Characterising elliptic solids of $Q(4,q)$, $q$ even

TL;DR

This paper characterizes specific sets of solids in projective 4-space over even finite fields, showing they are either disjoint from a hyperoval or intersect a non-singular quadric in an elliptic quadric.

Contribution

It provides a classification of elliptic solids in PG(4,q) for even q, identifying their structure relative to hyperovals and quadrics.

Findings

Sets of solids are either disjoint from a hyperoval or meet a non-singular quadric in an elliptic quadric.

Characterization applies to q even, q > 2.

Provides a structural understanding of elliptic solids in projective geometry.

Abstract

Let $E$ be a set of solids (hyperplanes) in $PG(4,q)$, $q$ even, $q>2$, such that every point of $PG(4,q)$ lies in either $0$, $\frac12q^3$ or $\frac12(q^3-q^2)$ solids of $E$, and every plane of $PG(4,q)$ lies in either $0$, $\frac12q$ or $q$ solids of $E$. This article shows that $E$ is either the set of solids that are disjoint from a hyperoval, or the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric.

Equations

$$\mathsf{e}=\frac{|\mathcal{E}|}{\frac{1}{2}{q}^2},$$

$$r.0+w.\frac{1}{2}{q}^3+b.\frac{1}{2}(q^3-q^2)$$

$$r.0.0+w.\frac{1}{2}{q}^3(\frac{1}{2}{q}^3-1)+b.\frac{1}{2}(q^3-q^2).(\frac{1}{2}(q^3-q^2)-1)$$

$$wq$$

$$s+q$$

$$t+s$$

$$t.(\frac{1}{2}{q}^3-1)+s.(\frac{1}{2}(q^3-q^2)-1)$$

$$s+q$$

$$s+q=x(q^2+q+1).$$

$$wq+b(q-1)$$

$$wq(q^3-2)+b(q-1)(q^3-q^2-2)$$

$$q^2(q-1)b=q(q-1)\left[(q^3-2)(q^3+q^2+q+1)-(q^4-q^3-2)(q^2+q+1)\right]$$

$$y+z=q^3+q^2+q+1.$$

$$y.(\frac{1}{2}q-1)+z(q-1)$$

$$u+v$$

$$(q+2).0+u.(\frac{1}{2}{q}^3)+v.(\frac{1}{2}(q^3-q^2))$$

$$(q+1-x).\frac{1}{2}{q}^3=\frac{1}{2}{q}^3(q-1).1$$

$$b.\frac{1}{2}(q^3-q^2)=\frac{1}{2}{q}^2(q^2-1).(q^2+1)$$

$$t+s$$

$$t.\frac{1}{2}{q}^3+s.\frac{1}{2}(q^3-q^2)$$

$$t+s$$

$$t.\frac{1}{2}{q}^3+s.\frac{1}{2}(q^3-q^2)$$

$$x.(q+1)+(q^3+q^2+q+1-x).1$$

$$y+(q^2+q+1)(q+1-y)=q^3+q^2+q+1.$$

$$q^3+q^2+q+1$$

$$q^2+1=(3+x)+(q+1)(q+1-(3+x)),$$

$$n+m$$

$$y+n.(q+1-y)+m.(2q+1-y)$$

Full text

AMS code: 51E20

Keywords: projective geometry, quadrics, hyperplanes

1 Introduction

In this article, we denote a non-singular quadric of $\textup{PG}(4,q)$ by $Q(4,q)$, and refer to the hyperplanes of $\textup{PG}(4,q)$ as solids. We prove the following characterisations, the second characterises the solids of $\textup{PG}(4,q)$ meeting a non-singular quadric $Q(4,q)$ in an elliptic quadric.

Theorem 1.1

Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, such that

• (I)

every point of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q^{3}$ or $\frac{1}{2}(q^{3}-q^{2})$ solids of $\mathcal{E}$, and

• (II)

every plane of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q$ or $q$ solids of $\mathcal{E}$.

Then either $\mathcal{E}$ is the set of solids that are disjoint from a hyperoval, or $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric.

Theorem 1.2

Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, satsifying conditions (I) and (II) of Theorem 1.1. If

• (III)

there exists a line of $\textup{PG}(4,q)$ that lies in exactly $\frac{1}{2}q(q+1)$ solids of $\mathcal{E}$,

then $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric.

For more details on non-singular quadrics, see [2]. Note that using techniques from [2], we can show that the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric satisfies conditions (I), (II) and (III).

There are several known characterisations of the points, lines, planes and solids relating to $Q(4,q)$; these are surveyed in [1].

2 Preliminary results

Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, such that

• (I)

every point of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q^{3}$ or $\frac{1}{2}(q^{3}-q^{2})$ solids of $\mathcal{E}$, and

• (II)

every plane of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q$ or $q$ solids of $\mathcal{E}$.

By assumption (I), there are three types of points; a point is called a

• •

red point if it lies in [math] solids of $\mathcal{E}$,

• •

white point if it lies in $\frac{1}{2}q^{3}$ solids of $\mathcal{E}$,

• •

black point if it lies in $\frac{1}{2}(q^{3}-q^{2})$ solids of $\mathcal{E}$.

Further, by assumption (I), a solid in $\mathcal{E}$ does not contain any red points, so we partition the solids of $\textup{PG}(4,q)$ into $\mathcal{E}$ and two further types of solids:

• •

let $\mathcal{T}$ be the set of solids containing at least one red point,

• •

let $\mathcal{H}$ be the set of solids not in $\mathcal{E}$ that contain no red points.

In the following, we will show that there are two possibilities. In case A, the red points form a hyperoval $\mathcal{O}$ in a plane $\pi$, the white points are the points in $\pi\setminus\mathcal{O}$, and the black points are the points of $\textup{PG}(4,q)\setminus\pi$. Moreover, in this case, $\mathcal{E}$ is the set of solids that do not meet $\mathcal{O}$, $\mathcal{T}$ is the set of solids that meet $\mathcal{O}$, and $\mathcal{H}=\emptyset$. In case B, the black points form a non-singular quadric $Q(4,q)$ whose nucleus is the unique red point. Moreover, in this case, $\mathcal{E}$, $\mathcal{H}$, $\mathcal{T}$ are the set of solids meeting $Q(4,q)$ in an elliptic quadric, a hyperbolic quadric and a conic cone respectively.

2.1 Determining $|\mathcal{E}|$

Lemma 2.1

Let

[TABLE]

then $\mathsf{e}$ is an integer and $\mathsf{e}$ is congruent to either [math] or $-1$ modulo $q$.

Proof   We first show that $\frac{1}{2}q^{2}$ divides $|\mathcal{E}|$. Let $r,w,b$ be the number of red, white and black points of $\textup{PG}(4,q)$ respectively. Count in two ways the incident pairs $(P,\Pi)$ where $\Pi$ is a solid in $\mathcal{E}$ and $P$ is a point of $\Pi$. We have

[TABLE]

As $\frac{1}{2}q^{2}$ divides the left hand side and gcd$(q^{2},q^{3}+q^{2}+q+1)=1$, it follows that $\frac{1}{2}q^{2}$ divides $|\mathcal{E}|$, so $\mathsf{e}$ is an integer.

Now we show that $\mathsf{e}$ is congruent to either 0 or $-1$ modulo $q$. Count in two ways incident triples $(P,\Pi,\Sigma)$ where $\Pi,\Sigma$ are distinct solids of $\mathcal{E}$ and $P$ is a point in $\Pi\cap\Sigma$. We have

[TABLE]

Substituting (1) into (2) to eliminate $b$ and then simplifying gives

[TABLE]

Consider the right hand side of this equation. If the first factor $\mathsf{e}$ is even, then the second factor is odd, and so $\mathsf{e}$ is congruent to 0 modulo $q$. On the other hand, if the first factor $\mathsf{e}$ is odd, then the second factor is congruent to 0 modulo $q$, and so $\mathsf{e}$ is congruent to $-1$ modulo $q$. $\square$

Lemma 2.2

Each solid in $\mathcal{E}$ contains $s$ black points where

[TABLE]

Proof   Let $\Sigma\in\mathcal{E}$, let $t$ be the number of white points in $\Sigma$ and let $s$ be the number of black points in $\Sigma$. As $\Sigma$ does not contain any red points, we have

[TABLE]

Count in two ways incident pairs $(P,\Pi)$ where $\Pi\in\mathcal{E}$, $\Sigma\neq\Pi$ and $P$ is a point in $\Sigma\cap\Pi$. We have

[TABLE]

Simplifying using (4) gives (3) as required. $\square$

Lemma 2.3

If $\mathsf{e}\equiv 0\pmod{q}$, then

$|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, 2. 2.

each solid in $\mathcal{E}$ contains $q^{3}+q^{2}$ black points and $q+1$ white points.

Proof   Suppose $\mathsf{e}\equiv 0\pmod{q}$, so $\mathsf{e}=kq$ for some integer $k$, and hence $|\mathcal{E}|=\tfrac{1}{2}kq^{3}$. By Lemma 2.2, a solid in $\mathcal{E}$ contains $s$ black points, with $s$ given in (3). Substituting $\mathsf{e}=kq$ into (3) gives

[TABLE]

So we have q^{2}+q+1\bigm{|}s+q and q\bigm{|}s+q. As gcd$(q^{2}+q+1,q)=1$, it follows that $s=xq(q^{2}+q+1)-q$ for some integer $x$. By (4), $0\leq s\leq q^{3}+q^{2}+q+1$, hence $x=1$ and so $s=q^{2}(q+1)$, $k=q-1$ and $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1).$ $\square$

Lemma 2.4

If $\mathsf{e}\equiv-1\pmod{q}$, then

$|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, 2. 2.

each solid in $\mathcal{E}$ contains $q^{2}+1$ black points and $q^{3}+q$ white points.

Proof   If $\mathsf{e}\equiv-1\pmod{q}$, then $\mathsf{e}=kq-1$ for some integer $k$, and hence $|\mathcal{E}|=\frac{1}{2}q^{2}(kq-1)$. We will show that $k=q$. By Lemma 2.2, a solid in $\mathcal{E}$ contains $s$ black points, with $s$ given in (3). Substituting $\mathsf{e}=kq-1$ into (3) gives

[TABLE]

So q^{2}+q+1\bigm{|}s+q, that is, for some integer $x$, we have

[TABLE]

Comparing (5) and (6) we get $x-1=q(q-k)$, so q\bigm{|}x-1. By (4), $0\leq s\leq q^{3}+q^{2}+q+1$, hence $0<x\leq q$. It follows that $x=1$ and $k=q$. This gives $s=q^{2}+1$ and $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, as required. $\square$

2.2 Case A: $\mathsf{e}\equiv 0\pmod{q}$, so $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$

Lemma 2.5

If $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, then in $\textup{PG}(4,q)$, there are $q+2$ red points, $q^{2}-1$ white points and $q^{4}+q^{3}$ black points.

Proof   Let $r$ be the number of red points, $w$ be the number of white points, and $b$ be the number of black points of $\textup{PG}(4,q)$. Substituting $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$ into (1) gives

[TABLE]

Substituting $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$ into (2) and dividing by $\frac{1}{4}q^{2}$ gives

[TABLE]

Multiplying (7) by $q^{3}-2$ and subtracting (8) gives

[TABLE]

giving $b=q^{4}+q^{3}$. Using (7) we get $w=q^{2}-1$. As $b+w+r=q^{4}+q^{3}+q^{2}+q+1$, we have $r=q+2$. $\square$

Lemma 2.6

If $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, then the $q+2$ red points of $\textup{PG}(4,q)$ lie in a plane.

Proof   By assumption (II), there are three types of planes in $\textup{PG}(4,q)$. Let $\Sigma\in\mathcal{E}$ and let $x,\ y,\ z$ denote the number of planes in $\Sigma$ that lie in $0,\ \frac{1}{2}q,\ q$ solids of $\mathcal{E}$ respectively. Noting that $x=0$, we have

[TABLE]

Counting in two ways the pairs $(\pi,\Pi)$ where $\pi$ is a plane in $\Sigma$, and $\Pi$ is a solid containing $\pi$, with $\Pi\in\mathcal{E}$ and $\Pi\neq\Sigma$ gives

[TABLE]

Using $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$ and (9) gives $z=q+1$ and $y=q^{3}+q^{2}$. Let $\alpha_{1},\ldots,\alpha_{q+1}$ be the $z=q+1$ planes in $\Sigma$ which lie in $q$ solids of $\mathcal{E}$ and so in one solid not in $\mathcal{E}$. For $i=1,\ldots,q+1$, let $\Gamma_{i}$ denote the unique solid not in $\mathcal{E}$ that contains $\alpha_{i}$. Recall that the solids in $\mathcal{E}$ contain no red points. For any fixed $i$, the solids containing $\alpha_{i}$ cover all the points of $\textup{PG}(4,q)$, so the $r=q+2$ red points lie in $\Gamma_{i}$. In particular, as $\alpha_{1},\alpha_{2}$ are distinct planes in $\Sigma$, $\pi=\Gamma_{1}\cap\Gamma_{2}$ is a plane which is not contained in $\Sigma$. The $r=q+2$ red points lie in $\Gamma_{1}$ and $\Gamma_{2}$, hence they lie in $\pi$. $\square$

Lemma 2.7

If $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, then the plane $\pi$ containing the $q+2$ red points contains no black points.

Proof   By Lemma 2.6, there are $q+2$ red points which lie in a plane $\pi$. Let $u$ be the number of white points in $\pi$ and $v$ the number of black points in $\pi$. As the number of red points in $\pi$ is $q+2$, we have $u+v=(q^{2}+q+1)-(q+2)$, that is,

[TABLE]

As solids in $\mathcal{E}$ contain no red points, $\pi$ does not lie in a solid of $\mathcal{E}$. Hence each solid in $\mathcal{E}$ meets $\pi$ in a line. We now count in two ways the incident pairs $(P,\Pi)$ where $\Pi$ is a solid in $\mathcal{E}$ and $P$ is a point in $\pi\cap\Pi$. Using assumption (I), we have

[TABLE]

Using $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$ and (10) gives $v=0$. $\square$

Lemma 2.8

If $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, then the $q+2$ red points of $\textup{PG}(4,q)$ form a hyperoval.

Proof   By Lemma 2.6, there are $q+2$ red points of $PG(4,q)$ and they lie in a plane $\pi$. Further, $\pi$ contains 0 black points and $q^{2}-1$ white points Lemma 2.7. Let $\ell$ be a line of $\pi$ which contains $x\geq 1$ red points, so $\ell$ contains $q+1-x$ white points. As $\ell$ contains a red point, $\ell$ does not lie in a solid of $\mathcal{E}$, so each solid of $\mathcal{E}$ meets $\ell$ in a white point. We count in two ways incident pairs $(P,\Pi)$ where $\Pi\in\mathcal{E}$ and $P$ is a white point in $\ell\cap\Pi$. Using assumption (I), we have

[TABLE]

and so $x=2$. Hence the $q+2$ points in $\pi$ are no three collinear, so they form a hyperoval. $\square$

Lemma 2.9

If $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$, then $\mathcal{E}$ is the set of solids that are disjoint from a hyperoval.

Proof   By Lemma 2.8, there are $q+2$ red points of $PG(4,q)$ and they form a hyperoval $\mathcal{O}$ in a plane $\pi$. Let $\Sigma$ be a solid in $\mathcal{E}$. As $\Sigma$ contains no red points, it does not contain $\pi$ and $\Sigma\cap\pi$ is a line that is disjoint from $\mathcal{O}$.

Conversely we count the solids in $\textup{PG}(4,q)$ that are disjoint from $\mathcal{O}$, that is, we count the number of solids of $\textup{PG}(4,q)$ that meet $\pi$ in a line containing 0 red points. The number of lines of $\pi$ that are external to $\mathcal{O}$ is $q^{2}+q+1-\frac{1}{2}(q+2)(q+1)=\frac{1}{2}(q^{2}-q)$. Through each of these external lines, there are $q^{2}+q+1$ solids of $\textup{PG}(4,q)$, $q+1$ of which contain $\pi$. Hence the number of solids of $\textup{PG}(4,q)$ that meet $\pi$ in a line containing 0 red points is $\frac{1}{2}(q^{2}-q).[(q^{2}+1+1)-(q+1)]=\frac{1}{2}(q^{2}-q).q^{2}$ which is equal to $|\mathcal{E}|$. Hence $\mathcal{E}$ is the set of solids that are disjoint from $\mathcal{O}$. $\square$

2.3 Case B: $\mathsf{e}\equiv-1\pmod{q}$ and so $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$

We now count the number of points of each colour in $\textup{PG}(4,q)$.

Lemma 2.10

If $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, then in $\textup{PG}(4,q)$, there are $q^{3}+q^{2}+q+1$ black points, $q^{4}-1$ white points and one red point.

Proof   In $\textup{PG}(4,q)$, let $r,w,b$ be the number of red, white and black points respectively. Count in two ways incident pairs $(P,\Pi)$ where $\Pi\in\mathcal{E}$ and $P$ is black point in $\Pi$. By Lemma 2.4, $\Pi$ contains $q^{2}+1$ black points, so using assumption (I), we have

[TABLE]

giving $b=q^{3}+q^{2}+q+1$. Substituting this and $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$ into (1) gives $w=q^{4}-1$. As $b+w+r=q^{4}+q^{3}+q^{2}+q+1$, we also have $r=1$. $\square$

Recall that we partition the solids of $\textup{PG}(4,q)$ into three sets: $\mathcal{E}$, $\mathcal{T}$ (the solids containing the red point), and $\mathcal{H}$ (the solids not in $\mathcal{E}$ which do not contain the red point). In Lemma 2.4 we counted the solids in $\mathcal{E}$. We now do the same for $\mathcal{T}$ and $\mathcal{H}$.

Lemma 2.11

If $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, then

Each solid in $\mathcal{H}$ contains $(q+1)^{2}$ black points, and $|\mathcal{H}|=\frac{1}{2}q^{2}(q^{2}+1)$; 2. 2.

Each solid in $\mathcal{T}$ contains $q^{2}+q+1$ black points, and $|\mathcal{T}|=q^{3}+q^{2}+q+1$.

Proof   By Lemma 2.10 there is a unique red point, so we have $|\mathcal{T}|=q^{3}+q^{2}+q+1$ (the number of solids through a point). Using $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$ and $|\mathcal{E}|+|\mathcal{H}|+|\mathcal{T}|=q^{4}+q^{3}+q^{2}+q+1$ gives $|\mathcal{H}|=\frac{1}{2}q^{2}(q^{2}+1)$ as required.

Let $\Sigma\in\mathcal{H}$, let $s$ be the number of black points in $\Sigma$ and let $t$ be the number of white points in $\Sigma$. As solids in $\mathcal{H}$ do not contain a red point, we have

[TABLE]

Count in two ways incident pairs $(P,\Pi)$ where $\Pi\in\mathcal{E}$, and $P$ is a point in $\Pi\cap\Sigma$. Using assumption (I), we have

[TABLE]

Simplifying using (11) gives $s=(q+1)^{2}$ proving part 1.

For part 2, let $\Sigma\in\mathcal{T}$, let $s$ be the number of black points in $\Sigma$ and let $t$ be the number of white points in $\Sigma$. As solids in $\mathcal{T}$ contain the unique red point, we have

[TABLE]

Count in two ways incident pairs $(P,\Pi)$ where $\Pi\in\mathcal{E}$, and $P$ is a point in $\Pi\cap\Sigma$, As $\Pi\in\mathcal{E}$, it does not contain the red point, so using assumption (I), we have

[TABLE]

Simplifying using (12) gives $s=q^{2}+q+1$ as required. $\square$

Lemma 2.12

If $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, then a plane containing the unique red point of $\textup{PG}(4,q)$ contains exactly $q+1$ black points.

Proof   Let $\pi$ be a plane through the red point, and let $x$ be the number of black points in $\pi$. By Lemma 2.10, there are $q^{3}+q^{2}+q+1-x$ black points not in $\pi$. Also, the $q+1$ solids containing $\pi$ lie in $\mathcal{T}$ and partition the remaining black points. Using this we count in two ways incident pairs $(P,\Pi)$ where $\Pi$ is a solid in $\mathcal{T}$ containing $\pi$ and $P$ is a black point in $\Pi$. By Lemma 2.11, we have

[TABLE]

and so $x=q+1$ as required. $\square$

Lemma 2.13

If $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, then every line through the unique red point of $\textup{PG}(4,q)$ contains exactly one black point.

Proof   Let $\ell$ be a line through the red point and suppose $\ell$ contains $y$ black points. By Lemma 2.12, each plane containing $\ell$ contains $q+1-y$ further black points. Counting the total number of black points of $\textup{PG}(4,q)$ in two ways: by looking at planes through $\ell$ and by using Lemma 2.10 gives

[TABLE]

Hence $y=1$ as required. $\square$

We now consider assumption (II), that a plane of $\textup{PG}(4,q)$ lies in [math], $\frac{1}{2}q$ or $q$ solids in $\mathcal{E}$, and determine properties of each class of planes.

Lemma 2.14

Suppose $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, and let $\pi$ be a plane of $\textup{PG}(4,q)$ not containing the red point. Then one of the following holds.

$\pi$* lies in [math] solids of $\mathcal{E}$, $q$ solids of $\mathcal{H}$ and $\pi$ contains $2q+1$ black points.* 2. 2.

$\pi$* lies in $\frac{1}{2}q$ solids of $\mathcal{E}$, $\frac{1}{2}q$ solids of $\mathcal{H}$ and $\pi$ contains $q+1$ black points.* 3. 3.

$\pi$* lies in $q$ solids of $\mathcal{E}$, [math] solids of $\mathcal{H}$ and $\pi$ contains $1$ black point.*

Proof   Let $\pi$ be a plane not containing the red point, and suppose $\pi$ contains $x$ black points. Let $e$ be the number of solids in $\mathcal{E}$ through $\pi$ and let $h$ be the number of solids in $\mathcal{H}$ through $\pi$. As each solid in $\mathcal{T}$ contains the red point, there is exactly one solid $\Sigma$ in $\mathcal{T}$ through $\pi$, so $e+h=q$. We count the black points in two ways, firstly using Lemma 2.10, and secondly via solids of $\mathcal{T},\mathcal{E},\mathcal{H}$ containing $\pi$ using Lemma 2.11. We have

[TABLE]

Simplifying gives $x=2q+1-2e$. By assumption (II), there are three possibilities for $e$, namely [math], $q/2$, and $q+1$. If $e=0$, then $x=2q+1$. If $e=q/2$, then $x=q+1$. If $e=q$, then $x=1$. $\square$

Hence there are three different types of planes of $\textup{PG}(4,q)$. We shall call a plane a 1-plane, $(q+1)$-plane or $(2q+1)$-plane according to the number of black points it contains.

Lemma 2.15

If $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$, and $\Pi\in\mathcal{E}$, then the black points in $\Pi$ form an ovoid.

Proof   Let $\Pi\in\mathcal{E}$ and suppose there exists a line $\ell$ in $\Pi$ containing $3+x$ black points for some $x\geq 0$. Let $\pi$ be a plane of $\Pi$ containing $\ell$. There are $q+1$ solids of $\textup{PG}(4,q)$ containing $\pi$, one contains the red point and one is $\Pi$. As $\pi$ contains at least three black points, by Lemma 2.14, $\pi$ is not a 1-plane. As $\pi$ lies in a solid of $\mathcal{E}$, by Lemma 2.14, $\pi$ is a $(q+1)$-plane. Counting the black points in $\Pi$ in two ways, firstly using Lemma 2.4 and secondly by counting the planes of $\Pi$ through $\ell$ gives

[TABLE]

giving a negative value of $x$, a contradiction. Thus the lines in $\Pi$ contain at most two black points. Hence the $q^{2}+1$ black points in $\Pi$ are no three collinear, and form an ovoid. $\square$

Lemma 2.16

Suppose $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$. Let $\Pi\in\mathcal{H}$ and let $\ell$ be a line in $\Pi$ which contains $y$ black points, so $0\leq y\leq q+1$. Then the number of $(q+1)$-planes and $(2q+1)$-planes of $\Pi$ containing $\ell$ is $q+1-y$ and $y$ respectively.

Proof   Let $\Pi\in\mathcal{H}$ and let $\ell$ be a line in $\Pi$ which contains $y$ black points, $0\leq y\leq q+1$. Let $n$ be the number of $(q+1)$-planes of $\Pi$ through $\ell$ and let $m$ be the number of $(2q+1)$-planes of $\Pi$ through $\ell$. It follows from Lemma 2.14 that there are no $1$-planes in $\Pi$, so

[TABLE]

By Lemma 2.11, $\Pi$ contains $(q+1)^{2}$ black points, so counting the black points of $\Pi$ gives

[TABLE]

Simplifying using (13) gives $m=y$ and $n=q+1-y$ as required. $\square$

Lemma 2.17

Suppose $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$. If $\Pi\in\mathcal{H}$, and $q>2$, then the black points in $\Pi$ form a hyperbolic quadric.

Proof   Let $\Pi\in\mathcal{H}$. We suppose $\ell$ is a line of $\Pi$ which contains $y$ black points with $3\leq y\leq q$, and work to a contradiction. By Lemma 2.16, the number of $(q+1)$-planes containing $\ell$ is $q+1-y\geq 1$. So there is a $(q+1)$-plane $\pi$ in $\Pi$ containing $\ell$. By Lemma 2.14, $\pi$ lies in a solid of $\mathcal{E}$. Hence by Lemma 2.15, the black points in $\pi$ form an oval. As $\ell$ is a line of $\pi$, this contradicts $3\leq y\leq q$. Hence every line of $\Pi$ contains either [math], $1$, $2$ or $q+1$ black points. By Lemma 2.10, there are $q^{2}+2q+1$ black points in $\Pi$. It follows from a characterisation of sets of type $(0,1,2,q+1)$ due to Tallini [3], that if $q>2$, the black points in $\Pi$ form a quadric containing $q^{2}+2q+1$ points, that is, a hyperbolic quadric. $\square$

Lemma 2.18

Suppose $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$. If $\Pi\in\mathcal{T}$, and $q>2$, then each line in $\Pi$ contains [math], $1$, $2$ or $q+1$ black points.

Proof   Let $\Pi\in\mathcal{T}$ and $\ell$ a line of $\Pi$. Recall from Lemma 2.10 that there is a unique red point which lies in each solid of $\mathcal{T}$. If $\ell$ contains the red point, then by Lemma 2.13, $\ell$ contains exactly one black point. If $\ell$ does not contain the red point, then $\ell$ lies in a plane $\pi$ not through the red point. By Lemma 2.14, $\pi$ lies in a solid of $\mathcal{H}$ or $\mathcal{E}$. Hence by Lemma 2.15 or Lemma 2.17, each line in $\pi$ contains [math], $1$, $2$ or $q+1$ black points. Thus every line in $\Pi$ contains [math], $1$, $2$ or $q+1$ black points as required. $\square$

3 Proof of main results

We now complete the proofs of the main results.

**Theorem 1.1 ** Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, such that

• (I)

every point of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q^{3}$ or $\frac{1}{2}(q^{3}-q^{2})$ solids of $\mathcal{E}$, and

• (II)

every plane of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q$ or $q$ solids of $\mathcal{E}$.

Then either $\mathcal{E}$ is the set of solids that are disjoint from a hyperoval, or $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric.

Proof   Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, satsifying satisfying Assumptions (I) and (II). By Lemma 2.1, $\mathsf{e}$ is congruent to [math] or $-1$ modulo $q$. We consider these two cases separately.

*Case A: *If $\mathsf{e}$ is congruent to [math] modulo $q$, then by Lemma 2.3, $|\mathcal{E}|=\frac{1}{2}q^{3}(q-1)$. By Lemma 2.8, the red points form a hyperoval $\mathcal{O}$. Further, by Lemma 2.9, $\mathcal{E}$ is the set of solids that are disjoint from $\mathcal{O}$.

*Case B: * If $\mathsf{e}$ is congruent to $-1$ modulo $q$, then by Lemma 2.4, $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$. Let $\mathcal{B}$ be the set of black points. Let $\ell$ be a line of $\textup{PG}(4,q)$ and let $\Sigma$ be a solid containing $\ell$, so $\Sigma$ lies in one of $\mathcal{E},\mathcal{H},\mathcal{T}$. Hence by one of Lemmas 2.15, 2.17, 2.18, $\ell$ contains [math], $1$, $2$ or $q+1$ black points. By Lemma 2.10, $|\mathcal{B}|=q^{3}+q^{2}+q+1$. It follows from a characterisation of sets of type $(0,1,2,q+1)$ due to Tallini [3] that as $q>2$, $\mathcal{B}$ is the set of points of either a non-singular quadric or a solid of $\textup{PG}(4,q)$. By Lemmas 2.15 and 2.17, the black points do not lie in a single solid, so the latter case does not occur. By Lemma 2.15, $\mathcal{E}$ is a set of solids of $\textup{PG}(4,q)$ that meet a non-singular quadric in an ovoid, which is necessarily an elliptic quadric. $\square$

Remark 3.1

• •

In case A, it follows that $\mathcal{T}$ is the set of solids that meet the hyperoval $\mathcal{O}$, and $\mathcal{H}=\emptyset.$ Furthermore, if $\pi$ is the plane containing $\mathcal{O}$, then the white points are the points in $\pi\setminus\mathcal{O}$, and the black points are the points of $\textup{PG}(4,q)\setminus\pi$.

• •

In case B, it follows that $\mathcal{T}$ is the set of solids that meet the non-singular quadric in a quadratic cone, and $\mathcal{H}$ is the set of solids that meet the non-singular quadric in a hyperbolic quadric.

Lemma 3.2

Let $\mathcal{E}$ be the set of solids of $\textup{PG}(4,q)$ that are disjoint from a fixed hyperoval. Then a line of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q(q-1)$, $\frac{1}{2}q^{2}$, or $q^{2}$ solids of $\mathcal{E}$. 2. 2.

Let $\mathcal{E}$ be the set of solids of $\textup{PG}(4,q)$ that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric. Then a line of $\textup{PG}(4,q)$ lies in either [math], $\frac{1}{2}q(q-1)$, $\frac{1}{2}q^{2}$, or $\frac{1}{2}q(q+1)$ solids of $\mathcal{E}$.

Proof   For part 1, let $\mathcal{E}$ be the set of solids of $\textup{PG}(4,q)$ that are disjoint from a fixed hyperoval $\mathcal{O}$ lying in a plane $\pi$. If $\ell$ is any line that contains one or two points of $\mathcal{O}$, then $\ell$ lies in [math] solids of $\mathcal{E}$. Suppose $\ell$ is a line of $\pi$ that does not meet $\mathcal{O}$. Then $\ell$ lies in $q+1$ solids that contain $\pi$, and $q^{2}$ solids that meet $\pi$ in $\ell$. That is, $\ell$ lies in $q^{2}$ solids of $\mathcal{E}$. Suppose $\ell$ is not a line of $\pi$, but $\ell$ meets $\pi$ in a point not in $\mathcal{O}$. Then $\ell$ lies in $1$ solid that contains $\pi$, $\frac{1}{2}(q+2)q$ solids that meet $\pi$ in a bisecant of $\mathcal{O}$, and $\frac{1}{2}q^{2}$ solids that meet $\pi$ in an exterior line of $\mathcal{O}$. That is, $\ell$ lies in $\frac{1}{2}q^{2}$ solids of $\mathcal{E}$. Finally, suppose $\ell$ is any line that does not meet $\pi$, so each solid through $\ell$ meets $\pi$ in a line. So $\frac{1}{2}(q+2)(q+1)$ solids through $\ell$ meet $\mathcal{O}$ in a secant line, and $\frac{1}{2}q(q-1)$ solids through $\ell$ meet $\mathcal{O}$ in an exterior line. That is, $\ell$ lies in $\frac{1}{2}q(q-1)$ solids of $\mathcal{E}$. Part 2 is straightforward counting in $Q(4,q)$. $\square$

Theorem 1.2 * Let $\mathcal{E}$ be a set of solids in $\textup{PG}(4,q)$, $q$ even, $q>2$, satsifying conditions (I) and (II) of Theorem 1.1. If*

• (III)

there exists a line of $\textup{PG}(4,q)$ that lies in exactly $\frac{1}{2}q(q+1)$ solids of $\mathcal{E}$,

then $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric.

Proof   As $\mathcal{E}$ satisfies conditions (I) and (II), by Theorem 1.1, $\mathcal{E}$ is either the set of solids that are disjoint from a hyperoval, or $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric. If $\ell$ is a line that lies in exactly $\frac{1}{2}q(q+1)$ solids of $\mathcal{E}$, then by Lemma 3.2, $\mathcal{E}$ is not the set of solids that are disjoint from a hyperoval, hence $\mathcal{E}$ is the set of solids that meet a non-singular quadric $Q(4,q)$ in an elliptic quadric. $\square$