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Chapter 23. Upper bounds using inclusion matrices

Chapter 23. Upper bounds using inclusion matrices

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142

23. Upper bounds using inclusion matrices

For a matrix M , the column space is a vector space spanned by its

column vectors. Let colsp M denote the column space of M . Then

(23.1) also shows the following.

Lemma 23.1. Let 0 ≤ j ≤ i ≤ k and F ⊂

colsp M (F,

[n]

j

[n]

k

) ⊂ colsp M (F,

. Then

[n]

i

).

[n]

We say that F ⊂ [n]

k is s-independent if the rows of M (F, s )

are linearly independent, that is, the inclusion matrix has full rowrank. In this case, |F| ≤ ns immediately follows.

Lemma 23.2. Let p be a prime and let f ∈ Q[x] be a polynomial of

degree s. Let F ⊂ [n]

k . Suppose that F satisﬁes

(23.2)

f (|F ∩ F |)

≡ 0 (mod p)

if F = F ,

≡ 0 (mod p)

if F = F

for F, F ∈ F. Then F is s-independent.

Proof. Since f is of degree s, it can be uniquely represented as

s

f (x) =

αi

i=0

x

,

i

where αi ∈ Q. Let Mi = M (F, [n]

i ) be the inclusion matrix. Then

T

the (F, F )-entry of Mi Mi counts the number of i-element subsets

|

. Let

contained in both F and F , that is, |F ∩F

i

s

(23.3)

αi Mi MiT .

A=

i=0

Then the (F, F )-entry of A is f (|F ∩F |), and A (mod p) is a diagonal

matrix with no zero diagonal entries. So A is non-singular because

det A ≡ 0 (mod p), and rank A = |F|. By Lemma 23.1 it follows that

colsp Mi ⊂ colsp Ms for 0 ≤ i ≤ s. This, together with (23.3), yields

colsp A ⊂ colsp Ms , and hence rank A ≤ rank Ms . Thus Ms has full

row-rank, that is, F is s-independent.

Theorem 23.3 (Frankl–Wilson [67]). Let p be a prime and let L ⊂

[0, p − 1] = {0, 1, . . . , p − 1} with |L| = s. Suppose that F ⊂ [n]

k

23.1. Bounds for s-independent families

143

satisﬁes

|F ∩ F |

∈L

(mod p)

if F = F ,

∈L

(mod p)

if F = F

for F, F ∈ F. Then F is s-independent and |F| ≤

n

s

.

Proof. Deﬁne a polynomial f ∈ Q[x] by

(x − l).

f (x) =

l∈L

Then f satisﬁes (23.2) in Lemma 23.2. Thus F is s-independent and

|F| ≤ ns .

Theorem 23.4 (Frankl–Wilson [67]). Let p be a prime and let q = pe

satisﬁes

be a prime power. Suppose that F ⊂ [n]

k

|F ∩ F | ≡ k

for distinct F, F ∈ F. Then F ≤

(mod q)

n

q−1

.

Proof. Choose rational numbers αi (0 ≤ i ≤ q − 1) so that

q−1

αi

i=0

x

i

=

x−k−1

,

q−1

and deﬁne a matrix

q−1

αi Mi MiT ,

A=

i=0

where Mi = M (F,

[n]

q−1

) is the inclusion matrix. Clearly we have

rank A ≤ rank Mi ≤

n

.

q−1

We show that A (mod p) is a diagonal matrix without zero diagonal entries. This would give rank A = |F|, completing the proof.

Recall that the (F, F )-entry of Mi MiT is xi , where x = |F ∩ F |.

=: a.

Thus the (F, F )-entry of A is given by x−k−1

q−1

−1

= (−1)(−2)···(−q+1)

If F = F then a = q−1

= (−1)q−1 . So p

(q−1)!

divides no diagonal entries of A. Next let F = F and we show that a

(mod p) ≡ 0. Since F = F implies x ≡ k (mod q), q does not divide

144

23. Upper bounds using inclusion matrices

x−k−1

= x−k

= (x−k)a

x − k. On the other hand, x−k

shows that

q

q

q

q−1

q divides (x − k)a. Thus p must divide a, as needed.

Exercise 23.5. Let n ≥ k ≥ 2l + 1, and let k − l be a prime power.

satisﬁes |F ∩ F | = l for all distinct F, F ∈

Suppose that F ⊂ [n]

k

n

. (Hint: Let q = k − l and apply

F. Show that |F| ≤ k−l−1

Theorem 23.4.)

We mention that Theorem 23.3 has been extended to non-uniform

families as follows by Babai and Frankl [9].

Theorem 23.6. Let p be a prime and let L ⊂ [0, p − 1] with |L| = s.

Let k be an integer with k ∈ L (mod p) and s + k ≤ n. Suppose that

F ⊂ 2[n] satisﬁes

|F ∩ F |

≡k

(mod p)

if F = F ,

∈L

(mod p)

if F = F

for F, F ∈ F. Then |F| ≤

n

s

.

We present a quick application of Theorem 23.3 due to Gottlieb

[75], which is also known as Kantor’s lemma.

Theorem 23.7. Let 0 ≤ i ≤ j ≤ n. Then it follows that

rank M (

[n]

i

,

[n]

j

) = min{

n

i

,

n

j

}.

Proof. First suppose that i + j ≥ n. Let F = [n]

j . Then, for

F, F ∈ F, |F ∩ F | ≥ 2j − n ≥ j − i. Thus F is an (n, j, L)-system,

where L = {j − i, . . . , j − 1}. Choose a prime p > n arbitrarily. By

Theorem 23.3 it follows that F is s-independent, where s = |L| = i.

[n]

n

Thus rank M ( [n]

i , j ) = i . Next suppose that i + j < n, and let

i = n − i and j = n − j. Note that M (

[n]

j

[n]

i

,

[n]

j

) is the transpose

[n]

i

,

) and i + j > n. So we apply the ﬁrst case and the

of M (

corresponding rank is jn = nj .

23.2. Shadows in a t-intersecting family

The exact lower bound for the size of u-shadows in a k-uniform family is given by the Kruskal–Katona Theorem (Theorem 6.2). But if

the family is t-intersecting, then the bound should be much larger.

23.2. Shadows in a t-intersecting family

145

Katona found a sharp inequality in this case. We will present a proof

of the result as an application of inclusion matrices.

Theorem 23.8 (Katona’s intersecting shadow theorem [81]). Let

1 ≤ t ≤ k ≤ n and let F ⊂ [n]

be t-intersecting. Then, for k − t ≤

k

u ≤ k, we have

(23.4)

|σu (F)|/|F| ≥

2k − t

2k − t

.

/

k

u

To prove the theorem we use the following lemma due to Frankl

and Fă

uredi [57].

Lemma 23.9. Let 0 ≤ s ≤ u ≤ k and let F ⊂

Then

(23.5)

|σu (F)|/|F| ≥

[n]

k

be s-independent.

k+s

k+s

/

.

u

k

Proof. For each x ∈ [n] let Wx = [n] \ {x} and

Fx = {G ∈

Wx

k−1

Claim 23.10. Fx ⊂

: {x} ∪ G ∈ F} = {F \ {x} : x ∈ F ∈ F}.

Wx

k−1

is s-independent, that is,

rank M (Fx ,

Wx

s

) = |Fx |.

We postpone the proof of Claim 23.10, and we ﬁrst prove the

lemma by induction on k assuming Claim 23.10. Inequality (23.5)

trivially holds for the following three cases: s = 0, u = s, and u = k.

So let 1 ≤ s < u < k and assume that (23.5) is true for k − 1. By

Wx

,

Claim 23.10 we can apply the induction hypothesis to Fx ⊂ k−1

and we get

(23.6)

|σu−1 (Fx )| ≥ |Fx |

(k−1)+s

u−1

(k−1)+s

(k−1)

.

By counting #{(x, F ) ∈ [n] × F : x ∈ F } in two ways, that is, by

counting the number of edges in the corresponding bipartite graph

from each side, we have

|Fx | = k |F|.

(23.7)

x∈[n]

146

23. Upper bounds using inclusion matrices

Similarly, by counting #{(x, G) ∈ [n] × σu (F) : x ∈ G}, we have

|σu−1 (Fx )| = u |σu (F)|.

(23.8)

x∈[n]

Using (23.8), (23.6), and (23.7), we get

1

|σu (F)| =

u

=

x∈[n]

k

|F|

u

1

|σu−1 (Fx )| ≥

u

k−1+s

u−1

k−1+s

k−1

= |F|

|Fx |

x∈[n]

k+s

u

k+s

k

k−1+s

u−1

k−1+s

k−1

.

This shows that (23.5) is true for k as well and completes the induction.

Now we prove Claim 23.10. Fix x ∈ [n]. Divide

parts C D, where C = Wsx and D = {{x} T : T ∈

F x = {F ∈ F : x ∈ F } ⊂

We divide the columns of M (F x ,

(23.9)

M (F x ,

[n]

s

[n]

s

[n]

k

[n]

s

Wx

s−1

into two

}. Let

.

) into two blocks:

) = M (F x , C) | M (F x , D) .

Wx

). Then, by Lemma 23.1,

By deﬁnition M (F x , D) = M (F x , s−1

Wx

x

x Wx

colsp M (F , s−1 ) ⊂ colsp M (F , s ), and

(23.10)

colsp M (F x , D) ⊂ colsp M (F x ,

Wx

s

By (23.9) and (23.10) we have colsp M (F x ,

and

(23.11)

rank M (F x ,

[n]

s

) = colsp M (F x , C).

[n]

s

) ⊂ colsp M (F x , C)

) ≤ rank M (F x , C).

The opposite inequality is trivial, so we have equality in (23.11). On

the other hand, since F is s-independent, F x is also s-independent

x

and rank M (F x , [n]

s ) = |F |. Thus equality in (23.11) gives us that

rank M (F x , C) = |F x |.

Finally, noting that |F x | = |Fx | and M (F x , C) = M (Fx , C), we

have rank M (Fx , C) = |Fx |, as needed. This completes the proof

of Claim 23.10 and Lemma 23.5.

23.2. Shadows in a t-intersecting family

147

Proof of Theorem 23.8. Choose a prime p > k and deﬁne a polynomial f ∈ Q[x] by

f (x) = (x − t)(x − t − 1) · · · (x − k + 1).

Let s = k − t and L = {t, t + 1, . . . , k − 1}, and apply Lemma 23.2.

Then F is s-independent. Now the inequality (23.4) follows from

Lemma 23.9.

One can apply the same proof technique to obtain a vector space

version of Theorem 23.8; see [65].

Chapter 24

Some algebraic

constructions for

L-systems

Designs, codes, and ﬁnite geometries are useful sources for constructing large L-systems. In this chapter we present some L-systems constructed by blowing up these structures.

24.1. Algebraic constructions

In this section we construct some L-systems related to Steiner systems

and codes. Here let us brieﬂy recall some basic deﬁnitions concerning

codes. Let V = Fnq be an n-dimensional vector space over the qelement ﬁeld. We say that a subset C ⊂ V is an [n, k, d]q -code if C is

a k-dimensional subspace of V and every x ∈ V has at least d nonzero

coordinates. An (n − k) × n matrix A is called a parity check matrix

of C if C = {x ∈ V : Ax = 0}.

As an illustrative example, let us construct a (7, {0, 1, 3})-system.

Let A be the following 3 × 7 matrix over F2 :

(24.1)

1 0 0 1 1 0 1

A = ⎣0 1 0 1 0 1 1⎦ .

0 0 1 0 1 1 1

149

150

24. Some algebraic constructions for L-systems

This is a parity check matrix of a [7, 4, 3]2 -code. This matrix has the

following properties.

(i) Any two columns are linearly independent over F2 .

(ii) For any two columns cp , cq of A, the subspace spanned by

cp , cq contains precisely three columns cp , cq , and cr of A,

where cr = cp + cq .

Note that (ii) implies that any four columns of A span the entire 3dimensional space F32 . The triples {p, q, r} in (ii) (the indices of three

columns corresponding to 2-dimensional subspaces) form the Steiner

triple system S(2, 3, 7), that is,

{{1, 2, 4}, {1, 3, 5}, {1, 6, 7}, {2, 3, 6}, {2, 5, 7}, {3, 4, 7}, {4, 5, 6}}.

We construct a {0, 1, 3}-system F ⊂ X7 using the matrix A. This

is a 7-partite hypergraph on the vertex set X = V1 · · · V7 , where

each partite set Vi is a distinct copy of Fd2 . So we may understand

F ⊂ (Fd2 )7 . Using A we deﬁne a linear map F : (Fd2 )3 → (Fd2 )7 as

follows. For a, b, c ∈ Fd2 , with an abuse of notation, let F (a, b, c) ∈

(Fd2 )7 be the ordered 7-tuple deﬁned by

F (a, b, c) = (a, b, c)A = (a, b, c, a + b, a + c, b + c, a + b + c).

Then deﬁne

(24.2)

F = {F (a, b, c) : a, b, c ∈ Fd2 }.

A key observation is as follows. Suppose that two members F and

F in F meet on at least two vertices, say they meet on V1 V2 . Then

they necessarily meet on V4 as well. Thus |F ∩ F | ≥ 3. Moreover, if

they meet on at least four vertices, then they necessarily coincide.

Exercise 24.1. Let n = |X| = 7 · 2d . Show that the family F deﬁned

by (24.2) is an (n, 7, {0, 1, 3})-system with |F| = (n/7)3 .

Let c ∈ Fsq and y ∈ (Fdq )s . We write y = (y1 , . . . , ys ), where

s

yj ∈ Fdq for 1 ≤ j ≤ s. Deﬁne c y = j=1 cj yj ∈ Fdq .

Exercise 24.2. Let ci ∈ F32 denote the ith column vector of A deﬁned

by (24.1). Show that if F (a, b, c) and F (a , b , c ) meet on Vi ∼

= Fd2

then ci y = 0, where y = (a − a , b − b , c − c ) ∈ (Fd2 )3 .

24.1. Algebraic constructions

151

Now we generalize the above construction. Let t < b < k. We

say that a (t + 1) × k matrix over Fq is a (t, b, k)q -matrix if it satisﬁes

the following two conditions concerning columns of the matrix:

(P1) Any t columns are linearly independent over Fq .

(P2) For any t columns, the t-dimensional subspace spanned by

these columns contains precisely b columns.

For a (t, b, k)q -matrix, the number of (t+1)×b minor matrices of rank

t is kt / bt . Each such minor matrix gives a b-element subset (block)

as indices of corresponding columns. The set of these blocks forms a

Steiner system S(t, b, k). In this case we say that a (t, b, k)q -matrix

supports S(t, b, k).

Theorem 24.3. If there exists a (t, b, k)q -matrix, then there exists

an (n, k, L)-system of size (n/k)t+1 where L = {0, 1, . . . , t − 1, b}.

Proof. Let A = (ai,j ) be a (t, b, k)q -matrix. We construct a (k, L)system F which is k-partite on the vertex set X = V1 · · · Vk , where

each Vi is a distinct copy of Fdq . To each (t + 1)-tuple (x1 , . . . , xt+1 ),

where xi ∈ Fdq , we assign a k-tuple

F (x1 , . . . , xt+1 ) ∈ V1 × · · · × Vk

by

t+1

F (x1 , . . . , xt+1 ) = (x1 , . . . , xt+1 )A =

t+1

ai1 xi , . . . ,

i=1

aik xi .

i=1

We understand this k-tuple as an edge of F; that is, we deﬁne

F = {F (x1 , . . . , xt+1 ) : (x1 , . . . , xt+1 ) ∈ (Fdq )t+1 }.

It is clear from the construction that F is k-partite and k-uniform.

We need to verify that F is an L-system. Let F = F (x1 , . . . , xt+1 )

and F = F (x1 , . . . , xt+1 ) be distinct edges of F. Let l = |F ∩ F |

be such that F and F meet on i∈I Vi . Let

and let I ∈ [k]

l

y = (x1 − x1 , . . . , xt+1 − xt+1 )

and let ci denote the ith column of A. Then F − F = yA and the

following four statements are equivalent: (i) i ∈ I; (ii) F and F meet

on Vi ; (iii) the ith entry of F − F is 0; (iv) ci y = 0.

152

24. Some algebraic constructions for L-systems

We want to show that l ∈ L. If l < t there is nothing to show.

First suppose that t ≤ l ≤ b. Let T ∈ It . Then ci y = 0 for i ∈ T .

such that for every j ∈ B

Moreover, by (P2), T determines B ∈ [k]

b

we can write cj = i∈T γij ci , where γij ∈ Fq . Thus for j ∈ B we

have

cj

y=

γij ci

i∈T

So F and F meet on

i∈B

y=

γij (ci

y) = 0.

i∈T

Vi , and l ≥ b. Thus l = b follows.

Next suppose that l ≥ b + 1. Again, by (P2), we can choose t + 1

columns from {ci : i ∈ I} so that the corresponding (t + 1) × (t + 1)

minor matrix C of A is non-singular. Then we have yC = 0, which

implies y = 0, i.e., F = F . This contradicts the assumption that F

and F are distinct. Thus we have l ∈ L.

Finally, this (k, L)-system F has n = k · q d vertices and |F| =

(q )

= (n/k)t+1 edges as required.

d t+1

Lemma 24.4. Let t + b < k. Every (t, b, k)q -matrix is a parity check

matrix of a [k, k − (t + 1), t + 1]q -code.

Proof. Let M be a (t, b, k)q -matrix and let C = {x ∈ Fkq : M x = 0}.

It follows from (P1), (P2), and t + k < b that M has full rank t + 1.

Thus dim C = k − (t + 1). We show that every x ∈ C has at least

t + 1 nonzero coordinates. Suppose, to the contrary, that there are

such that xi = 0 for all i ∈ T . Then

some x ∈ C and T ∈ [k]

t

x

c

=

0,

where

c

denotes

the ith column vector of M . This

i

i

i

i∈T

contradicts (P1). Therefore C is a [k, k − (t + 1), t + 1]q -code and M

is its parity check matrix.

To ﬁnd a (t, b, k)q -matrix it is tempting to look at a parity check

matrix of a corresponding code. In fact there are some successful

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