Counting H -free graphs for bipartite H Joint work with Asaf Ferber - PowerPoint PPT Presentation

Counting H -free graphs for bipartite H Joint work with Asaf Ferber and Wojciech Samotij Gwen McKinley July 26, 2017 Massachusetts Institute of Technology 1

Outline • Background & definitions • Conjecture, progress, & proof methods • Our result • Proof sketch & applications 2

Background & Definitions

Background & Definitions maximum possible number of edges in a graph G on n vertices that does not contain a H as a subgraph. • Call such a graph H -free . 3 • The extremal number ex( n , H ) for a graph H is the

Background & Definitions Classical result of Turán (1941) and Erdős-Stone (1946): Erdős-Stone Theorem 1 2 4 ( ) ( n ) ex( n , H ) = 1 − + o ( n 2 ) χ ( H ) − 1

Background & Definitions but what about bipartite graphs? • Answer: Very tricky! • See survey of Füredi and Simonovits, 2013 (97 pages!) 5 • This gives asymptotic behavior of ex( n , H ) when χ ( H ) ≥ 3,

Background & Definitions • Closely related problem: count H -free graphs. n vertices that do not contain H as a subgraph. 6 • Explicitly, find |F n ( H ) | , the number of (labeled) graphs on • How is this related to finding ex( n , H ) ?

Background & Definitions Trivial bounds 2 i 7 • Lower bound: |F n ( H ) | ≥ 2 ex( n , H ) ex( n , H ) (( n ) ) ∑ = 2 O ( ex ( n , H ) log( n )) • Upper bound: |F n ( H ) | ≤ i = 0 • Question: How to eliminate log ( n ) factor?

Background & Definitions Better bounds and Rödl in 1986. 8 • In general, |F n ( H ) | = 2 ex( n , H )+ o ( n 2 ) , proved by Erdős, Frankl, • If χ ( H ) ≥ 3, then this means |F n ( H ) | = 2 ( 1 + o ( 1 )) ex( n , H ) • But if H is a forest, |F n ( H ) | = 2 Θ(ex( n , H ) log( n ))

Background & Definitions Conjecture (Erdős, Frankl, and Rödl, 1986): For any H containing a cycle, • False! Morris and Saxton (2016). 9 |F n ( H ) | = 2 ( 1 + o ( 1 )) ex( n , H ) • Counterexample: |F n ( C 6 ) | ≥ 2 ( 1 + c ) ex( n , H ) for some c > 0;

The Problem

Background & Definitions New Conjecture For any H containing a cycle, 10 |F n ( H ) | = 2 O (ex( n , H ))

Progress • Known for C 4 , C 6 , and C 10 . (non-degenerate case). 11 • Known for K 2 , t , K 3 , t , and K s , t with t > ( s − 1 )! . • ”Almost” known for some others - e.g. |F n ( C 2 ℓ ) | = 2 O ( n 1 + 1 /ℓ ) . Known that ex( n , C 2 ℓ ) = O ( n 1 + 1 /ℓ ) , conjectured to be sharp. • Known for k -uniform hypergraphs with χ ( H ) > k

Methods • Main technique: hypergraph containers (Balogh, Morris, and Samotij, 2015; Saxton and Thomason, 2015) • Gives a way to count independent sets in hypergraphs. • Then H -free graphs on n vertices correspond to 12 • Application: create hypergraph Z whose vertices are the edges of K n and whose edges are all copies of H in K n . independent sets in Z .

Containers Method ”niceness” properties, there exists a family of containers 13 • Broad strokes : for a hypergraph Z satisfying certain C ⊆ P ( V ( Z )) so that each independent set in Z is contained in some C ∈ C . • So |F n ( H ) | ≤ (# containers ) · 2 size of max container

Containers • ”Niceness” : In general, in any graph with more than ”well-distributed” copies of H (a supersaturation condition) in order to apply containers. • Supersaturation results often very hard to prove • ”If only he had used his genius for niceness instead of evil” 14 ex( n , H ) edges, need to prove there are ”many” and

A New Hope • Question: Possible to prove supersaturation without • Answer: Maybe! Paper by Balogh, Liu, and Sharifzadeh (2016) counting k -arithmetic progression free subsets of • Sample smaller set of numbers, show they induce many 15 knowing ex( n , H ) ? [ n ] . k -APs, end up having to bound ratio of ex( m ) ex( n ) for m < n .

Our Contribution

Our result Main Theorem 16 If H is any graph containing a cycle, and ex( n , H ) = O ( n α ) for some α ∈ ( 1 , 2 ) , then |F n ( H ) | = 2 O ( n α ) In particular, if ex( n , H ) = Θ( n α ) , then |F n ( H ) | = 2 O (ex( n , H )) .

Proof Ideas • First: Inductive application of containers. Developed by • Second: Prove supersaturation result by bounding number of copies of H in small random subgraphs. 17 Morris and Saxton in paper on C 2 ℓ -free graphs (2016).

Notation Notation 18 • γ > 1 is a constant depending on H • v H = # vertices of H • e H = # edges of H

Supersaturation Condition Supersaturation Condition Let k be any constant depending only on H . If for every graph m 19 G on n vertices with m = γ t · k · n α edges, there exists a subset Z of all copies of H in G so that ) ℓ − 1 n α ( · |Z| ∆ ℓ ( Z ) ≤ m ( t + 1 ) 3 for all ℓ ∈ { 1 , . . . , e H } then |F n ( H ) | = 2 O ( n α ) . ∆ ℓ ( Z ) = maximum number of copies of H in Z that contain any subset of ℓ edges in G .

m Proof of Supersaturation reduces to Show only this case here - gives basic idea of the proof. 20 For ℓ = e H , the condition ) ℓ − 1 n α ( · |Z| ∆ ℓ ( Z ) ≤ m ( t + 1 ) 3 |Z| ≥ ( γ t · k ( t + 1 ) 3 ) e H − 1 · m .

Proof of Supersaturation • Strategy: show that random small subgraph of G gives many copies of H . 21 • Goal: given graph G with m = γ t · k · n α edges, want to show there at least ( γ t · k ( t + 1 ) 3 ) e H − 1 · m copies of H

Proof of Supersaturation Notation (random variable) • Z = total number of copies of H in G (what we’re trying to bound) 22 • R = uniformly random set of pn vertices in G • p ∈ ( 0 , 1 ) , yet to be chosen • X = number of copies of H in induced subgraph G [ R ]

Proof of Supersaturation pn Solve to get: pn Bounds 23 And • X ≥ e ( G [ R ]) − ex( pn , H ) • So E [ X ] ≥ E [ e ( G [ R ])] − ex( pn , H ) ( n − v H ) ( n ) • E [ X ] = Z · / ≈ Z · p v H pn − v H ( n − 2 ) ( n ) • E [ e ( G [ R ])] = m · / ≈ m · p 2 pn − 2 Z ≥ ( mp 2 − ex( pn , H )) p − v H .

Proof of Supersaturation many copies of H . Use this to bound total number Z of copies. Show: 24 Goal: Show that random subgraph G [ R ] of correct size pn has Z ≥ ( mp 2 − ex( pn , H )) p − v H ≥ ( γ t · k ( t + 1 ) 3 ) e H − 1 · m

Proof of Supersaturation Approach 1 25 • Do some algebra to get upper and lower bonds on p ≤ p α n α • Along the way, use fact that ex( pn , H ) = p α n α n α • End up with upper bound ≥ lower bound if and only if e H − 1 v H − 2 < 2 − α

Proof of Supersaturation Bohman and Keevash, 2009 1 1 For any H containing a cycle, 26 1 e H − 1 v H − 2 < 2 − α ? { } e ( F ) − 1 Definition: m 2 ( H ) = max v ( F ) − 2 : F ⊆ H with e ( F ) > 1 . ex( n , H ) ≥ n 2 − 1 / m 2 ( H ) · log( n ) eH − 1 . Since n α ≥ ex( n , H ) , n α ≥ n 2 − 1 / m 2 ( H ) · log( n ) eH − 1 .

Proof of Supersaturation 1 Know: 1 So And in particular, 27 e H − 1 v H − 2 < 2 − α ? n α − 2 + 1 / m 2 ( H ) > log( n ) eH − 1 α − 2 + 1 / m 2 ( H ) > 0 α − 2 + v H − 2 e H − 1 > 0

Summary Main Ideas in Proof of Supersaturation correct size has many copies of H . . there is a gap between upper and lower bounds. 28 • Probabilistic method: show that random subgraph G [ R ] of • Use assumption on growth rate of ex( n , H ) to bound ratio ex( pn , H ) n α • Use bound on ex( n , H ) in terms of 2-density m 2 ( H ) to show

Applications

Reproving Old Results Saxton (2016) • Hypergraphs: reproves recent result of Balogh, Nayaranan, • The list goes on! 29 • Reproves non-degenerate case (where χ ( H ) ≥ 3) • Reproves |F n ( H ) | = 2 O (ex( n , H )) for C 4 , C 6 , and C 10 , as well as K 2 , t , K 3 , t , and K s , t with t > ( s − 1 )! . • Reproves |F n ( C 2 ℓ ) | = 2 O ( n 1 + 1 /ℓ ) - result of Morris and k ) | = 2 O (ex( n , C r k )) and Skokan (2017) for linear cycles: |F n ( C r

New results Infinite Sequences for all i . Phillips, and Sarnak, 1988). 30 If there is a constant ε > 0 such that ex( n , H ) = Ω( n 2 − 1 / m 2 ( H )+ ε ) , then there exist an infinite sequence { n i } ⊆ N and a constant C > 0 such that |F n ( H ) | ≤ 2 C · ex( n , H ) In particular, this holds for all even cycles, C 2 ℓ . (Lubotzky,

Questions? 30