An exact result for hypergraphs and upper bounds for the Turán density of K r r+1

Transcription

1 A exact result for hypergraphs ad upper bouds for the Turá desity of K r r+1 Liyua Lu Departmet of Mathematics Uiversity of outh Carolia Columbia, C 908 Yi Zhao Departmet of Mathematics ad tatistics Georgia tate Uiversity Atlata, GA February 1, 009 Abstract We first aswer a questio of de Cae []: give r 3, if G is a r-uiform hypergraph o vertices such that every r + 1 vertices spa 1 or r + 1 edges, the G = K r or K 1, r assumig that > p 1r, where p is the smallest prime factor of r 1 We the show that the Turá desity πkr+1 r 1 1 r 1 1 r r p r+p r 1 + r+1, for all eve r 4, improvig a well-kow boud 1 1 r of de Cae [1] ad idoreko [15] 1 Itroductio Give a positive iteger ad a r-uiform hypergraph H or r-graph, for short, the Turá umber ex, H is the maximum umber of edges i a r-graph o vertices that does ot cotai H as a subgraph called H-free It is easy to see that f, H = ex, H/ r is a decreasig fuctio of Katoa et al [10] The limit πh = lim f, H, which always exists, is called the Turá desity of H Let Kk r deote the complete r-graph o k vertices Turá determied ex, Kk which implies that πk k = 1 1 k 1 for all k 3 However, o Turá desity πkk r is kow for ay k > r 3 The most well-kow case k = 4 ad r = 3 is a cojecture of Turá [0], claimig that πk4 3 = 5/9 Erdős [5] offered $500 for determiig Research supported i part by NF grat DM Research supported i part by NA grats H , H

2 ay πkk r with k > r 3 ad $1000 for aswerig it for all k ad r The best geeral kow upper boud is due to de Cae [1] ad the special case πk r k 1 1 k 1 πk r r r, 1 was also give by idoreko [15] For the lower boud, idoreko [18] showed that πkr+1 r 1 l r r 1 + o1 for large r ee survey papers of de Cae [] ad idoreko [17] for other bouds For odd r 3, Chug ad Lu [4] improved to πk r r+1 1 5r + 1 9r + 4r rr + 3 = 1 1 r 1 1 rr O r 3 3 Whe r = 3, this gives the best kow upper boud for πk 3 4 However, it seems more difficult to improve for eve r 4 For odd r, the edge desity f, Kr+1 r of extremal hypergraphs decreases quickly for small, for example, fr+, Kr r+1 < fr + 1, Kr+1 r It is o loger the case for eve r 4 It is easy to see that f8, K 4 5 = f7, K 4 5 = f6, K 4 5 = f5, K 4 5 = 4 5 Let G 4 8 be the 4-graph o [8] = {1,,, 8} with o-edges {1,, 3, 4}, {1,, 5, 6}, {1,, 7, 8}, {1, 3, 5, 7}, {1, 3, 6, 8}, {1, 4, 5, 8}, {1, 4, 6, 7}, {, 3, 5, 8}, {, 3, 6, 7}, {, 4, 5, 7}, {, 4, 6, 8}, {3, 4, 5, 6}, {3, 4, 7, 8}, {5, 6, 7, 8} It is easy to check that G 4 8 is K4 5 -free with edges, ad every 5 vertices o [8] cotai exactly 4 edges ad oe o-edge While studyig Turá umbers, De Cae [] asked the followig questio: Describe all r-graphs such that every r + 1 vertices spa 1 or r + 1 edges Frakl ad Füredi [7] solved a similar problem earlier: they described all 3-graphs such that every 4 vertices cotai 0 or edges the geeral case whe every r + 1 vertices spa 0 or edges is still ope Theorem 1 below aswers the questio of de Cae I fact, our uderlyig hypergraph G is the complemet of the r-graph i de Cae s questio: every r + 1 vertices cotai 0 or r edges of G 4 Fix a -vertex set V The empty r-graph is the oe with o edge; a complete star is a r-graph whose edge set cosists of all r-sets cotaiig some fixed vertex x V Clearly the empty graph ad complete stars satisfy 4, ad their complemets are K r ad K r 1

3 Theorem 1 Let r uppose that G is a r-graph o vertices satisfyig 4 1 If r =, the G is a complete bipartite graph If r 3 ad > rp 1, the G is either the empty graph or a complete star, where p is the smallest prime factor of r 1 The proof for the r = case is rather straightforward; the r 3 case requires more work, eg, we use the upper boud for ex, K r r+1 correspodig to Note that our costructio G4 8 suggests that a lower boud for is ecessary whe r 3 Let K4 3 deote the uique 3-graph with 4 vertices ad 3 edges It is ot hard to prove that ex, K eg, de Cae [1] Mubayi [13] used the result of Frakl ad Füredi [7] ad supersaturatio to obtai that ex, K4 3 < Talbot [19] recetly improved the boud to by cosiderig related Chromatic Turá problems Motivated by these works, we apply Lemma 4, the key step i the proof of Theorem 1, to slightly improve for all r 3 We ca also improve by Theorem 1 ad supersaturatio though the result turs out to be weaker tha Theorem ee the last sectio for details Theorem For r 3, let p be the smallest prime factor of r πkr+1 r r 8 + O 1 9r r, p =, r 1 1 r r p r+p + r+1, p 3 Whe r is odd thus p =, Theorem is slightly weaker tha 3 For eve r, Theorem gives a ew upper boud for πkr+1 r For example, whe r = 6k + 4 thus p = 3, it gives πkr+1 r 1 1 r 1 1 r 3 + O r 4 The rest of the paper is orgaized as follows After recallig the proof of i ectio, we prove Theorem 1 i ectio 3 ad Theorem i ectio 4 I the last sectio we give some cocludig remarks Prelimiaries I this sectio we recall the proof of, πkr+1 r 1 1 r, which idicates the source of our improvemet Let G be a Kr+1 r -free r-graph For 0 i r, let i deote the family of r + 1-sets that cotai exactly i edges of G ad δ i = i Lemma 3 For all r, ex, K r r r r r + 1 r 3

4 Moreover, if every Kr+1 r -free r-graph G o vertices satisfies δ i y r+1, the x = πkr+1 r satisfies x x 1 1 r 1y + r rr Proof Let G be a Kr+1 r -free r-graph G o vertices ad e edges For a -set V G, let d = {v : {v} EG} be the degree of We have d = er 7 d r i = δ i 8 r iδ i = e r 9 We derive that d = r = = i d + d δ i + er usig 7 ad 8 i i δ i i iδ i + er = i i δ i e r usig 9 r r iδ i e r = re r e r Usig the Cauchy-chwarz iequality er = d d ad 7, we coclude that er re r e r e r r 1 r r + 1 with equality holds if all d are the same ad δ i = 0 This proves 5 4

5 Whe δ i y r+1, we ca refie the estimate of d as d = = r r i δ i e r r r iδ i ir iδ i e r r iδ i r 1 δ i e r re r r 1y e r 10 r + 1 Cosequetly er re r y r+1 e r, ad 6 follows by lettig e = x r 3 Proof of Theorem 1 I this sectio we prove Theorem 1 We eed some otatios Let G = V, E be a r-graph ad T be a -subset of V The eighborhood N T is the set of x V such that {x} T E; ad let N T = V T N T We defie the degree d T = N T ad the o-degree d T = N T The followig lemma provides iformatio o the structures of all r-graphs satisfyig 4 Lemma 4 Let G be a r-graph with r such that every r + 1 vertices cotai 0 or r edges The followig are true for every r 1-vertex set T 1 T N T cotais o edge of G For every vertex x N T ad every r 1-subset D T N T, {x} D is a edge of G 3 If r 3, the either d T < or d T < p 1 Proof To prove Part 1, we eed to show that for i = 1,, r, every i-subset of N T ad every r i-subset of T together form a o-edge We prove this claim by iductio o i The defiitio of N T justifies the claim for i = 1 uppose the claim holds for some i 1 Cosider A N T ad B T with A = i + 1 ad B = r i 1 Pick a r i-set C such that B C T We kow that the r + 1 set A C cotais either 0 or r edges, ad by iductio hypothesis, oe of the i + 1 i-subsets of A together with C form a edge Therefore A C cotais o edge, i particular, A B is ot a edge To prove Part, we eed to show that for i = 0,,, every i-subset of N T, every i- subset of T ad x together form a edge We agai apply iductio o i The fact that x N T justifies the i = 0 case uppose that A N T ad B T with A = i + 1 ad B = r i for some i 0 Pick a set C such that B C T with C = r i 1 We kow that the 5

6 r + 1-set A C {x} cotais either 0 or r edges The iductio hypothesis says that all the i-subsets of A form edges together with C {x} O the other had, we kow from Part 1 that A C is ot a edge Therefore A C is the uique o-edge i A C {x}, which implies that A B {x} is a edge To prove Part 3, we assume istead that some r 1-set T satisfies d T ad d T p 1 Let x, y be two vertices i N T, ad be a r + p -subset of T N T By Part, every r 1-subset of together with x or y form a edge This forces that every r 1-subset of together with x, y cotai precisely r edges Defie a auxiliary r -graph H o i which a r -subset R of is a edge if ad oly if R {x, y} is a o-edge of G here we eed r 3 For k > t, we call a t-graph o vertices a [, k, t]-system if every k vertices cotai exactly oe edge Hece H is a [r + p, r 1, r ]-system The umber of edges of H, r+p /p is therefore a iteger This implies that p divides r+p = r+p, or p! divides r + p r Recall that p divides r 1 or r 1 mod p We thus obtai a cotradictio because r + p r p 1! 0 mod p Proof of Theorem 1 Whe r =, we apply Lemma 4 Parts 1 ad to ay T = {x} ad lear that A = {x} N {x} is a idepedet set ad ab EG for ay a A, b A We thus kow that B = V A is idepedet from the assumptio 4 Therefore G is a complete bipartite graph with partitio sets A ad B Now let r 3 By Lemma 4 Part 3, every r 1-subset T V G satisfies d T 1 or d T p We claim that if > p 1r, the there always exists a r 1-vertex set T 0 with d T0 1 I fact, suppose that d T p for all T The the umber of o-edges is at most p = p 1 < r r + 1 r r 1 r r + 1 r because > p 1r I other words, G cotais more tha 1 1 r + 1 r r+1 r edges By Lemma 3, there exists a r + 1-set which cotais r + 1 edges, cotradictig 4 Let U = T 0 N T0 The U 1 Lemma 4 Part 1 says that U cotais o edges of G If U =, G is the empty graph Otherwise, U = 1 Let {x} = V G U Lemma 4 Part implies that G is the complete star with ceter x 4 Proof of Theorem uppose that G = V, E is a Kr+1 r -free r-graph We call a r+1-set V bad if cotais i edges of G for some 1 i r 1 Let B be the family of all bad sets Let Q cosist of all r + p-subsets Q V such that some r 1-subset of Q has exactly two eighbors i Q ie, N Q = The i every iduced sub-hypergraph G[Q], Q Q, we have d = ad d = p 1 Lemma 4 Part 3 thus implies that every Q Q cotais some B B For a arbitrary r + p-set Q V, we defie fq as the umber of r 1-sets Q such that 6

7 N Q =, ad wq as the umber of bad sets i Q The wq 1 for all Q Q Let RG = max Q Q fq/wq The heart of the proof of Theorem is the lemma below Lemma 5 For r 3, let p be the smallest prime factor of r 1 ad R max{r 1, RG : G is a Kr+1 r -free r-graph } 11 The the Turá desity x = πk r r+1 satisfies x + r 1 R x1 x 1 1 r 1 The proof of Lemma 3 used the Cauchy-chwarz iequality, or the covexity of x I order to prove Lemma 5, we istead use the covexity of fx = x + cx 1 x with 0 < c 1/ ice fx > x, this covexity helps us to get a tighter boud for πk r r+1 Propositio 6 Let c, p be costats satisfyig 0 c 1/ ad p The the fuctio fx = x + cx 1 x is covex o [0, 1] Proof First let p = From fx = x + cx 1 x, we derive that f x = + c 6cx ice 1 3x ad 0 c 1/, we have f x 0 o [0, 1] Now assume that p 3 We have ad f x = x + cx1 x cp 1x 1 x p f x = + c1 x 4cp 1x1 x p + cp 1p x 1 x p 3 + c1 x 4cp 1x1 x p = + cgx, where gx = 1 x p 1x1 x p = 1 x p 1 p 1x It is ot hard to see that gx p 4 p > o [0, 1] Thus f x > + c 0 ad cosequetly fx is covex o [0, 1] Proof of Lemma 5 Let G = G be a Kr+1 r -free r-graph o vertices with x r = ex, Kr+1 r edges The fact that wq 1 for all Q Q ad the defiitio of R give d d = 1 = 1 p 1 wq = Q : Q =r+p, N Q =,Q B B:B Q = fq r + 1 wq R B p 1 B B Q Q:B Q We thus obtai a boud for the umber of bad sets: d d B = y 13 r + 1 R 7 r+1

8 We defie a fuctio fx = x + cx 1 x with c = R The defiitio of R forces c 1 By Propositio 6, fx is covex o [0, 1] For every r 1-set, let x = d r+1 The covexity of f gives ice x = O the other had, d r+1 = e f x fx r = x, we have f x fx = x + cx 1 x = d r + 1 = fx + c d d r + 1 p+1 Usig 10, d r + 1 re r r 1y r+1 e r r + 1 re r e r = r + 1 r 1 r + 1r y + O 1 ice d, d for all, we have d c d d r + 1 p+1 d = O p+ + p 1! d d ad = Op+ + cp 1! d d r + 1 p+1 R r+1 y r+1 cp 1! r + 1 p+1 + O 1 R = c r + 1r y + O 1 r 1 = r + 1r y + O 1 Puttig these together, we have fx fx = re r e r r O x + O 1 r The claim 1 follows by lettig ad substitutig fx = x + R x 1 x Trivially we ca choose R = r+p applyig it i 1 already yields a better boud for tha The followig techical lemma refies our boud for R πk r r+1 8

9 Lemma 7 For ay Kr+1 r -free r-graph G with at least r + p vertices, { 3r+ RG 4, p =, r+1 + r+p, p 3 Proof Fix Q Q Let q = r + p = Q We cosider a p-graph H o Q whose edges are the complemets of edges of G, amely, EH = {Q e : e EG, e Q} Therefore a p 1-subset Q has degree i i H if ad oly if the r + 1-set Q \ cotais exactly i edges of G Let W be the family of p 1-sets with degree i i H for some 1 i r 1 their complemetary sets i Q are bad sets for G, ad F be the set of p + 1-sets cotaiig exactly edges of H their complemetary sets i Q have degree i G[Q] Followig the defiitio of wq ad fq i the begiig of the sectio, we have W = wq ad F = fq Note that the degree of ay p 1-set i H is at most r sice G is Kr+1 r -free We also kow that W = wq 1 sice Q cotais at least oe bad set From ow o we cosider H as our uderlyig hypergraph istead of G Our goal is to obtai upper bouds for F / W The p = case: Partitio the vertex set Q ito W = {x : 1 d {x} r 1}, U = {x : d {x} = r}, ad Z = {x : d {x} = 0} ice each vertex i U misses at most oe vertex of Q, o-edges iside U form a matchig M Let U 1 = U V M cosist of vertices i U ot covered by M The family F cosists of all triples T = abc such that ab, ac EH ad bc EH Clearly T W U for all T F We partitio F = F 0 F 1 F such that F 0 cosists of all T F with T W, F 1 cosists of those T = abc with b, c U, ad F cosists of those abc with exactly oe of b, c i W We have F 0 W q, ad F1 M q ice M q U 1 W / ad W 1, W F 0 + F 1 q + 1 q U 1 W q 1 W q 1 W 1 + q U 1 W = 1 W q q U To estimate F, let us cosider a vertex x W By defiitio of M, x is adjacet to all the vertices i V M If x has t o-eighbors i U, the t U 1 ad the umber of triples of F cotaiig x is at most t U t { U 1 U U 1 if U 1 U /, U /4 if U 1 > U / First assume that U 1 U / ice U q 1, we have F W U 1 U U 1 W U 1 p 1 U 1 9

10 Together with 14 ad usig W 1, we derive that F W 1 q q U U 1 q 1 U 1 3q W 4 1, where equality holds whe U 1 = q/4 I cotrast, whe U 1 > U /, we have F W 1 q q U U W 1 q q 1 U U W q 1q 3, 1 4 where equality holds whe U = q 1 ad U 1 = U / It is easy to check that 3q q 1q 3 for ay q We therefore coclude that F / W 3q 4 1 = 3r+ 4 The p 3 case: Partitio F ito F 1 = { F : T for some T W } ad F = F F 1 Clearly F 1 W q p+1 We prove below that F W q Fix a p + 1-set F By the defiitio of F, all p 1-subsets of either have degree 0 or degree r i H equivaletly, o-degree 1, ad cotais exactly two edges p-sets of H We deote these two edges by T {x} ad T {y}, where T is a p 1-set We have two observatios Observatio 1 For ay a T, the p 1-set T {x} {a} must have degree r sice T {x} EH ad {a} EH This implies that T {x, u} {a} EH for ay vertex u Observatio For two distict vertices a, b T, the p 1-set {a, b} must have degree 0 sice two p-sets {a}, {b} are ot edges Note that we eed T or p 3 here We defie a fuctio g : F Q such that g = T If g is a oe-to-oe fuctio, the F q W q Otherwise, let T be a p 1-set with the maximum size of g 1 T = { F : g = T } Let M cosist of all -sets T such that g 1 T The M = g 1 T We first claim that M is a matchig, amely, ay two pairs i M are disjoit uppose istead xy, xz M For ay a T, the p 1-set {x} T {a} is cotaied i two oedges {x, y} T {a} ad {x, z} T {a}, cotradictig Observatio 1 We ext claim that {x, y} T {a, b} W for two pairs P, Q M, ay vertices x P, y Q, ad ay two vertices a, b T I fact, Observatio says that T P = P T {a, b} has degree 0, ad cosequetly {y} T P is a o-edge imilarly {x} T Q is also a o-edge with T Q = Q T {a, b} O the other had, we kow that {x, y} T {a} is a edge from Observatio 1 Hece {x, y} T {a, b} has degree at least 1 ad o-degree at least ad thus it belogs to W This implies that W M M sice M We thus coclude that T R g 1 T W q 10

11 Cosequetly F W q p+1 + W q ad F / W q p+1 + q Proof of Theorem : Let R be the upper boud for RG i Lemma 7: { 3r+ R = 4, p =, r+1 + r+p, p 3 It is easy to check that R r 1 ad cosequetly c = R 1 Let the Turá desity x = π, Kr+1 r = 1 1 r ε The iequality 1 from Lemma 5 gives ε c r ε r + ε > c r ε, r which implies that ε > > 1 1 c r c + r 1 1 c r r 1 c r We ow apply the defiitio of R to c = R p 3, we have ε > 1 1 r r 1 r+p r + r+1 1 c r > Whe p =, this gives ε > 8 9r + O1/r 3 Whe r p r+p r 1 + r r 5 Cocludig Remarks Followig the approach of Mubayi [13], we ca improve by Theorem 1 ad supersaturatio as follows By Theorem 1, every Kr+1 r -free r-graph o p 1r + 1 vertices either cotais a bad set or has at most r edges I fact, from a simple removig argumet ad the structure of the r-graphs that satisfy 4, we ca show that every Kr+1 r -free r-graph o 0 = p 1r + 1 vertices either cotais p 1 bad sets or has at most 0 1 edges By averagig ad 5, every K r r+1 -free r-graph with 0 vertices ad x r edges cotais at least x r r + 1 r 0 r+1 r p r+1 r + 1 bad sets this techique is usually called supersaturatio [6] By 6 i Lemma 3, as r, πkr+1 r 1 1 r 1 + o1 p r+1 r+1 rp 1, which is weaker tha Theorem This suggests that 13 gives a better estimate o the umber of bad sets tha the oe from supersaturatio 11

12 The boud > rp 1 i Theorem 1 is tight for odd r ad r = 4 but we do ot kow if it is tight for ay eve umber r > 4 The boud for R i Lemma 7 is ear tight for p =, tight up to a costat factor for p = 3, but we do ot kow if it is tight for p > 3 Let us cosider K5 4 The followig bouds are due to Giraud [8, 9] πk , 15 idoreko showed [15] that πk ad cojectured [17] that the lower boud above gives the correct value Applyig 1 with R = 13 the proof of R = 13 is ot very hard but ot simple either, we derive that πk Markström [1] recetly obtaied a much better boud πk by determiig ex, K4 5 for 16 ad applyig the simple boud πk5 4 ex, K4 5 / 4 ex, K 4 5 was oly kow [3] for 10 before For ay r-graph F with f edges, idoreko [16] used a aalytic approach to obtai πf 1 1 f 1 16 Whe F = Kr+1 r, f = r + 1 ad this gives A r-graph is called a forest if there is a orderig of its edges E 1,, E t such that for every i =,, t, there exists j i < i such that E i i 1 l=1 E l E ji Keevash [11] foud that the method of idoreko actually implies that π = πf satisfies π t 1 + f tπ 1 0, 17 where t is the size of a largest forest cotaied i F Whe t =, 17 reduces to 16; whe t >, 17 improves 16 I our case 17 does ot improve the 1 1/r boud for πkr+1 r because the largest forest i Kr r+1 has oly edges Ackowledgmets The authors thaks referees for their commets that improved the presetatio Refereces [1] D de Cae, Extesio of a theorem of Moo ad Moser o complete subgraphs Ars Combi , 5 10 [] D de Cae, The curret status of Turá s problem o hypergraphs Extremal problems for fiite sets Visegrád, 1991, , Bolyai oc Math tud, 3, Jáos Bolyai Math oc, Budapest,

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