We just finished the Erdős-Stone Theorem, and ex(n, F ) (1 1/(χ(F ) 1)) ( n

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1 Lecure 3 - Kövari-Sós-Turán Theorem Jacques Versraëe jacques@ucsd.edu We jus finished he Erdős-Sone Theorem, and ex(n, F ) ( /(χ(f ) )) ( n 2). So we have asympoics when χ(f ) 3 bu no when χ(f ) = 2 i.e. F is biparie. We focus on his case now, and his lecure looks a F = K s,, he complee biparie graph wih pars of sizes s and. Here is a warmup exercise. Exercise A. Deermine ex(n, K, ). The exremal problem for K s, is relaed o he Zarankiewicz problem (95). This problem asks for he maximum number z(m, n, s, ) of s in an m n 0- marix ha does no conain an s minor filled wih s. Observe ha if A is an m n 0- marix, hen A is he incidence marix of an m n biparie graph: he rows of he marix are labelled by he verices in he par of size m, and he columns are labelled by he verices in he par of size n. If A has no s minor filled wih s, hen he graph has no K s, wih s verices in he par of size m and verices in he par of size n. Theorem. (Kövari-Sós-Turán, 954) For m, n, s,, z(m, n, s, ) (s ) / nm / + ( )m. Proof. We wrie he proof in he language of graphs. Le A be an m n marix wih z = z(m, n, s, ) ones and no s minor filled wih s, and le G be he corresponding m n biparie graph, say wih pars R of size m and C of size n. They key observaion is d(v) C (s ). By convexiy, his implies v R E(G) / R C R (s ). Using (x + ) /! ( x ) x /! for x, we ge This gives he required bound. m(z/m + ) (s )n. The proof of he bound above also gives an upper bound on he Turán numbers ex(n, K s, ):

2 Theorem 2. For s,, ex(n, K s, ) 2 (s )/ n 2 / + ( )n. 2 Proof. If G = (V, E) is a K s, -free n-verex graph, hen d(v) n (s ). v V Since v V d(v) = 2e(G), he convexiy of binomial coefficiens gives n(2e(g)/n + ) (s )n which gives he heorem. In general, he order of magniude of ex(n, K s, ) and z(m, n, s, ) are no known when s, 4, and specifically, K 4.4. The bes lower bounds in general come from random graphs and random marices: Theorem 3. For all s,, ex(n, K s, ) = Ω(n 2 (s+ 2)/(s ) ). The same holds for z(n, n, s, ). Laer on we shall look a consrucions of K s, -free graphs. Righ now we concenrae on he case s = = 2. Quadrilaerals. The case s = = 2 of he Kővari-Sós-Turán heorem is exensively researched. From he las secion, we have: ex(n, C 4 ) 2 n n n and z(m, n, 2, 2) mn /2 + m. 4 Exremal consrucions come from projecive geomery. Recall a projecive plane of order q consiss of q 2 + q + poins and q 2 + q + lines, such ha each line is a se of q + poins, each poin is conained in q + lines, every pair of lines inersecs in exacly one poin, and every wo poins are conained in exacly one line. The exisence of projecive planes has been esablished only when q is a prime power, and remains one of he main open problems in projecive geomery. Direcly from projecive planes, by considering he incidence marix of poins and lines, one obains he exac value of z(n, n, 2, 2): Theorem 4. Le q be a posiive ineger and n = q 2 + q +. Then z(n, n, 2, 2) = (q + )n if and only if here exiss a projecive plane of order q. 2

3 Equivalenly, he biparie incidence graph of a projecive plane has (q + )(q 2 + q + ) edges, and so a projecive plane of order q exiss if and only if here exiss a (q + )-regular n by n biparie C 4 -free graph, where n = q 2 + q +. Exercise B. Prove his heorem. This heorem gives he asympoic formula z(n, n, 2, 2) n 3/2 as n, due o he exisence of projecive planes of all prime orders and he following lemma from number heory: Lemma. For all large enough n, here exiss a prime p beween n and n + n 2/40. Erdős, Rényi and Sós (966) were he firs o describe asympoically exremal C 4 -free graphs. The resuls of he las secion show ex(n, C 4 ) 2 n3/2 + o(n 3/2 ), and heir consrucion maches his upper bound asympoically. I is described as follows. Le V be a 3-dimensional vecor space over F q, and le ER q be he graph whose verex se is he se of -dimensional subspaces of V, and whose edge se consiss of pairs of orhogonal -dimensional subspaces of V. Then ER q is known as a polariy graph of a projecive plane, and one checks ha ER q is C 4 -free and has 2 q(q + )2 edges. Exercise C. Verify ha e(er q ) = 2 q(q + )2 and ha ER q is C 4 -free. In general, given a projecive plane wih poin se P and line se L, a polariy of he projecive plane is an involuion π : P L P L ha preserves incidences. We refer he reader o books on projecive geomery for more informaion on planes and polariies. This allows us in essense o fold he biparie incidence graph of a projecive plane ono one of is pars o obain a C 4 -free graph: if G q is he biparie incidence graph of a projecive plane, wih pars P and L corresponding o poins and lines, hen form a graph H q wih V (H q ) = P and where {p, p 2 } E(H q ) if p is adjacen o π(p 2 ). I is a challenging ask o prove ex(q 2 +q +, C 4 ) 2 q(q +)2 ; noe ha when n = q 2 +q +, he couning argumens of he las secion only give ex(n, C 4 ) (n/2) n 3/4 + n/4 = 2 (q + )(q2 + q + ). Füredi (988) showed ha in fac polariy graphs are exremal C 4 -free graphs, and showed ha any C 4 -free graph wih q 2 + q + verices where q > 3 is a prime power mus be he polariy graph of some projecive plane. In paricular, Theorem 5. Le q > 3 be a prime power and n = q 2 + q +. Then ex(n, C 4 ) = 2 q(q + )2. Once more, he disribuion of primes according o he lemma shows ha for all n, ex(n, C 4 ) 2 n3/2 + o(n 3/2 ). Deermining ex(n, C 4 ) when n is no of he form q 2 + q + seems o be a very difficul problem. There are many near exremal consrucions of C 4 -free graphs, we give an example here. Exercise D. Le G be he graph whose verex se is F q F q \{(0, 0)} where q is prime, and whose edge se consiss of pairs {(x, y), (a, b)} where b + y = xa. Verify ha G q is C 4 -free and has v(g) = q 2 and e(g) 2 (q )(q2 ). 3

4 Füredi (996) also deermined he asympoic value of ex(n, K 2, ) and z(n, n, 2, ), showing ex(n, K 2, ) 2 n 3/2 for all 2 and z(n, n, 2, ) n 3/2 for all 2. In addiion, Füredi (996) showed ex(n, K 3,3 ) 2 n5/3 + o(n 5/3 ). We end wih he following dense K 3,3 - free consrucion due o Brown (966) which gives he lower bound maching Füredi s upper bound on ex(n, K 3,3 ): Exercise E. Le G q be he graph whose verex se is F q F q F q. Join a riple (a, b, c) o (x, y, z) if (x a) 2 + (y b) 2 + (z c) 2 =. Prove ha G q conains no K 3,3 and deermine he number of edges in G q. Conclude ex(n, K 3,3 ) 2 n5/3 + o(n 5/3 ). Quick Applicaions. The firs quick applicaion is geomeric. The famous Erdős uni disance problem (946) asks for he maximum number of uni disances ha can be defined by n poins in he plane in oher words, if P is he se of poins, how many pairs {x, y} P have d(x, y) = in he Euclidean meric. Erdős conjecured ha he answer is n +o() as n, and his remains wide open despie a remendous amoun of work in rying o find good upper bounds. The saring poin is in he following heorem: Theorem 6. Any se of n poins in he plane defines a mos n 3/2 / 2+o(n 3/2 ) uni disances. Proof. We reduce o an exremal problem. Noe ha if x, y are poins in he plane, hen here exis a mos wo poins z such ha d(x, z) = d(y, z) =. Form he graph G whose verex se is he se of n poins in he plane, and where {x, y} E(G) if d(x, y) =. Then G is K 2,3 -free. By he Kövari-Sós-Turán Theorem, e(g) n 3/2 / 2 + o(n 3/2 ). This elemenary argumen is no very difficul o improve o O(n 4/3 ), however he conjecure sill appears ou of reach. Exercise F. Prove ha he number of uni disances beween n poins in R 3 is O(n 5/3 ). Wha lower bound can you give? The Szemerédi-Troer Theorem (984) on planar poin-line incidences saes ha m poins and n lines in he plane define a mos abou (mn) 2/3 + m + n poin-line incidences. This provides he saed upper bound of O(n 4/3 ) for uni disances in he plane. This resul is special o he plane, since in he projecive plane, we may have n poins and n lines and he oal number of poin-line incidences is abou n 3/2. We can prove a general upper bound as follows: Theorem 7. A se of m poins and n lines in he plane define a mos mn /2 + m + n incidences. 4

5 We leave he proof as an exercise. A final applicaion is o Sidon ses. A se S in an abelian group Γ is called a Sidon se if x + y = z + w wih x, y, z, w S implies {x, y} = {z, w}. The exremal size of Sidon ses has been sudied for a leas half a cenury, wih roos in projecive geomery. I urns ou ha a projecive plane may be defined by aking special Sidon ses S in Z q 2 +q+ hese are called Singer cycles or perfec difference ses and hey are Sidon ses of size q + in Z q 2 +q+. Their special propery is ha every non-zero ineger mod q 2 + q + can be represened uniquely as a difference of elemens in S. To ge a projecive plane, le he lines of he plane be he se of ranslaes of S, namely {S + λ : λ Z q 2 +q+}. Theorem 8. If Γ is an abelian group of order n, and S Γ is a Sidon se, hen S n /2 +. Proof. Form a biparie graph G wih pars X = Y = Γ, and join x X o y Y if x + a = y for some a S. One checks ha because S is a Sidon se, G is C 4 -free. Therefore if we wrie n = q 2 + q + for a posiive real q, hen e(g) z(n, n, 2, 2) (q + )n n 3/2 + n. On he oher hand, e(g) = S Γ = S n, and so S n /2 +. Exercise G. Le Γ = F q F q. Prove ha if q is odd, hen S = {(x, x 2 ) : x F q } is a Sidon se, and if q is even, hen S = {(x, x 3 ) : x F q } is a Sidon se. Exercise H. Le S be a Sidon se in an abelian group Γ. Le G be he graph wih V (G) = Γ and E(G) = {{x, y} : x + y S}. Deermine he number of edges in G, and wheher G conains a cycles of lengh hree and four. 5