Tải bản đầy đủ - 0 (trang)
Chapter 25. Oddtown and eventown problems

# Chapter 25. Oddtown and eventown problems

Tải bản đầy đủ - 0trang

156

25. Oddtown and eventown problems

Theorem 25.1. If F ⊂ 2[n] satisﬁes (E1) and (E2), then |F| ≤

2 n/2 .

For the proof we brieﬂy recall some basic facts from linear algebra.

For x, y ∈ Fn2 let x · y denote the standard inner product, and for a

vector space V ⊂ Fn2 let V ⊥ = {x ∈ V : v · x = 0 for all v ∈ V }.

Then it follows that

(25.1)

dim V + dim V ⊥ = n.

Proof of Theorem 25.1. Let F = {F1 , . . . , Fm } and let xi ∈ Fn2

be the characteristic vector of Fi . From (E1) and (E2) it follows that

xi · xj = 0 for all 1 ≤ i ≤ j ≤ m. Let V be the vector space over

F2 spanned by x1 , . . . , xm . Let r = dim V . We may assume that

x1 , . . . , xr is the basis of V . Then for every v ∈ V we can write

v = ri=1 αi xi , where αi ∈ F2 . Since v · xj = ri=1 αi xi · xj = 0

we have xj ∈ V ⊥ and V ⊂ V ⊥ . Thus we get r = dim V ≤ dim V ⊥ .

This, together with (25.1), gives us r ≤ n/2 and m ≤ |V | ≤ 2r ≤

2 n/2 .

Oddtown rules are the following:

(O1) |F | is odd for all F ∈ F, and

(O2) the same rule as (E2).

Theorem 25.2. If F ⊂ 2[n] satisﬁes (O1) and (O2), then |F| ≤ n.

Proof. Let F = {F1 , . . . , Fm } and let xi ∈ Fn2 be the characteristic vector of Fi . Then we have xi · xj = δij . Now we claim that

x1 , . . . , xm are linearly independent over F2 . Consider a linear combination

(25.2)

λ1 x1 + · · · + λm xm = 0.

By taking the inner product of each side of (25.2) with xi we get

xi · λi xi = xi · 0, that is, λi = 0. This holds for every i, and all the

coeﬃcients in the LHS of (25.2) vanish. Thus these m vectors in Fn2

are linearly independent, and m ≤ dim Fn2 = n follows.

These eventown/oddtown theorems were proved independently

by Berlekamp [10] and Graver [76]; see also Miniature 3 of [92].

25.2. Uniform eventown problems

157

25.2. Uniform eventown problems

Let n, k, and p be positive integers with n ≥ k and let L ⊂ Z/pZ. We

say that F ⊂ [n]

is an [n, k, L]p -system if |F ∩ F | ∈ L (mod p) for

k

all distinct F, F ∈ F. Let mp [n, k, L] denote the maximum size of

[n, k, L]p -systems. In this section we consider the case where |L| = 1.

By Theorem 23.3, k ≡ r (mod p) implies mp [n, k, {r}] ≤ n.

When p = 2 this result is a uniform hypergraph version of the oddtown theorem. If we replace the condition k ≡ r (mod p) with k ≡ r

(mod p), where p is not necessarily a prime, then the situation changes

completely. We may call the problem of ﬁnding m2 [n, k, {0}] a uniform eventown problem. More generally we show the following.

Theorem 25.3 ([63]). Let n, k, p, r be integers with 0 ≤ r < p and

p|k. If n > n0 (k), then

n/p

k/p

mp [n + r, k + r, {r}] =

.

Exercise 25.4. Deduce Theorem 25.3 for the case r = 0 and p|n

from Theorem 16.1. (Hint: In this case we can write n = pn and

k = pk , and we have

mp [n, k, {0}] = m(n, k, {0, p, 2p, . . . , (k − 1)p})

k −1

i=0

n − pi

=

k − pi

k −1

i=0

n −i

=

k −i

n

k

=

n/p

.

k/p

Recall that m(n, k, L) is deﬁned in Chapter 16.)

Proof of Theorem 25.3. Let n = pa + q and k = pb (0 ≤ q < p).

First we show the lower bound by construction. For the case

r = 0 let [n − q] = A1 A2 · · · Aa be a partition, where |Ai | = p

(1 ≤ i ≤ a), and deﬁne

Ai : I ∈

Fn,k =

[a]

b

.

i∈I

This is an [n, k, {0}]p -system of size

case r > 0 deﬁne

a

b

=

n/p

k/p

= Θ(nb ). For the

Gn+r,k+r = {F ∪ [n + 1, n + r] : F ∈ Fn,k }.

158

25. Oddtown and eventown problems

This is an [n + r, k + r, {r}]p -system of the same size as Fn,k . Consequently we have shown that

mp [n + r, k + r, {r}] ≥ |Fn,k | =

n/p

k/p

= Θ(nb )

for all 0 ≤ r < p.

Next we show the upper bound. We claim that it suﬃces to prove

the bound for the case r = 0:

(25.3)

|Fn,k | ≥ mp [n, k, {0}].

To see this let r > 0 and suppose that F is an [n+r, k+r, {r}]p -system

with |F| = mp [n + r, k + r, {r}]. Then the size of (distinct) pairwise

intersections is one of r, p + r, . . . , (b − 1)p + r. By Theorem 21.1

b

we have |F| ≤ n+r

b . Since |F| ≥ |Gn+r,k+r | we have |F| = Θ(n ).

Then, by (ii) of Theorem 18.1, there is an r-element set R which is

contained in all F ∈ F. In this case F = {F \ R : F ∈ F} is an

[n, k, {0}]p -system, and we have

(25.4)

mp [n, k, {0}] ≥ |F | = |F| = mp [n + r, k + r, {r}]

≥ |Gn+r,k+r | = |Fn,k |.

If (25.3) holds, then we have equalities everywhere in (25.4), and

mp [n, k, {0}] = mp [n + r, k + r, {r}] = |Fn,k |.

This would complete the proof of the theorem. Thus all we need is

to prove (25.3).

Now let F be one of the largest [n, k, {0}]p -systems, and we prove

(25.3) by showing that |F| ≤ |Fn,k |. We say that A ⊂ [n] with |A| ≥ p

is an atom of F if A ⊂ F or A ∩ F = ∅ for any F ∈ F and moreover

A is inclusion maximal with this property. Notice that atoms are

be the set of atoms

pairwise disjoint. Let A = {A1 , . . . , At } ⊂ [n]

p

of size p, and let X1 ⊂ [n] be the union of all atoms in A. Thus

X1 = A1 · · · At is a partition. Then t ≤ a because n = pa + q

and q < p. Let X0 = [n] \ X1 , FX1 = {F ∈ F : F ⊂ X1 }, and

FX0 = F \ FX1 . It follows from the deﬁnition of X1 that for every

F ∈ F, p divides |F ∩ X1 |, and hence p divides |F ∩ X0 | as well.

25.2. Uniform eventown problems

159

If FX0 = ∅, then |F| ≤ bt ≤ ab = |Fn,k |, and we are done. We

will show that FX0 = ∅ indeed holds. For each x ∈ X0 let

F(x) = {F \ {x} : x ∈ F ∈ F}.

This is an (n − 1, k − 1, L )-system, where

L = {p − 1, 2p − 2, . . . , (b − 1)p − 1}.

So |F(x)| = O(nb−1 ) follows from Theorem 18.1, but we can improve

the bound as follows.

Claim 25.5. |F(x)| = O(nb−2 ) for x ∈ X0 .

Proof. Let x ∈ X0 . First suppose that there is an atom A ⊂ X0

with x ∈ A. Let c = |A|. Since all atoms of size p are in X1 , we have

c > p. Let F(A) = {F \ A : A ⊂ F ∈ F}. Then |F(x)| = |F(A)|

and F(A) is an (n − c, k − c, {2p − c, 3p − c, . . . , (b − 1)p − c})-system.

Thus, by Theorem 18.1, the size is O(nb−2 ).

Next suppose that there is no atom containing x. Let Ix =

G∈F (x) G. If |Ix | < p − 1 = min L , then, by (ii) of Theorem 18.1,

we have |F(x)| = O(n|L |−1 ) = O(nb−2 ). Let |Ix | ≥ p − 1, and choose

Y ⊂ Ix with |Y | = p−1. Since Y {x} is a p-element subset in X0 , it is

not an atom. So there is an F1 ∈ F such that 0 < |F1 ∩(Y {x})| < p.

By deﬁnition Y ⊂ G for all G ∈ F(x), and x ∈ F1 implies Y ⊂ F1 .

In this case |F1 ∩ (Y {x})| = p, a contradiction. Thus we have

x ∈ F1 and F1 ∩ Y = ∅.

(25.5)

Fix Y and F1 . For z ∈ F1 \ Y deﬁne

G(z) = {G \ (Y

{z}) : Y

{z} ⊂ G ∈ F(x)}.

Then G(z) is an (n − p − 1, k − p − 1, {p − 1, 2p − 1, . . . , (b − 2)p − 1})system, and |G(z)| = O(nb−2 ) follows from Theorem 18.1.

Now suppose that for every G ∈ F(x),

(F1 \ Y ) ∩ G = ∅.

(25.6)

Then for each G there is a z ∈ F1 \ Y such that G \ (Y

Thus we get

{z}) ∈ G(z).

|G(z)| = |F1 \ Y | O(nb−2 ) = O(nb−2 ),

|F(x)| ≤

z∈F1 \Y

160

25. Oddtown and eventown problems

completing the proof provided (25.6) is true.

To prove (25.6) let G ∈ F(x). Since both F1 and F2 := G {x} are

in F, p divides |F1 ∩F2 |. Moreover, |F1 ∩F2 | = |F1 ∩G| ≥ |F1 ∩Y | > 0

by (25.5), so |F1 ∩ G| ≥ p. Since Y ⊂ Ix ⊂ G and |Y | = p − 1 we

have (F1 \ Y ) ∩ G = ∅.

By the claim we have

|FX0 | ≤

|F(x)| = O(|X0 |nb−2 ).

x∈X0

This, together with |FX1 | ≤ bt , implies that

(25.7)

Θ(nb ) = ab = |Fn,k | ≤ |F| = |FX0 | + |FX1 | ≤ O(|X0 |nb−2 ) +

Note that |X0 | = n − |X1 | = n − pt = O(n) and |X0 |nb−2 = O(n

If |X0 | = Ω(n), say n − pt = n for some

t = (1 − )n/p

and

a

b

t

b

t

b .

b−1

).

> 0, then

(n − q)/p = a,

If |X0 | = o(n), then it follows from (25.7) that

(25.8)

a

b

≤ o(nb−1 ) +

t

b

.

Compare the coeﬃcients of n

in both sides. Since t ≤ a = (n −

q)/p, (25.8) is possible only when t = a. Consequently |X1 | = pa and

|X0 | = n − pa = q < p. Thus |F ∩ X0 | = 0 for every F ∈ F because

p divides |F ∩ X0 |. This means that FX0 = ∅.

b−1

Chapter 26

Tensor product method

Suppose that we want to bound the size of a family F of subsets. If

we ﬁnd a map from F to a vector space V such that all members of

F are sent to linearly independent vectors, then we get |F| ≤ dim V .

We have seen such examples in earlier chapters. In this chapter we

repeat the same procedure, but this time V is not only a vector space

but also a tensor space which has richer structure and may reﬂect

some delicate combinatorial structures in F.

In the ﬁrst section we introduce some basic concepts and properties concerning tensor spaces. The remaining sections focus on applications to problems for two families with intersection constraints. We

mention that [9] and [92] also deal with similar topics and are highly

recommended. We also mention that Kalai used tensor products to

deﬁne algebraic shifting. We do not cover this topic here, but it is an

important tool for studying simplicial complexes; see, for example,

[80].

In this chapter, by a vector space we mean a vector space over R

for simplicity. See [9] for replacing the base ﬁeld R with a ﬁnite ﬁeld

of reasonably large characteristic.

161

162

26. Tensor product method

26.1. Some basic facts about tensor spaces

In this section we gather some basic properties of tensor products and

construct two structures called the symmetric algebra and the exterior

algebra, which are useful for our combinatorial purposes (and useful

in many other branches of mathematics, of course). We will need

only a small part of the mechanism, and we may introduce the facts

with proofs. In fact, the last miniature in the book [92] is written

in such a minimalistic way, and it is one nice way to prepare the

ingredients we need. On the other hand, the essence of considering

tensors is perhaps in an abstract conceptual framework rather than

in a concrete setting. So in this section we will show a moderate

“whole picture” of the mechanism without proofs. It is more than

we will need for our applications, but it will be helpful to understand

the motivations for using tensors. Most of the omitted proofs are

routine and one can ﬁnd them in any standard textbook dealing with

multilinear algebra, such as [113].

26.1.1. Multilinear map and tensor product. Let V1 , V2 , and

W be vector spaces. We say that a map

Φ : V1 × V2 → W

is bilinear if it is linear with respect to each variable, that is,

Φ(αv1 + βv1 , v2 ) = αΦ(v1 , v2 ) + βΦ(v1 , v2 ),

Φ(v1 , αv2 + βv2 ) = αΦ(v1 , v2 ) + βΦ(v1 , v2 )

hold for all v1 , v1 ∈ V1 , v2 , v2 ∈ V2 , and α, β ∈ R.

First we deﬁne a tensor product of V1 and V2 , which is based on

the following fact.

Theorem 26.1. Let V1 and V2 be vector spaces. Then there exist a

vector space W and a bilinear map Φ : V1 × V2 → W which satisfy

the following two conditions.

(i) W is spanned by Φ(V1 × V2 ) = {Φ(v1 , v2 ) : v1 ∈ V1 , v2 ∈

V2 }.

(ii) For every vector space U and every bilinear map

F : V1 × V2 → U

26.1. Some basic facts about tensor spaces

163

there exists a unique linear map f : W → U such that

F = f ◦ Φ.

We write V1 ⊗ V2 for W and call it a tensor product of V1 and V2 ,

and we call an element of W a tensor. Note that Φ(V1 ×V2 ) = V1 ⊗V2

in general. We also write v1 ⊗ v2 for Φ(v1 , v2 ), where v1 ∈ V1 and

v2 ∈ V2 . Then the condition (i) means that every tensor w ∈ W can

be represented in the form

v1 ⊗ v2 .

w=

v1 ∈V1 , v2 ∈V2

We remark that this representation is not necessarily unique in general. Let e1 , . . . , em (resp. f1 , . . . , fn ) be a basis for V1 (resp. V2 ).

Then

ei ⊗ fj

(1 ≤ i ≤ m, 1 ≤ j ≤ n)

is a basis for V1 ⊗ V2 . In particular,

dim V1 · dim V2 = dim V1 ⊗ V2 .

Using the basis, every w ∈ W can be uniquely written as

αi,j ei ⊗ fj ,

w=

i,j

where αi,j ∈ R and i and j run over 1 ≤ i ≤ m and 1 ≤ j ≤ n. The

condition (ii) is called universality of a tensor product.

We deﬁned a tensor product of two vector spaces, and we can

easily extend the concept to any ﬁnite number of vector spaces. So

let V1 , . . . , Vr and W be vector spaces. We say that a map

Φ : V1 × · · · × Vr → W

is r-linear if it is linear with respect to each variable. Then we have

the following.

Theorem 26.2. Let V1 × · · · × Vr be vector spaces. Then there exist

a vector space W and an r-linear map Φ : V1 × · · · × Vr → W which

satisfy the following two conditions.

(i) W is spanned by Φ(V1 × · · · × Vr ).

164

26. Tensor product method

(ii) For every vector space U and every r-linear map

F : V1 × · · · × Vr → U

there exists a unique linear map f : W → U such that

F = f ◦ Φ.

We write V1 ⊗ · · · ⊗ Vr for W and v1 ⊗ · · · ⊗ vr for Φ(v1 , . . . , vr ),

where vi ∈ Vi (1 ≤ i ≤ r).

26.1.2. Symmetric tensors and alternating tensors. Now we

consider the case where all Vi are the same, that is, V1 = · · · = Vr =

V . In this case let

T r (V ) = V ⊗ · · · ⊗ V

(r times),

and we sometimes write T r for T r (V ) for simplicity. We will deﬁne two important subspaces S r (V ) (symmetric tensors) and Ar (V )

(alternating tensors) of T r (V ). Let σ ∈ Sr , where Sr denotes the

symmetric group of order r, and apply Theorem 26.2 with U = T r

and Φ deﬁned by

Φ(v1 , . . . , vr ) = vσ(1) ⊗ · · · ⊗ vσ(r) .

Then we get a unique isomorphism fσ : T r → T r such that

(26.1)

fσ (v1 ⊗ · · · ⊗ vr ) = vσ(1) ⊗ · · · ⊗ vσ(r) .

We say that a tensor t ∈ T r is symmetric if

fσ (t) = t

for all σ ∈ Sr , and alternating if

fσ (t) = sgn(σ)t

for all σ ∈ Sr , where sgn(σ) is 1 if σ is an even permutation and −1

if σ is an odd permutation. Finally we deﬁne

S r (V ) = {t ∈ T r (V ) : t is symmetric},

Ar (V ) = {t ∈ T r (V ) : t is alternating}.

Then it is easy to verify that both S r (V ) and Ar (V ) are subspaces of

T r (V ). The dimensions of these subspaces are given as follows.

26.1. Some basic facts about tensor spaces

165

Theorem 26.3. Let n = dim V . Then we have

r+n−1

,

dim S r (V ) =

r

n

dim Ar (V ) =

.

r

In particular, Ar (V ) = {0} for r > n.

To explain the reason let us ﬁnd a basis for each subspace. We introduce two operators S r (symmetrizer) and Ar (alternatizer) deﬁned

by

1

S r (t) =

fσ (t),

r!

σ∈Sr

1

Ar (t) =

r!

sgn(σ)fσ (t),

σ∈Sr

where fσ satisﬁes (26.1). Note that each of them is a linear map on

T r . Then one can show that S r ◦ S r = S r and Ar ◦ Ar = Ar , from

which it follows that

S r (V ) = {S r (t) : t ∈ T r (V )},

Ar (V ) = {Ar (t) : t ∈ T r (V )}.

In other words, S r (resp. Ar ) is a projection from T r (V ) to S r (V )

(resp. Ar (V )). Now it is not too diﬃcult to show the following.

Claim 26.4. Let n = dim V and let e1 , . . . , en be a basis for V . Then

a basis for S r (V ) is given by

S r (ei1 ⊗ · · · ⊗ eir ),

1 ≤ i1 ≤ i2 ≤ · · · ≤ in ≤ r,

and a basis for Ar (V ) is given by

Ar (ei1 ⊗ · · · ⊗ eir ),

1 ≤ i1 < i2 < · · · < in ≤ r.

The dimension formulas immediately follow from the above claim.

Let V and U be vector spaces. We say that an r-linear map

F : V × ···× V → U

is symmetric if

F (v1 , . . . , vr ) = F (vσ(1) , . . . , vσ(r) )

166

26. Tensor product method

for all σ ∈ Sr , and alternating if

F (v1 , . . . , vr ) = sgn(σ) F (vσ(1) , . . . , vσ(r) )

for all σ ∈ Sr . Then we have the following universality.

Theorem 26.5. Let V and U be vector spaces and let F : V × · · · ×

V → U be an r-linear map. If F is symmetric, then there exists a

unique linear map f : S r (V ) → U such that

F = f ◦ S r ◦ Φ,

where Φ is deﬁned in Theorem 26.2. If F is alternating, then there

exists a unique linear map g : Ar (V ) → U such that

F = g ◦ Ar ◦ Φ.

Exercise 26.6. Let F be an alternating r-linear map.

(i) Show that F (. . . , vi , . . . , vj , . . .) = −F (. . . , vj , . . . , vi , . . .).

(ii) Let span{u1 , . . . , ur } = span{v1 , . . . , vr }. Show that there

is a real number λ = 0 such that

F (u1 , . . . , ur ) = λF (v1 , . . . , vr ).

26.1.3. Symmetric algebra and exterior algebra. Let S 0 (V ) =

R and let

S(V ) = S 0 (V ) ⊕ S 1 (V ) ⊕ S 2 (V ) ⊕ · · · .

The RHS is a set of formal sums v0 + v1 + · · · , where vi ∈ S i (V ) for

i = 0, 1, . . . , and we assume that the number of i such that vi = 0 is

ﬁnite. As usual, for v, v ∈ S(V ), where v =

vi and v =

vi , we

deﬁne v + v = (vi + vi ), and for α ∈ R we deﬁne αv = (αvi ).

Then S(V ) is a (possibly inﬁnite-dimensional) vector space.

Let us deﬁne a product in S(V ). For s ∈ S i (V ) and s ∈ S j (V )

we deﬁne the product by

s · s = S i+j (s ⊗ s ) ∈ S i+j (V ).

We extend the deﬁnition, and for s, s ∈ S(V ), where s =

s = j sj , we deﬁne

s·s =

si · sj .

i,j

i si

and

### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

Chapter 25. Oddtown and eventown problems

Tải bản đầy đủ ngay(0 tr)

×