2-Distance Problems. Combinatorics, 2016 Fall, USTC Week 16, Dec 20&22. Theorem 1. (Frankl-Wilson, 1981) If F is an L-intersecting family in 2 [n],

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1 Combinatorics, 206 Fall, USTC Week 6, Dec 20&22 2-Distance Problems Theorem (Frankl-Wilson, 98 If F is an L-intersecting family in 2 [n], then F L k=0 ( n k Proof Let F = {A, A 2,, A m } where A A 2 A m i [m], let f i (x in R n by For f i (x = l L,l< A i (x Ai l So f i (x is a polynomial with n variables and with degree L Claim : f, f 2,, f m are linearly independent Pf of Claim : Take A, A2,, Am, we have ˆ f i ( Ai = l L,l< A i ( A i l > 0 ˆ f i ( Aj = l L,l< A i ( A i A j l = 0 Because F is L-intersecting l L with l = A j A i and l < A i Observation: All vector Aj are 0/ vectors Thus, we can dene a new polynomial f i (x from f i (x by replacing all term x k i by x i So for all 0/ vectors v we still have f i (v = f i (v This also shows that f, f 2,, f m are linearly independent We see each f i (x is a linear combination of the monomials i I x i where I [n] and I L And clearly the

2 number of each monomials is L ( n k=0 k which is also is the dimension of the space containing f, f 2,, f m So L ( n F = m k k=0 Theorem 2 Let p be a prime and L Z p = {0,,, p } Let F 2 [n] be st ˆ A / L(mod P ˆ A B L(mod p for A B F Then F L ( n k=0 k Proof All operations are mod p Dene f i (x over Z n p for each set in F = {A,, A m } by f i (x = l L(x Ai l Then ˆ f i ( Ai = l L ( A i l 0(mod p ˆ f i ( Aj = l L ( A i A j l = 0(mod p for i j So f, f 2,, f m are linearly independent over Z n p The remaining proof is identical to the proof of Thm L ( n F = m k 2 k=0

3 Theorem 3 (Frankl-Wilson For any prime p, there is a graph G on n = p vertices st the size of minimum clique or maximum independent set 2 is p i=0 i Proof Let G = (V, E be as follows: ˆ V = ( [p 3 ] p 2 ˆ for A, B V, A G B i A B = p (mod p Consider the max clique with vertices sat A, A 2,, A m ( [p 3 ] p 2 Thus we have ˆ A i A j p (mod p, for i j ˆ A i = p 2 = p (mod p By Thm 2 with L = {0,, 2,, p 2} Z p we have m p i=0 Consider the maximum independent set, say B, B 2,, B s, then B i B j = p (mod p for i j So B i B j {p, 2p,, p(p } = L with L = p By Thm with L we have s p i=0 i i Corollary 4 R(k +, k + k Θ( log(k log(log(k Proof Let k = p i=0 i, n = p 2 ( p 3 k (p 2 p p 2p, n ( p3 p p 2 p2 p p2 3

4 log(k Θ(p log(p log(log(p log(p log(k p = Θ( log(log(k, n (p2p p/2 k Θ( log(k log(log(k Denition 5 Given a set S R n (bounded, the diameter of S is dened as Diam(S = sup{d(x, y : x, y S}(Euclidean distance between x and y in R n Borswk's Conjecture: Every bounded S R n can be partitioned into d + sets of strictly smaller diameter This was veried for all S R n with d 3 and for all S = sphere However, using Thm and 2 one show this is false! ( Lemma 6 For prime p, there is a set of 4p 2 2p vectors in {, } 4p st every subset of size 2 ( 4p p vectors contains an orthogonal pair of vectors Proof Let Q={I ( ( [4p] 2p : I}, then Q = 4p 2 2p For I Q, dene v I {, } 4p by, i I v i =, i / I Claim: v I v J i I J 0(modp Let F = {v I : I Q} with ( F = Q = 4p 2 2p Proof v I v J = I J I C J I J C + I C J C = 4p 2 I J So v I v J i I J = 2p = 4p 2 I J i I J = p 4

5 Claim: For any subset G F without orthogonal pairs, then G ( < 2 4p p k=0 ( 4p k p Proof Consider the corresponding subset Q Q of G, ie Q = {I Q : v I G By claim, Q' is a subfamily of ( [4p] 2p such that ˆ A = 2p 0(modp, A Q ˆ A B 0(modp, A B Q By thm 2, G = Q p ( 4p k=0 k = maximal subset without orthogonal pairs p k=0 ( 4p ( k < 2 4p p Theorem 7 For suciently large d, there exists a bounded set S R d (a nite set such that any partition of S into d subsets contains a subset of the same diameter Remark As d >> d + for large d, this disproves Borsuk's conj Denition 8 A tensor product of vectors v R n is w = v v R n2 w ij = v i v j for all i, j n by Proof Take the family F from the lemma, so F {, } n (where n=4p R n Let X = {v v : v F } st X R n2 For any w = v v X, w 2 = wij 2 = n vi 2 vj 2 = ( vi 2 ( vj 2 = n 2 j= i,j n i,j n i= = w = n 5

6 For w = v v, w = v v X, we have w w = w ij w ij = (v i v i(v j v j = ( v i v i 2 = (v v 2 i,j n i,j n This implies that w w v v Also, w w 2 = w 2 + w 2 2w w = 2n 2 2(v v 2 2n 2 Diam(X = 2n = X = F = ( [4p] 2 2p By the lemma, any subset of 2 ( 4p p vectors in F contains an orthogonal pair of vector v&v Thus, any subset of 2 ( 4p p vectors in X must contain a pair w = v v, w = v v with v v and thus of the maximum distance w w = 2n Thus, if we want to decrease the diameter, we must partition X into subsets, each of which has less than 2 ( 4p p vectors, so the number of subsets is at least X 2 ( 4p = p 2 ( 4p 2p 2 ( 4p p = 4 (3p + 92p + (2p (p where d = n 2 = 6p 2 is the dimension of X 4 (3 2 p+ C ( 3 2 d 4 d Bollobás Thm Recall: (Sperner's Thm Let F 2 [n] be: A B F, A B, B A, then F ( n n 2 LYM-Inequality: For such F, A F 6 ( n A

7 Theorem 9 (Bollobás Thm Let A, A 2,, A m and B, B 2,, B m be the sequences of sets in [n] st ˆ A i B j φ, i j ˆ A i B i = φ, i Then, where = A i, b i = B i m i= Remark Condition: A i B j φ, i j can't be weakened to i<j, or the base case doesn't hold any more Counter example: ˆ A = {} = B 2, A 2 = B = φ ˆ A = {} = B 2, A 2 = {3} = B, A 3 = {3}, B 3 = {, 2} Remark Bollobás = LY M = Sperner s Proof Let X = m i=(a i B i We prove by induction on n = X Base case: n = A = {}; B = φ, OK Assume this holds for X n For x X, dene I x = { i m : x / A i } Dene F x = {A i : i I x } {B i {x} : i I x } Note that any set of _x doesn't contain x, so F x has less than n elements Hence we apply induction hypothesis for each F x to get: ( Ai + B i {x} i I x A i 7 (

8 We summing up the above inequalities for all x X to get: ( Ai + B i {x} x X i I x A i n (2 For each i, it contributes either 0, or ( +b i or to each x The ( +b i term ( +b i corresponds to points x / A i B i, thus this term appears exactly (n b i times While, the term ( +b i corresponds to points x / A i&x B i, thus this term appears exactly b i times Since (k l ( k l (2 = m [(n b i i= = k l, we get k ( +b i = m [(n b i i= + b i ( +b i ] n ai+b i b i, plugging in, + + b i ] n m n i n m i= n Denition 0 Let F be a eld, a set A F n is general position if any n vectors in A are linearly independant over F Examples For a F, dene m(a = (, a, a 2,, a n F n (moment curve Then {m(a : a F is a general position 8

9 Next, we use the so-called "general position" argument to prove a version of Bollobás s Thm, which is weaker than the previous one But, on the other hand, the condition can be generalized to A i B j φ for i < j Theorem (Bollobás Thm(the skew version Let A,, A m be sets of size r and B,, B m be the sets of size s, such that : ˆ A i B j φ, i j ˆ A i B i = φ, i Then, ( r + s m s Proof (By Lovász: Let X= i (A i B i Take a set V R r+ of vectors v = (v 0, v,, v r such that ˆ V is in general position ˆ V = X Identify the elements of X with vectors in V Hence, we will view A i as a subset in V containing r vectors and B j as a subset in V containing s vectors For each B j, dene f j (x = v B j < v, x >= v B j (v 0 x v r x r For x R r+, note that f j (x = 0 iff < v, x >= 0 for some v B j (3 Consider the subspace span A i, which is spanned by the r vector in A i, since A i V R r+ and V is in general position, we see that all r vectors in 9

10 A i are linearly independent and thus dim(spana i = r So, (spana i has dimension Choose (spana i for i=,,m Then for each v V, < v, >= 0 iff v spana i iff v A i (4 (OW v / A i, {v} A i has r+ vectors in V, which must be linearly independent, contradicting to v spana i Combing (3&(4, f j ( = v B j < v, >= 0 i A i B j φ f j ( = 0, i < j = f j ( 0, j This shows that f,, f m are linearly independent Next, we give an upper bound on the dimension of the space containing f,, f m Recall: f j (x = v B j (v 0 x v r x r, it is homogeneous with degree s = B j and r+ variables (x 0, x,, x r So this polynomial space can be generated by all monomials of follows: x i 0 0 x i x ir r, where i 0 + i + + i r = s, i j 0 There are ( r+s r ( many solutions! So m the dimension= r+s s 0