8 Szemerédi Regularity Lemma

Informal statement

Let G be a graph, e(G)=αn2. We can partition V(G) into equal parts

V(G)=V1∪⋯∪Vk

where k≤L(𝜀) depends only on a “coarseness” parameter 𝜀>0, so that for all but 𝜀k2pairs 1≤i<j≤k, the graph between Vi, Vj “looks” quasi-random.

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So, in a sense, “every graph is somewhat random”.

Definition 8.1 (Edge density). Let G be a graph. Let U,V⊂V(G). First define

d(U,V)=e(U,V)|U||V|.

Definition 8.2 (eps-uniform). We say that (U,V) is 𝜀-uniform if ∀⁡U0⊂U and V0⊂V,

|d(U0,V0)−d(U,V)|≤𝜀

whenever |U0|≥𝜀|U| and |V0|≥𝜀|V|.

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Fact: Let G be a graph and let U,V⊂V(G) disjoint which are 𝜀-uniform. Let p=d(U,V). Then

|{x∈U:|N(x)∩V|>(p+𝜀)|V|}|≤𝜀|U||{x∈U:|N(x)∩V|<(p+𝜀)|V|}|≤𝜀|U|

Proof. Simply note that if

S={x∈U:|N(x)∩V|>(p+𝜀)|V|},

we have d(S,V)>(p+𝜀), hence |S|<𝜀|U| due to uniformity.

Second case is the same. □

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Lemma 8.3 (Counting lemma for triangles). Assuming that:

  • G a graph on vertex set V1∪V2∪V3 with Vi disjoint

  • |Vi|=n

  • 𝜀<p2

  • d(Vi,Vj)≥p and (Vi,Vj) is 𝜀-uniform

Then the number of triangles in G is
≥(1−2𝜀)(p−𝜀)3n3.

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Proof. Let

V′={x∈V1:|N(x)∩V2|,|N(x)∩V3|≥(p−𝜀)n}.

Note that |V′|≥(1−2𝜀)n. Now fix x∈V and note

e(N(x)∩V2,N(x)∩V3)≥(p−𝜀)|N(x)∩V2||N(x)∩V3|≥(p−𝜀)(p−𝜀)(p−𝜀)n2

using the 𝜀-uniform and p≥2𝜀. So summing over all x∈V′ gives us

(1−2𝜀)(p−𝜀)3n3

triangles. □

Definition 8.4 (Equipartition). Say that V(G)=V1∪⋯∪Vk is an equipartition if ||Vi|−|Vj||≤1 for all i,j.

Theorem 8.5 (Szemerédi Regularity Lemma). Assuming that:

  • 𝜀>0, l∈ℕ

Then there exists L=L(𝜀,l) such that the following holds: Let G be a graph. Then there is an equipartition of V(G)=V1∪⋯∪Vk, where l≤k≤L(l,𝜀) and such that all but 𝜀k2 pairs 1≤i<j≤k, we have that (Vi,Vj) is 𝜀-uniform.

Remark.

  • Technically, the Szemerédi Regularity Lemma also applies when e(G)=o(n2), but often it does not really tell us anything.

  • The dependence on 𝜀 of L(𝜀) is of tower type, i.e.

    L(𝜀)≤munder⏟𝜀−C.

    Gowers proved that

    L(𝜀)≥munder⏟𝜀c.

    So these tower bounds are necessary.

Theorem 8.6 (The triangle Removal Lemma). For all 𝜀>0, there exists δ>0 so that the following holds: Every graph G on n vertices with ≤δn3 triangles contains T⊂E(G) with |T|≤𝜀n2 and G−T⁄⊃K3.

Proof. Let 𝜀>0 be given, let L=L(10𝜀,𝜀4) and define δ=𝜀316L3. Let G be a graph with ≤δn3 triangles.

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Apply Szemerédi Regularity Lemma to G to find V(G)=V1∪⋯∪Vk, with k≤L(10𝜀,𝜀4). Now define

T={all edges between Vi, Vj whenever d(Vi,Vj)<𝜀2}∪{all edges between Vi, Vj such that (Vi,Vj) not 𝜀4-uniform}∪{all edges inside of the parts Vi}

We note

|T|≤k2⋅𝜀2(nk)2+𝜀4k2(nk)2+nk2k≤(𝜀2)k22(nk)2+𝜀8n2+n22k≤𝜀n2

Let’s see that G−T is triangle-free. If not, then x,y,z∈V(G) with x∈Vi, y∈Vj, z∈Vr. So we must have d(Vi,Vj)≥𝜀2 and (Vi,Vj) is 𝜀4-uniform.

So we can apply the “counting lemma” for triangles to find that the number of triangles is at least

(1−2𝜀)𝜀3(nk)3≥(12𝜀2L3)n3>δn3,

which contradicts the initial assumption. □

Remark.

  • This proof gives tower-type dependence

    δ=mfrac

    between δ and 𝜀.

  • It’s a big question to improve these bounds.

  • Best known is due to Fox, who proved that

    δ<mfrac

    is sufficient.

Theorem 8.7 (Roth). For 𝜀>0 there exists n(𝜀)∈ℕ such that if n≥n(𝜀) and A⊂[n] with |A|≥𝜀n, then A contains a,a+d,a+2d where d≠0.

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Proof. For 𝜀>0 given, let δ=δ(𝜀) be the δ from The triangle Removal Lemma. Let n(𝜀)≥1δ(𝜀). Now given a set A⊂[n], we define a graph G on V1∪V2∪V3 where Vi are disjoint copies of [3n] and:

E(V1,V2)={{x,x+a}:x∈V1,x+a∈V2,a∈A}E(V2,V3)={{y,y+a}:y∈V2,y+a∈V3,a∈A}E(V1,V3)={{x,x+2a}:x∈V1,x+3a∈V3,a∈A}

Say then x∈V1, y∈V2, z∈V3 form a triangle in G. So y=x+a, z=y+b, z=x+2c for some a,b,c∈A. Then

y+z+(x+2c)=(x+a)+(y+b)+z,

hence a+b=2c. If A is no 3-term arithmetic progressions, then G only contains triangles of the form {x,x+a,x+2a}, where a∈A. So G contains ≤3n|A|≤3n2 triangles.

So apply the The triangle Removal Lemma to find T⊂E(G) with |T|≤𝜀n2 such that G−T is triangle-free. This is possible since n>10δ. But now notice that

{{x,x+a,x+2a}:x∈[n],a∈A}

is a set of edge-disjoint triangles of size n|A| and hence

n|A|≤|T|≤𝜀n2,

so |A|≤𝜀n. □

Remark.

  • Again this gives bad bounds on our minimum density 𝜀.

  • Roth showed that if A⊂[n], |A|≥cnlog⁡ log ⁡n, then A contains a 3-term arithmetic progression.

  • Kelley–Meka showed that the same holds for |A|≥ne−c(log⁡n)112.

    There exist A⊂[n] with no 3-term arithmetic progression but

    |A|≥ne−clog⁡n.