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Extremal Graph Theory 2012-2013 Lecture Notes
Extremal Graph Theory
University of Cambridge
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Extremal Graph Theory
- Lent Term Lectured by A. Thomason- 1 The Erd ̋os-Stone Theorem
- 2 Stability
- 3 Supersaturation
- 4 Szemer ́edi’s Regularity Lemma
- 5 A couple of applications
- 6 Hypergraphs
- 7 The size of a hereditary property
- 8 Containers
- 9 The Local Lemma
- 10 Tail Estimation
- 11 Martingales Inequalities
- 12 The Chromatic Number of a Random Graph
- 13 The Semi-Random Method - Last updated: Sat 7thMay, Please let me know of corrections: glt1000@cam.ac
Course description
Extremal graph theory is an umbrella title for the study of graph properties and their dependence on the values of graph parameters. This course builds on the material introduced in the Part II Graph Theory course, in particular Tur ́an’s theorem and the Erd ̋os-Stone theorem, as well as developing the use of randomness in combinatorial proofs. Further techniques and extensions to hypergraphs will be discussed. It is intended to cover some reasonably large subset of the following.
The Erd ̋os-Stone theorem and stability. Supersaturation. Szemer ́edi’s Regularity Lemma, with applications. The number of complete subgraphs.
Hypergraphs. Erd ̋os’sr-partite theorem. Instability. The Fano plane. Razborov’s flag algebras. Hereditary properties and their sizes.
Probabilistic tools: the Local Lemma and concentration inequalities. The chromatic number of a random graph. The semi-random method, large independent sets and the Erd ̋os-Hanani problem. Dependent random choice.
Pre-requisite Mathematics
A knowledge of the basic concepts, techniques and results of graph theory, such as that afforded by the Part II Graph Theory course.
Literature
No book covers the course but the following can be helpful.
B. Bollob ́as,Modern graph theory, Graduate Texts in Mathematics 184 , Springer-Verlag, New York (1998), xiv+394 pp.
N. Alon and J. Spencer,The Probabilistic Method, Wiley, 3rd ed. (2008)
or εn 2
−|K| 6 |U|
(
1
r
−
ε 2
)
which, ifn 1 (r,ε) is large enough, implies
|U|
r
>
εn 3
.
Se we may assume that|U|> εrn 3
.
Now, each vertex inUis joined to at least ( 1 −
1
r
+
ε 2
)
|K|−(r−1)T =
(
1 −
1
r
+
ε 2
)
rT−(r−1)T = εrT 2
> t
vertices in each class ofK, and so is joined to someKr(t) inK. But there are only
(T
t
)r
manyKr(t) inK, and, recalling that
(n k
)
6 (enk)k, we have ( T t
)r 6
(
eT t
)tr 6
(
3 e εr
)rdlogn 6
εrn 3 t
6
|U|
t
ifd′(r,ε) is small andn 1 (r,ε) is large.
Hence there existsW⊂U, with|W|>t, joined to the sameKr(t) inK.
HenceKr+1(t)⊂G. 2
Lemma 1. Letc,ε >0. Then there existsn 2 =n 2 (c,ε) with the following property.
Suppose that|G|=n>n 2 ande(G)>(c+ε)
(n 2
)
. ThenGhas a subgraphHsuch that δ(H)>c|H|and|H|>ε 1 / 2 n.
Proof not, there is a sequenceG=Gn⊃Gn− 1 ⊃Gn− 2 ⊃ ··· ⊃Gs, wheres=⌊ε 1 / 2 n⌋, such that|Gj|=jand the only vertex inGjnot inGj− 1 has degree less thancj. Then
e(Gs)>(c+ε)
(
n 2
)
−
∑n
j=s+
cj= (c+ε)
(
n 2
)
−c
[(
n+ 1 2
)
−
(
s+ 1 2
)]
>
εn 2 2
>
(
s 2
)
providedn 2 is large enough. Contradiction. 2
Lecture 2 Theorem 1 (Erd ̋os-Stone, 1946).Letr>0 be an integer andε >0. Then there exist d =d(r,ε) andn 0 = n 0 (r,ε) such that, if|G|=n >n 0 and if forr >1 we have e(G)>
(
1 − 1 r+ε
)(n 2
)
, thenG⊃Kr+1(t) wheret=⌊dlogn⌋.
Proof 0 >n 2
(
1 − 1 r+ε 2 ,ε 2
)
, we may apply Lemma 1 toGto obtain a subgraph Hwithδ(H)>
(
1 − 1 r+ε 2
)
|H|and|H|>ε 1 / 2 n.
Providedn 0 >ε− 1 / 2 n 1 (r,ε 2 ), we may apply Lemma 1 toH to obtainKr+1(t) with t>
⌊
d 1 (r,ε 2 ) logε 1 / 2 n
⌋
.
Providedn 0 > 1 ε, if we taked(r,ε) = 12 d 1 (r,ε 2 ), we are done. 2
As observed by Erd ̋os and Simonovits in the mid-1960s, we can determineex(n,F) asymptoti- cally foreveryF.
Theorem 1.4 a fixed graph with chromatic numberr=χ(F). Then
lim n→∞
ex(n,F) (n 2
) = 1−
1
r− 1
Proofχ(Tr− 1 (n)) =r−1, we haveF6⊂Tr− 1 (n), hence
ex(n,F) >e(Tr− 1 (n)) >
(
1 −
1
r− 1
)(
n 2
)
.
On the other hand, givenε >0, if|G|=nande(G)>
(
1 −r− 11 +ε
)(n 2
)
, thenG⊃ Kr(|F|)⊃Fifnis large enough, by the Erd ̋os-Stone theorem.
Thus for everyε >0, we have lim sup
ex(n,F) (n 2
) 6
(
1 −
1
r− 1
+ε
)
. 2
Here’s a pretty consequence of Erd ̋os-Stone.
Define theupper densityof an infinite graph to be the supremum of densities of large finite subgraphs:
ud(G) = lim n→∞ sup
{
x:∃F⊂G,|F|>n,e(F)>x
(
|F|
2
)}
Corollary 1.5(G)∈{ 0 , 12 , 23 , 34 ,...}∪{ 1 }.
Can we strengthen Erd ̋os-Stone to obtain largert?
Theorem 1 Givenr∈N, there existsεr>0 such that, ifε < εr, there existsn(r,ε) such that for alln>n(r,ε) there is a graphGof ordernwithe(G)>
(
1 − 1 r+ε
)(n 2
)
and G6⊃Kr+1(t) wheret=
⌊3 logn log(1/ε)
⌋
.
Proof be a largest vertex class ofTr(n) with|W|=w=⌈n/r⌉. FormGby adding ε
(n 2
)
edges withinWso thatG[W]6⊃K 2 (t) and henceG6⊃Kr+1(t).
To see that this addition is possible, choose edges insideWindependently with probability p= 3εr 2. Takeεr= (3r 2 )− 6. Sop <1.
LetXbe the number of edges chosen andYthe number ofK 2 (t) formed by them. Then
E(X−Y) =EX−EY=p
(
w 2
)
−
1
2
(
w t
)(
w−t t
)
pt
2
Now, w 2 t− 2 pt
2 − 1 =
[
w 2 pt+
]t− 1 <
[
w 2 ε 56 (t+1)]t+ <
[
w 2 n− 5 / 2
]t− 1 <
1
2
.
HenceE(X−Y)>
p 2
(
w 2
)
. So there is a choice withX−Y >
p 2
(
w 2
)
.
Remove an edge from eachK 2 (t) to leave at leastX−Y > p 2
(
w 2
)
> ε
(
n 2
)
edges with no K 2 (t). 2
This shows dependence onnis correct. Our argument givesd(r,ε)>
ε 2 r(r−1)!
.
toy 1 ,...,ym. Thene(G∗)> e(G), soG∗⊃F. This copy ofFinG∗containsubut not xi, say. But then (F−u)∪xicontainsFinG. Contradiction. 2
Stability can sometimes be used as a bootstrapping device to obtain exact results. For example, ex(n,C 5 ) =t 2 (n) forn>6. In fact,ex(n,C 2 k+1) =t 2 (n) ifnis large. This is a special case of the following theorem.
Theorem 2 (Simonovits).LetFbe (r+1)-edge-critical, i.χ(F) =r+1 butχ(F−e) =r for all edgese∈E(F). Thenex(n,F) =tr(n) for largen, andTr(n) is the unique extremal graph.
Proof an extremal graph of ordern. Select, by Theorem 2(c) usingt=|F|, an r-partite subgraphHof minimum degree
(
1 − 1 r+o(1)
)
n. Necessarily, each part ofHis of order
( 1
r+o(1)
)
n. Assign theo(n) vertices ofG−Hto the parts ofHwith the fewest neighbours.
Lecture 4 Suppose a vertexxis joined toεnvertices in its own class. Then it hasεnneighbours in each part ofH. Thesersets ofεnvertices span
(
1 − 1 r+o(1)
)(rεn 2
)
edges because δ(H) =
(
1 − 1 r+o(1)
)
n.
By Erd ̋os-Stone the neighbours ofxspanKr(|F|) which containsF−vfor anyv∈V(F). HenceG⊃F, a contradiction.
So each vertex has onlyo(n) neighbours in its own part. By Corollary 2, we know δ(G) =
(
1 − 1 r+o(1)
)
n. So each vertex ofGis joined to all buto(n) vertices in each other class. Supposexyis an edge inside some class. Pick a setZof|F|vertices in this class, including{x,y}.
Then all buto(n) vertices of the other classes are common neighbours ofZ. Hence these common neighbours spanKr− 1 (|F|), either by Erd ̋os-Stone or directly. ButZtogether withKr− 1 (|F|) containsF, a contradiction.
HenceGisr-partite, and sinceTr(n) is the uniquer-partite graph of maximum size, we haveG=Tr(n). 2
Here is another example.
Theorem 2.4,sbe fixed. For largen, the unique extremal graph forsKr+1(i.,s disjoint copies ofKr+1) isKs− 1 +Tr(n−s+ 1) (where ‘+’ means ‘join all to all’).
Proof proof is similar to that of Theorem 2. Proceed by induction ons. The cases= 1 is Tur ́an’s theorem.
As before, chooseHand assignG−Hto the classes ofH. Once again, if some vertex xhasεnneighbours in its own class, then the neighbours ofxspan someKr(s(r+ 1)). ButG−xcannot contain (s−1)Kr+1, hencee(G−x) 6 e
(
Ks− 2 +Tr(n−s+ 1)
)
, so e(G) 6 e
(
Ks− 1 +Tr(n−s+ 1)
)
.
So equality holds in both cases, soG−x=Ks− 2 +Tr(n−s+1) by the induction hypothesis, soG=Ks− 1 +Tr(s−n+ 1).
Hence we may suppose that each vertex is joined too(n) in its own class. Suppose some class containssindependent edges. LetZbe the 2sendvertices of these edges. As in the previous proof, the common neighbours ofZspanKr− 1 (s(r−1)), in which caseG⊃sKr+1.
Thus no class containssindependent edges, so thejthclass contains a setAjof 2(s−1) vertices such that every edge in thejthclass meetsAj. But each vertex inAjhaso(n) neighbours in thejthclass.
Hencee(G) 62 r(s−1)o(n) +tr(n)< e
(
Ks− 1 +Tr(n−s+ 1)
)
. 2
3 Supersaturation
Supersaturation is the study of how many copies ofFmust exist inGife(G)>ex(n,F). The basic theorem holds in a general context.
Recall that anℓ-uniform hypergraphis a pairG= (V,E) where
E⊂V(ℓ)=
{
Y⊂V:|Y|=ℓ
}
.
For a classFofℓ-uniform hypergraphs, define
ex(n,F) = max
{
e(G) :|G|=n,Gisℓ-uniform,Gcontains noF∈F
}
and π(F) = lim n→∞
ex(n,F) (n ℓ
)
Exercise limit exists.
Theorem 3.1 anℓ-uniform hypergraph. Then for allε >0 there existsδ=δ(H,ε)> 0 such that everyℓ-uniform hypergraphGwith|G|=nande(G)>
(
π(H) +ε
)(n ℓ
)
⌊ contains δn|H|
⌋
copies ofH.
Lecture 5 Proof eachm-setM∈V(m), letG[M] be the hypergraph induced byGonM. If there areη
(n m
)
setsMwithe(G[M])>
(
π(H) +ε 2
)(m ℓ
)
then
(π(H) +ε)
(
n ℓ
)
6 e(G) =
∑
Me
(
G[M]
)
(n−ℓ m−ℓ
) 6
η
(n m
)(m ℓ
)
- (1−η)
(n m
)(
π(H) +ε 2
)(m ℓ
)
(n−ℓ m−ℓ
)
So, assumingn>m>ℓ, we have
π(H) +ε 6 η+ (1−η)
(
π(H) +ε 2
)
or η>
ε 2 1 −π(H)−ε 2
> 0.
Pickmso thatex(m,H)<
(
π(H) +ε 2
)(m ℓ
)
. ThenH⊂G[M] for each ofη
(n m
)
subsetsM.
SoGcontains>η
(
n m
)(
n−|H| m−|H|
)− 1
distinct copies ofH, i.,>η
(
n |H|
)(
m |H|
)− 1
copies.
Pickn 0 >mso that
(
n |H|
)
>
1
2
n|H| |H|!
.
Ifn>n 0 then pickδ 6
1
2
1
|H|!
(
m |H|
)− 1
small enough that
⌊
δn|H|
⌋
works forn 6 n 0. 2
Then, amongst graphsGof given order,f(G) is maximized on a complete multipartite graph. Moreover, ifαK 3 >0, there are no other maxima.
Proof may suppose thatαK 3 >0, the caseαK 3 = 0 following by a limiting argument.
Choose a graphGof ordernthat maximizesf(G). SupposeGis not complete multipartite. ThenGcontains two non-adjacent verticesx,ywhose neighbourhoodsX,Ydiffer.
The numberiF(α) contains contributions from four kinds ofF: those containing, respec- tively,xbut noty,ybut notx, bothxandy, neitherxnory. The first contribution depends only onX, and the second only onY. Moreover the third depends only onX∩Y andV−(X∪Y) whereV=V(G), sinceFis complete multipartite.
This fourfold partition ofiF(G) means we can write f(G) =g(X) +g(Y) +h(X∩Y,V−(X∪Y)) +C,
whereCis independent ofXandY.
Lecture 6 Note thath(A,B) 6 h(A′,B′) ifA⊂A′,B⊂B′, becauseFmakes no contribution tohif Fis complete, and otherwiseαF>0. Moreover, if alsoB 6 =B′, thenh(A,B)< h(A′,B′), butαK 3 >0 andiK 3 (G) contributes exactlyαK 3 |B|toh(A,B).
We may suppose thatg(X)>g(Y) and, ifg(X) 6 =g(Y), that|X| 6 |Y|and soX 6 =X∪Y. Therefore g(X) +h(X,V−X)> g(Y) +h(X∩Y,V−(X∪Y))
FormHfromGby removing all edgesytoYand inserting all edgesytoX. Then f(H) = 2g(X) +h(X,V−X) +C > f(G).
Contradiction. 2
Note that this transformation does not increase the clique size or the chromatic number.
Let 1 6 p 6 r. For 0 6 x 6
(n p
)
, letψ(x) be the maximal convex function defined byψ(0) = 0, ψ
(
kp(Tq(n))
)
=kr(Tq(n)). (Whereq=r− 1 ,r,r+ 1,...)
Theorem 3 (Bollob ́as, 1976).LetGbe a graph of ordern. Thenkr(G)>ψ(kp(G))
Proof(G) =kp(G)−ckr(G) for some realc >0. Sinceψis convex, it suffices to prove thatfis maximized on a Tur ́an graph. So letGbe a graph on whichfis maximized.
kp(Tr− 1 ) kp(Tr) kp(Tr+1)
q
q
q
✦✦
✦✦
✦★
★
★
★★
✓✓
q✓
✓
✓
✓
✓
✓
✓
✛
Suppose there exists G below ψ. Consider the straight line of gradient c >0. The intercept on the axis is kp(G)−ckr(G).
By Theorem 3, we know thatfis maximized on a complete-partite graph, withqclasses of sizes 0< a 16 a 26 ··· 6 aq. We may assume thatq>r, elseTr− 1 (n) maximizesf.
Now,f(G) =a 1 aqA−ca 1 aqB+C, whereA,B,Care non-zero rationals depending on a 2 ,a 3 ,...,aq− 1 anda 1 +aq.
We may assume without loss of generality thatcis irrational, soA−cB 6 = 0. IfA−cB <0, then replacea 1 by 0 andaqbya 1 +aqto increasef. IfA−cB >0 anda 16 aq−2, then replacea 1 bya 1 + 1 andaqbyaq−1 to increasef. Thusa 1 >aq−1 and soGis a Tur ́an graph. 2
The exact value of min
{
k 3 (G) :|G|=n,k 2 (G) =m
}
is unknown. It is conjectured to be given byr-partite graphs whereris minimal.
k 2 (T 2 (n)) k 2 (T 3 (n)) =n 2 / 4 =n 2 / 3
q
q
q
k 3 (T 4 (n))
k 3 (T 3 (n))
✦✦
✦✦
✦★
★
★
★★
✓✓
. ....................................
................................................
....................................................................................
.................................................................. ............................................................
............................ ✛ Cor. 3.
I think. I foolishly tried to draw this during the lecture, and so it might not be right...
The continuous envelope here forn 2 / 46 m 6 n 2 /3 was posed as a lower bound by Fisher (1989). The whole rangem 6
(n 2
)
was proved (again in the limit) by Razborov, who introduced the method of flag algebras (2002), a lowbrow view of this being a massive generalisation of Cauchy-Schwarz, using CSP to find optimal quadratic forms.
Nikiforov (2008) did likewise forn= 4. Reiher (2012) did likewise for allr, forp= 2.
Open problem inducibility: what is max{ilength 3 paths(G)}?
4 Szemer ́edi’s Regularity Lemma
A graph having the (large scale) property that its (induced) subgraphs all have roughly the same density, the graph itself can be regarded as “pseudo-random” in a sense that can be made precise. Consider the following bipartite version.
LetU,Wbe disjoint subsets of the vertex set of some graph. The number of edges betweenU andWis denoted bye(U,W).
Thedensityd(U,W) is
e(U,W) |U||W|
.
Definition 4. Let 0< ε <1. The pair (U,W) is said to beε-uniform(orε-regular) if for U′⊂U,W′⊂W, ∣ ∣d(U′,W′)−d(U,W)
∣
∣< ε whenever |U′|> ε|U|,|W′|> ε|W|
(Clearly a lower bound on|U′|is necessary – e.,|U′|= 1 is hopeless.)
Anε-uniform pair is roughly regular.
Lemma 4. Let (U,W) beε-uniform andd(U,W) =d. Then ∣ ∣
{
u∈U:|Γ(u)∩W|>(d−ε)|W|
}∣
∣>(1−ε)|U|
and ∣ ∣
{
u∈U:|Γ(u)∩W|<(d+ε)|W|
}∣
∣>(1−ε)|U|
Then there exist verticesx 1 ,...,xk(xi∈Vi) such that the mapvi7→xigives an isomor- phism betweenHandG[{x 1 ,...,xk}].
Proof that, by replacing the set ofVi−Vjedges by the complementary set ifvivj∈/E(H), we may assume thatHis complete andd(Vi,Vj)>λfor alli,j.
The result is then immediate from Lemma 4 on takingε=η,r=k,d=k−1, and s= 1. 2
It is a remarkable fact thateverygraph can be partitioned into aboundednumber of pieces, almost all pairs of which areε-uniform. This result is due to Szemer ́edi and is used in his proof that sets of integers with positive density contain arbitrarily long arithmetic progressions.
AnequipartitionofV(G) intokparts is a partitionV 1 ,...,Vksuch that⌊n/k⌋ 6 |Vi| 6 ⌈n/k⌉ for all 1 6 i 6 k, wheren=|V(G)|. The partition isε-uniformif (Vi,Vj) isε-uniform for all but at mostε
(k 2
)
pairs, for 1 6 i < j 6 k.
Theorem 4 (Szemer ́edi’s Regularity Lemma). Let 0< ε <1 and letℓbe a natural number. Then there existsL=L(ℓ,ε) such that every graph has anε-uniform equipartition intomparts, for someℓ 6 m 6 L.
Before proving the lemma, we establish two useful facts.
Lemma 4. LetU′⊂UandW′⊂Wsatisfy|U′|>(1−δ)|U|and|W′|>(1−δ)|W|.
Then
∣
∣d(U′,W′)−d(U,W)
∣
∣ 62 δ.
Proof=d(U,W) andd′=d(U′,W′). Then
d=d(U,W) =
e(U,W) |U||W|
>
e(U′,W′) |U||W|
=d(U′,W′)
|U′||W′|
|U||W|
>d′(1−δ) 2
Sod′−d 6 d′
(
1 −(1−δ) 2
)
62 δd′ 62 δ.
By considering the complementary graph, (1−d′)−(1−d) 62 δ, i.,d−d′ 62 δ. 2
Lemma 4. Letx 1 ,...,xnbe real numbers such thatX= 1 n
∑n i=
xi. Letx=m 1
∑m i=
xi. Then
1
n
∑n
i=
x 2 i>X 2 +
m n−m
(x−X) 2 >X 2 +
m n
(x−X) 2
Proof have
1 n
∑n
i=
x 2 i =
1
n
∑m
i=
x 2 i+
1
n
∑n
i=m+
x 2 i
>
m n
x 2 +
n−m n
(
nX−mx n−m
) 2
= X 2 +
m n−m
(x−X) 2
by two applications of Cauchy-Schwarz. 2
Proof of Theorem 4. Define theindexind(P) of an equipartitionPwithkpartsV 1 ,...,Vk to be 1 k 2
∑
i<j
d 2 (Vi,Vj).
We shall show that, ifn>k 16 kand 4kε 5 >100, andPis notε-uniform, then there exists an equipartitionQintok 4 kparts with ind(Q)>ind(P) +ε 5 /8.
This will be enough to prove the theorem. For, chooset>ℓwith 4tε 5 >100. Define f(0) =tandf(j+ 1) =f(j)4f(j). LetN=f(⌈ 4 ε− 5 ⌉). LetL=N 16 N. Then ifn 6 L, take an equipartition intonsingle vertices. Otherwise, begin with an equipartitionPinto tparts.
So long as the current partition intokparts is notε-uniform, replace it with one intok 4 k parts and larger index. Since ind(P) 612 , replacement can occur at most⌈ 4 ε− 5 ⌉times, so anε-uniform partition is found with 6 Lparts.
For each (Vi,Vj) that is notε-uniform, select witness setsXij⊂ViandXji⊂Vjsuch that|Xij|>ε|Xi|and|Xji|>ε|Xj|, and
∣
∣d(Xij,Xji)−d(Vi,Vj)
∣
∣>ε. For eachi, the sets XijpartitionViinto at most 2k− 1 atoms.
Letm=⌊n/k 4 k⌋and letn=k 4 km+ak+bwhere 0 6 a < 4 kand 0 6 b 6 k. Then ⌊n/k⌋= 4km+a, so the parts ofPhave size 4km+aor 4km+a+ 1, withBparts have the larger.
Lecture 9 Partition each part ofPinto 4ksets of sizemorm+ 1. Any such partition is a partition ofGintok 4 ksets of sizesmorm+ 1, i. an equipartition.
Choose such a partitionQwhose parts are as much as possible inside atoms: that is, every atom is a union of part ofQtogether with at mostmextra vertices. All that remains is to check that ind(Q)>ind(P) +ε 5 /8.
Let the setes ofQwithinVibeVi(s), for 1 6 s 6 q= 4k. That is,Vi=
⋃q s=1Vi(s). Note that
∑
16 s,t 6 qe
(
Vi(s),Vj(t)
)
=e(Vi,Vj) and|Vi|>q|Vi(s)|mm+1for eachs, and so
1
q 2
∑
16 s,t 6 q
d
(
Vi(s),Vj(t)
)
>
(
m m+ 1
) 2
d(Vi,Vj).
Sincen>k 16 k, we have ( m m+ 1
) 2
> 1 −
2
m
> 1 −
2
4 k
> 1 −
ε 5 50
.
Therefore
1 q 2
∑
16 s,t 6 q
d 2
(
Vi(s),Vj(t)
)C.
>
(
1
q 2
∑
16 s,t 6 q
d
(
Vi(s),Vj(t)
)
) 2
>d 2 (Vi,Vj)−
ε 5 25
The main point is that we can improve this bound if (Vi,Vj) is not uniform. LetXij∗ be the largest subset ofXijthat is a union of parts ofQ. Because we choose parts ofQas much as possible within atoms, and because|Vi| 6 ⌊nl⌋ 4 km+a> 4 km, we have
∣ ∣Xij∗
∣
∣>
∣
∣Xij
∣
∣− 2 k− 1 m>
∣
∣Xij
∣
∣
(
1 −
2 k− 1 m ε|Vi|
)
>
∣
∣Xij
∣
∣
(
1 −
1
ε 2 k
)
>
∣
∣Xij
∣
∣
(
1 −
ε 10
)
Theorem 5.1 an integerd, there exists a numberc(d) such thatr(G) 6 c(d)|G|if ∆(G) 6 d.
Remark same is conjectured to be true for every graph withe(H) 6 d|H|for allH⊂G.
It is known for a wider class than in the theorem, including planar graphs.
Proof=R(d+ 1). Chooseε 6 min
{ 1
t,
1 2 d(d+1)}.
Letℓ>t 2 and letL=L(ℓ,ε) be the number given by Szemer ́edi’s Regularity Lemma (SRL). Finally, letc=L/ε.
LetGbe a gaph a maximum degree 6 dand let the edges ofKn, wheren>c|G|, be coloured red/blue. Apply Szemer ́edi’s Lemma toR, the red subgraph ofKn, withℓ,εas above.
LetH be the graph with vertex set{V 1 ,...,Vm}, whereV 1 ,...,Vmis the partition of Rgiven by SRL. LetViVj ∈E(H) if (Vi,Vj) isε-uniform. Then|H|=m>t 2 and e(H) 6 ε
(m 2
)
. SoH⊃Kt. [Or else, by Tur ́an’s theorem, there exist integersm 1 ,...,mt− 1 with
∑
mi=mande(H)>
∑(mi 2
)
>(t−1)
(m/(t−1) 2
)
> ε
(m 2
)
whereε 61 tandm>t 2 .]
Thus we may assume that every pair (Vi,Vj), 1 6 i < j 6 tisε-uniform. Colour the edges ofKtgreen ifd(Vi,Vj)> 12 , and yellow otherwise.
Sincet=R(d+ 1), there exists a monochromaticKd+1, and we may assume that it is spanned byV 1 ,...,Vd+1. Therefore,d(Vi,Vj)> 12 for 1 6 i < j 6 d+ 1, ord(Vi,Vj)< 12 for 1 6 i < j 6 d+ 1.
We will show that in the first case we haveR⊃G, and that in the other case we obtain similarly a blueGinKn.
Take a vertex colouring ofGwithd+ 1 colours in which no colour is used more than |G|times. Lemma 4 applied with H(there) =G(here),G(there) = (subgraph ofR spanned byV 1 ,...,Vd+1),u=|Vi|>n/L>c|G|/L>|G|/ε,s=|G|,r=d+ 1,λ= 12 , (d+ 1)ε 61 / 2 d.
This completes the proof. 2
Theorem 5 was proved by Chvat ́al, R ̈odl, Szemer ́edi, Trotter in1983. It was extended to a larger class (using a modification of Lemma 4) by Chen and Schelp in 1993. Graham, R ̈odl,
Rucinski showed in 1999 thatc(d) 62 Cdlog
2 d (obviously without SRL).
Observe that the Erd ̋os-Stone theorem follows from SRL and Tur( ́an’s theorem. Indeed, ife(G)> 1 − 1 r+ε
)(n 2
)
, form the “reduced” graphHwhose edges correspond to pairs of positive density. Thene(H)>tr(|H|) soH⊃Kr+1, and soG⊃Kr+1by Lemma 4.
We make this precise, but, more interestingly, we recover stability too. We use an argument of Erd ̋os.
Theorem 5 (Erd ̋os, 1970).LetGbe aKr+1-free graph. Then there is a (complete)r- partite graphHwithV(H) =V(G) anddH(v)>dG(v) for allv∈V(G).
Proof induction onr. The caser= 1 is trivial.
In general, letxbe a vertex of maximum degree. ThenG′ =G[Γ(x)] isKr-free, so by induction there is an (r−1)-partite graphH′withV(H′) =V(G′) = Γ(x), and dH′(v)>dG′(v) for allv∈Γ(x).
FormH by joining every vertex ofV(G)\Γ(x) to every vertex in Γ(x). ThenHisr- partite and ifv /∈Γ(x) thendH(x) =|Γ(x)|=dG(x)>dG(v), while ifv∈Γ(x) then dH(v) =dH′(v) +|V(G)\Γ(x)|>dG′(v) +|V(G)\Γ(x)|>dG(v). 2
Lecture 11 Theorem 5 (F ̈uredi, 2010). LetGbe aKr+1-free graph. Then there is a completer-partite graphHwithV(H) =V(G) and|E(G)\E(H)| 612 |E(H)\E(G)|.
In particular, ife(G)>tr(|G|)−kthen|E(H)△E(G)| 63 k.
Proof in the previous proof. The proof is via induction, the caser= 1 being trivial.
Now|E(G′)\E(H′)| 612 |E(H′)\E(G′)|. Let there beeedges fromGinsideV(G)\Γ(x), andf edgesmissingfromGbetween Γ(x) andV(G)\Γ(x). Then|E(G)\E(H)|= |E(G′)\E(H′)|+e, and|E(H)\E(G)|=|E(H′)\E(G′)|+f, so it suffices to show that e 612 f.
For eachv∈V(G)\Γ(x), lete(v) be the number of edges ofGinsideV(G)\Γ(x) meeting v, and letf(v) be the number of edges missing fromGbetween Γ(x) andV(G)\Γ(x) meetingv. Thene= 12
∑
v
e(v) andf= 12
∑
v
f(v).
But for eachv,dG(x)>dG(v) =dG(x)−f(v) +e(v), soe(v) 6 f(v), and we are done.
Finally, ife(G)>tr(|G|)−k, thentr(|G|)>e(H) =e(G)−|E(G)\E(H)|+|E(H)\E(G)|> tr(G)−k+ 12 |E(H)\E(G)|, so|E(H)\E(G)| 62 k, hence|E(G)\E(H)| 6 k, and so |E(H)△E(G)| 63 k. 2
Definition 5. LetGbe a graph with anε-uniform partitionPintompartsV 1 ,...,Vm. Let λ∈R. Thereduced graphG(P,λ) is the graph of orderMwith vertex set{V 1 ,...,Vm} and edge set{ViVj: (Vi,Vj) isε-uniform andd(Vi,Vj) 6 λ}.
Lemma 5. LetG,Pbe as in this definition. Ife(G)>c
(|G|
2
)
then
e(G(P,λ)) 6
(
c−ε−λ−
1
m− 1
)(
m 2
)
,
provided|G|> 2 m 2.
Proof(G(P,λ)) =d
(m 2
)
andn=|G|. Then
e(G) 6 d
(
m 2
)(
n m
+ 1
) 2
+
(
m 2
)
ε
(
n m
+ 1
) 2
+
(
m 2
)
λ
(
n m
+ 1
) 2
+m
(n m+ 1 2
)
6
(
d+ε+λ+
1
m− 1
)(
m 2
)(
n m
+ 1
) 2
6
(
d+ε+λ+
1
m− 1
)(
n 2
)
2
Theorem 5.6,t∈Nand letε >0. Then there existsn 0 (r,t,ε) such that, if|G|=n>n 0 andG6⊃Kr+1(t), thenGhas a subgraphG′withe(G′)>e(G)−ε
(n 2
)
andG′6⊃Kr+1.
Tur ́an further conjecturedex(n,K 35 ) =
(n 3
)
−
(⌊n/ 2 ⌋ 3
)
−
(⌈n/ 2 ⌉ 3
)
, but this is false for oddn > 9 (Sidorenko, 1990s).
The situation appearsunstable.
π(F) is known only in one or two simple cases; more recently, in a couple of less trivial examples, e. the Fano plane. But stability holds here. (F ̈uredi, Simonovits, Pikhurko, Sudakov, Keevadi.)
Clearlyπ(Krℓ) 61 −
1
(r ℓ
). The best known general bound is as follows.
Lemma 6. LetGbe anℓ-uniform hypergraph withnrcopies ofKrl(forr >ℓ), where nℓ=e(G) andnℓ− 1 =
(n ℓ− 1
)
, andn=|G|. Then providednr− 1 >0,
nr+ nr
>
r 2 (r−ℓ+ 1)(r+ 1)
×
nr nr− 1
−
(ℓ−1)(n−r) +r (r−ℓ+ 1)(r+ 1)
Proof 1 ,...,Anr− 1 be an enumeration of theKr− 1 ’s, andB 1 ,...,Bnrenumerate theKr’s. Letaibe the number ofKr’s containingAiandbjbe the number ofKr+1’s containing Bj. Then
∑
ai=rnrand
∑
bj= (r+ 1)nr+1.
LetNbe the number of pairs (S,T) whereSis the vertex set of aKr,Tis anr-set not Kr, and|S∩T|=r−1. Clearly
N =
∑
ai(n−r+ 1−ai) = (n−r+ 1)rnr−
∑
a 2 i 6 (n−r+ 1)rnr− r 2 nr nr− 1
On the other hand, for eachBjwe can findn−r−bjverticesxsuch thatB∪{x}doesn’t spanKr+1. So there is an (ℓ−1)-setY⊂Bjsuch thatY∪{x}is not an edge.
Then eachz∈Bj\Ygives a pair (S,T) = (Bj,Bi{z}∪{x}). Thus
N >
∑
j
(n−r−bj)(r−ℓ+ 1) = (r−ℓ+ 1)
(
(n−r)nr−(r+ 1)nr+
)
Comparing bounds onNgives the result. 2
Corollary 6 (de Caen, 1983). ex(n,Krℓ) 6
(
n l
)(
1 −
1
(r− 1 ℓ− 1
)×
n−r+ 1 n−ℓ+ 1
)
.
In particular,π(Krℓ) 61 −
1
(r− 1 ℓ
).
Proof maximal with noKℓr. Thennℓ− 1 =
(n ℓ− 1
)
andns>0 forℓ 6 s < r.
We show by induction that
ns ns− 1
>
n−ℓ+ 1 s
(
s− 1 ℓ− 1
)(
nℓ (n ℓ
)−1 +
n−s+ 1 n−ℓ+ 1
×
1
(s− 1 ℓ− 1
)
)
holds forℓ 6 s < r, which proves the Corollary sincenr= 0.
The cases=ℓholds automatically sincens− 1 =
(n s− 1
)
.
Writingqs=ns/ns− 1 the desired inequality is
qs >
n−ℓ+ 1 s
(
s− 1 ℓ− 1
)(
nℓ (n ℓ
)− 1
)
+
n−s+ 1 s
which follows from
qs >
(s−1) 2 (s−ℓ) 2
qs− 1 −
(ℓ−1)(n−s+ 1) +s− 1 (s−ℓ)s
(from Lemma 6) and induction. 2
The next theorem (Erd ̋os, 1964) shows thatℓ-partite,ℓ-uniform hypergraphs have extremal functiono(nℓ), i.π= 0.
Anℓ-uniform hypergraphHisℓ-partiteifV(H) =V 1 ∪...∪VℓdisjointVi, and each edge has one vertex in each class. The completeℓ-partiteℓ-uniformKℓ(t 1 ,...,tℓ) has|Vi|=tiand all possible edges.
Lecture 13 Theorem 6.3 anℓ-uniform hypergraph of ordernand sizepnℓ/ℓ!. Lett 16 t 26 ... 6 tℓbe positive integers and suppose thatpt 1 ..ℓ− 1 > T 2 n− 1 , whereT=
∑
ti.
ThenGcontains at least 21 T!pt 1 ..ℓnTcopies ofKℓ(t 1 ,...,tℓ).
Remarks.
- Note that this quantity is similar to the expected number if the edges were chosen randomly.
- This implies thatex(n,Kℓ(t 1 ,...,tℓ)) 6 cTnℓ− 1 /t 1 ..ℓ.
Proofχ:Vℓ→{ 0 , 1 }be the characteristic (indicator) function of edges.
Then
∑
x 1 ,...,xℓ
χ(x 1 ,...,xl) =pnl.
Letf(t 1 ,...,tℓ) =
1
nT
∑
x 11 ,...,xt 11
···
∑
x 1 ℓ,...,xtℓℓ
∏
xi 11
···
∏
xiℓℓ
χ(xi 11 ···xiℓℓ).
Note thatnTf(t 1 ,...,tℓ) is the number of labelled homomorphic copies ofKℓ(t 1 ,...,tℓ) in G(copies where vertices in the same class might coincide). I.,f() is the probability that a randomly chosen collection ofTvertices spans a homomorphic copy ofKℓ(t 1 ,...,tℓ). Now
f(t 1 ,...,tℓ) =
1
nT−tℓ
∑
x 11 ,...,xt 11
···
∑
xℓ− 1 ,...,xtℓℓ−− 11
∏
xi 11
···
∏
xiℓℓ−− 11
∑
x 1 ℓ,...,xtℓℓ
∏
xℓ
χ(xi 11 ,...,xiℓℓ−− 11 ,xiℓℓ)
=
1
nT−tℓ
∑
x 11 ,...,xt 11
···
∑
xℓ− 1 ,...,xtℓℓ−− 11
∏
xi 11
···
∏
xiℓℓ−− 11
(
1
n
∑
xℓ
χ(xi 11 ,...,xiℓ−ℓ− 11 ,xℓ)
)tℓ
use Jensen’s inequality
>
1
nT−tℓ
∑
x 11 ,...,xt 11
···
∑
xℓ− 1 ,...,xtℓ−ℓ− 11
∏
xi 11
···
∏
xiℓℓ−− 11
1
n
∑
xℓ
χ(xi 11 ,...,xiℓ−ℓ− 11 ,xℓ)
tℓ
= f(t 1 ,...,tℓ− 1 ,1)tℓ > f(t 1 ,...,tℓ− 2 , 1 ,1)tℓ− 1 tℓ > ···> f(1,...,1)t 1 t 2 ..ℓ = pt 1 ..ℓ
Extremal Graph Theory 2012-2013 Lecture Notes
Module: Extremal Graph Theory
University: University of Cambridge
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