FIRST ORDER SENTENCES ON G(n, p), ZERO-ONE LAWS, ALMOST SURE AND COMPLETE THEORIES ON SPARSE RANDOM GRAPHS

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1 FIRST ORDER SENTENCES ON G(n, p), ZERO-ONE LAWS, ALMOST SURE AND COMPLETE THEORIES ON SPARSE RANDOM GRAPHS MOUMANTI PODDER 1. First order theory on G(n, p) We start with a very simple property of G(n, p), where n is the number of vertices in the vertex set V (G) = V of G G(n, p), and p = p(n) the edge probability. Let the property be that a graph contains a triangle. We denote this property by A and let µ n (A) = P [G(n, p) = A]. We shall for now consider p a constant not depending on n. In particular, let us look at p = 1/2. Theorem 1.1. lim µ n(a) = 1. Proof. We split the vertices of V (G) into s = n/3 many disjoint triples. A triple {i, j, k} forms a triangle with probability precisely 1/8. Because these involve different edges, these events are independent over all the s triples, and the probability that none of them forms a triangle is therefore (7/8) s. Now note that as n, this quantity also goes to 0, and as this is the upper bound on the probability of absence of a triangle in the graph, or in other words, an upper bound on P [G(n, 1/2) = A], hence this probability also goes to 0, which gives us the proof. Definition 1.2. When lim µ n (A) = 1 for some property A, we say that the property A occurs almost surely, i.e. almost all graphs have property A. When lim µ n (A) = 0, we say that A holds almost never, i.e. almost no graphs have property A. We can expect similar zero-one laws for the class of properties referred to as first order properties for graphs. These are formally defined as follows. Definition 1.3. The first order language consists of variables that are the vertices of the graph, denoted by x, y, z etc., binary predicates = (which means equality of vertices) and (which denotes the adjacency of two vertices), Boolean connectives,, =,, etc., existential ( ) and universal ( ) quantification over vertices (only). A first order sentence will be of finite length, and the number of nested quantifiers gives the quantifier depth of the sentence. Example 1.4. ( There exists an isolated vertex in the graph. This can be expressed as v[ u { (u v)}]. (i There exists a triangle in the graph. This can be expressed as x y z [{x y} {y z} {z x}]. Example 1.5. A property that is not first order: a graph is connected. Proof: using Ehrenfeucht games! Remark 1.6. When we study first order graph properties, it basically suffices to fix any possible arbitrary finite graph H and ask for the property A(H) that G contains a subgraph isomorphic to H. This basically exhausts all possible first order graph properties. 1

2 2 MOUMANTI PODDER Theorem 1.7 (Fagin-GKLT). Let A be a first order graph property. Then for constant p(n), we have lim µ n(a) {0, 1}. (1.1) That is, every first order sentence holds either almost surely or almost never. Definition 1.8 (Witness and (r, s) extension statement). Given r + s distinct vertices x 1,... x r, y 1,... y s, a point z (distinct from x1,... x r, y 1,... y s ) is called the witness with respect to these vertices if z x i for all 1 i r and z y j for all 1 j s. The (r, s) extension statement, denoted by A r,s, is that for all distinct vertices x 1,... x r and y 1,... y s, there exists a vertex z which is a witness with respect to these points. Lemma 1.9. For all r, s 0 and constant p(n), the extension statement A r,s holds almost surely. Proof. Fix r, s 0. Then choose distinct x 1,... x r, y 1,... y s and fix them. Now, for any z distinct from these points, we define the event Then G z = {z is a witness to x 1,... x r, y 1,... y s }. (1.2) P [G z ] = P [z x i 1 i r, z y j 1 j s] = p r (1 p) s. (1.3) Then if Bad(x 1,... x r, y 1,... y s ) denotes the event that there exists no z distinct from x 1,... x r, y 1,... y s such that G z holds, then P [Bad(x 1,... x r, y 1,... y s )] =P = z x 1,...x r,y 1,...y s G c z z x 1,...x r,y 1,...y s P [G c z] = [1 p r (1 p) s ] n r s. (1.4) We have used here the fact that when we consider two different z 1, z 2 and the events G z1 and G z2, they involve two mutually exclusive set of edges and hence are independent of each other. Now suppose Bad is the union of Bad(x 1,... x r, y 1,... y s ) over all x 1,... x r, y 1,... y s. Then if Bad happens then A r,s is not true. Hence [ ] P [ A r,s ] =P [Bad] = P Bad(x 1,... x r, y 1,... y s ) x 1,...x r,y 1,...y s P [Bad(x 1,... x r, y 1,... y s )] x 1,...x r,y 1,...y s ( )( ) n n r = [1 p r (1 p) s ] n r s r s Suppose we set then we have the upper bound for P [ A r,s ] as nr r! ns s! [1 pr (1 p) s ] n r s = nr+s r!s! [1 pr (1 p) s ] n r s. (1.5) ɛ = [1 p r (1 p) s ] < 1, P [ A r,s ] c n r+s ɛ n r s = exp [(r + s) log n + (n r s) log ɛ] 0 as n. (1.6) This completes the proof.

3 FIRST ORDER WORLD OF G(n, p) AND ZERO-ONE LAWS 3 This lemma is one of the key steps for the proof of the Fagin-GKLT theorem. The rest follows from creating a theory T that consists of A r,s as axioms for all r, s 0. This theory is complete, a notion we define later, and has a countable model (meaning a graph with countably infinite vertices that satisfy all the axioms. Because the theory is complete, for every first order sentence B, either B T or B T, i.e. either B is derivable from T or B is. This finishes the proof of the zero-one law. Let p(n) be any edge probability function. 2. Almost sure and complete theories Definition 2.1. The almost sure theory T relative to p(n) is the set of all first order sentences A that hold almost surely when p(n) is the edge function for G(n, p), i.e. lim P [G(n, p(n)) = A] = 1. We denote this theory by T α for the special case p(n) = n α, 0 < α < 1. Remark 2.2. A theory is a set of sentences which are closed under logical inferences in the first order language. Definition 2.3. A theory T is called complete when for every first order sentence B, either B T or B T. Definition 2.4. We say p(n) satisfies the zero-one law if relative to p(n), every first order prorperty A either holds almost surely or holds almost never. Lemma 2.5. Easy consequence: p(n) satisfies the zero-one law if the almost sure theory T corresponding to p(n) is complete. Remark 2.6. In the case where p(n) = n α, we generally talk about α irrational. Why? Because we get zero-one laws only when α is irrational. This fact is not easy to prove in full generality, but let me give you a flavour as to why this is kind of expected. Consider a fixed (though arbitrary) finite graph H, with v vertices and e edges. We consider the event that the graph G(n, p(n)) contains H as a subgraph. Set S = {S V : S = v}. Then for every S S, let X S be the indicator random variable for the event that the induced subgraph on S by G(n, p(n)) contains H. Suppose the number of automorphisms of H possible is a. This means that for each such automorphism, we require the presence of e specific edges, which happens with probability p(n) e. Finally, if X = S S X S, then E[X] = ( ) n E[X S ] = a p(n) e nv v v! a (n α) e a = v! nv αe. (2.1) S S What if we have α = v/e, which is a rational? We shall not get a zero-one law for that. A more specific example would be the presence of triangles. Here, if A denotes the property that a graph contains a triangle, then with p(n) 1/n, A holds almost never, and with p(n) 1/n, A holds almost always. But when p(n) = c/n, then we are within some kind of threshold and the probability of A is moving from 0 to 1 as c moves from 0 to 1, and in fact one can show that P [A] 1 e c3 /6. Hence we get no zero-one law in this case. A proof of this fact involves using Janson s inequality, a topic I talked about in the last to last student probability seminar.

4 4 MOUMANTI PODDER 3. Countable models Suppose p(n) satisfies the zero-one law and T be its almost sure theory. Then as mentioned above Remark 2.5, T will be a complete theory. A model of a theory T is simply a specific graph G that satisfies all the properties A T. We wish to find all possible models for this specific theory T. First note that there can be no finite models. For any k there exists a first order sentence stating that the graph has at least k vertices, i.e. x 1 x 2... x k {x i x j }. 1 i<j k This sentence holds almost surely as it holds with probability 1 for all n k, where n is the total number of vertices in G(n, p(n)). Thus G would have to satisfy it and hence have at least k elements, and this is true for all k N. Hence G cannot be finite. Does T necessarily have models? A deep theorem of Kurt Gödel says that any consistent theory will have a countable or finite model. A theory T is called N 0 -categorical, read as aleph-nought-categorical, if it has precisely one countable model up to isomorphism. One sufficient condition for T being complete is given by the following theorem: Theorem 3.1 (Skolem-Lowenheim). If T has no finite models and T is N 0 -categorical, then T is complete. Proof. Brief inspiration behind this theorem: Suppose T is not complete, then there exists some first order B such that B / T. Then we consider the new theory T + obtained by adding B to T, and the theory T obtained by adding B to T. Both are consistent and hence must have countable models G +, G. Clearly, G + and G cannot be isomorphic since they disagree on the truth value of B. But both G + and G are models for T. Hence T has two non-isomorphic countable models. Now take the contrapositive statement and get the theorem. 4. Very Sparse Random Graphs We will consider the evolution of G(n, p) as p increases from empty to full. We will consider various ranges of the function p(n). ( Suppose p(n) n 2. Then almost surely there are no edges. Proof. Suppose, for every pair of distinct vertices v i, v j, X i,j denotes the indicator random variable for the event that v i, v j are adjacent. Then if X is the total number of edges present, then E[X] = E X i,j = ( ) n E[X i,j ] = p(n) n2 p(n) 0, (4.1) i<j n 1 i<j n as n. Hence using Markov or Chebychev s inequality, P [X 1] E[X] 0 as n. (4.2) So, what should be the almost sure theory for p(n) in this case? It consists of the following axioms: (a) for all r N, P r = there exist at least r vertices, (b) no edge, i.e. x, y [ {x y}]. Hence the countable model for this theory consists of countably infinite vertices and no edges. This model is clearly N 0 -categorical and hence complete.

5 ( n i FIRST ORDER WORLD OF G(n, p) AND ZERO-ONE LAWS 5 (i When p reaches Θ(n 2 ), edges start to appear. They remain scattered until p reaches Θ(n 3/2 ). Then edges with common vertices appear - which we will consider as trees on three vertices. When p reaches Θ(n 1 1/k ), trees on k + 1 vertices appear. All of these occur well before cycles appear at Θ(n 1 ). Let us examine the theory when k p(n) n 1 1. (4.3) Are there cycles? Consider i vertices, for some i N. If these are to form a cycle, then there are (i 1)!/2 many automorphisms, and we require the presence of exactly i edges for each such automorphism. Finally these i vertices can be chosen out of n vertices in ( n ways. Hence the expected number of such cycles will be ( ) n (i 1)! p(n) i ni (i 1)! p(n) i 1 ( ) i p(n) i 2 i! 2 2i n 1 0, (4.4) as n. Are there components with more than k + 1 vertices? This means that we are asking for a subset S of V with S k +2, such that S is connected in the induced subgraph of G(n, p(n)). Let us say we fix an S of i vertices with i k+2. Then the probability that it will be connected is p(n) i 1 (since we need at least i 1 edges to make it connected, but since there can be no cycles, hence exactly i 1 edges). The number of such subsets is ( n. Hence the expected number of such structures is ) a p(n) i 1 ni i! a p(n)i 1 = C ( p(n) ) 1 i 1 n (i 1)( 1 i n ) = C ( p(n) ) i 1 i 1 n 1. (4.5) Because the exponent of n above is negative when i k + 2. Hence this expectation is going to 0. a is the number of automorphisms for the tree structure we are considering. For i k +1, there are ( n n i /i! many choices for a subset of i vertices out of n vertices, and then we have to make sure that there are i 1 many edges within that subset in the induced subgraph, which has probability p(n) i 1. Then the expected number of such structures is a constant times the following quantity ( p(n) n i p(n) i 1 = k ) i 1 n i n (i 1)( 1 1 k ), as n. (4.6) We now can use second moment method or Janson s inequality like argument to argue that duch structures are almost surely as n. These three properties (or observations) give us the almost sure theory T for this p(n). What are the countable models G of T? Each component (connected) can have at most k + 1 many vertices and must be a tree, including the possibility of an isolated vertex. But every such tree does occur as a component at least r times for all r N. So G consists of a countably infinite number of copies of every possible tree on i vertices for every i k + 1, and nothing else. Hence clearly G is N 0 -categorical and hence complete. Thus there is a zero-one law. References [1] The Strange Logic of Random Graphs, Joel Spencer, ISBN , Springer; [2] The Probabilistic Method, Noga Alon and Joel Spencer, ISBN-13: , Wiley Series in Discrete Mathematics and Optimization; Courant Institute of Mathematical Sciences, New York University