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Chapter 21. Upper bounds using multilinear polynomials

Chapter 21. Upper bounds using multilinear polynomials

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128



21. Upper bounds using multilinear polynomials



For I ⊂ [n] let xI =

for J ⊂ [n], it follows that



i∈I



xi ∈ RΩ ; in particular, x∅ = 1. Then,

1 if I ⊂ J,



xI (J) =



(21.1)



0 otherwise.



Lemma 21.2. Let f ∈ RΩ be such that f (I) = 0 for all I ⊂ [n] with

|I| ≤ r. Then the set {xI f : |I| ≤ r} ⊂ RΩ is linearly independent.

Proof. Suppose that there is a non-trivial linear combination

λI xI f = 0.

|I|≤r



Let I0 be an inclusion-minimal subset such that λI0 = 0. Substitute I0 into the above linear combination:

|I|≤r λI xI (I0 )f (I0 ) =

0. If I

I0 , then λI = 0 by the minimality. If I ⊂ I0 , then

xI (I0 ) = 0 by (21.1). Consequently we get λI0 xI0 (I0 )f (I0 ) = 0.

Since xI0 (I0 )f (I0 ) = 0 we have λI0 = 0, a contradiction.

Let F = {F1 , . . . , Fm } be an (n, k, L)-system. For x, y ∈ Rn let

x · y = ni=1 xi yi denote the standard inner product. For 1 ≤ i ≤ m

let vi ∈ Rn be the characteristic vector of Fi , and define fi ∈ RΩ by

fi (x) =



(vi · x − l).

l∈L



Since vi · vj = |Fi ∩ Fj |,

= 0 if i = j,



fi (vj )



(21.2)



= 0 otherwise.



This shows that the fi are linearly independent. We claim that we can

n

add s−1

i=0 i more functions while keeping the linear independency.

Claim 21.3. Let g =

(21.3)



n

j=1



xj − k ∈ RΩ . The set of functions



{fi : 1 ≤ i ≤ m} ∪ {xI g : I ⊂ [n], |I| ≤ s − 1}



in RΩ is linearly independent.

Proof. Consider a linear combination

m



(21.4)



λi fi +

i=1



μI xI g = 0.

|I|≤s−1



21.1. Ray-Chaudhuri–Wilson Theorem



129



n



Notice that the function g = j=1 xj − k vanishes on k-element sets.

So, for each j, substituting vj into (21.4) gives λj = 0 by (21.2). Thus

only the second term in (21.4) remains, that is,

μI xI g = 0. Since

s ≤ k and g vanishes only on a k-element set, it follows that g(I) = 0

for all I with |I| ≤ s − 1. Thus, by Lemma 21.2, the functions xI g

(|I| ≤ s − 1) are linearly independent, and therefore μI = 0 for all

I.

n

In (21.3) there are m+ s−1

i=0 i linearly independent polynomials

of degree at most s. On the other hand, si=0 ni is the dimension

of the space of polynomials in n variables of degree at most s. Thus

s

n

n

n

it follows that m + s−1

i=0 i ≤

i=0 i , that is, m ≤ s . This

completes the proof of Theorem 21.1.



Frankl and Wilson proved the modular version of Theorem 21.1.

We write a ∈ L (mod p) if a ≡ l (mod p) for some l ∈ L.

Theorem 21.4 (Frankl–Wilson [67]). Let n > k ≥ s be positive

integers, and let p be a prime. Let L ⊂ [0, p − 1] = {0, 1, . . . , p − 1}

satisfies the following

be a set of s integers. Suppose that F ⊂ [n]

k

conditions:

(i) k ∈ L (mod p).

(ii) If F, F ∈ F with F = F , then |F ∩ F | ∈ L (mod p).

Then |F| ≤



n

s



.



The condition that p is a prime cannot be dropped in general. We

will prove Theorem 21.4 in a slightly stronger form in Chapter 23; see

Theorem 23.3.

Exercise 21.5. Let G := {G ∈ [m]

11 : {1, 2, 3} ⊂ G}. Then we have

|G ∩ G | ∈ [3, 10] for distinct G, G ∈ G. Let F := { G

2 : G ∈ G} be

11

a k-uniform family on n := m

vertices,

where

k

=

2

2 = 55 ≡ 1

(mod 6). Verify that

|F ∩ F | ∈ {



i

2



: 3 ≤ i ≤ 10} ≡ {0, 3, 4} (mod 6)



for distinct F, F ∈ F, and that |F| = |G| =

that Theorem 21.4 fails if p = 6 and s = 3.



m−3

8



= Θ(n4 ). Show



130



21. Upper bounds using multilinear polynomials



Exercise 21.6. Prove Theorem 21.4 for the case s = 1 without

assuming that p is a prime. (Hint: Go through the proof of Theorem 21.1 using Z/pZ in place of R, where p is not necessarily a

prime.)

Exercise 21.7. Suppose that F satisfies the assumptions in Theorem 21.4 and, moreover, that

(iii) k ∈ [0, s − 1] (mod p).

Then the proof of Theorem 21.1 still works almost verbatim. Examine

the proof in this setting (using Fp in place of R) and show that |F| ≤

n

n

Ω

j=1 xj − k ∈ Fp satisfies

s . (Hint: It follows from (iii) that g =

g(I) = 0 for |I| ≤ s − 1. So we can apply Lemma 21.2.)

By taking p > n in the above exercise we can deduce Theorem 21.1.



21.2. Deza–Frankl–Singhi Theorem

Let p be a prime and let L ⊂ Fp . We say that a family of subsets H =

{C1 , . . . , Cm } ⊂ 2[n] is (p, L)-intersecting if it satisfies the following

two conditions:

• |Ci ∩ Cj | ∈ L (mod p) for 1 ≤ i < j ≤ m.

• |Ci | ∈ L (mod p) for 1 ≤ i ≤ m.

Theorem 21.8 (Deza–Frankl–Singhi Theorem [24]). Let s = |L|. If

a family H ⊂ 2[n] is (p, L)-intersecting, then

|H| ≤



n

n

n

+

+ ···+

.

0

1

s



For the proof we need some preparation. As usual we identify 2[n]

and {0, 1}n by the correspondence between a subset and its characteristic vector. A polynomial is said to be multilinear if it has degree

at most 1 in each variable. Let f be a polynomial in n variables of

degree s over Fp . Then there exists a unique multilinear polynomial

f¯ of degree at most s such that

f (x) = f¯(x)



21.2. Deza–Frankl–Singhi Theorem



131



for all x ∈ {0, 1}n . In fact, since x2 = x for x ∈ {0, 1} we get f¯

just by replacing monomials in f with the corresponding products of

distinct variables. For example, if

f (x, y) = x4 + 2x3 y 2 + 3xy 5 ,

then we have

f¯(x, y) = x + 5xy.

The set of multilinear polynomials in variables x1 , . . . , xn of degree s constitutes a vector space V over Fp :

xi : |I| ≤ s, I ⊂ [n] .



V = span

i∈I



We have

s



n

.

i



dim V =

i=0



For example, if n = 3 and s = 2, then

V = span{1, x1 , x2 , x3 , x1 x2 , x1 x3 , x2 x3 },

dim V =



3

3

3

+

+

0

1

2



= 7.



Proof of Theorem 21.8. Let X = {0, 1}n . To each subset Ci (1 ≤

i ≤ m) we assign a polynomial fi : X → Fp by

fi (x) :=



(x · vi − l),

l∈L



where vi ∈ X denotes the characteristic vector of Ci . Then, by the

(p, L)-intersecting property, we have

fi (vj )



=0



if i = j,



=0



if i = j.



Thus f1 , . . . , fm ∈ FX

p are independent over Fp . Here is the plan for

the proof. If we can find a vector space V ⊂ FX

p satisfying

• span{f1 , . . . , fm } ⊂ V , and

• dim V =



s

n

i=0 i



,



132



21. Upper bounds using multilinear polynomials



then it follows that

s



m = dim span{f1 , . . . , fm } ≤ dim V =

i=0



n

,

i



which will complete the proof.

By definition the polynomial fi can be written as

fi (x) := (x · vi − l1 )(x · vi − l2 ) · · · (x · vi − ls ),

where L = {l1 , . . . , ls }. Now let x = (x1 , . . . , xn ). Then fi is a

polynomial in n variables x1 , . . . , xn of degree s. This is a map from

X = {0, 1}n to Fp . Then we get the corresponding multilinear polynomial f¯i of degree s. So f¯i is an element of a vector space

xi : |I| ≤ s, I ⊂ [n] ,



V := span

i∈I



and the dimension is

s



dim V =

i=0



n

.

i



This is exactly what we needed.



21.3. Snevily’s Theorem

We have seen some simple applications of multilinear polynomials.

In this last section we present a more delicate usage of the method

due to Snevily to show the following strong result, which includes

some earlier results obtained by Fisher, Bose, de Bruijin and Erd˝

os,

Majumdar, Frankl and Wilson, Ramanan, and others.

Theorem 21.9 (Snevily’s Theorem [102]). Let L be a set of s positive

integers. If F ⊂ 2[n] satisfies |F ∩ F | ∈ L for all distinct F, F ∈ F,

s

.

then |F| ≤ i=0 n−1

i

The upper bound for |F| is sharp. We give two examples. The

first example has L = [s] and F = {F ⊂ [n] : 1 ∈ A, |A| ≤ s + 1}.

The second one consists of L = {1} and a Steiner system S(2, 3, 7),

that is, F ⊂ 2[7] is defined by

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



21.3. Snevily’s Theorem



133



The idea of the proof is as follows. First we assign a polynomial

to each member of the family F and show that these polynomials are

linearly independent. The dimension of the space of the polynomials

n−1

more

is si=0 ni . Then we show that we can still add s−1

i=0

i

polynomials to the same space without violating the linear indepenn−1

= si=0 n−1

.

dence. Thus we get |F| ≤ si=0 ni − s−1

i=0

i

i

Proof. To deal with multilinear polynomials we need some more notation. Let X ∗ = {x1 , x2 , . . . , xn } denote the set of n variables, and



let Xk denote the set of multilinear monomials of degree k, that is,

X∗

k



xi : I ∈



=

i∈I



[n]

k



,







where we understand X0 = {1}. Let σ(X ∗ , k) denote the basic symmetric function of degree k in X ∗ , that is, f ∈(X ∗ ) f . For example,

k



if X ∗ = {x1 , x2 , x3 } then σ(X ∗ , 2) = x1 x2 + x2 x3 + x1 x3 . Finally, for

a subset F ⊂ [n] let F ∗ = {xi : i ∈ F } and let v(F ) = (v1 , . . . , vn )

be the characteristic vector of F , that is, vi = 1 if i ∈ F and vi = 0

if i ∈ F . Notice that if we substitute v(F ) into σ(X ∗ , k) then we get

the value |Fk | . Also, if we substitute v(F ), where F ⊂ [n], into

|

σ(F ∗ , k) then we get the value |F ∩F

.

k

Let L be a set of s positive integers. We define a polynomial in

y of degree s by

(y − l).

g(y) =

l∈L



We can uniquely determine s + 1 real numbers c0 , c1 , . . . , cs by g(y) =

s

y

i=0 ci i . Then for x = (x1 , . . . , xn ) we define

s



g ∗ (x) =



ci σ(X ∗ , i),

i=0



and for F ⊂ [n] let

s



gF∗ (x)



ci σ(F ∗ , i),



=

i=0







where we understand σ(F , i) = 0 if |F | < i. By definition we get the

following.



134



21. Upper bounds using multilinear polynomials



Claim 21.10. For F, F ⊂ [n] we have that

gF∗ (v(F )) = g ∗ (v(F ∩ F )) = g(|F ∩ F |).

Now suppose that F ⊂ 2[n] satisfies |F ∩ F | ∈ L for all distinct

F, F ∈ F. Since g(|F ∩ F |) = 0 we get gF∗ (v(F )) = 0.

Claim 21.11. Polynomials in the set {gF∗ : F ∈ F} are linearly

independent.

Proof. Consider a linear combination

αF gF∗ = 0,



(21.5)

F ∈F



where αF ∈ R. By substituting v(F ) we obtain αF gF∗ (v(F )) = 0. So

if gF∗ (v(F )) = 0, or equivalently if |F | ∈ L, then we get αF = 0. In

particular, αF = 0 for all F ∈ F with |F | > max L. But we need to

show that all αF are zero. We prove this by induction on |L|.

The initial step is L = {l}. Let F ∈ F. If |F | = l then αF = 0. If

|F | = l, then F is the only l-element subset in F. Thus (21.5) reads

αF gF∗ = 0. Since gF∗ is not a zero polynomial, we have αF = 0.

We proceed to the induction step. Let L = {l1 , . . . , ls } with

l1 < · · · < ls , and let L = L \ {ls }. To apply the induction hypothesis

let F = F \ {F ∈ F : |F | > ls }. Then |F ∩ F | ∈ L for all distinct

F, F ∈ F . Since αF = 0 for all F ∈ F with |F | > ls , (21.5) can be

rewritten as F ∈F αF gF∗ = 0. By the hypothesis applied to F with

L we obtain that αF = 0 for all F ∈ F .

Let L = {l1 , . . . , ls } with 1 ≤ l1 < · · · < ls and F = {F1 , . . . , Fm }.

If |Fi | = l1 for some i, then Fi is the only l1 -element subset in F and

we may assume that n ∈ Fi by renaming the vertices if necessary. Let

r ≥ 0 be the number of edges in F not containing n. We may assume

that n ∈ Fi for 1 ≤ i ≤ r and n ∈ Fi for r + 1 ≤ i ≤ m. We may also

assume that l1 < |Fr+1 | ≤ |Fr+2 | ≤ · · · ≤ |Fm |.

We are going to assign a polynomial pi to Fi for 1 ≤ i ≤ m. To

this end let

fFi (x) =



(v(Fi ) · x − lj ),

lj <|Fi |



21.3. Snevily’s Theorem



135



and let f¯Fi be the corresponding multilinear polynomial. Then we

define

pi =



gF∗ i

f¯F



for 1 ≤ i ≤ r,

for r + 1 ≤ i ≤ m.



i



We note that all the pi sit in the vector space V of multilinear polys

nomials in n variables of degree at most s, and dim V = i=0 ni .

Claim 21.12. The polynomials p1 , . . . , pm are linearly independent.

Proof. By Claim 21.11, p1 , . . . , pr are linearly independent. For r +

1 ≤ j ≤ i ≤ m it follows that

= 0 if i = j,

f¯Fi (v(Fj ))

= 0 if i > j,

where we used |Fi | > l1 for the first case i = j. This means that

pr+1 , . . . , pm are linearly independent.

Now consider a linear combination

m



αi pi = 0.



(21.6)

i=1



Suppose that there is a nonzero scalar. Then there are at least two

of them: one is αi0 with i0 ≤ r and the other is αj0 with j0 > r.

So we may assume that αj0 = 0 and αj = 0 for r + 1 ≤ j < j0 .

Substituting v(Fj0 ) into (21.6), we get αj0 pj0 (v(Fj0 )) = 0 and so

αj0 = 0, a contradiction.

s



At this point we already have that |F| = m ≤ dim V = i=0

s−1

Next we will introduce N = i=0 n−1

more polynomials. Let

i

{B1 , B2 , . . . , BN } =



[n − 1]

0



[n − 1]

1



···



n

i



.



[n − 1]

.

s−1



We may assume that |B1 | ≤ |B2 | ≤ · · · ≤ |BN |. Then for 1 ≤ i ≤ N

we define

qi (x) = (xn − 1)



xj ,

j∈Bi



136



21. Upper bounds using multilinear polynomials



where we understand q1 (x) = xn − 1. For 1 ≤ j ≤ i ≤ N it follows

that

= 0 if i = j,

qi (v(Bj ))

= 0 if i > j.

This means that q1 , . . . , qN are linearly independent.

Claim 21.13. The m + N polynomials p1 , . . . , pm , q1 , . . . , qN are linearly independent.

Proof. Consider a linear combination

m



N



αi pi +



(21.7)

i=1



βj qj = 0.

j=1



First suppose that αi = 0 for all r + 1 ≤ i ≤ m. In this case let j0 be

such that βj0 = 0 and βj = 0 for 1 ≤ j < j0 , and let t = |Bj0 |. We

substitute y = (y1 , . . . , yn ) into (21.7), where yi = y if i ∈ Bj0 {n}

and yi = 0 otherwise. By the definition of gF∗ i it follows that pi (y)

(1 ≤ i ≤ r) is a polynomial in y of degree at most t. On the other

hand, qj0 (y) = (y − 1)y t , and this is the only term containing y t+1 ;

that is, the LHS of (21.7) reads βj0 y t+1 + O(y t ). This must be a zero

polynomial, so βj0 = 0, and this is a contradiction.

Next suppose that αi0 = 0 for some r + 1 ≤ i0 ≤ m. Since

v(Fi0 ) = (∗, . . . , ∗, 1) we get qj (v(Fi0 )) = 0 for all j. Then, by substituting v(Fi0 ) into (21.7), we have αi0 pi0 (v(Fi0 )) = 0 and hence

αi0 = 0, a contradiction.

We have found m + N linearly independent polynomials in V .

Consequently we obtain

s



|F| = m ≤ dim V − N =

i=0



as promised.



s−1



n

n−1



i

i

i=0



s



=

i=0



n−1

,

i



Chapter 22



Application to discrete

geometry



In some problems in discrete geometry one can express the geometric constraints in terms of intersections in hypergraphs. Then the

problems may be solved by applying the corresponding results on

L-systems. In this chapter we present some such examples.



22.1. Sylvester’s problem and Fisher’s inequality

The following result is called Fisher’s inequality, which can be obtained from Theorem 21.9 by setting s = 1.

Theorem 22.1 ([41]). Let C1 , C2 , . . . , Cm be distinct subsets of [n],

and let λ ≥ 1. If |Ci ∩ Cj | = λ for all i = j, then m ≤ n.

We present an application of Fisher’s inequality to a collinearity

problem in discrete geometry. We say that points are collinear if

they are on the same line. Two points in the plane determine a line.

So m points in the plane determine m

2 lines if those points are in

general position, that is, no three points are collinear. Of course, m

collinear points determine only one line. Now suppose that we have

m non-collinear points in the plane. Then at least how many lines are

determined by those points? This problem was posed by Sylvester,

137



138



22. Application to discrete geometry



solved by Gallai, and popularized by Erd˝

os through his article in the

American Mathematical Monthly [31].

Theorem 22.2. Any set of m non-collinear points in the plane determine at least m lines.

The idea of the following proof goes back to de Bruijn and Erd˝os [16].

Proof. Let p1 , . . . , pm be the given m points. Suppose that they

determine n lines, and let L = {l1 , l2 , . . . , ln } be the set of the lines.

Let Ci ⊂ L be the subset of lines passing through pi , and let H =

{C1 , . . . , Cm } ⊂ 2L . Then, for all i = j, Ci ∩Cj contains just one line,

that is, the line connecting two points pi and pj . Thus |Ci ∩ Cj | = 1,

and by the Fisher’s inequality we get |H| = m ≤ |L| = n.

Chv´

atal posed a generalization of this problem in a metric space.

We say that three points x, y, and z in a metric space with metric d

are collinear1 if they satisfy

d(x, z) = d(x, y) + d(y, z).

For example, a graph can be viewed as a metric space by the usual

shortest-path metric. Then we can define a line in the graph in the

sense above. Such lines behave rather differently from the usual lines

in Euclidean space. Nevertheless, Chen and Chv´

atal conjectured the

following.

Conjecture 22.3 (Chen–Chv´

atal [18]). In any metric space, m noncollinear points determine at least m lines.



22.2. Chromatic number of the unit-distance

graph

A family of subsets H ⊂ 2[n] is called t-avoiding if

|H ∩ H | = t

for all H, H ∈ H. Applying Theorem 21.8 to the case

n = 4p − 1, s = p − 1,

we get the following.

1



This is called collinearity by betweeness.



L = {0, . . . , p − 2},



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