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The main aim of this chapter is to use linear algebra in the construction of infinite families
of graphs which contain no copy of some specified subgraph. In nearly all cases these
families of graphs have been known for some time. However, by using linear algebra, and
more specifically bilinear forms, messy calculations with coordinates can be avoided and
in some cases one is able to extend these constructions to a larger family of graphs.
A graph H consists of a set of vertices and a set of edges which are subsets of the vertices
of size 2. Note that for some authors, this would be a loopless, undirected graph. We
shall assume throughout that H is finite, in other words the set of vertices is finite and
for our purposes we can assume that H is connected, that is that for all vertices x and
y, there is a sequence of intersecting edges e1 ,. . . ,er , with the property that x ∈ e1 and
y ∈ er .
The Turán number of a graph H is a function from N to N, denoted ex(n, H), and is
the maximum number of edges a graph with n vertices can have which contains no copy
of H as a subgraph. We shall be above all concerned with the asymptotic behaviour of
ex(n, H), that is, how it grows as n gets large.
1. The Erdős-Stone theorem
In a colouring of a graph H, each vertex is assigned a colour in such a way that no edge
contains two vertices assigned the same colour. The chromatic number χ(H) of a graph
H is the smallest number of colours required to colour the graph H.
The following theorem, the Erdős-Stone theorem, describes the asymptotic behaviour of
ex(n, H) for nearly all graphs.
Theorem 1. For all > 0, there is a n0 such that for all n ≥ n0 ,
1
1
1−
− 12 n2 < ex(n, H) < 1 −
+ 12 n2 .
χ(H) − 1
χ(H) − 1
An immediate consequence of the Erdős-Stone theorem is that we know the asymptotic
behaviour of ex(n, H) for all graph H where χ(H) ≥ 3. Therefore, we are only left with
the problem of determining the asymptotic behaviour of ex(n, H) when χ(H) = 2. Note
that when χ(H) = 2 the left-most term in this inequality is negative and so tells us
nothing. If χ(H) = 2 then H is called bi-partite, since we can partition the vertices into
Date: 9 April 2013.
1
2
SIMEON BALL
two disjoint subsets with the property that all edges contain one vertex from each of the
two disjoint subsets.
The following example is known as the Turán graph.
Example 2. Let Km,...,m denote the multi-partite graph with n = tm vertices, where the
set of vertices is the disjoint union of t subsets of size m and there is an edge between two
vertices belonging to different subsets in this partition of the vertex set. Since χ(Km,...,m ) =
t, the graph Km,...,m contains no graph H where χ(H) = t + 1. Furthermore, Km,...,m has
1
n(n
2
− m) = 12 n2 (1 − (1/t))
edges.
Now, with a small observation, similar to one which we shall use later, we can conclude
that Example 2 proves the lower bound in Theorem 1. Suppose that χ(H) = t + 1. For
all n, we have n − r = mt for some non-negative integer r ≤ t − 1. Example 2 with n − r
vertices has 12 (n − r)2 (1 − (1/t)) edges and since
1
(n
2
− r)2 (1 − (1/t)) ≥ 12 n2 (1 − (1/t)) − rn(1 − (1/t)) > 12 n2 (1 − (1/t) − ),
for n large enough, we are done.
2. Complete bipartite graphs
In view of Theorem 1 we shall from now on restrict our attention to the case that H is
bipartite.
Let Kt,s denote the complete bipartite graph with s + t vertices, where the vertices are
partitioned into a subset of size s and a subset of size t and where two vertices are joined
by an edge if and only if they belong to distinct subsets in the partition of the vertex set.
We start by proving an upper bound on ex(n, Kt,s ) using purely combinatorial counting.
Theorem 3. For all > 0, there is a n0 such that for all n ≥ n0 ,
ex(n, Kt,s ) < 21 (s − 1)1/t (1 + )n2−(1/t) ,
where t ≤ s.
Proof. Let G be a graph with n vertices and e edges which contains no Kt,s .
Let N be the number of copies of K1,t contained in G.
Since G contains no Kt,s for each subset S of t vertices, there are at most s − 1 common
neighbours of S. Hence,
n
nt
N≤
(s − 1) ≤ (s − 1)(1 + )t−1 .
t
t!
Let d(v) denote the degree of a vertex, that is the number of edges which contain the
vertex v and let δ = 2e/n denote the average degree of a vertex.
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By considering each vertex in turn,
X d(v)
δ
δt
N=
≥n
> n − 12 nδ t−1 ,
t!
t
t
v
for n large enough.
Suppose e > 12 (s − 1)1/t (1 + )n2−(1/t) . Comparing the upper and lower bounds on N we
have,
nt
nt
(s − 1)(1 + )t−1 > (s − 1)(1 + )t − 12 (s − 1)(t−1)/t (1 + )t−1 nt−(1/t) ,
t!
t!
which implies
n1/t
(t−1)/t
1
(s
−
1)
>
(s − 1),
2
t!
which is not true for n large enough.
3. Linear and bilinear forms
Let F be a field and let Vk (F) denote the k-dimensional vector space over F.
A linear form is a map α from Vk (F) to F with the property that
α(λu + µv) = λα(u) + µα(v),
for all λ, µ ∈ F and for all u, v ∈ Vk (F).
The set Vk (F)∗ of linear forms from V (F) to F forms a k-dimensional vector space over F,
if we define
(λα + µβ)(u) = λα(u) + µβ(u)
for all λ, µ ∈ F and for all α, β ∈ Vk (F)∗ .
Let αi ∈ Vk (F)∗ be a linear form, for i = 1, . . . , r.
Lemma 4. If α1 , . . . , αr are linearly independent then
r
\
dim
ker(αi ) = k − r.
i=1
Proof. Since α1 , . . . , αr are linearly independent, there is a basis for Vk (F)∗ of the form
{α1 , . . . , αr , αr+1 , . . . , αk } for some αr+1 , . . . , αk . With respect to this basis
u∈
r
\
ker(αi )
i=1
if and only if the first r coordinates of u are zero.
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SIMEON BALL
A bilinear form is a map b from Vk (F) × Vk (F) to F with the property that b(u, v) and
b(v, u) are both linear forms for all u ∈ Vk (F).
If there is a u ∈ Vk (F), u 6= 0, such that b(u, v) = 0, for all v ∈ Vk (F) then b is said to be
degenerate. If not then b is said to be non-degenerate.
Let U be a subspace of Vk (F) and define
U ⊥ = {v ∈ Vk (F) | b(u, v) = 0, for all u ∈ U }.
Lemma 5. If b is a non-degenerate bilinear form then
dim U + dim U ⊥ = k.
Proof. Let {u1 , . . . ur } be a basis for U .
Define linear maps αi , for i = 1, . . . r, by
αi (v) = b(ui , v).
If
Pr
i=1
λi αi = 0 then
Pr
i=1
0=
λi αi (v) = 0, for all v ∈ Vk (F). Therefore,
r
X
i=1
λi αi (v) =
r
X
λi b(ui , v) = b(
r
X
i=1
λi ui , v),
i=1
Pr
and so i=1 λi ui = 0, since b in non-degenerate. Thus, λ1 = · · · = λr = 0, which implies
that α1 , . . . , αr are linearly independent. The lemma now follows from Lemma 4 and the
observation that
r
\
⊥
U =
ker(αi ).
i=1
4. Subspaces of a finite vector space
Let Fq denote the finite field with q elements.
Lemma 6. The number of r-dimensional subspaces of Vk (Fq ) is
(q k − 1)(q k − q) · · · (q k − q r−1 )
.
(q r − 1)(q r − q) · · · (q r − q r−1 )
Proof. The numerator is the number of ordered sets of r linearly independent vectors
and the denominator is the number of ordered sets of r linearly independent vectors that
generate the same r-dimensional subspace.
Let U be an s-dimensional subspace of Vk (F). For all v ∈ Vk (F), the set
v + U = {u + v | u ∈ U }
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is a coset of U . The set of cosets
Vk (F)/U = {v + U | v ∈ Vk (F)}
forms a (k − s)-dimensional vector space over F, where we define
λ(v + U ) = λv + U
for all λ ∈ F and v ∈ Vk (F), and
v + U + w + U = v + w + U,
for all v, w ∈ Vk (F).
Lemma 7. The number of r-dimensional subspaces of Vk (F) containing a fixed s-dimensional
subspace is equal to the number of (r − s)-dimensional subspaces in Vk−s (F).
Hence in the case F = Fq , this number is
(q k−s − 1)(q k−s − q) · · · (q k−s − q r−s−1 )
.
(q r−s − 1)(q r−s − q) · · · (q r−s − q r−s−1 )
Proof. Let W be an r-dimensional subspace containing an s-dimensional subspace U .
Then W = U ⊕ W 0 where dim W 0 = r − s. Suppose that {w1 , . . . , wr−s } is a basis for W 0 .
The map α defined on the subspaces containing U by α(W ) = hw1 + U, . . . , wr−s + U i is
an isomorphism between the subspaces containing U and the subspaces of Vk (F)/U . 5. Graphs containing no K2,2
In this section we begin by determining the asymptotic behaviour of ex(n, K2,2 ).
Theorem 8. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
1
(1
2
− )n3/2 < ex(n, K2,2 ) < 21 (1 + )n3/2 .
Proof. The upper bound follows from Theorem 3.
To prove the lower bound we shall construct an infinite sequence of graphs on q 2 + q + 1
vertices, where q is a prime power, with roughly 12 q 3 edges. For a graph on n vertices we
take the graph in the infinite sequence with n − r = q 2 + q + 1, where r is minimised,
together with r vertices of degree zero. Now (q + 1)2 > n − r, so this graph has at least
− r)1/2 − 1)3 > 21 (n − r)3/2 − 32 (n − r)
√
edges. Bombieri’s theorem implies that r ≤ c n log n which gives
1
((n
2
1
(n
2
− r)3/2 − 23 (n − r) > 12 (1 − )n3/2 ,
for n large enough.
Let b be a non-degenerate symmetric bilinear form on V3 (Fq ), so
b(u, v) = b(v, u),
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SIMEON BALL
for all u, v ∈ V3 (Fq ).
Let G be the graph whose vertices are the one-dimensional subspaces of V3 (Fq ) and where
hui and hvi are joined by an edge if
b(u, v) = 0.
If hwi is a neighbour of both hui and hvi then b(u, w) = b(v, w) = 0, so w ∈ hu, vi⊥ . By
Lemma 5, dimhu, vi⊥ = 1, so hwi = hu, vi⊥ . Hence, the vertices hui and hvi have at most
one common neighbour and G contains no K2,2 .
By Lemma 6, the graph G has n = q 2 + q + 1 vertices. The neighbours of the vertex hui
are the 1-dimensional subspaces in hui⊥ and by Lemma 7, there are q + 1 of them (one
of which may be hui). So, counting edges through each vertex we conclude that G has at
least 21 nq edges and
1
nq > 12 n(n1/2 − 1) > 12 (1 − )n3/2 ,
2
for n large enough.
6. Projective planes
We can give a geometrical interpretation of the construction in Theorem 8 but first we
need to define some point and line geometries.
An incidence structure is a set of points and a set of lines, where a line is a subset of the
points of size at least two. If x ∈ `, where x is a point and ` is a line, we say x is incident
with ` and likewise ` is incident with x.
A projective plane is an incidence structure with the property that any two lines are
incident with a unique common point and any two points are incident with a unique
common line. To avoid degenerate example we add the axiom that there are four points,
no three of which are collinear.
A polarity π of an incidence structure is an incidence preserving bijection from the points
to the lines. In other words, x ∈ ` if and only if π −1 (`) ∈ π(x).
The construction of the graph in Theorem 8 is an example of the following construction.
Let G be a graph whose vertices are the points of a projective plane equipped with a
polarity π, where two vertices x and y are joined by an edge if and only if x ∈ π(y). The
common neighbours of x and z are the vertices of π(x) ∩ π(z), which is a singleton set
since two lines are incident with a unique point. Hence, this graph contains no K2,2 .
In Theorem 8, the projective plane we use is that projective plane whose points are the
one-dimensional subspaces of V3 (F) and whose lines are the two-dimensional subspaces of
V3 (F). This projective plane is denoted PG(2, F) and is called the Desarguesian projective
plane. A symmetric bilinear form on V3 (F) clearly induces a polarity on PG(2, F), where
we define
π(U ) = U ⊥ ,
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for any non-trivial subspace U .
It is worth noting that we can construct graphs on n vertices with cn3/2 edges from a
projective plane without using a polarity, albeit for a worse constant c.
The incidence graph of an incidence structure is a bi-partite graph whose vertices are the
points and lines of the incidence structure and where a point and a line are joined by an
edge if and only if they are incident. The incidence graph of PG(2, Fq ) has n = 2(q 2 +q+1)
vertices and 12 n(q + 1) ' 2√1 2 n3/2 edges.
7. Graphs containing no K2,s
We wish to extend Theorem 8 to deduce the asymptotic behaviour of ex(n, K2,s ) and to
do this we need to use the multiplicative structure of Fq . This we shall deduce in the
following lemma.
Let F∗q denote the non-zero elements of Fq .
Lemma 9. The elements of F∗q are roots of the polynomial
X q−1 − 1,
and form a multiplicative cyclic group.
Proof. Let x ∈ F∗q . The set {ax | a ∈ F∗q } consists of all elements of F∗q . Thus,
Y
Y
(ax) =
a,
a∈F∗q
a∈F∗q
which implies xq−1 = 1.
Let N (d) be the number of elements of F∗q of order d. There are at most d roots of the
polynomial
X d − 1.
If a is an element of order d then the roots of this polynomial are {1, a, . . . , ad−1 }. The
element ae has order d if and only if gcd(d, e) = 1. So N (d) = 0 or N (d) = φ(d), where φ
is Euler’s totient function.
Euler’s formula states
X
φ(d) = q − 1,
d|q−1
so we have
X
d|q−1
N (d) = q − 1 =
X
d|q−1
φ(d) ≥
X
N (d).
d|q−1
Therefore, we have equality throughout and N (q − 1) = φ(q − 1) 6= 0. Hence, F∗q has an
element of order q − 1 and is so cyclic.
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SIMEON BALL
Theorem 10. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
1
(s
2
− 1)1/2 (1 − )n3/2 < ex(n, K2,s ) < 21 (s − 1)1/2 (1 + )n3/2 .
Proof. The upper bound follows from Theorem 3.
To prove the lower bound we shall construct an infinite sequence of graphs on (q 2 −1)(s−1)
vertices, where q is a prime power congruent to 1 modulo s−1, with roughly 21 q 3 edges. For
a graph on n vertices we take the graph in the infinite sequence with n−r = (q 2 −1)(s−1),
where r is minimised, together with r vertices of degree zero. We then proceed to argue as
in the beginning of the proof of Theorem 8, and use a refiniement on Bombieri’s theorem,
the Huxley-Iwaniec theorem that states that there is a prime congruent to 1 modulo s − 1
with r < n2/3 .
Let q be a prime power congruent to 1 modulo s − 1, let S be a subgroup of F∗q with s − 1
elements and let R be a set of coset representatives for S.
Let b be a non-degenerate symmetric bilinear form on V3 (Fq ). Let v ∈ V3 (Fq ) be a vector
with the property that b(v, v) 6= 0, so v 6∈ hvi⊥ .
For each one-dimensional subspace U of hvi⊥ , fix a basis so that U = hxi and for all
ρ ∈ R, define
[x, ρ] = {ρx + λv | λ ∈ S}.
Let G be the graph with vertices [x, ρ], where [x, ρ] is joined to [x0 , ρ0 ] with an edge if and
only if
b(u, u0 ) = 0,
for some u ∈ [x, ρ] and u0 ∈ [x0 , ρ0 ].
Now,
b(ρx + λv, x0 ρ0 + λ0 v) = ρρ0 b(x, x0 ) + (λµ)(λ0 /µ)b(v, v),
so [x0 , ρ0 ] is a neighbour of [x, ρ] if for a fixed u ∈ [x, ρ], there is a u0 ∈ [x0 , ρ0 ] such that
b(u, u0 ) = 0,
or in other words u0 ∈ hui⊥ .
The subspace hui⊥ is two-dimensional, and since u 6∈ hvi, it intersects hvi⊥ in a onedimensional subspace, hx00 i say. For each one-dimensional subspace hx0 i of hvi⊥ , x0 6= x00 ,
the subspace hui⊥ intersects hx0 , vi in x0 + α0 v, for some α0 ∈ F∗q . Now α0 = λ0 ρ0 , for some
ρ0 ∈ R and λ0 ∈ S and so the vertex u has neighbour [x0 , (ρ0 )−1 ]. Therefore, each vertex
has at least q neighbours, by Lemma 6.
If for ρ, ρ0 ∈ R, λ, λ0 ∈ S and x, x0 ∈ hvi⊥ , there is a µ ∈ Fq such that
µ(ρx + λv) = ρ0 x0 + λ0 v
then µρ = ρ0 and µ = λ0 /λ. But λ0 /λ ∈ S and so µ ∈ S and µρ = ρ0 implies ρ = ρ0 .
So µ = 1, λ = λ0 and ρ = ρ0 . Hence, if ρx + λv 6= ρ0 x0 + λ0 v, then they are linearly
independent.
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To find the common neighbours of [x, ρ] and [x0 , ρ0 ], we can fix u ∈ [x, ρ] as above. For
each λ0 ∈ S, the common neighbours of u and ρ0 x0 + λ0 v intersect hu, ρ0 x0 + λ0 vi⊥ , which is
a one-dimensional subspace. Now, as we have seen in the previous paragraph two vectors
of the form ρx + λv, where ρ ∈ R and λ ∈ S, are linearly independent so these s − 1
one-dimensional subspaces intersect at most s − 1 vertices [x, ρ]. Hence, there are at most
s − 1 vertices common neighbours of [x, ρ] and [x0 , ρ0 ].
It only remains to count the number of vertices and the number of edges in G. The
number of vertices is
n = (q + 1)(q − 1)/(s − 1)
and since each vertex has at least q neighbours, the number of edges is at least
1
n((s
2
− 1)n + 1)1/2 > 12 (s − 1)1/2 n3/2 .
8. A probabilistic construction of graphs containing no Kt,s
Theorem 11. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
ex(n, Kt,s ) > c(1 − )n2−(s+t−2)/(st−1) ,
where
c = 2−2−2/(st−1) (t!s!)1/(st−1) .
Proof. Let G be a graph on n vertices where we join two vertices with an edge with
probability p, where p is to be determined.
Let Y be the random variable that counts the number of edges in G. The expected value
of Y is
n
E(Y ) =
p > c0 n2 p,
2
1
0
for any constant c < 2 , if n is large enough.
Let X be the random variable that counts the number of copies of Kt,s in G. The expected
value of X is
n n − s st
E(X) =
p < c00 ns+t pst ,
s
t
where c00 = 1/(s!t!).
By the linearity of expectation,
E(Y − X) > c0 n2 p − c00 ns+t pst .
If we put
p=
c0 1/(st−1)
n−(s+t−2)/(st−1) .
2c00
then
E(Y − X) ≥ 21 c0 pn2 = cn2−(s+t−2)/(st−1) ,
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SIMEON BALL
where
c = 12 c0
c0 1/(st−1)
< 2−2−2/(st−1) (t!s!)1/(st−1) .
2c00
We have already determined the asymptotic behaviour of ex(n, K2,s ), so let us consider
the upper and lower bounds which we have proved for ex(n, K3,s ).
By Theorem 3 and Theorem 11 we have
c(1 − )n3/2 < ex(n, K3,s ) < 12 (s − 1)1/3 (1 + )n5/3 .
The aim of the next section will be to improve the lower bound.
9. Graphs containing no K3,3
Let Q be the quadratic form on V4 (Fq ) defined by
Q(u) = u1 u2 + f (u3 , u4 ),
where f is an irreducible homogeneous polynomial of degree two.
Lemma 12. The set of vectors Z = {u ∈ V4 (Fq ) | Q(u) = 0} contains a subset O of
q 2 + 1 vectors with the property that every three vectors of S are linearly independent.
Proof. For any u, v ∈ V4 (Fq ) and λ ∈ Fq
Q(u + λv) = Q(u) + λ2 Q(v) + 2λb(u, v),
where b is a symmetric bilinear form.
Suppose u, v, u+µv ∈ Z, where v 6∈ hui and µ ∈ F∗q . Then b(u, v) = 0 and so u, v, u+λv ∈
Z for all λ ∈ Fq . In other words the two dimensional subspace U = hu, vi ⊂ Z. But then
U ∩ ker u1 is a one-dimensional subspace, whereas Z ∩ ker u1 = {0}, since f is irreducible.
This is a contradiction, which implies that any three vectors of Z which are pairwise
linearly independent are linearly independent.
Define a subset O ⊂ Z by
O = {(−f (u3 , u4 ), 1, u3 , u4 ) | u3 , u4 ∈ Fq } ∪ {(1, 0, 0, 0)}.
Then O is the required set.
Theorem 13. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
1
(1
2
− )n5/3 < ex(n, K3,3 ) < 12 21/3 (1 + )n5/3 .
Proof. The upper bound follows by Theorem 3.
To prove the lower bound we shall construct an infinite sequence of graphs on (q 2 +1)(q−1)
vertices, where q is a prime power, with roughly 21 q 5 edges. For a graph on n vertices we
take the graph in the sequence with n − r = (q 2 + 1)(q − 1), where r is minimised. Now
TURÁN NUMBERS
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q 3 > n − r, so the graph has at least 12 (n − r)5/3 edges. Again, Bombieri’s theorem is
enough to imply,
1
(n − r)5/3 > 21 (1 − )n5/3 ,
2
for n large enough.
Let b be a symmetric non-degenerate bilinear form on V5 (Fq ).
Let v ∈ V5 (Fq ) be such that b(v, v) 6= 0.
By Lemma 12, there is a subset O of hvi⊥ of q 2 + 1 vectors with the property that every
subset of three vectors of O are linearly independent.
Let G be the graph whose set of vertices is
{x + λv | x ∈ O, λ ∈ F∗q },
and where u and u0 are joined by an edge if
b(u, u0 ) = 0.
For any subspace U define U ⊥ as before and note that the common neighbours of the
vertices in U are the vertices in hU i⊥ .
Let A = {x1 + λ1 v, x2 + λ2 v, x3 + λ3 v} be a set of three vertices of G.
If xi = xj for some i 6= j, then v ∈ hAi and so hAi⊥ ⊆ hvi⊥ . Since hvi⊥ contains no
vertices of G, the vertices of A have no common neighbour.
If x1 , x2 , x3 are distinct then we shall show that the vectors in A are linearly independent.
If
3
X
0=
µi (xi + λi v)
i=1
then
3
3
X
X
(
µi λi )v = −
µ i xi .
i=1
i=1
⊥
Since v 6∈ v ,
3
X
µi λi = 0,
i=1
so
3
X
µi xi = 0,
i=1
which gives µi = 0, for i = 1, 2, 3, since the any three vectors in S are linearly independent.
Hence, dimhAi = 3 and so dimhAi⊥ = 2. The same argument as above implies A⊥ contains
at most two vertices of G and so G contains no K3,3 .
By Lemma 12, we can assume |O| = q 2 + 1, so G has n = (q 2 + 1)(q − 1) vertices.
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SIMEON BALL
For any vertex x + λv and x0 ∈ O \ hxi⊥ ,
b(x + λv, x0 − (b(x, x0 )/λb(v, v))v) = b(x, x0 ) − b(x, x0 )(b(v, v)/b(v, v)) = 0.
Now hxi⊥ ∩ hvi⊥ is a three-dimensional subspace which contains at most q + 2 vectors of
O (in fact it is at most q + 1), since if u ∈ hxi⊥ ∩ hvi⊥ ∩ O then each two-dimensional
subspace in hxi⊥ ∩ hvi⊥ containing u contains at most one other vector of O. There are
q + 1 such subspaces by Lemma 7.
Therefore, the number of edges in G is at least
1
n(q 2
2
− q − 1) > 12 (1 − )n5/3 ,
for n large enough.
In fact, the actual asymptotic behaviour of ex(n, K3,3 ) has been determined although we
shall not prove it here. It implies that the construction in Theorem 13 is asymptotically
best possible.
Theorem 14. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
1
(1
2
− )n5/3 < ex(n, K3,3 ) < 21 (1 + )n5/3 .
By mimicking the proof of Theorem 10, replacing hvi⊥ by the subset O of hvi⊥ used in
Theorem 13 one obtains the following lower bound on ex(n, K3,2r2 +1 ).
Theorem 15. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
ex(n, K3,2r2 +1 ) > 12 r2/3 (1 − )n5/3 .
It is fairly straightforward to generalise the construction in Theorem 13. Let O be a set
of vectors in V2t−2 (Fq ) with the property that every subset of t vectors of O are linearly
independent. Then consider the set O as a subset of v ⊥ where v ∈ V2t−1 (Fq ). The same
construction as in Theorem 13 will give a graph G with n vertices containing no Kt,t , and
by counting the number of edges we have that
ex(n, Kt,t ) > 21 (1 − )n2−1/(logq (|O|)+1) .
Unfortunately one can show that for t ≥ 6 this bound cannot improve on the probabilistic
lower bound in Theorem 11. Let M be the (2t − 2) × |O| matrix whose columns are the
vectors of O. The dual of the code generated by M has minimum distance at least t + 1
(since any t vectors of O are linearly independent) so by applying the sphere-packing
bound to the dual code, one sees that the above lower bound can only (possibly) improve
on Theorem 11 for t ≤ 5.
10. Projective spaces
To be able to give a geometrical interpretation of the construction in Theorem 13, we
need to introduce projective spaces.
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The projective space PG(k −1, F) consists of all non-trivial subspaces of Vk (F). The points
of PG(k−1, F) are the one-dimensional subspaces of Vk (F), the lines of PG(k−1, F) are the
two-dimensional subspaces of Vk (F), the planes of PG(k − 1, F) are the three-dimensional
subspaces of Vk (F), . . . and the hyperplanes of PG(k − 1, F) are the hyperplanes (the
(k − 1)-dimensional subspaces) of Vk (F). Note that in the geometry we think of the
subspace as the points it contains. In this way, incidence in the geometry comes from
the vector space, so for example a point is incident with a plane if the corresponding
one-dimensional subspace is contained in the corresponding three-dimensional subspace.
It is important to note the dimension shift here. The projective dimension of a subspace
in PG(k − 1, F) is one less than the dimension of the subspace in the vector space Vk (F).
An ovoid of PG(3, Fq ) is a set of q 2 + 1 points with the property that no three points are
collinear.
A polarity π of the projective space PG(k − 1, F) is an incidence preserving bijection from
the m-dimensional subspaces to the (k − 2 − m)-dimensional subspaces, m = 0, . . . , k − 2.
In other words, a subspace u is contained in a subspace u0 if and only if π −1 (u0 ) ⊂ π(u).
We now give a more general construction of a graph of which the graph constructed in
Theorem 13 is an example.
Let v be a point of PG(4, Fq ) equipped with a polarity π, such that v 6∈ π(v). Let O be
an ovoid of π(v). Let the vertices of a graph G be the set of points on the cone whose
vertex is v and whose base is O, but not the point v nor the points of O. In other words
the points on the lines hv, oi, where o ∈ O, not including v nor o. Two vertices x and y
are joined by an edge if x ∈ π(y).
Let A be a subset of three vertices of G.
If two of the points of A lie on a line of the cone then v ∈ hAi and so π(hAi) ⊂ π(v).
Since π(v) contains no vertices of G the vertices in A have no common neighbour.
If the points of A are incident with distinct lines of the cone then hAi is a plane, not
contained in π(v), so π(A) is a line not containing v. A line not containing v contains at
most 2 points of the cone and so the vertices in A have at most two common neighbours.
Hence the graph G contains no K3,3 .
11. The norm graph
The norm function N : Fqt−1 → Fq is defined by
N (x) = xq
t−2 +q t−3 +···+q+1
.
q
Note that N (x) = N (x), so the image is Fq .
The norm graph is a graph whose vertices are Fqt−1 × F∗q and where (x, λ) is joined by an
edge to (x0 , λ0 ) if and only if
N (x + x0 ) = λλ0 .
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SIMEON BALL
The norm graph has n = q t − q t−1 vertices and 12 nq t−1 edges.
Lemma 16. The norm graph on q t − q t−1 vertices contains no Kt,(t−1)!+1 .
We shall not prove the above lemma as it requires more algebra than we have studied
thus far but as a consequence of the lemma we have the following theorem.
Theorem 17. For all > 0, there is an n0 such that for all n ≥ n0
1
(1
2
− )n2−1/t < ex(n, Kt,(t−1)!+1 ) < 21 ((t − 1)!)1/t (1 + )n2−1/t .
Consider the construction of the graph G containing no K2,2 in Theorem 10. If we take
the symmetric bilinear form
b(u, v) = u1 v2 + v1 u2 − u3 v3
and v = (0, 0, 1), then the vertices of G are
{(x, 1, λ) | x, λ ∈ Fq , λ 6= 0}
0
and (x, 1, λ) is joined to (x , 1, λ0 ) if and only if
0 = b((x, 1, λ), (x0 , 1, λ0 )) = x + x0 − λλ0
which is if N (x + x0 ) = λλ0 . So this graph is the norm graph with t = 2.
Now, consider the construction of the graph G containing no K3,3 in Theorem 13. Let b
be the symmetric bilinear form,
b(u, v) = u1 v1q + uq1 v1 + u2 v3 + u3 v2 − u4 v4 ,
defined on V5 (Fq ) ∼
= Fq2 × F3q . Let v = (0, 0, 0, 1) ∈ Fq2 × F3q and let
S = {(x, xq+1 , 1, 0) | x ∈ Fq2 }.
The vertices of G are u + λv, where u ∈ S and λ ∈ F∗q , so u = (x, xq+1 , 1, λ) for some
x ∈ Fq2 . Two vertices (x, xq+1 , 1, λ) and (x0 , (x0 )q+1 , 1, λ0 ) are joined by an edge if and
only if
0 = x(x0 )q + xq x0 + xq+1 + (x0 )q+1 − λλ0 = N (x + x0 ) − λλ0 .
So, this is again the norm graph. To show that it is the same graph as in Theorem 13
we have to show that the vectors of S are the zeros of an equivalent quadratic form.
There are two non-equivalent quadratic forms on V4 (Fq ), the elliptic and the hyperbolic
forms. The set of zeros of a hyperbolic form intersect every hyperplane in at least q + 1
one-dimensional subspace, whereas S is contained in the zeros of the quadratic form
xq+1
− x2 x3 ,
1
which intersects ker(x2 ) in the one-dimensional subspace h(0, 0, 1)i. Therefore S is contained in the zeros of a quadratic form of elliptic type. In Theorem 13 we used the
quadratic form
x1 x2 + f (x3 , x4 )
whose set of zeros intersects ker(x2 ) in the one-dimensional subspace h(1, 0, 0, 0)i, so is
also of elliptic type.
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12. Graphs containing no K5,5
In this section we shall consider the norm graph with t = 4 in more detail. We shall aim
to prove that this graph, which contains no K4,7 , contains no K5,5 .
Define the following isomorphisms of V9 (Fq3 ),
σ((x1 , . . . , x8 , x9 )) = (x8 , x7 , x6 , x5 , x4 , x3 , x2 , x1 , x9 ),
and for each λ ∈ Fq3 ,
2
τλ ((x1 , . . . , x8 , x9 )) = (x1 , x2 + λx1 , x3 + λq x1 , x4 + λq x1 , x5 + λx3 + λq x2 + λq+1 x1 ,
2
x6 + λx4 + λq x2 + λq
2 +1
2
x1 , x7 + λq x4 + λq x3 + λq
q2
x8 + λx7 + λq x6 + λ x5 + λq+1 x4 + λ
q 2 +1
x3 + λ
q 2 +q
2 +q
x1 ,
q 2 +q+1
x2 + λ
x1 , x9 )
and
2
αλ ((x1 , . . . , x8 , x9 )) = (x1 , λx2 , λq x3 , λq x4 , λq+1 x5 , λq
2 +1
x6 , λq
2 +q
x7 , λq
2 +q+1
x8 , x9 ).
Let
2
u(a) = (1, a, aq , aq , aq+1 , aq
2 +1
, aq
2 +q
, aq
2 +q+1
, 0),
where a ∈ Fq3 and
u(∞) = (0, 0, 0, 0, 0, 0, 0, 1, 0).
Note
σ(u(a)) = aq
2 +q+1
u(a−1 ), σ(u(0)) = u(∞), σ(u(∞)) = u(0),
and
τλ (u(x)) = u(x + λ), τλ (u(∞)) = u(∞),
and
αλ (u(x)) = u(λx), αλ (u(∞)) = u(∞).
Define
S = {u(a) | a ∈ Fq3 ∪ {∞}}.
Lemma 18. If A ⊂ S and |A| = 4 then dimhAi = 4.
Proof. Suppose M is the 4 × 9 matrix whose rows are the vectors in A. We need to
show this matrix has rank 4. Suppose that the first row is u(a), for some a ∈ Fq3 . By
multiplying on the right, the matrix M by the matrix of the isomorphism τ−a and then σ,
we can assume the first row is u(∞). The second row of the matrix is now a multiple of
u(a) for some a ∈ Fq3 . By multiplying on the right by the matrix of the isomorphism τ−a ,
we can assume the second row is a multiple of u(0). The third row of the matrix is now a
multiple of u(a) for some a ∈ Fq3 \ {0}. By multiplying on the right by the matrix of the
isomorphism α1/a , we can assume the third row is u(1). The fourth row of the matrix is
now a multiple of u(a) for some a 6= 0, 1, ∞.
If
λ1 u(∞) + λ2 u(0) + λ3 u(1) + λ4 u(a) = 0,
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SIMEON BALL
then the second coordinate implies λ3 +λ4 a = 0 and the fifth coordinate gives λ3 +λ4 aq+1 =
0, so aq = 1 and a = 1, a contradiction implying that the four rows of the matrix M are
linearly independent.
Lemma 19. If A ⊂ S and |A| = 5 then either dimhAi ≥ 5 or |hAi ∩ S| ≥ q.
Proof. Suppose dimhAi ≤ 4. By Lemma 18, dimhAi = 4. As in the proof of Lemma 18, we
can apply suitable isomorphisms so that the five vectors in A are multiples of u(∞), u(0), u(1), u(a)
and u(b), where a, b 6= 0, 1, ∞ and a 6= b. Since dimhAi = 4 there exist λ1 , λ2 , λ3 , λ4 ∈ Fq3
such that
u(b) = λ1 u(∞) + λ2 u(0) + λ3 u(1) + λ4 u(a).
If λ4 = 0 then the second and fifth coordinates give λ3 = b = bq+1 , which implies b = 0, 1,
which it doesn’t.
If λ3 = 0 then the second and fifth coordinates give λ4 a = b and λ4 aq+1 = bq+1 , which
implies bq = aq and so b = a, which it doesn’t.
If λ3 λ4 6= 0 then the second, third and fourth coordinates give
2
b = λ3 + λ4 a, bq = λ3 + λ4 aq , bq = λ3 + λ4 aq
2
which imply
2
2
b − bq = λ4 (a − aq ), bq − bq = λ4 (aq − aq ),
and so
0 = (λq4 − λ4 )(a − aq ).
If a 6∈ Fq then λ4 ∈ Fq and the second (raised to the power q) and third coordinates give
bq = λq3 + λ4 aq , bq = λ3 + λ4 aq
and so λ3 ∈ Fq . The second and seventh coordinates
b = λ3 + λ4 a, bq
2 +q
= λ 3 + λ 4 aq
2 +q
combine to give
bq
which implies a + a
q 2 +q
2 +q+1
= (λ3 + λ4 a)(λ3 + λ4 aq
2 +q
) ∈ Fq ,
∈ Fq , since λ3 λ4 6= 0. Thus,
a + aq
2 +q
2
= aq + aq+q ,
2
and so (aq − a)(aq − 1) = 0 and so a ∈ Fq , a contradiction.
Hence, a ∈ Fq . For all b ∈ Fq the eight coordinates give just four equations
λ2 + λ3 + λ4 = 0, b = λ3 + λ4 a, b2 = λ3 + λ4 a2 , b3 = λ3 + λ4 a3 + λ1 ,
which have a solution for all b 6= 0, 1, a.
Theorem 20. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
ex(n, K5,5 ) > 21 (1 − )n7/4 .
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Proof. We shall prove that for q ≥ 7, the norm graph which contains no K4,7 , contains no
K5,5 , from which the result follows.
Let b be the non-degenerate symmetric bilinear form on V9 (Fq3 ) defined by
b(u, v) =
8
X
ui v9−i − u9 v9 .
i=1
Let G be the graph whose vertices are
{u(a) + λv | a ∈ Fq3 , λ ∈ F∗q },
where v = (0, 0, 0, 0, 0, 0, 0, 0, 1). A vertex u(a) + λv is joined to the vertex u(a0 ) + λ0 v if
and only if
2
0 = b(u(a) + λv, u(a0 ) + λ0 v = (a + a0 )q +q+1 − λλ0 .
The graph G is the norm graph defined in the Section 11 with t = 4, so by Lemma 16
contains no K4,7 .
Let B be a set of five vertices of G.
If v ∈ hBi then hBi⊥ ⊂ v ⊥ and since the hyperplane v ⊥ contains no vertices of G, the
vertices in B have no common neighbour. Hence we can assume that if u(a) + λv, u(a0 ) +
λ0 v ∈ B, then a 6= a0 .
If v 6∈ hBi then dimhv, Bi ∩ hvi⊥ = dimhBi. By Lemma 19, either dimhBi = 5 or
|hv, Bi ∩ hvi⊥ ∩ S| ≥ q + 1.
If |hv, Bi ∩ hvi⊥ ∩ S| ≥ q + 1 then there are at least q vectors in hBi of the form u(a) + λv
for some a, λ ∈ Fq3 . We want to show that λ ∈ Fq and hence conclude that there are at
least q vertices of G in hBi. We can assume that u(a0 ) + λ0 v, u(a00 ) + λ00 v ∈ hBi⊥ , for some
a0 , a00 ∈ Fq3 , a0 6= a00 , and λ0 , λ00 ∈ F∗q , since otherwise that vertices in B have at most one
common neighbour. Now u(a) + λv ∈ hBi and u(a0 ) + λ0 v ∈ hBi⊥ implies
0 = b(u(a) + λv, u(a0 ) + λ0 v) = (a + a0 )q
2 +q+1
− λλ0 .
Since λ0 ∈ F∗q we have λ ∈ Fq . If λ = 0 then a = −a0 and we can repeat the above
replacing u(a0 ) + λ0 v by u(a00 ) + λ00 v and conclude that λ 6= 0. Hence λ ∈ F∗q and there are
at least q vertices in hBi. By Lemma 16, G contains no K4,7 so there are at most three
vertices in hBi⊥ and so the vertices in B have at most three common neighbours.
If dimhBi = 5 then dimhBi⊥ = 4. Now v 6∈ hBi⊥ and so dimhv, B ⊥ i ∩ hvi⊥ = 4. By
Lemma 19, either |hv, B ⊥ i∩hvi⊥ ∩S| = 4 or |hv, B ⊥ i∩hvi⊥ ∩S| ≥ q +1. In the former case
the vertices in B have at most four common neighbours. In the latter case this implies
there are at least q vectors in hBi⊥ of the form u(a) + λv, where λ ∈ Fq3 . Since there are
at least two vertices in B we can argue as in the previous paragraph and again conclude
that λ ∈ F∗q . Hence there are at least q vertices in hBi⊥ and G contains a K5,q , which it
does not by Lemma 16.
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SIMEON BALL
13. Alternating forms
An alternating or symplectic form is a bilinear form b from Vk (F) × Vk (F) to F with the
property that
b(u, u) = 0,
for all u ∈ Vk (F).
Lemma 21. If b is an alternating bilinear form then
b(u, v) = −b(v, u),
for all u, v ∈ Vk (F).
Proof. Since
0 = b(u + v, u + v) = b(u, u) + b(u, v) + b(v, u) + b(v, v) = b(u, v) + b(v, u).
A subspace U of Vk (F) is called totally isotropic with respect to a bilinear form b if
b(u, v) = 0,
for all u, v ∈ Vk (F).
Lemma 22. A totally isotropic subspace with respect to a non-degenerate alternating form
rom Vk (F) × Vk (F) to F has dimension at most k/2.
Proof. If U is a totally isotropic subspace then U ⊆ U ⊥ . Hence,
dim U ≤ dim U ⊥ .
By Lemma 5,
dim U + dim U ⊥ = k,
so
2 dim U ≤ k.
14. A probabilistic construction of graphs containing no C2t
Let C2t denote the cyclic graph on 2t vertices. Since C2t is bipartite the Erdős-Stone
theorem, Theorem 1, does not tell us anything. In this case we have the following theorem
which we state without proof.
Theorem 23. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
ex(n, C2t ) < (t − 1)(1 + )n1+1/t .
We use a probabilistic construction to obtain a lower bound in the following theorem.
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Theorem 24. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
ex(n, C2t ) > c(1 − )n1+1/(2t−1) ,
where c = t1/(2t−1) 2−2−2(2t−1) .
Proof. Let G be a graph on n vertices where we join two vertices with an edge with
probability p, where p is to be determined.
Let Y be the random variable that counts the number of edges in G. The expected value
of Y is
n
E(Y ) =
p > c0 n2 p,
2
for any constant c0 < 21 , if n is large enough.
Let X be the random variable that counts the number of copies of C2t in G. The expected
value of X is
n
E(X) =
(2t − 1)!p2t < n2t p2t /(2t).
2t
By the linearity of expectation,
E(Y − X) > c0 n2 p − n2t p2t /(2t).
If we put
p = (c0 t)1/(2t−1) n−1+1/(2t−1) .
then
E(Y − X) ≥ c0 pn2 − 12 c0 pn2 = cn1+1/(2t−1) ,
where
c = 12 c0 (c0 t)1/(2t−1) .
For n large enough, we can put c0 = 12 if we replace c by c(1 − ).
Consider the bounds for small t. For t = 2, the graph C4 is the graph K2,2 so Theorem 8
says for all > 0, there exists an n0 , such that for all n ≥ n0 ,
1
(1
2
− )n3/2 < ex(n, C4 ) < 21 (1 + )n3/2 .
which indicates that the upper bound in Theorem 23 is good.
In the next section, we shall concentrate on providing a good lower bound for ex(n, C6 ).
15. Graphs containing no C6
The construction in the following theorem uses alternating bilinear forms, as opposed to
symmetric bilinear forms which we have been using until now.
Theorem 25. For all > 0, there exists an n0 , such that for all n ≥ n0 ,
2−4/3 (1 − )n4/3 < ex(n, C6 ) < 2(1 + )n4/3 .
20
SIMEON BALL
Proof. The upper bound comes form Theorem 23.
To prove the lower bound we shall construct a family of graphs on n = (q 4 − 1)/(q − 1)
vertices with 21 n(q + 1) edges which contains no C6 , for all prime powers q. Then, as in
Theorem 8, we can then use Bombieri’s theorem again.
Let b be a non-degenerate alternating form on V4 (Fq ).
Let G be the bipartite graph whose vertices are the totally isotropic one-dimensional subspaces and the totally isotropic two-dimensional subspaces and where a totally isotropic
one-dimensional subspace hui is joined by an edge to a totally isotropic two-dimensional
subspace U , if hui is contained in U .
Suppose that hui, hvi and hwi are three totally isotropic one-dimensional subspaces which
are part of a six cycle. Then hui and hvi have a common neighbour which implies that
hu, vi is a totally isotropic subsapce, and so
b(u, v) = 0.
Similarly, b(v, w) = 0 and b(u, w) = 0. Since b is alternating,
b(λ1 u + λ2 v + λ3 w, λ1 u + λ2 v + λ3 w) = 0,
for all λ1 , λ2 , λ3 ∈ Fq . Hence hu, v, wi is a totally isotropic subspace. By Lemma 22,
w ∈ hu, vi. Since b(u, v) = 0, the subspace hu, vi is totally isotropic, so u, v, w ∈ hu, vi⊥ .
Hence the vertices, hui, hvi and hwi have a common neighbour and do not form part of
a six cycle. Therefore, G contains no C6 .
Since, b is alternating, every one-dimensional subspace is totally isotropic, so by Lemma 6,
there are (q 4 − 1)/(q − 1) totally isotropic one-dimensional subspaces. For each non-zero
u ∈ V4 (Fq ), the subspace hu, vi is totally isotropic, for all v ∈ hui⊥ . By Lemma 7,
there are (q 2 − 1)/(q − 1) two-dimensional totally isotropic subspaces containing hui,
and so (q 4 − 1)/(q − 1) two-dimensional totally isotropic subspaces in all. Hence, G has
n = 2(q 4 − 1)/(q − 1) vertices and 21 n(q + 1) edges. The lower bound now follows since
1
n(q
2
+ 1) > (1 − )(n/2)4/3
for n large enough.
16. Generalised polygons
An ordinary n-gon in an incidence structure is a sequence of distinct points and lines
x1 , `1 , x2 , `2 , . . . , xn , `n ,
where the points xi , xi+1 are incident with the line `i , for i = 1, . . . , n, indices read modulo
n.
A generalised n-gon is an incidence structure that contains no ordinary r-gons for r ≤ n−1
and where any pair of points, or pair of lines or point-line pair is contained in an ordinary
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n-gon. To avoid degenerate examples, we add the axiom that every point and line is
contained in some ordinary (n + 1)-gon.
Note that the definition of a generalised 3-gon coincides with the previous definition of a
projective plane.
Lemma 26. A finite generalised n-gon has an order (s, t) with the property that every
line is incident with s + 1 points and every point is incident with t + 1 lines.
Proof. We provide proofs only for n = 3 and n = 4.
(n = 3). Let x be a point not incident with a line `. The axioms imply that for all y ∈ `
there is a unique line incident with both x and y. Moreover all the lines incident with
x are incident with some point of `. Hence, the number of lines incident with x is equal
to the number of points incident with `, s + 1 say. Now, varying ` we can conclude that
every line not incident with x contains s + 1 points. Using the non-degeneracy condition
we can find a point not on ` and distinct from x and conclude that the lines incident with
x also contain s + 1 points. We have also shown that there are s + 1 lines incident with
x and hence every point is incident with s + 1 points.
(n = 4) Let x be a point not incident with a line `. The axioms imply that there is a
unique lines incident with x and meeting `. If ` and `0 are skew lines then, since for all
x ∈ `0 there is a unique line joining x to `, the number of points on ` and `0 is the same. If
` and `0 are concurrent then it suffices to find a line skew to them both to conclude that
every line contains the same number of points, and this follows from the non-degeneracy
condition. Interchanging points and lines in this argument shows that every point is
incident with the same number of lines.
The following theorem is the Feit-Higman theorem.
Theorem 27. If s, t ≥ 2 then a finite generalised n-gon exists if and only if n = 3, 4, 6
or 8.
If n = 3 then s = t and the number of points is s(s + 1) + 1 and the number of lines is
also s(s + 1) + 1.
If n = 4 then t1/2 ≤ s ≤ t2 and the number of points is (s + 1)(st + 1) and the number of
lines is (t + 1)(st + 1).
If n = 6 then t1/3 ≤ s ≤ t3 , st is a square and the number of points is (s+1)((st)2 +st+1)
and the number of lines is (t + 1)((st)2 + st + 1).
If n = 8 then t1/2 ≤ s ≤ t2 , 2st is a square and the number of points is (s + 1)((st)3 +
(st)2 + st + 1) and the number of lines is (t + 1)((st)3 + (st)2 + st + 1).
For any prime power q, there is a finite generalised n-gon of order (q, q). We shall be
interested in when these generalised n-gons have polarities and that is stated in the
following lemmas.
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SIMEON BALL
Lemma 28. For all h ∈ N, there is a generalised 4-gon of order (22h+1 , 22h+1 ) which has
a polarity.
Lemma 29. For all h ∈ N, there is a generalised 6-gon of order (32h+1 , 32h+1 ) which has
a polarity.
17. Graphs containing no C2t , for t = 3, 5, 7
We can now extend the following bound in Theorem 25 to graphs containing no C2t , where
t = 3, 5 or 7, thanks to the existence of finite generalised (t + 1)-gons, for these values of
t.
Theorem 30. Let t = 2, 3, 5 or 7.
For all > 0, there exists an n0 , such that for all n ≥ n0 ,
2−(1+(1/t)) (1 − )n1+(1/t) < ex(n, C2t ) < (t − 1)(1 + )n1+(1/t) .
Proof. The upper bound comes from Theorem 23.
To prove the lower bound we construct a family of graphs on n = 2(q t+1 − 1)/(q + 1)
vertices with 12 n(q + 1) edges which contain no C2t .
Let G be the incidence graph of a generalised (t + 1)-gon Γ of order (q, q). In other words,
G is a bipartite graph whose vertices are the points and lines of a generalised (t + 1)-gon
of order (q, q) and where a point x is joined by an edge to a line ` in the graph G, if x is
incident with `. The number of vertices in G follows from Theroem 27 and the number
of edges follows from the fact that each point is incident with q + 1 lines.
By definition, Γ contains no ordinary r-gon for r ≤ t. Hence G contains no r-cycle of
length at most 2t. In particular, G contains no C2t .
In Theorem 8, we managed to prove a better contant in the lower bound than that of
Theorem 30 by using a polarity of the projective plane (generalised 3-gon), instead of
simply taking the incidence graph of the geometry.
This we can do for the polarities mentioned in Lemma 28 and Lemma 29, although we
cannot conclude that the lower bound is always bettered because the powers of two and
the power of three are not dense enough. However, we can conclude the following theorem.
Theorem 31. Let t = 3 or 5.
For all > 0, there exists an n0 , such that
sup ex(n, C2t ) > 12 (1 − )n1+(1/t) .
n≥n0
TURÁN NUMBERS
23
Proof. Let π be a polarity of a generalised (t + 1)-gon Γ of order (q, q). By Lemma 28
and Lemma 29, if t = 3 then q = 22h+1 and if t = 5 then q = 32h+1 , where h ∈ N.
Let G be graph whose vertices are the points of Γ and where x is joined to y by an edge
if and only if x ∈ π(y).
Suppose that
x1 , x2 , . . . , x2t
is a C2t in the graph G. Then
x1 , π(x2 ), x3 , π(x4 ), . . . , x2t−1 , π(x2t )
is an ordinary t-gon in Γ, contradicting the definition of a generalised (t + 1)-gon.
18. References
Theorem 1 comes from
P. Erdős and A. H. Stone, On the structure of linear graphs, Bull. Amer. Math. Soc., 52
(1946) 1087–1091.
Theorem 3 is from
T. Kővári, V. T. Sós and P. Turán, On a problem of K. Zarankiewicz, Colloq. Math., 3
(1954) 50–57.
Theorem 8 is due to
P. Erdős, A. Rényi and V. T. Sós, On a problem of graph theory, Studia, Sci. Math.
Hungar., 1 (1966) 215–235,
and the Bombieri’s theorem, used in the proof is from
E. Bombieri, Le Grand Crible dans la Th?orie Analytique des Nombres, Ast?risque, 18
(Seconde ed.), Paris (1987).
Theorem 10 is due to
Z. Füredi, New asymptotics for bipartite Turán numbers, J. Combin. Theory Ser. A, 75
(1996) 141–144,
and the Huxley-Ivaniec theorem used in the proof is from
M. N. Huxley and H. Iwaniec, Mathematika, 22 (1975) 188-194.
For more on the probabilistic constructions in Theorem 11 and Theorem 24, see
P. Erdős and J. Spencer, Probabilistic Methods in Combinatorics, Academic Press, London, New York, Akadémiai Kiadó, Budapest, 1974.
The lower bound in Theorem 13 is due to
W. G. Brown, On graphs that do not contain a Thomsen graph, Canad. Math. Bull., 9
(1966) 281–289.
24
SIMEON BALL
whereas improvement to the upper bound in Theorem 14 is from
Z. Füredi, An upper bound on Zarankiewicz’ problem, Combin. Probab. Comput., 5
(1996) 29–33.
Lemma 16 and Theorem 17 are from
N. Alon, L. Rónyai and T. Szabó, Norm-graphs: variations and applications, J. Combin.
Theory Ser. B, 76 (1999) 280–290,
although the main algebraic part of the proof is from
J. Kollár, L. Rónyai and T. Szabó, Norm-graphs and bipartite turán numbers, (Combinatorica, 16 (1996) 399–406.
Theorem 20 is from
S. Ball and V. Pepe, Asymptotic improvements to the lower bound of certain bipartite
Turán numbers, Combin. Probab. Comput., 21 (2012) 323–329.
Theorem 23 is from
O. Pikhurko, A note on the Turán function of even cycles, Proc. Amer. Math. Soc., 140
(2012) 3687–3692,
who improved the constant in the main theorem of
J. A. Bondy and M. Simonovits, Cycles of even length in graphs, J. Combin. Theory Ser.
B, 16 (1974) 87–105.
Theorem 25 and Theorem 30 for t = 5 are from
C. T. Benson, Minimal regular graphs of girth eight and twelve, Canad. J. Math, 18
(1966) 1091–1094.
Theorem 27 is from
W. Feit and G. Higman, The non-existence of certain generalized poly- gons, J. Algebra,
1 (1964) 114–131.
For details on Lemma 28, Lemma 29 and Theorem 31, see
F. Lazebnik, V.A. Ustimenko and A.J. Woldar, Polarities and 2k-cycle-free graphs, Discrete Math., 197/198 (1999) 503–513.