On a theorem of Lascar

Martin Ziegler ^{∗}

June 15, 2000

We denote by ℂ be a big saturated model (the monster model). If R, S are binary relations on ℂ the product RS is the class of all pairs (a, c) for which there is a b ∈ ℂ such that aRb and bSc. The smallest type–definable relation which contains R is denoted by by \overline{R}. The smallest invariant equivalence relation whith a bounded number of classes is E_{L}, the relation of having the same strong Lascar type. The smallest bounded type–definable equivalence relation E_{KP}, was introduced by Kim and Pillay.

The following theorem was proved by Lascar using the Lascar galois group:

Theorem 1 (D. Lascar)

\overline{E_{L}} E_{L} = E_{KP}

I will give another of this theorem.

A formula θ(x, y) is called thick if there is no infinite sequence of (a_{i}) such that ¬θ(a_{i}, a_{j}) for i < j. We denote by Θ(x, y) the relation which is defined by the set of all thick formulas.

It is well known that E_{L} is the transitive closure of Θ:

E_{L} = Θ ∪ Θ^{2} ∪ Θ^{3} ∪ ….

We will prove Theorem 1 in the following slightly stronger form.

Theorem 2 \overline{E_{L}} Θ = E_{KP}

Lemma 3 (Open mapping) Let A be a set of parameters, a an element and θ(x, y) a thick formula, possibly with parameters from A. Then there is an L_{A}– formula ϕ in tp(a/A) such that every type p ∈ S(A) which contains ϕ can be realized by an element b such that ℂ ⊧ θ(a, b).

Proof: Since every thick formula contains a symmetric thick formula we may assume that θ is symmetric. Also we may assume that A is empty.

∗I thank H. Casanovas and F. Wagner for a helpful discussion.

Let D be the class of all conjugates of a and D_{0} a finite subset of D such that D is contained in the definable class

B = {b ∈ ℂ | ℂ ⊧ θ(a_{0}, b) for some a_{0} ∈ D_{0}}.

The set of all types p ∈ S(∅) which can be realized by an element of ℂ \ B is closed. Therefore the set Φ of all p with only realizations in B is open. Since D ⊂ B, Φ contains tp(a). Thus we find a ϕ ∈ tp(a) such that every p which contains ϕ can only be realized by elements of B.

Fix a p ∈ S(∅) which contains ϕ. Choose a realization b ∈ B and a_{0} ∈ D_{0} such that ℂ ⊧ θ(a_{0}, b). Since a_{0} has the same type as a we find a b′ such that ab′ has the same type as a_{0}b. Then b′ realizes p and ℂ ⊧ θ(a, b′). This proves the Lemma.

Let R and S be two invariant relations on C. It is easy to see^{1} that \overline{RS} is always contained in \overline{R} \overline{S}. The converse inclusion is not generally true: Take as a model a set with a sequence of named elements 0, 1, 2, . . .. Take for R the set of all pairs (0, 1), (0, 3), (0, 5), … and for S the set of all pairs (2, 0), (4, 0), (6, 0), . . .. Then \overline{R} \overline{S} contains (0, 0) and \overline{RS} is empty.

Of course, R and S are not connected in the following sense:

Definition 4 Two invariant relation R and S are called connected if there is a complete type p over ∅ such that both, R(x, y) and S(y, z), imply p(y).

Example: Look at the group G = ℝ × ℝ with the lexicographical ordering. Forget everything except the ordering, addition with (1, 0) and addition with (0, 1). Define

R(x, y) ⇔ ⋁_{n∈ω} y < x + (0, n).

Since there is only one type over the empty set, R and R are connected. Also R is transitive, while

\overline{R}(x, y) ⇔ ⋀_{n∈ω} y < x + (1, −n)

is not. Whence \overline{R} \overline{R} ≠ \overline{R} = \overline{RR}.

Lemma 5 Assume the invariant relations R and S to be connected^{2}. Then

R\overline{S} ⊂ \overline{RS} Θ  (1)
\overline{R}S ⊂ Θ \overline{RS}  (2)
\overline{R}\overline{S} ⊂ Θ \overline{RS} Θ  (3)

1Note that the product of two type–definable relations is again type–definable.

2For (1) (and similarly for (2)) we only need that the first components of all pairs in S realize the same type.

Proof: We prove first (1). Assume (R\overline{S})(a, c). Since (\overline{RS} Θ)(x, z′) can be axiomatized

\{∃z (ψ(x, z) ∧ θ(z, z′)) | ψ ∈ \overline{RS}, θ thick\}

we have to show that for all ψ(x, z) ∈ \overline{RS} and all thick θ there is a c′ such that \overline{RS}(a, c′) and θ(c′, c). Let b be such that R(a, b) and \overline{S}(b, c). If we apply Lemma 3 to tp(c/b) we obtain a formula ϕ(z, b) such that every type over b which realizes ϕ can be realized by an element c′ which satisfies θ(c′, c). Since \overline{S}(b, c), and R and S are connected, there is a c′ which realizes ϕ and satisfies S(b, c′). By the choice of ϕ we can choose c′ in such a way that θ(c′, c).

The proof of (2) is symmetrical.

(3) follows from (1) and (2):

\overline{R} \overline{S} ⊂ \overline{\overline{R}S}Θ ⊂ \overline{Θ\overline{RS}}Θ = Θ \overline{RS} Θ

Proof of Theorem 2: Fix a complete type. First we prove the theorem restricted to the type p. We restrict the the meaning of E_{L}, \overline{E_{L}}, Θ and E_{KP} the the realization set of p. We can then apply the last lemma. Since \overline{E_{L}}Θ is type–definable it suffices to prove that \overline{E_{L}}Θ is transitive.

We have by the lemma E_{L}\overline{E_{L}} ⊂ \overline{E_{L}}Θ and \overline{E_{L}}E_{L} ⊂ Θ\overline{E_{L}}. This gives

\overline{E_{L}}Θ ⊂ \overline{E_{L}}E_{L} ⊂ Θ\overline{E_{L}} ⊂ E_{L}\overline{E_{L}} ⊂ \overline{E_{L}}Θ

and all four terms are equal. Part (3) of the Lemma gives

\overline{E_{L}} \overline{E_{L}} ⊂ Θ\overline{E_{L}}Θ = \overline{E_{L}}E_{L}.

Whence \overline{E_{L}}Θ is transitive:

\overline{E_{L}}Θ\overline{E_{L}}Θ = \overline{E_{L}} \overline{E_{L}}E_{L} = \overline{E_{L}}E_{L}^{2} = \overline{E_{L}}E_{L} = \overline{E_{L}}Θ.

Now the general case: Let for every complete type \overline{E_{L}(p)} be the closure of E_{L}∩p^{2}, Θ(p) = Θ ∩ p^{2} and E_{KP}(p) be the finest bounded type–definable equivalence relation on p. We have proved that

\overline{E_{L}(p)}Θ(p) = E_{KP}(p).

Since E_{KP}(p) = E_{KP} ∩ p^{2} this implies

⋃_{p} \overline{E_{L}(p)} Θ = E_{KP}.

But ⋃_{p} \overline{E_{L}(p)} ⊂ \overline{E_{L}} and the theorem is proved.