The Turán number of sparse spanning graphs

1. The Turánnumber of sparse spanning graphs Raphael Yuster joint work with Noga Alon Banff 2012

2. [Ore – 1961]A non-Hamiltonian graph of order n has at most edges. • ex(n,H)is the maximum number of edges in a graph of order nnot containing a subgraph isomorphic to H. • Ore’s result states that: • Recently, Ore's theorem has been generalized to the setting of Hamilton cycles in k-graphs: • Let Cn(k,t) denote the (k,t)-tight cycle of order n. • [Glebov, Person &Weps– 2012]determined ex(n,Cn(k,t)) (for nsuff.large).It is of the form where P is a specific fixed (k-1)-graph. t

3. It is natural to try to extend Ore's result to spanning structures other than just Hamilton cycles (in both the graph and hypergraphsettings). Suppose that His a k-graph of order nand with, say, bounded max degree. It is natural to suspect that for n sufficiently large,where L is aset of (k-1)-graphs that depending on neighborhoods in H. A conjecture raised in [GPW – 2012] asserts that it suffices to take Lto be the set of linksof H. (example: the links of a Cn(3,1) are L={K2 , 2K2}). Observe:the conjecture holds for both Ore's result and its aforementioned generalization to Hamilton cycles in hypergraphs (in fact, with equality).

4. In the graph-theoretic case, the link of a vertex is just a set of singletons whose cardinality is the degree of the vertex. In this case, the aforementioned conjecture states that: if His a graph of order nwith mindegδ>0and bounded maxdeg, then assuming nis sufficiently large (note: we trivially cannot do better). Main result: This is true in a strong sense (no need for bounded maxdeg): Theorem 1:For all nsufficiently large, if H is any graph of order n with no isolated vertices and , then

5. For all nsufficiently large, if H is any graph of order n with no isolated vertices and , then • Proof actually works for all n > 10000. • The constant 40 cannot be improved to less than .hence the bound on the maximum degree is optimal: • Take Hwith n=k(k+6)/2+1vertices, consisting of k disjoint cliques of size (n-1)/k each, and another vertex connected to δ≤ (n-1)/k-1vertices of the cliques. • Clearly, Δ(H)=(n-1)/kand δ(H)=δ. • H has no independent set of size k+2. • Hence, if Gis Kn - Kk+2, then His not a spanning subgraphof G. • However, Ghas more than edges.

6. A counter-example to the conjecture of [GPW – 2012], already for 3-graphs: Proposition 2:Let sbe a large integer, n=1+5sand let V1… Vs{x}where |Vi|=5. LetH be the 3-graph on V where each Vi forms a K5(3)andxis contained in a unique edge {x,u,v} with u,v in V1. Then ex(n-1,L(H)) = 0but: Proof: Take T to be U1U2  U3{x,y}where |Ui|=(n-2)/3.The edges are all the triples of U1U2U3 and all triples {x,ui,uj} ,{y,ui,uj}.T does not contain H because the links of x and y are 3-colorable so do not lie in a K5(3) . The result follows since T has edges.

7. Proof preliminaries We say that Gand Hof the same order pack, if H is a spanning subgraph of the complement of G. Let H=(W,F)be a graph with nvertices and with . Let G=(V,E)be any graph with nvertices and n-δ-1edges, where δ=δ(H). It suffices to prove that G and Hpack. Equivalently, a bijectionf : V  Wsuch that (u,v)E (f(u), f(v))F. Let V={v1,…,vn}where d(vi) ≥ d(vi+1). Observe: d(v1) ≤ n-δ-1 d(v2) ≤ n/2d(vi) ≤ 2n/i

8. We need the following independent setsof G, one for each vi : S1 consists of non-neighbors of v1that have small degree (less than 2n1/2) Si consists of non-neighbors of vithat have very small degree (at most 50) Each Siis chosen with maximum cardinality, under this restriction. It is not difficult to show that |S1|≥ δand |Si| ≥ n/7. vi Si N(vi) Random subsets of Si have whp some useful properties for our embedding: Let Bi be a random subset of Si where each vertex is chosen with prob. n-1/2 Lemma 1: Whp, all the Ciare relatively small (less than 4n1/2)the first few Diare relatively large (at least0.05n1/2fori=2,…,n1/2)

9. Proof outline The construction of the bijectionf : V  Wis done in four stages. At each point of the construction, some vertices of Vare matchedto some vertices of Wwhile the other vertices of V and Ware yet unmatched.Initially, all vertices are unmatched. We always maintain the packingproperty: for two matched vertices u,vVwith (u,v)Ewe have (f(u), f(v))F. Thus, once all vertices are matched, f defines a packing of Gand H.

10. Stage 1. We match v1 (a vertex with maximum degree in G) with a vertex wWhaving minimum degree δ in H. As N(w)= δand since|S1| ≥ δ , we may match an arbitrary subset B1 of δ vertices of S1 with N(w). Observe that the packing property is maintained since B1 is an independent set of non-neighbors of v1 . Note that after stage 1, precisely δ+1pairs are matched. v1 w |S1|≥ δ |N(w)|=δ N(v1) S1 G H Other vertices N(w)

11. Stage I1. This stage consists of iterations i=2,…,k where at iteration i we match vi and some subsets of Bi with a corresponding set of vertices of H.We do this as long as d(vi) ≥ 2n1/2(hence k ≤ n1/2).We make sure that after each iteration i, the following invariants are kept: After matching viwith some vertex w=f(vi) of H, we make sure that all neighbors of win H are matched to vertices of Bi. Any matched vertex of Gother than {v1,…,vi}is contained in some Bjwhere j ≤ i. The number of matched vertices after iteration i is at most i((H)+1). These invariants clearly hold after stage 1.

12. vi Observe: it is really yet unmatched Z X Y SiNon-neighbors G unmatched neighbors Other matched neighbors Bi matched vjj < i Di =Bi-j<iBj Y j<iBjY N(vi), Y Ci|Y|<5n1/2 |X|<i<k<n1/2 Lemma 1 |Di| ≥ n1/2/20, lemma 1 R Unmatchedneighbors of w |R|≤≤ n1/2/40 w non-neighbors of T The matches of X  Y in H H Q T |T|=|X|+|Y| <6n1/2 |Q| ≥ n-|T| Is there an unmatched vertex in Q? Yes! only (i-1)(|+1)matched so far

13. Stage I1I. • We are guaranteed that the unmatched vertices of Ghave degree ≤2n1/2. • By the third invariant of Stage 2, the number of unmatched vertices of Gis still linear in n (at least 19n/20). • As the unmatched vertices induce a subgraph with at least 19n/20 vertices and less than nedges, they contain an independent set of size at least n/4. • Let, therefore, J denote a maximum independent set of unmatched vertices of G. We have |J| ≥ n/4. • Let K be the remaining unmatched vertices of G. • The third stage consists of matching the vertices of Kone by one. • Details similar to those of Stage 2.

14. Stage IV. It remains to match the vertices of Jto the remaining unmatched vertices of H, denoting the latter by Q. Construct a bipartite graph Pwhose sides are Jand Q. Recall that |J|=|Q| ≥ n/4. We place an edge from v J to q Qif matching vto qis allowed. By this we mean that mapping vto qwill not violate the packing property. At the beginning of Stage 4, for each v J , there are at least 19n/20vertices of H that are non-neighbors of all vertices that are matches of matched neighbors of v. So, the degree of vin Pis at least 19n/20-(n-|J|) > |J|/2. It is not difficult to also show that the degree of each q Qis much larger than |J|/2. It follows by Hall's Theorem that Phas a perfect matching, completing the matching f.

15. Concluding remarks • The extremal graph in Ore's Theorem is unique (for all n>5).It is Kn-K1,n-2. • This is not the case in our more general Theorem 1: • Let H bea graph in which all vertices but one have degree at least 3, and one vertex v is of degree 2 and its two neighbors x and yare adjacent. • By Theorem 1, • One extremal graph is Kn-K1,n-2. • Another extremalgraph: The graph T obtained from Knby deleting a vertex-disjoint union of a star with n-3edges and a single edge.

16. Concluding remarks • all our counter-examples to the [GPW – 2012] conjecture regarding the extremal numbers ex(n,H)for hypergraphsH are based on a local obstruction. • It seems interesting to decide if these are all the possible examples: • Problem: • Is it true that for any k ≥ 2 and any Δ > 0there is an f = f(Δ) so that for any k-graph H on nvertices and with maximum degree at most Δ, any k-graph on n vertices which contains no copy of Hand with ex(n,H)edges, must contain a complete k-graph on at least n-fvertices?

17. Thanks