A characterisation of Baer subplanes

TL;DR

This paper characterizes certain point sets in projective spaces as tangent Baer subplanes, linking geometric intersection properties with algebraic structures in finite projective geometries.

Contribution

It provides a new characterization of tangent Baer subplanes in PG(2,q^2) through intersection properties with Baer-pencils.

Findings

Sets of points with specific 3-space intersection properties are ruled cubic surfaces.

Such sets correspond to tangent Baer subplanes via Bruck-Bose representation.

The characterization aids in identifying tangent Baer subplanes in finite projective planes.

Abstract

Let $K$ be a set of $q^2+2q+1$ points in $PG(4,q)$. We show that if every 3-space meets $K$ in either one, two or three lines, a line and a non-degenerate conic, or a twisted cubic, then $K$ is a ruled cubic surface. Moreover, $K$ corresponds via the Bruck-Bose representation to a tangent Baer subplane of $PG(2,q^2)$. We use this to prove a characterisation in $PG(2,q^2)$ of a set of points $B$ as a tangent Baer subplane by looking at the intersections of $B$ with Baer-pencils.

Full text

AMS code: 51E20

Keywords: projective geometry, ruled cubic surface, Baer subplanes, Bruck-Bose representation

1 Introduction

A ruled cubic surface in ${\mbox{{P}G}}(4,q)$ is a variety ${\cal V}^{3}_{2}$ which rules a line and a conic according to a projectivity (see Section 2.1 for more details). By [8, Lemma 2.1], a hyperplane of ${\mbox{{P}G}}(4,q)$ meets a ruled cubic surface in either one, two or three lines, a line and a non-degenerate conic, or a twisted cubic. We show that the converse holds, and prove the following characterisation.

Theorem 1

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$. Then ${\cal K}$ is a ruled cubic surface if and only if every 3-space meets ${\cal K}$ in either one, two or three lines, a line and a non-degenerate conic, or a twisted cubic.

Using known results on the Bruck-Bose representation gives the following corollary.

Corollary 2

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space meets ${\cal K}$ in either one, two or three lines, a line and a non-degenerate conic, or a twisted cubic. Then there is a regular spread ${\cal S}$ in a 3-space $\Sigma_{\infty}$ so that in the corresponding Bruck-Bose plane ${\cal P}({\cal S})\cong{\mbox{{P}G}}(2,q^{2})$, the set of points corresponding to ${\cal K}$ form a Baer subplane of ${\mbox{{P}G}}(2,q^{2})$ which is tangent to $\ell_{\infty}$.

We further prove a characterisation in ${\mbox{{P}G}}(2,q^{2})$ of a tangent Baer subplane by the intersection types with $\ell_{\infty}$-Baer pencils. A Baer pencil in ${\mbox{{P}G}}(2,q^{2})$ is the cone of $q+1$ lines joining a vertex point $P$ to a Baer subline base $b$, so contains $q^{2}(q+1)+1$ points. Every line of ${\mbox{{P}G}}(2,q^{2})$ not through the vertex of a Baer pencil meets the Baer pencil in a Baer subline. An $\ell_{\infty}$-Baer pencil is a Baer pencil containing $\ell_{\infty}$, so has vertex in $\ell_{\infty}$, and base $b$ meeting $\ell_{\infty}$ in a point. We will prove the following characterisation.

Theorem 3

Let ${\cal B}$ be a set of $q^{2}+q+1$ points in ${\mbox{{P}G}}(2,q^{2})$, $q\geq 5$, with $T={\cal B}\cap\ell_{\infty}$ a point. Then ${\cal B}$ is a Baer subplane if and only if every $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in either a point; one or two Baer sublines; or a non-degenerate $\mathbb{F}_{q}$-conic.

Section 2 contains background material. Section 3 contains a proof of Theorem 1 and Corollary 2. Section 4 contains a proof of Theorem 3.

2 Background

2.1 A ruled cubic surface of ${\mbox{{P}G}}(4,q)$

We consider a scroll in ${\mbox{{P}G}}(4,q)$ that rules a line and a non-degenerate conic according to a projectivity. That is, in ${\mbox{{P}G}}(4,q)$, let $\ell$ be a line and let ${\cal C}$ be a non-degenerate conic in a plane $\pi$ with $\ell\cap\pi=\emptyset$. Let $\theta,\phi$ be the non-homogeneous coordinates of $\ell$, ${\cal C}$ respectively. That is, without loss of generality, we can write $\ell=\{(1,\theta,\ 0,0,0)\,|\,\theta\in\mathbb{F}_{q}\cup\{\infty\}\}$ and ${\cal C}=\{(0,0,\ 1,\phi,\phi^{2})\,|\,\phi\in\mathbb{F}_{q}\cup\{\infty\}\}$. Let $\sigma$ be a projectivity in ${\mbox{PGL}}(2,q)$ that maps $(1,\theta)$ to $(1,\phi)$. Consider the set of $q+1$ lines of ${\mbox{{P}G}}(4,q)$ joining a point of $\ell$ to the corresponding point of ${\cal C}$ according to $\sigma$. Then the points on these lines form a scroll ruling a line and a conic according to a projectivity. The line $\ell$ is called the line directrix. The ruling lines are called generators and are pairwise disjoint. Further, the line directrix and the generators are the only lines on the scroll. The scroll contains $q^{2}$ conics, called conic directrices, which pairwise meet in a point. Each conic directrix contains one point of each generator and is disjoint from $\ell$. This scroll is a variety ${\cal V}^{3}_{2}$ of dimension 2 and order 3, so is a ruled cubic surface. For more details, see [4], in particular, all ruled cubic surfaces are projectively equivalent.

2.2 The Bruck-Bose representation of ${\mbox{{P}G}}(2,q^{2})$ in ${\mbox{{P}G}}(4,q)$

The Bruck-Bose representation of ${\mbox{{P}G}}(2,q^{2})$ in ${\mbox{{P}G}}(4,q)$ was introduced independently by André [1] and Bruck and Bose [5, 6]. Let $\Sigma_{\infty}$ be a hyperplane of ${\mbox{{P}G}}(4,q)$ and let ${\cal S}$ be a regular line spread of $\Sigma_{\infty}$. Consider the incidence structure ${\cal P}({\cal S})$ with points the points of ${\mbox{{P}G}}(4,q)\setminus\Sigma_{\infty}$ and the lines of ${\cal S}$; lines the planes of ${\mbox{{P}G}}(4,q)\setminus\Sigma_{\infty}$ that contain an element of ${\cal S}$, and a line at infinity $\ell_{\infty}$ whose points correspond to the lines of ${\cal S}$; and incidence induced by incidence in ${\mbox{{P}G}}(4,q)$. Then ${\cal P}({\cal S})\cong{\mbox{{P}G}}(2,q^{2})$.

Associated with a regular spread ${\cal S}$ in ${\mbox{{P}G}}(3,q)$ are a unique pair of transversal lines in the quadratic extension ${\mbox{{P}G}}(3,q^{2})$. These transversal lines are disjoint from ${\mbox{{P}G}}(3,q)$, and are conjugate with respect to the map $X=(x_{0},\ldots,x_{3})\mapsto X^{q}=(x_{0}^{q},\ldots,x_{3}^{q})$. We denote these transversal lines by $g,g^{q}$. The spread ${\cal S}$ is the set of $q^{2}+1$ lines $XX^{q}\cap{\mbox{{P}G}}(3,q)$ for $X\in g$, in particular, the points of $\ell_{\infty}$ are in 1-1 correspondence with the points of $g$. If ${\cal X}$ is a set of points in ${\mbox{{P}G}}(2,q^{2})$, then we denote the corresponding set of points in the Bruck-Bose representation in ${\mbox{{P}G}}(4,q)$ by $[{\cal X}]$. For more details on this representation, see [2]. In particular, the representation of Baer sublines and Baer subplanes is well known. In particular, a Baer subplane of ${\mbox{{P}G}}(2,q^{2})$ tangent to $\ell_{\infty}$ corresponds to a ruled cubic surface of ${\mbox{{P}G}}(4,q)$. The following result characterises the ruled cubic surfaces of ${\mbox{{P}G}}(4,q)$ which correspond to Baer subplanes of ${\mbox{{P}G}}(2,q^{2})$.

Result 4

[7]** Let ${\cal K}$ be a ruled cubic surface in ${\mbox{{P}G}}(4,q)$ with line directrix $b$. Let $\Sigma_{\infty}$ be a 3-space that meets ${\cal K}$ in the line $b$ and let ${\cal S}$ be a regular spread in $\Sigma_{\infty}$ containing $b$. Then ${\cal K}$ corresponds via the Bruck-Bose representation to a Baer subplane tangent to $\ell_{\infty}$ in the Bruck-Bose plane ${\cal P}({\cal S})\cong{\mbox{{P}G}}(2,q^{2})$ if and only if in the extension to ${\mbox{{P}G}}(4,q^{2})$, ${\cal K}$ contains the transversal lines of the regular spread ${\cal S}$.

3 Proof of Theorem 1

We begin by labelling the five different types of 3-spaces given in Theorem 1.

Definition 5

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$. Let $\Pi$ be a 3-space, we say $\Pi$ has type

• $T_{1}$:

if $\Pi\cap{\cal K}$ is 1 line,

• $T_{2}$:

if $\Pi\cap{\cal K}$ is 2 lines,

• $T_{3}$:

if $\Pi\cap{\cal K}$ is 3 lines,

• $T_{4}$:

if $\Pi\cap{\cal K}$ is a line and a non-degenerate conic,

• $T_{5}$:

if $\Pi\cap{\cal K}$ is a twisted cubic.

Throughout this section, we let ${\cal K}$ be a set of $q^{2}+2q+1$ points of ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{1}$, $T_{2}$, $T_{3}$, $T_{4}$ or $T_{5}$, that is, ${\cal K}$ satisfies the assumptions of Theorem 1.

We prove Theorem 1 with a series of lemmas, the structure of these lemmas is as follows. In Lemma 8 we show that ${\cal K}$ can be partitioned into $q+1$ mutually skew lines which we call sticks. In Lemma 9 we show that ${\cal K}$ contains a base line which meets each stick. In Lemma 10, we show that ${\cal K}$ contains $q^{2}$ non-degenerate conics. In Lemma 11, we show that the sticks rule the base line and a non-degenerate conic of ${\cal K}$ according to a projectivity. That is, ${\cal K}$ is a ruled cubic surface as defined in Section 2.1, with line directrix the base line, and generators the sticks. We then use [7, Theorem 2.1] to conclude that ${\cal K}$ corresponds to a tangent Baer subplane of ${\mbox{{P}G}}(2,q^{2})$.

Lemma 6

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$ such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Let $\Pi$ be a 3-space of Type $T_{4}$, so $\Pi\cap{\cal K}$ is a line $\ell$ and a non-degenerate conic ${\cal C}$. Then $\ell\cap{\cal C}$ is a point and $\langle\ell,{\cal C}\rangle=\Pi$.

Proof   Let $\Pi$ be a 3-space meeting ${\cal K}$ in a line $\ell$ and a non-degenerate conic ${\cal C}$, and let $\pi$ be the plane containing ${\cal C}$. Suppose $\ell$ is contained in $\pi$. Each 3-space containing $\pi$ has type $T_{i}$ for some $i\in\{1,\ldots,5\}$, so has type $T_{4}$ and hence contains no further point of ${\cal K}$. Every point of ${\mbox{{P}G}}(4,q)$ lies in a 3-space containing $\pi$, so if $\ell\subset\pi$, then $|{\cal K}|=|{\cal K}\cap\pi|\leq 2(q+1)$, a contradiction. Hence $\ell$ meets $\pi$ in a point. Suppose the point $P=\ell\cap\pi$ is not in ${\cal C}$. As the $q+1$ $3$-spaces through $\pi$, contain a non-degenerate conic of ${\cal K}$, they are of Type $T_{4}$. Hence each 3-space about $\pi$ also contains a line of ${\cal K}$, which necessarily contains the point $P$. Therefore the number of points of ${\cal K}$ is $|{\cal C}|+1+(q+1)\times q=q^{2}+2q+2$, a contradiction. Hence the line $\ell$ meets the non-degenerate conic ${\cal C}$ in a point. $\square$

Lemma 7

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Then ${\cal K}$ contains at least one non-degenerate conic.

Proof   Suppose that ${\cal K}$ does not contain a non-degenerate conic, that is, there are no 3-spaces of type $T_{4}$. We work to a contradiction. There are four cases for the lines of ${\cal K}$: (a) ${\cal K}$ contains no lines; (b) ${\cal K}$ contains exactly one line; (c) ${\cal K}$ contains two lines which meet; (d) ${\cal K}$ contains $s\geq 2$ lines which are pairwise skew.

Case (a): suppose ${\cal K}$ does not contain any lines, then every 3-space has type $T_{5}$. Let $\Pi$ be a 3-space, so there is a plane $\pi$ contained in $\Pi$ that contains 3 points of ${\cal K}$. Each $3$-space through $\pi$ has type $T_{5}$, so $|{\cal K}|=3+(q+1)(q-2)=q^{2}-q+1$, a contradiction. Hence case (a) does not occur.

Case (b): suppose that ${\cal K}$ contains exactly one line $\ell$. Let $P\in{\cal K}\setminus\ell$ and consider the plane $\alpha=\langle P,\ell\rangle$. The 3-spaces through $\alpha$ meet ${\cal K}$ in at least $P,\ell$, and so are of type $T_{2}$ or $T_{3}$, contradicting our assumption that ${\cal K}$ contains exactly one line. Hence case (b) does not occur.

Case (c): suppose ${\cal K}$ contains two lines $\ell,\ell^{\prime}$ which meet in a point. So $\pi=\langle\ell,\ell^{\prime}\rangle$ is a plane. The 3-spaces containing $\pi$ contain at least 2 lines of ${\cal K}$, so are of type $T_{2}$ or $T_{3}$. Suppose there is a 3-space $\Pi$ of type $T_{3}$ that contains $\pi$. If the points of $\Pi\cap{\cal K}$ all lie in $\pi$, then every 3-space containing $\pi$ has type $T_{3}$ and hence contains no further point of ${\cal K}$. So $|{\cal K}|=|\pi\cap{\cal K}|\leq 3q+1$, a contradiction as $q\geq 3$. Hence $\Pi$ contains a line $m$ which is not contained in $\pi$. Consider the point $m\cap\pi$. If $m\cap\pi$ is not in $\ell$ or $\ell^{\prime}$, then every 3-space that contains $\pi$ contains $\ell,\ell^{\prime}$ and a further point of ${\cal K}$, so has type $T_{3}$. In this case, the number of points of ${\cal K}$ is $|{\cal K}|=(2q+1)+1+(q+1).q=q^{2}+3q+2$, a contradiction. Hence the point $m\cap\pi$ lies in $\ell\cup\ell^{\prime}$. Let $x$ be the number of 3-spaces of type $T_{2}$ containing $\pi$ and let $y$ be the number of 3-spaces of type $T_{3}$ containing $\pi$. As every 3-space containing $\pi$ has type $T_{2}$ or $T_{3}$, $x+y=q+1$. Further, counting the points of ${\cal K}$ in planes about $\pi$ we have $2q+1+y.q=q^{2}+2q+1$. Hence $y=q$ and $x=1$.

Hence the points of ${\cal K}$ consist of the points on $q+2$ lines $\{\ell,\ell^{\prime},m_{1},\ldots,m_{q}\}$ where each $m_{i}$ meets $\pi$ in a point of $\ell\cup\ell^{\prime}$, and no two $m_{i}$ lies in a common 3-space about $\pi$. Suppose the lines $m_{1},\ldots,m_{q}$ are concurrent at the point $\ell\cap\ell^{\prime}$. Let $\Pi$ be a 3-space not through $\ell\cap\ell^{\prime}$, so $\Pi$ meets each of the $q+2$ lines $\{m_{1},\ldots,m_{q},\ell,\ell^{\prime}\}$ in distinct points, and $|{\cal K}\cap\Pi|=q+2$, contradicting $\Pi$ having type $T_{i}$ for some $i\in\{1,\ldots,5\}$. Thus at least one of the lines $m_{1},\ldots,m_{q}$ meets $\pi$ in a point of $\ell\cup\ell^{\prime}$ which is not the point $\ell\cap\ell^{\prime}$.

Without loss of generality, suppose $m_{1}$ meets $\pi$ in a point of $\ell$ distinct from $\ell\cap\ell^{\prime}$. Let $X\in{\cal K}$ with $X\notin\pi$ and $X\notin m_{1}$. Label the points on $\ell^{\prime}$ by $Y_{0}=\ell^{\prime}\cap\ell,Y_{1},\ldots,Y_{q}$. So for $i=1,\ldots,q$, the line $Y_{i}X$ is disjoint from $m_{1}$. Consider the 3-spaces $\Pi_{i}=\langle Y_{i},X,m_{1}\rangle$, $i=0,\ldots,q$. As there are no type $T_{4}$ 3-spaces; and $\Pi_{i}\cap{\cal K}$ contains the line $m_{1}$ and the two points $X$, $Y_{i}$; each $\Pi_{i}$ has type $T_{2}$ or $T_{3}$. Now $\Pi_{0}\cap{\cal K}$ contains $\ell,m_{1},X$, so $\Pi_{0}$ has type $T_{3}$. Hence $\Pi_{0}$ contains at least $q-1$ points of ${\cal K}$ not in $\ell\cup\ell^{\prime}\cup m_{1}\cup X$. If some $\Pi_{i}$ has type $T_{2}$, then the two lines in $\Pi_{i}\cap{\cal K}$ are $XY_{i}$ and $m_{1}$. As these two lines are disjoint, $\Pi_{i}\cap{\cal K}$ contains $q-1$ points not in $\ell\cup\ell^{\prime}\cup m_{1}\cup X$. If some $\Pi_{i}$, $i\neq 0$, has type $T_{3}$, then the three lines in $\Pi_{i}\cap{\cal K}$ are either $m_{1},XY_{i}$ and a further line; or $m_{1}$ and distinct lines $m^{\prime}$ through $X$ and $m^{\prime\prime}$ through $Y_{i}$. In the latter case, if $m^{\prime}$, $m^{\prime\prime}$ meet, then they meet in a point of $m_{1}$. As $XY_{i}$ and $m_{1}$ are disjoint, $\Pi_{i}\cap{\cal K}$ contains at least $2q-2$ points not in $\ell\cup\ell^{\prime}\cup m_{1}\cup X$. Let $a$ be the number of type $T_{2}$ 3-spaces in $\{\Pi_{1},\ldots,\Pi_{q}\}$ and let $b$ be the number of type $T_{3}$ 3-spaces in $\{\Pi_{1},\ldots,\Pi_{q}\}$, so $a+b=q$. Counting the points of ${\cal K}$ gives $|{\cal K}|\geq(q-1)+a(q-1)+b(2q-2)+(2q+1+q+1)$. Simplifying using $a+b=q$ gives $0\geq b(q-1)+q+2$, contradicting $b\geq 0$. Hence case (c) does not occur.

Case (d): suppose ${\cal K}$ contains at least two lines, and the lines contained in ${\cal K}$ are pairwise skew. Let $\ell,m$ be two skew lines of ${\cal K}$, then the 3-space $\langle\ell,m\rangle$ has type $T_{2}$ or $T_{3}$. Let $X\in\ell$ and consider the plane $\pi=\langle X,m\rangle$. By assumption, the lines of ${\cal K}$ are pairwise skew, so there is a unique line of ${\cal K}$ through $X$, namely $\ell$. Let $\Sigma$ be a 3-space through $\pi$, with $\ell$ not in $\Sigma$. As $\Sigma\cap{\cal K}$ contains the line $m$ and another point $X$, $\Sigma$ has type $T_{2}$ or type $T_{3}$. Hence $\Sigma\cap{\cal K}$ contains a line through $X$, so contains the unique line $\ell$ of ${\cal K}$ through $X$, a contradiction. That is, case (d) does not occur.

We have shown that if ${\cal K}$ does not contain a non-degenerate conic, then none of the cases (a), (b), (c), (d) can occur, a contradiction. Thus ${\cal K}$ contains at least one non-degenerate conic. $\square$

Lemma 8

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Then ${\cal K}$ can be partitioned into a set of $q+1$ lines called sticks. Further, no three sticks lie in a 3-space.

Proof   By Lemma 7, ${\cal K}$ contains a non-degenerate conic ${\cal C}$. Denote the plane containing ${\cal C}$ by $\pi$. Each 3-space which contains $\pi$ has type $T_{4}$ and so contains a further line of ${\cal K}$ meeting ${\cal C}$. Moreover, by Lemma 6, this line meets ${\cal C}$ and is not contained in $\pi$. Thus the $q+1$ 3-spaces containing $\pi$ give rise to a set of $q+1$ distinct lines of ${\cal K}$, we call these lines sticks. Suppose that two sticks $m,m^{\prime}$ meet in a point, which is necessarily a point of ${\cal C}$. As the 3-spaces containing $\pi$ partition the points of ${\mbox{{P}G}}(4,q)\setminus\pi$, there is a point $X\in{\cal C}$ through which there is no line of ${\cal K}$. Consider the 3-space $\Pi=\langle m,m^{\prime},X\rangle$. As $\Pi$ contains two lines and a further point of ${\cal K}$, $\Pi$ has type $T_{3}$ and so contains a line of ${\cal K}$ through $X$, a contradiction. Thus the $q+1$ sticks meets ${\cal C}$ in distinct points, and lie in distinct 3-spaces containing $\pi$, hence they are pairwise skew, and so partition the $(q+1)^{2}$ points of ${\cal K}$.

Finally suppose $\Pi$ is a 3-space containing three sticks $m_{1},m_{2},m_{3}$. Then $\Pi$ has type $T_{3}$, so does not contain a non-degenerate conic, so does not contain $\pi$. So $\Pi$ meets $\pi$ in a line $\ell$. Further, each stick meets $\pi$ in a point of ${\cal C}$, so $\ell$ contains the three collinear points $m_{i}\cap{\cal C}$, $i=1,2,3$, a contradiction. $\square$

Lemma 9

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Then ${\cal K}$ contains exactly $q+2$ lines; consisting of the $q+1$ sticks, and a baseline which intersects every stick.

Proof   By Lemma 8, ${\cal K}$ contains at least $q+1$ lines, namely the sticks. Suppose ${\cal K}$ contains no further line. Let $m,m^{\prime}$ be two sticks and let $X$ be any point on $m$. Consider the plane $\pi=\langle X,m^{\prime}\rangle$. Let $\Pi$ be a 3-space containing $\pi$, as $\Pi$ meets ${\cal K}$ in at least a line and a point, $\Pi$ must be type $T_{2}$, $T_{3}$ or $T_{4}$. If $\Pi$ is type $T_{3}$, then $\Pi$ contains three lines, which by assumption are all sticks, contradicting Lemma 8. As there is a unique line of ${\cal K}$ through $X$, there is a unique 3-space of type $T_{2}$ containing $\pi$. So the remaining $q$ 3-spaces containing $\pi$ have type $T_{4}$. Recalling Lemma 6, such a 3-space meets ${\cal K}$ in $2q+1$ points, $q+2$ of which lie in $\pi$. Hence counting the points of ${\cal K}$ yields $(q+2)+q+q(q-1)=q^{2}+q+2$, a contradiction. Hence ${\cal K}$ contains at least one line which is not a stick.

Now suppose ${\cal K}$ contained two lines, $\ell,\ell^{\prime}$ which are not sticks. As the sticks partition ${\cal K}$, and $\ell,\ell^{\prime}$ lie in ${\cal K}$ but are not sticks, they each meet every stick in a point. If $\ell,\ell^{\prime}$ do not meet, then the $q+1$ sticks all lie in the subspace $\Sigma=\langle\ell,\ell^{\prime}\rangle$. If $\ell,\ell^{\prime}$ meet in a point, then the $q$ sticks not containing the point $\ell\cap\ell^{\prime}$ lie in the subspace $\Sigma=\langle\ell,\ell^{\prime}\rangle$. In both cases, $\Sigma$ is either a 3-space, or is contained in a 3-space, that does not have type $T_{i}$, $i=1,\ldots,5$, a contradiction. Hence there is at most one line of ${\cal K}$ which is not a stick. Thus ${\cal K}$ contains exactly one line which is not a stick, and this line meets every stick. $\square$

Lemma 10

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Let $P,Q$ be any two points on distinct sticks, not lying on the baseline of ${\cal K}$. Then there exists a unique non-degenerate conic ${\cal C}$ contained in ${\cal K}$ through $P,Q$. Moreover, ${\cal C}$ meets each stick of ${\cal K}$ in a unique point, and does not meet the baseline.

Proof   If there were two non-degenerate conics of ${\cal K}$ containing two points $P$ and $Q$ of ${\cal K}$, then the two conics lie in a 3-space which is not type $T_{i}$, $i\in\{1,\ldots,5\}$, a contradiction. Hence $P,Q$ lie in at most one non-degenerate conic of ${\cal K}$.

Let $m,m^{\prime}$ be sticks of ${\cal K}$, and denote the baseline of ${\cal K}$ by $b$ (recall the baseline meets every stick). Let $P\in m$, $Q\in m^{\prime}$, with $P,Q\notin b$. We construct a non-degenerate conic containing $P$ and $Q$. Consider the plane $\pi=\langle m,Q\rangle$. As $\pi$ contains at least the line $m$ and point $Q$ of ${\cal K}$, each 3-space containing $\pi$ is type $T_{2}$, $T_{3}$ or $T_{4}$. The 3-space $\Pi_{0}=\langle\pi,b\rangle$ contains $m,m^{\prime},b$ and so has type $T_{3}$. As the sticks partition the points of ${\cal K}$, there are no type $T_{2}$ 3-spaces containing $\pi$, and a unique type $T_{3}$ 3-space containing $\pi$, namely $\Pi_{0}$. So the $q$ remaining 3-spaces $\Pi_{1},\ldots,\Pi_{q}$ through $\pi$ are type $T_{4}$. That is, $\Pi_{i}$, $i=1,\ldots,q$, meets ${\cal K}$ in the stick $m$ and a non-degenerate conic ${\cal C}_{i}$ which contains $Q$. Further, by Lemma 6, ${\cal C}_{i}$ meets the stick $m$ in a point, but ${\cal C}_{i}$ is not contained in $\pi$.

Suppose for some $i\neq j$, the non-degenerate conics ${\cal C}_{i},{\cal C}_{j}$ meet $m$ in the same point $X$, then ${\cal C}_{i},{\cal C}_{j}$ both contain the points $P,X$, contradicting the first paragraph. Hence each ${\cal C}_{i}$ meets $m$ in a distinct point. Now suppose some ${\cal C}_{i}$ contains the point $m\cap b$, Let $\Pi^{\prime}$ be the 3-space containing ${\cal C}_{i}$ and $b$. As $\Pi^{\prime}$ contains two points of the stick $m^{\prime}$ (namely $m^{\prime}\cap b$ and $Q\in{\cal C}_{i}$), $\Pi^{\prime}$ contains $m^{\prime}$. Hence $\Pi^{\prime}$ is a 3-space and $\Pi^{\prime}\cap{\cal K}$ contains two lines $b,m^{\prime}$ and a non-degenerate conic ${\cal C}_{i}$, a contradiction. Hence each ${\cal C}_{i}$ meets $m$ in a distinct point which is not $m\cap b$. One of these points of $m$ is $P$, so one of the ${\cal C}_{i}$ contains both $P$ and $Q$. That is, $P,Q$ lie in at least one non-degenerate conic of ${\cal K}$. We showed above that $P,Q$ lie in at most one non-degenerate conic of ${\cal K}$, hence they lie in a unique non-degenerate conic of ${\cal K}$. $\square$

Lemma 11

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. Then ${\cal K}$ is a scroll ruling the base line $b$ and a non-degenerate conic according to a projectivity.

Proof   Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. By Lemma 8, ${\cal K}$ contains $q+1$ sticks which we denote $\{m_{0},\ldots,m_{q}\}$. These sticks partition the points of ${\cal K}$ and are no three in a 3-space. By Lemma 9, ${\cal K}$ contains a base line $b$ which meets each stick in a distinct point. By Lemma 10, ${\cal K}$ contains a non-degenerate conic ${\cal C}$ that meets each stick in a point, and does not meet $b$. So the points of ${\cal K}$ lie on $q+1$ lines $\{m_{0},\ldots,m_{q}\}$ that are pairwise skew, each joining a point of $b$ to a point of ${\cal C}$. That is, ${\cal K}$ is a set of lines that rules $b$ and ${\cal C}$. We show that this ruling is a projectivity by: in step 1 we show that we can project ${\cal K}$ from a point of ${\cal K}$ onto a hyperbolic quadric in a 3-space; then in step 2 we use this hyperbolic quadric to demonstrate the required projectivity.

Step 1: Let $P$ be a point on the stick $m_{0}$ with $P\notin b$. By Lemma 10, there is a unique non-degenerate conic of ${\cal K}$ through $P$ and any point of ${\cal K}$ not on $m_{0}$ or $b$; moreover, this conic meets $m_{0}$ in $P$ and does not meet $b$. As $|{\cal K}|=q^{2}+2q+1$, we obtain a set $\{{\cal C}_{1},\ldots,{\cal C}_{q}\}$ of non-degenerate conics of ${\cal K}$ which each contain $P$, such that each point in ${\cal K}\backslash\{m_{0}\cup b\}$ lies in exactly one of these non-degenerate conics. Denote the plane containing ${\cal C}_{i}$ by $\pi_{i}$. If $\pi_{i}$ contains a point $X\in{\cal K}$ with $X\notin{\cal C}_{i}$, then $\pi_{i}$ contains the stick through $X$, contradicting Lemma 6. So $\pi_{i}\cap{\cal K}={\cal C}_{i}$, $i=1,\ldots,q$, so in particular, $b\cap\pi_{i}=\emptyset$. Let $t_{1}$ be any line of $\pi_{1}$ not through $P$, so $t_{1}$ is not a line of ${\cal K}$ and does not meet $b$. Consider the 3-space $\Pi=\langle b,t_{1}\rangle$. If $P\in\Pi$, then $\Pi$ contains the plane $\pi_{1}$, and so $b$ meets $\pi_{1}$, a contradiction, hence $P\notin\Pi$.

Let $\phi$ denote the projection of ${\cal K}$ from $P$ onto $\Pi$. We have $\phi(b)=b$. The stick $m_{0}$ contains $P$ and so $\phi(m_{0})$ is the point $\phi(m_{0})=M_{0}=m_{0}\cap b$. For $j=1,\ldots,q$, the stick $m_{j}$ is projected onto a line $s_{j}=\phi(m_{j})$. For $j=1,\ldots,q$, let $M_{j}=m_{j}\cap b$, and note that for $i\neq j$, we have $M_{i}\neq M_{j}$. Further, let $P_{j}={\cal C}_{1}\cap m_{j}$ and $Q_{j}=PP_{j}\cap t_{1}$, so we have $s_{j}=M_{j}Q_{j}$.

Suppose two lines $s_{i},s_{j}$, $i\neq j$ meet in a point $Y$. As $M_{i}\neq M_{j}$, we have $Y\notin b$. If $Y\in t_{1}$, then the line $PY$ contains three points of ${\cal C}_{1}$, namely $P,P_{i},P_{j}$, a contradiction as ${\cal C}_{1}$ is non-degenerate. So $Y\notin t_{1}$, and so $Q_{i}\neq Q_{j}$. Hence $\langle s_{i},s_{j}\rangle$ is a plane containing two points of $b$ (namely $M_{i},M_{j}$) and two points of $t_{1}$ (namely $Q_{i},Q_{j}$). Hence $b,t_{1}$ meet, a contradiction. Thus $s_{1},\ldots,s_{q}$ are pairwise skew lines.

Hence $\phi({\cal K})$ is a set of $q^{2}+q+1$ points, namely the points on the lines $s_{1},\ldots,s_{q}$, and the point $M_{0}$. Recall that the base line $b$ lies in $\phi(K)$ and contains the point $M_{0}$ and meets each of $s_{1},\ldots,s_{q}$ in a distinct point $M_{i}=s_{i}\cap b=m_{i}\cap b$.

Now consider the $q$-arcs ${\cal C}_{i}\setminus\{P\}$, $i=1,\ldots,q$. We have $\phi({\cal C}_{i}\setminus\{P\})=\langle P,\ {\cal C}_{i}\setminus\{P\}\rangle\cap\Pi$, and so $\phi({\cal C}_{i}\setminus\{P\})$ is a set of $q$ points on the line $t_{i}=\pi_{i}\cap\Pi$. By Lemma 10, ${\cal C}_{i}$, $i=1,\ldots,q$, meets the stick $m_{j}$, $j=1,\ldots,q$ in point not on $b$. Hence in $\phi({\cal K})$, the line $t_{i}$ meets the line $s_{j}$ in a point not on $b$, $i,j=1,\ldots,q$. The $q+1$ lines $\{b,t_{1},\ldots,t_{q}\}$ are mutually skew, and contain the $q^{2}+q+1$ points of $\phi({\cal K})$, and $q$ additional points, one on each line $t_{i}$. The $q$ lines $\{s_{1},\ldots,s_{q}\}$ are mutually skew, and cover the $q^{2}+q$ points of $\phi({\cal K})\setminus\{M_{0}\}$. Hence ${\cal R}=\{b,t_{1},\ldots,t_{q}\}$ is a regulus, with opposite regulus ${\cal R}^{\prime}$ containing $s_{1},\ldots,s_{q}$. As $q\geq 3$, there is a unique line $s_{0}$ in ${\cal R}^{\prime}$ through $M_{0}$ which meets $t_{1},\ldots,t_{q}$. Thus we can project ${\cal K}$ onto $q^{2}+q+1$ points of the hyperbolic quadric ${\cal R}\cup{\cal R}^{\prime}$.

Step 2: We showed in step 1 that we can project ${\cal K}$ onto $q^{2}+q+1$ points of the hyperbolic quadric ${\cal R}\cup{\cal R}^{\prime}$. For $i\in\{1,\ldots,q\}$, consider the stick $m_{i}$ which joins the point $M_{i}\in b$ with the point $m_{i}\cap{\cal C}_{1}$. This maps under $\phi$ to the line $\phi(m_{i})=s_{i}$, joining the point $\phi(M_{i})=M_{i}$ with the point $\phi(m_{i}\cap{\cal C}_{1})=s_{i}\cap t_{1}$. A hyperbolic quadric is ruled by a projectivity, in particular, it is ruled by a projectivity from $b$ to $t_{1}$. That is, there is a projectivity that maps the point $M_{i}\in b$ to the point $s_{i}\cap t_{1}$, $i=0,\ldots,q$. The projection of ${\cal K}$ onto $\Pi$ preserves this projectivity, so we have a projectivity mapping the points of $b\setminus\{M_{0}\}$ to the points of ${\cal C}_{1}\setminus\{P\}$ which maps $m_{i}\cap b$ to $m_{i}\cap{\cal C}_{1}$, $i=1,\ldots,q$. This projectivity will also map $M_{0}$ to $P$. That is, the sticks of ${\cal K}$ are lines of a scroll ruling the line $b$ and the non-degenerate conic ${\cal C}_{1}$ according to a projectivity. $\square$

Proof of Theorem 1 Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$. By Lemma 11, ${\cal K}$ is a scroll that rules a line and a non-degenerate conic according to a projectivity. Hence by [4], ${\cal K}$ is a ruled cubic surface. $\square$

Proof of Corollary 2 Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 3$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$, then by Theorem 1, ${\cal K}$ is a ruled cubic surface. Let $\Sigma_{\infty}$ be a 3-space that meets ${\cal K}$ in exactly the line directrix (baseline) $b$. For each conic ${\cal C}_{i}$, $i=1,\ldots,q^{2}$ contained in ${\cal K}$, let $\ell_{i}$ be the line in $\Sigma_{\infty}$ that lies in the plane containing ${\cal C}_{i}$. By [7, Theorem 2.1], the lines $\ell_{i}$, $i=1,\ldots,q^{2}$ together with $b$ form a regular spread ${\cal S}$, moreover ${\cal K}$ corresponds via the Bruck-Bose representation to a tangent Baer subplane of ${\cal P}({\cal S})\cong{\mbox{{P}G}}(2,q^{2})$. $\square$

4 Proof of Theorem 3

An $\ell_{\infty}$-Baer pencil of ${\mbox{{P}G}}(2,q^{2})$ has the following nice representation in the Bruck-Bose setting.

Result 12

[3]** An $\ell_{\infty}$-Baer pencil of ${\mbox{{P}G}}(2,q^{2})$ with vertex $L$ corresponds in the ${\mbox{{P}G}}(4,q)$ Bruck-Bose representation to a 3-space of ${\mbox{{P}G}}(4,q)$ containing the spread line $[L]$. Conversely if $\Pi$ is a $3$-space in ${\mbox{{P}G}}(4,q)$ distinct from $\Sigma_{\infty}$, then $\Pi$ corresponds in ${\mbox{{P}G}}(2,q^{2})$ to an $\ell_{\infty}$-Baer pencil with vertex corresponding to the unique spread line in $\Pi$.

We first show how an $\ell_{\infty}$-Baer pencil meets a Baer subplane of ${\mbox{{P}G}}(2,q^{2})$ tangent to $\ell_{\infty}$.

Lemma 13

Let ${\cal B}$ be a Baer subplane in ${\mbox{{P}G}}(2,q^{2})$ tangent to $\ell_{\infty}$ at the point $T={\cal B}\cap\ell_{\infty}$. Then an $\ell_{\infty}$-Baer pencil with vertex $L\in\ell_{\infty}$ meets ${\cal B}$ in either: the point $T$; one or two Baer sublines containing $T$; two Baer sublines with one containing $T$ and one contained in a line through $L$; or a non-degenerate $\mathbb{F}_{q}$-conic of ${\cal B}$.

Proof   Let ${\cal B}$ be a Baer subplane in ${\mbox{{P}G}}(2,q^{2})$ tangent to $\ell_{\infty}$ at the point $T={\cal B}\cap\ell_{\infty}$, so by Result 4, in ${\mbox{{P}G}}(4,q)$, $[{\cal B}]$ is a ruled cubic surface with line directrix the spread line $[T]$. By Result 12, an $\ell_{\infty}$-Baer pencil with vertex $L$ corresponds to a 3-space that meets $\Sigma_{\infty}$ in the spread line $[L]$. By [8, Lemma 2.1], a hyperplane of ${\mbox{{P}G}}(4,q)$ meets a ruled cubic surface in one of the following: (a) 1 line; (b) 2 lines, (c) 3 lines, (d) a line and a non-degenerate conic, (e) a twisted cubic. In case (a), the line is $[T]$, and so the corresponding $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in the point $T$. In case (b), the lines are $[T]$ and one generator which meets $[T]$, so the corresponding $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in a Baer subline through $T$. In case (c), the lines are $[T]$ and two generators which meet $[T]$, so the corresponding $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in two Baer sublines through $T$. In case (d), the line is a generator, and by [8, Lemma 2.1], the non-degenerate conic corresponds to a Baer subline disjoint from $\ell_{\infty}$, so the corresponding $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in two Baer sublines, one containing $T$, and one contained in a line through $L$. In case (e), by [8, Theorem 3.1], the twisted cubic corresponds to a non-degenerate conic in ${\cal B}$ which contains the point $T$. $\square$

We now prove that the converse holds.

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$. Recall that in Definition 5 we defined five types of 3-spaces with respect to ${\cal K}$. Here we define three further types of 3-spaces.

Definition 14

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$. Let $\Pi$ be a 3-space, we say $\Pi$ has type

• $T_{6}$:

if $\Pi\cap{\cal K}$ is one point,

• $T_{7}$:

if $\Pi\cap{\cal K}$ is a non-degenerate conic and two lines.

• $T_{8}$:

if $\Pi\cap{\cal K}$ is a line and a twisted cubic.

Lemma 15

Let ${\cal K}$ be a set of $q^{2}+2q+1$ points in ${\mbox{{P}G}}(4,q)$, $q\geq 5$, such that every 3-space has type $T_{i}$, $i\in\{1,\ldots,8\}$, then every 3-space has type $T_{i}$, $i\in\{1,\ldots,5\}$.

Proof   Suppose there is a 3-space $\Pi$ of type $T_{6}$, and let $P=\Pi\cap{\cal K}$. Let $\pi$ be a plane contained in $\Pi$ with $P\notin\pi$, so $|\pi\cap{\cal K}|=0$. Denote the $q+1$ 3-spaces containing $\pi$ by $\Pi,\Pi_{1},\ldots,\Pi_{q}$. As $|\pi\cap{\cal K}|=0$, $\Pi_{i}\cap{\cal K}$ cannot contain a line. Hence each $\Pi_{i}$, $i=1,\ldots,q$ has type $T_{5}$ or $T_{6}$. Let $x$ be the number of type $T_{5}$ 3-spaces containing $\pi$ and let $y$ be the number of type $T_{6}$ 3-spaces containing $\pi$, so $y\geq 1$. Then $x+y=q+1$, and counting the points in ${\cal K}$ we have $x(q+1)+y=q^{2}+2q+1$. These equations have solution $x=q+1$, $y=0$, a contradiction, hence there is no 3-space of type $T_{6}$.

Suppose there is a 3-space $\Pi$ of type $T_{7}$, so $\Pi\cap{\cal K}$ is a non-degenerate conic ${\cal C}$ and two lines $\ell,m$. Let $\pi$ be the plane of $\Pi$ containing the non-degenerate conic ${\cal C}$. If $\pi$ contains both $\ell$ and $m$, then every 3-space containing $\pi$ is of type $T_{7}$, and contains no further point of ${\cal K}$, hence $|{\cal K}|\leq(q+1)+(q+1)+q$, a contradiction. If $\pi$ contains $\ell$ and meets $m$ in a point, then every 3-space containing $\pi$ is of type $T_{7}$, and contains $q$ points of ${\cal K}$ not in $\pi$, so $|{\cal K}|\geq(q+1)+(q-1)+q(q+1)$, a contradiction. Hence both $\ell$ and $m$ meet $\pi$ in a point, and $|\pi\cap{\cal K}|=q+1+t$ for some $t\in\{0,1,2\}$. Denote the $q+1$ 3-spaces containing $\pi$ by $\Pi,\Pi_{1},\ldots,\Pi_{q}$. As $\pi$ contains a non-degenerate conic of ${\cal K}$, each of these 3-spaces has type $T_{4}$ or $T_{7}$. Let $x$ be the number of 3-spaces containing $\pi$ of type $T_{4}$, $y$ the number of 3-spaces containing $\pi$ of type $T_{7}$, so $x+y=q+1$ and $y\geq 1$. A 3-space of type $T_{4}$ containing $\pi$ contains $q$ points of ${\cal K}$ that are not in $\pi$. A 3-space of type $T_{7}$ containing $\pi$ contains at least $2q-1$ points of ${\cal K}$ that are not in $\pi$. Counting points of ${\cal K}$ gives $|{\cal K}|\geq xq+y(2q-1)+(q+1+t)$, hence $y(q-1)+t\leq 0$. As $q\geq 3$, we have $y=t=0$, contradicting $y\geq 1$. Hence there is no 3-space of type $T_{7}$.

Suppose there is a 3-space $\Pi$ of type $T_{8}$, so $\Pi\cap{\cal K}$ is a line $\ell$ and a twisted cubic ${\cal N}_{3}$. The line $\ell$ contains at most two points of ${\cal N}_{3}$, and each plane of $\Pi$ contains at most three points of ${\cal N}_{3}$. As $q\geq 4$, there are at least three points of ${\cal N}_{3}$ not on $\ell$. Let $\pi$ be the plane containing these three points, so $\pi\cap\ell\notin{\cal N}_{3}$. As $\pi\cap\ell$ is a point in ${\cal K}$, we have $|\pi\cap{\cal K}|=4$. Let $\Sigma$ be any 3-space containing $\pi$. As $|\pi\cap{\cal K}|=4$, $\Sigma$ has type $T_{7}$ or $T_{8}$. As we have shown that there are no type $T_{7}$ 3-spaces, each 3-space containing $\pi$ has type $T_{8}$. Hence $|\Sigma\cap{\cal K}|\geq 2q$. Counting points of ${\cal K}$ in the $q+1$ 3-spaces containing $\pi$ gives $(q^{2}+2q+1)\geq 4+(q+1)(2q-4)$, and so $q(q-4)\leq 1$, a contradiction as $q\geq 5$. Hence there is no 3-space of type $T_{8}$. $\square$

Lemma 16

Let ${\cal B}$ be a set of $q^{2}+q+1$ points in ${\mbox{{P}G}}(2,q^{2})$, $q\geq 5$, with $T={\cal B}\cap\ell_{\infty}$ a point. Suppose every $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in either a point; one or two Baer sublines; or a non-degenerate $\mathbb{F}_{q}$-conic. Then ${\cal B}$ is a Baer subplane.

Proof   Let ${\cal B}$ be a set of $q^{2}+q+1$ points in ${\mbox{{P}G}}(2,q^{2})$, $q\geq 5$, with $T={\cal B}\cap\ell_{\infty}$ a point. We work in the Bruck-Bose representation. So ${\mbox{{P}G}}(2,q^{2})$ is represented in ${\mbox{{P}G}}(4,q)$, and the representation involves a regular 1-spread ${\cal S}$ in the hyperplane at infinity $\Sigma_{\infty}$. In the Bruck-Bose representation in ${\mbox{{P}G}}(4,q)$, $[{\cal B}]$ is a set of $q^{2}+q$ affine points and the spread line $[T]$, so $[{\cal B}]$ is a set of $q^{2}+2q+1$ points.

Let $[{\cal D}]$ be any 3-space distinct from $\Sigma_{\infty}$. Then $[{\cal D}]$ contains a unique spread line $[P]$. By [3], $[{\cal D}]$ corresponds to an $\ell_{\infty}$-Baer pencil ${\cal D}$ in ${\mbox{{P}G}}(2,q^{2})$ with vertex $P$ (possibly $P=T$). As ${\cal D}$ contains $\ell_{\infty}$, $T\in{\cal D}$, so $T\in{\cal B}\cap{\cal D}$. So by assumption ${\cal B}\cap{\cal D}$ is either a point; one or two Baer sublines; or a non-degenerate conic. As each $\ell_{\infty}$-Baer pencil contains $T$, we have four cases: a $\ell_{\infty}$-Baer pencil meets ${\cal B}$ in either (a) the point $T$; (b) one Baer subline containing $T$; (c) two Baer sublines (at least one containing $T$); or (d) a non-degenerate $\mathbb{F}_{q}$-conic containing $T$. We show that in each case, $[{\cal B}]\cap[{\cal D}]$ is type $T_{i}$ for some $i\in\{1,\ldots,8\}$.

Case (a): suppose ${\cal B}\cap{\cal D}$ is the point $T$. If $P=T$, then in ${\mbox{{P}G}}(4,q)$, $[{\cal B}]$ meets $[{\cal D}]$ in the spread line $[T]$, and so $[{\cal D}]$ is a 3-space of type $T_{1}$. If $P\neq T$, then $[{\cal B}]$ meets $[{\cal D}]$ in one point of $[T]$, and so $[{\cal D}]$ is a 3-space of type $T_{6}$.

Case (b): suppose ${\cal B}\cap{\cal D}$ is a Baer subline $n$ through $T$. Then in ${\mbox{{P}G}}(4,q)$, $n$ corresponds to a line $[n]$ that meets $\Sigma_{\infty}$ in a point of $[T]$. Hence if $P=T$, then $[{\cal B}]$ meets $[{\cal D}]$ in two lines, namely the spread line $[T]$ and the line $[n]$. That is, $[{\cal D}]$ is a 3-space of type $T_{2}$. However, if $P\neq T$, then $[{\cal B}]$ meets $[{\cal D}]$ in one line, namely $[n]$, and so $[{\cal D}]$ is a 3-space of type $T_{1}$.

Case (c): suppose ${\cal B}\cap{\cal D}$ is two Baer sublines $m,n$, with $T\in m$. The line $m$ contains $T$ and so corresponds to a line $[m]$ of ${\mbox{{P}G}}(4,q)$ that meets $\Sigma_{\infty}$ in a point. If $T\in n$, then $[n]$ is a line of ${\mbox{{P}G}}(4,q)$ that meets $\Sigma_{\infty}$ in a point. If $T\notin n$, then $[n]$ is a conic of ${\mbox{{P}G}}(4,q)$. Hence if $P\neq T$, then $[{\cal B}]$ meets $[{\cal D}]$ in either two lines $[n],[m]$ (and so $[{\cal D}]$ is a 3-space of type $T_{2}$); or in a conic and a line $[m]$ (and so $[{\cal D}]$ is a 3-space of type $T_{4}$). If $P=T$, then $[{\cal B}]$ meets $[{\cal D}]$ in: either three lines, namely the spread line $[T]$ and the lines $[n],[m]$, and $[{\cal D}]$ is a 3-space of type $T_{3}$; or in a conic and two lines $[T]$, $[m]$, and $[{\cal D}]$ is a 3-space of type $T_{7}$.

Case (d): suppose ${\cal B}\cap{\cal D}$ is a non-degenerate $\mathbb{F}_{q}$-conic ${\cal C}$. Note that as $T\in{\cal D}$, we have $T\in{\cal C}$. By [8], in ${\mbox{{P}G}}(4,q)$, $[{\cal C}]$ is a twisted cubic that meets $[T]$ in a point. Hence in ${\mbox{{P}G}}(4,q)$, if $P=T$, $[{\cal B}]$ meets $[{\cal D}]$ in a line and a twisted cubic. That is, $[{\cal D}]$ is a 3-space of type $T_{8}$. If $P\neq T$, then $[{\cal B}]$ meets $[{\cal D}]$ in a twisted cubic, and so $[{\cal D}]$ is a 3-space of type $T_{5}$.

Hence we have shown that every 3-space of ${\mbox{{P}G}}(4,q)$ has type $T_{i}$ for some $i\in\{1,\ldots,8\}$. So by Lemma 15, every 3-space has type $T_{i}$ for some $i\in\{1,\ldots,5\}$. Hence by Theorem 1, $[{\cal B}]$ is a ruled cubic surface of ${\mbox{{P}G}}(4,q)$.

By Result 4, $[{\cal B}]$ corresponds to a Baer subplane of ${\mbox{{P}G}}(2,q^{2})$ if and only if the extension of $[{\cal B}]$ to ${\mbox{{P}G}}(4,q^{2})$ contains the transversal lines of the regular spread ${\cal S}$. Let $[L]$ be a spread line distinct from $[T]$, and $[{\cal D}]$ a 3-space containing $[L]$. Denote the transversals lines of ${\cal S}$ by $g,g^{q}$, so the extension of $[L]$ meets the transversal lines $g$ and $g^{q}$ in points we denote $g\cap[L]$, $g^{q}\cap[L]$. We need to show that these two points lie in the extension of $[{\cal B}]$.

By assumption, $[{\cal D}]$ does not contain $[T]$. So by [8, Lemma 2.1], $[{\cal D}]$ meets the ruled cubic surface $[{\cal B}]$ in either (i) a non-degenerate conic disjoint from $[T]$ and a generator line of $[{\cal B}]$, or (ii) a twisted cubic with a point in $[T]$. Correspondingly in ${\mbox{{P}G}}(2,q^{2})$, ${\cal D}$ is an $\ell_{\infty}$-Baer pencil with vertex $L\neq T$.

First we look at case (i). The generator line in $[{\cal D}]\cap[{\cal B}]$ corresponds in ${\mbox{{P}G}}(2,q^{2})$ to a Baer subline through $T$. So by assumption, ${\cal D}\cap{\cal B}$ is two Baer sublines. The conic in $[{\cal D}]\cap[{\cal B}]$ is a set of $q+1$ affine points, no two on a line with $T$, and so corresponds to a Baer subline $b$ that lies in a line $m$ with $m\cap\ell_{\infty}\neq T$. In ${\mbox{{P}G}}(4,q)$, $[b]$ is a non-degenerate conic lying in a plane about a spread line. As the only spread line in $[{\cal D}]$ is $[L]$, the conic $[b]$ lies in a plane about $[L]$. Moreover, the extension of the conic $[b]$ to ${\mbox{{P}G}}(4,q^{2})$ contains the two points $g\cap[L]$, $g^{q}\cap[L]$, see [2]. Hence the extension of the ruled cubic surface $[{\cal B}]$ contains the two points $g\cap[L]$, $g^{q}\cap[L]$.

In case (ii), the twisted cubic in $[{\cal D}]\cap[{\cal B}]$ contains a point of $[T]$, so contains $q$ affine points no two on a line through $T$. Hence by assumption, ${\cal D}\cap{\cal B}$ is an $\mathbb{F}_{q}$-conic ${\cal O}$ which contains the point $T$. By [3], $[{\cal O}]$ corresponds in ${\mbox{{P}G}}(4,q)$ to a twisted cubic $[{\cal O}]$ whose extension to ${\mbox{{P}G}}(4,q^{2})$ contains the two points $g\cap[L]$, $g^{q}\cap[L]$. Hence the extension of the ruled cubic surface contains the two points $g\cap[L]$, $g^{q}\cap[L]$.

Hence for each line $[L]$ in the spread ${\cal S}$ distinct from $[T]$, the points $g\cap[L]$, $g^{q}\cap[L]$ lie in the extension of the ruled cubic surface $[{\cal B}]$, and so $g$, $q^{q}$ are lines of the extended ruled cubic surface. Thus by Result 4, ${\cal B}$ is a tangent Baer subplane of ${\mbox{{P}G}}(2,q^{2})$, as required. $\square$

Theorem 3 now follows immediately from Lemmas 13 and 16.