This section, whose results are from joint papers with Lenz [27, 29] focuses on the problem of determiningρτr(Kt) forr ≥ 3, suggested by Erd˝os, Hajnal, S ´os, and Szemer´edi [50] .
Erd˝os, Hajnal, Simonovits, S ´os, and Szemer´edi [49] proved that ρτr(Kt) ≤
1
2 1−t−1r
and this is best possible for allt≡1 (mod r). This left open the ques-tion whent 6≡1 (mod r), where they made partial progress fors6min{5, r}. Theorem 78. For2≤s≤min{5, r},RTr(n, Kr+s, o(n))≤ s−14r n2 +o(n2).
Our main result is to construct for every2 ≤ s ≤ r an infinite graph family providing near-optimal lower bounds forρτr(n, Kr+s, o(n)). In particular, it in-cludes that Theorem 78 is sharp when4r/(s−1)is a power of2. Earlier the only sharp construction was by Bollob´as and Erd˝os [39] forr =s= 2.
Theorem 79. Let26s 6r. Let`be the largest positive integer such thatdr·2−`e< s.
Then
RTr(n, Kr+s, o(n))>2−`−2n2+o(n2).
For example, it yields that θ4(K6) = 1/16 and θ4(K7) = 1/8. We suspect that Theorem 79 should be best possible for all s when 4r/(s − 1) is a power of2; towards this direction we have only the following partial result extending Theorem 78.
Proposition 80. ρτ10(K16) =ρτ12(K19) = 18.
Our first approach in [29], where we provedρτr(Kr+2) > 0for every r ≥ 2. This resolved one of the main open questions from [49]. In [29] hypergraphs were constructed to estimate Ramsey-Tur´an numbers of some hypergraphs. Tak-ing the shadow graphs of the constructed hypergraphs implied the results for graphs.
In [27] we refined this method, and the proof of Theorem 79 considers directly graphs, it builds on the techniques developed in [39] and [29] combined with several new ideas.
The remainder of this Chapter is organized as follows: in the first subsection we describe the construction for the graphs used to prove Theorem 79, in second subsection we prove Theorem 79, and in Section 3.5 we list several open prob-lems. Here we omit the proof of Proposition 80, which is rather technical and uses only standard methods.
Construction
The construction for Theorem 79 builds on the Bollob´as-Erd˝os Graph [39].
First we briefly sketch a few properties of the unit sphere. Let µ be the Lebesgue measure on the k-dimensional unit sphere Sk ⊆ Rk+1 normalized so thatµ(Sk) = 1. Given anyα, β >0, it is possible to select >0small enough and thenk sufficiently large so that Properties (P1) and (P2) are satisfied.
(P1) Let C be a spherical cap in Sk with height h, where 2h = √
2−/√ k2
(this means all points of the spherical cap are within distance√
2−/√ kof the center). Thenµ(C)≥ 12 −α.
(P2) LetC be a spherical cap with diameter2−/(2r2√
k). Thenµ(C)≤β. To prove Theorem 79, it suffices to prove that for all integers n, r ≥ 2, every 2≤s ≤ r, and everyα, β > 0, there exists anN-vertex graphG =G(n, r, s, α, β) such thatGisKr+s-free,N ≥n, and
|E(G)| ≥ 2−`−2−α
N2 and αr(G)< βN, where`is the largest positive integer such that
r·2−`
< s.
Assumen,r,s,α,β, and`are given as above, we shall show how to construct G = G(n, r, s, α, β). For the givenα and β, there exists > 0 and k ≥ 3such that properties (P1) and (P2) hold. Defineθ = /√
k andz = 2n. Partition thek -dimensional unit sphereSkintozdomains having equal measures and diameter at mostθ/4. Choose a point from each set and let P be the set of these points.
Letφ :P → P(Sk)map points ofP to the corresponding domains of the sphere.
Before definingG, we construct some auxiliary bipartite graphsB1, . . . , B` and hypergraphsHandH0.
The vertex set of the auxiliary bipartite graphsB1, . . . , B`is[r], and the edges are built from the `-dimensional hypercube Q` as follows. Blow up Q` into Q0` so that each vertex is blown up into an independent set of sizes− 1. Discard vertices ofQ0`so thatQ0`has exactlyrvertices, discarding at most one vertex from each blow up class. (Note that`was chosen so thatQ`is the smallest hypercube with at leastr/(s−1)vertices.) Consider the vertices ofQ` as labeled by binary words of length`. If the(2i+ 1)-st discarded vertex is from the class labelled by (a1, . . . , a`), then the next, (2i+ 2)-nd vertex should be removed from the class labelled(1−a1, . . . ,1−a`). DenoteAi,0 the subset of vertices of Q0` which come from a blowup of a vertex with itsi-th coordinate zero. Similarly defineAi,1. The bipartite graphBiis the complete bipartite graph with partsAi,0 andAi,1.
Now we define anr-uniform hypergraphHwith vertex setP`, the family of ordered`-tuples of elements ofP. We let{¯x1, . . . ,x¯r} ⊆P`be a hyperedge ofHif for every16i < j 6rand16a6`ifij ∈E(Ba)thend(xia, xja)>2−θ. In other words, we form a hyperedge if the edges ofBa correspond to almost antipodal points on the sphere in thea-th vertex coordinate.
FromH, define a hypergraphH0by applying the following theorem toHwith γ =β andk=r3.
Theorem 81(Theorem 16 in [29]). LetH be an r-uniform hypergraph onn vertices.
Let0< γ <1and letkbe a positive integer. Then there exists at=t(H, k, γ, r)and an r-uniform hypergraphGwith vertex setV(H)×[t]with the following properties.
(i) For all {a1, . . . , ar} ∈ H and all sets Ui ⊆ {ai} × [t] with |Ui| ≥ γt for each 1≤i≤r, there exists at least one hyperedge ofG with one vertex in eachUi. (ii) G does not contain as a subhypergraph anyv-vertex hypergraph F withmedges
wherev ≤kandv+ (1 +γ−r)(m−1)< r.
We are finally ready to defineG. LetU andV be two distinct copies ofV(H0) and let the vertex set ofGbeU∪V˙ . We place a copy of the shadow graph of H0 on bothG[U]andG[V]. (Theshadow graphof a hypergraph has the same vertex set andxyforms an edge of the shadow graph ifxand yare contained together
in some hyperedge.) Lastly, foru¯ = hu1, . . . , u`i ∈ U andv¯=hv1, . . . , v`i ∈ V let
¯
u¯vbe an edge ifd(ui, vi)<√
2−θfor all1≤i≤`.
3.4 Verifying properties of G
To complete the proof of Theorem 79, we need to prove three properties of G: G has at least 2−`−2−α
N2 edges, G is Kr+s-free, and the Kr-independence number ofGis smaller thanβN.
The number of edges of G .
First we compute the number of vertices ofG. The hypergraphHhasz`vertices and each vertex inHis blown up into a set of sizet soH0 hastz` vertices. Thus Ghas 2tz` vertices. To estimate the number of edges of G, we fix some vertex x0 ∈ U; we will compute a lower bound on its degree inV. There exists a vertex x inH such that x0 is contained in the blowup of x. For y0 ∈ V to be adjacent tox0, we must have d(xi, yi) ≤ √
2−θ for all i. By Property (P1), there are at least 12 −α
|P| pointsyi that are within distance √
2−θ ofxi. Thus there are at least2−`|P| −Cα|P|choices fory whereC is some constant depending only on `. Since each y is blown up into a set of size t, the degree of x0 is at least 2−`tz`−Cαtz`. Thus
|E(G)| ≥ |V(G)|
2 2−`tz`−Cαtz` .
Sincetz` = |V(G)|2 ,
|E(G)| ≥2−`−2|V(G)|2−Cα|V(G)|2/2 = 2−`−2|V(G)|2(1−Cα/2).
Using that C was depending only on `, and α > 0 could be chosen arbitrarily small, it gives the required bound.
G is K
r+s-free
First we need a couple of short lemmas.
Lemma 82. α(∪Bi)< s.
Proof. Fix any two verticesx0, y0 ∈V(Q0`)and letxandybe the vertices ofQ`such thatx0 andy0 are contained in the blowups ofxandyrespectively. Ifx6=y, then their binary labels differ in at least one position so there will be someBiwherex0 andy0 appear in different classes of the bipartition ofBi. Thus the independent sets in∪Biare subset of the blowup of some vertex inQ`. Using that each vertex inQ`is blown up into a set of size at mosts−1, we the required bound.
Lemma 83. LetKw be a completew-vertex subgraph of G[U]. Then there exists a hy-peredgeEinH0 such thatV(Kw)⊆E.
Proof. LetKw ⊆G[U]andV(Kw) ={x1, . . . , xw}. SinceKwis complete, for every i, jthere exists some hyperedgeEi,j ofH0such thatEi,jcontains bothxiandxj. If theEi,j’s are not all the same hyperedge, then(ii)of Theorem 81 is violated.
Lemma 84. G[U](and similarlyG[V]) isKr+1-free.
Proof. This is an immediate corollary of Lemma 83. Hyperedges in H have size at mostr, soG[U]does not contain anyKr+1.
We now need the following property of the unit sphere observed by Bollob´as and Erd˝os [39].
Theorem 2(Bollob´as-Erd˝os Rombus Theorem). For any0 < γ < 14, it is impos-sible to havep1, p2, q1, q2 ∈ Sk such thatd(p1, p2) ≥ 2−γ, d(q1, q2) ≥ 2−γ, and d(pi, qj)≤√
2−γ for all1≤i, j ≤2.
Recall that from the hypergraphHwe formed the hypergraphH0 by blowing up each vertex in H into a strong independent set in H0. Also recall that the vertices inG[U]are vertices of H0, so vertices in G[U] correspond to blowups of vertices inH. We define a functionΞbetweenV(G)andV(H): forx ∈V(G), let Ξ(x)be the vertex ofV(H)such thatxis contained in the blowup ofΞ(x).
Lemma 85. GisKr+s-free.
Proof. For a contradiction, assume that K = Kr+s is a subgraph of G and let Ku =K[V(K)∩U]andKv =K[V(K)∩V]. SinceU andV are symmetric in the definition ofG, we may assume without loss of generality that|Ku| ≥ |Kv|. By Lemma 84 and sinces≥2,
dr/2e+ 1 ≤
r+s 2
≤ |V(Ku)| ≤r. (3.9)
This implies that
|V(Kv)|=r+s− |V(Ku)| ≥s. (3.10) By Lemma 83 and (3.10), there existx1, . . . , xs ∈V(Kv)and a hyperedgeE in H0such thatx1, . . . , xs ∈E. Sincex1, . . . , xsare all inEand edges ofHwere built from the auxiliary bipartite graphsB1, . . . , B`, we can think ofΞ(x1), . . . ,Ξ(xs)as vertices in∪Bi. By Lemma 82, there exists someBi and two vertices, say Ξ(x1) and Ξ(x2), such that the ith coordinate ofΞ(x1)and the ith coordinate ofΞ(x2) are almost antipodal. Fix thisifor the remainder of this proof.
By (3.9) there exist at leastdr/2e+ 1 vertices inV(Ku), sayy1, . . . , ydr/2e+1.By Lemma 83 there is a hyperedgeF inH0 containing them. Similarly as before, we can think ofΞ(y1), . . . ,Ξ(ydr/2e+1)as vertices in Bi (recall thatihas already been chosen.) The parts ofBi have size at mostdr/2e so there exist two vertices, say Ξ(y1)andΞ(y2), such that thei-th coordinates are almost antipodal.
Consider the i-th coordinates of Ξ(x1), Ξ(x2), Ξ(y1), and Ξ(y2). The cross-distances between thex’s andy’s are all at most√
2−θ, sincex1,x2,y1, andy2all came from the cliqueK. Hence we have four points violating Theorem 2.
The K
r-independence number of G
First we need an elementary statement about distances of points on a sphere.
Lemma 86. Let k ≥ 2and1 ≤ h ≤ bk/2c be any positive integers and fix a positive a < 1/(16h4). Letx1, . . . , xk ∈ Sk such that for everyi we haved(xi, xi+1) ≥ 2−a. Thend(x1, x2h)>2−4h2a.
Proof. For u ∈ Sk denote by u0 ∈ Sk the antipodal point to u. Note that for everyu, v triviallyd(u, u0) = 2andd(u, v) = d(u0, v0). First, we boundd(x0i, xi+1) for everyi. The points xi, x0i, andxi+1 form a right triangle sincexi and x0i are antipodal (the right angle is at the pointxi+1). Thus
d2(x0i, xi+1) =d2(xi, x0i)−d2(xi, xi+1)≤4−(2−a)2 ≤4a−a2 ≤4a.
Thusd(x0i, xi+1)≤2√
afor alli. Using the triangle inequality, we obtain d(x01, x2h)≤d(x01, x2) +d(x2, x03) +. . .+d(x02h−1, x2h)≤2(2h−1)√
a.
The points x1, x01, and x2h form a right triangle since x1 and x01 are antipodal.
Thus
d2(x1, x2h) = d2(x1, x01)−d(x01, x2h)≥4−4(2h−1)2a
= 4−16h2a+ 16ha−4a≥4−16h2a+a.
Sincea <1/(16h4)implies16h4a2 < awe have
d2(x1, x2h)≥4−16h2a+a >4−16h2a+ 16h4a2 = (2−4h2a)2.
We now need one lemma from [29]. There is a subtle point here: in [29] the statement of the lemma uses “d(pi, pj) ≥ 2−θ”. But the variable θ used in this chapter and the θ used in [29] are slightly different constants. The θ used in the statement of [29, Lemma 13] comes from the statement of [29, Property (P3)]
which matches our Property (P2). So theθin [29, Lemma 13] is replaced with the constant from our Property (P2) when we cite that lemma below.
Lemma 87. (Lemma 13 in [29]) If A1, . . . , Ar ⊆ P with |Ai| ≥ 2rβz andT is a tree on vertex set [r], then there exist p1 ∈ A1, . . . , pr ∈ Ar such that if ij ∈ E(T)then d(pi, pj)≥2−/
2r2√
k
.
One of the key improvements in this chapter compared to [29] is improving the above lemma by replacing trees with complete bipartite graphs.
Lemma 88. If A1, . . . , Ar ⊆ P with|Ai| ≥ 2rβz andB is a complete bipartite graph on vertex set[r], then there exist p1 ∈ A1, . . . , pr ∈ Ar such that if ij ∈ E(B)then d(pi, pj)≥2−θ.
Proof. LetT be a path on vertex set[r]. Apply Lemma 87 to findp1 ∈A1, . . . , pr∈ Arsuch that ifij ∈E(T)thend(pi, pj)≥2−θ. SinceT is a path, this implies that d(pi, pi+1)≥ 2−/
2r2√ k
= 2−θ/r2 for alli. We can then apply Lemma 86 to show thatd(p2i+1, p2j)>2−θfor alliandj (setx1 =p2i+1andx2h =p2j.)
Lemma 89. α(H)≤r`2`+rβz`.
Proof sketch. The proof is identical to the proof of [29, Lemma 14], except where [29, Lemma 14] uses Lemma 87 on trees, we instead use Lemma 88.
Lemma 90. αr(G)≤r`2`+r+2βz`t.
Proof sketch. The proof is identical to the proof of [29, Theorem 9(iv)].
3.5 Concluding Remarks
Remark. We conjecture that our construction is best possible when 4r/(s −1) is a power of2, we know this only when s ≤ 5 and for some additional cases (see Proposition 80). Probably, some mixture of more involved application of the Szemer´edi Regularity Lemma and some proof technique from weighted Tur´an theory could help to prove our conjecture.
Perhaps existing standard proof techniques can be improved to show our con-struction is best possible for alls.
It seems very hard to decide if our constructions are best possible when4r/(s−
1)is not a2-power. The simplest open cases are 1
16 ≤θ3(K5)6 1
12, 1
8 ≤θ3(K6)6 1
6, 1
8 ≤θ4(K8)6 3 16.
The upper bounds are from Theorem 78 and the lower bounds from Theorem 79.
Remark.Theorems 78 and 79 focus onθr(Kt)fort ≤2r. What happens whent >
2r? The construction from Theorem 79 can be easily extended to cliques larger
thanK2ras follows. For a lower bound onρτr(Kqr+s)with2≤s ≤r, letGbe the construction from Theorem 79 and join it to a complete(q−1)-partite graph with almost equal part sizes. Into each part insert aKr+1-free graph with small Kr -independence number (such a graph exists by the Erd˝os-Rogers Theorem [52].) Erd˝os, Hajnal, Simonovits, S ´os, and Szemer´edi [49] conjectured that this type of construction provides the correct answer (see also [120, Conjecture 18].) Can Theorem 78 be extended tot >2rand if so, does it match our lower bound?
Remark. In the area of the Ramsey-Tur´an theory, one of the major open prob-lems is to prove a generalization of the Erd˝os-Stone–Simonovits Theorem [55]
by proving that RT(n, H, o(n)) = RT(n, Ks, o(n)) where s = s(H) is equal to some parameter depending only on H. Erd˝os, Hajnal, S ´os, and Szemer´edi [50]
proved an upper bound using a parameter closely related to the arboricity. That is, one can takes to be the minimums such that V(H) can be partitioned into ds/2e setsV1, . . . , Vds/2e such thatV1, . . . , Vbs/2c span forests and if s is oddVds/2e
spans an independent set. This is known to be sharp for odd s. In several papers, Erd˝os mentioned the problem of solving the simplest open case when H = K2,2,2, where s(K2,2,2) = 4, i.e., one would like to have a lower bound of RT(n, K2,2,2, o(n))6 RT(n, K4, o(n)) = 18n2+o(n2). Even the question of deter-mining ifRT(n, K2,2,2, o(n)) = Ω(n2)is still open (see [120, Problem 4], [50, p. 72], [123, Problem 1.3] among others).
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