9 Ramsey–Turán

Theorem 9.1 (Turán). Assuming that:

  • G is an n-vertex graph with GKr+1

Then
e(G)(11r)n22.

In other words,

ex(n,Kr+1)=(11r)n22+o(n2).

Example:

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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, GHα(G)k}.

Example. RT(n,K3,k)(k1)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.

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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 ViVj if ij, 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={xVi:|N(x)Vj|𝜀|Vj|,|N(x)Vk|𝜀2|Vk|}.

We know

|Vi|(12𝜀)|Vi|>0,

so let xVi.

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Now let

{yVjN(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,vN(x)N(y)Vk such that uvG. 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 (12𝜀)nL>δn vertices that remain in Vi. So there exist u,vVi, uvG.

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Now note

|N(u)N(v)Vj|𝜀|Vj|.

Now use independence number property to find x,yN(u)N(v)Vi such that xyG. 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)2n28+4𝜀n2

Theorem 9.5 (Bollobás–Erdős, 1970s). There exists a graph G on n vertices with GK4, 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.

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“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 2𝜀d.

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A little more formally:

Let d slowly, and let 𝜀0 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

  • xR, yB: xy when |xy|<2𝜀d.

  • x,xR: xx when |xy|2𝜀d.

  • Same for y,yB

Properties:

  • e(G)=n28+o(n2).

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    xy iff x,yc𝜀d. Let’s imagine x=(1d,,1d).

    (1di=1dyic𝜀d).(∗)

    Varyi1d, Vari=1dyi1. So () is (Zc𝜀)=12Θ(𝜀), where ZN(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,zR and assume they form a K3. Then

    0x+y+z2=x+y+z,x+y+z=3+2(x,y+y,z+x,z)=9(xy2+yz2+xz2)=93(2𝜀d)2<0

    Contradiction.

    G is K4-free: consider x+xyy where x,xR and y,yB.

  • Finally, we need to check α(G)=o(n). Let IV(G) be independent. Let IR, and without loss of generality say |I||I|2. Let

    S=xIRx

    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 SSd with diameter

    <2cd.

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

    Hence |I|n0, 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 GG can be made triangle-free by removing 𝜀n2 edges.
  • (3) Every triangle-free graph on [n] is contained in some GG.

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

# of triangle-free graphsGCn,𝜀2HG1GC2n24n2𝜀2n22n24+𝜀n2

Proof of Lemma 9.7. Given a graph H on [k] and a partition [n]=V1Vk, we define G to be the blow up of H onte {Vi} by V(G)=[n] and include all of ViVj into G whenever ijH.

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 kl.
  • (2) Let u[n]=V1Vk be an equipartition, and blow up H onto {Vi}.
  • (3) Throw in 𝜀2n2 edges in any way.

Check |G| is small:

|G|2L2step 1(n!2k)step 2n2𝜀2n2step 32𝜀n2.

Each GG 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 GG. We apply Szemerédi Regularity Lemma with parameter l=10𝜀, 𝜀22, to get a partition [n]=V1Vk with kL. Consider the reduced graph R=R𝜀2,𝜀22. Note R is triangle-free. So use GR blown up to {V1} plus 𝜀2n extra edges.

Reminder of Szemerédi Regularity Lemma:

Theorem 9.8 (Regularity Lemma). For 𝜀>0, l, there exists L(l,𝜀) such that the following holds: Let G be a graph. Then there exists an equipartition V(G)=V1Vk with lkL(l,𝜀), where {Vi} equipartition and all but 𝜀k2 pairs (Vi,Vj) are 𝜀-uniform.

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

Start with V(G)=V1Vl, arbitrarily into an equipartition. At a general step, we are given

V(G)=V1Vk

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 WijVi, WjiVj, 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

1k2i,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

()=1r2s,t=1rd(Vis,Vjt)2d(Vi,Vj)2=1r2s,t(d(Vis,Vjt)d(Vi,Vj))2

Here we are using

1r2s,td(Vis,Vjt)=d(Vi,Vj).

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

S={s:VisWij}T={t:VjtWji}

So

()1r2sStT(d(Vis,Vjt)d(Vi,Vj))2|S||T|r21|S||T|sStT(d(Vis,Vjt)d(Vi,Vj))2𝜀2(1|S||T|sStTd(Vis,Vjt)d(Vi,Vj))2=𝜀2(d(Wij,Wji)d(Vi,Vj))2𝜀4

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 :

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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

(n4kk)2kk<n2k𝜀n.

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.