9 Ramsey–Turán

Theorem 9.1 (Turán). Assuming that:

  • G is an n-vertex graph with G⁄⊃Kr+1

Then
e(G)≤(1−1r)n22.

In other words,

ex⁡(n,Kr+1)=(1−1r)n22+o(n2).

Example:

PIC

Definition 9.2 (Ramsey–Turán number). Let H be a graph. Then we define

RT(n,H,k)=max⁡{e(G):G is on n vertices, G⁄⊃H, α(G)≤k}.

Example. RT(n,K3,k)≤(k−1)n2, since the neighbourhood of every vertex must be an independent set (using the fact that we are triangle-free).

Theorem 9.3 (Szemerédi). RT(n,K4,o(n))≤n28+o(n2). This was shown to be sharp by Bollobás and Erdős.

Theorem 9.4. For every 𝜀>0, there exists δ>0 such that if α(G)≤δn and G is K4-free, we have

e(G)≤n28+𝜀n2.

PIC

Proof. Given 𝜀>0, let L=L(10𝜀,𝜀2) and δ=𝜀22L and apply Szemerédi Regularity Lemma with parameters 𝜀2 and l=10𝜀. We define the graph on vertex set {V1,…,Vk}. Define the reduced graph R𝜀,𝜀2 on V(Rα,𝜀)={V1,…,Vk} and Vi∼Vj if i≠j, 𝜀2-uniform and d(Vi,Vj)≥𝜀.

Claim 1: R𝜀,𝜀2 is triangle-free.

If there exist Vi,Vj,Vk distinct with

d(Vi,Vj),d(Vj,Vk),d(Vi,Vk)≥𝜀

and are pairwise 𝜀2-uniform. Then consider

Vi′={x∈Vi:|N(x)∩Vj|≥𝜀|Vj|,|N(x)∩Vk|≥𝜀2|Vk|}.

We know

|Vi′|≥(1−2𝜀)|Vi|>0,

so let x∈Vi′.

PIC

Now let

{y∈Vj∩N(x):|(y)∩N(x)∩Vk|≥𝜀|N(x)∩Vk|}.

Again this set is non-empty, so choose y in it.

So now note,

|N(x)∩N(y)∩Vk|≥𝜀|N(x)∩Vk|≥𝜀2|Vk|≥𝜀2(nL)>δ.

So there exist u,v∈N(x)∩N(y)∩Vk such that uv∈G. But then x,y,u,v form a K4, contradiction.

Claim 2: d(Vi,Vj)≤12+2𝜀, for all i<j.

Throw away 𝜀|Vi| vertices in Vi with degree <12+𝜀 to Vj. There are at least (1−2𝜀)nL>δn vertices that remain in Vi. So there exist u,v∈Vi, uv∈G.

PIC

Now note

|N(u)∩N(v)∩Vj|≥𝜀|Vj|.

Now use independence number property to find x,y∈N(u)∩N(v)∩Vi such that xy∈G. Then x,y,u,v is a K4, contradiction.

We now estimate the number of edges in G using the claims.

e(G)≤e(R)(nk)2(12+2𝜀)+k2𝜀(nk)2+𝜀k2(nk)2≤n28+4𝜀n2□

Theorem 9.5 (Bollobás–Erdős, 1970s). There exists a graph G on n vertices with G⁄⊃K4, α(G)=o(n) and e(G)=n28+o(n2).

Sketch. Let d→∞ slowly as n→∞. We consider the sphere Sd in d+1 dimensions.

PIC

“join a red point to a blue point if they are reasonably close” (say having inner product bigger than some small quantity). “join red to red and join blue to blue if they have distance ≥2−𝜀d.

PIC

A little more formally:

Let d→∞ slowly, and let 𝜀→0 slowly.

  • Partition Sd into n2 regions of equal measure and with diameter ≤12d.

  • Let R={x1,…,xn2}, B={y1,…,yn2} be a set of points, with one from each of these regions.

We define G by

  • x∈R, y∈B: x∼y when |x−y|<2−𝜀d.

  • x,x′∈R: x∼x′ when |x−y|≥2−𝜀d.

  • Same for y,y′∈B

Properties:

  • e(G)=n28+o(n2).

    PIC

    x∼y iff ⟨x,y⟩≥c𝜀d. Let’s imagine x=(1d,…,1d).

    ℙ(1d∑i=1dyi≥c𝜀d).(∗)

    Var⁡yi∼1d, Var⁡∑⁡i=1dyi∼1. So (∗) is ≈ℙ(Z≥c𝜀)=12−Θ(𝜀), where Z∼N(0,1). So every vertex has degree ≈n4, so e(G)≈n28.

  • Now we check it is K4-free

    First note G[R],G[B] are triangle-free. Let x,y,z∈R and assume they form a K3. Then

    0≤∥x+y+z∥2=⟨x+y+z,x+y+z⟩=3+2(⟨x,y⟩+⟨y,z⟩+⟨x,z⟩)=9−(∥x−y∥2+∥y−z∥2+∥x−z∥2)=9−3(2−𝜀d)2<0

    Contradiction.

    G is K4-free: consider ∥x+x′−y−y′∥ where x,x′∈R and y,y′∈B.

  • Finally, we need to check α(G)=o(n). Let I⊂V(G) be independent. Let I′∩R, and without loss of generality say |I′|≥|I|2. Let

    S=⋃x∈I′Rx

    where Rx are the regions that we broke the sphere into at the start. We have μ(S)=2|I′|n (where μ is such that μ(Sd)=1). Note S⊂Sd with diameter

    <2−cd.

    Fact from life: this implies μ(S)→0 as d→∞.

    Hence |I|n→0, i.e. |I|=o(n).

□

What is the number of triangle-free graphs on n vertices?

Theorem 9.6 (Erdős, Klietman, Rothchild, 70s). The number of triangle-free graphs on n vertices is 2n24+o(n2).

Remark. “≥” comes from considering all subgraphs of Kn2,n2. Can sort of think of this as saying ‘almost all triangle-free graphs are bipartite’.

Lemma 9.7. For every 𝜀>0 and n large enough, there exists a family of graphs G=Gn,𝜀 on [n] with the following properties:

  • (1) |G|≤2𝜀n2.
  • (2) All G∈G can be made triangle-free by removing ≤𝜀n2 edges.
  • (3) Every triangle-free graph on [n] is contained in some G∈G.

Proof of Erdős, Klietman, Rothchild, 70s using Lemma 9.7. Let 𝜀>0 be given. Then for large enough n,

# of triangle-free graphs≤∑G∈Cn,𝜀2∑H⊂G1≤∑G∈C2n24⋅n2𝜀2n2≤2n24+𝜀n2□

Proof of Lemma 9.7. Given a graph H on [k] and a partition [n]=V1∪⋯∪Vk, we define G to be the blow up of H onte {Vi} by V(G)=[n] and include all of ViVj into G whenever ij∈H.

Let 𝜀>0 be given. We define G=Gn,𝜀 to be all graphs formed in the following steps. Let L=L(10𝜀,𝜀22).

  • (1) Let H be a triangle-free graph on [k] where k≤l.
  • (2) Let u[n]=V1∪⋯∪Vk be an equipartition, and blow up H onto {Vi}.
  • (3) Throw in 𝜀2n2 edges in any way.

Check |G| is small:

|G|≤2L2⏟step 1(n!2k)⏟step 2n2𝜀2n2⏟step 3≤2𝜀n2.

Each G∈G can be made triangle-free by removing ≤𝜀n2 edges: this is true by construction (in fact only need to remove 𝜀2n2 edges).

Finally, we need to check that every triangle-free graph on [n] is contained in some G∈G. We apply Szemerédi Regularity Lemma with parameter l=10𝜀, 𝜀22, to get a partition [n]=V1∪⋯∪Vk with k≤L. Consider the reduced graph R=R𝜀2,𝜀22. Note R is triangle-free. So use G⊂R blown up to {V1} plus 𝜀2n extra edges. □

Reminder of Szemerédi Regularity Lemma:

Theorem 9.8 (Regularity Lemma). For 𝜀>0, l∈ℕ, there exists L(l,𝜀) such that the following holds: Let G be a graph. Then there exists an equipartition V(G)=V1∪⋯∪Vk with l≤k≤L(l,𝜀), where {Vi} equipartition and all but 𝜀k2 pairs (Vi,Vj) are 𝜀-uniform.

PIC

Proof of Szemerédi Regularity Lemma. Given 𝜀,l and a graph G, we loop the following, iteratively refining a partition.

Start with V(G)=V1∪⋯∪Vl, arbitrarily into an equipartition. At a general step, we are given

V(G)=V1∪⋯∪Vk

an equipartition.

If there are ≤𝜀k2 non 𝜀-uniform pairs, then output {Vi}i=1k and done So assume >k2. For each non 𝜀-uniform pair Vi,Vj, we select Wij⊂Vi, Wji⊂Vj, Wij|,|Wji|≥𝜀|Vi| and |d(Wij,Wji)−d(Vi,Vj)|>𝜀.

Now for each Vi consider {Wij}i=1k. Let

Vi=⋃jUij

be a common refinement of these {Wij}. We now refine this partition further to an equipartition where each cell has size nk4k (this is a slight cheat ∗). Putting all these partitions together, we get

V(G)=⋃i,jVij.

Now loop again with this partition.

Next step: show that this loop eventually terminates.

Analysis of Algorithm: We define the index of a partition V1,…,Vk to be

1k2∑i,j(d(Vi,Vj))2.

We claim that as we go from V1,…,Vk to {Vij}, we increase the index by at least +𝜀52. Fix Vi, Vj and consider

(∗)=1r2∑s,t=1rd(Vis,Vjt)2−d(Vi,Vj)2=1r2∑s,t(d(Vis,Vjt)−d(Vi,Vj))2

Here we are using

1r2∑s,td(Vis,Vjt)=d(Vi,Vj).

PIC

Note (∗)≥0. Now assume that Vi, Vj are not 𝜀-uniform. Define

S={s:Vis⊂Wij}T={t:Vjt⊂Wji}

So

(∗)≥1r2∑s∈StT(d(Vis,Vjt)−d(Vi,Vj))2≥|S||T|r21|S||T|∑s∈St∈T(d(Vis,Vjt)−d(Vi,Vj))2≥𝜀2(1|S||T|∑s∈St∈Td(Vis,Vjt)−d(Vi,Vj))2=𝜀2(d(Wij,Wji)−d(Vi,Vj))2≥𝜀4

So since the number of non 𝜀-uniform pairs is ≥k2, we get an +𝜀4 boost from 𝜀 proportion of the parts. Also, the other pairs don’t hurt us. So index increases by ≥𝜀52.

How to fix the tiny lie at ∗:

PIC

We can assume that we start the algorithm with ≫1𝜀2 parts. Now each time we encounter a “rounding issue”, we just throw out the remainder of the cell to some bin I track. At a given step, we throw out at most

(n4k⋅k)⋅2k⋅k<n2k≪≪𝜀n.

Summing over all steps in algorithm, we still throw out at most ≪𝜀n. We just redistribute these vertices equally at the end.

  • This does not ruin the 𝜀-uniformity: your new partition is at most 2𝜀-uniform.

  • Also need to check that throwing out these bits does not affect the calculations by more than a small amount. □