On intriguing sets of the Penttila-Williford association scheme

TL;DR

This paper explores intriguing sets within a specific association scheme, providing examples and characterizations that deepen understanding of their structure and properties.

Contribution

It introduces new examples and characterizations of intriguing sets in the Penttila-Williford association scheme, advancing knowledge of non-P-polynomial schemes.

Findings

Identification of four types of intriguing sets

Examples illustrating the structure of these sets

Characterization results for each type

Abstract

We investigate intriguing sets of an association scheme introduced by Penttila and Williford (2011) that was the basis for their construction of primitive cometric association schemes that are not $P$-polynomial nor the dual of a $P$-polynomial scheme. In particular, we give examples and characterisation results for the four types of intriguing sets that arise in this scheme.

Tables

Table 1. Table 1. The Penttila-Williford Scheme

Relations	Description
$R_0$	equality
$R_1$	noncollinear but collinear to the conjugate of the other
$R_2$	noncollinear and noncollinear to the conjugate of the other
$R_3$	collinear and not conjugate
$R_4$	conjugate

Equations

$$\frac{k-\theta }{|X|}\cdot |Y|$$

$$\theta +\frac{k-\theta }{|X|}\cdot |Y|$$

$$R_0$$

$$R_1$$

$$R_2$$

$$R_3$$

$$R_4$$

$$P=$$

$$Q=$$

$$|Y\cap \mathcal{F}|=\frac{|Y|\cdot |\mathcal{F}|}{|X|}=\frac{|Y|}{q+1}.$$

$$(q+1{)}^2{E}^{-}=q{B}_0-B_1+\frac{1}{q}{B}_2$$

$$S{E}^{-}{S}^{\top }=$$

$$=$$

$$+\frac{1}{q^2(q+1{)}^2}\left(\frac{q(q^2+1)}{2}{A}_0-\frac{q(q-1)}{2}{A}_1+q{A}_2-\frac{q(q-1)}{2}{A}_3+\frac{q(q^2+1)}{2}{A}_4\right)$$

$$=$$

$$(\iota (v)E^{-})(\iota (v)E^{-}{)}^{\top }$$

$$=vS{E}^{-}{S}^{\top }{v}^{\top }=v(E_1+\frac{q-1}{q+1}{E}_4)v^{\top }$$

$$=0.$$

$$E_j{\chi }_{p^{\sigma }}=\frac{1}{|X|}\sum _i{Q}_{ij}{A}_i{\chi }_{p^{\sigma }}=\frac{1}{|X|}\sum _i(-1{)}^j{Q}_{ij}{A}_{4-i}{\chi }_p=(-1{)}^j{E}_j{\chi }_p.$$

$$({\chi }_p-{\chi }_{p^{\sigma }})\cdot {\chi }_S=0$$

$$|T|=\frac{q-1}{2}(q^3+q^2+q+1-|C|)+\frac{q-1}{2}(q^2+q+1-|m|)=\frac{1}{2}{q}^2(q^2-1).$$

$$|p^{\perp }\cap y^{\perp }\cap T|=|p^{\perp }\cap y^{\perp }\cap \mathcal{Q}|-|p^{\perp }\cap C|.$$

$$|p^{\perp }\cap {\ell }^{\perp }\cap T|=|p^{\perp }\cap {\ell }^{\perp }\cap \mathcal{Q}|-|p^{\perp }\cap m|.$$

$$|p^{\perp }\cap T|$$

$$=\sum _{y\in S_1}|p^{\perp }\cap y^{\perp }\cap \mathcal{Q}|+\sum _{\ell \in S_1}|p^{\perp }\cap {\ell }^{\perp }\cap \mathcal{Q}|-\frac{1}{2}(q-1)(|p^{\perp }\cap C|+|p^{\perp }\cap m|).$$

$$|p^{\perp }\cap T|=|S_1|(q^2+1)+|S_2|(q+1)-\frac{1}{2}(q-1)((q+1)+1)=\frac{1}{2}(q-1)q^2.$$

$$|p^{\perp }\cap T|=\{\begin{array}{ll}\frac{1}{2}(q-1)q^2& \text{if }p\notin T,\\ \frac{1}{2}(q-1)q^2+q& \text{if }p\in T.\end{array}$$

$$|Y\cap \mathcal{F}|=\frac{\alpha (q+1)\cdot q^2(q-1)}{q^2(q^2-1)}=\alpha .$$

$$\alpha :=\frac{k-\theta }{|L|}\cdot |Y|=\frac{1}{q}|Y|$$

$$|S\cap Y|=\frac{|S|\cdot |Y|}{|X|}=\frac{4(q+1)|Y|}{q^2(q^2-1)}=\frac{4|Y|}{q^2(q-1)}.$$

$$\sum _{p\in H\cap \mathcal{Q}}{s}_p(s_p-1)(q^2-q)=|Y|(q-1).$$

$$q^3+q+q{s}_C=\frac{1}{2}(q+1)(q^2-q)+\frac{1}{2}(q+1)(q^2+q)$$

$$q^3+q+q{s}_C=q(q+1-s_p)+q(q+1-s_r)+\frac{1}{2}(q-1)(q^2-q)+\frac{1}{2}(q-1)(q^2+q).$$

$$q^3+q+q{s}_C$$

$$q^3+q+(q^2+q)s_{\ell }$$

Full text

1. Introduction

A celebrated result of Beniamino Segre [13, n. 91] is that if there exists a nonempty proper subset $S$ of the point set $\mathcal{P}$ of the elliptic quadric $\mathsf{Q}^{-}(5,q)$, $q$ odd, such that every generator meets $S$ in a constant number $m$ of points, then $m=(q+1)/2$. Such a set $S$ is called a hemisystem, and by the work of Cossidente and Penttila [7], these configurations exist for every odd prime power $q$. Hemisystems garnered interest from the algebraic combinatorics community, as they give rise to cometric $Q$-antipodal association schemes [14]. Segre’s result was extended to $q$ even by Bruen and Hirschfeld [3], who showed that no such subsets $S$ can exist. Thus for $q$ even, there is an absence of such interesting configurations. However, Penttila and Williford [12] introduced the notion of a relative hemisystem of $\mathsf{Q}^{-}(5,q)$, $q>2$ even, which give rise to atypical and rare association schemes; primitive cometric association schemes that do not arise from distance regular graphs. Their idea was to consider a non-tangent hyperplane $H$ for $\mathsf{Q}^{-}(5,q)$ and to only regard the points $X$ that lie outside of $H$. Now, a nonempty proper subset $S$ of $X$ such that every generator meets $S$ in a constant number $m$ of points must have $m=q/2$, according to [1, Theorem 1]. In particular, $q$ is even. Such a configuration is a relative hemisystem.

In the background, there is a $4$-class cometric association scheme, which we call the Penttila-Williford scheme. It arises from taking the natural relations that are invariant under the stabiliser of $H$ in the full similarity group of $\mathsf{Q}^{-}(5,q)$, and it exists for all prime powers $q$, odd or even (but greater than $2$). Table 1 summarises the relations of this scheme, and full details will be given in Section 3.

A relative hemisystem, if it exists, provides an example of an intriguing set for this association scheme, in the language of De Bruyn and Suzuki [9]. Little is known about other intriguing sets for this association scheme, apart from the devices used in Melissa Lee’s MPhil thesis [10] and in [1]. In this paper, we begin an investigation into the various intriguing sets for this scheme and we derive characterisation results. Since there are four non-principal eigenspaces, there are four ‘types’ $i\in\{1,2,3,4\}$ of intriguing sets depending on the index $i$ for which its characteristic vector belongs to the sum $V_{0}\oplus V_{i}$ of eigenspaces for the Penttila-Williford scheme. There is a particular involution $\sigma$ that acts fixed-point-freely on $X$, and we say that two points $x,y\in X$ are conjugate if $x=y^{\sigma}$. A subset $S$ of $X$ is $\sigma$-invariant, if $x^{\sigma}\in X$ for all $x\in X$. If an intriguing set $S$ is not $\sigma$-invariant, then the image $S^{\sigma}$ of $S$ under $\sigma$ is disjoint $S$, and we shall see that $|S|=\tfrac{1}{2}|X|$.

Theorem 1.1** (Paraphrase of Theorem 3.3).**

Suppose that $Y$ is an intriguing set of type $i$, where $i\in\{1,2,3,4\}$.

1. (a)

If $i=1$ or $i=3$, then $|Y|=\frac{1}{2}|X|$ and $|\{p,p^{\sigma}\}\cap Y|=1$ for all $p\in X$. 2. (b)

If $i=2$ or $i=4$, then $Y^{\sigma}=Y$.

We show in Section 4 that an intriguing set of type 2 has size at least $4(q+1)$ (see Lemma 4.9), and this bound is sharp. In Theorem 4.12, we characterise the smallest examples for when $q\geqslant 59$. We also show in Section 5 that an intriguing set of type 4 has size at least $q^{2}(q-1)$, and this bound is sharp. The following result characterises the smallest examples.

Theorem 1.2** (Paraphrase of Theorem 5.5).**

Suppose $Y$ is an intriguing set of type 4 of the Penttila-Williford scheme. Then $|Y|\geqslant q^{2}(q-1)$, and in the case of equality, there exists a point $p\in H\cap\mathsf{Q}^{-}(5,q)$ such that $Y$ consists of the $q^{2}(q-1)$ points of $\mathsf{Q}^{-}(5,q)\setminus H$ that are collinear to $p$.

2. Preliminaries

Let $\Gamma=(X,R)$ be a finite connected regular graph, and we will also assume throughout that $\Gamma$ is nontrivial; neither empty nor complete. A subset $Y$ of $X$ is called an intriguing set if and only if there are integers $h_{1},h_{2}\geqslant 0$ such that every vertex of $Y$ is adjacent to exactly $h_{1}$ vertices of $Y$ and every vertex of $X\setminus Y$ is adjacent to exactly $h_{2}$ vertices of $Y$. We will use the symbol $j$ to denote the ‘all-ones’ row vector. We always assume a given numbering $X=\{x_{1},\dots,x_{n}\}$ of the vertices of $X$ where $n=|X|$. The adjacency matrix of $\Gamma$ is the real $(n\times n)$-matrix $A$ with $A_{ij}=1$ when $x_{i}$ and $x_{j}$ are adjacent and $A_{ij}=0$ otherwise. Notice that $A$ is a real symmetric matrix, so the row space $\mathbb{R}^{n}$ decomposes into the orthogonal sum of the eigenspaces of $A$, which we also call the eigenspaces of $\Gamma$. For every subset $Y$ of $X$ we denote by $\chi_{Y}$ its characteristic vector, that is the real row vector of length $n$ whose $i$-th entry is 1, if $x_{i}\in Y$, and [math] otherwise. The following theorem is from [9].

Result 2.1** ([9]).**

Let $\Gamma=(X,R)$ be a finite connected nontrivial regular graph with eigenspaces $V_{0},\dots,V_{s}$ where $V_{0}=\langle j\rangle$ and adjacency matrix $A$. Let $Y$ be a non-empty proper subset of $X$.

1. (i)

$Y$* is an intriguing set of $\Gamma$ if and only if $\chi_{Y}\in V_{0}\oplus V_{i}$ for some integer $1\leqslant i\leqslant s$.* 2. (ii)

If $Y$ is an intriguing set, $i\geqslant 1$ is the integer with $\chi_{Y}\in V_{0}\oplus V_{i}$, and if $\theta$ is the eigenvalue of $A$ on $V_{i}$, then every vertex of $X\setminus Y$ is adjacent to exactly

[TABLE]

elements of $Y$ and every vertex of $Y$ is adjacent to exactly

[TABLE]

elements of $Y$.

If $Y$ is an intriguing set with $Y\neq\varnothing,X$ and $i$ is the unique index such that $\chi_{Y}\in V_{0}\oplus V_{i}$, then we say that $Y$ is an intriguing set of type $i$, or for the eigenspace $V_{i}$. Notice that $\chi_{Y}=cj+v$ with $c:=|Y|/|X|$ and $v\in V_{i}$. Since eigenvectors for distinct eigenvalues have inner product zero, the following well-known result follows easily.

Result 2.2**.**

Let $\Gamma=(X,R)$ be a finite connected regular graph. Let $Y_{1}$ and $Y_{2}$ be intriguing sets of $\Gamma$ for different eigenspaces. Then $|Y_{1}\cap Y_{2}|\cdot|X|=|Y_{1}|\cdot|Y_{2}|$.

We give an example that will be used later.

Example 2.3**.**

The collinearity graph of $\mathsf{Q}^{-}(5,q)$ is a strongly regular graph with eigenvalues $(q^{2}+1)q$, $q-1$ and $-(q^{2}+1)$ where $(q^{2}+1)q$ has eigenspace $V_{0}:=\langle j\rangle$. Intriguing sets for the eigenvalue $-(q^{2}+1)$ are hemisystems (by Segre’s theorem), that is, sets of points of size $(q^{3}+1)(q+1)/2$ such that every generator meets in $(q+1)/2$ elements. The intriguing sets for the eigenvalue $q-1$ are usually called ‘tight sets’ in the literature. It follows from the above results that for any tight set $Y$ we have $|Y|=c(q+1)$, where $c$ is the number of points in $Y$ collinear to a given point not in $Y$. There exist many different examples of tight sets of $\mathsf{Q}^{-}(5,q)$. The $q+1$ points of a line form a tight set. The union of two perpendicular conics is a tight set. Moreover, any union of disjoint tight sets is a tight set, so a large number of examples can be obtained by taking a disjoint union of lines and pairs of perpendicular conics. If $c$ is sufficiently small, it was proven in [11, Theorem 2.15] that this construction describes all tight sets of size $c(q+1)$.

3. The Penttila–Williford scheme

We consider a finite $5$-dimensional projective space $\mathcal{P}=\mathsf{PG}(5,q)$, $q>2$, and in there, an elliptic quadric $\mathcal{Q}=\mathsf{Q}^{-}(5,q)$. We let $\perp$ be the associated polarity of $\mathcal{P}$ whose absolute points are the points of $\mathcal{Q}$. Let $H$ be a hyperplane of $\mathcal{P}$ meeting $\mathcal{Q}$ in a parabolic quadric $\mathsf{Q}(4,q)$.

We consider the point set $X:=\mathcal{Q}\setminus(\mathcal{Q}\cap H)$ obtained from $\mathcal{Q}$ by removing the points of $H\cap\mathcal{Q}$. Note that $|X|=q^{2}(q^{2}-1)$. We define $\sigma\colon X\to X$ as follows. Let $p$ be a point of $\mathcal{Q}$ not in $H$. Then the line joining $H^{\perp}$ and $p$ is a hyperbolic line and so contains a unique second point $p^{\prime}$ of $\mathcal{Q}$. Let $\sigma$ be the central collineation of $\mathsf{PG}(5,q)$ having axis $H$ and centre $H^{\perp}$, mapping $p$ to $p^{\prime}$. Then $\sigma$ commutes with the polarity $\perp$ and so stabilises $\mathcal{Q}$. Moreover, since $pp^{\prime}$ has only two points of $\mathcal{Q}$ on it, we have $\sigma(p^{\prime})=p$ and hence $\sigma^{2}$ is the identity. Via the Klein correspondence, we can map the points of $X$ to lines of the Hermitian surface $\mathsf{H}(3,q^{2})$ that lie outside of a symplectic subgeometry $\mathsf{W}(3,q)$. This subgeometry can alternatively be defined as the fixed subspaces of a Baer involution, and this involution corresponds to our map $\sigma$.

Two points $p,p^{\prime}$ of $\mathcal{Q}$ are said to be collinear when $p\neq p^{\prime}$ and $p^{\prime}\in p^{\perp}$, that is, $p$ and $p^{\prime}$ are distinct points that span a line of $\mathsf{PG}(5,q)$ contained in $\mathcal{Q}$. In this case we write $p\sim p^{\prime}$. On $X$ the following symmetric relations $R_{i}$, $0\leqslant i\leqslant 4$, define a cometric association scheme on $X$ (see [12, Theorem 1] and [10, Section 5.4]).

[TABLE]

We will frequently refer to the ‘association scheme $X$’, which will mean the association scheme on $X$ given by the above relations. The matrix of eigenvalues $P$ and matrix of dual eigenvalues $Q$ of this association scheme are as follows:

[TABLE]

If $A_{i}$ denotes the adjacency matrix of the graph $(X,R_{i})$, this means that we can number the eigenspaces $V_{0},\dots,V_{4}$ of the association scheme in such a way that $A_{j}v=P_{ij}v$ for $v\in V_{i}$, where the entries of $P$ are numbered $P_{ij}$ with $i,j=0,\dots,4$. Notice that this implies that $V_{0}=\langle\chi_{X}\rangle$.

The reason to exclude the case when $q=2$ is that the relation $R_{2}$ is empty when $q=2$ as can be seen from the middle entry in the first line of $P$.

Let $Y$ be a non-empty proper subset of $X$. We call $Y$ an intriguing set of this association scheme, if $\chi_{Y}\in V_{0}\oplus V_{i}$ for some $i\in\{1,2,3,4\}$. Result 2.1 shows that an intriguing set $Y$ of the association scheme is an intriguing set of each graph $(V,R_{i})$.

Example 3.1** (Intriguing sets of type 1).**

It was shown in [1] that an intriguing set of type 1, if it exists, is a relative hemisystem and $q$ is even. Moreover, Penttila and Williford [12] provided an infinite family of examples that exist for each even prime power $q>2$. Further examples were constructed in [2, 4, 5, 6].

Example 3.2** (Some intriguing sets of type 4).**

Let $w$ be a point of $H\cap\mathcal{Q}$ and let $Y$ be the set consisting of all points of $X$ that are in $\mathcal{Q}$ collinear to $w$. Then $|Y|=(q^{2}-q)q$, every point of $Y$ is collinear to $q-1$ points of $Y$ and every point of $X\setminus Y$ is collinear to $q^{2}-q$ points of $Y$. Hence $Y$ is an intriguing set of $(X,R_{3})$ corresponding to the eigenvalue $q-1-(q^{2}-q)=-(q-1)^{2}$. Since $V_{4}$ is the eigenspace of this graph for the eigenvalue $-(q-1)^{2}$, it follows that $\chi_{Y}\in V_{0}\oplus V_{4}$, so $Y$ is also an intriguing set of the association scheme $X$.

Examples of intriguing sets of types 2 and 3 will appear in Section 4. In the following result, we show that an intriguing set is either $\sigma$-invariant or has its image under $\sigma$ disjoint from it.

Theorem 3.3**.**

Suppose that $Y$ is a subset of $X$ such that $\chi_{Y}\in V_{0}\perp V_{i}$ for some $i\in\{1,2,3,4\}$.

1. (a)

If $i=1$ or $i=3$, then $|Y|=\frac{1}{2}|X|$ and $|\{p,p^{\sigma}\}\cap Y|=1$ for all $p\in X$. 2. (b)

If $i=2$ or $i=4$, then $Y^{\sigma}=Y$.

Proof.

Let $\theta$ be the eigenvalue of the relation $R_{3}$ on $V_{i}$. Result 2.1 shows that every point of $X\setminus Y$ is in relation $R_{3}$ to $a_{3}:=\frac{|Y|}{|X|}(k-\theta)$ points of $Y$ (where $k$ is the valency of $R_{3}$), and every point of $Y$ is in relation $R_{3}$ to $b_{3}:=a_{3}+\theta$ points of $Y$. Inspection of the first matrix of eigenvalues shows that $(-1)^{i}\theta$ is the eigenvalue of $R_{1}$ on $V_{i}$. Result 2.1 thus shows that every point of $X\setminus Y$ is in relation $R_{1}$ to $a_{1}:=\frac{|Y|}{|X|}(k-(-1)^{i}\theta)$ points of $Y$, and every point of $Y$ is in relation $R_{1}$ to $b_{1}:=a_{1}+(-1)^{i}\theta$ points of $Y$. Hence, if $p\in X\setminus Y$, then $p^{\sigma}$ is collinear to $a_{1}$ points of $Y$, and for every point $p\in Y$, the point $p^{\sigma}$ is collinear to exactly $b_{1}$ points of $Y$. Since $\theta\neq 0$, we have $a_{1}\neq b_{1}$ and $a_{3}\neq b_{3}$.

If $i$ is even, that is $i=2$ or $i=4$, then $a_{1}=a_{3}$ and $b_{1}=b_{3}$, and it follows immediately that for every point $p$, both points $p$ and $p^{\sigma}$ are in relation $R_{3}$ to the same number of points of $Y$, hence both points lie in $Y$ or both points do not. Hence $Y=Y^{\sigma}$ in this case.

If $i$ is odd, then $\theta=-(q^{2}+1)$ or $\theta=q-1$ and an easy calculation gives $a_{1}=b_{3}$ and $a_{3}=b_{1}$. This time it follows for every point $p\in X$ that $p$ and $p^{\sigma}$ are in relation $R_{3}$ to a different number of points of $Y$ and hence, exactly one of these two points lies in $Y$. ∎

As a corollary, we obtain the following result of [1].

Theorem 3.4**.**

If $Y$ is a non-trivial $m$-cover of $H(3,q^{2})\setminus W(3,q)$, then $q$ is even, $m=\frac{q}{2}$, and from any pair of conjugate lines exactly one lies in $Y$.

Proof.

We do this in the dual setting, that is we consider a set $Y$ of points of $\mathsf{Q}^{-}(5,q)\setminus\mathsf{Q}(4,q)$ such that every line of $\mathsf{Q}^{-}(5,q)$ that is not contained in $\mathsf{Q}(4,q)$ has exactly $m$ points in $Y$. Hence, in the graph $(X,R_{3})$, every point of $Y$ is collinear to exactly $(q^{2}+1)(m-1)$ points of $Y$, and every point of $X\setminus Y$ is collinear to exactly $(q^{2}+1)m$ points of $Y$. Therefore Result 2.1 implies that $Y$ is an intriguing set for the eigenvalue $-(q^{2}+1)$ of $(X,R_{3})$. Since the eigenspace of the graph $(X,R_{3})$ for this eigenvalue is $V_{1}$, which is an eigenspace of the association scheme, it follows that $Y$ is an intriguing set of the scheme. Theorem 3.3 shows that $|Y|=|X|/2$ and that $Y$ contains exactly one point of every pair of conjugate points. Since each point of $X\setminus Y$ is collinear to $m(q^{2}+1)$ points of $Y$, part (ii) of Result 2.1 gives $m=q/2$. ∎

4. A connection to tight sets of $\mathsf{Q}^{-}(5,q)$

In this section, we consider intriguing sets of the graph $(X,R_{3})$ corresponding to the eigenvalue $q-1$. Notice that the eigenspace for this eigenvalue of $(X,R_{3})$ is $V_{2}\oplus V_{3}$. Hence a subset $Y$ of $X$ is an intriguing set of $(X,R_{3})$ if and only if $\chi_{Y}\in V_{0}\oplus V_{2}\oplus V_{3}$. We will investigate when $Y$ is also an intriguing set of the association scheme $X$, that is when $\chi_{Y}\in V_{0}\oplus V_{2}$ or $\chi_{Y}\in V_{0}\oplus V_{3}$.

By Result 2.1, every point $p\in X\setminus Y$ is collinear to exactly $\alpha:=|Y|/(q+1)$ points of $X$ and every point of $Y$ is collinear to exactly $q-1+\alpha$ points of $Y$. Recall that $X$ is the set of points of the elliptic quadric $\mathcal{Q}$ that lie outside the non-tangent hyperplane $H$ of the ambient projective space.

Lemma 4.1**.**

If $Y$ is an intriguing set of the graph $(X,R_{3})$ corresponding to the eigenvalue $q-1$, then $Y$ is a tight set of $\mathsf{Q}^{-}(5,q)$.

Proof.

Let $p$ be a point of $H\cap\mathcal{Q}$. We have seen in Example 3.2 that the set $\mathcal{F}$ consisting of the $q^{2}(q-1)$ points of $X$ that are in $\mathcal{Q}$ collinear to $p$ is an intriguing set of the association scheme with $\chi_{\mathcal{F}}\in V_{0}\oplus V_{4}$. Applying Result 2.2 to the intriguing sets $Y$ and $\mathcal{F}$ of $(X,R_{3})$, it follows that

[TABLE]

Hence $p$ is collinear to exactly $|Y|/(q+1)$ points of $Y$. Since this is true for all points of $H\cap\mathcal{Q}$ and since the points of $X\setminus Y$ are collinear to the same number $|Y|/(q+1)$ points of $Y$, it follows that $Y$ is an intriguing set of the collinearity graph of $\mathsf{Q}^{-}(5,q)$ for the eigenvalue $q-1$. Hence $Y$ is a tight set of $\mathsf{Q}^{-}(5,q)$; see Example 2.3. ∎

Remark 4.2**.**

The algebraic reason behind this phenomenon is the following. If $\mathcal{P}$ is the set of points of $\mathsf{Q}^{-}(5,q)$, then we have the following decomposition into eigenspaces of the collinearity relation:

[TABLE]

Whereas for the association scheme on $X$, we have

[TABLE]

We have two linear maps: (1) the natural inclusion map $\iota\colon\mathbb{C}X\to\mathbb{C}\mathcal{P}$; (2) the projection map $\rho\colon\mathbb{C}\mathcal{P}\to\mathbb{C}X$ which maps $\sum_{x\in\mathcal{P}}\alpha_{x}\chi_{x}$ to $\sum_{x\in X}\alpha_{x}\chi_{x}$. Note that $\iota$ is injective, $\rho$ is surjective, and $\rho\circ\iota$ is the identity on $\mathbb{C}X$. The kernel of $\rho$ is clearly $K:=\langle\chi_{h}\colon h\in H\cap\mathcal{P}\rangle$. Let $S$ be the $|X|\times|\mathcal{P}|$ inclusion matrix for $X$ in $\mathcal{P}$. Let $(A_{0},A_{1},A_{2},A_{3},A_{4})$ be the adjacency matrices for the association scheme on $X$, and let $(B_{0},B_{1},B_{2})$ be the adjacency matrices for the association scheme on $\mathcal{P}$, where $B_{1}$ represents the collinearity relation. So $SB_{0}S^{\top}=A_{0}$, $SB_{1}S^{\top}=A_{3}$, and $SB_{2}S^{\top}=A_{1}+A_{2}+A_{4}$.

From the matrix of dual eigenvalues111which is $\begin{bmatrix}1&q^{2}(q^{2}+1)&q(q^{2}-q+1)\\ 1&q(q-1)&-(q^{2}-q+1)\\ 1&-\tfrac{1}{q}(q^{2}+1)&\tfrac{1}{q}(q^{2}-q+1)\\ \end{bmatrix}$ for the association scheme on $\mathbb{C}\mathcal{P}$, we have

[TABLE]

and hence

[TABLE]

Therefore, $\rho(V^{-})$ contains the rowspace of $E_{1}+\tfrac{q-1}{q+1}E_{4}$. Since $E_{1}E_{4}=0$, the rowspace of $E_{1}+\tfrac{q-1}{q+1}E_{4}$ is the sum of the rowspaces of $E_{1}$ and $E_{4}$; that is, $V_{1}\perp V_{4}$. Hence $\rho(V^{-})$ contains $V_{1}\perp V_{4}$. As $\dim(V^{-})=\dim(V_{1})+\dim(V_{4})$, it follows that $\rho$ restricted to $V^{-}$ is a bijective map from $V^{-}$ onto $V_{1}\perp V_{4}$.

For each $v\in\langle\chi_{X}\rangle\perp V_{2}\perp V_{3}$, we have

[TABLE]

Therefore, $\iota(v)E^{-}=0$ and consequently $\iota(v)\in\langle\chi_{\mathcal{P}}\rangle\perp V^{+}$. Hence $\iota$ maps $\langle\chi_{X}\rangle\perp V_{2}\perp V_{3}$ into $\langle\chi_{\mathcal{P}}\rangle\perp V^{+}$, which affirms Lemma 4.1. So we have the following diagram:

V_{1}$$V_{2}$$V_{3}$$V_{4}$$V^{-}$$V^{+}$$\iota$$\rho

Now we investigate when $Y$ is an intriguing set of the association scheme $X$, that is when $\chi_{Y}\in V_{0}\perp V_{2}$ or $\chi_{Y}\in V_{0}\perp V_{3}$. We already know that $\chi_{Y}\in V_{0}\oplus V_{2}$ implies $Y=Y^{\sigma}$ and that $\chi_{Y}\in V_{0}\oplus V_{3}$ implies $Y\cap Y^{\sigma}=\varnothing$. The following observation is crucial in proving that the converse is also true.

Lemma 4.3**.**

We have $\langle\chi_{p}-\chi_{p^{\sigma}}\mid p\in X\rangle=V_{1}\oplus V_{3}$ and $\langle\chi_{p}+\chi_{p^{\sigma}}\mid p\in X\rangle=V_{0}\oplus V_{2}\oplus V_{4}$.

Proof.

We number the rows and columns of $Q$ from [math] to $4$. Inspection of $Q$ shows that $Q_{ij}=(-1)^{j}Q_{4-i,j}$ for all $i,j=0,\dots,4$. The definition of the relations $R_{i}$ shows that $A_{i}\chi_{p^{\sigma}}=A_{4-i}\chi_{p}$. It follows for each $j\in\{0,1,2,3,4\}$ that

[TABLE]

Hence, for each $p\in X$ we have $E_{j}(\chi_{p}-\chi_{p^{\sigma}})=0$ for $j=0,2,4$ and $E_{j}(\chi_{p}+\chi_{p^{\sigma}})=0$ for $j=1,3$. Hence $\langle\chi_{p}-\chi_{p^{\sigma}}\mid p\in X\rangle\subseteq V_{1}\oplus V_{3}$ and $\langle\chi_{p}+\chi_{p^{\sigma}}\mid p\in X\rangle\subseteq V_{0}\oplus V_{2}\oplus V_{4}$. Since $\{\chi_{p}\mid p\in X\}$ forms a basis of $V_{0}\oplus V_{1}\oplus V_{2}\oplus V_{3}\oplus V_{4}$, we have equality. ∎

Lemma 4.4**.**

If $S\subseteq X$, then $S=S^{\sigma}$ if and only if $\chi_{S}\in V_{0}\oplus V_{2}\oplus V_{4}$.

Proof.

We have $S=S^{\sigma}$ if and only if

[TABLE]

for all $p\in\mathcal{P}$. Therefore the statement follows from Lemma 4.3. ∎

Theorem 4.5**.**

Let $Y$ be a proper nonempty subset of $X$ with $\chi_{Y}\in V_{0}\oplus V_{2}\oplus V_{3}$, that is, $Y$ is an intriguing set of $(X,R_{3})$. Then $\chi_{Y}\in V_{0}\oplus V_{2}$ if and only if $Y^{\sigma}=Y$, and $\chi_{Y}\in V_{0}\oplus V_{3}$ if and only if $Y$ contains exactly one of every two conjugate points of $X$.

Proof.

If we assume that $\chi_{Y}\in V_{0}\oplus V_{2}$ or $\chi_{Y}\in V_{0}\oplus V_{3}$, then Theorem 3.3 proves the statement. If we assume that $Y^{\sigma}=Y$, then $\chi_{Y}\in V_{0}\oplus V_{2}\oplus V_{4}$ by Lemma 4.4 and hence $\chi_{Y}\in V_{0}\oplus V_{2}$. Finally assume that $Y$ contains one point of any pair of conjugate points. Then $(\chi_{p}+\chi_{p^{\sigma}})\cdot(2\chi_{Y}-j)=0$ for every point $p\in X$. Therefore Lemma 4.3 shows that $2\chi_{Y}-j\in V_{1}\oplus V_{3}$. Hence $2\chi_{Y}-j\in V_{3}$, which shows that $\chi_{Y}\in V_{0}\oplus V_{3}$. ∎

Intriguing sets of type $3$

We construct an intriguing set $Y$ of the association scheme $X$ with $\chi_{Y}\in V_{0}\oplus V_{3}$. Recall that $X=\mathcal{Q}\setminus(H\cap\mathcal{Q})$ where $\mathcal{Q}$ is an elliptic quadric $\mathcal{Q}$ in $\mathsf{PG}(5,q)$ and $H$ is a non-tangent hyperplane of $\mathsf{PG}(5,q)$, that is a hyperplane that meets $\mathcal{Q}$ in a parabolic quadric $\mathsf{Q}(4,q)$.

Example 4.6** (Intriguing sets of type $3$, $q$ even).**

Assume $q$ is even and $q>2$. Consider a solid $S$ of $H$ such that $S\cap\mathcal{Q}$ is a hyperbolic quadric. Then $S^{\perp}$ is an elliptic line of $\mathsf{PG}(5,q)$ and $H^{\perp}$ is a point on this line. Since $q$ is even, $H^{\perp}\in H$ and $\sigma$ has $q/2$ orbits of length two in its action on the remaining $q$ points of $S^{\perp}$.

Consider $p\in S^{\perp}$ with $p\neq H^{\perp}$. Then $p^{\perp}$ is a non-tangent hyperplane to $\mathcal{Q}$ and hence $p^{\perp}\cap\mathcal{Q}$ is a parabolic quadric $Q(4,q)$ of $\mathcal{Q}$. This implies that $p^{\perp}\cap\mathcal{Q}$ is a tight set of $\mathcal{Q}$. As $p\in S^{\perp}$ we have $p^{\perp}\cap H=S$. As $S\cap\mathcal{Q}$ is a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$, which is the union of its lines of each of its reguli, the set $S\cap\mathcal{Q}$ is also a tight set of $\mathcal{Q}$ and it is contained in $p^{\perp}\cap\mathcal{Q}$. Hence $T_{p}:=(p^{\perp}\setminus S)\cap\mathcal{Q}$ is a tight set of $\mathcal{Q}$. For different points $p$ the different hyperplanes $p^{\perp}$ meet within $S$ and hence the corresponding tight sets $T_{p}$ are disjoint. As $\sigma$ has order two and $H^{\perp}$ is the only point of $S^{\perp}$ fixed by $\sigma$, we can partition the set of $q$ points of $S^{\perp}$ into two sets $M$ and $M^{\prime}$ of cardinality $q/2$ such that $M^{\sigma}=M^{\prime}$. Then the union $T_{M}$ of the tight sets $T_{p}$ with $p\in M$ is a tight set of $\mathcal{Q}$ and the union $T_{M^{\prime}}$ of the tight sets $T_{p}$ with $p\in M^{\prime}$ is a tight set of $\mathcal{Q}$. Moreover every point of $X$ is contained in exactly one of the two sets $T_{M}$ and $T_{M^{\prime}}$. Since $M^{\sigma}=M^{\prime}$, we have $(T_{M})^{\sigma}=T_{M^{\prime}}$. Theorem 4.5 now shows that $T_{M}$ is an intriguing set of $X$ with $\chi_{T_{M}}\in V_{0}\oplus V_{3}$.

Example 4.7** (Intriguing sets of type $3$, $q$ odd).**

Here we assume that $q$ is odd, and hence $H^{\perp}\notin H$. Let $m$ be a line of $H\cap\mathcal{Q}$. Then $H^{\perp}$ and $m$ generate a plane $\pi$ that meets the quadric $\mathcal{Q}:=\mathcal{Q}$ only in the line $m$. Set $z=H^{\perp}$ and let $u$ be a point of $m$. Then $(zu)^{\perp}$ is a 3-space that is contained in $H$ and meets $\mathcal{Q}$ in a cone $C$ with vertex $u$ over a conic. In particular $|C|=q^{2}+q+1$.

Let $S_{1}$ be any set consisting of $\frac{1}{2}(q-1)$ points of the line $uz$ such that $u,z\notin S_{1}$, let $S_{2}$ be any set consisting of $\frac{1}{2}(q-1)$ lines of $\pi$ on $u$ such that $uz,m\notin S_{2}$, and let $T$ be the set consisting of all points of $\mathcal{Q}\setminus H$ that are perpendicular to a point in $S_{1}$ or to a line in $S_{2}$. For $y\in S_{1}$, the hyperplane $y^{\perp}$ meets $\mathcal{Q}$ in a quadric $\mathsf{Q}(4,q)$, and $y^{\perp}\cap H$ meets $\mathcal{Q}$ in the cone $C$. For $\ell\in S_{2}$, the solid $\ell^{\perp}$ meets $\mathcal{Q}$ in a cone with vertex $u$ over a conic, and $\ell^{\perp}\cap H$ meets $\mathcal{Q}$ in the line $m$. It follows that

[TABLE]

Let $p$ be a point of $X$. We want to determine $|p^{\perp}\cap T|$. First notice that $p^{\perp}$ meets the line $uz$ in exactly one point and this point is different from $z$. Also, if $p^{\perp}\cap uz$ is the point $u$, then $p^{\perp}\cap\pi$ is one of the lines of $\pi$ on $u$ that is different from $m$ and $uz$. For $y\in S_{1}$ we have $y^{\perp}\cap H\cap\mathcal{Q}=C$ and thus

[TABLE]

For $\ell\in S_{2}$ we have $\ell^{\perp}\cap H\cap\mathcal{Q}=m$ and thus

[TABLE]

As $|S_{1}|=|S_{2}|=\frac{1}{2}(q-1)$, it follows that

[TABLE]

**Case 1. $p^{\perp}\cap uz\notin S_{1}\cup\{u\}$: **

Then $p^{\perp}\cap C$ is a conic and $p^{\perp}\cap m$ is a single point. If $y\in S_{1}$, then $y^{\perp}\cap\mathcal{Q}$ is a $\mathsf{Q}(4,q)$ not containing $p$, so $p^{\perp}$ meets this $\mathsf{Q}(4,q)$ in a $\mathsf{Q}^{-}(3,q)$. If $\ell\in S_{2}$, then $\ell^{\perp}\cap\mathcal{Q}$ is a cone with vertex $u$ over a conic and since $p\notin u^{\perp}$, the hyperplane $p^{\perp}$ meets this cone in a conic. Hence

[TABLE]

**Case 2. $p^{\perp}\cap uz$ is a point $y_{0}$ of $S_{1}$: **

The only difference to Case 1 occurs for the point $y_{0}\in S_{1}$, namely $p^{\perp}\cap y_{0}^{\perp}$ is a solid that meets $\mathcal{Q}$ in a cone with vertex $p$ over a conic, so $|p^{\perp}\cap y_{0}^{\perp}\cap\mathcal{Q}|=q^{2}+q+1$. It follows that $|p^{\perp}\cap T|=\frac{1}{2}(q-1)q^{2}+q$.

**Case 3. $p^{\perp}\cap uz=\{u\}$ and the line $p^{\perp}\cap\pi$ is not in $S_{2}$: **

Then $p^{\perp}\cap C=p^{\perp}\cap m=\{u\}$. If $y\in S_{1}$, then $p^{\perp}\cap y^{\perp}\cap\mathcal{Q}$ is a $\mathsf{Q}^{-}(3,q)$ as before. As $pu$ is a line of $\mathcal{Q}$ we have $(pu)^{\perp}\cap\mathcal{Q}=pu$. If $\ell\in S_{2}$, then $\ell$ and $p$ are not perpendicular and therefore $\ell^{\perp}\cap p^{\perp}\cap\mathcal{Q}=\ell^{\perp}\cap(pu)^{\perp}\cap\mathcal{Q}=\ell^{\perp}\cap pu=\{u\}$. Hence $|p^{\perp}\cap T|=\frac{1}{2}(q-1)q^{2}$.

**Case 4. $p^{\perp}\cap uz=\{u\}$ and the line $\ell_{0}:=p^{\perp}\cap\pi$ is a line of $S_{2}$: **

The only difference to Case 3 is that now $p^{\perp}\cap\ell_{0}^{\perp}$ meets $\mathcal{Q}$ in the line $pu$, so that $|p^{\perp}\cap\ell_{0}^{\perp}\cap\mathcal{Q}|=q+1$. Hence $|p^{\perp}\cap T|=\frac{1}{2}(q-1)q^{2}+q$.

Notice that Case 2 or 4 occurs if and only if $p\in T$. Hence

[TABLE]

This shows that $T$ is a tight set of the graph $(X,R_{3})$. Since $\sigma$ has orbits of length two on the points not equal to $u,z$ of $uz$, and orbits of length two on the lines not equal to $m,uz$ of $\pi$ on $u$, we can choose $S_{1}$ and $S_{2}$ in such a way that $S_{1}^{\sigma}\cap S_{1}=\varnothing$ and $S_{2}^{\sigma}\cap S_{2}=\varnothing$. As the polarity of $\mathcal{Q}$ commutes with $\sigma$, this implies that $T^{\sigma}\cap T=\varnothing$. Since $|T|=\frac{1}{2}|X|$, it follows that $|T\cap\{p,p^{\sigma}\}|=1$ for every point $p\in X$. Therefore Theorem 4.5 shows that $\chi_{T}\in V_{0}\oplus V_{3}$.

Intriguing sets of type $2$

Now we focus on the intriguing sets $Y$ with $\chi_{Y}\in V_{0}\oplus V_{2}$. We will prove a lower bound on the size of $Y$ and provide an example that shows that the bound is sharp.

Lemma 4.8**.**

If $S$ is the union of two perpendicular conics of $\mathsf{Q}^{-}(5,q)$ with $S\cap H=\varnothing$, then $S^{\sigma}\neq S$.

Proof.

Assume $S$ is the union of conics $C_{1}=\pi_{1}\cap\mathsf{Q}^{-}(5,q)$ and $C_{2}=\pi_{2}\cap\mathsf{Q}^{-}(5,q)$ where $\pi_{1}$ and $\pi_{2}$ are perpendicular planes. Then $\pi_{1}\cap\pi_{2}=\varnothing$. Also $\pi_{1}\cap H$ and $\pi_{2}\cap H$ are elliptic lines. Suppose that $S^{\sigma}=S$. As $\pi_{1}^{\sigma}$ is a plane on $\pi_{1}\cap H$, then either $\pi_{1}^{\sigma}=\pi_{1}$ or $\pi_{1}^{\sigma}\cap S\subseteq\pi_{1}^{\sigma}\cap\pi_{2}$. As $\pi_{1}^{\sigma}\cap S=\pi_{1}^{\sigma}\cap S^{\sigma}=(\pi_{1}\cap S)^{\sigma}$ is a conic, the second case is impossible. Hence $\pi_{1}^{\sigma}=\pi_{1}$. As $\sigma$ is a non-trivial central collineation with centre $H^{\perp}$, it follows that $H^{\perp}\in\pi_{1}$. The same argument shows $H^{\perp}\in\pi_{2}$. But $\pi_{1}\cap\pi_{2}=\varnothing$, a contradiction. ∎

Lemma 4.9**.**

If $Y$ is a non-empty subset of $X$ with $\chi_{Y}\in V_{0}\oplus V_{2}$, then $|Y|=\alpha(q+1)$ where $\alpha$ is an even integer such that $\alpha\geqslant 4$.

Proof.

The set $Y$ is a tight set of $\mathsf{Q}^{-}(5,q)$ (by Lemma 4.1), and so $|Y|=\alpha(q+1)$ for some positive integer $\alpha$. Let $\mathcal{F}$ be an intriguing set of type 4 of size $q^{2}(q-1)$ as in Example 3.2. Then by Result 2.1,

[TABLE]

Now $Y$ and $\mathcal{F}$ are both $\sigma$-invariant (by Theorem 3.3) and so $Y\cap\mathcal{F}$ is $\sigma$-invariant. So $Y\cap\mathcal{F}$ is a union of $\sigma$-orbits, each of size 2. Therefore, $|Y\cap\mathcal{F}|=\alpha$ is even.

A $2$-tight set is either a union of two lines or two perpendicular conics (c.f., [11, Theorem 1.1]). As $Y$ is disjoint from the hyperplane $H$, we have that $Y$ contains no line. As $Y^{\sigma}=Y$, Lemma 4.8 shows that $Y$ is not a union of two perpendicular conics. Hence $\alpha\neq 2$ and thus $\alpha\geqslant 4$. ∎

Example 4.10** (Intriguing sets of type 2).**

Recall that $X=\mathcal{Q}\setminus H$. Consider a solid $S$ of $H$ such that $S\cap\mathcal{Q}$ is a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$. Then $S^{\perp}$ is an elliptic line of $\mathsf{PG}(5,q)$, that is a line of $\mathsf{PG}(5,q)$ with no point in $\mathcal{Q}$, and $H^{\perp}$ is a point on this line. Consider an elliptic line $\ell_{1}$ of $S$. Then $\ell_{2}:=\ell_{1}^{\perp}\cap S$ is also an elliptic line. Let $p_{1}$ and $p_{2}$ be two perpendicular points of the line $S^{\perp}$ that are different from $H^{\perp}$ (and hence of $S^{\perp}\cap H$). Then $\langle p_{1},\ell_{1}\rangle$ and $\langle p_{2},\ell_{2}\rangle$ are perpendicular conic planes that have all their conic points outside $H$. Let $Y$ be the set consisting of the $2(q+1)$ conic points in these two planes. As the conics are perpendicular, $Y$ is a $2$-tight set of $\mathcal{Q}$ (cf. Example 2.3). The points $p_{1}^{\sigma}$ and $p_{2}^{\sigma}$ are perpendicular points of $S^{\perp}$ and as before, $\langle p_{1}^{\sigma},\ell_{1}\rangle$ and $\langle p_{2}^{\sigma},\ell_{2}\rangle$ are conic planes and the $2(q+1)$ points of their conics form a $2$-tight set $Y^{\prime}$ of $\mathcal{Q}$. We have $Y^{\prime}=Y^{\sigma}$ and $Y=(Y^{\prime})^{\sigma}$, so $Y\cup Y^{\prime}$ is $\sigma$-invariant. Notice that either $p_{1}^{\sigma}=p_{2}$ and $p_{2}^{\sigma}=p_{1}$ or otherwise the points $p_{1}$, $p_{2}$, $p_{1}^{\sigma}$ and $p_{2}^{\sigma}$ are four distinct points. In both cases, it is easy to see that no four of the used conic planes share a point of $\mathcal{Q}$. Hence $Y\cap Y^{\prime}=\varnothing$, which implies that $Y\cup Y^{\prime}$ is a $4$-tight set of $\mathcal{Q}$. As it is invariant under $\sigma$ and since $Y,Y^{\prime}\subseteq X$, it follows from Theorem 4.5 that $\chi_{Y\cup Y^{\prime}}\in V_{0}+V_{2}$. Hence $Y\cup Y^{\prime}$ is an intriguing set of $X$ size $4(q+1)$.

Lemma 4.11**.**

Let $\mathcal{Q}$ be a five-dimensional elliptic quadric of $\mathsf{PG}(5,q)$, and suppose $\ell_{1}$ and $\ell_{2}$ are two elliptic lines (with respect to $\mathcal{Q}$) that are disjoint and perpendicular. Then the solid $\langle\ell_{1},\ell_{2}\rangle$ meets $\mathcal{Q}$ in a hyperbolic quadric.

Proof.

Let $S$ be the solid spanned by $\ell_{1}$ and $\ell_{2}$. Now $\ell_{2}^{\perp}$ is equal to $\langle\ell_{1},S^{\perp}\rangle$ because $\ell_{1}$ and $S^{\perp}$ are both contained in $\ell_{2}^{\perp}$, and $\ell_{1}\cap S^{\perp}=\ell_{1}\cap(\ell_{1}^{\perp}\cap\ell_{2}^{\perp})=\varnothing$. Note that $\ell_{2}^{\perp}$ is a solid of hyperbolic type. Within the space $\ell_{2}^{\perp}$, if we consider it just as the hyperbolic quadric $\mathsf{Q}^{+}(3,q)$, the line $S^{\perp}$ is simply the polar image of $\ell_{1}$, and therefore, $m$ is elliptic. Hence it follows that $S$ is hyperbolic. ∎

Theorem 4.12**.**

For $q\geqslant 59$, every intriguing set of $Z$ with $\chi_{Z}\in V_{0}+V_{2}$ and $|Z|=4(q+1)$ is of the form described in Example 4.10.

Proof.

Let $Z$ be such an intriguing set (of type 2). Now $Z$ does not contain a line because $Z\cap H=\varnothing$. Since $q\geqslant 59$, it follows therefore from [11, Theorem 2.15] that $Z$ is the union of four conics $C_{1}$, $C_{2}$, $C_{3}$ and $C_{4}$ where $C_{1}$ and $C_{2}$, as well as $C_{3}$ and $C_{4}$, are perpendicular. Let $\pi_{i}$ be the plane spanned by $C_{i}$. Since $Z\subseteq X$ and hence $Z\cap H=\varnothing$, the lines $\ell_{i}:=\pi_{i}\cap H$, $i=1,2,3,4$, are elliptic lines. Since $\pi_{1}$ and $\pi_{2}$ are perpendicular and hence skew, the lines $\ell_{1}$ and $\ell_{2}$ are skew and perpendicular. Hence these two lines span a solid $S$ that meets $\mathcal{Q}$ in a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$ (by Lemma 4.11). We have $S\subseteq H$ and $S^{\perp}$ is an elliptic line. Since $S^{\perp}$ and $\pi_{1}$ are contained in $\ell_{2}^{\perp}$, it follows that $\pi_{1}$ and $S^{\perp}$ meet in a point $p_{1}$, and similarly $\pi_{2}$ and $S^{\perp}$ meet in a point $p_{2}$. Lemma 4.8 shows that $C_{1}\cup C_{2}$ is not invariant under $\sigma$, and Theorem 4.5 shows that $Z^{\sigma}=Z$. Since $C_{1}\cup C_{2}\subset Z$, it follows that $\pi_{1}^{\sigma}\neq\pi_{1}$ or $\pi_{2}^{\sigma}\neq\pi_{2}$. We may assume without loss of generality that $\pi_{1}^{\sigma}\neq\pi_{1}$. Then $\pi_{1}^{\sigma}$ is a conic plane on $\ell_{1}$. Since the conic $\pi_{1}^{\sigma}\cap\mathcal{Q}$ is contained in $Z^{\sigma}=Z$ and hence in the union of the conics $C_{2}$, $C_{3}$ and $C_{4}$, it follows from $q\geqslant 59$ that $\pi_{1}^{\sigma}$ must be one of the planes $\pi_{2}$, $\pi_{3}$ or $\pi_{4}$. Now $\ell_{1}\subseteq\pi_{1}^{\sigma}$, and so $\pi_{1}^{\sigma}\neq\pi_{2}$ and hence we may assume that $\pi_{1}^{\sigma}=\pi_{3}$. Therefore, $\pi_{2}^{\sigma}=(\pi_{1}^{\perp})^{\sigma}=(\pi_{1}^{\sigma})^{\perp}=\pi_{3}^{\perp}=\pi_{4}$. Thus with $Y:=C_{1}\cup C_{2}$ and $Y^{\prime}=C_{3}\cup C_{4}$ we have $Y^{\sigma}=Y^{\prime}$. So, $Z=Y\cup Y^{\prime}$ has the structure described in Example 4.10. ∎

5. Intriguing sets that are not tight; type 4

Let $Y$ be an intriguing set of the association scheme $X$ corresponding to $V_{4}$. Since the relation $R_{3}$ has the eigenvalue $\theta:=-(q-1)^{2}$ on $V_{4}$, Result 2.1 shows that every point of $X\setminus Y$ is collinear to

[TABLE]

points of $Y$, and every point of $Y$ is collinear to exactly $\theta+\alpha$ points of $Y$. Since $\alpha+\theta\geqslant 0$, we have $\alpha\geqslant-\theta=(q-1)^{2}$ and hence $|Y|\geqslant q(q-1)^{2}$.

Lemma 5.1**.**

Let $Y$ be an intriguing set of type $4$. Then $|Y|\geqslant q^{2}(q-1)$.

Proof.

We have seen in Example 4.10 that there exists an intriguing set $S$ of size $4(q+1)$ with $\chi_{S}\in V_{0}\oplus V_{2}$. Hence

[TABLE]

that is $q^{2}(q-1)$ divides $4|Y|$. We also have $|Y|\geqslant q(q-1)^{2}$. As $q\geqslant 3$, it follows that either $|Y|\geqslant q^{2}(q-1)$ or otherwise that $q=4$ and $|Y|=q(q-1)^{2}=36$. However, $|Y|=q^{2}(q-1)$ and $q=4$ implies that $\alpha+\theta=0$, that is $Y$ is a partial ovoid. It is known that $\mathsf{Q}^{-}(5,4)$ has no partial ovoids of size larger than $35$ [8, Theorem 4.2]. ∎

We have seen in Example 3.2 that this bound is tight. In the rest of this section we show that every intriguing set $Y$ of type 4 with $|Y|=q^{2}(q-1)$ has the form described in Example 3.2. Recall that $H$ is the hyperplane of $\mathsf{PG}(5,q)$ such that $X=\mathcal{Q}\setminus H$, and that $H\cap\mathcal{Q}$ is a parabolic quadric $\mathsf{Q}(4,q)$.

Lemma 5.2**.**

Suppose that $Y$ is an intriguing set of $X$ with $q^{2}(q-1)$ points such that every point in $Y$ is collinear to $q-1$ points of $Y$, and such that every point of $X\setminus Y$ is collinear to $q^{2}-q$ points of $Y$. Then we have.

1. (a)

For each $p\in H\cap\mathcal{Q}$ there exists an integer $s_{p}$ such that $|\ell\cap Y|=s_{p}$ for every line of $\mathcal{Q}$ that contains $p$ and is not contained in $H$. For every subset $M$ of $H\cap\mathcal{Q}$ we set $s_{M}:=\sum_{p\in M}s_{p}$. 2. (b)

$s_{H\cap\mathcal{Q}}=q^{3}+q$* and $\sum_{p\in H\cap\mathcal{Q}}s_{p}(s_{p}-1)=q(q-1)$.* 3. (c)

If $\ell$ is a line of $H\cap\mathcal{Q}$, then $s_{\ell}=q$. 4. (d)

Let $U$ be a solid of $H$ such that $\mathcal{E}:=U\cap\mathcal{Q}$ is an elliptic quadric $\mathsf{Q}^{-}(3,q)$ and suppose that $s_{p}=0$ for at least one point $p\in\mathcal{E}$. Then $s_{\mathcal{E}}=q^{2}-q$. 5. (e)

Let $U$ be a solid of $H$ such that $\mathcal{H}:=U\cap\mathcal{Q}$ is a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$. Then $s_{\mathcal{H}}=q^{2}+q$. 6. (f)

Suppose that $q$ is odd. Let $\pi$ be a plane of $H$ such that $C:=\pi\cap\mathcal{Q}$ is a conic and suppose that $s_{x}=0$ for at least one point $x$ of $C$. If the line $\ell:=\pi^{\perp}\cap H$ is external to $\mathcal{Q}$, then $s_{C}=q-1$, and if $\ell$ meets $\mathcal{Q}$ in two points $p$ and $r$, then $s_{C}=q+1-s_{p}-s_{r}$. 7. (g)

Suppose that $q$ is even. Let $\pi$ be a plane of $H$ such that $C:=\pi\cap\mathcal{Q}$ is a conic.

If $H^{\perp}\notin\pi$ and $s_{p}=0$ for at least one point $p$ of $C$, then $\pi^{\perp}\cap H$ is a line on $H^{\perp}$ meeting $\mathcal{Q}$ in exactly one point $u$ and we have $s_{C}+s_{u}=q$.

If $H^{\perp}\in\pi$, then $\pi^{\perp}\cap H$ is a plane that meets $\mathcal{Q}$ in a conic $C^{\prime}$ and we have $s_{C}+s_{C^{\prime}}=2q$.

Proof.

(a) Let $\ell$ be a line of $\mathcal{Q}$ that contains $p$ and is not contained in $H$. Let $s$ be the number of points of $Y$ on $\ell$. Let $r$ be a point of $\ell$ with $r$ not equal to $p$. If $r\in Y$, then $r$ is collinear to $q-1$ points of $Y$, of which $s-1$ are on $\ell$, and if $r\notin Y$, then $r$ is collinear to $q^{2}-q$ points of $Y$, of which $s$ are on $\ell$. Hence there exist exactly $s(q-s)+(q-s)(q^{2}-q-s)=(q-s)(q^{2}-q)$ points in $Y$ that are collinear to exactly one of the points $\neq p$ of $\ell$. The remaining points of $Y$ must therefore be collinear to $p$. This shows that $s$ depends only on $p$ but not on the choice of $\ell$.

(b) Each point $p\in H\cap\mathcal{Q}$ is collinear to $(q^{2}-q)s_{p}$ points of $Y$. On the other hand, every point of $Y$ is collinear to $q^{2}+1$ points of $H\cap\mathcal{Q}$. A double counting argument thus gives $\sum_{p\in H\cap\mathcal{Q}}s_{p}(q^{2}-q)=|Y|(q^{2}+1)$. Since $|Y|=q^{2}(q-1)$, this gives $\sum_{p\in H\cap\mathcal{Q}}s_{p}=q^{3}+q$.

For the second equality in (b) we count triples $(p,r_{1},r_{2})\in(H\cap\mathcal{Q})\times Y\times Y$ of collinear points with $r_{1}\neq r_{2}$. Each point $r_{1}\in Y$ is collinear to $q-1$ points of $Y$ and thus occurs in $q-1$ such triples in the middle position. Each point $p\in H\cap\mathcal{Q}$ occurs in $(q^{2}-q)s_{p}(s_{p}-1)$ such triples. It follows that

[TABLE]

Since $|Y|=q^{2}(q-1)$, this implies the second equation of statement (b).

(c) A point $p$ of $\ell$ is collinear to $s_{p}(q^{2}-q)$ points of $Y$. Since every point of $Y$ is collinear to exactly one point of $\ell$, it follows that $s_{\ell}(q^{2}-q)=|Y|$, hence $s_{\ell}=q$.

(d) The line $U^{\perp}$ meets $\mathcal{Q}$ in two points $r$ and $r^{\prime}$ and these points do not lie in $H$. Since $s_{p}=0$ for some point of $\mathcal{E}$, we see that $r,r^{\prime}\notin Y$. Thus $r$ is collinear to exactly $q^{2}-q$ points of $Y$ and these are exactly the points of $Y$ on the lines $rz$ with $z\in\mathcal{E}$. Since $r\notin Y$, it follows that $s_{\mathcal{E}}=q^{2}-q$.

(e) Because a hyperbolic quadric is the union of the $q+1$ lines of any of its two reguli, this follows from part (c).

(f) As $q$ is odd, $\pi^{\perp}\cap H$ is in fact a line and moreover this line is skew to $\pi$. First consider the case that $\ell$ has no point on the quadric. Then each solid $T$ of $H$ on $\pi$ meets $\mathcal{Q}$ in an elliptic quadric $\mathsf{Q}^{-}(3,q)$ or a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$, and there are $\frac{1}{2}(q+1)$ solids of each kind. Counting the sum $\sum_{p\in H\cap\mathcal{Q}}s_{p}$ using the solids on $\pi$, it follows from (b), (d) and (e) that

[TABLE]

This implies that $s_{C}=q-1$.

Now consider the case that the $\pi^{\perp}\cap H$ is a secant line to the quadric, and let $p$ and $r$ be the two points of $\mathcal{Q}$ on this line. The solids $\langle\pi,p\rangle$ and $\langle\pi,r\rangle$ meet $\mathcal{Q}$ in a cone $C_{p}$ resp. $C_{r}$ with vertex $p$ resp. $r$ over the conic $C$. From the remaining $q-1$ solids of $H$ through $\pi$ one half meets $\mathcal{Q}$ in an elliptic quadric $\mathsf{Q}^{-}(3,q)$ and one half meets $\mathcal{Q}$ in a hyperbolic quadric $\mathsf{Q}^{+}(3,q)$. It follows from (c) that $s_{C_{p}}=q(q+1-s_{p})$ and $s_{C_{r}}=q(q+1-s_{r})$. Then a similar counting as in the first case gives

[TABLE]

This implies that $s_{C}=q+1-s_{p}-s_{r}$.

(g) First consider the case when $\pi$ does not contain $H^{\perp}$. Then $\langle\pi,H^{\perp}\rangle$ meets $\mathcal{Q}$ in a cone $U$ with vertex a point $u$ over the conic $C$. The vertex $u$ is the unique point on the line joining $H^{\perp}$ to the nucleus of the conic in $\pi$. Part (c) shows that $s_{U}=q(q+1-s_{u})$. Every other solid on $\pi$ meets $\mathcal{Q}$ in a $\mathsf{Q}^{-}(3,q)$ or $\mathsf{Q}^{+}(3,q)$ and there are $q/2$ solids of each type. If $C$ has a point $p$ with $s_{p}=0$, then the previous parts imply similarly as in the proof of (f) that

[TABLE]

and this gives $s_{C}+s_{u}=q$.

Now consider the case when $\pi$ contains $H^{\perp}$. Then $\tau:=\pi^{\perp}\cap H$ is a conic plane with $\pi\cap\tau=H^{\perp}$. Let $C^{\prime}$ be the conic $\tau\cap\mathcal{Q}$. The solids on $\pi$ are spanned by $\pi$ and a point of $C^{\prime}$. Hence all these solids meet $\mathcal{Q}$ in cones with vertex a point of $C^{\prime}$. If $u$ is a point of $C^{\prime}$, then the solid $T:=\langle\pi,u\rangle$ meets $\mathcal{Q}$ in the union of $q+1$ lines on $u$, so part (c) shows that $s_{T}=q(q+1-s_{u})$. It follows that

[TABLE]

which gives $s_{C}+s_{C^{\prime}}=2q$. ∎

Lemma 5.3**.**

Suppose that $Y$ is an intriguing set of $X$ with $q^{2}(q-1)$ points such that every point in $Y$ is collinear to $q-1$ points of $Y$, and every point of $X\setminus Y$ is collinear to $q^{2}-q$ points of $Y$. Let $\ell$ be a line of $H$ that meets $\mathcal{Q}$ in exactly two points $p$ and $r$, and let $C$ be the conic $\ell^{\perp}\cap H\cap\mathcal{Q}$. Suppose that $s_{p}=0$. Then $s_{C}=q(2-s_{r})$.

Proof.

Put $C=\{z_{0},\dots,z_{q}\}$ and $\pi:=\ell^{\perp}\cap H$. Then $C=\pi\cap\mathcal{Q}$.

Case 1. We assume that $q$ is odd.

The planes of $H$ on $\ell$ are the planes $\tau$ spanned by $\ell$ and a point $z$ of $\pi$. For $z=z_{i}\in C$, the plane $\tau:=\langle\ell,z_{i}\rangle$ meets $\mathcal{Q}$ in the union of the lines $z_{i}p$ and $z_{i}r$, so we have $s_{\tau}=2q-s_{z_{i}}$. Now consider one of the $q^{2}$ points $z$ of $\pi\setminus C$ and the plane $\tau_{z}:=\langle\ell,z\rangle$. Then $\tau_{z}$ meets $\mathcal{Q}$ in a conic. The line $\tau_{z}^{\perp}\cap H=z^{\perp}\cap\pi$ is either an exterior line or a secant line of the conic $\pi\cap\mathcal{Q}$, depending on whether or not $z$ is an interior or exterior point of the conic $C$. Thus, exactly ${q\choose 2}$ planes $\tau_{z}$ have the property that $\tau_{z}^{\perp}\cap H$ is an exterior line of the quadric and these planes satisfy $s_{\tau_{z}}=q-1$ by Lemma 5.2 (f). The remaining ${q+1\choose 2}$ planes $\tau_{z}$ of $H$ on $\ell$ can be indexed $\tau_{ij}$ with $i<j$ where $z_{i}$ and $z_{j}$ are the two points of $C$ that lie on the line $\langle z,\ell\rangle^{\perp}\cap\pi$. Then $s_{\tau_{ij}}=q+1-s_{z_{i}}-s_{z_{j}}$ again by Lemma 5.2 (f). Counting the sum $\sum_{p\in H\cap\mathcal{Q}}s_{p}$ by considering the planes of $H$ on $\ell$ we thus find

[TABLE]

For each $i$ the term $s_{z_{i}}$ occurs $q+1$ times in this equation. It follows thus that

[TABLE]

As $s_{C}=\sum_{i}s_{z_{i}}$ this gives $qs_{\ell}=2q-s_{C}$. By Lemma 5.2(f), we have

[TABLE]

and hence $s_{C}=q$ and $s_{\ell}=1$. Therefore, $s_{r}=s_{p}+s_{r}=s_{\ell}=1$ and $s_{C}=q(2-s_{r})$.

Case 2. We assume that $q$ is even.

As before the planes of $H$ on $\ell$ are the planes spanned by $\ell$ and a point of $\pi$. Let $z$ be a point of $\pi$ and $\tau=\langle\ell,z\rangle$. If $z=z_{i}\in C$, then $s_{\tau}=2q-s_{z_{i}}$ as in Case 1. If $z=H^{\perp}$, then $\tau^{\perp}\cap H=\pi$, so $s_{\tau}=2q-s_{\pi}$ by Lemma 5.2 (g). If $z$ is one of the $q^{2}-1$ points $z$ with $z\neq H^{\perp}$ and $z\notin C$, then the line on $H^{\perp}$ and $z$ meets $C$ in a unique point $z_{i}$ and Lemma 5.2 (g) shows $s_{\tau}=q-s_{z_{i}}$. Notice that every point $z_{i}$ of $C$ occurs for exactly $q-1$ choices for $z$ of $\pi$ with $z\neq H^{\perp}$ and $z\notin C$. Counting the sum $\sum_{p\in H\cap\mathcal{Q}}s_{p}$ by considering the planes of $H$ on $\ell$ we thus find

[TABLE]

Using $s_{C}=\sum_{i}s_{z_{i}}$, this simplifies to $(q^{2}+q)s_{\ell}=2q(q+1)-(q+1)s_{C}$. Since $s_{\ell}=s_{p}+s_{r}=s_{r}$, this proves the statement. ∎

Lemma 5.4**.**

Suppose that $q$ is odd and that $Y$ is an intriguing set with $q^{2}(q-1)$ points such that every point in $Y$ is collinear to $q-1$ points of $Y$, and every point of $X\setminus Y$ is collinear to $q^{2}-q$ points of $Y$. Then $s_{r}\in\{0,1,q\}$ for every point $r\in H\cap\mathcal{Q}$.

Proof.

Since $\sum_{w\in H\cap\mathcal{Q}}s_{w}(s_{w}-1)=q(q-1)$, we have $s_{r}\leqslant q$. We may assume that $0<s_{r}<q$. Let $\ell$ be a line of $H\cap\mathcal{Q}$ on $r$. It follows from Lemma 5.2 (c) that $\ell$ contains a point $z$ with $z\neq r$ and $s_{z}>0$. Let $h$ be a line of $H\cap\mathcal{Q}$ on $z$ with $h\neq\ell$. Lemma 5.2 (c) shows that $h$ contains a point $p$ with $s_{p}=0$. Then $pr$ is a secant line and $z$ is a point of the conic $C:=(pr)^{\perp}\cap H\cap\mathcal{Q}$. Hence $s_{C}\geqslant s_{z}>0$. Lemma 5.3 shows that $s_{C}=q(2-s_{r})$. Hence $s_{r}<2$, that is $s_{r}=1$. ∎

Theorem 5.5**.**

Suppose $Y$ is an intriguing set of the Pentilla-Williford scheme $(X,\{R_{0},\dots,R_{4}\})$ with $|Y|=q^{2}(q-1)$ such that every point in $Y$ is collinear to $q-1$ points of $Y$, and every point of $X\setminus Y$ is collinear to $q^{2}-q$ points of $Y$. Then there exists a point $p\in H\cap\mathcal{Q}$ such that $Y$ consists of the $(q^{2}-q)q$ points of $X=\mathcal{Q}\setminus H$ that are collinear to $p$.

Proof.

Since $s_{p}\in\{0,1,q\}$ for all points $p\in H\cap\mathcal{Q}$, Lemma 5.2 (b) implies that there exists a unique point $p$ with $s_{p}=q$. Then each of the $q^{2}-q$ lines of $\mathcal{Q}$ on $p$ that is not contained in $H$ meets $Y$ in $q$ points. Hence, $Y$ consists of the points of $X$ that are collinear to $p$. ∎