Lecture 5. Symmetric Shearer s Lemma

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1 Stanfor University Spring 208 Math 233: Non-constructive methos in combinatorics Instructor: Jan Vonrák Lecture ate: January 23, 208 Original scribe: Erik Bates Lecture 5 Symmetric Shearer s Lemma Here we iscuss a corollary of Shearer s Lemma that consiers the symmetric case, in which all events are given the same probability boun Theorem 5 (Symmetric Shearer s Lemma) Suppose there is a collection of events {E i } n i= such that each E i is inepenent of all but other events ( 2), an Then P(E i ) ( ) =: p Shearer i =, 2,, n (5) ( n ) P E i > 0 i= Proof: Let G be a epenency graph for {E i } with maximum egree, an let p = ( ) / We may assume that G is connecte, since otherwise the problem reuces to a collection of inepenent problems For G connecte, we can fin an orering of vertices (v,, v n ) such that each v i, i 2, has egree at most among {v i,, v n } (However, this is not possible to arrange for v if G is -regular Therefore, we nee to hanle this case separately later) By inuction on S we claim that a > for a S where S Γ(a) The base case of the inuction is satisfie as q {a} q = q {a} = p = ( ) > = For the general case, we will use an ientity establishe in the proof of the asymmetric case (see Lecture 5): a = p \Γ+ (a) a (52) Assume that a S is such that S Γ(a), an write S Γ + (a) = {a, a, a 2,, a k }, k Since each a i has egree at most insie S \ {a}, the inuctive hypothesis gives \Γ + (a) = \Γ+ (a) q (S\Γ + (a))+a k a a a q (S\Γ + (a))+a k q (S\Γ + (a))+a k +a k a }{{} at most terms < (53) ( /)

2 From (62), we get = p \Γ+ (a) > p a a ( ) = which finishes the inuctive claim Finally let us hanle the case where S = {v,, v n } an v has egree We can still use (62), but now the telescoping prouct in (63) may involve terms, giving q [n] = p q [n]\γ+ (v ) > p q [n] v q [n] v ( ) = We note that coul be 0 (for = 2) but the strict inequality ensures that the ratio is still positive We conclue that ( n ) P E i q [n] = q [n] q [n] v q ( v n > ) ( n 0, q [n] v v 2 q ) i= q [n] a completing the proof Let us compare Symmetric Shearer s Lemma to the Lovász Local Lemma In the LLL, assuming that all events get the same parameter x, it is require that p x( x) (54) for some x (0, ) The optimal choice here can be shown to be x = +, which gives p ( + ) + =: p LLL (55) Comparing (65) to (6), we see that the threshol probability in Shearer s lemma, p Shearer := ( ), has the benefit of aitional epenency over the LLL Further, the inequalities show that ( + ) < e < ( ) e( + ) < p LLL < e < p Shearer < e( ) Of course, as grows large, p LLL an p Shearer are asymptotically the same 5 Worst instance: -regular trees We woul like to emonstrate that p Shearer is optimal, in the sense that Theorem 6 fails if p Shearer is taken any larger The extreme case is when each E i is epenent on exactly other events an moreover the epenency graph is a (large) -regular tree Begin with a root vertex r, by itself calle T 0 A root with chilren is calle T Constructe recursively, T l is the tree obtaine by taking a root with subtrees, each of which is T l Note that all vertices in levels through

3 Figure : A binary tree ( = 3) Here, T 3 is shown (levels 0 through 3) l have egree ; the root has egree, an level l consists of leaves We call this a -regular tree of epth l (ignoring the slightly ifferent egree at the root) For consistency, we also efine T to be the empty tree Suppose that the probability of each event is p From (62), we have q Tl = q Tl \r p q Tl \Γ + (r) But T l \ r is the union of isjoint copies of T l Similarly, T l \ Γ + (r) is the union of ( ) 2 isjoint copies of T l 2 Hence q Tl = ( q Tl ) p ( qtl 2 ) ( ) 2 Let us efine b l := q Tl ( q Tl ) = p ( ( qtl 2 ) q Tl ) That is, ( ) b l = p b l If Shearer s positivity conitions are satisfie for an arbitrarily large -regular tree, then b l > 0 for all l 0, an also the sequence is ecreasing by inuction: b 0 = p, b = b 0, an if b l b l, then b l+ = p/b l which must satisfy λ = which gives ( ) ( p Inee, p Shearer is optimal p/b l = b l Hence there is a limit, λ := lim l b l, p ( p) p λ, an hence p = λ λ The maximum is attaine at λ =, ) = ( ) = p Shearer 52 Application of Shearer s Lemma: the multipartite Turán problem Consier an r-partite graph G on V V 2 V r Suppose we have at least a certain ensity ρ between any two parts: e(v i, V j ) ρ V i V j i j

4 How large must ρ be to guarantee the existence of a clique K r in G? More generally, given a graph H on r vertices, assume {i, j} E(H) e(v i, V j ) ρ V i V j How large must ρ be to guarantee the existence of a copy of H in G? Following (Csikvári an Nagy, 202), we show how to apply Shearer s Lemma Pick x i V i inepenently an uniformly at ranom For each (i, j) E(H), efine an event E ij = {{x i, x j } / E(G)}, so that if all E ij are avoie, then a copy of H is present Note that the probability of each event is at most ρ by assumption A epenency graph for the events E ij is the line graph of H, which we call D: The vertices of D are the eges in H, an two of these vertices are ajacent if an only if the corresponing eges in H share a vertex So inepenent sets in D are exactly matchings in H First, consier Symmetric Shearer s Lemma The egrees in D are at most 2( (H) ) where (H) is the maximum egree in H Hence, if the probability of each event is at most 2e( (H) ), then by Theorem 6, P[ (i,j) E(H) E ij] > 0 Equivalently, if ρ 2e( (H) ) then G contains a copy of H This problem is actually a rare setting where we can apply Shearer s Lemma irectly an obtain a stronger result Consier the polynomial q (p) = I In(D)( ) I p I = ( ) M p M M H Mmatching This last sum is a variant of the matching polynomial of K r It is most commonly efine in the following form, which we refer to as the matching efect polynomial: M H (x) = M H Mmatching ( ) M x r 2 M (Recall that r = V (H) ) A simple calculation gives M H (x) = x r q ( x 2 ) It is useful in this setting to consier Property 4 of Shearer s Lemma, state in Lecture 5 In particular, we ask, for which p is it true that q (λp) > 0 λ [0, ]? To answer this question, it suffices to locate the minimum positive root of q, or equivalently the maximum positive root of M H Here we appeal to the following theorem (which we will prove later in this course) Theorem 52 (Heilmann-Lieb) For any graph H, the roots of the matching efect polynomial are all real an the maximum root is at most 2 (H)

5 It follows that the minimum positive root of q (p) for H is at least 4( (H) ) Consequently, if ρ 4( (H) ) then G contains a copy of H, which improves the boun from above (2e has been improve to 4) For H = K r, which is perhaps the most interesting special case here, we obtain that ensity ρ 4(r 2) is sufficient to guarantee a copy of K r In fact, here we can go one step further a obtain a slightly tighter boun M Kr is known to be the Hermite polynomial of egree r Recall that the Hermite polynomials are efine recursively, corresponing to the recursion in the context of matchings: H 0 (x) = H r+ (x) = xh r (x) rh r (x) For this special case, more accurate bouns are known In particular, the maximum root of M Kr is known to be 2 r Θ(r /6 ) Hence, the minimum positive root of q is (2 r Θ(r /6 )) 2 = 4r Θ(r /3 ) Consequently, if ρ 4r Θ(r /3 ) then G contains a copy of K r, a slight improvement over the boun of 4(r 2) from the Heilmann-Lieb theorem We conclue by mentioning that it is easy to construct an r-partitite graph of ensity ρ = r which oes not contain a K r (an exercise) A better counterexample which can be foun in [Csikvári-Nagy 2] implies that ρ = (2 ɛ)r is not sufficient to guarantee a copy of K r for any ɛ > 0 The gap between 2r an 4r remains open References Péter Csikvári an Zoltán Lóránt Nagy The ensity Turán problem Combinatorics, Probability an Computing, 2(04):53 553, 202