Maximal co-cliques in the Kneser graph on plane-solid flags in $PG(6,q)$

TL;DR

This paper determines the independence number of a Kneser graph on plane-solid flags in projective space PG(6,q) for large q, describing large maximal independent sets and bounding smaller ones.

Contribution

It precisely characterizes all large maximal independent sets in the Kneser graph on plane-solid flags in PG(6,q) for q > 27.

Findings

All maximal independent sets of size at least q^{11} are described.

Other maximal independent sets have size at most a constant times q^{10}.

The independence number is determined for q > 27.

Abstract

For $q>27$ we determine the independence number $\alpha(\Gamma)$ of the Kneser graph $\Gamma$ on plane-solid flags in $PG(6,q)$. More precisely we describe all maximal independent sets of size at least $q^{11}$ and show that every other maximal example has cardinality at most a constant times $q^{10}$.

Equations

$${\left[\genfrac{}{}{0ex}{}{6}{4}\right]}_q\cdot {\left[\genfrac{}{}{0ex}{}{4}{3}\right]}_q+{\left[\genfrac{}{}{0ex}{}{5}{3}\right]}_q\cdot q^3$$

$${\left[\genfrac{}{}{0ex}{}{n}{d}\right]}_q:=\{\begin{array}{ll}{\prod }_{i=1}^d\frac{q^{n+1-i}-1}{q^i-1}& \text{if 0≤d≤n,}\\ 0& \text{otherwise.}\end{array}$$

$$s_q(l,k,d,n):=q^{(l+1)(d-k)}\cdot {\left[\genfrac{}{}{0ex}{}{n-k-l-1}{d-k}\right]}_q.$$

$$\alpha (\Gamma )=s(3,5)=q^8+q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1.$$

$$\begin{gathered} |\Lambda (H,\mathcal{E})|=s(3,5)\cdot s(3)+|\mathcal{E}|\cdot q^3 \\ \le q^{11}+2q^{10}+5q^9+7q^8+10q^7+11q^6+11q^5+9q^4+7q^3+4q^2+2q+1 \end{gathered}$$

$$s(2)\cdot s(1,4)\cdot \xi =(q^8+2q^7+4q^6+5q^5+6q^4+5q^3+4q^2+2q+1)\cdot \xi$$

$$|{\Delta }_{P_1}(C)|>(q+1)(6q^6+10q^5+17q^4+15q^3+15q^2+9q+5),$$

$$|{\Delta }_{P_2}(C)|>(q+1)(6q^6+10q^5+17q^4+18q^3+15q^2+9q+5),$$

$$\begin{gathered} (q+1)(2\cdot s(1,4)-s(3)+2\cdot s(2)\cdot s(4)) \\ =(q+1)(4q^6+6q^5+10q^4+9q^3+9q^2+5q+3) \end{gathered}$$

$$\begin{gathered} (q+1)(2\cdot s(1,4)-s(3)+4\cdot s(2)\cdot s(4)+q^4) \\ =(q+1)(6q^6+10q^5+17q^4+15q^3+15q^2+9q+5) \end{gathered}$$

$$s(1,3,6)+(q^2+q{)}^3+|\mathcal{E}|\cdot s(0,2,3,6).$$

$$|{\Delta }_Q(C)|\le 3q^8+12q^7+21q^6+28q^5+26q^4+18q^3+12q^2+8q+4.$$

$$(s(3)-s(u){)}^2+s(0,1,u+1)(s(1,2,6)-s(1,2,u+1))+s(0,2,u+1)$$

$$m:=3q^6+6q^5+7q^4+4q^3+2q^2+q+1.$$

$$n:=2q^6+2q^5+3q^4+2q^3+2q^2+q+1$$

$$|C|\le 24q^{10}+79q^9+155q^8+210q^7+216q^6+187q^5+140q^4+93q^3+51q^2+22q+5.$$

$$|C|$$

$$=q^{10}+3q^9+5q^8+7q^7+8q^6+8q^5+7q^4+5q^3+3q^2+2q+1$$

$$(q^8+2q^7+4q^6+5q^5+6q^4+5q^3+4q^2+2q+1)\cdot s(1)$$

$$|{\Delta }_P(C)|\le |{\Delta }_{P_2}(C)|\le 6q^7+16q^6+27q^5+35q^4+33q^3+24q^2+14q+5$$

$$|{\Delta }_{P_1}(C)|\ge |{\Delta }_{P_2}(C)|>6q^7+16q^6+27q^5+35q^4+33q^3+24q^2+14q+5.$$

$$|{\Delta }_P(C)|\le 24q^7+54q^6+71q^5+67q^4+48q^3+33q^2+22q+11$$

$$|{\Delta }_{\hat{P}}(C)|\le 3q^8+12q^7+21q^6+28q^5+26q^4+18q^3+12q^2+8q+4,$$

$$\begin{gathered} (s(3)-1)\cdot (24q^7+54q^6+71q^5+67q^4+48q^3+33q^2+22q+11) \\ +3q^8+12q^7+21q^6+28q^5+26q^4+18q^3+12q^2+8q+4 \\ =24q^{10}+78q^9+152q^8+204q^7+207q^6+176q^5+129q^4+84q^3+45q^2+19q+4. \end{gathered}$$

$$|C|\le 24q^{10}+79q^9+155q^8+210q^7+218q^6+189q^5+142q^4+95q^3+53q^2+22q+5.$$

$$2\cdot s(1,4)\cdot s(2)\cdot (s(2)+1)$$

$$n:=2q^8+4q^7+6q^6+4q^5+4q^4+3q^3+2q^2+q+1.$$

$$|\mathcal{S}|\le s(3,5)-s(1,-1,3,5)=q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1,$$

$$|C|>26q^{10}+83q^9+159q^8+216q^7+222q^6+193q^5+144q^4+97q^3+53q^2+22q+5.$$

$$|C|\le 2(q^7+2q^6+2q^5+3q^4+2q^3+2q^2+q+1)\cdot q^3+|C^{\prime }|.$$

Full text

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disposition

1 Introduction

For integers $n\geq 2$ and prime powers $q$ we denote by $\text{PG}(n,q)$ the $n$-dimensional projective space over the finite field $\mathbb{F}_{q}$. A flag $F$ of $\text{PG}(n,q)$ is a set of non-trivial subspaces of $\text{PG}(n,q)$ such that $U\subseteq U^{\prime}$ or $U^{\prime}\subseteq U$ for all $U,U^{\prime}\in F$. Here non-trivial means different from $\emptyset$ and $\text{PG}(n,q)$. The set $\{\dim(U)\mid U\in F\}$ is called the type of the flag $F$. Two flags $F_{1}$ and $F_{2}$ of $\text{PG}(n,q)$ are said to be in general position, if for all subspaces $U_{1}\in F_{1}$ and $U_{2}\in F_{2}$ we have $U_{1}\cap U_{2}=\emptyset$ or $\langle U_{1},U_{2}\rangle=\text{PG}(n,q)$.

If $S$ is a non-empty subset of $\{0,1,\dots,n-1\}$, then the Kneser graph of flags of type $S$ is the simple graph whose vertices are the flags of type $S$ of $\text{PG}(n,q)$ with two flags $F$ and $F^{\prime}$ adjacent if and only if they are in general position. If $S=\{i\}$ for some $i\in\{0,\dots,n-1\}$ then we also speak of the Kneser graph of $i$-dimensional subspaces. Note that this graph, among other generalizations of Kneser graphs, has already been defined by C. Güven [6].

This graph and maximal co-cliques therein were studied in [3] for $S=\{1,2\}$ and $n=4$, in [4] for $S=\{0,n-1\}$ and $n\geq 2$, and in [5] for $S=\{0,2\}$ and $n=4$. The result given in [4] was already conjectured by Mussche in his thesis [8] and is also given by C. Güven (one of the authors of [4]) in her thesis [6].

In this paper we determine the independence number $\alpha(\Gamma)$ for the Kneser graph $\Gamma$ of type $\{2,3\}$ in $\text{PG}(6,q)$ for $q>27$. We point out that a flag of type $\{2,3\}$ of $\text{PG}(6,q)$ is a self-dual object, hence any independent set of $\Gamma$ can also be seen as an independent set of the Kneser graph of the same type in the dual space of $\text{PG}(6,q)$. To simplify notation, we also denote a flag $\{E,S\}$ of type $\{2,3\}$ by $(E,S)$ where $E$ is a plane and $S$ is a solid. We first provide some examples.

1.1 Example**.**

For a hyperplane $H$ of $\text{PG}(6,q)$ and a maximal set $\mathcal{E}$ of mutually intersecting planes of $H$, we denote by $\Lambda(H,\mathcal{E})$ the set of all flags $(E,S)$ of type $\{2,3\}$ of $\text{PG}(6,q)$ that satisfy $S\subseteq H$ or $E\in\mathcal{E}$. Dually, for a point $P$ of $\text{PG}(6,q)$ and a maximal set $\mathcal{S}$ of $3$-dimensional subspaces on $P$ any two of which share at least a line, denote by $\Lambda(P,\mathcal{S})$, the set of all flags $(E,S)$ of type $\{2,3\}$ of $\text{PG}(6,q)$ that satisfy $P\in E$ or $S\in\mathcal{S}$.

Indeed, the following special case of this example was already covered in a more general setting in [3].

1.2 Example**.**

For an incident point-hyperplane pair $(P,H)$ of $\text{PG}(6,q)$, let $\Lambda(P,H)$ be the set of all flags $(E,S)$ of type $\{2,3\}$ that satisfy $P\in E$ or $P\in S\subseteq H$ and let $\Lambda(H,P)$ be the set of all flags $(E,S)$ of type $\{2,3\}$ that satisfy $S\subseteq H$ or $P\in E\subseteq H$.

For an incident point-line pair $(P,l)$ of $\text{PG}(6,q)$, let $\Lambda(P,l)$ be the set of all flags $(E,S)$ of type $\{2,3\}$ that satisfy $P\in E$ or $l\subseteq S$.

For an incident pair $(U,H)$ of a $4$-dimensional space $U$ and a hyperplane $H$ of $\text{PG}(6,q)$, let $\Lambda(H,U)$ be the set of all flags $(E,S)$ of type $\{2,3\}$ that satisfy $S\subseteq H$ or $E\subseteq U$.

We shall show in Proposition 3.2 that the sets described in Example 1.1 are maximal independent sets in the Kneser graph of flags of type $\{2,3\}$ in $\text{PG}(6,q)$. Notice that the condition on $\mathcal{E}$ means that $\mathcal{E}$ is an independent set of the Kneser graph of planes of $H\simeq\text{PG}(5,q)$ and that the condition on $\mathcal{S}$ means that $\mathcal{S}$ is an independent set of the Kneser graph of solids on $PG(6,q)$.

The sets constructed in Example 1.2 are special cases of the ones in Example 1.1 and hence also maximal independent sets. Notice that the first and second examples as well as the third and forth examples in Example 1.2 are dual to each other.

In order to state our first theorem, we need the Gaussian coefficients $\genfrac{[}{]}{0.0pt}{}{n}{k}_{q}$ which will be defined in Section 2.

1.3 Theorem**.**

For $q>27$, the independence number of the Kneser graph of flags of type $\{2,3\}$ of $\text{PG}(6,q)$ is

[TABLE]

and the independent sets attaining this bound are precisely the four sets described in Example 1.2.

The independence number is of order $q^{11}$. As our proof of this theorem is geometric it also provides a stability result for independence sets. Essentially it says that, for large values of prime powers $q$, Example 1.1 describes all maximal independent sets with at least $27q^{10}$ elements. A precise formulation is given in Theorem 6.5.

1.4 Remarks**.**

For a Hilton-Milner type result for the Kneser graph of plane-solid flags in $\text{PG}(6,q)$ one needs to determine the second largest cardinality of a maximal set $\mathcal{E}$ of mutually intersecting planes of $\text{PG}(5,q)$ and use this in Example 1.1. However, this number is unknown as it is one of the cases that is not covered by the Hilton-Milner type theorem in **[2]**. 2. 2.

Every upper bound $b$ for the independence number of a graph with $n$ vertices leads to the lower bound $\chi\geq n/b$ for its chromatic number $\chi$. In our situation this shows that the chromatic number of the Kneser graph of plane-solid flags of $\text{PG}(6,q)$, $q>27$, has chromatic number at least $q^{4}-q^{2}+2q+1$. On the other hand, if $U$ is a $4$-space, then the sets $\Lambda(P,\emptyset)$ with $P\in U$ are independent sets whose union covers every vertex, so the chromatic number is at most $q^{4}+q^{3}+q^{2}+q+1$. Using independent sets of the form $\Lambda(P,l)$, we show that this upper bound can be improved. 3. 3.

We keep all estimations in this paper as easy as possible. A more detailed approach, especially in Lemma 4.2, shows that Theorem 1.3 holds also for $q=27$. This will appear in the Ph.D. thesis of the second author.

2 Preliminaries

Let $q$ be a prime power and $F_{q}$ the finite field of order $q$. For integers $n$ and $d$ we define the Gaussian coefficient

[TABLE]

For $n\geq 0$ this is the number of $d$-dimensional subspaces of an $F_{q}$-vector space of dimension $n$. If $0\leq d\leq n$, and if $D$ is a $d$-dimensional subspace of an $n$-dimensional $F_{q}$ vector space $V$ then $D$ has exactly $q^{d(n-d)}$ complements in $V$. These two facts can be found in Section 3.1 of [7]. We define

[TABLE]

We also set $s_{q}(k,d,n):=s_{q}(-1,k,d,n)$, $s_{q}(k,n):=s_{q}(-1,d,n)$ and $s_{q}(n):=s_{q}(0,n)$ and omit the subscript $q$ in the following.

2.1 Lemma**.**

Given two skew subspaces in $\text{PG}(n,q)$ of dimensions $k$ and $l$ respectively and any integer $d$ the number of $d$-subspaces of $\text{PG}(n,q)$ that contain the $k$-subspace and are skew to the $l$-subspace is $s(l,k,d,n)$.

Proof.

We prove this for the underlying $F_{q}$-vector space $V$ of dimension $n+1$ and two skew subspaces $K$ and $L$ of dimension $k+1$ and $l+1$ respectively, where we have to count the number of subspaces $D$ of dimension $d+1$ that contain $K$ and are skew to $L$. Every such subspace $D$ gives rise to a subspace $D+L$ of dimension $d+l+2$ of $V$. Going to the factor space $V/(K+L)$, we see that $V$ has $\genfrac{[}{]}{0.0pt}{}{n-k-l-1}{d-k}_{q}$ subspaces $U$ of dimension $d+l+2$ that contain $K+L$. For such a subspace $U$ we see in the quotient space $U/K$ that $U$ has $q^{(d-k)(l+1)}$ subspaces $D$ of dimension $d+1$ with $U=L+D$. ∎

2.2 Lemma**.**

If $n\geq 5$ and if $\mathcal{E}$ is a set of planes of $\text{PG}(n,q)$ such that any two distinct planes of $\mathcal{E}$ meet in a line, then $|\mathcal{E}|\leq s(n-2)$.

Proof.

If there exists a line contained in all planes of $\mathcal{E}$, then $|\mathcal{E}|\leq s(1,2,n)=s(n-2)$. Otherwise there exist planes $E_{1},E_{2},E_{3}\in\mathcal{E}$ such that $E_{1}\cap E_{3}$ and $E_{2}\cap E_{3}$ are distinct lines, which implies that $E_{3}$ is contained in the 3-space $U:=\langle E_{1},E_{2}\rangle$. In this case, for every further plane $E$ of $\mathcal{E}$ at least two of the lines $E\cap E_{1}$, $E\cap E_{2}$ and $E\cap E_{3}$ are distinct, so $E$ is contained in $U$. Thus, in this case, every plane of $\mathcal{E}$ is one of the $s(2,3)$ planes of $U$. ∎

The following result has been proven in Theorem 1.4 of [2], where it was formulated in its dual version.

2.3 Result**.**

For $q\geq 3$ the independence number $\alpha(\Gamma)$ of the Kneser graph $\Gamma$ of type 3 in $\text{PG}(6,q)$ is given by

[TABLE]

For each hyperplane $H$ of $\text{PG}(6,q)$ the set consisting of all solids of $H$ is an independent set of $\Gamma$ with $\alpha(\Gamma)$ vertices. Every other maximal independent set has cardinality at most $q^{6}+2q^{5}+3q^{4}+3q^{3}+2q^{2}+q+1$.

3 Sets of flags of type $\{2,3\}$

In this section we study sets of flags of type $\{2,3\}$ of $\text{PG}(6,q)$. Recall that we also denote a flag $\{E,S\}$ of type $\{2,3\}$ as the ordered pair $(E,S)$ where $E$ is the plane and $S$ the solid of the flag. Note that two distinct such flags $(E,S)$ and $(E^{\prime},S^{\prime})$ are adjacent in $\Gamma$ if and only if $E\cap S^{\prime}=\emptyset=E^{\prime}\cap S$. Let $\pi_{2}$ and $\pi_{3}$ be the maps from the set of all flags of type $\{2,3\}$ to the set of subspaces of $\text{PG}(6,q)$ with $\pi_{2}(f):=E$ and $\pi_{3}(f):=S$ for all flags $f=(E,S)$ of type $\{2,3\}$. For any set $C$ of such flags, we define $\pi_{i}(C):=\{\pi_{i}(f):f\in C\}$, $i=2,3$.

3.1 Lemma**.**

Let $\Gamma$ be the Kneser graph of flags of type $\{2,3\}$ of $\text{PG}(6,q)$, let $C$ be an independent set of $\Gamma$, let $H$ be a hyperplane and let $P$ be a point of $\text{PG}(6,q)$.

1. i)

Let $\mathcal{E}$ be the set whose elements are the planes $E$ of $H$ for which there exists a solid $S$ with $(E,S)\in C$ and $E=H\cap S$. Then $E\cap E^{\prime}\neq\emptyset$ for all $E,E^{\prime}\in\mathcal{E}$, that is, $\mathcal{E}$ is an independent set of the Kneser graph of planes of $H$. Hence $|\mathcal{E}|\leq s(1,4)$. 2. ii)

Let $\mathcal{S}$ be the set whose elements are the solids $S$ for which there exists a flag $(E,S)\in C$ with $P\in S\setminus E$. Then $|S|\leq s(1,4)$.

Proof.

1. i)

For $E,E^{\prime}\in\mathcal{E}$ let $(E,S)$ and $(E^{\prime},S^{\prime})$ be flags of $C$ with $S\cap H=E$ and $S^{\prime}\cap H=E^{\prime}$. Then $S^{\prime}\cap E=E^{\prime}\cap E$ and $S\cap E^{\prime}=E^{\prime}\cap E$. Since $C$ is independent, it follows that $E\cap E^{\prime}\not=\emptyset$. Thus $\mathcal{E}$ is an independent set of the Kneser graph of planes of $H$. The main result of [1] implies that $|\mathcal{E}|\leq s(2,4)=s(1,4)$. 2. ii)

This is a special case of the dual statement of i).∎

In the following proposition we investigate the sets constructed in Example 1.1 up to duality.

3.2 Proposition**.**

Let $H$ be a hyperplane of $\text{PG}(6,q)$ and let $\Gamma$ be the Kneser graph of flags of type $\{2,3\}$ of $\text{PG}(6,q)$.

1. i)

$\Lambda(H,\emptyset)$* is an independent set of $\Gamma$.* 2. ii)

The maximal independent sets of $\Gamma$ that contain $\Lambda(H,\emptyset)$ are the sets $\Lambda(H,\mathcal{E})$ for maximal independent sets $\mathcal{E}$ of planes of $H$. 3. iii)

For every maximal independent set $\mathcal{E}$ of the Kneser graph of planes of $H$ we have

[TABLE]

If equality holds, that is if $|\mathcal{E}|=s(1,4)$, then either there exists a point $P$ in $H$ such that $\mathcal{E}$ consists of all planes of $H$ that contain $P$, or there exists a $4$-dimensional subspace of $H$ such that $\mathcal{E}$ consists of all planes of this $4$-dimensional subspace.

Proof.

1. i)

This follows from the fact that every solid of $H$ meets every plane of $H$. 2. ii)

If $\mathcal{E}$ is an independent set of planes in the Kneser graph of planes of $H$, then every solid of $H$ meets every plane of $\mathcal{E}$ non-trivially and every two planes of $\mathcal{E}$ meet non-trivially. Therefore $\Lambda(H,\mathcal{E})$ is an independent set of $\Gamma$. In order to prove the assertion, it therefore suffices to consider an independent set $C$ of $\Gamma$ with $\Lambda(H,\emptyset)\subseteq C$ and to show that $C$ is contained in $\Lambda(H,\mathcal{E})$ for a set $\mathcal{E}$ of mutually intersecting planes of $H$.

Let $C$ be an independent set with $\Lambda(H,\emptyset)\subseteq C$. Let $\mathcal{E}$ be the set of all planes $E$ of $H$ such that $C$ contains a flag $(E,S)$ with $E=S\cap H$. Lemma 3.1 shows that the planes of $\mathcal{E}$ are mutually intersecting. It remains to show that $C\subseteq\Lambda(H,\mathcal{E})$. Suppose on the contrary that there exists a flag $(E,S)\in C$ with $S\not\subseteq H$ and $H\cap S\not=E$. Then $S\cap H$ is a plane and $E\cap H$ is a line of this plane and $H$ contains a solid $S^{\prime}$ that is skew to the line $E\cap H$. This implies that $S^{\prime}$ meets the plane $S\cap H$ in a point and therefore $S^{\prime}$ contains a plane $E^{\prime}$ with $E^{\prime}\cap S\cap H=\emptyset$. Then $(E^{\prime},S^{\prime})\in\Lambda(H,\emptyset)\subseteq C$ with $S^{\prime}\cap E=\emptyset=S\cap E^{\prime}$ and since $C$ is independent this is a contradiction. 3. iii)

Since $H$ contains $s(3,5)$ solids all of which contain $s(2,3)=s(3)$ planes, we have $|\Lambda(H,\emptyset)|=s(3,5)\cdot s(3)$. Every plane $E$ of $H$ lies on $s(2,3,6)-s(2,3,5)=q^{3}$ solids $S$ with $S\cap H=E$. Hence $|\Lambda(H,\mathcal{E})|=|\Lambda(H,\emptyset)|+|\mathcal{E}|\cdot q^{3}$. A special case of a result of [1] shows that $|\mathcal{E}|\leq s(2,4)=s(1,4)$ with equality if and only if all planes of $\mathcal{E}$ contain a common point of $H$ or lie in a common $4$-subspace of $H$.∎

3.3 Lemma**.**

Let $\mathcal{C}$ be an independent set of the Kneser graph of type $\{2,3\}$ in $\text{PG}(6,q)$ and let $\xi\in\mathbb{N}$ be such that every solid of $\text{PG}(6,q)$ occurs in at most $\xi$ flags of $C$. Let $(E,S)$ be an element of $C$. Then there are at most

[TABLE]

flags $(E^{\prime},S^{\prime})\in\mathcal{C}$ with $E^{\prime}\cap E=\emptyset$ and $S^{\prime}\cap E\neq\emptyset$.

Proof.

Since $C$ is independent, every flag $(E^{\prime},S^{\prime})\in\mathcal{C}$ with $E^{\prime}\cap E=\emptyset$ and $S^{\prime}\cap E\neq\emptyset$ has the property that $S^{\prime}\cap E$ is a point $P$ with $P\notin E^{\prime}$. Hence for every such flag there exists a point $P$ in $E$ with $P\in S^{\prime}\setminus E^{\prime}$. Since $E$ has $s(2)$ points and since every solid occurs in at most $\xi$ flags of $C$ Lemma 3.1 ii) proves the statement. ∎

We now proceed to prove our theorem in three steps, where we consider two special cases in the first two steps: In the first step we only consider independent sets $C$ in which no plane or solid occurs in more than $s(1)$ flags of $C$ and in the second step we consider independent sets $C$ in which no plane or solid occurs in more than $s(2)$ flags of $C$.

4 The first special case

In this section we consider an independent set $C$ of the Kneser graph of type $\{2,3\}$ in $\text{PG}(6,q)$ that has the property that every plane and every solid of $\text{PG}(6,q)$ occurs in at most $q+1$ flags of $C$. Our aim is to prove an upper bound for $|C|$. For every point $P$ we denote the set of all flags $(E,S)\in C$ with $P\in E$ by $\Delta_{P}(C)$.

4.1 Lemma**.**

Let $P_{1}$, $P_{2}$ and $P_{3}$ be non-collinear points of $\text{PG}(6,q)$.

1. i)

If

[TABLE]

then there are flags $f_{i}=(E_{i},S_{i})\in C$ for $i\in\{1,2,3\}$ with $\dim(\langle E_{1},E_{2},E_{3}\rangle)\geq 5$, $P_{2},P_{3}\notin S_{1},S_{2},S_{3}$ as well as $E_{i}\cap E_{j}=P_{1}$ and $P_{2},P_{3}\notin\langle E_{i},E_{j}\rangle$ for all distinct $i,j\in\{1,2,3\}$. 2. ii)

If there are flags $f_{1}$, $f_{2}$ and $f_{3}$ with the properties stated in i) and if

[TABLE]

then there are flags $f_{i}^{\prime}=(E_{i}^{\prime},S_{i}^{\prime})\in C$ for $i\in\{1,2,3\}$ with $\dim(\langle E_{1}^{\prime},E_{2}^{\prime},E_{3}^{\prime}\rangle)\geq 5$, $P_{1},P_{3}\notin S_{1}^{\prime},S_{2}^{\prime},S_{3}^{\prime}$, $\dim(S_{i}\cap S_{j}^{\prime})\leq 1$ for all $i,j\in\{1,2,3\}$ as well as $E_{i}^{\prime}\cap E_{j}^{\prime}=P_{2}$ and $P_{1},P_{3}\notin\langle E_{i}^{\prime},E_{j}^{\prime}\rangle$ for all distinct $i,j\in\{1,2,3\}$.

Proof.

1. i)

We frequently make use of the fact that every plane and every solid occurs in at most $q+1$ flags of $C$. We also make use of the following properties:

1. $(Q1)$

There are $2\cdot s(1,3,6)-s(2,3,6)=2\cdot s(1,4)-s(3)$ solids that contain $P_{1}$ and a point of $\{P_{2},P_{3}\}$. 2. $(Q2)$

If $E$ is a plane on $P_{1}$ and $P$ is a point not contained in $E$, then every plane $E^{\prime}$ on $P_{1}$ with $E^{\prime}\cap E\not=P_{1}$ or $P\in\langle E,E^{\prime}\rangle$ meets the solid $\langle P,E\rangle$ in at least a line and hence there are at most $s(0,1,3)\cdot s(1,2,6)=s(2)\cdot s(4)$ such planes $E^{\prime}$. 3. $(Q3)$

If $E_{1}$ and $E_{2}$ are planes with $E_{1}\cap E_{2}=P_{1}$, then there exist less than $s(1,0,2,4)=q^{4}$ planes in $\langle E_{1},E_{2}\rangle$ with $E\cap E_{1}=E\cap E_{2}=P_{1}$.

From i)$(Q1)$ and the bound in equation (1) we see that there exists a flag $(E_{1},S_{1})$ in $C$ with $P_{1}\in E_{1}$ and $P_{2},P_{3}\notin S_{1}$. According to i)$(Q1)$ and i)$(Q2)$ the number of flags $(E,S)\in\Delta_{P_{1}}(C)$ for which $E_{1}\cap E\not=P_{1}$ or for which $\langle E_{1},E\rangle$ or $S$ contains a point of $\{P_{2},P_{3}\}$ is at most

[TABLE]

which is smaller than the right hand side of equation (1). Therefore, we find a flag $(E_{2},S_{2})\in\Delta_{P_{1}}(C)$ such that $E_{1}\cap E_{2}=P_{1}$ and neither of the spaces $\langle E_{1},E_{2}\rangle$ or $S_{2}$ contains one of the points $P_{2}$ and $P_{3}$. Notice that $\dim(\langle E_{1},E_{2}\rangle)=4$, so for the remaining flag $(E_{3},S_{3})$ we need that $E_{3}$ is not contained in $\langle E_{1},E_{2}\rangle$. Using i)$(Q1)$, i)$(Q2)$ and i)$(Q3)$ a similar argument shows that at most

[TABLE]

flags of $\Delta_{P_{1}}(C)$ do not satisfy all of the properties we want for the final flag $(E_{3},S_{3})$. Since this is the right hand side of (1) and thus smaller than $|\Delta_{P_{1}}(C)|$ we find a flag $(E_{3},S_{3})$ with the desired properties. 2. ii)

We can argue analogously to the proof of i). However, when choosing the flags $(E^{\prime}_{i},S^{\prime}_{i})$ for $i\in\{1,2,3\}$ we additionally have to avoid all flags $(E,S)\in\Delta_{P_{2}}(C)$ for which $S$ meets one of the solids $S_{1}$, $S_{2}$ and $S_{3}$ in a plane $\pi$ with $P_{1}\notin\pi$. For $j\in\{1,2,3\}$ each $S_{j}$ has $q^{3}$ planes that do not contain $P_{1}$, so in total there are at most $3q^{3}$ solids $S$ that must not appear in any of our desired flags $(E^{\prime}_{i},S^{\prime}_{i})$ for $i\in\{1,2,3\}$ and were not considered before. Therefore, it is sufficient to check that the sum of the number in equation (2) and the number $3q^{3}(q+1)$ is the right hand side of equation (1) and thus smaller than $|\Delta_{P_{2}}(C)|$, which is obviously true.∎

4.2 Lemma**.**

Let $P_{1}$ and $P_{2}$ be two distinct points of $\text{PG}(6,q)$ and let $E_{1}$, $E_{2}$ and $E_{3}$ be planes such that $E_{i}\cap E_{j}=P_{1}$ and $P_{2}\notin\langle E_{i},E_{j}\rangle$ for all distinct $i,j\in\{1,2,3\}$. Furthermore, let $\mathcal{S}$ be the set of all solids of $\text{PG}(6,q)$ with $P_{2}\in S$ and $S\cap E_{i}\neq\emptyset$ for all $i\in\{1,2,3\}$. Then we have $|\mathcal{S}|\leq 3q^{6}+6q^{5}+7q^{4}+4q^{3}+2q^{2}+q+1$.

Proof.

Let $\mathcal{E}$ be the set of all planes that contain $P_{2}$ but not $P_{1}$ and that meet all the planes $E_{1}$, $E_{2}$ and $E_{3}$. There are $s(1,3,6)$ solids on $P_{2}$ that contain $P_{1}$. If a solid on $P_{2}$ does not contain $P_{1}$ nor any plane of $\mathcal{E}$, then it meets all the planes $E_{1}$, $E_{2}$ and $E_{3}$ in unique points (different from $P_{1}$) and these three intersection points together with $P_{2}$ span the solid. Hence, there are at most $(q^{2}+q)^{3}$ such solids. Finally, each plane of $\mathcal{E}$ lies in at most $s(0,2,3,6)$ solids that do not contain $P_{1}$, which shows that the number of solids on $P_{2}$ that meet $E_{1}$, $E_{2}$ and $E_{3}$ is at most

[TABLE]

It remains to determine an upper bound on $|\mathcal{E}|$. We put $U:=\langle E_{1},E_{2}\rangle$ with $P_{2}\notin U$ and if $E\in\mathcal{E}$ we know from $P_{2}\in E$ and $P_{1}\notin E$ that $E\cap U$ is a line. We show that every point of $E_{3}\setminus\{P_{1}\}$ lies on at most $q$ planes of $\mathcal{E}$. To see this, let $Q$ be a point of $E_{3}\setminus\{P_{1}\}$ and suppose that $Q$ lies on at least one plane $E$ of $\mathcal{E}$. Since the lines $\langle P_{2},Q\rangle$ and $E\cap U$ of $E$ are distinct, they meet in a unique point $R$, and $E\cap U$ is a line on $R$. Since $P_{2}$ is not contained in $\langle E_{1},E_{3}\rangle$ nor in $\langle E_{2},E_{3}\rangle$ we have $R\notin E_{1}$ and $R\notin E_{2}$. This implies that $R$ lies on exactly $q$ lines of $U$ that meet $E_{1}$ and $E_{2}$ but do not contain $P_{1}$. Since every plane of $\mathcal{E}$ on $Q$ is generated by $P_{2}$ and such a line, we see that $Q$ lies on at most $q$ planes of $\mathcal{E}$. As there are $q^{2}+q$ choices for $Q$, we find $|\mathcal{E}|\leq(q^{2}+q)q$. Using this upper bound for $|\mathcal{E}|$, the statement follows from equation (3). ∎

4.3 Lemma**.**

Let $P$ be a point and suppose that there are flags $(E_{i},S_{i})\in\Delta_{P}(C)$, for $i\in\{1,2,3\}$, such that $E_{i}\cap E_{j}=P$ for distinct $i,j\in\{1,2,3\}$. Then every point $Q$ with $Q\notin\langle E_{i},E_{j}\rangle$ and $Q\notin S_{i}$ for all $i,j\in\{1,2,3\}$ satisfies

[TABLE]

Proof.

For $i\in\{1,2,3\}$ exactly $n:=s(0,2,6)-s(3,0,2,6)$ planes on $Q$ meet $S_{i}$. Since every plane lies in at most $q+1$ flags of $C$, it follows that there exists at most $3n(q+1)$ flags $(E,S)\in\Delta_{Q}(C)$ such that $E$ has non-empty intersection with at least one of the solids $S_{1}$, $S_{2}$ or $S_{3}$.

Every other flag $f=(E,S)\in\Delta_{Q}(C)$ has the property that its solid $S$ meets $E_{1}$, $E_{2}$ and $E_{3}$. Lemma 4.2 shows that there at most $n^{\prime}:=3q^{6}+6q^{5}+7q^{4}+4q^{3}+2q^{2}+q+1$ such solids. Since each solid lies in at most $q+1$ flags of $C$, there are at most $n^{\prime}(q+1)$ such flags. Therefore $|\Delta_{Q}(C)|\leq(3n+n^{\prime})(q+1)$ proving the desired bound. ∎

4.4 Lemma**.**

Let $S_{1}$ and $S_{2}$ be solids of $\text{PG}(6,q)$ with $\dim(S_{1}\cap S_{2})\leq 1$ and let $P$ be a point that is not contained in $S_{1}\cup S_{2}$. Then the number of planes that contain $P$ and meet $S_{1}$ and $S_{2}$ non-trivially is at most $2q^{6}+2q^{5}+3q^{4}+2q^{3}+2q^{2}+q+1$.

Proof.

We have $d:=\dim(S_{1}\cap S_{2})\in\{0,1\}$. A line through $P$ meets $S_{1}$ and $S_{2}$ if and only if it meets one and hence both of the subspaces $U_{1}:=\langle P,S_{2}\rangle\cap S_{1}$ and $U_{2}:=\langle P,S_{1}\rangle\cap S_{2}$, that is, if the line is contained in the subspace $V:=\langle U_{1},P\rangle$. The subspaces $U_{1}$ and $U_{2}$ have the same dimension $u$ where $u=1$ if $d=0$ and $u\in\{1,2\}$ when $d=1$. We have $\dim(V)=u+1$.

A plane on $P$ that meets $V$ only in $P$ is spanned by $P$, a point of $S_{1}\setminus U_{1}$ and a point of $S_{2}\setminus U_{2}$, so there are $(s(3)-s(u))^{2}$ such planes. The number of planes on $P$ that meet $V$ in a line is equal to the number $s(0,1,u+1)$ of lines of $V$ on $P$ multiplied with the number $s(1,2,6)-s(1,2,u+1)$ of planes that meet $V$ in a given line. Finally there are $s(0,2,u+1)$ planes on $P$ that are contained in $V$. Hence, the total number of planes on $P$ that meet $S_{1}$ and $S_{2}$ non-trivially is equal to

[TABLE]

The larger value occurs for $u=2$ and gives the bound in the lemma. ∎

4.5 Lemma**.**

Let $P_{1}$, $P_{2}$ and $P_{3}$ be non-collinear points of $\text{PG}(6,q)$. Suppose that for $i\in\{1,2\}$ and $r\in\{1,2,3\}$ there exist flags $f_{i,r}=(E_{i,r},S_{i,r})\in\Delta_{P_{i}}(C)$ such that

1. $\bullet$

$\forall r,s\in\{1,2,3\}:\dim(S_{1,r}\cap S_{2,s})\leq 1$* and* 2. $\bullet$

$\forall i\in\{1,2\},\forall\{r,s,t\}=\{1,2,3\}:P_{3-i},P_{3}\notin\langle E_{i,r},E_{i,s}\rangle\cup S_{i,r}\text{ and }E_{i,r}\cap E_{i,s}=P_{i}$.

Then $|\Delta_{P_{3}}(C)|\leq 24q^{7}+54q^{6}+71q^{5}+67q^{4}+48q^{3}+33q^{2}+22q+11$.

Proof.

Because $C$ is independent we know that for every $(E,S)\in C$ and every $i\in\{1,2\}$ we have $S\cap E_{i,r}\neq\emptyset$ for all $r\in\{1,2,3\}$ or $E\cap S_{i,r}\neq\emptyset$ for at least one $r\in\{1,2,3\}$. For $i\in\{1,2\}$ Lemma 4.2 shows that the number of solids of $\text{PG}(6,q)$ that contain $P_{3}$ and meet $E_{i,1}$, $E_{i,2}$ and $E_{i,3}$ is at most

[TABLE]

For every flag $(E,S)\in\Delta_{P_{3}}(C)$ for which $S$ is not such a solid we know that $E$ is a plane that meets $S_{1,r}$ and $S_{2,s}$ for some $r,s\in\{1,2,3\}$. For any choice of $r,s\in\{1,2,3\}$, Lemma 4.4 shows that there exist at most

[TABLE]

planes on $P_{3}$ that meet $S_{1,r}$ and $S_{2,s}$. Since every plane and every solid occurs in at most $q+1$ flags of $C$, it follows that $|\Delta_{P_{3}}(C)|\leq(2m+9n)(q+1)$, as claimed. ∎

4.6 Proposition**.**

Let $C$ be an independent set of the Kneser graph of type $\{2,3\}$ in $\text{PG}(6,q)$ with $q\geq 7$ that has the property that every plane and every solid of $\text{PG}(6,q)$ is contained in at most $s(1)$ flags of $C$. Then

[TABLE]

Proof.

Let $P_{1}$ and $P_{2}$ be distinct points of $\text{PG}(6,q)$ such that $|\Delta_{P_{1}}(C)|,|\Delta_{P_{2}}(C)|\geq|\Delta_{P}(C)|$ for all points $P\neq P_{1}$. If every flag $(E,S)\in C$ satisfies $\langle P_{1},P_{2}\rangle\cap E\neq\emptyset$, then

[TABLE]

since there are $s(2,6)-s(1,-1,2,6)$ planes that meet the line $\langle P_{1},P_{2}\rangle$ and since every plane lies in at most $q+1$ flags of $C$. Therefore, we may assume that $C$ contains a flag $f=(E,S)$ with $\langle P_{1},P_{2}\rangle\cap E=\emptyset$ and thus $\dim(S\cap\langle P_{1},P_{2}\rangle)\leq 0$. Every flag $(E^{\prime},S^{\prime})\in C$ either satisfies $E^{\prime}\cap S\neq\emptyset$ or $E^{\prime}\cap S=\emptyset\neq S^{\prime}\cap E$. Lemma 3.3 shows that at most

[TABLE]

flags $(E^{\prime},S^{\prime})$ of $C$ satisfy $E^{\prime}\cap S=\emptyset\neq S^{\prime}\cap E$. Before we count all flags $f^{\prime}=(E^{\prime},S^{\prime})$ with $E^{\prime}\cap S\neq\emptyset$ we note that we either have

[TABLE]

for all $P\in\text{PG}(6,q)\setminus\langle P_{1},P_{2}\rangle$ or

[TABLE]

If the second situation occurs, then Lemma 4.1 provides flags $f_{i,j}\in C$ for $i\in\{1,2\}$ and $j\in\{1,2,3\}$ required to apply Lemma 4.5 proving

[TABLE]

for all $P\in\text{PG}(6,q)\setminus\langle P_{1},P_{2}\rangle$. Since the bound in equation (7) is weaker than the bound given in equation (6) we know that it also holds in the first case. In particular, equation (7) holds for all $P\in S\setminus(S\cap\langle P_{1},P_{2}\rangle)$. Note that we chose $f$ such that $S\cap\langle P_{1},P_{2}\rangle$ is at most a point. Now, if $\widehat{P}:=S\cap\langle P_{1},P_{2}\rangle\neq\emptyset$, then, since $P_{1}$ and $P_{2}$ are distinct, there is an index $i\in\{1,2\}$ such that $\widehat{P}\neq P_{i}$ and, using the flags $f_{i,1}$, $f_{i,2}$ and $f_{i,3}$, we may apply Lemma 4.3 to see that

[TABLE]

which is weaker than the bound in equation (7) for $q\geq 7$. Therefore, the number of all flags $(E^{\prime},S^{\prime})$ of $C$ with $E^{\prime}\cap S\neq\emptyset$ is at most

[TABLE]

Together with the upper bound in equation (5) for the remaining flags of $C$, this provides the claimed upper bound on the cardinality of $C$. ∎

5 The second special case

In this section we study independent sets of the Kneser graph of type $\{2,3\}$ in $\text{PG}(6,q)$ with the property that every plane and every solid of $\text{PG}(6,q)$ occurs in at most $q^{2}+q+1$ flags of $C$.

5.1 Lemma**.**

Let $E$ be a plane and suppose that the solids $S$ with $(E,S)\in C$ span a subspace $H$ of dimension at least $5$. Suppose also that every plane of $\text{PG}(6,q)$ occurs in at most $s(2)=q^{2}+q+1$ flags of $C$. Then the number of flags $(E^{\prime},S^{\prime})\in C$ with $E^{\prime}\cap E=\emptyset$ is at most $s(1,4)\cdot s(2)\cdot(s(2)+1)$.

Proof.

Let $M$ be the set consisting of all flags $(E^{\prime},S^{\prime})$ of $C$ such that $S^{\prime}\cap E=\emptyset$ and let $N$ be the set consisting of all flags $(E^{\prime},S^{\prime})$ of $C$ such that $S^{\prime}\cap E\not=\emptyset$ and $E^{\prime}\cap E=\emptyset$. Every flag $(E^{\prime},S^{\prime})\in C$ with $E^{\prime}\cap E=\emptyset$ lies in $M\cup N$. Lemma 3.3 applied with $\xi=s(2)$ shows that $|N|\leq s(1,4)\cdot s(2)^{2}$. For an upper bound on the number of flags in $M$, we let $\mathcal{E}$ denote the set of all planes that occur in a flag of $M$. The hypothesis of this lemma shows that $|M|\leq|\mathcal{E}|\cdot s(2)$. In order to prove the statement, it remains to show that $|\mathcal{E}|\leq s(1,4)$.

Consider $E^{\prime}\in\mathcal{E}$. Let $S^{\prime}$ be a solid with $(E^{\prime},S^{\prime})\in M$. Then $S^{\prime}\cap E=E^{\prime}\cap E=\emptyset$. Since $C$ is independent and since $S^{\prime}\cap E=\emptyset$, then $E^{\prime}$ meets every solid $S$ for which $(E,S)\in C$. Then every such solid $S$ is spanned by $E$ and a point of $E^{\prime}$, so $H\subseteq\langle E,E^{\prime}\rangle$. Since $H$ has dimension at least $5$, it follows that $H$ has dimension 5 and that $H=\langle E,E^{\prime}\rangle$ for all $E^{\prime}\in\mathcal{E}$. Lemma 3.1 shows that $|\mathcal{E}|\leq s(1,4)$. ∎

5.2 Proposition**.**

Let $C$ be an independent set of the Kneser graph of type $\{2,3\}$ in $\text{PG}(6,q)$ with $q\geq 8$ that has the property that every plane and every solid of $\text{PG}(6,q)$ occurs in at most $s(2)$ flags of $C$. Then

[TABLE]

Proof.

Let $\mathcal{E}$ be the set consisting of all planes of $\text{PG}(6,q)$ that lie in at least $q+2$ flags of $C$, and let $\mathcal{S}$ be the set consisting of all solids of $\text{PG}(6,q)$ that lie in at least $q+2$ flags of $C$. We distinguish three cases.

Case 1. We assume that $|\mathcal{E}|\leq s(4)$ and $|\mathcal{S}|\leq s(4)$. In this case we choose a subset $C^{\prime}$ of $C$ such that every plane and every solid of $C^{\prime}$ lies in at most $q+1$ flags of $C^{\prime}$. Since every plane and solid lies in at most $q^{2}+q+1$ flags of $C$, we can find such a subset with $|C^{\prime}|\geq|C|-(|\mathcal{E}|+|\mathcal{S}|)q^{2}$ and then $|C|\leq|C^{\prime}|+2\cdot s(4)\cdot q^{2}$. Now the statement follows by applying Proposition 4.6 to $C^{\prime}$.

Case 2. We assume that $|\mathcal{E}|>s(4)$. Lemma 2.2 proves the existence of planes $E_{1},E_{2}\in\mathcal{E}$ satisfying $\dim(E_{1}\cap E_{2})\leq 0$. From Lemma 5.1 we know that at most

[TABLE]

flags $(E,S)\in C$ satisfy $E\cap E_{1}=\emptyset$ or $E\cap E_{2}=\emptyset$. It remains to find an upper bound on the number of flags in $C$ whose planes meet both $E_{1}$ and $E_{2}$. Therefore, we count the number of planes of $\text{PG}(6,q)$ that meet $E_{1}$ and $E_{2}$. First consider the case that $E_{1}\cap E_{2}$ is a point $Q$. In this case there are $s(0,2,6)$ planes on $Q$, there are $(s(2)-1)^{2}(s(1,2,6)-(2\cdot s(0,1,2)-1))$ planes that do not contain $Q$ and meet both $E_{1}$ and $E_{2}$ in exactly one point and there are $2\cdot s(0,-1,1,2)(s(2)-1)$ planes that do not contain $Q$ and meet $E_{1}$ or $E_{2}$ in a line and the other plane in a point. Thus, in this case the number of planes that meet $E_{1}$ and $E_{2}$ is equal to

[TABLE]

If $E_{1}$ and $E_{2}$ are skew than a similar calculation shows that there are even less than $n$ planes that meet $E_{1}$ and $E_{2}$, so that $n$ is an upper bound for the number of planes that meet $E_{1}$ and $E_{2}$ in both situations. Since every plane lies in at most $s(2)$ flags of $C$, it follows that there are at most $n\cdot s(2)$ flags $(E,S)\in C$ such that $E$ meets $E_{1}$ and $E_{2}$. Together with the count in equation (8) we find $|C|\leq n\cdot s(2)+2\cdot s(1,4)\cdot s(2)\cdot(s(2)+1)$ and this bound is better than the one in the statement.

Case 3. We assume that $|\mathcal{S}|>s(4)$. This is dual to Case 2. ∎

6 Proof of the theorem

In this section, $\Gamma$ denotes the Kneser graph of plane-solid flags in $\text{PG}(6,q)$ and $C$ denotes a maximal independent set of $\Gamma$.

6.1 Lemma**.**

1. i)

Every solid $S$ of $\text{PG}(6,q)$ has a subspace $U$ with the following property: For every plane $E$ of $S$ we have $(E,S)\in C$ if and only if $U\subseteq E$. 2. ii)

For every plane $E$ of $\text{PG}(6,q)$ there exists a subspace $U$ containing $E$ with the following property: For every solid $S$ on $E$ we have $(E,S)\in C$ if and only if $S\subseteq U$.

Proof.

Since the two statements are dual to each other, it suffices to prove the first statement. Thus consider a plane $E$ and let $\mathcal{S}$ be the set of solids $S$ satisfying $E\subseteq S$ and $(E,S)\in C$. In the quotient space $\text{PG}(6,q)/E$ the set $\{S/E\mid S\in\mathcal{S}\}$ is a set of points and we have to show that this set is a subspace of $\text{PG}(6,q)/E$. In that regard, it is sufficient to show for any two distinct solids $S_{1},S_{2}\in\mathcal{S}$ and every solid $S$ with $E\subseteq S\subseteq\langle S_{1},S_{2}\rangle$ we have $S\in\mathcal{S}$. Let $S$ be such a solid. If $(E^{\prime},S^{\prime})$ is any flag of $C$ then either $S^{\prime}\cap E\not=\emptyset$ or $E^{\prime}$ meets $S_{1}\setminus E$ and $S_{2}\setminus E$. In the second case $E^{\prime}$ meets $\langle S_{1},S_{2}\rangle$ in a line and hence $E^{\prime}$ meets $S$. Thus for every $(E^{\prime},S^{\prime})\in C$ we have $E\cap S^{\prime}\not=\emptyset$ or $E^{\prime}\cap S\not=\emptyset$. This shows that $C\cup\{(E,S)\}$ is an independent set of $\Gamma$ and since $C$ is a maximal independent set we have $(E,S)\in C$, that is, $S\in\mathcal{S}$. ∎

6.2 Definition**.**

A plane $E$ will be called saturated (for $C$) if $(E,S)\in C$ for all solids $S$ of $\text{PG}(6,q)$ that contain $E$. Dually, a solid $S$ will be called saturated (for $C$), if $(E,S)\in C$ for all planes $E$ of $S$.

6.3 Lemma**.**

1. i)

For every saturated solid $S$ and every flag $(E^{\prime},S^{\prime})\in C$ we have $E^{\prime}\cap S\neq\emptyset$. 2. ii)

If $S$ is a solid with $S\cap E^{\prime}\neq\emptyset$ for all flags $(E^{\prime},C^{\prime})$ of $C$, then $S$ is saturated. 3. iii)

If $S$ and $S^{\prime}$ are saturated solids, then $\dim(S\cap S^{\prime})\geq 1$. 4. iv)

Let $H$ be a hyperplane of $\text{PG}(6,q)$ and suppose that $E\subseteq H$ for all flags $(E,S)\in C$. Then every solid of $H$ is saturated.

Proof.

1. i)

Suppose that there is a flag $(E^{\prime},S^{\prime})\in C$ with $E^{\prime}\cap S=\emptyset$. Since $\text{PG}(6,q)$ has dimension $6$, it follows $S^{\prime}\cap S$ is a point $P$ with $P\notin E^{\prime}$. Let $E$ be a plane of $S$ with $P\notin E$. Then $E\cap S^{\prime}=\emptyset$. As $S$ is a saturated solid we have $(E,S)\in C$. But then $(E,S)$ and $(E^{\prime},S^{\prime})$ are flags of the independent set $C$ with $E\cap S^{\prime}=\emptyset$ and $E^{\prime}\cap S=\emptyset$, a contradiction. 2. ii)

Let $E$ be a plane of $S$. We have to show that $(E,S)\in C$. Since $S\cap E^{\prime}\not=\emptyset$ for every flag $(E^{\prime},S)$ of $C$, the set $C\cup\{(E,S)\}$ is independent. Maximality of $C$ implies $(E,S)\in C$. 3. iii)

Assume to the contrary that $S$ and $S^{\prime}$ only meet in a point $P$. Choose planes $E$ of $S$ and $E^{\prime}$ of $S^{\prime}$ with $P\notin E,E^{\prime}$. Then $S\cap E^{\prime}=\emptyset=S^{\prime}\cap E$. Hence $(E,S)$ and $(E^{\prime},S^{\prime})$ are adjacent elements of the Kneser graph $\Gamma$. As $C$ is independent, this is a contradiction. 4. iv)

Let $S$ be a solid of $H$. The dimension formula shows that $S\cap E\not=\emptyset$ for all planes $E$ of $H$. Therefore part ii) shows that $S$ is saturated.∎

6.4 Lemma**.**

Let $C$ be a maximal independent set of $\Gamma$. If there are more than $c:=q^{7}+2q^{6}+2q^{5}+3q^{4}+2q^{3}+2q^{2}+q+1$ saturated solids for $C$, then $C=\Lambda(H,\mathcal{E})$ for some hyperplane $H$ and some maximal independent set $\mathcal{E}$ of the Kneser graph of planes of $H$ (cf. Example 1.1).

Proof.

Let $\mathcal{S}$ be the set of saturated solids in $\Pi_{3}(C)$. We have $c>q^{6}+2q^{5}+3q^{4}+3q^{3}+2q^{2}+q+1$ and according to Corollary 6.3 iii) we have $\dim\left(S_{1}\cap S_{2}\right)\geq 1$ for all $S_{1},S_{2}\in\mathcal{S}$. Result 2.3 shows that there exists a hyperplane $H$ containing all saturated solids. If there would be a flag $(E,S)\in C$ such that $E\not\leq H$, then according to Lemma 6.3 i) all solids of $\mathcal{S}$ would have non-empty intersection with the line $E\cap H$ and thus

[TABLE]

which is a contradiction. This shows that $E\subseteq H$ for all planes $E\in\Pi_{2}(C)$. Lemma 6.3 iv) shows that all solids of $H$ are saturated. This means that $\Lambda(H,\emptyset)\subseteq C$. Proposition 3.2 now proves the statement. ∎

6.5 Theorem**.**

Suppose that $q\geq 8$ and that $C$ is a maximal independent set in $\Gamma$ with

[TABLE]

Then $C=\Lambda(H,\mathcal{E})$ for a hyperplane $H$ and a maximal set $\mathcal{E}$ of mutually intersecting planes of $H$, or $C=\Lambda(P,\mathcal{S})$ for a point $P$ and a maximal set $\mathcal{S}$ of solids on $P$ any two of which share at least a line.

Proof.

The class of examples described in Example 1.1 is closed under duality. In view of Lemma 6.4 we may assume that there exists at most $c:=q^{7}+2q^{6}+2q^{5}+3q^{4}+2q^{3}+2q^{2}+q+1$ saturated solids, and, dually, that there are at most $c$ saturated planes. For every saturated plane $E$ choose one hyperplane $H_{E}$ on $E$, and for every saturated solid $S$ choose a point $P_{S}$ of $S$. Let $C^{\prime}$ be the subset of $C$ that is obtained from $C$ by removing all flags $(E,S)$ such that $E$ is saturated and $S$ is not contained in $H_{E}$ and by removing all flags $(E,S)$ such that $S$ is saturated and $E$ does not contain $P_{S}$. Then $|C^{\prime}|\geq|C|-2cq^{3}$, that is

[TABLE]

Lemma 6.1 shows for every plane $E$ that is not saturated for $C$ has the property that the solids $S$ with $(E,S)\in C$ span a proper subspace of $\text{PG}(6,q)$. Therefore the construction of $C^{\prime}$ implies that every plane $E$ has the property that the solids $S$ with $(E,S)\in C^{\prime}$ span a proper subspace of $\text{PG}(6,q)$. Consequently every plane of $\text{PG}(6,q)$ lies in at most $q^{2}+q+1$ flags of $C^{\prime}$. Dually, every solid $S$ of $\text{PG}(6,q)$ lies in at most $q^{2}+q+1$ flags of $C^{\prime}$. Therefore Proposition 5.2 proves an upper bound for $|C^{\prime}|$. Now (9) proves the bound for $|C|$ that is given in the statement. ∎

6.6 Corollary**.**

For $q>27$, the maximal independent set in the Kneser graph of flags of type $\{2,3\}$ in $\text{PG}(6,q)$ with $|C|\geq q^{11}+2q^{10}$ are the independent sets described in Example 1.1.

Theorem 1.3 follows from this Corollary and Proposition 3.2 iii).

7 Bounds on the chromatic number of $\Gamma$

Let $\Gamma$ be the Kneser graph of flags of type $\{2,3\}$ in $\text{PG}(6,q)$. The chromatic number of $\Gamma$ is the smallest number $\chi$ such that the vertex set can be represented as the union of $\chi$ independent sets. Using the upper bound $\alpha$ for the size of such an independent set this immediately gives the bound $\chi\geq\frac{n}{\alpha}$. With the upper bound from Theorem 1.3 we find

7.1 Proposition**.**

For $q>27$, the chromatic number of $\Gamma$ is at least $q^{4}-q^{2}+2q+1$.

On the other hand, if $V$ is a subspace of dimension $4$ of $\text{PG}(6,q)$, then the independent sets $\Lambda(P,\emptyset)$ with $P\in V$ comprise all vertices of $\Gamma$, so we have the trivial upper bound $\chi\leq s(4)=q^{4}+q^{3}+q^{2}+q+1$. We can slightly improve this bound.

7.2 Proposition**.**

The chromatic number $\chi$ of $\Gamma$ satisfies $\chi\leq q^{4}+q^{3}+q^{2}+1$.

Proof.

Consider a point $P$, a line $l$, a plane $E$ and a $4$-space $V$ that are mutually incident. Let $Q$ be a point of $V$ that is not in $E$. Let $l_{1},\dots,l_{q}$ be the lines of plane $\langle l,Q\rangle$ with $P\in l_{i}$ and $Q\notin l_{i}$, let $E_{1},\dots,E_{q}$ be the planes of $\langle E,Q\rangle$ with $l\subseteq E_{i}$ and $Q\notin E_{i}$, and let $S_{1},\dots,S_{q}$ be the solids of $V$ with $E\subseteq S_{i}$ and $Q\notin S_{i}$. For $i\in\{1,\dots,q\}$ put $M_{i}:=l_{i}\cup(E_{i}\setminus l)\cup(S_{i}\setminus E)$. Then $|M_{i}|=q^{3}+q^{2}+q+1$ with $M_{i}\cap M_{j}=P$ for distinct $i,j\in\{1,\dots,q\}$ and the union of the sets $M_{1},\dots,M_{q}$ is $\{P\}\cup V\setminus\langle P,Q\rangle$. Let $\{Q_{1},\dots,Q_{q}\}=\langle P,Q\rangle\setminus\{P\}$ and consider the independent set $\Lambda(X,\langle X,Q_{i}\rangle)$ for $X\in M_{i}$ and $i\in\{1,\dots,q\}$. Then for $i\in\{1,\dots,q\}$ all lines of $V$ on $Q_{i}$ occur in one of these sets and every solid that contains $Q_{i}$ contains a line $\langle X,Q_{i}\rangle$ with $X\in M_{i}$. Therefore the union of the sets $\Lambda(X,\langle X,Q_{i}\rangle)$ for $i\in\{1,\dots,q\}$ covers all vertices of $\Gamma$. ∎