A ruled quintic surface in $PG(6,q)$

TL;DR

This paper investigates a specific ruled quintic surface in projective 6-space, exploring its geometric properties and its connection to subplanes in finite projective geometry.

Contribution

It introduces and analyzes a ruled quintic surface in PG(6,q), linking it to subplanes in PG(2,q^3) through the Bruck-Bose representation.

Findings

Identifies the surface as a ruled quintic $ ext{V}^5_2$

Studies its geometric properties

Establishes its relationship with an $ extbf{F}_q$-subplane

Abstract

In this article we look at a scroll of $PG(6,q)$ that uses a projectivity to rule a conic and a twisted cubic. We show this scroll is a ruled quintic surface $\mathcal V^5_2$, and study its geometric properties. The motivation in studying this scroll lies in its relationship with an $\mathbb F_q$-subplane of $PG(2,q^3)$ via the Bruck-Bose representation.

Equations

$$\begin{array}[]{rccl}\sigma:&\pi&\longrightarrow&{\cal V}\\ &(x,y,z,0)&\longmapsto&(x^{3},x^{2}y,xy^{2},y^{3},x^{2}z,xyz,y^{2}z).\end{array}$$

$$5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1.$$

$$r_3=\frac{q^3-q}{6},r_2=q^2+q,r_1=\frac{q^3}{2}+\frac{q}{2}+1,r_0=\frac{q^3-q}{3}.$$

$$r_3=\left(\genfrac{}{}{0ex}{}{q+1}{3}\right)$$

$$r_2=2\times \left(\genfrac{}{}{0ex}{}{q+1}{2}\right)=(q+1)q.$$

$$(q+1)(q^2+q+1)=3r_3+2r_2+r_1.$$

$$s_2=\frac{q^2+q}{2},s_1=q+1,s_0=\frac{q^2-q}{2}.$$

$$x=(q+1{)}^2\times (q^5+q^4+q^3+q^2+q+1)=q^7+3q^6+4q^5+4q^4+4q^3+4q^2+3q+1.$$

$$x=n_2+n_3+n_4+n_5.$$

$$n_2=\sum _{i=0}^3{r}_i(iq+q+1)=2q^4+4q^3+4q^2+3q+1.$$

$$n_3=q^2\sum _{i=0}^2{s}_i(iq+q+1)=2q^5+4q^4+3q^3+q^2.$$

$$n_4=(q^5+q^4-q^3-q^2)\times (2q+1)=2q^6+3q^5-q^4-3q^3-q^2.$$

$$n_5=y\times (q+1).$$

Full text

AMS code: 51E20

keywords: projective space, varieties, scroll, Bruck-Bose representation

1 Introduction

In this article we consider a scroll of ${\rm PG}(6,q)$ that rules a conic and a twisted cubic according to a projectivity. The motivation in studying this scroll lies in its relationship with an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ via the Bruck-Bose representation as described in Section 3. In ${\rm PG}(6,q)$, let ${\cal C}$ be a non-degenerate conic in a plane $\alpha$, ${\cal C}$ is called the conic directrix. Let ${\cal N}_{3}$ be a twisted cubic in a 3-space $\Pi_{3}$ with $\alpha\cap\Pi_{3}=\emptyset$, ${\cal N}_{3}$ is called the twisted cubic directrix. Let $\phi$ be a projectivity from the points of ${\cal C}$ to the points of ${\cal N}_{3}$. By this we mean that if we write the points of ${\cal C}$ and ${\cal N}_{3}$ using a non-homogeneous parameter, so ${\cal C}=\{C_{\theta}=(1,\theta,\theta^{2})\,|\,\theta\in\mathbb{F}_{q}\cup\{\infty\}\}$, and ${\cal N}_{3}=\{N_{\epsilon}=(1,\epsilon,\epsilon^{2},\epsilon^{3})\,|\,\epsilon\in\mathbb{F}_{q}\cup\{\infty\}\}$, then $\phi\in{\rm PGL}(2,q)$ is a projectivity mapping $(1,\theta)$ to $(1,\epsilon)$. Let ${\cal V}$ be the set of points of ${\rm PG}(6,q)$ lying on the $q+1$ lines joining each point of ${\cal C}$ to the corresponding point (under $\phi$) of ${\cal N}_{3}$. These $q+1$ lines are called the generators of ${\cal V}$. As the two subspaces $\alpha$ and $\Pi_{3}$ are disjoint, ${\cal V}$ is not contained in a 5-space. We note that this construction generalises the ruled cubic surface ${\cal V}^{3}_{2}$ in ${\rm PG}(4,q)$, a variety that has been well studied, see [8].

We will be working with normal rational curves in ${\rm PG}(6,q)$, if ${\cal N}$ is a normal rational curve that generates an $i$-dimensional space, then we call ${\cal N}$ an $i$-dim nrc, and often use the notation ${\cal N}_{i}$. See [6] for details on normal rational curves. As we will be looking at 5-dim nrcs contained in ${\cal V}$, we will assume $q\geq 6$ throughout.

This article studies the geometric structure of ${\cal V}$. In Section 2, we show that ${\cal V}$ is a variety ${{\mathcal{V}}^{5}_{2}}$ of order 5 and dimension 2, and that all such scrolls are projectively equivalent. Further, we show that ${\cal V}$ contains exactly $q+1$ lines and one non-degenerate conic. In Section 3, we describe the Bruck-Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(6,q)$, and discuss how ${\cal V}$ corresponds to an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$. We use the Bruck-Bose setting to show that ${\cal V}$ contains exactly $q^{2}$ twisted cubics, and that each can act as a directrix of ${\cal V}$. In Section 4, we count the number of 4 and 5-dim nrcs contained in ${\cal V}$. Further, we determine how 5-spaces meet ${\cal V}$, and count the number of 5-spaces of each intersection type. The main result is Theorem 4.8. In Section 5, we determine how 5-spaces meet ${\cal V}$ in relation to the regular 2-spread in the Bruck-Bose setting.

2 Simple properties of ${\cal V}$

Theorem 2.1

Let ${\cal V}$ be a scroll of ${\rm PG}(6,q)$ that rules a conic and a twisted cubic according to a projectivity. Then ${\cal V}$ is a variety of dimension 2 and order 5, denoted ${{\mathcal{V}}^{5}_{2}}$ and called a ruled quintic surface. Further, any two ruled quintic surfaces are projectively equivalent.

Proof   Let ${\cal V}$ be a scroll of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ in a plane $\alpha$, twisted cubic directrix ${\cal N}_{3}$ in a 3-space $\Pi_{3}$, and ruled by a projectivity as described in Section 1. The group of collineations of ${\rm PG}(6,q)$ is transitive on planes; and transitive on 3-spaces. Further, all non-degenerate conics in a projective plane are projectively equivalent; and all twisted cubics in a 3-space are projectively equivalent. Hence without loss of generality, we can coordinatise ${\cal V}$ as follows. Let $\alpha$ be the plane which is the intersection of the four hyperplanes $x_{0}=0,\ x_{1}=0,\ x_{2}=0,\ x_{3}=0.$ Let ${\cal C}$ be the non-degenerate conic in $\alpha$ with points $C_{\theta}=(0,0,0,0,1,\theta,\theta^{2})$ for $\theta\in\mathbb{F}_{q}\cup\{\infty\}$. Note that the points of ${\cal C}$ are the exact intersection of $\alpha$ with the quadric of equation $x_{5}^{2}=x_{4}x_{6}.$ Let $\Pi_{3}$ be the 3-space which is the intersection of the three hyperplanes $x_{4}=0,\ x_{5}=0,\ x_{6}=0.$ Let ${\cal N}_{3}$ be the twisted cubic in $\Pi_{3}$ with points $N_{\theta}=(1,\theta,\theta^{2},\theta^{3},0,0,0)$ for $\theta\in\mathbb{F}_{q}\cup\{\infty\}$. Note that the points of ${\cal N}_{3}$ are the exact intersection of $\Pi_{3}$ with the three quadrics with equations $x_{1}^{2}=x_{0}x_{2},\ x_{2}^{2}=x_{1}x_{3},\ x_{0}x_{3}=x_{1}x_{2}.$ A projectivity in ${\rm PGL}(2,q)$ is uniquely determined by the image of three points, so without loss of generality, let ${\cal V}$ have generator lines $\ell_{\theta}=\{V_{\theta,t}=N_{\theta}+tC_{\theta},\ t\in\mathbb{F}_{q}\cup\{\infty\}\}$ for $\theta\in\mathbb{F}_{q}\cup\{\infty\}$. That is, $V_{\theta,t}=(1,\theta,\theta^{2},\theta^{3},t,t\theta,t\theta^{2})$. Equivalently, ${\cal V}$ consists of the points $V_{x,y,z}=(x^{3},\ x^{2}y,\ xy^{2},\ y^{3},\ zx^{2},\ zxy,\ zy^{2})$ for $x,y\in\mathbb{F}_{q}$, not both 0, $z\in\mathbb{F}_{q}\cup\{\infty\}.$ It is straightforward to verify that the pointset of ${\cal V}$ is the exact intersection of the following ten quadrics, $x_{0}x_{5}=x_{1}x_{4},\ x_{0}x_{6}=x_{1}x_{5}=x_{2}x_{4},\ x_{1}x_{6}=x_{2}x_{5}=x_{3}x_{4},\ x_{2}x_{6}=x_{3}x_{5},\ x_{1}^{2}=x_{0}x_{2},\ x_{2}^{2}=x_{1}x_{3},\ x_{5}^{2}=x_{4}x_{6},\ x_{0}x_{3}=x_{1}x_{2}.$ Hence the points of ${\cal V}$ form a variety.

We follow [7] to calculate the dimension and order of ${\cal V}$. The following map defines an algebraic one to one correspondence between the plane $\pi$ of $PG(3,q)$ with points $(x,y,z,0)$, $x,y,z\in\mathbb{F}_{q}$ not all 0, and the points of ${\cal V}$.

[TABLE]

Thus ${\cal V}$ is an absolutely irreducible variety of dimension two and so we are justified in calling it a surface. Now consider a generic 4-space of ${\rm PG}(6,q)$ with equation given by the two hyperplanes $\Sigma_{1}:a_{0}x_{0}+\cdots+a_{6}x_{6}=0$ and $\Sigma_{2}:b_{0}x_{0}+\cdots+b_{6}x_{6}=0$, $a_{i},b_{i}\in\mathbb{F}_{q}$. The point $V_{x,y,z}=(x^{3},x^{2}y,xy^{2},y^{3},x^{2}z,xyz,y^{2}z)$ lies on $\Sigma_{1}$ if $a_{0}x^{3}+a_{1}x^{2}y+a_{2}xy^{2}+a_{3}y^{3}+a_{4}x^{2}z+a_{5}xyz+a_{6}y^{2}z=0$. This corresponds to a cubic $\mathcal{K}$ in the plane $\pi$, moreover $\mathcal{K}$ contains the point $P=(0,0,1,0)$, and $P$ is a double point of $\mathcal{K}$. Similarly $V_{x,y,z}\in\Sigma_{2}$ corresponds to a cubic in $\pi$ with a double point $(0,0,1,0)$. Two cubics in a plane meet generically in nine points. As $(0,0,1,0)$ lies in the kernel of $\sigma$, in ${\rm PG}(6,q)$ the 4-space $\Sigma_{1}\cap\Sigma_{2}$ meets ${\cal V}$ in five points, and so ${\cal V}$ has order 5. $\square$

Theorem 2.2

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface in ${\rm PG}(6,q)$.

No two generators of ${{\mathcal{V}}^{5}_{2}}$ lie in a plane. 2. 2.

No three generators of ${{\mathcal{V}}^{5}_{2}}$ lie in a 4-space. 3. 3.

No four generators of ${{\mathcal{V}}^{5}_{2}}$ lie in a 5-space.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ in a plane $\alpha$, and twisted cubic directrix ${\cal N}_{3}$ lying in a 3-space $\Pi_{3}$. Suppose two generator lines $\ell_{0},\ell_{1}$ of ${{\mathcal{V}}^{5}_{2}}$ lie in a plane. Let $m$ be the line in $\alpha$ joining the distinct points $\ell_{0}\cap\alpha$, $\ell_{1}\cap\alpha$. Let $m^{\prime}$ be the line in $\Pi_{3}$ joining the distinct points $\ell_{0}\cap\Pi_{3}$, $\ell_{1}\cap\Pi_{3}$. The lines $m,m^{\prime}$ lie in the plane $\langle\ell_{0},\ell_{1}\rangle$ and so meet in a point, contradicting $\alpha$, $\Pi_{3}$ being disjoint. Hence the generator lines of ${{\mathcal{V}}^{5}_{2}}$ are pairwise skew.

For part 2, suppose a 4-space $\Pi_{4}$ contains three distinct generators of ${{\mathcal{V}}^{5}_{2}}$. As distinct generators meet ${\cal C}$ in distinct points, $\Pi_{4}$ contains three distinct points of ${\cal C}$, and so contains the plane $\alpha$. Further, distinct generators meet ${\cal N}_{3}$ in distinct points, hence $\Pi_{4}$ contains three points of ${\cal N}_{3}$, and so $\Pi_{4}\cap\Pi_{3}$ has dimension at least two. Hence $\langle\Pi_{4},\Pi_{3}\rangle$ has dimension at most $4+3-2=5$. However, ${{\mathcal{V}}^{5}_{2}}\subseteq\langle\Pi_{4},\Pi_{3}\rangle$, a contradiction as ${{\mathcal{V}}^{5}_{2}}$ is not contained in a 5-space.

For part 3, suppose a 5-space $\Pi_{5}$ contains four distinct generators of ${{\mathcal{V}}^{5}_{2}}$. Distinct generators meet $\Pi_{3}$ in distinct points of ${\cal N}_{3}$, so $\Pi_{5}$ contains four points of ${\cal N}_{3}$, which do not lie in a plane. Hence $\Pi_{5}$ contains $\Pi_{3}$. Similarly $\Pi_{5}$ contains $\alpha$, and so $\Pi_{5}$ contains ${{\mathcal{V}}^{5}_{2}}$, a contradiction as ${{\mathcal{V}}^{5}_{2}}$ is not contained in a 5-space. $\square$

Corollary 2.3

No two generators of ${{\mathcal{V}}^{5}_{2}}$ lie in a 3-space containing $\alpha$.

Proof   Suppose a 3-space $\Pi_{3}$ contained $\alpha$ and two generators of ${{\mathcal{V}}^{5}_{2}}$. Let $P$ be a point of ${{\mathcal{V}}^{5}_{2}}$ not in $\Pi_{3}$, and let $\ell$ be the generator of ${{\mathcal{V}}^{5}_{2}}$ through $P$. Then $\Pi_{4}=\langle\Pi_{3},P\rangle$ contains two distinct points of $\ell$, namely $P$ and $\ell\cap{\cal C}$, and so $\Pi_{4}$ contains $\ell$. That is, $\Pi_{4}$ is a 4-space containing three generators, contradicting Theorem 2.2. $\square$

We now show that the only lines on ${{\mathcal{V}}^{5}_{2}}$ are the generators, and the only non-degenerate conic on ${{\mathcal{V}}^{5}_{2}}$ is the conic directrix. We show later in Theorem 3.2 that there are exactly $q^{2}$ twisted cubics on ${{\mathcal{V}}^{5}_{2}}$, and that each is a directrix.

Theorem 2.4

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface in ${\rm PG}(6,q)$. A line of ${\rm PG}(6,q)$ meets ${{\mathcal{V}}^{5}_{2}}$ in [math], $1$, $2$ or $q+1$ points. Further, ${{\mathcal{V}}^{5}_{2}}$ contains exactly $q+1$ lines, namely the generator lines.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ lying in a plane $\alpha$, and twisted cubic directrix ${\cal N}_{3}$ lying in the 3-space $\Pi_{3}$. Let $m$ be a line of ${\rm PG}(6,q)$ that is not a generator of ${{\mathcal{V}}^{5}_{2}}$, and suppose $m$ meets ${{\mathcal{V}}^{5}_{2}}$ in three points $P,Q,R$. As $m$ is not a generator of ${{\mathcal{V}}^{5}_{2}}$, the points $P,Q,R$ lie on distinct generator lines denoted $\ell_{P},\ell_{Q},\ell_{R}$ respectively. As ${\cal C}$ is a non-degenerate conic, $m$ is not a line of $\alpha$ and so at most one of the points $P,Q,R$ lie in ${\cal C}$. Suppose firstly that $P,Q,R\notin{\cal C}$. Then $\langle\alpha,m\rangle$ is a 3- or 4-space that contains the three generators $\ell_{P},\ell_{Q},\ell_{R}$, contradicting Theorem 2.2. Now suppose $P\in{\cal C}$ and $Q,R\notin{\cal C}$. Then $\Sigma_{3}=\langle\alpha,m\rangle$ is a 3-space which contains the two generator lines $\ell_{Q},\ell_{R}$. So $\Sigma_{3}\cap\Pi_{3}$ contains the distinct points $\ell_{R}\cap{\cal N}_{3}$, $\ell_{Q}\cap{\cal N}_{3}$, and so has dimension at least one. Hence $\langle\Sigma_{3},\Pi_{3}\rangle$ has dimension at most $3+3-1=5$, a contradiction as ${{\mathcal{V}}^{5}_{2}}\subset\langle\Sigma_{3},\Pi_{3}\rangle$, but ${{\mathcal{V}}^{5}_{2}}$ is not contained in a 5-space. Hence a line of ${\rm PG}(6,q)$ is either a generator line of ${{\mathcal{V}}^{5}_{2}}$, or meets ${{\mathcal{V}}^{5}_{2}}$ in 0, 1 or 2 points. $\square$

Theorem 2.5

The ruled quintic surface ${{\mathcal{V}}^{5}_{2}}$ contains exactly one non-degenerate conic.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface with conic directrix ${\cal C}$ in a plane $\alpha$. Suppose ${{\mathcal{V}}^{5}_{2}}$ contains another non-degenerate conic ${\cal C}^{\prime}$ in a plane $\alpha^{\prime}\neq\alpha$. If ${\cal C}^{\prime}$ contains two points on a generator $\ell$ of ${{\mathcal{V}}^{5}_{2}}$, then $\alpha^{\prime}\cap{{\mathcal{V}}^{5}_{2}}$ contains ${\cal C}^{\prime}$ and $\ell$. However, by the proof of Theorem 2.1, ${{\mathcal{V}}^{5}_{2}}$ is the intersection of quadrics, and the configuration ${\cal C}^{\prime}\cup\ell$ is not contained in any planar quadric. Hence ${\cal C}^{\prime}$ contains exactly one point on each generator of ${{\mathcal{V}}^{5}_{2}}$. We consider the three cases where $\alpha\cap\alpha^{\prime}$ is either empty, a point or a line. Suppose $\alpha\cap\alpha^{\prime}=\emptyset$, then $\langle\alpha,\alpha^{\prime}\rangle$ is a 5-space that contains ${\cal C}$ and ${\cal C}^{\prime}$, and so contains two distinct points on each generator of ${{\mathcal{V}}^{5}_{2}}$. Hence $\langle\alpha,\alpha^{\prime}\rangle$ contains each generator of ${{\mathcal{V}}^{5}_{2}}$ and so contains ${{\mathcal{V}}^{5}_{2}}$, a contradiction as ${{\mathcal{V}}^{5}_{2}}$ is not contained in a 5-space. Suppose $\alpha\cap\alpha^{\prime}$ is a point $P$, then $\langle\alpha,\alpha^{\prime}\rangle$ is a 4-space that contains at least $q$ generators of ${{\mathcal{V}}^{5}_{2}}$, contradicting Theorem 2.2 as $q\geq 6$. Finally, suppose $\alpha\cap\alpha^{\prime}$ is a line, then $\langle\alpha,\alpha^{\prime}\rangle$ is a 3-space that contains at least $q-1$ generators, contradicting Theorem 2.2 as $q\geq 6$. So ${{\mathcal{V}}^{5}_{2}}$ contains exactly one non-degenerate conic. $\square$

We aim to classify how 5-spaces meet ${{\mathcal{V}}^{5}_{2}}$, so we begin with a simple description.

Remark 2.6

Let $\Pi_{5}$ be a 5-space, then $\Pi_{5}\cap{{\mathcal{V}}^{5}_{2}}$ contains a set of $q+1$ points, one on each generator.

Lemma 2.7

A $5$-space meets ${{\mathcal{V}}^{5}_{2}}$ in either (a) a 5-dim nrc, (b) a 4-dim nrc and 0 or 1 generators, (c) a 3-dim nrc and 0, 1 or 2 generators, (d) the conic directrix and 0, 1, 2 or 3 generators.

Proof   Using properties of varieties (see for example [7]) we have ${{\mathcal{V}}^{5}_{2}}\cap{\cal V}^{1}_{5}={\cal V}^{5}_{1}$, that is, the variety ${{\mathcal{V}}^{5}_{2}}$ meets a 5-space ${\cal V}^{1}_{5}$ in a curve of degree five. Denote this curve of ${\rm PG}(6,q)$ by $\mathcal{K}$. The degree of $\mathcal{K}$ can be partitioned as

[TABLE]

By Theorem 2.4, the only lines on ${{\mathcal{V}}^{5}_{2}}$ are the generators. By Theorem 2.2, $\mathcal{K}$ does not contain more than 3 generators. By Remark 2.6, $\mathcal{K}$ contains at least one point on each generator. Hence $\mathcal{K}$ is not empty, and is not the union of 1, 2 or 3 generators, so the partition $1+1+1+1+1$ for the degree of $\mathcal{K}$ does not occur.

Suppose that the degree of $\mathcal{K}$ is partitioned as either (a) $2+2+1$ or (b) $2+1+1+1$. By Remark 2.6, $\mathcal{K}$ contains a point on each generator, so $\mathcal{K}$ contains an irreducible conic. By Theorem 2.5, this conic is the conic directrix ${\cal C}$ of ${{\mathcal{V}}^{5}_{2}}$, and case (a) does not occur. Hence $\mathcal{K}$ consists of ${\cal C}$ and 0, 1, 2 or 3 generators of ${{\mathcal{V}}^{5}_{2}}$.

Suppose that the degree of $\mathcal{K}$ is partitioned as $3+1+1$. So $\mathcal{K}$ consists of at most 2 generators, and an irreducible cubic $\mathcal{K}^{\prime}$. By Remark 2.6, $\mathcal{K}$ contains a point on each generator, so $\mathcal{K}^{\prime}$ contains a point on at least $q-1$ generators. If $\mathcal{K}^{\prime}$ generates a 3-space, then it is a 3-dim nrc of ${\rm PG}(6,q)$. If not, $\mathcal{K}^{\prime}$ is an irreducible cubic contained in a plane $\Pi_{2}$. By the proof of Theorem 2.1, $\mathcal{K}^{\prime}$ is contained in a quadric, so $\mathcal{K}^{\prime}$ is not an irreducible planar cubic. Thus $\mathcal{K}^{\prime}$ is a 3-dim nrc of ${\rm PG}(6,q)$. Hence $\mathcal{K}$ consists of a 3-dim nrc and 0, 1 or 2 generators of ${{\mathcal{V}}^{5}_{2}}$.

Suppose that the degree of $\mathcal{K}$ is partitioned as $2+3$. By Remark 2.6, $\mathcal{K}$ contains a point on each generator. As argued above, $\mathcal{K}$ does not contain an irreducible planar cubic. Suppose $\mathcal{K}$ contained both an irreducible conic ${\cal C}$ and a twisted cubic ${\cal N}_{3}$, then there is at least one generator $\ell$ that meet ${\cal C}$ and ${\cal N}_{3}$ in distinct points. In this case $\ell$ lies in the 5-space and so lies in $\mathcal{K}$, a contradiction. So $\mathcal{K}$ is not the union of an irreducible conic and a twisted cubic.

Suppose that the degree of $\mathcal{K}$ is partitioned as $4+1$. So $\mathcal{K}$ consists of at most 1 generator, and an irreducible quartic $\mathcal{K}^{\prime}$. By Remark 2.6, $\mathcal{K}$ contains a point on each generator, so $\mathcal{K}^{\prime}$ contains a point on at least $q$ generators. If $\mathcal{K}^{\prime}$ generates a 4-space, then it is a 4-dim nrc of ${\rm PG}(6,q)$. If not, $\mathcal{K}^{\prime}$ is an irreducible quartic contained in a 3-space $\Pi_{3}$. Let $\ell,m$ be two generators not in $\mathcal{K}$, then by Remark 2.6 they meet $\mathcal{K}^{\prime}$. So $\langle\Pi_{3},\ell,m\rangle$ has dimension at most 5, and meets ${{\mathcal{V}}^{5}_{2}}$ in a irreducible quartic and 2 lines, which is a curve of degree 6, a contradiction. Thus $\mathcal{K}^{\prime}$ is a 4-dim nrc of ${\rm PG}(6,q)$. That is, $\mathcal{K}$ consists of a 4-dim nrc and 0 or 1 generators of ${{\mathcal{V}}^{5}_{2}}$.

Suppose the curve $\mathcal{K}$ is irreducible. By Remark 2.6, $\mathcal{K}$ contains a point on each generator. So either $\mathcal{K}$ is a 5-dim nrc of ${\rm PG}(6,q)$, or $\mathcal{K}$ lies in a 4-space. Suppose $\mathcal{K}$ lies in a 4-space $\Pi_{4}$, and let $\ell$ be a generator, then $\langle\Pi_{4},\ell\rangle$ has dimension at most 5 and meets ${{\mathcal{V}}^{5}_{2}}$ in a curve of degree 6, a contradiction. So $\mathcal{K}$ is a 5-dim nrc of ${\rm PG}(6,q)$. $\square$

Corollary 2.8

Let $\Pi_{r}$ be an $r$-space, $r=3,4,5$ that contains an $r$-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. Then $\Pi_{r}$ contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$.

Proof   First suppose $r=3$, by Lemma 2.7, a 5-space containing a twisted cubic ${\cal N}_{3}$ of ${{\mathcal{V}}^{5}_{2}}$ contains at most two generators of ${{\mathcal{V}}^{5}_{2}}$. Hence a 4-space containing ${\cal N}_{3}$ contains at most one generator of ${{\mathcal{V}}^{5}_{2}}$. Hence the 3-space $\Pi_{3}$ containing ${\cal N}_{3}$ contains no generator of ${{\mathcal{V}}^{5}_{2}}$. If $r=4$, by Lemma 2.7, a 5-space containing a 4-dim nrc ${\cal N}_{4}$ of ${{\mathcal{V}}^{5}_{2}}$ contains at most one generator of ${{\mathcal{V}}^{5}_{2}}$. Hence the 4-space $\Pi_{4}$ containing ${\cal N}_{4}$ contains no generators of ${{\mathcal{V}}^{5}_{2}}$. If $r=5$, then by Lemma 2.7, $\Pi_{5}$ contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$. $\square$

Theorem 2.9

Let ${\cal N}_{r}$ be an $r$-dim nrc lying on ${{\mathcal{V}}^{5}_{2}}$, $r=3,4,5$. Then ${\cal N}_{r}$ contains exactly one point on each generator of ${{\mathcal{V}}^{5}_{2}}$.

Proof   Let ${\cal N}_{r}$ be an $r$-dim nrc lying on ${{\mathcal{V}}^{5}_{2}}$, $r=3,4,5$, and denote the $r$-space containing ${\cal N}_{r}$ by $\Pi_{r}$. If $\Pi_{r}$ contained 2 points of a generator of ${{\mathcal{V}}^{5}_{2}}$, then it contains the whole generator, so by Corollary 2.8, the $q+1$ points of ${\cal N}_{r}$ are one on each generator of ${{\mathcal{V}}^{5}_{2}}$. $\square$

3 ${{\mathcal{V}}^{5}_{2}}$ and $\mathbb{F}_{q}$-subplanes of ${\rm PG}(2,q^{3})$

To study ${{\mathcal{V}}^{5}_{2}}$ in more detail, we use the linear representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(6,q)$ developed independently by André and Bruck and Bose in [1, 4, 5]. Let $\mathcal{S}$ be a regular 2-spread of ${\rm PG}(6,q)$ in a 5-space $\Sigma_{\infty}$. Let $\mathscr{I}$ be the incidence structure with: points the points of ${\rm PG}(6,q)\backslash\Sigma_{\infty}$; lines the 3-spaces of ${\rm PG}(6,q)$ that contain a plane of $\mathcal{S}$ and are not in $\Sigma_{\infty}$; and incidence is inclusion. Then $\mathscr{I}$ is isomorphic to ${\rm AG}(2,q^{3})$. We can uniquely complete $\mathscr{I}$ to ${\rm PG}(2,q^{3})$, the points on $\ell_{\infty}$ correspond to the planes of $\mathcal{S}$. We call this the Bruck-Bose representation of ${\rm PG}(2,q^{3})$ in ${\rm PG}(6,q)$, see [2] for a detailed discussion on this representation. Of particular interest is the relationship between the ruled quintic surface of ${\rm PG}(6,q)$ and the $\mathbb{F}_{q}$-subplanes of ${\rm PG}(2,q^{3})$.

To describe this relationship, we need to use the cubic extension of ${\rm PG}(6,q)$ to ${\rm PG}(6,q^{3})$. The regular 2-spread $\mathcal{S}$ has a unique set of three conjugate transversal lines in this cubic extension, denoted $g,g^{q},g^{q^{2}}$, which meet each extended plane of $\mathcal{S}$, for more details on regular spreads and transversals, see [6, Section 25.6]. An $r$-space $\Pi_{r}$ of ${\rm PG}(6,q)$ lies in a unique $r$-space of ${\rm PG}(6,q^{3})$, denoted $\Pi_{r}^{{}^{\mbox{\tiny\char 73}}}$. A nrc ${\cal N}$ of ${\rm PG}(6,q)$ lies in a unique nrc of ${\rm PG}(6,q^{3})$, denoted ${\cal N}^{{}^{\mbox{\tiny\char 73}}}$. Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface with conic directrix ${\cal C}$, twisted cubic directrix ${\cal N}_{3}$, and associated projectivity $\phi$. Then we can extend ${{\mathcal{V}}^{5}_{2}}$ to a unique ruled quintic surface ${{\mathcal{V}}^{5}_{2}}^{{}^{\mbox{\tiny\char 73}}}$ of ${\rm PG}(6,q^{3})$ with conic directrix ${\cal C}^{{}^{\mbox{\tiny\char 73}}}$, twisted cubic directrix ${\cal N}_{3}^{{}^{\mbox{\tiny\char 73}}}$, with the same associated projectivity, that is, extend $\phi$ from acting on ${\rm PG}(1,q)$ to acting on ${\rm PG}(1,q^{3})$. We need the following characterisations.

Result 3.1

[2, 3]** Let $\mathcal{S}$ be a regular 2-spread in a $5$-space $\Sigma_{\infty}$ in ${\rm PG}(6,q)$ and consider the Bruck-Bose plane ${\rm PG}(2,q^{3})$.

An $\mathbb{F}_{q}$-subline of ${\rm PG}(2,q^{3})$ that meets $\ell_{\infty}$ in a point corresponds in ${\rm PG}(6,q)$ to a line not in $\Sigma_{\infty}$. 2. 2.

An $\mathbb{F}_{q}$-subline of ${\rm PG}(2,q^{3})$ that is disjoint from $\ell_{\infty}$ corresponds in ${\rm PG}(6,q)$ to a twisted cubic ${\cal N}_{3}$ lying in a 3-space about a plane of $\mathcal{S}$, such that the extension ${\cal N}_{3}^{{}^{\mbox{\tiny\char 73}}}$ to ${\rm PG}(6,q^{3})$ meets each transversal of $\mathcal{S}$ in a point. 3. 3.

An $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ tangent to $\ell_{\infty}$ at the point $T$ corresponds in ${\rm PG}(6,q)$ to a ruled quintic surface ${{\mathcal{V}}^{5}_{2}}$ with conic directrix in the spread plane corresponding to $T$, such that in the cubic extension ${\rm PG}(6,q^{3})$, the transversals $g,g^{q},g^{q^{2}}$ of $\mathcal{S}$ are generators of ${{{\mathcal{V}}^{5}_{2}}}{{}^{\mbox{\tiny\char 73}}}$.

Moreover, the converse of each is true.

We use this characterisation to show that ${{\mathcal{V}}^{5}_{2}}$ contains exactly $q^{2}$ twisted cubics.

Theorem 3.2

The ruled quintic surface ${{\mathcal{V}}^{5}_{2}}$ contains exactly $q^{2}$ twisted cubics, each is a directrix of ${{\mathcal{V}}^{5}_{2}}$.

Proof   By Theorem 2.1, all ruled quintic surfaces are projectively equivalent. So without loss of generality, we can position a ruled quintic surface so that it corresponds to an $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$ which we denote by ${\mathscr{B}}$. That is, by Result 3.1, $\mathcal{S}$ is a regular 2-spread in a hyperplane $\Sigma_{\infty}$, ${{\mathcal{V}}^{5}_{2}}\cap\Sigma_{\infty}$ is the conic directrix ${\cal C}$ of ${{\mathcal{V}}^{5}_{2}}$, ${\cal C}$ lies in a plane of $\mathcal{S}$, and in the cubic extension ${\rm PG}(6,q^{3})$, the transversals $g,g^{q},g^{q^{2}}$ of $\mathcal{S}$ are generators of ${{{\mathcal{V}}^{5}_{2}}}^{{}^{\mbox{\tiny\char 73}}}$.

Let ${\cal N}_{3}$ be a twisted cubic contained in ${{\mathcal{V}}^{5}_{2}}$, and denote the 3-space containing ${\cal N}_{3}$ by $\Pi_{3}$. As ${{\mathcal{V}}^{5}_{2}}\cap\Sigma_{\infty}={\cal C}$, $\Pi_{3}$ meets $\Sigma_{\infty}$ in a plane, we show this is a plane of $\mathcal{S}$. In ${\rm PG}(6,q^{3})$, ${{\mathcal{V}}^{5}_{2}}^{{}^{\mbox{\tiny\char 73}}}$ is a ruled quintic surface that contains the twisted cubic ${\cal N}_{3}^{{}^{\mbox{\tiny\char 73}}}$, moreover, the transversals $g,g^{q},g^{q^{2}}$ of $\mathcal{S}$ are generators of ${{\mathcal{V}}^{5}_{2}}^{{}^{\mbox{\tiny\char 73}}}$. So by Theorem 2.9, ${\cal N}_{3}^{{}^{\mbox{\tiny\char 73}}}$ contains one point on each of $g,g^{q}$ and $g^{q^{2}}$. Hence the 3-space $\Pi_{3}^{{}^{\mbox{\tiny\char 73}}}$ contains an extended plane of $\mathcal{S}$, and so $\Pi_{3}$ meets $\Sigma_{\infty}$ in a plane of $\mathcal{S}$. Hence $\Pi_{3}\cap\alpha=\emptyset$, further, by Theorem 2.9, ${\cal N}_{3}$ contains one point on each generator of ${{\mathcal{V}}^{5}_{2}}$, thus ${\cal N}_{3}$ is a directrix of ${{\mathcal{V}}^{5}_{2}}$.

By Result 3.1, ${\cal N}_{3}$ corresponds in ${\rm PG}(2,q^{3})$ to an $\mathbb{F}_{q}$-subline of ${\mathscr{B}}$ disjoint from $\ell_{\infty}$. Conversely, every $\mathbb{F}_{q}$-subline of ${\mathscr{B}}$ disjoint from $\ell_{\infty}$ corresponds to a twisted cubic on ${{\mathcal{V}}^{5}_{2}}$. Thus the twisted cubics in ${{\mathcal{V}}^{5}_{2}}$ are in 1-1 correspondence with the $\mathbb{F}_{q}$-sublines of ${\mathscr{B}}$ that are disjoint from $\ell_{\infty}$. As there are $q^{2}$ such $\mathbb{F}_{q}$-sublines, there are $q^{2}$ twisted cubics on ${{\mathcal{V}}^{5}_{2}}$. $\square$

Suppose we position ${{\mathcal{V}}^{5}_{2}}$ so that it corresponds via the Bruck-Bose representation to a tangent $\mathbb{F}_{q}$-subplane ${\mathscr{B}}$ of ${\rm PG}(2,q^{3})$. So we have a regular 2-spread $\mathcal{S}$ in a hyperplane $\Sigma_{\infty}$, and the conic directrix of ${{\mathcal{V}}^{5}_{2}}$ lies in a plane $\alpha\in\mathcal{S}$. We define the splash of ${\mathscr{B}}$ to be the set of $q^{2}+1$ points on $\ell_{\infty}$ that lie on an extended line of ${\mathscr{B}}$. The splash of ${{\mathcal{V}}^{5}_{2}}$ is defined to be the corresponding set of $q^{2}+1$ planes of $\mathcal{S}$. We denote the splash of ${{\mathcal{V}}^{5}_{2}}$ by $\mathbb{S}$. Note that $\alpha$ is a plane of $\mathbb{S}$. We show that the remaining $q^{2}$ planes of $\mathbb{S}$ are related to the $q^{2}$ twisted cubics of ${{\mathcal{V}}^{5}_{2}}$.

Corollary 3.3

Let $\mathcal{S}$ be a regular $2$-spread in a hyperplane $\Sigma_{\infty}$ of ${\rm PG}(6,q)$. Without loss of generality, we can position ${{\mathcal{V}}^{5}_{2}}$ so that it corresponds via the Bruck-Bose representation to a tangent $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$. Then the conic directrix of ${{\mathcal{V}}^{5}_{2}}$ lies in a plane $\alpha\in\mathcal{S}$, the $q^{2}$ 3-spaces containing a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$ meet $\Sigma_{\infty}$ in distinct planes of $\mathcal{S}$, and these planes together with $\alpha$ form the splash $\mathbb{S}$ of ${{\mathcal{V}}^{5}_{2}}$.

Proof   By Theorem 2.1, all ruled quintic surfaces are projectively equivalent, so without loss of generality, let ${{\mathcal{V}}^{5}_{2}}$ be positioned so that it corresponds to an $\mathbb{F}_{q}$-subplane ${\mathscr{B}}$ of ${\rm PG}(2,q^{3})$ which is tangent to $\ell_{\infty}$. Let $b$ be an $\mathbb{F}_{q}$-subline of ${\mathscr{B}}$ disjoint from $\ell_{\infty}$, so the extension of $b$ meets $\ell_{\infty}$ in a point $R$ which lies in the splash of ${\mathscr{B}}$. By Result 3.1, $b$ corresponds in ${\rm PG}(6,q)$ to a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$ which lies in a 3-space that meets $\Sigma_{\infty}$ in the plane of $\mathbb{S}$ corresponding to the point $R$. $\square$

Using this Bruck-Bose setting, we describe the 3-spaces of ${\rm PG}(6,q)$ that contain a plane of the regular 2-spread $\mathcal{S}$.

Corollary 3.4

Position ${{\mathcal{V}}^{5}_{2}}$ as in Corollary 3.3, so $\mathcal{S}$ is a regular 2-spread in the hyperplane $\Sigma_{\infty}$, and the conic directrix of ${{\mathcal{V}}^{5}_{2}}$ lies in a plane $\alpha$ contained in the splash $\mathbb{S}\subset\mathcal{S}$ of ${{\mathcal{V}}^{5}_{2}}$.

Let $\beta\in\mathbb{S}\backslash\alpha$, then there exists a unique 3-space containing $\beta$ that meets ${{\mathcal{V}}^{5}_{2}}$ in a twisted cubic. The remaining $3$-spaces containing $\beta$ (and not in $\Sigma_{\infty}$) meet ${{\mathcal{V}}^{5}_{2}}$ in 0 or 1 point. 2. 2.

Let $\gamma\in\mathcal{S}\backslash\mathbb{S}$, then each 3-space containing $\gamma$ and not in $\Sigma_{\infty}$ meets ${{\mathcal{V}}^{5}_{2}}$ in 0 or 1 point.

Proof   By Corollary 3.3, we can position ${{\mathcal{V}}^{5}_{2}}$ so that it corresponds to an $\mathbb{F}_{q}$-subplane ${\mathscr{B}}$ of ${\rm PG}(2,q^{3})$ which is tangent to $\ell_{\infty}$. The 3-spaces that contain a plane of $\mathcal{S}$ (and do not lie in $\Sigma_{\infty}$) correspond to lines of ${\rm PG}(2,q^{3})$. Each point on $\ell_{\infty}$ not in ${\mathscr{B}}$ but in the splash of ${\mathscr{B}}$ lies on a unique line that meets ${\mathscr{B}}$ in an $\mathbb{F}_{q}$-subline. By Result 3.1, this corresponds to a twisted cubic in ${{\mathcal{V}}^{5}_{2}}$. The remaining lines meet ${\mathscr{B}}$ in 0 or 1 point, so the remaining 3-spaces meet ${{\mathcal{V}}^{5}_{2}}$ in 0 or 1 point. $\square$

As ${{\mathcal{V}}^{5}_{2}}$ corresponds to an $\mathbb{F}_{q}$-subplane, we have the following result.

Theorem 3.5

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface in ${\rm PG}(6,q)$.

Two twisted cubics on ${{\mathcal{V}}^{5}_{2}}$ meet in a unique point. 2. 2.

Let $P,Q$ be points lying on different generators of ${{\mathcal{V}}^{5}_{2}}$, and not in the conic directrix. Then $P,Q$ lie on a unique twisted cubic of ${{\mathcal{V}}^{5}_{2}}$.

Proof   Without loss of generality let ${{\mathcal{V}}^{5}_{2}}$ be positioned as described in Corollary 3.3. So the conic directrix lies in a plane $\alpha$ contained in a regular 2-spread $\mathcal{S}$ in $\Sigma_{\infty}$, and ${{\mathcal{V}}^{5}_{2}}$ corresponds to a $\mathbb{F}_{q}$-subplane ${\mathscr{B}}$ of ${\rm PG}(2,q^{3})$ tangent to $\ell_{\infty}$. Let ${\cal N}_{1},{\cal N}_{2}$ be two twisted cubics contained in ${{\mathcal{V}}^{5}_{2}}$. By Result 3.1, they correspond in ${\rm PG}(2,q^{3})$ to two $\mathbb{F}_{q}$-sublines of ${\mathscr{B}}$ not containing ${\mathscr{B}}\cap\ell_{\infty}$, and so meet in a unique affine point $P$. This corresponds to a unique point $P\in{{\mathcal{V}}^{5}_{2}}\backslash\alpha$ lying in both ${\cal N}_{1}$ and ${\cal N}_{2}$, proving part 1.

For part 2, let $P,Q$ be points lying on distinct generators of ${{\mathcal{V}}^{5}_{2}}$, $P,Q\notin{\cal C}$. If the line $PQ$ met $\alpha$, then $\langle\alpha,P,Q\rangle$ is a 3-space that contains $\alpha$ and the generators of ${{\mathcal{V}}^{5}_{2}}$ containing $P$ and $Q$, contradicting Corollary 2.3. Hence the line $PQ$ is skew to $\alpha$. In ${\rm PG}(2,q^{3})$, $P,Q$ correspond to two affine points in the tangent $\mathbb{F}_{q}$-subplane ${\mathscr{B}}$, so they lie on a unique $\mathbb{F}_{q}$-subline $b$ of ${\mathscr{B}}$. By Result 3.1, the generators of ${{\mathcal{V}}^{5}_{2}}$ correspond to the $\mathbb{F}_{q}$-sublines of ${\mathscr{B}}$ through the point ${\mathscr{B}}\cap\ell_{\infty}$. As $PQ$ is skew to $\alpha$, we have $b\cap\ell_{\infty}=\emptyset$. Hence by Result 3.1, in ${\rm PG}(6,q)$, $P,Q$ lie on a unique twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. $\square$

4 $5$-spaces meeting ${{\mathcal{V}}^{5}_{2}}$

In this section we determine how 5-spaces meet ${{\mathcal{V}}^{5}_{2}}$ and count the different intersection types. A series of lemmas is used to prove the main result which is stated in Theorem 4.8.

Lemma 4.1

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$. Of the $q^{3}+q^{2}+q+1$ $5$-spaces of ${\rm PG}(6,q)$ containing ${\cal C}$, $r_{i}$ meet ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal C}$ and $i$ generators, where

[TABLE]

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ lying in a plane $\alpha$. By Lemma 2.7, a 5-space containing ${\cal C}$ contains at most three generator lines of ${{\mathcal{V}}^{5}_{2}}$. By Theorem 2.2, three generators of ${{\mathcal{V}}^{5}_{2}}$ lie in a unique $5$-space. Hence there are

[TABLE]

5-spaces that contain three generators of ${{\mathcal{V}}^{5}_{2}}$. Such a 5-space contains three points of ${\cal C}$, and so contains ${\cal C}$ and $\alpha$.

Denote the generator lines of ${{\mathcal{V}}^{5}_{2}}$ by $\ell_{0},\ldots,\ell_{q}$ and consider two generators, $\ell_{0},\ell_{1}$ say. By Corollary 2.3, $\Sigma_{4}=\langle\alpha,\ell_{0},\ell_{1}\rangle$ is a 4-space. By Theorem 2.2, $\langle\Sigma_{4},\ell_{i}\rangle$ $i=2,\ldots,q$ are distinct 5-spaces. That is, $q-1$ of the 5-spaces about $\Sigma_{4}$ contain 3 generators, and hence the remaining two contain $\ell_{0}$, $\ell_{1}$ and no further generator of ${{\mathcal{V}}^{5}_{2}}$. Hence by Lemma 2.7, $q-1$ of the 5-spaces about $\Sigma_{4}$ meet ${{\mathcal{V}}^{5}_{2}}$ in exactly ${\cal C}$ and 3 generators; and the remaining two 5-spaces about $\Sigma_{4}$ meet ${{\mathcal{V}}^{5}_{2}}$ in exactly ${\cal C}$ and two generators. There are ${q+1\choose 2}$ choices for $\Sigma_{4}$, hence the number of 5-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal C}$ and two generators is

[TABLE]

Next, let $r_{1}$ be the number of $5$-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal C}$ and one generator. We count in two ways ordered pairs $(\ell,\Pi_{5})$ where $\ell$ is a generator of ${{\mathcal{V}}^{5}_{2}}$, and $\Pi_{5}$ is a 5-space that contains $\ell$ and $\alpha$, giving

[TABLE]

Hence $r_{1}=q^{3}/2+q/2+1$. Finally, the number of 5-spaces containing ${\cal C}$ and zero generators is $r_{0}=(q^{3}+q^{2}+q+1)-r_{3}-r_{2}-r_{1}=(q^{3}-q)/3$, as required. $\square$

Lemma 4.2

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ and let ${\cal N}_{3}$ be a twisted cubic directrix of ${{\mathcal{V}}^{5}_{2}}$.

Of the $q^{2}+q+1$ $5$-spaces of ${\rm PG}(6,q)$ containing ${\cal N}_{3}$, $s_{i}$ meet ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal N}_{3}$ and $i$ generators, where

[TABLE] 2. 2.

The total number of 5-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in a twisted cubic and $i$ generators is $q^{2}s_{i}$, $i=0,1,2$.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with a twisted cubic directrix ${\cal N}_{3}$ lying in the 3-space $\Pi_{3}$. By Lemma 2.7, a 5-space containing ${\cal N}_{3}$ contains at most two generators of ${{\mathcal{V}}^{5}_{2}}$, so the number of 5-spaces that contain $\Pi_{3}$ and exactly two generator lines is $s_{2}={q+1\choose 2}$. Let $\ell$ be a generator of ${{\mathcal{V}}^{5}_{2}}$ and consider the 4-space $\Pi_{4}=\langle\Pi_{3},\ell\rangle$. For each generator $m\neq\ell$, $\langle\Pi_{4},m\rangle$ is a 5-space about $\Pi_{4}$ that meets ${{\mathcal{V}}^{5}_{2}}$ in ${\cal N}_{3}$, $\ell$ and $m$, and in no further point by Lemma 2.7. This accounts for $q$ of the 5-spaces containing $\Pi_{4}$. Hence the remaining 5-space containing $\Pi_{4}$ meets ${{\mathcal{V}}^{5}_{2}}$ in exactly ${\cal N}_{3}$ and $\ell$. That is, exactly one of the 5-spaces about $\Pi_{4}=\langle\Pi_{3},\ell\rangle$ meets ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal N}_{3}$ and $\ell$. There are $q+1$ choices for the generator $\ell$, hence $s_{1}=q+1$. Finally $s_{0}=(q^{2}+q+1)-s_{2}-s_{1}=(q^{2}-q)/2$, as required. For part 2, by Lemma 3.2, ${{\mathcal{V}}^{5}_{2}}$ contains $q^{2}$ twisted cubics, so the total number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in a twisted cubic and $i$ generators is $q^{2}s_{i}$, $i=0,1,2$. $\square$

The next result looks at properties of 4-dim nrcs contained in ${{\mathcal{V}}^{5}_{2}}$. In particular, we show that there are no 5-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in a 4-dim nrc and 0 generator lines.

Lemma 4.3

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ in the plane $\alpha$, and let ${\cal N}_{4}$ be a $4$-dim nrc contained in ${{\mathcal{V}}^{5}_{2}}$.

The $q+1$ $5$-spaces containing ${\cal N}_{4}$ each contain a distinct generator line of ${{\mathcal{V}}^{5}_{2}}$. 2. 2.

The $4$-space containing ${\cal N}_{4}$ meets $\alpha$ in a point $P$, and either $P={\cal C}\cap{\cal N}_{4}$ or $q$ is even and $P$ is the nucleus of ${\cal C}$.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface in ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ lying in a plane $\alpha$. Let ${\cal N}_{4}$ be a 4-dim nrc contained in ${{\mathcal{V}}^{5}_{2}}$, so ${\cal N}_{4}$ lies in a 4-space which we denote $\Pi_{4}$. By Corollary 2.8, $\Pi_{4}$ does not contain a generator of ${{\mathcal{V}}^{5}_{2}}$. By Lemma 2.7, a 5-space containing ${\cal N}_{4}$ can contain at most one generator of ${{\mathcal{V}}^{5}_{2}}$. Hence each of the $q+1$ 5-spaces containing ${\cal N}_{4}$ contains a distinct generator. In particular, if we label the points of ${\cal C}$ by $Q_{0},\ldots,Q_{q}$, and the generator through $Q_{i}$ by $\ell_{Q_{i}}$, then the $q+1$ 5-spaces containing ${\cal N}_{4}$ are $\Sigma_{i}=\langle\Pi_{4},\ell_{Q_{i}}\rangle$, $i=0,\ldots,q$.

If $\Pi_{4}$ met the plane $\alpha$ in a line, then $\langle\Pi_{4},\alpha\rangle$ is a 5-space whose intersection with ${{\mathcal{V}}^{5}_{2}}$ contains ${\cal N}_{4}$ and ${\cal C}$, contradicting Lemma 2.7. Hence $\Pi_{4}$ meets $\alpha$ in a point $P$. There are three possibilities for the point $P=\Pi_{4}\cap\alpha$, namely $P\in{\cal C}$; $q$ even and $P$ the nucleus of ${\cal C}$; or $q$ even, $P\notin{\cal C}$, and $P$ not the nucleus of ${\cal C}$.

Case 1, suppose $P\in{\cal C}$. For $i=0,\ldots,q$, the 5-space $\Sigma_{i}=\langle\Pi_{4},\ell_{Q_{i}}\rangle$ meets $\alpha$ in a line $m_{i}$. Label ${\cal C}$ so that $P=Q_{0}$, so the line $m_{0}$ is the tangent to ${\cal C}$ at $P$, and $m_{i}$, $i=1,\ldots,q$, is the secant line $PQ_{i}$. We now show that $P=Q_{0}$ is a point of ${\cal N}_{4}$. Let $i\in\{1,\ldots,q\}$, then by Lemma 2.7, $\Sigma_{i}$ meets ${{\mathcal{V}}^{5}_{2}}$ in precisely ${\cal N}_{4}\cup\ell_{Q_{i}}$, and $\Sigma_{i}\cap{{\mathcal{V}}^{5}_{2}}\cap\alpha$ is the two points $P,Q_{i}$. As $P\notin\ell_{Q_{i}}$ we have $P\in{\cal N}_{4}$. That is, $P={\cal C}\cap{\cal N}_{4}$.

Case 2, suppose $q$ is even and $P=\Pi_{4}\cap\alpha$ is the nucleus of ${\cal C}$. For $i=0,\ldots,q$, the 5-space $\Sigma_{i}=\langle\Pi_{4},\ell_{Q_{i}}\rangle$ meets $\alpha$ in the tangent to ${\cal C}$ through $Q_{i}$. In this case, ${\cal C}\cap{\cal N}_{4}=\emptyset$.

Case 3, suppose $P=\Pi_{4}\cap\alpha$ is not in ${\cal C}$, and $P$ is not the nucleus of ${\cal C}$. Now $P$ lies on some secant $m=QR$ of ${\cal C}$, for some points $Q,R\in{\cal C}$. The intersection of the 5-space $\langle\Pi_{4},m\rangle$ with ${{\mathcal{V}}^{5}_{2}}$ contains ${\cal N}_{4}$ and two points $R,Q$ of ${\cal C}$. As $R,Q$ lie on distinct generators and are not in ${\cal N}_{4}$, this contradicts Lemma 2.7. Hence this case cannot occur. $\square$

We can now describe how a nrc of ${{\mathcal{V}}^{5}_{2}}$ meets the conic directrix, and note that Theorem 5.1 shows that each possibility in part 3 can occur.

Corollary 4.4

Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$.

A twisted cubic ${\cal N}_{3}\subseteq{{\mathcal{V}}^{5}_{2}}$ contains 0 points of ${\cal C}$. 2. 2.

A $4$-dim nrc ${\cal N}_{4}\subseteq{{\mathcal{V}}^{5}_{2}}$ either (i) contains 1 point of ${\cal C}$; or (ii) contains 0 points of ${\cal C}$, in which case $q$ is even and the 4-space containing ${\cal N}_{4}$ contains the nucleus of ${\cal C}$. 3. 3.

A $5$-dim nrc ${\cal N}_{5}\subseteq{{\mathcal{V}}^{5}_{2}}$ contains 0 or 1 or 2 points of ${\cal C}$.

Proof   Let ${{\mathcal{V}}^{5}_{2}}$ be a ruled quintic surface of ${\rm PG}(6,q)$ with conic directrix ${\cal C}$ in a plane $\alpha$. Let ${\cal N}_{3}$ be a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$, so by Lemma 3.2, ${\cal N}_{3}$ is a directrix of ${{\mathcal{V}}^{5}_{2}}$, and so is disjoint from $\alpha$, proving part 1. Next let ${\cal N}_{4}$ be a 4-dim nrc on ${{\mathcal{V}}^{5}_{2}}$, and let $\Pi_{4}$ be the 4-space containing ${\cal N}_{4}$. By Lemma 4.3, $\Pi_{4}\cap\alpha$ is a point $P$, and either $P={\cal C}\cap{\cal N}_{4}$; or $q$ is even and $P$ is the nucleus of ${\cal C}$, and so $P\notin{{\mathcal{V}}^{5}_{2}}$, hence $P\notin{\cal N}_{4}$, proving part 2. Let $\Pi_{5}$ be a 5-space containing a 5-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. By Lemma 2.7, $\Pi_{5}$ cannot contain $\alpha$. Hence $\Pi_{5}$ meets $\alpha$ in a line, and so contains at most two points of ${\cal C}$, proving part 3. $\square$

We now use the Bruck-Bose setting to count the 4-dim nrcs contained in ${{\mathcal{V}}^{5}_{2}}$.

Lemma 4.5

Let $\mathcal{S}$ be a regular 2-spread in a 5-space $\Sigma_{\infty}$ in ${\rm PG}(6,q)$. Position ${{\mathcal{V}}^{5}_{2}}$ as in Corollary 3.3, so ${{\mathcal{V}}^{5}_{2}}$ has splash $\mathbb{S}\subset\mathcal{S}$. Then a 5-space/4-space about a plane $\beta\in\mathbb{S}$ cannot contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$.

Proof   Position ${{\mathcal{V}}^{5}_{2}}$ as described in Corollary 3.3, so $\mathcal{S}$ is a regular 2-spread in a 5-space $\Sigma_{\infty}$, the conic directrix of ${{\mathcal{V}}^{5}_{2}}$ lies in a plane $\alpha\in\mathcal{S}$, and $\mathbb{S}\subset\mathcal{S}$ denotes the splash of ${{\mathcal{V}}^{5}_{2}}$. By Lemma 2.7, a 4-space containing $\alpha$ cannot contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. Let $\beta\in\mathbb{S}\backslash\alpha$, then by Corollary 3.4, $\beta$ lies in exactly one 3-space that contains a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$, denote these by $\Pi_{3}$ and ${\cal N}_{3}$ respectively. By Theorem 3.2, ${\cal N}_{3}$ is a directrix of ${{\mathcal{V}}^{5}_{2}}$, and so $\Pi_{3}$ is disjoint from $\alpha$. So if $\ell_{P}$ is a generator of ${{\mathcal{V}}^{5}_{2}}$, then $\Pi_{4}=\langle\Pi_{3},\ell_{P}\rangle$ is a 4-space and $\Pi_{4}\cap\alpha$ is the point $P=\ell_{P}\cap{\cal C}$. Let $\ell$ be a line of $\alpha$ through $P$ and let $\Pi_{5}=\langle\Pi_{3},\ell\rangle$. If $\ell$ is tangent to ${\cal C}$, then $\Pi_{5}\cap{{\mathcal{V}}^{5}_{2}}$ is exactly ${\cal N}_{3}\cup\ell_{P}$. If $\ell$ is a secant of ${\cal C}$, so $\ell\cap{\cal C}=\{P,Q\}$, then $\Pi_{5}\cap{{\mathcal{V}}^{5}_{2}}$ consists of ${\cal N}_{3},\ell_{P}$ and the generator $\ell_{Q}$ through $Q$. Varying $\ell_{P}$ and $\ell$, we get all the 5-spaces that contain $\beta$ and contain 1 or 2 generators of ${{\mathcal{V}}^{5}_{2}}$. That is, each 5-space containing $\beta$ and 1 or 2 generators of ${{\mathcal{V}}^{5}_{2}}$ also contains ${\cal N}_{3}$. The remaining 5-spaces about $\beta$ hence contain 0 generators of ${{\mathcal{V}}^{5}_{2}}$ and meet $\alpha$ in an exterior line of ${\cal C}$. Hence by Lemma 4.3, none of the 5-spaces about $\beta$ contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. $\square$

Lemma 4.6

The number of 4-dim nrcs contained in ${{\mathcal{V}}^{5}_{2}}$ is $q^{4}-q^{2}$. 2. 2.

The number of 5-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in a 4-dim nrc and one generator is $q^{5}+q^{4}-q^{3}-q^{2}$.

Proof   Without loss of generality position ${{\mathcal{V}}^{5}_{2}}$ as described in Corollary 3.3. That is, let $\mathcal{S}$ be a regular 2-spread in a 5-space $\Sigma_{\infty}$, let the conic directrix of ${{\mathcal{V}}^{5}_{2}}$ lie in a plane $\alpha\in\mathcal{S}$, and let $\mathbb{S}\subset\mathcal{S}$ be the splash of ${{\mathcal{V}}^{5}_{2}}$. Straightforward counting shows that a 5-space distinct from $\Sigma_{\infty}$ contains a unique spread plane. If this plane is in the splash $\mathbb{S}$, then by Lemma 4.5, the 5-space does not contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. So a 5-space containing a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$ contains a unique plane of $\mathcal{S}\backslash\mathbb{S}$. Consider a plane $\gamma\in\mathcal{S}\backslash\mathbb{S}$. Let $P\in{\cal C}$, let $\ell_{P}$ be the generator of ${{\mathcal{V}}^{5}_{2}}$ through $P$, and consider the 4-space $\Pi_{4}=\langle\gamma,\ell_{P}\rangle$. Suppose first that $\Pi_{4}$ contains two generators of ${{\mathcal{V}}^{5}_{2}}$, then there is a 5-space $\Pi_{5}$ containing $\gamma$ and two generators. By Lemma 2.7, $\Pi_{5}$ contains either ${\cal C}$ or a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. A 5-space distinct from $\Sigma_{\infty}$ cannot contain two planes of $\mathcal{S}$, so $\Pi_{5}$ does not contain ${\cal C}$. Moreover, by Corollary 3.3, $\Pi_{5}$ does not contain a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. Hence $\Pi_{4}$ contains exactly one generator of ${{\mathcal{V}}^{5}_{2}}$. If every generator of ${{\mathcal{V}}^{5}_{2}}$ contained at least one point of $\Pi_{4}$, then the intersection of $\Pi_{4}$ with ${{\mathcal{V}}^{5}_{2}}$ contains at least $\ell_{P}$ and $q$ further points, one on each generator. By Lemma 2.7 and Corollary 2.8, the only possibility is that $\Pi_{4}\cap{{\mathcal{V}}^{5}_{2}}$ contains a twisted cubic, which is not possible by Corollary 3.3. Hence there is at least one generator which is disjoint from $\Pi_{4}$, denote this $\ell_{Q}$. Label the points of $\ell_{Q}$ by $X_{0},\ldots,X_{q}$, then the $q+1$ 5-spaces containing $\Pi_{4}$ are $\Sigma_{i}=\langle\gamma,\ell_{P},X_{i}\rangle$. For each $i=0,\ldots,q$, the intersection of $\Sigma_{i}$ with ${{\mathcal{V}}^{5}_{2}}$ contains the generator $\ell_{P}$ and the point $X_{i}$. By Corollary 3.3, $\Sigma_{i}$ does not contain a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. Hence by Lemma 2.7, $\Sigma_{i}\cap{{\mathcal{V}}^{5}_{2}}$ is $\ell_{P}$ and a 4-dim nrc.

That is, there are $(q+1)^{2}$ 5-spaces containing $\gamma$ and one generator of ${{\mathcal{V}}^{5}_{2}}$, each contains a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. Further, if $\Pi_{5}$ is a 5-space containing $\gamma$ and zero generators of ${{\mathcal{V}}^{5}_{2}}$, then by Lemma 4.3, $\Pi_{5}$ does not contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. Hence as there are $q^{3}-q^{2}$ choices for $\gamma$, there are $(q+1)^{2}\times(q^{3}-q^{2})=q^{5}+q^{4}-q^{3}-q^{2}$ 5-spaces that meet ${{\mathcal{V}}^{5}_{2}}$ in one generator and a 4-dim nrc. By Lemma 4.3, every 4-dim nrc in ${{\mathcal{V}}^{5}_{2}}$ lies in $q+1$ such 5-spaces. Hence the number of 4-dim nrcs contained in ${{\mathcal{V}}^{5}_{2}}$ is $(q^{5}+q^{4}-q^{3}-q^{2})/(q+1)$ as required. $\square$

We now count the number of 5-dim nrcs contained in ${{\mathcal{V}}^{5}_{2}}$.

Lemma 4.7

The number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc is $q^{6}-q^{4}$.

Proof   We show that the number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc is $q^{6}-q^{4}$ by counting in two ways the number $x$ of incident pairs $(A,\Pi_{5})$ where $A$ is a point of ${{\mathcal{V}}^{5}_{2}}$, and $\Pi_{5}$ is a 5-space containing $A$. The number of ways to choose a point $A$ of ${{\mathcal{V}}^{5}_{2}}$ is $(q+1)^{2}.$ The point $A$ lies in $q^{5}+q^{4}+q^{3}+q^{2}+q+1$ 5-spaces. So

[TABLE]

Alternatively, we count the 5-spaces first; there are several possibilities for $\Pi_{5}$. By Lemmas 2.7, $\Pi_{5}\cap{{\mathcal{V}}^{5}_{2}}$ is either empty, or contains an $r$-dim nrc, for some $r\in\{2,\ldots,5\}$. Let $n_{r}$ be the number of pairs $(A,\Pi_{5})$ with $A\in{{\mathcal{V}}^{5}_{2}}\cap\Pi_{5}$ and $\Pi_{5}$ containing an $r$-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. Note that

[TABLE]

We now calculate $n_{2}$, $n_{3}$ and $n_{4}$, and then use (1) to determine the number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc.

Consider a 5-space $\Pi_{5}$ that contains the conic directrix ${\cal C}$, so by Lemma 4.1, $\Pi_{5}$ contains 0, 1, 2 or 3 generators of ${{\mathcal{V}}^{5}_{2}}$, and the number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in exactly the conic directrix and $i$ generators is $r_{i}$. In this case the number of ways to pick a point of $\Pi_{5}\cap{{\mathcal{V}}^{5}_{2}}$ is $iq+q+1$. Hence the total number of pairs $(A,\Pi_{5})$ with $\Pi_{5}$ containing the conic directrix is

[TABLE] 2. 2.

Consider a 5-space $\Pi_{5}$ that contains a twisted cubic, then by Lemma 4.2, $\Pi_{5}$ contains 0, 1 or 2 generators of ${{\mathcal{V}}^{5}_{2}}$, and the number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in a given twisted cubic and $i$ generators is $s_{i}$. In this case the number of ways to pick $A$ in ${{\mathcal{V}}^{5}_{2}}\cap\Pi_{5}$ is $iq+q+1$. Hence the number of pairs $(A,\Pi_{5})$ with $\Pi_{5}$ containing a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$ is

[TABLE] 3. 3.

Consider a 5-space $\Pi_{5}$ that contains a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. By Lemma 4.3, $\Pi_{5}$ contains 1 generator of ${{\mathcal{V}}^{5}_{2}}$. By Lemma 4.6, the number of 5-spaces meeting ${{\mathcal{V}}^{5}_{2}}$ in exactly a 4-dim nrc and one generator is $q^{5}+q^{4}-q^{3}-q^{2}$. The number of ways to pick $A$ in ${{\mathcal{V}}^{5}_{2}}\cap\Pi_{5}$ is $2q+1$. So

[TABLE] 4. 4.

Denote the number of $5$-spaces containing a 5-dim nrc of ${{\mathcal{V}}^{5}_{2}}$ by $y$. Then the number of pairs $(A,\Pi_{5})$ with $\Pi_{5}$ containing a 5-dim nrc of ${{\mathcal{V}}^{5}_{2}}$ is

[TABLE]

Substituting the calculated values for $x,n_{2},n_{3},n_{4},n_{5}$ into (1) and rearranging gives $y=q^{6}-q^{4}$ as required. $\square$

Summarising the preceding lemmas gives the following theorem describing ${{\mathcal{V}}^{5}_{2}}$.

Theorem 4.8

Let ${{\mathcal{V}}^{5}_{2}}$ be the ruled quintic surface in ${\rm PG}(6,q)$, $q\geq 6$.

${{\mathcal{V}}^{5}_{2}}$* contains exactly*

[TABLE] 2. 2.

A 5-space meets ${{\mathcal{V}}^{5}_{2}}$ in one of the following configurations

[TABLE]

5 $5$-spaces and the Bruck-Bose spread

Let $\mathcal{S}$ be a regular 2-spread in a 5-space $\Sigma_{\infty}$ in ${\rm PG}(6,q)$, and position ${{\mathcal{V}}^{5}_{2}}$ so that it corresponds to a tangent $\mathbb{F}_{q}$-subplane of ${\rm PG}(2,q^{3})$. So ${{\mathcal{V}}^{5}_{2}}$ has splash $\mathbb{S}\subset\mathcal{S}$, the conic directrix ${\cal C}$ lies in a plane $\alpha\in\mathbb{S}$, and each of the $q^{2}$ 3-spaces containing a twisted cubic directrix of ${{\mathcal{V}}^{5}_{2}}$ meets $\Sigma_{\infty}$ in a distinct plane of $\mathbb{S}\backslash\alpha$. In Corollary 3.4, we looked at how 3-spaces containing a plane of $\mathcal{S}$ meet ${{\mathcal{V}}^{5}_{2}}$. In Lemma 4.5, we looked at how 4-spaces containing a plane of $\mathcal{S}$ meet ${{\mathcal{V}}^{5}_{2}}$. Next we look at how 5-spaces containing a plane of $\mathcal{S}$ meet ${{\mathcal{V}}^{5}_{2}}$. Note that straightforward counting shows that a 5-space distinct from $\Sigma_{\infty}$ contains a unique plane $\pi$ of $\mathcal{S}$, and meets every other plane of $\mathcal{S}$ in a line. If $\pi=\alpha$, then Lemma 4.1 describes the possible intersections with ${{\mathcal{V}}^{5}_{2}}$. The next theorem describes the possible intersections with ${{\mathcal{V}}^{5}_{2}}$ for the remaining cases $\pi\in\mathbb{S}\backslash\alpha$ and $\pi\in\mathcal{S}\backslash\mathbb{S}$.

Theorem 5.1

Position ${{\mathcal{V}}^{5}_{2}}$ as in Corollary 3.3, so $\mathcal{S}$ is a regular 2-spread in a hyperplane $\Sigma_{\infty}$, the conic directrix ${\cal C}$ lies in a plane $\alpha\in\mathcal{S}$ and ${{\mathcal{V}}^{5}_{2}}$ has splash $\mathbb{S}\subset\mathcal{S}$. Let $\ell$ be a line of $\alpha$ with $|\ell\cap{\cal C}|=i$ and let $\pi\in\mathcal{S}$, $\pi\neq\alpha$. Then the $q$ 5-spaces containing $\pi,\ell$ and distinct from $\Sigma_{\infty}$ meet ${{\mathcal{V}}^{5}_{2}}$ as follows.

If $\pi\in\mathbb{S}\backslash\alpha$, then $q-1$ meet ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, and 1 meets ${{\mathcal{V}}^{5}_{2}}$ in a twisted cubic and $i$ generators. 2. 2.

If $\pi\in\mathcal{S}\backslash\mathbb{S}$, then $q-i$ meet ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, and $i$ meet ${{\mathcal{V}}^{5}_{2}}$ in a 4-dim nrc and $1$ generator.

Proof   By [2], the group of collineations of ${\rm PG}(6,q)$ fixing $\mathcal{S}$ and ${{\mathcal{V}}^{5}_{2}}$ is transitive on the planes of $\mathbb{S}\backslash\alpha$ and on the planes of $\mathcal{S}\backslash\mathbb{S}$. As this group fixes the conic directrix ${\cal C}$, it is transitive on the lines of $\alpha$ tangent to ${\cal C}$, the lines of $\alpha$ secant to ${\cal C}$ and the lines of $\alpha$ exterior to ${\cal C}$. So without loss of generality let $\ell_{0}$ be a line of $\alpha$ exterior to ${\cal C}$, let $\ell_{1}$ be a line of $\alpha$ tangent to ${\cal C}$, let $\ell_{2}$ be a line of $\alpha$ secant to ${\cal C}$, let $\beta$ be a plane in $\mathbb{S}\backslash\alpha$, and let $\gamma$ be a plane of $\mathcal{S}\backslash\mathbb{S}$. For $i=0,1,2$, label the 4-spaces $\Sigma_{4,i}=\langle\beta,\ell_{i}\rangle$ and $\Pi_{4,i}=\langle\gamma,\ell_{i}\rangle$. By Corollary 3.4, as $\beta\in\mathbb{S}\backslash\alpha$, there is a unique twisted cubic of ${{\mathcal{V}}^{5}_{2}}$ that lies in a 3-space about $\beta$, denote this 3-space by $\Pi_{3}$. Hence for $i=0,1,2$, there is a unique 5-space containing $\Sigma_{4,i}$ whose intersection with ${{\mathcal{V}}^{5}_{2}}$ contains a twisted cubic, namely the 5-space $\langle\Pi_{3},\ell_{i}\rangle$.

First consider the line $\ell_{0}$ which is exterior to ${\cal C}$. A 5-space meeting $\alpha$ in $\ell_{0}$ contains 0 points of ${\cal C}$, and so contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$. The 4-space $\Sigma_{4,0}=\langle\beta,\ell_{0}\rangle$ lies in $q$ 5-spaces distinct from $\Sigma_{\infty}$, each containing 0 generators of ${{\mathcal{V}}^{5}_{2}}$. Exactly one of these 5-spaces, namely $\langle\Pi_{3},\ell_{0}\rangle$, contains a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. The remaining $q-1$ 5-spaces about $\Sigma_{4,0}$ contain 0 generators, and do not contain a conic or twisted cubic of ${{\mathcal{V}}^{5}_{2}}$, so by Theorem 4.8, they meet ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 1 for $i=0$. For part 2, let $\Pi_{5}\neq\Sigma_{\infty}$ be any 5-space containing $\Pi_{4,0}=\langle\gamma,\ell_{0}\rangle$. As $\gamma\notin\mathbb{S}$, by Corollary 3.3, $\Pi_{5}$ cannot contain a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. As $\Pi_{5}$ contains 0 generator lines of ${{\mathcal{V}}^{5}_{2}}$ and does not contain a conic or twisted cubic of ${{\mathcal{V}}^{5}_{2}}$, by Theorem 4.8, $\Pi_{5}$ meets ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc. That is, the $q$ 5-spaces (distinct from $\Sigma_{\infty}$) containing $\Pi_{4,0}$ meet ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 2 for $i=0$.

Next consider the line $\ell_{1}$ which is tangent to ${\cal C}$. Let $P=\ell_{1}\cap{\cal C}$ and denote the generator of ${{\mathcal{V}}^{5}_{2}}$ through $P$ by $\ell_{P}$. A 5-space meeting $\alpha$ in a tangent line contains 1 point of ${\cal C}$, and so contains at most one generator of ${{\mathcal{V}}^{5}_{2}}$. So exactly one 5-space contains $\Sigma_{4,1}$ and a generator, namely the 5-space $\langle\Sigma_{4,1},\ell_{P}\rangle$. Consider the 5-space $\langle\Pi_{3},\ell_{1}\rangle$, it contains $P$ and a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$ which by Corollary 4.4 is disjoint from $\alpha$, hence $\langle\Pi_{3},\ell_{1}\rangle$ contains the generator $\ell_{P}$. That is, $\langle\Pi_{3},\ell_{1}\rangle$ contains $\beta$, $\ell_{1}$, $\ell_{P}$ and so $\langle\Pi_{3},\ell_{1}\rangle=\langle\Sigma_{4,1},\ell_{P}\rangle$. That is, the intersection of $\langle\Sigma_{4,1},\ell_{P}\rangle$ with ${{\mathcal{V}}^{5}_{2}}$ is a twisted cubic and one generator. Let $\Pi_{5}\neq\Sigma_{\infty}$ be one of the remaining $q-1$ 5-spaces (distinct from $\Sigma_{\infty}$) that contains $\Sigma_{4,1}$, so $\Pi_{5}$ contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$ and does not contain a conic or twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. So by Theorem 4.8, $\Pi_{5}$ meets ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 1 for $i=1$. For part 2, we consider $\Pi_{4,1}=\langle\gamma,\ell_{1}\rangle$. By Corollary 3.3, as $\gamma\notin\mathbb{S}$, no 5-space containing $\Pi_{4,1}$ contains a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$. The 5-space $\langle\Pi_{4,1},\ell_{P}\rangle$ contains one generator of ${{\mathcal{V}}^{5}_{2}}$, so by Theorem 4.8, it meets ${{\mathcal{V}}^{5}_{2}}$ in exactly a 4-dim nrc and the generator $\ell_{P}$. Let $\Pi_{5}\neq\Sigma_{\infty}$ be one of the remaining $q-1$ 5-spaces containing $\Pi_{4,1}$, then $\Pi_{5}$ contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$. So by Theorem 4.8, $\Pi_{5}$ meets ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 2 for $i=1$.

Finally, consider the line $\ell_{2}$ which is secant to ${\cal C}$. Let ${\cal C}\cap\ell_{2}=\{P,Q\}$ and let $\ell_{P},\ell_{Q}$ be the generators of ${{\mathcal{V}}^{5}_{2}}$ through $P,Q$ respectively. The intersection of the 5-space $\langle\Pi_{3},\ell_{2}\rangle$ and ${{\mathcal{V}}^{5}_{2}}$ contains a twisted cubic, and $P$ and $Q$. By Corollary 4.4, this twisted cubic is disjoint from $\alpha$, so $\langle\Pi_{3},\ell_{2}\rangle$ contains the two generators $\ell_{P},\ell_{Q}$. Thus $\langle\Pi_{3},\ell_{2}\rangle=\langle\Sigma_{4,2},\ell_{P}\rangle=\langle\Sigma_{4,2},\ell_{Q}\rangle=\langle\Sigma_{4,2},\ell_{P},\ell_{Q}\rangle$. The remaining $q-1$ 5-spaces (distinct from $\Sigma_{\infty}$) about $\Sigma_{4,2}$ contain 0 generators and two points of ${\cal C}$. By Lemma 4.3 they cannot contain a 4-dim nrc of ${{\mathcal{V}}^{5}_{2}}$. So by Theorem 4.8, they meet ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 1 for $i=2$. For part 2, let $\Pi_{5}\neq\Sigma_{\infty}$ be a 5-space containing $\Pi_{4,2}=\langle\gamma,\ell_{2}\rangle$. By Corollary 3.3, $\Pi_{5}$ does not contain a twisted cubic of ${{\mathcal{V}}^{5}_{2}}$, as $\gamma\notin\mathbb{S}$. So by Theorem 4.8, $\Pi_{5}$ contains at most one generator of ${{\mathcal{V}}^{5}_{2}}$. Hence $\langle\Pi_{4,2},\ell_{P}\rangle$, $\langle\Pi_{4,2},\ell_{Q}\rangle$ are distinct 5-spaces about $\Pi_{4,2}$, and by Theorem 4.8, they each meet ${{\mathcal{V}}^{5}_{2}}$ in a 4-dim nrc and one generator. Let $\Sigma_{5}\neq\Sigma_{\infty}$ be one of the remaining $q-2$ 5-spaces about $\Pi_{4,2}$. Then $\Sigma_{5}$ contains 0 generators of ${{\mathcal{V}}^{5}_{2}}$, and so by Theorem 4.8, meets ${{\mathcal{V}}^{5}_{2}}$ in a 5-dim nrc, proving part 2 for $i=2$. $\square$