Characterising hyperbolic hyperplanes of a non-singular quadric in $PG(4,q)$

TL;DR

This paper characterizes a specific set of hyperplanes in projective 4-space over a finite field, showing they correspond to hyperplanes intersecting a non-singular quadric in a hyperbolic quadric, based on combinatorial properties.

Contribution

It provides a geometric characterization of hyperplanes related to a non-singular quadric in PG(4,q) using combinatorial intersection properties.

Findings

The set of hyperplanes meeting a non-singular quadric in a hyperbolic quadric is uniquely characterized by specific point and plane intersection counts.

The characterization applies to even q and relies on combinatorial conditions of hyperplanes and planes in PG(4,q).

This result links geometric configurations with combinatorial intersection properties in finite projective spaces.

Abstract

Let $H$ be a non-empty set of hyperplanes in $PG(4,q)$, $q$ even, such that every point of $PG(4,q)$ lies in either $0$, $\frac12q^3$ or $\frac12(q^3+q^2)$ hyperplanes of $ H$, and every plane of $PG(4,q)$ lies in $0$ or at least $\frac12q$ hyperplanes of $H$. Then $H$ is the set of all hyperplanes which meet a given non-singular quadric $Q(4,q)$ in a hyperbolic quadric.

Equations

$$H(q^2+2q)+(q^3-H)q^2=(q^3+q^2+q)q^2.$$

$$T(q^2+2q+1)+(q^3-T)(q^2+1)=(q^3+q^2+q)q^2+\mathrm{1.0.}$$

$$\mathsf{h}=\frac{|\mathcal{H}|}{\frac{1}{2}{q}^2},$$

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

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

$$\begin{array}{rlr}\mathsf{h}\left((q+1-\mathsf{h})(q^2+q+1)+q(q^3+q^2-2)\right)& =& wq.\end{array}$$

$$s+t$$

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

$$t-q$$

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

$$s+t$$

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

$$s+t$$

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

$$t(q^2+q+1)+s(q^2+2q+1)+(q+1-s-t)(q^2+1)=q^3+q^2+q+1+qx.$$

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

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

$$\sum _i{x}_i^2=(q+1{)}^2[q^3+4q^2+3q+1].$$

$$\sum _i(x_i-(q+1))(x_i-(2q+1))=\sum _i{x}_i^2-(3q+2)\sum _i{x}_i+(2q+1)(q+1)(q^2+1)(q+1)=0.$$

Full text

AMS code: 51E20

Keywords: projective geometry, quadrics, hyperplanes

1 Introduction

This article gives a characterisation of the hyperbolic hyperplanes of a non-singular quadric $Q(4,q)$ of $\textup{PG}(4,q)$, see [8] for background on non-singular quadrics. Note that we refer to the hyperplanes of $\textup{PG}(4,q)$ as solids. There are a number of known characterisations of points, lines, planes and solids relating to the non-singular quadric $Q(4,q)$. In 1956, Tallini [10] characterised sets of points in $\textup{PG}(n,q)$ by their intersection numbers with lines, the following is a version specialised to $\textup{PG}(4,q)$.

Result 1.1

[10]** Let $\mathcal{A}$ be a set of points in $\textup{PG}(4,q)$, $q\neq 2$, with $|\mathcal{A}|\geq q^{3}+q^{2}+q+1$. Suppose every line of $\textup{PG}(4,q)$ contains [math], $1$, $2$ or $q+1$ points of $\mathcal{A}$. Then $\mathcal{A}$ is either all of $\textup{PG}(4,q)$; a quadric; the union of a quadric and its nucleus ($q$ even); or the union of a solid and a subspace.

Buekenhout [2] characterised a set of points in $\textup{PG}(n,q)$ by looking at its plane intersections; and a modification specific to $\textup{PG}(4,q)$ was given by Butler in [3]. Ferri and Tallini [7] proved a characterisation of the pointset of $Q(4,q)$ using plane and solid intersections. The fourth author proved a corollary of this result in [9] (see below), which we will use in our proof. In joint work with De Winter [6], he characterised non-singular polar spaces by their intersection numbers with solids and codimension 2 spaces, which in particular reproves the following result without using Tallini’s result. It is not possible to characterise quadrics by solid intersection numbers alone due to the existence of quasi-quadrics [4].

Result 1.2

[9]** Let ${\mathcal{K}}$ be a set of points in $\textup{PG}(4,q)$, such that each plane of $\textup{PG}(4,q)$ meets $\mathcal{K}$ in either $1$, $q+1$ or $2q+1$ points, and each solid meets $\mathcal{K}$ in either $q^{2}+1$, $q^{2}+q+1$ or $(q+1)^{2}$ points, then $\mathcal{K}$ is the set of points of a non-singular quadric $Q(4,q)$.

When $q$ is odd, there is a polarity associated with $Q(4,q)$ which interchanges planes and lines. Using this polarity, we can dualise these characterisations when is $q$ odd. For example, dualising Results 1.1 and 1.2 give characterisations of the tangent solids of a non-singular quadric $Q(4,q)$, $q$ odd. As a counterpoint to these, this article gives a characterisation involving solids when $q$ is even.

There are also characterisations known of sets of lines and planes related to $Q(4,q)$. For $q$ odd, [5], [1] and [3] give characterisations of the set of tangents and generator lines of $Q(4,q)$. These results can be dualised using the polarity of $Q(4,q)$, $q$ odd, to give characterisations of a set of planes that meet $Q(4,q)$ in a conic. The following result of Butler [3] gives a combinatorial characterisation of a set of planes in $\textup{PG}(4,q)$ that meet $Q(4,q)$ in a conic which holds for all $q$.

Result 1.3

[3]** Let $\mathcal{S}$ be a set of planes in $\textup{PG}(4,q)$, such that

• (I)

every point lies on $q^{4}$ or $q^{4}+q^{2}$ planes of $\mathcal{S}$;

• (II)

every line lies on [math] or $q^{2}$ planes of $\mathcal{S}$;

• (III)

every solid contains at least one plane of $\mathcal{S}$.

Then $\mathcal{S}$ is the set of planes meeting a non-singular quadric $Q(4,q)$ in a conic.

This article proves a combinatorial characterisation of solids in the style of Result 1.3. We first count hyperbolic solids incidences in $Q(4,q)$.

Lemma 1.4

Let $Q(4,q)$ be a non-singular quadric in $\textup{PG}(4,q)$, $q$ even, and let $\mathcal{H}$ be the set of solids that meet $Q(4,q)$ in a hyperbolic quadric. Then

• (a)

every point of $\textup{PG}(4,q)$ lies on [math] or $\frac{1}{2}q^{3}$ or $\frac{1}{2}(q^{3}+q^{2})$ solids of $\mathcal{H}$;

• (b)

every line of $\textup{PG}(4,q)$ lies in [math], $\frac{1}{2}q(q-1)$, $\frac{1}{2}q^{2}$, $\frac{1}{2}q(q+1)$ or $q^{2}$ solids of $\mathcal{H}$; and

• (c)

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

Proof   All references in this proof refer to [8]. By Corollary 1.8, $Q(4,q)$ has a nucleus. By Lemma 1.13(ii) and Theorem 1.54, a solid is a tangent solid if and only if it contains the nucleus. Moreover Theorem 1.57 counts how many hyperbolic, elliptic and tangent solids there are. By Theorem 1.46 the subgroup of $\mathrm{PGL}(5,q)$ which fixes $Q(4,q)$ acts transitively on the set of points on $Q(4,q)$ and by Theorem 1.49 it has two orbits on the points of $\textup{PG}(4,q)\setminus Q(4,q)$, one of which is a singleton, namely the nucleus. Combining the above facts with the exact numbers from [8] it follows that the nucleus is not contained in any hyperbolic solid, the other points not in $Q(4,q)$ are contained in exactly $\frac{1}{2}q^{3}$ hyperbolic solids, and the points of $Q(4,q)$ are contained in exactly $\frac{1}{2}(q^{3}+q^{2})$ hyperbolic solids, proving part (a).

Part (c) can be proved similarly by counting the incident plane-solid pairs $(\pi,\Pi)$ where $\pi\subset\Pi$, (where $\Pi$ is a hyperbolic solid) for different types of planes $\pi$ using Theorem 1.57, Lemma 1.13 and Corollary 1.68, see also Example 1.70. More precisely, any plane not containing the nucleus which meets $Q(4,q)$ in a conic is contained in exactly $\frac{1}{2}q$ hyperbolic solids, any plane meeting $Q(4,q)$ in 2 lines is contained in exactly $q$ hyperbolic solids, all other planes are not contained in any hyperbolic solids. Part (b) can be proved similarly by counting incident line-solid pairs. $\square$

In this article, we assume part (a) and a weaker version of part (c) to prove a characterisation of the hyperbolic solids of $Q(4,q)$, $q$ even. The main result of this article is the following.

Theorem 1.5

Let $\mathcal{H}$ be a non-empty set of solids in $\textup{PG}(4,q)$, $q$ even, 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 $\mathcal{H}$, and

• (II)

a plane which is contained in a solid of $\mathcal{H}$ is contained in at least $\frac{1}{2}q$ solids of $\mathcal{H}$.

Then the set of points contained in $\frac{1}{2}(q^{3}+q^{2})$ solids of $\mathcal{H}$ forms a non-singular quadric $Q=Q(4,q)$ and $\mathcal{H}$ is the set of solids that meets $Q$ in a non-singular hyperbolic quadric $Q^{+}(3,q)$.

Consider a quasi-quadric as defined in [4], more precisely: Let $n=2m$ and let $q$ be even. A parabolic quasi-quadric with nucleus the point $N$ in $\mathrm{PG}(n,q)$ is a set $\mathcal{K}$ of $\frac{q^{n}-1}{q-1}$ points such that each line on $N$ contains a unique point of $\mathcal{K}$ and each hyperplane not on $N$ meets $\mathcal{K}$ in $\frac{(q^{m}+1)(q^{m-1}-1)}{q-1}$ or $\frac{(q^{m}-1)(q^{m-1}+1)}{q-1}$ points. (Note for $n=4$ these values are $q^{2}+1$ and $(q+1)^{2}$). These are constructed in Theorem 10 of [4] using a pivoting method and for all $q>2$ even there exist quasi-quadrics which are not quadrics.

An auxiliary result leading to Theorem 1.5 is

Theorem 1.6

Let $\mathcal{H}$ be a non-empty set of solids in $\textup{PG}(4,q)$, $q$ even, satisfying condition (I). Then $\mathcal{H}$ is the set of solids intersecting a quasi-quadric in $\mathrm{PG}(4,q)$ (which is formed by the points contained in $\frac{1}{2}(q^{3}+q^{2})$ elements of $\mathcal{H}$) in $q^{2}+2q+1$ points.

We show the necessity of Condition (II) in Theorem 1.5 by showing:

Lemma 1.7

The set of solids intersecting a quasi-quadric $\mathcal{Q}$ in $\mathrm{PG}(4,q)$ in $(q+1)^{2}$ points satisfies Condition (I). In particular, every quasi-quadric in $\mathrm{PG}(4,2)$ is a non-singular quadric.

Proof   Let $\mathcal{H}$ be the set of hyperplanes which intersect $\mathcal{Q}$ in $(q+1)^{2}$ points and let $N$ denote the nucleus. Fix a point $X\in\mathcal{Q}$ and let $H$ be the number of elements of $\mathcal{H}$ containing $X$. We count in two ways the pairs $(P,\Pi)$ where $P\in\mathcal{Q},P\neq X$ where $\Pi$ is a solid containing $P,X$ but not $N$. Firstly we count the solids through $X$, and then we look at the points $P\in\mathcal{Q}$, $P\neq X$. Note that from the definition of the quasi-quadric the line $PX$ does not contain $N$, so there are $q^{2}$ solids through $PX$ which do not contain $N$.

[TABLE]

We use the same argument for a point $Y$ not belonging to the quadric and different from $N$ and let $T$ be the number of elements of $\mathcal{H}$ containing $Y$. Note that the line $NY$ has one point $Z$ of $\mathcal{Q}$ and so in this case there are no solids in $\mathcal{H}$ containing $Y$ and $Z$.

[TABLE]

Hence $H=\frac{1}{2}(q^{3}+q^{2})$ and $T=\frac{1}{2}q^{3}$. $\square$

2 Proof of Theorem 1.5

Let $\mathcal{H}$ be a non-empty set of solids in $\textup{PG}(4,q)$, $q$ even, such that

• (I)

every point of $\textup{PG}(4,q)$ lies on [math] or $\frac{1}{2}q^{3}$ or $\frac{1}{2}(q^{3}+q^{2})$ solids of $\mathcal{H}$, and

• (II)

a plane which is contained in a solid of $\mathcal{H}$ is contained in at least $\frac{1}{2}q$ solids of $\mathcal{H}$.

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

• •

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

• •

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

• •

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

Throughout, we will denote the number of red points of $\textup{PG}(4,q)$ by $r$, the number of white points of $\textup{PG}(4,q)$ by $w$, and the number of black points of $\textup{PG}(4,q)$ by $b$.

A solid in $\mathcal{H}$ does not contain any red points, and we partition the solids of $\textup{PG}(4,q)$ into $\mathcal{H}$ and two further types of solids:

• •

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

• •

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

In the following, we will show that the black points form a non-singular quadric $Q(4,q)$ whose nucleus is the unique red point. Further, we show that $\mathcal{H}$ is the set of solids meeting $Q(4,q)$ in a hyperbolic quadric. Indeed, $\mathcal{E}$ is the set of solids meeting $Q(4,q)$ in an elliptic quadric, and $\mathcal{T}$ is the set of solids meeting $Q(4,q)$ in a conic cone.

We proceed with a series of lemmas.

Lemma 2.1

Let

[TABLE]

then ${\mathsf{h}}$ is an integer congruent to either 0 or 1 modulo $q$, and $q+1$ divides ${\mathsf{h}}({\mathsf{h}}-2)$.

Proof   We first show that $\frac{1}{2}q^{2}$ divides $|\mathcal{H}|$. Count in two ways the incident pairs $(P,\Pi)$ where $\Pi$ is a solid in $\mathcal{H}$ and $P$ is a point of $\Pi$.

[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 $\frac{1}{2}q^{2}$ divides $|\mathcal{H}|$.

Now we show that ${\mathsf{h}}=\dfrac{|\mathcal{H}|}{\frac{1}{2}q^{2}}$ 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 $\mathcal{H}$ and $P$ is a point in $\Pi\cap\Sigma$. We have

[TABLE]

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

[TABLE]

Hence ${\mathsf{h}}({\mathsf{h}}-1)\equiv 0\pmod{q}$. As $q$ is a power of 2, it follows that either ${\mathsf{h}}\equiv 0\pmod{q}$ or ${\mathsf{h}}\equiv 1\pmod{q}$. It follows from (1) that $q+1|w$, hence using (3), we deduce that $q+1|{\mathsf{h}}({\mathsf{h}}-2)$. $\square$

Lemma 2.2

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

Proof   Let $\Sigma\in\mathcal{H}$, let $s$ be the number of white points in $\Sigma$ and let $t$ 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 $\Pi\in\mathcal{H}$, $\Sigma\neq\Pi$ and point $P\in\Sigma\cap\Pi$. We have

[TABLE]

Simplifying using (4) using the notation of Lemma 2.1 yields

[TABLE]

Since $0\leq t\leq q^{3}+q^{2}+q+1$, from (5) we have $q^{2}\leq{\mathsf{h}}<q^{2}+q$. It follows from Lemma 2.1 that ${\mathsf{h}}=q^{2}+1$ and $t=(q+1)^{2}$ as required. $\square$

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

Lemma 2.3

There are $q^{3}+q^{2}+q+1$ black points, $q^{4}-1$ white points and one red point.

Proof   Count in two ways incident pairs $(P,\Sigma)$ where $\Sigma\in\mathcal{H}$ and $P$ is black point in $\Sigma$. By Lemma 2.2, $|\mathcal{H}|=\frac{1}{2}q^{2}(q^{2}+1)$ and $\Sigma$ contains $(q+1)^{2}$ black points, so we have

[TABLE]

giving $b=q^{3}+q^{2}+q+1$. Substituting this into (1) gives $w=q^{4}-1$ and so $r=1$, as required. $\square$

Recall that we partition the solids of $\textup{PG}(4,q)$ into three sets: $\mathcal{H}$, $\mathcal{T}$ (the solids containing the red point), and $\mathcal{E}$ (the solids not in $\mathcal{H}$ which do not contain the red point). In Lemma 2.2 we counted the solids in $\mathcal{H}$. We now do the same for $\mathcal{T}$ and $\mathcal{E}$.

Lemma 2.4

Each solid in $\mathcal{E}$ contains $q^{2}+1$ black points, and $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$. 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 2.3 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 Lemma 2.2 and $|\mathcal{H}|+|\mathcal{E}|+|\mathcal{T}|=q^{4}+q^{3}+q^{2}+q+1$ gives $|\mathcal{E}|=\frac{1}{2}q^{2}(q^{2}-1)$ as required.

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

[TABLE]

Count in two ways incident pairs $(P,\Sigma)$ where $\Sigma\in\mathcal{H}$, and $P\in\Pi\cap\Sigma$. We have

[TABLE]

Simplifying using (6) gives $t=q^{2}+1$ proving part 1.

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

[TABLE]

Count in two ways incident pairs $(P,\Sigma)$ where $\Sigma\in\mathcal{H}$, and $P\in\Pi\cap\Sigma$, noting that as $\Sigma\in\mathcal{H}$, it does not contain the red point. We have

[TABLE]

Simplifying using (7) gives $t=q^{2}+q+1$ as required. $\square$

Remark 2.5

Notice that so far we have only used Condition (I) and to prove that the set of black points is a quasi-quadric, it remains to show that each line through the red point contains exactly one black point. Standard counting arguments show first that every plane through the red point contains exactly $q+1$ black points and then that every line through the red point contains exactly 1 black point.

Lemma 2.6

All planes contain $1,q+1$ or $2q+1$ black points.

Proof   Let $\pi$ be a plane, and denote the number of black points in $\pi$ by $x$. Further, let $t$ be the number of solids of $\mathcal{T}$ containing $\pi$ and $s$ the number of solids of $\mathcal{H}$ containing $\pi$, so there are $q+1-s-t$ solids of $\mathcal{E}$ containing $\pi$. Counting the black points in $\textup{PG}(4,q)$ by counting in solids about $\pi$, we have (using Lemmas 2.2, 2.3 and 2.4)

[TABLE]

If $\pi$ contains the red point, then $t=q+1$ and $s=0$, hence $x=q+1$. If $\pi$ does not contain the red point, then $t=1$, and so $x=2s+1$. If $s=0$, then $x=1$. If $s>0$, then using (II), we have $q/2\leq s\leq q+1$ and so $q+1\leq x\leq 2q+1$. We now show that in this case (where $\pi$ lies in a solid of $\mathcal{H}$) we have $x=q+1$ or $2q+1$.

Consider a fixed solid $\Sigma\in\mathcal{H}$. We will first count pairs $(P,\pi)$ where $P$ is a black point in a plane $\pi$ belonging to $\Sigma$ and then triples $(P,P^{\prime},\pi)$ with $P,P^{\prime}$ black points, $\pi\in\Sigma$ where $P\in\pi$, $P^{\prime}\in\pi$, $P\neq P^{\prime}$. If $\pi_{i}$ is a plane belonging to $\Sigma$, let $x_{i}$ denote the number of black points in $\pi_{i}$. We have

[TABLE]

[TABLE]

Hence we obtain

[TABLE]

So we obtain that

[TABLE]

As $q+1\leq x_{i}\leq 2q+1$, we have $x_{i}=q+1$ or $2q+1$ as required. $\square$

Proof of Theorem 1.5 Let $\mathcal{B}$ be the set of black points. By Lemma 2.6, planes meet $\mathcal{B}$ in either $1$, $q+1$ or $2q+1$ points. By Lemmas 2.2 and 2.4, solids meet $\mathcal{B}$ in either $q^{2}+1$, $q^{2}+q+1$ or $q^{2}+2q+1$ points. Hence by Result 1.2, $\mathcal{B}$ is the set of points of points of a non-singular quadric of $\textup{PG}(4,q)$, and so $\mathcal{H}$ is the set of solids of $\textup{PG}(4,q)$ that meet a non-singular quadric in a hyperbolic quadric as required. $\square$

We note that it follows that $\mathcal{E}$ is the set of solids that meet the non-singular quadric in an elliptic quadric, and $\mathcal{T}$ is the set of solids that meet the non-singular quadric in a quadratic cone.