More on invariant types in NIP theories

More on invariant types in NIP theories Pierre Simon June 22, 2013 This note is not intended for publication in its present form. I wrote it to be used as a reference for other papers (mainly [CPS13]) and because I think that the ideas explored here might be of interest to others. We develop here some ideas hinted in [Sim12] concerning the study of global invariant types in NIP theories. There are two ideas or intuitions: the first one is that the relation p commutes with q, where p and q are global invariant types, is very meaningful and expresses in some way that p and q are far away from each other (think of DLO). It is shown in [Sim13] that two types which commute behave with respect to each other as in a stable theory. For example p has a unique non-forking extension to a realization of q and the limit type of any Morley sequence of p over a realization of q is equal to the invariant extension of p. We recall some of those facts below. The second idea is that amongst M-invariant types, definable types and M- finitely satisfiable types are two opposite extremes, and other invariant types sit somehow in between. We showed in [Sim12] that this intuition is justified in the dp-minimal case: a dp-minimal M-invariant types is either definable or finitely satisfiable in M. Our hope is that given an M-invariant type p, one can somehow decompose it into an M-finitely satisfiable type and a definable quotient. In this note, we make a first step towards such a result. Fixing a model M, we construct a continuous retraction F M from the set of M-invariant types to that of M- finitely satisfiable types. The idea is that F M (p) gives the finitely-satisfiable part of p. In all that follows, T is an NIP theory in a language L and U a monster model of T. For basic facts about NIP theories, we refer the reader to [Sim]. The notation M + N means that N is M + -saturated. 1

By an M-invariant type, we mean a global type p S(U) which is invariant under automorphisms of U fixing M. Equivalently (by NIP) it is a global type which does not fork (resp. divide) over M. A type is M-finitely satisfiable if it is finitely satisfiable in M. If p is M-invariant and finitely satisfiable in some small model, then it is finitely satisfiable in M; similarly for definable. If p(x) and q(y) are two M-invariant types, then p q denotes the M- invariant type tp(a, b/u) in variables (x, y) where b = q and a = p Ub. We recall some facts from previous papers: Proposition 0.1 ([Sim12] Lemma 6). An M-invariant type p is definable if and only if for every M-finitely satisfiable type q, p q M = q p M. Proposition 0.2 ([Sim13] Corollary 3.24). Let p and q be two global M- invariant types which commute. Assume that N M is M + -saturated, I = (a i : i < ω) = p (ω) and b = q, then lim(i/nb) = p Nb. Lemma 0.3. Let p, q be M-invariant types. Assume that p (ω) q (ω) M = q (ω) p (ω) M, then p and q commute. Proof. Build a sequence (a n, b n : n < ω) such that (a n, b n ) = p x q y Ua <n b <n when n is even and (a n, b n ) = q y p x Ua <n b <n when n is odd. We show by induction that this is an indiscernible sequence. If p and q do not commute, then we get infinite alternation. 1 The construction Let M = T and let M + M. Consider the pair (M, M): the expansion of M obtained by adding a new predicate P(x) naming M. Call L P the expanded language. Let (N, N) be an M + -saturated elementary extension of (M, M). Let p be a global M-invariant type (in the language L). Claim: The partial L P -type p N P(x) implies a complete L-type over M. Assume not. Then for some L-formula φ(x; b) L(M ), p N P(x) is consistent with φ(x; b) and with φ(x; b). We construct by induction a sequence (a i : i < ω) of points of N such that: a 2i = p N P(x) φ(x; b); a 2i+1 = p N P(x) φ(x; b). The sequence (a i : i < ω) is L-indiscernible, but the formula φ(x; b) alternates infinitely often on it. This contradicts NIP. 2

Therefore there is a unique type q S(M ) such that p N P(x) q is consistent. As q is consistent with P(x) it must be finitely satisfiable in M. As M is M + -saturated, q extends uniquely to a global M-finitely satisfiable L-type q. We now define F M (p) to be equal to q. It is not hard to check that this is well defined, i.e., does not depend on neither M nor (N, N). Lemma 1.1. Let M = T, then the map F M from M-invariant types to M- finitely satisfiable types has the following properties: (i) F M (p) M = p M ; (ii) F M is continuous; (iii) if p is finitely satisfiable, then F M (p) = p; (iv) for any M-definable function f, f (F M (p)) = F M (f (p)). Proof. Point (i) is clear. Point (ii) is by compactness: given a formula φ(x; b) over M, if F M (p) = φ(x; b), then there is some finite part p 0 of p N such that p 0 P(x) implies φ(x; b). If p is finitely satisfiable in M, then p N P(x) is a consistent L P -type, hence F M (p) = p. Point (iv) is easy to check. Remark 1.2. Beware that in general we do not have F M (p q) = F M (p) F M (q). 1.1 The reverse type We present another construction which associates to an invariant type, a finitely satisfiable type, but this time over a larger base. It already appeared in [Sim13]. Let p be an M-invariant type and let M + N. Given any b U, there is some B N of size M such that any two realizations of p B in N have the same type over Mb. To see this, assume not. Then we can build a sequence (a i,0 a i,1 : i < M + ) in N such that a i,0, a i,1 = p Ma<i,0 a <i,1, but a i,0, a i,1 do not have the same type over Mb. Then for any η : M + {0, 1}, the sequence (a i,η(i) : i < M + ) is indiscernible and we easily contradict NIP. This fact allows us to associate canonically to any b U a type over Mb which is compatible with p M. Letting b vary and pasting together the different type, we obtain a type R M (p) over U. This type is clearly N-finitely satisfiable and its restriction to N coincides with p N. Lemma 1.3. The following facts hold: 3

(i) For any M-invariant type q, we have (R N (p) q) N = (q p) N. (ii) The type R N (p) commutes with any M-invariant type (in particular with p). Proof. (i) Let q be M-invariant and b = q N. Let B N containing M and a = p B, then tp(b, a/b) = q p B. (iii) Let q be M-invariant; take some d U, a = R N (p) Nd and b = q Na d. Let a N such that tp(a/mdb) = R N (p) Mdb. Then (a, b) = (R N (p) x q y ) Md. Also a Md a, hence by M-invariance of p, (a, b) = (q y R N (p) x ) Md. 2 Commutativity properties We want to argue that F M (p) somehow captures the finitely satisfiable part of p. We do not define the finitely satisfiable part of p, but we have in mind something like {q M-finitely satisfiable p does not commute with q}. Lemma 2.1. Let p, q be M-invariant types, q being finitely satisfiable in M. Then we have F M (q p) = q F M (p). Proof. Let (M, M) + (N, N) be as in the definition of F M. Take yet another extension (N, N) + (N 1, N 1). Let a N 1 realize p N. As q is finitely satisfiable in M, q N a P(x) is consistent and is realized by some b N 1. Then by definition F M (q p) is given by tp(b, a/m ). But tp(b/m a) = q M a, hence the result. We know that a type is definable if and only if it commutes with all finitely satisfiable types. We show now that it is enough to check commutativity with one specific finitely satisfiable. Proposition 2.2. Let p be an M-invariant type. If p commutes with F M (p), then p is definable. Proof. If s 1 and s 2 are M-invariant types and M + N, saying that a 1 = s 1 N and a 2 = s 2 Na1 is equivalent to saying that a 2 = s 2 N and a 1 = R N (s 1 ) Na2. We will use this in the proof without explicit mention. Assume that p is not definable. Then there is some type q finitely satisfiable over M such that p q M q p M. Take N M, M + -saturated. Let b = q N and build a maximal sequence (c 0 i, c1 i : i < α) such that: 4

tp(b + (c 1 i, c0 i : i < α)/n) is finitely satisfiable in M; there is some M-finitely satisfiable type r i such that (c 1 i, c0 i ) = R N(r i ) (2) c 0 <i c1 <i N; the types r i commute with p; tp(bc 0 i /M) tp(bc1 i /M). Let a 0 = R N (p) over everything and a 1 = p over everything. Then tp(ba 0 /M) tp(ba 1 /M). Let s be the type of b+(c 1 i, c0 i ) i<α +a 0 a 1 over N. This type is M-invariant. If we apply F M to it, then the restriction to the variables corresponding to b + (c 1 i, c0 i ) i<α does not change since this type is M-finitely satisfiable. Thus we can find (g 0, g 1 ) such that b + (c 1 i, c0 i ) i<α + g 0 g 1 realizes F M (s) over N. By Lemma 2.1 and because the types r i commute with p, we have (g 1, g 0 ) = R N (F M (p)) (2) c 0 <αc 1 <αn. Also, g 0 = R N (F M (p)) Mb. We also still have tp(bg 0 /M) tp(bg 1 /M). So g 1 does not realize R N (F M (p)) over Mb. Then take g satisfying R N (F M (p)) over everything. Set c 0 α = g and c 1 α = g 1. This contradicts maximality of the sequence (c 0 i c1 i : i < α). Lemma 2.3. Let p be an M-invariant, not finitely satisfiable in M. Let I be a Morley sequence of p over M, then p MI is not finitely satisfiable in M. Proof. Otherwise, let D be an ultrafilter on M whose limit type q has the same restriction as p over MI. Then p and q have the same Morley sequence, hence p = q. Lemma 2.4. If p is M-invariant, but not finitely satisfiable in M, then F N (p) does not commute with R N (p) (where as before M + N). Proof. Let I = (a i : i < ω) be a Morley sequence of p over N. We know that p NI is not finitely satisfiable in N. Let φ(x; a 1,..., a n ) be a formula witnessing it. We have (a 1,..., a n ) = R N (p) (n) N. As φ(n; a 1,..., a n ) =, necessarily F N (p) R N (p) (n) = φ(x; y 1,..., y n ). On the other hand, R N (p) (n) F N (p) and R N (p) (n) p have the same restriction to N, hence R N (p) (n) F N (p) = φ(x; y 1,..., y n ). So F N (p) and R N (p) (n) do not commute, hence F N (p) and R N (p) do not commute. Lemma 2.5. The type p is definable if and only if F N (p) = R N (p). Proof. Assume that p is definable. Then we can certainly check directly from the definition that F N (p) = R N (p), but here is an easier way: we know that a definable type has a unique coheir over any model. (Proof: show that the 5

Morley sequence of a coheir is determined by reading it backwords and using the fact that p has a unique heir.) As both R N (p) and F N (p) are coheirs of p N, they must be equal. Conversely, if p is not definable, then p does not commute with F N (p). But R N (q) does commute with p, hence F N (p) R N (p). Proposition 2.6. Let p be M-invariant and let q be a type finitely satisfiable in M. Then if q commutes with F M (p), it commutes with p. Proof. Let M + N. Assume that q does not commute with p. Let r = p q (ω) p and let (a 1, b, a 0 ) = F M (r) N. Then for every k < ω, tp(a 1, b k /M) tp(a 0, b k /M). On the other hand by Lemma 2.1, tp( b, a 0 /N) = q (ω) F M (p) N. Thus for every k, (b k, a 0 ) does not realize q F M (p) over M, and tp(b k, a 0 /M) is constant as k varies. If q commutes with F M (p), this contradicts Proposition 0.2. 2.1 Application to dp-minimal types Let p be M-invariant and M + N. The situation is as follows: in any case, p and R N (p) commute; F N (p) commutes with R N (p) if and only if p is finitely satisfiable; F N (p) commutes with p if and only if p is definable. Recall that a type p is dp-minimal if for any A and any two sequences I and J mutually indiscernible over A, for any a = p, either I or J is indiscernible over Aa. In particular if p is M-invariant and dp-minimal, q and r are two M-invariant types which commute, then p commutes with either q or r. To see this let M + N and build ( b, c, a, b, c ) = q (ω) r (ω) p q (ω) r (ω) over N. The sequences b + b and c + c are mutually indiscernible but none is indiscernible over Na. Proposition 2.7. Let p be M-invariant and dp-minimal, then p is either finitely satisfiable in M or definable. Proof. If p is dp-minimal, then already p N is dp-minimal. Hence F N (p) is dpminimal. Since R N (p) and p commute, we know that F N (p) must commute with one or the other. We deduce that p is either finitely satisfiable or definable. 6

2.2 One last result Proposition 2.8. Let p be some global M-finitely satisfiable type orthogonal to all types finitely satisfiable in a set of size < κ. Then p is orthogonal to every global type invariant over a set of size < κ. Proof. Let p as in the statement. Let M of size < κ. There is e U such that for a = p, we have tp(a/e) tp(a/m). More precisely, for every φ(x; y), there are d 0 φ (y; e) and d1 φ (y; e) such that tp x(a/e) d l φ (y; e) φl (x; y) and d 0 φ (y; e) d1 φ (y; e) cover M. Let p be some type finitely satisfiable in a small N, N M. Let r be an heir of tp(e/m) over N and let e realize r. Let σ be an automorphism sending e to e. Then p is orthogonal to every type finitely satisfiable in σ(n) because every such type satisfies φ d0 φ (y; e) d1 φ (y; e). In particular, p is orthogonal to σ(p ). Let b = q and a 0 = p such that tp(a 0, b/u) p q. Let σ as in the previous paragraph for p = p. As q is M-invariant, it is also non-orthogonal to p 1 := σ(p). Hence we can find a 1 = p 1 such that tp(a 1, b/u) p 1 q. By orthogonality of p and p 1, we have tp(a 1, a 0 /U) = p 1 p. Now we iterate the construction. At stage α, we have defined (a i a i : i < α) = i<α p(2) i where each p i is orthogonal to j<i p j. Take σ such that p is orthogonal to σ( i<α p(2) i ). Set p α = σ 1 (p). Let a α = p α be such that tp(a α, b/u) p α q. Then let a α = p α Uba <α a <α. After T + steps, we contradict NIP. Question 2.9. Does this still hold if we replace orthogonal by commuting? References [CPS13] Artem Chernikov, Anand Pillay, and Pierre Simon. External definability and groups in nip theories. preprint, 2013. [Sim] Pierre Simon. Lecture notes on nip theories. preprint. [Sim12] Pierre Simon. Invariant types in dp-minimal theories. preprint, 2012. [Sim13] Pierre Simon. Distal and non-distal theories. Annals of Pure and Applied Logic, 164(3):294 318, 2013. 7