Uniformly X Intersecting Families. Noga Alon and Eyal Lubetzky

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1 Uniformly X Intersecting Families Noga Alon and Eyal Lubetzky April 2007

2 X Intersecting Families Let A, B denote two families of subsets of [n]. The pair (A,B) is called iff `-cross-intersecting A B =` for all A A, B B. Q: What is P`(n), the maximum value of A B over all `-X-intersecting pairs (A,B)?

3 Previous work: single family What is the maximal size of F 2 [n] with given pair-wise intersections? Erdős-Ko-Rado 61: F F t and F =k for all F,F F, then: F ³ n t k t Katona s Thm 64: no restriction on F. Additional examples: Ray-Chaudhuri-Wilson 75, Frankl-Wilson 81, Frankl-Füredi 85.

4 Previous work: two families Conj (Erdős 75): if F 2 [n] has no pair-wise b n 4 c intersection of then F (2 ε) n Settled by studying pairs of families: Frankl-Rödl 87: if A,B 2 [n] have a forbidden X-intersection ηn l ( 1 2 η)n then A B (4 ε(η)) n η < ¼ Frankl-Rödl 87 studied several notions of X-intersecting pairs, including P`(n).

5 Previous work: P`(n) Frankl-Rödl 87: P 0 (n) 2 n, and for all ` 1, P`(n) 2 n -1. Ahlswede-Cai-Zhang 89: lower bound: take n 2`, and: Linear algebra over (Z p ) n One set A 1 of 2` elements A A 1 =[2`] B A 1 B =` All sets containing ` elements of A 1 A B = ³ 2` 2n 2` ` =(1+o(1)) 2n π`

6 Previous work: P`(n) Conj (Ahlswede-Cai-Zhang 89): the above construction maximizes P`(n). True for `=0 and `=1. For general `: Gap = [ Θ( `) 2n, Θ(2 n )] Q (Sgall 99): does P`(n) decrease with `? Keevash-Sudakov 06: the above conj is true for `=2 as well.

7 Our results Confirmed the conj of Ahlswede-Cai-Zhang 89 for any sufficiently large `. Characterized all the extremal pairs A,B which attain the maximum of A B = ³ 2` This also provides a positive answer to the Q of Sgall 99. ` 2n 2`

8 Main Theorem There exists some `0, such that, for all ` `0, every `-X-intersecting pair A,B 2 [n] satisfies: Furthermore, equality holds iff the pair A,B is w.l.o.g. as follows:

9 The construction of Ahlswede et al. fits the special case τ =0,κ =2` A B = ³ κ ` Extremal pairs A,B 2n κ = ³ 2` 2n 2` ` τ κ, κ {2` 1, 2`} A A A = ³ κ ` 2M ` objects any subset τ 1 1 κ +1 τ... 2 κ +2 κ + τ 1 κ + τ τ +1 τ +2 x 1 x 2 x M κ 1 κ y 1 y 2 y N. 1 from each object any subset B B B =2 τ+n n = τ + κ + M + N

10 Ideas used in the Proof Tools from Linear Algebra: study the vector spaces of the characteristic vectors of the sets in A,B over R n. Techniques from Extremal Combinatorics, including: The Littlewood-Offord Lemma. Extensions of Sperner s Theorem Large deviation estimates. Prove: Upper bound up to a constant Asymp. tight upper bound Main result

11 A weaker result Upper bound tight up to a constant: There exists some `0, such that, for all ` `0, every `-X-intersecting pair A,B 2 [n] satisfies:

12 Vector spaces over R n Define: F A =span ({χ A :A A}) over R, F B =span ({χ B :B B}) over R. Set: F B =span ({χ B χ B1 :B B}) over R, k=dim(f A ), h=dim(f B ). A,B are `-X-inter F A F B. k + h n.

13 Vector spaces over R n Let M A and M B denote the matrices of bases for F A and F B after performing Gauss elimination: M A = ³ I k, M B = ³ I h Since target vectors are in {0,1} n : A B 2 k+h 2 n. If, say, M A can produce at most A < 8 n 2 k sets, we are done.

14 What do M A and M B look like? Can we indeed produce 2 d n d legal char. vectors from? M = ³ I d We get constraints if there are: Columns with many non-zero entry. Rows not in {0, ±1} n \{0, 1} n. The Littlewood-Offord Lemma (1D) Families are antichains by induction

15 The Littlewood-Offord Lemma (1D) Q (Littlewood-Offord 43): Let a 1,...,a n R with a i >1 for all i. P What is the max num of sub-sums i I a i, I [n], which lie in a unit interval? Lemma (Erdős 45): Let a 1,...,a n R \{ δ, δ} and let U denote an interval of length δ. P Then the number of sub-sums i I a i, I [n], which belong to U, is at most ³ n bn/2c

16 Erdős s Pf of the L-O Lemma (1D) Without loss of generality, all the a i -s are positive (o/w, shift the interval U). A sub-sum which belongs to U is an antichain of [n] and the result follows from Sperner s Thm.

17 What do M A and M B look like? Either, or w.l.o.g. : M A = M B = I k 0 I k 0 0 I h 0 I h 0 0 k 0 = 2 5n O(log n) h 0 = 13 40n O(log n)

18 Completing the proof of the Thm Recall: each row of M A is orthogonal to each row of M B. Two (1,-1,0,,0) rows are orthogonal only if the (1,-1) indices are disjoint. M A gives M B gives n O(log n) n O(log n) pairs of indices. pairs of indices. 4 5 n n>n contradiction. Qed

19 Proof of Main result some ideas If M A is far from a structure which produces 2 k sets for A, M B must be close to a structure producing 2 h sets for B. Clean the matrices gradually, using orthogonality to switch back and forth between M A and M B. An easy scenario to illustrate this:

20 Scenario: Ω(n) rows of M A in {0,1} n k {2` 1, 2`} M A = I h 0 I h 0 I k h 0 A A: `-subset of pairs and singles. A = ³ k ` M B = I h 0 χ B1 = I h 1,...,1 1,...,1 0,..., 0 Optimal family with κ = k, τ = h, n = κ + τ. B B: One of each pair and single. B =2 h

21 Thank you!