Cycles of length 3 and 4 in tournaments Timothy F. N. Chan Joint - PowerPoint PPT Presentation

Cycles of length 3 and 4 in tournaments Timothy F. N. Chan Joint work with: Andrzej Grzesik (Krak´ ow) Daniel Kr´ al’(Brno) Jonathan A. Noel (Warwick) March 18, 2019

Context Mantel 1907: Any graph with more than ⌊ n 2 / 4 ⌋ copies of K 2 contains a copy of K 3 . os-Rademacher problem: If a graph exceeds ⌊ n 2 / 4 ⌋ copies of Erd˝ K 2 , how many copies of K 3 are forced? A : Asymptotically solved by Razborov 2008, using flag algebras. Topic of this talk: Analogous problem for tournaments .

b b b b u u u u u u Tournaments Complete graph with every edge given a direction. e.g. random tournament, transitive tournament R

b b b b b b b b b b b b Main question Q : What is the minimum number of K 3 ’s in a graph with a given number of K 2 ’s?

b b b b b b b b b b b b Main question Q : What is the minimum number of K 3 ’s in a tournament with a given number of K 2 ’s?

b b b b b b b b b b b b Main question Q : What is the minimum number of K 3 ’s in a tournament with a given number of K 2 ’s? Q : What is the minimum number of C 4 ’s in a tournament with a given number of C 3 ’s?

b b b b b b b b b b b b Main question Q : What is the minimum number of K 3 ’s in a tournament with a given number of K 2 ’s? Q : What is the minimum number of C 4 ’s in a tournament with a given number of C 3 ’s? density: c ℓ ( T ) := probability that a random mapping from V ( C ℓ ) to V ( T ) is a homomorphism i.e. arcs of C ℓ map to arcs of T . c 3 ( T ) = (3 + 3) / 4 3 = 3 / 32 Q : Given c 3 ( T ), asymptotically minimise c 4 ( T ).

An extremal construction? Fix z ∈ [0 , 1]. Create as many blocks of vertices of size zn as possible, and put the remaining ≤ zn vertices in a single block. Edges within blocks behave randomly, edges between blocks go to the right. ”random blow-up of a transitive tournament”

An extremal construction? Fix z ∈ [0 , 1]. Create as many blocks of vertices of size zn as possible, and put the remaining ≤ zn vertices in a single block. Edges within blocks behave randomly, edges between blocks go to the right. ”random blow-up of a transitive tournament” c 3 ( T ) = 1 � ⌊ z − 1 ⌋ z 3 + � 3 � � 1 − ⌊ z − 1 ⌋ z + o (1) 8 c 4 ( T ) = 1 � ⌊ z − 1 ⌋ z 4 + � 4 � 1 − ⌊ z − 1 ⌋ z � + o (1) 16

An extremal construction? Fix z ∈ [0 , 1]. Create as many blocks of vertices of size zn as possible, and put the remaining ≤ zn vertices in a single block. Edges within blocks behave randomly, edges between blocks go to the right. ”random blow-up of a transitive tournament” Conjecture (Linial & Morgenstern 2016) For every tournament T , c 4 ( T ) ≥ g ( c 3 ( T )) + o (1) .

Main result Conjecture (Linial & Morgenstern 2016) For every tournament T , c 4 ( T ) ≥ g ( c 3 ( T )) + o (1) .

Main result Conjecture (Linial & Morgenstern 2016) For every tournament T , c 4 ( T ) ≥ g ( c 3 ( T )) + o (1) . Theorem (C., Grzesik, Kr´ al’, Noel 2018) The above conjecture is true for c 3 ( T ) ≥ 1 / 72 . Furthermore, we characterise the extremal tournaments when c 3 ( T ) ≥ 1 / 32 .

Main result Conjecture (Linial & Morgenstern 2016) For every tournament T , c 4 ( T ) ≥ g ( c 3 ( T )) + o (1) . Theorem (C., Grzesik, Kr´ al’, Noel 2018) The above conjecture is true for c 3 ( T ) ≥ 1 / 72 . Furthermore, we characterise the extremal tournaments when c 3 ( T ) ≥ 1 / 32 . Notes: Behaviour appears similar to the Razborov result Proof uses spectral methods instead of flag algebras The space of extremal tournaments is surprisingly large!

The c 3 - c 4 profile t ( C 4 , T ) 1 12 1 16 1 128 1 432 0 t ( C 3 , T ) 1 1 1 72 32 8

Aside: The upper bound Upper bound is c 4 ( T ) ≤ 2 3 c 3 ( T ) + o (1). Bottom left construction ( c 3 = 0 , c 4 = 0): transitive tournament Upper right construction ( c 3 = 1 / 8 , c 4 = 1 / 12): the “circular” tournament, edges directed from v i to v i +1 , . . . v i + n / 2 for each i (indices modulo n )

b b b The spectral approach tournament matrix: non-negative square matrix satisfying A + A T = matrix of ones. tournament �→ tournament matrix by taking the usual (directed) adjacency matrix and replacing the diagonal entries with 1 / 2. B A C   1 / 2 1 0 0 1 / 2 0   1 1 1 / 2

The spectral approach tournament matrix: non-negative square matrix satisfying A + A T = matrix of ones. tournament �→ tournament matrix by taking the usual (directed) adjacency matrix and replacing the diagonal entries with 1 / 2. Fact : If A is the tournament matrix corresponding to a T of order n and ℓ ≥ 3, then the number of homomorphisms from C ℓ to T is Tr( A ℓ ) + O ( n ℓ − 1 ). density: σ ℓ ( A ) := 1 n ℓ Tr A ℓ ↔ c ℓ ( T ) Fact : n Tr ( A ℓ ) = � λ ℓ i , i =1 where the λ i are the eigenvalues of A .

Rephrasing the problem Minimise c 4 ( T ) for fixed c 3 ( T ) ⇒ Minimise Tr( A 4 ) for fixed Tr( A 3 ) ⇐ ⇐ ⇒ Minimise the sum of 4th powers of the eigenvalues of A , given a fixed the sum of 3rd powers The main property of A that we know is that the sum of eigenvalues is n / 2.

Rephrasing the problem Minimise c 4 ( T ) for fixed c 3 ( T ) ⇒ Minimise Tr( A 4 ) for fixed Tr( A 3 ) ⇐ ⇐ ⇒ Minimise the sum of 4th powers of the eigenvalues of A , given a fixed the sum of 3rd powers The main property of A that we know is that the sum of eigenvalues is n / 2. Lemma (Linial & Morgenstern) Let x 1 , . . . , x n be non-negative real numbers summing to 1 / 2. Then x 4 1 + · · · + x 4 n ≥ g ( x 3 1 + . . . x 3 n ) .

Rephrasing the problem Minimise c 4 ( T ) for fixed c 3 ( T ) ⇒ Minimise Tr( A 4 ) for fixed Tr( A 3 ) ⇐ ⇐ ⇒ Minimise the sum of 4th powers of the eigenvalues of A , given a fixed the sum of 3rd powers The main property of A that we know is that the sum of eigenvalues is n / 2. Lemma (Linial & Morgenstern) Let x 1 , . . . , x n be non-negative real numbers summing to 1 / 2. Then x 4 1 + · · · + x 4 n ≥ g ( x 3 1 + . . . x 3 n ) . Problem: What if the eigenvalues are complex?

Taking a step back general case: A has eigenvalues ρ n , the spectral radius r 1 n , . . . , r k n , the remaining real eigenvalues ( a 1 ± ι b 1 ) n , . . . , ( a ℓ ± ι b ℓ ) n , conjugate pairs of complex eigenvalues

Taking a step back general case: A has eigenvalues ρ n , the spectral radius r 1 n , . . . , r k n , the remaining real eigenvalues ( a 1 ± ι b 1 ) n , . . . , ( a ℓ ± ι b ℓ ) n , conjugate pairs of complex eigenvalues Optimization problem Spectrum Parameters: reals c 3 ∈ [0 , 1 / 8] and ρ ∈ [0 , 1 / 2] non-negative integers k and ℓ Variables: real numbers r 1 , . . . , r k , a 1 , . . . , a ℓ and b 1 , . . . , b ℓ Constraints: 0 ≤ r 1 , . . . , r k ≤ ρ 0 ≤ a 1 , . . . , a ℓ k ℓ ρ + � r i + 2 � a i = 1 / 2 i =1 i =1 ℓ k ρ 3 + r 3 � a 3 i − 3 a i b 2 � � i + 2 � = c 3 i i =1 i =1 ℓ k minimize ρ 4 + r 4 � a 4 i − 6 a 2 i b 2 i + b 4 � Objective: � i + 2 � i i =1 i =1

Structure of optimal solutions Key lemma Let r 1 , . . . r k , a 1 , . . . a ℓ , b 1 , . . . , b ℓ be an optimal solution to the optimisation problem. Then one of the following holds: 1 There exist positive reals r ′ and r ′′ such that r 1 , . . . , r k ∈ { 0 , r ′ , r ′′ , ρ } and ( a 1 , b 1 ) , . . . , ( a ℓ , b ℓ ) ∈ { (0 , 0) , ( r ′ , 0) , ( r ′′ , 0) } . 2 There exist reals a ′ and b ′ � = 0 such that r 1 , . . . , r k ∈ { 0 , ρ } and ( a 1 , b 1 ) , . . . , ( a ℓ , b ℓ ) ∈ { (0 , 0) , ( a ′ , b ′ ) , ( a ′ , − b ′ ) } .

Structure of optimal solutions Key lemma Let r 1 , . . . r k , a 1 , . . . a ℓ , b 1 , . . . , b ℓ be an optimal solution to the optimisation problem. Then one of the following holds: 1 All eigenvalues are real 2 There exist reals a ′ and b ′ � = 0 such that r 1 , . . . , r k ∈ { 0 , ρ } and ( a 1 , b 1 ) , . . . , ( a ℓ , b ℓ ) ∈ { (0 , 0) , ( a ′ , b ′ ) , ( a ′ , − b ′ ) } .

Structure of optimal solutions Key lemma Let r 1 , . . . r k , a 1 , . . . a ℓ , b 1 , . . . , b ℓ be an optimal solution to the optimisation problem. Then one of the following holds: 1 All eigenvalues are real 2 Up to multiplicity, there’s only real eigenvalue and one pair of complex eigenvalues

A key ingredient Our optimisation problem involves: An objective function f (sum of 4th powers of e-values) Two constraint functions g 1 = 1 / 2 and g 2 = c 3 (sum of e-values and 3rd powers of e-values) Some boundary conditions on the r i and a i

A key ingredient Our optimisation problem involves: An objective function f (sum of 4th powers of e-values) Two constraint functions g 1 = 1 / 2 and g 2 = c 3 (sum of e-values and 3rd powers of e-values) Some boundary conditions on the r i and a i The method of Lagrange multipliers tells us that the extrema of f in the feasible set occur at the boundary of the feasible set, or where ∇ f = λ 1 ∇ g 1 + λ 2 ∇ g 2 for some constants λ 1 , λ 2 .

Lagrange multipliers The extrema of f in the feasible set occur at the boundary of the feasible set, or where ∇ f = λ 1 ∇ g 1 + λ 2 ∇ g 2 for some constants λ 1 , λ 2 .

Theorem revisited Theorem (C., Grzesik, Kr´ al’, Noel 2018) If c 3 ( T ) ≥ 1 / 72 , then c 4 ( T ) ≥ g ( c 3 ( T )) + o (1) .