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Extremal combinatorics in generalized Kneser graphs

Citation for published version (APA):Mussche, T. J. J. (2009). Extremal combinatorics in generalized Kneser graphs. Eindhoven: TechnischeUniversiteit Eindhoven. https://doi.org/10.6100/IR642440

DOI:10.6100/IR642440

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https://doi.org/10.6100/IR642440

https://doi.org/10.6100/IR642440

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Extremal combinatorics in generalized Kneser graphs

PROEFSCHRIFT

ter verkrijging van de graad van doctor aan deTechnische Universiteit Eindhoven, op gezag van de

Rector Magnificus, prof.dr.ir. C.J. van Duijn, voor eencommissie aangewezen door het College voor

Promoties in het openbaar te verdedigenop donderdag 16 april 2009 om 16.00 uur

door

Tim Joris Jacqueline Mussche

geboren te Eeklo, Belgie

Dit proefschrift is goedgekeurd door de promotoren:

prof.dr. A.E. Brouwerenprof.dr. A.M. Cohen

Copromotor:dr. A. Blokhuis

Contents

Contents 3

1 Introduction 71.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.1 Basic definitions . . . . . . . . . . . . . . . . . . . . . 91.2.2 Graph colorings and chromatic numbers . . . . . . . . 111.2.3 Graph homomorphisms . . . . . . . . . . . . . . . . . 13

1.3 Finite projective spaces . . . . . . . . . . . . . . . . . . . . . 151.3.1 Finite fields . . . . . . . . . . . . . . . . . . . . . . . . 151.3.2 Projective spaces . . . . . . . . . . . . . . . . . . . . . 171.3.3 The projective space PG(n, q) . . . . . . . . . . . . . . 221.3.4 Some counting in PG(n, q) . . . . . . . . . . . . . . . 24

1.4 Finite polar spaces . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Combinatorics and q-analogues 272.1 An easy example . . . . . . . . . . . . . . . . . . . . . . . . . 272.2 Sperner’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 28

2.2.1 Original problem . . . . . . . . . . . . . . . . . . . . . 282.2.2 q-analogue . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.3 Erdos-Ko-Rado theorem . . . . . . . . . . . . . . . . . . . . . 312.3.1 Original problem . . . . . . . . . . . . . . . . . . . . . 312.3.2 q-Analogue . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Bollobas’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 332.4.1 Set version . . . . . . . . . . . . . . . . . . . . . . . . 332.4.2 Set vs. subspace version . . . . . . . . . . . . . . . . . 362.4.3 Subspace version . . . . . . . . . . . . . . . . . . . . . 36

2.5 The Hilton-Milner theorem . . . . . . . . . . . . . . . . . . . 362.5.1 Original problem . . . . . . . . . . . . . . . . . . . . . 36

3

4 CONTENTS

2.5.2 The q-analogue . . . . . . . . . . . . . . . . . . . . . . 382.6 Small maximal cliques . . . . . . . . . . . . . . . . . . . . . . 53

3 The Kneser and q-Kneser graphs 593.1 Definitions and properties . . . . . . . . . . . . . . . . . . . . 593.2 Homomorphisms. . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.2.1 Homomorphisms between Kneser graphs . . . . . . . . 623.2.2 Homomorphisms between q-Kneser graphs . . . . . . . 643.2.3 A homomorphism from a q-Kneser graph into a Kneser

graph . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.3 Chromatic numbers . . . . . . . . . . . . . . . . . . . . . . . 65

3.3.1 The Kneser graphs . . . . . . . . . . . . . . . . . . . . 653.3.2 The q-Kneser graphs . . . . . . . . . . . . . . . . . . . 68

4 A family of point-hyperplane graphs 774.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.2 PH(2, q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 834.3 PH(3, q) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5 Polar versions of the q-Kneser graphs 875.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875.2 Chromatic numbers . . . . . . . . . . . . . . . . . . . . . . . 89

5.2.1 KQ+

q (2m+ 2,m+ 1), m ≥ 2 even, a trivial case . . . . 895.2.2 KPq (n, 1) . . . . . . . . . . . . . . . . . . . . . . . . . . 905.2.3 ΓPq (n, 2) and KPq (n, 2), where P has rank 2 . . . . . . 94

6 Generalized Kneser graphs 996.1 Chevalley groups . . . . . . . . . . . . . . . . . . . . . . . . . 99

6.1.1 Tits systems . . . . . . . . . . . . . . . . . . . . . . . 996.1.2 Coxeter systems . . . . . . . . . . . . . . . . . . . . . 1016.1.3 Root systems . . . . . . . . . . . . . . . . . . . . . . . 1016.1.4 Chevalley groups . . . . . . . . . . . . . . . . . . . . . 103

6.2 Generalized Kneser graphs . . . . . . . . . . . . . . . . . . . . 1056.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 1056.2.2 Parameters . . . . . . . . . . . . . . . . . . . . . . . . 105

6.3 Chromatic numbers in the thin An(1) case . . . . . . . . . . . 1146.4 Chromatic numbers in the thick An case . . . . . . . . . . . . 121

CONTENTS 5

A Parameters of generalized Kneser graphs 123A.1 The case Jw0 = J . . . . . . . . . . . . . . . . . . . . . . . . . 123A.2 The case Jw0 6= J . . . . . . . . . . . . . . . . . . . . . . . . . 143

Bibliography 148

Index 153

Samenvatting 157

Summary 159

Acknowledgements 161

6 CONTENTS

Chapter 1

Introduction

In this first chapter we will introduce the basic notions used in this thesis.It is not the intention to give a complete, self contained description of everytopic. Instead the most important notions are defined and the most usefulresults are mentioned, often without proof. We refer the reader to the ci-tations with each topic for more background information and proofs of theresults.

First we introduce some notions of set theory. Then some graph theo-retic topics are mentioned. Since this thesis describes a way of translatingproblems in set theory and graph theory into finite geometry we need tointroduce the basic facts about projective spaces. In Chapter 5 we general-ize the construction over projective spaces of Chapter 3 to work over polarspaces, so we define here what polar spaces are and give some properties.

Unless stated otherwise all the objects used here are finite.

1.1 Sets

Take a set X with n elements. Such a set is also called an n-set. We canask ourselves how many subsets of X with k elements (k-subsets) there are,where 0 ≤ k ≤ n. This is of course a very easy calculation, but in a latersection we will use the same method to calculate some numbers about vectorspaces, so we will give the calculation anyway. To form such a k-subset wehave to choose k elements from X. For the first element we have n choices,for the second element n−1 choices, and so on. Finally for the k-th element

7

8 CHAPTER 1. INTRODUCTION

we have n− k+ 1 choices left. But now there are many choices that lead tothe same k-subset; indeed all choices of the same k points but in a differentorder give rise to the same subset. So, to get the correct number we have todivide by the number of permutations of k points, which is k!. So we have:

(the number of k-subsets of an n-set) =n(n− 1)(n− 2) · · · (n− k + 1)

k!

=n!

k!(n− k)!

=(n

k

)

Definition 1.1. The number of k-subsets of an n-set (with 1 ≤ k ≤ n) isgiven by the binomial coefficient

(nk

).

Definition 1.2. A set of sets F is called a set system or a family of sets.If all the elements of F are subsets of some set X, then F is called a familyover X, and X is called the universe of F . If all the members of F have sizek, then F is called a k-uniform family. F is called uniform if it is k-uniformfor some integer k.

We give two examples of special set systems.

Definition 1.3. A set system F is called a chain when for any two elementsone is a proper subset of the other.

Definition 1.4. A family of sets F is called an antichain if no set in Fis a proper subset of another set in F . Easy examples of such families areuniform families.Antichains are also named Sperner families after E. Sperner, who proved atight upper bound on the size of an antichain (see Theorem 2.2).

A set system can be viewed as a poset (or partially ordered set). Todefine this notion we need to define a partial order.

1.2. GRAPHS 9

Definition 1.5. A relation � on a set X is called a partial order on X if itsatisfies:

• Reflexivity: x � x for all x ∈ X,

• Antisymmetry: if x � y and y � x then x = y for all x, y ∈ X, and

• Transitivity: if x � y and y � z then x � z for all x, y, z ∈ X.

If x � y or y � x then x and y are called comparable with respect to �,otherwise they are called incomparable.

Definition 1.6. A poset (or partially ordered set) is an ordered pair (X,�)where X is a set, called the ground set, and � a partial order on X.

A chain in a poset is a subset of X whose elements are pairwise compa-rable. An antichain is a subset whose elements are pairwise incomparable.A maximal chain is a chain that cannot be extended to a larger chain.

It is clear to see that the relation ⊆ (“is subset of”) is a partial order.So the pair (F ,⊆), where F is a set system is a poset. It is easy to see thata(n) (anti-)chain in this poset is exactly what we called a(n) (anti-)chainbefore. The poset (2X ,⊆), where 2X is the set system containing all subsetsof X, is called the subset poset of X.

1.2 Graphs

1.2.1 Basic definitions

Definition 1.7. A (simple) graph Γ = (V,E) is a pair consisting of a set Vof vertices and a 2-uniform family E, called the edge set, with universe V ,whose members are called edges. Two vertices u, v ∈ V are called adjacentif {u, v} ∈ E and this edge is said to join the vertices u and v. In that case,u and v are also called the end points of the edge {u, v}. The neighbors of avertex are the vertices adjacent to that vertex. The degree of a vertex is thenumber of its neighbors. A graph in which each vertex has the same degreeis called regular .

We will sometimes use |Γ| as a notation for the number of vertices of Γ.


Definition 1.8. Two vertices u, v ∈ V are called joined if there are verticesu = u0, u1, . . . , um = v in V for a certain m ≥ 0 such that {ui, ui+1} is anedge for all 0 ≤ i ≤ m − 1. Such a sequence of adjacent vertices is calleda path between u and v. If all the ui’s are different, this is called a simplepath. If u and v are connected, the length of the shortest path betweenthem is called the distance and is denoted by d(u, v). A path of length atleast three without repeated vertices except that the two endpoints are thesame is called a cycle.

Being joined by a path is an equivalence relation on the set of vertices.The equivalence classes are called the connected components of the graph.A graph is called connected when it has precisely one connected component.Thus the empty graph (the graph with no vertices) is disconnected becauseit has no connected components.

Some subsets of the vertex set have special properties:

Definition 1.9. An independent set is a subset of the vertex set in whichno two vertices are adjacent. A clique is a subset of the vertex set in whichall pairs of vertices are adjacent. A complete graph is a graph where theedge set E contains all pairs of vertices in V . This means that all verticesare pairwise adjacent. The complete graph on n vertices is denoted by Kn.

Starting from one graph, there are many ways to construct other graphs.The most straightforward ways are taking subgraphs and taking the com-plement:

Definition 1.10. A graph Γ′ = (V ′, E′) is called a subgraph of a graphΓ = (V,E) if V ′ is a subset of V and all edges in E′ are also in E. A sub-graph is called induced if all unordered vertex pairs of Γ′ that are adjacentin Γ are also adjacent in Γ′.

Definition 1.11. The complement of a graph Γ = (V,E) is the graphΓ = (V,E) where the edge set E consists of all pairs of vertices that are notelements of E. In other words, adjacent pairs in Γ are not adjacent in Γ andvice versa.

It is clear that the complement of a clique is an independent set andthat the complement of an independent set is a clique. For that reason, anindependent set is also called a coclique. The complement of the complete

1.2. GRAPHS 11

graph Kn has n vertices and no edges and is called an edgeless graph.

To each graph we can associate certain numbers, called graph parameters.Some well-known examples are the independence number and the cliquenumber.

Definition 1.12. The independence number α(Γ) of a graph Γ is the sizeof the largest independent set in Γ. The clique number ω(Γ) is the size ofthe largest clique in Γ. Again it is clear that α(Γ) = ω(Γ) and ω(Γ) = α(Γ).

1.2.2 Graph colorings and chromatic numbers

Definition 1.13 (Graph coloring). A vertex coloring of a graph Γ = (V,E)is a partition of the vertex set. The color of a vertex is determined by theblock of the partition that the vertex is in. A proper (vertex) coloring is avertex coloring such that the end points of an edge have a different color.From now on, if the term (graph) coloring is used without further qualifi-cation, we are referring to a proper vertex coloring of a graph. Colors willusually be denoted by integers.

We can now define some other graph parameters:

Definition 1.14. The chromatic number χ(Γ) of a graph Γ is the minimumnumber of colors needed to color Γ. A coloring using only χ(Γ) colors iscalled a minimal coloring of Γ.

Because all vertices of the same color form an independent set, there is avery easy connection between the independence number and the chromaticnumber of a graph:

Proposition 1.15. For each graph Γ we have that

|Γ| ≤ α(Γ)χ(Γ).

Graph colorings are generalized by multiple graph colorings:


Definition 1.16 (Multiple coloring). A k-fold coloring of a graph Γ for apositive integer k is an assignment of exactly k colors to each vertex of Γsuch that two adjacent vertices have no colors in common.

A connected graph Γ with at least two vertices must have at least oneedge. In a k-fold coloring of Γ with c colors, the endpoints of this edgecannot have a color in common, so we have:

Proposition 1.17. If a k-fold coloring of a connected graph Γ exists withc colors, then:

• c = mk if Γ is an m-clique,

• c ≥ mk if Γ contains an m-clique.

Associated with these multiple colorings is the multiple chromatic num-ber:

Definition 1.18. The k-fold chromatic number χk(Γ) of a graph Γ for apositive integer k is the minimum number of colors needed for a k-fold col-oring of Γ. Such a coloring is called a minimal k-fold coloring of Γ.

It is obvious that a 1-fold coloring of a graph is just a proper vertexcoloring.

It is easy to see that χk(Γ) is subadditive:

Proposition 1.19. For each graph Γ and positive integers k1, k2 we have:

χk1+k2(Γ) ≤ χk1(Γ) + χk2(Γ).

Proof. Let C1 be a minimal k1-fold coloring of Γ and C2 a minimal k2-foldcoloring such that C1 and C2 have no colors in common. Now each vertex ofΓ is colored by exactly k1 colors of C1 and exactly k2 colors of C2, so we havea (k1 + k2)-fold coloring of Γ and hence an upper bound on χk1+k2(Γ).

Definition 1.20 (Fractional chromatic number). The fractional chromaticnumber χF (Γ) of a graph Γ is defined as:

χF (Γ) = limk→∞

χk(Γ)k

1.2. GRAPHS 13

The existence of this limit is guaranteed by the previous proposition andthe following lemma by M. Fekete:

Lemma 1.21 (M. Fekete (1923) [20]). If a sequence of real numbers {an}satisfies the subadditivity condition, then

limn→∞

ann

= infn

ann.

For the fractional chromatic number of a graph, this means:

χF (Γ) = infk

χk(Γ)k

.

We state some properties regarding fractional chromatic numbers:

Proposition 1.22 ([43]). Given a graph Γ we have that:

(i) χF (Γ) is a rational number,

(ii) there is a positive integer k such that χF (Γ) = χk(Γ)k , and

(iii) χF (Γ) ≥ |Γ|α(Γ) , with equality if Γ is vertex-transitive1.

Note that the second property means that the fractional chromatic num-ber of a graph Γ is actually the minimum of all χk(Γ)

k .

1.2.3 Graph homomorphisms

Homomorphisms from a graph Γ into certain other graphs can give infor-mation about the parameters of Γ. We will give the definition of a graphhomomorphism and show what “target graphs” give information about thevarious chromatic numbers.

1A graph is vertex-transitive if the automorphism group of the graph is transitive onits vertices


Definition 1.23 (Graph Homomorphism). Consider the graphs Γ = (V,E)and Γ′ = (V ′, E′). A (graph) homomorphism η : Γ → Γ′ is a map from Vto V ′ such that adjacent vertices of Γ are mapped to adjacent vertices of Γ′.This implies that the fibers2 over the vertices in Γ′ are independent sets of Γ.

If the map is injective, we call the homomorphism an embedding . Asurjective embedding whose inverse is also a homomorphism is called anisomorphism. Two graphs that have an isomorphism between them arecalled isomorphic.

If there is a homomorphism from Γ to Γ′ we will note this as follows:Γ → Γ′. If there is an embedding from Γ into Γ′ the notation becomes:Γ ↪→ Γ′. We will write Γ ∼= Γ′ if Γ is isomorphic to Γ′.

Graph homomorphisms can be used to obtain bounds on the variouschromatic numbers of a graph. Suppose there is a homomorphism η : Γ →Γ′, and take a (k-fold) coloring of Γ′ (for a positive integer k). Now give thevertices of Γ the same color(s) as their images under η. It is clear that thisyields a proper (k-fold) coloring of Γ. This shows the following proposition:

Proposition 1.24. If Γ→ Γ′, then χk(Γ) ≤ χk(Γ′) for all positive integersk.

A coloring of a graph Γ with c colors can be seen as a homomorphismΓ → Kc. Indeed, number the colors m1,m2, . . . ,mc and the vertices of Kc

v1, v2, . . . , vc, and define for all vertices x of Γ:

η(x) = vi where mi is the color of x.

Two adjacent vertices of Γ must have a different color, so they will bemapped to different vertices of Kc, and those are adjacent, so η defines ahomomorphism.

Conversely every homomorphism of a graph into a complete graph canbe seen as a coloring of the graph with as many colors as vertices of thecomplete graph. So we have:

Proposition 1.25. A coloring of a graph Γ with c colors is equivalent to ahomomorphism from Γ to the complete graph Kc.

2inverse images

1.3. FINITE PROJECTIVE SPACES 15

With this in mind we can give an alternative definition of the chromaticnumber of a graph:

Proposition 1.26. χ(Γ) = min{c such that Γ→ Kc}.

A natural question is now whether we can do the same for a multiplecoloring. Take a k-fold coloring of a graph Γ with n colors. Now each vertexis colored by exactly k out of n colors, so consider the graph we will denoteK(n, k), whose vertices are all the k-subsets out of the n-set of used colors.We know that two adjacent vertices of Γ cannot share a color, so if we makeall pairs of vertices of K(n, k) that are disjoint as subsets adjacent, there isa natural homomorphism from Γ into K(n, k).

Again, the converse is also clear, so we have the following equivalence:

Proposition 1.27. A k-fold coloring of a graph Γ with n colors is equiva-lent to a homomorphism from Γ into the graph K(n, k).

As a consequence:

Proposition 1.28. For each positive integer k we have:

χk(Γ) = min{n such that Γ→ K(n, k)}.

This graph K(n, k) is called the Kneser graph, and we will say moreabout it in Chapter 3.

1.3 Finite projective spaces

1.3.1 Finite fields

Definition 1.29 (Group). A group (G, �) is a set G with a binary operation� such that

(i) for all a, b ∈ G: a � b is an element in G,

(ii) for all a, b, c ∈ G: (a � b) � c = a � (b � c),


(iii) there is an element e ∈ G such that a� e = e�a = a for all a ∈ G (thiselement is called the identity element), and

(iv) for each a ∈ G, there is an inverse element a′ such that a�a′ = a′�a = e.

A group G for which a � b = b � a for all a, b ∈ G is called a commutativegroup.The number of elements of the group is called the order of the group.

Definition 1.30 (Field). A field (F,+, ·) is a set F together with two binaryoperations + and · such that

(i) (F,+) is a commutative group with identity element 0 (called theadditive group),

(ii) (F∗, ·) is a commutative group with identity element 1 (called the mul-tiplicative group), and

(iii) for all a, b, c ∈ F: a · (b+ c) = a · b+ a · c and (a+ b) · c = a · c+ b · c

where F∗ = F \ {0}.The number of elements of the field is called the order of the field.

Remark 1.31. A set F with binary operations + and · that has the sameproperties as above, except that the group (F∗, ·) is not commutative, iscalled a division ring .

Most of the times we will write just F instead of (F,+, ·).

Some well-known fields are the field of rationals Q, the field of real num-bers R and the field of complex numbers C. Note that they all have infiniteorder.

A field with finite order is called a finite field . The following theorem isa well-known fact:

Theorem 1.32. If F is a finite field with order q, then q = ph where p is aprime number and h ≥ 1. Moreover this field is unique (up to isomorphism)with this order.

We will denote the finite field of order q by GF(q).


Remark 1.33. If p is a prime then GF(p) is actually the set of integersmodulo p, with the operations defined by performing the operations in Zand taking the result modulo p. This field is also denoted by Z/pZ or Fp.

Remark 1.34. The (little) theorem of Wedderburn states that all finitedivision rings are in fact fields.

Definition 1.35. The characteristic of a field F is the smallest positive in-teger k, if it exists, such that a+ a+ . . .+ a = 0 (k repeated terms) for alla ∈ F and 0 otherwise. It is denoted by char(F ).

It is known that char(Q) = char(R) = char(C) = 0 and char(GF(q)) = pwhere q = ph with p a prime.

Definition 1.36. A subfield (F′,+, ·) of a field (F,+, ·) is a subset F′ ofF that, together with the additive and multiplicative operations of F againforms a field.

For example: the field of rationals is a subfield of the field of reals, whichis again a subfield of the complex numbers. For the finite fields we have thefollowing:

Proposition 1.37. GF(q1) is a subfield of GF(q2) if and only if q1 = ph1

and q2 = ph2, for some prime p and h1|h2.

1.3.2 Projective spaces

We will introduce projective spaces using their axiomatic description. Afterthat, we will introduce the models of projective spaces that we will workwith for the rest of this thesis.

First we need to know what a point-line incidence structure is:

Definition 1.38. A point-line incidence structure (sometimes also called apoint-line geometry) (P,B, I) consists of two disjoint sets P and B. Theelements of P are called points, and P is called the point set. The elementsof B are called lines, and B is called the line set. I is a symmetric relationI ⊂ (P × B) ∪ (B × P), called the incidence relation.


If (p, L) ∈ I for a point p and a line L we can say that p is incident withL (and because I is symmetric, also the other way around). Most of thetime we will use the more intuitive expressions “the point p lies on the lineL” and “the line L goes through the point p”. Similarly, we will say thattwo lines intersect in a point if they are both incident with that point andthat a line connects two points if those points are both incident with the line.

With some axioms, we can define a projective space:

Definition 1.39. A projective space is an incidence structure S = (P,B, I)that satisfies the following axioms:

(i) Any two distinct points p and q are connected by exactly one line (andwe can denote this line by pq),

(ii) for any four distinct points a, b, c and d such that the line ab intersectscd we have that the line ac intersects bd, and

(iii) any line is incident with at least three points.

Now consider a line L of a projective space S = (P,B, I) and the setP ′, the set of all points of P incident with L. Define the incidence structureL = (P ′, {L}, I ′), where I ′ is the restriction of I to (P ′×{L})∪ ({L}×P ′).It is clear that L is also a projective space, called a projective line. A pro-jective space that is no projective line is called a non-degenerate projectivespace.

A projective plane is a projective space S = (P,B, I) that also satisfiesa stronger version of axiom (ii) and a fourth axiom:

(i) Any two distinct points p and q are connected by exactly one line (andwe can denote this line by pq),

(ii∗) any two lines intersect in at least one point,

(iii) any line is incident with at least three points, and

(iv) there are at least two lines.

Note that axiom (ii*) also holds for projective lines, that is why we needthe extra axiom (iv).

Remark 1.40. There is an equivalent axiom system for projective planes:


(i) Any two distinct points are connected by exactly one line,

(ii′) any two distinct lines intersect in exactly one point, and

(iii′) there exist four distinct points such that no three of those points areincident with the same line.

Now take a projective space S = (P,B, I). Because two points deter-mine exactly one line, we can identify a line with the set of points incidentwith the line. Hence we can think of lines as subsets of P.

Definition 1.41. A set A ⊂ P is called linear if every line meeting A inat least two points is completely contained in A. A linear set containing atleast two points must therefore contain at least one line. Denote the set oflines contained in A by B′, then it is not hard to see that S(A) := (A,B′, I ′)is a non-degenerate projective space or a line (here I ′ is the incidence re-lation I restricted to the set A). The incidence structure S(A) is called a(linear) subspace of S.

Like we identified lines with their point sets we can also identify everysubspace by its point set and refer to both of them as subspace. It is obviousthat the empty set, a singleton of P, a line and the set P itself are examplesof subspaces. As every subset of P is clearly contained in at least onesubspace, we can define the span of an arbitrary subset of P:

Definition 1.42. The span of a set B ⊆ P, denoted by 〈B〉 is defined as:

〈B〉 =⋂{C|B ⊆ C,C is a linear set}.

It is clear that 〈B〉 is always a subspace.

Note that for two sets A,B ∈ P we will use 〈A,B〉 as a notation for〈A ∪ B〉.

A set of points A ⊂ P is called linearly independent if for any subsetA′ ⊂ A and point p ∈ A \A′, we have p 6∈ 〈A′〉. In other words: 〈A′〉

⋂A =

A′.


Definition 1.43. A basis of a projective space S is a linearly independentset of points that spans the entire space. It is not hard to show that everybasis has the same number of points. This number is called the rank of Sand will be denoted by rk(S).

One can define the rank of a subspace in exactly the same way. Takea point p. It is obvious that 〈{p}〉 = {p}, therefore {p} is a basis for thesubspace with {p} as point set, hence a point is a rank-1 subspace. A lineis spanned by two distinct points and the set consisting of those two pointsis linearly independent, so the rank of a line is 2. A projective plane can beshown to have rank 3 and every rank-3 subspace of a projective space canbe shown to be a projective plane. The empty set will be considered as asubspace of rank 0.

We will define the projective dimension of a projective (sub)space S, tobe the rank of S minus one3. We denote this by dim(S). An m-dimensionalsubspace (that is, a space of rank m + 1) will also be called an m-space.In that way, the empty set is a (−1)-space, a point is a 0-space, a line a1-space and plane is a 2-space. If the projective space has dimension n, an(n− 1)-space in that space will be called a hyperplane.

The following formula is very useful to determine the dimension of thespan or the intersection of two subspaces:

Theorem 1.44 (Dimension formula). If U and V are two subspaces of aprojective space S, then

dim(〈U, V 〉) + dim(U ∩ V ) = dim(U) + dim(V ).

Now we will state two theorems that are important in characterizingprojective spaces. Consider a projective space S = (P,B, I) of dimensionat least 2.

Take any distinct points p1, p2, p3 and r1, r2, r3 of S for which the linesp1r1, p2r2 and p3r3 are concurrent in a point s, and such that no linepipj or rirj (for i, j ∈ {1, 2, 3} and i 6= j) contains s. Define the points

3The reason for that we want to conserve the intuition that a point has dimension 0,a line dimension 1, a plane dimension 2, etc.


tij := pipj ∩ rirj for i, j = 1, 2, 3, i < j. Now the projective space S iscalled Desarguesian if and only if t12, t13 and t23 are collinear for all possiblechoices of pi and ri. This configuration in a Desarguesian projective spaceis called a Desargues configuration. See Figure 1.1.

Figure 1.1: Desargues configuration

This criterion classifies all projective spaces of dimension at least 3:

Theorem 1.45. An n-dimensional projective space with n ≥ 3 is Desargue-sian.

A proof of this theorem is given in [2].

Now take two distinct lines L and M and points li ∈ L and mi ∈ Mfor i = 1, 2, 3 all different from L ∩M . Define the points tij := limj ∩ ljmi

for i, j = 1, 2, 3, i < j. The projective space S is called Pappian if andonly if t12, t13 and t23 are collinear for all possible choices of li and mi. Theresulting configuration in a Pappian projective space is called a Pappus con-figuration. See Figure 1.2.


Figure 1.2: Pappus configuration

A connection between Desarguesian and Pappian projective spaces isgiven in the following theorem which is proved in [26]:

Theorem 1.46. All Pappian projective spaces are also Desarguesian.

We will end the axiomatic description here and introduce the model wewill work with for the rest of this thesis.

1.3.3 The projective space PG(n, q)

Consider the (n + 1)-dimensional vector space V(n + 1,K) over the finitefield K. More generally one can take a left vector space over a divisionring. Define P as the set of 1-dimensional subspaces of V(n + 1,K) and Bas the set of 2-dimensional subspaces of V(n + 1,K). Define I to be thesymmetrised set theoretic inclusion. Now it is easy to check that (P,B, I)is a projective space of dimension n. This projective space will be denoted


by PG(n,K). It is clear that an (r+1)-dimensional subspace of V(n+1,K)becomes an r-dimensional subspace of PG(n, q). Thus the rank of a projec-tive subspace of PG(n,K) is equal to the dimension of the correspondingsubspace of the underlying vector space, whereas the (geometric) dimensionof a projective subspace is one less than the (vectorial) dimension of thecorresponding subspace of the underlying vector space. For that reason wewill also refer to the (vectorial) dimension of a subspace of the underlyingvector space as the rank of that subspace.

One can define PG(n,K) in a more practical but equivalent way. Twovectors x, y of V(n+ 1,K) \ {0} are called equivalent if and only x = ky forsome k ∈ K \{0}. Now the point set of PG(n,K) is just the set of all equiv-alence classes under this equivalence relation. The point that is the equiva-lence class of a vector x will be denoted by P (x), and x is called a coordinatevector of the point P (x). Points P (x1), . . . , P (xr) are linearly independentif the corresponding set of vectors is linearly independent in V(n+ 1,K). Asubspace of PG(n,K) of dimension r is a set of all points whose correspond-ing vectors form a (r + 1)-dimensional subspace of V(n+ 1, q).

The reason that PG(n,K) is highlighted here is given by the followingtheorem:

Theorem 1.47. Let S be a projective space. Then

(i) S = PG(n,K) for some division ring K if and only if S is Desargue-sian.

(ii) S = PG(n,K) for some field K if and only if S is Pappian.

From Theorem 1.45 it now follows that:

Theorem 1.48. If S = (P,B, I) is a projective space of dimension at least3, then S = PG(n,K) for some division ring K.

Note that there exist a lot of projective planes (projective spaces of di-mension two) that are not isomorphic to PG(n,K) for some division ring K.

If K is finite, it must be a field, according to the (little) Wedderburntheorem, hence K = GF(q) for some prime power q. In this case, theprojective space PG(n,K) is also denoted by PG(n, q). But now Theorem1.47 states that for finite projective spaces being Pappian is equivalent withthem being Desarguesian. From this fact and Theorem 1.48 one can concludethe following:


Theorem 1.49. If S = (P,B, I) is a finite projective space of dimensionat least 3, then S = PG(n, q) for some prime power q.

For the rest of this thesis we will only use finite projective spaces (andhence, over a finite field).

1.3.4 Some counting in PG(n, q)

In this section we will determine some numbers that will play a role in therest of this thesis.

A first useful question is determining the number of k-dimensional sub-spaces in PG(n, q) (where −1 ≤ k ≤ n). This is the number of rank-(k + 1)subspaces of a rank-(n+ 1) vector space.

So let us count the number of rank-k subspaces of V(n, q). We can usethe same method that we used in Section 1.1 to calculate the number ofk-subsets in an n-set. A rank-k subspace is spanned by k linearly indepen-dent vectors. For the first vector we have qn − 1 choices (the all-zero vectorcannot be chosen). The second vector cannot lie in the rank 1-subspacedefined by the first, so we have qn − q choices. For the third vector we haveqn − q2. And so on. Finally for the k-th vector we have qn − qk−1. But thek chosen vectors are not unique to span this rank-k subspace. So we haveto divide by the number of choices of vectors that span the same subspace.Using the same argument this is (qk−1)(qk−q) · · · (qk−qk−1). We thus havethat the number of rank-k subspaces in a rank-n vector space over GF(q) is:

(qn − 1)(qn − q) · · · (qn − qk−1)(qk − 1)(qk − q) · · · (qk − qk−1)

=[nk

]q

Definition 1.50 (Gaussian coefficient). The number of rank k-subspacesin a rank n-vector space over GF(q) is given by the Gaussian coefficient

[nk

]q.

We can restate this in terms of (projective) dimensions:The number of k-subspaces in an n-dimensional projective space over

GF(q) is given by[n+1k+1

]q.

1.4. FINITE POLAR SPACES 25

Another useful piece of combinatorial information is the number of k-dimensional subspaces containing a fixed l-dimensional subspace in PG(n, q),for −1 ≤ l < k < n. This is the same as the number of rank-(k+1) subspacescontaining a fixed rank-(l+ 1) subspace in V(n+ 1, q). Now this is the sameas the number of rank-(k − l) subspaces in the quotient V(n + 1, q)/V(l +1, q) which is isomorphic to V(n − l, q). So we have that the number of k-dimensional subspaces containing a fixed l-dimensional subspace in PG(n, q)is the same as the number of (k − l − 1)-dimensional subspaces in PG(n −l − 1, q), which is

[n−lk−l

]q.

1.4 Finite polar spaces

In the previous section we encountered a first kind of incidence structure,namely the projective spaces (and planes). In this section we will defineanother type: the polar spaces.

Definition 1.51. A (non degenerate) polar space of rank n (n ≥ 2) consistsof a point set P together with a family of subsets of P, called the singularsubspaces, satisfying the following axioms:

(i) A singular subspace, together with the singular subspaces it containsis a k-dimensional projective space, for some −1 ≤ k ≤ n − 1. Thisdimension is by definition also the dimension of the singular subspaceof the polar space.

(ii) The intersection of two singular subspaces is again a singular subspace.

(iii) Take a singular subspace U of dimension n− 1 and a point p ∈ P \U .There is exactly one singular subspace V such that p ∈ V and U ∩ Vhas dimension n− 2. The singular subspace V contains all the pointsof U that lie on a common line with p.

(iv) There are at least two disjoint singular subspaces of dimension n− 1.

Note that an incidence structure that satisfies axioms (i),(ii) and (iii) butnot (iv) is called a degenerate polar space.The singular subspaces of (maximal) dimension n− 1 are called the gener-ators of the polar space.A finite polar space is a polar space with a finite point set.

As with the projective spaces, all polar spaces of rank at least 3 havebeen classified. In the (thick) finite case all polar spaces of rank at least 3


are classical polar spaces. Here is a list of all the finite classical polar spaceswith their ranks:

• Q+(2n+1, q) non-singular hyperbolic quadric in PG(2n+1, q) for somen ≥ 1, giving a hyperbolic polar space of rank n+ 1.

• Q(2n, q): non-singular parabolic quadric in PG(2n, q) for some n ≥ 2,giving a parabolic polar space of rank n.

• Q−(2n+ 1, q): non-singular elliptic quadric in PG(2n+ 1, q) for somen ≥ 2, giving a hyperbolic polar space of rank n.

• W (2n + 1, q): polar space consisting the points of PG(2n + 1, q) to-gether with the totally isotropic subspaces of a non-singular symplecticpolarity of PG(2n+ 1, q), giving a symplectic polar space of rank n.

• H(2n, q2) : non-singular Hermitian variety in PG(2n, q2) for somen ≥ 2, giving a Hermitian polar space of rank n.

• H(2n + 1, q2): non-singular Hermitian variety in PG(2n + 1, q2) forsome n ≥ 1, giving a Hermitian polar space of rank n+ 1.

The following theorem (see e.g. [28]) gives the size (number of points)of those finite classical spaces:

Theorem 1.52. The numbers of points of the finite classical polar spacesare given by:

|Q+(2n+ 1, q)| = (qn + 1)(qn+1 − 1)/(q − 1),|Q(2n, q)| = (q2n − 1)/(q − 1),

|Q−(2n+ 1, q)| = (qn − 1)(qn+1 + 1)/(q − 1),|W (2n+ 1, q)| = (q2n+1 − 1)/(q − 1),|H(2n, q2)| = (q2n+1 + 1)(q2n − 1)/(q2 − 1),

|H(2n+ 1, q2)| = (q2n+2 − 1)(q2n+1 + 1)/(q2 − 1).

In the case of rank 2, there are a lot of non-classical polar spaces known.

Chapter 2

Combinatorics andq-analogues

A natural way of generalizing problems from extremal combinatorics to finitegeometry is the q-analogue. This is done by changing the definition of theproblem as follows: replace all occurrences of the words “set of size n” with“vector space over the field GF(q) with or rank n”. Of course some otherwords in the definition should also be changed accordingly, for example ifthe original problem talks about disjoint subsets, the q-analogous problemtalks about subspaces that intersect trivially. In this chapter we give someexamples of such generalizations by takingq-analogues of some classical problems in extremal combinatorics.

2.1 An easy example

In Section 1.1 we counted the number of k-subsets in an n-set. Theq-analogue of this problem is counting the number of rank-k subspaces inV(n, q) and that is what we did in Section 1.3.4. Let us take a closer lookat both results.

The number of projective points in PG(n− 1, q) is given by[n1

]q. If we

now recursively define [n+ 1]q! =[n+1

1

]q

[n]q! and [0]q! = 1 we can write

[nk

]q

=[n]q!

[k]q![n− k]q!.

27

28 CHAPTER 2. COMBINATORICS AND Q-ANALOGUES

Note that this notation looks a lot like the definition of the binomialcoefficients. In that light it is not so strange that Gaussian coefficients arealso called q-binomial coefficients.

An even stronger validation of this name is a phenomenon we see witha lot of q-analogues. Because we work with GF(q), the q’s used are primepowers. But if we look at the formulas as functions on R in the variable q,we can take the limit of these functions for q → 1+. For example the limit(by using de l’Hopital’s limit rule) of the number of rank 1-subspaces in arank-n vector space is:

limq→1+

[n1

]q

= limq→1+

qn − 1q − 1

= n

which is of course the number of elements in an n-set. Using this it isstraightforward that

limq→1+

[nk

]q

=(n

k

).

We will see that, in a lot of cases, taking the limit for q → 1+ of a valuein the q-analogue of a problem gives us the value from the original problem.Note that this is not always the case; see, for example, Remark 3.9 in thenext chapter.

2.2 Sperner’s Theorem

An important subdomain of extremal combinatorics is extremal set the-ory. The goal in this domain is to find the maximum size of a family of setssatisfying certain assumptions. In this section, we consider Sperner families.

2.2.1 Original problem

A classical theorem in extremal set theory is Sperner’s theorem. If we fix auniverse X with size n, the largest uniform family, that is a family in whichall sets have the same size, is the family of all k-subsets where k =

⌊n2

⌋(or

2.2. SPERNER’S THEOREM 29

k =⌈n2

⌉). This uniform family has size

(nbn2 c

). In 1928 Sperner proved

([44]) that this is the best possible upper bound for the size of an antichainin general (for both uniform and non-uniform Sperner families) and thatequality occurs exactly in the case just described.

There are several proofs known of this upper bound. Here we give onethat uses the stronger LYM inequality. This inequality is named after D.Lubell [37], K. Yamamoto [48] and L.D. Meshalkin [39]. Here Lubell’s Per-mutation Method [37] is used to prove the inequality:

Lemma 2.1 (LYM inequality). If F is a Sperner family of subsets of a setX with |X| = n, then ∑

A∈F

1(n|A|

) ≤ 1.

Proof. Consider the subset poset of X: (2X ,⊆). It is easy to see that thenumber of maximal chains in this poset is n! and that the number of maxi-mal chains containing a certain subset A is |A|!(n− |A|)!.

Now count the pairs (A, C) where A ∈ F and C is a maximal chain thatcontains A. If we take a subset A in F , we have |A|!(n−|A|)! maximal chainsthat contain A. On the other hand, a maximal chain C can contain at mostone element of F (because F is an antichain). Thus

∑A∈F|A|!(n− |A|)! ≤ n!.

Dividing by n! gives the required result.

For other applications of the LYM inequality, see eg. [33].

Now we can prove Sperner’s Theorem:

Theorem 2.2. If F is a Sperner family of subsets of an n-set X, then

|F| ≤(

n⌊n2

⌋).Proof. Because (

n⌊n2

⌋) ≥ ( n

|A|

)for all A ∈ F


we have, using the LYM inequality:

1 ≥∑A∈F

1(n|A|

) ≥ |F|(nbn2 c

)

2.2.2 q-analogue

Sperner’s theorem gives the maximum size of an antichain in the subsetposet of a finite set. The natural q-analogue question therefore is: what isthe maximum size of an antichain in the subspace poset of a finite vectorspace V(n, q)?

It turns out that the answer is what one would expect, and the proof is acomplete q-analogue of the set case. We now state and prove the q-analogueof the LYM-inequality.

Lemma 2.3. If F is an antichain in the subset poset of V(n, q), then∑A∈F

1[n

rk(A)

]q

≤ 1.

Proof. The number of maximal chains in this poset is [n]q! and the numberof chains containing a subspace A is [rk(A)]q![n − rk(A)]q!. Counting thepairs (A, C) where A ∈ F and C is an antichain containing A yields:∑

A∈F[rk(A)]q![n− rk(A)]q! ≤ [n]q!.

Dividing both sides of this inequality by [n]q! gives the stated result.

If we now use this lemma together with the fact that[nk

]q≤[

nbn2 c

]q

for

all 0 ≤ k ≤ n we find the q-analogue sought for.

Theorem 2.4. If F is an antichain in the subset poset of V(n, q), then

|F| ≤[n⌊n2

⌋]q

.

In this case equality holds if and only if F is the antichain consisting ofall subspaces of rank

⌊n2

⌋(or

⌈n2

⌉).

2.3. ERDOS-KO-RADO THEOREM 31

2.3 Erdos-Ko-Rado theorem

Another classic result in extremal set theory is the Erdos-Ko-Rado Theorem.


Theorem 2.5 (Erdos-Ko-Rado Theorem (1961) [18]). If F is a k-uniformfamily with a universe X of size n, where k ≤ n

2 , and every pair of membersof F intersect, then

|F| ≤(n− 1k − 1

).

A family of mutually intersecting k-subsets of an n-set is called an (n, k)-EKR family. An EKR family that cannot be extended to another EKRfamily by adding subsets is called a maximal EKR family (or maximal in-tersecting family).

We do not give the original proof here, but we sketch a much more elegantproof by G. Katona in 1972 [31], which is inspired by Lubell’s PermutationMethod.

Proof. Consider a labeling of an n-cycle by the elements of X. A path oflength k in this labeled cycle corresponds to a k-subset of X. A collectionof paths that pairwise overlap1 has size at most k, and in total there are npaths of length k on this cycle. So at most a fraction k/n of the k-subsets“occurring” in this cycle as paths mutually intersect. Each subset occurs thesame number of times as a path among all possible labelings of an n-cycleby elements of X. So at most a fraction k/n of the

(nk

)k-subsets mutually

intersect.

This bound is tight. We give some examples of families attaining thisbound. There is an obvious example of a family of size

(n−1k−1

), namely a

family that consists of all k-subsets containing a common element of X.Such a family is called a point pencil and the common element is called thecenter of the point pencil. In Section 2.5 we will show that this is the onlytype of family attaining this upper bound for n > 2k.

1Here to overlap means to have at least a vertex in common.


If n = 2k, there are other examples besides point pencils. Each k-subsethas exactly one complementary k-subset. An intersecting family can containat most one k-subset out of each complementary pair. A family consistingof exactly one k-subset out of each complementary pair has size 1

2

(2kk

)which

is the maximum size.

2.3.2 q-Analogue

In the original Erdos-Ko-Rado theorem, F is a family of k-subsets of ann-set (with 2k ≤ n) such that no two members are disjoint. The statementof the theorem can be viewed in two ways:

1.(n−1k−1

)is an upper bound on the size of F , or

2. |F||Fk| ≤

kn where Fk is the family of all k-subsets of this n-set (this is

the statement that was proved by Katona).

In a q-analogue version of the Erdos-Ko-Rado theorem, F will be afamily of rank-k subspaces of V(n, q) with 2k ≤ n in which no two membersintersect trivially. We will call such a family an [n, k]q-EKR family. So nowwe can expect the following two statements:

(1q) |F| ≤[n−1k−1

]q, and

(2q)|F||Fk| ≤

kn where Fk is the family of all rank k subspaces of V(n, q).

Because the number of rank-k subspaces containing the same rank-1subspace is

[n−1k−1

]q, the inequality in (1q) is best possible. Furthermore, we

have that [n−1k−1

]q[

nk

]q

=qk − 1qn − 1

� k

n.

Therefore (1q) imposes a much stronger bound than (2q).

W.H. Hsieh [29] generalized Katona’s permutation method in 1973 toprove statement (2q), and, using a lot of counting arguments, he proved thestronger statement (1q) for n ≥ 2k + 1.

2.4. BOLLOBAS’S THEOREM 33

The case n = 2k remained open until 1986 when P. Frankl and R.M.Wilson [21] also proved the statement in this case.

Examples of families reaching this upper bound are easily found. Ifn > 2k, a family consisting of all rank-k spaces that contain the same rank-1 subspace has this size. In the case n = 2k we have the same type offamilies and also the families that consist of all the rank-k spaces in a givenhyperplane (rank-(2k − 1) subspace).

In [21], Frankl and Wilson state they have a proof that those are theonly families attaining the upper bound. In a paper, [24], of M.W. Newmanand C. Godsil a short proof of this statement is given.

Remark 2.6. In [29] and [21], Hsieh, Frankl and Wilson actually prove thefollowing, more general result.

Theorem 2.7. If F is a [n, k]q-EKR family such that the rank of the inter-section of each pair of members is at least t (with n ≥ 2k − t), then

|F| ≤ max

{[n− tk − t

]q

,

[2k − tk

]q

}.

This result is used to improve some bounds on a similar generalizationof the original Erdos-Ko-Rado problem in [18].

2.4 Bollobas’s theorem

For all of the problems we encountered in the previous sections, the answerin the original version was the limit case for q → 1+ of the answer in thevector space version. We already noted that this is not always true. For thefollowing problem we see an even stronger connection between the differentversions.

2.4.1 Set version

Helly’s Theorem gives a characterization of the dimension of a linear spaceover R in terms of intersection properties of convex sets:


Theorem 2.8. If C1, C2, . . . , Cm ⊆ Rn are convex sets such that any n+ 1intersect, then all of them intersect.

One can wonder if other objects than convex sets obey similar laws. Forexample, it is easy to see that, if for each set of at most three edges of agraph we have that those edges have nonempty intersection, then all edgesof the graph have nonempty intersection. This statement even generalizesover r-uniform set systems (note that a graph is a 2-uniform set system):

Proposition 2.9. If each family of at most r+ 1 members of an r-uniformset system intersects, then all members intersect.

The statement about graphs can also be generalized in a different direc-tion:

Theorem 2.10 ([17]). If each family of at most(s+2

2

)edges of a graph can

be covered by s vertices, then all edges can.

This was proved by P. Erdos, H. Hajnal and J. W. Moon in 1964. Wewill not give this proof but prove a more general result later in this section.The complete graph on s+ 2 vertices shows that this result is sharp.

In 1965 B. Bollobas proved ([4]) the following generalization of thoseproblems:

Theorem 2.11. If each family of at most(r+ss

)members of an r-uniform

set system can be covered by s points, then all members can.

We will not prove this theorem, but a slightly more powerful theorem,also by Bollobas in [4]:

Theorem 2.12. Let A1, . . . , Am and B1, . . . , Bm be subsets of size s of aset of size r such that

(i) Ai and Bi are disjoint for i = 1, . . .m, and

(ii) Ai and Bj intersect if i 6= j (1 ≤ i, j ≤ m).

Then m ≤(r+sr

).

There exist a lot of different proofs of this theorem, for example byJaeger-Payan ([30]) and Katona ([32]). We will give a proof by L. Lovasz in[35]. First we need a proposition called the diagonal criterion:

2.4. BOLLOBAS’S THEOREM 35

Proposition 2.13. For i = 1, . . . ,m, let fi : X → F, (where X is a set andF a field) be functions and ai ∈ X elements such that

fi(aj){6= 0 if i = j;= 0 if i 6= j.

Then f1, . . . fm are linearly independent in the space FX .

Proof. Suppose that∑m

i=1 λifi = 0 is a linear relation on the fi’s. Substi-tuting aj for some 1 ≤ j ≤ m in the functions gives λjfj(aj) = 0, whichimplies that λj = 0. This is true for all j = 1, . . . ,m, hence the functionsare linearly independent.

Now we can give Lovasz’s proof of Theorem 2.12:

Proof. Define V =⋃mi=1(Ai ∪Bi) and associate vectors

p(v) = (p0(v), p1(v), . . . , pr(v)) ∈ Rr+1

to each v ∈ V such that any r + 1 of those vectors are linearly independent(note that there are constructive ways of doing this). Now we associate apolynomial fW (x) in r + 1 variables x = (x0, . . . , xr) to every W ⊆ V :

fW (x) :=∏v∈W

(p0(v)x0 + p1(v)x1 + . . .+ pr(v)xr).

This is a homogeneous polynomial of degree |W | with:

fW (x){6= 0 if x is orthogonal to none of the p(v), v ∈W ;= 0 otherwise.

Now take a vector aj orthogonal to the subspace spanned by the vectorscorresponding to the elements of Aj for each j = 1, . . . ,m. This guaranteesthat aj is orthogonal to p(v) if and only if v ∈ Aj . We can see that fBi(aj) =0 if and only if Aj and Bi intersect, hence if and only if i 6= j. By thediagonal criterion this means that the polynomials fB1 , . . . , fBm are linearlyindependent in the space of homogeneous polynomials of degree s in r + 1variables. Since this space has dimension

((r+1)+s−1

s

), we have that m ≤(

r+ss

).


2.4.2 Set vs. subspace version

In the same paper where he proves Theorem 2.12, [35], Lovasz also provesthe following result using a similar technique:

Theorem 2.14. Let U1, . . . , Um be r-dimensional subspaces and B1, . . . , Bmbe s-elements sets in a linear space W over the field F such that

(i) Ui and Bi are disjoint for i = 1, . . .m, and

(ii) Ui and Bj intersect if i 6= j (1 ≤ i, j ≤ m).

Then m ≤(r+sr

).

2.4.3 Subspace version

Using tensor product methods Lovasz gives another proof of theorem 2.12in [35] and generalizes this proof to show the following result:

Theorem 2.15. Let U1, . . . , Um be r-dimensional subspaces and V1, . . . , Vmbe s-dimensional subspaces of a linear space W over the field F such that

(i) Ui ∩ Vi = 0 for i = 1, . . .m, and

(ii) Ui ∩ Vj 6= 0 if i 6= j (1 ≤ i, j ≤ m).

Then m ≤(r+sr

)We see that in all three versions (set-set, set-subspace and subspace-

subspace) the answer of the problem is the same. Our observation that inmany cases the answer in the set version of a problem is a limit case of theanswer in the vector space version is trivially true here.

2.5 The Hilton-Milner theorem


In Section 2.3, a tight upper bound on the size of an (n, k)-EKR family wasshown. We saw that a point pencil is a family that attains this bound. In[27], A.J.W. Hilton and E.C. Milner give two examples of maximal (n, k)-EKR families that are not point pencils.

2.5. THE HILTON-MILNER THEOREM 37

Example 2.16. Take the set X = {1, 2, . . . n} and A = {1, 2, . . . k} with2 ≤ k ≤ n

2 . The family F1 consisting of all k-subsets of X containing theelement n and intersecting A, together with A, is clearly a maximal (n, k)-EKR family. Suppose that this is a point pencil, say with center x. Becausex ∈ A we have that 1 ≤ x ≤ k, but the set {1, 2, . . . , k, n}\{x} is an elementof F1 that does not contain x, a contradiction. Hence F1 cannot be a pointpencil. The size of F1 is

(n−1k−1

)−(n−k−1k−1

)+ 1.

Example 2.17. The family F2, consisting of all k-subsets of X containingat least two elements of {1, 2, 3}, is a maximal intersecting family that is nopoint pencil. If k = 2, F2 has the same structure as F1. If k ≥ 3, it has size3(n−3k−2

)+(n−3k−3

), which is the same as |F1| if k = 3 and smaller than |F1|

if k > 4.

In the same paper Hilton and Milner prove that F1 (and F2 if k = 2, 3)has the maximum size a maximal (n, k)-EKR family can have if it is not apoint pencil:

Theorem 2.18. If F is a (n, k)-EKR family, with 2 ≤ k ≤ n2 , that is no

point pencil and not contained in one, then

|F| ≤(n− 1k − 1

)−(n− k − 1k − 1

)+ 1.

If n > 2k and this upper bound is attained, then F has the structure de-scribed in Example 2.16

Note that if n = 2k this is bound is the same bound as in the EKRtheorem, because in that case, there were other maximal EKR families thanpoint pencils that attain the EKR bound, too.

If n > 2k, this theorem is of course enough to conclude the following:

Corollary 2.19. If F is an (n, k)-EKR family, with n > 2k and

|F| >(n− 1k − 1

)−(n− k − 1k − 1

)+ 1,

then F is a point-pencil and hence

|F| =(n− 1k − 1

).


2.5.2 The q-analogue

We are looking for maximal [n, k]q-EKR families that are no point pencils(and no hyperplanes if n = 2k).

For k = 2, the only possibility is a family consisting of all rank-2 spaces(projective lines) in a rank-3 space (projective plane):

Proposition 2.20. Every [n, 2]q-EKR family, with n ≥ 4 F is contained ina point pencil or a projective plane

Proof. Suppose F is an [n, 2]q-EKR family that is not contained in a plane.Take two lines l1 and l2, they intersect, say in the point x and span a plane,say α. Take another line l3 of F , not in α. This line intersects l1 and l2in respectively y1 and y2. Now y1 and y2 are the same point, otherwise l3would lie in α. So the point y1 = y2 lies both on l1 and l2, hence it mustbe x. Now take another line l4 in F . Because l1, l2 and l3 are not coplanar(they do not lie on the same plane), at least one pair of them spans a planethat does not contain l4. Now repeat the same reasoning as above withthose two lines and l4 to conclude that x is on l4. Continue this reasoningto conclude that all lines of F go through x, so F is contained in the pointpencil with center x.

For k ≥ 3 the situation is much more complex. There are lots and lotsof types of maximal EKR-families that are no point pencil (or hyperplaneif n = 2k). In what follows we develop a description of the structure of thelargest of those families (if n ≥ 2k + 1).

First, note that a point pencil has[n− 1k − 1

]q

≈ q(n−k)(k−1)

elements.

Consider the following large examples.

Example 2.21. Let F3 be the family of all rank-k spaces (projective (k−1)-spaces) on a fixed rank-2 space (projective line). This [n, k]q-EKR familyhas [

n− 2k − 2

]q

≈ q(n−k)(k−2)


elements. Note that this family is not maximal, it can, for instance, beextended to a point pencil or to a family described in the next example.

Example 2.22. Let F4 be the family of all (k − 1)-spaces that intersect agiven plane in at least a line. This family has size approximately

(q2 + q + 1)[n− 2k − 2

]q

≈ q(n−k)(k−2)+2.

We can still find larger examples

Example 2.23. Let F5 be the family of all (k − 1)-spaces that contain afixed point p and meet a fixed (k − 1)-space π (that does not contain p),together with π. This family has size approximately[

k

1

]q

[n− 2k − 2

]q

≈ q(n−k)(k−2)+k−1.

Note that this is not a maximal family, but if we add all the (k − 1)-spacesin 〈p, π〉 that do not contain p, it is maximal. Doing that only adds ap-proximately qk elements to F5. So the order of magnitude of the size doesnot change. Note that this is the q-analogue of the Hilton-Milner example(Example 2.16). We will call this example the q-Hilton-Milner example.

Example 2.24. Take two (k − 1)-spaces π1 and π2 and a point p not con-tained in them such that π2 6⊂ 〈p, π1〉. In that case both π1 \ π2 and π2 \ π1

contain approximately qk−1 points. Let F6 be the family of all (k−1)-spacesthat contain p and meet both π1 \ π2 and π2 \ π1, together with π1 and π2.This family has approximate size

q2(k−1)

[n− 3k − 3

]q

≈ q(n−k)(k−2)−(n−k)+2k−2.

Again, here we can add all (k − 1)-spaces that contain p and meet π1 ∩ π2,and they add at most approximately[

k − 11

]q

[n− 2k − 2

]q

≈ q(n−2)(k−2)+k−2

to the previous number. Note that if n = 2k + 1, q(n−k)(k−2)−(n−k)+2k−2 =q(n−k)(k−2)+k−3, so in that case, if π1 and π2 intersect in a (k−2)-space, thesecond term dominates and we have an example that is of almost the sameorder of magnitude than Example 2.23. Note that also here we can add the(k − 1)-spaces that are contained in 〈p, π1〉 ∪ 〈p, π2〉 but do not contain pwithout changing the order of magnitude of the size of this family.


Remark 2.25. In both Examples 2.23 and 2.24 (the two largest we havefound so far) there is a point that contains almost all (k − 1)-sets of thefamily.

The following lemma assures the existence of points with a high degreeif there is a subspace that meets all elements of an EKR-family (the degreeof a subspace is the number of elements of F it is contained in).

Lemma 2.26. Let F be an [n, k]q EKR-family. If there is an (l − 1)-spaceΛ that meets F (by which we mean that it meets all elements of F), thenthere is a point contained in at least |F|/

[l1

]q

elements of F and this point

is contained in Λ.

Proof. Label the points of Λ as p1, p2, . . . , p[ l1 ]q

and define di as the number

of elements of F that contain the point pi. Counting the pairs (p, π), wherep is a point of Λ and π and element of F that contains p, in two ways yields

[ l1 ]q∑

i=1

di ≥ |F|.

Define d as the maximum of all di, then we have the required result for allpoints that attain this maximum.

We would like to prove a q-analogue of Theorem 2.18. We propose, forreasons that will become clear later, a lower bound for the families F wewant to consider:

|F| ≥[n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n.

For n = 2k + 1, this bound has a size of the same order of magnitude asthe size of the q-Hilton-Milner example. For n > 2k + 1, the size is at leastone order of magnitude smaller than the size of the q-Hilton-Milner example.

Depending on the value of q and n, this bound is larger or smaller thanthe size of the q-Hilton-Milner example. Let us consider some cases.

• Case 1: If q ≥ 4 and n ≥ 2k + 2; q = 3 and n ≥ 2k + 3 or q = 2 andn ≥ 2k + 4, the bound on F is smaller than the size of the q-Hilton-Milner example.


• Case 2: If q ≥ 4 and n = 2k + 1; q = 3 and n = 2k + 2 or q = 2 andn = 2k+ 3, the order of magnitudes are equal but the bound is largerthan the size of the q-Hilton-Milner example.

• Case 3: If q = 3 and n = 2k+ 1 or q = 2 and 2k+ 1 ≤ n ≤ 2k+ 2, theorder of magnitue of the bound is larger than that of the size of theq-Hilton-Milner example.

Let us start with the first case.

Case 1

So define a type 1 family as a maximal [n, k]q-EKR family F , with k ≥ 3and the given bounds on n and q, such that

|F| ≥[n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n.

The following lemma shows that there is always a line meeting such F .

Lemma 2.27. If there is an (l−1)-subspace meeting a type 1 family F and3 ≤ l ≤ k, then there is an (l − 2)-subspace meeting F .

Proof. Lemma 2.26 guarantees a point p with degree at least |F|/[l1

]q. Now

either all elements of F contain this point or there is an F1 ∈ F that doesnot contain this point. In the former case F is a point pencil and all (l−2)-spaces on the center of this point pencil meet F . In the latter case we canrepeat the argument in the proof of Lemma 2.26 for p and F and find thatthere is a point q in F1, and hence a line L = pq, with degree at least|F|/

[l1

] [k1

].

Now either all elements of F intersect L or there is an F2 ∈ F that isdisjoint from L. In the former case all (l − 2)-spaces containing this linemeet F . In the latter case we can find a plane π with degree at least

|F|/[l1

]q

[k1

]2

q. Continuing this argument we either find at some point an

(l − 2)-space meeting F , or we get an (l − 1)-subspace with degree at least

|F|/[l1

]q

[k1

]l−1

q. Since an (l−1)-space is contained in

[n−lk−l

]q

(k−1)-spaces

of PG(n, q), we have a contradiction if[n− lk − l

]q

<|F|[

l1

]q

[k1

]l−1

q

.


This is equivalent with

l−1∏i=2

(qn−i − 1

(qk−i − 1)(qk − 1)

)(q − 1)l(q2 + q + 1)q3k−n

(ql − 1)(qk − 1)(q − 1)2> 1.

Now for 2 ≤ i ≤ l − 1 ≤ k − 1 we have

qn−2k(qk−i − 1)(qk − 1) = qn−i − qn−k − qn−k−i + qn−2k

< qn−i − qn−k

< qn−i − 1

so we get a contradiction if

q(n−2k)(l−2) (q − 1)l(q2 + q + 1)q3k−n

(ql − 1)(qk − 1)(q − 1)2> 1. (2.1)

Since substituting l by l + 1 in the right hand side of this inequality isequivalent with multiplying this expression by

qn−2k(q − 1)(ql − 1)ql+1 − 1

> 1,

we only need to get a contradiction for the smallest l, which is l = 3.Substitute for l = 3 in the right hand side of inequality 2.1 yields:

qn−2k(q − 1)3(q2 + q + 1)q3k−n

(q3 − 1)(qk − 1)(q − 1)2=

qk

qk − 1> 1.

So for l = 3, inequality 2.1 holds and we obtain a contradiction.

Since all elements of F satisfy the criterion of this lemma for l = k, thereis indeed a line meeting F . Such a line that meets F will be called a hittingline.

With this knowledge we can improve our bound if n ≥ 3k.

Corollary 2.28. If F is a type 1 family, then

|F| ≥[n− 2k − 2

]q

.

Proof. Lemma 2.27 guarantees the existence of a hitting line of F . Since Fis maximal, all (k− 1)-spaces through that hitting line must be elements ofF .


Note that for n ≥ 3k this bound is indeed larger than our assumed lowerbound.

We prove some results about hitting lines.

Lemma 2.29. If F is a [n, k]q-EKR family, and L1 and L2 are hitting linesof F that intersect in a point p. Then all lines L through p in the plane〈L1, L2〉 are also hitting lines.

Proof. Take such a line L. Since each element of F intersects both L1 andL2, it intersects the plane 〈L1, L2〉 in a line, and that line intersects L. Notethat if L were not to lie on p then it would not intersect with the elementsof F that intersect both L1 and L2 in p.

If there are three hitting lines in a triangle, we can characterize F .

Lemma 2.30. If L1, L2, L3 are non concurrent hitting lines of an [n, k]q-EKR family F in a plane α, then F is a family of (k − 1)-spaces that meeta fixed plane (in this case α) in at least a line (see Example 2.22).

Proof. Each element of F has either 2 or 3 intersection points with thetriple (L1, L2, L3). In the former case those two points determine the line inwhich the element intersects α and in the latter case α is contained in theelement.

Define a type 1’ family F as a type 1 family that allows no plane suchthat all elements of F intersect it in at least a line.

The following lemma gives a condition for a line to be a hitting line.

Lemma 2.31. Let F be an [n, k]q-EKR family. If a line L has a degree

larger than[k1

]q

[n−3k−3

]q

then it intersects F .

Proof. Suppose that there is an element F ∈ F that does not intersect L.That means that there is a point p on F , and hence a plane 〈p, L〉, thatintersects more than

[n−3k−3

]q

elements of F , a contradiction.

The following lemma states that if F is a type 1 family that is largeenough, unless F is (part of) a point pencil, a hitting line always intersectsat least one other hitting line.


Lemma 2.32. Let F be a type 1 family. If L is a hitting line and no linemeeting L has degree more than

[k1

]q

[n−3k−3

]q, then either

|F| ≤[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

,

or F is contained in a point pencil.

Proof. Let F ∈ F be such that F ∩ L = {p0}. If such an element doesnot exist then L is contained in all elements and F is contained in a pointpencil. Now count the number of elements of F not containing L. For apoint p 6= p0, p ∈ L and any point q ∈ F , the line pq has degree at most[k1

]q

[n−3k−3

]q. Since every element of F meets F we have that there are at

most([

k1

]q− 1)[

k1

]q

[n−3k−3

]q

elements of F through p, for all points p on

L different from p0.Now if all elements of F contain p0, F is contained in a point pencil. Soassume that there is an element F ′ of F that does not contain p0. The

same argument yields that there are also at most([

k1

]q− 1)[

k1

]q

[n−3k−3

]q

elements of F through p0. Finally, there are at most[n−2k−2

]q

elements of Fthat contain L. Adding this up yields the desired result.

Remark 2.33. The bound in the previous lemma is approximately

q(n−k)(n−2) + q(n−k)(k−2)−n+3k−1.

• If n ≥ 3k, the first term of this bound dominates. Because of Corollary2.28 we cannot hope for a smaller bound, since we can simply take all(k − 1)-spaces through a hitting line. So at this point, we need

|F| >[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

.

• If n = 3k − 1, this bound roughly doubles the previously assumedbound.

• If n < 3k − 1, the second term dominates and the new bound is a lotsmaller than the previous bound.


An intersection point of two hitting lines of a type 1’ family has largedegree.

Lemma 2.34. Let F be a type 1’ family and let L1, L2 be two hitting lineswith L1 ∩ L2 = {p}, then there are at most q2

[k1

]q

[n−3k−3

]q

elements of Fthat do not contain p.

Proof. Suppose there is a line L in 〈L1, L2〉 that does not contain p andhas degree more than

[k1

]q

[n−3k−3

]q. By Lemma 2.31 L is a hitting line, but

we assumed that such a configuration does not exist. That means that allsuch lines have degree at most this number. Now we count the number ofelements of F that do not contain p. Each such element intersects the plane〈L1, L2〉 in a line not through p. We just saw that such lines have a degreeat most

[k1

]q

[n−3k−3

]q. This yields the result.

A point with this property we will call a point with large degree. Theprevious lemmas give us a lower bound on F for the existence of such points.

Corollary 2.35. If F is a type 1’ family such that

|F| >[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

,

then there is a point of large degree.

Proof. Lemma 2.32, together with Lemma 2.27 guarantee the existence oftwo intersecting hitting lines. Now Lemma 2.34 guarantees that the inter-section point of those two lines has a large degree.

The Lemma 2.34 has strong implications on the structure of hitting lines.

Lemma 2.36. Two hitting lines of a type 1’ family F always meet.

Proof. Suppose that L1 and L2 are two disjoint hitting lines. If there is ahitting line meeting L1, by Lemma 2.34 the intersection point has degree atleast[

n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n − q2

[k

1

]q

[n− 3k − 3

]q

≈ q(n−k)(k−2)+3k−n.

If there is also a hitting line intersecting L2, this intersection point has alsoat least this degree. But then we have that

|F| ≥ 2|F | − 2q2

[k

1

]q

[n− 3k − 3

]q

−[n− 2k − 2

]q

,


or

|F| ≤ 13

(2q2

[k

1

]q

[n− 3k − 3

]q

+[n− 2k − 2

]q

).

Since n ≥ 2k + 1, the second term dominates. If n < 3k, this bound issmaller than the lower bound we assumed for F , a contradiction. If n ≥ 3k,Corollary 2.28 gives a lower bound on F that is approximately 3 times higherthan the upper bound we find here, which is also impossible.So we can assume that all lines meeting L2 are non hitting lines and hencehave degree less than

[k1

]q

[n−3k−3

]q. So we can apply Lemma 2.32 and find

that

|F| ≤[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

.

Actually, since all elements of F containing L2 must also intersect L1, thefirst term of the left hand side can be reduced to (q + 1)

[n−3k−3

]q. But now

the second term is always dominating and since this is always some ordersof magnitudes smaller than the assumed lower bound on F we have a con-tradiction.

Let us now consider the next case.

Case 2

In this case the lower bound we assumed on F is larger than the size of theq-Hilton Milner example. That is why we will propose another bound thatis slightly smaller than that example.Define a type 2 family F as a maximal [n, k]q-EKR family for the givenbounds on n and q, such that

|F| ≥[k

1

]q

[n− 2k − 2

]q

− q[k

2

]q

[n− 3k − 3

]q

.

In this case we can also prove the existence of a hitting line (or a hittingplane in the case q = 2) in the same way as before.

Lemma 2.37. Let F be a type 2 family. If there is an (l − 1)-subspacemeeting F and 3 ≤ l ≤ k (or 4 ≤ l ≤ k if q = 2), then there is an (l − 2)-subspace meeting F .


Proof. Using the same arguments as in Lemma 2.27 we find a contraditionif

|F|[n−lk−l

]q

[l1

]q

[k1

]l−1

q

> 1,

or ifl−1∏i=2

qn−i − 1(qk−i − 1)(qk − 1)

([k

1

]q

(q − 1)l

(ql − 1)(qk − 1)

)

−l−1∏i=3

qn−i − 1(qk−i − 1)(qk − 1)

(q

[k

2

]q

(q − 1)l

(ql − 1)(qk − 1)2

)> 1.

By using inequality 2.1 from Lemma 2.27 we can simplify this to obtain acontradiction if

q(n−2k)(l−2)

([k

1

]q

(q − 1)l

(ql − 1)(qk − 1)

)1−q[k2

]q

(qk−2 − 1)[k1

]q

(qn−2 − 1)

> 1,

or if

q(n−2k)(l−2)

((q − 1)l−1

ql − 1

)1−

([k1

]q− 1)

(qk−2 − 1)

(q + 1)(qn−2 − 1)

> 1.

Now since [k

1

]q

− 1 <qk − 1q − 1

the following inequality is needed for a contradiction:

q(n−2k)(l−2)

((q − 1)l−1

ql − 1

)(1− (qk − 1)(qk−2 − 1)

(q + 1)(q − 1)(qn−2 − 1)

)> 1. (2.2)

1. If n = 2k + 1, and hence q ≥ 4, substituting l + 1 for l is equivalentwith multiplying the right hand side of inequality 2.2 with

q(q − 1)(ql − 1)ql+1 − 1

> 1.

Therefore it is enough to obtain a contradiction for l = 3. In this casea contradiction is obtained if

q(q − 1)2

q3 − 1

(1− (qk − 1)(qk−2 − 1)

(q + 1)(q − 1)(q2k−1 − 1)

)> 1.

Carefully checking this shows that this inequality is true for q ≥ 4.


2. If n = 2k + 2, and hence q = 3, substituting l + 1 for l is equivalentwith multiplying the right hand side of inequality 2.2 with

q2(q − 1)(ql − 1)ql+1 − 1

> 1.


q2(q − 1)2

q3 − 1

(1− (qk − 1)(qk−2 − 1)

(q + 1)(q − 1)(q2k−1 − 1)

)> 1.

Carefully checking this shows that this inequality is true for q = 3.

3. If n = 2k+3 and hence q = 2 substituting l+1 for l is equivalent withmultiplying the right hand side of inequality 2.2 with

q3(q − 1)(ql − 1)ql+1 − 1

> 1.


q3(q − 1)2

q3 − 1

(1− (qk − 1)(qk−2 − 1)

(q + 1)(q − 1)(q2k−1 − 1)

)> 1.

Carefully checking this shows that this inequality is not true for q = 2.So we look at the case l = 4. In this case a contradiction is obtained if

q6(q − 1)3

q4 − 1

(1− (qk − 1)(qk−2 − 1)

(q + 1)(q − 1)(q2k−1 − 1)

)> 1.

Carefully checking this shows that this inequality is not true for q = 2.

In the same way as Corollary 2.28 one proves the following.

Corollary 2.38. If F is a type 2 family with q > 2, then

|F| ≥[n− 2k − 2

]q

.

Now since Lemmas 2.29, 2.30 and 2.31 are stated for general [n, k]q-EKR families, they still hold (for q > 2). A quick check shows that theother results in Case 1 still hold in Case 2 for q > 2.

Let us now continue with the last case.


Case 3

Here we will use the same lower bound for F as in Case 2.Define a type 3 family F as a maximal [n, k]q-EKR family for the givenbounds on n and q, such that

|F| ≥[k

1

]q

[n− 2k − 2

]q

− q[k

2

]q

[n− 3k − 3

]q

.

In this case we can only prove the existence of a hitting plane for q = 3.

Lemma 2.39. Let F be a type 3 family with q = 3. If there is an (l − 1)-subspace meeting F and 4 ≤ l ≤ k, then there is an (l− 2)-subspace meetingF .

Proof. The proof is completely analoguous to the proof of Lemmas 2.27 and2.37 with l = 4.

We summarize everything.

Main theorem

Remark 2.33 shows that the cases n ≥ 3k − 1 and n < 3k − 1 should betreated a little bit different.

The following lemma summarizes what we know about the existence ofhitting lines in the different cases.

Lemma 2.40. Hitting lines have the following properties:

1. Let F be a maximal [n, k]q-EKR family with n ≥ 3k − 1.

(a) If

|F| >[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

there exist at least two hitting lines.

(b) If [n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n ≤ |F|


and

|F| ≤[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

there exists at least one hitting line.

2. Let F be a maximal [n, k]q-EKR family with n = 2k + d, for 1 ≤ d <k − 1 (such that n < 3k − 1).

(a) If q ≥ 4 and n ≥ 2k + 2; q = 3 and n ≥ 2k + 3; or q = 2 andn ≥ 2k + 4, and

|F| ≥[n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n


(b) If q ≥ 4 and n = 2k + 1 or q = 3 and 2k + 2, and

|F| ≥[k

1

]q

[n− 2k − 2

]q

− q[k

2

]q

[n− 3k − 3

]q


Now we investigate the structure of F if there are at least two hittinglines. Because of Lemma 2.36 we know that all hitting lines pairwise inter-sect. If there is no plane such that all elements of F intersect this plane inat least a line, then this means that all hitting lines intersect in the samepoint p, and this is a point of high degree. Now all the hitting lines span anm-dimensional subspace H for some m. Lemma 2.29 now guarantees thatall lines in H that contain p are also hitting lines.So the existence of two hitting lines gives us the following structure: if thereis no plane such that all elements of F intersect that plane in at least a line,there is an m-dimensional subspace H, with m ≥ 2, and a point p in H, suchthat all lines through p in H are exactly all the hitting lines of F . If all ele-ments of F contain p, F is a point pencil. Otherwise m ≤ k and an elementF of F that does not contain p hits H in at least an (m − 1)-dimensionalsubspace, since it must intersect all hitting lines. That means that two suchelements intersect in at least an (m− 2)-subspace. Define F∗ as the subsetof F of elements that do not contain p. The previous argument shows that

|F∗| ≤ qm[n−mk −m

]q

.


If m ≥ 3, this is a lot smaller than the size of F , if l = 2 Lemma 2.34 showsthat F∗ is a lot smaller than F . In both cases the size of F∗ is negligiblecompared to the size of F . Now we have enough information to determinea new upper bound on the size of F :

|F| ≤[m

1

]q

[n− 2k − 2

]q

+ qm[n−mk −m

]q

if m ≥ 3, and

|F| ≤[m

1

]q

[n− 2k − 2

]q

+ q2

[k

1

]q

[n− 3k − 3

]q

if m = 2.

Note that if m = k, this structure is the structure of the q-Hilton-Milnerexample. If m < k, consider 〈F, p〉 for an element F ∈ F that does notcontain p such that 〈F, p〉 contains H. It is clear that F meets H in an(m − 1)-subspace L. If now all elements of F are also contained in 〈F, p〉,than m = k, so there is another element F ′ ∈ F not contained in 〈F, p〉.Now F ′ also meets H in an (m − 1)-subspace L′ and F ∩ F ′ is at most an(m− 2)-space. If the intersection is an (m− 2)-space, the point p can havedegree at most [

k − 11

]q

[n− 2k − 2

]q

+[k

1

]2

q

[n− 3k − 3

]q

.

If n < 3k−1, write n = 2k+d, for 1 ≤ d < k−1. Then we assumed thatthe order of magnitude of the size of F is at least q(n−k)(k−2)+k−d. We foundthat the size of F must have an order of magnitude at most q(n−k)(k−2)+m−1.Therefore k−d ≤ m− 1, or m ≥ k−d+ 1. In particular, if n = 2k+ 1, thatmeans that m = k and we have the q-Hilton-Milner example.

We summarize everything in the following theorem.

Theorem 2.41. Let F is a maximal [n, k]q-EKR family and one of thefollowing holds:

1. n ≥ 3k − 1 and

|F| >[n− 2k − 2

]q

+ (q + 1)

([k

1

]q

− 1

)[k

1

]q

[n− 3k − 3

]q

,

2. n = 2k + d, for 1 ≤ d < k − 1 (such that n < 3k − 1),

|F| ≥[n− 2k − 2

]q

q2 + q + 1(q − 1)2

q3k−n,

and one of the following hold:


(a) q ≥ 4 and n ≥ 2k + 2,

(b) q = 3 and n ≥ 2k + 3, or

(c) q = 2 and n ≥ 2k + 4,

3. n = 2k + d, for 1 ≤ d < k − 1 (such that n < 3k − 1),

|F| ≥[k

1

]q

[n− 2k − 2

]q

− q[k

2

]q

[n− 3k − 3

]q

,

and one of the following hold:

(a) q ≥ 4 and n = 2k + 1, or

(b) q = 3 and n = 2k + 2.

Then F is one of the following:

• F is a point pencil, and

|F| =[n− 1k − 1

]q

.

• There is a plane such that all elements of F intersect that plane in atleast a line, in that case F has approximate size

(q2 + q + 1)[n− 2k − 2

]q

.

• F has the structure described above for some 2 ≤ m ≤ k, and

|F| ≤[m

1

]q

[n− 2k − 2

]q

+ qm[n−mk −m

]q

if m ≥ 3, and

|F| ≤ (q + 1)[n− 2k − 2

]q

+ q2

[k

1

]q

[n− 3k − 3

]q

if m = 2.

If m = k, F is a q-Hilton-Milner family.

In the following corollary we state the situation for n = 2k+ 1 explicitlysince we need this in the next chapter.

Corollary 2.42. Let F be a maximal [2k + 1, k]q-EKR family, for q ≥ 4,such that

|F| ≥[k

1

]q

[2k − 1k − 2

]q

− q[k

2

]q

[2k − 2k − 3

]q

,

Then F is one of the following:

2.6. SMALL MAXIMAL CLIQUES 53

• F is a point pencil, and

|F| =[

2kk − 1

]q

.

• There is a plane such that all elements of F intersect that plane in atleast a line, in that case F has approximate size

(q2 + q + 1)[

2k − 1k − 2

]q

.

• F is a q-Hilton-Milner family and

|F| ≤[k

1

]q

[2k − 1k − 2

]q

+ qk.

2.6 Small maximal cliques

In a previous section we studied the maximal size of maximal intersectingfamilies of subsets of a set, and the q-analogous question of the maximalsize of a maximal intersecting family of subspaces of a vector space. In thissection we will study the other end of the spectrum. The main question hereis what is the minimal size of a maximal intersecting family of finite sets (orsubspaces of some finite vector space).

A k-clique is a collection of pairwise intersecting k-sets. A k-clique ismaximal if it is not possible to add a new k-set that intersects all of thek-sets in the clique. The support of a clique is the union of all elements ofthe sets in the clique. A blocking set of a k-clique is a set that intersects allthe k-sets of that clique.

It’s easy to see that if each blocking set of a k-clique either is an elementof the clique or its size is larger than k, then the clique is maximal.

The minimal number of k-sets in a maximal k-clique will be denoted bym(k).

It is trivial to determine m(2). Take two intersecting 2-sets, say {0, 1}and {0, 2}; it is clear that this is not maximal. To add a third 2-set, we have2 possibilities: we can add a new element 3 and take the pair {0, 3}, or take


the pair {1, 2}. It is clear that in the former case, we can add more newsets by adding new elements, while in the latter case, the clique is maximal.Therefore m(2) = 3.

The following case, m(3), is a bit more work, but can also be determined:

Proposition 2.43. m(3) = 7.

Proof. Consider the Fano plane PG(2, 2), this is a configuration of 7 pair-wise intersecting lines of size 3 and hence a 3-clique. Because the smallestblocking sets of a projective plane are lines, this clique is maximal. Thisgives m(3) ≤ 7. Explicit checking of all cases shows that smaller examplesdo not exist.

We saw that in this case the projective plane of order 2 gives an upperbound for m(3). In the same way a projective plane of order n gives a max-imal n + 1-clique. This bound, and the other known bounds for m(k) aregiven in the following overview:

Theorem 2.44. 1. m(k) ≥ 83k − 3 (Erdos and Lovasz [19]),

2. m(k) ≤ k2 − k + 1 if a projective plane of order k − 1 exists (Meyer[41]),

3. m(2k) ≤ 3k2 (Furedi [22]),

4. m(k2 +k) ≤ k4 +k3 +k2 if a projective plane of order k exists (Furedi[22]),

5. m(kn + kn−1) ≤ k2n + k2n−1 + k2n−2 for all n, if a projective plane oforder k exists (Drake and Sane [15]),

6. m(k) ≥ 3k if k ≥ 4 (Dow, Drake, Furedi and Larson [14],

7. m(k) < kf(k), where f(k) = c · k712 (Furedi [22]),

8. m(k) ≤ k5 (Blokhuis [3]),

9. m(k) ≤ 34k

2 + 32k − 1 if k − 1 is an odd prime power ≥ 7 (Blokhuis

[3]),

10. m(k) ≤ k2

2 + 5k + o(k) if k − 1 is a prime power (Boros, Furedi andKahn [5]).


With those bounds, we can determine m(4): the third bound gives:m(4) ≤ 12, while the sixth gives m(4) ≥ 12. A maximal 4-clique with 12sets can be constructed as follows: take the sets {0 + i, 1 + i, 4 + i, 6 + i} fori = 0, . . . , 11 where addition is modulo 12. It is easy to see that any two ofthose sets intersect. Explicit checking of all cases shows that this clique ismaximal.

Now consider the q-analogue problem: what is the size (denoted mq(k))of the smallest maximal intersecting family of rank-k subspaces of some vec-tor space V(n, q). Just as in the set case, where the support of the familywas not a parameter of the problem, here the dimension n is not important.

Can we determine mq(2)? We are looking for small maximal intersect-ing families of rank-2 spaces (projective lines). We saw that the only twopossible types of maximal intersecting families of projective lines were pointpencils or hyperplanes (all the lines of a hyperplane). Because we can alwaysadd more lines to a point pencil, this type will never be maximal. So thesmallest maximal intersecting family consists of all the projective lines of aprojective plane. Hence mq(2) = q2 + q+ 1. Note that, again, the limit casefor q → 1+ is 3 = m(2).

Now what about mq(k) for k ≥ 3?

Theorem 2.45. If k − 1 is a prime power, then mq(k) ≤ k2 − k + 1.

Proof. Let k − 1 be a prime power and consider the maximal k-clique Karising from the lines of the projective plane PG(2, k − 1). We showed thatthis was a maximal k-clique of size k2 − k + 1. From K we will constructan intersecting family K of rank-k spaces, that has the same size, and showthat it is maximal.

Take a basis {e1, e2, . . . , ek2−k+1} of V(k2−k+1, k−1). For each elementK = {i1, i2, . . . , ik} of K, consider the rank-k space K =< ei1 , ei2 , . . . , eik >

and let K be the family of all those spaces. It is obvious that any two ofthose spaces intersect, so K is an intersecting family of k2 − k + 1 rank-kspaces.

Now suppose that K is not maximal, that means we can add a rank-kspace U that intersects all elements of K. Now as each rank-k space, U isthe row space of a unique k× (k2− k+ 1) matrix M in row reduced echelon


form. Without loss of generality we can assume that (maybe after relabelingthe basis vectors, and thus relabeling the elements of X) M has this form:

1 0 · · · 0 ? · · · ?

0 1... ? · · · ?

.... . .

... ? · · · ?0 · · · · · · 1 ? · · · ?

With this labeling, the space < e1, e2, . . . , ek > must be an element of

K. Suppose that {1, 2, . . . , k} 6∈ K, that means there is a k-set K in K thatis disjoint from it, so K ⊆ {k + 1, . . . , k2 − k + 1}. But then U is disjointfrom K, a contradiction.

Now consider a matrix element Mi,j where 1 ≤ i ≤ k and k + 1 ≤ j ≤k2−k+1. Because K arises from a projective plane, there is a k-set Ki,j ∈ Kthat intersects {1, 2, . . . , k} in the element i and does not contain the elementj. Because U is a blocking set of K, U has to intersect Ki,j . Each vectorin this intersection must be a multiple of the i-th row of M , but if Mi,j

is a non-zero entry, then none of these vectors belong to Ki,j . ThereforeMi,j = 0 for each 1 ≤ i ≤ k and k+ 1 ≤ j ≤ k2− k+ 1. That means that Uis spanned by the first k basis vectors, but then U =< e1, e2, . . . ek >. Thisis in contradiction with the assumption that U was not an element of K.

In the proof of this theorem we constructed a maximal intersecting fam-ily of k-spaces from a maximal k-clique arising from a projective plane. Wecan apply this construction to any maximal k-cliques, also those not arisingfrom a projective plane, but the resulting intersecting family of k-spaces willnot always be maximal, as the following example shows.

Example 2.46. The 3-clique consisting of the ten 3-subsets of a 5-set ismaximal. If we apply the construction to this clique, we get an intersectingfamily of 3-spaces. This family is not maximal because (once a basis is cho-sen) we can add the3-space < e1 + e2, e3, e4 > for example.

In the proof of Theorem 2.45, we only use the following property of aprojective plane: For each line L, for each point p on L and for each pointq not on L, there is a line that intersects L in p and does not contain q. We


will call this property in the setting of k-sets in a k-clique property P . Solet K be a k-clique with support X, then property P is satisfied in K if:

For each k-set K ∈ K, for each x ∈ K and for each y ∈ X \K there is ak-set K ′ ∈ K intersecting K in x and not containing y.

So now we can use the same construction to prove:

Theorem 2.47. If property P is satisfied for a maximal k-clique of size m,then there is a maximal intersecting family of k-spaces with the same size.

Chapter 3

The Kneser and q-Knesergraphs

Besides extremal set theory, extremal graph theory is another importantsubdomain of extremal combinatorics. A typical problem in here is deter-mining the extremal values of a graph parameter of a certain type of graphs.In this chapter we will look at some graph parameters of the Kneser graph,which we already encountered in Section 1.2.3, and its q-analogue. Theultimate goal is to find the chromatic number of the q-Kneser graph.

We already saw that a coloring of a graph is nothing else than a ho-momorphism of this graph into a complete graph and that in the sameway homomorphisms into Kneser graphs are equivalent to multiple color-ings. Another reason why the Kneser graph is interesting is because it is anexample of a graph with high chromatic number and no short odd cycles.Those objects are much desired in graph theory, but we will come back tothis later in this chapter.

3.1 Definitions and properties

Definition 3.1 (Kneser graph). The Kneser graph K(n, k), with 1 ≤ k ≤ n,is a graph with vertex set the k-subsets of an n-set, in which two verticesare adjacent if they are disjoint as k-subsets. It is obvious that K(n, k) isan edgeless graph if 2k > n and that K(n, 1) is a complete graph. That iswhy we will restrict our definition to k ≥ 2 and 2k ≤ n.

59

60 CHAPTER 3. THE KNESER AND Q-KNESER GRAPHS

Without loss of generality we can take {1, 2, . . . , n} for this n-set in thedefinition of the Kneser graph and when we write out a vertex as a k-subsetwe will write the elements in increasing order.

Remark 3.2. • The Johnson scheme J(n, k) (k ≤ n), named afterS. M. Johnson, is the association scheme of which the elements arek-subsets of an n-set and the relation between two k-sets (also calledthe Johnson distance) is half the size of their symmetric distance. Forany 2 ≤ k ≤ n

2 , the Kneser graph K(n, k) is the graph that describesthe distance-k relation in the Johnson scheme J(n, k). Another graphassociated with the Johnson scheme is the Johnson graph, it describesthe distance-1 relation in the Johnson scheme.

• Odd graphs are a special case of Kneser graphs: the odd graph Om+1

is the graph with vertices the m-subsets of an 2m + 1-set where twom-subsets are adjacent if and only if they are disjoint, hence Om+1 =K(2m+ 1,m).

We have the following properties for the Kneser graph:

Proposition 3.3. For n ≥ 2k we have:

(i) If n > 2k, the automorphism group of K(n, k) is the symmetric groupon n symbols, Sym(n);

(ii) K(n, k) is vertex and edge transitive;

(iii) K(n, k) is regular with degree(n−kk

);

(iv) K(n, k) has 12

(n−kk

)(nk

)edges.

If n = 2k, each vertex has degree 1. It is easy to see that K(2k, k) isa graph with

(2kk

)vertices and 1

2

(2kk

)edges that are pairwise disjoint (this

configuration is sometimes called a perfect matching).

The smallest non-trivial Kneser graph which is not a perfect matchingis K(5, 2), which is the Petersen graph (see figure 3.1).

The q-analogous definition of the Kneser graphs goes as follows:

3.2. HOMOMORPHISMS. 61

Figure 3.1: K(5, 2) also known as the Petersen graph.

The q-Kneser graph Kq(n, k) (n ≥ 2k) is the graph with vertex set allrank-k subspaces of V(n, q). Two vertices are adjacent if they intersect triv-ially as subspaces. Or stated projectively: Kq(n, k) (n ≥ 2k) is the graphwith vertex set all (k − 1)-spaces of PG(n − 1, q) where two vertices areadjacent if the correspondent subspaces are disjoint.

We have similar properties as those for the Kneser graph:

Proposition 3.4. For n > 2k and prime power q we have:

(i) If n > 2k, the automorphism group of Kq(n, k) is isomorphic to PΓL(n, q),

(ii) Kq(n, k) is vertex and edge transitive,

(iii) Kq(n, k) is regular with degree qk2[n−kk

]q,

(iv) Kq(n, k) has 12qk2[n−kk

]q

[nk

]q

edges.

Here the case n = 2k is not so trivial anymore. We will notice that whenwe determine the chromatic number.

3.2 Homomorphisms.

In this section we will determine some homomorphisms between Knesergraphs, which will help us to determine the chromatic number later.


3.2.1 Homomorphisms between Kneser graphs

We give three different homomorphisms between Kneser graphs:

Extension map

A k-subset of an n-set is of course also a k-subset of that same n-set withone extra element added. Therefore the vertex set of K(n, k) is a subsetof the vertex set of K(n + 1, k). It’s clear that adjacent vertices in K(n, k)are also adjacent in K(n + 1, k) and that all edges in K(n + 1, k) betweenvertices of K(n, k) are also edges in K(n, k). Therefore K(n, k) is an inducedsubgraph of K(n + 1, k) and there is a embedding K(n, k) ↪→ K(n + 1, k).This homomorphism is called the extension map.

Multiplication map

Take a positive integer t and define the map µt : K(n, k) → K(tn, tk) asfollows:

µt(a) =t−1⋃i=0

{a1 + in, . . . , ak + in}

for every vertex a = {a1, . . . , ak} of K(n, k). This clearly defines an embed-ding which is called the multiplication map.

Coloring map

For K(n, k) with n > 2k, a vertex a = {a1, a2, . . . , ak} is said to be l-regularif al = l and al+1 > l + 1 for some 1 ≤ l ≤ k. A vertex that is not l-regularfor any l is called irregular. Because we write the elements of a k-subset inincreasing order, l-regularity means that ai = i for 1 ≤ i ≤ l and ai > i forl + 1 ≤ i ≤ k. If a is irregular, this means that ai > i for all i.

In [45] S. Stahl defines the map η : K(n, k)→ K(n− 2, k− 1) as follows:

• if a is l-regular for some l:

η(a) = {a2 − 1, a3 − 1, . . . , al − 1, al+1 − 2, . . . ak − 2},

3.2. HOMOMORPHISMS. 63

• if a is irregular:

η(a) = {a2 − 2, a3 − 2, . . . , ak − 2}.

This is clearly a map into K(n−2, k−1) but is it also a homomorphism?Take two adjacent vertices a = {a1, a2, . . . , ak} and b = {b1, b2, . . . , bk} ofK(n, k). This means that a and b are disjoint as sets, so they cannot bothbe regular, since then they would both contain the element 1. So we haveto check two cases:

1. Both a and b are irregular. In that case η(a) = {a2−2, a3−2, . . . , ak−2} and η(b) = {b2− 2, b3− 2, . . . , bk− 2}. If those sets are not disjoint,there is a ai−2 equal to a bj−2 for some 2 ≤ i, j ≤ k, but then ai = bjand the sets were not disjoint, a contradiction.

2. One vertex, say a, is l-regular for some l and the other, b, is irregular.So we have:

η(a) = {a2 − 1, a3 − 1, . . . , al − 1, al+1 − 2, . . . ak − 2},

η(b) = {b2 − 2, b3 − 2, . . . , bk − 2}.If those sets are not disjoint we have that for some 2 ≤ j ≤ k bj ∈ η(a),so there are two possibilities:

(i) bj − 2 = ai − 2 for some l + 1 ≤ i ≤ k. But then ai = bj and wehave a contradiction.

(ii) bj − 2 = ai − 1 for some 2 ≤ i ≤ l. Because 1, 2, . . . , l ∈ a allelements of b must be at least l+ 1. Together with bj = ai + 1 =i+ 1 ≤ l+ 1 this means that bj = l+ 1. But l+ 1 is the smallestvalue an element of b can have, so j must be equal to 1, but b1 isnot an element of η(b), a contradiction.

In each case the assumption that η(a) and η(b) are not disjoint yieldsa contradiction, so they must be disjoint and hence adjacent as vertices inK(n− 2, k − 1), so η is a homomorphism.

For reasons that will become clear in Section 3.3.1 this homomorphismis called the coloring map.

In the same paper ([45]) Stahl conjectures that any homomorphism fromK(n, k) to K(n′, k′) is a composition of extension, multiplication and color-ing maps.


3.2.2 Homomorphisms between q-Kneser graphs

In [9] the following homomorphisms are given:

q-Extension map

There is a q analogue of the extension map between ordinary Kneser graphs,this is an embedding Kq(n, k) ↪→ Kq(n+ 1, k) which is also called the exten-sion map.

q-Multiplication map

A rank-k subspace of V(n, qr) for some r ≥ 1 can be viewed as a rank-rksubspace of V(rn, q). This yields an embedding Kqr(n, k) ↪→ Kq(rn, rk).This can be seen as a q-analogue of the multiplication map, hence the nameq-multiplication map.

Note that there is no q-analogue of the coloring map. We will give anindication of the reason for this in Section 3.3.2. The following two homo-morphisms have no analogue for the ordinary Kneser graphs.

Subfield map

Because GF(q) is a subfield of GF(qr) for all r ≥ 1, there is a natural ho-momorphism Kq(n, k)→ Kqr(n, k), called the subfield map.

Deletion map

Each rank-k space of V(n, q) (once a basis is chosen) is the row space ofa unique k × n matrix in row reduced echelon form. When the first lineis deleted from this matrix, the first column only contains zeroes, so thiscolumn can be deleted too. What is left is a (k− 1)× (n− 1) matrix in rowreduced echelon form, with row space a rank-(k−1) subspace of V(n−1, q).This defines a map γ : Kq(n, k) → Kq(n − 1, k − 1). The matrices of twoadjacent vertices in Kq(n, k) put underneath each other give a 2k×n matrixof row rank 2k. The matrices of the images under this map underneath eachother give a (2k−2)× (n−1) matrix. If those images are not adjacent, they

3.3. CHROMATIC NUMBERS 65

are not disjoint as subspaces and the row rank of this (2k−2)×(n−1) matrixis smaller than 2k − 2, but then the row rank of the original 2k × n matrixcannot be 2k. So this map defines a homomorphism, called the deletion map.

3.2.3 A homomorphism from a q-Kneser graph into a Knesergraph

If we forget the vector space structure on V(n, q), this is just a set of[n1

]q

rank-1 subspaces and a rank-k subset of V(n, q) is just a[k1

]q-subset of this

set. This yields an embedding Kq(n, k) ↪→ K([

n1

]q,[k1

]q

).

3.3 Chromatic numbers

In this section we will determine the chromatic number of the Kneser graphand present new results on the chromatic number of the q- Kneser graph.

3.3.1 The Kneser graphs

We saw that K(2k, k) is a perfect matching. As a consequence the chromaticnumber is known: χ(K(2k, k)) = 2.

If we write n = 2k + r, we can find an upper bound for the chromaticnumber using the coloring map defined in the previous chapter. By usinginduction there is a homomorphism

K(2k + r, k)→ K(2k + r − 2, k − 1)→ . . .→ K(r + 2, 1).

The last graph in this chain is of course the complete graph Kr+2 so thereis a coloring of K(2k+ r, k) with r+ 2 colors. Hence χ(K(2k+ r, k)) ≤ r+ 2.

We can easily make this coloring explicit: each vertex that is a k-subsetof {1, 2, . . . , 2k − 1} is given the color 1. Next, all vertices containing thenumber 2k + r are colored with the color r + 2, all vertices containing thenumber 2k+ r− 1 that are not colored already are given the color r+ 1, theuncolored vertices containing 2k+r−2 are colored r, and so on. Finally, thevertices containing the element 2k that are not colored yet are colored with 2.


In 1955 M. Kneser [34] conjectured that r + 2 was also a lower bound.

For k = 2 and k = 3 there are nice combinatorial proofs of this conjec-ture that use the Hilton-Milner theorem (theorem 2.5):

Proposition 3.5. For n ≥ 4 we have that χ(K(n, 2)) = n− 2.

Proof. The proof goes by induction on n. For n = 4, K(4, 2) is bipartite, soit has chromatic number 2 = 4− 2.

Suppose for contradiction that K(n, 2) has a coloring with n − 3 colorsfor some n > 4. From the Hilton-Milner theorem, it follows that, if a colorclass contains more than three vertices, the sets lie in a point pencil. If thisis the case for some color class, recolor the graph such that this color classforms an entire point pencil. By removing all vertices in this color class weget a coloring of K(n− 1, 2) with n− 4 < (n− 1)− 2 colors, a contradictionwith the induction hypothesis. So each color class contains at most threevertices. Therefore,

3(n− 3) ≥ |K(n, 2)| = n(n− 1)2

.

This implies that n2−7n+ 18 ≤ 0, which is impossible. Hence, the assump-tion that K(n, 2) can be colored with n− 3 colors is false.

The proof for k = 3 is completely analogous. In that case the Hilton-Milner theorem states that a color class with more than 3n−8 vertices mustlie in a point pencil. The calculation in the proof leads to the inequalityn3 − 6n2 + 25n− 40 ≤ 0 that has no solutions.

Unfortunately for k ≥ 4 the Hilton-Milner bound does not lead to acontradiction anymore in the method used above, so we need another prooffor the general case.

In 1978, L. Lovasz proved the general conjecture ([36]) using clever topo-logical constructions and Borsuk’s theorem. This was one of the first andmost spectacular applications of topological methods in combinatorics. Afew weeks after this I. Barany [1] found a much shorter proof also usingBorsuk’s theorem and a theorem of D. Gale. Here we state Gale’s theorem,Borsuk’s theorem (actually an equivalent statement) and show Barany’s


proof of Kneser’s conjecture.

Theorem 3.6 (D. Gale (1956) [23]). For every m, r ≥ 0 there exists anarrangement of 2m+ r points on the r-dimensional unit sphere Sr in Rr+1

such that every open hemisphere contains at least m of them.

We will call such a set of 2m+ r points a Gale set.

Two points x and y on the unit sphere Sd−1 in Rd are called antipodal ifx+ y = 0. For an ε > 0 we call two points ε-nearly antipodal if ||x+ y|| < ε.Borsuk’s graph B(d, ε) is an infinite graph with vertex set the points of Sd−1

and two points are adjacent if they are ε-nearly antipodal.

Borsuk’s theorem is equivalent with:

Theorem 3.7 (K. Borsuk (1933) [6]). For any ε > 0 the chromatic numberof Borsuk’s graph B(d, ε) is at least d+ 1.

Using this we can now show Barany’s proof of Kneser’s Conjecture:

Theorem 3.8 (Kneser’s Conjecture). The chromatic number of Kneser’sgraph K(2k + r, k) (r ≥ 0) is r + 2.

Proof. It is clear that we only have to prove that r + 2 is a lower bound.Take a Gale set G ⊂ Sr of 2k + r points. From Gale’s theorem it followsthat every open hemisphere contains at least k points of G. It is easy toshow that for some δ > 0 this property remains valid if we exclude the rimof width δ of each hemisphere. The property thus becomes: there are atleast k points at distance at most

√2− δ from each point in Sr.

Now take the Kneser graph K(2k + r, k) with vertex set the k-subsetsof G and suppose we have a coloring of this graph with c colors. Now weconstruct a coloring of all the points of Sr. Take a point x ∈ Sr. There is ak-subset of G in which all the points have distance at most

√2− δ from x.

We give this point the same color as the k-subset (as vertex in the Knesergraph) we just mentioned.

To show that this coloring is also a coloring of the Borsuk graph B(r+1, ε)for some ε > 0 we have to show that two ε-nearly antipodal points have adifferent color. Take two points x, y ∈ Sr with the same color. That meansthat their corresponding k-subsets must intersect (because otherwise they


are adjacent as vertices in the Kneser graph and can not have the samecolor). So both x and y are at distance at most

√2 − δ from a point z in

this intersection. If δ is small enough and we take ε = δ it is impossible thatx and y are ε-nearly antipodal.

Because we have a coloring with c colors of B(r + 1, ε), it follows fromTheorem 3.7 that c is at least r + 2.

After 1987 other proofs of this conjecture appeared (by V.L. Dol’nikov[13, 12] and K.S. Sarkaria [42] among others) but all of them using topolog-ical methods. In 2004 J. Matousek [38] gave a purely combinatorial proof(by specializing a combinatorial proof of Tucker’s lemma [47], a topologicaltheorem).

An interesting problem in graph theory is finding graphs with large chro-matic number and some extra constraint. A graph containing a completegraph of size k as a subgraph has chromatic number at least k, but such agraph clearly contains a lot of short cycles. Are there graphs with high chro-matic number and no short cycles? It turns out that constructing graphswith high chromatic number and no short even cycles is not so easy. Butavoiding short odd cycles is now easier. Suppose we want a graph withchromatic number c, choosing r = c − 2 gives a Kneser graph K(2k + r, k)for each k with this chromatic number. A 3-cycle in a Kneser graph cor-responds with three mutually disjoint k-subsets of the (2k + r)-set, whichis impossible if 3k > 2k + r. Hence if k > r there are no 3-cycles. Usingsimilar arguments it can be shown that making k even bigger relative to r wecan avoid also larger odd cycles, while the chromatic number stays the same.

3.3.2 The q-Kneser graphs

The case n > 2k

In this case largest independent sets are the families consisting of all rank-ksubspaces that have a given rank-1 subspace in common. Such a family hassize

[n−1k−1

]q. A rather easy upper bound here is given as follows: a rank-

(n−k+ 1) subspace has a nontrivial intersection with each rank-k space, sothe rank-1 spaces of this subspace color all-k spaces. Therefore:

χ(Kq(n, k)) ≤[n− k + 1

1

]q

.


For k = 2 A. Chowdhury, C. Godsil and G. Royle [9] prove that this isthe correct value:

χ(Kq(n, 2)) =[n− 1

1

]q

.

Remark 3.9. Note that if we use the parameters from the original Knesergraphs this translates to:

χ(Kq(4 + r, 2)) =[r + 3

1

]q

.

Taking the limit for q → 1+ we get r+ 3 instead of r+ 2. This is a casewhere the q-analogue does not correspond with the original problem if wetake the limit.

What remains to prove is the chromatic number of Kq(2k+r, k) for k > 2and r > 0.

The following lemma shows that if we can prove that for r = 1 the upperbound mentioned above is the right number, we can prove that this upperbound is the chromatic number for all r:

Lemma 3.10. If

χ(Kq(2k + 1, k)) =[k + 2

1

]q

for k ≥ 3 and some values of q, then

χ(Kq(n, k)) =[n− k + 1

1

]q

for all n > 2k + 1 and the same values of q.

Proof. Using the deletion map, a homomorphism Kq(n, k)→ Kq(n−1, k−1),we can construct a homomorphism Kq(2n − 2k − 1, n − k − 1) → . . . →Kq(n, k). , Because of our assumption, we have that

χ(Kq(2(n− k − 1) + 1, n− k − 1)) =[n− k + 1

1

]q

.


This, together with the upper bound we get from the coloring describedabove gives us:[

n− k + 11

]q

≤ χ(Kq(n, k)) ≤[n− k + 1

1

]q

.

Now to continue we have to know what is the size of a maximal indepen-dent set of Kq(2k+ 1, k) that is no point pencil. In the previous chapter wesaw that the largest such families (in the case of n = 2k+ 1) were q-Hilton-Milner type families, so from Corollary 2.42 we can conclude the followingproposition.

Proposition 3.11. The size of a maximal independent set of Kq(2k+ 1, k)(k ≥ 3, q ≥ 4) that is no point-pencil is at most cqk

2−3 for some positiveconstant c.

Now we can prove that Kq(2k+1, k) has the asymptotic chromatic num-ber we think it has for k ≥ 3:

Lemma 3.12. For all k ≥ 3, there is a prime power qk such that for allq ≥ qk we have the following. If there is a coloring of Kq(2k + 1, k) with at

most[k+2

1

]q

colors, then all colors are point pencils.

Proof. Suppose there is a coloring of Kq(2k + 1, k) with[k+2

1

]q

colors. Let

G be the set of centers of the point pencils used in the coloring and B beset of colors that are no point pencil (let’s call those the bad colors), hence|G|+ |B| ≤

[k+2

1

]q.

There are exactly[k+2

1

]q

(k − 1)-dimensional subspaces that contain a

fixed (k− 2)-dimensional subspace π. If π does not contain a point of G, atmost |G| of the (k−1)-spaces containing π are colored by point pencils. Thismeans that at least |B| of the (k − 1)-subspaces containing π are coloredwith a bad color.

Now count the pairs (π, ρ), where π is a (k− 2)-subspace not containingelements of G, ρ is a (k− 1)-subspace colored by a bad color and ρ containsπ. This counting yields:

|B|[

2k + 1k − 1

]q

≤ cqk2−3|B|[k + 1

1

]q

.


If |B| 6= 0 we can divide both sides of this inequality by |B|. In that casethe left hand side of this inequality is a monic polynomial in q of degreek2 + k − 2 and the right hand side a polynomial of degree k2 + k − 4 withleading coefficient c > 0. That means that for each k ≥ 3 there is a primepower qk such that this inequality does not hold anymore for all q ≥ qk.

In those cases there cannot be any bad colors.

To complete the theorem we need a theorem by R.C. Bose and R.C. Bur-ton of 1966 that says that we need at least an (n− k)-dimensional subspaceto color Kq(n, k) with point pencils only.

Theorem 3.13 ([7]). If F is a set of points in PG(n− 1, q) that intersectsall (k − 1)-subspaces then:

|F | ≥[n− k + 1

1

]q

.

Equality holds if, and only if, F is an (n− k)-dimensional subspace.

Now we can prove the chromatic number for (almost) all cases.

Theorem 3.14. For all k ≥ 3 and n > 2k there is a prime power qk suchthat for all q ≥ qk we have that

χ(Kq(n, k)) =[n− k + 1

1

]q

.

The colors in a minimal coloring of χ(Kq(n, k)) are point pencils with centerson a (n− k)-dimensional subspace.

Proof. Lemma 3.12 together with Theorem 3.13 tells us the chromatic num-ber in the case n = 2k+1. Lemma 3.10 completes the proof of the chromaticnumber for all n > 2k.

Theorem 3.13 shows that all minimal colorings come from an (n − k)-subspace.

The case n = 2k

This case was very easy and not so interesting in the set case. Here on thecontrary it is not so straightforward. An independent set in this graph isa collection of rank-k subspaces that pairwise intersect non-trivially. The


largest independent sets in this graph are the two types described in theq-analogue of the Erdos-Ko-Rado problem: all rank-k subspaces containinga given rank-1 subspace (type I, point pencils) and all rank-k subspacescontained in a hyperplane (type II). We can identify sets of type I with therank-1 subspace contained in all rank-k subspaces of this set and the sets oftype II with the hyperplane that contains all the rank-k subspaces of thisset. Both types have size

[2k−1k−1

]q

and this is also the independence number

of the q-Kneser graph.

We try to find an upper bound for the chromatic number. A colorcorresponds to an independent set in the graph and by using the largestindependent sets to color the graph we get a small chromatic number. Anidea is to take a rank-(k + 1) subspace U in V(n, q). Because all the rank-ksubspaces intersect this subspace in at least a rank-1 space, taking all therank-1 spaces (as sets of type I) in this rank-(k + 1) space as colors, all thevertices are colored1. That way we get:

α(Kq(2k, k)) =[k + 1

1

]q

.

Unfortunately, taking the limit for q → 1+ gives k+1, which is much higherthan the chromatic number of the Kneser graph. But we can do better. Westill consider the rank-(k+1) space, let us call it U but now we take a rank-kspace inside of it, say V. We take qk type I colors: the rank-1 spaces in Uthat are not in V; and qk−1 type II colors: the hyperplanes that intersectU in V. It is easy to see that all rank-k spaces are colored that way. Thisgives us:

α(Kq(2k, k)) ≤ qk + qk−1

which, in the limit to the set case, gives us the expected value 2. So maybewe have found the right value?

In a 2001 paper J. Eisfeld, L. Storme and P. Sziklai [16] prove usinglengthy counting arguments that for k = 2 this is the chromatic number. Inthis case we are talking about projective lines in PG(3, q). They also provethat a minimal coloring is of the following form: take a plane and a pointon this plane, now take s (with 1 ≤ s ≤ q) lines on that plane through that

1For a vertex meeting U in at least a rank 2 space, we can choose one of the colors inthis subspace.


point. As type I colors take the points on the plane but not on one of the slines and as type II colors take the planes that intersect the given plane inone of the s lines.

Until now the only lower bound known in the case k > 2 was the one weget from proposition 1.15:

χ(Kq(2k, k)) ≥

[2kk

]q[

k+11

]q

= qk + 1.

If we only use point pencils and hyperplanes as colors we can show thatyou need at least qk + qk−1 colors:

Theorem 3.15. A minimal cover of the rank-k subspaces of V(2k, q), withk ≥ 2, using only point pencils and hyperplanes has size at least qk + qk−1.

Proof. Let P be the set of centers of the point pencils used in the coverand let H be the set of hyperplanes used in the cover. We will show that|P |+ |H| ≥ qk + qk−1.

Take a rank-k subspace π that is covered by a point pencil with centera projective point p. We can assume that π is not contained in one of theelements of H, because if each element of P is contained in an element of Hthen the size of the cover is certainly larger than qk + qk−1

The number of rank-k-subspaces that intersect π in a rank-(k − 1) sub-space not containing p is

qk−1([k + 1

1

]q

− 1).

Indeed, there are[k1

]q−[k−1

1

]q

= qk−1 rank-(k−1) subspaces on π that

do not contain p, and each of those subspaces is contained in[k+1

1

]q− 1

rank-k subspaces different from π.

All those subspaces need to be covered by point pencils or hyperplanes.Take a point pencil with center not on π, this covers at most qk−1 rank-kspaces, one for each rank-(k− 1) space on π not containing p. A hyperplane


intersecting π in a rank-(k− 1) space, say πk−1, covers at most[k1

]q

rank-k

spaces. But each other hyperplane intersecting π in πk−1 covers at most anadditional qk−1 rank-k spaces. That means that the hyperplanes in H coverat most |H|qk−1 plus an additional qk−1(

[k1

]q− qk−1) = qk−1(

[k−1

1

]q) of

those rank-k spaces. So we have:

(|P | − 1)qk−1 + |H|qk−1 + qk−1

[k − 1

1

]q

≥ qk−1([k + 1

1

]q

− 1).

And this reduces to what we wanted to prove.

For k = 2 this is already the proof for the chromatic number becauseall maximal independent sets are point pencils and hyperplanes (see Section2.5.2). For k ≥ 3 there are also other maximal independent sets that aresmaller in size, but we do have to worry about them.

For k = 3, we can show that the maximal independent sets that are nopoint pencils or hyperplanes have a size that, as a polynomial in q, has adegree smaller than that of the size of a point pencil or hyperplane (this sizeis[

52

]q∼ q6:

Lemma 3.16. The size of a maximal independent set of Kq(6, 3) that is nopoint pencil or hyperplane is at most 6q5.

To show that q3 + q2 really is the chromatic number we need an upperbound on the amount of ‘bad’ colors:

Lemma 3.17. There is a prime power q3 such that for all prime powersq ≥ q3 we have the following. In a coloring of Kq(6, 3) with at most q3 + q2

colors, the number of colors that are not point pencils or hyperplanes isbounded above by q2 + 6q + 30.

Proof. Assume we have a coloring of Kq(6, 3) with at most q3 + q2 colors.Let G be the set of point pencils and hyperplanes used as a color (we willcall these the ’good’ colors) and B be the set of other colors (’bad’ colors).So we have that

|G|+ |B| ≤ q3 + q2.

Because those colors have to color all rank-3 spaces we have:


[63

]q

≤ |G|[

52

]q

+ |B|6q5

≤ (q3 + q2)[

52

]q

− |B|

([52

]q

− 6q5

)

Which leads to:

|B| ≤ q8 + q7 + q6 + q5 − q3 − q2 − q − 1q6 − 5q5 + 2q4 + 2q3 + 2q2 + q + 1

= q2 + 6q + 29 +132q5 − 72q4 − 72q3 − 66q2 − 36q − 30q6 − 5q5 + 2q4 + 2q3 + 2q2 + q + 1

Since this last fraction becomes strictly smaller than 1 when q is largeenough (say q ≥ q3 for some prime power q3) and |B| is a non-negativeinteger, we have that for q ≥ q3:

|B| ≤ q2 + 6q + 30.

With this upper bound we can prove what we wanted:

Theorem 3.18. There is a positive integer q3 such that for q ≥ q3 thechromatic number of the q-Kneser graph Kq(6, 3) is q3 + q2.

Proof. Assume we have a coloring of Kq(6, 3) with at most q3 + q2 colorsand consider again the sets G and B as in lemma 3.17. Furthermore, calla rank-3 space colored with a color from G ‘good’ and one colored with acolor from B ‘bad’.

Now we will proceed as in theorem 3.15: take a rank-3 space π (this is aprojective plane) colored by a point pencil with center the projective pointp. Again the number of planes intersecting π in a line not containing p isq2(q3 + q2 + q). But now, a number of those planes, say bπ will be ”bad”. Ifwe count the number of those planes colored by an other point pencil andby a hyperplane, we get the same results as in theorem 3.15. This meansthat:

(|G| − 1)q2 ≥ q2(q3 + q2 + q)− bπ − q2(q + 1)


This leads to:

|G| ≥ 1 +q2(q3 + q2 + q)− bπ − q2(q + 1)

q2

= q3 + q2 − bπq2.

Hence, |B| ≤ bπ/q2, or bπ ≥ |B|q2. Of course this holds for all planes

colored by a point, and all planes colored by a hyperplane (this situation iscompletely dual). So if we denote b = min{bπ|π is a good plane}, we havethat b ≥ |B|q2.

A double counting of the pairs (π, ρ) where π is a good plane, ρ a badone, and their intersection is a line gives us:

b([

63

]q

− |B|6q5) ≤ q(q2 + q + 1)2|B|6q5.

This reduces to

|B|([

63

]q

− |B|6q5) ≤ |B|(q2 + q + 1)26q4.

If we assume that |B| > 0 we can divide both sides of this inequality by|B|. For the remaining |B| in this inequality we can use the bound of lemma3.17 and assume that q is large enough.

Now we see that the left hand side of this equation has degree 9 and theright hand side only degree 8, so if q is large enough then this inequality willbe false.

This means that our assumption |B| > 0 is false for q large enough, soin those cases we only have good colors.

We believe the same can be done for k > 3.

Chapter 4

A family of point-hyperplanegraphs

4.1 Definition

In the previous chapter we generalized Kneser graphs over projective spaces.This generalization was quite literally: we replaced (sub)sets by (sub)spaces.Two subsets or subspaces were adjacent if they are disjoint. We remarkedthat this relation was the distance-k relation, or maximal distance relation,in the Johnson scheme J(n, k). In this chapter we will consider a graph withpairs of points and hyperplanes of some projective space as vertices and theadjacency relation will be represented by the maximal distance such twopairs can have.

Definition 4.1. Define the graph PH(n + 1, q) as follows: the vertex setis the set of incident point hyperplane pairs of PG(n, q) and two vertices(p,H) and (q,K) are adjacent if p 6∈ K and q 6∈ H.

Note that the discussion in this chapter is inspired by a particular po-lar q-Kneser graph defined in the next chapter. Indeed, consider the graphKQ

+

q (6, 2) (for a definition and properties see Chapter 5). It has as vertexset the set L of the totally isotropic projective lines of Q+(5, q) and twolines L,K ∈ L are adjacent in the graph if L ∩ K⊥ = ∅ (or equivalentlyK ∩ L⊥ = ∅).

Using the Klein correspondence we can give another description of thisgraph. The set of totally isotropic lines of Q+(5, q) is mapped under the

77

78 CHAPTER 4. A FAMILY OF POINT-HYPERPLANE GRAPHS

Klein correspondence to the set of incident point plane pairs of PG(3, q).For two such pairs (p,H) and (q,K), being at ‘maximal distance’ in thecollinearity relation means that p 6∈ K and q 6∈ H. The graphs PH(n+ 1, q)are a generalization of this graph over PG(n, q).

Proposition 4.2. The graph PH(n+1, q) as defined above has the followingproperties:

(i) PH(n+ 1, q) has[n+1

1

]q

[n1

]q

vertices,

(ii) PH(n+ 1, q) is regular with degree q2n−1.

Proof. (i) There are[n+1

1

]q

hyperplanes in PG(n, q) and each hyperplaneis incident with

[n1

]q

points.

(ii) Take a vertex (p,H). The hyperplane of an adjacent vertex cannotcontain p so there are qn possibilities for this hyperplane. The point ofan adjacent vertex cannot lie in H but has to lie in its own hyperplane,so there are qn−1 possibilities.

To find the chromatic number of PH(n+ 1, q) it is useful to study largecocliques. The following proposition shows that a maximal coclique has anice “linear” properties:

Proposition 4.3. Let C be a maximal coclique of PH(n+ 1, q).

(i) For each hyperplane H of PG(n, q), there is a subspace UH of H suchthat (p,H) ∈ C if and only if p ∈ UH . Dually, for each point p thereis a subspace Up containing p such that (p,H) ∈ C if and only if Up iscontained in H.

(ii) Take a hyperplane Y of PG(n, q) and a vertex (p,G) of PH(n, q) de-fined over the projective space Y (so G is an (n− 2)-dimensional pro-jective space contained in Y and p is a point of Y ). There are 0, 1 orq hyperplanes Hi of PG(n, q) such that Hi ∩ Y = G and (p,Hi) ∈ Cfor all i. Moreover, if it extends in q ways, then (p, Y ) ∈ C.

(iii) Take a hyperplane Y of PG(n, q) and suppose that both vertices (p,G′),(p,G′′) of PH(n, q) defined over Y extend in q ways into vertices ofC. Then all vertices (p,G) of PH(n, q) for which G′ ∩G′′ ⊆ G extendin q ways into vertices of C. Dually if both vertices (p′, G), (p′′, G) of

4.1. DEFINITION 79

PH(n, q) extend in q ways, then all vertices (p,G) for which p ∈ p′p′′extend in q ways.

Proof. (i) Take a hyperplane H and consider the set UH := {p | (p,H) ∈C}. We have to prove that this set is a subspace of H. This set isclearly a subset of H, so if we take two points p, q ∈ UH and provethat all points on the line pq are in UH we are done. So take a pointr on the line pq and suppose (r,H) 6∈ C. This is only possible if thereis a vertex (s,K) ∈ C such that both r 6∈ K and s 6∈ H. But becauses 6∈ H, both p and q must lie in K, but then pq and hence also r liesin K, a contradiction.

(ii) Denote by Hi, i = 1, . . . , q, the q hyperplanes that, intersected with Y ,give G. Suppose (p,G) extends in at least 2 ways into a vertex of C:without loss of generality we can assume that (p,H1), (p,H2) ∈ C. LetUp be the subspace mentioned in (i). Clearly Up ⊆ H1 ∩H2 = G, soUp ⊆ Hi for i = 3, . . . , q and Up ⊆ Y . Hence (p,Hi) ∈ C for i = 3, . . . , qand (p, Y ) ∈ C.

(iii) Denote by Hi, H′i, H

′′i, i = 1, . . . , q, the q hyperplanes that, inter-

sected with Y , give respectively G,G′ and G′′ for a G ⊇ G′ ∩G′′. Weknow that (p,H ′i), (p,H ′′i) ∈ C for i = 1, . . . , q. Now suppose that(p,H1) 6∈ C, this means that there is a (q,K) ∈ C such that p 6∈ Kand q 6∈ H1. But H ′1 ∩H ′2 ∩H ′′1 ∩H ′′2 = G′ ∩G′′ ⊆ G ⊆ H1, so atleast one of the four hyperplanes H ′1, H ′2, H ′′1, H ′′2 does not containq. Hence p ∈ K, a contradiction, so (p,Hi) ∈ C for i = 1, . . . , q.

If we represent a point p of PG(n, q) by a column vector p and ahyperplane H by a row vector H, then a point hyperplane pair (p,H) canbe represented by the rank-1 (n + 1) × (n + 1) matrix pH, and this is anincident pair if and only if Hp = 0, which is equivalent with tr(pH) = 0.Now take two vertices x = pH and y = qK, this means that Hp = Kq = 0.Those vertices are adjacent if and only if tr(xy) 6= 0. Indeed, if x andy are non-adjacent then Hq = 0 or Kp = 0, which is equivalent with(Hq)(Kp) = 0. If (Hq)(Kp) = 0 then 0 = tr(HqKp) = tr(pHqK) = tr(xy).If tr(xy) = 0, then 0 = tr(pHqK) = (Hq)tr(pK), which means that Hq = 0or Kp = tr(pK) = 0, hence x and y are non-adjacent.

Using this representation we can show another linearity property for amaximal coclique:


Proposition 4.4. Let C be a maximal coclique of PH(n+1, q). Viewed as aset of matrices C = 〈C〉∩X where X ⊆ sln+1(q) is the set of (n+1)×(n+1)rank 1-matrices with trace 0.

Proof. It is obvious that C ⊆ 〈C〉 ∩X. Now take an x ∈ 〈C〉 ∩X and takea basis {x1, . . . , xm} ⊆ C of 〈C〉, so x is a linear combination of the basiselements xi. Because the trace function is linear, tr(x) = 0, so x = pH isa vertex of PH(n + 1, q). Take a vertex y ∈ C. The linearity of the tracefunction and the fact that tr(xiy) = 0 for each i guarantees that x and yare non-adjacent. This means that x ∈ C and hence 〈C〉 ∩X ⊆ C.

Now take U1, U2 subspaces of PG(n, q) and consider the sets CUi :={(p,H) | p ∈ Ui, Ui ⊆ H} for i = 1, 2. Clearly both sets are cocliques ofPH(n+1, q) and it is not hard to see that if U1 ⊆ U2 then CU1 ∪CU2 is also acoclique. Using this fact we can construct a large coclique. Take a chamber(maximal flag) F of PG(n, q). This means that F = {U0, U1, . . . , Un−1},where, for each i, Ui is an i-dimensional projective subspace of PG(n, q),and Ui ⊆ Uj if i ≤ j. For this chamber we construct the following set:

CF := {(p,H) | ∃i ∈ {0, 1 . . . , n− 1} : p ∈ Ui ⊆ H} =⋃Ui∈F

CUi .

The observation above guarantees that this set is a coclique. One checksthat this coclique is maximal. We will call a maximal coclique of this typea chamber-type coclique.

Proposition 4.5. For each maximal flag F of PG(n, q), we have that

|CF | = 1 + 2q + 3q2 + . . .+ nqn−1 =d

dn

[n+ 1

1

]q

:= fq(n+ 1)

Proof. Suppose that F = {U0, U1, . . . , Un−1}. We prove that for each i =0, . . . n − 1 the number of incident point hyperplane pairs (p,H) such thatp ∈ Ui ⊆ H, and p 6∈ Uj or Uj 6⊆ H for all j < i, is qi

[n−i

1

]q. Then summing

this up for all i gives f(n).Take an i ∈ {0, 1, . . . n − 1} and a incident point hyperplane (p,H) suchthat p ∈ Ui ⊆ H. If p ∈ Ui−1, this pair was already counted before, sop ∈ Ui \ Ui−1. Hence there are qi possibilities for p. Each of those points,together with a hyperplane H containing Ui, is a pair we want to count.Now there are

[n−i

1

]q

such hyperplanes, so in total we have qi[n−i

1

]q

of

pairs.

4.1. DEFINITION 81

The following theorem justifies calling a chamber-type coclique large:

Theorem 4.6. For each n > 0 there is a prime power qn such that for allprime powers q ≥ qn we have that

|C| ≤ fq(n+ 1)

for all maximal cocliques C of PH(n+ 1, q).

Proof. Take a maximal coclique C of PH(n + 1, q). We prove by inductionthat |C| ≤ fq(n+ 1).

For n = 1, the vertices of PH(2, q) are just the projective points ofPG(1, q) and all vertices are adjacent. A chamber here is also just a point,and so is a chamber-type coclique. Hence |C| ≤ 1 = fq(2) for all maximalcocliques C.

Now take n > 1. Suppose there is a k-dimensional subspace U ofPG(n, q) such that C contains all pairs (q,K) for which q ∈ U ⊆ K. Inthis case there are three types of pairs (p,H) in C:

• p ∈ U ⊆ H: at most[k+1

1

]q

[n−k

1

]q

pairs of this type.

• p 6∈ U : Then U ⊆ H, and we can look at the quotient V(n+ 1, q)/U ,by mapping (p,H) onto ((p+U)/U,H/U). Those pairs form a cocliqueC′ in PH(n − k) (the graph defined over the (n − k − 1)-dimensionalprojective quotient space). By induction we know that |C′| ≤ fq(n−k).Now each “point” (p + U)/U in this quotient has qk+1 preimages inPG(n, q), and each “hyperplane” H/U has only one preimage. So wehave at most qk+1fq(n− k) pairs of this type.

• U 6⊆ H: Then p ∈ U and we can look at the pairs (p,H ∩ U) in thek-dimensional projective space U . Again those pairs form a cocliqueC′, and by induction, |C′| ≤ fq(k + 1). Now each “hyperplane” H ∩ Uextends in qn−k ways to a hyperplane of PG(n, q) not containing U .So we have at most qn−kfq(k + 1).

In total we have that

|C| ≤[k + 1

1

]q

[n− k

1

]q

+ qk+1fq(n− k) + qn−kfq(k + 1) = fq(n+ 1).

Now suppose that there is a subspace U such that for all pairs (p,H) ∈ C


we have that p ∈ U or U ⊆ H. Take a incident point hyperplane pair (q,K)such that q ∈ U ⊆ K. If q 6∈ H for a pair (p,H) ∈ C then U 6⊆ H, but thenp ∈ U ⊆ K, so (q,K) ∈ C and we are in the previous case.

So now we can assume that for all subspaces U of PG(n, q) there is apair (q,K) ∈ C such that q 6∈ U and U 6⊆ K. Take a basis for the span of C.Because of Proposition 4.3 this is a maximal set of incident point hyperplanepairs {(pi, Hi)|i ∈ I} ⊆ C that are linearly independent as rank-1 matrices.It is clear that |I| ≤ (n+ 1)2 − 1.Suppose that the pi’s do not span PG(n, q), then they lie in a hyperplaneH. Now take a vertex (q,K) ∈ C. Because of our assumption, there is avertex (r, L) ∈ C such that r 6∈ K and L 6= K. In terms of matrices thismeans that the scalar Kr 6= 0. But then

(Hq)(Kr) = H(qK)r = H

(∑i∈I

ai(piHi)

)r =

∑i∈I

ai(Hpi)(Hir) = 0

for some ai ∈ GF(q), and hence Hq = 0 or q ∈ H. This means that q ∈ Hfor all (q,K) ∈ C, a contradiction. So the pi must span PG(n, q). In thesame way we can prove that the intersection of all the Hi’s is empty.

Now for each pair (p,H) ∈ C we can define J(p,H) := {j ∈ I|pj ∈ H}.It is clear that J(p,H) 6= I, because otherwise all pi are contained in H, butthen they cannot span PG(n, q). So take an i 6∈ J(p,H), this means thatpi 6∈ H, but then p ∈ Hi. Now it is clear that J(p,H) is not empty becausethen p would be in the intersection of all Hi’s and we assumed that this wasempty.

Take a vertex (p,H) ∈ C, and define U(p,H) := ∩i∈I\J(p,H)Hi. Suppose

that pj ∈ U(p,H) for each j ∈ J(p,H). Now we can show that U(p,H) is asubspace we assumed did not exist. Indeed take a (q,K) ∈ C such thatq 6∈ U(p,H) and U(p,H) 6⊆ K. The first condition on (q,K) guarantees thatthere is an i ∈ I \J(p,H) such that q 6∈ Hi, while the second gives a j ∈ J(p,H)

for which pj ∈ U(p,H) \K. So as a matrix product (Hiq)(Kpj) 6= 0. But

(Hiq)(Kpj) = Hi(qK)pj = Hi

(∑k∈I

ak(pkHk)

)pj =

∑k∈I

ak(Hipk)(Hkpj) = 0

for some ak ∈ GF(q), because if k ∈ J(p,H), then Hipk = 0, and if k ∈I \ J(p,H), then Hkpj = 0. So we have q ∈ U(p,H) or U(p,H) ⊆ K for each

4.2. PH(2, Q) 83

(q,K) ∈ C, a contradiction.

So we can assume that for each J ⊆ I, there is a j ∈ J and an i ∈ I \ Jsuch that pj 6∈ Hi. Take a J ⊆ I and the corresponding j ∈ J and i ∈ I \ Jsuch that pj 6∈ Hi, and consider all the pairs (p,H) such that J(p,H) = J .We can map those pairs to the quotient space PG(n, q)/〈pj〉: p is mapped toppj/〈pj〉 and H is mapped to H/〈pj〉. Because i 6∈ J , we have that pi 6∈ Hand hence p ∈ Hi. Because pj 6∈ Hi, p is the intersection point of ppj andHi. This means that the mapping is one to one. In the quotient space thereare at most fq(n) vertices in the coclique. Hence for each J ⊆ I there areat most fq(n) vertices in C, so

|C| ≤ 2(n+1)2−1fq(n).

For large enough q this is smaller than fq(n+ 1).

Let us try to determine the chromatic number for small n:

4.2 PH(2, q)

In this case hyperplanes are points, so the only hyperplane incident with apoint, is the point itself. The vertices of the graph PH(1, q) are the points ofthe projective line PG(1, q), and two vertices are always adjacent. So PH1

is the complete graph Kq+1, and the chromatic number is q + 1.

4.3 PH(3, q)

In the planar case, besides the chamber-type coclique of size 2q + 1, thereis another obvious maximal coclique. Take three points p1, p2, p3 not on aline and consider the vertices (p1, p1p2), (p2, p2p3) and (p3, p3p1). This isclearly a coclique and it is trivial to check that it is maximal. This cocliqueis called the triangular coclique.

Now we can show that there are no other types of maximal cocliques.

Proposition 4.7. If C is a maximal coclique in PH(3, q), then it is ofchamber-type or it is a triangular coclique.

Proof. Take a maximal coclique C and an incident point line pair (p, L)contained in C. Let M be another line through p.


1. Suppose that (q,M) is in C for a q 6= p. Now take another vertex (r,N)in C. This vertex cannot be adjacent to (p, L) so p ∈ N or r ∈ L.

(a) p ∈ N : now q ∈ N , or if not, r ∈M .

i. M = N : now all other vertices in C must have p as theirpoint or M as their line, so C is of chamber-type.

ii. M 6= N : now q 6∈ N , so r = p. Again C can only be ofchamber-type with chamber {p,M}.

(b) p 6∈ N , so r ∈ L: now q ∈ N , otherwise (q,M) ∼ (r,N), so wehave a triangular coclique.

2. Now (p,M) is the only vertex with M as line. We can assume thatthe same holds for all other lines N through p (except for L), becauseotherwise consider case 1 with N instead of M . Now take anothervertex (q,K) in C, with q 6= p. This vertex has to be non-adjacentwith all vertices (p,X) where X is a line through p, so p ∈ K, and wehave that C is of chamber-type with chamber {p,K}.

It follows that, in order to find a minimal coloring, only those two typesof cocliques need to be considered. But we can even do better:

Lemma 4.8. If there is a coloring of PH(3, q) using some triangular co-cliques, we can find a coloring with the same number of colors that uses onlychamber-type cocliques.

Proof. Since each proper subset of a triangular coclique can be colored witha chamber-type coclique, we have that if a triangular coclique cannot bereplaced by a chamber-type coclique, all three point-line pairs of the trian-gular coclique are colored by the triangular coclique only. But that meansthat also the three point-line pairs in the opposite triangular coclique arecolored by that opposite triangular coclique. Now let p be a point that is in-cident with m lines ppi (for 1 ≤ i ≤ m) such that the point line pairs (p, ppi)are each colored by a different triangular coclique. Because our previousargument, m ≥ 2. Now replace each of those m triangular cocliques by them chamber-type cocliques CFi where Fi{pi, ppi}. This yields a new coloringwith the same number of colors and m less triangular cocliques. Now repeatthis procedure for each such point p.

Using this lemma, we find an upper bound for the chromatic number:

4.3. PH(3, Q) 85

Proposition 4.9.χ(PH(2, q)) ≤ q2 + 1.

Proof. To color this graph we are looking for as little chamber-type cocliquesas possible whose union contains all vertices. Look at the incidence graphof PG(2, q). This is the graph with the points and lines as vertices and twovertices are adjacent if and only if they are an incident point line pair. Thisis a regular bipartite graph of degree q + 1 with q2 + q + 1 vertices in eachclass. A regular bipartite graph admits perfect matchings (this is a specialcase of Hall’s theorem). It is clear that a perfect matching of the incidencegraph of PG(2, q) is a coloring of PH(2, q) with q2 + q + 1 colors. Of coursethis coloring is trivial: for all points p take a arbitrary line Lp through itand use the flag-type coclique.But we can do a bit better. Take a flag (p, L) and delete all lines throughp and all points on L. In the geometry that is left, all points lie on exactlyq lines and all lines contain exactly q points. So the incidence graph of thisgeometry now is a regular bipartite graph of degree q with q2 vertices ineach class. In this case we get a perfect matching (and hence a coloring) ofsize q2. Together with the flag-type coclique of (p, L) we have a coloring ofPH(2, q) with q2 + 1 colors. This gives the upper bound.

Note that for q = 2 this is bound is tight: PH(2, 2) has 21 vertices and aflag-type coclique has size 5, so a coloring must have at least 5 colors. Thecoloring constructed above has exactly 5 colors, so χ(PH(2, 2)) = 5 = q2 +1.

For q = 4, and actually for all q = r2 for some prime power r, we can doeven better. But to do this we need the following definition and result:

Definition 4.10. A non-degenerate maximal k-arc of the projective planePG(2, q) for some 1 ≤ k < q + 1 is a set of points such that each lineintersects this set in either 0 or k points. It can be shown that maximalk-arcs can only exist for k a divisor of q.

In 1969 Denniston proved the existence of maximal k-arcs for all divisorsk of q if q is even:

Theorem 4.11 (R.H.F Denniston [11]). There exist maximal k-arcs inPG(2, q) for all divisors k of q if q is even.

Using these maximal arcs, we can construct a coloring with less colors ifq is an even prime power.


Proposition 4.12. If q = r2 for some prime power r, then

χ(PH(3, q)) ≤ (q + 1)(q + 1−√q).

Proof. Take a maximal r-arc A in PG(2, q). It is not hard to count that|A| = r3−r2+r. Indeed take a point on the arc. This point is on q+1 = r2+1lines that all intersect the arc in r−1 other points. This means that there are(r−1)(r2−1)+1 = r3−r2+r points on A. Now count the pairs (p, L), wherep is a point not on the arc and L a line on p that does not intersect the arc, intwo ways. We have q2+q+1−r3+r2−r = (r2+1)(r2−r+1) choices for p, andfor each p there are r choices for L. Fixing L we have q+1 = r2+1 choices forp. That means that there are exactly r3− r2 + r lines disjoint from A. Nowit is clear that the set of those lines is a dual (0, r)-arc AD: each point lies oneither 0 or r lines of this set. Now delete all points of A and all lines of AD.The incidence graph of the resulting geometry is a regular bipartite graphof degree r2−r+1 with r4−r3 +2r2−r+1 vertices in each class. A perfectmatching gives a coloring with r4−r3 +2r2−r+1 = q2 +2q+1− (q+1)

√q

colors, which is the desired result.

The next proposition gives a lower bound for all q and shows that forq = r2 for some prime power r the coloring above is the best possible.

Proposition 4.13.

χ(PH(3, q)) ≥ (q + 1)(q + 1−√q).

Proof. Take a coloring of PH(3, q) consisting of chamber-type cocliques onlywith t colors. That is, we have a collection of incident point line pairs (pi, Li)(1 ≤ i ≤ t) such that for each point line pair (q,K) we have that q = pi forsome 1 ≤ i ≤ t or K = Lj for some 1 ≤ j ≤ t (or both). Now let X be theset of points of PG(2, q) that do not occur as pi and Y the set of lines ofPG(2, q) that do not occur as Lj . It is clear that |X| = |Y | = q2 + q+ 1− t.In the incidence graph of PG(2, q) there are no edges between the sets Xand Y .A standard interlacing result (see e.g. [25], Corollary 5.3) gives that (q +1)2|X||Y | ≤ q(q2 + q + 1 − |X|)(q2 + q + 1 − |Y |). This means that (q +1)2(q2 + q + 1− t)2 ≤ qt2 or t2 − 2(q + 1)2t+ (q + 1)2(q2 + q + 1) ≤ 0. Thisinequality results in either t ≥ (q+ 1)(q+ 1 +

√q) or t ≥ (q+ 1)(q+ 1−√q).

In both cases we have the bound we tried to prove.

Note that for q = 3 this lower bound gives χ(PH(3, q)) ≥ 10 and theupper bound gives χ(PH(3, q)) ≤ 10.

Chapter 5

Polar versions of theq-Kneser graphs

In Chapter 3 we constructed a q-analogue of the Kneser graph using a fi-nite vector space instead of a finite set as a starting point. Of course theconstruction of a Kneser graph can be generalized to other spaces than justvector spaces. One generalization was considered in Chapter 4. In this chap-ter we will use (finite) polar spaces as a basis for the construction and lookat some variations of the construction.

5.1 Definition

We defined the q-Kneser graph in Chapter 3 as a generalization of the ordi-nary Kneser graph by replacing sets by vector spaces (or projective spaces).We can generalize this graph in another way using polar spaces instead ofprojective spaces. Since the condition for adjacency was disjointness, we cando the same here and define the polar disjointness graph as follows:

Definition 5.1. Consider a finite, rank r, polar space P (n − 1, q) (whichhas a natural embedding in PG(n−1, q)) with n ≥ 3. The polar disjointnessgraph ΓPq (n, k) associated to the polar space P (n − 1, q) with 1 ≤ k ≤ r isthe graph with

• vertex set the totally isotropic (k − 1)-spaces of P (n− 1, q), and

• two vertices are adjacent if their corresponding (k− 1)-spaces are dis-joint in P (n− 1, q).

87

88 CHAPTER 5. POLAR VERSIONS OF THE Q-KNESER GRAPHS

For the same reason as before we will only consider polar disjointnessgraphs with k ≥ 2, because the case k = 1 yields a complete graph. Be-cause 2r ≤ n in P (n− 1, q), and k is bounded from above by r (all maximalsubspaces have dimension r − 1), the condition n ≥ 2k to exclude edgelessgraphs is satisfied automatically.

We can immediately compare the chromatic number of those graphs withthe chromatic number of the corresponding q-Kneser graphs.

Proposition 5.2. For all k ≥ 2, n ≥ 2k, prime powers q and projectivespaces P (n− 1, q) we have that

χ(ΓPq (n, k)) ≤ χ(Kq(n, k)).

Proof. The (k − 1)-spaces of P (n − 1, q) are the totally isotropic (k − 1)-subspaces of PG(n − 1, q), that means that the vertex set of ΓPq (n, k) is asubset of the vertex set of Kq(n, k). Two disjoint (k−1)-spaces in P (n−1, q)are also disjoint in PG(n−1, q) and any two disjoint totally isotropic (k−1)-spaces of PG(n − 1, q) are disjoint in P (n − 1, q). Therefore ΓPq (n, k) is aninduced subgraph of Kq(n, k). Hence the chromatic number of the polarq-Kneser graph is upper bounded by the chromatic number of the q-Knesergraph.

The polar disjointness graphs defined here are a generalization of theq-Kneser graphs over polar spaces. Two vertices of such a graph are adja-cent if there corresponding subspaces are disjoint, just as in the q-Knesergraphs. In the q-Kneser graphs over projective spaces two subspaces of acertain rank are disjoint if and only if they lie at maximal distance in thecollinearity graph of the subspaces of this rank. In this sense two verticesof the q-Kneser graphs (and also the original Kneser graphs) are adjacent ifthey are ”at maximal distance” from each other in the vector space (or setrespectively).

We use this notion of adjacency to define the polar q-Kneser graphs:

Definition 5.3. Consider a finite, rank r, polar space P (n − 1, q) (whichhas a natural embedding in PG(n − 1, q)) with n ≥ 3. The polar q-Knesergraph KPq (n, k) associated to the polar space P (n − 1, q) with 1 ≤ k ≤ r isthe graph with

• vertex set the totally isotropic (k − 1)-spaces of P (n− 1, q), and


• two vertices U and V and are adjacent if and only if U ∩ V ⊥ = ∅,where V ⊥ is the pole of V (the image of V under the polarity thatdefines the polar space P (n− 1, q)).

Note that the condition U ∩ V ⊥ = ∅ is equivalent with V ∩ U⊥ = ∅.

In this case the graph is no longer trivial for k = 1 since all totallyisotropic points of a polar space are not necessarily mutually non-orthogonal.The condition n ≥ 2k remains for the same reasons as before.

Remark 5.4. If 2k = n a totally isotropic rank-k subspace is its own pole.In that case the adjacency condition for the polar disjointness graph andthe polar q-Kneser graph are the same and hence the two graphs are equal.

Here again we can give an immediate upper bound for the chromaticnumber of these graphs.

Proposition 5.5. For all k ≥ 2, n ≥ 2k, prime powers q and projectivespaces P (n− 1, q) we have that

χ(KPq (n, k)) ≤ χ(ΓPq (n, k)).

Proof. By definition both the polar disjointness graph ΓPq (n, k) and the polarq-Kneser graph KPq (n, k) have the same vertex set but two vertices thatare adjacent in ΓPq (n, k), and hence disjoint in the polar space, are notnecessarily adjacent in KPq (n, k). Two adjacent vertices in KPq (n, k) arecertainly disjoint as subspaces and hence adjacent in ΓPq (n, k).

Together with Proposition 5.2 this gives the following inequality:

χ(KPq (n, k)) ≤ χ(ΓPq (n, k)) ≤ χ(Kq(n, k)).

In the rest of this chapter we will try to determine the chromatic numberof some polar disjointness graphs and polar q-Kneser graphs associated tothe classical polar spaces. Table 5.1 gives an overview of those classical polarspaces, their rank and the associated polar q-Kneser graph.

5.2 Chromatic numbers

5.2.1 KQ+

q (2m + 2, m + 1), m ≥ 2 even, a trivial case

There is one case where determining the chromatic number is trivial, namelyKQ

+

q (2m + 2,m + 1) with m ≥ 2 even. Note that in this case n = 2k and


Polar space P (n, q) Rank KPq (n+ 1, k)Hyperbolic quadric Q+(2m+ 1, q), m ≥ 1 m+ 1 KQ

+

q (2m+ 2, k)Elliptic quadric Q−(2m+ 1, q), m ≥ 2 m KQ

−q (2m+ 2, k)

Parabolic quadric Q(2m, q), m ≥ 2 m KQq (2m+ 1, k)Symplectic space W (2m+ 1, q), m ≥ 2 m KWq (2m+ 2, k)Hermitian variety H(2m, q2), m ≥ 2 m KHq2(2m+ 1, k)Hermitian variety H(2m+ 1, q2), m ≥ 1 m+ 1 KHq2(2m+ 2, k).

Table 5.1: The classical polar spaces and their corresponding polar q-Knesergraphs.

hence KQ+

q (2m+ 2,m+ 1) = ΓQ+

q (2m+ 2,m+ 1). The vertices are the gen-erators (maximal totally isotropic m-spaces) of the hyperbolic polar spaceQ+(2m + 1, q). This polar space has two families of generators. Whethertwo generators belong to the same family is determined by the dimension oftheir intersection: two generators belong to different families if and only ifthe parity of the dimension of the intersection of both is equal to the parityof the rank of the polar space (m+ 1).In the case that m is even, the rank is odd, so two disjoint generators belongto a different family. This means that there are two families of vertices ofKQ

+

q (2m + 2,m + 1) and two vertices in the same family cannot have anedge between them. Therefore KQ

+

q (2m+ 2,m+ 1) is a bipartite graph and:

χ(KQ+

q (2m+ 2,m+ 1)) = 2 if m is even.

5.2.2 KPq (n, 1)

If k = 1, the polar disjointness graph is a complete graph so we only considerthe polar q-Kneser graph KPq (n, 1) here. The vertices of this graph are thetotally isotropic points of the polar space P . Two points are adjacent in thegraph if they are not perpendicular (in other words, if one point is not in thepole of the other), that means that a coclique is a set of mutual orthogonalpoints. The largest sets of mutual orthogonal points, and hence the largestcocliques, are the maximal totally isotropic subspaces of the polar space.

Using Proposition 1.15 this already gives a lower bound for the chromaticnumber. If we could partition the points of the polar space by maximal


totally isotropic subspaces, then this lower bound is the chromatic number.Such a partition is called a spread of the polar space:

Definition 5.6. A spread of a polar space is a partition of the points of thepolar space in maximal totally isotropic subspaces.

The following theorem (see e.g. [46]) gives an overview of the existenceof spreads in the classical polar spaces:

Theorem 5.7. The following is known for the existence of spreads of clas-sical polar spaces:

1. Q+(2m+ 1, q), m ≥ 1:

(a) Q+(2m+ 1, q) has no spread if m is even,

(b) Q+(2m+ 1, q), m odd has a spread if q is even,

(c) Q+(3, q) has a spread for all prime powers q,

(d) Q+(7, q) has a spread if q is even, q is an odd prime or q ≡ 0 or2 mod 3.

2. Q−(2m+ 1, q), m ≥ 2:

(a) Q−(2m+ 1, q) has a spread if q is even,

(b) Q−(5, q) has a spread for all q.

3. Q(2m, q), m ≥ 2:

(a) Q(2m, q), q even has spreads for all m ≥ 2,

(b) Q(2m, q) with q odd and m even has no spreads,

(c) Q(6, q), has a spread if q is even, q is an odd prime or q ≡ 0 or2 mod 3.

4. W (2m+ 1, q), contains a spread for all m ≥ 1 and all prime powers q.

5. H(2m, q2), m ≥ 2: H(4, 4) has no spreads.

6. H(2m+ 1, q2), m ≥ 1 has no spreads.

The following cases are still open problems:

• Q+(2m+ 1, q), m ≥ 5 odd and q odd,

• Q+(7, q), for q odd, with q ≡ 0 mod 3 and not a prime,


• Q−(2m+ 1, q), for q > 2 and q odd,

• Q(2m, q), m ≥ 5 odd and q odd,

• Q(6, q), for q odd, with q ≡ 0 mod 3 and not a prime,

• H(2m, q2) for m > 2,

• H(4, q2) for q > 2.

Let us now look at the chromatic number of each case in more detail.

1. KQ+

q (2m+ 2, 1), m ≥ 1: This graph has

(qm + 1)(qm+1 − 1)q − 1

vertices. A maximal totally isotropic subspace has rank m + 1, so itcontains

[m+1

1

]q

points, and hence

χ(KQ+

q (2m+ 2, 1)) ≥ qm + 1.

Furthermore, by Theorem 5.7 we know that:

(a) χ(KQ+

q (2m+ 2, 1)) > qm + 1 if m is even,

(b) χ(KQ+

q (2m+ 2, 1)) = qm + 1 if m is odd and q even,

(c) χ(KQ+

q (4, 1)) = q + 1 for all q,

(d) χ(KQ+

q (8, 1)) = q3 + 1 if q is even, q is an odd prime or q ≡ 0 or2 mod 3.

2. KQ−

q (2m+ 2, 1), m ≥ 2: This graph has

(qm − 1)(qm+1 + 1)q − 1

vertices. A maximal totally isotropic subspace has rank m, so it con-tains

[m1

]q

points, and hence

χ(KQ−q (2m+ 2, 1)) ≥ qm+1 + 1.

Furthermore, by Theorem 5.7 we know that


(a) χ(KQ−

q (2m+ 2, 1)) = qm+1 + 1 if q is even,

(b) χ(KQ−

q (6, 1)) = q3 + 1 for all q.

3. KQq (2m + 1, 1), m ≥ 2: This graph has (q2m − 1)/(q − 1) vertices. Amaximal totally isotropic subspace has rank m, so it contains

[m1

]q

points, and hence

χ(KQq (2m+ 1, 1)) ≥ qm + 1.

Furthermore, by Theorem 5.7 we know that:

(a) χ(KQq (2m+ 1, 1)) = qm + 1 if q is even,

(b) χ(KQq (2m+ 1, 1)) > qm + 1 if q is odd and m even,

(c) χ(KQq (7, 1)) = q3 + 1 if q is even, q is an odd prime or q ≡ 0 or 2mod 3.

4. KWq (2m+ 2, 1), m ≥ 2: This graph has[

2m+21

]q

vertices. A maximaltotally isotropic subspace has rank m, so it contains

[m1

]q

points, andhence

χ(KWq (2m+ 2, 1)) ≥ qm + 1.

Furthermore, by Theorem 5.7 we know that in all cases there is aspread, so

χ(KWq (2m+ 2, 1)) = qm + 1.

5. KHq2(2m+ 1, 1), m ≥ 2: This graph has

(q2m+1 + 1)(q2m − 1)q2 − 1

vertices. A maximal totally isotropic subspace has rank m, so it con-tains

[m1

]q2

points, and hence

χ(KHq2(2m+ 1, 1)) ≥ q2m+1 + 1.


χ(KHq2(5, 1)) = q5 + 1.


6. KHq2(2m+ 2, 1), m ≥ 1: This graph has

(q2m+2 − 1)(q2m+1 + 1)q2 − 1

vertices. A maximal totally isotropic subspace has rank m + 1, so itcontains

[m+1

1

]q2

points, and hence

χ(KHq2(2m+ 2, 1)) ≥ q2m+1 + 1.


χ(KHq2(2m+ 2, 1)) > q2m+1 + 1 for all q and m ≥ 1.

5.2.3 ΓPq (n, 2) and KPq (n, 2), where P has rank 2

If we consider the definition of a polar space (Definition 1.51) for rank 2 weget a point set P, together with a collection B of subspaces of dimensionone, the lines, that satisfy the following axioms:

(i) A line, together with the points it contains is (isomorphic to) a pro-jective line.

(ii) The intersection of two lines, or the intersection of a point and a lineis again a subspace. In the former case this subspace can be a line, apoint or the empty subspace, in the latter case this intersection is apoint or the empty subspace.

(iii) Take a line L and a point p not on L. There is exactly one line Mthat contains p and intersects L in a point.

(iv) There are at least two distinct lines.

The incidence structure described here is a special case of a generalizedquadrangle. Indeed, a generalized quadrangle is defined as follows:

Definition 5.8. A generalized quadrangle GQ(s, t) (with s, t ≥ 1) is anincidence structure (P,B, I) that satisfies the following axioms:

(i) On every line there are exactly s + 1 points. Two distinct lines haveat most one point in common.

(ii) On every point there are exactly t+ 1 lines. There is at most one linethrough two distinct points.


(iii) For every line L and point p not on L there is exactly one line M andone point q such that p is on M and q is the intersection of L and M .

The numbers s and t are called the parameters of the generalized quad-rangles. An easy counting argument shows that a generalized quadrangleGQ(s, t) has (st+ 1)(s+ 1) points and (st+ 1)(t+ 1) lines.

Note that since the lines of a polar space of rank 2 are projective linesover GF(q) (or GF(q2) in the Hermitian case) we always have s = q (ors = q2 resp.).

The classical generalized quadrangles are the generalized quadranglesthat arise from a classical rank-2 polar space. We list them here, togetherwith their parameters:

• Q+(3, q), the hyperbolic quadric in PG(3, q). Here s = q and t = 1.This is a grid.

• Q(4, q), the parabolic quadric in PG(4, q). Here s = t = q.

• Q−(5, q), the elliptic quadric in PG(5, q). Here s = q and t = q2.

• W (3, q), the symplectic polar space in PG(3, q). Here s = t = q.

• H(3, q2), the Hermitian variety in PG(3, q2). Here s = q2 and t = q.

• H(4, q2), the Hermitian variety in PG(4, q2). There s = q2 and t = q3.

Let us consider the polar disjointness graphs ΓPq (n, 2). If the generalizedquadrangle is embedded in PG(3, q), the pole of a line is again a line. Sothe adjacency condition for both Kq(4, 2)P and ΓPq (4, 2) are the same, soboth graphs are equal. For the polar disjointness graphs ΓPq (n, 2) a cocliquein the graph is a set of lines in the generalized quadrangle that mutuallyintersect. Because of axiom (iii), such a set is always a point pencil. Thatmeans that the maximal size of a coclique is t + 1 and that the chromaticnumber is at least st + 1. To color the graph we look for a set of pointswhose point pencils together contain all lines. Such a set is called a blockingset of the generalized quandrangle. A blocking set is called minimal if theset that remains after removing any point is no longer a blocking set. Thesize of the smallest minimal blocking set clearly gives the chromatic numberof the graph.


If a blocking set partitions the set of all lines into point pencils, thisblocking set is also called an ovoid . In other words, an ovoid is a set ofpoints of a generalized quadrangle such that a line on one of the pointsof this set does not contain any other point of this set. It is clear that ifan ovoid exists in a generalized quadrangle, then it is the smallest minimalblocking set of that generalized quadrangle. In this case the chromatic num-ber is exactly st+ 1, the size of the ovoid.

The following theorem (see e.g. [46],[40]) gives an overview of the exis-tence of ovoids and in the classical generalized quadrangles. If no ovoidsexist, we give the size of the smallest minimal blocking set.

Theorem 5.9. For all prime powers q we have the following:

1. Q+(3, q) has ovoids,

2. Q(4, q) has ovoids,

3. Q−(5, q) has no ovoids. The smallest minimal blocking set in this casehas size q3 + q.

4. W (3, q) has ovoids if q is even,

5. W (3, q) has no ovoids if q is odd. The smallest minimal blocking setis not known in this case.

6. H(3, q2) has ovoids,

7. H(4, q2) has no ovoids. The smallest minimal blocking set in this casehas size q5 + q2.

This theorem gives us the chromatic numbers of most polar disjointnessgraphs and some polar q-Kneser graphs for k = 2 and polar rank 2, thosenumbers are summarized in Table 5.2. The table also shows where theminimal coloring arises from an ovoid (OV) and where it arises from thesmallest minimal blocking set (BS).


P (n− 1, q) ΓPq (n, 2) χ(ΓPq (n, 2))Q+(3, q) ΓQ

+

q (4, 2) = KQ+

q (4, 2) q + 1 (OV)Q−(5, q) ΓQ

−q (6, 2) q3 + q (BS)

Q(4, q) ΓQq (5, 2) q2 + 1 (OV)W (3, q), q even ΓWq (4, 2) = KWq (4, 2) q2 + 1 (OV)W (3, q), q odd ΓWq (4, 2) = KWq (4, 2) ? (BS)

H(4, q2) ΓHq2(5, 2) q5 + q2 (BS)H(3, q2) ΓHq2(4, 2) = KHq2(4, 2) q3 + 1 (OV)

Table 5.2: The chromatic numbers of the disjointness graphs

Chapter 6

Generalized Kneser graphs

In Chapter 3 we defined Kneser graphs and generalized them by definingthem over finite projective spaces instead of sets. In Chapter 5 we sawthat we could also define Kneser graphs over finite polar spaces. In thischapter we will generalize this idea even further. We will define Chevalleygroups and see how it is possible to define Kneser graphs over quotients ofChevalley groups. The graphs defined in Chapters 3 and 5 are special casesof the graphs we will define in this chapter.

6.1 Chevalley groups

6.1.1 Tits systems

Definition 6.1. Let G be a group with a (B,N)-pair . This means thatthere are subgroups B and N of G with the following properties:

• G is generated by B and N ,

• T := B ∩N is a normal subgroup of N ,

• W := N/T is a group generated by a set R of order 2 elements,

• for any r ∈ R and w ∈W we have that

rBw ⊆ BwB ∪BrwB

andrBr−1 6⊆ B.

99

100 CHAPTER 6. GENERALIZED KNESER GRAPHS

A 4-tuple (B,N,W,R) with these properties is called a Tits system. Thegroup B of the Tits system (B,N,W,R) is sometimes called the Borel sub-group, T can be called the Cartan subgroup or maximal split torus, and Wis called the Weyl group.

Note that from the second property of the definition it follows thatN = NG(T ). If J ⊆ R, we denote by WJ = 〈J〉 the subgroup of Wgenerated by J .

A large class of groups having (B,N)-pairs are the groups of Lie type.

Definition 6.2. A group G(K) is said to be of Lie type if it is a group ofrational points of a reductive linear algebraic group G with values in thefield K. In the case K = GF(q), this means that G(K) is a subgroup of agroup of invertible matrices over a finite field, hence G ≤ GLn(q) for somen ≥ 2 and prime power q.

So take a group G of Lie type and the corresponding Tits system(B,N,W,R). For w ∈ W the length of w, denoted by l(w), is the lengthof a shortest expression of w as a product of elements in R. One can provethat (see for example [8], Theorem 10.1.2 (v)) if W is finite (which is thecase here), that there is a unique longest element w0 in W and that elementis an involution.

Proposition 6.3. For all r ∈ R, w ∈W we have that:

BrBwB ={BrwB if l(rw) > l(w),BwB ∪BrwB otherwise.

Note that if l(rw) > l(w) then l(rw) = l(w) + 1 and otherwise l(rw) =l(w)− 1, because r ∈ R and hence l(r) = 1.

Using this, G can be written as a disjoint union as follows:

Theorem 6.4.G =

∐w∈W

BwB.

This decomposition is called the Bruhat decomposition of G.

6.1. CHEVALLEY GROUPS 101

6.1.2 Coxeter systems

If G is any group with a Tits system (this includes groups of Lie type), onecan prove (see e.g. [8], Theorem 10.5.1 (i)) that (W,R) is a Coxeter system.That is, there is a symmetric matrix M = (mr,t)r,t∈R, the Coxeter matrix ,with entries in N ∪ {∞} and diagonal entries 1, such that

(rt)mr,t = 1 for all r, t ∈ R.

For this reason, W is also called the Coxeter group of the Tits system. Withthis Coxeter system, one can associate the Coxeter diagram (R,M). SinceM is symmetric this diagram can be viewed as a graph with labeled edgesby letting {r, t} for two distinct r, t ∈ R be an edge labeled mr,t whenevermr,t ≥ 3. Labels 3 are usually omitted and edges with label 4 drawn asdouble edges.

Theorem 6.5. Let (W,R) be a Coxeter system

(i) The choice of the matrix M = (mr,t)r,t∈R associated with (W,R) isuniquely determined by the fact that mr,t is the order of rt for allr, t ∈ R.

(ii) If I ⊆ R, then (WI , I) is a Coxeter system with WI ∩R = I.

(iii) The system (W,R) is finite if and only if R is finite and every con-nected component of its Coxeter diagram appears in Figure 6.1

As a corollary any Coxeter system has a decomposition into a directproduct of Coxeter groups whose diagrams correspond to the connectedcomponents of the diagram of the Coxeter system. A Coxeter system thathas a connected diagram is called irreducible. If the Tits system has a Cox-eter system with diagram Xn, the Tits system is said to be of type Xn.Hence, Figure 6.1 gives a complete list of the finite irreducible Coxeter sys-tems.

6.1.3 Root systems

Now most of the finite Coxeter systems can be described in terms of rootsystems.

Definition 6.6. A (reduced) root system is a finite collection Φ of nonzerovectors spanning Rl, for some l ≥ 1, such that


Figure 6.1: The finite irreducible Coxeter diagrams

• if α ∈ Φ, then Rα ∩ Φ = {α,−α},

• if α, β ∈ Φ, then 2(β,α)(α,α) ∈ Z, and

• if α, β ∈ Φ, then wα(β) := β − 2(β,α)(α,α) α ∈ Φ.

It is an easy exercise to see that the wα are involutions. In fact, theyare reflections in the hyperplane perpendicular to α. Now the group W :=〈wα|α ∈ Φ〉 is called the Weyl group of the root system.

A root system Φ is called irreducible when it cannot be nontriviallydecomposed as Φ1 ∪ Φ2 where all vectors in Φ1 are perpendicular to allvectors in Φ2. It can be shown that each irreducible root system Φ containsa fundamental subsystem of roots ∆.

Definition 6.7. A fundamental subsystem of roots ∆ of a root system Φ is

6.1. CHEVALLEY GROUPS 103

a subcollection of vectors in Φ such that

• ∆ is a basis of Rl,

• each root, when written as a linear combination of vectors in ∆, haseither only nonnegative or only nonpositive coefficients.

Now one can check that (W,R := {wα|α ∈ ∆}) is a Coxeter system. Itis known that each irreducible finite Coxeter system, except for H3, H4 andIm2 (m = 5 or m ≥ 7), arises in this way. The set R is called the set offundamental reflexions.

6.1.4 Chevalley groups

Chevalley groups are a special kind of Lie type groups:

Definition 6.8. The group G = Xl(q) is an untwisted Chevalley group oftype Xl over the finite field GF(q) if G is a finite simple group of Lie type,and the following properties hold:

• The Tits system (B,N,W,R) associated to G is of type Xl,

• there is a U CB with B = UT and U ∩ T = {1},

•⋂w∈W

wBw−1 = T .

Because of the second property the Tits system is called split, and becauseof the third it is called saturated.

The root system of a Chevalley group has a special property:

Proposition 6.9. The root system Φ corresponding to the Coxeter system(W,R) of a Chevalley group G has the property that every root α ∈ Φ is anintegral combination of the fundamental roots such that either all coefficientsare nonnegative (α is called a positive root) or all coefficients are nonpositive(α is called a negative root).

We denote by Φ+ and Φ− respectively the sets of positive and negativeroots.

Given the root system and the Cartan subgroup T of a Chevalley group,we can find back the groups N,W,U and B of the Tits system of G: denote


by Xα (for α ∈ Φ) the root subgroup with respect to α. Then T normalizeseach Xα. Now, because N := NG(T ) and W := N/T , we have that W per-mutes the Xα: wXαw

−1 = Xwα for w ∈W and α ∈ Φ. Now U := Πα∈Φ+Xα

is a subgroup of G normalized by T , so that B := UT is a subgroup of Gwith B ∩N = T .

Take w ∈W , and define Φεw, with ε ∈ {+,−} as follows:

Φεw := {α ∈ Φ+|w−1α ∈ Φε}

Using this notation we can define for w ∈W and ε ∈ {+,−}:

U εw :=∏α∈Φεw

Xα.

Now it can be shown that U = U−wU+w and |U−w | = ql(w).

Proposition 6.10. For a Chevalley group G we have the following decom-position:

G =∐w∈W

U−wwB

Proof. For a w ∈ W we first show that w−1U+ww ⊆ B. Take an element u

of U+w , then u = uα1uα2 · · ·uαN , with uαi ∈ Xαi and αi ∈ Φ+

w . This meansthat w−1αi ∈ Φ+ for each i. Now we have that w−1Xαiw = Xw−1αi , hence

w−1uw ∈ Xw = Xwα1Xwα2· · ·Xw

αN= Xw−1α1

Xw−1α2· · ·Xw−1αN ⊆ U ⊆ B.

Now for w ∈W we have that

BwB = UTwB

= UwTB (because w = nT for some n ∈ N = NG(T ))= UwB (because T ⊆ B)= U−wU

+wwB

= U−www−1U+

wwB

= U−wwB (because w−1U+ww ⊆ B).

Substituting this in the Bruhat decomposition

G =∐w∈W

BwB.

gives the desired decomposition of G.

6.2. GENERALIZED KNESER GRAPHS 105

Definition 6.11. A standard parabolic subgroup P of G is a subgroup thatcontains B.

It can be shown that P = GJ := BWJB for some J ⊆ R. Using areasoning as before, one can show that

G =∐

w∈WJ

U−wwP (6.1)

where W J is the set of shortest representatives of left cosets of WJ in W .

6.2 Generalized Kneser graphs

6.2.1 Definition

For G = Xl(q), a Chevalley group of type Xl over the field GF(q), and P , aparabolic subgroup of G, we can define the generalized Kneser graph Γ overG/P :

• The vertex set V Γ of Γ is G/P , that is, the left cosets of P in G, and

• two vertices gP and hP are adjacent if and only if they are opposite,that is, if and only if h−1g ∈ Pw0P , where w0 is the unique longestelement of the Coxeter system (W,R).

As G acts as a transitive group of automorphisms on Γ, it is clear thatthis graph is regular.

6.2.2 Parameters

To calculate the size of a generalized Kneser graph we can use the followingformulas for the size of a Chevalley group and a parabolic subgroup:

|G| = q|Φ+|(qd1 − 1) · · · (qdl − 1)

and|P | = q|Φ

+|(q − 1)|R\J |(qe1 − 1) · · · (qem − 1)

where the di (i = 1, . . . , l) are the degrees of the diagram of G, and ej(j = 1, . . . ,m = |J |) the degrees of the diagram of P . We do not define herewhat the degrees of a diagram are but list them in Table 6.1. Note that theformula for |G| is just a special case of the expression for |P | where J = R.


Xl d1, d2, . . . , dlAl 2, 3, . . . , l + 1

Bl, Cl 2, 4, . . . , 2lDl 2, 4, . . . , 2l − 2, lE6 2, 5, 6, 8, 9, 12E7 2, 6, 8, 10, 12, 14, 18E8 2, 8, 12, 14, 18, 20, 24, 30F4 2, 6, 8, 12G2 2, 6

Table 6.1: Degrees of Chevalley groups

Proposition 6.12. If the diagram of G has degrees di (i = 1, . . . , l) and theconnected components of the diagram of P are (X1)l1 , (X2)l2 , . . . , (Xm)lm,where (Xj)lj has degrees ej i (i = 1, . . . , lj), then:

|V Γ| = |G/P | = |G||P |

=(qd1 − 1) · · · (qdl − 1)

(q − 1)|R\J |m∏j=1

(qej1 − 1) · · · (qej lj − 1)

. (6.2)

This is a monic polynomial in q of degree∑di −

∑li + |R \ J |.

The following proposition determines the valency of Γ:

Proposition 6.13. The valency of a generalized Kneser graph Γ over G/Pis given by

k = ql(v0)∑u∈VJ

ql(u)

where v0 is the unique word of smallest length in WJw0WJ and VJ = {u ∈WJ |uv0 ∈W J}.

Proof. To calculate the valency of Γ, take the vertex P . Another vertex gPis adjacent with P if g ∈ Pw0P . Now {gP |g ∈ Pw0P} = {gP |g ∈ Pv0P},where v0 is the unique word of smallest length in WJw0WJ . This is becausethere are aJ , bJ ∈ WJ ⊆ P such that w0 = aJv0bJ , and hence Pw0P =PaJa

−1J w0b

−1J bJP = Pa−1

J w0b−1J P = Pv0P . Now by Proposition 6.3,


Pv0P = BWJBv0P

= BWJv0BP

= BWJv0P

=⋃

u∈WJ

Buv0P

=⋃

u∈WJ

U−uv0uv0P.

Note that this union is not a disjoint union. Because of Proposition 6.10this is a disjoint union if we only consider the u ∈WJ such that uv0 is rightreduced w.r.t. J . Or in other words, uv0 ∈W J .

This means that

|{gP |g ∈ Pw0P}| = |{gP |g ∈ Pv0P}|= |

∐u∈VJ

U−uv0uv0P |

=∑u∈VJ

|U−uv0 |

=∑u∈VJ

ql(uv0)

= ql(v0)∑u∈VJ

ql(u)

where VJ = {u ∈WJ |uv0 ∈W J}.

The following proposition shows that a generalised Kneser graph has alarge valency compared to its size. As q grows to infinity, the ratio of thesize to the valency of the graph tends to 1.

Proposition 6.14. Let Γ be the generalized Kneser graph over G/P , whereG is a Chevalley group and P a parabolic subgroup of G. The degree of thesize of Γ, as polynomial in q, is the same as the degree of the valency of Γ.Moreover, both polynomials are monic.

Proof. Formula 6.2 uses the degree of the diagrams of G and the degreesthe components of the diagram of P to calculate the size of the generalized


Kneser graph over G/P , but there is another way of calculating this size.Using the decomposition given by formula 6.1.4 we have that

|G/P | =∑w∈WJ

|U−w | =∑w∈WJ

ql(w).

The (unique) longest element in W J is clearly the right J reduced form ofw0, so the length of this word determines the degree of the size of Γ. On theother hand, one sees that the unique longest element of W J is uv0, whereu is the longest element in WJ such that uv0 ∈ W J , and this is also thelongest element of VJ , by definition of VJ . So this uv0 also determines thedegree of the valency of Γ. The uniqueness of this element guarantees thatboth polynomials are monic.

We will see that in a lot of the cases, the valency is actually only one term.

Proposition 6.15. The following statements are equivalent:

(i) Jw0 = J ,

(ii) Jv0 = J ,

(iii) Ww0J = WJ ,

(iv) W v0J = WJ .

Theorem 6.16. Let G = Xl(q) be a Chevalley group and P = BWJB aparabolic subgroup of G. Then Jw0 = J , where w0 is the unique longest wordin the Weyl group W , if and only if the valency of the generalized Knesergraph Γ over G/P is ql(v0), where v0 is the unique word of smallest lengthin WJw0WJ .

Proof. Take a u ∈WJ and consider

uv0 = v0v0−1uv0

= v0uv0 .

Now if Jw0 = J , we have because of Proposition 6.15 that W v0J = WJ . This

means that uv0 ∈ WJ and hence uv0 ∈ W J if and only if uv0 , and hence u,is the empty word. So in that case the valency of Γ is

ql(v0)∑u∈VJ

ql(u) = ql(v0).


Now let us look in what cases the equality Jw0 = J holds:

Al: For a fundamental reflection i we have that iw0 = l − i + 1. So iffor every i ∈ J (J is symmetric) it holds that l − i + 1 ∈ J we haveJw0 = J .

Bl: iw0 = i for all i ∈ {1, 2, . . . , l}, so Jw0 = J for all J ⊆ R.

D2m: iw0 = i for all i ∈ {1, 2, . . . , 2m}, so Jw0 = J for all J ⊆ R.

D2m+1: iw0 = i for all i ∈ {1, 2, . . . , 2m− 1} and (2m)w0 = 2m+ 1, so Jw0 = Jfor all J ⊆ R such that {2m, 2m+ 1} ⊆ J or {2m, 2m+ 1} ∩ J = ∅.

E6: 1w0 = 6, 3w0 = 5, 2w0 = 2 and 4w0 = 4, so if J is symmetric, thenJw0 = J .

E7: iw0 = i for all i ∈ {1, 2, . . . , l}, so Jw0 = J for all J ⊆ R.

E8: iw0 = i for all i ∈ {1, 2, . . . , l}, so Jw0 = J for all J ⊆ R.

F4: iw0 = i for all i ∈ {1, 2, . . . , l}, so Jw0 = J for all J ⊆ R.

G2: iw0 = i for all i ∈ {1, 2, . . . , l}, so Jw0 = J for all J ⊆ R.

We conclude that only in the case of Al, D2m+1 and E6 we can have va-lencies that are a sum of different powers of q, rather than a single power of q.

We start with the cases where Jw0 = J . To calculate the size of thegeneralized Kneser graph, formula 6.2 is used. The valency is just the leadingterm of the polynomial in q representing the number of vertices.

In Appendix A.1 those values are listed for small diagram sizes. As anextra check, we explicitly calculated the valencies in this list to comparethem to the corresponding sizes. This calculation was done using the freeand open source computer algebra package LiE ([10]). First we have toset the group in which we are working with the command setdefault andthen define the set J (note that sets are not supported by LiE as datasets so we use a vector to represent J). Next we use the built in functionslong_word, lr_reduce and length to calculate w0, v0 and l(v0) in each case.The following example calculates the valency of the generalized Kneser graphfor G of type A4 and J = {1, 4}:

> setdefault A4> j=[1,4]> w_0=long_word


> v_0=lr_reduce(j,w_0,j)> length(v_0)

8

In the cases where Jw0 6= J we need to determine for which u ∈ WJ wehave that uv0 ∈ W J . In LiE it is not so hard to list the elements of W J .But we have not found an easy way to list the elements of WJ , so we canuse the following proposition:

Proposition 6.17. Consider the subsets of WJ and W J , resp. given by

VJ = {u ∈WJ |uv0 ∈W J}

andV J = {x ∈W J |xv−1

0 ∈WJ}.

The mapα : VJ → V J , v 7→ vv0

is a bijection between VJ and V J .

Proof. The map α is clearly injective. Because for all x ∈ V J we have thatxv−1

0 ∈WJ and xv−10 v0 = x ∈W J the map is also surjective.

Hence, ∑u∈VJ

ql(u) =∑x∈V J

ql(xv−10 ).

The function r_cosets(j) lists the elements of W J ; this is not a builtin function but it can be found in a file distributed with the LiE package:

r_cosets(vec s)=for wt row W_orbit(char_v(s)) do print(W_word(wt)) od

Note that some other functions used in the r_cosets function are also foundin the same file. We modified this function to, in stead of listing the elements,test if the elements (multiplied by the inverse of v0) are in WJ :

val_exp (vec s) ={loc ret = [];loc v = lr_reduce(s,long_word,s);for wt row W_orbit(char_v(s)) doloc u = canonical(W_word(wt)^inverse(v));


if test(u,s) thenret=ret^[length(u)];fiod;ret}

The testing happens in the seventh line: test(u,s), where u and sare both vectors, tests whether each component of u (which represents aWeyl word, each component is an element of R) are equal to at least onecomponent of the vector s (which represents a subset of R):

test (int x; vec t) = #test if at least one component of t is#equal to x{loc ans = 0;for i = 1 to size(t) doif !ans && x==t[i] thenans = 1;fiod;ans}

test (vec s,t) =#test if each component of s is equal to at least one component#of t{loc ans = 1;for i = 1 to size(s) doif ans && !test(s[i],t) thenans = 0;fiod;ans}

So val_exp(j) returns a vector that has the lengths of all u ∈ VJ ascomponents.

In Tables 6.2 and 6.3 we list the valencies k in all cases (for small pa-rameters) where Jw0 6= J .


G J P k

A2 {1}, {2} A1 q(q + 1)A3 {1}, {3} A1 q4(q + 1)

{1, 2}, {2, 3} A2 q(q2 + q + 1)A4 {1}, {2}, {3}, {4} A1 q8(q + 1)

{1, 2}, {3, 4} A2 q4(q3 + 2q2 + 2q + 1){1, 3}, {2, 4} A1A1 q6(q2 + 2q + 1){1, 2, 3}, {2, 3, 4} A3 q(q3 + q2 + q + 1){1, 2, 4}, {1, 3, 4} A2A1 q4(q2 + q + 1)

A5 {1}, {2}, {4}, {5} A1 q13(q + 1){1, 2}, {4, 5} A2 q9(q3 + 2q2 + 2q + 1){1, 3}, {3, 5} A1A1 q12(q + 1){1, 4}, {2, 5} A1A1 q11(q2 + 2q + 1){2, 3}, {3, 4} A2 q10(q2 + q + 1){1, 2, 3}, {3, 4, 5} A3 q4(q5 + 2q4 + 3q3 + 3q2

+2q + 1){1, 2, 4}, {1, 2, 5}, {1, 4, 5}, {2, 4, 5} A2A1 q9(q2 + q + 1){1, 3, 4}, {2, 3, 5} A2A1 q8(q3 + 2q2 + 2q + 1){1, 2, 3, 4}, {2, 3, 4, 5} A4 q(q4 + q3 + q3 + q + 1){1, 2, 3, 5}, {1, 3, 4, 5} A3A1 q4(q4 + q3 + 2q2 + q + 1)

Table 6.2: Valencies for G = Al(q) (l ≤ 5) in the case Jw0 6= J .


G J P k

D5 {4}, {5} A1 q18(q + 1){1, 4}, {1, 5}, {2, 4}, {2, 5} A1A1 q17(q + 1){3, 4}, {3, 5} A2 q15(q2 + q + 1){1, 2, 4}, {1, 2, 5} A2A1 q15(q + 1){1, 3, 4}, {1, 3, 5} A2A1 q14(q2 + q + 1){2, 3, 4}, {2, 3, 5} A3 q11(q3 + q2 + q + 1){1, 2, 3, 4}, {1, 2, 3, 5} A4 q6(q4 + q3 + q2 + q + 1)

E6 {1}, {3}, {5}, {6} A1 q34(q + 1){1, 2}, {1, 4}, {2, 3}, A1A1 q33(q + 1){2, 5}, {2, 6}, {4, 6}{1, 3}, {5, 6} A2 q30(q3 + 2q2 + 2q + 1){1, 5}, {3, 6} A1A1 q32(q2 + 2q + 1){3, 4}, {4, 5} A2 q31(q2 + q + 1){1, 2, 3}, {1, 4, 5}, A2A1 q29(q3 + 2q2 + 2q + 1){2, 5, 6}, {3, 4, 6}{1, 2, 4}, {2, 4, 6} A2A1 q31(q + 1){1, 2, 5}, {2, 3, 6} A1A1A1 q31(q2 + 2q + 1){1, 3, 4}, {4, 5, 6} A3 q25(q5 + 2q4 + 3q3 + 3q2

+2q + 1){1, 3, 5}, {1, 3, 6}, A2A1 q30(q2 + q + 1){1, 5, 6}, {3, 5, 6}{2, 3, 4}, {2, 4, 5} A3 q27(q3 + q2 + q + 1){1, 2, 3, 4}, {2, 4, 5, 6} A4 q19(q7 + 2q6 + 3q5 + 4q4 + 4q3

+3q2 + 2q + 1){1, 2, 3, 5}, {1, 2, 3, 6}, A2A1A1 q29(q2 + q + 1){1, 2, 5, 6}, {2, 3, 5, 6}{1, 2, 4, 5}, {2, 3, 4, 6} A3A1 q25(q4 + 2q3 + 2q2 + 2q + 1){1, 3, 4, 5}, {3, 4, 5, 6} A4 q22(q4 + q3 + q2 + q + 1){1, 3, 4, 6}, {1, 4, 5, 6} A3A1 q25(q4 + q3 + 2q2 + q + 1){1, 2, 3, 4, 5}, {2, 3, 4, 5, 6} D5 q8(q8 + q7 + q6 + q5 + 2q4 + q3

+q2 + q + 1){1, 2, 3, 4, 6}, {1, 2, 4, 5, 6} A4A1 q19(q6 + q5 + 2q4 + 2q3

+2q2 + q + 1)

Table 6.3: Valencies for G = D5, E6 in the case Jw0 6= J .


A similar list with sizes included is found in Appendix A.2.

In the first case (Jw0 = J), for a fixed Chevalley group G, the valencyof the graph depends only on the size, and hence on the type of P . In thecase where Jw0 6= J this is no longer true. For example consider G = E6(q)and J = {1, 2, 3}, J = {1, 2, 4} and J = {1, 3, 5}, respectively. In all threecases P has type A2A1. But in the first case only the A1 component is fixedunder w0, in the second case the A2 component is fixed and in the last case,nothing is fixed. So when Jw0 6= J , the valency also depends on the actionof w0 on the components of the type of P .

6.3 Chromatic numbers in the thin An(1) case

The construction of generalized Kneser graphs can be transferred to thethin case (q = 1). Indeed, take a thin Chevalley group Xl(1). In this casethe Weyl group W = 〈R〉 of Xl(1) is actually the whole group Xl(1). Aparabolic subgroup in this case is of the form P = B〈J〉B = 〈J〉 for someJ ⊆ R.

The generalized Kneser graph over Xl(1)/P where P = 〈J〉 for someJ ⊆ R is the graph with vertex set Xl(1)/P . Two vertices g〈J〉 and h〈J〉are adjacent if and only if g = hjw0j

′ for some j, j′ ∈ 〈J〉 and w0 the uniquelongest word of W .

Using the same reasoning as in Theorem 6.16 we can show that in thecase where Jw0 = J (where w0 is the longest word in the Weyl group) thevalency of the generalized Kneser graph is 1. That means that only theother cases are interesting here.

The thin Chevalley group of type An, noted as An(1) is Sym(n + 1),the symmetric group on n+ 1 letters, say on the set {1, 2, . . . , n, n+ 1}. Aparabolic subgroup in this case is a subgroup P = 〈ri1 , . . . , rim〉 where ri isthe transposition (i, i + 1). The unique longest word w0 in this case is thereverse permutation.

Let us consider some examples:

Example 6.18. Take n = 3, so the group is A3(1) = Sym(4) and take P =〈r3〉. The vertices of this graph are of the form g〈(3, 4)〉 where g ∈ Sym(4).

6.3. CHROMATIC NUMBERS IN THE THIN AN (1) CASE 115

We know that 〈(3, 4)〉 must be a stabilizer of the set {g〈(3, 4)〉 | g ∈ Sym(4)}.Therefore we can represent the vertices of this graph as ordered pairs outof {1, 2, 3, 4}, so v = 12. Note that this is also the result of substitutingq = 1 in the formula for the valency of this graph in the non-thin case (seeAppendix A.2. Two vertices v and v′ in this representation are adjacent inthe thin case if there is a w ∈ 〈(3, 4)〉w0〈(3, 4)〉 = 〈(3, 4)〉(1, 4)(2, 3)〈(3, 4)〉such that vw = v′. This means that k = |〈(3, 4)〉| = 2. Again substitutingq = 1 in the value for k in the non-thin case gives the same result.Take the vertex [1, 2] of this graph. This vertex is adjacent with [1, 2](1,4)(2,3) =[4, 3] and [1, 2](1,4)(2,3)(3,4) = [3, 4] and those two vertices are both also ad-jacent with [2, 1]. Since k = 2 this results in a 4-cycle. Hence this graphis isomorphic to 3 · C4, the disjoint union of three copies of the cycle withlength 4. The chromatic number is trivially 2.

We can generalize this example as follows:

Proposition 6.19. Let Γ be the generalized Kneser graph over An(1)/P forn ≥ 2 and P = 〈rn〉. Then

Γ ∼=(n+ 1)!

8· C4

and χ(Γ) = 2.

Proof. Take a vertex of Γ, this is an ordered (n−1)-tuple out of {1, 2, . . . , n, n+1}. Without loss of generality we can take [1, 2, . . . , n − 1]. This vertex isadjacent with the vertices [n+1, n, n−1, . . . , 3] and [n, n+1, n−1, . . . , 3] andthose vertices are both also adjacent with [2, 1, . . . , n−1]. Since the valencyis |〈(n, n+1)〉| = 2 this is a 4-cycle and since this graph is transitive, it mustbe a disjoint union of 4-cycles. The number of vertices of Γ is

v =(n+ 1n− 1

)(n− 1)! =

(n+ 1)!2

so Γ consists of the asserted number of 4-cycles.

Example 6.20. The thin Chevalley group of type A2 is Sym(3), so w0 =(1, 3). Take P = 〈(2, 3)〉. The generalized Kneser graph in this case has asvertex set {1, 2, 3}, so v = 3. It is clear that all elements are adjacent, sothis graph is a triangle.Now take A3(1) with P = 〈(2, 3), (3, 4)〉 the vertices of this graph are theelements of {1, 2, 3, 4} and all elements are mutually adjacent, so this graphis K4.


For general n this gives:

Proposition 6.21. Let Γ be the generalized Kneser graph over An(1)/P forn ≥ 2 and P = 〈r2, . . . , rn〉. Then Γ ∼= Kn+1 and hence χ(Γ) = n+ 1.

Let us study another example:

Example 6.22. Take A4(1) with P = 〈(3, 4), (4, 5)〉 and consider the gen-eralized Kneser graph Γ over A4(1)/P . The vertex set is the set of orderedpairs of {1, 2, 3, 4, 5}, so v = 20. Consider the vertex [1, 2], this vertex is ad-jacent with the following six vertices: [3, 4], [4, 3], [3, 5], [5, 3], [4, 5] and [5, 4].Note that those are all the ordered pairs that as a set are disjoint from{1, 2}. This reminds us of the definition of the original Kneser graph, butinstead of pairs we look at ordered pairs.

To continue this example we need some definitions and results:

Definition 6.23. Define the tensor product Γ× Γ′ of two graphs Γ and Γ′

as the graph with vertex set the Cartesian product of the two vertex setsand two vertices (u, u′) and (v, v′) (with u and v vertices of Γ and u′ and v′

vertices of Γ′) are adjacent if and only if u ∼ v in Γ and u′ ∼ v′ in Γ′.

Definition 6.24. We define a loop on a vertex v of a graph Γ as the edgethat has v as both endpoints. Note that we didn’t allow this in our originaldefinition of graphs. In what follows we will still hold on to this originaldefinition, so unless stated otherwise graphs have no loops.Now we can consider the complete graph Kn. Take the graph that is theresult of adding loops to all vertices of the complete graph. We will denotethis graph by K◦n.

The following result shows that taking a tensor product of a graph withthe complete graph with loops doesn’t change the chromatic number:

Proposition 6.25. For all graphs Γ and all n ≥ 1 we have that

χ(Γ×K◦n) = χ(Γ).

Proof. First we will show that each coloring of Γ yields a valid coloring ofΓ × K◦n with the same number of colors. Indeed, color each vertex (v, i),where v is a vertex of Γ and i a vertex ofK◦n, with the color of v in the originalcoloring of Γ. To show that this is a valid coloring we take two adjacentvertices (v, i) and (w, j). Now suppose that those vertices are colored withthe same color, that means that v and w are colored with the same color.


But for (v, i) and (w, j) to be adjacent, v and w have to be adjacent in Γ, butthen they could not have had the same color in Γ. So χ(Γ×K◦n) ≤ χ(Γ).Now since Γ ∼= Γ × K◦1 ≤ Γ × K◦n we need at least χ(Γ) colors to colorΓ×K◦n.

Now let us continue with Example 6.22:

Example 6.26. The vertices of this thin generalized Kneser graph Γ overA4(1)/〈(3, 4), (4, 5)〉 are the ordered pairs out of {1, 2, 3, 4, 5} and two ver-tices are adjacent if and only if they are disjoint. Consider the Kneser graphK(5, 2) with vertex set only those ordered pairs out of {1, 2, 3, 4, 5} that arein lexicographical order instead of the unordered pairs. This is clearly aninduced subgraph of Γ of half the size of Γ. Now we can write a vertex of Γas ([a, b], σ) := [a, b]σ with a < b and σ ∈ Sym(2), so the vertex set of Γ is thedirect product of the vertex set of K(5, 2) and a set with two elements. Thecondition for adjacency in both Γ and K(5, 2) is the same. One easily checksthat Γ ∼= K(5, 2)×K◦2 . Proposition 6.25 tells us that χ(Γ) = χ(K(5, 2)) = 3.

Again this example can be generalized:

Proposition 6.27. Let Γ be the generalized Kneser graph over An(1)/P forn ≥ 2 and

P = 〈R \ {r1 . . . , ri−1, ri}〉 = 〈ri+1, . . . , rn〉

for 1 ≤ i ≤ (n+ 1)/2, then

Γ ∼= K(n+ 1, i)×K◦i!

and χ(Γ) = χ(K(n+ 1, i)) = n− 2i+ 3.

Proof. The vertices of Γ are the ordered i-tuples out of {1, . . . , n, n+1} thatare adjacent if they are disjoint. Each vertex can be written as ([t1, . . . , ti], σ):= [t1, . . . , ti]σ, where 1 ≤ t1 < . . . < ti ≤ n+ 1 and σ ∈ Sym(i).

Another example shows that we encounter already known cases:

Example 6.28. Let Γ be the generalized Kneser graph overA4(1)/〈r1, r3, r4〉.In the previous example we saw that if P = 〈r3, r4〉 we get ordered pairs(2-tuples) as vertices. Now because r1 = (1, 2) is also in P , a permutationof the first two elements of the tuple (in this case there are only two ele-ments in the tuple) does not change the vertex. That means that insteadof ordered pairs, the vertices are unordered pairs of {1, 2, 3, 4, 5}. ThereforeΓ ∼= K(5, 2).


As another example, let ∆ be the generalized Kneser graphA7(1)/〈r1, r2, r5, r6, r7〉.If P would have been 〈r5, r6, r7〉, the vertex set would have been the set ofordered 4-tuples of {1, . . . , 8}. Now that r1 and r2 are also in P a permu-tation of the first three elements of the tuple does not change the vertex.Therefore each vertex (4-set) of K(8, 4) only gives rise to 4 vertices of ∆instead of 4!. So ∆ ∼= K(8, 4)×K◦4 .

These examples show how we can describe all generalized q-Kneser graphsover An(1)/P where P = 〈R \ S〉 and Q = 〈S〉 is of type Al for some(0 ≤ l ≤ n):

Proposition 6.29. Let Γ be the generalized Kneser graph over An(1)/P forn ≥ 2 and

P = 〈R \ {rj , rj+1 . . . , ri−1, ri}〉 = 〈r1, r2, . . . , rj−1, ri+1, . . . , rn〉

for some 1 ≤ j ≤ i ≤ n, where ri is the transposition (i, i+ 1).

(i) If 1 ≤ j ≤ i ≤ (n+ 1)/2, then

Γ ∼= K(n+ 1, i)×K◦i!j!

andχ(Γ) = χ(K(n+ 1, i)) = n− 2i+ 3.

(ii) If (n+ 1)/2 < i ≤ n and 1 ≤ j < (n+ 1)/2, then

Γ ∼=(

n+ 12i− n− 1

)(2i−n−1)!·

(K(2(n− i+ 1), n− i+ 1)×K◦(n−i+1)!

j!

),

which is a disjoint union of(

n+12i−n−1

)(2i− n− 1)! copies of the graph

K(2(n− i+ 1), n− i+ 1)×K◦(n−i+1)!j!

. In that case

χ(Γ) = χ(K(2(n− i+ 1), n− i+ 1)) = 2.

Proof. (i) The vertices of Γ are ordered i-tuples of {1, . . . , n, n+1} wherea permutation of the first j doesn’t change the vertex. Two verticesare adjacent if and only if they are disjoint.

(ii) We first consider the case where j = 1. We will demonstrate this withan example that completely generalizes. Take for Γ the generalizedKneser graph over A7(1)/〈r6, r7〉 (so n = 7 and i = 5). The vertices


are ordered 5-tuples with elements from 1 to 8. Take for example thevertex [1, 2, 3, 4, 5], this is adjacent with the vertex [8, 7, 6, 5, 4] and allvertices obtained by permuting the first three (= n− i+ 1) elements.All those vertices are also adjacent with the vertices that are obtainedby permuting the first three elements in [1, 2, 3, 4, 5]. Notice that inall those vertices the last two (= 2i − n − 1) elements are the same(although not necessarily in the same order). In fact two adjacentvertices have the same elements in the last two places but in reversedorder. That means that this graph cannot be connected.Now if we delete the last two (= 2i−n−1) elements in all those vertices,then we get the graph with vertices the ordered 3-tuples (where 3 =n − i + 1) of {1, 2, 3, 6, 7, 8} and two vertices are adjacent if they aredisjoint. This is the graph K(6, 3)×K◦3!, or in general

K(2(n− i+ 1), n− i+ 1)×K◦(n−i+1)!.

So we deleted the two (= 2i − n − 1) elements 4 and 5. But we have(82

)(=(

n+12i−n−1

)) choices for those elements. And the order in which

they appear in the last two places in the i-tuple is important, so foreach choice we have another two (= (2i−n− 1)!) possibilities. In thisway we get that

Γ ∼=(

n+ 12i− n− 1)

)(2i−n−1)!·

[K(2(n− i+ 1), n− i+ 1)×K◦(n−i+1)!

].

Now if j > 1 the same reasoning as before gives the correct result. Notethat since j < (n+ 1)/2 and (n+ 1)/2 < i, we have that j ≤ n− i+ 1.

Now let us consider an example where the complement of J is not con-nected:

Example 6.30. Let Γ be the generalized Kneser graph over A4(1)/〈r2, r4〉.The vertices of this graph need to be stabilized by P = 〈(2, 3), (4, 5)〉,so they can be represented by a point-pair pair {a, {b, c}}, for distincta, b, c ∈ {1, 2, 3, 4, 5}. This means that v = 30. Take the vertex {1, {2, 3}}, itis adjacent to the vertices {5, {3, 4}}, {5, {2, 4}}, {4, {3, 5}} and {4, {2, 5}},so k = 4. Note that if we take a vertex, say {1, {2, 3}} which is deter-mined by the point 1 and the pair {2, 3}, the disjoint pair {4, 5} is alsodetermined. Or in other words, this vertex is also fully determined by theordered pair [{4, 5}, {2, 3}]. Note that, because of this last representation,


we can represent the vertices of Γ by the directed edges in the Petersen graphK(5, 2). What is adjacency in this representation? Take the two adjacentvertices {1, {2, 3}} and {5, {3, 4}}, they are represented as [{4, 5}, {2, 3}] and[{1, 2}, {3, 4}]. Note that {4, 5} is adjacent with {1, 2} in K(5, 2) and {2, 3}is not adjacent with {3, 4} in K(5, 2). So two vertices of Γ are adjacent ifand only if as directed edges of the Petersen graph, the starting vertices ofthe directed edges are adjacent and the end vertices of the directed edges arenon-adjacent. Now take a maximal coclique in the Petersen graph, this hassize 4, and consider the set of vertices of Γ that have their starting vertexin this maximal coclique. This set has size 12 and one can easily check thatthis is a maximal coclique of Γ. This yields that χ(Γ) ≥ 3. Now each color-ing of K(5, 2) yields a coloring of Γ. Indeed, color each vertex of Γ with thecolor of its starting vertex as a directed graph of the Petersen graph. Taketwo vertices of Γ and suppose that they have the same color. That meansthat their starting vertices have the same color in the Petersen graph, butthose are adjacent in the Petersen graph, clearly a contradiction. Thereforeχ(Γ) = χ(K(5, 2)) = 3.

This example generalizes as follows:

Proposition 6.31. Let Γ be the generalized Kneser graph over A2k(1)/Pfor k ≥ 1 and P = 〈R \ {r1, rk+1}〉, then Γ is isomorphic with the graphthat has the directed edges of K(2k + 1, k) as vertices, and two verticesare adjacent if and only if their starting vertex are adjacent. Thereforeχ(Γ) = χ(K(2k + 1, k)) = 3.

Proof. The vertices of Γ are pairs that consist of an element and an un-ordered k-tuple that does not contain that first element. Take a vertex, say{1, {2, 3, . . . , k + 1}}. This vertex is adjacent to all vertices {j, S}, wherej ≥ k+2, 1, j 6∈ S, |S| = k and |S∩{2, . . . , k+2}| = 1. We can represent thevertex {1, {2, 3, . . . , k+ 1}} by [{k + 2, . . . , 2k + 1}, {2, . . . , k + 1}], which isa directed edge of K(2k + 1, k). It is clear that two vertices are adjacent ifand only if their starting vertices are adjacent (and hence their end verticesare not). Now take a maximal coclique C of K(2k+ 1, k), by Theorem 2.3.1this coclique has size

(2kk−1

). The set C′ consisting of all vertices of Γ that

have their starting vertices in C is clearly a maximal coclique of Γ with sizek(

2kk−1

). This yields that

χ(Γ) ≥

(2k+1k

)(k+1k

)k(

2kk−1

) =2k + 1k

= 2 +1k.

6.4. CHROMATIC NUMBERS IN THE THICK AN CASE 121

Since χ(Γ) is an integer we have χ(Γ) ≥ 3.Now a coloring of K(2k + 1, k) yields a coloring of Γ. Indeed color eachvertex of Γ with the color of its starting vertex in K(2k + 1, k).

6.4 Chromatic numbers in the thick An case

Finally we observe that Proposition 6.29 can be generalized almost literally.

Proposition 6.32. Let Γ be the generalized Kneser graph over An(q)/Pwhere n ≥ 2 and P is the parabolic subgroup P = B〈J〉B for

J = R \ {rj , rj+1 . . . , ri−1, ri} = {r1, r2, . . . , rj−1, ri+1, . . . , rn}

for some 1 ≤ j ≤ i ≤ n.

(i) If 1 ≤ j ≤ i ≤ (n+ 1)/2, then

Γ ∼= Kq(n+ 1, i)×K◦[i]q ![j]q !

andχ(Γ) = χ(Kq(n+ 1, i)).

(ii) If (n+ 1)/2 < i ≤ n and 1 ≤ j < (n+ 1)/2, then

Γ ∼=[

n+ 12(n− i+ 1)

]q

[2i−n−1]q!·(Kq(2(n− i+ 1), n− i+ 1)×K◦[n−i+1]q !

[j]q !

).

In that case

χ(Γ) = χ(Kq(2(n− i+ 1), n− i+ 1)).

Proof. (i) The vertices of Γ are the flags of PG(n, q) of type (Uj , Uj+1, . . . , Ui)where Ul (for j ≤ l ≤ i) is an (l− 1)-dimensional subspace of PG(n, q)and two vertices are adjacent if their (i − 1) dimensional subspacesare disjoint. This means that, just as in the thin case, Γ is a tensorproduct of Kq(n+1, i) and some complete graph with loops. Now each(i− 1)-space in Kq(n+ 1, i) gives rise to[

i

i− 1

]q

[i− 1i− 2

]q

. . .

[j + 1j

]q

=[i

1

]q

[i− 1

1

]q

. . .

[j + 1

1

]q

=[i]q![j]q!

flags of the right type. So this is the number of vertices of the completegraph with loops we are looking for.


(ii) Again, the vertices of Γ are the flags of PG(n, q) of type (Uj , Uj+1, . . . , Ui)where Ul (for j ≤ l ≤ i) is an (l−1)-dimensional subspace of PG(n, q).But now two vertices (Uj , Uj+1, . . . , Ui) and (U ′j , U

′j+1, . . . , U

′i) are ad-

jacent if and only if Ul ∩ U ′l = ∅ for all j ≤ l ≤ n/2 and Ul ∩ U ′l is a2l − n− 1 space for all n/2 < l ≤ i.Now consider the projective space PG(n, q)/(Ui∩U ′i). Since Ui∩U ′i is(2i−n−2)-dimensional, the vertices of whose (i−1)-dimensional sub-space contain Ui∩U ′i in this quotient become flags of PG(2n−2i+1, q)of the type (Vj , . . . Vn−i+1) and now two vertices are adjacent if andonly if their (n − i)-dimensional subspaces are disjoint. So the sub-graph of Γ on the vertices whose (i− 1)-dimensional subspace containUi ∩ U ′i is isomorphic to

Kq(2(n− i+ 1), n− i+ 1)×K◦[n−i+1]q !

[j]q !

.

Now the intersection of the flag (Uj , Uj+1, . . . , Ui) with a (2i− n− 2)-dimensional subspace contained in Ui is a maximal flag (V1, . . . V2i−n−1).So for each choice of such a flag we get a copy of this subgraph. Nowthere are

[n+1

2i−n−1

]q

[2i−n−1]q! ways of choosing such a flag and each

choice gives a different copy.

Appendix A

Parameters of generalizedKneser graphs

In Chapter 6 we defined generalized Kneser graphs defined over Chevalleygroups. We gave a formula for their size and a method of determiningtheir valency. In this appendix we give a list of those parameters for thegeneralized Kneser graphs over An, Bn, Dn for n ≤ 5 and E6, E7, E8, F4 andG2. Note that for clarity we will denote the ri by i.

A.1 The case Jw0 = J

A3 – J = {2}; type of P : A1

v = q5 + 2q4 + 3q3 + 3q2 + 2q + 1k = q5

A4 – J = {1, 4}; type of P : A1A1

v = q8 + 2q7 + 4q6 + 5q5 + 6q4 + 5q3 + 4q2 + 2q + 1k = q8

– J = {2, 3}; type of P : A2

v = q7 + 2q6 + 3q5 + 4q4 + 4q3 + 3q2 + 2q + 1k = q7

A5 – J = {3}; type of P : A1

v = q14 + 4q13 + 10q12 + 19q11 + 30q10 + 41q9 + 49q8 + 52q7 + 49q6 + 41q5 + 30q4 +19q3 + 10q2 + 4q + 1k = q14

– J = {1, 5}, {2, 4}; type of P : A1A1

v = q13 + 3q12 + 7q11 + 12q10 + 18q9 + 23q8 + 26q7 + 26q6 + 23q5 + 18q4 + 12q3 +7q2 + 3q + 1k = q13

– J = {1, 3, 5}; type of P : A1A1A1

v = q12 + 2q11 + 5q10 + 7q9 + 11q8 + 12q7 + 14q6 + 12q5 + 11q4 + 7q3 + 5q2 + 2q+ 1k = q12

123

124APPENDIX A. PARAMETERS OF GENERALIZED KNESER GRAPHS

– J = {2, 3, 4}; type of P : A3

v = q9 + 2q8 + 3q7 + 4q6 + 5q5 + 5q4 + 4q3 + 3q2 + 2q + 1k = q9

– J = {1, 2, 4, 5}; type of P : A2A2

v = q9 + q8 + 2q7 + 3q6 + 3q5 + 3q4 + 3q3 + 2q2 + q + 1k = q9

B2 – J = {1}, {2}; type of P : A1

v = q3 + q2 + q + 1k = q3

B3 – J = {1}, {2}, {3}; type of P : A1

v = q8 + 2q7 + 3q6 + 4q5 + 4q4 + 4q3 + 3q2 + 2q + 1k = q8

– J = {1, 2}; type of P : A2

v = q6 + q5 + q4 + 2q3 + q2 + q + 1k = q6

– J = {1, 3}; type of P : A1A1

v = q7 + q6 + 2q5 + 2q4 + 2q3 + 2q2 + q + 1k = q7

– J = {2, 3}; type of P : B2

v = q5 + q4 + q3 + q2 + q + 1k = q5

B4 – J = {1}, {2}, {3}, {4}; type of P : A1

v = q15 + 3q14 + 6q13 + 10q12 + 14q11 + 18q10 + 21q9 + 23q8 + 23q7 + 21q6 + 18q5 +14q4 + 10q3 + 6q2 + 3q + 1k = q15

– J = {1, 2}, {2, 3}; type of P : A2

v = q13 + 2q12 + 3q11 + 5q10 + 6q9 + 7q8 + 8q7 + 8q6 + 7q5 + 6q4 + 5q3 + 3q2 + 2q+ 1k = q13

– J = {1, 3}, {1, 4}, {2, 4}; type of P : A1A1

v = q14 + 2q13 + 4q12 + 6q11 + 8q10 + 10q9 + 11q8 + 12q7 + 11q6 + 10q5 + 8q4 +6q3 + 4q2 + 2q + 1k = q14

– J = {3, 4}; type of P : B2

v = q12 + 2q11 + 3q10 + 4q9 + 5q8 + 6q7 + 6q6 + 6q5 + 5q4 + 4q3 + 3q2 + 2q + 1k = q12

– J = {1, 2, 3}; type of P : A3

v = q10 + q9 + q8 + 2q7 + 2q6 + 2q5 + 2q4 + 2q3 + q2 + q + 1k = q10

– J = {1, 2, 4}; type of P : A2A1

v = q12 + q11 + 2q10 + 3q9 + 3q8 + 4q7 + 4q6 + 4q5 + 3q4 + 3q3 + 2q2 + q + 1k = q12

– J = {1, 3, 4}; type of P : B2A1

v = q11 + q10 + 2q9 + 2q8 + 3q7 + 3q6 + 3q5 + 3q4 + 2q3 + 2q2 + q + 1k = q11

– J = {2, 3, 4}; type of P : B3

v = q7 + q6 + q5 + q4 + q3 + q2 + q + 1k = q7

A.1. THE CASE JW0 = J 125

B5 – J = {1}, {2}, {3}, {4}, {5}; type of P : A1

v = q24 + 4q23 + 10q22 + 20q21 + 34q20 + 52q19 + 73q18 + 96q17 + 119q16 + 140q15 +157q14 + 168q13 + 172q12 + 168q11 + 157q10 + 140q9 + 119q8 + 96q7 + 73q6 + 52q5 +34q4 + 20q3 + 10q2 + 4q + 1k = q24

– J = {1, 2}, {2, 3}, {3, 4}; type of P : A2

v = q22 + 3q21 + 6q20 + 11q19 + 17q18 + 24q17 + 32q16 + 40q15 + 47q14 + 53q13 +57q12 +58q11 +57q10 +53q9 +47q8 +40q7 +32q6 +24q5 +17q4 +11q3 +6q2 +3q+1k = q22

– J = {1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {3, 5}; type of P : A1A1

v = q23+3q22+7q21+13q20+21q19+31q18+42q17+54q16+65q15+75q14+82q13+86q12 +86q11 +82q10 +75q9 +65q8 +54q7 +42q6 +31q5 +21q4 +13q3 +7q2 +3q+1k = q23

– J = {4, 5}; type of P : B2

v = q21 + 3q20 + 6q19 + 10q18 + 15q17 + 21q16 + 27q15 + 33q14 + 38q13 + 42q12 +44q11 + 44q10 + 42q9 + 38q8 + 33q7 + 27q6 + 21q5 + 15q4 + 10q3 + 6q2 + 3q + 1k = q21

– J = {1, 2, 3}, {2, 3, 4}; type of P : A3

v = q19 + 2q18 + 3q17 + 5q16 + 7q15 + 9q14 + 11q13 + 13q12 + 14q11 + 15q10 + 15q9 +14q8 + 13q7 + 11q6 + 9q5 + 7q4 + 5q3 + 3q2 + 2q + 1k = q19

– J = {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {2, 3, 5}; type of P : A2A1

v = q21 + 2q20 + 4q19 + 7q18 + 10q17 + 14q16 + 18q15 + 22q14 + 25q13 + 28q12 +29q11 + 29q10 + 28q9 + 25q8 + 22q7 + 18q6 + 14q5 + 10q4 + 7q3 + 4q2 + 2q + 1k = q21

– J = {1, 3, 5}; type of P : A1A1A1

v = q22 + 2q21 + 5q20 + 8q19 + 13q18 + 18q17 + 24q16 + 30q15 + 35q14 + 40q13 +42q12 + 44q11 + 42q10 + 40q9 + 35q8 + 30q7 + 24q6 + 18q5 + 13q4 + 8q3 + 5q2 + 2q+ 1k = q22

– J = {1, 4, 5}, {2, 4, 5}; type of P : B2A1

v = q20 +2q19 +4q18 +6q17 +9q16 +12q15 +15q14 +18q13 +20q12 +22q11 +22q10 +22q9 + 20q8 + 18q7 + 15q6 + 12q5 + 9q4 + 6q3 + 4q2 + 2q + 1k = q20

– J = {3, 4, 5}; type of P : B3

v = q16 + 2q15 + 3q14 + 4q13 + 5q12 + 6q11 + 7q10 + 8q9 + 8q8 + 8q7 + 7q6 + 6q5 +5q4 + 4q3 + 3q2 + 2q + 1k = q16

– J = {1, 2, 3, 4}; type of P : A4

v = q15+q14+q13+2q12+2q11+3q10+3q9+3q8+3q7+3q6+3q5+2q4+2q3+q2+q+1k = q15

– J = {1, 2, 3, 5}; type of P : A3A1

v = q18 + q17 + 2q16 + 3q15 + 4q14 + 5q13 + 6q12 + 7q11 + 7q10 + 8q9 + 7q8 + 7q7 +6q6 + 5q5 + 4q4 + 3q3 + 2q2 + q + 1k = q18

– J = {1, 2, 4, 5}; type of P : B2A2

v = q18 + q17 + 2q16 + 3q15 + 4q14 + 5q13 + 6q12 + 7q11 + 7q10 + 8q9 + 7q8 + 7q7 +6q6 + 5q5 + 4q4 + 3q3 + 2q2 + q + 1k = q18

– J = {1, 3, 4, 5}; type of P : B3A1

v = q15 + q14 + 2q13 + 2q12 + 3q11 + 3q10 + 4q9 + 4q8 + 4q7 + 4q6 + 3q5 + 3q4 +2q3 + 2q2 + q + 1k = q15


– J = {2, 3, 4, 5}; type of P : B4

v = q9 + q8 + q7 + q6 + q5 + q4 + q3 + q2 + q + 1k = q9

D4 – J = {1}, {2}, {3}, {4}; type of P : A1

v = q11 + 3q10 + 6q9 + 10q8 + 13q7 + 15q6 + 15q5 + 13q4 + 10q3 + 6q2 + 3q + 1k = q11

– J = {1, 2}, {2, 3}, {2, 4}; type of P : A2

v = q9 + 2q8 + 3q7 + 5q6 + 5q5 + 5q4 + 5q3 + 3q2 + 2q + 1k = q9

– J = {1, 3}, {1, 4}, {3, 4}; type of P : A1A1

v = q10 + 2q9 + 4q8 + 6q7 + 7q6 + 8q5 + 7q4 + 6q3 + 4q2 + 2q + 1k = q10

– J = {1, 2, 3}, {1, 2, 4}, {2, 3, 4}; type of P : A3

v = q6 + q5 + q4 + 2q3 + q2 + q + 1k = q6

– J = {1, 3, 4}; type of P : A1A1A1

v = q9 + q8 + 3q7 + 3q6 + 4q5 + 4q4 + 3q3 + 3q2 + q + 1k = q9

D5 – J = {1}, {2}, {3}; type of P : A1

v = q19 + 4q18 + 10q17 + 20q16 + 34q15 + 51q14 + 69q13 + 86q12 + 99q11 + 106q10 +106q9 + 99q8 + 86q7 + 69q6 + 51q5 + 34q4 + 20q3 + 10q2 + 4q + 1k = q19

– J = {1, 2}, {2, 3}; type of P : A2

v = q17 +3q16 +6q15 +11q14 +17q13 +23q12 +29q11 +34q10 +36q9 +36q8 +34q7 +29q6 + 23q5 + 17q4 + 11q3 + 6q2 + 3q + 1k = q17

– J = {1, 3}, {4, 5}; type of P : A1A1

v = q18 + 3q17 + 7q16 + 13q15 + 21q14 + 30q13 + 39q12 + 47q11 + 52q10 + 54q9 +52q8 + 47q7 + 39q6 + 30q5 + 21q4 + 13q3 + 7q2 + 3q + 1k = q18

– J = {1, 2, 3}, {3, 4, 5}; type of P : A3

v = q14+2q13+3q12+5q11+7q10+8q9+9q8+10q7+9q6+8q5+7q4+5q3+3q2+2q+1k = q14

– J = {1, 4, 5}, {2, 4, 5}; type of P : A1A1A1

v = q17 + 2q16 + 5q15 + 8q14 + 13q13 + 17q12 + 22q11 + 25q10 + 27q9 + 27q8 + 25q7 +22q6 + 17q5 + 13q4 + 8q3 + 5q2 + 2q + 1k = q17

– J = {1, 2, 4, 5}; type of P : A2A1A1

v = q15 + q14 + 3q13 + 4q12 + 6q11 + 7q10 + 9q9 + 9q8 + 9q7 + 9q6 + 7q5 + 6q4 +4q3 + 3q2 + q + 1k = q15

– J = {1, 3, 4, 5}; type of P : A3A1

v = q13 + q12 + 2q11 + 3q10 + 4q9 + 4q8 + 5q7 + 5q6 + 4q5 + 4q4 + 3q3 + 2q2 + q+ 1k = q13

– J = {2, 3, 4, 5}; type of P : D4

v = q8 + q7 + q6 + q5 + 2q4 + q3 + q2 + q + 1k = q8

E6 – J = {2}, {4}; type of P : A1

v = q35 +5q34 +15q33 +35q32 +70q31 +125q30 +204q29 +310q28 +444q27 +604q26 +


785q25 + 980q24 + 1179q23 + 1370q22 + 1541q21 + 1681q20 + 1780q19 + 1831q18 +1831q17 + 1780q16 + 1681q15 + 1541q14 + 1370q13 + 1179q12 + 980q11 + 785q10 +604q9 + 444q8 + 310q7 + 204q6 + 125q5 + 70q4 + 35q3 + 15q2 + 5q + 1k = q35

– J = {1, 6}, {3, 5}; type of P : A1, A1

v = q34 +4q33 +11q32 +24q31 +46q30 +79q29 +125q28 +185q27 +259q26 +345q25 +440q24 + 540q23 + 639q22 + 731q21 + 810q20 + 871q19 + 909q18 + 922q17 + 909q16 +871q15 + 810q14 + 731q13 + 639q12 + 540q11 + 440q10 + 345q9 + 259q8 + 185q7 +125q6 + 79q5 + 46q4 + 24q3 + 11q2 + 4q + 1k = q34

– J = {2, 4}; type of P : A2

v = q33 +4q32 +10q31 +21q30 +39q29 +65q28 +100q27 +145q26 +199q25 +260q24 +326q23 + 394q22 + 459q21 + 517q20 + 565q19 + 599q18 + 616q17 + 616q16 + 599q15 +565q14 +517q13 +459q12 +394q11 +326q10 +260q9 +199q8 +145q7 +100q6 +65q5 +39q4 + 21q3 + 10q2 + 4q + 1k = q33

– J = {1, 2, 6}, {1, 4, 6}, {2, 3, 5}; type of P : A1A1A1

v = q33 + 3q32 + 8q31 + 16q30 + 30q29 + 49q28 + 76q27 + 109q26 + 150q25 + 195q24 +245q23 + 295q22 + 344q21 + 387q20 + 423q19 + 448q18 + 461q17 + 461q16 + 448q15 +423q14 + 387q13 + 344q12 + 295q11 + 245q10 + 195q9 + 150q8 + 109q7 + 76q6 + 49q5 +30q4 + 16q3 + 8q2 + 3q + 1k = q33

– J = {3, 4, 5}; type of P : A3

v = q30 + 3q29 + 6q28 + 11q27 + 19q26 + 29q25 + 41q24 + 56q23 + 73q22 + 90q21 +107q20 + 124q19 + 138q18 + 148q17 + 155q16 + 158q15 + 155q14 + 148q13 + 138q12 +124q11 + 107q10 + 90q9 + 73q8 + 56q7 + 41q6 + 29q5 + 19q4 + 11q3 + 6q2 + 3q + 1k = q30

– J = {1, 2, 4, 6}; type of P : A2A1A1

v = q31 + 2q30 + 5q29 + 9q28 + 16q27 + 24q26 + 36q25 + 49q24 + 65q23 + 81q22 +99q21 + 115q20 + 130q19 + 142q18 + 151q17 + 155q16 + 155q15 + 151q14 + 142q13 +130q12 +115q11 +99q10 +81q9 +65q8 +49q7 +36q6 +24q5 +16q4 +9q3 +5q2 +2q+1k = q31

– J = {1, 3, 5, 6}; type of P : A2A2

v = q30 + 2q29 + 4q28 + 8q27 + 13q26 + 19q25 + 28q24 + 38q23 + 48q22 + 60q21 +72q20 + 82q19 + 91q18 + 99q17 + 103q16 + 104q15 + 103q14 + 99q13 + 91q12 + 82q11 +72q10 + 60q9 + 48q8 + 38q7 + 28q6 + 19q5 + 13q4 + 8q3 + 4q2 + 2q + 1k = q30

– J = {2, 3, 4, 5}; type of P : D4

v = q24 + 2q23 + 3q22 + 4q21 + 7q20 + 9q19 + 11q18 + 13q17 + 17q16 + 18q15 + 19q14 +20q13+22q12+20q11+19q10+18q9+17q8+13q7+11q6+9q5+7q4+4q3+3q2+2q+1k = q24

– J = {1, 2, 3, 5, 6}; type of P : A2A2A1

v = q29 + q28 + 3q27 + 5q26 + 8q25 + 11q24 + 17q23 + 21q22 + 27q21 + 33q20 + 39q19 +43q18 + 48q17 + 51q16 + 52q15 + 52q14 + 51q13 + 48q12 + 43q11 + 39q10 + 33q9 +27q8 + 21q7 + 17q6 + 11q5 + 8q4 + 5q3 + 3q2 + q + 1k = q29

– J = {1, 3, 4, 5, 6}; type of P : A5

v = q21 + q20 + q19 + 2q18 + 3q17 + 3q16 + 4q15 + 5q14 + 5q13 + 5q12 + 6q11 + 6q10 +5q9 + 5q8 + 5q7 + 4q6 + 3q5 + 3q4 + 2q3 + q2 + q + 1k = q21

E7 – J = {1}, {2}, {3}, {4}, {5}, {6}, {7}; type of P :v = q62 + 6q61 + 21q60 + 56q59 + 126q58 + 252q57 + 461q56 + 786q55 + 1265q54 +


1940q53 +2855q52 +4054q51 +5578q50 +7462q49 +9732q48 +12402q47 +15472q46 +18926q45 + 22731q44 + 26836q43 + 31173q42 + 35658q41 + 40194q40 + 44674q39 +48985q38 + 53012q37 + 56643q36 + 59774q35 + 62313q34 + 64184q33 + 65330q32 +65716q31 + 65330q30 + 64184q29 + 62313q28 + 59774q27 + 56643q26 + 53012q25 +48985q24 + 44674q23 + 40194q22 + 35658q21 + 31173q20 + 26836q19 + 22731q18 +18926q17 +15472q16 +12402q15 +9732q14 +7462q13 +5578q12 +4054q11 +2855q10 +1940q9 + 1265q8 + 786q7 + 461q6 + 252q5 + 126q4 + 56q3 + 21q2 + 6q + 1k = q62

– J = {1, 2}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {2, 3}, {2, 5}, {2, 6}, {2, 7}, {3, 5}, {3, 6}, {3, 7},{4, 6}, {4, 7}, {5, 7}; type of P : A1A1

v = q61 + 5q60 + 16q59 + 40q58 + 86q57 + 166q56 + 295q55 + 491q54 + 774q53 +1166q52 + 1689q51 + 2365q50 + 3213q49 + 4249q48 + 5483q47 + 6919q46 + 8553q45 +10373q44 + 12358q43 + 14478q42 + 16695q41 + 18963q40 + 21231q39 + 23443q38 +25542q37 + 27470q36 + 29173q35 + 30601q34 + 31712q33 + 32472q32 + 32858q31 +32858q30 + 32472q29 + 31712q28 + 30601q27 + 29173q26 + 27470q25 + 25542q24 +23443q23 + 21231q22 + 18963q21 + 16695q20 + 14478q19 + 12358q18 + 10373q17 +8553q16 + 6919q15 + 5483q14 + 4249q13 + 3213q12 + 2365q11 + 1689q10 + 1166q9 +774q8 + 491q7 + 295q6 + 166q5 + 86q4 + 40q3 + 16q2 + 5q + 1k = q61

– J = {1, 3}, {2, 4}, {3, 4}, {4, 5}, {5, 6}, {6, 7}; type of P : A2

v = q60 +5q59 +15q58 +36q57 +75q56 +141q55 +245q54 +400q53 +620q52 +920q51 +1315q50 + 1819q49 + 2444q48 + 3199q47 + 4089q46 + 5114q45 + 6269q44 + 7543q43 +8919q42 + 10374q41 + 11880q40 + 13404q39 + 14910q38 + 16360q37 + 17715q36 +18937q35 + 19991q34 + 20846q33 + 21476q32 + 21862q31 + 21992q30 + 21862q29 +21476q28 + 20846q27 + 19991q26 + 18937q25 + 17715q24 + 16360q23 + 14910q22 +13404q21 +11880q20 +10374q19 +8919q18 +7543q17 +6269q16 +5114q15 +4089q14 +3199q13 + 2444q12 + 1819q11 + 1315q10 + 920q9 + 620q8 + 400q7 + 245q6 + 141q5 +75q4 + 36q3 + 15q2 + 5q + 1k = q60

– J = {1, 2, 3}, {1, 2, 4}, {1, 3, 5}, {1, 3, 6}, {1, 3, 7}, {1, 4, 5}, {1, 5, 6}, {1, 6, 7}, {2, 4, 6},{2, 4, 7}, {2, 5, 6}, {2, 6, 7}, {3, 4, 6}, {3, 4, 7}, {3, 5, 6}, {3, 6, 7}, {4, 5, 7}, {4, 6, 7}; typeof P : A2A1

v = q59 +4q58 +11q57 +25q56 +50q55 +91q54 +154q53 +246q52 +374q51 +546q50 +769q49 + 1050q48 + 1394q47 + 1805q46 + 2284q45 + 2830q44 + 3439q43 + 4104q42 +4815q41 + 5559q40 + 6321q39 + 7083q38 + 7827q37 + 8533q36 + 9182q35 + 9755q34 +10236q33 + 10610q32 + 10866q31 + 10996q30 + 10996q29 + 10866q28 + 10610q27 +10236q26 + 9755q25 + 9182q24 + 8533q23 + 7827q22 + 7083q21 + 6321q20 + 5559q19 +4815q18 + 4104q17 + 3439q16 + 2830q15 + 2284q14 + 1805q13 + 1394q12 + 1050q11 +769q10 + 546q9 + 374q8 + 246q7 + 154q6 + 91q5 + 50q4 + 25q3 + 11q2 + 4q + 1k = q59

– J = {1, 2, 5}, {1, 2, 6}, {1, 2, 7}, {1, 4, 6}, {1, 4, 7}, {1, 5, 7}, {2, 3, 5}, {2, 3, 6}, {2, 3, 7},{2, 5, 7}, {3, 5, 7}; type of P : A1A1A1

v = q60 + 4q59 + 12q58 + 28q57 + 58q56 + 108q55 + 187q54 + 304q53 + 470q52 +696q51 + 993q50 + 1372q49 + 1841q48 + 2408q47 + 3075q46 + 3844q45 + 4709q44 +5664q43+6694q42+7784q41+8911q40+10052q39+11179q38+12264q37+13278q36+14192q35 + 14981q34 + 15620q33 + 16092q32 + 16380q31 + 16478q30 + 16380q29 +16092q28 + 15620q27 + 14981q26 + 14192q25 + 13278q24 + 12264q23 + 11179q22 +10052q21 + 8911q20 + 7784q19 + 6694q18 + 5664q17 + 4709q16 + 3844q15 + 3075q14 +2408q13 + 1841q12 + 1372q11 + 993q10 + 696q9 + 470q8 + 304q7 + 187q6 + 108q5 +58q4 + 28q3 + 12q2 + 4q + 1k = q60

– J = {1, 3, 4}, {2, 3, 4}, {2, 4, 5}, {3, 4, 5}, {4, 5, 6}, {5, 6, 7}; type of P : A3

v = q57 +4q56 +10q55 +21q54 +40q53 +70q52 +114q51 +176q50 +260q49 +370q48 +509q47 + 680q46 + 885q45 + 1125q44 + 1399q43 + 1705q42 + 2040q41 + 2399q40 +2775q39 + 3160q38 + 3546q37 + 3923q36 + 4281q35 + 4610q34 + 4901q33 + 5145q32 +


5335q31 + 5465q30 + 5531q29 + 5531q28 + 5465q27 + 5335q26 + 5145q25 + 4901q24 +4610q23 + 4281q22 + 3923q21 + 3546q20 + 3160q19 + 2775q18 + 2399q17 + 2040q16 +1705q15 + 1399q14 + 1125q13 + 885q12 + 680q11 + 509q10 + 370q9 + 260q8 + 176q7 +114q6 + 70q5 + 40q4 + 21q3 + 10q2 + 4q + 1k = q57

– J = {1, 2, 3, 4}, {1, 3, 4, 5}, {2, 4, 5, 6}, {3, 4, 5, 6}, {4, 5, 6, 7}; type of P : A4

v = q53 + 3q52 + 6q51 + 11q50 + 19q49 + 31q48 + 47q47 + 68q46 + 95q45 + 129q44 +170q43 + 218q42 + 273q41 + 335q40 + 403q39 + 476q38 + 553q37 + 632q36 + 711q35 +788q34+862q33+930q32+990q31+1040q30+1079q29+1106q28+1120q27+1120q26+1106q25 +1079q24 +1040q23 +990q22 +930q21 +862q20 +788q19 +711q18 +632q17 +553q16 + 476q15 + 403q14 + 335q13 + 273q12 + 218q11 + 170q10 + 129q9 + 95q8 +68q7 + 47q6 + 31q5 + 19q4 + 11q3 + 6q2 + 3q + 1k = q53

– J = {1, 2, 3, 5}, {1, 2, 3, 6}, {1, 2, 3, 7}, {1, 2, 4, 6}, {1, 2, 4, 7}, {1, 2, 5, 6}, {1, 2, 6, 7},{1, 3, 5, 7}, {1, 4, 5, 7}, {1, 4, 6, 7}, {2, 3, 5, 6}, {2, 3, 6, 7}; type of P : A2A1A1

v = q58 + 3q57 + 8q56 + 17q55 + 33q54 + 58q53 + 96q52 + 150q51 + 224q50 + 322q49 +447q48 + 603q47 + 791q46 + 1014q45 + 1270q44 + 1560q43 + 1879q42 + 2225q41 +2590q40 + 2969q39 + 3352q38 + 3731q37 + 4096q36 + 4437q35 + 4745q34 + 5010q33 +5226q32 + 5384q31 + 5482q30 + 5514q29 + 5482q28 + 5384q27 + 5226q26 + 5010q25 +4745q24 + 4437q23 + 4096q22 + 3731q21 + 3352q20 + 2969q19 + 2590q18 + 2225q17 +1879q16 +1560q15 +1270q14 +1014q13 +791q12 +603q11 +447q10 +322q9 +224q8 +150q7 + 96q6 + 58q5 + 33q4 + 17q3 + 8q2 + 3q + 1k = q58

– J = {1, 2, 4, 5}, {1, 3, 4, 6}, {1, 3, 4, 7}, {1, 4, 5, 6}, {1, 5, 6, 7}, {2, 3, 4, 6}, {2, 3, 4, 7},{2, 4, 5, 7}, {2, 5, 6, 7}, {3, 4, 5, 7}, {3, 5, 6, 7}; type of P : A3A1

v = q56 + 3q55 + 7q54 + 14q53 + 26q52 + 44q51 + 70q50 + 106q49 + 154q48 + 216q47 +293q46 +387q45 +498q44 +627q43 +772q42 +933q41 +1107q40 +1292q39 +1483q38 +1677q37 + 1869q36 + 2054q35 + 2227q34 + 2383q33 + 2518q32 + 2627q31 + 2708q30 +2757q29 + 2774q28 + 2757q27 + 2708q26 + 2627q25 + 2518q24 + 2383q23 + 2227q22 +2054q21+1869q20+1677q19+1483q18+1292q17+1107q16+933q15+772q14+627q13+498q12+387q11+293q10+216q9+154q8+106q7+70q6+44q5+26q4+14q3+7q2+3q+1k = q56

– J = {1, 2, 5, 7}, {2, 3, 5, 7}; type of P : A1A1A1A1

v = q59 + 3q58 + 9q57 + 19q56 + 39q55 + 69q54 + 118q53 + 186q52 + 284q51 + 412q50 +581q49 + 791q48 + 1050q47 + 1358q46 + 1717q45 + 2127q44 + 2582q43 + 3082q42 +3612q41 + 4172q40 + 4739q39 + 5313q38 + 5866q37 + 6398q36 + 6880q35 + 7312q34 +7669q33 + 7951q32 + 8141q31 + 8239q30 + 8239q29 + 8141q28 + 7951q27 + 7669q26 +7312q25 + 6880q24 + 6398q23 + 5866q22 + 5313q21 + 4739q20 + 4172q19 + 3612q18 +3082q17 + 2582q16 + 2127q15 + 1717q14 + 1358q13 + 1050q12 + 791q11 + 581q10 +412q9 + 284q8 + 186q7 + 118q6 + 69q5 + 39q4 + 19q3 + 9q2 + 3q + 1k = q59

– J = {1, 3, 5, 6}, {1, 3, 6, 7}, {2, 4, 6, 7}, {3, 4, 6, 7}; type of P : A2A2

v = q57 + 3q56 + 7q55 + 15q54 + 28q53 + 48q52 + 78q51 + 120q50 + 176q49 + 250q48 +343q47+457q46+594q45+754q44+936q43+1140q42+1363q41+1601q40+1851q39+2107q38 + 2363q37 + 2613q36 + 2851q35 + 3069q34 + 3262q33 + 3424q32 + 3550q31 +3636q30 + 3680q29 + 3680q28 + 3636q27 + 3550q26 + 3424q25 + 3262q24 + 3069q23 +2851q22 + 2613q21 + 2363q20 + 2107q19 + 1851q18 + 1601q17 + 1363q16 + 1140q15 +936q14 + 754q13 + 594q12 + 457q11 + 343q10 + 250q9 + 176q8 + 120q7 + 78q6 + 48q5 +28q4 + 15q3 + 7q2 + 3q + 1k = q57

– J = {2, 3, 4, 5}; type of P : D4

v = q51 + 3q50 + 6q49 + 10q48 + 17q47 + 27q46 + 40q45 + 56q44 + 77q43 + 103q42 +133q41 + 167q40 + 206q39 + 250q38 + 296q37 + 344q36 + 394q35 + 446q34 + 495q33 +541q32 + 584q31 + 624q30 + 656q29 + 680q28 + 697q27 + 707q26 + 707q25 + 697q24 +


680q23 + 656q22 + 624q21 + 584q20 + 541q19 + 495q18 + 446q17 + 394q16 + 344q15 +296q14 + 250q13 + 206q12 + 167q11 + 133q10 + 103q9 + 77q8 + 56q7 + 40q6 + 27q5 +17q4 + 10q3 + 6q2 + 3q + 1k = q51

– J = {1, 2, 3, 4, 5}, {2, 3, 4, 5, 6}; type of P : D5

v = q43 + 2q42 + 3q41 + 4q40 + 6q39 + 9q38 + 12q37 + 15q36 + 19q35 + 24q34 + 29q33 +34q32 + 39q31 + 45q30 + 50q29 + 55q28 + 60q27 + 65q26 + 68q25 + 70q24 + 72q23 +74q22 + 74q21 + 72q20 + 70q19 + 68q18 + 65q17 + 60q16 + 55q15 + 50q14 + 45q13 +39q12 + 34q11 + 29q10 + 24q9 + 19q8 + 15q7 + 12q6 + 9q5 + 6q4 + 4q3 + 3q2 + 2q+ 1k = q43

– J = {1, 2, 3, 4, 6}, {1, 2, 3, 4, 7}, {1, 2, 4, 5, 6}, {1, 3, 4, 5, 7}, {1, 4, 5, 6, 7}; type of P :A4A1

v = q52 + 2q51 + 4q50 + 7q49 + 12q48 + 19q47 + 28q46 + 40q45 + 55q44 + 74q43 +96q42 + 122q41 + 151q40 + 184q39 + 219q38 + 257q37 + 296q36 + 336q35 + 375q34 +413q33 + 449q32 + 481q31 + 509q30 + 531q29 + 548q28 + 558q27 + 562q26 + 558q25 +548q24 + 531q23 + 509q22 + 481q21 + 449q20 + 413q19 + 375q18 + 336q17 + 296q16 +257q15 + 219q14 + 184q13 + 151q12 + 122q11 + 96q10 + 74q9 + 55q8 + 40q7 + 28q6 +19q5 + 12q4 + 7q3 + 4q2 + 2q + 1k = q52

– J = {1, 2, 3, 5, 6}, {1, 2, 3, 6, 7}, {1, 2, 4, 6, 7}; type of P : A2A2A1

v = q56 + 2q55 + 5q54 + 10q53 + 18q52 + 30q51 + 48q50 + 72q49 + 104q48 + 146q47 +197q46 + 260q45 + 334q44 + 420q43 + 516q42 + 624q41 + 739q40 + 862q39 + 989q38 +1118q37 + 1245q36 + 1368q35 + 1483q34 + 1586q33 + 1676q32 + 1748q31 + 1802q30 +1834q29 + 1846q28 + 1834q27 + 1802q26 + 1748q25 + 1676q24 + 1586q23 + 1483q22 +1368q21 +1245q20 +1118q19 +989q18 +862q17 +739q16 +624q15 +516q14 +420q13 +334q12+260q11+197q10+146q9+104q8+72q7+48q6+30q5+18q4+10q3+5q2+2q+1k = q56

– J = {1, 2, 3, 5, 7}; type of P : A2A1A1A1

v = q57 + 2q56 + 6q55 + 11q54 + 22q53 + 36q52 + 60q51 + 90q50 + 134q49 + 188q48 +259q47 +344q46 +447q45 +567q44 +703q43 +857q42 +1022q41 +1203q40 +1387q39 +1582q38 + 1770q37 + 1961q36 + 2135q35 + 2302q34 + 2443q33 + 2567q32 + 2659q31 +2725q30 + 2757q29 + 2757q28 + 2725q27 + 2659q26 + 2567q25 + 2443q24 + 2302q23 +2135q22 + 1961q21 + 1770q20 + 1582q19 + 1387q18 + 1203q17 + 1022q16 + 857q15 +703q14 + 567q13 + 447q12 + 344q11 + 259q10 + 188q9 + 134q8 + 90q7 + 60q6 + 36q5 +22q4 + 11q3 + 6q2 + 2q + 1k = q57

– J = {1, 2, 4, 5, 7}, {1, 2, 5, 6, 7}; type of P : A3A1A1

v = q55 + 2q54 + 5q53 + 9q52 + 17q51 + 27q50 + 43q49 + 63q48 + 91q47 + 125q46 +168q45 + 219q44 + 279q43 + 348q42 + 424q41 + 509q40 + 598q39 + 694q38 + 789q37 +888q36 + 981q35 + 1073q34 + 1154q33 + 1229q32 + 1289q31 + 1338q30 + 1370q29 +1387q28 + 1387q27 + 1370q26 + 1338q25 + 1289q24 + 1229q23 + 1154q22 + 1073q21 +981q20 + 888q19 + 789q18 + 694q17 + 598q16 + 509q15 + 424q14 + 348q13 + 279q12 +219q11 + 168q10 + 125q9 + 91q8 + 63q7 + 43q6 + 27q5 + 17q4 + 9q3 + 5q2 + 2q + 1k = q55

– J = {1, 3, 4, 5, 6}, {2, 4, 5, 6, 7}, {3, 4, 5, 6, 7}; type of P : A5

v = q48 +2q47 +3q46 +5q45 +8q44 +12q43 +17q42 +23q41 +30q40 +39q39 +49q38 +60q37+72q36+85q35+98q34+112q33+126q32+139q31+151q30+162q29+172q28+180q27 + 186q26 + 189q25 + 190q24 + 189q23 + 186q22 + 180q21 + 172q20 + 162q19 +151q18 + 139q17 + 126q16 + 112q15 + 98q14 + 85q13 + 72q12 + 60q11 + 49q10 + 39q9 +30q8 + 23q7 + 17q6 + 12q5 + 8q4 + 5q3 + 3q2 + 2q + 1k = q48

– J = {1, 3, 4, 6, 7}, {1, 3, 5, 6, 7}, {2, 3, 4, 6, 7}, {2, 3, 5, 6, 7}; type of P : A3A2

v = q54 + 2q53 + 4q52 + 8q51 + 14q50 + 22q49 + 34q48 + 50q47 + 70q46 + 96q45 +127q44 + 164q43 + 207q42 + 256q41 + 309q40 + 368q39 + 430q38 + 494q37 + 559q36 +


624q35 + 686q34 + 744q33 + 797q32 + 842q31 + 879q30 + 906q29 + 923q28 + 928q27 +923q26 + 906q25 + 879q24 + 842q23 + 797q22 + 744q21 + 686q20 + 624q19 + 559q18 +494q17 + 430q16 + 368q15 + 309q14 + 256q13 + 207q12 + 164q11 + 127q10 + 96q9 +70q8 + 50q7 + 34q6 + 22q5 + 14q4 + 8q3 + 4q2 + 2q + 1k = q54

– J = {2, 3, 4, 5, 7}; type of P : D4A1

v = q50 +2q49 +4q48 +6q47 +11q46 +16q45 +24q44 +32q43 +45q42 +58q41 +75q40 +92q39 + 114q38 + 136q37 + 160q36 + 184q35 + 210q34 + 236q33 + 259q32 + 282q31 +302q30 + 322q29 + 334q28 + 346q27 + 351q26 + 356q25 + 351q24 + 346q23 + 334q22 +322q21 + 302q20 + 282q19 + 259q18 + 236q17 + 210q16 + 184q15 + 160q14 + 136q13 +114q12 +92q11 +75q10 +58q9 +45q8 +32q7 +24q6 +16q5 +11q4 +6q3 +4q2 +2q+1k = q50

– J = {1, 2, 3, 4, 5, 6}; type of P : E6

v = q27 +q26 +q25 +q24 +q23 +2q22 +2q21 +2q20 +2q19 +3q18 +3q17 +3q16 +3q15 +3q14 + 3q13 + 3q12 + 3q11 + 3q10 + 3q9 + 2q8 + 2q7 + 2q6 + 2q5 + q4 + q3 + q2 + q+ 1k = q27

– J = {1, 2, 3, 4, 5, 7}; type of P : D5A1

v = q42 + q41 + 2q40 + 2q39 + 4q38 + 5q37 + 7q36 + 8q35 + 11q34 + 13q33 + 16q32 +18q31 + 21q30 + 24q29 + 26q28 + 29q27 + 31q26 + 34q25 + 34q24 + 36q23 + 36q22 +38q21 + 36q20 + 36q19 + 34q18 + 34q17 + 31q16 + 29q15 + 26q14 + 24q13 + 21q12 +18q11 + 16q10 + 13q9 + 11q8 + 8q7 + 7q6 + 5q5 + 4q4 + 2q3 + 2q2 + q + 1k = q42

– J = {1, 2, 3, 4, 6, 7}; type of P : A4A2

v = q50 + q49 + 2q48 + 4q47 + 6q46 + 9q45 + 13q44 + 18q43 + 24q42 + 32q41 + 40q40 +50q39 +61q38 +73q37 +85q36 +99q35 +112q34 +125q33 +138q32 +150q31 +161q30 +170q29 + 178q28 + 183q27 + 187q26 + 188q25 + 187q24 + 183q23 + 178q22 + 170q21 +161q20 +150q19 +138q18 +125q17 +112q16 +99q15 +85q14 +73q13 +61q12 +50q11 +40q10 + 32q9 + 24q8 + 18q7 + 13q6 + 9q5 + 6q4 + 4q3 + 2q2 + q + 1k = q50

– J = {1, 2, 3, 5, 6, 7}; type of P : A3A2A1

v = q53 + q52 + 3q51 + 5q50 + 9q49 + 13q48 + 21q47 + 29q46 + 41q45 + 55q44 + 72q43 +92q42 + 115q41 + 141q40 + 168q39 + 200q38 + 230q37 + 264q36 + 295q35 + 329q34 +357q33 + 387q32 + 410q31 + 432q30 + 447q29 + 459q28 + 464q27 + 464q26 + 459q25 +447q24 + 432q23 + 410q22 + 387q21 + 357q20 + 329q19 + 295q18 + 264q17 + 230q16 +200q15 + 168q14 + 141q13 + 115q12 + 92q11 + 72q10 + 55q9 + 41q8 + 29q7 + 21q6 +13q5 + 9q4 + 5q3 + 3q2 + q + 1k = q53

– J = {1, 2, 4, 5, 6, 7}; type of P : A5A1

v = q47+q46+2q45+3q44+5q43+7q42+10q41+13q40+17q39+22q38+27q37+33q36+39q35+46q34+52q33+60q32+66q31+73q30+78q29+84q28+88q27+92q26+94q25+95q24+95q23+94q22+92q21+88q20+84q19+78q18+73q17+66q16+60q15+52q14+46q13+39q12+33q11+27q10+22q9+17q8+13q7+10q6+7q5+5q4+3q3+2q2+q+1k = q47

– J = {1, 3, 4, 5, 6, 7}; type of P : A6

v = q42 + q41 + q40 + 2q39 + 3q38 + 4q37 + 5q36 + 7q35 + 8q34 + 10q33 + 12q32 +14q31 + 16q30 + 18q29 + 20q28 + 22q27 + 24q26 + 25q25 + 26q24 + 27q23 + 28q22 +28q21 + 28q20 + 27q19 + 26q18 + 25q17 + 24q16 + 22q15 + 20q14 + 18q13 + 16q12 +14q11 + 12q10 + 10q9 + 8q8 + 7q7 + 5q6 + 4q5 + 3q4 + 2q3 + q2 + q + 1k = q42

– J = {2, 3, 4, 5, 6, 7}; type of P : D6

v = q33 + q32 + q31 + q30 + 2q29 + 2q28 + 3q27 + 3q26 + 4q25 + 4q24 + 5q23 + 5q22 +6q21 + 6q20 + 6q19 + 6q18 + 7q17 + 7q16 + 6q15 + 6q14 + 6q13 + 6q12 + 5q11 + 5q10 +4q9 + 4q8 + 3q7 + 3q6 + 2q5 + 2q4 + q3 + q2 + q + 1k = q33


E8 – J = {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}; type of P : A1

v = q119 + 7q118 + 28q117 + 84q116 + 210q115 + 462q114 + 924q113 + 1716q112 +3002q111 + 4998q110 + 7980q109 + 12292q108 + 18353q107 + 26663q106 + 37807q105 +52457q104+71372q103+95396q102+125453q101+162539q100+207711q99+262073q98+326760q97+402920q96+491693q95+594187q94+711453q93+844459q92+994064q91+1160992q90 + 1345806q89 + 1548882q88 + 1770386q87 + 2010254q86 + 2268175q85 +2543577q84 + 2835617q83 + 3143175q82 + 3464854q81 + 3798986q80 + 4143642q79 +4496646q78 + 4855594q77 + 5217878q76 + 5580715q75 + 5941181q74 + 6296247q73 +6642817q72 + 6977769q71 + 7297999q70 + 7600465q69 + 7882231q68 + 8140509q67 +8372699q66 + 8576428q65 + 8749588q64 + 8890369q63 + 8997287q62 + 9069207q61 +9105361q60 + 9105361q59 + 9069207q58 + 8997287q57 + 8890369q56 + 8749588q55 +8576428q54 + 8372699q53 + 8140509q52 + 7882231q51 + 7600465q50 + 7297999q49 +6977769q48 + 6642817q47 + 6296247q46 + 5941181q45 + 5580715q44 + 5217878q43 +4855594q42 + 4496646q41 + 4143642q40 + 3798986q39 + 3464854q38 + 3143175q37 +2835617q36 + 2543577q35 + 2268175q34 + 2010254q33 + 1770386q32 + 1548882q31 +1345806q30+1160992q29+994064q28+844459q27+711453q26+594187q25+491693q24+402920q23+326760q22+262073q21+207711q20+162539q19+125453q18+95396q17+71372q16+52457q15+37807q14+26663q13+18353q12+12292q11+7980q10+4998q9+3002q8 + 1716q7 + 924q6 + 462q5 + 210q4 + 84q3 + 28q2 + 7q + 1k = q119

– J = {1, 2}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {2, 3}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {3, 5},{3, 6}, {3, 7}, {3, 8}, {4, 6}, {4, 7}, {4, 8}, {5, 7}, {5, 8}, {6, 8}; type of P : A1A1

v = q118 + 6q117 + 22q116 + 62q115 + 148q114 + 314q113 + 610q112 + 1106q111 +1896q110 + 3102q109 + 4878q108 + 7414q107 + 10939q106 + 15724q105 + 22083q104 +30374q103+40998q102+54398q101+71055q100+91484q99+116227q98+145846q97+180914q96+222006q95+269687q94+324500q93+386953q92+457506q91+536558q90+624434q89+721372q88+827510q87+942876q86+1067378q85+1200797q84+1342780q83+1492837q82 + 1650338q81 + 1814516q80 + 1984470q79 + 2159172q78 + 2337474q77 +2518120q76 + 2699758q75 + 2880957q74 + 3060224q73 + 3236023q72 + 3406794q71 +3570975q70 + 3727024q69 + 3873441q68 + 4008790q67 + 4131719q66 + 4240980q65 +4335448q64 + 4414140q63 + 4476229q62 + 4521058q61 + 4548149q60 + 4557212q59 +4548149q58 + 4521058q57 + 4476229q56 + 4414140q55 + 4335448q54 + 4240980q53 +4131719q52 + 4008790q51 + 3873441q50 + 3727024q49 + 3570975q48 + 3406794q47 +3236023q46 + 3060224q45 + 2880957q44 + 2699758q43 + 2518120q42 + 2337474q41 +2159172q40 + 1984470q39 + 1814516q38 + 1650338q37 + 1492837q36 + 1342780q35 +1200797q34+1067378q33+942876q32+827510q31+721372q30+624434q29+536558q28+457506q27+386953q26+324500q25+269687q24+222006q23+180914q22+145846q21+116227q20 + 91484q19 + 71055q18 + 54398q17 + 40998q16 + 30374q15 + 22083q14 +15724q13 + 10939q12 + 7414q11 + 4878q10 + 3102q9 + 1896q8 + 1106q7 + 610q6 +314q5 + 148q4 + 62q3 + 22q2 + 6q + 1k = q118

– J = {1, 3}, {2, 4}, {3, 4}, {4, 5}, {5, 6}, {6, 7}, {7, 8}; type of P : A2

v = q117 + 6q116 + 21q115 + 57q114 + 132q113 + 273q112 + 519q111 + 924q110 +1559q109 + 2515q108 + 3906q107 + 5871q106 + 8576q105 + 12216q104 + 17015q103 +23226q102 +31131q101 +41039q100 +53283q99 +68217q98 +86211q97 +107645q96 +132904q95+162371q94+196418q93+235398q92+279637q91+329424q90+385003q89+446565q88+514238q87+588079q86+668069q85+754106q84+846000q83+943471q82+1046146q81 + 1153558q80 + 1265150q79 + 1380278q78 + 1498214q77 + 1618154q76 +1739226q75 + 1860498q74 + 1980991q73 + 2099692q72 + 2215564q71 + 2327561q70 +2434644q69 + 2535794q68 + 2630027q67 + 2716410q66 + 2794072q65 + 2862217q64 +2920139q63 + 2967232q62 + 3002998q61 + 3027057q60 + 3039152q59 + 3039152q58 +3027057q57 + 3002998q56 + 2967232q55 + 2920139q54 + 2862217q53 + 2794072q52 +2716410q51 + 2630027q50 + 2535794q49 + 2434644q48 + 2327561q47 + 2215564q46 +2099692q45 + 1980991q44 + 1860498q43 + 1739226q42 + 1618154q41 + 1498214q40 +1380278q39 + 1265150q38 + 1153558q37 + 1046146q36 + 943471q35 + 846000q34 +754106q33+668069q32+588079q31+514238q30+446565q29+385003q28+329424q27+


279637q26+235398q25+196418q24+162371q23+132904q22+107645q21+86211q20+68217q19 + 53283q18 + 41039q17 + 31131q16 + 23226q15 + 17015q14 + 12216q13 +8576q12 + 5871q11 + 3906q10 + 2515q9 + 1559q8 + 924q7 + 519q6 + 273q5 + 132q4 +57q3 + 21q2 + 6q + 1k = q117

– J = {1, 2, 3}, . . . ; type of P : A2A1

v = q116 +5q115 +16q114 +41q113 +91q112 +182q111 +337q110 +587q109 +972q108 +1543q107 + 2363q106 + 3508q105 + 5068q104 + 7148q103 + 9867q102 + 13359q101 +17772q100 + 23267q99 + 30016q98 + 38201q97 + 48010q96 + 59635q95 + 73269q94 +89102q93+107316q92+128082q91+151555q90+177869q89+207134q88+239431q87+274807q86+313272q85+354797q84+399309q83+446691q82+496780q81+549366q80+604192q79+660958q78+719320q77+778894q76+839260q75+899966q74+960532q73+1020459q72 + 1079233q71 + 1136331q70 + 1191230q69 + 1243414q68 + 1292380q67 +1337647q66 + 1378763q65 + 1415309q64 + 1446908q63 + 1473231q62 + 1494001q61 +1508997q60 + 1518060q59 + 1521092q58 + 1518060q57 + 1508997q56 + 1494001q55 +1473231q54 + 1446908q53 + 1415309q52 + 1378763q51 + 1337647q50 + 1292380q49 +1243414q48 + 1191230q47 + 1136331q46 + 1079233q45 + 1020459q44 + 960532q43 +899966q42+839260q41+778894q40+719320q39+660958q38+604192q37+549366q36+496780q35+446691q34+399309q33+354797q32+313272q31+274807q30+239431q29+207134q28+177869q27+151555q26+128082q25+107316q24+89102q23+73269q22+59635q21 + 48010q20 + 38201q19 + 30016q18 + 23267q17 + 17772q16 + 13359q15 +9867q14 +7148q13 +5068q12 +3508q11 +2363q10 +1543q9 +972q8 +587q7 +337q6 +182q5 + 91q4 + 41q3 + 16q2 + 5q + 1k = q116

– J = {1, 2, 5}, . . . ; type of P : A1A1A1

v = q117+5q116+17q115+45q114+103q113+211q112+399q111+707q110+1189q109+1913q108 + 2965q107 + 4449q106 + 6490q105 + 9234q104 + 12849q103 + 17525q102 +23473q101 + 30925q100 + 40130q99 + 51354q98 + 64873q97 + 80973q96 + 99941q95 +122065q94+147622q93+176878q92+210075q91+247431q90+289127q89+335307q88+386065q87+441445q86+501431q85+565947q84+634850q83+707930q82+784907q81+865431q80 + 949085q79 + 1035385q78 + 1123787q77 + 1213687q76 + 1304433q75 +1395325q74 + 1485632q73 + 1574592q72 + 1661431q71 + 1745363q70 + 1825612q69 +1901412q68 + 1972029q67 + 2036761q66 + 2094958q65 + 2146022q64 + 2189426q63 +2224714q62 + 2251515q61 + 2269543q60 + 2278606q59 + 2278606q58 + 2269543q57 +2251515q56 + 2224714q55 + 2189426q54 + 2146022q53 + 2094958q52 + 2036761q51 +1972029q50 + 1901412q49 + 1825612q48 + 1745363q47 + 1661431q46 + 1574592q45 +1485632q44 + 1395325q43 + 1304433q42 + 1213687q41 + 1123787q40 + 1035385q39 +949085q38+865431q37+784907q36+707930q35+634850q34+565947q33+501431q32+441445q31+386065q30+335307q29+289127q28+247431q27+210075q26+176878q25+147622q24 + 122065q23 + 99941q22 + 80973q21 + 64873q20 + 51354q19 + 40130q18 +30925q17+23473q16+17525q15+12849q14+9234q13+6490q12+4449q11+2965q10+1913q9 + 1189q8 + 707q7 + 399q6 + 211q5 + 103q4 + 45q3 + 17q2 + 5q + 1k = q117

– J = {1, 3, 4}, . . . ; type of P : A3

v = q114 + 5q113 + 15q112 + 36q111 + 76q110 + 146q109 + 261q108 + 441q107 +711q106+1102q105+1652q104+2406q103+3416q102+4742q101+6451q100+8617q99+11321q98 + 14650q97 + 18695q96 + 23551q95 + 29315q94 + 36084q93 + 43954q92 +53018q91 + 63362q90 + 75064q89 + 88193q88 + 102805q87 + 118941q86 + 136626q85 +155866q84+176646q83+198931q82+222663q81+247760q80+274117q79+301606q78+330075q77+359352q76+389245q75+419542q74+450015q73+480424q72+510517q71+540035q70+568716q69+596296q68+622514q67+647118q66+669866q65+690529q64+708897q63+724780q62+738011q61+748451q60+755990q59+760546q58+762070q57+760546q56+755990q55+748451q54+738011q53+724780q52+708897q51+690529q50+669866q49+647118q48+622514q47+596296q46+568716q45+540035q44+510517q43+480424q42+450015q41+419542q40+389245q39+359352q38+330075q37+301606q36+274117q35+247760q34+222663q33+198931q32+176646q31+155866q30+136626q29+


118941q28 + 102805q27 + 88193q26 + 75064q25 + 63362q24 + 53018q23 + 43954q22 +36084q21 + 29315q20 + 23551q19 + 18695q18 + 14650q17 + 11321q16 + 8617q15 +6451q14 +4742q13 +3416q12 +2406q11 +1652q10 +1102q9 +711q8 +441q7 +261q6 +146q5 + 76q4 + 36q3 + 15q2 + 5q + 1k = q114

– J = {1, 2, 3, 4}, . . . ; type of P : A4

v = q110 +4q109 +10q108 +21q107 +40q106 +71q105 +119q104 +190q103 +291q102 +431q101 + 621q100 + 873q99 + 1200q98 + 1617q97 + 2140q96 + 2787q95 + 3577q94 +4529q93+5662q92+6996q91+8551q90+10346q89+12399q88+14726q87+17340q86+20253q85 + 23475q84 + 27011q83 + 30862q82 + 35025q81 + 39493q80 + 44255q79 +49296q78 + 54594q77 + 60122q76 + 65850q75 + 71744q74 + 77765q73 + 83871q72 +90015q71+96147q70+102217q69+108174q68+113964q67+119533q66+124828q65+129797q64+134392q63+138568q62+142281q61+145491q60+148165q59+150275q58+151799q57+152721q56+153030q55+152721q54+151799q53+150275q52+148165q51+145491q50+142281q49+138568q48+134392q47+129797q46+124828q45+119533q44+113964q43 + 108174q42 + 102217q41 + 96147q40 + 90015q39 + 83871q38 + 77765q37 +71744q36 + 65850q35 + 60122q34 + 54594q33 + 49296q32 + 44255q31 + 39493q30 +35025q29 + 30862q28 + 27011q27 + 23475q26 + 20253q25 + 17340q24 + 14726q23 +12399q22 +10346q21 +8551q20 +6996q19 +5662q18 +4529q17 +3577q16 +2787q15 +2140q14 + 1617q13 + 1200q12 + 873q11 + 621q10 + 431q9 + 291q8 + 190q7 + 119q6 +71q5 + 40q4 + 21q3 + 10q2 + 4q + 1k = q110

– J = {1, 2, 3, 5}, . . . ; type of P : A2A1A1

v = q115 + 4q114 + 12q113 + 29q112 + 62q111 + 120q110 + 217q109 + 370q108 +602q107+941q106+1422q105+2086q104+2982q103+4166q102+5701q101+7658q100+10114q99 + 13153q98 + 16863q97 + 21338q96 + 26672q95 + 32963q94 + 40306q93 +48796q92 + 58520q91 + 69562q90 + 81993q89 + 95876q88 + 111258q87 + 128173q86 +146634q85+166638q84+188159q83+211150q82+235541q81+261239q80+288127q79+316065q78+344893q77+374427q76+404467q75+434793q74+465173q73+495359q72+525100q71+554133q70+582198q69+609032q68+634382q67+657998q66+679649q65+699114q64+716195q63+730713q62+742518q61+751483q60+757514q59+760546q58+760546q57+757514q56+751483q55+742518q54+730713q53+716195q52+699114q51+679649q50+657998q49+634382q48+609032q47+582198q46+554133q45+525100q44+495359q43+465173q42+434793q41+404467q40+374427q39+344893q38+316065q37+288127q36+261239q35+235541q34+211150q33+188159q32+166638q31+146634q30+128173q29 + 111258q28 + 95876q27 + 81993q26 + 69562q25 + 58520q24 + 48796q23 +40306q22 + 32963q21 + 26672q20 + 21338q19 + 16863q18 + 13153q17 + 10114q16 +7658q15 +5701q14 +4166q13 +2982q12 +2086q11 +1422q10 +941q9 +602q8 +370q7 +217q6 + 120q5 + 62q4 + 29q3 + 12q2 + 4q + 1k = q115

– J = {1, 2, 4, 5}, . . . ; type of P : A3A1

v = q113 +4q112 +11q111 +25q110 +51q109 +95q108 +166q107 +275q106 +436q105 +666q104 +986q103 +1420q102 +1996q101 +2746q100 +3705q99 +4912q98 +6409q97 +8241q96 + 10454q95 + 13097q94 + 16218q93 + 19866q92 + 24088q91 + 28930q90 +34432q89 + 40632q88 + 47561q87 + 55244q86 + 63697q85 + 72929q84 + 82937q83 +93709q82+105222q81+117441q80+130319q79+143798q78+157808q77+172267q76+187085q75+202160q74+217382q73+232633q72+247791q71+262726q70+277309q69+291407q68+304889q67+317625q66+329493q65+340373q64+350156q63+358741q62+366039q61+371972q60+376479q59+379511q58+381035q57+381035q56+379511q55+376479q54+371972q53+366039q52+358741q51+350156q50+340373q49+329493q48+317625q47+304889q46+291407q45+277309q44+262726q43+247791q42+232633q41+217382q40+202160q39+187085q38+172267q37+157808q36+143798q35+130319q34+117441q33 + 105222q32 + 93709q31 + 82937q30 + 72929q29 + 63697q28 + 55244q27 +47561q26 + 40632q25 + 34432q24 + 28930q23 + 24088q22 + 19866q21 + 16218q20 +13097q19 +10454q18 +8241q17 +6409q16 +4912q15 +3705q14 +2746q13 +1996q12 +


1420q11 +986q10 +666q9 +436q8 +275q7 +166q6 +95q5 +51q4 +25q3 +11q2 +4q+1k = q113

– J = {1, 2, 5, 7}, . . . ; type of P : A1A1A1A1

v = q116 + 4q115 + 13q114 + 32q113 + 71q112 + 140q111 + 259q110 + 448q109 +741q108 + 1172q107 + 1793q106 + 2656q105 + 3834q104 + 5400q103 + 7449q102 +10076q101 + 13397q100 + 17528q99 + 22602q98 + 28752q97 + 36121q96 + 44852q95 +55089q94 + 66976q93 + 80646q92 + 96232q91 + 113843q90 + 133588q89 + 155539q88 +179768q87+206297q86+235148q85+266283q84+299664q83+335186q82+372744q81+412163q80+453268q79+495817q78+539568q77+584219q76+629468q75+674965q74+720360q73+765272q72+809320q71+852111q70+893252q69+932360q68+969052q67+1002977q66 + 1033784q65 + 1061174q64 + 1084848q63 + 1104578q62 + 1120136q61 +1131379q60 + 1138164q59 + 1140442q58 + 1138164q57 + 1131379q56 + 1120136q55 +1104578q54 + 1084848q53 + 1061174q52 + 1033784q51 + 1002977q50 + 969052q49 +932360q48+893252q47+852111q46+809320q45+765272q44+720360q43+674965q42+629468q41+584219q40+539568q39+495817q38+453268q37+412163q36+372744q35+335186q34+299664q33+266283q32+235148q31+206297q30+179768q29+155539q28+133588q27 + 113843q26 + 96232q25 + 80646q24 + 66976q23 + 55089q22 + 44852q21 +36121q20 + 28752q19 + 22602q18 + 17528q17 + 13397q16 + 10076q15 + 7449q14 +5400q13 + 3834q12 + 2656q11 + 1793q10 + 1172q9 + 741q8 + 448q7 + 259q6 + 140q5 +71q4 + 32q3 + 13q2 + 4q + 1k = q116

– J = {1, 3, 5, 6}, . . . ; type of P : A2A2

v = q114 +4q113 +11q112 +26q111 +54q110 +102q109 +181q108 +304q107 +487q106 +752q105+1124q104+1632q103+2312q102+3204q101+4351q100+5804q99+7617q98+9846q97 + 12553q96 + 15802q95 + 19655q94 + 24178q93 + 29436q92 + 35488q91 +42392q90 + 50202q89 + 58961q88 + 68706q87 + 79467q86 + 91258q85 + 104082q84 +117932q83+132783q82+148594q81+165314q80+182872q79+201180q78+220140q77+239638q76+259542q75+279714q74+300004q73+320248q72+340280q71+359931q70+379022q69+397378q68+414830q67+431206q66+446344q65+460097q64+472322q63+482890q62+491696q61+498645q60+503660q59+506692q58+507708q57+506692q56+503660q55+498645q54+491696q53+482890q52+472322q51+460097q50+446344q49+431206q48+414830q47+397378q46+379022q45+359931q44+340280q43+320248q42+300004q41+279714q40+259542q39+239638q38+220140q37+201180q36+182872q35+165314q34+148594q33+132783q32+117932q31+104082q30+91258q29+79467q28+68706q27 + 58961q26 + 50202q25 + 42392q24 + 35488q23 + 29436q22 + 24178q21 +19655q20 +15802q19 +12553q18 +9846q17 +7617q16 +5804q15 +4351q14 +3204q13 +2312q12 +1632q11 +1124q10 +752q9 +487q8 +304q7 +181q6 +102q5 +54q4 +26q3 +11q2 + 4q + 1k = q114

– J = {2, 3, 4, 5}; type of P : D4

v = q108 +4q107 +10q106 +20q105 +37q104 +64q103 +105q102 +164q101 +247q100 +360q99+511q98+708q97+961q96+1280q95+1676q94+2160q93+2745q92+3444q91+4269q90 +5232q89 +6345q88 +7620q87 +9067q86 +10696q85 +12513q84 +14524q83 +16732q82 + 19140q81 + 21745q80 + 24544q79 + 27528q78 + 30688q77 + 34010q76 +37480q75 + 41077q74 + 44780q73 + 48563q72 + 52400q71 + 56262q70 + 60120q69 +63940q68 + 67688q67 + 71331q66 + 74836q65 + 78170q64 + 81300q63 + 84195q62 +86824q61 + 89162q60 + 91184q59 + 92870q58 + 94200q57 + 95161q56 + 95740q55 +95934q54 + 95740q53 + 95161q52 + 94200q51 + 92870q50 + 91184q49 + 89162q48 +86824q47 + 84195q46 + 81300q45 + 78170q44 + 74836q43 + 71331q42 + 67688q41 +63940q40 + 60120q39 + 56262q38 + 52400q37 + 48563q36 + 44780q35 + 41077q34 +37480q33 + 34010q32 + 30688q31 + 27528q30 + 24544q29 + 21745q28 + 19140q27 +16732q26+14524q25+12513q24+10696q23+9067q22+7620q21+6345q20+5232q19+4269q18 + 3444q17 + 2745q16 + 2160q15 + 1676q14 + 1280q13 + 961q12 + 708q11 +511q10 + 360q9 + 247q8 + 164q7 + 105q6 + 64q5 + 37q4 + 20q3 + 10q2 + 4q + 1k = q108


– J = {1, 2, 3, 4, 5}, {2, 3, 4, 5, 6}; type of P : D5

v = q100 + 3q99 + 6q98 + 10q97 + 16q96 + 25q95 + 38q94 + 55q93 + 77q92 + 105q91 +141q90 + 186q89 + 241q88 + 307q87 + 385q86 + 477q85 + 585q84 + 710q83 + 852q82 +1012q81 + 1191q80 + 1391q79 + 1612q78 + 1854q77 + 2115q76 + 2396q75 + 2697q74 +3018q73 + 3356q72 + 3709q71 + 4074q70 + 4451q69 + 4838q68 + 5232q67 + 5628q66 +6023q65 + 6414q64 + 6799q63 + 7175q62 + 7537q61 + 7880q60 + 8201q59 + 8499q58 +8771q57 + 9014q56 + 9223q55 + 9396q54 + 9532q53 + 9632q52 + 9693q51 + 9714q50 +9693q49 + 9632q48 + 9532q47 + 9396q46 + 9223q45 + 9014q44 + 8771q43 + 8499q42 +8201q41 + 7880q40 + 7537q39 + 7175q38 + 6799q37 + 6414q36 + 6023q35 + 5628q34 +5232q33 + 4838q32 + 4451q31 + 4074q30 + 3709q29 + 3356q28 + 3018q27 + 2697q26 +2396q25 + 2115q24 + 1854q23 + 1612q22 + 1391q21 + 1191q20 + 1012q19 + 852q18 +710q17 + 585q16 + 477q15 + 385q14 + 307q13 + 241q12 + 186q11 + 141q10 + 105q9 +77q8 + 55q7 + 38q6 + 25q5 + 16q4 + 10q3 + 6q2 + 3q + 1k = q100

– J = {1, 2, 3, 4, 6}, . . . ; type of P : A4A1

v = q109 + 3q108 + 7q107 + 14q106 + 26q105 + 45q104 + 74q103 + 116q102 + 175q101 +256q100+365q99+508q98+692q97+925q96+1215q95+1572q94+2005q93+2524q92+3138q91 + 3858q90 + 4693q89 + 5653q88 + 6746q87 + 7980q86 + 9360q85 + 10893q84 +12582q83 + 14429q82 + 16433q81 + 18592q80 + 20901q79 + 23354q78 + 25942q77 +28652q76 + 31470q75 + 34380q74 + 37364q73 + 40401q72 + 43470q71 + 46545q70 +49602q69 + 52615q68 + 55559q67 + 58405q66 + 61128q65 + 63700q64 + 66097q63 +68295q62 + 70273q61 + 72008q60 + 73483q59 + 74682q58 + 75593q57 + 76206q56 +76515q55 + 76515q54 + 76206q53 + 75593q52 + 74682q51 + 73483q50 + 72008q49 +70273q48 + 68295q47 + 66097q46 + 63700q45 + 61128q44 + 58405q43 + 55559q42 +52615q41 + 49602q40 + 46545q39 + 43470q38 + 40401q37 + 37364q36 + 34380q35 +31470q34 + 28652q33 + 25942q32 + 23354q31 + 20901q30 + 18592q29 + 16433q28 +14429q27 +12582q26 +10893q25 +9360q24 +7980q23 +6746q22 +5653q21 +4693q20 +3858q19 + 3138q18 + 2524q17 + 2005q16 + 1572q15 + 1215q14 + 925q13 + 692q12 +508q11 + 365q10 + 256q9 + 175q8 + 116q7 + 74q6 + 45q5 + 26q4 + 14q3 + 7q2 + 3q+ 1k = q109

– J = {1, 2, 3, 5, 6}, . . . ; type of P : A2A2A1

v = q113 + 3q112 + 8q111 + 18q110 + 36q109 + 66q108 + 115q107 + 189q106 + 298q105 +454q104 + 670q103 + 962q102 + 1350q101 + 1854q100 + 2497q99 + 3307q98 + 4310q97 +5536q96+7017q95+8785q94+10870q93+13308q92+16128q91+19360q90+23032q89+27170q88 + 31791q87 + 36915q86 + 42552q85 + 48706q84 + 55376q83 + 62556q82 +70227q81 + 78367q80 + 86947q79 + 95925q78 + 105255q77 + 114885q76 + 124753q75 +134789q74+144925q73+155079q72+165169q71+175111q70+184820q69+194202q68+203176q67+211654q66+219552q65+226792q64+233305q63+239017q62+243873q61+247823q60+250822q59+252838q58+253854q57+253854q56+252838q55+250822q54+247823q53+243873q52+239017q51+233305q50+226792q49+219552q48+211654q47+203176q46+194202q45+184820q44+175111q43+165169q42+155079q41+144925q40+134789q39 +124753q38 +114885q37 +105255q36 +95925q35 +86947q34 +78367q33 +70227q32 + 62556q31 + 55376q30 + 48706q29 + 42552q28 + 36915q27 + 31791q26 +27170q25 + 23032q24 + 19360q23 + 16128q22 + 13308q21 + 10870q20 + 8785q19 +7017q18 + 5536q17 + 4310q16 + 3307q15 + 2497q14 + 1854q13 + 1350q12 + 962q11 +670q10 + 454q9 + 298q8 + 189q7 + 115q6 + 66q5 + 36q4 + 18q3 + 8q2 + 3q + 1k = q113

– J = {1, 2, 3, 5, 7}, . . . ; type of P : A2A1A1A1

v = q114 + 3q113 + 9q112 + 20q111 + 42q110 + 78q109 + 139q108 + 231q107 + 371q106 +570q105 +852q104 +1234q103 +1748q102 +2418q101 +3283q100 +4375q99 +5739q98 +7414q97 + 9449q96 + 11889q95 + 14783q94 + 18180q93 + 22126q92 + 26670q91 +31850q90 + 37712q89 + 44281q88 + 51595q87 + 59663q86 + 68510q85 + 78124q84 +88514q83+99645q82+111505q81+124036q80+137203q79+150924q78+165141q77+179752q76+194675q75+209792q74+225001q73+240172q72+255187q71+269913q70+284220q69+297978q68+311054q67+323328q66+334670q65+344979q64+354135q63+362060q62+368653q61+373865q60+377618q59+379896q58+380650q57+379896q56+


377618q55+373865q54+368653q53+362060q52+354135q51+344979q50+334670q49+323328q48+311054q47+297978q46+284220q45+269913q44+255187q43+240172q42+225001q41+209792q40+194675q39+179752q38+165141q37+150924q36+137203q35+124036q34 + 111505q33 + 99645q32 + 88514q31 + 78124q30 + 68510q29 + 59663q28 +51595q27 + 44281q26 + 37712q25 + 31850q24 + 26670q23 + 22126q22 + 18180q21 +14783q20 +11889q19 +9449q18 +7414q17 +5739q16 +4375q15 +3283q14 +2418q13 +1748q12 + 1234q11 + 852q10 + 570q9 + 371q8 + 231q7 + 139q6 + 78q5 + 42q4 + 20q3 +9q2 + 3q + 1k = q114

– J = {1, 2, 4, 5, 7}, . . . ; type of P : A3A1A1

v = q112 + 3q111 + 8q110 + 17q109 + 34q108 + 61q107 + 105q106 + 170q105 + 266q104 +400q103 + 586q102 + 834q101 + 1162q100 + 1584q99 + 2121q98 + 2791q97 + 3618q96 +4623q95+5831q94+7266q93+8952q92+10914q91+13174q90+15756q89+18676q88+21956q87 + 25605q86 + 29639q85 + 34058q84 + 38871q83 + 44066q82 + 49643q81 +55579q80 + 61862q79 + 68457q78 + 75341q77 + 82467q76 + 89800q75 + 97285q74 +104875q73+112507q72+120126q71+127665q70+135061q69+142248q68+149159q67+155730q66+161895q65+167598q64+172775q63+177381q62+181360q61+184679q60+187293q59+189186q58+190325q57+190710q56+190325q55+189186q54+187293q53+184679q52+181360q51+177381q50+172775q49+167598q48+161895q47+155730q46+149159q45+142248q44+135061q43+127665q42+120126q41+112507q40+104875q39+97285q38 + 89800q37 + 82467q36 + 75341q35 + 68457q34 + 61862q33 + 55579q32 +49643q31 + 44066q30 + 38871q29 + 34058q28 + 29639q27 + 25605q26 + 21956q25 +18676q24+15756q23+13174q22+10914q21+8952q20+7266q19+5831q18+4623q17+3618q16 +2791q15 +2121q14 +1584q13 +1162q12 +834q11 +586q10 +400q9 +266q8 +170q7 + 105q6 + 61q5 + 34q4 + 17q3 + 8q2 + 3q + 1k = q112

– J = {1, 3, 4, 5, 6}, . . . ; type of P : A5

v = q105+3q104+6q103+11q102+19q101+31q100+49q99+74q98+107q97+151q96+209q95 +283q94 +376q93 +491q92 +630q91 +798q90 +999q89 +1235q88 +1509q87 +1825q86 + 2185q85 + 2593q84 + 3052q83 + 3562q82 + 4123q81 + 4738q80 + 5407q79 +6129q78 +6903q77 +7725q76 +8591q75 +9500q74 +10448q73 +11427q72 +12431q71 +13453q70 + 14485q69 + 15521q68 + 16554q67 + 17571q66 + 18563q65 + 19523q64 +20442q63 + 21311q62 + 22123q61 + 22866q60 + 23532q59 + 24118q58 + 24618q57 +25024q56 + 25333q55 + 25540q54 + 25642q53 + 25642q52 + 25540q51 + 25333q50 +25024q49 + 24618q48 + 24118q47 + 23532q46 + 22866q45 + 22123q44 + 21311q43 +20442q42 + 19523q41 + 18563q40 + 17571q39 + 16554q38 + 15521q37 + 14485q36 +13453q35+12431q34+11427q33+10448q32+9500q31+8591q30+7725q29+6903q28+6129q27 + 5407q26 + 4738q25 + 4123q24 + 3562q23 + 3052q22 + 2593q21 + 2185q20 +1825q19 +1509q18 +1235q17 +999q16 +798q15 +630q14 +491q13 +376q12 +283q11 +209q10 + 151q9 + 107q8 + 74q7 + 49q6 + 31q5 + 19q4 + 11q3 + 6q2 + 3q + 1k = q105

– J = {1, 3, 4, 6, 7}, . . . ; type of P : A3A2

v = q111 + 3q110 + 7q109 + 15q108 + 29q107 + 51q106 + 86q105 + 138q104 + 212q103 +316q102 + 458q101 + 646q100 + 892q99 + 1208q98 + 1605q97 + 2099q96 + 2705q95 +3437q94 +4312q93 +5348q92 +6558q91 +7960q90 +9570q89 +11400q88 +13462q87 +15770q86 + 18329q85 + 21145q84 + 24223q83 + 27561q82 + 31153q81 + 34995q80 +39074q79 + 43372q78 + 47873q77 + 52553q76 + 57382q75 + 62332q74 + 67371q73 +72457q72 + 77554q71 + 82622q70 + 87615q69 + 92489q68 + 97205q67 + 101713q66 +105971q65+109941q64+113581q63+116851q62+119724q61+122166q60+124149q59+125657q58+126673q57+127181q56+127181q55+126673q54+125657q53+124149q52+122166q51+119724q50+116851q49+113581q48+109941q47+105971q46+101713q45+97205q44 + 92489q43 + 87615q42 + 82622q41 + 77554q40 + 72457q39 + 67371q38 +62332q37 + 57382q36 + 52553q35 + 47873q34 + 43372q33 + 39074q32 + 34995q31 +31153q30 + 27561q29 + 24223q28 + 21145q27 + 18329q26 + 15770q25 + 13462q24 +11400q23 + 9570q22 + 7960q21 + 6558q20 + 5348q19 + 4312q18 + 3437q17 + 2705q16 +2099q15 + 1605q14 + 1208q13 + 892q12 + 646q11 + 458q10 + 316q9 + 212q8 + 138q7 +


86q6 + 51q5 + 29q4 + 15q3 + 7q2 + 3q + 1k = q111

– J = {2, 3, 4, 5, 7}, . . . ; type of P : D4A1

v = q107 + 3q106 + 7q105 + 13q104 + 24q103 + 40q102 + 65q101 + 99q100 + 148q99 +212q98 +299q97 +409q96 +552q95 +728q94 +948q93 +1212q92 +1533q91 +1911q90 +2358q89 + 2874q88 + 3471q87 + 4149q86 + 4918q85 + 5778q84 + 6735q83 + 7789q82 +8943q81 + 10197q80 + 11548q79 + 12996q78 + 14532q77 + 16156q76 + 17854q75 +19626q74 + 21451q73 + 23329q72 + 25234q71 + 27166q70 + 29096q69 + 31024q68 +32916q67 + 34772q66 + 36559q65 + 38277q64 + 39893q63 + 41407q62 + 42788q61 +44036q60 + 45126q59 + 46058q58 + 46812q57 + 47388q56 + 47773q55 + 47967q54 +47967q53 + 47773q52 + 47388q51 + 46812q50 + 46058q49 + 45126q48 + 44036q47 +42788q46 + 41407q45 + 39893q44 + 38277q43 + 36559q42 + 34772q41 + 32916q40 +31024q39 + 29096q38 + 27166q37 + 25234q36 + 23329q35 + 21451q34 + 19626q33 +17854q32 + 16156q31 + 14532q30 + 12996q29 + 11548q28 + 10197q27 + 8943q26 +7789q25 + 6735q24 + 5778q23 + 4918q22 + 4149q21 + 3471q20 + 2874q19 + 2358q18 +1911q17 +1533q16 +1212q15 +948q14 +728q13 +552q12 +409q11 +299q10 +212q9 +148q8 + 99q7 + 65q6 + 40q5 + 24q4 + 13q3 + 7q2 + 3q + 1k = q107

– J = {1, 2, 3, 4, 5, 6}; type of P : E6

v = q84 + 2q83 + 3q82 + 4q81 + 5q80 + 7q79 + 10q78 + 13q77 + 16q76 + 20q75 + 25q74 +31q73 + 38q72 + 45q71 + 52q70 + 61q69 + 71q68 + 82q67 + 94q66 + 106q65 + 118q64 +132q63 + 147q62 + 162q61 + 177q60 + 192q59 + 207q58 + 224q57 + 241q56 + 256q55 +270q54 + 284q53 + 298q52 + 312q51 + 325q50 + 335q49 + 344q48 + 353q47 + 361q46 +367q45 + 371q44 + 372q43 + 372q42 + 372q41 + 371q40 + 367q39 + 361q38 + 353q37 +344q36 + 335q35 + 325q34 + 312q33 + 298q32 + 284q31 + 270q30 + 256q29 + 241q28 +224q27 + 207q26 + 192q25 + 177q24 + 162q23 + 147q22 + 132q21 + 118q20 + 106q19 +94q18 + 82q17 + 71q16 + 61q15 + 52q14 + 45q13 + 38q12 + 31q11 + 25q10 + 20q9 +16q8 + 13q7 + 10q6 + 7q5 + 5q4 + 4q3 + 3q2 + 2q + 1k = q84

– J = {1, 2, 3, 4, 5, 7}, . . . ; type of P : D5A1

v = q99 + 2q98 + 4q97 + 6q96 + 10q95 + 15q94 + 23q93 + 32q92 + 45q91 + 60q90 +81q89 + 105q88 + 136q87 + 171q86 + 214q85 + 263q84 + 322q83 + 388q82 + 464q81 +548q80+643q79+748q78+864q77+990q76+1125q75+1271q74+1426q73+1592q72+1764q71 + 1945q70 + 2129q69 + 2322q68 + 2516q67 + 2716q66 + 2912q65 + 3111q64 +3303q63 + 3496q62 + 3679q61 + 3858q60 + 4022q59 + 4179q58 + 4320q57 + 4451q56 +4563q55 + 4660q54 + 4736q53 + 4796q52 + 4836q51 + 4857q50 + 4857q49 + 4836q48 +4796q47 + 4736q46 + 4660q45 + 4563q44 + 4451q43 + 4320q42 + 4179q41 + 4022q40 +3858q39 + 3679q38 + 3496q37 + 3303q36 + 3111q35 + 2912q34 + 2716q33 + 2516q32 +2322q31 + 2129q30 + 1945q29 + 1764q28 + 1592q27 + 1426q26 + 1271q25 + 1125q24 +990q23 + 864q22 + 748q21 + 643q20 + 548q19 + 464q18 + 388q17 + 322q16 + 263q15 +214q14 + 171q13 + 136q12 + 105q11 + 81q10 + 60q9 + 45q8 + 32q7 + 23q6 + 15q5 +10q4 + 6q3 + 4q2 + 2q + 1k = q99

– J = {1, 2, 3, 4, 6, 7}, . . . ; type of P : A4A2

v = q107+2q106+4q105+8q104+14q103+23q102+37q101+56q100+82q99+118q98+165q97 +225q96 +302q95 +398q94 +515q93 +659q92 +831q91 +1034q90 +1273q89 +1551q88 + 1869q87 + 2233q86 + 2644q85 + 3103q84 + 3613q83 + 4177q82 + 4792q81 +5460q80 +6181q79 +6951q78 +7769q77 +8634q76 +9539q75 +10479q74 +11452q73 +12449q72 + 13463q71 + 14489q70 + 15518q69 + 16538q68 + 17546q67 + 18531q66 +19482q65 + 20392q64 + 21254q63 + 22054q62 + 22789q61 + 23452q60 + 24032q59 +24524q58 + 24927q57 + 25231q56 + 25435q55 + 25540q54 + 25540q53 + 25435q52 +25231q51 + 24927q50 + 24524q49 + 24032q48 + 23452q47 + 22789q46 + 22054q45 +21254q44 + 20392q43 + 19482q42 + 18531q41 + 17546q40 + 16538q39 + 15518q38 +14489q37+13463q36+12449q35+11452q34+10479q33+9539q32+8634q31+7769q30+6951q29 + 6181q28 + 5460q27 + 4792q26 + 4177q25 + 3613q24 + 3103q23 + 2644q22 +


2233q21+1869q20+1551q19+1273q18+1034q17+831q16+659q15+515q14+398q13+302q12+225q11+165q10+118q9+82q8+56q7+37q6+23q5+14q4+8q3+4q2+2q+1k = q107

– J = {1, 2, 3, 4, 6, 8}, . . . ; type of P : A4A1A1

v = q108 + 2q107 + 5q106 + 9q105 + 17q104 + 28q103 + 46q102 + 70q101 + 105q100 +151q99 +214q98 +294q97 +398q96 +527q95 +688q94 +884q93 +1121q92 +1403q91 +1735q90 + 2123q89 + 2570q88 + 3083q87 + 3663q86 + 4317q85 + 5043q84 + 5850q83 +6732q82+7697q81+8736q80+9856q79+11045q78+12309q77+13633q76+15019q75+16451q74 + 17929q73 + 19435q72 + 20966q71 + 22504q70 + 24041q69 + 25561q68 +27054q67 + 28505q66 + 29900q65 + 31228q64 + 32472q63 + 33625q62 + 34670q61 +35603q60 + 36405q59 + 37078q58 + 37604q57 + 37989q56 + 38217q55 + 38298q54 +38217q53 + 37989q52 + 37604q51 + 37078q50 + 36405q49 + 35603q48 + 34670q47 +33625q46 + 32472q45 + 31228q44 + 29900q43 + 28505q42 + 27054q41 + 25561q40 +24041q39 + 22504q38 + 20966q37 + 19435q36 + 17929q35 + 16451q34 + 15019q33 +13633q32 +12309q31 +11045q30 +9856q29 +8736q28 +7697q27 +6732q26 +5850q25 +5043q24 + 4317q23 + 3663q22 + 3083q21 + 2570q20 + 2123q19 + 1735q18 + 1403q17 +1121q16 + 884q15 + 688q14 + 527q13 + 398q12 + 294q11 + 214q10 + 151q9 + 105q8 +70q7 + 46q6 + 28q5 + 17q4 + 9q3 + 5q2 + 2q + 1k = q108

– J = {1, 2, 3, 5, 6, 7}, . . . ; type of P : A3A2A1

v = q110 + 2q109 + 5q108 + 10q107 + 19q106 + 32q105 + 54q104 + 84q103 + 128q102 +188q101+270q100+376q99+516q98+692q97+913q96+1186q95+1519q94+1918q93+2394q92 + 2954q91 + 3604q90 + 4356q89 + 5214q88 + 6186q87 + 7276q86 + 8494q85 +9835q84 + 11310q83 + 12913q82 + 14648q81 + 16505q80 + 18490q79 + 20584q78 +22788q77 + 25085q76 + 27468q75 + 29914q74 + 32418q73 + 34953q72 + 37504q71 +40050q70 + 42572q69 + 45043q68 + 47446q67 + 49759q66 + 51954q65 + 54017q64 +55924q63 + 57657q62 + 59194q61 + 60530q60 + 61636q59 + 62513q58 + 63144q57 +63529q56 + 63652q55 + 63529q54 + 63144q53 + 62513q52 + 61636q51 + 60530q50 +59194q49 + 57657q48 + 55924q47 + 54017q46 + 51954q45 + 49759q44 + 47446q43 +45043q42 + 42572q41 + 40050q40 + 37504q39 + 34953q38 + 32418q37 + 29914q36 +27468q35 + 25085q34 + 22788q33 + 20584q32 + 18490q31 + 16505q30 + 14648q29 +12913q28 +11310q27 +9835q26 +8494q25 +7276q24 +6186q23 +5214q22 +4356q21 +3604q20 + 2954q19 + 2394q18 + 1918q17 + 1519q16 + 1186q15 + 913q14 + 692q13 +516q12+376q11+270q10+188q9+128q8+84q7+54q6+32q5+19q4+10q3+5q2+2q+1k = q110

– J = {1, 2, 3, 5, 6, 8}, . . . ; type of P : A2A2A1A1

v = q112 + 2q111 + 6q110 + 12q109 + 24q108 + 42q107 + 73q106 + 116q105 + 182q104 +272q103 + 398q102 + 564q101 + 786q100 + 1068q99 + 1429q98 + 1878q97 + 2432q96 +3104q95 +3913q94 +4872q93 +5998q92 +7310q91 +8818q90 +10542q89 +12490q88 +14680q87 + 17111q86 + 19804q85 + 22748q84 + 25958q83 + 29418q82 + 33138q81 +37089q80 + 41278q79 + 45669q78 + 50256q77 + 54999q76 + 59886q75 + 64867q74 +69922q73 + 75003q72 + 80076q71 + 85093q70 + 90018q69 + 94802q68 + 99400q67 +103776q66+107878q65+111674q64+115118q63+118187q62+120830q61+123043q60+124780q59+126042q58+126796q57+127058q56+126796q55+126042q54+124780q53+123043q52+120830q51+118187q50+115118q49+111674q48+107878q47+103776q46+99400q45 + 94802q44 + 90018q43 + 85093q42 + 80076q41 + 75003q40 + 69922q39 +64867q38 + 59886q37 + 54999q36 + 50256q35 + 45669q34 + 41278q33 + 37089q32 +33138q31 + 29418q30 + 25958q29 + 22748q28 + 19804q27 + 17111q26 + 14680q25 +12490q24 +10542q23 +8818q22 +7310q21 +5998q20 +4872q19 +3913q18 +3104q17 +2432q16 +1878q15 +1429q14 +1068q13 +786q12 +564q11 +398q10 +272q9 +182q8 +116q7 + 73q6 + 42q5 + 24q4 + 12q3 + 6q2 + 2q + 1k = q112

– J = {1, 2, 4, 5, 6, 7}, . . . ; type of P : A5A1

v = q104 + 2q103 + 4q102 + 7q101 + 12q100 + 19q99 + 30q98 + 44q97 + 63q96 + 88q95 +121q94 + 162q93 + 214q92 + 277q91 + 353q90 + 445q89 + 554q88 + 681q87 + 828q86 +


997q85 + 1188q84 + 1405q83 + 1647q82 + 1915q81 + 2208q80 + 2530q79 + 2877q78 +3252q77 + 3651q76 + 4074q75 + 4517q74 + 4983q73 + 5465q72 + 5962q71 + 6469q70 +6984q69 + 7501q68 + 8020q67 + 8534q66 + 9037q65 + 9526q64 + 9997q63 + 10445q62 +10866q61 + 11257q60 + 11609q59 + 11923q58 + 12195q57 + 12423q56 + 12601q55 +12732q54 + 12808q53 + 12834q52 + 12808q51 + 12732q50 + 12601q49 + 12423q48 +12195q47 + 11923q46 + 11609q45 + 11257q44 + 10866q43 + 10445q42 + 9997q41 +9526q40 + 9037q39 + 8534q38 + 8020q37 + 7501q36 + 6984q35 + 6469q34 + 5962q33 +5465q32 + 4983q31 + 4517q30 + 4074q29 + 3651q28 + 3252q27 + 2877q26 + 2530q25 +2208q24 + 1915q23 + 1647q22 + 1405q21 + 1188q20 + 997q19 + 828q18 + 681q17 +554q16 +445q15 +353q14 +277q13 +214q12 +162q11 +121q10 +88q9 +63q8 +44q7 +30q6 + 19q5 + 12q4 + 7q3 + 4q2 + 2q + 1k = q104

– J = {1, 3, 4, 5, 6, 7}, . . . ; type of P : A6

v = q99 +2q98 +3q97 +5q96 +8q95 +12q94 +18q93 +26q92 +35q91 +47q90 +63q89 +82q88 + 105q87 + 133q86 + 165q85 + 203q84 + 248q83 + 299q82 + 356q81 + 421q80 +493q79 +573q78 +662q77 +758q76 +860q75 +971q74 +1090q73 +1215q72 +1347q71 +1484q70 + 1624q69 + 1769q68 + 1919q67 + 2069q66 + 2219q65 + 2369q64 + 2516q63 +2660q62 + 2802q61 + 2936q60 + 3061q59 + 3179q58 + 3288q57 + 3385q56 + 3472q55 +3545q54 + 3602q53 + 3647q52 + 3679q51 + 3694q50 + 3694q49 + 3679q48 + 3647q47 +3602q46 + 3545q45 + 3472q44 + 3385q43 + 3288q42 + 3179q41 + 3061q40 + 2936q39 +2802q38 + 2660q37 + 2516q36 + 2369q35 + 2219q34 + 2069q33 + 1919q32 + 1769q31 +1624q30+1484q29+1347q28+1215q27+1090q26+971q25+860q24+758q23+662q22+573q21 + 493q20 + 421q19 + 356q18 + 299q17 + 248q16 + 203q15 + 165q14 + 133q13 +105q12 + 82q11 + 63q10 + 47q9 + 35q8 + 26q7 + 18q6 + 12q5 + 8q4 + 5q3 + 3q2 + 2q+ 1k = q99

– J = {1, 3, 4, 6, 7, 8}, . . . ; type of P : A3A3

v = q108 + 2q107 + 4q106 + 8q105 + 15q104 + 24q103 + 39q102 + 60q101 + 89q100 +128q99 + 181q98 + 248q97 + 335q96 + 444q95 + 578q94 + 742q93 + 941q92 + 1176q91 +1453q90 + 1778q89 + 2151q88 + 2578q87 + 3063q86 + 3608q85 + 4213q84 + 4886q83 +5622q82 +6424q81 +7291q80 +8224q79 +9214q78 +10266q77 +11370q76 +12522q75 +13715q74 + 14946q73 + 16199q72 + 17472q71 + 18754q70 + 20032q69 + 21296q68 +22540q67 + 23747q66 + 24906q65 + 26012q64 + 27048q63 + 28005q62 + 28876q61 +29652q60 + 30318q59 + 30878q58 + 31318q57 + 31635q56 + 31826q55 + 31894q54 +31826q53 + 31635q52 + 31318q51 + 30878q50 + 30318q49 + 29652q48 + 28876q47 +28005q46 + 27048q45 + 26012q44 + 24906q43 + 23747q42 + 22540q41 + 21296q40 +20032q39 + 18754q38 + 17472q37 + 16199q36 + 14946q35 + 13715q34 + 12522q33 +11370q32 +10266q31 +9214q30 +8224q29 +7291q28 +6424q27 +5622q26 +4886q25 +4213q24 + 3608q23 + 3063q22 + 2578q21 + 2151q20 + 1778q19 + 1453q18 + 1176q17 +941q16 + 742q15 + 578q14 + 444q13 + 335q12 + 248q11 + 181q10 + 128q9 + 89q8 +60q7 + 39q6 + 24q5 + 15q4 + 8q3 + 4q2 + 2q + 1k = q108

– J = {2, 3, 4, 5, 6, 7}; type of P : D6

v = q90 + 2q89 + 3q88 + 4q87 + 6q86 + 8q85 + 12q84 + 16q83 + 21q82 + 26q81 + 34q80 +42q79 +53q78 +64q77 +77q76 +90q75 +108q74 +126q73 +147q72 +168q71 +192q70 +216q69 + 245q68 + 274q67 + 304q66 + 334q65 + 367q64 + 400q63 + 435q62 + 470q61 +503q60 + 536q59 + 570q58 + 604q57 + 635q56 + 666q55 + 692q54 + 718q53 + 742q52 +766q51 + 782q50 + 798q49 + 808q48 + 818q47 + 823q46 + 828q45 + 823q44 + 818q43 +808q42 + 798q41 + 782q40 + 766q39 + 742q38 + 718q37 + 692q36 + 666q35 + 635q34 +604q33 + 570q32 + 536q31 + 503q30 + 470q29 + 435q28 + 400q27 + 367q26 + 334q25 +304q24 + 274q23 + 245q22 + 216q21 + 192q20 + 168q19 + 147q18 + 126q17 + 108q16 +90q15 + 77q14 + 64q13 + 53q12 + 42q11 + 34q10 + 26q9 + 21q8 + 16q7 + 12q6 + 8q5 +6q4 + 4q3 + 3q2 + 2q + 1k = q90

– J = {2, 3, 4, 5, 7, 8}; type of P : D4A2

v = q105 +2q104 +4q103 +7q102 +13q101 +20q100 +32q99 +47q98 +69q97 +96q96 +


134q95 + 179q94 + 239q93 + 310q92 + 399q91 + 503q90 + 631q89 + 777q88 + 950q87 +1147q86 + 1374q85 + 1628q84 + 1916q83 + 2234q82 + 2585q81 + 2970q80 + 3388q79 +3839q78 + 4321q77 + 4836q76 + 5375q75 + 5945q74 + 6534q73 + 7147q72 + 7770q71 +8412q70+9052q69+9702q68+10342q67+10980q66+11594q65+12198q64+12767q63+13312q62 + 13814q61 + 14281q60 + 14693q59 + 15062q58 + 15371q57 + 15625q56 +15816q55 + 15947q54 + 16010q53 + 16010q52 + 15947q51 + 15816q50 + 15625q49 +15371q48 + 15062q47 + 14693q46 + 14281q45 + 13814q44 + 13312q43 + 12767q42 +12198q41+11594q40+10980q39+10342q38+9702q37+9052q36+8412q35+7770q34+7147q33 + 6534q32 + 5945q31 + 5375q30 + 4836q29 + 4321q28 + 3839q27 + 3388q26 +2970q25 + 2585q24 + 2234q23 + 1916q22 + 1628q21 + 1374q20 + 1147q19 + 950q18 +777q17 + 631q16 + 503q15 + 399q14 + 310q13 + 239q12 + 179q11 + 134q10 + 96q9 +69q8 + 47q7 + 32q6 + 20q5 + 13q4 + 7q3 + 4q2 + 2q + 1k = q105

– J = {1, 2, 3, 4, 5, 6, 7}; type of P : E7

v = q57 + q56 + q55 + q54 + q53 + q52 + 2q51 + 2q50 + 2q49 + 2q48 + 3q47 + 3q46 +4q45 + 4q44 + 4q43 + 4q42 + 5q41 + 5q40 + 6q39 + 6q38 + 6q37 + 6q36 + 7q35 + 7q34 +7q33 + 7q32 + 7q31 + 7q30 + 8q29 + 8q28 + 7q27 + 7q26 + 7q25 + 7q24 + 7q23 + 7q22 +6q21 + 6q20 + 6q19 + 6q18 + 5q17 + 5q16 + 4q15 + 4q14 + 4q13 + 4q12 + 3q11 + 3q10 +2q9 + 2q8 + 2q7 + 2q6 + q5 + q4 + q3 + q2 + q + 1k = q57

– J = {1, 2, 3, 4, 5, 6, 8}; type of P : D6A1

v = q83+q82+2q81+2q80+3q79+4q78+6q77+7q76+9q75+11q74+14q73+17q72+21q71+24q70+28q69+33q68+38q67+44q66+50q65+56q64+62q63+70q62+77q61+85q60 + 92q59 + 100q58 + 107q57 + 117q56 + 124q55 + 132q54 + 138q53 + 146q52 +152q51 + 160q50 + 165q49 + 170q48 + 174q47 + 179q46 + 182q45 + 185q44 + 186q43 +186q42 + 186q41 + 186q40 + 185q39 + 182q38 + 179q37 + 174q36 + 170q35 + 165q34 +160q33 + 152q32 + 146q31 + 138q30 + 132q29 + 124q28 + 117q27 + 107q26 + 100q25 +92q24+85q23+77q22+70q21+62q20+56q19+50q18+44q17+38q16+33q15+28q14+24q13 +21q12 +17q11 +14q10 +11q9 +9q8 +7q7 +6q6 +4q5 +3q4 +2q3 +2q2 +q+1k = q83

– J = {1, 2, 3, 4, 5, 7, 8}; type of P : D5A2

v = q97 + q96 + 2q95 + 3q94 + 5q93 + 7q92 + 11q91 + 14q90 + 20q89 + 26q88 + 35q87 +44q86 + 57q85 + 70q84 + 87q83 + 106q82 + 129q81 + 153q80 + 182q79 + 213q78 +248q77 + 287q76 + 329q75 + 374q74 + 422q73 + 475q72 + 529q71 + 588q70 + 647q69 +710q68+772q67+840q66+904q65+972q64+1036q63+1103q62+1164q61+1229q60+1286q59 + 1343q58 + 1393q57 + 1443q56 + 1484q55 + 1524q54 + 1555q53 + 1581q52 +1600q51 + 1615q50 + 1621q49 + 1621q48 + 1615q47 + 1600q46 + 1581q45 + 1555q44 +1524q43 + 1484q42 + 1443q41 + 1393q40 + 1343q39 + 1286q38 + 1229q37 + 1164q36 +1103q35 +1036q34 +972q33 +904q32 +840q31 +772q30 +710q29 +647q28 +588q27 +529q26 + 475q25 + 422q24 + 374q23 + 329q22 + 287q21 + 248q20 + 213q19 + 182q18 +153q17 + 129q16 + 106q15 + 87q14 + 70q13 + 57q12 + 44q11 + 35q10 + 26q9 + 20q8 +14q7 + 11q6 + 7q5 + 5q4 + 3q3 + 2q2 + q + 1k = q97

– J = {1, 2, 3, 4, 6, 7, 8}; type of P : A4A3

v = q104 + q103 + 2q102 + 4q101 + 7q100 + 10q99 + 16q98 + 23q97 + 33q96 + 46q95 +63q94 + 83q93 + 110q92 + 142q91 + 180q90 + 227q89 + 282q88 + 345q87 + 419q86 +505q85+600q84+709q83+830q82+964q81+1110q80+1273q79+1445q78+1632q77+1831q76 + 2043q75 + 2263q74 + 2497q73 + 2736q72 + 2983q71 + 3236q70 + 3494q69 +3750q68 + 4009q67 + 4265q66 + 4514q65 + 4758q64 + 4994q63 + 5216q62 + 5424q61 +5620q60 + 5794q59 + 5951q58 + 6087q57 + 6200q56 + 6286q55 + 6354q54 + 6391q53 +6404q52 + 6391q51 + 6354q50 + 6286q49 + 6200q48 + 6087q47 + 5951q46 + 5794q45 +5620q44 + 5424q43 + 5216q42 + 4994q41 + 4758q40 + 4514q39 + 4265q38 + 4009q37 +3750q36 + 3494q35 + 3236q34 + 2983q33 + 2736q32 + 2497q31 + 2263q30 + 2043q29 +1831q28 + 1632q27 + 1445q26 + 1273q25 + 1110q24 + 964q23 + 830q22 + 709q21 +600q20 + 505q19 + 419q18 + 345q17 + 282q16 + 227q15 + 180q14 + 142q13 + 110q12 +


83q11 + 63q10 + 46q9 + 33q8 + 23q7 + 16q6 + 10q5 + 7q4 + 4q3 + 2q2 + q + 1k = q104

– J = {1, 2, 3, 5, 6, 7, 8}; type of P : A4A2A1

v = q106 + q105 + 3q104 + 5q103 + 9q102 + 14q101 + 23q100 + 33q99 + 49q98 + 69q97 +96q96 + 129q95 + 173q94 + 225q93 + 290q92 + 369q91 + 462q90 + 572q89 + 701q88 +850q87 + 1019q86 + 1214q85 + 1430q84 + 1673q83 + 1940q82 + 2237q81 + 2555q80 +2905q79 + 3276q78 + 3675q77 + 4094q76 + 4540q75 + 4999q74 + 5480q73 + 5972q72 +6477q71 + 6986q70 + 7503q69 + 8015q68 + 8523q67 + 9023q66 + 9508q65 + 9974q64 +10418q63 + 10836q62 + 11218q61 + 11571q60 + 11881q59 + 12151q58 + 12373q57 +12554q56 + 12677q55 + 12758q54 + 12782q53 + 12758q52 + 12677q51 + 12554q50 +12373q49 + 12151q48 + 11881q47 + 11571q46 + 11218q45 + 10836q44 + 10418q43 +9974q42 + 9508q41 + 9023q40 + 8523q39 + 8015q38 + 7503q37 + 6986q36 + 6477q35 +5972q34 + 5480q33 + 4999q32 + 4540q31 + 4094q30 + 3675q29 + 3276q28 + 2905q27 +2555q26 + 2237q25 + 1940q24 + 1673q23 + 1430q22 + 1214q21 + 1019q20 + 850q19 +701q18 + 572q17 + 462q16 + 369q15 + 290q14 + 225q13 + 173q12 + 129q11 + 96q10 +69q9 + 49q8 + 33q7 + 23q6 + 14q5 + 9q4 + 5q3 + 3q2 + q + 1k = q106

– J = {1, 2, 4, 5, 6, 7, 8}; type of P : A6A1

v = q98 + q97 + 2q96 + 3q95 + 5q94 + 7q93 + 11q92 + 15q91 + 20q90 + 27q89 + 36q88 +46q87+59q86+74q85+91q84+112q83+136q82+163q81+193q80+228q79+265q78+308q77 + 354q76 + 404q75 + 456q74 + 515q73 + 575q72 + 640q71 + 707q70 + 777q69 +847q68 + 922q67 + 997q66 + 1072q65 + 1147q64 + 1222q63 + 1294q62 + 1366q61 +1436q60 + 1500q59 + 1561q58 + 1618q57 + 1670q56 + 1715q55 + 1757q54 + 1788q53 +1814q52 + 1833q51 + 1846q50 + 1848q49 + 1846q48 + 1833q47 + 1814q46 + 1788q45 +1757q44 + 1715q43 + 1670q42 + 1618q41 + 1561q40 + 1500q39 + 1436q38 + 1366q37 +1294q36+1222q35+1147q34+1072q33+997q32+922q31+847q30+777q29+707q28+640q27 + 575q26 + 515q25 + 456q24 + 404q23 + 354q22 + 308q21 + 265q20 + 228q19 +193q18 + 163q17 + 136q16 + 112q15 + 91q14 + 74q13 + 59q12 + 46q11 + 36q10 + 27q9 +20q8 + 15q7 + 11q6 + 7q5 + 5q4 + 3q3 + 2q2 + q + 1k = q98

– J = {1, 3, 4, 5, 6, 7, 8}; type of P : A7

v = q92+q91+q90+2q89+3q88+4q87+6q86+8q85+10q84+13q83+17q82+21q81+26q80+32q79+38q78+46q77+55q76+64q75+74q74+86q73+98q72+112q71+127q70+142q69 + 157q68 + 175q67 + 193q66 + 211q65 + 230q64 + 249q63 + 267q62 + 287q61 +307q60 + 325q59 + 343q58 + 361q57 + 377q56 + 393q55 + 409q54 + 421q53 + 432q52 +443q51 + 452q50 + 458q49 + 464q48 + 466q47 + 466q46 + 466q45 + 464q44 + 458q43 +452q42 + 443q41 + 432q40 + 421q39 + 409q38 + 393q37 + 377q36 + 361q35 + 343q34 +325q33 + 307q32 + 287q31 + 267q30 + 249q29 + 230q28 + 211q27 + 193q26 + 175q25 +157q24 +142q23 +127q22 +112q21 +98q20 +86q19 +74q18 +64q17 +55q16 +46q15 +38q14+32q13+26q12+21q11+17q10+13q9+10q8+8q7+6q6+4q5+3q4+2q3+q2+q+1k = q92

– J = {2, 3, 4, 5, 6, 7, 8}; type of P : D7

v = q78 + q77 + q76 + q75 + 2q74 + 2q73 + 3q72 + 4q71 + 5q70 + 5q69 + 7q68 + 8q67 +10q66 + 11q65 + 13q64 + 14q63 + 17q62 + 19q61 + 21q60 + 23q59 + 26q58 + 28q57 +31q56 + 34q55 + 36q54 + 38q53 + 41q52 + 44q51 + 46q50 + 49q49 + 50q48 + 52q47 +54q46 + 57q45 + 57q44 + 59q43 + 59q42 + 60q41 + 60q40 + 62q39 + 60q38 + 60q37 +59q36+59q35+57q34+57q33+54q32+52q31+50q30+49q29+46q28+44q27+41q26+38q25+36q24+34q23+31q22+28q21+26q20+23q19+21q18+19q17+17q16+14q15+13q14 +11q13 +10q12 +8q11 +7q10 +5q9 +5q8 +4q7 +3q6 +2q5 +2q4 +q3 +q2 +q+1k = q78

F8 – J = {1}, {2}, {3}, {4}; type of P : A1

v = q23+3q22+6q21+10q20+15q19+21q18+27q17+33q16+38q15+42q14+45q13+47q12 +47q11 +45q10 +42q9 +38q8 +33q7 +27q6 +21q5 +15q4 +10q3 +6q2 +3q+1k = q23

A.2. THE CASE JW0 6= J 143

– J = {1, 2}, {3, 4}; type of P : A2

v = q21 + 2q20 + 3q19 + 5q18 + 7q17 + 9q16 + 11q15 + 13q14 + 14q13 + 15q12 + 16q11 +16q10 + 15q9 + 14q8 + 13q7 + 11q6 + 9q5 + 7q4 + 5q3 + 3q2 + 2q + 1k = q21

– J = {1, 3}, {1, 4}, {2, 4}; type of P : A1A1

v = q22 +2q21 +4q20 +6q19 +9q18 +12q17 +15q16 +18q15 +20q14 +22q13 +23q12 +24q11 + 23q10 + 22q9 + 20q8 + 18q7 + 15q6 + 12q5 + 9q4 + 6q3 + 4q2 + 2q + 1k = q22

– J = {2, 3}; type of P : B2

v = q20 + 2q19 + 3q18 + 4q17 + 6q16 + 8q15 + 9q14 + 10q13 + 11q12 + 12q11 + 12q10 +12q9 + 11q8 + 10q7 + 9q6 + 8q5 + 6q4 + 4q3 + 3q2 + 2q + 1k = q20

– J = {1, 2, 3}, {2, 3, 4}; type of P : B3

v = q15+q14+q13+q12+2q11+2q10+2q9+2q8+2q7+2q6+2q5+2q4+q3+q2+q+1k = q15

– J = {1, 2, 4}, {1, 3, 4}; type of P : A2A1

v = q20 + q19 + 2q18 + 3q17 + 4q16 + 5q15 + 6q14 + 7q13 + 7q12 + 8q11 + 8q10 + 8q9 +7q8 + 7q7 + 6q6 + 5q5 + 4q4 + 3q3 + 2q2 + q + 1k = q20

G2 – J = {1}, {2}; type of P : A1

v = q5 + q4 + q3 + q2 + q + 1k = q5

A.2 The case Jw0 6= J

A2 – J = {1}, {2}; type of P : A1

v = q2 + q + 1k = q(q + 1)

A3 – J = {1}, {3}; type of P : A1

v = q5 + 2q4 + 3q3 + 3q2 + 2q + 1k = q4(q + 1)

– J = {1, 2}, {2, 3}; type of P : A2

v = q3 + q2 + q + 1k = q(q2 + q + 1)

A4 – J = {1}, {2}, {3}, {4}; type of P : A1

v = q9 + 3q8 + 6q7 + 9q6 + 11q5 + 11q4 + 9q3 + 6q2 + 3q + 1k = q8(q + 1)

– J = {1, 2}, {3, 4}; type of P : A2

v = q7 + 2q6 + 3q5 + 4q4 + 4q3 + 3q2 + 2q + 1k = q4(q3 + 2q2 + 2q + 1)

– J = {1, 3}, {2, 4}; type of P : A1A1

v = q8 + 2q7 + 4q6 + 5q5 + 6q4 + 5q3 + 4q2 + 2q + 1k = q6(q2 + 2q + 1)

– J = {1, 2, 3}, {2, 3, 4}; type of P : A3

v = q4 + q3 + q2 + q + 1k = q(q3 + q2 + q + 1)

– J = {1, 2, 4}, {1, 3, 4}; type of P : A2A1

v = q6 + q5 + 2q4 + 2q3 + 2q2 + q + 1k = q4(q2 + q + 1)


A5 – J = {1}, {2}, {4}, {5}; type of P : A1

v = q14 + 4q13 + 10q12 + 19q11 + 30q10 + 41q9 + 49q8 + 52q7 + 49q6 + 41q5 + 30q4 +19q3 + 10q2 + 4q + 1k = q13(q + 1)

– J = {1, 2}, {4, 5}; type of P : A2

v = q12 +3q11 +6q10 +10q9 +14q8 +17q7 +18q6 +17q5 +14q4 +10q3 +6q2 +3q+1k = q9(q3 + 2q2 + 2q + 1)

– J = {1, 3}, {3, 5}; type of P : A1A1

v = q13 + 3q12 + 7q11 + 12q10 + 18q9 + 23q8 + 26q7 + 26q6 + 23q5 + 18q4 + 12q3 +7q2 + 3q + 1k = q12(q + 1)

– J = {1, 4}, {2, 5}; type of P : A1A1

v = q13 + 3q12 + 7q11 + 12q10 + 18q9 + 23q8 + 26q7 + 26q6 + 23q5 + 18q4 + 12q3 +7q2 + 3q + 1k = q11(q2 + 2q + 1)

– J = {2, 3}, {3, 4}; type of P : A2

v = q12 +3q11 +6q10 +10q9 +14q8 +17q7 +18q6 +17q5 +14q4 +10q3 +6q2 +3q+1k = q10(q2 + q + 1)

– J = {1, 2, 3}, {3, 4, 5}; type of P : A3

v = q9 + 2q8 + 3q7 + 4q6 + 5q5 + 5q4 + 4q3 + 3q2 + 2q + 1k = q4(q5 + 2q4 + 3q3 + 3q2 + 2q + 1)

– J = {1, 2, 4}, {1, 2, 5}, {1, 4, 5}, {2, 4, 5}; type of P : A2A1

v = q11 + 2q10 + 4q9 + 6q8 + 8q7 + 9q6 + 9q5 + 8q4 + 6q3 + 4q2 + 2q + 1k = q9(q2 + q + 1)

– J = {1, 3, 4}, {2, 3, 5}; type of P : A2A1

v = q11 + 2q10 + 4q9 + 6q8 + 8q7 + 9q6 + 9q5 + 8q4 + 6q3 + 4q2 + 2q + 1k = q8(q3 + 2q2 + 2q + 1)

– J = {1, 2, 3, 4}, {2, 3, 4, 5}; type of P : A4

v = q5 + q4 + q3 + q2 + q + 1k = q(q4 + q3 + q3 + q + 1)

– J = {1, 2, 3, 5}, {1, 3, 4, 5}; type of P : A3A1

v = q8 + q7 + 2q6 + 2q5 + 3q4 + 2q3 + 2q2 + q + 1k = q4(q4 + q3 + 2q2 + q + 1)

D5 – J = {4}, {5}; type of P : A1

v = q19 + 4q18 + 10q17 + 20q16 + 34q15 + 51q14 + 69q13 + 86q12 + 99q11 + 106q10 +106q9 + 99q8 + 86q7 + 69q6 + 51q5 + 34q4 + 20q3 + 10q2 + 4q + 1k = q18(q + 1)

– J = {1, 4}, {1, 5}, {2, 4}, {2, 5}; type of P : A1A1

v = q18 + 3q17 + 7q16 + 13q15 + 21q14 + 30q13 + 39q12 + 47q11 + 52q10 + 54q9 +52q8 + 47q7 + 39q6 + 30q5 + 21q4 + 13q3 + 7q2 + 3q + 1k = q17(q + 1)

– J = {3, 4}, {3, 5}; type of P : A2

v = q17 +3q16 +6q15 +11q14 +17q13 +23q12 +29q11 +34q10 +36q9 +36q8 +34q7 +29q6 + 23q5 + 17q4 + 11q3 + 6q2 + 3q + 1k = q15(q2 + q + 1)

– J = {1, 2, 4}, {1, 2, 5}; type of P : A2A1

v = q16 + 2q15 + 4q14 + 7q13 + 10q12 + 13q11 + 16q10 + 18q9 + 18q8 + 18q7 + 16q6 +13q5 + 10q4 + 7q3 + 4q2 + 2q + 1k = q15(q + 1)


– J = {1, 3, 4}, {1, 3, 5}; type of P : A2A1

v = q16 + 2q15 + 4q14 + 7q13 + 10q12 + 13q11 + 16q10 + 18q9 + 18q8 + 18q7 + 16q6 +13q5 + 10q4 + 7q3 + 4q2 + 2q + 1k = q14(q2 + q + 1)

– J = {2, 3, 4}, {2, 3, 5}; type of P : A3

v = q14+2q13+3q12+5q11+7q10+8q9+9q8+10q7+9q6+8q5+7q4+5q3+3q2+2q+1k = q11(q3 + q2 + q + 1)

– J = {1, 2, 3, 4}, {1, 2, 3, 5}; type of P : A4

v = q10 + q9 + q8 + 2q7 + 2q6 + 2q5 + 2q4 + 2q3 + q2 + q + 1k = q6(q4 + q3 + q2 + q + 1)

E6 – J = {1}, {3}, {5}, {6}; type of P : A1

v = q35 +5q34 +15q33 +35q32 +70q31 +125q30 +204q29 +310q28 +444q27 +604q26 +785q25 + 980q24 + 1179q23 + 1370q22 + 1541q21 + 1681q20 + 1780q19 + 1831q18 +1831q17 + 1780q16 + 1681q15 + 1541q14 + 1370q13 + 1179q12 + 980q11 + 785q10 +604q9 + 444q8 + 310q7 + 204q6 + 125q5 + 70q4 + 35q3 + 15q2 + 5q + 1k = q34(q + 1)

– J = {1, 2}, {1, 4}, {2, 3}, {2, 5}, {2, 6}, {4, 6}; type of P : A1A1

v = q34 +4q33 +11q32 +24q31 +46q30 +79q29 +125q28 +185q27 +259q26 +345q25 +440q24 + 540q23 + 639q22 + 731q21 + 810q20 + 871q19 + 909q18 + 922q17 + 909q16 +871q15 + 810q14 + 731q13 + 639q12 + 540q11 + 440q10 + 345q9 + 259q8 + 185q7 +125q6 + 79q5 + 46q4 + 24q3 + 11q2 + 4q + 1k = q33(q + 1)

– J = {1, 3}, {5, 6}; type of P : A2

v = q33 +4q32 +10q31 +21q30 +39q29 +65q28 +100q27 +145q26 +199q25 +260q24 +326q23 + 394q22 + 459q21 + 517q20 + 565q19 + 599q18 + 616q17 + 616q16 + 599q15 +565q14 +517q13 +459q12 +394q11 +326q10 +260q9 +199q8 +145q7 +100q6 +65q5 +39q4 + 21q3 + 10q2 + 4q + 1k = q30(q3 + 2q2 + 2q + 1)

– J = {1, 5}, {3, 6}; type of P : A1A1

v = q34 +4q33 +11q32 +24q31 +46q30 +79q29 +125q28 +185q27 +259q26 +345q25 +440q24 + 540q23 + 639q22 + 731q21 + 810q20 + 871q19 + 909q18 + 922q17 + 909q16 +871q15 + 810q14 + 731q13 + 639q12 + 540q11 + 440q10 + 345q9 + 259q8 + 185q7 +125q6 + 79q5 + 46q4 + 24q3 + 11q2 + 4q + 1k = q32(q2 + 2q + 1)

– J = {3, 4}, {4, 5}; type of P : A2

v = q33 +4q32 +10q31 +21q30 +39q29 +65q28 +100q27 +145q26 +199q25 +260q24 +326q23 + 394q22 + 459q21 + 517q20 + 565q19 + 599q18 + 616q17 + 616q16 + 599q15 +565q14 +517q13 +459q12 +394q11 +326q10 +260q9 +199q8 +145q7 +100q6 +65q5 +39q4 + 21q3 + 10q2 + 4q + 1k = q31(q2 + q + 1)

– J = {1, 2, 3}, {1, 4, 5}, {2, 5, 6}, {3, 4, 6}; type of P : A2A1

v = q32 + 3q31 + 7q30 + 14q29 + 25q28 + 40q27 + 60q26 + 85q25 + 114q24 + 146q23 +180q22 + 214q21 + 245q20 + 272q19 + 293q18 + 306q17 + 310q16 + 306q15 + 293q14 +272q13 + 245q12 + 214q11 + 180q10 + 146q9 + 114q8 + 85q7 + 60q6 + 40q5 + 25q4 +14q3 + 7q2 + 3q + 1k = q29(q3 + 2q2 + 2q + 1)

– J = {1, 2, 4}, {2, 4, 6}; type of P : A2A1

v = q32 + 3q31 + 7q30 + 14q29 + 25q28 + 40q27 + 60q26 + 85q25 + 114q24 + 146q23 +180q22 + 214q21 + 245q20 + 272q19 + 293q18 + 306q17 + 310q16 + 306q15 + 293q14 +272q13 + 245q12 + 214q11 + 180q10 + 146q9 + 114q8 + 85q7 + 60q6 + 40q5 + 25q4 +14q3 + 7q2 + 3q + 1k = q31(q + 1)


– J = {1, 2, 5}, {2, 3, 6}; type of P : A1A1A1

v = q33 + 3q32 + 8q31 + 16q30 + 30q29 + 49q28 + 76q27 + 109q26 + 150q25 + 195q24 +245q23 + 295q22 + 344q21 + 387q20 + 423q19 + 448q18 + 461q17 + 461q16 + 448q15 +423q14 + 387q13 + 344q12 + 295q11 + 245q10 + 195q9 + 150q8 + 109q7 + 76q6 + 49q5 +30q4 + 16q3 + 8q2 + 3q + 1k = q31(q2 + 2q + 1)

– J = {1, 3, 4}, {4, 5, 6}; type of P : A3

v = q30 + 3q29 + 6q28 + 11q27 + 19q26 + 29q25 + 41q24 + 56q23 + 73q22 + 90q21 +107q20 + 124q19 + 138q18 + 148q17 + 155q16 + 158q15 + 155q14 + 148q13 + 138q12 +124q11 + 107q10 + 90q9 + 73q8 + 56q7 + 41q6 + 29q5 + 19q4 + 11q3 + 6q2 + 3q + 1k = q25(q5 + 2q4 + 3q3 + 3q2 + 2q + 1)

– J = {1, 3, 5}, {1, 3, 6}, {1, 5, 6}, {3, 5, 6}; type of P : A2A1

v = q32 + 3q31 + 7q30 + 14q29 + 25q28 + 40q27 + 60q26 + 85q25 + 114q24 + 146q23 +180q22 + 214q21 + 245q20 + 272q19 + 293q18 + 306q17 + 310q16 + 306q15 + 293q14 +272q13 + 245q12 + 214q11 + 180q10 + 146q9 + 114q8 + 85q7 + 60q6 + 40q5 + 25q4 +14q3 + 7q2 + 3q + 1k = q30(q2 + q + 1)

– J = {2, 3, 4}, {2, 4, 5}; type of P : A3

v = q30 + 3q29 + 6q28 + 11q27 + 19q26 + 29q25 + 41q24 + 56q23 + 73q22 + 90q21 +107q20 + 124q19 + 138q18 + 148q17 + 155q16 + 158q15 + 155q14 + 148q13 + 138q12 +124q11 + 107q10 + 90q9 + 73q8 + 56q7 + 41q6 + 29q5 + 19q4 + 11q3 + 6q2 + 3q + 1k = q27(q3 + q2 + q + 1)

– J = {1, 2, 3, 4}, {2, 4, 5, 6}; type of P : A4

v = q26 +2q25 +3q24 +5q23 +8q22 +11q21 +14q20 +18q19 +22q18 +25q17 +28q16 +31q15 + 32q14 + 32q13 + 32q12 + 31q11 + 28q10 + 25q9 + 22q8 + 18q7 + 14q6 + 11q5 +8q4 + 5q3 + 3q2 + 2q + 1k = q19(q7 + 2q6 + 3q5 + 4q4 + 4q3 + 3q2 + 2q + 1)

– J = {1, 2, 3, 5}, {1, 2, 3, 6}, {1, 2, 5, 6}, {2, 3, 5, 6}; type of P : A2A1A1

v = q31 + 2q30 + 5q29 + 9q28 + 16q27 + 24q26 + 36q25 + 49q24 + 65q23 + 81q22 +99q21 + 115q20 + 130q19 + 142q18 + 151q17 + 155q16 + 155q15 + 151q14 + 142q13 +130q12 +115q11 +99q10 +81q9 +65q8 +49q7 +36q6 +24q5 +16q4 +9q3 +5q2 +2q+1k = q29(q2 + q + 1)

– J = {1, 2, 4, 5}, {2, 3, 4, 6}; type of P : A3A1

v = q29 + 2q28 + 4q27 + 7q26 + 12q25 + 17q24 + 24q23 + 32q22 + 41q21 + 49q20 +58q19 + 66q18 + 72q17 + 76q16 + 79q15 + 79q14 + 76q13 + 72q12 + 66q11 + 58q10 +49q9 + 41q8 + 32q7 + 24q6 + 17q5 + 12q4 + 7q3 + 4q2 + 2q + 1k = q25(q4 + 2q3 + 2q2 + 2q + 1)

– J = {1, 3, 4, 5}, {3, 4, 5, 6}; type of P : A4

v = q26 +2q25 +3q24 +5q23 +8q22 +11q21 +14q20 +18q19 +22q18 +25q17 +28q16 +31q15 + 32q14 + 32q13 + 32q12 + 31q11 + 28q10 + 25q9 + 22q8 + 18q7 + 14q6 + 11q5 +8q4 + 5q3 + 3q2 + 2q + 1k = q22(q4 + q3 + q2 + q + 1)

– J = {1, 3, 4, 6}, {1, 4, 5, 6}; type of P : A3A1

v = q29 + 2q28 + 4q27 + 7q26 + 12q25 + 17q24 + 24q23 + 32q22 + 41q21 + 49q20 +58q19 + 66q18 + 72q17 + 76q16 + 79q15 + 79q14 + 76q13 + 72q12 + 66q11 + 58q10 +49q9 + 41q8 + 32q7 + 24q6 + 17q5 + 12q4 + 7q3 + 4q2 + 2q + 1k = q25(q4 + q3 + 2q2 + q + 1)

– J = {1, 2, 3, 4, 5}, {2, 3, 4, 5, 6}; type of P : D5

v = q16 + q15 + q14 + q13 + 2q12 + 2q11 + 2q10 + 2q9 + 3q8 + 2q7 + 2q6 + 2q5 + 2q4 +q3 + q2 + q + 1k = q8(q8 + q7 + q6 + q5 + 2q4 + q3 + q2 + q + 1)

– J = {1, 2, 3, 4, 6}, {1, 2, 4, 5, 6}; type of P : A4A1

v = q25+q24+2q23+3q22+5q21+6q20+8q19+10q18+12q17+13q16+15q15+16q14+


16q13 +16q12 +16q11 +15q10 +13q9 +12q8 +10q7 +8q6 +6q5 +5q4 +3q3 +2q2 +q+1k = q19(q6 + q5 + 2q4 + 2q3 + 2q2 + q + 1)

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[1] I. Barany. A short proof of Kneser’s conjecture. Journal of Combina-torial Theory, Series A, 25:325–326, 1978.

[2] A. Beutelspacher and U. Rosenbaum. Projective geometry: from foun-dations to applications. Cambridge University Press, 1998.

[3] A. Blokhuis. More on maximal intersecting families of finite sets. Jour-nal of Combinatorial theory, series A, 44(2):299–303, 1987.

[4] B. Bollobas. On generalized graphs. Acta Math. Acad. Sci. Hungar.,16:447–452, 1965.

[5] E. Boros, Z. Furedi, and J. Kahn. Maximal intersecting families andaffine regular polygons in PG(2, q). J. Combin. Theory Ser. A, 52(1):1–9, 1989.

[6] K. Borsuk. Drei Satze uber die n-dimensionale euklidische Sphare.Fundamenta Mathematicae, 20:177–190, 1933.

[7] R. C. Bose and R. C. Burton. A characterization of flat spaces in afinite geometry and the uniqueness of the Hamming and the MacDonaldcodes. J. Combinatorial Theory, 1:96–104, 1966.

[8] A. E. Brouwer, A. M. Cohen, and A. Neumaier. Distance-regular graphs.Springer-Verlag, 1989.

[9] A. Chowdhury, C. Godsil, and G. Royle. Colouring lines in projectivespace. Journal of Combinatorial Theory, Series A, 113:39–52, 2006.

[10] A. M. Cohen, M. van Leeuwen, B. Lisser, R. Sommeling, B. de Smit, andB. Ruitenburg. LiE. http://www-math.univ-poitiers.fr/~maavl/LiE/.

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[11] R. H. F. Denniston. Some maximal arcs in finite projective planes. J.Combinatorial Theory, 6:317–319, 1969.

[12] V. L. Dol’nikov. Transversals of families of sets, pages 30–36,109.Yaroslav. Gos. Univ., 1981.

[13] V. L. Dol’nikov. Transversals of families of sets in Rn and a connectionbetween the Helly and Borsuk theorems. Russian Acad. Sci. Sb. Math.,79:93–107, 1994.

[14] S. J. Dow, D. A. Drake, Z. Furedi, and J. A. Larson. A lower boundfor the cardinality of a maximal family of mutually intersecting sets ofequal size. In Proceedings of the sixteenth Southeastern internationalconference on combinatorics, graph theory and computing (Boca Raton,Fla., 1985), volume 48, pages 47–48, 1985.

[15] D. A. Drake and S. S. Sane. Maximal intersecting families of finite setsand n-uniform Hjelmslev planes. Proc. Amer. Maht. Soc., 86:358–362,1982.

[16] J. Eisfeld, L. Storme, and P. Sziklai. Minimal covers of the Kleinquadric. Journal of Combinatorial Theory, Series A, 95(1):145–157,2001.

[17] P. Erdos, H. Hajnal, and J. W. Moon. A problem in graph theory.Amer. Math. Monthly, 1964.

[18] P. Erdos, C. Ko, and R. Rado. Intersection theorems for systems offinite sets. Quarterly Journal of Mathematics, Oxford Series, series 2,12:313–320, 1961.

[19] P. Erdos and L. Lovasz. Problems and results on 3-chromatic hyper-graphs and some related questions. Colloq. Math. Janos Bolyai, 10:609–627, 1974.

[20] M. Fekete. Uber die Verteilung der Wurzeln bei gewissen algebraischenGleichungen mit ganzzahligen Koeffizienten. Mathematische Zeitschrift,17:228–249, 1923.

[21] P. Frankl and R. M. Wilson. The Erdos-Ko-Rado theorem for vectorspaces. Journal of Combinatorial Theory, Series A, 46:228–236, 1986.

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[24] C. Godsil and M. W. Newman. Independent sets in association schemes.Combinatorica, 4:431–443, 2006.

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[27] A. J. W. Hilton and E. C. Milner. Some intersection theorems forsystems of finite sets. Quart. J. Math. Oxford Ser. (2), 18:369–384,1967.

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[29] W. N. Hsieh. Intersection theorems for systems of finite vector spaces.Discrete Mathematics, 12:1–16, 1975.

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[31] G. O. H. Katona. A simple proof of the Erdos-Chao Ko-Rado theorem.Journal of Combinatorial Theory, Series B, 13(2):183–184, 1972.

[32] G. O. H. Katona. Solution of a problem of Ehrenfeucht and Mycielski.J. Combin. Th. B, 13:265–266, 1974.

[33] D. J. Kleitman. On an extremal property of antichains in partial orders.The LYM property and some of its implications and applications. InCombinatorics (Proc. NATO Advanced Study Inst., Breukelen, 1974),Part 2: Graph theory; foundations, partitions and combinatorial ge-ometry, pages 77–90. Math. Centre Tracts, No. 56. Math. Centrum,Amsterdam, 1974.

[34] M. Kneser. Aufgabe 300. Jahresbericht der Deutschen Mathematiker-Vereinigung, 58:27, 1955.

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[35] L. Lovasz. Flats in matroids and geometric graphs. In P. J. Cameron,editor, Combinatorial surveys, Proc. 6th British Comb. Conf., Egham,pages 45–86. Acad. Press, London, 1977.

[36] L. Lovasz. Kneser’s conjecture, chromatic number, and homotopy.Journal of Combinatorial Theory, Series A, 25:319–324, 1978.

[37] D. Lubell. A short proof of Sperner’s lemma. Journal of CombinatorialTheory, 1:299, 1966.

[38] J. Matousek. A combinatorial proof of Kneser’s conjecture. Combina-torica, 24(1):163–170, 2004.

[39] L. D. Meshalkin. Generalization of Sperner’s theorem on the numberof subsets of a finite set. Theory of Probability and it’s Applications,8:203–204, 1963.

[40] K. Metsch. Blocking sets in projective spaces and polar spaces. J.Geom., 76(1-2):216–232, 2003. Combinatorics, 2002 (Maratea).

[41] J. C. Meyer. Quelques problemes concernant les cliques des hyper-graphes h-complets et q-parti h-complets. In Hypergraph Seminar(Proc. First Working Sem., Ohio State Univ., Columbus, Ohio, 1972;dedicated to Arnold Ross), pages 127–139. Lecture Notes in Math., Vol.411. Springer, Berlin, 1974.

[42] K. S. Sarkaria. A generalized Kneser conjecture. Journal of Combina-torial Theory, Series B, 49:236–240, 1990.

[43] E. R. Scheinerman and D. H. Ullman. Fractional Graph Theory. Wiley-Interscience, 1997.

[44] E. Sperner. Ein Satz uber Untermengen einer endlichen Menge. Math-ematische Zeitschrift, 27(1):544–548, 1928.

[45] S. Stahl. n-tuple colourings and associated graphs. Journal of Combi-natorial Theory, Series B, 20:185–203, 1976.

[46] J. A. Thas. Ovoids, spreads and m-systems of finite classical polarspaces. In Surveys in combinatorics, 2001 (Sussex), volume 288 ofLondon Math. Soc. Lecture Note Ser., pages 241–267. Cambridge Univ.Press, Cambridge, 2001.

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[47] A. W. Tucker. Some topological properties of disk and sphere. In Proc.First Canadian Math. Congr., Montreal 1945, pages 285–309. Univ. ofToronto Press, 1946.

[48] K. Yamamoto. Logarithmic order of free distributive lattice. Journalof the Mathematical Society of Japan, 6:343–353, 1954.

Index

antichain, 8

Barany, I., 66basis, 20binomial coefficient, 8

q-binomial coefficient, 28blocking set, 95

minimal, 95BN-pair, 99Borel subgroup, 100Borsuk, K., 66Bose, R.C., 71Burton, R.C., 71

Cartan subgroup, 100chain, 8Chowdhury, A., 69chromatic number, 11

fractional chromatic number,12

multiple chromatic number, 12clique, 10

clique number, 11coclique, 10

chamber-type coclique, 80connected component, 10Coxeter diagram, 101Coxeter group, 101Coxeter matrix, 101Coxeter system, 101cycle, 10

Desargues configuration, 21diagonal criterion, 34dimension

projective dimension, 20distance, 10division ring, 16Dol’nikov, V.L., 68

edge, 9join, 9

Eisfeld, J., 72EKR family, 31

maximal, 31embedding, 14Erdos-Ko-Rado theorem, 31extension map, 62

family (of sets), 8Fekete, M., 13field, 16

characteristic, 17finite field, 16order, 16

Frankl, P., 33

Gale, D., 66Gaussian coefficient, 24generalized quadrangle, 94geometry, see incidence structureGodsil C., 69Godsil, C., 33graph, 9

154

INDEX 155

coloring, 11k-fold coloring, 12minimal coloring, 11minimal multiple coloring, 12

complement, 10complete graph, 10connected, 10edgeless graph, 11empty graph, 10homomorphism, 14regular, 9subgraph, 10

induced subgraph, 10tensor product, 116

group, 15commutative, 16of Lie type, 100order, 16

Hilton, A.J.W., 36Hsieh, W.H., 32hyperplane, 20

incidence relation, 17incidence structure, 17independence number, 11independent set, 10isomorphism, 14

Katona, G., 31Kneser, E.

q-Kneser graph, 61Kneser graph, 15, 59Polar q-Kneser graph, 88

Kneser, M., 66

line, 17linearly independent set, 19loop, 116Lovasz, L., 66Lubell, D., 29

LYM inequality, 29

maximal arc, 85maximal intersecting family, 31maximal split torus, 100Meshalkin, L.D., 29Milner, E.C., 36multiplication map, 62

neighbor, 9Newman, M.W., 33

ovoid, 96

Pappus configuration, 21parabolic subgroup, 105partial order, 9partially ordered set, 9path, 10

simple path, 10point, 17point pencil, 31

center, 31point-line geometry, see incidence

structurepoint-line incidence structure, see

incidence structurepolar disjointness graph, 87polar space, 25

generator, 25poset, 9projective line, 18projective plane, 18projective space, 18

Desarguesian projective space,21

Pappian projective space, 21

rank, 20root system, 101Royle G., 69

156 INDEX

Sarkaria, K.S., 68set system, 8span, 19Sperner, E., 29

Sperner family, 8Sperner’s theorem, 28

spread, 91Stahl, S., 62Storme, L., 72subfield, 17subspace, 19Sziklai, P., 72

Tits system, 100Tucker, A.W., 68

vertex, 9adjacent, 9coloring, 11degree, 9joined, 10

Weyl group, 100Wilson, R.M., 33

Yamamoto, K., 29

Samenvatting

Dit proefschrift behandelt de kruisbestuiving van extremale combinatorieken eindige meetkunde. Combinatoriek beschrijft discrete (en doorgaanseindige) objecten. Extremale combinatoriek bestudeert hoe groot of hoeklein een collectie eindige objecten kan zijn onder bepaalde voorwaarden.Deze objecten kunnen verzamelingen, grafen, vectoren en dergelijke meerzijn. De vragen in dit gebied komen vaak uit de informatietheorie en comput-erwetenschap. Een andere tak van de combinatoriek is eindige meetkundeover lichamen van orde q. Sommige deelstructuren in een eindige meetkundekunnen beschouwd worden als versies van projectieve vlakken voor q = 1,zelfs al bestaan er geen lichamen van orde 1. Een driehoek in een projectiefvlak is een goed voorbeeld van dit fenomeen.

Na een inleidend hoofdstuk geeft Hoofdstuk 2 een overzicht van een aan-tal klassieke problemen in de extremale combinatoriek en hun q-analoga.De meest recente resultaten voor de q-analoga van de stelling van Sperner,de Erdos-Ko-Rado stelling en verschillende versies van de stelling van Bol-lobas worden beschreven. We tonen een aantal nieuwe resultaten voorhet q-analogon van de Hilton-Milner stelling. Verder beschrijven we eennieuwe grens voor de minimale grootte van het q-analogon van kleine max-imale klieken. We besluiten dat, in sommige gevallen, resultaten voor eenq-analogon kunnen gevonden worden door gebruik te maken van dezelfdemethoden als in het origineel geval. In sommige gevallen is het antwoordzelfs identiek als in het origineel probleem. In andere gevallen kunnen detechieken die gebruikt worden voor het q-analogon ook gebruikt worden omgrenzen in het origineel probleem te verbeteren.

Het derde hoofdstuk beschrijft de reeds gekende resultaten voor hetkleuringsgetal van de Knesergraaf en geeft nieuwe grenzen voor het kleur-ingsgetal van het q-analogon. De knopen van een Knesergraaf zijn deelverza-melingen (van een vaste grootte) van een gekozen verzameling. Twee knopen

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158 INDEX

zijn verbonden als ze disjunct zijn in hun voorstelling als deelverzameling.M. Kneser formuleerde in 1955 een vermoeden over het kleuringsgetal vandeze grafen. Dit vermoeden werd in 1987 bewezen door L. Lovasz. Hetkleuringsgetal van twee kleine gevallen in het q-analogon werd gevondendoor J. Eisfeld, L. Storme en P. Sziklai in 2001 en door A. Chowdhury, C.Godsil en G. Royle in 2006. We beschrijven een asymptotisch resultaat vooralle gevallen (behalve voor een parameterfamilie) dat gebruik maakt van hetq-analogon van de Hilton-Milner stelling.

Hoofdstukken 4, 5 en 6 beschrijven andere q-analoga van de Knesergrafen. In Hoofdstuk 4 definieren we een familie Kneser grafen over parenvan incidente punten en hypervlakken van een projectieve ruimte. Webeschrijven grote maximale coklieken in deze grafen en bewijzen dat dezecoklieken de grootst mogelijke zijn in kleine dimensies. We formuleren hetvermoeen dat dit het geval is in alle dimensies. In Hoofdstuk 5 veralgemenenwe de Kneser grafen over eindige polaire ruimten in plaats van projectieveruimten. We geven beschrijvingen en kleuringsgetallen van deze grafen in eenaantal gevallen. Hoofdstuk 6 beschrijft de veralgemening van Kneser grafenover nevenklassen ten opzichte van parabolische subgroepen van Chevalleygroepen (geparametriseerd door q). We beschrijven deze grafen en bepalenkleuringsgetallen in een aantal gevallen waar q = 1. Vervolgens beschouwenwe deze grafen voor algemene q en bekijken we welke resultaten in het q = 1geval kunnen vertaald worden naar het algemene geval.

We vonden in dit proefschrift verschillende mogelijkheden voor het ver-band tussen de grenzen voor het q = 1 geval en het algemene geval. Ineen aantal voorbeelden zijn de grenzen voor het algemene geval identiek aandeze in het q = 1 geval. In andere voorbeelden bekomen we de grenzen voorhet q = 1 geval door de limiet voor q → 1 te nemen in de grenzen in hetalgemeen geval. In sommige voorbeelden tenslotte zijn de resultaten in hetq = 1 geval totaal verschillend van deze in het algemeen geval.

Summary

This thesis focuses on the interplay of extremal combinatorics and finitegeometry. Combinatorics is concerned with discrete (and usually finite) ob-jects. Extremal combinatorics studies how large or how small a collection offinite objects can be under certain restrictions. Those objects can be sets,graphs, vectors, etc. These questions are often motivated by problems ininformation theory and computer science. Another branch of combinatoricsis finite geometry over finite fields of order q. Although there is no fieldof order 1, certain substructures in finite geometry can be interpreted asversions of projective spaces for q = 1. A triangle in a projective plane is agood example of this phenomenon.

Following an introductory chapter, Chapter 2 gives an overview of someclassical problems and their q-analogues in extremal combinatorics. Themost recent results for the q-analogues of Sperner’s theorem, the Erdos-Ko-Rado theorem and several versions of Bollobas’s theorem are described. Forthe q-analogue of the Hilton-Milner theorem we give some new results. Wealso give a new bound for the minimum size of the q-analogue of small max-imal cliques. We conclude that sometimes results for a q-analogue can beobtained by using the same technique as in the original problem. In somecases the answer to the problem is even identical to that of the originalproblem. In other cases techniques used for the q-analogue could be usedfor improving bounds for the original problem.

The third chapter describes the known results for the chromatic numberof the Kneser graphs and gives new bounds for the chromatic number of theq-analogue. The vertices of a Kneser graph are subsets (of a fixed size) of aset, whereas two vertices are adjacent if they are disjoint in their subset rep-resentation. In 1955 M. Kneser conjectured the chromatic number of thosegraphs. In 1978 this was proven correct by L. Lovasz. Two small cases inthe projective space q-analogue were solved in 2001 by J. Eisfeld, L. Storme

159

160 INDEX

and P. Sziklai and in 2006 by A. Chowdhury, C. Godsil and G. Royle. Wedescribe an asymptotic result for all cases (except for one parameter family,where we give a partial proof) using the bounds from the q-analogue of theHilton-Milner theorem.

Chapters 4, 5 and 6 describe other q-analogues of the Kneser graphs. InChapter 4 we define a family of Kneser type graphs over pairs of incidentpoints and hyperplanes of a projective space. We describe large maximalcocliques in these graphs and prove that for small dimensions these are thelargest possible cocliques. We conjecture that this is the case for all dimen-sions. In Chapter 5 we extend the Kneser graphs to the case of finite polarspaces instead of finite projective spaces and in some cases give descriptionsand chromatic numbers of these graphs. Chapter 6 describes the generaliza-tion of Kneser graphs over coset spaces of Chevalley groups (parameterizedby q) with respect to parabolic subgroups. This encompasses all previouscases and extends to some interesting new cases. First we describe thesegraphs when q = 1 and give chromatic numbers for some families. Then weconsider the graphs defined over coset spaces of Chevalley groups for generalq with respect to parabolic subgroups and study what results of the q = 1case can be translated to the general case.

In this thesis we found different possibilities for the connection betweenthe bounds for the q = 1 case and the case for general q. In some casesthe bounds for the general case are identical to the bounds for q = 1. Inother cases we obtain the bounds for q = 1 by taking the limit for q → 1 inthe general bound. Other cases show a completely different bound in bothcases.

Acknowledgements

First of all, I would like to thank my co-supervisor Aart Blokhuis. It wasAart who introduced me to the beautiful cross-fertilization between extremalcombinatorics and finite geometry. Furthermore this thesis would not havebeen here without the help of my supervisors Andries Brouwer and ArjehCohen. Andries continually came up with new ideas and improvements andArjeh made sure the research was kept on track and helped me with theresults in the last chapter of this thesis.

I would also like to thank Tamas Szonyi, who made a great contributionto the Hilton-Milner results in Chapter 2. The other members of my readingcommittee are Hans Cuypers and Leo Storme. Their work in reading mythesis and commenting on it is greatly appreciated.

I also owe a great deal to the Incidence Geometry research group in Gent,where I completed my Master thesis. Especially I would like to thank FrankDe Clerk who introduced me to Finite Geometry, Hendrik Van Maldeghemwho introduced me to the Discrete Algebra group in Eindhoven and Jan DeBeule, with whom I had a lot of interesting discussions.

There are a lot of people who made the four years in Eindhoven verypleasant. First of all, thanks to the other Ph.D. students of the DiscreteAlgebra group and Coding and Crypto group, lunches were never boring.Furthermore, Andrey was a great room mate during those years, alwaysready to have some fun or discuss something serious. I’m very grateful Imet Ellen during my first month in Eindhoven. She became a good friendwho was always there to have a good talk or share some good advice with me.

Finally I would like to thank my family. My parents, grandparents, andsister Isabelle for their support, encouragement and understanding. Andlast but not least my wife Inge for her endless love.

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Curriculum Vitae

Tim Mussche was born on October 12, 1981 in Eeklo, Belgium. Between1993 and 1999 he was a high school student at the Sint Vincentius collegein Eeklo.

In 1999 he started studying at the University of Ghent where he tooka variety of courses in mathematics, physics, computer science and philoso-phy. He received his bachelor’s degree in mathematics in 2001. Tim wrotehis Master’s thesis on model theory and generalized quadrangles under thesupervision of prof.dr. Hendrik Van Maldeghem. In 2003 he received hisMaster’s degree in pure mathematics.

Thanks to professor Van Maldeghem Tim came in contact with the Dis-crete Algebra and Geometry group at the Technical University of Eind-hoven. In 2003 he joined this group where he was a Ph.D student un-der the supervision of prof.dr. Andries Brouwer, prof.dr. Arjeh Cohen anddr. Aart Blokhuis. During this time he researched the connections betweenextremal combinatorics and finite geometry in the context of generalizedKneser graphs. This thesis is the result of this research.

Tim is currently employed in Belgium as business process managementconsultant at the consultancy firm MOBIUS.

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