The small Kakeya sets in $T^{*}_{2}(\mathcal{C})$, $\mathcal{C}$ a conic

0.19 MB·

by  Maarten De Boeck

T
The small Kakeya sets in ( ), a conic
2∗
C C
Maarten De Boeck∗
6 Abstract
1
0 A Kakeya set in the linear representation T∗(C), C a non-singular conic, is the point set
2
2 covered by a set of q+1 lines, one through each point of C. In this article we classify the
n small Kakeya sets in T2∗(C). The smallest Kakeya sets have size j3q24+2qk, and all Kakeya
Ja sets with weight less than j3(q24−1)k+q are classified: thereare approximately pq2 types.

4
1 Keywords: linear representation, Kakeya sets, edge-disjoint maximal cliques, bipartite
graphs

]
O
MSC 2010 codes: 05B25, 51E20, 51E26, 05C35
C
.
h
1 Introduction
t
a
m
A Kakeya set in the affine space AG(n,q) is the point set covered by a set of lines, precisely one
[ in every direction, i.e. one affine line through every point in the hyperplane at infinity. The finite

1 field Kakeya problem asks for the size of the smallest Kakeya sets in AG(n,q). This problem was
v stated by Wolff in [13] as the finite field analogue of the classical Euclidean Kakeya problem. We
9 refer to [11, Section 1.3]for a survey on the EuclideanKakeyaproblem. Important results on the
3
finite field Kakeya problem were obtained in [4, 5, 10]. In [4] it was proved that a Kakeya set
5 contains at least q+n−1 points.
3 n
0 The moststud(cid:0)iedfin(cid:1)ite Kakeyasets arethose inAG(2,q). They arisefroma setofq+1lines.
. Not only is the minimal size of these Kakeya sets known, but there is also a classification of the
1
small examples. We refer to [1, 2, 3] for the results. The problem we will discuss in this article
0
6 was inspired by the Kakeya problem for AG(2,q).
1 A linear representation T∗( ) is a point-line geometry which is embedded in AG(3,q), with
2 K
: a point set in π = PG(2,q), the projective plane at infinity of AG(3,q). It is defined in the
v ∞
K
Xi following way: the points of T2∗(K) are the points of AG(3,q), the lines of T2∗(K) are the lines
of AG(3,q) whose direction is a point of , and the incidence is the incidence inherited from
r AG(3,q). K
a
SinceAG(3,q)togetherwithπ =PG(2,q)formsaprojectivegeometryPG(3,q)wewillhave
∞
a look at some substructures of PG(3,q). A quadric of PG(n,q) is an algebraic variety described
by a homogeneous polynomial of degree 2 in the variables X ,...,X . It is called singular if a
0 n
projective transformationcan be found such that it can be written with less than n+1 variables,
and non-singular otherwise. A quadric in PG(2,q) is called a conic. All non-singular conics are
projectively equivalent. In PG(3,q) there are two types of non-singular quadrics: the hyperbolic
quadrics and the elliptic quadrics. In this article we focus on hyperbolic quadrics. Up to a
projective transformation a hyperbolic quadric is described by the equation X X +X X = 0.
0 1 2 3
Ona hyperbolicquadricthere are2(q+1)lines,whichcanbe dividedintwogroupsofq+1lines,
called reguli, such that two lines from the same regulus are disjoint, and two lines from different
reguli meet each other. The set of lines meeting three disjoint lines ℓ ,ℓ ,ℓ , is a regulus,and the
1 2 3
∗Address: UGent,DepartmentofMathematics,Krijgslaan281-S22,9000Gent, Flanders,Belgium.
Emailaddress: [email protected]
1
set of lines meeting all lines of this regulus is also a regulus containing ℓ ,ℓ ,ℓ , which is called
1 2 3
the opposite regulus. These two reguli form a hyperbolic quadric. More background information
on quadrics and reguli can be found in [7, 8].
Inthis article we consider the linear representationT∗( ) of a non-singularconic in π , the
2 C C ∞
plane at infinity of AG(3,q). We will use the notation T∗( ) throughout. We study the Kakeya
2 C
sets in T∗( ), which we define analogouslyto Kakeyasets in AG(2,q): a Kakeyaline set in T∗( )
2 C 2 C
is a set of lines, precisely one through each point of . The corresponding Kakeya set is the set
C
of points covered by this line set. A. Blokhuis raised the question of finding the minimal size of
these Kakeya sets, and to classify the small examples. The Kakeya sets in T∗( ) are similar to
2 C
the Kakeya sets in AG(2,q) in the sense that they arise both from a set of q+1 lines.
In Section 2 we study this problem by looking at a related graph-theoretical problem, so we
recallbriefly somegraph-theoreticalconcepts. Agraph Γisapair(V,E)whichconsistsofasetV
of vertices and a set E of edges, which are 2-subsets of V; since we consider E to be a set, graphs
will be consideredto be simple. If u,v is an edge, then the vertices u and v are called adjacent.
{ }
A clique is a set of vertices together with the edges connecting them, such that any two vertices
in this set are adjacent. A clique is maximal if it is not contained in a larger clique.
A graph is complete if any two vertices are adjacent. The complete graph on a vertex set V
is denoted by K . By K , r N, we denote a complete graph on a set of r vertices, without
V r
∈
specifying the vertex set. Note that cliques are complete subgraphs. A graph is called bipartite
if there is a partition V ,V of the vertex set V such that two vertices in V are not adjacent,
1 2 i
{ }
i = 1,2. It is called complete bipartite if there is a partition V ,V of the vertex set V such
1 2
{ }
that two vertices are adjacent if and only if they belong to different parts of the partition. The
completebipartitegraphonthevertexsetsV′ andV′′ isdenotedbyKV′,V′′. ByKr,s,r,s N,we
∈
denote a complete bipartite graphon a set of r vertices and a set of s vertices, without specifying
the vertex sets.
2 The graph theory approach
In this section we will discuss a particular kind of graph. This discussion is motivated by the
following definition and the two succeding lemmata.
Definition 2.1. Let be a Kakeya line set in T∗( ). The graph Γ( ) is the graph with vertex
L 2 C L
set and such that two vertices are adjacent if the corresponding lines have a point in common.
L
Lemma 2.2. Let be aKakeyaline set inT∗( ). The maximalcliques ofΓ( ) areedge-disjoint
L 2 C L
and correspond to subsets of through a common point.
L
Proof. Letℓ , ℓ andℓ bethreelinesof ,havingpairwiseapointincommon. Theplane ℓ ,ℓ
1 2 3 1 2
L h i
inPG(3,q)meetsπ inasecantlineto containingP andP ,thepointsatinfinityofℓ andℓ .
∞ 1 2 1 2
C
The line ℓ passes through a point of , different from P and P , hence meets ℓ ,ℓ in a point.
3 1 2 1 2
C h i
Since ℓ meets both ℓ and ℓ , it contains the intersection point ℓ ℓ . Consequently, maximal
3 1 2 1 2
∩
cliques of Γ( ) correspond to subsets of the line set through a common point.
L L
So, two different maximal cliques C and C of Γ( ) correspond to two different subsets
1 2 1
L L
and of , which are two line sets, each through a common point, say P and P . The points
2 1 2
L L
P and P are different since the maximal cliques C and C are different. Hence, and can
1 2 1 2 1 2
L L
have at most one line in common. So, C and C have at most a vertex in common and are thus
1 2
edge-disjoint.
Lemma 2.3. Let be a Kakeya line set in T∗( ), and let k be the number of maximal cliques
with i vertices in ΓL( ). Then, the Kakeya set 2 (C ) contains qi(q+1) q+1k (i 1) points.
L K L − i=1 i −
P
Proof. We know that =q+1 and everyline contains q affine points. A point oni lines, which
|L|
corresponds to a maximal clique with i vertices in Γ( ), needs to be counted only once. So, we
L
2
substract for each of these points, i 1 from q(q+1). This proves the result
−
q+1
( ) =q(q+1) k (i 1).
i
|K L | − −
Xi=1
To find small Kakeya sets, we need the value q+1k (i 1) to be as large as possible, using
i=1 i −
the notation from Lemma 2.3. The next lemmataPdeal with this question.
In this discussion we use some Tur´an-style graph theoretical results. The next theorem about
triangle-free graphs is due to Mantel.
Theorem 2.4 ([9]). A triangle-freegraphon n verticescontains at most n2 edges. A triangle-
4
j k
free graph on n vertices with n2 edges, is the complete bipartite graph K .
4 n , n
j k ⌈2⌉⌊2⌋
This result was later generalised to the famous theorem by Tur´an on K -free graphs ([12]). A
r
stability result for Mantel’s theorem was proved by Hanson and Toft.
Theorem 2.5 ([6]). A triangle-free graph on n vertices with n2 l edges, l < n 1, is a
bipartite graph. j 4 k− (cid:4)2(cid:5)−
Note that all maximal cliques of a triangle-free graph are trivially edge-disjoint. We restate
the previous theorem in the following way:
Corollary 2.6. Let Γ be a triangle-free graph on n vertices.
1. If n is even, and Γ contains n2 ε edges, ε< n 1, then Γ arisesfrom a complete bipartite
graph Kn+δ,n−δ by removing4 ε− δ2 edges, for2s−ome δ √ε, δ N.
2 2 − ≤ ∈
2. If n is odd, and Γ contains n2−1 ε edges, ε < n−1 1, then Γ arises from a complete
4 − 2 −
δbipaNrt.ite graph Kn+21+δ,n−21−δ by removing ε −δ2 −δ edges, for some δ ≤ qε+ 14 − 21,
∈
We present the main lemma of this section, which generalises the previous result.
Lemma 2.7. LetΓbe agraphonnverticeswithedge-disjointmaximalcliques,andletk be the
i
number of maximal cliques with i vertices in Γ.
1. If n is even, and n k (i 1)> n2 n+1, then Γ arisesfrom a complete bipartite graph
i=1 i − 4 − 2
Kn2+δ,n2−δ by remPoving ε−δ2 edges, for some δ ≤√ε, δ ∈N, with ε= n42 − ni=1ki(i−1).
P
2. If n is odd, and n k (i 1) > n2−1 n−1 +1, then Γ arises from a complete bipartite
i=1 i − 4 − 2
graph Kn+21+δ,n−2P1−δ by removing ε−δ2−δ edges, for some δ ≤ qε+ 14 − 21, δ ∈ N, with
ε= n2−1 n k (i 1).
4 − i=1 i −
P
In both cases Γ is a bipartite graph.
Proof. For a graph G on n vertices, we denote the value n kG(i 1) by C(G), with kG the
i=1 i − i
number of maximal cliques with i vertices in G. P
Denote the vertex set of Γ by V. Let C ,...,C be the maximal cliques of Γ with at least
1 m
three vertices, and let V be the vertex set of C . We know that V V 1 if i = j. Also, if x
i i i j
| ∩ | ≤ 6
and y are adjacent vertices in Γ, then there is a maximal clique C such that x,y V .
i i
∈
For each i=1,...,m, we choose a partition of V with two subsets V′ and V” such that the
i i i
sizes V′ and V′′ differ at most one. Let Γ be the graph on the vertex set V with the edges
| i| | i |
of the complete bipartite subgraphs KV′,V′′ and the edges of Γ that are not contained in any of
i i
the maximal cliques C . In other words, the graph Γ is obtained by replacing the edges of the
i
3
maximalcliquesC ,...,C (completesubgraphsK )inΓbythe edgesofthecompletebipartite
1 m Vi
subgraphs KV′,V′′, i=1,...,m.
i i
We note that Γ arises from Γ by deleting k s(s 1)+ k s2 edges. We denote
s≥1 2s − s≥1 2s+1
the set of edges that are in Γ but not in Γ byP. Moreover,we knPow that
E
C(Γ) C(Γ)= k (s 1)2+ k s(s 1) 0.
2s 2s+1
− − − ≥
Xs≥1 Xs≥1
Consequently,
n2 n
C(Γ) C(Γ)> +1.
≥ (cid:22) 4 (cid:23)−j2k
The graphΓ is triangle-free since any triangle of Γ is contained in a unique maximal clique of
size at least3. Hence, C(Γ) equals the number of edgesin Γ, which we denote by n2 ε′. Note
j 4 k−
that
n2 n
ε′+C(Γ)= <C(Γ)+ 1 ,
(cid:22) 4 (cid:23) (cid:16)j2k− (cid:17)
hence
n
ε′ < 1 C(Γ) C(Γ) .
(cid:16)j2k− (cid:17)−(cid:0) − (cid:1)
WecanapplyCorollary2.6onΓsinceitistriangle-freeandwefindthatΓisabipartitegraph,
arisingfromthebipartitegraphK bytheremovaloffewedges;wedenotethissetofremoved
V1,V2
edgesby ′. Here, V andV formapartitionofthevertexsetV. Ifniseven,then V = n δ,
E 1 2 | 1| 2−
V = n +δ, δ N and the number of removed edges equals
| 2| 2 ∈
n
′ =ε′ δ2 < 1 δ2 C(Γ) C(Γ) .
|E | − 2 − − − −
(cid:0) (cid:1)
If n is odd, then V = n+1 +δ, V = n−1 δ, δ N and the number of removed edges equals
| 1| 2 | 2| 2 − ∈
′ =ε′ δ2 δ < n−1 1 δ2 δ C(Γ) C(Γ) .
|E | − − 2 − − − − −
Letxandy be twoverticesinV suc(cid:0)hthatx,y V(cid:1) , j =1,2. We knowthatthe edge(x,y)is
i j
∈
in . So,theverticesxandy arebothcontainedinV′ orarebothcontainedinV′′. Consequently,
E i i
V V V′ or V V V′′, j =1,2. Since V =(V V ) (V V ) and both V′ and V′′ are
i∩ j ⊆ i i∩ j ⊆ i i i∩ 1 ∪ i∩ 2 i i
nonempty,the partitions V V ,V V and V′,V′′ areequal. Without lossofgeneralitywe
{ i∩ 1 i∩ 2} { i i }
can assume V V =V′ and V V =V′′.
i∩ 1 i i∩ 2 i
So, Γ arises from the bipartite graph K by first deleting the edges of ′ (necessarily
V1,V2 E
connecting a vertex in V and a vertex in V ), and then adding the edges of , each of which
1 2
E
connects two vertices of V or two vertices of V .
1 2
We distinguish now between two cases. First we assume n is even. We consider the maximal
clique C of Γ. Two vertices in V′ cannot be adjacent to the same vertex of V V′′ since the
i i 2 \ i
maximal cliques of Γ are edge-disjoint. If V = 2s, and thus V′ = V′′ = s, then ′ should
contain at least (s 1) n +δ s edges. |Wi|e also know that|Ci(|Γ) |Ci(|Γ) (s 1E)2 by the
− 2 − − ≥ −
existence of Ci. Hence, (cid:0) (cid:1)
n n
(s 1) +δ s ′ < 1 δ2 (s 1)2 .
− (cid:16)2 − (cid:17)≤|E | 2 − − − −
This inequality is equivalent to
n
1 (s 2)+δ(δ+s 1)<0,
(cid:16)2 − (cid:17) − −
which is false since s 2, n 2 and δ 0. If V =2s+1, s 2, and thus V′ =s, V′′ =s+1
or V′ = s+1, V′′ ≥= s, th≥en ′ sho≥uld con|taii|n at least (s≥ 1) n +δ |si|1 ed|geis.| In this
| i| | i | E − 2 − −
case we know that C(Γ) C(Γ) s(s 1). We find the inequality(cid:0) (cid:1)
− ≥ −
n n n
(s 1) +δ s 1 ′ < 1 δ2 s(s 1) 1 (s 2)+δ(δ+s 1)<0,
− (cid:16)2 − − (cid:17)≤|E | 2 − − − − ⇔ (cid:16)2 − (cid:17) − −
4
which is false. If V = 3, then either V′ = 1, V′′ = 2 or else V′ = 2, V′′ = 1. In the first
case ′ should con|tai|in at least n δ | 1i|edges,|ini t|he second ca|seia|t least| ni |+δ 1 edges. In
generEal, ′ should contain at lea2st−n −δ 1 edges. We find the inequality 2 −
E 2 − −
n n
δ 1 ′ < 1 δ2 δ(δ 1)<0,
2 − − ≤|E | 2 − − ⇔ −
which is also false for all δ N. We conclude that Γ does not contain maximal cliques of size at
∈
least three. Hence, Γ is equal to Γ and the theorem follows.
Secondly,weassumenisodd. Weproceedinthesamewayasinthecaseneven. For V =2s
i
| |
we find the inequality
n+1 n 1
(s 1) +δ s ′ < − 1 δ2 δ (s 1)2
− (cid:18) 2 − (cid:19)≤|E | 2 − − − − −
n 1
− (s 2)+δ(δ+s)+1<0.
⇔ 2 −
For V =2s+1, s 2, we find the inequality
i
| | ≥
n+1 n 1
(s 1) +δ s 1 ′ < − 1 δ2 δ s(s 1)
− (cid:18) 2 − − (cid:19)≤|E | 2 − − − − −
n 1
− (s 2)+δ(δ+s)+1<0.
⇔ 2 −
For V =3 we again need to look at two different situations. Analogously,we find the inequality
i
| |
n 1 n 1 n 1
− δ 1 ′ < − 1 δ2 δ δ(δ 1)+ − 1 <0.
2 − − ≤|E | 2 − − − ⇔ − (cid:18) 2 − (cid:19)
In all three cases we find a contradiction.
We describe a particular graph,whose existence shows that the bound in Lemma 2.7 is sharp.
Example 2.8. Let n be odd. Consider the vertex sets W , W and W , with W = W = n−3
1 2 3 | 1| | 2| 2
and W = x,y,z . The sets W′ and W′′ form a partition of W . The graph G on the vertex set
3 { } 2 2 2
W W W is defined by the following adjacencies. Any vertexof W is adjacentto any vertex
1 2 3 1
∪ ∪
of W x , the vertex y is adjacent to all vertices of W′ x , the vertex z is adjacent to all
vertic2e∪s o{f }W′′ x and the vertices y and z are adjacen2t∪. {Th}is graph has n2−1 n−5 edges.
2 ∪{ } 4 − 2
It contains one maximal clique of size 3, namely W ; all other maximal cliques are edges. Hence,
3
n k (i 1)= n2−1 n−3 with k the number of maximal cliques with i vertices in G.
i=1 i − 4 − 2 i
P
3 The classification result
We give two examples of a small Kakeya set in T∗( ).
2 C
Example 3.1. We consider the linear representation T∗( ) embedded in the projective space
2 C
PG(3,q), with π the plane containing the non-singular conic . We denote the points of by
C C
P ,...,P . Let be a hyperbolic quadric in PG(3,q) meeting the plane π in the conic , and let
0 q
Q C
and ′ be the two reguli of . So, through each point P there is a unique line ℓ and a
i i
R R Q ∈ R
uniquelineℓ′ ′. Weconsiderapartitionofthepointsof intwosets,withoutlossofgenerality
P ,...,P i ∈aRnd P ,...,P ,0 k q+1. Let be thCe Kakeyalineset k−1ℓ ( q ℓ′).
{ 0 k−1} { k q} ≤ ≤ L ∪i=0 i ∪ ∪i=k i
Then, the corresponding Kakeya set ( ) covers (cid:0) (cid:1)
K L
3q2+2q 1 q+1 2
kq+(q+1 k)(q k)= − + k
− − 4 (cid:18) 2 − (cid:19)
affine points.
Note that and ′ can be interchanged, hence, it is sufficient to look at the examples with
0 k q+1. ARlso noRte that Γ( ) is a complete bipartite graph.
≤ ≤ 2 L
5
Example3.2. Weusethesamenotationasinthepreviousexample. Now,weconsiderapartition
ofthe pointsof P intwosets,withoutlossofgenerality P ,...,P and P ,...,P ,
q 0 k−1 k q−1
C\{ } { } { }
0 k q. Let m be a secantline meeting in P and a point on a line ℓ , 0 i<k or a line ℓ′,
≤ ≤ Q q i ≤ i
k i q 1. Let be the Kakeyaline set k−1ℓ q−1ℓ′ m . Then, the corresponding
Ka≤key≤a se−t ( ) coLvers (cid:0)∪i=0 i(cid:1)∪(cid:16)∪i=k i(cid:17)∪{ }
K L
3 q 2
kq+(q k)(q k)+(q 1)= q2+q 1+ k
− − − 4 − (cid:16)2 − (cid:17)
affine points.
Also here, we note that and ′ can be interchanged. Hence, it is sufficient to look at the
examples with 0 k q. RThe graRph Γ( ) depends on the second intersection point of m and
≤ ≤ 2 L
, the point P being the first. If it is a point on a line ℓ , 0 i < k but not on a line ℓ′,
Q q i ≤ i
k i q 1 (or vice versa), then Γ( ) is a bipartite graph arising from a complete bipartite
≤ ≤ − L
graph by removing all edges but one through a given vertex. If this second intersection point is
a point ℓ ℓ′, 0 i < k and k j q 1, then Γ( ) is the graph constructed in Example 2.8
i∩ j ≤ ≤ ≤ − L
with W′ =W .
2 2
Intheproofofthemaintheoremswewillusethefollowinglemmawhichshowsthatahyperbolic
quadric is determined by a conic and two disjoint lines.
Lemma 3.3. Let C be a non-singular conic in a plane π of PG(3,q) and let m and m′ be two
disjoint lines in PG(3,q), not in π, both meeting C. Then, there is a unique hyperbolic quadric
containing C, m and m′.
Proof. Let P be the point m C and P be the point m′ C, and let P be the intersection
1 2 3
∩ ∩
point ofthe tangentlines in P andP to the conic C, inthe plane π. Let P be a point m P
1 2 4 1
\{ }
and let P be a point m P . Now we choose a frame of PG(3,q) such that P = (1,0,0,0),
5 2 1
\{ }
P = (0,1,0,0), P = (0,0,1,0), P = (0,0,0,1) and P = (1,1,1,1). The plane π is given by
2 3 4 5
X =0 and the non-singular conic C is given by (X =0) X X +dX2 =0 for some d F∗.
3 3 ∩ 0 1 2 ∈ q
TheequationofaquadriccontainingC isX0X1+dX22+X(cid:0)3(a0X0+a1X1+a(cid:1)2X2+a3X3)=0.
All points on the line m = P ,P are on this quadric, hence a = a = 0. Also all points
1 4 0 3
h i
on the line m′ = P ,P are on this quadric, hence a = 1 and a = d. Consequently,
2 5 1 2
h i − −
there is only one quadric containing C, m and m′, namely the quadric given by the equation
X X +dX2 X X dX X =0. This quadric is hyperbolic: its equation can be rewritten as
0 1 2 − 1 3− 2 3
(X X )X +dX (X X )=0.
0 3 1 2 2 3
− −
Thetwofollowingtheoremspresentthe mainresultofthisarticle. Aclassificationofthesmall
Kakeya sets in T∗( ) is proved.
2 C
Theorem 3.4. Let be a Kakeya line set in T∗( ), embedded in PG(3,q), q odd, with ( ) its
L 2 C K L
corresponding Kakeya set. If ( ) < 3 q2 1 +q, then ( ) is a Kakeya set as described in
|K L | 4 − K L
Example 3.1. (cid:0) (cid:1)
Proof. InDefinition2.1wedefinedthegraphΓ( )correspondingtotheKakeyalineset . Denote
L L
its number of maximal cliques with i vertices by k . By Lemma 2.3 and the assumption we know
i
that
q+1
3
q(q+1) k (i 1)= ( ) < q2 1 +q
i
− − |K L | 4 −
Xi=1 (cid:0) (cid:1)
and thus
q+1 (q+1)2 q+1
k (i 1)> +1.
i
− 4 − 2
Xi=1
Since Γ( ) is a graph on q+1 vertices with edge-disjoint maximal cliques (by Lemma 2.2), we
L
can apply Lemma 2.7(1.). Since q is odd the number of vertices of Γ( ) is even.
L
6
We denote (q+1)2 q+1k (i 1) by ε. Now, we know that Γ( ) arises from a complete
bipartitegraphK4q+21+−δ,qP+21i−=δ1byi re−movingε−δ2 edges,forsomeδ ≤√Lε, δ ∈N. LetV1 betheset
ofvertices ofthe partitioncontaining q+1+δ verticesandlet be the setof lines corresponding
2 L1
to it; let V be the set of vertices of the partition containing q+1 δ vertices and let be the
2 2 − L2
set of lines corresponding to it. Since the number of edges removed from the complete bipartite
graph equals ε δ2 ε < q+1 1, we know there are two vertices in V that are adjacent to all
− ≤ 2 − 1
vertices in V , hence there are two lines, say ℓ and ℓ , in meeting all lines of .
2 0 1 1 2
L L
The lines ℓ and ℓ are disjoint. By Lemma 3.3 they define together with the non-singular
0 1
conic a unique hyperbolic quadric . Let be the regulus containing ℓ and ℓ , and let ′ be
0 1
C Q R R
the opposite regulus. All lines in meet the lines ℓ and ℓ . Since they also contain a point of
2 0 1
L
, they are lines on , necessarily belonging to ′.
C Q R
A line of contains a point of , and hence it is a tangent line to , a secant line to or
1
L C Q Q
a line on . So, it meets zero, one or all lines of . If contains a line meeting at most one
2 1
Q L L
line of , then there is a vertex in V adjacent to at most one vertex of V . Consequently, Γ( )
2 1 2
arisesfrLom the complete bipartite graphK by removingatleast q+1 δ 1 edges. HoweveLr,
V1,V2 2 − −
we know that the number of removed edges equals ε δ2 < q+1 δ 1, a contradiction. So, all
− 2 − −
lines of are lines meeting all lines of , and therefore all lines of belong to .
1 2 1
L Q L L R
We conclude that ( ) is a Kakeya set as described in Example 3.1.
K L
Theorem 3.5. Let be a Kakeya line set in T∗( ), embedded in PG(3,q), q even, with ( )
L 2 C K L
its corresponding Kakeya set. If ( ) < 3q2+q 1, then ( ) is a Kakeya set as described in
|K L | 4 − K L
Example 3.1.
Proof. We proceed as in the proof of Theorem 3.4. We look at the graph Γ( ) on q+1 vertices.
L
By Lemma 2.3 and the assumption we know
q+1
3
q(q+1) k (i 1)= ( ) < q2+q 1,
i
− − |K L | 4 −
Xi=1
hence
q+1 (q+1)2 1 (q+1) 1
k (i 1)> − − +1.
i
− 4 − 2
Xi=1
Wedenote (q+1)2−1 q+1k (i 1)byε. ApplyingLemma2.7(2.) weknownowthatΓ( )arises
4 − i=1 i − L
fLreotmVacboemtphleetseetbiopfarvtPeirtteicgersapohf tKheq2+p1a+rδt,i2qti−oδnbcyonretaminovininggqε+−δ12+−δδveedrgteics,esδa≤ndqlεet+ 41−be12,tδhe∈sNet.
of line1s corresponding to it; let V be the set of vertices o2f the partition containing Lq1 δ vertices
2 2 −
and let be the set of lines corresponding to it. Since the number of edges removed from the
2
completeLbipartite graph equals ε δ2 δ ε < q 1, we know also in this cases there are two
− − ≤ 2 −
vertices in V that are adjacent to all vertices in V , hence there are two lines, say ℓ and ℓ , in
1 2 0 1
meeting all lines of .
1 2
L L
Now,weintroducethehyperbolicquadric andthereguli and ′asintheproofofTheorem
Q R R
3.4. We know that all lines of belong to ′. We also know that the lines of meet zero, one
2 1
L R L
or all lines of .
2
L
If there is a line in meeting at most one line of , then Γ( ) arises from the complete
1 2
bipartite graph K bLy removing at least q δ 1 edLges. We knoLw however that the number
of removed edges Ve1q,uV2als ε δ2 δ < q δ 2 −1. S−o, we find a contradiction. As in the proof of
− − 2 − −
Theorem 3.4, we conclude that ( ) is a Kakeya set as described in Example 3.1.
K L
We look at the previous results in some more detail.
Remark 3.6. We firstconsider the case q odd. We know that all smallKakeyasets in T∗( ) are
ofthe type describedExample3.1. Amongthemthe smallestexample,correspondingtok2=Cq+1,
2
has γ(q)= 3q2+2q−1 points. The corresponding Kakeya line set has q+1 lines in each regulus.
4 2
7
Theorem 3.4 allows us in general to classify the q−1 smallest types of Kakeya sets in
(cid:24)q 2 (cid:25)
T∗( ). Thesecond-smallestexamplehasγ(q)+1points. ItarisesfromaKakeyalinesetwith q+3
2 C 2
linesinoneregulusand q−1 linesintheoppositeregulus. Ingeneralthe(m+1)-smallestexample
2
has γ(q)+m2 points and arises from a Kakeya line set with q+1 +m lines in one regulus and
2
q+1 m lines in the opposite regulus, m q−1 1. This corresponds to the construction in
2 − ≤(cid:24)q 2 (cid:25)−
Example 3.1 with parameter k = q+1 m.
2 −
NotethattheresultinTheorem3.4issharp. TheKakeyasetofthetypedescribedinExample
3.2, with parameter k = q−1, contains precisely 3 q2 1 +q points.
2 4 −
Now we consider the case q even. We know t(cid:0)hat all(cid:1)small Kakeya sets in T∗( ) are of the
type described Example 3.1. Among them the smallest example, corresponding w2itCh k = q, has
2
γ(q)= 3q2+2q = 3q2+1q points. ThecorrespondingKakeyalinesethas q+1linesinoneregulus
4 4 2 2
and q in the opposite.
2
Theorem 3.5 allows us in general to classify the q 3 1 smallest types of Kakeya sets
lq2 − 4 − 2m
in T∗( ). The second-smallest example has γ(q)+2 points. It arises from a Kakeya line set with
q+22liCnesinoneregulusand q 1linesintheoppositeregulus. Ingeneralthem-smallestexample
h2as γ(q)+m(m 1) points an2d−arisesfrom a Kakeyaline set with q +m lines in one regulus and
− 2
q m+1linesinthe oppositeregulus,m q 3 1 . Thiscorrespondstothe construction
i2n−Example 3.1 with parameter k = q +1≤lmq. 2 − 4 − 2m
2 −
NotethattheresultinTheorem3.5issharp. TheKakeyasetofthetypedescribedinExample
3.2, with parameter k = q, contains precisely 3q2+q 1 points.
2 4 −
For small q values, the results from Theorem 3.5, Theorem 3.4 and Remark 3.6 give little
information. For q = 2 it yields no classification result, for q = 3,4 only the smallest example
has been classified. For q 5 at least the smallest and the second-smallest example have been
≥
classified. Therefore we study in the next remarks the small q values, namely q =2,3,4. Given a
Kakeya line set , we know the size of ( ) through the graph Γ( ) by Lemma 2.3. Therefore,
L K L L
we study the different possibilities for Γ( ).
L
Remark 3.7. We first look at the case q = 2. We already noted that the previous theorems do
notyielda classificationresultin this case. Inthis remarkhoweverwe willpresentaclassification
of all Kakeya sets in the case q =2. A Kakeya line set contains 3 lines, hence corresponds to a
L
simple graphs on 3 vertices. There are four nonisomorphic simple graphs on 3 vertices.
f f f f
1 2 3 4
Figure 1: The simple graphs on 3 vertices.
In each of the graphs in Figure 1, the maximal cliques are edge-disjoint. By Lemma 2.3 the
Kakeya sets corresponding to the graphs f and f have size 4, the Kakeya sets corresponding to
3 4
the graph f have size 5 and the Kakeya sets corresponding to the graph f have size 6. Recall
2 1
that the structure of the graph gives directly a description of the Kakeya line set.
A Kakeya line set whose graph Γ( ) equals f , is a cone, a set of three lines through a
• L L 4
common affine point. There is a unique hyperbolic quadric in PG(3,2) containing the conic
inthe plane atinfinity andtwoofthese three lines. So,anysuchconeofthree lines through
a common affine point can also be seen as a Kakeya line set described in Example 3.2 with
k=1 and m passing through ℓ ℓ′.
0∩ 1
A Kakeya line set whose graph Γ( ) equals f , consists of two disjoint lines, and a third
• L L 3
linemeetingboth. ByLemma3.3weknowthatwecanfindahyperbolicquadriccontaining
8
the conic in the plane at infinity and the two disjoint lines. Hence, any Kakeya line set
whose graph Γ( ) equals f can be described by Example 3.1, with k =1. L
L 3
A Kakeya line set whose graph Γ( ) equals f , consists of two intersecting lines, and a
• L L 2
third line meeting none. Again by Lemma 3.3 we can see that any such Kakeyaline set can
be described by Example 3.2, with k =0.
A Kakeya line set whose graph Γ( ) equals f , consists of three disjoint lines. These
• L L 1
three lines determine a unique hyperbolic quadric, which necessarily meets π in the conic
∞
. Hence, this is a Kakeya line set described in Example 3.1, with k =0.
C
We observe that all Kakeya sets for q =2 are described by Examples 3.1 and 3.2.
Remark 3.8. Now, we look at the case q = 3. Then a Kakeya line set contains 4 lines. We
L
already classified the smallest Kakeya sets for q =3 in Theorem 3.4: they have size 8, arise from
the construction in Example 3.1 with k = 2. In this remark we classify all Kakeya sets of size
9. Recall that the structure of a Kakeya line set is described by its graph Γ( ). There are 11
L L
nonisomorphic simple graphs on 4 vertices. One of them has two maximal cliques of size 3 which
haveanedgeincommon,socannotbeagraphΓ( ). ThetenothergraphsarepresentedinFigure
L
2.
m m m m m
1 2 3 4 5
m m m m m
6 7 8 9 10
Figure 2: The simple graphs on 4 vertices with edge-disjoint maximal cliques.
The graphm correspondsto a Kakeyasetofsize 8, the graphsm , i=2,...,5, correspondto
1 i
a Kakeyaset of size 9, the graphs m , i=6,7,8, correspondto a Kakeyaset of size 10, the graph
i
m corresponds to a Kakeyaset of size 11, and the graph m corresponds to a Kakeyaset of size
9 10
12.
The smallest Kakeya sets for q = 3, which have size 8, correspond to the graph m . We look
1
at the second-smallest Kakeya sets, those of size 9.
AKakeyaline set whose graphΓ( ) equalsm , is a cone, the set ofallfour lines in T∗( )
• L L 2 2 C
through a common affine point.
A Kakeyaline set whose graphΓ( ) equals m , consists of three lines through a common
• L L 3
affine point and a line meeting one of these three lines in a different point. E.g. the Kakeya
linesetdescribedinExample3.2withk =1andmpassingthroughℓ ℓ′ yieldsthisgraph.
0∩ 2
Moreover, by Lemma 3.3 we can see that any Kakeya line set whose graph Γ( ) equals
m can be described by Example 3.2 with k =1 and m passingLthrough ℓ ℓ′. L
3 0∩ 2
A Kakeya line set whose graph Γ( ) equals m , consists of two pairs of two intersecting
• L L 4
lines, such that one line of the first pair meets precisely one line of the other pair and the
other line of the first pair does not meet a line of the secondpair. Again by Lemma 3.3 any
suchKakeyaline set can be described by Example 3.2, with k =1 and m passing througha
point of ℓ′ (or ℓ′) not on ℓ .
1 2 0
AKakeyalineset whosegraphΓ( )equalsm ,consistsofthreedisjointlines,andafourth
• L L 5
line meeting all three. All these Kakeya line sets are described by Example 3.1, with k =1.
9
Remark 3.9. Finally, we look at the case q = 4. Theorem 3.5 characterizes the Kakeya sets in
T∗( ) of size 14, and we characterize the Kakeya sets of size 15 here. A Kakeya line set consists
2 C
of 5 lines in this case. There are 34 nonisomorphic simple graphs on 5 vertices, 9 of which have
maximalcliquesthatarenotedge-disjoint. WementionedabovethevalueC(G)= q+1kG(i 1)
i=1 i −
for a graph G, with kiG the number of maximal cliques with i vertices of G. ForPeach of the 25
remaining graphs we can calculate this value. There is one graph with C(G) = 0 and one graph
with C(G)=1. There are 3 graphs with C(G)=2, 6 graphs with C(G) =3 and 10 graphs with
C(G)=4. InFigure3weshowthethreegraphswithC(G)=5andtheonegraphwithC(G)=6.
These are the only graphs that can correspond to Kakeyasets of size at most 15 by Lemma 2.3.
d d d d
1 2 3 4
Figure 3: The simple graphs on 5 vertices that can correspond to Kakeya sets of size at most 15.
We already classified the smallest Kakeya sets for q = 4: they have size 14, arise from the
constructionin Example 3.1 with k =2, and correspondto the graphd . We look at the second-
1
smallest Kakeya sets, those of size 15. Recall once more that the graphdescribes the structure of
the Kakeya line set.
A Kakeya line set whose graph Γ( ) equals d consists of five lines m ,m ,m ,m′,m′
• L L 2 1 2 3 1 2
with m ,m ,m pairwise disjoint, m′,m′ disjoint, m′ meeting m , m and m and m′
1 2 3 1 2 1 1 2 3 2
meeting m and m but not m . By applying Lemma 3.3 we find a hyperbolic quadric
1 2 3
containing and the lines m′ and m′. It follows that is a Kakeya line set that can be
C 1 2 L
described by Example 3.2, with k = 2 and m passing through a point of ℓ′ or ℓ′ not on ℓ
2 3 0
or ℓ or through a point on ℓ or ℓ not on ℓ′ or ℓ′. Consequently, any Kakeya line set
who1se graph Γ( ) equals d i0s give1n by Exam2ple 3.32 in the described way. L
L 2
It can be proved analogously, again using Lemma 3.3, that any Kakeya line set whose
• graphΓ( ) equals d canbe seen as a Kakeyaline set described in Example 3.2 wiLth k =2
L 3
and m passing through ℓ ℓ′, i=0,1 and j =2,3.
i∩ j
AKakeyalineset whosegraphΓ( )equalsd ,consistsoffivelinesm ,...,m ,suchthat
• L L 4 0 4
m meets m and m , but not m and m , 0 i 4, whereby the addition in the
i i+1 i+4 i+2 i+3
≤ ≤
indices is takenmodulo 5. We will show that such a Kakeyaline set cannotexist, hence the
graph d is not admissible as graph of a Kakeya line set.
4
Let P be the point m and let be the hyperbolic quadric defined by the disjoint lines
i i
C∩ Q
m and m and the conic . Let be the regulus of containing m , and let ′ be the
0 2 0
C R Q R
opposite regulus. We denote the line of through P by ℓ and the line of ′ through P
i i i
R R
by ℓ′, 0 i 4. We know that m =ℓ , m =ℓ′ and m =ℓ . We denote the point ℓ ℓ′
i ≤ ≤ 0 0 1 1 2 2 i∩ j
by P .
i,j
The lines m and m are both secants to the quadric . Since m and m meet each other,
3 4 3 4
Q
V = m ,m is a plane through the line P ,P . There are five planes through P ,P ,
3 4 3 4 3 4
h i h i h i
one of which is π . The four remaining planes π ,...,π meet in the union of two
∞ 1 4
Q
lines ℓ ℓ′, the union of two lines ℓ ℓ′, the conic P ,P ,P ,P ,P and the conic
3∪ 4 4∪ 3 { 3 4 0,1 1,2 2,0}
P ,P ,P ,P ,P ,respectively. Thelinem hastobe asecantmeetingthelineℓ and
3 4 1,0 2,1 0,2 3 2
{ }
the line m has to be a secant meeting the line ℓ . In each of the four planes there is only
4 0
onepointonℓ andonepointonℓ . Thelineinπ throughP meetingℓ isℓ′,whichisnot
0 2 1 4 0 4
a secant line, a contradiction, so V = π . Analogously, starting from P , also V = π . The
1 3 2
6 6
unique line in π through P meeting ℓ is P ,P . However this line also meets ℓ′ =m ,
3 4 0 h 4 0,1i 1 1
a contradiction, so V =π . Analogously, considering P ,P , also V =π .
3 3 2,1 4
6 h i 6
We conclude that there are only two types of Kakeya sets of size 15 for q = 4, both arising from
the construction in Example 3.2.
10