Flag Algebra Methods (more formal approach) Bernard Lidick y 6th - - PowerPoint PPT Presentation

Flag Algebra Methods (more formal approach)

Bernard Lidick´ y

6th Lake Michigan Workshop on Combinatorics and Graph Theory

Apr 7, 2019

Outline

• Flag Algebras “definitions”
• First try for Mantel’s theorem
• More automatic approach
• Additional constraints
• maybe break
• Define flag algebras
• Graph sequences and homomorphisms
• Tur´

an’s Theorem (in limit)

• Finally Mantel’s Theorem (for real)
• mega break
• Applications

2

Building an algebra

Let F denote the set of all graphs (up to isomorphism). Let RF be the set of all finite formal linear combinations of graphs. 2 ·

• 4 ·

− 1 3 · + 2 3 · + π

• It comes with natural addition and multiplication by a real number

(think of a vector ∈ RF.) How to define multiplication of two elements in RF?

3

Towards multiplication

Let F denote the set of all graphs. Let F1, F2, F ∈ F such that |F1| + |F2| ≤ |F|.

Definition

Let X1, X2 ⊆ V (F) be random disjoint of sizes |F1| and |F2|. P(F1, F2; F) is the probability F[Xi] ∼ = Fi for both i. P

• ,

;

• = 2

3 P

• ,

;

• = 1

2 P

• ,

;

• = 1

6 P

• ,

;

• = 1

6

4

Towards multiplication

Let F denote the set of all graphs. Let F1, F2, F ∈ F such that |F1| + |F2| ≤ |F|.

Definition

Let X1, X2 ⊆ V (F) be random disjoint of sizes |F1| and |F2|. P(F1, F2; F) is the probability F[Xi] ∼ = Fi for both i. P

• ,

;

• = 2

3 P

• ,

;

• = 1

2 P

• ,

;

• = 1

6 P

• ,

;

• = 1

6 −2

4

Multiplication

Let F denote the set of all graphs. Let F1, F2 ∈ F. We need F1 · F2 ∈ RF. Define F1 · F2 :=

• F∈Fℓ

P(F1, F2; F)F, where |F1| + |F2| = ℓ. Notice there is NO +o(1). · = P

• ,

;

• ·

+ P

• ,

;

• ·

= 0 + 1 6 + 1 3 + 1 2 + 1 2 + · · · This extends to a · b for a, b ∈ RF.

5

Factorizing

Recall from previous presentation we had identities like = 0 + 1 3 + 2 3 + 1 Let K be linear subspace generated by F −

• F ′∈Fℓ

P(F, F ′)F ′ (= 0), where F ∈ F and |F| ≤ ℓ. Algebra A is RF factorized by K. a, b ∈ RF are in one equivalence class if a = b + c for some c ∈ K Note: Think of RF and A as Z × Z and Q. (correctness of definitions proved by Razborov)

6

Algebra A

• F is the set of all graphs.
• Fℓ is on ℓ vertices.
• RF formal linear combinations
• K := span(F −

F ′∈Fℓ P(F, F ′)F ′)

• A is RF factorized by K.
• addition in A comes from RF
• F1 · F2 =

F∈Fℓ P(F1, F2; F)F,

• F ∈ F is called a flag.
• informally called unlabeled flags

7

Convergent Graph Sequence

Let F denote the set of all graphs.

Definition

A sequence of graphs (Gn)n∈N is convergent if for every finite graph H, limn→∞ P(H, Gn) exists. Examples: lim

n→∞ P(H, Kn) =

• 1

if H ∼ = Km

• therwise

lim

n→∞ P(H, Kn,n) =

       1/2 if H ∼ = K2 3/4 if H ∼ = P2 . . . lim

n→∞ P(H, Pn) =

• 1

is |E(H)| = 0

• therwise

Gives map F → [0, 1] or a point in [0, 1]F.

8

Convergent Graph Sequence

Let F denote the set of all graphs.

Definition

A sequence of graphs (Gn)n∈N is convergent if for every finite graph H, limn→∞ P(H, Gn) exists. For all H ∈ F lim

n→∞ P(H, Kn,n) = lim n→∞ P(H, Kn,n ∪ K1) = lim n→∞ P(H, Kn,n + K1)

In particular for H ∈

• ,
• .

Small changes in (Gn)n∈N are not noticeable.

9

Positive Homomorphisms

Let Hom(A, R) be the set of all homomorphisms from A to R. I.e. for any φ ∈ Hom(A, R) and a, b ∈ A:

• φ(a + b) = φ(a) + φ(b)
• φ(a · b) = φ(a) · φ(b).

Since p(H, G) ∈ [0, 1], we consider only Hom+(A, R), i.e. φ(H) ≥ 0 for all H ∈ F Notice φ(∅) = 1.

Theorem (Razborov)

Hom+(A, R) corresponds exactly to the convergent sequences.

Theorem (Felix; Podolski; (only for graphs))

Let a ∈ RF. φ(a) = 0 for all φ ∈ Hom+(A, R) iff a ∈ K.

10

Theorem (Mantel)

Every triangle-free graph on n vertices has at most n2/4 edges.

Theorem (Mantel from previous presentation)

For a large graph if = 0 then ≤ 1

2 + o(1).

Theorem (Mantel with Homomorphisms)

For every φ ∈ Hom+(A, R) holds that if φ

• = 0 then φ
• ≤ 1

2.

11

Example from last time 0 ≤ 3 · − − + 3 · +o(1) We want to find a ∈ A such that for EVERY φ ∈ Hom+(A, R) we have φ(a) ≥ 0. We write a ≥ 0.

Theorem (Hatami and Norin 11)

Determining if a ≥ 0 is not algorithmically decidable. Norin: “Extremal combinatorics remains an art”. But we can still generate a lot of them!

12

Algebra Aσ

• vertices of σ ∈ F are labeled by 1, . . . , |σ|.
• Fσ is the set of all graphs each containing a fixed induced

labeled copy of σ.

• Fσ

ℓ is on ℓ vertices.

• RFσ formal linear combinations
• Kσ := span(F −

F ′∈Fσ

ℓ P(F, F ′)F ′)

• Aσ is RFσ factorized by Kσ.
• addition in Aσ comes from RFσ
• F1 · F2 =

F∈F σ

ℓ P(F1, F2; F)F,

• F ∈ Fσ is called a σ-flag.
• σ is called a type

1 2 1 2 3 1 2 1 2

13

F1 · F2 =

F∈F σ

ℓ P(F1, F2; F)F

1 2 · · · |σ| X1 X2 F

Pick randomly X1, X2 ⊂ V (F) such that X1 ∩ X2 are exactly all labeled vertices and |Xi| = |Fi|. P(F1, F2; F) is the probability that X1 ∼ = F1 and X2 ∼ = F2.

14

Averaging (unlabeling) operator

Let F = (G, θ) be a σ-flag, where θ : 1, . . . , |σ| → V (G). Define Fσ = qθ(F) · G, where qθ(F) is the probability that (G, θ′) is isomorphic to F for a random injective θ′ : 1, . . . , |σ| → V (G).

• 1

2

• σ

= 6 20 Linear extension gives the averaging operator ·σ : RFσ → RF. It is a linear mapping, not a homomorphism. F1 + F2σ = F1σ + F2σ F1 · F2σ might NOT be F1σ · F2σ

15

Aσ, A and Hom+

For any φ ∈ Hom+(A, R) and any a ∈ Aσ φ(a · aσ) ≥ 0.

Lemma (Razborov)

Cauchy-Schwarz inequality for all φ ∈ Hom+(A, R) φ(a2σ · b2σ) ≥ φ(ab2

σ).

Special cases: φ(a2σ) ≥ φ(a2

σ)

φ(σσ) φ(a2 ) ≥ φ(a2) Uses special probability distribution on Hom+(Aσ, R) related to φ.

16

Let σ and Hom+(A, R) be fixed. This gives G1, G2, G3, . . . To make φσ ∈ Hom+(Aσ, R), we need a sequence of labeled graphs. Incorrect: Take a copy of σ in each Gi and get a convergent G σ

1 , G σ 2 , G σ 3 , . . .

Correct: Fix Gi. Randomly label a copy of σ and get G σ

i . This

gives P(F, G σ

n ) for all F ∈ Fσ.

By a random σ, we get a probability distribution Pσ

Gn on the

functions P(., G σ

n ).

These Pσ

Gn then weakly converge to a (unique) probability

distribution Pσ

φ on φσ ∈ Hom+(Aσ, R).

A crucial feature: if a ∈ Aσ and φ ∈ Hom+(A, R), then φ(σσ) · EPσ

φ [φσ(a)] = φ(aσ).

This can be viewed as an analogue of P(B) · P(A|B) = P(A ∧ B). Skip this. Hard to follow.

17

A crucial feature: if a ∈ Aσ and φ ∈ Hom+(A, R), then φ(σσ) · EPσ

φ [φσ(a)] = φ(aσ).

(1) Bonus: For a given φ ∈ Hom+(A, R), exists a unique probability distribution Pσ

φ satisfying (1).

(1) is especially useful when φσ(a) ≥ 0 with probability one. It gives φ(aσ) ≥ 0. In particular, for any φ ∈ Hom+(A, R) and any a ∈ Aσ φ(a · aσ) ≥ 0. Simplified notation is just a · aσ ≥ 0. Skip this. Hard to follow.

18

Consider sequence (Kn,n + K1)n with φ ∈ Hom+(A, R) and σ = 1.

1 1

There exists φ′ ∈ Hom+(Aσ, R) such that φ′

• 1
• = 1.

But φ

• 1
• σ
• = φ(σσ) · EPσ

φ

• φσ
• 1
• = 1

2 Flag algebras do not see ’rare vertices’. Skip this. Hard to follow.

19

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

1

2

1

• ≤ φ

1 2 1

• a2

1 ≤ a21

20

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

• 2

= φ

1

2

1

• ≤ φ

1 2 1

• = φ

1

• 1
• a2

1 ≤ a21

20

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

• 2

= φ

1

2

1

• ≤ φ

1 2 1

• = φ

1

• 1
• φ
• 2

≤ φ 1 3

• a2

1 ≤ a21

20

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

• 2

= φ

1

2

1

• ≤ φ

1 2 1

• = φ

1

• 1
• φ
• 2

≤ φ 1 3

• φ

1 3 · + 2 3 ·

• = φ
• a2

1 ≤ a21

20

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

• 2

= φ

1

2

1

• ≤ φ

1 2 1

• = φ

1

• 1
• φ
• 2

≤ φ 1 3

• φ

2 3 ·

• ≤ φ

1 3 · + 2 3 ·

• = φ
• a2

1 ≤ a21

20

Mantel’s Theorem again

For all φ ∈ Hom+(A, R) σ is K1 denoted by 1. Over triangle-free graphs i.e. φ(K3) = 0 φ

• 2

= φ

1

2

1

• ≤ φ

1 2 1

• = φ

1

• 1
• φ
• 2

≤ φ 1 3

• φ

2 3 ·

• ≤ φ

1 3 · + 2 3 ·

• = φ
• φ
• 2

≤ φ 1 3

• ≤ 1

2φ

• hence

φ

• ≤ 1

2 a2

1 ≤ a21

20

A and forbidden graphs

In previous example, we used = 1 3 · + 2 3 · rather than = 1 3 · + 2 3 · + 1 This can be done formally by defining A∇ over F∇, which are all triangle free graphs. Works for any forbidden graphs.

21

Forbidding triangles can be done formally by defining A∇ over F∇, which are all triangle free graphs. Works for any forbidden graphs.

Theorem (Mantel - Homomorphisms from A)

For every φ ∈ Hom+(A, R) holds that if φ

• = 0 then φ
• ≤ 1

2.

Theorem (Mantel - Homomorphisms from A∇)

For every φ ∈ Hom+(A∇, R) holds φ

• ≤ 1

2.

Formulation with A∇ is particularly useful in practical calculations if |F∇

ℓ | < |Fℓ|.

22

Forbidding triangles can be done formally by defining A∇ over F∇, which are all triangle free graphs. Works for any forbidden graphs.

Theorem (Mantel - Homomorphisms from A)

For every φ ∈ Hom+(A, R) holds that if φ

• = 0 then φ
• ≤ 1

2.

Theorem (Mantel - Homomorphisms from A∇)

For every φ ∈ Hom+(A∇, R) holds φ

• ≤ 1

2.

Formulation with A∇ is particularly useful in practical calculations if |F∇

ℓ | < |Fℓ|.

ℓ |Fℓ| |F∇

ℓ |

3 4 3 4 11 7 5 34 14 6 156 38 7 1,044 107 8 12,346 410 9 274,668 1,897 10 12,005,168 12,172 11 1,018,997,864 105,071

22

Let a ∈ Aσ. Recall a · aσ ≥ 0. Let a = X Tv, where X ∈ (Fσ)m and v ∈ Rm. a =

1

+ 2

1

X T =

• 1

,

1

• v =

1 2

• a2 = a · a = X Tv · (X Tv) = X TvvTX = X TMX
• 1

+ 2

1

2 =

• 1

,

1

1 2 2 4

1

,

1

T

Observation

Let σ be fixed. If M 0 and X is a vector (Fσ

ℓ )n, then

∀φ ∈ Hom+(A, R) φ

• X TMXσ
• ≥ 0. Similar to φ
• a∈Aσ

a2σ

• ≥ 0.

23

Last time - Density approach

0 ≤ 1 n

• v
• v

,

v

a c c b

v

,

v

T = 1 n

• v

a

v ?

+ b

v ?

+ 1 2c

v ?

+ 1 2c

v ?

= a + a + 2c 3 + b + 2c 3 + b

24

Homomorphism approach

For all φ ∈ Hom+(A, R) 0 ≤ φ

1

,

1

a c c b

1

,

1

T

σ

• 25

Homomorphism approach

For all φ ∈ Hom+(A, R) 0 ≤ φ

1

,

1

a c c b

1

,

1

T

σ

• = φ

a

• 1

+ 1

• + c
• 1

+ 1

• + b
• 1

+ 1

• σ
• 25

Homomorphism approach

For all φ ∈ Hom+(A, R) 0 ≤ φ

1

,

1

a c c b

1

,

1

T

σ

• = φ

a

• 1

+ 1

• + c
• 1

+ 1

• + b
• 1

+ 1

• σ
• = φ
• a
• 1
• σ

+

• 1
• σ
• + c
• 1
• σ

+

• 1
• σ
• + b
• 1

25

Homomorphism approach

For all φ ∈ Hom+(A, R) 0 ≤ φ

1

,

1

a c c b

1

,

1

T

σ

• = φ

a

• 1

+ 1

• + c
• 1

+ 1

• + b
• 1

+ 1

• σ
• = φ
• a
• 1
• σ

+

• 1
• σ
• + c
• 1
• σ

+

• 1
• σ
• + b
• 1

= φ

• a

1 3 +

• + c

2 3 + 2 3

• + b
• + 1

3

25

Homomorphism approach

For all φ ∈ Hom+(A, R) 0 ≤ φ

1

,

1

a c c b

1

,

1

T

σ

• = φ

a

• 1

+ 1

• + c
• 1

+ 1

• + b
• 1

+ 1

• σ
• = φ
• a
• 1
• σ

+

• 1
• σ
• + c
• 1
• σ

+

• 1
• σ
• + b
• 1

= φ

• a

1 3 +

• + c

2 3 + 2 3

• + b
• + 1

3 = φ

• a ·

+ a + 2c 3 · + b + 2c 3 · + b ·

• 25

Tur´ an Theorem

Tur´ an question

What is the maximum number of edges in Kt-free n-vertex graphs? We seek answer as ≤ t−2

t−1.

Computer can do at most t = 8. We do by hand induction on t. Main tool a2σ ≥ a2

σ

σσ . Our first goal is to prove ≥ 2 ·

• − 1

2

• We do not write φ(. . .) any more but it is there!

26

Fill the rest of the following 2

• 1

+ 1

• 1

= 2

• 1

2 1

≥ 2

• 1

2

1

= 2

2

2 + 2 3 ≥ 2

2

a21 ≥ a2

1

27

Fill the rest of the following 2

• 1

+ 1

• 1

= 2

• 1

2 1

≥ 2

• 1

2

1

= 2

2

2 + 2 3 ≥ 2

2

Recall + 2 3 · + 1 3 · = . By subtracting, we obtain ≥ − 1 3 · ≥ 2

2

− = 2 ·

• − 1

2

• a21 ≥ a2

1

27

We got the base of induction. ≥ 2 ·

• − 1

2

• (2)

The real goal is Kt+1 ≥ t · Kt ·

• − t − 1

t

• .

(3) Inequality (2) is just (3) with t = 2. Why is (3) proving the Tur´ ans theorem?

28

We got the base of induction. ≥ 2 ·

• − 1

2

• (2)

The real goal is Kt+1 ≥ t · Kt ·

• − t − 1

t

• .

(3) Inequality (2) is just (3) with t = 2. Why is (3) proving the Tur´ ans theorem? Let (Gn)n∈N is Kt+1 free sequence with φ ∈ Hom+(A, R) . Then φ(Kt+1) = 0. This gives in (3) that 0 ≥ t · φ(Kt) ·

• φ (K2) − t − 1

t

• .

If φ(Kt) > 0 then φ(K2) ≤ t−1

t .

If φ(Kt) = 0 induction on t gives φ(K2) ≤ t−2

t−1.

28

Induction step for t = 4. Goal is ≥ 3 · ·

• − 2

3

• 1

2

+

1 2

• σ

=

• 1

2

2

• σ

≥

• 1

2

• 2

σ

+ 1 6 ≥

2

a2σ ≥ a2

σ

σσ

29

+ 1 6 ≥

2

and ≥ 2 ·

• − 1

2

• When we use induction, which means (2) and we get

2

= · ≥ · 2

• − 1

2

• .

Hence we get after scaling by 3

2

3 2 + 1 4 ≥ · 3

• − 1

2

• 30

3 2 + 1 4 ≥ · 3

• − 1

2

• Add 1

2 times the following

= + 1 2 + · · · ≥ + 1 2 and get ≥ · 3

• − 1

2

• − 1

2 = · 3

• − 2

3

• In a similar way we get also

Kt+1 ≥ t · Kt ·

• − t − 1

t

• 31

Actual proof of Mantel’s theorem

Theorem (Mantel for all n)

If G is a triangle free graph on n vertices, then |E(G)| ≤ n2/4. For contradiction let H be a triangle-free graph on n0 vertices, with |E(H)| = cn2

0, where c > 1 4.

32

Actual proof of Mantel’s theorem

Theorem (Mantel for all n)

If G is a triangle free graph on n vertices, then |E(G)| ≤ n2/4. For contradiction let H be a triangle-free graph on n0 vertices, with |E(H)| = cn2

0, where c > 1 4.

Let B(H, k) be a blow-up of H by k vertices. (B(H, k))k∈N is convergent, gives φH ∈ Hom+(A, R).

32

(B(H, k))k∈N is convergent, gives φH ∈ Hom+(A, R).

Observation

B(H, k) is a triangle-free graph. Hence φH(K3) = 0 |E(B(H, k))| = cn2k2. Hence φH(K2) = lim

k→∞

|E(B(H, k)| |V ((B(H,k)|

2

= lim

k→∞

cn2

0k2

n0k

2

= 2c > 1 2 Before, we proved that φ(K2) ≤ 1

2 whenever φ(K3) = 0,

contradiction. Tur´ an’s theorem can be proved with the same trick.

33

See you in the afternoon! (maybe)

• Flag Algebras “definitions”
• First try for Mantel’s theorem
• More automatic approach
• Additional constraints
• maybe break
• Define flag algebras
• Graph sequences and homomorphisms
• Tur´

an’s Theorem (in limit)

• Finally Mantel’s Theorem (for real)
• mega break
• Applications

34