The Schr oder-Bernstein property and a-saturated...

Introduction to SB SB for saturated models SB over a base

The Schroder-Bernstein property and a-saturatedmodels

John Goodrick

Universidad de los Andes

XV Simposio Latinoamericano de Logica MatematicaBogota, 8 de junio de 2012

John Goodrick Universidad de los Andes

The Schroder-Bernstein property and a-saturated models

The Schroder-Bernstein property

Theorem

(Schroder-Bernstein) If there are injective functions f : X → Yand g : Y → X , then there is a bijection between the two sets Xand Y .

K is a (concrete) category;

Mor is a distinguished class of injective morphisms in K.

(K,Mor) has the Schroder-Bernstein property or SB property ifany two Mor-biembeddable objects in K are isomorphic.

The Schroder-Bernstein property

Theorem

(Schroder-Bernstein) If there are injective functions f : X → Yand g : Y → X , then there is a bijection between the two sets Xand Y .

K is a (concrete) category;

Mor is a distinguished class of injective morphisms in K.

(K,Mor) has the Schroder-Bernstein property or SB property ifany two Mor-biembeddable objects in K are isomorphic.

The SB property for theories

T will always be a complete, first-order theory.

All embeddings in this talk will be elementary.

Mod(T ) is the class of models of T , Elem is the class ofelementary embeddings.

We say T has the SB property if (Mod(T ),Elem) has the SBproperty.

Example

T = the theory of dense linear orderings without endpoints (in thelanguage L = {<}) does not have the SB property: all injectiveembeddings are elementary (by q.e.), and consider (R, <) and(R \ {0}, <).

Example

SB and classifiability of T

It turns out that (for T a countable theory)

T is ℵ1 − categorical ⇒ T has SB ⇒ T is classifiable.

Classifiable ⇔ [superstable + NOTOP + NDOP + shallow]⇔ all models are prime over a well-founded, independent tree ofcountable elementary submodels.

All implications are strict.

Example

T = the theory of algebraically closed fields of characteristic 0 inthe language {+, ·}. Any model of T is determined by itstranscendence degree over Qalg , a cardinal number. So T isuncountably categorical and SB.

Example

SB and cardinal invariants

Idea: We should expect (K,Mor) to have SB whenever the theobjects of K can be classified by a bounded collection of cardinalinvariants which are preserved by Mor.

Example

If T is any countable, ℵ1-categorical theory, the models of T areclassified by a single cardinal invariant: the dimension of the set ofrealizations of a strongly minimal set. (As in Hans Adler’s tutorial.)

SB and cardinal invariants

Idea: We should expect (K,Mor) to have SB whenever the theobjects of K can be classified by a bounded collection of cardinalinvariants which are preserved by Mor.

Example

If T is any countable, ℵ1-categorical theory, the models of T areclassified by a single cardinal invariant: the dimension of the set ofrealizations of a strongly minimal set. (As in Hans Adler’s tutorial.)

Categories of models

Pipe dream: Use stability theory techniques to describe all possiblecategories (Mod(T ),Elem) as T varies.

We expect dichotomies or “gaps” of the form:

If (Mod(T ),Elem) has a small subcategory isomorphic to a certainC, then must also have a subcategory isomorphic to D.

Categories of models

Pipe dream: Use stability theory techniques to describe all possiblecategories (Mod(T ),Elem) as T varies.

We expect dichotomies or “gaps” of the form:

If (Mod(T ),Elem) has a small subcategory isomorphic to a certainC, then must also have a subcategory isomorphic to D.

SB for T : some special cases

Theorem

(Nurmagambetov) If T is totally transcendental, then T has SB ifand only if T is non-multidimensional.

Theorem

(G., Laskowski) If T is countable and weakly minimal, then T hasSB if and only if (†) for every minimal type p ∈ S(acleq(∅)) andevery f ∈ Aut(C), there is some n < ω such that f n(p) is notalmost orthogonal to p ⊗ f (p)⊗ . . .⊗ f n−1(p).

Conjecture

T is SB if and only if T is classifiable and every regular typesatisfies a condition like (†) for every regular type.

Theorem

Conjecture

Theorem

Conjecture

SB and saturation

Characterizing the SB property turns out to be easier if we restrictourselves to the sufficiently-saturated models.

Example: T1 = Th(Z; +). T1 is classifiable, weakly minimal, butnot ℵ1-categorical and not totally transcendental.

All models of T1 are submodels of Z⊕Qκ for some κ, whereZ =

∏p prime Z(p).

T1 does not have SB (fun exercise!).

The set ℵ1-saturated (or even a-saturated) models are of the formZ⊕Qκ. Thus the ℵ1-saturated models do have SB.

SB and saturation

∏p prime Z(p).

SB and saturation

∏p prime Z(p).

SB and saturation

∏p prime Z(p).

SB and saturation

∏p prime Z(p).

SB and saturation

∏p prime Z(p).

Stability

The most basic Shelahian dichotomy is stability vs. instability:

Definition

T is unstable if there is a model M |= T , a formula ϕ(x ; y), andtuples 〈ai : i < ω〉 in M such that

M |= ϕ(ai ; aj)⇔ (i < j).

T is stable if T is not unstable.

Theorem

(Shelah) If T is unstable, then for every κ > |T | it has 2κ modelsof size κ which are pairwise nonisomorphic.

Stability

Definition

M |= ϕ(ai ; aj)⇔ (i < j).

Theorem

Stability

Definition

M |= ϕ(ai ; aj)⇔ (i < j).

Theorem

Stability and SB

Definition

M |= ϕ(ai ; aj)⇔ (i < j).

Theorem

(G.) If T is unstable, then T is not SB.

In fact, for any cardinal κ, there is a collection of κ non-isomorphicand pairwise bi-embeddable models of T , each of which isκ-saturated.

Superstability: examples

Definition

T is superstable if it has no infinite forking chain of types: so thereis no

p0(x) ⊆ p1(x) ⊆ . . . ⊆ pn(x) ⊆ . . .

such that pi ∈ S(Mi ) and pi+1 forks over Mi .

Equivalently (for T countable): T is superstable if for every modelM |= T such that |M| ≥ 2ℵ0 , |S(M)| = |M|.

Some superstable theories: Any ℵ1-categorical countable T .Any weakly minimal theory (e.g. Th(Z; +) is). Differentially closeddifference fields of characteristic 0.

Definition

p0(x) ⊆ p1(x) ⊆ . . . ⊆ pn(x) ⊆ . . .

Definition

p0(x) ⊆ p1(x) ⊆ . . . ⊆ pn(x) ⊆ . . .

Superstability: non-examples

Definition

p0(x) ⊆ p1(x) ⊆ . . . ⊆ pn(x) ⊆ . . .

Stable but not superstable: Separably closed fields ofcharacteristic p > 0 and fixed degree of imperfection e. The freegroup with more than one generator. Infinitely manyinfinitely-refining equivalence relations 〈Ei : i < ω〉.

Superstability and SB

Theorem

(G.) If T is not superstable, then T does not have the SB property.

However, there do exist stable, not superstable T such that thecollection of all ℵ1-saturated models has the SB property.

Superstability and SB

Theorem

(G.) If T is not superstable, then T does not have the SB property.

However, there do exist stable, not superstable T such that thecollection of all ℵ1-saturated models has the SB property.

a-saturated models

Definition

A model M is a-saturated (or an a-model) if for every finite A ⊆ Mand every A-indiscernible sequence I ⊆ N for some N � M, thereis an A-indiscernible extension J with I ⊆ J ⊆ N and J ∩M 6= ∅.

For T countable,

M is ℵ1-saturated ⇒ M is an a-model ⇒ M is ℵ0-saturated.

(These implications are strict.)

a-saturated models

Definition

A model M is a-saturated (or an a-model) if for every finite A ⊆ Mand every A-indiscernible sequence I ⊆ N for some N � M, thereis an A-indiscernible extension J with I ⊆ J ⊆ N and J ∩M 6= ∅.

For T countable,

M is ℵ1-saturated ⇒ M is an a-model ⇒ M is ℵ0-saturated.

(These implications are strict.)

Orthogonality

From now on, we assume T is stable.

Definition

If p and q are nonalgebraic types over an a-model M, then we saythat p is orthogonal to q (or p ⊥ q) if there are a-modelsMp,Mq � M such that:

1 In Mp there is a realization of p but not of q; and

2 in Mq there is a realization of q but not of p.

Intuition: p ⊥ q basically means that the number of realizations ofp in a model is independent of the number of realizations of q in amodel (as in Jouko Vaananen’s talk!).

This can be extended to p ∈ S(M) and q ∈ S(N) using nonforkingextensions.

Orthogonality

Definition

Orthogonality

Definition

Nonmultidimensionality and nomadic types

Definition

(Shelah?) T is multidimensional if for every cardinal κ, there is ana-model M and a collection of κ pairwise-orthogonal types over M.

Definition

(G.) A type p ∈ S(M) is nomadic if there is an automorphism f ofthe monster model (any sufficiently saturated C � M) such that{p, f (p), f 2(p), . . .} are pairwise orthogonal.

Note: T nonmultidimensional ⇒ T has a nomadic type.

(Standard forking calculus...)

Definition

Non-multidimensional examples

Suppose that T2 is the {E}-theory saying: E is an equivalencerelation with infinitely many classes, each of which is infinite.

T2 is complete, has q.e., and is superstable.

Any model of this T2 is an a-model!

If M |= T2 and a ∈ M, let pa(x) ∈ S(M) be the type that says:“E (x , a)” but “x 6= c” for every c ∈ M.

pa ⊥ pb iff M |= ¬E (a, b). Therefore, if |M/E | = κ, then there areκ pairwise orthogonal types in S(M).

Therefore, T2 is multidimensional (or “T2 is not nmd”).

Not hard to show: T2 does not have SB.

More examples

Other stable multidimensional theories: Separably closed fields;differentiably closed fields.

Nomadic types, but nonmultidimensional

There is a superstable T3 which is non-multidimensional (nmd) butwhich has a nomadic type.

Let T3 be the theory of refining equivalence relations {Ei : i ∈ N}(so Ei+1 ⇒ Ei ) such that E0 has two classes and each Ei -classsplits into two Ei+1-classes.

T3 is complete, has q.e., and is superstable and nmd.

If p(x) and q(x) are nonalgebraic 1-types, p ⊥ q iff they belong todifferent E∞-classes (E∞ =

⋂i∈N Ei ), i.e. iff p 6= q.

If p is any nonalgebraic 1-type and f ∈ Aut(C) permutes theEi -classes in a single cycle of order 2i+1, thenp ⊥ f (p) ⊥ f 2(p) ⊥ ...., so p is nomadic.

SB for a-saturated models

Theorem

(G., Laskowski) If T is superstable, then the following areequivalent:

1 The a-models of T have the SB property,

2 T has no nomadic types,

3 there is no infinite collection of a-models of T which arepairwise bi-embeddable but pairwise non-isomorphic.

The only case this leaves out is when T is “strictly stable” (stablebut not superstable), since we already know that unstable T doesnot have SB for κ-saturated models.

SB for a-saturated models

Theorem

(G., Laskowski) If T is superstable, then the following areequivalent:

1 The a-models of T have the SB property,

2 T has no nomadic types,

3 there is no infinite collection of a-models of T which arepairwise bi-embeddable but pairwise non-isomorphic.

The only case this leaves out is when T is “strictly stable” (stablebut not superstable), since we already know that unstable T doesnot have SB for κ-saturated models.

No nomadic types ⇔ SB for a-models

To characterize SB for saturated models in superstable theories, weuse a-prime models and regular types.

Regular types have a nice dimension theory (pregeometry ormatroid), just like the strongly minimal sets in Adler’s tutorial,given by nonforking.

We for p a regular type and M ≺ N two a-models, we can definedim(p,M,N), the dimension of the realizations of p in N over M,and this has the expected properties.

a-prime models

If A ⊆ M |= T , we say that M is a-prime over A if:

1 M is an a-model; and

2 Any embedding f : A→ N into an a-model of T has can beextended to a map g : M → N.

Theorem

(Shelah; T stable) If A ⊆ M |= T is an a-model, then there is amodel N such that A ⊆ N ≺ M and N es a-prime over A.Furthermore, N is unique up to isomorphisms fixing A.

Main lemma

If T is superstable and nmd, then there are two a-prime modelsM0 ≺ M and a family 〈Ij : J ∈ S〉 of infinite M0-indiscerniblesequences such that:

1 M is a-prime over M0 ∪⋃

j∈S Ij ;

2 For any a-model N � M, there are M0-indiscernible sequencesI ′j such that Ij ⊆ I ′j ⊆ N and N is a-prime over M0 ∪

⋃j∈S I ′j ;

3 If pj = tp(aj/M0) for every aj ∈ Ij , then the types 〈pj : j ∈ S〉are pairwise orthogonal.

Note: the cardinals 〈|I ′j | : j ∈ S〉 determine the isomorphism typeof N.

Nomadic types ⇒ failure of SB

Suppose T is stable with a nomadic type p ∈ S(M), witnessed byp ⊥ f (p) ⊥ f 2(p) ⊥ . . ..

We can construct an infinite collection of pairwise bi-embeddablea-models M1,M2, . . . using properties of a-prime models (similar tothe example of infinitely many infinite equivalence relations).

Nomadic types ⇒ failure of SB

Suppose T is stable with a nomadic type p ∈ S(M), witnessed byp ⊥ f (p) ⊥ f 2(p) ⊥ . . ..

We can construct an infinite collection of pairwise bi-embeddablea-models M1,M2, . . . using properties of a-prime models (similar tothe example of infinitely many infinite equivalence relations).

No nomadic types ⇒ SB for a-saturated models

Suppose T is superstable with no nomadic types.

If f : N1 → N2 and g : N2 → N1 are embeddings of a-models, wewant to just say (for any regular type p):

dim(p,N1) ≤ dim((g ◦ f )(p),N1) ≤ dim((g ◦ f )2(p),N1) ≤ . . . ;

...then, since (g ◦ f )n(p) is nonorthogonal to p for some n (“nonomadic types”), these dimensions must all be equal.

Technicality Important detail: g ◦ f might move the “base model”M0 ≺ N1.

This argument needs the “ubiquity of regular types” insuperstable theories.

Fixing a base model

If T is superstable and nmd, then any elementary extension of ana-saturated model is a-saturated.

Theorem

The following are equivalent:

1 There is some expansion of T by constants which is SB;

2 If M |= T is a-saturated, T has SB after adding constants forelements in M;

3 T is superstable and nmd.

Proof.

Use the above lemma and the fact that an nmd T will not havenomadic types after we add constants for M.

Fixing a base model

Theorem

Proof.

Fixing a base model

Theorem

Proof.

Fixing a base model

Theorem

Proof.

Which constants do we need to add?

Conjecture

If T is superstable, countable, nmd, and NOTOP, then T has SBafter adding constants for any model.

Interesting questions

Question

If T is strictly stable and has no nomadic types, then do itsℵ1-saturated models have SB?

Question

If T has SB, then is it true that any expansion of T by constantsstill has SB?

Question

If T does not have SB, then there is an infinite collection ofnon-isomorphic, pairwise-bi-embeddable models of T ?

Question

Embarassing questions

Question

If T is stable, non necessarily superstable, M is |T |+-saturated,and p ∈ S(M) is a nonalgebraic type, then is there a regular typeq ∈ S(M) nonorthogonal to p?

Question

Is the SB property absolute over forcing extensions of the universeof set theory V ?

Yes for some special cases, e.g. if T is weakly minimal, T is totallytranscendental...

Question

References

John Goodrick, When are elementarily bi-embeddable structuresisomorphic?, 2007, Berkeley.

John Goodrick and Michael Laskowski, “The Schroder-Bernsteinproperty for a-saturated models,” to appear in The Proceedings of theAMS, 2012, arxiv:1202.6535.

John Goodrick and Michael Laskowski, “The Schroder-Bernsteinproperty for weakly minimal theories,” The Israel Journal ofMathematics vol. 188 (2012) no. 1, 91-110, arXiv:math/0912.1363v1.

T. A. Nurmagambetov, “The mutual embeddability of models,” fromThe Theory of Algebraic Structures (in Russian), Karagand. Gos.Univ. 1985.

Saharon Shelah, Classification Theory (second edition), 1990,North-Holland.

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