# Projective Rectangles:
A New Kind of Incidence Structure

Rigoberto Flórez  Dept. of Mathematical Sciences, The Citadel, Charleston,
South Carolina 29409<EMAIL_ADDRESS>and  Thomas Zaslavsky  Dept. of
Mathematics and Statistics, Binghamton University, Binghamton, New York
13902-6000<EMAIL_ADDRESS>
###### Abstract.

A projective rectangle is like a projective plane that has different lengths
in two directions. We develop the basic theory of projective rectangles
including incidence properties, projective subplanes, configuration counts, a
partial Desargues’s theorem, a construction, and alternative formulations. In
sequels we study harmonic conjugation and the graphs of lines and subplanes.

###### Key words and phrases:

Projective rectangle; incidence geometry; Pasch axiom; projective plane;
Desargues’s theorem; orthogonal array

###### 2010 Mathematics Subject Classification:

Primary 51E26; Secondary 05B15, 05B35, 05C22, 51A30, 51E20

###### Contents

  1. 1 Introduction
  2. 2 Projective rectangles
  3. 3 Properties of projective rectangles
    1. 3.1 Fundamental properties
    2. 3.2 Projective subplanes
    3. 3.3 No Vamos configuration
  4. 4 Finite projective rectangles
    1. 4.1 Counts
    2. 4.2 Constraints on the parameters
  5. 5 Axial and central Desargues’s theorems
  6. 6 Construction of projective rectangles
    1. 6.1 An abstract construction
    2. 6.2 A subplane construction
  7. 7 Narrow rectangles
  8. 8 Orthogonal arrays from projective rectangles
    1. 8.1 An orthogonal array from points and lines
    2. 8.2 An orthogonal array from points and planes
  9. 9 The dual incidence structure
  10. 10 Open problems

##  1\. Introduction

A projective rectangle is like a projective plane, but narrower than it is
tall. More precisely, it is like the set of points on a certain kind of family
of lines in a projective plane, with their induced lines. Very precisely, it
is an axiomatic incidence structure based on adapting axioms of projective
geometry, inspired by harmonic conjugation in matroids.

Projective rectangles, regarded as rank-3 matroids, are found in all known
harmonic matroids, such as full algebraic matroids. Harmonic matroids are
matroids within which there is harmonic conjugation [3]; their definition was
inspired by Lindström’s article [11] about abstract harmonic conjugation.
Harmonic conjugation applied to complete lift matroids of group expansions
[15, Example 6.7] of a triangle (for instance, $L_{2}^{k}$, Example 2.1) led
us to structures that looked like vertical strips in projective planes—whence
the name “projective rectangle” and the impulse to find a general theory of
this idea in terms of incidence geometry. Projective rectangles themselves are
almost examples of harmonic matroids, seemingly falling short only in special
lines, as we prove in a sequel [5].

An indication of what we accomplish in this article: First, the axioms
(Section 2) and basic consequences for incidence geometry (Section 3) and
counting (Section 4). Especially, we see that a projective rectangle, if it is
not a projective plane, contains a multitude of maximal projective planes; we
call them its “planes”. Section 5 develops partial Desarguesian properties of
projective rectangles, which satisfy limited versions of the two halves of
Desargues’s Theorem. In Section 6 we show that the construction based on a
subplane and a special point, alluded to above, actually works to produce
projective rectangles in planes that are Pappian, i.e., coordinatized by a
field; we do not know how far that subplane construction generalizes. The
following section treats the narrowest projective rectangles, which are the
simplest and best understood. Next are two sections that give alternative
viewpoints: in Section 8 we see that a projective rectangle is essentially a
Paschian transversal design and thus is equivalent to a special kind of
orthogonal array, and in Section 9 we take the approach of projective duality
by interchanging points and lines, which may suggest new properties but which
we have not studied deeply. We have only an elementary understanding of
projective rectangles in general, as is shown by the list of significant open
problems in Section 10.

In sequels we treat adjacency graphs and harmonic conjugation. Two sequels
explore the graphs of adjacency of lines and of planes in finite projective
rectangles [6, 7]. Notably, the graph of (ordinary) lines, where adjacency
means having a point in common, is strongly regular; and in projective
rectangles that are not projective planes the graph of planes, where adjacency
means having an ordinary line in common, has striking internal structure that
presents a tantalizing vision of higher dimensionality. The other sequel [5]
explores abstract harmonic conjugation as a theme linking harmonic matroids
and projective rectangles. In one direction, a projective rectangle is almost
a harmonic matroid. In the other direction, a harmonic matroid contains a
projective rectangle if it contains a matroid of a finite-field expansion of a
triangle, in particular if it contains a Reid cycle matroid.

Our personal interest is mainly in finite systems, but many results apply to
infinite projective rectangles. For instance, Section 3 encompasses infinite
systems, while Section 4 requires finiteness. Our viewpoint is influenced by
matroid theory but is largely that of incidence geometry; matroid theory is
not needed to read this paper.

We wish to acknowledge the inspiration of the elegant and deep short papers
[10, 11] of Bernt Lindström. Lindström’s ideas, as further developed by the
first author in his doctoral dissertation and [3], led to this study of
projective rectangles.

### Acknowledgement

We are grateful to a referee of a previous version, who read the paper with
great care and warned us of errors, gaps, and infelicities. Those observations
led us to major improvements.

##  2\. Projective rectangles

An _incidence structure_ is a triple $({\mathcal{P}},\mathcal{L},\mathcal{I})$
of sets with $\mathcal{I}\subseteq{\mathcal{P}}\times\mathcal{L}$. The
elements of ${\mathcal{P}}$ are _points_ , the elements of $\mathcal{L}$ are
_lines_. A point $p$ and a line $l$ are _incident_ if $(p,l)\in\mathcal{I}$. A
set $P$ of points is said to be _collinear_ if all points in $P$ are in the
same line. We say that two distinct lines _intersect in a point_ if they are
incident with the same point.

A _projective rectangle_ is an incidence structure
$({\mathcal{P}},\mathcal{L},\mathcal{I})$ that satisfies the following axioms:

  1. (A1)

Every two distinct points are incident with exactly one line.

  2. (A2)

There exist four points with no three of them collinear.

  3. (A3)

Every line is incident with at least three distinct points.

  4. (A4)

There is a _special point_ $D$. A line incident with $D$ is called _special_.
A line that is not incident with $D$ is called _ordinary_ , and a point that
is not $D$ is called _ordinary_.

  5. (A5)

Each special line intersects every other line in exactly one point.

  6. (A6)

Let $l_{1}$ and $l_{2}$ be two ordinary lines that intersect in a point. If
$l_{3}$ and $l_{4}$ are distinct lines that intersect $l_{1}$ and $l_{2}$ in
four distinct points, then $l_{3}$ and $l_{4}$ intersect in a point. (We
sometimes will call $l_{1}$ and $l_{2}$ the “intersecting lines”, and $l_{3}$
and $l_{4}$ the “crossing lines” because they cross the first two.)

A _complete quadrilateral_ is an incidence structure that consists of four
lines, no three concurrent, and their six points of intersection. A _nearly
complete quadrilateral_ is like a complete quadrilateral but with only five of
the intersection points; the sixth intersection point may or may not exist.
Axiom (A6) states that almost every nearly complete quadrilateral in a
projective rectangle is complete. This is a partial Pasch axiom (e.g., see
[14, page 314]), not the full Pasch axiom because it has an exception when
either of the first two lines is special; then the remaining two lines may or
may not be concurrent. This exception is what admits projective rectangles
that are not projective planes. Section 8 has more discussion of the
significance of Axiom (A6).

_Notation_ : $\mathbb{PR}$ denotes a projective rectangle. Axiom (A3) lets us
treat lines as sets of points; thus for a point $p$ and line $l$, the
statements “$p$ is a point of $l$”, “$p\in l$”, “$p$ is on $l$”, “$l$ contains
$p$”, etc., mean the same. We write $\overline{pq}$ for the unique line that
contains two points $p$ and $q$. After we establish the existence of
projective planes in $\mathbb{PR}$, we use the notation $\overline{abc\dots}$
to mean the unique line (if $abc\dots$ are collinear) or plane (if they are
coplanar but not collinear) that contains the points $abc\dots$.

The projective planes are some familiar examples of projective rectangles. A
projective plane is called a _trivial projective rectangle_. In particular the
Fano plane $F_{7}$ is the smallest projective rectangle (see Theorem 3.1 Part
(b)). The non-Fano configuration is not a projective rectangle; it fails Axiom
(A6).

In a trivial projective rectangle $\mathbb{PR}$ the special point $D$,
although selected as part of the definition, may be chosen to be any point. We
show in Theorem 3.1 that every special line has the same number $n$ of
ordinary points and every ordinary line has the same number $m+1$ of points,
and $m\leq n$. If $m\neq n$, then the special lines are distinguished from the
ordinary lines by cardinality; in that case $D$ is determined by lying on all
special lines. If $m<n$, therefore, the special point is unique. In general we
regard the specification of $D$ as part of the definition of a projective
rectangle.

A projective rectangle $\mathbb{PR}$ is a rank-3 matroid; the elements are the
points of $\mathbb{PR}$ and the rank-2 flats are the lines of $\mathbb{PR}$.
(The matroid is, of course, infinite if the number of points in a special line
is not finite.) The simplest such matroid is that of the following example.

######  Example 2.1.

The matroid $L_{2}^{k}$ is another example of a projective rectangle (see
Figure 2.1). It has $m+1=3$ special lines. Let $A:=\left\\{a_{g}\mid
g\in\mathbb{Z}_{2}^{k}\right\\}\cup\\{D\\}$, $B:=\left\\{b_{g}\mid
g\in\mathbb{Z}_{2}^{k}\right\\}\cup\\{D\\}$ and $C:=\left\\{c_{g}\mid
g\in\mathbb{Z}_{2}^{k}\right\\}\cup\\{D\\}$. Let $L_{2}^{k}$ be the simple
matroid of rank 3 defined on the ground set $E:=A\cup B\cup C$ by its rank-2
flats, which are the special lines $A$, $B$, $C$ and the sets
$\\{a_{g},b_{g+h},c_{h}\\}$ with $g$ and $h$ in $\mathbb{Z}_{2}^{k}$, which
are the ordinary lines.

We note that $L_{2}^{k}$ is the complete lift matroid of the group expansion
of a triangle, i.e., $L_{0}(\mathbb{Z}_{2}^{k})$ in the language of [15, 16].
We say more about projective rectangles with $m=2$ in Section 7.

Figure 2.1. The matroid $L_{2}^{2}$ with group the Klein 4-group,
$\mathfrak{V}_{4}=\\{1,g,h,g+h\\}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.

##  3\. Properties of projective rectangles

In this section we study essential properties of projective rectangles. We
begin with basic facts; then we prove that the projective rectangle contains
projective planes and we conclude with a section of counting formulas for
later use.

###  3.1. Fundamental properties

If a projective rectangle $\mathbb{PR}$ with exactly $m+1$ special lines has
one of them with $n+1$ points, then we say that the _order_ of $\mathbb{PR}$
is $(m,n)$. We do not assume $m$ or $n$ is finite unless we so state. In
Theorem 3.1 we prove $m\leq n$; we also prove that every special line has the
same number of points, that every ordinary line has the same number of points,
and many other elementary facts about points and lines.

The following result states basic properties of a projective rectangle.

######  Theorem 3.1.

If $\mathbb{PR}$ is a projective rectangle of order $(m,n)$, then the
following hold in $\mathbb{PR}$:

  1. (a)

The point set of $\mathbb{PR}\smallsetminus D$ is partitioned by all special
lines deleting $D$.

  2. (b)

There are at least three special lines and four ordinary lines. Moreover,
there are at least seven points.

  3. (c)

If $l$ is a line and $p$ is a point not in $l$, then the number of distinct
lines incident with $p$ intersecting $l$ equals the number of points on $l$.

  4. (d)

Through each ordinary point there passes exactly one special line.

  5. (e)

All ordinary lines have the same number of points. The number of points in an
ordinary line is equal to the number of special lines, that is, $m+1$.

  6. (f)

All special lines have the same number of points, i.e., $n+1$ points, and the
same number of ordinary points, i.e., $n$.

  7. (g)

There are exactly $(m+1)n$ ordinary points.

  8. (h)

The number of lines incident with an ordinary point is equal to the number of
points in a special line, that is, $n+1$. The number of ordinary lines that
contain each ordinary point is $n$.

  9. (i)

The number of points in a special line is at least the number of points in an
ordinary line; that is, $n\geq m$.

  10. (j)

There are exactly $n^{2}$ ordinary lines.

  11. (k)

For a given point $p$ in an ordinary line $l$, there are $n-1$ ordinary lines
intersecting $l$ at $p$.

###### Proof.

Proof of Part (a). By Axiom (A1), every point $p\in\mathbb{PR}\smallsetminus
D$ belongs to the unique special line $\overline{pD}$.

Proof of Part (b). From Axiom (A2) we know that in $\mathbb{PR}$ there are
four points, no three of them collinear. If one is $D$, each other one with
$D$ generates a special line, all of which are distinct by noncollinearity. If
none of them is $D$, the points generate six distinct lines, of which at most
two can contain $D$ because no three of the four points are collinear. Thus,
the four remaining lines are ordinary lines. Since in one of the ordinary
lines there are at least three points, these points form with $D$ three
special lines. We have proved that in $\mathbb{PR}$ there are at least three
special lines and three ordinary lines. By Axiom (A3), each special line
contains at least two ordinary points, so there are at least seven points.

Now consider two special lines $s,s^{\prime}$ and two ordinary points
$p_{1},p_{2}$ on $s$ and $p_{1}^{\prime},p_{2}^{\prime}$ on $s^{\prime}$. The
lines $\overline{p_{i}p^{\prime}_{j}}$ are four distinct ordinary lines.

We prove Part (c). Let $q\in l$ and $p\notin l$. From (A1) there is exactly
one line incident with $p$ that intersects $l$ at $q$, and all such lines are
distinct.

We prove Parts (d) and (e). Given an arbitrary ordinary line $l$, we know by
(A1) that each point in $l$ together with $D$ determines a unique special
line. Every special line is generated in this way, by (A5). Thus, there is a
bijection between the special lines and the points in $l$. This implies the
number of points in any ordinary line equals the number of special lines.

We prove Parts (f) and (h). We suppose that $l_{1}$ and $l_{2}$ are special
lines in $\mathbb{PR}$ with $n_{1}+1$ and $n_{2}+1$ points, respectively. Let
$p$ be a point non-incident with either of those lines. Part (c) implies that
there are $n_{1}+1$ distinct lines intersecting $l_{1}$ that are incident with
$p$. Those $n_{1}+1$ lines also intersect $l_{2}$. Indeed, one of those lines
is special and the remaining $n_{1}$ lines intersect $l_{2}$ because they are
ordinary. Therefore, $n_{1}\leq n_{2}$. Similarly, $n_{2}\leq n_{1}$. This
proves that all special lines have the same number of points. Deducting 1 for
the special point $D$ gives the number of ordinary points on a special line.

Proof of Part (g). The number of special lines is $m+1$, Part (f) says the
number of ordinary points in each special line equals $n$ and Part (a) says
the special lines partition the ordinary points.

Proof of Part (i). We suppose that $l$ is an ordinary line and $s$ is a
special line. We produce an injection of the point set of $l$ into the point
set of $s$. Let $p$ be a point not in either $l$ or $s$; it exists because any
special line other than $s$ contains at least three points by Axiom (A3), one
of which is not $D$ and not in $l$. For each $q\in l$, the line
$\overline{pq}$ intersects $s$ in a point $r_{q}$ by Axiom (A5). For distinct
$q,q^{\prime}\in l$, $r_{q}$ and $r_{q^{\prime}}$ must be distinct because the
lines $\overline{pr_{q}}=\overline{pq}$ and
$\overline{pr_{q^{\prime}}}=\overline{pq^{\prime}}$ are distinct and intersect
only at $p$ by Axiom (A1). Thus, the mapping $l\to s$ by $q\mapsto r_{q}$ is
an injection. This proves that $s$ has at least as many points as $l$, i.e.,
$n+1\geq m+1$.

By Part (f) every special line has $n$ ordinary points, and by definition
there are $m+1$ special lines.

Proof of Part (j). Let $p$ be a point in an ordinary line. Two ordinary points
in two special lines give rise to a unique ordinary line. Since every special
line has $n+1$ points and one of them is $D$, it is easy to see that the two
special lines give rise to $n^{2}$ ordinary lines. Those are all the ordinary
lines that intersect the two special lines. Since every ordinary line
intersects every special line, we conclude that there are no more ordinary
lines in $\mathbb{PR}$.

Proof of Part (k). Since $p$ is a point in an ordinary line $l$, from Part (h)
there are $n+1$ lines incident with $p$. Only one of those $n+1$ lines is
special; the other $n$ are not. This implies that there are $n-1$ ordinary
lines intersecting $l$ at $p$. ∎

###  3.2. Projective subplanes

We show that a projective rectangle is a combination of projective planes, in
the strong sense that every two intersecting ordinary lines are lines of a
substructure that is a projective plane. Before our results, though, we have
to clarify the notion of substructure of an incidence structure
$({\mathcal{P}},{\mathcal{L}},{\mathcal{I}})$.

An _incidence substructure_ of $({\mathcal{P}},{\mathcal{L}},{\mathcal{I}})$
is an incidence structure
$({\mathcal{P}}^{\prime},{\mathcal{L}}^{\prime},{\mathcal{I}}^{\prime})$ in
which ${\mathcal{P}}^{\prime}\subseteq{\mathcal{P}}$,
${\mathcal{L}}^{\prime}\subseteq{\mathcal{L}}$, and
${\mathcal{I}}^{\prime}={\mathcal{I}}|{\mathcal{P}}^{\prime}\times{\mathcal{L}}^{\prime}$,
i.e., the incidence relation is the same as in the superstructure but
restricted to the elements of the substructure. In particular, if
$({\mathcal{P}}^{\prime},{\mathcal{L}}^{\prime},{\mathcal{I}}^{\prime})$ is a
projective plane, we call it a _subplane_ of
$({\mathcal{P}},{\mathcal{L}},{\mathcal{I}})$.

In a projective rectangle a subplane may contain an ordinary line and all its
points; we call that kind _full_. A full subplane necessarily has order $m$. A
subplane need not be full; it also need not be a maximal subplane, for
instance if it is a proper subplane of a full subplane. In fact, that is the
only way a subplane can fail to be maximal, as we will see in Theorem 3.7.

The special point $D$ is very special, as are the special lines.

######  Proposition 3.2.

In a projective rectangle $\mathbb{PR}$, the special point $D$ is a point of
every full subplane. Also, for every special line $s$ and every full subplane
$\pi$, $s\cap\pi$ is a line of $\pi$.

###### Proof.

A full subplane $\pi$ contains at least two lines, $l$ and $l^{\prime}$, which
intersect at a point $p\in\pi$, and at least one is ordinary, say $l$.

If $l^{\prime}$ is ordinary, then every special line $s$ intersects both $l$
and $l^{\prime}$ at different points, unless $s$ is the special line $s_{p}$
on $p$. These two points of $s$ determine a line of $\pi$, which is the
intersection of $s$ with $\pi$. Thus, for every special line except possibly
$s_{p}$, $s\cap\pi$ is a line of $\pi$.

If $l^{\prime}$ is special, or rather if $l^{\prime}=s^{\prime}\cap\pi$ for
some special line $s^{\prime}$, then there is at least one point $p^{\prime}$
on $l^{\prime}$ that is neither $p$ nor $D$. Let $q$ be a point in
$l\smallsetminus p$; then $\pi$ has a line $l^{\prime\prime}$ determined by
$p^{\prime}$ and $q$, which is ordinary since it contains not only $p\in
s_{p}$ but also $q\notin s_{p}$. Then we can replace $l^{\prime}$ by
$l^{\prime\prime}$ and have the case of two ordinary lines, so we may as well
assume $l^{\prime}$ is ordinary.

Let $s_{1}$ and $s_{2}$ be two special lines that are not $s_{p}$. Then
$s_{1}\cap\pi$ and $s_{2}\cap\pi$ are lines of $\pi$ whose intersection is a
point $d\in\pi$. Since $\\{d\\}=(s_{1}\cap\pi)\cap(s_{2}\cap\pi)=(s_{1}\cap
s_{2})\cap\pi=\\{D\\}\cap\pi$, we conclude that $D=d\in\pi$.

Let $p_{1}$ be the intersection of $l$ with $s_{1}$ and let $p_{2}$ be the
intersection of $l^{\prime}$ with $s_{2}$. Since $p_{1}\notin l^{\prime}$ and
$p_{2}\notin l$, the line $l^{\prime\prime}$ of $\pi$ determined by $p_{1}$
and $p_{2}$ does not contain $p$. Since the points $p_{1},p_{2}$ are not $D$
and are not in the same special line, $l^{\prime\prime}$ is ordinary, hence it
is contained in $\pi$. Being ordinary, by Axiom (A5) $l^{\prime\prime}$
intersects $s_{p}$ in a point $p_{12}$, which cannot be $p$, so $p$ and
$p_{12}$ determine a line of $\pi$, which must be $s_{p}\cap\pi$. That is,
$s_{p}\cap\pi$ is a line of $\pi$. ∎

Now we present the fundamental result about subplanes.

######  Theorem 3.3 (Planes in $\mathbb{PR}$).

Let $\mathbb{PR}$ be a projective rectangle. If two ordinary lines in
$\mathbb{PR}$ intersect in a point, then both lines are lines of a unique full
projective plane in $\mathbb{PR}$.

First we state the construction that gives the projective plane.

######  Construction 3.4.

Let $l_{0}$ and $l_{1}$ be ordinary lines in $\mathbb{PR}$ with exactly one
point $q$ in common. (See Figure 3.1.) Let $a_{0s}\in l_{0}\cap s$ and
$a_{1s}\in l_{1}\cap s$, where $s$ ranges over the set ${\mathcal{S}}$ of
special lines in $\mathbb{PR}$, and pick three special lines to be called $x$,
$y$, and $z$ such that $q\in x$. Thus, $q=a_{0x}=a_{1x}$. (We know there are
three special lines by Theorem 3.1 Part (b).) Let $b_{1s}\in n_{1}\cap s$,
where $n_{1}$ is the ordinary line that passes through $a_{0y}$ and $a_{1z}$.

Figure 3.1. Construction 3.4.

Suppose that $s$ and $t$ denote two special lines. We denote by $l_{st}$ the
ordinary line passing through $a_{0s}$ and $a_{1t}$ with $s,t\neq x$ and we
denote by $n_{st}$ the ordinary line passing through $a_{0s}$ and $b_{1t}$
with $s,t\neq y$. Let

$L=\\{l_{st}:s,t\in{\mathcal{S}},\ s,t\neq x\text{ and }s\neq t\\}$

and

$N=\\{n_{st}:s,t\in{\mathcal{S}},\ s,t\neq y\text{ and }s\neq t\\}.$

Note that $n_{1}=l_{yz}\in L$ and $l_{1}=n_{xz}\in N$, but $l_{0}=n_{xy}\notin
N$.

We set $\Pi:=({\mathcal{P}}_{\Pi},\mathcal{L}_{\Pi},\mathcal{I}_{\Pi})$, where
$\mathcal{I}_{\Pi}$ is the incidence relation defined in $\mathbb{PR}$ and

$\begin{array}[]{lcl}{\mathcal{P}}_{\Pi}&:=&{(\bigcup_{l\in
N}l)\cup(\bigcup_{l\in L}l)\cup l_{0}\cup\\{D\\}},\\\\[3.0pt]
{\mathcal{L}}_{1}&:=&\left\\{s\cap{\mathcal{P}}_{\Pi}:s\in{\mathcal{S}}\right\\},\\\\[3.0pt]
{\mathcal{L}}_{2}&:=&L\cup N\cup\\{l_{0}\\},\\\\[3.0pt]
{\mathcal{L}}_{\Pi}&:=&{\mathcal{L}}_{1}\cup{\mathcal{L}}_{2}.\end{array}$

###### ”New” New Proof of Theorem 3.3.

We begin with the incidence structure $\Pi$ given by Construction 3.4. With
the notation there, we prove that $\Pi$ is a projective plane.

First of all, we note that one of the defining properties of a projective
plane, that there are four points in ${\mathcal{P}}_{\Pi}$ with no three of
them collinear, is satisfied by $a_{0y}$, $a_{1z}$, $q$, and $D$.

Figure 3.2. For the proof of Theorem 3.3: lines $\lambda=l_{st}$ and
$\eta=n_{xt^{\prime}}$ intersect (general case).

We next prove that given two lines in $\Pi$, they intersect. We begin with a
lemma that simplifies the list of lines in $\Pi$.

######  Lemma 3.5.

Every line $n_{st}$ with $s,t\neq x$ is a line $l_{st^{\prime}}$. Each line
$n_{xt}$ is $\overline{qb_{1t}}$.

###### Proof.

By definition of $n_{st}$ (with $s,t\neq x$), we have $s,t\neq y$. The lines
$l_{1}$ and $n_{st}$ are in $\Pi$ so they intersect in a point
$a_{1t^{\prime}}$. We know $a_{1t^{\prime}}\neq q$ since $q\notin n_{st}$,
therefore $t^{\prime}\neq x$. Because $a_{0s}\in n_{st}$ and $\notin l_{1}$,
we know that $a_{1t^{\prime}}\neq a_{0s}$, so $t^{\prime}\neq s$ and the line
$l_{st^{\prime}}$ is well defined.

A line $n_{xt}=\overline{a_{0x}b_{1t}}=\overline{qb_{1t}}$. ∎

Suppose that the two given lines are in $L$ (so they are ordinary). If they
intersect in a point in $l_{0}$ or in a point in $l_{1}$, there is nothing to
prove. Suppose that neither of those two cases holds. So, they are two
ordinary lines that intersect $l_{0}$ and $l_{1}$ in four different points.
Therefore, by Axiom (A6) the two given lines intersect. By a similar argument
we conclude that if the two given lines are in $N$, then they intersect. It is
clear that any two lines in ${\mathcal{L}}_{1}$ intersect in $D$ and that a
line in ${\mathcal{L}}_{2}$ intersects every line in ${\mathcal{L}}_{1}$.

Suppose the two given lines are $\lambda$ and $\eta$ with $\lambda=l_{st}\in
L$ and $\eta\in N$. We may assume $\eta\neq l_{1}$ because $\lambda$ and
$l_{1}$ intersect by the definition of $L$. By definition, $\lambda=l_{st}$
for $s\neq t$ and both $\neq x$. By Lemma 3.5 we may assume
$\eta=n_{xt^{\prime}}$ for some $t^{\prime}\neq x,y,z$.

Suppose that $a_{0y}\neq a_{0s}$ and $a_{1t}\neq a_{1z}$. Since $l_{st}$ and
$n_{1}$ intersect $l_{0}$ and $l_{1}$ in four distinct points, by (A6) we know
that $l_{st}$ intersects $n_{1}$ at a point $b_{1u}$. Then $l_{st}$ and
$n_{xt^{\prime}}$ intersect $l_{0}$ and $n_{1}$ in four distinct points
(because $n_{1}$ intersects $l_{0}$ at $a_{0y}\notin l_{st}$), unless
$b_{1u}=b{1t^{\prime}}$. That, with (A6), implies that $l_{st}$ and
$n_{xt^{\prime}}$ intersect in a point. In the exceptional case that
$b_{1u}=b{1t^{\prime}}$, that is the intersection point.

If $a_{0y}=a_{0s}$, then $l_{st}$ and $n_{xt^{\prime}}$ intersect $l_{0}$ and
$n_{1}$ in four distinct points unless $a_{1t}=a_{1z}$. Since $l_{0}$ and
$n_{1}$ intersect in $a_{0y}$, by (A6) we conclude that $l_{st}$ and
$n_{xt^{\prime}}$ intersect. If $a_{1t}=a_{1z}$, $l_{st}=n_{1}$ and the
intersection point is $b_{1t^{\prime}}$.

In the remaining special case $a_{1t}=a_{1z}$ and $a_{0y}\neq a_{0s}$. Here
$l_{st}$ and $n_{xt^{\prime}}$ intersect $l_{0}$ and $n_{1}$ in four distinct
points so by (A6) we obtain the desired intersection.

Finally, suppose one line is $l=s\cap{\mathcal{P}}_{\Pi}$ for a special line
$s$. The other is either $t\cap{\mathcal{P}}_{\Pi}$ for a special line $t$,
and intersects $l$ at $D$, or is an ordinary line $l^{\prime}=l_{st}$ or
$n_{st}$, in which case it intersects $l$ at a point in
$l^{\prime}\subseteq{\mathcal{P}}_{\Pi}$.

This completes the proof that any two lines in $\Pi$ intersect in $\Pi$.

Figure 3.3. For the proof of Theorem 3.3: the framework for intersections of
ordinary lines of $\Pi$, with the fundamental ordinary lines
$l_{0},l_{1},n_{1}$ and important special lines $x,y,z$.

We now prove that given two points $p_{0},p_{1}\in{\mathcal{P}}_{\Pi}$, they
are in a line in $\Pi$. (If they are in one line, they cannot be in two,
because the lines of $\Pi$ are ordinary lines or restrictions of special lines
of $\mathbb{PR}$, and every line in $\mathbb{PR}$ is determined by two of its
points.)

Our objective is to prove that the line $\overline{p_{0}p_{1}}$ is either a
line in $\Pi$ or is special. We divide the proof into cases depending on how
$p_{0}$ and $p_{1}$ are located with respect to the fundamental lines
$l_{0},l_{1},n_{1}$ of Construction 3.4. The cases are:

  1. Case 1.

$p_{0},p_{1}$ are both in a special line or both in any one of $l_{0}$,
$l_{1}$, or $n_{1}$.

  2. Case 2.

$p_{0}=q$. By symmetry this covers the case $p_{1}=q$.

  3. Case 3.

$p_{0}\in x\smallsetminus q$. This covers the case $p_{1}\in x\smallsetminus
q$.

  4. Case 4.

The generic case, where there are no collinear threesomes among
$q,p_{0},p_{1},D$ and neither point is in one of the fundamental lines
$l_{0},l_{1},n_{1}$. By the definition of ${\mathcal{P}}_{\Pi}$ and Lemma 3.5,
each of $p_{0}$ and $p_{1}$ must belong to an ordinary line $l_{st}$ or
$n_{xt}$.

  5. Case 5.

$p_{0}\in l_{0}\smallsetminus q$ and $p_{1}\notin x,l_{0}$.

  6. Case 6.

$p_{0}\in n_{1}$ but is not in any of $x,l_{0},l_{1}$, and $p_{1}\notin
x,l_{0},l_{1},n_{1}$.

Now we prove each case. We provide figures for the more complicated cases and
some subcases.

Case 1. If $p_{0},p_{1}$ are both in a special line, its intersection with
${\mathcal{P}}_{\Pi}$ is in ${\mathcal{L}}_{1}\subseteq{\mathcal{L}}_{\Pi}$.
If $p_{0},p_{1}$ are both in any one of $l_{0}$, $l_{1}$, or $n_{1}$, then
$\overline{p_{0}p_{1}}$ is in ${\mathcal{L}}_{\Pi}$ by definition.

Case 2. Suppose $p_{0}=q$. By Case 1 we may assume $p_{1}\notin x\cup
l_{0}\cup l_{1}$. Now, $p_{1}$ is in a line $n_{xt}$ or $l_{st}$. In the
former case $\overline{p_{0}p_{1}}=n_{xt}$. In the latter case there is a
point $b_{1t^{\prime}}$ common to $l_{st}$ and $n_{1}$. If $s\neq y$ (so
$t^{\prime}\neq y$), Axiom (A6) with intersecting lines $l_{0},l_{st}$ and
crossing lines $n_{1}$ and $\overline{p_{0}p_{1}}$ produces an intersection
point $b_{1t^{\prime\prime}}$; and if $s\neq z$ (so $t^{\prime}\neq z$), lines
$l_{1},l_{st}$ and $n_{1},\overline{p_{0}p_{1}}$ produce an intersection point
$b_{1t^{\prime\prime}}$. Either way,
$\overline{p_{0}p_{1}}=n_{qx}\in{\mathcal{L}}_{\Pi}$ where $x$ is the special
line that contains $b_{1t^{\prime\prime}}$.

Case 3. Suppose $p_{0}\in x$ but is not $q$; then $p_{0}\in l_{st}$ for some
$s,t\neq x$. We may assume $p_{1}\notin x,l_{st}$; then $p_{1}\in
l_{s^{\prime}t^{\prime}}$ or $n_{xt^{\prime}}$.

If the former, (A6) with intersecting lines $l_{st},l_{s^{\prime}t^{\prime}}$
and crossing lines $l_{0},\overline{p_{0}p_{1}}$ imply an intersection
$a_{0,s^{\prime\prime}}$ if $a_{0s}\neq a_{0s^{\prime}}$ (i.e., $s\neq
s^{\prime}$), and replacing $l_{0}$ by $l_{1}$ implies an intersection point
$a_{1t^{\prime\prime}}$ if $a_{1t}\neq a_{1t^{\prime}}$ (i.e., $t\neq
t^{\prime}$); note that one of these must hold or $p_{0},p_{1}\in
l_{st}=l_{s^{\prime}t^{\prime}}$ and we are done. If both hold, then
$\overline{p_{0}p_{1}}=l_{s^{\prime\prime}t^{\prime\prime}}$ and we are done.

If $s=s^{\prime}$, then the intersecting lines
$l_{st},l_{s^{\prime}t^{\prime}}$ and crossing lines
$l_{1},\overline{p_{0}p_{1}}$ imply a point $a_{1t^{\prime\prime}}\in
l_{1}\cap\overline{p_{0}p_{1}}$. Then intersecting $l_{1},l_{st}$ and crossing
$l_{0},\overline{p_{0}p_{1}}$ generate a point $a_{0s^{\prime\prime}}\in
l_{0}\cap\overline{p_{0}p_{1}}$. Thus,
$\overline{p_{0}p_{1}}=l_{s^{\prime\prime}t^{\prime\prime}}$ and we are done.
The case $t=t^{\prime}$ is similar.

Case 4. We suppose $p_{0},p_{1}\notin l_{0},l_{1},n_{1}$, no three of
$q,p_{0},p_{1},D$ are collinear, and there are ordinary lines $k_{0},k_{1}$ in
$\Pi$ such that $p_{0}\in k_{0}$ and $p_{1}\in k_{1}$. Since
$\overline{p_{0}p_{1}}\not\ni D$, it is an ordinary line. We show that
$\overline{p_{0}p_{1}}$ is an $l_{st}$, hence a line of $\Pi$. First, $k_{0}$
and $k_{1}$ intersect because they are lines of $\Pi$. Let $r$ be their
intersection.

(a) (b)
(c)

Figure 3.4. For the proof of Case 4, with the three possible types of pair
$k_{0},k_{1}$.

Case 4(a). Assume $k_{0},k_{1}\in L$. There are two subcases.

If $r\notin l_{0}\cup l_{1}$, then (for $i=0,1$)
$l_{i}\cap\overline{p_{0}p_{1}}$ is a point, by Axiom (A6) applied to the
intersecting lines $k_{0},k_{1}$ and the crossing lines $l_{i}$ and
$\overline{p_{0}p_{1}}$. Call these points $a_{0s}$ and $a_{1t}$. Because
$q\notin\overline{p_{0}p_{1}}$, they are different points and $s,t\neq x$.
Because $D\notin\overline{p_{0}p_{1}}$, $s\neq t$. So
$\overline{p_{0}p_{1}}=l_{st}$.

If $r\in l_{0}$ (or $l_{1}$, but they are similar), then (A6) applied to the
intersecting lines $k_{0},k_{1}$ and the crossing lines $l_{1}$ and
$\overline{p_{0}p_{1}}$ implies an intersection point
$l_{1}\cap\overline{p_{0}p_{1}}$, which is not $q$ so it is $a_{1t}$ for some
$t\neq x$. Applying the axiom again to intersecting lines $k_{1}$ and $l_{1}$
and crossing lines $l_{0}$ and $\overline{p_{0}p_{1}}$ implies an intersection
point of the latter two lines. This point cannot be $q$ because it is in
$l_{0}$ but not $l_{1}$, so it is an $a_{0s}$ for some $s\neq x$. We have
found points $a_{0s},a_{1t}\in\overline{p_{0}p_{1}}$, so that line is
$l_{st}$.

Case 4(b). Assume $k_{0}\in L$ and $k_{1}=n_{xt^{\prime}}\in N$. The
intersecting lines $k_{0},k_{1}$ with crossing lines
$l_{0},\overline{p_{0}p_{1}}$ generate a crossing point $a_{0s}\in
l_{0}\cap\overline{p_{0}p_{1}}$ and with crossing lines
$l_{1},\overline{p_{0}p_{1}}$ they generate $a_{1t}\in
l_{1}\cap\overline{p_{0}p_{1}}$. Thus, $\overline{p_{0}p_{1}}=l_{st}$.

Case 4(c). Last, assume $k_{0},k_{1}\in N$. We need two steps. Axiom (A6)
applied to intersecting $k_{0},k_{1}$ and crossing
$n_{1},\overline{p_{0}p_{1}}$ gives a point $r\in
n_{1}\cap\overline{p_{0}p_{1}}$. Now we use the intersecting lines
$k_{0},n_{1}$ and crossing lines $l_{i},\overline{p_{0}p_{1}}$ for each of
$l_{0}$ and $l_{1}$ to get points $a_{0s},a_{1t}\in\overline{p_{0}p_{1}}$,
proving that $\overline{p_{0}p_{1}}=l_{st}$.

Case 5. Let $p_{0}$ in $l_{0}\smallsetminus q$, say $p_{0}=a_{0s}$. Since
$p_{0}$ is in $l_{0}$, it cannot be in a line $n_{xt}$ (other than $l_{0}$
itself); thus, it is in a line $l_{st}$ for some $t\neq x,s$. If $p_{1}\in x$
or $l_{0}$ or $s$ we are in a previous case. If $p_{1}\in l_{1}\smallsetminus
q$, then $p_{1}=a_{1t^{\prime}}$ and $\overline{p_{0}p_{1}}=l_{st^{\prime}}$
for some $t^{\prime}$. So, we assume $p_{1}$ is not in $l_{1}$. If $p_{1}\in
l_{st}$ then $\overline{p_{0}p_{1}}=l_{st}$, so we assume $p_{1}\notin
l_{st}$. We know it is in either an $l_{s^{\prime}t^{\prime}}$ or an
$n_{xt^{\prime}}$.

(a)  (b)

Figure 3.5. For the proof of Cases 5(a) and (b).

Case 5(a). Suppose $p_{1}\in l_{s^{\prime}t^{\prime}}$. If $s=s^{\prime}$,
then $\overline{p_{0}p_{1}}=l_{s^{\prime}t^{\prime}}$ and we are done, so we
assume $s\neq s^{\prime}$. If $t\neq t^{\prime}$, then
$l_{st},l_{s^{\prime}t^{\prime}}$ form the intersecting lines (intersecting at
a point $r\neq p_{1}$) and $l_{1},\overline{p_{0}p_{1}}$ the crossing lines
for (A6); this implies an intersection point
$a_{1t^{\prime\prime}}\in\overline{p_{0}p_{1}}$ and then
$\overline{p_{0}p_{1}}=l_{st^{\prime\prime}}$. We consider the case
$t=t^{\prime}$. Then (A6) applies to the intersecting pair
$l_{0},l_{s^{\prime}t^{\prime}}$ and the crossing pair
$l_{1},\overline{p_{0}p_{1}}$, giving a point $a_{1,t^{\prime\prime}}\in
l_{1}\cap\overline{p_{0}p_{1}}$, so
$\overline{p_{0}p_{1}}=l_{st^{\prime\prime}}$, and we are done.

Case 5(b). Suppose, however, that $p_{1}\in n_{xt^{\prime}}$ (but not in
$x,l_{0},l_{1}$). Here we take as intersecting lines $l_{0}$ and
$n_{xt^{\prime}}$ with crossing lines $l_{1}$ and $\overline{p_{0}p_{1}}$.
This gives us a point $a_{1t^{\prime\prime}}\in
l_{1}\cap\overline{p_{0}p_{1}}$ so that
$\overline{p_{0}p_{1}}=l_{st^{\prime\prime}}\in{\mathcal{L}}_{\Pi}$, provided
that $p_{0},p_{1}\notin n_{1}$ so there are four distinct crossing points. For
the exceptional cases, If both are in $n_{1}$ we are done:
$\overline{p_{0}p_{1}}=n_{1}$. If $p_{0}\in n_{1}$ and $p_{1}\notin n_{1}$,
then intersecting lines $n_{1},n_{xt^{\prime}}$ and crossing lines
$l_{1},\overline{p_{0}p_{1}}$ give us an intersection point
$a_{1t^{\prime\prime}}$ so that $\overline{p_{0}p_{1}}=l_{tt^{\prime\prime}}$.
If the reverse, where $p_{1}\in n_{1}$, so $p_{1}=b_{1t^{\prime}}$, and
$p_{0}\notin n_{1}$, the intersecting lines should be $l_{0}$ and $n_{1}$ and
the crossing lines $l_{1},\overline{p_{0}p_{1}}$, giving an intersection point
$a_{1t^{\prime\prime}}$ and making
$\overline{p_{0}p_{1}}=l_{st^{\prime\prime}}$—but with the one exception when
$b_{1z}\in\overline{p_{0}p_{1}}$ since then the crossing lines make only three
crossing points. In that case $\overline{p_{0}p_{1}}=l_{sz}=l_{st}$, so we are
done.

Case 6. We assume $p_{0}\in n_{1}$ but not in any of $x,l_{0},l_{1}$, and
$p_{1}\notin x,l_{0},l_{1},n_{1}$. As with $p_{1}$ in Case 5, $p_{0}$ is
either in a line $l_{st}$ or a line $n_{xt}$. Also as with Case 5, there are
exceptional subcases.

(a)  (b)

Figure 3.6. For the proof of Cases 6(a) and (b).

Case 6a. Suppose $p_{1}$ is in an $n_{xt^{\prime}}$. We assume $p_{0}\notin
n_{xt^{\prime}}$, since otherwise $\overline{p_{0}p_{1}}=n_{xt^{\prime}}$.
There are two steps. First, if $p_{0}\in l_{0}$, then $p_{0}=a_{0y}$; let
$s^{\prime\prime}=y$ and go to the second step. Otherwise, apply (A6) to
intersecting lines $n_{1},n_{st^{\prime}}$ and crossing lines
$l_{0},\overline{p_{0}p_{1}}$ to get a point $a_{0s^{\prime\prime}}\in
l_{0}\cap\overline{p_{0}p_{1}}$. In the second step, the intersecting lines
are the same and the crossing lines are $l_{1},\overline{p_{0}p_{1}}$, giving
a point $a_{1t^{\prime\prime}}$ and a line
$l_{s^{\prime\prime}t^{\prime\prime}}=\overline{p_{0}p_{1}}$.

Case 6b. Suppose $p_{1}\in l_{s^{\prime}t^{\prime}}$. If $p_{0}\in
l_{s^{\prime}t^{\prime}}$, that is the line $\overline{p_{0}p_{1}}$. Otherwise
we apply (A6) simultaneously to two situations. The intersecting lines are
$n_{1},l_{s^{\prime}t^{\prime}}$ and the crossing lines are
$l_{i},\overline{p_{0}p_{1}}$ for $i=0,1$. This gives points
$a_{0s^{\prime\prime}},a_{1t^{\prime\prime}}\in\overline{p_{0}p_{1}}$ so
$\overline{p_{0}p_{1}}=l_{s^{\prime\prime}t^{\prime\prime}}\in{\mathcal{L}}_{\Pi}$,
and we are done.

That concludes the cases. As they are exhaustive, every pair of points of
$\Pi$ is joined by a line of $\Pi$, which establishes the existence of a
projective plane on $l_{0}$ and $l_{1}$. The plane is unique because by Axiom
(A1) the line joining any two points in $\mathbb{PR}$ is unique so, given
$l_{0}$ and $l_{1}$, there is no choice in constructing $\Pi$. ∎

An interpretation of Theorem 3.3 is the following corollary.

######  Corollary 3.6.

Given three noncollinear ordinary points in a projective rectangle
$\mathbb{PR}$, there is a unique full projective plane in $\mathbb{PR}$ that
contains all three points.

Given an ordinary line $l$ and an ordinary point $p$ not in $l$, there is a
unique full projective plane in $\mathbb{PR}$ that contains both.

###### Proof.

For the first part, let the three points be $p,q,r$. No special line contains
all three, so there is one, say $p$, that is not in a special line through the
others. The lines $pq$ and $pr$ are ordinary lines, they are distinct by
noncollinearity of the three points, and they intersect, so by Theorem 3.3
there is a unique full projective plane that contains them and the three
points.

The second part follows by taking $q,r\in l$. ∎

######  Theorem 3.7.

In a projective rectangle, every maximal subplane is full.

###### Proof.

The line set of an incidence subplane $\pi$ contains two ordinary lines
$l_{1},l_{2}$ and its point set contains their intersection point. It follows
from Theorem 3.3 that $\pi$ is a subplane of the full subplane determined by
$l_{1}$ and $l_{2}$. ∎

Thus, maximality and fullness are equivalent for projective subplanes of a
projective rectangle. From now on, when we refer to a _plane_ in a projective
rectangle, we mean a full projective subplane. Also, when we say several lines
are _coplanar_ , we mean there is a plane $\pi$ such that each of the lines
that is ordinary is a line of $\pi$ and for each line $s$ that is special,
$s\cap\pi$ is a line of $\pi$.

We can now characterize a nontrivial projective rectangle as a projective
rectangle that contains more than one maximal projective subplane. Such
projective rectangles have properties not common to all projective planes;
e.g., they satisfy the dual half of Desargues’s Theorem (see Theorem 5.2) and
they are harmonic matroids (see [5]).

######  Corollary 3.8.

Let $\mathbb{PR}$ be a projective rectangle. Every ordinary line in
$\mathbb{PR}$ is a line of a plane in $\mathbb{PR}$. If $\mathbb{PR}$ is
nontrivial, then every ordinary line $l$ is a line of at least three planes
that contain $l$.

###### Proof.

Let $l$ be an ordinary line in $\mathbb{PR}$. From Theorem 3.1 Part (c) we
know that there is another ordinary line $l^{\prime}$ that intersects $l$ at
exactly one point. This and Theorem 3.3 imply that $l$ is in a plane $\pi$.

If $\mathbb{PR}$ is nontrivial, there is a point $q$ not in $\pi$. Let
$p_{1},p_{2}\in\pi$ be points in $l$ that are not in the special line that
contains $q$. Then the plane $\overline{p_{1}p_{2}q}$ that contains both
ordinary lines $\overline{p_{1}q}$ and $\overline{p_{2}q}$, which exists and
is unique by Theorem 3.3, is a plane containing $l$ that is different from
$\pi$. To find a third plane, let $p_{1}\in\pi_{1}$ and $p_{2}\in\pi_{2}$ be
ordinary points not in $l$. There is an ordinary line $\overline{p_{1}p_{2}}$
that must contain a third point $p_{3}$ since $m\geq 2$ by Theorem 3.1. By
Corollary 3.6 there is a unique plane $\pi_{3}$ that contains $l$ and $p_{3}$.
∎

######  Corollary 3.9.

If $s$ is a special line in the projective rectangle $\mathbb{PR}$ and $\pi$
is a plane in $\mathbb{PR}$, then $s\cap\pi$ is a line of $\pi$.

###### Proof.

Let $p_{1}$ and $p_{2}$ be points in distinct special lines that are not $s$.
Then by Axiom (A6) there is an ordinary line $l$ that contains both $p_{1}$
and $p_{2}$, and by Corollary 3.8 there is a plane $\pi$ that contains $l$. In
$\pi$ there is another line $l^{\prime}$ that intersects $l$ at $p_{1}$; then
$q=l\cap s$ and $q^{\prime}=l^{\prime}\cap s$ are two points in $s\cap\pi$,
which determine a line in $\pi$ that is contained in the unique line $s$ of
$\mathbb{PR}$ that contains $q$ and $q^{\prime}$. Thus, $s\cap\pi$ is a line
of $\pi$. ∎

Now we prove a generalization of Theorem 3.3 to all lines, although we lose
uniqueness of the containing plane.

######  Corollary 3.10.

Let $\mathbb{PR}$ be a projective rectangle. If two lines $l_{1}$ and $l_{2}$
intersect in a point $p$, then they are coplanar.

###### Proof.

Suppose $l_{1}$ is a special line. There are points $p_{1}$ in
$l_{1}\smallsetminus l_{2}\smallsetminus D$ and $p_{2}$ in
$l_{2}\smallsetminus l_{1}$. By Axiom (A6) there is an ordinary line $l_{3}$
determined by $p_{1}$ and $p_{2}$.

If $l_{2}$ is ordinary, by Theorem 3.3 there is a unique plane $\pi$ that
contains $l_{2}$ and $l_{3}$. By Proposition 3.9 the restriction of $l_{1}$ to
$\pi$ is a line of $\pi$, so $l_{1}$ and $l_{2}$ are coplanar.

If $l_{2}$ is special, then $l_{3}$ is ordinary. By Proposition 3.8 there is a
plane $\pi$ that contains $l_{3}$, and by Proposition 3.9 both $l_{1}\cap\pi$
and $l_{2}\cap\pi$ are lines of $\pi$. Thus, $l_{1}$ and $l_{2}$ are coplanar.
∎

Next is an intersection property of lines that has a consequence for the
matroid structure of a projective rectangle.

######  Proposition 3.11.

Suppose three lines in a projective rectangle $\mathbb{PR}$ intersect pairwise
in three different points. Then they are a coplanar triple.

Equivalently, if three lines intersect pairwise (i.e., are pairwise coplanar)
but are not a coplanar triple, then they all intersect in the same point.

###### Proof.

Suppose two ordinary lines $l_{1},l_{2}$ intersect in a point $p$ and lie in a
common plane $\pi$, and suppose a third line $l_{3}$, possibly special,
intersects $l_{1}$ and $l_{2}$ in points different from $p$. Choosing any
points $q_{1}\in l_{1}\smallsetminus p$ and $q_{2}\in l_{2}\smallsetminus p$
determines a line of $\pi$ through $q_{1}$ and $q_{2}$. By Construction 3.4
and Theorem 3.7, this line is either an ordinary line of $\mathbb{PR}$ or the
restriction to $\pi$ of a special line of $\mathbb{PR}$. In particular, this
applies to $l_{3}$, hence $l_{1}$, $l_{2}$ and $l_{3}$ are a coplanar triple
of lines of $\mathbb{PR}$.

In case $l_{1}$ is ordinary while $l_{2}$ and $l_{3}$ are special, by
Corollary 3.10 $l_{1}$ and $l_{2}$ are coplanar in a plane $\pi$ and by
Proposition 3.2 $l_{3}\cap\pi$ is a line of $\pi$, so the three lines are
coplanar.

The second statement, which is the contrapositive of the first (and see
Corollary 3.10), is a useful restatement. ∎

######  Proposition 3.12.

If a finite projective rectangle has order $(n,n)$, then it is a projective
plane.

###### Proof.

Because $n=m$, the projective plane of Corollary 3.8 is the whole projective
rectangle. ∎

This proposition does not apply to the infinite case; see Example 6.6.

###  3.3. No Vamos configuration

Figure 3.7. The Vamos matroid and incidence structure. The vertical lines are
coplanar in pairs $l_{1}l_{2}$, $l_{2}l_{3}$, $l_{1}l_{3}$, $l_{1}l_{4}$, and
$l_{3}l_{4}$, but not $l_{2}l_{4}$. No three of them are coplanar. The top and
bottom squares may be planes or not; that does not affect Corollary 3.13.

The Vamos matroid is the matroid of eight points in Figure 3.7. It is one of
the smallest matroids that cannot be represented in a projective geometry; for
that reason it is one of the fundamental matroid examples. However, we shall
not think of it as a matroid but as an incidence structure with eight points
as well as lines and planes. The lines are the solid lines in Figure 3.7 and
the planes are the ones composed of pairs of lines as described in the
caption. (As a matroid a projective rectangle has rank 3 while the Vamos
matroid has rank 4 and therefore it is trivial that it cannot be a submatroid
of a projective rectangle. That is why it is important to think of the Vamos
incidence structure instead of the Vamos matroid, even though they look the
same in a diagram.)

######  Corollary 3.13.

The Vamos incidence structure is not a substructure of any projective
rectangle.

###### Proof.

Suppose a configuration of this kind exists in a projective rectangle. By
Proposition 3.11 the lines $l_{1},l_{2},l_{3}$ are concurrent in a point and
the lines $l_{2},l_{3},l_{4}$ are also concurrent in a point. Clearly, these
points are one point, so $l_{1}$ and $l_{3}$ contain a common point and hence
are coplanar, contrary to the structure of the Vamos matroid. That proves the
corollary. ∎

##  4\. Finite projective rectangles

In finite projective rectangles there are many possibilities for counting
elements and configurations. They are the topic of this section.

###  4.1. Counts

We extend the counts of points, lines, etc. in Section 3.1 to planes and
various kinds of incidence.

######  Theorem 4.1.

Let $\mathbb{PR}$ be a finite projective rectangle of order $(m,n)$.

  1. (a)

The number of ordinary lines that are concurrent with each ordinary line is
$(m+1)(n-1)$.

  2. (b)

There are $m(m+1)$ ordinary points and $m^{2}$ ordinary lines in each plane.

  3. (c)

The number of pairs $(p,l)$ that consist of an ordinary point $p$ and an
ordinary line $l$ that contains $p$ is $(m+1)n^{2}$.

  4. (d)

The number of planes that contain each ordinary line is $\dfrac{n-1}{m-1}.$

  5. (e)

The number of pairs $(l,\pi)$ such that $l$ is an ordinary line and $\pi$ is a
plane that contains $l$ is $n^{2}\dfrac{n-1}{m-1}.$

  6. (f)

The number of planes in $\mathbb{PR}$ is $\dfrac{n^{2}(n-1)}{m^{2}(m-1)}.$

  7. (g)

For a fixed ordinary point $p$, the number of triples $(p,l,\pi)$ such that
$l$ is an ordinary line incident with $p$ and $\pi$ is a plane that contains
$l$ is $n\dfrac{n-1}{m-1}.$

  8. (h)

The number of triples $(p,l,\pi)$ such that $p$ is an ordinary point, $l$ is
an ordinary line, and $\pi$ is a plane that contains $l$ is
$(m+1)n^{2}\dfrac{n-1}{m-1}.$

  9. (i)

The number of pairs $(p,\pi)$ such that $p$ is an ordinary point and $\pi$ is
a plane that is incident with $p$ is
$\dfrac{(m+1)n^{2}}{m}\,\dfrac{n-1}{m-1}.$

  10. (j)

The number of planes that are incident with each ordinary point is
$\dfrac{n}{m}\,\dfrac{n-1}{m-1}.$

###### Proof.

Proof of (a). Let $l$ be an ordinary line. From Part (g)) there are $m+1$
points on $l$. From Theorem 3.1 Part (k) we know there are $n-1$ ordinary
lines that intersect $l$ at each point. All those lines are distinct.

Proof of (b). This follows from the fact that the plane is projective of order
$m$. We exclude the one special point $D$ and the $m+1$ special lines in the
plane.

Proof of (c). Each of the $n^{2}$ ordinary lines (Theorem 3.1 Part (j))
contains $m+1$ ordinary points (Part (e)).

Proof of (d). Let $l$ be an ordinary line. From Part (a) there are
$(m+1)(n-1)$ ordinary lines $l^{\prime}$ that intersect $l$ at exactly one
point. Theorem 3.3 guarantees the existence of a unique plane $\pi$ that
contains both $l$ and $l^{\prime}$. By Part (b) the number of ordinary lines
in $\pi$ that intersect $l$ is $m^{2}-1=(m+1)(m-1)$. Thus, the number of
planes on $l$ is the quotient, $(m+1)(n-1)/(m+1)(m-1)=(n-1)/(m-1)$.

Proof of (e). The number of ordinary lines should be multiplied by the number
of planes on each line.

Proof of (f). The number of incident line-plane pairs should be divided by the
number of ordinary lines in a plane.

Proof of (g). The number of incident line-plane pairs should be multiplied by
the number of points in an ordinary line.

Proof of (h). The number of triples in Part (g) should be multiplied by the
number of ordinary points from Part (g).

Proof of (i). The number of triples in Part (h) should be divided by the
number of ordinary lines in $pi$ that contain $p$, which is $m$.

Proof of (j). Either divide the number of triples in Part (h) by $m$, the
number of ordinary lines on $p$ in $\pi$, or divide the number in Part (i) by
$(m+1)n$, the whole number of ordinary lines on $p$. ∎

Two lines are _skew_ if they have no point in common. A _skew class_ of lines
is a maximal set of lines, in which every pair is skew. If a line has no skew
mate, it is a skew class of one. A line may belong to more than one skew
class. Two lines that are skew to the same line may intersect.

######  Theorem 4.2.

If $\mathbb{PR}$ is a projective rectangle of finite order $(m,n)$, then the
following hold in $\mathbb{PR}$:

  1. (a)

Given an ordinary point $p$ and given any ordinary line $l$ that does not
contain $p$, there are exactly $n-m$ ordinary lines containing $p$ that are
skew to $l$.

  2. (b)

If $l$ is an ordinary line, then there are $(n-1)(n-m)$ lines that are skew to
$l$.

  3. (c)

If $l_{1}$ is skew to $l$, there are $(m+1)(n-m)$ lines skew to $l$ that are
concurrent with $l_{1}$.

###### Proof.

Proof of Part (a). From Theorem 3.1 Part (h) we know that there are exactly
$n+1$ lines passing through $p$ (including a special line). From Theorem 3.1
Part (c) we also know that there are exactly $m+1$ lines passing through $p$
that intersect $l$ (including a special line). Therefore, there are exactly
$n-m$ ordinary lines passing through $p$ and skew to $l$.

Part (b) follows by subtracting from the number of ordinary lines, $n^{2}$
(Theorem 3.1 Part (j)), the number that are concurrent with $l$, which is
$(m+1)(n-1)$ (Theorem 4.1 Part (a)), and the number that are $l$, which is
$1$.

Part (c) follows from Part (a). ∎

######  Proposition 4.3.

Suppose that $\mathbb{PR}$ is a nontrivial projective rectangle of finite
order $(m,n)$. Let $l$ be an ordinary line in $\mathbb{PR}$. Tthere is a skew
line class containing $l$ that has at least $m+1$ lines in it.

Let $M=\lceil(n-m)/m\rceil-m+1$, the largest integer such that $n/m>m+1+M$.
Then there is a skew class containing $l$ that has at least $m+1+M$ lines in
it.

###### Proof.

Let $l$ be an ordinary line and let $l_{1}\neq l$ be an ordinary line passing
though $q\in l$. Let $p\neq q$ be a second point in $l$. By Theorem 4.2 Part
(a), since $n>m$ there is an ordinary line $l_{2}$ passing through $p$ skew to
$l_{1}$. Let $a_{i}$ and $b_{i}^{\prime}$ be the points in $l_{1}$ and $l_{2}$
for $i=1,2,\dots,m+1$, labeled so that the line $a_{i}b_{i}^{\prime}$ is
special. Lines $a_{i}b_{i}$ and $a_{j}b_{j}$ for $i,j\in\\{1,2,\dots,m+1\\}$
with $i\neq j$, $b_{i}\neq b_{j}$, $b_{i}\neq b_{i}^{\prime}$, and $b_{j}\neq
b_{j}^{\prime}$ are ordinary and are skew to each other, because if they
intersect, then by Axiom (A6), $l_{1}$ intersects $l_{2}$, which is a
contradiction. Note that it is easy to choose all $b_{i}\neq b_{i}^{\prime}$
since $m\geq 1$. Also, we can suppose that $l$ is the line $a_{1}b_{1}$.

Now we suppose that $n/m-m-1>0$ and $M$ is the largest integer such that
$n/m>m+1+M$. (Thus, $n>m+M$.) Let $s$ be a special line with points
$s_{1},s_{2},\dots,s_{m},\dots,s_{n},D$. Suppose that $s\cap a_{i}b_{i}=s_{i}$
for $i=1,\dots,m+1$. We prove by induction that there are lines
$h_{1},h_{2},\dots,h_{M}$, skew to one other and to all lines of the form
$a_{i}b_{i}$. Assume we have $k$ lines $h_{1},h_{2},\dots,h_{k}$ that are skew
to one other and to all lines of the form $a_{i}b_{i}$ for some
$k\in\\{0,1,\dots,M-1\\}$, where $s_{m+1+t}\in h_{t}$ for $t=1,2,\dots,k$.
First note that neither $h_{t}$ nor $a_{i}b_{i}$ contains the point
$s_{m+1+k+1}$ and that $m(m+1+k)$ is the number of points in
$(\bigcup_{t=1}^{k}h_{t}\cup\bigcup_{i=1}^{m+1}a_{i}b_{i})\smallsetminus s$.
Thus, the maximum number of ordinary lines passing through $s_{m+1+k+1}$
intersecting a line of the form $a_{i}b_{i}$ and the lines $h_{1},\dots,h_{k}$
is $m(m+1+k)$. Since $s_{m+1+k+1}$ is an ordinary point, by Theorem 3.1 Part
(3.1) we know there are $n$ ordinary lines passing through this point. Since
$n>m(m+1+k)$ there must be at least one ordinary line $h_{k+1}$ passing
through $s_{m+1+k+1}$ that is skew to all lines of the form $a_{i}b_{i}$ and
the lines $h_{1},\dots,h_{k}$. This proves the induction, completing the
proof. ∎

In the notation of Theorem 4.5, $M=(\tau-1)(m+1)-2\tau$. This is negative or
zero if $\tau=1$, or if $\tau=2$ and $m\leq 3$, and positive otherwise, so in
the “otherwise” case the second bound on the maximum size of the skew class is
the better one.

###  4.2. Constraints on the parameters

We have found some integers in Theorem 4.1, namely,

$\rho=\dfrac{n-1}{m-1},\quad\dfrac{n}{m}\,\dfrac{n-1}{m-1},\text{ and
}\dfrac{n^{2}}{m^{2}}\,\dfrac{n-1}{m-1}.$

These integral fractions imply relationships between $m$ and $n$. Theorem 4.5
is a constraint on $n$, given a value of $m$. By Section 3.2 $m$ must be the
order of a projective plane; that is the only constraint we know on $m$.
Recall that ${\mathcal{P}}$ is the point set of the projective rectangle.

######  Proposition 4.4.

Let $p,p^{\prime}$ be two ordinary points in a special line $s$ of a finite or
infinite projective rectangle. The planes $\pi$ that contain both $p$ and
$p^{\prime}$ partition ${\mathcal{P}}\smallsetminus s$ into sets
$\pi\smallsetminus s$ of size $m^{2}$. For each other special line
$s^{\prime}\neq s$, the same planes partition $s^{\prime}\smallsetminus D$
into sets $\pi\cap(s^{\prime}\smallsetminus D)$ of size $m$, and each such set
is in a unique plane that contains $p$ and $p^{\prime}$. When $m,n$ are finite
there are $n/m$ such planes.

###### Proof.

For an ordinary point $q\in s^{\prime}$ let $\pi(q)$ denote the plane that
contains $p,p^{\prime},q$. This plane is unique, by Theorem 3.3, because it is
determined by the intersecting ordinary lines $pq$ and $p^{\prime}q$. Choose
another ordinary point $q^{\prime}\in s^{\prime}\smallsetminus\pi(q)$ and
suppose $\pi(q)$ and $\pi(q^{\prime})$ contain a common point $r\notin s$.
Then both planes contain the intersecting ordinary lines $pr$ and
$p^{\prime}r$, so they must be the same plane. It follows that the distinct
planes $\pi(q)$ for $q\in s^{\prime}\smallsetminus D$ partition the points not
in $s$ and in particular those of $s^{\prime}\smallsetminus D$. The
intersection $\pi(q)\cap s^{\prime}$ is a line of $\pi(q)$ that contains $D$,
so the number of ordinary points in it is $m$ and the total number of points
in $\pi(q)\smallsetminus s$ is $m^{2}$. In the finite case the number of sets
into which $s^{\prime}\smallsetminus D$ is partitioned is therefore equal to
$n/m$, and this is the number of planes that contain both $p$ and
$p^{\prime}$. ∎

######  Theorem 4.5.

For a projective rectangle $\mathbb{PR}$ of finite order $(m,n)$, there is an
integer $\tau\geq 0$ such that $n=m+\tau m(m-1)$. If $\mathbb{PR}$ is
nontrivial, then $\tau\geq 1$.

###### Proof.

Integrality of $(n-1)/(m-1)$ implies that there is an integer $\rho\geq 1$
such that $n=1+\rho(m-1)$. Proposition 4.4 implies that $n=\sigma m$ for some
positive integer $\sigma$. Therefore, $n=\rho(m-1)+1=\sigma m$. It follows
that $(\rho-\sigma)m=\rho-1$, so $\rho-1$ is a multiple of $m$, say $\rho=\tau
m+1$ where $\tau\geq 0$. Then substituting for $\rho$ gives $(\tau
m+1-\tau)m=\sigma m$, and upon division by $m$ we find that
$\sigma=\tau(m-1)+1$. This implies $n=\tau m(m-1)+m$, so $n-m=n-m=\tau
m(m-1)$. ∎

We infer the expressions

$\begin{gathered}\dfrac{n-1}{m-1}=\tau m+1,\quad\dfrac{n}{m}=\tau(m-1)+1,\\\
\dfrac{n}{m}\,\dfrac{n-1}{m-1}=[\tau(m-1)+1][\tau m+1],\\\
\dfrac{n^{2}}{m^{2}}\,\dfrac{n-1}{m-1}=[\tau(m-1)+1]^{2}[\tau
m+1].\end{gathered}$ (4.1)

######  Corollary 4.6.

If the projective rectangle is nontrivial, $n\geq m^{2}$ and $\rho\geq m+1$.

######  Example 4.7.

If the projective rectangle has $m=2$, then $n=2+2\tau$, where $\tau\geq 0$.
The value $\tau=0$ gives the Fano plane and $\tau=1$ gives $n=4$ as with the
$L_{2}^{2}$ projective rectangle of Example 2.1. However, not all those values
of $\tau$ admit a projective rectangle with $m=2$; there are examples only for
$n=2^{k}$, that is, for $\tau=2^{k-1}-1$ (see Section 7). Our numerical
constraints need strengthening.

##  5\. Axial and central Desargues’s theorems

Consider two triangles in a projective rectangle, $A=\triangle
a_{1}a_{2}a_{3}$ and $B=\triangle b_{1}b_{2}b_{3}$. (A triangle consists of
three points, not all collinear, and the three lines joining the points in
pairs.) We always assume the six vertices $a_{i}$ and $b_{i}$ are distinct.
There are three lines $l_{i}=\overline{a_{i}b_{i}}$; if they concur in a point
$p$ we say the triangles are _centrally perspective from center $p$_. (The
three lines are determined by the subscript correspondence of the two
triangles’ vertices.) If each of the three pairs of lines
$\overline{a_{i}a_{j}}$ and $\overline{b_{i}b_{j}}$ meets in a point $p_{ij}$
and the points $p_{12},p_{13},p_{23}$ are collinear in a line $l$, we say $A$
and $B$ are _axially perspective from axis $l$_. The Central Desargues’s
Theorem says that, if two triangles are centrally perspective, then they are
axially perspective. The converse is the Axial Desargues’s theorem. The two
together are generally known as Desargues’s Theorem.

In a projective plane the points $p_{ij}$ always exist. However, neither half
of Desargues’s Theorem is valid in every projective plane; in fact the
validity of Desargues’s Theorem is equivalent to the existence of plane
coordinates in a division ring. Thus, for any plane, knowing whether
Desargues’s theorem holds true is a fundamental question.

Every projective plane is a projective rectangle, so we cannot say that
Desargues’s Theorem holds true in every projective rectangle; but eliminating
projective planes from consideration changes the situation. We first establish
that each triangle in the axial configuration is contained in a plane.

######  Lemma 5.1.

If $A=\triangle a_{1}a_{2}a_{3}$ is a triangle and $l$ is a line that
intersects the three lines $\overline{a_{i}a_{j}}$ in three points $p_{ij}$,
then all six points and the four lines are contained in a unique plane.

###### Proof.

There are four lines in the configuration of six points: $l$ and the lines
$l_{ij}=\overline{a_{i}a_{j}}$. At most two can be special, so two are
ordinary, say $l^{\prime}$ and $l^{\prime\prime}$. Any two of the four lines
intersect, so $l^{\prime}$ and $l^{\prime\prime}$ intersect; this implies they
are in a unique plane $\pi$ (by Theorem 3.3). The other two lines of the four
are each determined by one point in $l$ and one in $l^{\prime}$, so each is a
line of $\pi$, or if special the intersection with $\pi$ is a line of $\pi$. ∎

First we prove that a nontrivial projective rectangle $\mathbb{PR}$ satisfies
the Axial Desargues’s Theorem when the axis is an ordinary line. We do not
know whether the assumption that the axis is ordinary can be removed.

######  Theorem 5.2 (Ordinary Axial Desargues’s Theorem).

In a nontrivial projective rectangle $\mathbb{PR}$, if two triangles are
axially perspective and the axis is an ordinary line, then they are centrally
perspective.

Figure 5.1. Triangles $A$ and $B$ that are axially perspective, with the extra
triangle $C$. $A$ and $B$ may or may not be in the same plane.

###### Proof.

We begin by assuming triangles $A=\triangle a_{1}a_{2}a_{3}$ and $B=\triangle
b_{1}b_{2}b_{3}$ are in planes $\pi_{A}$ and $\pi_{B}$, respectively, and are
axially perspective from an ordinary line $l$ with intersection points
$p_{ij}$, as in Figure 5.1. The two planes may be the same or different; if
they are different, $l$ is their intersection.

If $\overline{a_{1}b_{1}},\overline{a_{2}b_{2}},\overline{a_{3}b_{3}}$ are not
all coplanar, they are coplanar in pairs, since
$a_{i},b_{i},a_{j},b_{j}\in\overline{p_{ij}a_{i}a_{j}}$. Hence, by Proposition
3.11 there is a point $q$ at which all three lines are concurrent; therefore,
$q$ is a center of perspectivity for $A$ and $B$. Thus, we assume henceforth
that $\overline{a_{1}b_{1}},\overline{a_{2}b_{2}},\overline{a_{3}b_{3}}$ are
all in one plane, so that $\pi_{A}=\pi_{B}$.

There is another plane $\pi_{C}$ on $l$ because $\mathbb{PR}$ is nontrivial
and $l$ is ordinary (by Corollary 3.8), and in this plane we can find a
triangle $C=\triangle c_{1}c_{2}c_{3}$ that is axially perspective from $l$
with the same intersection points $p_{ij}=l\cap\overline{c_{i}c_{j}}$.

The lines $\overline{b_{i}c_{i}}$ and $\overline{b_{j}c_{j}}$ are coplanar in
a plane $\overline{p_{ij}b_{i}c_{j}}=\overline{b_{i}c_{i}b_{j}c_{j}}$.
Therefore, they intersect in a point $s_{ij}$. The pairwise coplanar lines
$\overline{b_{1}c_{1}}$, $\overline{b_{2}c_{2}}$, and $\overline{b_{3}c_{3}}$
are not all coplanar because $\overline{c_{1}c_{2}c_{3}}=\pi_{C}\not\ni
b_{1},b_{2},b_{3}$. By Proposition 3.11, those three lines have a common point
$s=s_{12}=s_{13}=s_{23}$. See Figure 5.2.

Similarly, there is a point
$r=\overline{a_{1}c_{1}}\cap\overline{a_{2}c_{2}}\cap\overline{a_{3}c_{3}}$.

Figure 5.2. The three planes $\overline{a_{i}b_{i}c_{i}}$ and the points $r$
and $s$.

We prove that $r\neq s$ and $r,s\notin\pi_{A}$. If $r=s$, then
$\overline{a_{i}c_{i}}=\overline{ra_{i}c_{i}}=\overline{rc_{i}}$ and
$\overline{b_{i}c_{i}}=\overline{sb_{i}c_{i}}=\overline{rc_{i}}$, so
$\overline{ra_{i}c_{i}}$ and $\overline{rb_{i}c_{i}}$ are the same line; that
is, $a_{i},b_{i},c_{i}$ are collinear; but this is impossible. If $r$ or
$s\in\pi_{A}$ then $c_{1},c_{2},c_{3}\in\pi_{A}$, which contradicts
$\pi_{C}\neq\pi_{A}$.

Each plane $\overline{a_{i}b_{i}c_{i}}$ contains $r$ and $s$ so the lines
$\overline{a_{i}b_{i}}$ and $\overline{rs}$ are coplanar. We know that
$r,s\notin\overline{a_{i}b_{i}}\subset\pi_{A}$. Hence, we have three triples
$\overline{a_{i}b_{i}},\overline{a_{j}b_{j}},\overline{rs}$ of lines that are
coplanar in pairs but not all coplanar. By Proposition 3.11 there is a point
$q_{ij}$ at which each triple is concurrent. Then taking $i=1$ and $j=2,3$, we
have $q_{12}=\overline{rs}\cap\overline{a_{1}b_{1}}=q_{13}$, so
$q_{12}=q_{13}$ is a point on all three lines
$\overline{a_{1}b_{1}},\overline{a_{2}b_{2}},\overline{a_{3}b_{3}}$ and a
center of perspectivity for $A$ and $B$.

That completes the proof. ∎

The case in which $A$ and $B$ are not coplanar is reminiscent of the higher-
dimensional Desargues’s Theorem for projective geometries. That suggests a
central Desargues’s Theorem for noncoplanar triangles. We are able to prove a
mildly limited theorem.

######  Theorem 5.3 (Ordinary Higher Central Desargues’s Theorem).

Let $\mathbb{PR}$ be a nontrivial projective rectangle. If two triangles in
different planes are centrally perspective by ordinary lines, then they are
axially perspective.

###### Proof.

We begin by assuming triangles $A$ and $B$ are in two different planes,
$\pi_{A}$ and $\pi_{B}$ respectively, and are centrally perspective from a
point $p$, such that all the lines $\overline{pa_{i}b_{i}}$ of perspectivity
are ordinary.

OLD
Let $l_{i}:=\overline{a_{i}b_{i}}$ (which exists and contains $p$ by central
perspectivity), $p_{ij}:=\overline{a_{i}a_{j}}\cap\overline{b_{i}b_{j}}$
(which exists because $a_{i},b_{i},a_{j},b_{j},p$ are coplanar and any
noncollinear three of them, excluding $D$ if one of them is not ordinary,
determine the plane), and $\lambda_{ij}:=\overline{p_{ik}p_{jk}}$ where
$\\{i,j,k\\}=\\{1,2,3\\}$. The lines $\lambda_{ij}$ exist because if
$p_{ij}=p_{ik}$ ($i,j,k$ all different), then this point is the intersection
of $\overline{a_{i}a_{j}}$ and $\overline{a_{i}a_{k}}$ but that intersection
is $a_{i}$, and it is also the intersection of $\overline{b_{i}b_{j}}$ and
$\overline{b_{i}b_{k}}$ but that intersection is $b_{i}$, from which it
follows that $a_{i}=b_{i}$, contrary to our standing assumption that all six
vertices are distinct.

Now we observe that all points $p_{ij}\in\pi_{A}\cap\pi_{B}$, so all lines
$\lambda_{ij}\subseteq\pi_{A}\cap\pi_{B}$. But as we assumed
$\pi_{A}\neq\pi_{B}$, their intersection cannot consist of more than one line.
It follows that $\lambda_{12}=\lambda_{13}=\lambda_{23}$ and this is the
required axis of perspectivity.

NEW, ?
Let $l_{i}:=\overline{a_{i}b_{i}}$, which exists and contains $p$ by central
perspectivity. Then $l_{i}\cup l_{j}$, which are ordinary lines that intersect
in $p$, lie in a plane $\pi_{ij}$ and
$p_{ij}:=\overline{a_{i}a_{j}}\cap\overline{b_{i}b_{j}}$, which exists because
$a_{i},b_{i},a_{j},b_{j},p$ are coplanar in $\pi_{ij}$ and any noncollinear
three of them, excluding $D$ if one of them is not ordinary, determine the
plane. All points $p_{ij}\in\pi_{A}\cap\pi_{B}$. As we assumed
$\pi_{A}\neq\pi_{B}$, their intersection cannot consist of more than one line;
it is either an ordinary line or a subset of a special line. This line is the
required axis of perspectivity. ∎

Theorem 5.3 reinforces our belief that a nontrivial projective rectangle
should be regarded as, in a strange way, nonplanar. Unfortunately, we were not
yet able to make this intuition precise.

##  6\. Construction of projective rectangles

We begin with a general formula for constructing projective rectangles inside
projective planes, which we then put to use to prove that many Desarguesian
planes contain nontrivial projective rectangles.

###  6.1. An abstract construction

We found a general construction method for projective rectangles, which
however is too abstract to be readily applicable.

######  Theorem 6.1.

Let $\Pi=({\mathcal{P}},{\mathcal{L}},{\mathcal{I}})$ be a projective plane,
$D$ a point in $\Pi$, and ${\mathcal{S}}$ any subset of the lines on $D$. Let
$R=({\mathcal{P}}_{R},{\mathcal{L}}_{R},{\mathcal{I}}_{R})$ be the incidence
structure with point set ${\mathcal{P}}_{R}=\bigcup_{s\in{\mathcal{S}}}s$,
line set ${\mathcal{L}}_{R}={\mathcal{S}}\cup{\mathcal{O}}_{R}$ where
${\mathcal{O}}_{R}=\\{l\cap{\mathcal{P}}_{R}:l\in{\mathcal{L}}\smallsetminus{\mathcal{S}}\\}$,
and incidence relation as in $\Pi$. Then $R$ is a projective rectangle if and
only if it satisfies the following properties:

  1. (R1)

${\mathcal{S}}$ contains at least $2$ lines.

  2. (R2)

If two lines $l_{1},l_{2}\in{\mathcal{O}}_{R}$ have a common point, then $R$
contains a projective plane of which $l_{1}$ and $l_{2}$ are lines.

First, we show that (R1) can be weakened.

######  Lemma 6.2.

Property (R1) can be replaced by

  1. (R1′)

${\mathcal{S}}$ contains at least $3$ lines.

###### Proof.

Suppose ${\mathcal{S}}$ contains (at least) two lines on $D$. Let $p$ be in
one line of ${\mathcal{S}}$ and $q_{1},q_{2}$ in another, neither one the
special point $D$; then the lines
$l_{1}=\overline{pq_{1}}\cap{\mathcal{P}}_{R}$ and
$l_{2}=\overline{pq_{2}}\cap{\mathcal{P}}_{R}$ are in ${\mathcal{O}}_{R}$ with
a common point. By assumption, $R$ contains a projective plane $\pi$ in which
$l_{1}$ is a line. A line in a projective plane has at least three points.
$\overline{pq_{1}}$ intersects each line $s\in{\mathcal{S}}$ in a point of
$\Pi$, each of which is in ${\mathcal{P}}_{R}$ hence in $l_{1}$, so
${\mathcal{S}}$ must contain at least three lines. ∎

###### Proof of Theorem 6.1..

We review the axioms.

  1. (A1)

Two points in ${\mathcal{P}}_{R}$ generate a line $l$ of $\Pi$. If
$l\in{\mathcal{S}}$, the points are collinear in $R$. If
$l\notin{\mathcal{S}}$, the points are collinear in
$l\cap{\mathcal{P}}\in{\mathcal{O}}_{R}$.

  2. (A2)

Two lines $s_{1},s_{2}\in{\mathcal{S}}$ each contain two points other than
$D$, which make four points of which no three are collinear.

  3. (A3)

is implied by Lemma 6.2.

  4. (A4)

By its definition, ${\mathcal{P}}_{R}$ contains $D$.

  5. (A5)

is implicit in the proof of Lemma 6.2.

  6. (A6)

Let $\pi$ be the projective plane contained in ${\mathcal{P}}_{R}$ of which
$l_{1}$ and $l_{2}$ are lines. If $l_{3}$ or $l_{4}$ is in ${\mathcal{S}}$,
the conclusion of (A6) follows from (A5). Otherwise, $l_{3}$ and $l_{4}$ are
lines of $\pi$ and therefore intersect in a point of
$\pi\subseteq{\mathcal{P}}_{R}$, which is the conclusion of (A6). ∎

The difficulty of applying Theorem 6.1 is that (R2) is hard to verify in
examples.

###  6.2. A subplane construction

Given a projective plane $\pi$ and a subplane $\pi^{\prime}$, we wish to get a
projective rectangle by taking a point $D$, all the lines joining it to points
of $\pi^{\prime}$, all the points on those lines, and all the restrictions to
this point set ${\mathcal{P}}_{R}$ of the lines in $\pi$ that are generated by
our point set ${\mathcal{P}}_{R}$ (i.e., contain at least two of our points).

######  Lemma 6.3.

$D$ must be taken in the subplane.

###### Proof.

Suppose $D$ is not in $\pi^{\prime}$. Take a point $p\in\pi^{\prime}$ and the
line $l=\overline{pD}$ in $\pi$. This is supposed to be a special line so it
must be a line of any plane in the projective rectangle; the proof is that
every line of a projective rectangle, thus every line of $\pi^{\prime}$,
intersects every special line (Axiom (A5)), so $l\cap\pi^{\prime}$ cannot be
one point. Therefore $l\cap\pi^{\prime}$ must be a line of $\pi^{\prime}$. Now
consider a second point $p^{\prime}\in\pi^{\prime}\smallsetminus l$. Then $l$
and $l^{\prime}=\overline{p^{\prime}D}$ are both extensions of lines of
$\pi^{\prime}$ so they intersect in $\pi^{\prime}$, but in the projective
rectangle they intersect at $D$; this means $D\in\pi^{\prime}$. ∎

We could simplify the construction: Take a subplane $\pi^{\prime}$ and one
line $l_{0}$ in it, and any point $D$ in $\pi^{\prime}\smallsetminus l_{0}$.
For the projective rectangle, take all lines of $\pi$ that join $D$ to $l_{0}$
and for ${\mathcal{P}}_{R}$ take all points of $\pi$ on those lines. This
gives precisely the subplane construction, because already it gives all the
points of $\pi^{\prime}$ and then only the points generated from $D$ and
$\pi^{\prime}$ in that construction.

We want the subplane construction to give rise to a projective rectangle.
Unfortunately, we have not found a direct proof of this (and it may not be
always true), so our proof calls upon a theorem from the sequel [5] about
harmonic conjugation. This proof is only valid in a plane that has harmonic
conjugation and when the subplane is prime, i.e., isomorphic to
PP$(\mathbb{Z}_{p})$ for a prime number $p$. The planes with harmonic
conjugation are the Moufang planes [13, p. 202]. Since all finite Moufang
planes are Pappian, that is, coordinatized by a (commutative) field [13], any
finite subplane of a Moufang plane is Pappian and a minimal finite subplane of
a Moufang plane is prime.

######  Theorem 6.4.

The subplane construction from a prime subplane in a Moufang projective plane
produces a projective rectangle.

######  Example 6.5.

For use in the proof we define a matroid generalization of $L_{2}^{k}$ that we
call $L_{p}^{\mathbf{V}}$, where $p$ is a prime number and $\mathbf{V}$ is a
vector space of any dimension over $\mathbb{Z}_{p}$. Let

$A:=\\{a_{g}\mid g\in\mathbf{V}\\}\cup\\{D\\},\ B:=\\{b_{g}\mid
g\in\mathbf{V}\\}\cup\\{D\\},\text{ and }C:=\\{c_{g}\mid
g\in\mathbf{V}\\}\cup\\{D\\}.$

All points $a_{g},b_{g},c_{g},D$ are distinct. $L_{p}^{\mathbf{V}}$ is the
simple matroid of rank 3 on the ground set $A\cup B\cup C$ whose rank-2 flats
are the $3$ lines $A$, $B$, $C$ and the $p^{2\dim\mathbf{V}}$ lines
$\\{a_{g},b_{g+h},c_{h}\\}$ with $g$ and $h$ in $\mathbf{V}$.

(This matroid is the complete lift matroid $L_{0}(\mathbf{V}{\cdot}K_{3})$ in
[16, Section 3].)

###### Proof.

The subplane construction produces a family of lines through $D$ whose point
set ${\mathcal{P}}_{R}$ constitutes the putative projective rectangle. The
Moufang plane $\pi$ is coordinatized by an alternative ring that is an algebra
over the coordinate field $\mathbb{Z}_{p}$ of $\pi^{\prime}$; we let
$\mathbf{V}$ denote the $\mathbb{Z}_{p}$-vector space of that ring. Any three
lines through $D$ in the subplane $\pi^{\prime}$ form a matroid isomorphic to
$L_{p}^{\mathbf{V}}$ ([17, Section 4]). According to [5, Theorem 5.1], by
harmonic conjugation in $\pi$ we obtain a projective rectangle $\mathbb{PR}$
whose point set is the harmonic closure of $A\cup B\cup C$, whose short lines
have order $p$ (that is, $p+1$ points) and whose long lines are the lines
$\overline{pD}$ of $\pi$ for $p\in\pi^{\prime}\smallsetminus D$. ∎

######  Example 6.6 (Some Infinite Projective Rectangles).

Consider an infinite projective plane that contains a countably infinite
subplane, such as $\mathbb{PP}(\mathbb{R})$ with $\mathbb{PP}(\mathbf{F})$ for
any countably infinite field $\mathbf{F}$. Theorem 6.6 of [5] tells us that
the harmonic closure of $\mathbb{PP}(\mathbb{Q})$ is a projective rectangle
$\mathbb{PR}$ of order $(|\mathbf{F}|,|\mathbb{R}|)$. The order of
$\mathbb{PR}$ is independent of the choice of $\mathbf{F}$ but the projective
planes in $\mathbb{PR}$ are Pappian with coordinate field $\mathbf{F}$, which
is certainly not independent of $\mathbf{F}$. These examples demonstrate that
the order of an infinite projective rectangle does not determine the
rectangle.

A similar conclusion holds more generally for any uncountably infinite Moufang
plane that contains $\mathbb{PP}(\mathbf{F})$ as a subplane.

##  7\. Narrow rectangles

The smallest allowed value of $m+1$ is 3. We call a projective rectangle
_narrow_ if it has $m=2$. We classify the narrow projective rectangles using
some matroid theory.

A matroid like $L_{2}^{k}$ of Example 2.1 is defined for any nontrivial
quasigroup $\mathfrak{G}$, simply replacing $\mathbb{Z}_{2}^{k}$ by
$\mathfrak{G}$; it is the complete lift matroid $L_{0}(\mathfrak{G}K_{3})$
from [16] or [4]). We define it in a way compatible with Example 2.1. The
ground set is $E:=A\cup B\cup C$ where $A:=\left\\{a_{g}\mid
g\in\mathfrak{G}\right\\}\cup\\{D\\}$, $B:=\left\\{b_{g}\mid
g\in\mathfrak{G}\right\\}\cup\\{D\\}$ and $C:=\left\\{c_{g}\mid
g\in\mathfrak{G}\right\\}\cup\\{D\\}$. The lines (rank-2 flats of the matroid)
are $A$, $B$, and $C$ and the sets $\\{a_{g},b_{gh},c_{h}\\}$ with
$g,h\in\mathfrak{G}$. If this is a projective rectangle, $A$, $B$, and $C$ are
the special lines and the other lines are the ordinary lines. If
$L_{0}(\mathfrak{G}K_{3})$ is a projective rectangle, it is narrow, and
$\mathfrak{G}$ can only be certain groups.

######  Proposition 7.1.

Every narrow projective rectangle $\mathbb{PR}$ has the form
$L_{0}(\mathfrak{G}K_{3})$ where $\mathfrak{G}$ is a nontrivial group with
exponent $2$, and conversely. If $\mathbb{PR}$ is finite the group is
$\mathbb{Z}_{2}^{k}$ with $k\geq 1$ and its parameters are $(m,n)=(3,2^{k}+1)$
with $k\geq 1$.

This proposition includes infinite groups.

###### Proof.

First we note that every narrow projective rectangle $\mathbb{PR}$ is an
$L_{0}(\mathfrak{G}K_{3})$ where $\mathfrak{G}$ is a quasigroup of order
greater than 1. There are three special lines, which we call $A$, $B$, and
$C$. We label the elements of each line, except $D$, by a set $\mathfrak{G}$
of labels and we define an operation on $\mathfrak{G}$ by $gh=k$ such that
$a_{g}c_{h}b_{k}$ is an ordinary line of $\mathbb{PR}$. It is clear that this
is well defined and that any two of $g,h,k$ determine the third, so
$\mathfrak{G}$ is a quasigroup. Then $\mathbb{PR}$ is the same as
$L_{0}(\mathfrak{G}K_{3})$.

Now let $\mathfrak{G}$ be a quasigroup, and assume that
$L_{0}(\mathfrak{G}K_{3})$ is a projective rectangle. We prove that
$\mathfrak{G}$ satisfies the following fundamental property:

$gh=ef\implies gf=eh.$ (7.1)

Consider the lines $l_{1}=\\{a_{g},b_{gh},c_{h}\\}$ and
$l_{2}=\\{a_{e},b_{ef},c_{f}$ in Axiom (A6), and two other lines,
$l=\\{a_{g},b_{gf},c_{f}\\}$ and $l^{\prime}=\\{a_{e},b_{eh},c_{h}\\}$.
According to Axiom (A6) the lines $l$ and $l^{\prime}$ should have a common
point, so $b_{gf}=b_{eh}$, which means $gf=eh$.

The matroid structure of $L_{0}(\mathfrak{G}K_{3})$, hence the structure of
the corresponding projective rectangle, is not affected by isotopy of
$\mathfrak{G}$ because it is entirely determined by the 3-point lines
$\\{a_{g},b_{gh},c_{h}\\}$, in which isotopy only permutes subscript names
without changing their algebraic relation. Thus, we may replace $\mathfrak{G}$
by any convenient isotope of itself. In particular, any quasigroup is isotopic
to a loop (a quasigroup with identity element, 1), so we may assume
$\mathfrak{G}$ is a loop. Suppose $h=e=1$ in Equation (7.1). Then $g=f\implies
gf=1$; in other words, $gg=1$ for every element of $\mathfrak{G}$. Suppose
$g=h$ and $e=f$. Then $1=1\implies ge=eg$; that is, $\mathfrak{G}$ is
commutative. A property that characterizes a quasigroup that is isotopic to a
group is the Quadrangle Criterion [1], which is

$\left.\begin{array}[]{l}a_{1}c_{1}=a_{2}c_{2}\\\ a_{1}d_{1}=a_{2}d_{2}\\\
b_{1}c_{1}=b_{2}c_{2}\end{array}\right\\}\implies b_{1}d_{1}=b_{2}d_{2}.$

We prove the Quadrangle Criterion for $\mathfrak{G}$ by means of Equation
(7.1).

$\displaystyle a_{1}c_{1}=a_{2}c_{2}$ $\displaystyle\implies
a_{1}a_{2}=c_{1}c_{2},$ $\displaystyle a_{1}d_{1}=a_{2}d_{2}$
$\displaystyle\implies a_{1}a_{2}=d_{1}d_{2},$ $\displaystyle
b_{1}c_{1}=b_{2}c_{2}$ $\displaystyle\implies b_{1}b_{2}=c_{1}c_{2}.$

The first two lines imply that $c_{1}c_{2}=d_{1}d_{2}$ and combined with the
third line we deduce that $b_{1}b_{2}=d_{1}d_{2}$, proving the Quadrangle
Criterion. Hence, $\mathfrak{G}$ is isotopic to a group. By isotopy we may
assume $\mathfrak{G}$ is a group, and since a group is a loop, the group is
abelian and has exponent 2 (hence may be written additively as in Example
2.1). If $\mathfrak{G}$ is finite, it is $\mathbb{Z}_{2}^{k}$ for some
positive integer $k$ as in Example 2.1. These necessary properties of
$\mathfrak{G}$ are sufficient for $L_{0}(\mathfrak{G}K_{3})$ to be a
projective rectangle, because exponent 2 implies Axiom (A6), as is easy to
verify. ∎

The geometry of a narrow projective rectangle is determined by the isotopy
type of its quasigroup. Thus, the finite such rectangles are obtained from a
finite Pappian projective plane of 2-power order by the subplane construction
of Section 6 using a Fano subplane.

##  8\. Orthogonal arrays from projective rectangles

A _transversal design_ is a partition of a set ${\mathcal{P}}_{T}$ of $(m+1)n$
points into $m+1$ special sets of size $n$ together with a family of
$m+1$-subsets of ${\mathcal{P}}_{T}$ such that each such $m+1$-set intersects
each special set exactly once and each pair of points not contained in a
special set lies in exactly one $m+1$-set. A projective rectangle with $D$
deleted is exactly a transversal design with the extra partial Pasch property
Axiom (A6). A dual concept to transversal designs is that of orthogonal
arrays; the corresponding dual to projective rectangles is orthogonal arrays
with a dual property to (A6). We explore that dual concept in this
section.111We thank Douglas Stinson for drawing our attention to transversal
designs.

An orthogonal array (OA) is a generalization of orthogonal latin squares. We
adopt the notation for orthogonal arrays used in [8]. An $N\times k$ array $A$
with entries from $S$ (a set of size $s$) is said to be an _orthogonal array_
, $OA_{\lambda}(N,k,s,t)$, with $s$ symbols, strength $0\leq t\leq k$, and
index $\lambda$ if every $N\times t$ subarray of $A$ contains each $t$-tuple
based on $S$ exactly $\lambda$ times as a row. We write $a(r,c)$ for the label
that appears in row $r$ and column $c$.

###  8.1. An orthogonal array from points and lines

In order to represent a projective rectangle $\mathbb{PR}$ as an orthogonal
array of points and lines, we formulate a special property for an orthogonal
array of type $OA_{1}(n^{2},m+1,n,2)$.

  1. (OA6)

If four rows in the orthogonal array appear like the first five columns
$c_{ij}$ in this table,

| $c_{12}$ | $c_{13}$ | $c_{24}$ | $c_{14}$ | $c_{23}$ | $c_{34}$
---|---|---|---|---|---|---
$r_{1}$ | $a_{12}$ | $a_{13}$ |  | $a_{14}$ |  |
$r_{2}$ | $a_{12}$ |  | $a_{24}$ |  | $a_{23}$ |
$r_{3}$ |  | $a_{13}$ |  |  | $a_{23}$ | $a_{34}$
$r_{4}$ |  |  | $a_{24}$ | $a_{14}$ |  | $a_{34}$

where it is possible that $c_{13}=c_{24}$ or $c_{14}=c_{23}$, then there is a
sixth column that appears like $c_{34}$. (The empty cells are arbitrary.)

The property (OA6) does not follow from the definition of an orthogonal array.
We are not aware that it has been considered in the theory of orthogonal
arrays or dually in transversal designs. Its contrary, that the sixth column
of (OA6) never appears, arises (in the language of transversal designs) as the
“anti-Pasch configuration” in [2] (whose “Pasch configuration” is slightly
stricter than ours).222We are very grateful to Charles Colbourn for hunting in
the literature and communicating these facts.

######  Theorem 8.1.

Let $n\geq m\geq 2$.

  1. (i)

A projective rectangle $\mathbb{PR}$ of order $(m,n)$ gives rise to an
orthogonal array $OA_{1}(n^{2},m+1,n,2)$ with property (OA6).

  2. (ii)

An orthogonal array $OA_{1}(n^{2},m+1,n,2)$ gives rise to a projective
rectangle $\mathbb{PR}$ of order $(m,n)$ if, and only if, it satisfies the
additional property (OA6).

We note that Part (ii) is a strengthening of the converse of Part (i).

###### Proof.

We begin by proving Part (i). In $\mathbb{PR}\smallsetminus D$ we have $m+1$
special lines partitioning all the points, and $n^{2}$ ordinary lines. By
Theorem 3.1, every ordinary line intersects every special line exactly once
and every pair of points in different special lines lie in exactly one
ordinary line. Each ordinary line will give a row of the orthogonal array and
each special line will give a column. We label the points in each special line
by the numbers $1,\dots,n$ and we write $a(p)$ for the label of the point $p$.
The entries in a row are the labels of the points that appear in that ordinary
line, arranged in the column of the special line that contains the point.
Thus, each pair of labels appears once in each pair of columns. That is a
2-$(n,m+1,1)$ orthogonal array in standard notation. In the notation used in
[8], it is an $OA_{1}(n^{2},m+1,n,2)$.

Property (OA6) is the interpretation of Axiom (A6) for an
$OA_{1}(n^{2},m+1,n,2)$. In Axiom (A6) let $l_{3}$ and $l_{4}$ be the two
lines besides $l_{1}$ and $l_{2}$. The assumption in the axiom is that points
$p_{ij}=l_{i}\cup l_{j}$ exist for $(i,j)=(1,2),(1,3),(2,4),(1,4),(2,3)$. Let
$s_{ij}$ be the special line that contains $p_{ij}$; we note that the special
lines are distinct except that $s_{13}$ may be the same as $s_{24}$ and
$s_{14}$ may be the same as $s_{23}$. In the orthogonal array derived from
$\mathbb{PR}$, the row of line $l_{i}$ is $r_{i}$, the column of line $s_{ij}$
is $c_{ij}$, and the label of $p_{ij}$ is $a(r_{i},c_{ij})=a(r_{j},c_{ij})$.
Therefore, the array looks as in Property (OA6), except for the last column.

The conclusion of Axiom (A6) is that there is a point $p_{34}$ that is
incident with both lines $l_{3}$ and $l_{4}$. That translates to the existence
of a final column as in (OA6) with $a_{34}=a(p_{34})$. Hence, Property (OA6)
is satisfied by the array derived from the projective rectangle $\mathbb{PR}$.

Proof of Part (ii). Suppose we have an $OA_{1}(n^{2},m+1,n,2)$. Let $C$ be the
set of $m+1$ columns, let $R$ be the set of rows, let $L$ be the set of $n$
labels in the array, and write $a(r,c)$ for the entry in row $r$, column $c$.
We form an incidence structure whose point set is $(C\times L)\cup D$. The
lines of this structure are special lines, of the form $s_{c}=\\{(c,a):a\in
L\\}\cup D$, for each $c\in C$, and ordinary lines, of the form
$l_{r}=\\{(c,a):c\in C\text{ and }a=a(r,c)\\}$, for each $r\in R$.

We prove this incidence structure satisfies Axioms (A1)–(A5) of a projective
rectangle. We assumed $n\geq m\geq 2$ so in the orthogonal array there are at
least two distinct labels, which we call $a_{1}$ and $a_{2}$, and at least $3$
columns, of which three are $c_{1},c_{2},c_{3}$. There are also at least
$2^{3}$ rows.

Proof of Axiom (A1). We consider two points $p_{1}=(r_{1},a_{1})$ and
$p_{2}=(r_{2},a_{2})$ where $a_{1}=a(r_{1},c_{1})$ and $a_{2}=a(r_{2},c_{2})$.
The points belong to the same special line if and only if $c_{1}=c_{2}$. The
special line is $s_{c_{1}}$. Otherwise, there is exactly one row $r$ where the
entry in column $c_{1}$ is $a_{1}$ and the entry in column $c_{2}$ is $a_{2}$.
Then $p_{1}$ and $p_{2}$ belong to the ordinary line $l_{r}$.

Proof of Axiom (A2). Among the three pairs $a(r_{1},c_{j}),a(r_{2},c_{j})$ for
$j=1,2,3$, only one can be the same label, $a(r_{1},c_{j})=a(r_{2},c_{j})$,
because each ordered pair of labels appears only once in the same two columns.
Say $a(r_{1},c_{1})\neq a(r_{2},c_{1})$ and $a(r_{1},c_{2})\neq
a(r_{2},c_{2})$. Then
$(c_{1},a(r_{1},c_{1})),(c_{1},a(r_{2},c_{1})),(c_{2},a(r_{1},c_{2})),(c_{2},a(r_{2},c_{2}))$
are four points, no three collinear.

Proof of Axiom (A3). The special line $s_{c}$ contains at least the three
points $D,(c,a_{1}),(c,a_{2})$. The ordinary line $l_{r}$ contains the points
$(c_{1},a(r,c_{1})),(c_{2},a(r,c_{2})),(c_{3},a(r,c_{3}))$.

Proof of Axiom (A4). This follows by the definition of the incidence
structure.

Proof of Axiom (A5). Two special lines intersect only in $D$. A special line
$s_{c}$ and an ordinary line $l_{r}$ intersect only in the point $(c,a(r,c))$.

Finally, we prove Axiom (A6) from Property (OA6). Let $r_{1},r_{2}$ be the
rows of the array that correspond to the lines $l_{1},l_{2}$ in this axiom and
let $l_{3},l_{4}$ be the two other lines with corresponding rows
$r_{3},r_{4}$. The hypotheses of intersection imply that the diagram in
Property (OA6) is satisfied, possibly except for the last column. By the
assumption of Property (OA6), the final column does exist. This implies that
$l_{3}\cap l_{4}$ is the point $p_{34}$ in the special line $s_{34}$ that
corresponds to column $c_{34}$ and has the label $a(p_{34}=a_{34}$. Therefore,
the conclusion of Axiom (A6) is satisfied.

If on the contrary there is a failure of Property (OA6), then the final column
fails to exist in at least one instance and the corresponding configuration in
the rectangle fails to have the intersection point promised by Axiom (A6), so
it is not a projective rectangle. ∎

###  8.2. An orthogonal array from points and planes

Ryser gives a nice construction of an orthogonal array from a projective plane
[12, p. 92]. We extend Ryser’s ideas to construct an orthogonal array from
points and planes of a projective rectangle by partitioning the ordinary
points outside a given ordinary line by means of the separate planes that
contain that line. The proof is based on the proof that Ryser gives for
projective planes, adapted to the existence of multiple planes.

######  Lemma 8.2.

Let $l$ be an ordinary line in a finite $\mathbb{PR}$. The family of sets
$\pi\smallsetminus(l\cup D)$ for all planes $\pi$ that contain $l$ is a
partition of the points in $\mathbb{PR}\smallsetminus(l\cup D)$ into
$(n-1)/(m-1)$ parts of $(m+1)(m-1)$ points each.

###### Proof.

We observe that every plane in $\mathbb{PR}$ containing $l$ also contains the
special point $D$. If $p\not\in l\cup D$, then by Corollary 3.6 there is a
unique plane on $l$ that contains $p$; thus, the planes on $l$ partition the
points in $\mathbb{PR}\smallsetminus(l\cup D)$. The number of such planes is
given by Theorem 4.1 Part (d). The number of parts of the resulting partition
equals the number of planes that contain the line $l$. ∎

######  Theorem 8.3.

Suppose that $(m,n)$ is the order of the projective rectangle $\mathbb{PR}$.
Let $l\in\mathbb{PR}$ be an ordinary line and let
$\pi_{1},\pi_{2},\dots,\pi_{w}$ be all the planes in $\mathbb{PR}$ that
contain $l$, where $w=(n-1)/(m-1)$. Then $\mathbb{PR}$ gives rise to an
orthogonal array of the form $OA_{w}(m^{2}w,m+1,m,2)$.

###### Proof.

Let $p_{1},p_{2},\dots,p_{m}$ be the points of $l$. We label the points in
$\pi_{i}\smallsetminus l$ by $q_{1}^{i},q_{2}^{i},\dots,q_{k}^{i}$ where
$k=m^{2}$ ($D$ is one of these points) and label the lines on $p_{r}$ in
$\pi_{i}\smallsetminus l$ with $1,2,\dots,m-1$ for each $r=1,2,\dots,m$. We
write $a_{st}^{i}$ to record the label of the line $q_{s}^{i}p_{t}\in\pi_{i}$.

We claim that the matrix $A_{i}=[a_{st}^{i}]_{s,t}$ is an orthogonal array of
the form $OA_{1}(m^{2},m,m-1,2)$. We prove this by contradiction. Suppose that
there two ordered pairs in the rows of $A_{i}$ that are equal; that is,
$(a_{s_{1}t_{1}}^{i},a_{s_{1}t_{2}}^{i})=(a_{s_{2}t_{1}}^{i},a_{s_{2}t_{2}}^{i})$
with $s_{1}\neq s_{2}$. Therefore, $a_{s_{1}t_{1}}^{i}=a_{s_{2}t_{1}}^{i}$ and
$a_{s_{1}t_{2}}^{i}=a_{s_{2}t_{2}}^{i}$. The equality of these labels implies
that the points $q_{s_{1}}^{i}$, $q_{s_{2}}^{i}$, and $p_{t_{1}}$ are
collinear and that $q_{s_{1}}^{i}$, $q_{s_{2}}^{i}$, and $p_{t_{2}}$ are also
collinear. Thus, each $p_{t_{j}}$ is the unique point of $l$ on the same line
$\overline{q_{s_{1}}^{i}q_{s_{2}}^{i}}$. Therefore, $p_{t_{1}}=p_{t_{2}}$, but
that is impossible because $t_{1}\neq t_{2}$.

Now let $B=\begin{bmatrix}A_{1}\\\ A_{2}\\\ \vdots\\\ A_{w}\end{bmatrix}$.
This matrix is an orthogonal array of the form $OA_{\lambda}(m^{2}w,m+1,m,2)$
where $\lambda=\sum_{i=1}^{w}1=w$. That completes the proof. ∎

######  Example 8.4.

We give an example for Theorem 8.3 using the projective rectangle $L_{2}^{2}$
depicted in Figure 2.1. For the sake of simplicity we pick the line
$l=\\{a_{1},b_{1},c_{1}\\}$. We recall that for an ordinary line in
$L_{2}^{2}$, there are exactly $\lambda=3$ planes having that line in common.
Figure 8.1 shows the three planes embedded in $L_{2}^{2}$ with $l$ as common
line.

For the first plane, let’s say $\pi_{1}$, we distinguish the points $a_{1}$,
$a_{g}$, $b_{1}$, $b_{g}$, $c_{1}$, $c_{g}$ and $D_{1}:=D$. For a fixed point
in $l$ theres two lines in $\pi_{1}\smallsetminus l$ passing by the fixed
point; from the set $\\{1,2\\}$ we assign labels to these lines. For the lines
$\\{a_{1},a_{g},D_{1}\\}$ and $\\{a_{1},b_{g},c_{g}\\}$, which intersect $l$
at $a_{1}$, we assign $1$ and $2$ to them, respectively. We arbitrarily assign
$1$ and $2$ to $\\{b_{1},b_{g},D_{1}\\}$ and $\\{a_{g},b_{g},c_{g}\\}$,
respectively, and also to $\\{a_{g},b_{g},c_{1}\\}$ and
$\\{c_{g},c_{1},D_{1}\\}$. With these labels we construct the first four rows
of the rectangular array in Table 8.1. The columns of the array are labeled on
top with the points in the line $l$ and the rows are labeled on the left with
the points in each plane that are not in $l$. In this case the first four rows
are labeled with the points in $\pi_{1}\smallsetminus l$. The entries of the
rectangular array are the labels of the lines passing through the point in the
column label and the point in the row label. For instance, the first entry of
the first row in Table 8.1 is $1$, because the line passing through $a_{1}$
and $a_{g}$ has label $1$. The first entry of the fourth row is $1$, because
the line passing through $a_{1}$ and $D$ has label $1$.

The second plane in Figure 8.1, $\pi_{2}$, has the points $a_{1}$, $a_{h}$,
$b_{1}$, $b_{h}$, $c_{1}$, $c_{h}$ and $D_{2}:=D$. As in $\pi_{1}$, we assign
arbitrary labels from $\\{1,2\\}$. We choose $1$ to be the label of
$\\{a_{1},b_{h},c_{h}\\}$, $\\{a_{h},b_{1},c_{h}\\}$, and
$\\{c_{1},c_{h},D_{2}\\}$ and $2$ as the label of $\\{a_{1},a_{h},D_{2}\\}$,
$\\{b_{1},b_{h},D_{2}\\}$, and $\\{a_{h},b_{h},c_{1}\\}$.

For the third plane in Figure 8.1, $\pi_{3}$ with points $a_{1}$, $a_{g+h}$,
$b_{1}$, $b_{g+h}$, $c_{1}$, $c_{g+h}$ and $D_{3}:=D$, we also assign
arbitrary labels from $\\{1,2\\}$. So, for example, $1$ will be the label of
$\\{a_{1},a_{g+h},D_{3}\\}$, $\\{a_{g+h},b_{1},c_{g+h}\\}$, and
$\\{a_{g+h},b_{b+h},c_{1}\\}$ and $2$ will be the label of
$\\{a_{1},b_{g+h},c_{g+h}\\}$, $\\{b_{1},b_{g+h},D_{3}\\}$, and
$\\{c_{1},c_{g+h},c_{1}\\}$.

These give the orthogonal array $OA_{3}(12,3,2,2)$. This is a $12\times 3$
array filled with $2$ symbols, such that in any $2$ columns there are 4
different ordered pairs, each repeated $\lambda=3$ times.

| $a_{1}$ | $b_{1}$ | $c_{1}$
---|---|---|---
$a_{g}$ | $1$ | $2$ | $1$
$b_{g}$ | $2$ | $1$ | $1$
$c_{g}$ | $2$ | $2$ | $2$
$D_{1}$ | $1$ | $1$ | $2$
$a_{h}$ | $2$ | $1$ | $2$
$b_{h}$ | $1$ | $2$ | $2$
$c_{h}$ | $1$ | $1$ | $1$
$D_{2}$ | $2$ | $2$ | $1$
$a_{g+h}$ | $1$ | $1$ | $1$
$b_{g+h}$ | $2$ | $2$ | $1$
$c_{g+h}$ | $2$ | $1$ | $2$
$D_{3}$ | $1$ | $2$ | $2$
Table 8.1. Orthogonal array $OA_{3}(12,3,2,2)$ for $L_{2}^{2}$.

$a_{1}$ $b_{1}$ $c_{1}$ $a_{g}$ $1$ $2$ $1$ $b_{g}$ $2$ $1$ $1$ $c_{g}$ $2$
$2$ $2$ $D_{1}$ $1$ $1$ $2$

$a_{1}$ $b_{1}$ $c_{1}$ $a_{h}$ $2$ $1$ $2$ $b_{h}$ $1$ $2$ $2$ $c_{h}$ $1$
$1$ $1$ $D_{2}$ $2$ $2$ $1$

$a_{1}$ $b_{1}$ $c_{1}$ $a_{g+h}$ $1$ $1$ $1$ $b_{g+h}$ $2$ $2$ $1$ $c_{g+h}$
$2$ $1$ $2$ $D_{3}$ $1$ $2$ $2$

Figure 8.1. Three views of the projective rectangle $L_{2}^{2}$. In each view
the lines of one plane $\pi_{i}$ are highlighted. Next to that view is the OA
$A_{i}$ corresponding to that plane with a small diagram showing the line
labels.

##  9\. The dual incidence structure

The dual structure is obtained by interchanging the roles of points and lines.
It is interesting in its own right, as it connects projective rectangles with
incidence geometry in a different way. The dual is essentially a net with a
complete quadrangle property. Being a dual projective rectangle, it contains
all the dual projective planes of the planes of the original projective
rectangle.

A _net_ $\mathcal{N}$ is an incidence structure
$({\mathcal{P}},{\mathcal{L}},\mathcal{I})$ which consists of a set
${\mathcal{P}}$ of points and a set ${\mathcal{L}}$ of parallel classes
${\mathcal{L}}_{i}$ ($i\in$ an index set) of lines, such that each line is a
set of points, every point belongs to exactly one line of each parallel class,
and any two lines of different parallel classes have exactly one point in
common. The theory of nets is extensive. It is easy to prove that every
parallel class has the same number of lines and that the number of points on
every line is the same.

We call these points and lines _ordinary_. By adding a _special point_ for
each parallel class, which is defined to belong to all lines of that class and
no other ordinary lines, and adding one _special line_ that contains all the
special points, we get a _projectively extended net_. (“Projectively” refers
to the existence of the special line.)

Two points might not be in any common line. They are called _collinear_ if
they are in a line. They cannot be in more than one line.

We state the dualized rectangle axioms.

  1. (A1*)

Every two distinct lines contain exactly one point in common.

  2. (A2*)

There exist four lines in the extended net with no three of them concurrent.

  3. (A3*)

Every point is in at least three distinct lines.

  4. (A4*)

There is a _special line_ $D^{*}$. (A point in $D^{*}$ is called _special_. A
point that is not in $D*$ and a line that is not $D^{*}$ are called
_ordinary_.)

  5. (A5*)

Each special point belongs to exactly one line with each other point.

  6. (A6*)

If two ordinary points $P_{1}$ and $P_{2}$ are collinear, then any two other
points $P_{1}^{\prime}$ and $P_{2}^{\prime}$ that are collinear with $P_{1}$
and $P_{2}$ through four distinct lines (i.e., there are four distinct lines
$P_{i}P_{j}^{\prime}$ for $i,j=1,2$), are themselves collinear.

A _complete quadrangle_ in a net consists of 4 points, no three collinear, and
6 lines determined by them. A _nearly complete quadrangle_ consists of the
same 4 points and 5 of the 6 lines, the 6th line possibly existing or not
existing. The dual Axiom (A6*) can be stated (more elegantly) in terms of
quadrangles:

  1. (CQP)

(Complete Quadrangle Property) Every nearly complete quadrangle is complete.

######  Lemma 9.1.

A net has the complete quadrangle property if and only if its projective
extension satisfies (A6*).

###### Proof.

Let $\mathcal{N}$ be the net and $\overline{\mathcal{N}}$ its projective
extension.

Suppose $\mathcal{N}$ satisfies (CQP). In Axiom (A6*), if $P_{1}^{\prime}$ is
special, it is in a line with every other point including $P_{2}^{\prime}$, so
we only need to consider ordinary points
$P_{1},P_{2},P_{1}^{\prime},P_{2}^{\prime}$. Then (CQP) implies the line
$\overline{P_{1}^{\prime}P_{2}^{\prime}}$ so (A6*) is satisfied by
$\overline{\mathcal{N}}$.

Conversely, assume $\overline{\mathcal{N}}$ satisfies Axiom (A6*). Let
$P_{1},P_{2},P_{1}^{\prime},P_{2}^{\prime}$ be any four points in
$\mathcal{N}$ of which five pairs, including $P_{1}$ and $P_{2}$, are
collinear. (A6*) implies the sixth pair is collinear, thus verifying (CQP) in
$\mathcal{N}$. ∎

######  Theorem 9.2.

The dual of a projective rectangle is a projective extension of a net that has
the complete quadrangle property, at least three parallel classes, and at
least two lines in each parallel class; and conversely such a projective
extension is the dual of a projective rectangle.

###### Proof.

First, we show that the dualized rectangle axioms imply a projective extension
of a net $\mathcal{N}$ with the three stated properties. Each point of
$\mathbb{PR}$ becomes a line in $\mathcal{N}$ and each line becomes a point. A
parallel class in $\mathcal{N}$ is the set of lines dual to the ordinary
points of a special line $s$ of $\mathbb{PR}$; thus it is the set of ordinary
dual lines that contain a fixed special point $s^{*}$; the lines are parallel
because of Axiom (A1*). There are at least three parallel classes because
there is one for each special point and $\mathbb{PR}$ has at least three
special points by Theorem 3.1 Part (b). There are at least two lines in a
parallel class because, by Theorem 3.1 Part (i), every special line in
$\mathbb{PR}$ has at least three points, two of which are ordinary and
correspond to parallel lines in the net.

Second, we consider how the dualized rectangle axioms apply to a projective
extension of a net that satisfies the (CQP) and the other properties in the
theorem.

(A1*) is true by definition if one of the lines is the special line. It is
valid in the net except when the lines are parallel. Parallel lines have a
common point in the extension.

(A2*) To find four lines in the extended net with no three concurrent, take
the special line, three special points, and one ordinary line on each of the
special points. If the three ordinary lines are concurrent, replace one of
them by a parallel line. Alternatively, take two lines from each of two
parallel classes.

(A3*) The existence of three distinct lines on each point is equivalent for an
ordinary point to the existence of at least 3 parallel classes, and for a
special point to the existence of a parallel to each ordinary line.

(A4*) is part of the definition of a projectively extended net.

(A5*) is part of the definition of a projectively extended net.

(A6*) Lemma 9.1 shows that the (CQP) of $\mathcal{N}$ implies this axiom for
the extended net.

∎

##  10\. Open problems

Our work on nontrivial projective rectangles leaves many unanswered questions.
Here are some to add those in the body of the paper.

  1. Q1.

All our examples of projective rectangles are substructures of Pappian
projective planes that can be obtained by the subplane construction. Are there
other examples?

  2. Q2.

We are ignorant of how a special line compares in its intersections with two
planes $\pi$ and $\pi^{\prime}$. Two questions stand out.

    1. (a)

If a plane $\pi$ has an ordinary line $l$, there are many other planes in
which $l$ is a line. However, if $l$ is special, i.e., $l=s\cap\pi$ for a
special line $s$, we have no idea whether even one other plane has $l$ as a
line.

    2. (b)

We do not know whether there may be another plane $\pi^{\prime}$ such that
$s\cap\pi\cap\pi^{\prime}$ has a specific cardinality (not greater than
$m+1$), what the possible values of $|s\cap\pi\cap\pi^{\prime}|$ may be,
whether 0 is a possible value in every nontrivial $\mathbb{PR}$ (aside from
$L_{2}^{2}$, where it is not), or in the infinite case whether it is even
possible that $s\cap\pi^{\prime}$ may properly contain $s\cap\pi$.

  3. Q3.

We proved the subplane construction of Section 6 only for Pappian planes,
coordinatizable by a field.

    1. (a)

Is there an analytic proof for skew fields?

    2. (b)

Does an analytic proof using alternative algebras succeed in planes with
weaker coordinate algebras such as near fields and alternative algebras?

    3. (c)

Is there a synthetic proof for Pappian or Desarguesian or other projective
planes?

    4. (d)

Does the construction exist in non-Desarguesian, or non-Moufang, planes?

  4. Q4.

Are all planes in a projective rectangle isomorphic? We were unable to find a
proof or a counterexample.

  5. Q5.

What do the partial Desargues’s theorems in Section 5 imply about
automorphisms and coordinatizations?

  6. Q6.

Is there a rigorous sense in which a projective rectangle is higher-
dimensional, as suggested in Section 5 and [7]?

  7. Q7.

If every plane in $\mathbb{PR}$ is Moufang, it has coordinates in an
alternative ring. If all such rings are isomorphic, does $\mathbb{PR}$ extend
to a Moufang plane with an alternative ring that extends that of the planes in
$\mathbb{PR}$?

  8. Q8.

Given a projective rectangle, in what projective planes can it be embedded? In
particular, our constructions by subplanes and harmonic extension give
projective rectangles embedded in a Pappian plane but the same rectangles may
possibly be isomorphically embeddable in planes that are not Pappian, not
Desarguesian, maybe not even Moufang, in a nontrivial way, i.e., not by
finding the Pappian plane as a subplane of a non-Pappian plane.

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