The Bose representation of PG(2,q^3) in PG(8,q)

TL;DR

This paper explores the Bose representation of PG(2,q^3) within PG(8,q), revealing how sublines, subplanes, and conics correspond to geometric varieties like 2-reguli and Segre varieties, and analyzing their extensions.

Contribution

It establishes the correspondence between substructures in PG(2,q^3) and geometric varieties in PG(8,q), and determines their extensions to larger projective spaces.

Findings

Subline corresponds to a 2-regulus.

Subplane corresponds to a Segre variety S_{2;2}.

Extension of varieties to PG(8,q^3) and PG(8,q^6) analyzed.

Abstract

This article looks at the Bose representation of $PG(2,q^3)$ as a 2-spread of $PG(8,q)$. It is shown that an $\mathbb F_q$-subline of $PG(2,q^3)$ corresponds to a 2-regulus, and an $\mathbb F_q$-subplane corresponds to a Segre variety $S_{2;2}$. Moreover, the extension of these varieties to $PG(8,q^3)$ and $PG(8,q^6)$ is determined. These are used to determine the structure of an $\mathbb F_q$-conic of $PG(2,q^3)$ in the Bose representation in $PG(8,q)$.

Equations

$$\begin{array}[]{ccccccccccc}{\rm PG}(2,q^{3})&\cong&\Gamma&\cong&{\cal I}_{\mbox{\tiny{Bose}}}\\ \bar{X}&\longleftrightarrow&X&\longleftrightarrow&\mbox{$\llbracket X\rrbracket$}=\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q).\end{array}$$

$$D=r_{0}A+r_{1}B+r_{2}C=s_0{D}_0+s_1{D}_1+s_2{D}_2$$

$$(r_0{a}_0,r_0{a}_1,r_0{a}_2,r_1{b}_0,r_1{b}_1,r_1{b}_2,r_2{c}_0,r_2{c}_1,r_2{c}_2)=(s_0,s_1,s_2,s_0,s_1,s_2,s_0,s_1,s_2)$$

$${\mathcal{V}}_i=\{(s^{n_i},s^{n_i-1}t,\dots ,s{t}^{n_i-1},t^{n_i})|s,t\in {\mathbb{F}}_q\}$$

$${\psi }_i:(s,t,0)⟼(s^{n_i},s^{n_i-1}t,\dots ,s{t}^{n_i-1},t^{n_i})$$

$$\{(a{s}^{n_0},a{s}^{n_0-1}t,\dots ,as{t}^{n_0-1},a{t}^{n_0},b{s}^{n_1},b{s}^{n_1-1}t,\dots ,bs{t}^{n_1-1},b{t}^{n_1})|a,b\in {\mathbb{F}}_q\}$$

$$(a{s}^{n_1},a{s}^{n_1-1}t,\dots ,a{s}^{n_1-n_0+1}{t}^{n_0-1},a{s}^{n_1-n_0}{t}^{n_0},b{s}^{n_1},b{s}^{n_1-1}t,\dots ,bs{t}^{n_1-1},b{t}^{n_1})$$

$$V(x_0{x}_3-x_1{x}_2)$$

$${\rm PG}(1,q)\times{\rm PG}(1,q)=\{\big{(}(s,t),(a,b)\big{)}|s,t,a,b\in\mathbb{F}_{q}\}.$$

$$\mathcal{C}$$

$${\mathcal{C}}^{\prime }$$

$${\mathcal{C}}^{\prime \prime }$$

$$\sigma (y_0,y_1,y_2,y_3,0)$$

$$(y_0^3,y_0^2{y}_1,y_0{y}_1^2,y_0^2{y}_2,y_0{y}_1{y}_2,y_1^2{y}_2,y_1^2{y}_2,y_0^2{y}_3,y_0{y}_1{y}_3,y_1^2{y}_3)$$

$$F_i=a_{i0}{\theta }^2+a_{i1}\theta +a_{i2},G_i=a_{i3}{\theta }^2+a_{i4}\theta +a_{i5},H_i=a_{i6}{\theta }^2+a_{i7}\theta +a_{i8}$$

$$P_{\theta ,\varphi }=(\theta H_1,H_1,\varphi H_1,-\theta F_1-\varphi G_1).$$

$$H_{1}^{2}\big{(}\theta H_{1}F_{2}+\phi H_{1}G_{2}-\theta H_{2}F_{1}-\phi H_{2}G_{1}\big{)}=0.$$

$$\varphi =\theta \frac{H_1{F}_2-H_2{F}_1}{H_2{G}_1-H_1{G}_2}$$

$$Q_{\theta}=P_{\theta,\phi}=\big{(}\theta(H_{2}G_{1}-H_{1}G_{2}),\ (H_{2}G_{1}-H_{1}G_{2}),\ \theta(H_{1}F_{2}-H_{2}F_{1}),\ \theta(F_{1}G_{2}-F_{2}G_{1})\big{)}$$

$$(G_{2}H_{1}-G_{1}H_{2})^{2}\,\theta\Big{(}(H_{2}G_{1}-H_{1}G_{2})F_{3}+(H_{1}F_{2}-H_{2}F_{1})G_{3}+(F_{1}G_{2}-F_{2}G_{1})H_{3}\Big{)}=0.$$

$$x^3-t_2{x}^2-t_{1}x-t_0$$

$$\bar{X}=(x,y,z)=\big{(}x_{0}+x_{1}\tau+x_{2}\tau^{2},\ y_{0}+y_{1}\tau+y_{2}\tau^{2},\ z_{0}+z_{1}\tau+z_{2}\tau^{2}\big{)}$$

$$X_0=(x_0,x_1,x_2,y_0,y_1,y_2,z_0,z_1,z_2).$$

$$\tau x$$

$$\displaystyle X_{1}=\Big{(}x_{2}t_{0},\ x_{0}+x_{2}t_{1},\ x_{1}+x_{2}t_{2},\ \ \ldots\ldots\Big{)}.$$

$$t_{0}(x_{1}+t_{2}x_{2})\,+\,\big{(}t_{0}x_{2}+t_{1}(x_{1}+t_{2}x_{2})\big{)}\tau\,+\,\big{(}x_{0}+t_{1}x_{2}+t_{2}(x_{1}+t_{2}x_{2})\big{)}\tau^{2}.$$

$$\displaystyle X_{2}=\Big{(}t_{0}(x_{1}+t_{2}x_{2}),\ t_{0}x_{2}+t_{1}(x_{1}+t_{2}x_{2}),\ x_{0}+t_{1}x_{2}+t_{2}(x_{1}+t_{2}x_{2}),\ \ \ldots\ldots\Big{)}.$$

$$\mbox{$\llbracket X\rrbracket$}=\langle X_{0},X_{1},X_{2}\rangle$$

$${\mathsf{a}}_0$$

$${\mathsf{a}}_1$$

$${\mathsf{a}}_2$$

$$A_0$$

$$A_1$$

Full text

Keywords: Bose representation, $\mathbb{F}_{q}$-subplanes, $\mathbb{F}_{q}$-sublines, conics, $\mathbb{F}_{q}$-conics

AMS code: 51E20

1 Introduction

In [4], Bose gave a representation of ${\rm PG}(2,q^{2})$ as a regular 1-spread in ${\rm PG}(5,q)$. It is straightforward to generalise this to represent ${\rm PG}(2,q^{3})$ as a regular 2-spread in ${\rm PG}(8,q)$, and we examine this representation in detail. In particular, we determine the representation of conics, $\mathbb{F}_{q}$-sublines, $\mathbb{F}_{q}$-subplanes and $\mathbb{F}_{q}$-conics of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$. Moreover, we determine the extensions of the corresponding varieties to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$. Our motivation in looking at these extensions is to study the representation of $\mathbb{F}_{q}$-conics of ${\rm PG}(2,q^{3})$ in the Bruck-Bose ${\rm PG}(6,q)$ representation. In particular, [3] uses these extensions to characterise $\mathbb{F}_{q}$-conics of ${\rm PG}(2,q^{3})$ as corresponding to certain normal rational curves of ${\rm PG}(6,q)$.

The article is set out as follows. In Section 2, we introduce the background and notation we use. In Section 3, we prove some preliminary results. First we need several results relating to disjoint planes in ${\rm PG}(8,q)$. Secondly, in Section 3.3, we carefully define a notion of a scroll in ${\rm PG}(8,q)$ ruled by three planar varieties. Section 3.4 is devoted to the variety associated with a scroll ruled by three conics, and we determine the order and dimension of this variety.

In Section 4, we investigate coordinates for the Bose representation. We calculate coordinates for the transversal planes of the regular 2-spread. We need a suitable description of certain points in ${\rm PG}(8,q^{3})$, and we determine these by looking at the conjugacy map with respect to an $\mathbb{F}_{q}$-subplane.

Section 5 looks at a variety of ${\rm PG}(2,q^{3})$, and uses coordinates to describe the corresponding variety in ${\rm PG}(8,q)$. A geometric description of this variety is given. An application to the representation of a non-degenerate conic of ${\rm PG}(2,q^{3})$ in the Bose representation is given in Theorem 5.7.

The machinery that has been developed in the article is then used to look at $\mathbb{F}_{q}$-structures of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$; and to determine the extension of the resulting varieties to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$. Section 6 determines the representation of $\mathbb{F}_{q}$-sublines and $\mathbb{F}_{q}$-subplanes of ${\rm PG}(2,q^{3})$ in the Bose representation, and Section 7 determines the representation of $\mathbb{F}_{q}$-conics of ${\rm PG}(2,q^{3})$ in the Bose representation.

2 Background and Notation

2.1 Background

We denote the unique finite field of prime power order $q$ by $\mathbb{F}_{q}$. An $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ is a subplane that is projectively equivalent to ${\rm PG}(2,q)$. Similarly, an $\mathbb{F}_{q}$-subline is a subline that is projectively equivalent to ${\rm PG}(1,q)$.

The Frobenius map $x\mapsto x^{q}$ for $x\in\mathbb{F}_{q^{h}}$ gives rise to an automorphic collineation in {\mbox{P\GammaL}}(n,q^{h}) of order $h$ acting on points of ${\rm PG}(n,q^{h})$ that fixes the points of ${\rm PG}(n,q)$ pointwise, that is $X=(x_{0},\ldots,x_{n})\ \longmapsto X^{q}=(x_{0}^{q},\ldots,x_{n}^{q}).$ We say the points $X,X^{q}\ldots,X^{q^{h}-1}$ are conjugate points with respect to the conjugacy map $X\mapsto X^{q}$.

We work with Segre varieties, see [6, Section 4.5] for details. In particular, the Segre variety $\mathcal{S}_{2;2}$ in ${\rm PG}(8,q)$ contains two systems of maximal subspaces (planes), denoted $\mathcal{R}$ and $\mathcal{R}^{\prime}$, such that every plane in $\mathcal{R}$ meets every plane in $\mathcal{R}^{\prime}$ in a point. A 2-regulus of ${\rm PG}(5,q)$ is the system of maximal 2-spaces of a Segre variety ${\mathscr{S}}_{1;2}$, see [6, Section 4.6].

A 2-spread of ${\rm PG}(8,q)$ is a set of planes that partition the points of ${\rm PG}(8,q)$. We use the following construction of a regular 2-spread of ${\rm PG}(3s+2,q)$, see [5]. Embed ${\rm PG}(8,q)$ in ${\rm PG}(8,q^{3})$ and consider the collineation $X=(x_{0},\ldots,x_{8})\longmapsto X^{q}=(x_{0}^{q},\ldots,x_{8}^{q})$ acting on ${\rm PG}(8,q^{3})$. Let $\Gamma$ be a plane in ${\rm PG}(8,q^{3})$ which is disjoint from ${\rm PG}(8,q)$, such that $\Gamma,\Gamma^{q},\Gamma^{q^{2}}$ span ${\rm PG}(8,q^{3})$ (so any two span a 5-space which is disjoint from the third). For a point $X\in\Gamma$, the plane $\langle X,X^{q},X^{q^{2}}\rangle$ of ${\rm PG}(8,q^{3})$ meets ${\rm PG}(8,q)$ in a plane. The planes $\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q)$ for $X\in\Gamma$ form a regular 2-spread of ${\rm PG}(8,q)$. The planes $\Gamma$, $\Gamma^{q}$ and $\Gamma^{q^{2}}$ are called the three transversal spaces of the 2-spread. Conversely, any regular 2-spread of ${\rm PG}(8,q)$ has a unique set of three transversal $s$-spaces in ${\rm PG}(8,q^{3})$, and can be constructed in this way, see [5, Theorem 6.1].

2.2 Variety-extensions

A variety in ${\rm PG}(n,q)$ has a natural extension to a variety in the cubic extension ${\rm PG}(n,q^{3})$ and to a variety in ${\rm PG}(n,q^{6})$ (a sextic extension of ${\rm PG}(n,q)$, and a quadratic extension of ${\rm PG}(n,q^{3})$). If $\mathcal{K}$ is a variety of ${\rm PG}(n,q)$, then the pointset of $\mathcal{K}$ is the set of points of ${\rm PG}(n,q)$ satisfying the set ${\cal F}=\{f_{i}(x_{0},\ldots,x_{n})=0,\ i=1,\ldots,k\}$ of $k$ homogeneous $\mathbb{F}_{q}$-equations $f_{i}$. We define the variety-extension ${\mathcal{K}^{\mbox{\tiny\char 73}}}$ of $\mathcal{K}$ to ${\rm PG}(n,q^{3})$, to be the set of points of ${\rm PG}(n,q^{3})$ that satisfy the same set ${\cal F}$ of homogeneous equations as $\mathcal{K}$. Similarly, we can define the variety-extension ${\mathcal{K}^{\mbox{\tiny\char 72}}}$ of $\mathcal{K}$ to ${\rm PG}(n,q^{6})$. So if $\Pi_{r}$ is an $r$-dimensional subspace of ${\rm PG}(n,q)$, then ${\Pi}^{\mbox{\tiny\char 73}}_{r}$ is the natural extension to an $r$-dimensional subspace of ${\rm PG}(n,q^{3})$, and ${\Pi}^{\mbox{\tiny\char 72}}_{r}$ is the extension to ${\rm PG}(n,q^{6})$. Moreover, if $\Sigma_{r}$ is an $r$-dimensional subspace of ${\rm PG}(n,q^{3})$ (possibly disjoint from ${\rm PG}(n,q)$), then ${\Sigma}^{\mbox{\tiny\char 72}}_{r}$ denotes the extension to ${\rm PG}(n,q^{6})$. In this article we use the ${{}^{\mbox{\tiny\char 73}}}$ and ${{}^{\mbox{\tiny\char 72}}}$ notation for varieties in the Bose representations, that is, when $n=8$. We do not use the ${{}^{\mbox{\tiny\char 73}}}$ and ${{}^{\mbox{\tiny\char 72}}}$ notation in ${\rm PG}(2,q^{3})$.

2.3 The Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$

Bose [4] gave a construction to represent ${\rm PG}(2,q^{2})$ using a regular 1-spread in ${\rm PG}(5,q)$. This construction generalises to the Bose representation of ${\rm PG}(2,q^{h})$ using a regular $(h-1)$-spread in ${\rm PG}(3h-1)$. We consider $h=3$, that is, the Bose representation of ${\rm PG}(2,q^{3})$ using a regular 2-spread $\mathbb{S}$ in ${\rm PG}(8,q)$. Let ${\cal I}_{\mbox{\tiny{Bose}}}$ be the incidence structure with points the $q^{6}+q^{3}+1$ planes of ${\mathbb{S}}$; lines the 5-spaces of ${\rm PG}(8,q)$ that meet ${\mathbb{S}}$ in $q^{3}+1$ planes; and incidence is inclusion. The 5-spaces of ${\rm PG}(8,q)$ that meet ${\mathbb{S}}$ in $q^{3}+1$ planes form a dual spread $\mathbb{H}$ (that is, each 7-space of ${\rm PG}(8,q)$ contains a unique 5-space in $\mathbb{H}$). Then ${\cal I}_{\mbox{\tiny{Bose}}}\cong{\rm PG}(2,q^{3})$, and this representation is called the Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$. The regular 2-spread $\mathbb{S}$ has three conjugate transversal planes which we denote throughout this article by $\Gamma$, $\Gamma^{q}$, $\Gamma^{q^{2}}$. Note that ${\cal I}_{\mbox{\tiny{Bose}}}\cong\Gamma\cong{\rm PG}(2,q^{3})$.

We use the following notation. Let $\bar{X}$ be a point of ${\rm PG}(2,q^{3})$, then the Bose representation of $\bar{X}$ is a plane of $\mathbb{S}$ denoted by $\llbracket X\rrbracket$. In ${\rm PG}(8,q^{3})$, we have \mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}}\cap\Gamma=X and \mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}}=\langle X,X^{q},X^{q^{2}}\rangle. Thus $\bar{X}$ corresponds to a unique point $X$ of $\Gamma$, and the points of $\Gamma$ and ${\rm PG}(2,q^{3})$ are in one-to-one correspondence.

More generally, if $\bar{\mathcal{K}}$ is a set of points of ${\rm PG}(2,q^{3})$, then \llbracket\mathcal{K}\rrbracket=\{\mbox{\llbracket X\rrbracket}\,|\,\bar{X}\in\bar{\mathcal{K}}\} denotes the corresponding set of planes in the Bose representation in ${\rm PG}(8,q)$, and \mathcal{K}=\{\mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}}\cap\Gamma\,|\,\bar{X}\in\bar{\mathcal{K}}\} denotes the corresponding set of points of $\Gamma$ .

So we have the following correspondences:

[TABLE]

2.4 Notation summary

• $\bullet$

For a variety $\mathcal{K}$ in ${\rm PG}(n,q)$, $n>2$, denote the variety-extension to ${\rm PG}(n,q^{3})$ by ${\mathcal{K}^{\mbox{\tiny\char 73}}}$, and the extension to ${\rm PG}(n,q^{6})$ by $\mathcal{K}{{}^{\mbox{\tiny\char 72}}}$.

• $\bullet$

For a point $X=(x_{0},\ldots,x_{n})$ of ${\rm PG}(n,q^{t})$, let $X^{q}=(x_{0}^{q},\ldots,x_{n}^{q}).$

• $\bullet$

$\mathbb{S}$ is a regular 2-spread in ${\rm PG}(8,q)$.

• $\bullet$

$\mathbb{S}$ has three transversal planes in ${\rm PG}(8,q^{3})$, denoted $\Gamma$, $\Gamma^{q}$, $\Gamma^{q^{2}}$.

• $\bullet$

A point $\bar{X}$ in ${\rm PG}(2,q^{3})$ corresponds to a point $X$ in the transversal plane $\Gamma$, and to a plane \mbox{\llbracket X\rrbracket}=\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q) of $\mathbb{S}$.

3 Preliminary Results

3.1 Planes of ${\rm PG}(8,q)$

Lemma 3.1

Let $\alpha,\beta,\gamma$ be three planes which span ${\rm PG}(8,q)$. Let $P$ be a point of ${\rm PG}(8,q)$ not on a line meeting two of $\alpha,\beta,\gamma$. Then $P$ lies on a unique plane that meets each of $\alpha,\beta,\gamma$.

Proof   Let $P$ be a point of ${\rm PG}(8,q)$ which is not on a line joining two of $\alpha,\beta,\gamma$. So $\langle P,\alpha\rangle$ is a 3-space that does not meet $\beta$ or $\gamma$. Hence $\Sigma_{6}=\langle P,\alpha,\beta\rangle$ is a 6-space. As $\alpha,\beta,\gamma$ span ${\rm PG}(8,q)$, $\Sigma_{6}$ meets $\gamma$ in a point we denote by $Q$. So $\Sigma_{4}=\langle P,Q,\alpha\rangle$ is a 4-space contained in $\Sigma_{6}$. As $\alpha,\beta$ span a 5-space contained in $\Sigma_{6}$, $\Sigma_{4}$ meets $\beta$ in a point denoted $R$. The two planes $\pi=\langle P,Q,R\rangle$ and $\alpha$ lie in the 4-space $\Sigma_{4}$, and so meet in a point $S$. That is, $\pi$ meets $\alpha$ in the point $S$, meets $\beta$ in the point $R$ and meets $\gamma$ in the point $Q$. That is, $P$ lies on at least one plane that meets each of $\alpha,\beta,\gamma$.

Suppose $P$ lies in two distinct planes $\pi_{1},\pi_{2}$ that meet each of $\alpha,\beta,\gamma$, so $\langle\pi_{1},\pi_{1}\rangle$ has dimension 3 or 4. Consider the set of six (possibly repeated) points $\mathcal{K}=\{\pi_{i}\cap\alpha,\,\pi_{i}\cap\beta,\,\pi_{i}\cap\gamma\,|\,i=1,2\}$. Suppose $\langle\pi_{1},\pi_{2}\rangle$ is a 3-space, so $\pi_{1}\cap\pi_{2}=\ell$ is a line. As $P\in\ell$, by assumption at most one of the planes $\alpha,\beta,\gamma$ meets $\ell$. Hence $|\mathcal{K}|\geq 5$, so the 3-space $\langle\pi_{1},\pi_{2}\rangle$ meets two of the planes $\alpha,\beta,\gamma$ in a line, and meets the other in at least a point. This contradicts the three planes $\alpha,\beta,\gamma$ spanning ${\rm PG}(8,q)$. If $\langle\pi_{1},\pi_{2}\rangle$ is a 4-space, then $|\mathcal{K}|=6$, and so $\langle\pi_{1},\pi_{2}\rangle$ meets each of $\alpha,\beta,\gamma$ in a line, contradicting the three planes $\alpha,\beta,\gamma$ spanning ${\rm PG}(8,q)$. Hence $P$ lies on at most one plane that meets each of $\alpha,\beta,\gamma$. We conclude that $P$ lies on exactly one plane that meets each of $\alpha,\beta,\gamma$. $\square$

Lemma 3.2

Let $\alpha,\beta,\gamma,\delta$ be four planes of ${\rm PG}(8,q)$, such that any three span ${\rm PG}(8,q)$. Then $\alpha,\beta,\gamma,\delta$ are contained in a unique Segre variety $\mathcal{S}_{2;2}$.

Proof   By Lemma 3.1, through each point $P_{i}\in\alpha$, there is a plane $\pi_{i}$ that meets each of $\beta,\gamma,\delta$ in a point. As $\beta,\gamma,\delta$ span ${\rm PG}(8,q)$, the planes $\pi_{i}$, $i=1,\ldots,q^{2}+q+1$ are pairwise disjoint. Hence there is a unique set $\mathcal{R}=\{\pi_{1},\ldots,\pi_{q^{2}+q+1}\}$ of $q^{2}+q+1$ pairwise disjoint planes that each meet $\alpha,\beta,\gamma,\delta$ in a point. We show that $\mathcal{R}$ is one set of maximal spaces of a Segre variety $\mathcal{S}_{2;2}$.

Let $U_{i}$, $i=0,\ldots,8$, be the point of ${\rm PG}(8,q)$ with a 1 in entry $i$, and zeroes elsewhere. A generalisation of the proof that three pairwise disjoint lines in ${\rm PG}(3,q)$ lie in a unique regulus (see for example [1, Theorem 3.5]) shows that without loss of generality, we can coordinatise so that $\alpha=\langle U_{0},U_{1},U_{2}\rangle$, $\beta=\langle U_{3},U_{4},U_{5}\rangle$, $\gamma=\langle U_{6},U_{7},U_{8}\rangle$, and $\delta=\langle D_{0},D_{1},D_{2}\rangle$ with $D_{0}=(1,0,0,1,0,0,1,0,0)$, $D_{1}=(0,1,0,0,1,0,0,1,0)$, $D_{2}=(0,0,1,0,0,1,0,0,1).$ Let $A\in\alpha$, so $A=(a_{0},a_{1},a_{2},0,0,0,0,0,0)$ for some $a_{0},a_{1},a_{2}\in\mathbb{F}_{q}$, not all zero. Let $\pi_{A}$ be the unique plane of $\mathcal{R}$ through $A$. Let $B=\pi_{A}\cap\beta$, $C=\pi_{A}\cap\gamma$ and $D=\pi_{A}\cap\delta$. So $B=(0,0,0,b_{0},b_{1},b_{2},0,0,0)$ for some $b_{0},b_{1},b_{2}\in\mathbb{F}_{q}$, not all zero, and $C=\pi_{A}\cap\gamma=(0,0,0,0,0,0,c_{0},c_{1},c_{2})$ for some $c_{0},c_{1},c_{2}\in\mathbb{F}_{q}$, not all zero. As $D=\pi_{A}\cap\delta$, we have

[TABLE]

for some $r_{0},r_{1},r_{2},s_{0},s_{1},s_{2}\in\mathbb{F}_{q}$. That is,

[TABLE]

and so $B=(0,0,0,a_{0},a_{1},a_{2},0,0,0)$ and $D=(r_{0}a_{0},r_{0}a_{1},r_{0}a_{2},\ r_{1}a_{0},r_{1}a_{1},r_{1}a_{2},\ r_{2}a_{0},r_{2}a_{1},r_{2}a_{2})$. Hence as in [6, page 209], $\pi_{A}=\langle A,B,D\rangle$ is a plane in one maximal system of a Segre variety $\mathcal{S}_{2;2}$. $\square$

3.2 T-points, T-lines and T-planes

We need some properties of planes of ${\rm PG}(8,q^{3})$ that meet all three transversal planes $\Gamma$, $\Gamma^{q}$, $\Gamma^{q^{2}}$ of the Bose spread $\mathbb{S}$. We call a plane of ${\rm PG}(8,q^{3})$ that meets all three transversal planes a T-plane; call a line of ${\rm PG}(8,q^{3})$ that meets two transversal planes a T-line; and call a point of ${\rm PG}(8,q^{3})$ that lies in a transversal plane a T-point.

Lemma 3.3

Two T-planes of  ${\rm PG}(8,q^{3})$ are either equal, disjoint, meet in a T-point, or meet in a T-line.

Proof   Let $P$ be a point of ${\rm PG}(8,q^{3})$ which is not a T-point, and does not lie on a T-line. As $P$ is not on a T-line, by Lemma 3.1, $P$ lies on a unique T-plane. The result follows. $\square$

3.3 Scrolls

The notion of a scroll that rules two normal rational curves according to a projectivity in ${\rm PGL}(2,q)$ is well known. Here, we are interested in scrolls which rule three planar varieties, that is, scrolls that are ruled by planes. We give a general definition of scrolls.

Let $r,d$ be positive integers. For $i=0,\dots,r$, let $n_{i}$ be a positive integer and let ${\mathscr{V}}_{i}$ be a variety of ${\rm PG}(n_{i},q)$ of dimension $d$. Suppose that there exists a primal ${\mathscr{V}}$ of ${\rm PG}(d+1,q)$ such that, for $i=0,\dots,r$, ${\mathscr{V}}_{i}$ is in one-to-one correspondence with ${\mathscr{V}}$ under a birational mapping $\psi_{i}\colon{\mathscr{V}}\rightarrow{\mathscr{V}}_{i}$. Let $\phi_{i}$, $i=1,\dots,r$ be projectivities of ${\rm PG}(d+1,q)$ fixing ${\mathscr{V}}$. Consider ${\rm PG}(n,q)$ with $n=n_{0}+\cdots+n_{r}+r-1$ and suppose that ${\mathscr{V}}_{0},\dots,{\mathscr{V}}_{r}$ are varieties contained in mutually disjoint subspaces of ${\rm PG}(n,q)$. For each point $P\in{\mathscr{V}}_{0}$ let $\Pi_{P}$ be the $r$-dimensional subspace $\langle P,P^{\psi_{0}^{-1}\phi_{1}\psi_{1}},\dots,P^{\psi_{0}^{-1}\phi_{r}\psi_{r}}\rangle$ (that is, the span of the point of each subspace corresponding to $P$). Define $\mathscr{S}=\mathscr{S}({\mathscr{V}}_{0},\dots,{\mathscr{V}}_{r})$ to be the set of subspaces $\{\Pi_{P}|P\in{\mathscr{V}}_{0}\}$. We say $\mathscr{S}$ is a scroll of type $(r,d)$ or an $(r,d)$-scroll.

Example 1. Normal Rational Curves Let $r=1,d=1$. Suppose that ${\mathscr{V}}_{0}$ and ${\mathscr{V}}_{1}$ are normal rational curves of ${\rm PG}(n_{0},q)$ and ${\rm PG}(n_{1},q)$ respectively. For $i=0,1$, let

[TABLE]

and let $\psi_{i}$ be the mapping from the line $z=0$ of ${\rm PG}(2,q)$ to ${\mathscr{V}}_{i}$ given by

[TABLE]

for $s,t\in\mathbb{F}_{q}$. Embedding ${\mathscr{V}}_{0}$ and ${\mathscr{V}}_{1}$ in independent subspaces of ${\rm PG}(n_{0}+n_{1}+1,q)$ and taking $\phi$ to be the identity we have that the collection $\mathscr{S}({\mathscr{V}}_{0},{\mathscr{V}}_{1})$ of lines

[TABLE]

for $s,t\in\mathbb{F}_{q}$, $(s,t)\not=(0,0)$, is a scroll of type $(1,1)$. It is well known that $\mathscr{S}({\mathscr{V}}_{0},{\mathscr{V}}_{1})$ is a variety of dimension $2$. We can see this using techniques of [7] as follows. Supposing $n_{1}\geq n_{0}$, the pointset of $\mathscr{S}({\mathscr{V}}_{0},{\mathscr{V}}_{1})$ may also be written as the set of points

[TABLE]

for $a,b,s,t\in\mathbb{F}_{q},s\not=0$, together with $(0,\dots,0,at^{n_{1}},0,\dots,0,bt^{n_{1}})$ for $t,a,b\in\mathbb{F}_{q},t\not=0.$ This pointset is in one-to-one algebraic correspondence with the hyperbolic quadric

[TABLE]

of ${\rm PG}(3,q)$ since they are both in one-to-one correspondence with

[TABLE]

Hence the pointset is a variety of dimension 2.

3.4 Scroll-extensions

We will be working with $(2,2)$-scrolls in ${\rm PG}(8,q)$, and define a natural extension to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$.

First consider a variety ${\mathscr{V}}$ in a plane $\Pi_{\mathscr{V}}$ of ${\rm PG}(8,q)$, so the points of ${\mathscr{V}}$ satisfy a set ${\cal F}$ of equations in $\Pi_{\mathscr{V}}$. Note that ${\mathscr{V}}$ is also a variety of ${\rm PG}(8,q)$, it consists of the points that satisfy both the equations in ${\cal F}$ and the equations defining $\Pi_{\mathscr{V}}$. We define the extension of ${\mathscr{V}}$ to ${\rm PG}(8,q^{3})$, denoted ${\mathscr{V}}{{}^{\mbox{\tiny\char 73}}}$, to be the set of points in the extended plane ${\Pi}^{\mbox{\tiny\char 73}}_{{\mathscr{V}}}$ that satisfy the same set of equations that define ${\mathscr{V}}$.

Suppose $\mathscr{S}({\mathscr{V}}_{1},{\mathscr{V}}_{2},{\mathscr{V}}_{3})$ is a $(2,2)$-scroll in ${\rm PG}(8,q)$. The birational mappings $\psi_{1},\psi_{2},\psi_{3}$ and the projectivities $\phi_{1}$, $\phi_{2}$ associated with $\mathscr{S}({\mathscr{V}}_{1},{\mathscr{V}}_{2},{\mathscr{V}}_{3})$ act on points of the form $(x,y,z)$, $x,y,z\in\mathbb{F}_{q}$. They have unique extensions to act on points of the form $(x,y,z)$, $x,y,z\in\mathbb{F}_{q^{3}}$. For $P\in{{\mathscr{V}}}^{\mbox{\tiny\char 73}}_{1}$, let {\Pi}^{\mbox{\tiny\char 73}}_{\mbox{\raisebox{0.4pt}{\scalebox{0.5}{P}}}} be the plane $\langle P,P^{\psi_{0}^{-1}\phi_{1}\psi_{1}},P^{\psi_{0}^{-1}\phi_{2}\psi_{2}}\rangle$. The scroll-extension of $\mathscr{S}({\mathscr{V}}_{1},{\mathscr{V}}_{2},{\mathscr{V}}_{3})$ to ${\rm PG}(8,q^{3})$ is a $(2,2)$-scroll $\mathscr{S}({{\mathscr{V}}}^{\mbox{\tiny\char 73}}_{1},{{\mathscr{V}}}^{\mbox{\tiny\char 73}}_{2},{{\mathscr{V}}}^{\mbox{\tiny\char 73}}_{3})$ consisting of the set of subspaces \{{\Pi}^{\mbox{\tiny\char 73}}_{\mbox{\raisebox{0.4pt}{\scalebox{0.5}{P}}}}|P\in{{\mathscr{V}}}^{\mbox{\tiny\char 73}}_{1}\}. Similarly, we can define the scroll-extension to ${\rm PG}(8,q^{6})$ (and similarly can define the scroll-extension of a scroll from ${\rm PG}(5,q)$ to ${\rm PG}(5,q^{3})$ and ${\rm PG}(5,q^{6})$).

3.5 A scroll ruling three conics

We now show that the points of a scroll of ${\rm PG}(8,q)$ that rules three conics forms a variety, and determine the order and dimension using techniques given in Semple and Roth [7, I.4].

Lemma 3.4

In ${\rm PG}(8,q)$, let $\Pi$, $\Pi^{\prime}$, $\Pi^{\prime\prime}$ be three disjoint planes which together span ${\rm PG}(8,q)$. Let ${\cal C}$, ${\cal C}^{\prime}$, ${\cal C}^{\prime\prime}$ be non-degenerate conics in $\Pi,\Pi^{\prime},\Pi^{\prime\prime}$ respectively. Then the pointset of any scroll $\mathscr{S}({\cal C},{\cal C}^{\prime},{\cal C}^{\prime\prime})$ form a variety of dimension 3 and order 6.

Proof   Denote the set of points on the scroll $\mathscr{S}({\cal C},{\cal C}^{\prime},{\cal C}^{\prime\prime})$ by $\mathscr{S}$. Without loss of generality, we can coordinatise so that

[TABLE]

and so that the homographies $\phi_{i}$ are essentially the identity. That is, without loss of generality we coordinatise so that $\mathscr{S}$ contains the $q+1$ planes $\langle P_{r,s},\ P^{\prime}_{r,s},\ P^{\prime\prime}_{r,s}\rangle$ for $r,s\in\mathbb{F}_{q}$ (not both 0). Thus the points of $\mathscr{S}$ form a variety denoted ${\mathscr{V}}(\mathscr{S})$. We determine the order and dimension of ${\mathscr{V}}(\mathscr{S})$ using techniques given in Semple and Roth [7, I.4]. We show that ${\mathscr{V}}(\mathscr{S})$ has dimension 3 by showing that the points of $\mathscr{S}$ are in one-to-one algebraic correspondence with the points of a 3-space in ${\rm PG}(4,q)$.

Using the notation of [7], let $M(y_{0},y_{1},y_{2},y_{3},y_{4})=y_{4}$, the points of ${\rm PG}(4,q)$ satisfying $M=0$ form a 3-space which we denote $M_{3}$. Note that $M_{3}$ is an irreducible primal of dimension 3 and order 1. Consider the following nine homogeneous cubic equations $F_{i}=F_{i}(y_{0},y_{1},y_{2},y_{3},y_{4})=0$ where $F_{0}=y_{0}^{3}$, $F_{1}=y_{0}^{2}y_{1}$, $F_{2}=y_{0}y_{1}^{2}$, $F_{3}=y_{0}^{2}y_{2}$, $F_{4}=y_{0}y_{1}y_{2},$ $F_{5}=y_{1}^{2}y_{2}$, $F_{6}=y_{0}^{2}y_{3}$, $F_{7}=y_{0}y_{1}y_{3}$, $F_{8}=y_{1}^{2}y_{3}.$ Consider the map $\sigma\colon{\rm PG}(4,q)\rightarrow{\rm PG}(8,q)$ such that

[TABLE]

The map $\sigma$ is an algebraic one-to-one correspondence between the points of the 3-space $M_{3}$ and the points of $\mathscr{S}$. Hence the points of $\mathcal{S}$ form a variety ${\mathscr{V}}(\mathscr{S})$ which has dimension 3. Note that the kernel of $\sigma$ is $\ker\sigma=\{(0,1,0,0,0),(0,0,a,b,0)\,|\,a,b\in\mathbb{F}_{q}\}$.

Next we show that ${\mathscr{V}}(\mathscr{S})$ has order $6$, that is, we show that a generic 5-space of ${\rm PG}(8,q)$ meets the pointset $\mathscr{S}$ in six points. A 5-space in ${\rm PG}(8,q)$ is determined by the intersection of three hyperplanes. Consider three hyperplane $\Pi_{i}$, $i=1,2,3$ of ${\rm PG}(8,q)$ of equation $a_{i0}x_{0}+\ldots+a_{i8}x_{8}=0$ respectively. The point

[TABLE]

of $\mathscr{S}$ lies on the hyperplane $\Pi_{i}$ if $a_{i0}y_{0}^{3}+a_{i1}y_{0}^{2}y_{1}+a_{i2}y_{0}y_{1}^{2}+a_{i3}y_{2}y_{0}^{2}+a_{i4}y_{2}y_{0}y_{1}+a_{i5}y_{2}y_{1}^{2}+a_{i6}y_{0}^{2}y_{3}+a_{i7}y_{0}y_{1}y_{3}+a_{i8}y_{1}^{2}y_{3}=0,$ and we denote this equation by $g_{i}(y_{0},y_{1},y_{2},y_{3})=0$. The set of points of $\mathscr{S}$ satisfying $g_{i}=0$ for some $i\in\{1,2,3\}$ corresponds to a set of points $\mathcal{K}_{i}$ in the 3-space $M_{3}$. So $\mathcal{K}_{i}$ is a cubic primal in ${\rm PG}(3,q)$, that is a variety ${\mathscr{V}}^{3}_{2}$. Note that each primal $\mathcal{K}_{i}$, $i=1,2,3$ contains the kernel of $\sigma$. Further, for each $i\in\{1,2,3\}$ the four partial derivatives of the equation $g_{i}$ vanish for the points $(0,0,a,b)$, $a,b\in\mathbb{F}_{q}$. That is, the line $\ell=\{(0,0,a,b)\,|\,a,b\in\mathbb{F}_{q}\}$ is a singular line of each primal $\mathcal{K}_{1},\mathcal{K}_{2},\mathcal{K}_{3}$. As each plane meets a cubic primal ${\mathscr{V}}^{3}_{2}$ in a cubic curve ${\mathscr{V}}^{3}_{1}$, each plane through $\ell$ meets $\mathcal{K}_{i}$ in either the line $\ell$ counted thrice, or $\ell$ counted twice and another line.

To determine the number of points of $\mathscr{S}$ in a generic 5-space of ${\rm PG}(8,q)$, we determine the number of points in $\mathcal{K}_{1}\cap\mathcal{K}_{2}\cap\mathcal{K}_{3}$ which are not in the kernel of $\sigma$, with $\mathcal{K}_{1},\mathcal{K}_{2},\mathcal{K}_{3}$ in general position in relation to each other. To abbreviate the notation, we define nine functions $F_{i}=F_{i}(\theta)$, $G_{i}=G_{i}(\theta)$, $H_{i}=H_{i}(\theta)$, for $i=1,2,3$, where

[TABLE]

As we are looking at the generic case, we can assume $y_{1}\neq 0$ (since the plane $y_{1}=0$ containing a point of $\mathcal{K}_{1}\cap\mathcal{K}_{2}\cap\mathcal{K}_{3}$ imposes a condition on the $a_{ij}$, contradicting the surfaces being in general position). We can parameterise the points of $\mathcal{K}_{1}$ with $y_{1}\neq 0$ by letting $y_{1}=1$, $y_{0}=\theta$, $y_{2}=\phi$, so $g_{1}(\theta,1,\phi,w)=\theta F_{1}+\phi G_{1}+y_{3}H_{1}$, and $-y_{3}H_{1}=\theta F_{1}+\phi G_{1}$. If $H_{1}(\theta)=0$, this either gives the point $(0,1,0,1)$ which lies in the kernel of $\sigma$, and so is not of interest to us; or it forces a linear relationship between the coefficients $a_{ij}$, contradicting the surfaces being in general position. That is, we can assume $H_{1}(\theta)\neq 0$, and so the points on $\mathcal{K}_{1}$ with $y_{1}\neq 0$ are

[TABLE]

To find the intersection of $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$, we substitute the point $P_{\theta,\phi}$ into the equation $g_{2}=0$, giving the following curve of order nine,

[TABLE]

As $H_{1}\neq 0$ for points on $\mathcal{K}_{1}$, we have $\phi(H_{2}G_{1}-H_{1}G_{2})=\theta(H_{1}F_{2}-H_{2}F_{1}).$ So provided $H_{2}G_{1}-H_{1}G_{2}\neq 0$,

[TABLE]

So points on $\mathcal{K}_{1}\cap\mathcal{K}_{2}$ have form:

[TABLE]

Note that $F_{i},G_{i},H_{i}$ are quadratics in $\theta$, so the point $Q_{\theta}$ contains coordinates which are degree at most five in $\theta$. This corresponds to the fact the ${\mathscr{V}}^{3}_{2}\cap{\mathscr{V}}^{3}_{2}={\mathscr{V}}^{9}_{1}$, and as $\ell$ is a double line in both $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$, it counts four times in the intersection, leaving a quintic ${\mathscr{V}}^{5}_{1}$ which contains the point $(0,1,0,0)$.

Finally, we want to find points of the form $Q_{\theta}$ which lie on $\mathcal{K}_{3}$. Substituting $Q_{\theta}$ into the equation $g_{3}=0$ gives

[TABLE]

This is an equation of degree 15 in $\theta$ (recall ${\mathscr{V}}^{9}_{1}\cap{\mathscr{V}}^{3}_{2}={\mathscr{V}}^{15}_{1}$). Recall that $(G_{2}H_{1}-G_{1}H_{2})\neq 0$ for points on $\mathcal{K}_{1}\cap\mathcal{K}_{2}$, so eight of the roots do not contribute to a point in the general intersection. If $\theta=0$, then we have a point on the plane $y_{0}=0$, which lies in the kernel of the map $\sigma$, so is not of interest to us. The remaining term is degree six in $\theta$, so there are in general six solutions. That is, excluding points with $y_{0}=0$, our three cubic surfaces generically meet in six points. That is, the number of points of $\mathscr{S}$ in a generic 5-space is six, and so the variety ${\mathscr{V}}(\mathscr{S})$ has order six as required. $\square$

4 Coordinates in the Bose representation

4.1 Coordinates for a Bose plane $\llbracket X\rrbracket$

Let $\tau$ be a primitive element of $\mathbb{F}_{q}$ with primitive polynomial

[TABLE]

for $t_{0},t_{1},t_{2}\in\mathbb{F}_{q}$. Let $\bar{X}=(x,y,z)\in{\rm PG}(2,q^{3})$. Using homogeneous coordinates we can write $\bar{X}=\rho(x,y,z)=(\rho x,\rho y,\rho z)$ for any $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$. As $\rho$ varies, we generate a related point of ${\rm PG}(8,q)$, giving us the $(q^{3}-1)/(q-1)$ points of the plane $\llbracket X\rrbracket$. To describe this plane, we determine the coordinates of three non-collinear points in it.

Firstly, consider the representation of the coordinates of $\bar{X}$ with $\rho=1$. We can write

[TABLE]

for unique $x_{i},y_{i},z_{i}\in\mathbb{F}_{q}$. Related to this representation of $\bar{X}$ is the point $X_{0}$ of ${\rm PG}(8,q)$ where

[TABLE]

Secondly, consider the representation of the coordinates of $\bar{X}$ with $\rho=\tau$, so $\bar{X}=\tau(x,y,z)=(\tau x,\tau y,\tau z)$. The first coordinate expands as

[TABLE]

Similarly, we can expand the second and third coordinates $\tau y$ and $\tau z$. Corresponding to this representation of $\bar{X}$ is the point $X_{1}\in{\rm PG}(8,q)$ whose first three coordinates are

[TABLE]

Thirdly, consider the representation $\bar{X}=\tau^{2}(x,y,z)$. The first coordinate is $\tau^{2}x=\tau^{2}(x_{0}+x_{1}\tau+x_{2}\tau^{2})$. Expanding and simplifying yields

[TABLE]

Similarly, we expand the other two coordinates $\tau^{2}y$ and $\tau^{2}z$. Corresponding to this representation of $\bar{X}$ is the point $X_{2}\in{\rm PG}(8,q)$ whose first three coordinates are

[TABLE]

More generally, consider the representation $\bar{X}=\rho(x,y,z)=(\rho x,\rho y,\rho z)$, where $\rho=p_{0}+p_{1}\tau+p_{2}\tau^{2}$ for unique $p_{0},p_{1},p_{2}\in\mathbb{F}_{q}$. Straightforward expanding and simplifying yields the point of ${\rm PG}(8,q)$ corresponding to this representation of $\bar{X}$ is the point $p_{0}X_{0}+p_{1}X_{1}+p_{2}X_{2}$ (where $X_{0},X_{1},X_{2}$ are given in (1), (2), (3)). We have proved the following result.

Lemma 4.1

Let $\bar{X}$ be a point in ${\rm PG}(2,q^{3})$, then the Bose representation of $\bar{X}$ in ${\rm PG}(8,q)$ is the plane

[TABLE]

where $X_{0},X_{1},X_{2}$ are given in (1), (2) and (3).

4.2 The transversal planes of the Bose 2-spread $\mathbb{S}$

We determine the coordinates of the three transversal planes of the Bose 2-spread $\mathbb{S}$. First, we define three points in ${\rm PG}(8,q^{3})$, we state this as a definition for easy reference.

Definition 4.2

Denote the following constants ${\mathsf{a}}_{0},{\mathsf{a}}_{1},{\mathsf{a}}_{2}\in\mathbb{F}_{q^{3}}$ and the points $A_{0},A_{1},A_{2}\in{\rm PG}(8,q^{3})$, as

[TABLE]

Lemma 4.3

Using the notation of Definition 4.2, the three transversal planes of the Bose 2-spread $\mathbb{S}$ are

[TABLE]

Moreover, the point $\bar{X}=(x,y,z)\in{\rm PG}(2,q^{3})$ corresponds to the point $X$ in $\Gamma$ where X=\mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}}\cap\Gamma and

[TABLE]

where $X_{0},X_{1},X_{2}$ are given in (1), (2), (3).

Proof   Let $\bar{X}=(x,y,z)\in{\rm PG}(2,q^{3})$, so \mbox{\llbracket X\rrbracket}=\langle X_{0},X_{1},X_{2}\rangle as calculated above. Straightforward calculations show that in ${\rm PG}(8,q^{3})$ we have ${\mathsf{a}}_{0}X_{0}+{\mathsf{a}}_{1}X_{1}+{\mathsf{a}}_{2}X_{2}=xA_{0}+yA_{1}+zA_{2}.$ Hence these are the homogeneous coordinates of a point that lies in the plane \mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}} and in the plane $\langle A_{0},A_{1},A_{2}\rangle$. Hence the extension of the Bose spread element $\llbracket X\rrbracket$ meets the plane $\langle A_{0},A_{1},A_{2}\rangle$. As every extended Bose spread element meets the plane $\langle A_{0},A_{1},A_{2}\rangle$, it is one of the transversal planes, which we denote by $\Gamma$. The other two transversal planes are hence $\Gamma^{q}$ and $\Gamma^{q^{2}}$. $\square$

Note that the points $A_{0},A_{1},A_{2}$ of the transversal plane $\Gamma$ correspond to the fundamental triangle of ${\rm PG}(2,q^{2})$, namely $\bar{A}_{0}=(1,0,0)$, $\bar{A}_{1}=(0,1,0)$, $\bar{A}_{2}=(0,0,1)$.

We can now write $\llbracket X\rrbracket$ in terms of the point $X\in\Gamma$, namely

[TABLE]

where

[TABLE]

In order to study $\mathbb{F}_{q}$-subplanes later in this article, we need to develop a description of the plane in ${\rm PG}(8,q^{3})$ which contains the three points $xA_{0}+yA_{1}+zA_{2}$, $xA_{0}^{q}+yA_{1}^{q}+zA_{2}^{q}$ and $xA_{0}^{q^{2}}+yA_{1}^{q^{2}}+zA_{2}^{q^{2}}$. The next two subsections are devoted to carefully calculating a useful description of these points.

4.3 Conjugacy with respect to an $\mathbb{F}_{q}$-subplane

Let $\mathbb{F}_{q}$ denote the finite field of order $q$. An $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ is a subplane of ${\rm PG}(2,q^{3})$ which has order $q$, that is, a subplane isomorphic to ${\rm PG}(2,q)$. An $\mathbb{F}_{q}$-subline is a line of an $\mathbb{F}_{q}$-subplane, that is, isomorphic to ${\rm PG}(1,q)$. We will define conjugacy with respect to an $\mathbb{F}_{q}$-subplane or subline.

First consider the $\mathbb{F}_{q}$-subplane $\bar{\pi}_{0}={\rm PG}(2,q)$ of ${\rm PG}(2,q^{3})$. There are two collineations in {\mbox{P\GammaL}}(3,q^{3}) which have order 3 and fix $\bar{\pi}_{0}$ pointwise, namely $\bar{\mathsf{c}}$ and $\bar{\mathsf{c}}^{2}$ where

[TABLE]

In fact, ${\bar{G}}_{\pi_{0}}=\langle\bar{\mathsf{c}}\rangle$ is the unique subgroup of {\mbox{P\GammaL}}(3,q^{3}) which fixes $\bar{\pi}_{0}$ pointwise and has order 3. For $\bar{X}\in{\rm PG}(2,q^{3})\backslash\bar{\pi}_{0}$, the three points $\bar{X}$, $\bar{X}^{\bar{\mathsf{c}}}$, $\bar{X}^{\bar{\mathsf{c}}^{2}}$ are called conjugate with respect to $\bar{\pi}_{0}$. The collineation $\bar{\mathsf{c}}$ can be extended to act on points of ${\rm PG}(2,q^{6})$, and we denote the extension of $\bar{\mathsf{c}}$ to {\mbox{P\GammaL}}(3,q^{6}) by $\bar{\mathsf{c}}$ as well. That is, acting on points of ${\rm PG}(2,q^{6})$, we have $\bar{\mathsf{c}}(\bar{X})=\bar{X}^{q}$, so $\bar{\mathsf{c}}$ has order 6 when acting on ${\rm PG}(2,q^{6})$. Under the collineation $\bar{\mathsf{c}}$, a point $\bar{X}\in{\rm PG}(2,q^{6})$ lies in an orbit of size: $1$ if $\bar{X}\in\bar{\pi}_{0}$; $3$ if $\bar{X}\in{\rm PG}(2,q^{3})\backslash\bar{\pi}_{0}$; $2$ or $6$ if $\bar{X}\in{\rm PG}(2,q^{6})\backslash{\rm PG}(2,q^{3})$, depending on whether $\bar{X}$ belongs to the $\mathbb{F}_{q^{2}}$-subplane ${\rm PG}(2,q^{2})$ that contains $\bar{\pi}_{0}={\rm PG}(2,q)$, or not.

More generally, let $\bar{\pi}$ be an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$. Acting on the points of ${\rm PG}(2,q^{3})$ is a unique collineation group {\bar{G}}_{\pi}\subseteq{\mbox{P\GammaL}}(3,q^{3}) which fixes $\bar{\pi}$ pointwise and has order 3. We wish to distinguish between the two non-identity collineations in ${\bar{G}}_{\pi}$, and do so as follows. Consider any homography that maps $\bar{\pi}$ to $\bar{\pi}_{0}$, and denote its $3\times 3$ non-singular matrix over $\mathbb{F}_{q^{3}}$ by $A$, so if $\bar{X}\in\bar{\pi}$, then $A\bar{X}\in\bar{\pi}_{0}$. Let $\bar{\mathsf{c}}_{\pi}(\bar{X})=A^{-1}\,\bar{\mathsf{c}}(A\bar{X}).$ As $\bar{\mathsf{c}}_{\pi}$ has order 3 and fixes $\bar{\pi}$ pointwise, we have ${\bar{G}}_{\pi}=\langle\bar{\mathsf{c}}_{\pi}\rangle$. We expand $\bar{\mathsf{c}}_{\pi}$, and to avoid confusion use the following notation. For a $3\times 3$ matrix $A=(a_{ij})$, $i,j=1,2,3$, we let the matrix $A^{\sigma}=(a_{ij}^{q})$, $i,j=1,2,3$. Thus $\bar{\mathsf{c}}_{\pi}(\bar{X})=A^{-1}A^{\sigma}\bar{X}^{q}$, or writing $B=A^{-1}A^{\sigma}$, we have $\bar{\mathsf{c}}_{\pi}(\bar{X})=B\bar{X}^{q}$. That is, we can without loss of generality write ${\bar{G}}_{\pi}=\langle\bar{\mathsf{c}}_{\pi}\rangle$ with

[TABLE]

with $B$ a $3\times 3$ non-singular matrix over $\mathbb{F}_{q^{3}}$. For $\bar{X}\in{\rm PG}(2,q^{3})\backslash\bar{\pi}$, the three points $\bar{X}$, $\bar{X}^{\bar{\mathsf{c}}_{\pi}}$, $\bar{X}^{\bar{\mathsf{c}}^{2}_{\pi}}$ are called conjugate with respect to $\bar{\pi}$. Note that $\bar{X}$, $\bar{X}^{\bar{\mathsf{c}}_{\pi}}$, $\bar{X}^{\bar{\mathsf{c}}^{2}_{\pi}}$ are collinear if and only if $\bar{X}$ lies on an extended line of $\bar{\pi}$.

Suppose we extend the plane ${\rm PG}(2,q^{3})$ to ${\rm PG}(2,q^{6})$. Then the collineation \bar{\mathsf{c}}_{\pi}\in{\mbox{P\GammaL}}(3,q^{3}) has a natural extension to a collineation of {\mbox{P\GammaL}}(3,q^{6}) acting on points of ${\rm PG}(2,q^{6})$. We also denote the extended collineation by $\bar{\mathsf{c}}_{\pi}$, so

[TABLE]

The collineation $\bar{\mathsf{c}}_{\pi}$ has order 3 when acting on ${\rm PG}(2,q^{3})$, and order 6 when acting on ${\rm PG}(2,q^{6})$. Under the collineation $\bar{\mathsf{c}}_{\pi}$, a point $\bar{X}\in{\rm PG}(2,q^{6})$ lies in an orbit of size: $1$ if $\bar{X}\in\bar{\pi}$; $3$ if $\bar{X}\in{\rm PG}(2,q^{3})\backslash\bar{\pi}$; $2$ or $6$ if $\bar{X}\in{\rm PG}(2,q^{6})\backslash{\rm PG}(2,q^{3})$.

Similarly, if $\bar{b}$ is an $\mathbb{F}_{q}$-subline of a line $\bar{\ell}_{b}$ of ${\rm PG}(2,q^{3})$, then acting on the points of $\bar{\ell}_{b}$ is a unique collineation group \bar{G}_{b}\subseteq{\mbox{P\GammaL}}(2,q^{3}) of order 3 which fixes $\bar{b}$ pointwise. Moreover, ${\bar{G}}_{\pi}$ restricted to acting on $\bar{\ell}_{b}$ is isomorphic to $\bar{G}_{b}$ if and only if $\bar{b}$ is a line of $\bar{\pi}$. Without loss of generality we can write ${\bar{G}}_{b}=\langle\bar{\mathsf{c}}_{b}\rangle$ where for a point $\bar{X}\in\bar{\ell}_{b}$,

[TABLE]

with $D$ a non-singular matrix over $\mathbb{F}_{q^{3}}$.

4.4 Conjugacy with respect to the $\mathbb{F}_{q}$-subplane $\bar{\pi}_{0}={\rm PG}(2,q)$

We now return to the $\mathbb{F}_{q}$-subplane $\bar{\pi}_{0}={\rm PG}(2,q)$ of ${\rm PG}(2,q^{3})$ and look in more detail at the collineation \bar{\mathsf{c}}\in{\mbox{P\GammaL}}(3,q^{3}) defined by

[TABLE]

We have ${\bar{G}}_{\pi_{0}}=\{id,{\bar{\mathsf{c}}},{\bar{\mathsf{c}}}^{2}\}$ where for $\bar{X}=(x,y,z)\in{\rm PG}(2,q^{3})$,

[TABLE]

The collineation ${\bar{\mathsf{c}}}$ induces a map denoted $\mathsf{c}$ which acts on the transversal plane $\Gamma$ in the Bose representation. In ${\rm PG}(8,q^{3})$, $\pi$ is an $\mathbb{F}_{q}$-subplane of the transversal plane $\Gamma$, and ${\mathsf{c}}$ acts on the points of $\Gamma$, note that ${\mathsf{c}}$ is not a collineation of ${\rm PG}(8,q^{3})$. By Lemma 4.3, a point $X\in\Gamma$ has form

[TABLE]

for some $x,y,z\in\mathbb{F}_{q^{3}}$. So

[TABLE]

Moreover, we will be interested in the images of these points under the map $X\mapsto X^{q}$ of ${\rm PG}(8,q^{3})$, in particular,

[TABLE]

For $X=xA_{0}+yA_{1}+zA_{2}\in\Gamma$, we will be interested in the plane

[TABLE]

Consider the extension to ${\rm PG}(8,q^{6})$. Let {\mathsf{e}}\in{\mbox{P\GammaL}}(9,q^{6}) be the unique involution acting on points of ${\rm PG}(8,q^{6})$ fixing ${\rm PG}(8,q^{3})$ pointwise. We say the points $X,X^{\mathsf{e}}$ are conjugate with respect to the quadratic extension from ${\rm PG}(8,q^{3})$ to ${\rm PG}(8,q^{6})$, and have

[TABLE]

In particular, for $x,y,z\in\mathbb{F}_{q^{6}}$, not all zero, we have

[TABLE]

As $\mathsf{e}$ fixes ${\rm PG}(8,q^{3})$ pointwise, it fixes $\Gamma$ pointwise, hence $\mathsf{e}$ induces an automorphism of the plane $\Gamma$. Moreover, for a point $X\in\Gamma{{}^{\mbox{\tiny\char 72}}}\backslash\Gamma$, we have $X^{\mathsf{e}}=X^{q^{3}}=X^{\mathsf{c}^{3}}$. It is straightforward to verify that

[TABLE]

For $X=xA_{0}+yA_{1}+zA_{2}\in\Gamma{{}^{\mbox{\tiny\char 72}}}$, we will be interested in the plane

[TABLE]

Note that if $X\in\Gamma$, then this equation agrees with (6).

4.5 Coordinates in the Bruck-Bose representation

We can construct the Bruck-Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(6,q)$ by intersecting the Bose representation with a 6-space $\Sigma_{6,q}$ of ${\rm PG}(8,q)$ which contains a unique 5-space that meets $\mathbb{S}$ in $q^{3}+1$ planes. To obtain the same coordinates for the Bruck-Bose representation used in [2, Section 2.2], we take the line at infinity in ${\rm PG}(2,q^{3})$ to have equation $z=0$, which contains the points $\bar{X}=(1,0,0),\bar{Y}=(0,1,0)$. In ${\rm PG}(8,q^{3})$, this corresponds to the line $g$ in the transversal plane $\Gamma$ where $g=\langle A_{0},A_{1}\rangle$. Take $\Sigma_{6,q}$ as the 6-space of ${\rm PG}(8,q)$ consisting of the points $(x_{0},x_{1},x_{2},x_{3},x_{4},x_{5},x_{6},0,0)$, $x_{i}\in\mathbb{F}_{q}$, not all 0. Then {\Sigma}^{\mbox{\tiny\char 73}}_{\mbox{\raisebox{0.4pt}{\scalebox{0.6}{\infty}}}}=\langle g,g^{q},g^{q^{2}}\rangle and contains all points of form $(x_{0},x_{1},x_{2},x_{3},x_{4},x_{5},0,0,0)$. Further, $g,g^{q},g^{q^{2}}$ are the transversal lines of the regular 2-spread $\mathcal{S}$ in $\Sigma_{\infty}$. An affine point $\bar{P}=(x,y,1)=(x_{0}+x_{1}\tau+x_{2}\tau^{2},\ y_{0}+y_{1}\tau+y_{2}\tau^{2},\ 1)$ of ${\rm PG}(2,q^{3})$ corresponds to the affine point of $\Sigma_{6,q}\backslash\Sigma_{\infty}$ denoted [P]=\mbox{\llbracket P\rrbracket}\cap\Sigma_{6,q}, with coordinates $[P]=(x_{0},x_{1},x_{2},\ y_{0},y_{1},y_{2},\ 1,0,0)$.

5 Varieties in the Bose representation

In this section we use coordinates to show how to convert a variety of ${\rm PG}(2,q^{3})$ into a variety in the Bose representation ${\rm PG}(8,q)$. This generalises known techniques from the Bose representation of ${\rm PG}(2,q^{2})$ in ${\rm PG}(5,q)$, and can be used more generally to convert a primal of ${\rm PG}(2,q^{h})$ to the intersection of varieties in the Bose representation ${\rm PG}(3h-1,q)$. Of particular interest is the variety-extensions to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$.

We use the following notation. Let $f(x_{0},\ldots,x_{n})$ be a homogeneous form over $\mathbb{F}_{q^{h}}$ of degree $d$ in indeterminates $x_{0},\ldots,x_{n}$. Then $V(f)$ denotes the set of points $X\in{\rm PG}(n,q^{h})$ such that $f(X)=0$. If $f_{1},\ldots,f_{k}$ are homogeneous forms over $\mathbb{F}_{q^{h}}$ in indeterminates $x_{0},\ldots,x_{n}$, then $V(f_{1},\ldots,f_{k})=V(f_{1})\cap\cdots\cap V(f_{k})$ denotes the set of points $X\in{\rm PG}(n,q^{h})$ such that $f_{i}(X)=0$, $i=1,\ldots,k$.

5.1 Varieties of ${\rm PG}(2,q^{3})$

We study varieties of ${\rm PG}(2,q^{3})$ and their corresponding structure in the Bose representation. This study proceeds as follows. We first look in Lemma 5.1 at a homogeneous equation in ${\rm PG}(2,q^{3})$ and convert it to a homogeneous equation in ${\rm PG}(8,q)$. Next, in Lemma 5.2, we look at a variety $\overline{\mathcal{K}}$ in ${\rm PG}(2,q^{3})$, and use the calculations from Lemma 5.1 to show that the corresponding set of points in the Bose representation in ${\rm PG}(8,q)$, forms a variety denoted {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}). We determine the extension {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}){{}^{\mbox{\tiny\char 73}}} to ${\rm PG}(8,q^{3})$ in Lemma 5.3. Lemma 5.4 is a stepping stone to Lemma 5.5 which gives a geometric description of {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}){{}^{\mbox{\tiny\char 73}}}. The results of these lemmas is then summarised in Theorem 5.6. In Section 5.3 we then use this result to study conics of ${\rm PG}(2,q^{3})$ in the Bose representation.

Lemma 5.1

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$. Using $\tau$ as in Section 4.1, we write $x=x_{0}+\tau x_{1}+\tau^{2}x_{2}$, $y=y_{0}+\tau y_{1}+\tau^{2}y_{2}$, $z=z_{0}+\tau z_{1}+\tau^{2}z_{2}$ for unique $x_{i},y_{i},z_{i}\in\mathbb{F}_{q}$. Expanding and simplifying yields

[TABLE]

where $f_{i}=f_{i}(x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2})$ is a homogeneous equation of degree $k$ over $\mathbb{F}_{q}$, $i=0,1,2$.

Proof   Recall from Section 4.1, $\tau$ is a primitive element in $\mathbb{F}_{q^{3}}$ satisfying $\tau^{3}=t_{0}+t_{1}\tau+t_{2}\tau^{2}$ for some $t_{0},t_{1},t_{2}\in\mathbb{F}_{q}$. Let $\bar{F}=\bar{F}(x,y,z)$ be a homogeneous form over $\mathbb{F}_{q^{3}}$ of degree $k$ in indeterminates $x,y,z$. For indeterminates $x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2}$, substitute $x=x_{0}+x_{1}\tau+x_{2}\tau^{2}$, $y=y_{0}+y_{1}\tau+y_{2}\tau^{2}$, $z=z_{0}+z_{1}\tau+z_{2}\tau^{2}$ in $\bar{F}$ to obtain the homogeneous form $G(x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2})$ of degree $k$. For each coefficient $a\in\mathbb{F}_{q^{3}}$ of $G$, rewrite as $a=a_{0}+a_{1}\tau+a_{2}\tau^{2}$ for unique $a_{0},a_{1},a_{2}\in\mathbb{F}_{q}$. Then using $\tau^{3}=t_{0}+t_{1}\tau+t_{2}\tau^{2}$, we may write

[TABLE]

for unique homogeneous forms $f_{0},f_{1},f_{2}$ over $\mathbb{F}_{q}$ of degree $k$ in indeterminates $x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2}$. $\square$

Note that $G$ is a homogeneous equation with coefficients in $\mathbb{F}_{q^{3}}$, so it does not make sense to talk about a variety in ${\rm PG}(8,q)$ corresponding to $G$. However, $f_{0}$, $f_{1}$ and $f_{2}$ are homogeneous equations with coefficients in $\mathbb{F}_{q}$, so we have the corresponding varieties $V(f_{0})$, $V(f_{1})$ and $V(f_{2})$ in ${\rm PG}(8,q)$. Moreover, a point $P\in{\rm PG}(8,q)$ satisfies $G(P)=0$ if and only if $f_{0}(P)=f_{1}(P)=f_{2}(P)=0$ if and only if $P\in V(f_{0})\cap V(f_{1})\cap V(f_{2})=V(f_{0},f_{1},f_{2})$.

Lemma 5.2

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$, and let $\bar{\mathcal{K}}=V(\bar{F})$ be the corresponding variety in ${\rm PG}(2,q^{3})$. Let $G$, $f_{0}$, $f_{1}$ and $f_{2}$ be derived from $\bar{F}$ as in Lemma 5.1. In the Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$, the pointset of $\llbracket\mathcal{K}\rrbracket$ coincides with the pointset of the variety $V(f_{0},f_{1},f_{2})=V(f_{0})\cap V(f_{1})\cap V(f_{2})$. We define {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket})=V(f_{0},f_{1},f_{2}).

Proof   Using the coordinates described in Section 4.1, a point $Q$ in ${\rm PG}(8,q)$ corresponds to a unique point of ${\rm PG}(2,q^{3})$ which we denote by $\bar{X}_{Q}$ as follows. Let $Q$ have homogeneous coordinates

[TABLE]

for any $\lambda\in\mathbb{F}_{q}\backslash\{0\}$. The corresponding point in ${\rm PG}(2,q^{3})$ has coordinates

[TABLE]

By Lemma 5.1, $G(Q)=\bar{F}(\bar{X}_{Q})$, so $G(Q)=0$ if and only if $\bar{F}(\bar{X}_{Q})=0.$ As noted above, $G(Q)=0$ if and only if $Q\in V(f_{0},f_{1},f_{2})$. Hence a point $Q\in{\rm PG}(8,q)$ lies in $V(f_{0},f_{1},f_{2})$ if and only if the corresponding point $\bar{X}_{Q}$ lies in $V(\bar{F})=\bar{\mathcal{K}}$.

We now consider the converse, a point $\bar{P}$ in ${\rm PG}(2,q^{3})$ corresponds to $q^{2}+q+1$ points in ${\rm PG}(8,q)$, namely the $q^{2}+q+1$ points of the plane $\llbracket P\rrbracket$. We want to show that if $\bar{P}\in\bar{\mathcal{K}}$, then in ${\rm PG}(8,q)$, every point Y\in\mbox{\llbracket P\rrbracket} lies in the variety $V(f_{0},f_{1},f_{2})$. The point $\bar{P}$ has homogeneous coordinates $\bar{P}=(x,y,z)\equiv\rho(x,y,z)$, for any $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$. By Lemma 4.1, we have \mbox{\llbracket P\rrbracket}=\langle X_{0},X_{1},X_{2}\rangle with points $X_{0},X_{1},X_{2}$ defined in Section 4.1. Moreover, writing $\rho=p_{0}+p_{1}\tau+p_{2}\tau^{2}$, for $p_{0},p_{1},p_{2}\in\mathbb{F}_{q}$, then by Section 4.1, the point in $\llbracket P\rrbracket$ corresponding to the coordinates $\rho(x,y,z)$ is the point

[TABLE]

So $\bar{P}\in\bar{K}$ if and only if \bar{F}(\bar{P})=\bar{F}\big{(}\rho(x,y,z)\big{)}=0 for all $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$ if and only if $G(Y_{\rho})=0$ for all $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$ if and only if $Y_{\rho}\in V(f_{0},f_{1},f_{2})$ for all $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$. Note that the points $Y_{\rho}$, $\rho\in\mathbb{F}_{q^{3}}\backslash\{0\}$ are exactly the points of the plane $\llbracket P\rrbracket$. Hence $\bar{P}\in\bar{\mathcal{K}}$ if and only if \mbox{\llbracket P\rrbracket}\subseteq V(f_{0},f_{1},f_{2}). It follows that \mbox{\llbracket\mathcal{K}\rrbracket}=V(f_{0},f_{1},f_{2}) as required. $\square$

As noted above, $G$ is a homogeneous equation with coefficients in $\mathbb{F}_{q^{3}}$, so $G$ corresponds to a variety in ${\rm PG}(8,q^{3})$. That is, we let $V(G)$ denote the variety of ${\rm PG}(8,q^{3})$ consisting of the points $X$ of ${\rm PG}(8,q^{3})$ satisfying $G(X)=0$. Further, we can define the varieties $V(G^{q})$ and $V(G^{q^{2}})$ in ${\rm PG}(8,q^{3})$.

Lemma 5.3

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$, and let $\bar{\mathcal{K}}=V(\bar{F})$ be the corresponding variety in ${\rm PG}(2,q^{3})$. Let $G$, $f_{0}$, $f_{1}$ and $f_{2}$ be derived from $\bar{F}$ as in Lemma 5.1, and let {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket})=V(f_{0},f_{1},f_{2}). The variety-extension of {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}) to ${\rm PG}(8,q^{3})$ is

[TABLE]

Proof   As {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket})=V(f_{0})\cap V(f_{1})\cap V(f_{2}), we have

[TABLE]

The set of points of ${\rm PG}(8,q^{3})$ satisfying the three equations $f_{0}=0$, $f_{1}=0$, $f_{2}=0$ is equivalent to the set of points satisfying any three linearly independent equations of form $\lambda_{0}f_{0}+\lambda_{1}f_{1}+\lambda_{2}f_{2}$ where $\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{F}_{q^{3}}$. Recalling that $f_{0},f_{1},f_{2}$ are equations over $\mathbb{F}_{q}$, we consider the three linearly independent equations

[TABLE]

So a point $P\in{\rm PG}(8,q^{3})$ satisfies $f_{0}(P)=f_{1}(P)=f_{2}(P)=0$ if and only if $G(P)=G^{q}(P)=G^{q^{2}}(P)=0$. Hence

[TABLE]

as required. $\square$

Lemma 5.4

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$, and let $\bar{\mathcal{K}}=V(\bar{F})$ be the corresponding variety in ${\rm PG}(2,q^{3})$. Let $G$, $f_{0}$, $f_{1}$ and $f_{2}$ be derived from $\bar{F}$ as in Lemma 5.1, and let {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket})=V(f_{0},f_{1},f_{2}). Consider the Bose representation of ${\rm PG}(2,q^{3})$ with transversal planes $\Gamma,\Gamma^{q},\Gamma^{q^{2}}$ in ${\rm PG}(8,q^{3})$. Then the variety $V(G)$ of ${\rm PG}(8,q^{3})$ is a cone with base $V(G)\cap\Gamma=\mathcal{K}$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$.

Proof   We first determine how the variety $V(G)$ meets the transversal plane $\Gamma$. By Lemma 4.3, a point $Q$ in the transversal plane $\Gamma$ has coordinates

[TABLE]

for some $x,y,z\in\mathbb{F}_{q^{3}}$, with $A_{i}$ and $\mathsf{a}_{i}$ as in Definition 4.2. Moreover, the points of $\Gamma$ are in one-to-one correspondence with the points of ${\rm PG}(2,q^{3})$ and $Q$ corresponds to the point $\bar{Q}$ of ${\rm PG}(2,q^{3})$ where

[TABLE]

By Lemma 5.1, $G(Q)=\bar{F}(\bar{Q})$. So the point $Q$ of $\Gamma$ lies in the variety $V(G)$ if and only if the point $\bar{Q}$ of ${\rm PG}(2,q^{3})$ lies in $V(\bar{F})=\bar{\mathcal{K}}$. That is, $Q\in V(G)\cap\Gamma$ if and only if $\bar{Q}\in\bar{\mathcal{K}}$, and so $V(G)\cap\Gamma=\mathcal{K}$ as required.

We now determine the base of $V(G)$. As $V(G)$ does not contain the plane $\Gamma$, the maximum dimension of the singular space of $V(G)$ is five. We show that the 5-space $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$ is the singular space of $V(G)$ by showing that every point of $V(G)$ lies on a line joining a point $Q\in\mathcal{K}$ to a point $R\in\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$. Let $Q\in\mathcal{K}$, $\ R\in\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$ and $P\in QR$. By Lemma 4.3, $\Gamma=\langle A_{0},A_{1},A_{2}\rangle$, $\Gamma^{q}=\langle A_{0}^{q},A_{1}^{q},A_{2}^{q}\rangle$ and $\Gamma^{q^{2}}=\langle A_{0}^{q^{2}},A_{1}^{q^{2}},A_{2}^{q^{2}}\rangle$. So $P$ has homogeneous coordinates of form

[TABLE]

for some $x,y,z,r,s,t,u,v,w\in\mathbb{F}_{q^{3}}$. Simplifying $P$ (using the coordinates for $A_{i}$ and $\mathsf{a}_{i}$ as in Definition 4.2) we calculate the first three coordinates of $P$ are

[TABLE]

As in the proof of Lemma 5.2, recall that $P$ corresponds to a unique point of ${\rm PG}(2,q^{3})$ which we denote by $\bar{X}_{P}$, and the first coordinate of $\bar{X}_{P}$ is

[TABLE]

Straightforward manipulation shows that

[TABLE]

and using this we simplify the first coordinate of $\bar{X}_{P}$ to $({\mathsf{a}}_{0}+\tau{\mathsf{a}}_{1}+\tau^{2}{\mathsf{a}}_{2})\,x$. Similarly we calculate the other coordinates of $P$, and the coordinates of $\bar{X}_{P}$ are

[TABLE]

By Lemma 5.1, $G(P)=\bar{F}(\bar{X}_{P})$, so $P$ lies in the variety $V(G)$ if and only if the point $\bar{X}_{P}=\bar{Q}$ lies in $\bar{\mathcal{K}}$. That is, if $P$ is on a line joining $Q\in\Gamma$ with a point $R$ of $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$, then $G(P)=0$ if and only if $\bar{F}(\bar{Q})=0$. That is, $P\in V(G)$ if and only if $\bar{Q}\in\bar{\mathcal{K}}$ if and only if $Q\in\mathcal{K}$. Hence $V(G)$ is a cone with base $\mathcal{K}$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$. $\square$

Lemma 5.5

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$, and let $\bar{\mathcal{K}}=V(\bar{F})$ be the corresponding variety in ${\rm PG}(2,q^{3})$ and let {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}) be the corresponding variety in ${\rm PG}(8,q)$ as defined in Lemma 5.2. Then in ${\rm PG}(8,q^{3})$, the pointset of {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}){{}^{\mbox{\tiny\char 73}}} is equivalent to the pointset of the planes $\{\langle X,Y^{q},Z^{q^{2}}\rangle\,|\,X,Y,Z\in\mathcal{K}\}$.

Proof   By Lemma 5.4, $V(G)$ is a cone with base $V(G)\cap\Gamma=\mathcal{K}$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$. Hence $V(G^{q})$ is a cone with base $\mathcal{K}^{q^{2}}$ in $\Gamma^{q^{2}}$ and vertex $\langle\Gamma^{q},\Gamma\rangle$; and $V(G^{q^{2}})$ is a cone with base $\mathcal{K}^{q}$ and vertex $\langle\Gamma,\Gamma^{q^{2}}\rangle$. Note that each of the three cones is a set of T-planes. Hence the intersection of the three cones $V(G)$, $V(G^{q})$, $V(G^{q^{2}})$ is the set of points lying on the T-planes

[TABLE]

as required. $\square$

Summarising the results of this section we have proved the following Theorem.

Theorem 5.6

Let $\bar{F}(x,y,z)$ be a homogeneous equation of degree $k$ over $\mathbb{F}_{q^{3}}$, and let $\bar{\mathcal{K}}=V(\bar{F})$ be the corresponding variety in ${\rm PG}(2,q^{3})$. Using $\tau$ as in Section 4.1, we write $x=x_{0}+\tau x_{1}+\tau^{2}x_{2}$, $y=y_{0}+\tau y_{1}+\tau^{2}y_{2}$, $z=z_{0}+\tau z_{1}+\tau^{2}z_{2}$ for unique $x_{i},y_{i},z_{i}\in\mathbb{F}_{q}$. Expanding and simplifying yields

[TABLE]

where $f_{i}=f_{i}(x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2})$ is a homogeneous equation of degree $k$ over $\mathbb{F}_{q}$, $i=0,1,2$. Consider the Bose representation of ${\rm PG}(2,q^{3})$.

In ${\rm PG}(8,q)$, the pointset of $\llbracket\mathcal{K}\rrbracket$ forms a variety $V(f_{0},f_{1},f_{2})$ which we denote by {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}). 2. 2.

In ${\rm PG}(8,q^{3})$,

[TABLE]

and the pointset of {\mathscr{V}}(\mbox{\llbracket\mathcal{K}\rrbracket}){{}^{\mbox{\tiny\char 73}}} is equivalent to the pointset of the planes $\{\langle X,Y^{q},Z^{q^{2}}\rangle\,|\,X,Y,Z\in\mathcal{K}\}$.

5.2 A Bose representation convention regarding variety-extensions

Using the notation of Theorem 5.6, we have the pointset of the Bose representation of the variety $V(\bar{F})$ of ${\rm PG}(2,q^{3})$ is the variety $V(f_{0},f_{1},f_{2})$ of ${\rm PG}(8,q)$. Note that this pointset may be the pointset of more than one variety of ${\rm PG}(8,q)$. In particular, as $a^{q+r}=a^{1+r}$, for all $a\in\mathbb{F}_{q}$, where $r\geq 0$, if the form $f_{i}$ of degree $d$ only contains terms having an indeterminate raised to a power greater than or equal to $q$, then reducing these exponents by $q-1$ yields a homogeneous form $g_{i}$ of degree $d-q+1$ such that the evaluations of $f_{i}$ and $g_{i}$ over $\mathbb{F}_{q}$ are equal. Thus the pointsets of the varieties $V(f_{0},f_{1},f_{2})$ and $V(g_{0},g_{1},g_{2})$ will be equal in ${\rm PG}(8,q)$. Note, however, that these evaluations may not be equal over the extension $\mathbb{F}_{q^{3}}$ and consequently these varieties may not yield the same pointset when extended to the extended projective space ${\rm PG}(8,q^{3})$.

As this is an important notion, we illustrate this with two examples in the well known Bose representation of ${\rm PG}(2,q^{2})$ in ${\rm PG}(5,q)$.

Example 1 Let $\bar{\cal U}$ be a classical unital of ${\rm PG}(2,q^{2})$. So $\bar{\cal U}$ is projectively equivalent to the variety $V(\bar{F})$ where $\bar{F}(x,y,z)=x^{q+1}+y^{q+1}+z^{q+1}$. Let $\tau\in\mathbb{F}_{q^{2}}$ have minimal polynomial $x^{2}-t_{1}x-t_{0}$. Let $x_{0},x_{1},y_{0},y_{1},z_{0},z_{1}$ be indeterminates and substitute $x=x_{0}+x_{1}\tau,y=y_{0}+y_{1}\tau,z=z_{0}+z_{1}\tau$ in $\bar{F}$ to obtain the form

[TABLE]

As $\tau^{q}=t_{1}-\tau$ and $\tau\tau^{q}=-t_{0}$, this becomes $G(x_{0},x_{1},y_{0},y_{1},z_{0},z_{1})=x_{0}^{q+1}+\tau x_{0}^{q}x_{1}+(t_{1}-\tau)x_{0}x_{1}^{q}-t_{0}x_{1}^{q+1}+y_{0}^{q+1}+\tau y_{0}^{q}y_{1}+(t_{1}-\tau)y_{0}y_{1}^{q})-t_{0}y_{1}^{q+1}+z_{0}^{q+1}+\tau z_{0}^{q}z_{1}+(t_{1}-\tau)z_{0}z_{1}^{q})-t_{0}z_{1}^{q+1}.$ It follows that

[TABLE]

and

[TABLE]

We now consider the corresponding forms $g_{i}$ where the exponents of the forms $f_{i}$ are reduced by $q-1$, giving

[TABLE]

and

[TABLE]

The forms $f_{0},f_{1}$ have degree $q+1$, while the form $g_{0}$ has degree $2$ and $g_{1}$ is identically zero. As we have two varieties $\mathcal{Q}=V(g_{0},g_{1})$ and ${\mathscr{V}}=V(f_{0},f_{1})$ covering the same pointset of ${\rm PG}(5,q)$, there are two variety-extensions ${\mathcal{Q}^{\mbox{\tiny\char 73}}}$ and ${{\mathscr{V}}^{\mbox{\tiny\char 73}}}$ to ${\rm PG}(5,q^{2})$.

There are two possible ways to consider the extension. The standard way used in the literature to extend (see [1]) is to consider the variety of ${\rm PG}(5,q)$ corresponding to a unital to be the elliptic quadric $\mathcal{Q}=V(g_{0},g_{1})=V(g_{0})$. The variety-extension of this elliptic quadric to ${\rm PG}(5,q^{2})$ is a hyperbolic quadric that contains the transversal plane $\Gamma$ of the Bose representation. Alternatively, we can consider the variety of ${\rm PG}(5,q)$ corresponding to a unital to be the variety ${\mathscr{V}}=V(f_{0},f_{1})$. By Theorem 5.6, the extension of the variety ${\mathscr{V}}=V(f_{0},f_{1})$ to ${\rm PG}(5,q^{2})$ is a variety which consists of the points on the lines $XY^{q}$ for points $X,Y\in{\cal U}$ (where ${\cal U}$ is the classical unital in the transversal plane $\Gamma$ that corresponds to $\bar{\cal U}$). That is, the variety-extension meets the transversal plane $\Gamma$ in a unital.

Example 2 Consider the Bose representation of the subplane $\bar{\pi}_{0}={\rm PG}(2,q)$ of ${\rm PG}(2,q^{2})$. Let $\bar{F}_{0}(x,y,z)=xy^{q}-x^{q}y$, $\bar{F}_{1}(x,y,z)=yz^{q}-y^{q}z$, $\bar{F}_{2}(x,y,z)=zx^{q}-z^{q}x$, then $\bar{\pi}_{0}$ may be described as the variety $V(\bar{F}_{0},\bar{F}_{1},\bar{F}_{2})$. A similar analysis to Example 1 yields varieties $V(f_{i,j})$, $i=0,1,2$, $j=0,1$, with $f_{0,0}=x_{0}y_{0}^{q}-x_{0}^{q}y_{0}+t_{1}(x_{0}y_{1}^{q}-x_{1}y^{q}y_{0})-t_{0}(x_{1}y_{1}^{q}-x_{1}^{q}y_{1})$, $f_{0,1}=x_{0}y_{1}^{q}+x_{1}^{q}y_{0}+x_{1}y_{0}^{q}-x_{0}^{q}y_{1}$, and so on. Hence reducing the exponents gives varieties $V(g_{i,j})$, $i=0,1,2$, $j=0,1$, with $g_{0,0}=t_{1}(x_{0}y_{1}-x_{1}y_{0})$, $g_{0,1}=-2(x_{0}y_{1}-x_{1}y_{0})$, and so on. The forms defining the variety $\mathcal{K}=V(f_{0,0},f_{0,1},f_{1,0},f_{1,1},f_{2,0},f_{2,1})$ have degree $q+1$, and the forms defining the variety $\mathcal{K}^{\prime}=V(g_{0,0},g_{0,1},g_{1,0},g_{1,1},g_{2,0},g_{2,1})$ are quadrics. The two varieties $\mathcal{K},\mathcal{K}^{\prime}$ have identical pointsets in ${\rm PG}(5,q)$, but their extensions to ${\rm PG}(5,q^{2})$ differ. The extension of $\mathcal{K}$ to ${\rm PG}(5,q^{2})$ meets the transversal plane $\Gamma$ in the Baer subplane $\pi_{0}$, but the extension of $\mathcal{K}^{\prime}$ (a Segre variety) to ${\rm PG}(5,q^{2})$ contains the transversal plane $\Gamma$. The standard way used in the literature to extend (see [1]) is using the variety $\mathcal{K}^{\prime}=V(g_{0,0},g_{0,1},g_{1,0},g_{1,1},g_{2,0},g_{2,1})$.

In Section 6 we consider the Bose representation of an $\mathbb{F}_{q}$-subplane and its relation to the Segre variety and its extension. In Section 7, we consider an $\mathbb{F}_{q}$-conic of an $\mathbb{F}_{q}$-subplane, and its representation on the Segre variety and the extended Segre variety. In light of this discussion, we are careful in this work to describe the variety of ${\rm PG}(8,q)$ that we extend.

5.3 Application to conics in the Bose representation

We use Theorem 5.6 to show that a non-degenerate conic of ${\rm PG}(2,q^{3})$ corresponds to the intersection of three quadrics in $PG(8,q)$. Further, we show that the points of the variety-extension lie on a set of planes.

Theorem 5.7

Let $\bar{\mathcal{O}}$ be a non-degenerate conic in ${\rm PG}(2,q^{3})$. Consider the Bose representation.

In ${\rm PG}(8,q)$, the pointset of $\llbracket{\mathcal{O}}\rrbracket$ coincides with the pointset of a variety {\mathscr{V}}(\mbox{\llbracket\mathcal{O}\rrbracket})=\mathcal{Q}_{0}\cap\mathcal{Q}_{1}\cap\mathcal{Q}_{2}, where $\mathcal{Q}_{1},\mathcal{Q}_{2},\mathcal{Q}_{3}$ are quadrics of ${\rm PG}(8,q)$. 2. 2.

In ${\rm PG}(8,q^{3})$, the variety-extension {\mathscr{V}}(\mbox{\llbracket\mathcal{O}\rrbracket}){{}^{\mbox{\tiny\char 73}}}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{0}\cap{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}\cap{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{2} has pointset which coincides with the points on the planes $\{\langle X,Y^{q},Z^{q^{2}}\rangle\,|\,X,Y,Z\in\mathcal{O}\}.$ 3. 3.

In ${\rm PG}(8,q^{6})$, the variety-extension {\mathscr{V}}(\mbox{\llbracket\mathcal{O}\rrbracket}){{}^{\mbox{\tiny\char 72}}}={\mathcal{Q}}^{\mbox{\tiny\char 72}}_{0}\cap{\mathcal{Q}}^{\mbox{\tiny\char 72}}_{1}\cap{\mathcal{Q}}^{\mbox{\tiny\char 72}}_{2} has pointset which coincides with the points on the planes $\{\langle X,Y^{q},Z^{q^{2}}\rangle\,|\,X,Y,Z\in\mathcal{O}{{}^{\mbox{\tiny\char 72}}}\}$, where $\mathcal{O}{{}^{\mbox{\tiny\char 72}}}$ is the quadratic extension of the conic $\mathcal{O}$ to $\Gamma{{}^{\mbox{\tiny\char 72}}}\subset{\rm PG}(8,q^{6})$.

Proof   Let $\bar{\mathcal{O}}$ be a non-degenerate conic in ${\rm PG}(2,q^{3})$, so $\bar{\mathcal{O}}$ is the set of points satisfying a homogeneous equation $F(x,y,z)=0$ of degree 2 over $\mathbb{F}_{q^{3}}$. As in Theorem 5.6, we can write $F(x,y,z)=f(x_{0},x_{1},x_{2},y_{0},y_{1},y_{2},z_{0},z_{1},z_{2})=f_{0}+f_{1}\tau+f_{2}\tau^{2}$ where $f_{0},f_{1},f_{2}$ are homogeneous equations of degree 2 over $\mathbb{F}_{q}$. Hence the set of points of ${\rm PG}(8,q)$ satisfying $f_{i}=0$ form a quadric denoted $\mathcal{Q}_{i}$, $i=0,1,2$. Further, by Theorem 5.6(1), in ${\rm PG}(8,q)$, the pointset of $\llbracket\mathcal{O}\rrbracket$ form a variety ${\mathscr{V}}(\mathcal{O})=\mathcal{Q}_{0}\cap\mathcal{Q}_{1}\cap\mathcal{Q}_{2}$. The variety-extension to ${\rm PG}(8,q^{3})$ is ${\mathscr{V}}(\mathcal{O}){{}^{\mbox{\tiny\char 73}}}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{0}\cap{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}\cap{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{2}$, where $\mathcal{Q}_{i}{{}^{\mbox{\tiny\char 73}}}$ is now a quadric of ${\rm PG}(8,q^{3})$. Moreover, ${\mathscr{V}}(\mathcal{O}){{}^{\mbox{\tiny\char 73}}}$ is also defined by any three independent quadrics in the set $\lambda_{0}\mathcal{Q}_{0}+\lambda_{1}\mathcal{Q}_{1}+\lambda_{2}\mathcal{Q}_{2}$ with $\lambda_{0},\lambda_{1},\lambda_{2}\in\mathbb{F}_{q^{3}}$ (where $\lambda_{0}\mathcal{Q}_{0}+\lambda_{1}\mathcal{Q}_{1}+\lambda_{2}\mathcal{Q}_{2}$ denotes the quadric with homogeneous equation $\lambda_{0}f_{0}+\lambda_{1}f_{1}+\lambda_{2}f_{2}$). Consider the following three independent quadrics of ${\rm PG}(8,q^{3})$, ${\cal T}_{0}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{0}+\tau{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}+\tau^{2}{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{2}$, ${\cal T}_{1}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{0}+\tau^{q}{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}+\tau^{2q}{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{2}$, ${\cal T}_{2}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{0}+\tau^{q^{2}}{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}+\tau^{2q^{2}}{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{2}$. If we denote the equation of ${\cal T}_{0}$ as $g=f_{0}+\tau f_{1}+\tau^{2}f_{2}$, then ${\cal T}_{1}$ has equation $g^{q}$ and ${\cal T}_{2}$ has equation $g^{q^{2}}$. By Theorem 5.6, ${\cal T}_{0}$ is a cone with base $\mathcal{O}$ in $\ \Gamma$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$. Similarly ${\cal T}_{1}$ is a cone with base $\mathcal{O}^{q}$ and vertex $\langle\Gamma,\Gamma^{q^{2}}\rangle$, and ${\cal T}_{2}$ is a cone with base $\mathcal{O}^{q^{2}}$ and vertex $\langle\Gamma,\Gamma^{q}\rangle$. The intersection of these three cones is the set of T-planes that contain a point of $\mathcal{O}$, a point of $\mathcal{O}^{q}$ and a point of $\mathcal{O}^{q^{2}}$, as in part 2.

For part 3, the extension of ${\mathscr{V}}(\mathcal{O})$ to ${\rm PG}(8,q^{6})$ is ${\mathscr{V}}(\mathcal{O}){{}^{\mbox{\tiny\char 72}}}={\cal T}_{0}{{}^{\mbox{\tiny\char 72}}}\cap{\cal T}_{1}{{}^{\mbox{\tiny\char 72}}}\cap{\cal T}_{2}{{}^{\mbox{\tiny\char 72}}}$. In ${\rm PG}(8,q^{3})$, the quadric ${\cal T}_{0}$ is a cone with base $\mathcal{O}$ in $\ \Gamma$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle$. Hence the quadric extension ${\cal T}_{0}{{}^{\mbox{\tiny\char 72}}}$ is a cone with base $\mathcal{O}{{}^{\mbox{\tiny\char 72}}}$ in $\ \Gamma{{}^{\mbox{\tiny\char 72}}}$ and vertex $\langle\Gamma^{q},\Gamma^{q^{2}}\rangle{{}^{\mbox{\tiny\char 72}}}=\langle(\Gamma^{q}){{}^{\mbox{\tiny\char 72}}},(\Gamma^{q^{2}}){{}^{\mbox{\tiny\char 72}}}\rangle$.

We can similarly describe the quadrics ${\cal T}_{1}{{}^{\mbox{\tiny\char 72}}},{\cal T}_{2}{{}^{\mbox{\tiny\char 72}}}$ as cones. The intersection of these three cones is the set of T-planes that contain a point of $\mathcal{O}{{}^{\mbox{\tiny\char 72}}}$, a point of $(\mathcal{O}^{q}){{}^{\mbox{\tiny\char 72}}}=(\mathcal{O}{{}^{\mbox{\tiny\char 72}}})^{q}$ and a point of $(\mathcal{O}^{q^{2}}){{}^{\mbox{\tiny\char 72}}}=(\mathcal{O}{{}^{\mbox{\tiny\char 72}}})^{q^{2}}$. That is, the set of planes of form $\langle X,Y^{q},Z^{q^{2}}\rangle$ where $X,Y,Z\in\mathcal{O}{{}^{\mbox{\tiny\char 72}}}$, proving part 3. $\square$

6 Sublines and

subplanes in the Bose representation

In this section we determine the Bose representation of $\mathbb{F}_{q}$-sublines and $\mathbb{F}_{q}$-subplanes of ${\rm PG}(2,q^{3})$. We show in Section 6.1 that an $\mathbb{F}_{q}$-subline of ${\rm PG}(2,q^{3})$ corresponds to a 2-regulus contained in a 5-space; and that an $\mathbb{F}_{q}$-subplane corresponds to a Segre variety $\mathcal{S}_{2;2}$. Moreover, we later wish to characterise $\mathbb{F}_{q}$-conics, and in order to do so, we describe in Section 6.2 the extensions of a 2-regulus and a $\mathcal{S}_{2;2}$ to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$.

6.1 $\mathbb{F}_{q}$-sublines and subplanes in ${\rm PG}(8,q)$

Let $\bar{\pi}$ be an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$, we show that the Bose representation is a Segre variety.

Theorem 6.1

Let $\bar{\pi}$ be a $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ then in ${\rm PG}(8,q)$, the planes \{\mbox{\llbracket X\rrbracket}\,|\,X\in\pi\} of the Bose representation $\llbracket\pi\rrbracket$ form one system of maximal spaces of a Segre variety $\mathcal{S}_{2;2}$.

Proof   The $\mathbb{F}_{q}$-subplane $\bar{\pi}$ corresponds to an $\mathbb{F}_{q}$-subplane $\pi$ of the transversal plane $\Gamma$ in ${\rm PG}(8,q^{3})$. Let $P_{1},P_{2},P_{3},P_{4}$ be a quadrangle of $\pi$, then the corresponding Bose planes \mbox{\llbracket P_{i}\rrbracket}=\langle P_{i},P_{i}^{q},P_{i}^{q^{2}}\rangle\cap{\rm PG}(8,q), $i=1,\ldots,4$ are four planes of ${\rm PG}(8,q)$, any three of which generate ${\rm PG}(8,q)$. By Lemma 3.2 there is a unique Segre variety ${\mathscr{V}}=\mathcal{S}_{2;2}$. Denote the two systems of maximal spaces by $\mathcal{R},\mathcal{R}^{\prime}$, with \mbox{\llbracket P_{1}\rrbracket},\ldots,\mbox{\llbracket P_{4}\rrbracket}\in\mathcal{R}^{\prime}. We will show that \mathcal{R}^{\prime}=\{\mbox{\llbracket X\rrbracket}\,|\,X\in\pi\}.

Consider the extension of the Segre variety ${\mathscr{V}}$ to a Segre variety ${{\mathscr{V}}^{\mbox{\tiny\char 73}}}$ of ${\rm PG}(8,q^{3})$. Denote the two systems of maximal spaces (planes) of ${\mathscr{V}}{{}^{\mbox{\tiny\char 73}}}$ by $\mathcal{R}{{}^{\mbox{\tiny\char 73}}}$ and $(\mathcal{R}^{\prime}){{}^{\mbox{\tiny\char 73}}}$. So \mbox{\llbracket P_{i}\rrbracket}{{}^{\mbox{\tiny\char 73}}}\in(\mathcal{R}^{\prime}){{}^{\mbox{\tiny\char 73}}}, $i=1,\ldots,4$, and for $\pi_{j}\in\mathcal{R}$, we have $\pi_{j}{{}^{\mbox{\tiny\char 73}}}\in\mathcal{R}{{}^{\mbox{\tiny\char 73}}}$. As $\Gamma$ is a plane of ${\rm PG}(8,q^{3})$ that meets \mbox{\llbracket P_{i}\rrbracket}{{}^{\mbox{\tiny\char 73}}}\in(\mathcal{R}^{\prime}){{}^{\mbox{\tiny\char 73}}}, $i=1,\ldots,4$, we have $\Gamma\in\mathcal{R}{{}^{\mbox{\tiny\char 73}}}$. Let $\alpha$ be a plane of $\mathcal{R}^{\prime}$, then in ${\rm PG}(8,q^{3})$, $\alpha{{}^{\mbox{\tiny\char 73}}}$ is a plane of $(\mathcal{R}^{\prime}){{}^{\mbox{\tiny\char 73}}}$, and so $\alpha$ meets $\Gamma$ in a point $A$. As $\alpha$ is a plane of ${\rm PG}(8,q)$, the points $A^{q}$ and $A^{q^{2}}$ lie in $\alpha$, and so \alpha=\langle A,A^{q},A^{q^{2}}\rangle\cap{\rm PG}(8,q)=\mbox{\llbracket A\rrbracket}.

As ${\mathscr{V}}$ is a Segre variety, the planes in $\mathcal{R}^{\prime}$ are a scroll ruled by a homography, hence the set of planes $\{\alpha{{}^{\mbox{\tiny\char 73}}}\,|\,\alpha\in\mathcal{R}^{\prime}\}$ of ${\rm PG}(8,q^{3})$ are ruled by a homography. Thus the set of points $\{\alpha{{}^{\mbox{\tiny\char 73}}}\cap\Gamma\,|\,\alpha\in\mathcal{R}^{\prime}\}$ form an $\mathbb{F}_{q}$-subplane $\pi^{\prime}$ of $\Gamma$. As both $\pi$ and $\pi^{\prime}$ contain the quadrangle $P_{1},\ldots,P_{4}$, we have $\pi^{\prime}=\pi$. Hence $\mathcal{R}^{\prime}$ is the set of Bose planes $\llbracket X\rrbracket$ for $X\in\pi$. $\square$

We use this to determine the Bose representation of an $\mathbb{F}_{q}$-subline. Let $\bar{b}$ be an $\mathbb{F}_{q}$-subline of the line $\bar{\ell_{b}}$ in ${\rm PG}(2,q^{3})$. In the Bose representation in ${\rm PG}(8,q)$, we have \mbox{\llbracket b\rrbracket}=\{\langle Y,Y^{q},Y^{q^{2}}\rangle\,|\,Y\in b\}\cap{\rm PG}(8,q). We show that this set of $q+1$ planes forms a 2-regulus.

Theorem 6.2

Let $\bar{b}$ be a $\mathbb{F}_{q}$-subline lying on the line $\bar{\ell_{b}}$ of ${\rm PG}(2,q^{3})$, then the planes of $\llbracket b\rrbracket$ form a 2-regulus of the 5-space $\Pi_{b}=\langle\ell_{b},\ell_{b}^{q},\ell_{b}^{q^{2}}\rangle\cap{\rm PG}(8,q)$.

Proof   Let $\bar{\pi}$ be any $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ which contains $\bar{b}$ as a line. In the Bose representation, $\pi$ is an $\mathbb{F}_{q}$-subplane of $\Gamma$ that contains the $\mathbb{F}_{q}$-subline $b$. As $b=\pi\cap\ell_{b}$, \mbox{\llbracket b\rrbracket}=\mbox{\llbracket\pi\rrbracket}\cap\mbox{\llbracket\ell_{b}\rrbracket}. These $q+1$ planes all lie in the 5-space $\Pi_{b}=\langle\ell_{b},\ell_{b}^{q},\ell_{b}^{q^{2}}\rangle\cap{\rm PG}(8,q)$, so \mbox{\llbracket b\rrbracket}=\mbox{\llbracket\pi\rrbracket}\cap\Pi_{b}. By Theorem 6.1, the planes of $\llbracket\pi\rrbracket$ form one system of maximal planes of a Segre variety $\mathcal{S}_{2;2}$. By [6, Thm 4.109], the 5-space $\Pi_{b}$ meets the $\mathcal{S}_{2;2}$ in a Segre variety $\mathcal{S}_{1;2}$. That is, the planes of $\llbracket b\rrbracket$ are the system of maximal planes of a Segre variety $\mathcal{S}_{1;2}$, and so are a 2-regulus. $\square$

6.2 Extending to ${\rm PG}(8,q^{3})$

We now look at the extension of the varieties in Theorem 6.1 and Theorem 6.2 to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$. We recall two collineations of ${\rm PG}(8,q^{6})$, namely $X=(x_{0},\ldots,x_{8})\longmapsto X^{q}=(x_{0}^{q},\ldots,x_{8}^{q})$ and $\mathsf{e}\colon X=(x_{0},\ldots,x_{8})\mapsto X^{\mathsf{e}}=(x_{0}^{q^{3}},\ldots,x_{8}^{q^{3}})$.

Theorem 6.3

Let $\bar{\pi}$ be an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$. In the Bose representation, let ${\mathsf{c}_{\pi}}$ be the collineation of order 3 acting on the points of $\Gamma$ which fixes $\pi$ pointwise as defined in (4.3). Let {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}) denote the Segre variety $\mathcal{S}_{2;2}$ whose pointset coincides with the pointset of $\llbracket\pi\rrbracket$.

In ${\rm PG}(8,q^{3})$, {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} is a Segre variety, with one system of maximal spaces the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in\Gamma\}$ where

[TABLE] 2. 2.

In ${\rm PG}(8,q^{6})$, {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 72}}} is a Segre variety, with one system of maximal spaces the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in\Gamma{{}^{\mbox{\tiny\char 72}}}\}$ where

[TABLE]

Proof   We first show that the planes of $\llbracket\pi\rrbracket$ form a scroll. Without loss of generality, let $\bar{\pi}={\rm PG}(2,q)=\{(x,y,z)\,|\,x,y,z\in\mathbb{F}_{q},\ \textup{not all }0\}$. In the Bose representation, $\pi=\{xA_{0}+yA_{1}+zA_{2}\,|\,x,y,z\in\mathbb{F}_{q},\ \textup{not all }0\}$ is an $\mathbb{F}_{q}$-subplane of the transversal plane $\Gamma$. For a point $X=xA_{0}+yA_{1}+zA_{2}\in\pi$, (so $x,y,z\in\mathbb{F}_{q}$) we have \mbox{\llbracket X\rrbracket}=\langle\ X,\ X^{q},\ X^{q^{2}}\ \rangle\cap{\rm PG}(8,q), that is,

[TABLE]

So the planes of $\llbracket\pi\rrbracket$ form a scroll denoted $\mathscr{S}_{\pi}$, where the associated homographies are the identity.

By Theorem 6.1, in ${\rm PG}(8,q)$, the planes of $\llbracket\pi\rrbracket$ form one system of maximal spaces of a Segre variety $\mathcal{S}_{2;2}$, denoted by {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}). The variety-extension of {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}) to {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} in ${\rm PG}(8,q^{3})$ is also a Segre variety $\mathcal{S}_{2;2}$. Hence {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} consists of two systems of maximal subspaces, each ruled by a homography, that is, each system of planes of {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} forms a scroll. As a homography is uniquely determined by a quadrangle, a scroll has a unique scroll-extension. Hence the Segre variety {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} has one system of maximal subspaces being the same set of planes as the scroll-extension of $\mathscr{S}_{\pi}$.

We now determine the planes of the scroll-extension of $\mathscr{S}_{\pi}$ to ${\rm PG}(8,q^{3})$. As the homographies of $\mathscr{S}_{\pi}$ are the identity, the scroll-extension is the set of planes

[TABLE]

Using the calculations from (6) in Section 4.3, this is the set of planes

[TABLE]

Hence the Segre variety {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} has as one system of maximal spaces the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in\Gamma\}$, with \llparenthesis X\rrparenthesis_{\pi}=\langle\,X,\ ({X^{\mathsf{c}^{2}_{\pi}}})^{q},\ ({X^{\mathsf{c}_{\scalebox{0.5}{\mbox{\pi}}}}})^{q^{2}}\,\rangle, proving part 1.

The proof of part 2 is similar. The scroll-extension of $\mathscr{S}_{\pi}$ to ${\rm PG}(8,q^{6})$ is the set of planes

[TABLE]

Using the calculations from (7) in Section 4.3, this is the set of planes

[TABLE]

Hence the Segre variety {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 72}}} has as one system of maximal spaces the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in\Gamma{{}^{\mbox{\tiny\char 72}}}\}$ with $\llparenthesis X\rrparenthesis_{\pi}=\langle\,X,(X^{{\mathsf{c}}^{2}_{\pi}\mathsf{e}})^{q},(X^{\mathsf{c}_{\pi}\mathsf{e}})^{q^{2}}\,\rangle$, proving part 2. $\square$

Note that for a point $X\in\pi$, {X^{\mathsf{c}_{\scalebox{0.5}{\mbox{\pi}}}}}=X and \llparenthesis X\rrparenthesis_{\pi}=\mbox{\llbracket X\rrbracket}{{}^{\mbox{\tiny\char 73}}}, which meets ${\rm PG}(8,q)$ in a plane $\llbracket X\rrbracket$. However, if $X\in\Gamma\backslash\pi$, then by Lemma 3.3, the plane $\llparenthesis X\rrparenthesis_{\pi}$ does not meet any element of $\{\llbracket X\rrbracket\,|\,X\in\Gamma\}$. As the planes in the set $\{\llbracket X\rrbracket\,|\,X\in\Gamma\}$ partition the points of ${\rm PG}(8,q)$, it follows that $\llparenthesis X\rrparenthesis_{\pi}$ is disjoint from ${\rm PG}(8,q)$ for $X\in\Gamma\backslash\pi$.

We use Theorem 6.3 to look at an $\mathbb{F}_{q}$-subline $\bar{b}$, and determine the planes of the extension of the 2-regulus $\llbracket b\rrbracket$ to ${\rm PG}(8,q^{3})$ and ${\rm PG}(8,q^{6})$.

Corollary 6.4

Let $\bar{b}$ be an $\mathbb{F}_{q}$-subline lying on the line $\bar{\ell_{b}}$ of ${\rm PG}(2,q^{3})$. In the Bose representation, let ${\mathsf{c}_{b}}$ be the collineation of order 3 acting on the points of $\ell_{b}$ which fixes $b$ pointwise as defined in (5). The 2-regulus $\llbracket b\rrbracket$ can be extended to a unique 2-regulus of ${\rm PG}(8,q^{3})$ with planes $\{\llparenthesis X\rrparenthesis_{b}\,|\,X\in\ell_{b}\}$; and to a unique 2-regulus of ${\rm PG}(8,q^{6})$ with planes $\{\llparenthesis X\rrparenthesis_{b}\,|\,X\in\ell^{{{}^{\mbox{\tiny\char 72}}}}_{b}\}$ where

[TABLE]

Proof   Note that if $X\in\ell_{b}$, then $X^{{\mathsf{c}}^{5}_{b}}=X^{{\mathsf{c}}^{2}_{b}}$ and $X^{\mathsf{c}^{4}_{b}}=X^{\mathsf{c}_{b}}$. Further, if $X\in b$, then $X^{{\mathsf{c}}^{5}_{b}}=X^{\mathsf{c}^{4}_{b}}=X$. If $\pi$ is an $\mathbb{F}_{q}$-subplane of $\Gamma$, then $\mathsf{c}_{b}$ coincides with $\mathsf{c}_{\pi}$ restricted to $\ell_{b}$ if and only if $b$ is a line of $\pi$. Hence letting $\pi$ be an $\mathbb{F}_{q}$-subplane of $\Gamma$ such that $b$ is a line of $\pi$, we can intersect the results of Theorem 6.3 with the 5-space $\Pi_{b}=\langle\ell_{b},\ell_{b}^{q},\ell_{b}^{q^{2}}\rangle$ to obtain the required result. $\square$

7 $\mathbb{F}_{q}$-conics in the Bose representation

We define an $\mathbb{F}_{q}$-conic of ${\rm PG}(2,q^{2})$ to be a non-degenerate conic in an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{2})$. That is, an $\mathbb{F}_{q}$-conic is projectively equivalent to a set of points in ${\rm PG}(2,q)$ that satisfy a non-degenerate homogeneous quadratic equation over $\mathbb{F}_{q}$. We determine the Bose representation of an $\mathbb{F}_{q}$-conic $\bar{\cal C}$ of ${\rm PG}(2,q^{3})$. An $\mathbb{F}_{q}$-conic $\bar{\cal C}$ of ${\rm PG}(2,q^{3})$ corresponds to an $\mathbb{F}_{q}$-conic in the transversal plane $\Gamma$ denoted ${\cal C}$, and ${\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}$ denotes the unique $\mathbb{F}_{q^{3}}$-conic of $\Gamma$ containing ${\cal C}$. The quadratic extension of the non-degenerate conic ${\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\subset\Gamma$ to the extended transversal plane $\Gamma{{}^{\mbox{\tiny\char 72}}}\cong{\rm PG}(2,q^{6})$ is a non-degenerate conic which we denote by ${\cal C}^{{\rm{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}\!{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}$.

Theorem 7.1

Let $\bar{\cal C}$ be an $\mathbb{F}_{q}$-conic in the $\mathbb{F}_{q}$-subplane $\bar{\pi}$ of ${\rm PG}(2,q^{3})$, and consider the Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(8,q)$.

In ${\rm PG}(8,q)$, the planes of $\llbracket{\cal C}\rrbracket$ form a scroll of ${\rm PG}(8,q)$, and the pointset of $\llbracket{\cal C}\rrbracket$ forms a variety {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket})={\mathscr{V}}^{6}_{3}. 2. 2.

In ${\rm PG}(8,q^{3})$, the points of the variety {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}} coincide with the points on the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\},$ which form a scroll. 3. 3.

In ${\rm PG}(8,q^{6})$, the points of the variety {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 72}}} coincide with the points on the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in{\cal C}^{{\rm{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}\!{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}\}$, which form a scroll.

Proof   Let ${\cal C}$ be an $\mathbb{F}_{q}$-conic in an $\mathbb{F}_{q}$-subplane $\pi$ of $\Gamma$, so ${\cal C}=\pi\cap{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}$. By definition, in ${\rm PG}(8,q)$, \mbox{\llbracket{\cal C}\rrbracket}=\{\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q)\,|\,X\in{\cal C}\}, \mbox{\llbracket\pi\rrbracket}=\{\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q)\,|\,X\in\pi\} and \mbox{\llbracket{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\rrbracket}=\{\langle X,X^{q},X^{q^{2}}\rangle\cap{\rm PG}(8,q)\,|\,X\in{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\}, so

[TABLE]

By Theorem 6.1, the planes of $\llbracket\pi\rrbracket$ form one system of maximal subspaces of a Segre variety $\mathcal{S}_{2;2}$, and so form a scroll. As the planes of $\llbracket{\cal C}\rrbracket$ are a subset of the planes of $\llbracket\pi\rrbracket$, the planes of $\llbracket{\cal C}\rrbracket$ form a scroll, ruled by the same homography as for the scroll $\llbracket\pi\rrbracket$. Hence by Lemma 3.4, the pointset of $\llbracket{\cal C}\rrbracket$ forms a variety ${\mathscr{V}}^{6}_{3}$.

To prove part 2, we first need to describe this variety ${\mathscr{V}}^{6}_{3}$ in more detail. By Theorem 5.7, the pointset of $\llbracket{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\rrbracket$ forms a variety {\mathscr{V}}(\mbox{\llbracket{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\rrbracket})=\mathcal{Q}_{1}\cap\mathcal{Q}_{2}\cap\mathcal{Q}_{3} where $\mathcal{Q}_{i}$ is a quadric with homogenous equation $f_{i}=0$ of degree two over $\mathbb{F}_{q}$, $i=1,2,3$. By [6], a Segre variety $\mathcal{S}_{2;2}$ is the intersection of six quadrics, so by Theorem 6.1, the pointset of $\llbracket\pi\rrbracket$ forms a variety {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket})=\mathcal{Q}_{4}\cap\cdots\cap\mathcal{Q}_{9} where $\mathcal{Q}_{i}$ is a quadric with homogenous equation $f_{i}=0$ of degree two over $\mathbb{F}_{q}$, $i=4,\ldots,9$. So by (8), the pointset of $\llbracket{\cal C}\rrbracket$ coincides with the pointset of a variety {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}) which is the intersection of nine quadrics, namely {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket})=(\mathcal{Q}_{1}\cap\mathcal{Q}_{2}\cap\mathcal{Q}_{3})\cap(\mathcal{Q}_{4}\cap\cdots\cap\mathcal{Q}_{9}).

The variety extension of {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}) is {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}}={\mathcal{Q}}^{\mbox{\tiny\char 73}}_{1}\cap\cdots\cap{\mathcal{Q}}^{\mbox{\tiny\char 73}}_{9}, so in particular,

[TABLE]

We now determine the points of {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}}. By Theorem 5.7, the points of {\mathscr{V}}(\mbox{\llbracket{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\rrbracket}){{}^{\mbox{\tiny\char 73}}} are the points of ${\rm PG}(8,q^{3})$ on the planes

[TABLE]

By Theorem 6.3, the points of {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} are the points of ${\rm PG}(8,q^{3})$ on the planes

[TABLE]

The planes in (10) and (11) are T-planes, so by Lemma 3.3, two planes in (10) and (11) either coincide, are disjoint, or meet in a T-point or a T-line. Thus by (9), {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}} consists of points on the set of planes which are in both (10) and (11). That is, {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}} consists of the points of ${\rm PG}(8,q^{3})$ on the planes $\{\llparenthesis X\rrparenthesis_{\pi}\,|\,X\in{\cal C}^{{\rm\boldsymbol{{\raisebox{0.2pt}{\scalebox{0.5}{{+}}}}}}}\}$.

By Theorem 6.3, the points of the variety-extension {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}} are one system of maximal subspaces of a Segre variety and so form a scroll. As the planes of {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}} are a subset of the planes of {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}}, the planes of {\mathscr{V}}(\mbox{\llbracket{\cal C}\rrbracket}){{}^{\mbox{\tiny\char 73}}} form a scroll, ruled by the same homography as for the scroll {\mathscr{V}}(\mbox{\llbracket\pi\rrbracket}){{}^{\mbox{\tiny\char 73}}}. This completes the proof of part 2. Part 3 is similar. $\square$