Hedgehogs are not colour blind

Transcription

1 Hedgehogs are no colour blind David Conlon Jacob Fox Vojěch Rödl Absrac We exhibi a family of 3-uniform hypergraphs wih he propery ha heir 2-colour Ramsey numbers grow polynomially in he number of verices, while heir 4-colour Ramsey numbers grow exponenially. This is he firs example of a class of hypergraphs whose Ramsey numbers show a srong dependence on he number of colours. 1 Inroducion The Ramsey number r k (H) of a k-uniform hypergraph H is he smalles n such ha any 2-colouring of he edges of he complee k-uniform hypergraph K n (k) conains a monochromaic copy of H. Similarly, for any q 2, we may define a q-colour Ramsey number r k (H; q). One of he main ousanding problems in Ramsey heory is o decide wheher he Ramsey number for complee 3-uniform hypergraphs is double exponenial. The bes known bounds, due o Erdős, Hajnal and Rado [5, 6], sae ha here are posiive consans c and c such ha 2 c2 r 3 (K (3) ) 2 2c. Paul Erdős has offered $500 for a proof ha he upper bound is correc, ha is, ha here exiss a posiive consan c such ha r 3 (K (3) ) 2 2c. Some evidence ha his may be rue was given by Erdős and Hajnal (see, for example, [9]), who showed ha he analogous bound holds for 4 colours, ha is, ha here exiss a posiive consan c such ha r 3 (K (3) ; 4) 2 2c. In his paper, we show ha his evidence may no be so compelling by finding a naural class of hypergraphs, which we call hedgehogs, whose Ramsey numbers show a srong dependence on he number of colours. The hedgehog H is he 3-uniform hypergraph wih verex se [ + ( ] such ha for every pair (i, j) wih 1 i < j here is a unique verex k > such ha ijk is an edge. We will someimes refer o he se {1, 2,..., } as he body of he hedgehog. Our main resul is ha he 2-colour Ramsey number r 3 (H ) grows as a polynomial in, while he 4-colour Ramsey number r 3 (H ; 4) grows as an exponenial in. Theorem 1 If H is he 3-uniform hedgehog wih body of order, hen (i) r 3 (H ) 4 3, Mahemaical Insiue, Oxford OX2 6GG, Unied Kingdom. david.conlon@mahs.ox.ac.uk. Research suppored by a Royal Sociey Universiy Research Fellowship and by ERC Saring Gran Deparmen of Mahemaics, Sanford Universiy, Sanford, CA 94305, USA. jacobfox@sanford.edu. Research suppored by a Packard Fellowship, by NSF Career Award DMS and by an Alfred P. Sloan Fellowship. Deparmen of Mahemaics and Compuer Science, Emory Universiy, Alana, GA 30322, USA. rodl@mahcs.emory.edu. Research parially suppored by NSF grans DMS and DMS

2 (ii) here exiss a posiive consan c such ha r 3 (H ; 4) 2 c. For he inermediae 3-colour case, we show ha he answer is inimaely conneced wih a special case of he mulicolour Erdős Hajnal conjecure [4]. This conjecure saes ha for any complee graph K wih a fixed q-colouring of is edges, here exiss a posiive consan c(k) such ha any q-colouring (wih he same q colours) of he edges of he complee graph on n verices wih no copy of K conains a clique of order n c(k) which receives only q 1 colours. Though his conjecure is known o hold in a number of special cases (see, for example, Secion 3.3 of [3]), he bes known general resul, due o Erdős and Hajnal hemselves, says ha here exiss a posiive consan c(k) such ha any q-colouring of he edges of he complee graph on n verices wih no copy of K conains a clique of order e c(k) log n which receives only q 1 colours. We will be concerned wih he paricular case where q = 4 and he banned configuraion K is a rainbow riangle wih one edge in each of he firs hree colours. Definiion. Le F () be he smalles n such ha every 4-colouring of he edges of K n, in red, blue, green and yellow, conains eiher a rainbow riangle K, wih one edge in each of red, blue and green, or a clique of order wih a mos 3 colours. We will show ha r 3 (H ; 3) is bounded above and below by polynomials in F () (sricly speaking, he upper bound is a polynomial in F ( 3 ), bu, provided F () does no jump pahologically, his will be a mos polynomial in F ()). Since he resul of Erdős and Hajnal menioned in he previous paragraph implies ha F () c log for some consan c, his in urn shows ha r 3 (H ; 3) c log for some consan c. Moreover, he Erdős Hajnal conjecure holds in his case if and only if here is a polynomial upper bound for r 3 (H ; 3). Theorem 2 If H is he 3-uniform hedgehog wih body of order, hen (i) r 3 (H ; 3) = O( 4 F ( 3 ) 2 ), (ii) r 3 (H ; 3) F (). In paricular, here exiss a consan c such ha r 3 (H ; 3) c log. We will prove Theorem 1 in he nex secion and Theorem 2 in Secion 3. We conclude by discussing a number of ineresing quesions ha arose from our work. 2 The basic dichoomy In his secion, we prove Theorem 1. We begin by proving ha he 2-colour Ramsey number of H is a mos 4 3. Proof of Theorem 1(i): Le n = 4 3. We will show ha every red/blue-colouring of he complee 3-uniform hypergraph on n verices conains a monochromaic copy of H. To begin, we define a parial colouring of he edges of he complee graph on he same verex se. We will colour an edge uv red if here are fewer han ( + red riples conaining u and v. Similarly, we colour uv blue if here are fewer han ( + blue riples conaining u and v. To find a monochromaic H, i will clearly 2

3 suffice o find a subse of order conaining no red edge or no blue edge, since we can consider his se as he body of he hedgehog and embed he spines greedily. We claim ha no verex is conained in 2 2 red edges and 2 2 blue edges. Suppose, on he conrary, ha u is such a verex and le V R and V B be he verices which are conneced o u in red and blue, respecively. Since i is easy o see ha no edge can be coloured boh red and blue, V R and V B are disjoin. Moreover, for each verex v in V R, since uv is conained in fewer han ( + red riples, here are a leas ( ) V B > V B 2 2 verices w in V B such ha uvw is blue. This implies ha more han half of he riples uvw wih v V R and w V B are blue. However, by firs considering verices w in V B, he same argumen also shows ha more han half of hese riples are red, a conradicion. We now assign a colour o each verex in he graph, colouring i red if i is conained in fewer han 2 2 red edges and blue oherwise. In he laer case, he claim of he las paragraph shows ha i will be conained in fewer han 2 2 blue edges. By he pigeonhole principle, a leas half he verices in he graph have he same colour, say red. Tha is, we have a subse of order a leas n/2 such ha every verex is conained in fewer han 2 2 red edges. By Brooks heorem, we conclude ha his se conains a subse of order n/4 2 conaining no red edge. Since n/4 2, his is he required se. We will now show ha he 4-colour Ramsey number of H is a leas 2 c for some posiive consan c. This is clearly sharp up o he consan in he exponen. Proof of Theorem 1(ii): A sandard applicaion of he firs momen mehod gives a posiive consan c such ha, for every ineger 4, here is a 4-colouring χ of he edges of he complee graph on 2 c verices wih he propery ha every clique of order conains all 4 colours. We now 4-colour he edges of he complee 3-uniform hypergraph on he same verex se by colouring he riple uvw wih any colour which is no conained wihin he se {χ(u, v), χ(v, w), χ(w, u)}. Suppose now ha here is a monochromaic copy of H wih colour 1, say, and le u 1, u 2,..., u be he body of his copy. Then, in he original graph colouring χ, none of he edges u i u j wih 1 i < j received he colour 1. However, his conradics he propery ha every se of order conains all 4 colours. 3 Three colours and he Erdős Hajnal conjecure To prove Theorem 2(i), we require wo lemmas. The firs is a resul of Spencer [13] which says ha any 3-uniform hypergraph wih few edges conains a large independen se. Lemma 1 If H is a 3-uniform hypergraph wih n verices and e edges, hen α(h) = Ω(n 3/2 /e 1/2 ). The second lemma we require is a resul of Fox, Grinshpun and Pach [7] saying ha he mulicolour Erdős Hajnal conjecure holds for 3-colourings of K n wih no rainbow riangle. The resul we use is somewha weaker han he main resul in [7], bu will be more han sufficien for our purposes. Lemma 2 Suppose ha he edges of he complee graph K n have been 3-coloured, in red, blue and green, so ha here are no rainbow riangles wih one edge in each of red, blue and green. Then here is a clique of order n 1/3 conaining a mos wo of he hree colours. 3

4 We are now ready o prove Theorem 2(i), ha r 3 (H ; 3) = O( 4 F ( 3 ) 2 ). Proof of Theorem 2(i): Suppose ha he edges of he complee 3-uniform hypergraph on n = c 4 F ( 3 ) 2 verices have been 3-coloured, in red, blue and green, where c is a sufficienly large consan o be chosen laer. We will 4-colour he edges of he graph on he same verex se as follows: if u and v are conained in fewer han ( + riples of a given colour, hen we give he edge uv ha colour, noing ha an edge may receive more han one colour (bu a mos wo). On he oher hand, if an edge is no coloured wih any of red, blue or green, we colour i yellow. We claim ha his colouring has a mos 2 n 2 riangles conaining all hree of he colours red, blue and green (where we include he possibiliy ha wo of hese colours may appear on he same edge). To see his, noe ha here are a mos ( ( ( + ) n red riples conaining a red edge. In paricular, since he riangles we wish o coun always conain a red edge, here are a mos ( ( ( +) n of hese riangles in he graph corresponding o a red riple. Since we may similarly bound he number of hese riangles corresponding o blue or green riples, we see ha, for 3, here are a mos 3( ( ( + ) n ) 2 2 n 2 riangles in he graph which conain all hree of he colours red, blue and green, as required. If we le H be he 3-uniform hypergraph on n verices whose edges correspond o riangles conaing all hree of he colours red, blue and green, Lemma 1 now yields a subse U of order Ω(n 1/2 /) conaining no such riangle. By aking c o be sufficienly large, we may assume ha U has order a leas F ( 3 ). We now consider he graph G on verex se U whose edge se consiss of all hose edges which received wo colours in he 4-colouring defined above. If we fix a verex u U, hen each of he edges in G ha conain u mus have received he same wo colours in he original colouring. Oherwise, we would have a riangle conaining all hree of he colours red, blue and green. Suppose, herefore, ha every edge in G ha conains u received he colours red and blue in he original colouring. Then, again using he propery ha every riangle conains a mos wo of he colours red, blue and green, we see ha he neighbourhood of u in G conains no green edges. Therefore, if u had neighbours in G, we could use his neighbourhood o find a green copy of H. Since a similar argumen holds if he edges conaining u correspond o blue and green or o red and green, we may assume ha every verex u U is conained in fewer han edges in he graph G. By Brooks heorem, i follows ha U conains a subse V of order a leas U / F ( 3 ) conaining no edges from G, ha is, such ha every edge received a mos one colour in he original colouring. Since V is a 4-coloured graph of order a leas F ( 3 ) conaining no rainbow riangle in red, blue and green, here is a subse of order a leas 3 wih a mos hree colours. If he missing colour is red, we may easily find a red copy of H and similar conclusions hold if he missing colour is eiher blue or green. On he oher hand, if he missing colour is yellow, we have a 3-colouring, in red, blue and green, of a se of order a leas 3 conaining no rainbow riangle, so Lemma 2 ells us ha here is a subse of order a leas wih a mos wo colours. If we again consider he missing colour, i is easy o find a monochromaic copy of H in ha colour. The lower bound r 3 (H ; 3) F () follows from a simple adapaion of he proof of Theorem 1(ii). Proof of Theorem 2(ii): By he definiion of F (), here exiss a 4-colouring χ, in red, blue, green and yellow, say, of he edges of he complee graph on F () 1 verices conaining no rainbow riangle wih one edge in each of red, blue and green and such ha every clique of order conains all 4 colours. We now 3-colour he complee 3-uniform hypergraph on he same verex se in red, blue and green, colouring he riple uvw wih any colour which is no conained wihin he se {χ(u, v), χ(v, w), χ(w, u)}. 4

5 Since here are no rainbow riangles in red, blue and green, his colouring is well-defined. Suppose now ha here is a monochromaic copy of H in red, say, and le u 1, u 2,..., u be he body of his copy. Then, in he original graph colouring χ, none of he edges u i u j wih 1 i < j are red. However, his conradics he propery ha every se of order conains all 4 colours. 4 Concluding remarks The resuls of his paper raise a number of ineresing quesions, some of which we describe below. 4.1 Higher-uniformiy hedgehogs The k-uniform hedgehog H (k) is he hypergraph wih verex se [ + ( k 1) ] such ha for every (k 1)- uple (i 1,..., i k 1 ) wih 1 i 1 < < i k 1 here is a unique verex i k > such ha i 1... i k is an edge. A sraighforward generalisaion of he proof of Theorem 1(i) gives he following resul. Theorem 3 For every ineger k 4, here exiss a consan c k such ha if H (k) hedgehog wih body of order, hen r k (H (k) ) k 2 (c k ), where he ower funcion i (x) is defined by 1 (x) = x and i+1 (x) = 2 i(x). is he k-uniform A consrucion due o Kosochka and Rödl [10] shows ha his resul is igh for k = 4, ha is, ha here exiss a posiive consan c such ha r 4 (H (4) ) 2 c. Since he consrucion is simple, we describe i in full. To begin, ake a colouring of he edges of he complee graph on 2 c verices such ha every se of order conains boh a red riangle and a blue riangle. We hen colour he edges of he 4-uniform hypergraph on he same verex se by colouring a 4-uple red if i conains a red riangle, blue if i conains a blue riangle and arbirarily oherwise. I is easy o check ha his 2-colouring conains no monochromaic copy of H (4). Already for k = 5, we were unable o prove a maching lower bound, since i seems ha one would firs need o know how o prove a double-exponenial lower bound for r 3 (K ). We were also unable o prove an analogue of Theorem 1(ii) for k = 4. Again, his is because of a basic gap in our undersanding of hypergraph Ramsey problems. While we know ha here are 4-colourings of he 3-uniform hypergraph on 2 2c verices such ha every subse of order receives a leas wo colours, we do no know if he following varian holds. Problem 1 Is here an ineger q, a posiive consan c and a q-colouring of he 3-uniform hypergraph on 2 2c verices such ha every subse of order receives a leas hree colours? A posiive answer o he analogous quesion where we ask ha every subse of order receives a leas five colours would allow us o prove ha here exiss an ineger q such ha r 4 (H (4) ; q) 2 2c. The proof of his saemen is a varian of he proof of Theorem 1(ii). Indeed, suppose ha we have a q-colouring χ of he edges of he 3-uniform hypergraph K n (3) such ha every subse of order receives a leas five colours. Then we define a colouring of he complee 4-uniform hypergraph K n (4) wih a mos q + ( q ( + q ) ( 3 + q ) 4 colours by colouring he edge uvwx wih he se {χ(uvw), χ(vwx), χ(wxu), χ(xuv)}. I is now easy o check ha if here is a monochromaic H (4) in his colouring, hen, in he original colouring 5

6 χ, he body of he hedgehog is a subse of order which receives a mos 4 colours, conradicing our choice of χ. Our moivaion for invesigaing higher-uniformiy hedgehogs was he hope ha hey migh allow us o show ha here are families of hypergraphs for which here is an even wider separaion beween he 2-colour and q-colour Ramsey numbers. However, i seems likely ha for hedgehogs he separaion beween he ower heighs is a mos one for any uniformiy. This leaves he following problem open. Problem 2 For any ineger h 3, do here exis inegers k and q and a family of k-uniform hypergraphs for which he 2-colour Ramsey number grows as a polynomial in he number of verices, while he q-colour Ramsey number grows as a ower of heigh h? 4.2 Burr Erdős in hypergraphs The degeneracy of a graph H is he minimum d such ha every induced subse conains a verex of degree a mos d. Building on work of Kosochka and Sudakov [11] and Fox and Sudakov [8], Lee [12] recenly proved he famous Burr Erdős conjecure [1], ha graphs of bounded degeneracy have linear Ramsey numbers. Tha is, he showed ha for every posiive ineger d here exiss a consan c(d) such ha he Ramsey number of any graph H wih n verices and degeneracy d saisfies r(h) c(d)n. If we define he degeneracy of a hypergraph H in a similar way, ha is, as he minimum d such ha every induced subse conains a verex of degree a mos d, we may ask wheher he analogous saemen holds in hypergraphs. Unforunaely, as firs observed by Kosochka and Rödl [10], he 4- uniform analogue of he Burr Erdős conjecure is false, since H (4) is 1-degenerae and r 4 (H (4) ) 2 c. Since H is a 1-degenerae hypergraph, he resuls of his paper show ha he Burr Erdős conjecure also fails for 3-uniform hypergraphs and 3 or more colours. For 4 colours, his follows immediaely from Theorem 1(ii). For 3 colours, i follows from Theorem 2(ii) and he observaion ha F () = Ω( 3 / log 6 ). To show his, we amend a consrucion of Fox, Grinshpun and Pach [7], aking he lexicographic produc of hree 3-colourings of he complee graph on /16 log 2 verices, one for each riple of colours from he se {red, blue, green, yellow} ha conains yellow, each having he propery ha he union of any wo colours conains no clique of order 4 log. This colouring will conain no rainbow riangle wih one edge in each of red, blue and green and no clique of order wih a mos 3 colours. For furher deails, we refer he reader o Theorem 3.1 of [7]. While i is also unlikely ha an analogue of he Burr Erdős conjecure holds in he 2-colour case, i may sill be he case ha r 3 (H ) is linear in he number of verices, ha is, ha r 3 (H ) = O( 2 ). I would already be ineresing o prove an approximae version of his saemen. Problem 3 Show ha r 3 (H ) = 2+o(1). 4.3 Mulicolour Erdős Hajnal I is somewha curious ha our upper bound for r 3 (H ; 3) mirrors he bes known lower bound for r 3 (K ; 3), due o Conlon, Fox and Sudakov [2], which says ha here exiss a posiive consan c such ha r 3 (K ; 3) 2 c log. 6

7 However, i seems likely ha his is mere coincidence and ha he funcion F () defined in he inroducion is polynomial in. Phrasing he quesion in a more radiional fashion, we would very much like o know he answer o he following special case of he mulicolour Erdős Hajnal conjecure. Problem 4 Show ha here exiss a posiive consan c such ha if he edges of K n are 4-coloured, in red, blue, green and yellow, so ha here are no rainbow riangles wih one edge in each of red, blue and green, hen here is a clique of order n c conaining a mos hree of he four colours. Tha being said, if F () were superpolynomial, i would no only disprove he mulicolour Erdős Hajnal conjecure, i would also srenghen he curious correspondence beween he bounds for r 3 (H ; q) and r 3 (K ; q). This would cerainly be he more ineresing oucome. References [1] S. A. Burr and P. Erdős, On he magniude of generalized Ramsey numbers for graphs, in Infinie and Finie Ses, Vol. 1 (Keszhely, 1973), , Colloq. Mah. Soc. János Bolyai, Vol. 10, Norh- Holland, Amserdam, [2] D. Conlon, J. Fox and B. Sudakov, Hypergraph Ramsey numbers, J. Amer. Mah. Soc. 23 (2010), [3] D. Conlon, J. Fox and B. Sudakov, Recen developmens in graph Ramsey heory, in Surveys in Combinaorics 2015, London Mah. Soc. Lecure Noe Ser., Vol. 424, , Cambridge Universiy Press, Cambridge, [4] P. Erdős and A. Hajnal, Ramsey-ype heorems, Discree Appl. Mah. 25 (1989), [5] P. Erdős, A. Hajnal and R. Rado, Pariion relaions for cardinal numbers, Aca Mah. Acad. Sci. Hungar. 16 (1965), [6] P. Erdős and R. Rado, Combinaorial heorems on classificaions of subses of a given se, Proc. London Mah. Soc. 3 (195, [7] J. Fox, A. Grinshpun and J. Pach, The Erdős Hajnal conjecure for rainbow riangles, J. Combin. Theory Ser. B 111 (2015), [8] J. Fox and B. Sudakov, Two remarks on he Burr Erdős conjecure, European J. Combin. 30 (2009), [9] R. L. Graham, B. L. Rohschild and J. H. Spencer, Ramsey heory, 2nd ediion, John Wiley & Sons, [10] A. V. Kosochka and V. Rödl, On Ramsey numbers of uniform hypergraphs wih given maximum degree, J. Combin. Theory Ser. A 113 (2006), [11] A. V. Kosochka and B. Sudakov, On Ramsey numbers of sparse graphs, Combin. Probab. Compu. 12 (2003), [12] C. Lee, Ramsey numbers of degenerae graphs, o appear in Ann. of Mah. [13] J. Spencer, Turán s heorem for k-graphs, Discree Mah. 2 (197,