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Most(?) theories have a Borel complete reduct
Chris LaskowskiUniversity of Maryland
Joint work withDouglas Ulrich
29 October, 2020
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Teasers:
(Z,+) has a Borel complete reduct.Theory of ‘independent unary predicates’ has a Borel completereduct. [L = {Un(x) : n ∈ ω}, T says every booleancombination is consistent.]
(Sanity check:) Theory of ‘infinite sets’ in language of equalitydoes not have a Borel complete reduct. ACF does not either.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Borel complexity (to be defined later) is a measure of thecomplexity of a theory T in a countable language.
Borel complete means of maximal complexity, i.e., there is a Borelembedding of countable graphs into countable models of T .
Examples: Real closed fields (RCF) and differentially closed fields(DCF0) are Borel complete, but algebraically closed fields (ACF) isnot Borel complete.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Reducts
In model theory, a reduct of an L-structure M is like a ‘forgetfulfunctor’.
Example: (Z,+) is a reduct of (Z,+,×).
Definition
An L0-structure M0 is a reduct of an L-structure M if M0 and Mhave the same universes and every L0-definable subset of M0(without parameters) is L-definable in M (without parameters).
Equivalently, an L0-structure M0 is a reduct of an L-structure M ifthere is an expansion by definitions of M to an L0 ∪ L-structure M∗and M0 is obtained by deleting symbols from M
∗ as above.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Some reducts of (Z,+)
Note that for each n ≥ 1,
Dn = {x ∈ Z : ∃y(x = y + y + · · ·+ y) [n times]}
is definable in (Z,+).
Thus, (Z,Un)n≥1 (with unary predicates Un interpreted as Dn) is areduct of (Z,+).
Moreover, (Z,Up)p prime is a reduct of (Z,Un)n≥1, and hence of(Z,+).
Note: Th(Z,Up)p prime is ‘independent unary predicates’. We willsee that it has a Borel complete reduct.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Reducts (generally) have lower complexity
If M is decidable, then every reduct M0 (to a reasonablelanguage) is decidable. e.g., (Z,+) is a reduct of(Z,+, 0, 1,
Spaces of countable models
Fix L a countable language. Let
StrL = {L-structures M with universe ω}
Topologize: Basic open sets
Uφ(n1,...,nk ) = {M ∈ StrL : M |= φ(n1, . . . , nk)}
StrL is a standard Borel space (separable, completely metrizable ofsize continuum).
For T any L-theory, Mod(T ) = {M ∈ StrL : M |= T} is a Borelsubset.
Sym(ω) acts on StrL by σ.M = the L-structure M′ formed by
permuting ω. Note: (Mod(T ),∼=) is invariant under this action.Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
How complicated is (Mod(T ),∼=)?
Friedman-Stanley Given two theories T ,S (possibly in differentcountable languages L, L′) we say (Mod(T ),∼=) is Borel reducibleto (Mod(S),∼=), T ≤B S , if there is a Borel f : StrL → StrL′ suchthat for all M,N ∈ Mod(T ), f (M), f (N) ∈ Mod(S) and
M ∼= N ⇐⇒ f (M) ∼= f (N)
There is a maximal ≡B -class, containing graphs, linear orders,RCF, DCF0. A theory T is Borel complete if (Mod(T ),∼=) is inthis maximal class.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Families of equivalence relations
Let L = {En : n ∈ ω}, T asserts each En is an equivalence relation.
Old: [Ulrich-Rast-L] REF(bin)=‘binary splitting, refiningequivalence relations’ is not Borel complete, but the isomorphismrelation ⊆ Mod(T )×Mod(T ) is not Borel.
Theorem (L-Ulrich)
Let h : ω → ω \ {0} be any strictly increasing function. Let Th bethe L-theory asserting that:
Each En is an equivalence relation with h(n) classes;
The En’s cross-cut i.e., for any finite F ⊆ ω,EF (x , y) :=
∧n∈F En(x , y) has Πn∈Fh(n) classes.
Then Th is Borel complete.
Pf: www.math.umd.edu/∼laskow/Crosscutting.pdfChris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
(Z,+) has a Borel complete reduct
Corollary
(Z,Up)p prime (and hence (Z,+)) has a Borel complete reduct.
Pf: Partition Primes=⊔{Fn : n ∈ ω}, where each Fn is finite, but
|Fn+1| > |Fn| for all n. For each n, let
En(x , y) :=∧p∈Fn
Up(x)↔ Up(y)
So (Z,En)n∈ω is a reduct of (Z,Up)p prime (and hence of (Z,+)).Each En is an equivalence relation with 2
|Fn| classes, and the family{En} is cross-cutting. By Theorem, (Z,En)n∈ω is Borel complete.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Extensions
Theorem
Let M be any structure (in a countable language) withuncountably many 1-types consistent with Th(M). Then M has aBorel complete reduct.
Corollary
Suppose Th(M) is not small (i.e., for some n ≥ 1, there areuncountably many n-types). Then Meq has a Borel completereduct.
Corollary
If Th(M) is not ω-stable, then the elementary diagram ElDiag(M)has a Borel complete reduct.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Further extensions
The previous results are not characterizations:
REF(bin) is small and not Borel complete, but has a Borelcomplete reduct. [Build a tree of strong types.]
There are extremely nice 2-dimensional, ω-stable structures Mthat have Borel complete reducts. [Start with the ‘standardeni-DOP example’ M0. Form an expansion M by adding afamily of commuting functions enforcing that any two boxeshave the same cardinality.]
But... if M is ω-stable of Morley rank 1 [e.g., (C,+,×)] thenevery reduct has only countably many countable models(hence is not Borel complete).
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
λ-Borel sets and functions
Not all Borel complete theories have the same complexity!
Fix L a countable language and λ an infinite cardinal. Let
StrλL = {L-structures M with universe λ}
Topologize: Basic open sets
Uφ(α1,...,αk ) = {M ∈ StrλL : M |= φ(α1, . . . , αk)}(α1, . . . , αk) ∈ λk
Call a set X λ-Borel if it is in the (≤ λ)-algebra generated by theopen sets, and a function f : X → Y λ-Borel if the inverse imageof every basic open set is λ-Borel.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
λ-Borel reducibility
Given two theories T , S (possibly in different countable languagesL, L′) we say T is λ-Borel reducible to S if there is a λ-Borelf : StrλL → StrλL′ such that for all M,N |= T with universe λ,f (M), f (N) |= S and
M ≡λ+,ω N ⇐⇒ f (M) ≡λ+,ω f (N)
There is a maximal ≡λB -class, containing graphs, linear orders,RCF, DCF0. A theory T is λ-Borel complete if T is in thismaximal class.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Borel complete does not imply λ-Borel complete for large λ
Fact: If T is weakly minimal, then T is not λ-Borel complete forλ ≥ i2. [Every model N is back-and-forth equivalent to asubmodel of size at most 2ℵ0 .]
Corollary
Suppose h : ω → ω \ {0} is strictly increasing and Th is the theoryof cross-cutting equivalence relations where En has h(n) classes.Then Th is Borel complete, but not λ-Borel complete for λ ≥ i2.
Note: A more complicated example of a Borel complete T that isnot λ-Borel complete for large λ appears in [Ulrich-Rast-L].
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Classifying cross-cutting equivalence relations
Let L = {En : n ∈ ω} and let h : ω → (ω \ {0, 1}) ∪ {∞}.Th := Theory of cross-cutting equivalence relations, where En hash(n) classes.Model theory facts:
Every Th admits QE and is stable.
If h(n)
Three test questions for Th
A. Isomorphism on Mod(Th)×Mod(Th) is not Borel?
⇑
B. Th is Borel complete?
⇑
C. Th is λ-Borel complete for all infinite λ?
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
All h(n)
h(n) =∞ for exactly one n.
Case: h(0) =∞, but h(n) 0. [Th is superstable ofR∞-rank 2.]
Theorem
1 Iso on Mod(Th) is not Borel iff h(n) > 2 for infinitely many n.
2 Th is Borel complete iff {h(n) : n > 0} is unbounded.3 Th is not λ-Borel complete for large λ.
Exactly the same result!
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
h(n) =∞ for at least two n.
Case: h(n) =∞ for at least two n’s.
Theorem
Th is λ-Borel complete for all λ.
Pf: Use two relations with infinitely many classes to code eni-DOP.
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct
Thanks for listening!
Proof of main theorem atwww.math.umd.edu/∼laskow/Crosscutting.pdf
Chris Laskowski University of Maryland Joint work with Douglas Ulrich
Most(?) theories have a Borel complete reduct