We proved: a = b is type-definable over A, say by E(x,y, A). For x, define E'(x', z, y, w) as z = w ∧ (E(x,y,z) ∨ x = y). Then E' is a type-definable equivalence relation. Let c = (a, A)/E'. What does the type of a/c say? Let p(x) := tp(a/c); It says: ∃B ∃t. c = (b, B)/E' which implies B = A ∧ x = a i.e. B = A ∧ x = a In other words tp(a/c) = lstp(a/A). You can do even better than this: Let bdd(A) = all hyperimaginaries with small orbits over bounded closure of A. Then c ∈ bdd(A). Cheating since is a proper class but can get around But then tp(a/c) ≠ lstp(a/A) = tp(a/bdd(A)) = tp(a/bdd(A)) = tp(a/A) since an automorphism fixing A pointwise fixes bdd(A) setwise. So we can conclude lstp(a/A) = tp(a/bdd(A)). (Analogous to stp(φ/A) = φ(A/acl(A))) Defn: A type-definable equivalence relation E(x,y) is small if |{t : |t| ≤ |E(x,t)| ≤ |T|}| ≤ |T|. IF E is small then the hyperimaginary sort z/E is also called small. Exercise/Remark: Every type-definable eq. reln E(x,y) can be written as E(x,y) = ∧ Eᵢ(x,yᵢ) where z(x,y) can be small and Eᵢ is small. This remark implies every hyper-imaginary is interdefinable with a tuple of small ones. Defn: Where U = all small h.l. sorts. Fact 2 (Hrushovski): U is simple ⇔ Wher U is not simple. Proof (Sketchy) If U is not simple neither is Wher U Conversely, ∃ hᵢ sorts xᵢ, yᵢ and formulas φᵢ(xᵢ,yᵢ), ψᵢ(yᵢ,xᵢ) s.t. ψᵢ(yᵢ,xᵢ) is a k-indiscernibility witness for φᵢ(xᵢ,yᵢ) and D(xᵢ = xᵢ) contains arbitrarily long sequences (t₁,t₂), (t₂,t₃), ... Defn Let p∈S(a), q∈S(b) be amalgamation bases. p ⊥⊥₁ q (p is 1-parallel to q) if they have a common nd. exth, i.e., ∃ t∈S(p,q), t↓a, t↓b. ||n (n-parallel) := n-iterate of ⊥⊥₁. ||∞ (parallel) := tr cl. Eg if p∈S(a) is an amal base, q∈S(a,b) is a nd. exth of p ∈ S is an amal base then, p ⊥⊥₁ q. → for discussion of coincidences From now on all types are amalgamation bases (a,L). "p(x,a)" etc are complete types over a etc. Lemma Assume p(x,a) ⊥⊥₁ q(x,b), r(x,c) and a↓c then p(x,a) ⊥⊥₁ r(x,c). Proof same as similar in last lecture (slightly less general). Defn Fix p(x,a) (actually fix p(x,a) - a will vary) aRb := p(x,a) ⊥⊥₁ p(x,b).

18.996A Simplicity Theory: Lecture 8 Notes

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