1. 1 Quasi-randomness is determined by the distribution of copies of a graph in equicardinal large sets Raphael Yuster University of Haifa

2. 2 Quasi-random graphs – background The theory of quasi-random graphs asks the following fundamental question: Which graph properties are such that any graph that satisfies them, resembles an appropriate random graph? The theory was initiated by Thomason [1987] and by Chung, Graham and Wilson [1989] who proved the fundamental theorem of quasi-random graphs.

3. 3 Fundamental theorem of quasi-random graphs For a set of vertices U  V(G) denote by : H[U] # of labeled copies of H induced by U e(U) # of edges of G induced by U Example: U H H[U] =16 A graph sequence (G n ) is an infinite sequence of graphs {G 1,G 2,…} where G n has n vertices. The following result shows that many properties of different nature are equivalent to the notion of quasi-randomness, defined using edge distribution.

4. 4 Fundamental theorem of quasi-random graphs Fix any 0 < p < 1. For any graph sequence (G n ) the following properties are equivalent: P 1 : For an even t ≥ 4, let C t denote the t-cycle. Then e(G n ) = ½pn 2 +o(n 2 ) and C t [V(G n )] = p t n t + o(n t ). P 2 : For any U  V(G n ) we have e(U) = ½p|U| 2 + o(n 2 ). P 3 : For any U  V(G n ), |U|=n/2, e(U) = ½p|U| 2 + o(n 2 ). P 4 : Fix an α  (0,½). For any U  V(G n ), |U|= αn we have e(U,V-U)=pα(1- α)n 2 +o(n 2 ). Chung, Graham and Wilson [1989]

5. 5 Fundamental theorem of quasi-random graphs A graph property is p-quasi-random if it is equivalent to any (and thus all) of the properties in the theorem. Each of the properties in the theorem is a property we would expect G(n,p) to satisfy with high probability. But it is far from true, however, that any property that almost surely holds for G(n,p) is p-quasi-random. Example: Having vertex degrees np(1+o(1)) is not a quasi- random property (just take vertex-disjoint cliques of size roughly np each).

6. 6 The property The property P H An important family of non-quasi-random properties are those requiring the graphs in the sequence to have the correct number of copies of a fixed graph H. Note that P 1 guarantees that for any even t, if a graph sequence has the correct number of edges as well as the correct number of copies of H=C t then the sequence is quasi-random. This is not true for all graphs H. Already for H=K 3 there are simple constructions showing that this is false. Namely: having ½pn 2 +o(n 2 ) edges and p 3 n 3 + o(n 3 ) labeled triangles does not suffice. For an even t ≥ 4, let C t denote the t-cycle. Then e(G n ) = ½pn 2 +o(n 2 ) and C t [V(G n )] = p t n t + o(n t ).

7. 7 The property The property P H As quasi-randomness is a hereditary property (a sub- structure of a random-like object is expected to be random-like), Simonovits and Sós introduced a variant of P 1 where now we require all subsets of vertices to contains the “correct” number of H. P H : For any U  V(G n ) we have H[U] = p e(H) |U| h + o(n h ). For any fixed graph H that has edges, property P H is p-quasi-random. Simonovits and Sós [1997]

8. 8 The property The property P H We can view P H as a generalization of P 2 since P 2 is just the special case P K 2. But property P 3 shows that it is enough to require that sets of vertices of size n/2 contain the “correct” number of K 2 (namely edges). An open problem of Simonovits and Sós and, in a stronger form, of Shapira is that the analogous condition also holds for any H. Namely: In order to infer that a sequence is quasi-random it is enough to require only the sets of vertices of size n/2 to contain the correct number of copies of H.

9. 9 The property The property P H Shapira (Combinatorica, to appear) proved it is enough to consider sets of size n/(h+1). Hence, in his result, the cardinality of the sets depends on h. We completely settle the above mentioned open problem. More generally, we show that for any H, it is enough to check subsets of size αn for any fixed α  (0,1) P H,α : For any U  V(G n ) with |U| = αn we have H[U] = p e(H) |U| h + o(n h ). For any fixed graph H and any fixed α  (0,1) property P H,α is p-quasi-random. Main result

10. 10 Proof of the main result Fix H with h vertices and r edges and fix α  (0,1). We will prove that: (G n ) satisfies P H,α  (G n ) satisfies P H. This suffices since by the Simonovits-Sós Theorem (G n ) satisfies P H  (G n ) is p-quasi random. Thus we need to prove the following lemma: If G satisfies that for all U  V(G) with |U|= αn we have |H[U] - p r |U| h | = o(n h ) then also for all W  V(G) we have |H[W]- p r |W| h | < o(n h ). Main lemma

11. 11 Proof of the main lemma For simplicity we will demonstrate the proof for H=K 3. So let G be a graph with n vertices satisfying that for all U  V(G) with |U|= αn we have |K 3 [U] – p 3 |U| 3 | = o(n 3 ). Consider any subset W  V(G). We need to prove that |K 3 [W]- p 3 |W| 3 | = o(n 3 ). This is trivial if |W| = o(n). This is also very easy if |W| > αn (counting argument; consider all subsets of U  W of size precisely αn and divide by appropriate multiplicity since each triangle is counted many times as it belongs to many different U).

12. 12 Proof of the main lemma So we can assume that |W| =  n < αn. The triangles of G are partitioned into 4 types H 0 H 1 H 2 H 3 according to the # of vertices they have outside W. By definition, H 0 =K 3 [W]. For convenience, denote w j = |H j | /n 3. Since n = |V(G)| > αn we already know that w 0 + w 1 + w 2 + w 3 = p 3 + o(1). W This triangle belongs to H 1 This triangle belongs to H 0

13. Fix reals α 3, α 2, α 1 so that 1 > α 3 > α 2 > α 1 > α. Let L i be a random subset of vertices of G that contains W and has size α i n. What’s the probability p j,i that an element of H j “survives” in L i ? 13 Proof of the main lemma W L i – chosen uniformly from all sets of size α i n that contain W V(G)V(G) Two elements of H 2 One survived The other didn’t

14. 14 Proof of the main lemma Clearly p j,i = (x i ) j +o(1). where x i = (|L i |-|W|)/(n-|W|) = (α i -  )/(1-  ). Hence the expected number of triangles of G that survived in L i is: E[K 3 [L i ]] = K 3 [W] + p 1,i |H 1 |+ p 2,i |H 2 | + p 3,i |H 3 |. But |L i | = α i n > αn so we already know that the number of triangles in L i is p 3 (α i n) 3 + o(n 3 ) hence K 3 [W] + p 1,i |H 1 |+ p 2,i |H 2 | + p 3,i |H 3 | = p 3 (α i n) 3 + o(n 3 ).

15. 15 Proof of the main lemma Dividing by n 3 we get the following equations: w 0 + x 1 w 1 + (x 1 ) 2 w 2 + (x 1 ) 3 w 3 = p 3 α 1 3 + o(1). w 0 + x 2 w 1 + (x 2 ) 2 w 2 + (x 2 ) 3 w 3 = p 3 α 2 3 + o(1). w 0 + x 3 w 1 + (x 3 ) 2 w 2 + (x 3 ) 3 w 3 = p 3 α 3 3 + o(1). Recalling that also w 0 + w 1 + w 2 + w 3 = p 3 + o(1) we get a system of 4 linear equations in the variables w 0, w 1, w 2, w 3 which is just the Vandermonde matrix A[1, x 1, x 2, x 3 ]. As the x i are distinct the system has a unique solution which, for w 0 gives p 3  3 + o(1). Recalling that w 0 n 3 = K 3 [W] we get |K 3 [W]- p 3 |W| 3 | = o(n 3 ).

16. 16 Thanks