Regularity Szemeredis Lemma
One of the most important tools in dense combinatorics. Message: every graph G is the approximate union of constantly many random-like bipartite graph. The number of parts depends only on the error of the approximation constant but not the size of G! For disjoint subsets X, Y V , |E(X, Y )| d(X, Y ) := |X| |Y |
is the density of the pair (X, Y ).
A pair (A, B) of disjoint subsets A, B V is called -regular pair for some > 0 if all X A, and Y B with |X| |A| and |Y | |B| satisfy |d(X, Y ) d(A, B)| . Remark Just like in a random bipartite graph...
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Szemeredis Regularity Lemma
A partition {V0 , V1, . . . , Vk } of V is called an -regular partition if (i) |V0| |V |
k (iii) all but at most 2 of the pairs (Vi , Vj ), with 1
(ii) |V1| = = |Vk |
i < j k2 , are -regular
V0 is the exceptional set Regularity Lemma (Szemeredi) > 0 and integer m 1 integer M = M (, m) such that every graph of order at least m admits an -regular partition {V0, V1 , . . . , Vk } with m k M . Was devised to prove that dense sets of integers contain an arithmetic progression of arbitrary length.
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History of Szemeredis Theorem
Szemeredis Theorem (1975) For any integer k 1 and > 0 there is an integer N = N (k, ) such that any subset S {1, . . . , N } with |S| N contains an arithmetic progression of length k. Was conjectured by Erdos and Turan (1936). Featured problem in mathematics, inspired lots of great new ideas and research in various elds; Case of k = 3: analytic number theory (Roth, 1953; Fields medal) First proof for arbitrary k: combinatorial (Szemeredi, 1975) Second proof: ergodic theory (Furstenberg, 1977) Third proof: analytic number theory (Gowers, 2001; Fields medal) Fourth proof: fully combinatorial (Rodl-Schacht, Gowers, 2007) Fifth proof: measure theory (Elek-Szegedy, 2007+) One of the ingredients in the proof of Green and Tao: primes contain arbitrary long arithmetic progression
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Proof of the Erdos-Stone Thm
Erdos-Stone Theorem. (Reformulation) For any > 0 and integers r 2, t 1 there exists an integer N = N (r, t, ), such that any graph G on n N ver1 tices with more than 1 r1 + n edges con2 tains Trt,r . Proof strategy: Based on an -regular partition, build a regularity graph R of G. (Regularity Lemma) Show that R contains a Kr (Turans Theorem) Show that Kr R Trt,r G
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Regularity graph
Given -regular partition P = {V0, V1 , . . . , Vk } of G, m k M (, m), dene the regularity graph R = R(P, d) V (R) = {V1, . . . , Vk } ViVj E(R) if (Vi , Vj ) is -regular pair with density d(Vi , Vj ) d Goal Choose , m, d such that most edges of G go between the sets Vi and Vj with ViVj E(R) How many edges are not at the right place?
n2 < n2 < k m # of edges incident to V0 : at most n n = n2
# of edges inside Vi: at most k n/k 2
# of edges between non-regular pairs:
k at most 2 n 2 < n2 k
# of edges between pairs of density < d: at most
2 k d n dn2 2 k
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Regularity graph contains an r-clique
Conclusion: If , m, and d is chosen such that 1 < m 2 then most edges of G go between sets Vi and Vj with ViVj E(R). d + 2 +
1 most means at least 1 r1 n + 2 n2 2
On the other hand: # of edges of G going between sets Vi and Vj with ViVj E(R): Hence 1 1 r1 1 1 r1
2 at most |E(R)| n k
n 2 n 2 + n |E(R)| 2 2 k k + k2 |E(R)| 2 2
Choose m = m() such that
1 ex(m, Kr ) 1 r1 m + 2 m2 2
Then Turans Theorem R contains a Kr
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Finding Trt,r
There are r classes Vi1 , . . . , Vir such that (Vij , Vi ) is an -regular pair of density at least d, for every 1 j < r. Let n = |Vij |. Then n n 1 n. k k We nd a Trt,r in G[Vi1 Vir ]. Lemma Let (A, B) be an -regular pair with d(A, B) d Let Y B be a subset with |Y | |B|. Then |{v A : dY (v) < (d )|Y |}| < |A|. Proof. Otherwise the subsets Y B and {v A : dY (v) < (d )|Y |} A would contradict the -regularity of (A, B). 2 For a set S V let (S) = vS N (v) denote the set of common neighbors of the vertices of S.
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Finding Trt,r
(d )t1 n n (r 1)t n t n S1 V1, |S1| = t |Vi (S1 )| (d )t n for i = 2, 3, . . . , r (d )2t1 n n (r 2)t (d )t n t n S2 V2, |S2| = t |Vi (S1 S2)| (d )2tn for i = 3, . . . , r . . . (d )(r1)t1 n n t (d )(r2)tn t n Sr1 Vr1, |Sr1| = t r1 |Vr (i=1 Si)| (d )(r1)tn
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Finding Trt,r
r1 Sr NVr (i=1 Si), |Sr | = t and thus G[S1 Sr ] contains a Trt,r provided
(d )(r1)t n t
Strongest of the blue conditions: (d )(r1)t1 Lets not forget: 1 d + 2 + < m 2 Choose, for example: 6 m
d=6
= min
d t(r1) , 1 2 t(r1)
Green conditions are satised by choosing a large enough threshold vertex number N = N (r, t, ). r, t, ; m, d, ; N
We
also needed large m earlier for using Turans Theorem.
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The Erdos-Turan conjecture
A set S of positive integers is k-AP-free if {a, a + d, a + 2d, . . . , a + (k 1)d} S implies d = 0. sk (n) = max{|S| : S [n] is k-AP-free} How large is sk (n)? Could it be linear in n? Erdos-Turan Conjecture (Szemeredis Theorem) For every constant k, we have sk (n) = o(n). Construction (Erdos-Turan, 1936) s3 (n)
log 2 log 3 . n
S = {s : there is no 2 in the ternary expansion of s} S is 3-AP-free. For n = 3l , |S [n]| = 2l Roths Theorem (1953) s3(n) = o(n).
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Applications of the Regularity Lemma
Removal Lemma For > 0 = () such that the following holds. Let G be an n-vertex graph such that at least n edges has to be deleted from G 2 to make it triangle-free. Then G has at least n tri3 angles. Proof. Apply Regularity Lemma (Homework). Roths Theorem For > 0 N = N ( ) such that for any n N and S [n], |S| n, there is a three-element arithmetic progression in S. Proof. Create a tri-partite graph H = H(S) from S. V (H) = {(i, 1) : i [n]} {(j, 2) : j [2n]} {(k, 3) : k [3n]} (i, 1) and (j, 2) are adjacent if j i S (j, 2) and (k, 3) are adjacent if k j S (i, 1) and (k, 3) are adjacent if k i 2S
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Roths Theorem Proof contd
(i, 1), (i + x, 2), (i + 2x, 3) form a triangle for every i [n], x S. These |S|n triangles are pairwise edge-disjoint. At least n2 18 |V (H)| edges must be removed 2 from H to make it triangle-free. Let = 18 provided by the Removal Lemma. There are at least |V (H)| triangles in H. 3 S has no three term arithmetic progression {(i, 1), (j, 2), (k, 3)} is a triangle iff ji = kj S. Hence the number of triangles in H is equal to n|S| n2 < 6n , provided n > N ( ) := 1 . 2 3
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Behrends Construction
Construction (Behrend, 1946) s3(n) n
1O 1
log N
.
Construct set of vectors = (a0 , a1, . . . , al1): a Vk = { Z a Z where = a
l
:
a
2
d = k, 0 ai < for all i < q}, 2
l1 2 i=0 ai .
Interpret a vector {0, 1, . . . , d 1}l as an integer a written in d-ary:
l1
n = a
i=0
ai di .
Let Sk = {n : Vk } a a
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Claim Sk [dl ] is 3-AP-free for every k. Proof. Assume n + n = 2n. a c b Then ai + bi = 2ci for every i < l, because ai + bi and 2ci are both < d (so there is no carry-over) So + = 2. But a b c 2 = 2 = 2 k = + + , c c a b a b and equality happens only if and are parallel. Since a b they are of the same length, we conclude = . 2 a b Take the largest Sk . Bound its size by averaging: {0, 1, . . . , d 1}l 2 < ld2 , a a so there is a k for which | ...