\perp}$ (in
particular, $N^{\perp}=\langle N,\\{a,b\\}^{\perp\perp}\rangle$). We say that
$\\{a,b\\}$ is regular if all generators of $\\{a,b\\}^{\perp}$ are
$\perp$-minimal. If all pairs of opposite points of ${\cal S}$ are regular,
then ${\cal S}$ is said to be regular.

###### Lemma 3.2

We have $\langle
X,\\{a,b\\}^{\perp\perp}\rangle=\cup_{x\in\\{a,b\\}^{\perp\perp}}\langle
X,x\rangle$ for every singular subspace $X$ of $\\{a,b\\}^{\perp}$.

Proof. Let $x,y$ be distinct points of $\\{a,b\\}^{\perp\perp}$. Then
$x\not\perp y$. Consequently, no point of $\langle X,x\rangle\setminus X$ can
be collinear with a point of $\langle X,y\rangle\setminus X$. The conclusion
follows from this remark. $\Box$

In particular, if $N$ is a generator of $\\{a,b\\}^{\perp}$ then $\langle
N,\\{a,b\\}^{\perp\perp}\rangle$ is the union of the generators of ${\cal S}$
which contain $N$ and meet $\\{a,b\\}^{\perp\perp}$ non-trivally. Accordingly,
$\\{N,N\\}$ is $\perp$-minimal if and only if all generators of ${\cal S}$
containing $N$ meet $\\{a,b\\}^{\perp\perp}$ non-trivially.

###### Theorem 3.3

Let $N$ be a generator of $\\{a,b\\}^{\perp}$. If the pair $\\{N,N\\}$ is
$\perp$-minimal then $N$ is $\perp$-minimal.

Proof. Let $\\{N,N\\}$ be $\perp$-minimal, namely all generators of ${\cal S}$
containing $N$ can be obtained as $\langle N,c\rangle$ for a point
$c\in\\{a,b\\}^{\perp\perp}$ (Lemma 3.2). Given another generator $N^{\prime}$
of $\\{a,b\\}^{\perp}$, let $x\in\\{N,N^{\prime}\\}^{\perp}$ and suppose that
$x\not\in N\cap N^{\prime}$. Therefore $x\not\in N$ (otherwise $\langle
x,N^{\prime}\rangle$ is a singular subspace contained in $\\{a,b\\}^{\perp}$
and properly containing $N^{\prime}$, a contradiction with $N^{\prime}$ being
a generator of $\\{a,b\\}^{\perp}$). Accordingly, $M:=\langle N,x\rangle$ is a
generator of ${\cal S}$. As $\\{N,N\\}$ is $\perp$-minimal, $M$ contains a
point $c\in\\{a,b\\}^{\perp}$. If $x=c$ then
$x\in\\{a,b\\}^{\perp\perp}\subseteq\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$. Otherwise, the line $\langle
x,c\rangle$ meets $N$ in a point, say $y$. We have $\langle
x,c\rangle\subseteq N^{\prime\perp}$, since both $x$ and $c$ belong to
$N^{\prime\perp}$. However, $N\cap N^{\prime\perp}=N\cap N^{\prime}$.
Therefore $y\in N\cap N^{\prime}$. Hence $x\in\langle
y,c\rangle\subseteq\langle N\cap N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$.

We have proved that $\\{N,N^{\prime}\\}^{\perp}\subseteq\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$. Hence
$\\{N,N^{\prime}\\}^{\perp}=\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$ since $(N\cap
N^{\prime})\cup\\{a,b\\}^{\perp\perp}\subseteq\\{N,N^{\prime}\\}^{\perp}$. So,
$\\{N,N^{\prime}\\}$ is $\perp$-minimal. As $N^{\prime}$ is an arbitrary
generator of $\\{a,b\\}^{\perp}$, $N$ is $\perp$-minimal. $\Box$

Theorem 1.1 immediately follows from Theorem 3.3. Indeed in property
$({\mathrm{R}}3)$ it is assumed that every pair of generators of
$\\{a,b\\}^{\perp}$ is $\perp$-minimal while $({\mathrm{R}}2)$ is equivalent
to $\\{N,N\\}$ being $\perp$-minimal for every generator $N$ of
$\\{a,b\\}^{\perp}$. Trivially, $({\mathrm{R}}3)$ implies $({\mathrm{R}}2)$.
Conversely, by Theorem 3.3, property $({\mathrm{R}}2)$ implies
$({\mathrm{R}}3)$.

###### Corollary 3.4

Assume that $\mathrm{rank}({\cal S})<\infty$ and let $N$ be a generator of
$\\{a,b\\}^{\perp}$. If there exists a generator $N^{\prime}$ of
$\\{a,b\\}^{\perp}$ such that $\\{N,N^{\prime}\\}$ is $\perp$-minimal then $N$
is $\perp$-minimal.

Proof. In view of Theorem 3.3, we only need to prove that, if a generator
$N^{\prime}$ of $\\{a,b\\}^{\perp}$ exists such that $\\{N,N^{\prime}\\}$ is
$\perp$-minimal, then $\\{N,N\\}$ is $\perp$-minimal. If $N^{\prime}=N$ there
is nothing to prove. So, assume that $N^{\prime}\neq N$. Hence $N\cap
N^{\prime\perp}=N\cap N^{\prime}$. Let $M$ be a generator of ${\cal S}$
containing $N$. Since $\dim(N^{\prime})$ is finite and
$\dim(M)=\dim(N^{\prime})+1$, we have $\dim(M\cap
N^{\prime\perp})=1+\dim(N\cap N^{\prime})$ (recall that
$\dim(N^{\prime})=\dim(N)$ as $\mathrm{rank}({\cal S})<\infty$). Accordingly,
$M\cap N^{\prime\perp}$ is not contained in $N\cap N^{\prime}$. Let
$x\in(M\cap N^{\prime\perp})\setminus(N\cap N^{\prime})$. Clearly
$x\in\\{N,N^{\prime}\\}^{\perp}$. Therefore $x\in\langle N\cap
N^{\prime},\\{a,b\\}^{\perp}\rangle$ since by assumption $\\{N,N^{\prime}\\}$
is $\perp$-minimal. By lemma 3.2, the point $x$ belongs to a line $\ell$
ioining a point $c\in\\{a,b\\}^{\perp\perp}$ with a point $y\in N\cap
N^{\prime}$. We have $x\neq y$, since $x\not\in N\cap N^{\prime}$. Therefore
$\ell=\langle x,y\rangle$. Consequently $\ell\subseteq M$, since both $x$ and
$y$ belong to $M$. However $c\in\ell$. Hence $M$ meets
$\\{a,b\\}^{\perp\perp}$ in $c$. We have proved that all generators of ${\cal
S}$ which contain $N$ meet $\\{a,b\\}^{\perp}$ non-trivially, namely
$\\{N,N\\}$ is $\perp$-minimal. $\Box$

###### Corollary 3.5

Still assuming that $\mathrm{rank}({\cal S})<\infty$, when
$\mathrm{rank}({\cal S})=2$ we also assume that ${\cal S}$ is embeddable.
Suppose that there exist two opposite points $a$ and $b$ of ${\cal S}$ such
that $\\{a,b\\}^{\perp}$ admits a $\perp$-minimal pair of generators. Then
${\cal S}$ is regular.

Proof. If ${\cal S}$ is non-embeddable of rank $3$ then ${\cal S}$ is regular,
as proved in [12, Proposition 5.9.4] (also [3, Result 3.3]). Let ${\cal S}$ be
embeddable of finite rank. Then $\mathrm{Aut}({\cal S})$ satisfies both the
following transitivity properties:

  * $(\mathrm{T}1)$

$\mathrm{Aut}({\cal S})$ acts transitively on the set of pairs of opposite
points of ${\cal S}$;

  * $(\mathrm{T}2)$

for any two opposite points $x$ and $y$ of ${\cal S}$, the stabilizer of
$\\{x,y\\}$ in $\mathrm{Aut}({\cal S})$ acts transitively on the set of
generators of $\\{a,b\\}^{\perp}$.

By assumption, two opposite points $a$ and $b$ exist in ${\cal S}$ such that
$\\{a,b\\}^{\perp}$ admits a $\perp$-minimal pair of generators
$\\{N,N^{\prime}\\}$. By Corollary 3.4, both $N$ and $N^{\prime}$ are
$\perp$-minimal. Hence $\\{a,b\\}$ is regular by $(\mathrm{T}2)$ and ${\cal
S}$ is regular by $(\mathrm{T}1)$. $\Box$

###### Remark 2

When $\mathrm{rank}({\cal S})=2$ and $(\mathrm{T}1)$ fails to hold (hence
${\cal S}$ is non-embeddable) it can happen that some but not all of the pairs
of opposite points of ${\cal S}$ are regular (see [3, Remark 6]).

###### Remark 3

When $\mathrm{rank}({\cal S})$ is infinite condition $(\mathrm{T}1)$ holds
true but in general $(\mathrm{T}2)$ fails to hold. As we shall see in Section
5.4, when $\mathrm{rank}({\cal S})=\infty$ this case it can happen that, for
any two opposite points $a$ and $b$ of ${\cal S}$, some but not all of the
generators of $\\{a,b\\}^{\perp}$ are $\perp$-minimal.

###  3.2 An improvement of Definition 3.1

Our definition of $\perp$-minimality as stated in Definition 3.1 is relative
to (the hyperbolic line spanned by) a given pair $\\{a,b\\}$ of opposite
points, but we have dropped the reference to $\\{a,b\\}$ in our terminology
presuming that this is implicit in choosing $N$ and $N^{\prime}$ among the
generators of $\\{a,b\\}^{\perp}$. One might believe that, if we regard $N$
and $N^{\prime}$ as sub-generators of ${\cal S}$ without choosing in advance a
pair of opposite points $\\{a,b\\}$ such that $\\{a,b\\}^{\perp}$ contains $N$
and $N^{\prime}$, then whether $\\{N,N^{\prime}\\}$ is or is not
$\perp$-minimal depends on the choice of a pair of opposite points in
$\\{N,N^{\prime}\\}^{\perp}$. However, as stated in the next theorem, this is
not the case, except possibly when $\mathrm{rank}({\cal S})=2$ and ${\cal S}$
is non-embeddable. So, all in all, the terminology adopted in Definition 3.1
is not so deceiving as it looks.

###### Theorem 3.6

Let $N$ and $N^{\prime}$ be sub-generators of ${\cal S}$ and let $\\{a,b\\}$
and $\\{c,d\\}$ be pairs of opposite points contained in
$\\{N,N^{\prime}\\}^{\perp}$. Then the pair $\\{N,N^{\prime}\\}$ is
$\perp$-minimal with respect to $\\{a,b\\}$ if and only if it is
$\perp$-minimal with respect to $\\{c,d\\}$, except possibly when ${\cal S}$
is non-embeddable of rank $2$, $N=N^{\prime}$ and the stabilizer in
$\mathrm{Aut}({\cal S})$ of the point $p:=N=N^{\prime}$ acts intransitively on
the set of hyperbolic lines of ${\cal S}$ contained in $p^{\perp}$.

Proof. Suppose firstly that ${\cal S}$ is embeddable and let
$\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be an embedding of ${\cal S}$.
We can assume to have chosen $\varepsilon$ in such a way that it is minimal,
namely $\pi_{\varepsilon}$ is a polarity. Then
$\\{a,b\\}^{\perp\perp}=\varepsilon^{-1}(\langle\varepsilon(a),\varepsilon(b)\rangle)$
and
$\\{c,d\\}^{\perp\perp}=\varepsilon^{-1}(\langle\varepsilon(c),\varepsilon(d)\rangle)$.
Moreover, by Lemma 3.2 the subpspace $\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$ is the union of the subspaces
$\langle N\cap N^{\prime},x\rangle$ for $x\in\\{a,b\\}^{\perp\perp}$, A
similar description holds for $\langle N\cap
N^{\prime},\\{c,d\\}^{\perp}\rangle$.

Suppose that $\\{N,N^{\prime}\\}$ is $\perp$-minimal with respect to
$\\{a,b\\}$, namely:

$\\{N,N^{\prime}\\}^{\perp}~{}=~{}\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle.$ (8)

By assumption, $c,d\in\\{N,N^{\prime}\\}^{\perp}$ and $c\not\perp d$.
Therefore, by (8) and the previous paragraph, $c\in\langle N\cap
N^{\prime},x\rangle\setminus N\cap N^{\prime}$ and $d\in\langle N\cap
N^{\prime},y\rangle\setminus N\cap N^{\prime}$ for suitable points
$x,y\in\\{a,b\\}^{\perp}$. Without loss, we can assume that $x=a$ and $y=b$.
It follows that $\langle\varepsilon(N\cap
N^{\prime}),\varepsilon(a),\varepsilon(b)\rangle=\langle\varepsilon(N\cap
N^{\prime}),\varepsilon(c),\varepsilon(d)\rangle$. Accordingly,
$\langle\varepsilon(c),\varepsilon(d)\rangle$ meets every subspace
$\langle\varepsilon(N\cap N^{\prime}),\varepsilon(x)\rangle$ non-trivially,
for every $x\in\\{a,b\\}^{\perp\perp}$. Similarly,
$\langle\varepsilon(a),\varepsilon(b)\rangle$ meets $\langle\varepsilon(N\cap
N^{\prime}),\varepsilon(y)\rangle$ non-trivially for every
$y\in\\{c,d\\}^{\perp\perp}$. It follows that

$\langle N\cap N^{\prime},\\{a,b\\}^{\perp\perp}\rangle~{}=~{}\langle N\cap
N^{\prime},\\{c,d\\}^{\perp\perp}\rangle.$ (9)

By (8) and (9) we get $\\{N,N^{\prime}\\}^{\perp}=\langle N\cap
N^{\prime},\\{c,d\\}^{\perp\perp}\rangle$, namely $\\{N,N^{\prime}\\}$ is also
$\perp$-minimal with respect to $\\{c,d\\}$.

When ${\cal S}$ is non-embeddable of rank 3 then ${\cal S}$ is regular [12,
Proposition 5.9.4]; also [3, Result 3.3]). In this case there is nothing to
prove. Finally, suppose that $\mathrm{rank}({\cal S})=2$. Then $N^{\prime}$
and $N^{\prime}$ are points, say $p=N$ and $p^{\prime}=N^{\prime}$. If $p\neq
p^{\prime}$ and $\\{p,p^{\prime}\\}$ is $\perp$-minimal with respect to
$\\{a,b\\}$ then $\\{p,p^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$.
Consequently, since $c,d\in\\{p,p^{\prime}\\}^{\perp}$ and $c\not\perp d$, we
have $c,d\in\\{a,b\\}^{\perp\perp}$, namely
$\\{c,d\\}^{\perp\perp}=\\{a,b\\}^{\perp\perp}$. Again, there is nothing to
prove. Finally, let $p^{\prime}=p$. If the stabilizer of $p$ in
$\mathrm{Aut}({\cal S})$ acts transitively on the set of hyperbolic lines
contained in $p^{\perp}$ (as it is the case when ${\cal S}$ is embeddable),
then the conclusion follows. $\Box$

When either $\mathrm{rank}({\cal S})>2$ or ${\cal S}$ is embeddable of rank 2
Theorem 3.6 allows us to replace Definition 3.1 with the following.

###### Definition 3.7

Let $N$ and $N^{\prime}$ be sub-generators of ${\cal S}$, possibly
$N=N^{\prime}$. We say that $N$ and $N^{\prime}$ form a $\perp$-minimal pair
if either $\\{N,N^{\prime}\\}^{\perp}$ is a singular subspace or
$\\{N,N^{\prime}\\}^{\perp}=\langle N\cap
N^{\prime},\\{a,b\\}^{\perp\perp}\rangle$ for every (equivalently, at least
one) choice of opposite points $a,b\in\\{N,N^{\prime}\\}^{\perp}$. A sub-
generator $N$ is said to be $\perp$-minimal if $\\{N,N\\}$ is $\perp$-minimal,
namely either $N^{\perp}$ is a generator of ${\cal S}$ or $N^{\perp}=\langle
N,\\{a,b\\}^{\perp\perp}\rangle$ for some (equivalently, every) pair
$\\{a,b\\}$ of opposite points of $N^{\perp}$. A pair $\\{a,b\\}$ of opposite
points is regular if all generators of $\\{a,b\\}^{\perp}$ are
$\perp$-minimal. The polar space ${\cal S}$ is regular if all of its sub-
generators are $\perp$-minimal (equivalently, all pairs of opposite points of
${\cal S}$ are regular).

Note that in Definition 3.7 when defining $\perp$-minimal sub-generators
Theorem 3.3 is also taken into consideration. Note also that, if $N$ and
$N^{\prime}$ are opposite sub-generators, then not two distinct points of
$\\{N,N^{\prime}\\}^{\perp}$ are colliear. Hence $\\{N,N^{\prime}\\}^{\perp}$
is a singular subspace if and only it is either empty or a singleton. When
$\mathrm{rank}({\cal S})<\infty$ no sub-generator is contained in one single
generator and, if $N$ and $N^{\prime}$ are opposite sub-generators, then
$\\{N,N^{\prime}\\}^{\perp}$ has the same cardinality as the set of generators
containing a given sub-generator; hence $\\{N,N^{\prime}\\}^{\perp}$ is never
a singular subspace.

In contrast, in general a polar space of infinite rank admits sub-generators
which are contained in just one generator and pairs of opposite generators
$\\{N,N^{\prime}\\}$ such that $|\\{N,N^{\prime}\\}^{\perp}|\leq 1$.

###### Remark 4

When $\mathrm{rank}({\cal S})=2$, if $a$ and $b$ are opposite points and $c$
and $d$ are distinct (hence opposite) points of $\\{a,b\\}^{\perp}$, then
$\\{c,d\\}^{\perp}=\\{a,b\\}^{\perp\perp}$ if and only if
$\\{a,b\\}^{\perp}=\\{c,d\\}^{\perp\perp}$. Accordingly, $\\{a,b\\}$ is
regular if and only if $\\{c,d\\}$ is regular. This is a special case of the
following general fact: for any choice of singular subspaces
$X_{1},X_{2},Y_{1},Y_{2}$, we have
$\\{Y_{1},Y_{2}\\}^{\perp}=\\{X_{1},X_{2}\\}^{\perp\perp}$ if and only if
$\\{X_{1},X_{2}\\}^{\perp}=\\{Y_{1},Y_{2}\\}^{\perp\perp}$.

###  3.3 Proof of Theorem 1.4 and more on tight embeddings

Throughout this subsection the polar space ${\cal S}$ is assumed to be
embeddable. We firstly prove Theorem 1.4. With the terminology introduced so
far, we rephrase Theorem 1.4 as follows.

###### Theorem 3.8

The polar space ${\cal S}$ admits a tight embedding if and only if

  * $({\mathrm{R}}4)$

${\cal S}$ admits a $\perp$-minimal pair of opposite sub-generators $N$ and
$N^{\prime}$ such that $|\\{N,N^{\prime}\\}^{\perp}|>1$ and $\langle
N,N^{\prime}\rangle$ is optimally embeddable.

Proof. Let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be a minimal
embedding of ${\cal S}$. So, $\pi_{\varepsilon}$ is a polarity. Accordingly,
$\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}\perp_{\varepsilon}}=\langle\varepsilon(a),\varepsilon(b)\rangle$
for every pair of opposite points $a,b$ of ${\cal S}$.

Suppose that $({\mathrm{R}}4)$ holds. By assumption,
$\\{N,N^{\prime}\\}^{\perp}$ contains at least two points and, since $N$ and
$N^{\prime}$ are opposite sub-generators, all points of
$\\{N,N^{\prime}\\}^{\perp}$ are mutually opposite. Let $a$ and $b$ be
distinct points of $\\{N,N^{\prime}\\}^{\perp}$. Then
$\\{N,N^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$ since $\\{N,N^{\prime}\\}$
is $\perp$-minimal. Moreover $\langle N,N^{\prime}\rangle$ is optimally
embeddable. Therefore

$\begin{array}[]{l}\langle\varepsilon(N),\varepsilon(N^{\prime}),\varepsilon(a),\varepsilon(b)\rangle=\langle\varepsilon(N),\varepsilon(N^{\prime}),\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\\\
=~{}\langle\varepsilon(N),\varepsilon(N^{\prime}),\varepsilon(\\{N,N^{\prime}\\}^{\perp})\rangle=\langle\varepsilon(N),\varepsilon(N^{\prime}),\\{\varepsilon(N),\varepsilon(N)\\}^{\perp_{\varepsilon}}\rangle=\\\
=\mathrm{PG}(V).\end{array}$

Hence

$\langle\varepsilon(N),\varepsilon(N^{\prime}),\varepsilon(a),\varepsilon(b)\rangle~{}=~{}\mathrm{PG}(V).$
(10)

Put $M=\langle N,a\rangle$ and $M^{\prime}=\langle N^{\prime},b\rangle$. Then
$M$ and $M^{\prime}$ are opposite generators of ${\cal S}$. Equation (10)
shows that
$\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle=\mathrm{PG}(V)$. So,
$\varepsilon$ is tight.

Conversely, let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be a tight
embedding of ${\cal S}$ and let $M$ and $M^{\prime}$ be generators of ${\cal
S}$ such that
$\mathrm{PG}(V)=\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle$. Then
$M\cap M^{\prime}=\emptyset$, namely $M$ and $M^{\prime}$ are opposite. The
equality $\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle=\mathrm{PG}(V)$
and the fact that $M\cap M^{\prime}=\emptyset$ force $\pi_{\varepsilon}$ to be
a polarity. Choose $a\in M$ and $b\in M^{\prime}\setminus a^{\perp}$ and put
$N=b^{\perp}\cap M$ and $N^{\prime}=a^{\perp}\cap N^{\prime}$. Then $N$ and
$N^{\prime}$ are opposite generators of $\\{a,b\\}^{\perp}$. Clearly,
$\langle\langle\varepsilon(N\cup
N^{\prime})\rangle,\langle\varepsilon(a),\varepsilon(b)\rangle\rangle=\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle$.
However $\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle=\mathrm{PG}(V)$
by assumption and $\\{N,N^{\prime}\\}^{\perp}\supseteq\\{a,b\\}^{\perp\perp}$.
Therefore

$\langle\varepsilon(N),\varepsilon(N^{\prime}),\\{\varepsilon(N),\varepsilon(N^{\prime}\\}^{\perp_{\varepsilon}}\rangle~{}=~{}\mathrm{PG}(V)$
(11)

and, since
$\langle\varepsilon(N),\varepsilon(N^{\prime})\rangle\cap\\{\varepsilon(N),\varepsilon(N^{\prime})\\}^{\perp_{\varepsilon}}=\emptyset$,
necessarily

$\\{\varepsilon(N),\varepsilon(N^{\prime})\\}^{\perp_{\varepsilon}}~{}=~{}\langle\varepsilon(a),\varepsilon(b)\rangle~{}=~{}\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle.$
(12)

Equality (11) shows that $\langle N,N^{\prime}\rangle$ is optimally embeddale
while (12) is equivalent to
$\\{N,N^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$, namely
$\\{N,N^{\prime}\\}$ is a $\perp$-minimal pair. $\Box$

As previously said, the condition called $({\mathrm{R}}4)$ in Theorem 3.8 is
just a rephrasing of condition $({\mathrm{R}}4)$ of Theorem 1.4. The subspaces
called $X$ and $Y$ in the statement of Theorem 1.4 are called $N$ and
$N^{\prime}$ in Theorem 3.8, condition (1) of $({\mathrm{R}}4)$ as stated in
Theorem 1.4 says that $\\{X,Y\\}$ is $\perp$-minimal and $\\{X,Y\\}^{\perp}$
is not a singular subspace and condition (2) is equivalent to $\langle
X,Y\rangle$ being optimally embeddable. Indeed, since
$\\{X,Y\\}^{\perp}\not\subseteq\\{X,Y\\}^{\perp\perp}$, we have
$\varepsilon(X\cup Y)^{\perp_{\varepsilon}}=\langle\varepsilon((X\cup
Y)^{\perp})\rangle$ by Proposition 2.1. So, $\langle\varepsilon(X\cup
Y\cup(X\cup Y)^{\perp})\rangle=\langle\varepsilon(\langle
X,Y\rangle),\varepsilon(\langle X,Y\rangle)^{\perp_{\varepsilon}}\rangle$.

The next Corollary is just the same as Proposition 1.3. We add its proof here
in order to show that Proposition 1.3 indeed follows from Theorem 1.4.

###### Corollary 3.9

Let ${\cal S}$ be embeddable of finite rank. Then ${\cal S}$ is regular if and
only if it admits a tight embedding.

Proof. Let $\mathrm{rank}({\cal S})<\infty$. Then the subspace spanned by two
opposite sub-generators is optimally embeddable by Proposition 2.7.
Accordingly, we can drop the condition that $\langle N,N^{\prime}\rangle$ is
optimally embeddable from $({\mathrm{R}}4)$ of Theorem 3.8. However, if we
remove that condition then, in view of Corollary 3.5, what remains of
$({\mathrm{R}}4)$ amounts to say that ${\cal S}$ is regular. $\Box$

###### Lemma 3.10

Suppose that $\mathrm{rank}({\cal S})\geq 3$ and ${\cal S}$ admits a unique
embedding. Let $N$ and $N^{\prime}$ be opposite sub-generators of ${\cal S}$
such that the pair $\\{N,N^{\prime}\\}$ is $\perp$-minimal and
$|\\{N,N^{\prime}\\}^{\perp}|>1$. Then $\langle N,N^{\prime}\rangle$ is
optimally embeddable if and only if $\langle
N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$ for every choice of distinct (hence
opposite) points $a,b\in\\{N,N^{\prime}\\}^{\perp}$.

Proof. The ‘if’ part of the statement immediately follows from the fact that
$\\{a,b\\}^{\perp}$ is optimally embeddable for every choice of opposite
points $a$ and $b$ (Proposition 2.9). Note that the hypotheses that
$\mathrm{rank}({\cal S})>2$ and ${\cal S}$ admits a unique embedding play no
role in this implication. Turning to the ‘only if’ part, let
$\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be the (unique) embedding of
${\cal S}$ and suppose that $\langle N,N^{\prime}\rangle$ is optimally
embeddable. Then

$\langle\varepsilon(\langle N,N^{\prime}\rangle),\varepsilon(N\cup
N^{\prime})^{\perp_{\varepsilon}}\rangle~{}=~{}\varepsilon(\langle
N,N^{\prime}\rangle\cap\\{N,N^{\prime}\\}^{\perp})^{\perp_{\varepsilon}}.$

Hence, given two opposite points $a,b\in\\{N,N^{\prime}\\}^{\perp}$ (which
exist because $\\{N,N^{\prime}\\}^{\perp}$ is non-singular by assumption), we
have

$\langle\varepsilon(\langle N\cup
N^{\prime}\rangle),\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}\rangle~{}=~{}\mathrm{PG}(V)$
(13)

because $\\{N,N^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$ by assumption,
$\\{a,b\\}^{\perp}\cap\\{a,b\\}^{\perp\perp}=\emptyset$ and
$\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$
by Corollary 2.3.

Recall that
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}\cup\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$
spans $\mathrm{PG}(V)$. The embedding $\varepsilon$ is both absolutely
universal and minimal, since by assumption $\varepsilon$ is the unique
embedding of ${\cal S}$. As $\varepsilon$ is minimal,
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}$ and
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$ are disjoint
by Lemma 2.8. So, $\mathrm{PG}(V)$ is the direct sum of
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}$ and
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}\perp_{\varepsilon}}$. Moreover
$\varepsilon(\\{a,b\\})^{\perp_{\varepsilon}}=\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$
by Corollary 2.3 and $\langle\varepsilon(\langle
N,N^{\prime}\rangle)\rangle\subseteq\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$,
since $N\cup N^{\prime}\subseteq\\{a,b\\}^{\perp}$. By comparing all this
information with equation (13) we see that $\langle\varepsilon(\langle
N,N^{\prime}\rangle)\rangle=\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$.

So far we have made no use of the hypothesis that $\mathrm{rank}({\cal S})>2$.
We shall use it now. Since $\mathrm{rank}({\cal S})>2$, the subspace $\langle
N,N^{\prime}\rangle$ is not a rosette. Therefore it arises from an embedding
of ${\cal S}$, by Proposition 2.5. However $\varepsilon$ is the unique
embedding of ${\cal S}$. Hence $\langle N,N^{\prime}\rangle$ arises from
$\varepsilon$, namely $\langle
N,N^{\prime}\rangle=\varepsilon^{-1}(\langle\varepsilon(\langle
N,N^{\prime}\rangle)\rangle)$. Similarly,
$\\{a,b\\}^{\perp}=\varepsilon^{-1}(\langle\varepsilon(\\{a,b\\}^{\perp})\rangle)$.
The equality $\langle\varepsilon(\langle
N,N^{\prime}\rangle)\rangle=\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$ now
forces $\langle N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$. $\Box$

###### Theorem 3.11

Suppose that $\mathrm{rank}({\cal S})\geq 3$ and ${\cal S}$ admits a unique
embedding. Then the following are equivalent:

  * $(1)$

the unique embedding of ${\cal S}$ is tight;

  * $(2)$

${\cal S}$ admits admits a pair opposite sub-generators $N$ and $N^{\prime}$
such that $\langle N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$ for two opposite
points $a$ and $b$ of ${\cal S}$;

  * $(3)$

${\cal S}$ admits two opposite generators $M$ and $M^{\prime}$ such that
$\langle M,M^{\prime}\rangle={\cal S}$.

Proof. Trivially (3) implies (1) while (1) implies (2) by Theorem 3.8 and
Lemma 3.10. It remains to show that (2) implies (3). Given $N,N^{\prime},a$
and $b$ as in (2), let $M=\langle N,a\rangle$ and $M^{\prime}=\langle
N^{\prime},b\rangle$. Then $M$ and $M^{\prime}$ are opposite generators of
${\cal S}$. Moreover $\langle M,M^{\prime}\rangle=\langle
N,N^{\prime},a,b\rangle=\langle\\{a,b\\}^{\perp},a,b\rangle$, since $\langle
N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$. However
$\langle\\{a,b\\}^{\perp},a,b\rangle={\cal S}$. Hence $\langle
M,M^{\prime}\rangle={\cal S}$. $\Box$

###### Remark 5

In Lemma 3.10 and Theorem 3.11 we cannot drop the hypothesis that
$\mathrm{rank}({\cal S})>2$. Indeed the sub-generators of a generalized
quadrangle are its points. Hence, with $N,N^{\prime},a$ and $b$ as in the
hypotheses of Lemma 3.10, when $\mathrm{rank}({\cal S})=2$ we have $\langle
N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$ only if ${\cal S}$ is a grid. When
$\mathrm{rank}({\cal S})=2$ we should replace the equality $\langle
N,N^{\prime}\rangle=\\{a,b\\}^{\perp}$ with $\langle\varepsilon(\langle
N,N\rangle)\rangle=\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$, for a
minimal embedding $\varepsilon$ of ${\cal S}$, but this condition amounts to
$\langle N,N^{\prime}\rangle$ being optimally embeddable.

The hypothesis that ${\cal S}$ admits a unique embedding is also necessary,
both in Lemma 3.10 and in Theorem 3.11. Indeed suppose that ${\cal S}$ admits
a tight embedding $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ and let $M$
and $M^{\prime}$ be opposite generators of ${\cal S}$ such that
$\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle=\mathrm{PG}(V)$. If
$\varepsilon$, which is minimal, is not universal, let
$\tilde{\varepsilon}:{\cal S}\rightarrow\mathrm{PG}(\widetilde{V})$ be a
proper cover of $\varepsilon$. Then
$\langle\tilde{\varepsilon}(M),\tilde{\varepsilon}(M^{\prime})\rangle\subset\mathrm{PG}(\widetilde{V})$,
hence $\langle M,M^{\prime}\rangle\subset{\cal S}$; similarly, $\langle\
X,Y\rangle\subset\\{a,b\\}^{\perp}$ for any two singular subspaces
$X,Y\subseteq\\{a,b\\}^{\perp}$, even if $X$ and $Y$ are such that
$\langle\varepsilon(\langle
X,Y\rangle)\rangle=\langle\varepsilon(\\{a,b\\}^{\perp})\rangle$.

###### Remark 6

When $\mathrm{rank}({\cal S})<\infty$, if ${\cal S}=\langle
M,M^{\prime}\rangle$ for two (necessarily opposite) generators $M$ and
$M^{\prime}$ then ${\cal S}=\langle X,X^{\prime}\rangle$ for any two opposite
generators $X$ and $X^{\prime}$. In general, as we shall see in Section 5,
when the rank of ${\cal S}$ is infinite ${\cal S}$ can admit pairs of
generators $\\{X,X^{\prime}\\}$ such that $\langle X,X^{\prime}\rangle={\cal
S}$ as well as opposite generators $Y$ and $Y^{\prime}$ such that $\langle
Y,Y^{\prime}\rangle\subset{\cal S}$.

###  3.4 Complementary generators and regularity

In this subsection ${\cal S}$ is supposed to admit the absolutely universal
embedding. Hence it also admits the minimum embedding. Throughout this
subsection $\varepsilon:{\cal S}:\rightarrow\mathrm{PG}(V)$ is the minimum
embedding of ${\cal S}$.

The following definitions, which complete previously stated definitions, will
help us to make our exposition more concise. Let $M$ and $M^{\prime}$ be
opposite generators of ${\cal S}$. If
$\langle\varepsilon(M),\varepsilon(M^{\prime})\rangle=\mathrm{PG}(V)$ then we
say that $M$ and $M^{\prime}$ are complementary, also that each of them is a
complement of the other one. Obviously, ${\cal S}$ admits a pair of
complementary generators if and only if $\varepsilon$ is tight.

A pair $\\{N,N^{\prime}\\}$ of sub-generators of ${\cal S}$ is said to be
optimally embeddable if $\langle N,N^{\prime}\rangle$ is optimally embeddable
(Definition 2.6); it is said to be $\perp$-degenerate if
$\\{N,N^{\prime}\\}^{\perp}$ is a singular subspace. Note that, according to
Definition 3.7, all $\perp$-degenerate pairs of opposite sub-generators are
$\perp$-minimal. Note also that, if $N$ and $N^{\prime}$ are opposite sub-
generators, then $\\{N,N^{\prime}\\}$ is $\perp$-degenerate precisely when
$|\\{N,N^{\prime}\\}^{\perp}|\leq 1$.

###### Lemma 3.12

No optimally embeddable pair of opposite sub-generators is $\perp$-degenerate.

Proof. Let $\\{N,N^{\prime}\\}$ be an optimally embeddable pair of opposite
sub-generators of ${\cal S}$. By way of contradiction, let
$|\\{N,N^{\prime}\\}^{\perp}|\leq 1$. Suppose first that
$\\{N,N^{\prime}\\}^{\perp}=\\{c\\}$ for a point $c$. Let
$\tilde{\varepsilon}:{\cal S}\rightarrow\mathrm{PG}(\widetilde{V})$ be the
universal embedding of ${\cal S}$. Since $\\{N,N^{\prime}\\}$ is optimally
embeddable, we have
$\langle\tilde{\varepsilon}(N),\tilde{\varepsilon}(N^{\prime}),\\{\tilde{\varepsilon}(N),\tilde{\varepsilon}(N^{\prime})\\}^{\perp_{\tilde{\varepsilon}}}\rangle=\mathrm{PG}(\widetilde{V})$.
However, $\tilde{\varepsilon}(c)$ is the unique singular point of
$\\{\tilde{\varepsilon}(N),\tilde{\varepsilon}(N^{\prime})\\}^{\perp_{\tilde{\varepsilon}}}$.
Then
$\\{\tilde{\varepsilon}(N),\tilde{\varepsilon}(N^{\prime})\\}^{\perp_{\tilde{\varepsilon}}}\subseteq\tilde{\varepsilon}(c)^{\perp_{\tilde{\varepsilon}}}$.
This forces
$\tilde{\varepsilon}(c)^{\perp_{\tilde{\varepsilon}}}=PG(\widetilde{V})$,
hence $c^{\perp}={\cal S}$. We have reached a contradiction.

Therefore $\\{N,N^{\prime}\\}^{\perp}=\emptyset$. Hence
$\langle\tilde{\varepsilon}(N),\tilde{\varepsilon}(N^{\prime})\rangle=\mathrm{PG}(\widetilde{V})$,
since $\\{N,N^{\prime}\\}$ is optimally embedded by assumption. Consequently,
if $M$ is a generator of ${\cal S}$ contaning $N$, then
$\tilde{\varepsilon}(M)$ meets $\tilde{\varepsilon}(N^{\prime})$ non-
trivially. Equivalently $M$ meets $N^{\prime}$ non-trivially. This contradicts
the assumption that $N$ and $N^{\prime}$ are opposite. $\Box$

###### Lemma 3.13

Let $\\{N,N^{\prime}\\}$ be an optimally embeddable pair of opposite sub-
generators of ${\cal S}$ and suppose that $\\{N,N^{\prime}\\}$ is
$\perp$-minimal. Then both $N$ and $N^{\prime}$ are $\perp$-minimal.

Proof. By Lemma 3.12, $\\{N,N^{\prime}\\}^{\perp}$ contains at least two
distinct (hence opposite) points $a$ and $b$. So, $N$ and $N^{\prime}$ are
opposite generators of $\\{a,b\\}^{\perp}$ and
$\\{N,N^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$ since $\\{N,N^{\prime}\\}$
is $\perp$-minimal.

By way of contradiction, suppose that $N$ is not $\perp$-minimal. So,
$N^{\perp}$ contains a point $c\not\in N\cup\\{a,b\\}^{\perp\perp}$. However
$\\{N,N^{\prime}\\}^{\perp}=\\{a,b\\}^{\perp\perp}$ by assumption. Moreover
$\\{\varepsilon(a),\varepsilon(b)\\}^{\perp_{\varepsilon}\perp_{\varepsilon}}=\langle\varepsilon(a),\varepsilon(b)\rangle$
since $\varepsilon$ is minimal and
$\mathrm{PG}(V)=\langle\varepsilon(N),\varepsilon(N^{\prime}),\varepsilon(a),\varepsilon(b)\rangle$,
since $\langle N,N^{\prime}\rangle$ is optimally embeddable. It follows that
the projective plane
$P=\langle\varepsilon(a),\varepsilon(b),\varepsilon(c)\rangle$ meets
$\langle\varepsilon(N),\varepsilon(N^{\prime})\rangle$ in a point, say $p$.
However $P\subseteq\varepsilon(N)^{\perp_{\varepsilon}}$. Hence $p\in
X:=\varepsilon(N)^{\perp_{\varepsilon}}\cap\langle\varepsilon(N),\varepsilon(N^{\prime})\rangle$.
However $p\not\in\varepsilon(N)$. Therefore $X$ properly contains
$\varepsilon(N)$. Consequently, $X\cap\varepsilon(N^{\prime})\neq\emptyset$.
So, $\varepsilon(N^{\prime})$ meets $X$ non-trivially. Equivalently,
$N^{\prime}$ meets $N^{\perp}$ non-trivially. This contradicts the hypothesis
that $N$ and $N^{\prime}$ are opposite. $\Box$

In the finite rank case the next lemma is absolutely trivial and its
hypotheses are redundant. However $\mathrm{rank}(S)=\infty$ is allowed here.
We recall that, so far, nobody has been able to prove that in every polar
space of infinite rank every generator admits an opposite, although no
counterexample is known which refutes this conjecture. So, the hypothesis that
$M$ admits an opposite cannot be dropped from the next lemma and a proof is
required for the conclusions.

###### Lemma 3.14

Let $M$ be a generator of ${\cal S}$ and $p$ a point exterior to $M$. If $M$
is opposite to at least one generators of ${\cal S}$ then there exists a
generator of ${\cal S}$ opposite to $M$ and containing $p$. If moreover $M$
admits a complement, then $M$ also admits a complement which contains $p$.

Proof. We firstly prove that if $M$ admits an opposite generator then a
generator $M^{\prime}$ also exists which is opposite to $M$ and contains $p$.

Let $M_{1}$ be a generator of ${\cal S}$ opposite to $M$. If $M_{1}$ contains
$p$ then there is nothing to prove. Otherwise, let $M_{2}:=\langle
p^{\perp}\cap M_{1},p\rangle$. If $M_{2}\cap M=\emptyset$ then again we are
done: $M^{\prime}=M_{2}$ does the job. Assume that $M_{2}\cap M\neq\emptyset$.
As $M_{2}$ contains a hyperplane $p^{\perp}\cap M_{1}$ which is disjoint from
$M$, necessarily $M_{2}\cap M=\\{a\\}$ for a point $a$. Clearly, $a\neq p$.
Put $N:=M\cap p^{\perp}$. As both $p$ and $a$ belong to $N^{\perp}$, the line
$\langle p,a\rangle$ is contained in $N^{\perp}$. Therefore this line meets
$N$ in a point, since $N$ is a sub-generator. It follows that $a\in N$. Now,
$p^{\perp}\not\subseteq a^{\perp}$ (indeed ${\cal S}$ is non-degenerate and
$p^{\perp}\neq a^{\perp}$ since $p\neq a$). Hence there exists a point $b\in
p^{\perp}\setminus a^{\perp}$. Put $M_{3}:=\langle b,b^{\perp}\cap
M_{2}\rangle$. The generator $M_{3}$ contains $p$.

We claim that $M_{3}\cap M=\emptyset$. Indeed, suppose the contrary and let
$c\in M_{3}\cap M$. Then $c\neq b$, as $b\not\perp a$ while $c\perp a$. The
line $\langle b,c\rangle$ meets $b^{\perp}\cap M_{2}$ in a point $d$. If
$c\neq d$ then $b\in\langle c,d\rangle$. However, $\langle c,d\rangle\subseteq
a^{\perp}$, since $c\perp a$ (because $c,a\in M$) and $d\perp a$ (because
$d,a\in M_{2})$. It follows that $b\perp a$. This conclusion contradicts the
choice of $b$. Therefore $d=c$. However $d\in b^{\perp}\cap M_{2}$, which is
disjont from $M$ since $b\not\perp a$ and $a$ is the unique point of
$M_{2}\cap M$, while $c\in M$. Again a contradiction. Hence $M_{3}\cap
M=\emptyset$ and $M^{\prime}=M_{3}$ is the required generator opposite to $M$
and containing $p$.

Turning to the second claim of the lemma, suppose that $M_{1}$ is a complement
of $M$. If $p\in M_{1}$ there is nothing to prove. Otherwise, let $M_{2}$ be
constructed as above. If $M_{2}\cap M=\emptyset$ then $M_{2}$ is the required
complement. Otherwise $\varepsilon(M)\cup\varepsilon(M_{2})$ spans a
hyperplane of $\mathrm{PG}(V)$. In this case the generator $M_{3}$ constructed
as above is a complement of $M$. $\Box$

###### Theorem 3.15

The following are equivalent:

  * $(1)$

${\cal S}$ is regular and, for every sub-generator $N$ of ${\cal S}$, if
$N^{\perp}$ is not a generator of ${\cal S}$ then a sub-generator $N^{\prime}$
of ${\cal S}$ exists opposite to $N$ and such that $\\{N,N^{\prime}\\}$ is
optimally embeddable.

  * $(2)$

Every generator of ${\cal S}$ admits a complement.

Proof. Assume (1). Let $M$ be a generator of ${\cal S}$ and, given a point
$a\in M$, let $b$ be a point opposite to $a$. Put $N=b^{\perp}\cap M$. Then
$a,b\in N^{\perp}$, hence $N^{\perp}$ is not a generator. By (1), there exists
a sub-generator $N^{\prime}$ such that $\\{N,N^{\prime}\\}$ is optimally
embeddable, hence it is not $\perp$-degenerate by Lemma 3.12. However, ${\cal
S}$ is regular by (1). Hence $N^{\perp}=\langle
N,\\{a,b\\}^{\perp\perp}\rangle$ and, since $\\{N,N^{\prime}\\}$ is not
$\perp$-degenerate, $N^{\prime}\subset\\{c,d\\}^{\perp}$ for two opposite
points $c,d\in N^{\perp}$. Moreover, $\\{N,N^{\prime}\\}$ is $\perp$-minimal,
namely $\\{N,N^{\prime}\\}^{\perp}=\\{c,d\\}^{\perp\perp}$, because ${\cal S}$
is regular. The hyperbolic line $\\{c,d\\}^{\perp\perp}$ meets every generator
$\langle N,x\rangle$ with $x\in\\{a,b\\}^{\perp\perp}$. So, we can assume that
$c\in\langle N,a\rangle$ and $d\in\langle N,b\rangle$. Accordingly, we can
safely replace $a$ with $c$ and $b$ with $d$. In other words, we can assume
that $a=c$ and $b=d$. Put $M^{\prime}:=\langle N^{\prime},b\rangle$. Then $M$
and $M^{\prime}$ are opposite. Recalling that $\\{N,N^{\prime}\\}$ is perp-
minimal and optimally embeddable, we obtain that
$\varepsilon(M)\cup\varepsilon(M^{\prime})$ spans $\mathrm{PG}(V)$ as in the
proof of Theorem 3.8. So, $M^{\prime}$ is a complement of $M$.

Conversely, assume (2). We firstly prove that every sub-generator of ${\cal
S}$ is $\perp$-minimal. Let $N$ be a sub-generator of ${\cal S}$. If
$N^{\perp}$ is a generator then $N$ is $\perp$-minimal (Definition 3.7).
Suppose that $N$ contains at least two distinct points, say $a$ and $b$.
Clearly, $a\not\perp b$. Put $M:=\langle N,a\rangle$. In view of (2), $M$
admits a complement. By Lemma 3.14, there exists a complement $M^{\prime}$ of
$M$ which contains $b$. Put $N^{\prime}:=a^{\perp}\cap M^{\prime}$. As in the
proof of Theorem 3.8, we see that $\\{N,N^{\prime}\\}$ is $\perp$-minimal and
optimally embeddable. By Lemma 3.13, the sub-generator $N$ is $\perp$-minimal.
So we have proved that every sub-generator of ${\cal S}$ is $\perp$-minimal.
Namely, ${\cal S}$ is regular. Morover, $\\{N,N^{\prime}\\}$ is optimally
embeddable, as required in (1). $\Box$

All polar spaces constructed in Sections 5.1, 5.2 and 5.3 are regular and, in
view of Proposition 5.3 and its analogues for the hyperbolic and hermitian
case, they satisfy condition (2) (hence (1) too) of Theorem 3.15.

###### Remark 7

When $\mathrm{rank}({\cal S})=\infty$, the existence of a pair of
complementary generators (namely the minimal embedding being tight) is not
sufficient for ${\cal S}$ to be regular and, conversely, ${\cal S}$ being
regular does not imply that all opposite generators are complementary. For
instance, the quadric ${\cal S}_{q^{\prime}}$ described in Section 5.4 is not
regular but it admits a pair of complementary generators, namely $[V]$ and
$[V^{\prime}]$. Conversely, the polar spaces described in Sections 5.1, 5.2
and 5.3 are regular but each of them admits pairs of opposite generators which
are not complementary. For instance, in each of these spaces, the generators
called $[V]$ and $[V^{*}]$ are complementary. Generators also exist which are
opposite to both $[V]$ and $[V^{*}]$. Each of them is a complement of
$[V^{*}]$ but not of $[V]$. (See also Lemma 4.13.)

##  4 The three generators game

Throughout this section ${\cal S}$ is assumed to be embeddable. For ease of
exposition, when $\mathrm{rank}({\cal S})=2$ we assume that ${\cal S}$ is
neither a grid of order $s>3$ nor a generalized quadrangle as in [17,
8.6(II)(a)]. So, ${\cal S}$ admits both the absolutely universal embedding and
a unique minimum embedding.

We shall deal with triples of mutually opposite generators. Recall that a
polar space ${\cal S}$ of finite rank admits triples of opposite generators if
and only if either ${\cal S}$ is thick or $\mathrm{rank}({\cal S})$ is even.
We cannot hope for such a sharp picture for polar spaces of infinite rank, all
the more that we do not even know if every generator of a polar space of
infinite rank admits an opposite. However polar spaces of infinite rank exist
which admit triples of mutually opposite generators. For instance, in each of
the spaces described in Sections 5.1, 5.2 and 5.3, the generators $[V]$ and
$[V^{*}]$ are opposite and generators exist which are opposite to both $[V]$
and $[V^{*}]$. So, the theory we are going to develope in this section is not
vacuous in the infinite rank case.

###  4.1 More on opposite generators

Let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be the minimum embedding
of ${\cal S}$. Given two opposite generators $M_{1}$ and $M_{2}$ of ${\cal
S}$, put ${\cal
S}_{M_{1},M_{2}}:=\varepsilon^{-1}(\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$.
Clearly, ${\cal S}_{M_{1},M_{2}}\supseteq\langle M_{1},M_{2}\rangle$. We have
${\cal S}_{M_{1},M_{2}}=\langle M_{1},M_{2}\rangle$ if and only if
$\varepsilon$ is the unique embedding of ${\cal S}$.

###### Lemma 4.1

The following holds for $\\{i,j\\}=\\{1,2\\}$: all hyperplanes of $M_{i}$ are
$\perp$-minimal as sub-generators of ${\cal S}_{M_{1},M_{2}}$ and, if $N$ is a
hyperplane of $M_{i}$ such that $N^{\perp}\cap{\cal S}_{M_{1},M_{2}}\neq
M_{i}$, then $N^{\perp}\cap M_{j}=\\{x_{j}\\}$ for a point $x_{j}\in M_{j}$
and $N^{\perp}\cap{\cal S}_{M_{1},M_{2}}=\langle M_{i},x_{j}\rangle=\langle
N,x_{i},x_{j}\rangle$ for any $x_{i}\in M_{i}\setminus N$.

Proof. Put ${\cal S}^{\prime}:={\cal S}_{M_{1},M_{2}}$ for short and, denoted
by $V^{\prime}$ the subspace of $V$ corresponding to
$\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$, let
$\varepsilon^{\prime}:{\cal
S}^{\prime}\rightarrow\mathrm{PG}(V^{\prime})=\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$
be the embedding of ${\cal S}^{\prime}$ induced by $\varepsilon$. The
embedding $\varepsilon^{\prime}$ is minimal and $\pi_{\varepsilon}$ induces
$\pi_{\varepsilon^{\prime}}$ on $\mathrm{PG}(V^{\prime})$. Explicitly,
$X^{\perp_{\varepsilon^{\prime}}}=X^{\perp_{\varepsilon}}\cap\mathrm{PG}(V^{\prime})$
for every subset $X$ of $\mathrm{PG}(V^{\prime})$.

Let $N$ be a hyperplane of $M_{1}$ such that $M_{1}\subset N^{\perp}\cap{\cal
S}^{\prime}$. Then $\varepsilon(N)^{\perp_{\varepsilon^{\prime}}}$ properly
contains $\varepsilon(M_{1})$. Hence it meets $\varepsilon(M_{2})$ non-
trivially. The intersection
$\varepsilon(N)^{\perp_{\varepsilon^{\prime}}}\cap\varepsilon(M_{2})$ cannot
contain a line, otherwise $N^{\perp}\cap M_{2}$ contains a line and $\langle
N,N^{\perp}\cap M_{2}\rangle$ is a singular subspace containg $N$ as a
subspace of codimension at least $2$, contradcting the fact that $N$ is a sub-
generator. Therefore
$\varepsilon(N)^{\perp_{\varepsilon^{\prime}}}\cap\varepsilon(M_{2})=\\{\varepsilon(x_{2})\\}$
for a point $x_{2}\in M_{2}$ and $\varepsilon(N)$ has codimension 2 in
$\varepsilon(N)^{\perp_{\varepsilon^{\prime}}}$. This implies that
$\varepsilon(N)^{\perp_{\varepsilon^{\prime}}}=\langle\varepsilon(N),\varepsilon(x_{1}),\varepsilon(x_{2})\rangle=\langle\varepsilon(N),\\{\varepsilon(x_{1}),\varepsilon(x_{2})\\}^{\perp_{\varepsilon^{\prime}}\perp_{\varepsilon^{\prime}}}\rangle$
for $x_{1}\in M_{1}\setminus N$. Accordingly, $N^{\perp}=\langle
N,x_{1},x_{2}\rangle=\langle N,(\\{x_{1},x_{2}\\}^{\perp}\cap{\cal
S}^{\prime})^{\perp}\cap{\cal S}^{\prime}\rangle$. $\Box$

###### Corollary 4.2

The generators $M_{1}$ and $M_{2}$ are complementary if and only if the
following holds for $\\{i,j\\}=\\{1,2\\}$:

  * $(\ast)$

all hyerplanes of $M_{i}$ are $\perp$-minimal as sub-generators of ${\cal S}$
and, if $N$ is a hyperplane of $M_{i}$ such that $N^{\perp}\neq M_{i}$, then
$N^{\perp}\cap M_{j}\neq\emptyset$ and $N^{\perp}=\langle M_{i},N^{\perp}\cap
M_{j}\rangle$.

Proof. The ‘only if’ part is Lemma 4.1 (see also the final part of the proof
of that lemma). Turning to the ‘if’ part, assume $(\ast)$ and let $z$ be any
point of ${\cal S}\setminus M_{1}$. Put $N:=z^{\perp}\cap M_{1}$. Then
$N^{\perp}\supset M_{1}$, since $z\not\in M_{1}$. Hypothesis $(\ast)$ now
implies that $N^{\perp}\cap M=\\{y\\}$ for a point $y\in M_{2}$ and
$N^{\perp}=\langle N,\\{x,y\\}^{\perp\perp}\rangle$ for $x\in M_{1}\setminus
N$. In the first case $z\in M_{2}$. In the second case, $x\in\langle
N,z^{\prime}\rangle$ for some $z^{\prime}\in\\{x,y\\}^{\perp\perp}$. However
$\varepsilon(\\{x,y\\}^{\perp\perp})\subseteq\langle\varepsilon(x),\varepsilon(y)\rangle$
because $\varepsilon$ is minimal and
$\langle\varepsilon(x),\varepsilon(y)\rangle\subseteq\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$.
Hence $z\in\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$ for every
point $z\in{\cal S}$. Therefore
$\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle=\mathrm{PG}(V)$. $\Box$

###### Remark 8

In the proof of the ‘if’ part of Corollary 4.2 we have exploited only the
hypothesis that $(\ast)$ holds for $(i,j)=(1,2)$. This is enough to obtain
that $M_{1}$ and $M_{2}$ are complementary. Hence if $(\ast)$ holds for
$(i,j)=(1,2)$ then it also holds for $(i,j)=(2,1)$.

###  4.2 Deep and hyperbolic sub-generators

Let $X$ be a sub-sub-generator of ${\cal S}$, namely a singular subspace of
${\cal S}$ of codimension 2 in at least one (hence all) of the generators of
${\cal S}$ which contain it. The star $\mathrm{St}(X)$ of $X$ is either a line
(if $X$ is contained in just one generator) or a pencil of lines (if just one
of the sub-generators containing $X$ is contained in at least two generators)
or a non-degenerate generalized quadrangle (if none of the sub-generators
containing $X$ is contained in just one generator).

###### Lemma 4.3

If a generator $M$ of ${\cal S}$ contains a sub-generator which is contained
in just two generators, then all sub-generators contained in $M$ are contained
in at most two generators.

Proof. By way of contradiction, suppose that $M$ contains two sub-generators
$N$ and $N^{\prime}$ with $N$ contained in just two generators and
$N^{\prime}$ in at least three generators. Put $X=N\cap N^{\prime}$. Then
$\mathrm{St}(X)$ is a non-degenerate thick-lined generalized quadrangle.
However, the point of $\mathrm{St}(X)$ corresponding to $N$ belongs to just
two lines of $\mathrm{St}(X)$ while the point corresponding to $N^{\prime}$
belongs to at least three lines. This is impossible in any thick-lined
generalized quadrangle. $\Box$

###### Definition 4.4

Let $N$ be a sub-generator of ${\cal S}$. We say that $N$ is deep if
$N^{\perp}$ is a generator, namely $N$ is contained in just one generator; in
other words, $\\{N,N\\}$ is $\perp$-degenerate. If $N$ is contained in exactly
two generators then we say that $N$ is hyperbolic. So, by Lemma 4.3, if a
generator contains a hyperbolic sub-generator then it is hyperbolic
(Definition 1.9). If all non-deep sub-generators of ${\cal S}$ are hyperbolic
(equivalently, all of its generators are hyperbolic) then we say that ${\cal
S}$ is hyperbolic.

Obviously, hyperbolic sub-generators are $\perp$-minimal. If ${\cal S}$ admits
a hyperbolic sub-generator, then at least one of the hyperbolic lines of
${\cal S}$ has size 2. Hence all of them have size 2, since
$\mathrm{Aut}({\cal S})$ is transitive on the set of pairs of opposite points
of ${\cal S}$. So, if ${\cal S}$ is hyperbolic then it is regular.

If $M$ is a generator and $p$ a point exterior to $M$ then $p^{\perp}\cap M$
is a non-deep sub-generator. So, every generator contains non-deep sub-
generators. Accordingly, a generator is non-hyperbolic if and only if all of
its non-deep hyperplanes are non-hyperbolic.

We recall that if $\mathrm{rank}({\cal S})<\infty$ then ${\cal S}$ admits no
deep sub-generators. In this case either ${\cal S}$ is hyperbolic or none of
its sub-generators is hyperbolic (hence ${\cal S}$ is thick). In contrast, in
general, when $\mathrm{rank}({\cal S})=\infty$ deep sub-generators exist in
${\cal S}$. So, in general, a generator of ${\cal S}$ contains both deep and
non-deep sub-generators. Moreover, polar spaces of infinite rank also exist
where some but not all of the non-deep sub-generators are hyperbolic. The
quadrics discussed in Section 5.4 have this property.

###### Remark 9

When $\mathrm{rank}({\cal S})=\infty$ the stabilizer in $\mathrm{Aut}({\cal
S})$ of a generator $M$ of ${\cal S}$ acts intransitively on the set of
hyperplanes of $M$. Indeed $M$ is an infinite-dimensional projective space and
the automorphism group of an infinite-dimensional projective space is never
transitive on the set of hyperplanes of that space. In view of this fact, the
existence of generators containing both deep and non-deep hyperplanes is not
so surprising.

###  4.3 An existence result for triples of opposite generators

The next theorem is an anlogue of Lemma 3.14, with triples of opposite
generators instead of pairs.

###### Theorem 4.5

Let $M$ and $M^{\prime}$ be opposite generators of ${\cal S}$ such that a
generator of ${\cal S}$ also exists which is opposite to both $M$ and
$M^{\prime}$. Suppose moreover that neither $M$ nor $M^{\prime}$ are
hyperbolic. Then for every point $p\not\in M\cup M^{\prime}$ there exists a
generator of ${\cal S}$ opposite to both $M$ and $M^{\prime}$ and containing
$p$.

Proof. Let $M_{1}$ be a generator opposite to both $M$ and $M^{\prime}$. If
$p\in M_{1}$ there is nothig to prove. Otherwise, put $M_{2}:=\langle
p^{\perp}\cap M_{1},p\rangle$. If $M_{2}$ is opposite to both $M$ and
$M^{\prime}$ then we are done. Suppose firstly that $M_{2}$ is opposite to
neither $M$ nor $M^{\prime}$. Then $M\cap M_{2}=\\{a\\}$ and $M^{\prime}\cap
M_{2}=a^{\prime}$ for distinct points $a,a^{\prime}\in
M_{2}\setminus(p^{\perp}\cap M_{1})$. Pick a point $b\in
p^{\perp}\setminus(a^{\perp}\cup a^{\prime\perp})$ and put $M_{3}:=\langle
b^{\perp}\cap M_{2},b\rangle$. Then $M_{3}$ is a generator and, as in the
proof of Lemma 3.14, we can see that $M_{3}$ is opposite to both $M$ and
$M^{\prime}$.

Assume now that $M_{2}$ is opposite to just one of $M$ and $M^{\prime}$, say
$M_{2}\cap M=\emptyset$ but $M_{2}\cap M^{\prime}:=\\{a\\}$ for a point $a\in
M_{2}\setminus(p^{\perp}\cap M_{1})$. Put $X:=M\cap\\{p,a\\}^{\perp}$. Then
$X$ has codimension 2 in $M$, $\langle X,p,a\rangle$ is a generator and
$M\cap\langle X,p,a\rangle=X$. The star $\mathrm{St}(X)$ of $X$ in ${\cal S}$
is a non-degenerate polar space of rank 2 and $M$ and $\langle X,p,a\rangle$
are opposite lines in it. The star $\mathrm{St}(X)$ is not a grid, since $M$
is non-hyperbolic by assumption. Hence there exists a point $b\in X^{\perp}$
such that $b\perp p$ but $b\not\perp a$ and $\langle X,p,b\rangle\cap M=X$,
namely $\\{p,b\\}^{\perp}\cap M=X$.

Put $M_{3}:=\langle b^{\perp}\cap M_{2},b\rangle$. Then $M_{3}$ is a
generator, it contains $p$ and, as in the proof of Lemma 3.14, we can see that
it is opposite to $M^{\prime}$. By way of contradiction, suppose that $M\cap
M_{3}\neq\emptyset$. So $M\cap M_{3}=\\{c\\}$ for a point $c\in
M_{3}\setminus(b^{\perp}\cap M_{2})$. So, $c\in\\{p,b\\}^{\perp}$. However,
$\\{p,b\\}^{\perp}\cap M=X$. Therefore $c\in X$. Clearly, $c\neq b$. Hence
$\langle b,c\rangle$ is a line and meets $b^{\perp}\cap M_{2}$ in a point, say
$d$. We have $c\neq d$ because $d\in M_{2}$ and $M_{2}\cap M=\emptyset$.
Therefore $b\in\langle c,d\rangle$. However $d\perp a$ because both $d$ and
$a$ belong to $M_{2}$ while $c\perp a$ because $c\in X\subset M\cap
a^{\perp}$. It follows that $\langle c,d\rangle\subseteq a^{\perp}$. Therefore
$b\in a^{\perp}$. This contradicts the choice of $b$. $\Box$

###### Remark 10

When $\mathrm{rank}({\cal S})<\infty$ the hypothesis that $M$ and $M^{\prime}$
are non-hyperbolic can be dropped from Theorem 4.5. Indeed in this case either
${\cal S}$ admits no triples of opposite generators (when ${\cal S}$ is
hyperbolic of odd rank) or for any two opposite generators $M$ and
$M^{\prime}$ of ${\cal S}$ and every point $p\not\in M\cup M^{\prime}$ there
exists a generator $M^{\prime\prime}$ opposite to both $M$ and $M^{\prime}$
and containing $p$. So, either the main hypothesis of Theorem 4.5 fails to
hold (hence the statement of Theorem 4.5 is logically valid) or the conclusion
of Theorem 4.5 holds true.

###  4.4 Partial dualities defined by pairs of opposite generators

Let $M_{1}$ and $M_{2}$ be opposite generators of ${\cal S}$. The proof of the
next lemma is easy. We leave it to the reader.

###### Lemma 4.6

Let $X$ be a subspace of $M_{1}$ of codimension $2$ in $M_{1}$. Let
$M_{1}^{*}(X)$ be the set of hyperplanes of $M_{1}$ containing $X$ (a line of
the dual $M_{1}^{*}$ of $M_{1}$). Then one of the following occurs:

  * $(1)$

We have $N^{\perp}\cap M_{2}\neq\emptyset$ for every $N\in M_{1}^{*}(X)$, the
subspace $X^{\perp}\cap M_{2}$ is a line of $M_{2}$ and the mapping $N\mapsto
N^{\perp}\cap M_{2}$ is a bijection between $M_{1}^{*}(X)$ and $X^{\perp}\cap
M_{2}$.

  * $(2)$

We have $N^{\perp}\cap M_{2}=\emptyset$ for all but at most one of the
hyperplanes $N\in M_{1}^{*}(N)$; moreover $X^{\perp}\cap M_{2}$ is either
empty of a single point, according to whether $N^{\perp}\cap M_{2}=\emptyset$
for either all or all but one of the hyperplanes $N\in M_{1}^{*}(X)$.

Let now $M$ be a generator of ${\cal S}$ opposite to both $M_{1}$ and $M_{2}$.
We define a partial duality $\pi_{1,2}$ on $M$ as follows: for $x\in M$ let
$X_{1}:=x^{\perp}\cap M_{1}$. If $X_{1}^{\perp}\cap M_{2}=\emptyset$ then we
put $\pi_{1,2}(x)=M$. Otherwise, $X_{1}^{\perp}\cap M_{1}=\\{x_{2}\\}$ for a
point $x_{2}\in M_{2}$; in this case we put $\pi_{1,2}(x)=x_{2}^{\perp}\cap
M$. It follows from Lemma 4.6 that $\pi_{1,2}$ is a partial duality of $M$.
The partial duality $\pi_{2,1}$ is defined in the same way as $\pi_{1,2}$ but
for permuting the roles of $M_{1}$ and $M_{2}$.

###### Lemma 4.7

Suppose that both $\pi_{1,2}$ and $\pi_{2,1}$ are non-degenerate. Then
$\pi_{1,2}$ is a polarity if and only if $\pi_{1,2}=\pi_{2,1}$ (if and only if
$\pi_{2,1}$ is a polarity).

Proof. Suppose that $\pi_{1,2}$ is a polarity. Then $x\in\pi_{1,2}(y)$ for
every $y\in\pi_{1,2}(x)$. We have $\pi_{2,1}(x)=x_{1}^{\perp}\cap M$ where
$x_{1}=X_{2}^{\perp}\cap M_{1}$ and $X_{2}=x^{\perp}\cap M_{2}$. Also,
$\pi_{1,2}(y)=y_{2}^{\perp}\cap M$ with $y_{2}=Y_{1}^{\perp}\cap M_{2}$ and
$Y_{1}=y^{\perp}\cap M_{1}$. However $x\in\pi_{1,2}^{M}(y)$. Therefore
$y_{2}\in X_{2}$. Accordingly, $x_{1}\in Y_{1}$ and therefore
$y\in\pi_{2,1}(x)$. Hence $\pi_{1,2}(x)\subseteq\pi_{2,1}(x)$. However both
$\pi_{1,2}(x)$ and $\pi_{2,1}(x)$ are hyperplanes of $M$. Consequently,
$\pi_{1,2}(x)=\pi_{2,1}(x)$. As this equality holds for every point $x\in M$,
we have $\pi_{1,2}=\pi_{2,1}$. By reversing the previous argument we can also
see that if $\pi_{1,2}=\pi_{2,1}$ then $\pi_{1,2}$ and $\pi_{2,1}$ are
polarites. $\Box$

###### Lemma 4.8

For $x\in M$, we have $x\in\pi_{1,2}(x)\neq M$ if and only if $x$ belongs to a
line of ${\cal S}$ which meets both $M_{1}$ and $M_{2}$ non-trivially.
Moreover, if $x\in\pi_{1,2}(x)\neq M$ then $\pi_{2,1}(x)=\pi_{1,2}(x)$.

Proof. Suppose $x\in\pi_{1,2}(x)\neq M$ and let $X_{1}=x^{\perp}\cap M_{1}$.
Then $X_{1}^{\perp}\cap M_{2}=\\{x_{2}\\}$ for a point $x_{2}\in M_{2}$
(because $\pi_{1,2}(x)\neq M$) and $x_{2}\perp x$ (because
$x\in\pi_{1,2}(x)$). It follows that $\langle X_{1},x,x_{2}\rangle$ is
singular. However $X_{1}$ is a sub-generator. Therefore the line
$\ell:=\langle x,x_{2}\rangle$ meets $X_{1}$ in a point. So, $\ell$ meets both
$M_{1}$ and $M_{2}$ non-trivially. Conversely, if there exists a line $\ell$
of ${\cal S}$ through $x$ which meets both $M_{1}$ and $M_{2}$ non trivially,
then $x\in\pi_{1,2}(x)=\pi_{2,1}(x)=\ell^{\perp}\cap M$. $\Box$

As in Section 4.1, let $\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ be the
minimum embedding of ${\cal S}$ and put ${\cal
S}_{M_{1},M_{2}}=\varepsilon^{-1}(\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$.

###### Lemma 4.9

Suppose that $M\subseteq{\cal S}_{M_{1},M_{2}}$. Then $\pi_{1,2}=\pi_{2,1}$
and $\pi_{1,2}$ is a polarity.

Proof. Let $\varepsilon(M)\subseteq{\cal S}_{M_{1},M_{2}}$. Lemma 4.1 implies
that neither $\pi_{1,2}$ nor $\pi_{2,1}$ is degenerate. We shall prove that
$\pi_{1,2}=\pi_{2,1}$. By Lemma 4.7 this will be enough to obtain also that
$\pi_{1,2}$ is a polarity. As in the proof of Lemma 4.1, we put ${\cal
S}^{\prime}:={\cal S}_{M_{1},M_{2}}$. Moreover, for a subset $X\subseteq{\cal
S}^{\prime}$, we put $X^{\perp^{\prime}}:=X^{\perp}\cap{\cal S}^{\prime}$.

Let $x\in M$, $X_{1}=x^{\perp}\cap M_{1}$ and $x_{2}\in M_{2}$ be such that
$X_{1}^{\perp}\cap M_{2}=\\{x_{2}\\}$. If $x_{2}\perp x$ then
$x\in\pi_{1,2}(x)$. In this case the equality $\pi_{1,2}(x)=\pi_{2,1}(x)$
follows from Lemma 4.8. Suppose that $x_{2}\not\perp x$. Then by (the proof
of) Lemma 4.1 we have that $X_{1}^{\perp^{\prime}}=\langle X_{1},h\rangle$
with $h=\\{x,x_{2}\\}^{\perp^{\prime}\perp^{\prime}}$ (a hyperbolic line of
${\cal S}^{\prime}$). Accordingly, and since $M_{1}\subseteq X_{1}^{\perp}$,
the hyperbolic line $h$ meets $M_{1}$ in a point, say $x_{1}$. The sub-
generator $X_{2}:=x^{\perp}\cap M_{2}$ is contained in $h^{\perp^{\prime}}$.
Hence $X_{2}\perp x_{1}$, namely $X_{2}^{\perp}\cap M_{1}=\\{x_{1}\\}$.
However $\pi_{1,2}(x)=x_{2}^{\perp}\cap M=h^{\perp}\cap M=x_{1}^{\perp}\cap
M=\pi_{2,1}(x)$. Hence $\pi_{1,2}(x)=\pi_{2,1}(x)$. $\Box$

###### Lemma 4.10

Suppose that $\pi_{1,2}$ is a polarity and admits at least one absolute point.
Suppose morever that ${\cal S}$ is defined over a division ring of
characteristic different from $2$. Then $M\subseteq{\cal S}_{M_{1},M_{2}}$.

Proof. Let $\pi_{1,2}$ be a polarity. Then, denoted by $V_{M}$ the inderlying
vector space of $M$ and by $\mathbb{K}$ the underlying division ring of ${\cal
S}$, the polarity $\pi_{1,2}$ is defined by a reflexive sesquilinear form
$f_{M}:V_{M}\times V_{M}\rightarrow\mathbb{K}$. The form $f_{M}$ is non-
degenerate, since polarities are non-degenerate reflexive partial dualities,
by defnition. The form $f_{M}$ is trace-valued because
$\mathrm{char}(\mathbb{K})\neq 2$ by assumption [17, §8.1.5]. Hence either $f$
admits no non-zero isotropic vector or $V_{M}$ is spanned by the isotropic
vectors of $f_{M}$. Therefore, if $\pi_{1,2}$ admits at least one absolute
point, then the absolute points of $\pi_{1,2}$ generate $M$. By Lemma 4.8, all
abosulte points of $\pi_{1,2}$ belong to ${\cal S}_{M_{1},M_{2}}$. Hence
$M\subseteq{\cal S}_{M_{1},M_{2}}$. $\Box$

###### Remark 11

When ${\cal S}$ is defined over a division ring of characteristic 2 it can
happen that the form $f_{M}$ considered in the proof of Lemma 4.10 is not
trace-valued, namely the absolute points of $\pi_{1,2}$ do not span $M$. If
this is the case then it might happen that $M\not\subseteq{\cal
S}_{M_{1},M_{2}}$.

###  4.5 Proof of Theorem 1.10

With $M$, $M_{1}$ and $M_{2}$ as in the previous subsection, in order to keep
a record of $M$ in our notation henceforth we write $\pi_{1,2}^{M}$ and
$\pi_{2,1}^{M}$ instead of $\pi_{1,2}$ and $\pi_{2,1}$. In the sequel we will
often refer to the three generators property by the acronym $(3\mathrm{G})$.

The next theorem is the same as Theorem 1.10.

###### Theorem 4.11

Suppose that ${\cal S}$ is defined over a division ring $\mathbb{K}$ of
characteristic $\mathrm{char}(\mathbb{K})\neq 2$. Let $M_{1}$ and $M_{2}$ be
opposite generators of ${\cal S}$ such that at least one generator of ${\cal
S}$ exists which is opposite both $M_{1}$ and $M_{2}$. Assume moreover that
neither $M_{1}$ nor $M_{2}$ is hyperbolic. Then $\\{M_{1},M_{2}\\}$ enjoys the
three generators property if and only if $\langle M_{1},M_{2}\rangle={\cal
S}$.

Proof. The ‘if’ part is Lemma 4.9. Conversely, suppose that
$\\{M_{1},M_{2}\\}$ enjoys $(3\mathrm{G})$. Choose a line $\ell$ of ${\cal S}$
meeting both $M_{1}$ and $M_{2}$ non-trivially and let $x$ be a point of
$\ell$ different from $\ell\cap M_{1}$ and $\ell\cap M_{2}$. By Theorem 4.5,
there exists a generator $M$ of ${\cal S}$ containing $x$ and opposite to both
$M_{1}$ and $M_{2}$. The point $x$ is absolute for $p_{1,2}^{M}$ by Lemma 4.8.
Hence $M\subseteq{\cal S}_{M_{1},M_{2}}$ by Lemma 4.10 and because
$p_{1,2}^{M}$ is a polarity by $(3\mathrm{G})$ on $\\{M_{1},M_{2}\\}$. Let now
$y$ be any point of $x^{\perp}\setminus{\cal S}_{M_{1},M_{2}}$ and put
$M^{\prime}:=y^{\perp}\cap M$. Then $M^{\prime}$ is opposite to both $M_{1}$
and $M_{2}$ and $\pi_{1,2}^{M^{\prime}}$ is a polarity by $(3\mathrm{G})$ on
$\\{M_{1}.M_{2}\\}$. Moreover $x\in M^{\prime}$ is absolute for
$p_{1,2}^{M^{\prime}}$ by Lemma 4.8. Lemma 4.10 forces
$M^{\prime}\subseteq{\cal S}_{M_{1},M_{2}}$. In particular $y\in{\cal
S}_{M_{1},M_{2}}$, a contradiction with the choice of $y$. It follows that
$x^{\perp}\subseteq{\cal S}_{M_{1},M_{2}}$.

So far, we have proved that $x^{\perp}\subseteq{\cal S}_{M_{1},M_{2}}$ for
every point $x\not\in M_{1}\cup M_{2}$ belonging to a line which meets both
$M_{1}$ and $M_{2}$ non trivially. Let $X$ be the set of points $x$ as above
and put $\overline{X}:=\cup_{x\in X}x^{\perp}$. In order to finish the proof
it is sufficient to prove that $\overline{X}\supseteq{\cal
S}\setminus(M_{1}\cup M_{2})$. Let $y$ be a point exterior to $M_{1}\cup
M_{2}$ and put $N_{1}=y^{\perp}\cap M_{1}$ and $N_{2}=y^{\perp}\cap M_{2}$. At
most one of the points of $M_{1}\setminus N_{1}$ is collinear with $N_{2}$.
Let $x_{1}\in M_{1}\setminus N_{1}$ be such that $x_{1}^{\perp}\cap M_{2}\neq
N_{2}$ and let $\ell$ be a line through $x_{1}$ meeting $M_{2}$ in a point
$x_{2}\not\in N_{2}$. Then $y^{\perp}$ meets $\ell$ in a point $x\not\in
M_{1}\cup M_{2}$. So $x\in X$ and therefore $y\in\overline{X}$. $\Box$

As said in Conjecture 1.11, we believe that the hypothesis that
$\mathrm{char}(\mathbb{K})\neq 2$ can be removed from Theorem 4.11, provided
that the conclusion is weakened as follows: the pair $\\{M_{1},M_{2}\\}$
satisfies the three-generator property if and only if $M_{1}$ and $M_{2}$ are
complementary. The next theorem offers a (faint) clue in favour of this
conjecture.

###### Theorem 4.12

Suppose that ${\cal S}$ is regular and let $M_{1}$ and $M_{2}$ be opposite
generators of ${\cal S}$ such that at least one generator of ${\cal S}$ exists
which is opposite to both $M_{1}$ and $M_{2}$. Suppose moreover that neither
$M_{1}$ nor $M_{2}$ is hyperbolic. Then $\\{M_{1},M_{2}\\}$ satisfies the
three-generator property if and only if $M_{1}$ and $M_{2}$ are complementary.

Proof. Lemma 4.9 provides the ‘if’ part of the theorem. Put ${\cal
S}^{\prime}:={\cal S}_{M_{1},M_{2}}$. By definition, $M_{1}$ and $M_{2}$ are
complementary if and only if ${\cal S}^{\prime}={\cal S}$. We shall prove that
if ${\cal S}^{\prime}\subset{\cal S}$ then $(3\mathrm{G})$ fails to hold for
$\\{M_{1},M_{2}\\}$, thus proving also the ‘only if’ part.

Suppose that ${\cal S}^{\prime}\subset{\cal S}$. Pick a point $a\in{\cal
S}\setminus{\cal S}^{\prime}$ and let $N:=a^{\perp}\cap M_{1}$. Suppose that
$N^{\perp}\cap M_{2}\neq\emptyset$, say $N^{\perp}\cap M_{2}=\\{b\\}$. As
${\cal S}$ is regular, $N$ is $\perp$-minimal. Hence $N^{\perp}=\langle
N,\\{a,b\\}^{\perp\perp}\rangle$ and $\\{a,b\\}^{\perp\perp}$ meets $M_{1}$ in
a point $c$. Recall now that the embedding $\varepsilon$ of ${\cal S}$ which
we use to define the subspace ${\cal
S}_{M_{1},M_{2}}=\varepsilon^{-1}(\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle_{V})$
is the minimum one. Therefore
$\langle\varepsilon(\\{a,b\\}^{\perp\perp})\rangle=\langle\varepsilon(a),\varepsilon(b)\rangle$.
It follows that the line $L:=\langle\varepsilon(a),\varepsilon(b)\rangle$
meets $\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$ in two distinct
points, namely $\varepsilon(b)$ and $\varepsilon(c)$. Hence
$L\subseteq\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$, namely
$\\{a,b\\}^{\perp\perp}\subseteq{\cal S}^{\prime}={\cal S}_{M_{1},M_{2}}$. So,
$a\in{\cal S}^{\prime}$. This contradicts the choice of $a$.

Therefore $N^{\perp}\cap M_{2}=\emptyset$. Consequently, if $M$ is a generator
of ${\cal S}$ containing $a$ and opposite both $M_{1}$ and $M_{2}$ (which
exists by Theorem 4.5), then $\pi_{1,2}^{M}(a)=M$. Property $(3\mathrm{G})$
fails to hold for $\\{M_{1},M_{2}\\}$. $\Box$

Theorems 4.11 and 4.12 show that, under suitable hypotheses on ${\cal S}$, the
three generators property boils down to forming a complementary pair. We shall
see now that, given three mutually opposite generators, if they have not the
same dimension (hence $\mathrm{rank}({\cal S})=\infty$) then at least one of
the three pairs we can form with them is not a complementary pair.

###### Lemma 4.13

Let $\\{M_{1},M_{2}\\}$ be a complementary pair of generators of ${\cal S}$
with $\dim(M_{1})\geq\dim(M_{2})$. Then $\dim(M_{3})\leq\dim(M_{2})$ for every
generator $M_{3}$ of ${\cal S}$ opposite to both $M_{1}$ and $M_{2}$. If
moreover $M_{3}$ is a complemet of $M_{1}$ then $\dim(M_{3})=\dim(M_{2})$.

Proof. Suppose that $\mathrm{rank}({\cal S})=\infty$ (otherwise there is
nothing to prove). So, the hupothesis that $M_{2}$ is a complement of $M_{1}$
combined with the hypothesis that $\dim(M_{1})\geq\dim(M_{2})$ imply that
$\dim(M_{1})=\dim(\varepsilon)=\mathrm{rank}({\cal S})$, with
$\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ the minimum embedding of
${\cal S}$. Hence $\dim(M_{3})\leq\dim(M_{1})$. Therefore, if
$\dim(M_{2})=\dim(M_{1})$ then $\dim(M_{3})\leq\dim(M_{2})$.

Let now $\dim(M_{2})<\dim(M_{1})$ and, for a contradiction, suppose that
$\dim(M_{3})>\dim(M_{2})$. Recall
$\mathrm{PG}(V)=\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle$ since
$M_{1}$ and $M_{2}$ are complementary. Let
$p^{M_{1}}_{M_{2}}:\mathrm{PG}(V)\rightarrow\varepsilon(M_{2})$ be the
projection of $\mathrm{PG}(V)$ onto $\varepsilon(M_{2})$ with
$\varepsilon(M_{1})$ as the kernel. As $\dim(M_{3})>\dim(M_{2})$,
$p^{M_{1}}_{M_{2}}$ induces a non-injective mapping on $\varepsilon(M_{3})$.
Hence there exist points $z\in M_{2}$ and $x,y\in M_{1}$ such that $x\neq y$
and both lines $\ell_{x}:=\langle\varepsilon(z),\varepsilon(x)\rangle$ and
$\ell_{y}:=\langle\varepsilon(z),\varepsilon(y)\rangle$ meet
$\varepsilon(M_{3})$ non-trivially, say
$\ell_{x}\cap\varepsilon(M_{3})=\\{p_{x}\\}$ and
$\ell_{y}\cap\varepsilon(M_{3})=p_{y}$. Clearly $p_{x}\neq p_{y}$, since
otherwise $p_{x}=p_{y}=\varepsilon(z)$, which is impossible since $M_{3}\cap
M_{2}=\emptyset$. Therefore $\ell:=\langle p_{x},p_{y}\rangle$ is a line and
$\ell$ meets $\langle\varepsilon(x),\varepsilon(y)\rangle$ in a point, say
$p$. This point belongs to both $\varepsilon(M_{3})$ and $\varepsilon(M_{1})$.
This conclusion contradict the hypothesis that $M_{3}$ is opposite to $M_{1}$.
Therefore $\dim(M_{3})\leq\dim(M_{2})$.

When $M_{3}$ is also a complement of $M_{1}$ we can permute $M_{2}$ and
$M_{3}$ in the previous paragraph, thus obtaining that
$\dim(M_{2})\leq\dim(M_{3})$. In this case $\dim(M_{3})=\dim(M_{2})$. $\Box$

###### Corollary 4.14

Let $M_{1},M_{2}$ and $M_{3}$ be three mutually opposite generators of ${\cal
S}$ and suppose that they have not the same dimension. Then at least one of
$\\{M_{1},M_{2}\\},\\{M_{1},M_{3}\\}$ or $\\{M_{2},M_{3}\\}$ is not a
complementary pair.

Proof. Since $M_{1},M_{2},M_{3}$ have not the same dimension, necessarily
$\mathrm{rank}({\cal S})=\infty$. To fix ideas, suppose that both $M_{2}$ and
$M_{3}$ are complements of $M_{1}$. Suppose firstly that
$\dim(M_{1})\geq\dim(M_{2})$. Then $\dim(M_{2})=\dim(M_{3})$ by Lemma 4.13. By
assumption, $M_{1},M_{2}$ and $M_{3}$ have not the same dimension. Hence
$\dim(M_{1})>\dim(M_{2})=\dim(M_{3})$. It follows that
$\dim(M_{1})=\dim(\varepsilon)=\mathrm{rank}({\cal S})$, where
$\varepsilon:{\cal S}\rightarrow\mathrm{PG}(V)$ is the minimum embedding of
${\cal S}$. Hence
$\dim\langle\varepsilon(M_{2}),\varepsilon(M_{3})\rangle=\dim(M_{2})=\dim(M_{3})<\dim(M_{1})=\dim(\varepsilon)$.
So,
$\langle\varepsilon(M_{1}),\varepsilon(M_{2})\rangle\subset\mathrm{PG}(V)$.
The generators $M_{3}$ is not a compement of $M_{2}$.

On the other hand, let $\dim(M_{1})$ be less than $\dim(M_{2})$ or
$\dim(M_{3})$, say $\dim(M_{1})<\dim(M_{2})$. Then
$\dim(\varepsilon)=\dim(M_{2})=\mathrm{max}(\dim(M_{1}),\dim(M_{3}))$, since
both pairs $\\{M_{1},M_{2}\\}$ and $\\{M_{1},M_{3}\\}$ are complementary. As
$\dim(M_{1})<\dim(M_{2})$, the equality
$\dim(M_{2})=\mathrm{max}(\dim(M_{1}),\dim(M_{3}))$ implies that
$\dim(M_{3})=\dim(M_{2})$. So, $\dim(M_{1})<\dim(M_{2})=\dim(M_{3})$. If
$M_{3}$ is a complement of $M_{2}$ then, by permuting $M_{1}$ and $M_{3}$ in
the statement of Lemma 4.13, we obtain that $\dim(M_{1})=\dim(M_{2})$, while
$\dim(M_{1})<\dim(M_{2})$ by assumption. Therefore $\\{M_{2},M_{3}\\}$ is not
a complementary pair. (Indeed now we would obtain that $M_{2}\cap
M_{3}\neq\emptyset$, thus contradicting the hypothesis that $M_{1},M_{2}$ and
$M_{3}$ are mutually opposite.) $\Box$

##  5 A few examples of infinite rank

Throughout this section $\mathbb{F}$ is a given field, $V$ is a infinite
dimensional $\mathbb{F}$-vector space, $V^{*}$ is the dual of $V$ and
$\overline{V}:=V\oplus V^{*}$. Recall that $\dim(V)<\dim(V^{*})$. So
$\dim(\overline{V})=\dim(V^{*})$.

We will adopt the following notation. We denote by $[v]$ the point of
$\mathrm{PG}(\overline{V})$ represented by a non-zero vector $v$ of
$\overline{V}$ and by $[W]$ the subspace of $\mathrm{PG}(\overline{V})$
corresponding to a subspace $W$ of $\overline{V}$.

The kernel of a linear functional $\xi\in V^{*}$ is denoted by $\ker(\xi)$
and, for a subspace $\Xi$ of $V^{*}$, we put
$\ker(\Xi)=\cap_{\xi\in\Xi}\ker(\xi)$. Given $x\in V$, we define
$\ker^{*}(x):=\\{\xi\in V^{*}~{}|~{}\xi(x)=0\\}$ (a subspace of $V^{*}$) and,
for a subspace $X$ of $V$, we put $\ker^{*}(X):=\cap_{x\in X}\ker^{*}(x)$.

Finally, $p_{V}$ and $p_{V^{*}}$ are the projections of $\overline{V}$ onto
$V$ and $V^{*}$ respectively with respectively $V^{*}$ and $V$ as the kernels
and, for a subspace $W$ of $\overline{V}$, $p_{V}^{W}$ and $p_{V^{*}}^{W}$ are
their restrictions to $W$.

All polar spaces to be considered in this section are full subgeometries of
$\mathrm{PG}(\overline{V})$. In order to avoid confusions between generation
in a polar space ${\cal S}\subseteq\mathrm{PG}(\overline{V})$ and spans in
$\mathrm{PG}(\overline{V})$, henceforth we renounce to use the symbol
$\langle.\rangle$ when referring to spans in $\mathrm{PG}(\overline{V})$,
keeping it for generation in ${\cal S}$ and using phrases as “the span of … in
$\mathrm{PG}(\overline{V})$” or switching to spans in $\overline{V}$ when
dealing with spans in $\mathrm{PG}(\overline{V})$. However we keep the symbol
$\langle.\rangle$ for spans in $\overline{V}$. No ambiguity arises from doing
so, since the vectors of $\overline{V}$ are not points of
$\mathrm{PG}(\overline{V})$, let alone points of ${\cal S}$. Thus, if
$X\subseteq\overline{V}$ then $\langle X\rangle$ is the span of $X$ on
$\overline{V}$, $[\langle X\rangle]$ is the span of $[X]:=\\{[x]\\}_{x\in
X\setminus\\{0\\}}$ in $\mathrm{PG}(\overline{V})$ and, if $[X]\subseteq{\cal
S}$, then $\langle[X]\rangle$ is the subspace of ${\cal S}$ generated by
$[X]$.

Turning to perps, let $f$ be the bilinear or hermitian form which defines the
quasi-polarity of $\mathrm{PG}(\overline{V})$ associated to ${\cal S}$ (or the
quasi-polarity of $\mathrm{PG}(\overline{V}^{\prime})$ associated to ${\cal
S}$ if ${\cal S}$ lives in $\mathrm{PG}(\overline{V}^{\prime})$ for a subspace
$\overline{V}^{\prime}$ of $\overline{V}$). We use the symbol $\perp_{f}$ for
orthogonality with respect to $f$ both in $\mathrm{PG}(\overline{V})$ and
$\overline{V}$ (respectively, in $\mathrm{PG}(\overline{V}^{\prime})$ and
$\overline{V}^{\prime}$), keeping the symbol $\perp$ with no subscript for
collinearity in ${\cal S}$.

###  5.1 A symplectic polar space

As in [9], we define a (non-degenerate) alternating form
$f:\overline{V}\times\overline{V}\rightarrow\mathbb{F}$ as follows:

$f(a\oplus\alpha,b\oplus\beta)~{}=~{}\alpha(a)-\beta(b),\hskip
14.22636pt\forall a,b\in V,~{}\alpha,\beta\in V^{*}.$ (14)

Let ${\cal S}_{f}$ be the symplectic polar space defined by $f$ in
$\mathrm{PG}(\overline{V})$. Of course, the inclusion maping of ${\cal S}_{f}$
into $\mathrm{PG}(\overline{V})$ is the minimum embedding of ${\cal S}_{f}$
(the unique embedding of ${\cal S}_{f}$ when $\mathrm{char}(\mathbb{F})\neq
2$.). This embedding is tight. Indeed the subspaces $V=V\oplus\\{0\\}$ and
$V^{*}=\\{0\\}\oplus V^{*}$ of $\overline{V}$ yield opposite generators $[V]$
and $[V^{*}]$ of ${\cal S}_{f}$ and $V+V^{*}=\overline{V}$ (but if
$\mathrm{char}(\mathbb{F})=2$ then $\langle[V],[V^{*}]\rangle\subset{\cal
S}_{f}$). Clearly, $\mathrm{rank}({\cal S}_{f})=\dim(V^{*})$.

As proved in [3], all symplectic polar spaces are regular. Hence ${\cal
S}_{f}$ is regular. So, if $N$ is a non-deep sub-generator of ${\cal S}_{f}$
then $N^{\perp}$ contains a hyperbolic line $h$ and $\\{\langle
N,x\rangle\\}_{x\in h}$ is the set of generators containing $N$. Deep sub-
generators also exist in ${\cal S}_{f}$. For instance, $V^{*}$ admits
infinitely many hyperplanes $H$ such that $H\neq\ker^{*}(x)$ for any $x\in V$.
If $N=[H]$ for such a hyperplane $H$ then $N^{\perp}=[V^{*}]$.

####  5.1.1 Singular subspaces and generators of ${\cal S}_{f}$

Let $W$ be a subspace of $\overline{V}$. Put $K_{1}:=\ker(p_{V^{*}}^{W})=W\cap
V$ and $K_{2}:=\ker(p_{V}^{W})=W\cap V^{*}$ and let $C$ be a complement of
$K_{1}\oplus K_{2}$ in $W$. Then $p_{V}^{C}$ and $p_{V^{*}}^{C}$ are
injective. Accordingly, if $C_{1}:=p_{V}^{C}(C)$ and
$C_{2}:=p_{V^{*}}^{C}(C)$, then $C_{1}\cong C\cong C_{2}$ and the mappings
$p_{1,2}:=p_{V^{*}}^{C}\circ(p_{V}^{C})^{-1}$ and
$p_{2,1}=p_{V^{*}}^{C}\circ(p_{V}^{C})^{-1}$ are isomorphisms from $C_{1}$ to
$C_{2}$ and from $C_{2}$ to $C_{1}$ respectively. Also,
$p_{2,1}=p_{1,2}^{-1}$. So, $C=\\{x\oplus p_{1,2}(x)\\}_{x\in
C_{1}}=\\{p_{2,1}(\xi)\oplus\xi\\}_{\xi\in C_{2}}$.

###### Lemma 5.1

With $C_{1},C_{2},K_{1},K_{2}$ as above, $C_{i}\cap K_{i}=\\{0\\}$ for
$i=1,2$.

Proof. If $x\in C_{1}\cap K_{1}$ then $W$ contains $(x\oplus
p_{1,2}(x))-x=p_{1,2}(x)$. Therefore $p_{1,2}(x)\in W\cap V^{*}=K_{2}$, namely
$x\oplus p_{1,2}(x)\in K_{1}\oplus K_{2}$. Hence $x\oplus p_{1,2}(x)=0$,
namely $x=0$. $\Box$

Clearly, $[W]$ is a singular subspace of ${\cal S}_{f}$ if and only if

$\left.\begin{array}[]{lll}K_{1}\subseteq\ker(C_{2}+K_{2}),&K_{2}\subseteq\ker^{*}(C_{1}+K_{1})&\mbox{and}\\\
p_{1,2}(x)(y)=p_{1,2}(y)(x),&\forall x,y\in C_{1}.&\end{array}\right\\}$ (15)

where for subspaces $X\subseteq V$ and $\Xi\subseteq V^{*}$ we put

$\ker^{*}(X):=\\{\xi\in V^{*}~{}|~{}\ker(\xi)\supseteq X\\}\mbox{ and
}\ker(\Xi):=\cap_{\xi\in\Xi}\ker(\xi).$

###### Lemma 5.2

Assume that $[W]$ is a singular subspace of ${\cal S}_{f}$. Then $[W]$ is a
generator of ${\cal S}_{f}$ if and only if

$\ker^{*}(C_{1}+K_{1})=K_{2}~{}\mbox{ and }~{}\ker(C_{2}+K_{2})=K_{1}.$ (16)

Proof. Suppose that $[W]$ is a generator of ${\cal S}_{f}$. Clearly
$W^{\perp_{f}}$ contains both $\ker(C_{2}+K_{2})$ and $\ker^{*}(C_{1}+K_{1})$.
However $W$ is maximal among the totally $f$-isotropic subspaces of
$\overline{V}$. Hence both $\ker(C_{2}+K_{2})$ and $\ker^{*}(C_{1}+K_{1})$ are
contained in $W$, namely $\ker(C_{2}+K_{2})=K_{1}$ and
$\ker^{*}(C_{1}+K_{1})=K_{2}$, as claimed in (16).

Conversely, assume (16). Note that, if $X$ is a subspace of $V$, then for
every vector $x\in V\setminus X$ there exists a hyperplane $H$ of $V$ which
contains $X$ but does not contain $x$. Consequently,

$\ker(\ker^{*}(X))~{}=~{}X.$ (17)

By (17) and the first equality of (16) we obtain

$C_{1}+K_{1}~{}=~{}\ker(K_{2}).$ (18)

Let now $a\oplus\alpha\in W^{\perp_{f}}$. Then
$0=f(a\oplus\alpha,\kappa)=\kappa(a)$ for every $\kappa\in K_{2}$. Therefore
$a\in\ker(K_{2})$. Hence $a=c+k$ for some $c\in C_{1}$ and some $k\in K_{1}$
by (18). So, we can replace $a\oplus\alpha$ with
$c\oplus\alpha=(a\oplus\alpha)-(k\oplus 0)$. In other words, we can assume to
have chosen $a\oplus\alpha$ with $a\in C_{1}$. However $W$ contains $a\oplus
p_{1,2}(a)$. Hence we can replace $a\oplus\alpha$ with $a\oplus\alpha-a\oplus
p_{1,2}(a)=0\oplus(\alpha-p_{1,2}(a))=\alpha-p_{1,2}(a)$. So, we can assume to
have chosen $a\oplus\alpha$ with $a=0$. Thus, we are reduced to the case of
$\alpha\in V^{*}$ such that $\alpha\perp_{f}W$. This condition forces
$\alpha\in\ker^{*}(C_{1}+K_{1})$. Hence $\alpha\in K_{2}$ by the first
equality of (16). We have proved that $W^{\perp_{f}}=W$. Thus $[W$] is a
generator of ${\cal S}_{f}$. $\Box$

###### Proposition 5.3

Every generator of ${\cal S}_{f}$ admits a complement.

Proof. Let $[W]$ be a generator of ${\cal S}_{f}$. With
$C_{1},C_{2},K_{1},K_{2}$ defined as in the first paragraph of this
subsection, choose a complement $H_{1}$ of $C_{1}+K_{1}$ in $V$ and put
$H_{2}:=\ker^{*}(H_{1})$. Let $W^{\prime}:=H_{1}\oplus H_{2}$. The two
relations of (16), referred to $W^{\prime}=H_{1}\oplus H_{2}$, amount to the
equality $\ker^{*}(H_{1})=H_{2}$, which holds by definition of $H_{2}$, and
the equality $\ker(H_{2})=H_{1}$, namely $\ker(\ker^{*}(H_{1}))=H_{1}$, which
is a special case of (17). So, $W^{\prime}$ satisfies both equations of (16).
Hence $[W^{\prime}]$ is a generator.

Let $x\oplus\xi\in W\cap W^{\prime}$. Then $x\in H_{1}$, $\xi\in H_{2}$ and
$x\oplus\xi=(k+c)\oplus(\kappa+p_{1,2}(c))$ for suitable $k\in K_{1}$,
$\kappa\in K_{2}$ and $c\in C_{1}$. However $H_{1}\cap(K_{1}+C_{1})=0$.
Therefore $x=0$. Accordingly, $k+c=0$, namely $c=-k$. Lemma 5.1 now forces
$c=k=0$. Hence $p_{1,2}(c)=0$. In the end, $\xi=\kappa$. Recall that $\xi\in
H_{2}=\ker^{*}(H_{1})$ while $\kappa\in K_{2}=\ker^{*}(K_{1}+C_{1})$. Moreover

$\ker^{*}(H_{1})\cap\ker^{*}(K_{1}+C_{1})~{}=~{}\ker^{*}(H_{1}+(K_{1}+C_{1})).$

However $H_{1}+K_{1}+C_{1}=V$ by our choice of $H_{1}$. Hence $H_{2}\cap
K_{2}=\ker^{*}(V)=0$. It follows that $\xi=\kappa=0$. Therefore $W\cap
W^{\prime}=0$.

We shall now prove that $W+W^{\prime}=\overline{V}$. The subspace
$W+W^{\prime}$ contains $K_{2}+H_{2}$. Let $\xi\in V^{*}$. If $\ker(\xi)$
contains $H_{1}$ then $\xi$ belongs to $H_{2}$. Suppose that $\ker(\xi)\cap
H_{1}$ is a hyperplane of $H_{1}$. As $H_{1}$ is a complement of $K_{1}+H_{1}$
and $K_{2}=\ker^{*}(K_{1}+H_{1})$, there exists $\kappa\in K_{2}$ such that
$\kappa$ and $\xi$ induce the same linear functional on $H_{1}$. Put
$\chi:=\xi-\kappa$. Then $\chi\in\ker^{*}(H_{1})=H_{2}$. So,
$\xi=\kappa+\chi\in K_{2}+H_{2}$.

We have proved that $W+W^{\prime}$ contains $V^{*}$. Consequently
$(W+W^{\prime})\cap V$ contains $(K_{1}+C_{1})+H_{1}=V$. So, $W+W^{\prime}$
contains both $V$ and $V^{*}$. Hence $W+W^{\prime}=\overline{V}$. $\Box$

The space ${\cal S}_{f}$ also admits opposite generators which are not
complementary, as in the following example.

###### Example 5.4

Given a basis $(e_{i})_{i\in\mathfrak{I}}$ of $V$, for every
$i\in\mathfrak{I}$ define a linear functional $\eta_{i}\in V^{*}$ by the
clause that $\eta_{i}(e_{j})=\delta_{i,j}$ (Kronecker symbol) for any
$j\in\mathfrak{I}$. The linear functionals $\eta_{i}$ defined in this way span
a proper subspace $V^{\prime}$ of $V^{*}$ isomorphic to $V$. Note that
$\ker(V^{\prime})=0$. Let $W$ be the subspace of $\overline{V}$ spanned by the
vectors $e_{i}\oplus\eta_{i}$ for $i\in\mathfrak{I}$. It is easy to see that
$[W]$ is a generator of ${\cal S}_{f}$. Clearly, $V\cap W=0$. However
$\dim(V+W)=\dim(V)<\dim(V^{*})=\dim(\overline{V})$. Hence
$V+W\subset\overline{V}$. So, $[W]$ is opposite to $[V]$ but it is not a
complement of $[V]$.

####  5.1.2 The characteristic two case

The natural embeding of ${\cal S}_{f}$ in $\mathrm{PG}(\overline{V})$ is
universal if and only if $\mathrm{char}(\mathbb{F})\neq 2$. Accordingly, when
$\mathrm{char}(\mathbb{F})\neq 2$ the union of any two complementary
generators generates ${\cal S}_{f}$. In contrast, when
$\mathrm{char}(\mathbb{F})=2$ no two generators of ${\cal S}_{f}$ generate
${\cal S}_{f}$, even when they are complementary. For instance, when
$\mathrm{char}(\mathbb{F})=2$ the subspace ${\cal
Q}:=\langle[V],[V^{*}]\rangle$ of ${\cal S}_{f}$ is the set of points of
$\mathrm{PG}(\overline{V})$ represented by the non-zero vectors
$a\oplus\alpha$ of $\overline{V}$ such that $\alpha(a)=0$. The set ${\cal Q}$
is a proper subspace of ${\cal S}_{f}$. In fact ${\cal Q}$ is an infinite
dimensional analogue of a hyperbolic quadric, defined by the quadratic form
which maps every vector $a\oplus\alpha\in\overline{V}$ onto the scalar
$\alpha(a)\in\mathbb{F}$. Referring to the next subsection (Section 5.2) for a
discussion of these quadrics, we devote the next paragraph to a description
the universal embedding of ${\cal S}_{f}$ when $\mathrm{char}(\mathbb{F})=2$.

Assuming that $\mathrm{char}(\mathbb{F})=2$, let
$\mathbb{F}^{2}:=\\{t^{2}\\}_{t\in\mathbb{F}}$ be the square subfield of
$\mathbb{F}$ and let $\mathbb{F}^{1/2}$ be the extension of $\mathbb{F}$
obtained by adding a square root $t^{1/2}$ of $t$ for every
$t\in\mathbb{F}\setminus\mathbb{F}^{2}$. Of course,
$\mathbb{F}^{2}\subseteq\mathbb{F}\subseteq\mathbb{F}^{1/2}$, with
$\mathbb{F}^{2}=\mathbb{F}=\mathbb{F}^{1/2}$ if and only if $\mathbb{F}$ is
perfect. As $\mathbb{F}^{1/2}$ contains $\mathbb{F}$, the field
$\mathbb{F}^{1/2}$ is naturally equipped with an $\mathbb{F}$-vector space
structure. Put $\widetilde{V}:=V\oplus\overline{V}\oplus\mathbb{F}^{1/2}$ and
define a quadratic form $q:\widetilde{V}\rightarrow\mathbb{F}$ as follows: