18.996A Simplicity Theory: Lecture 6 Notes

University:
Massachusetts Institute of Technology
Course:
18.996A | Simplicity Theory
Academic year:

2024
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75

Pages:

17
Author:

pakuonaqy

2/5 Defn let a & b be tuples of the same length & let A be a set of parameters. We say that a & b are the same Lascar strong type over A denoted a =b, if for each A-invariant equivalence relation E on types of same length as a (24) that has a bounded number of equiv classes then we have E(a, b). Note that the finest equality of Lascar strong type over A is the finest bounded biinvariant equivalence relation on tuples of the relevant length. Notation: Define da(a, b) = least n∈ω st. there are a = a₀, a₁, ..., an = b st. for each i∈n there is an infinite A-indiscernible sequence shifting with (aᵢ, aᵢ₊₁); or ∞ if it doesn't exist. Lemma a =b iff da(a, b) < ∞ Proof ⇒: First assume that da(a, b) ≤ 1 ∃ a = a₀, a₁, a₂ st. a₀ = a, a₂ = b, a₁ is an A-indiscernible sequence If a₁ ≠ b, extend the sequence to an arbitrary length (aᵢ)ᵢ∈ω By invariance, da(aᵢ, aⱼ) ≤ 1 (contradicting boundedness) ⇐: If da(a, b) = n ∈ ω, then ∃ a₀, a₁, ..., an = b st. so a = a₀ = a₁ = a₂ = ... = an = b. Clearly da(x, y) < ∞ is an A-invariant equiv reln. (symmetric since we can always extend an indiscernible sequence to negative indices ... a₂ ... a₁ a₀ a₁ a₂ ...) ⇒ x = y. If it is bounded, then by defn, da(x, y) < ∞ ⇒ x = y. If not, then for I sufficiently big, ∃ i ∈ I st. da(aᵢ, aⱼ) = ∞ Extract an A-indiscernible sequence (aᵢ : i ∈ ω) ∃ i ∈ I st. a₀ = a, a₁ = b ⇒ da(a₁, a₀) = 1. But da(a₁, a₀) ≤ 1. □ Lemma (Extension for Strong types) (Tsimple) Let A ≤ B & let a be a n-tuple. Then there exists (aᵢ : i ∈ ω) a large Morley sequence in tpl(a/A) with a₀ = a (need bound: cf >> f(t) + H(a)) By simplicity, there exists I ∈ X ⊆ B Let j ≤ sup I. So aⱼ ∈ B Thus aⱼ ∈ B → aᵢ ∈ B a₁ ∈ B But aᵢ, a₂ is a Morley sequence ⇒ a₁ = a₂ Now use this corollary... Cor a =b ⇒ da(a, b) ≤ 2. Assume a =b. Find cst. c ∈ clb such that a = c = b ⇒ da(a, b) ≤ 1 Cor Equality of Lascar strong types is type-definable, i.e. ∃ E(x,y) over A st. ∀ E(a, b) ↔ a =b. Take p(x) = E(x, a) then p(x) is logically equivalent to 1stp(a/A). IOU From Previous lecture If a ≤ b then D(a/c,=) = D(a/bc,=) Proof D(a/c,=) = D(a/bc,=) V We prove by induction on x: If x ∈ D(a/c,=) ⇒ x ∈ D(a/bc,=) x=0: ✓ x=limit: ✓ x=B+H: ∃ d ∈ D(x,y) s.t. p(x, d) divides,wrt ψ Then ∃ e ∈ tp(a/c) s.t. p(x, d) = e(x) Write e = tp(a/c)₄(x, d) = e₁(x) This means that div(b, e(x)) ∧ p(x, d) ∧ e₁(x, d) is consistent. Let a' realize it, so find d' st. a' ∈ d = ad'. Furthermore, find d'' ∈ d = ad' st d'' ∈ b Then (i) e ∈ D(a'/cd') so e ∈ DE(a'/cd'), ⇒ e(a'/cd') = 0 since d' ∈ c, ψ(x,d') divides/c. (3) d" ∈ b & a = a' ⇒ d" ∈ b & a ∈ b ⇒ d" ∈ a 3a: a ≤ b ∧ e ∈ D(a'/c,=) ⇒ e ∈ D(a'/bc,=) ⇒ e ∈ D(a'/bc,=) 3b: d'' ∈ b ⇒ ψ(x,d'') div/bc wrt ψ List: ∃ c-indiscernible sequence (di) in tp(d'/c) Since d'' ∈ b, this sequence has an automorphism image in tp(d'/bc), thus ψ(x,d') divides/bc wrt ψ. All together: ∃ e ∈ ψ(x,d') ∈ D(a'/bc,=) □ (Corollary) a ≤ b iff for some e ∈ D(a/c,=) maximal, if D(a/bc,=) = D(a/c,=). □ Important Corollary: Given any a ∈ c, and of there is a partial type p(y, a, c) such that for b ∈ at the appropriate length, b ⊨ p(y, a, c). Proof Let g ∈ D(a/c,=) be maximal. Let p(y, a, c) = div(g, a) Then a ≤ b ⇒ g ∈ tp(a/bc,=) ⇒ ∃ div(g, b) ⇒ b ⊨ p(y, a, c). □

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