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 sindependent 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 sindependent.
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 ielement 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 nonsingular 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
rowrank, that is, F is sindependent.
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 sindependent families
143
satisﬁes
F ∩ F 
∈L
(mod p)
if F = F ,
∈L
(mod p)
if F = F
for F, F ∈ F. Then F is sindependent 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 sindependent 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 nonuniform
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 sindependent, 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 tintersecting family
The exact lower bound for the size of ushadows in a kuniform family is given by the Kruskal–Katona Theorem (Theorem 6.2). But if
the family is tintersecting, then the bound should be much larger.
23.2. Shadows in a tintersecting 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 tintersecting. 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 sindependent.
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 sindependent, 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 sindependent, F x is also sindependent
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 tintersecting 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 sindependent. 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
Lsystems
Designs, codes, and ﬁnite geometries are useful sources for constructing large Lsystems. In this chapter we present some Lsystems constructed by blowing up these structures.
24.1. Algebraic constructions
In this section we construct some Lsystems related to Steiner systems
and codes. Here let us brieﬂy recall some basic deﬁnitions concerning
codes. Let V = Fnq be an ndimensional vector space over the qelement ﬁeld. We say that a subset C ⊂ V is an [n, k, d]q code if C is
a kdimensional 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 Lsystems
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 2dimensional 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 7partite 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 7tuple 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 tdimensional 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 belement 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 kpartite 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 ktuple
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 ktuple 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 kpartite and kuniform.
We need to verify that F is an Lsystem. 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 Lsystems
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 nonsingular. 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