The classification problem for extensions of torsion abelian groups
aa r X i v : . [ m a t h . G R ] F e b THE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS
MARTINO LUPINI
Abstract.
Given countable abelian groups
C, A , with C torsion, we compute the potential complexity class of theclassification problem for extensions of C by A . In particular, we show that such a problem can have arbitrarily highpotential complexity. Toward this goal, we further develop the theory of groups with a Polish cover, in particularby showing that they form an abelian category. Introduction
In this paper we study the complexity of classifying extensions of two given countable abelian groups up toequivalence. (In what follows, we assume all the groups to be abelian and additively denoted.) Suppose that A and C are countable groups. An extension of C by A is a short exact sequence0 → A → X → C → → A → X → C → → A → X ′ → C → equivalent if there exists a group isomorphism ψ : X → X ′ that makes the following diagram commute. A X CA X ′ C ψ We will explain below that one can regard the space
Ext ( C, A ) of extensions of C by A as a Polish space. Thus,the relation of equivalence of extensions of C by A is an equivalence relation R Ext ( C,A ) on Ext ( C, A ).We will study the relations R Ext ( C,A ) from the perspective of Borel complexity theory. This framework allowsone to compare the complexity of different classification problems in mathematics. A classification problem isidentified with a pair ( X, R ) where X is a Polish space and R is an equivalence relation on X , which we will assumeto be Borel as a subset of X × X . The notion of Borel reducibility captures the idea that a classification problemis at most as complicated as another one.
Definition 1.1.
Suppose that ( X, R ) and ( Y, S ) are Borel equivalence relations on Polish spaces. A Borel reduction from R to S is a Borel function f : X → Y satisfying f ( x ) S f ( y ) ⇔ x R y for every x, y ∈ X . We say that R is Borel reducible to S , and write R ≤ S , if there is a Borel reduction from R to S . We say that R is Borel bireducible to S if R ≤ S and
S ≤ R .While Borel reducibility provides a way to compare different classification problems, some important Borelequivalence relations serve as benchmarks to calibrate the complexity of other classification problems. Particularly,an equivalence relation R is: • smooth if it is Borel reducible to the relation of equality on some Polish space; • essentially hyperfinite if it is Borel reducible to the relation E of tail equivalence of binary sequences; Date : February 9, 2021.2000
Mathematics Subject Classification.
Primary 20K10, 20K35, 54H05; Secondary 20K40, 20K45.
Key words and phrases.
Group extension, Polish group, pure extension, Borel complexity theory, potential complexity, Borel re-ducibility, cotorsion functor, group with a Polish cover.The author was partially supported by a Marsden Fund Fast-Start Grant VUW1816 from the Royal Society Te Ap¯arangi. • essentially countable if it is Borel reducible to a Borel equivalence relation whose equivalence classes arecountable.The relation E is by definition the Borel equivalence relation on C := { , } ω defined by setting, for ( x i ) , ( y i ) ∈ C ,( x i ) E ( y i ) ⇔ ∃ n ∀ i ≥ n , x i = y i . We then let E ω to be the equivalence relation on C ω defined by setting, for( x i ) , ( y i ) ∈ C ω , ( x i ) E ω ( y i ) ⇔ ∀ i ∈ ω ( x i E y i ).As R Ext ( C,A ) is a Borel equivalence relation, by the Glimm–Effros Dichotomy [HKL90], R Ext ( C,A ) is not smoothif and only if E ≤ R Ext ( C,A ) . Furthermore, R Ext ( C,A ) is the orbit equivalence relation induced by a continuousaction of a non-Archimedean abelian Polish group. Therefore, by [DG17, Theorem 6.1, Corollary 6.3] and [HK97,Theorem 8.1], R Ext ( C,A ) is not essentially hyperfinite if and only if it is not essentially countable if and only if E ω ≤ R Ext ( C,A ) . In the following we will completely characterize the pairs A, C of countable abelian groups with C torsion such that R Ext ( C,A ) is smooth, bireducible with E , and bireducible with E ω , respectively.Recall that an abelian group A has a largest divisible subgroup D ( A ), which is a direct summand of A . We saythat A is reduced if D ( A ) = 0, and bounded if nA = 0 for some n ≥
1, where nA = { nx : x ∈ A } . For a prime p ,the p - primary subgroup of A is the subgroup consisting of elements of order p n for some n ∈ ω . The Ulm subgroups u α ( A ) = A α for α < ω are defined recursively by setting A = A and A α = T β<α T n ∈ ω nA β for 0 < α < ω . Theorem 1.2.
Suppose that
C, A are countable abelian groups, with C torsion. For a prime number p , we let A p ⊆ A/D ( A ) be the p -primary subgroup of A/D ( A ) and C p ⊆ C be the p -primary subgroup of C .(1) R Ext ( C,A ) is smooth if and only if, for every prime number p , either u ( C p ) = 0 or A p is bounded;(2) R Ext ( C,A ) is essentially hyperfinite if and only if for every prime number p , either u ( C p ) = 0 or u ( A p ) =0 , and the sum of | u ( C p ) | ranging over all prime numbers p such that A p is unbounded is finite;(3) R Ext ( C,A ) ≤ E ω if and only if, for every prime number p , one of the following holds: (a) u ( C p ) = 0 ; (b) u ( A p ) = 0 ; (c) u ( C p ) = 0 and u ( A p ) is bounded. We will obtain a generalization of Theorem 1.2 that completely determines the potential complexity of the relation R Ext ( C,A ) . The notion of potential complexity of a Borel equivalence relation on a Polish space was introduced byLouveau in [Lou94]; see also [Kec02, HK96, HKL98] and [Gao09, Definition 12.5.1]. A complexity class of sets Γ isa function X Γ ( X ) that assigns to each Polish space X a collection Γ ( X ) of Borel subsets of X , such that if f : X → Y is a continuous function and A ∈ Γ ( Y ) then f − ( A ) ∈ Γ ( X ). If A ∈ Γ ( X ) we also say that A is Γ in X .Following [HKL98], for a complexity class Γ, we let D (Γ) be the complexity class consisting of differences betweensets in Γ, and ˇΓ be the dual complexity class consisting of complements of elements of Γ. We denote by ˇ D (Γ) thedual class of D (Γ). We will mainly be interested in the complexity classes Σ α , Π α , and D (cid:0) Π α (cid:1) for α ∈ ω ; see[Kec95, Section 11.B]. Definition 1.3.
Let Γ be a complexity class. Suppose that ( X, R ) is a Borel equivalence relation on a Polish space.Then R is potentially Γ if there exists a Borel equivalence relation ( Y, S ) such that S ∈
Γ ( Y × Y ) and R is Borelreducible to S .The concept of potential complexity affords us to measure the complexity of an equivalence relation. Thebenchmarks considered above can be expressed in terms of potential complexity as follows. Let ( X, R ) be a Borelequivalence relation on a Polish space. Then R is smooth if and only if it is potentially Π if and only if itis potentially Π [Gao09, Lemma 12.5.3]. Suppose that R is the orbit equivalence relation associated with acontinuous action of a non-Archimedean abelian Polish group. Then R is essentially hyperfinite if and only if R ispotentially Σ if and only if R is potentially Σ by [HK96, Theorem 3.8], [HKL98, Theorem 4.1(ii)], and [DG17,Theorem 6.1]; see also [Gao09, Theorem 12.5.7]. Furthermore, R ≤ E ω if and only if R is potentially Π [All20,Corollary 6.11]. Definition 1.4.
Let Γ be one of the classes Π µ , Σ µ , D ( Π µ ), D ( Σ µ ), ˇ D ( Π µ ), ˇ D ( Σ µ ) for some countable ordinal µ ≥
1. Let ( X, R ) be a Borel equivalence relation on a Polish space. Then Γ is the potential class of R if R ispotentially Γ and R is not potentially ˇΓ.Theorem 1 in [HKL98] imposes restrictions on the possible values of the potential class of an orbit equivalencerelation associated with a continuous action of a non-Archimedean Polish group. HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 3
In view of the above remarks, the following result can be seen as an extension of Theorem 1.2. To simplify thestatement, we will assume that, for some prime number p , u ( C p ) is nonzero and A p is unbounded. If this fails,then Theorem 1.2 applies. We denote by P the set of primes. Theorem 1.5.
Suppose that
C, A are countable abelian groups, where C is torsion. For p ∈ P , let A p ⊆ A/D ( A ) be the p -primary subgroup of A/D ( A ) and C p ⊆ C be the p -primary subgroup of C . Assume that, for some p ∈ P , u ( C p ) is nonzero and A p is unbounded. For every p ∈ P , let µ p be the least countable ordinal such that either C µ p = 0 or µ p is the successor of µ p − and A µ p − = 0 . Set µ := sup p ∈ P µ p . If µ is the successor of µ − , set P max = (cid:8) p ∈ P : µ p = µ (cid:9) and W := X (cid:8)(cid:12)(cid:12) C µ − p (cid:12)(cid:12) : p ∈ P max (cid:9) .If furthermore µ − is the successor of µ − , set w := X (cid:8)(cid:12)(cid:12) C µ − p (cid:12)(cid:12) : p ∈ P max , A µ − p is unbounded (cid:9) .Then we have that:(1) Π µ is the complexity class of R Ext ( C,A ) if and only if either (a) µ is a limit ordinal, or (b) µ = 1 + λ + n where λ is either zero or limit, ≤ n < ω , and w = 0 ;(2) Σ µ is the complexity class of R Ext ( C,A ) if and only if µ = 1 + λ + 1 where λ is either zero or limit and W < ∞ ;(3) D (cid:0) Π µ (cid:1) is the complexity class of R Ext ( C,A ) if and only if µ = 1 + λ + n where λ is either zero or limit, ≤ n < ω , and < w < ∞ ;(4) Π µ +1 is the complexity class of R Ext ( C,A ) if and only if either (a) µ = 1 + λ + 1 where λ is either zero orlimit and W = ∞ , or (b) µ = 1 + λ + n where λ is either zero or limit and ≤ n < ω and w = ∞ .In particular, R Ext ( C,A ) is Π µ +1 and not Π α for α < µ . Corollary 1.6.
Let Z ( p ∞ ) be the Pr¨ufer p -group. Suppose that A is a countable unbounded reduced p -group of Ulmrank λ + n , where λ is either zero or limit and n ∈ ω . Then we have that:(1) Π λ + n +1 is the complexity class of R Ext ( Z ( p ∞ ) ,A ) if and only if n ≥ and A λ + n − is bounded;(2) Π λ + n +2 is the complexity class of R Ext ( Z ( p ∞ ) ,A ) if and only if either n = 0 , or n ≥ and A λ + n − isunbounded. Towards the proof of Theorem 1.5 and Theorem 1.2, we further develop the theory of groups with a Polishcover, originally introduced in [BLP20]. In particular, we show that groups with a Polish cover form an abeliancategory, whose morphisms are the definable group homomorphisms. We define canonical Polishable subgroups ofa group with a Polish cover indexed by countable ordinals, which we call Solecki subgroups as they correspond tothe canonical approximation of a Polishable subgroup of a Polish group introduced by Solecki in [Sol99]. We alsostudy the complexity of Polishable subgroups of groups with a Polish cover, and in particular of Solecki subgroups,building on results from [FS06, HKL98].For countable abelian groups
C, A , the group Ext (
C, A ) classifying extensions of C by A can be seen as a groupwith a Polish cover as in [BLP20]. The main ingredient in the proof of Theorem 1.5 and Theorem 1.2 is that,for groups C and A as in the statement, the (1 + α )-th Ulm subgroup of Ext ( C, A ) is equal to the α -th Soleckisubgroup of Ext ( C, A ) for α < ω . This is obtained by applying homological results about the functors C C α due to Nunke [Nun67].The rest of this paper is divided into three sections. Section 2 presents the notions of definable set and definablegroup, group with a Polish cover, and introduces in this context the notions of Polishable subgroup, Solecki subgroup,and their complexity. Section 3 recalls the definition of the group Ext and how it can be regarded as a group with aPolish cover, and presents in this context the definable content of the results about cotorsion functors from [Nun67].These results are used in Section 4 to describe the Solecki subgroups of Ext in terms of the Ulm subgroups, andthus establish Theorem 1.5 and Theorem 1.2. Acknowledgments.
We are grateful to Alexander Kechris and Andr´e Nies for their comments on a preliminaryversion of this paper.
MARTINO LUPINI Groups with a Polish cover and their Solecki subgroups
In this section, we begin with recalling the notion of definable set and definable group from [Lup20b, Lup20a],and the notion of group with a Polish cover from [BLP20]. We introduce the notion of Polishable subgroup of agroup with a Polish cover, and show that groups with a Polish cover form an abelian category. We then describe acanonical chain of Polishable subgroups of a group with a Polish cover, which we call
Solecki subgroups , originallydefined by Solecki in [Sol99] and also considered in [FS06, Sol09]. Building on [FS06], we present some resultsconcerning the complexity of such subgroups. We also consider the notion of non-Archimedean group with a Polishcover, and show that non-Archimedean groups with a Polish cover form an additive subcategory of the category ofgroups with a Polish cover. We conclude with discussing the
Ulm subgroups of a group with a Polish cover. Recallthat we assume all the groups to be abelian and additively denoted.2.1.
Definable sets and groups.
We present here the notion of definable set and definable group as in [Lup20b,Lup20a]. We begin with recalling the notion of idealistic equivalence relation from [Kec94]; see also [Gao09,Definition 5.4.9] and [KM16]. We will consider as in [Lup20b] a slightly more generous notion. Recall that a σ -filteron a set C is a nonempty family F of nonempty subsets of C that is closed under countable intersections and suchthat A ⊆ B ⊆ C and A ∈ F implies B ∈ F . Definition 2.1.
Suppose that X is a standard Borel space, and E is an equivalence relation on X . We say E is idealistic if there exist a Borel function s : X → X and an assignment C → F C mapping each E -class C to a σ -filter F C on C such that, for every Borel subset A ⊆ X × X , (cid:8) x ∈ X : { x ′ ∈ X : ( s ( x ) , x ′ ) ∈ A } ∈ F [ x ] E (cid:9) is a Borel subset of X .The term idealistic is due to the fact that it can be equivalently defined in terms of σ -ideals, in view of theduality between σ -ideals and σ -filters. If X is a Polish space endowed with a continuous action of a Polish group G , then the corresponding orbit equivalence relation E XG is idealistic [Gao09, Proposition 5.4.10].A definable set is a pair ( ˆ X, E ) where ˆ X is a standard Borel space and E is a Borel and idealistic equivalencerelation on ˆ X . We denote such a definable set by X = ˆ X/E , as we think of it as an explicit presentation of theset X as a quotient of the Polish space ˆ X by the “well-behaved” equivalence relation E . We identify a standardBorel space ˆ X with the definable set X = ˆ X/E where E is the identity relation on ˆ X . A subset Z of a definableset X = ˆ X/E is Borel if ˆ Z := { x ∈ ˆ X : [ x ] E ∈ Z } is a Borel subset of ˆ X , in which case Z is itself a definable set.Suppose that X = ˆ X/E and Y = ˆ Y /F are definable sets, and f : X → Y is a function. A lift ˆ f of f is a functionˆ f : ˆ X → ˆ Y such that f ([ x ] E ) = [ ˆ f ( x )] F for every x ∈ X . In this case, we also say that f is induced by ˆ f . Definition 2.2.
Suppose that X = ˆ X/E and Y = ˆ Y /F are definable sets, and f : X → Y is a function. Then f is Borel-definable (or, briefly, definable ) if it has a lift that is a Borel function.The category of definable sets has definable functions as morphisms. This category contains the category ofstandard Borel spaces and Borel functions as a full subcategory, and it satisfies natural generalizations of severalgood properties of the latter. We recall here the most salient ones; see [Lup20b, Proposition 1.10] and referencestherein.
Proposition 2.3 (Kechris–Macdonald [KM16]) . Suppose that X and Y are definable sets. If f : X → Y is adefinable injection, and A ⊆ X is Borel, then f ( A ) ⊆ Y is Borel, and the inverse function f − : f ( A ) → Y isdefinable. Proposition 2.4 (Motto Ros [MR12]) . Suppose that X and Y are definable sets. If there exist a definable injection X → Y and a definable injection Y → X , then there exists a definable bijection X ↔ Y . If X = ˆ X/E and Y = ˆ Y /F are definable sets, then the product X × Y in the category of definable sets is thedefinable set ( ˆ X × ˆ Y ) / ( E × F ), where ( x, y ) ( E × F ) ( x ′ , y ′ ) ⇔ xEx ′ and yF y ′ . (One can verify that E × F is Boreland idealistic whenever both E and F are Borel and idealistic.)A definable group is simply a group object in the category of definable sets in the sense of [ML98, Section III.6].Explicitly, a definable group is a definable set G = ˆ X/E that is also a group, and such that the group operation
HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 5 G × G → G and the function G → G mapping each element to its inverse are definable. Notice that every standardBorel group is, in particular, a definable group. Definition 2.5.
Suppose that X = ˆ X/E is a definable set, and Γ is a complexity class. Fix n ≥ n -aryBorel relation R ⊆ X n . We say that R is Γ-definable if there exists a Polish topology τ on ˆ X that induces its Borelstructure, and such that { ( x , . . . , x n ) ∈ ˆ X : ([ x ] E , . . . , [ x n ] E ) ∈ R } belongs to Γ( ˆ X n ), where ˆ X n is endowed withthe product topology induced by τ . Remark 2.6.
By definition, the relation of equality = X on X = ˆ X/E is Γ-definable if and only if the equivalencerelation E on ˆ X is potentially Γ in the sense of Definition 1.3. Remark 2.7.
Suppose that
X, Y are definable sets, R X ⊆ X n and R Y ⊆ Y n are Borel relations, and f : X → Y is a definable function such that, for x , . . . , x n ∈ X , ( x , . . . , x n ) ∈ R X ⇔ ( f ( x ) , . . . , f ( x n )) ∈ R Y . The sameproof as [Gao09, Lemma 12.5.4] shows that if R Y is Γ-definable, then R X is Γ-definable.More generally, one can consider sets that are presented as X = ˆ X/E where ˆ X is a standard Borel space and E is a analytic equivalence relation on ˆ X that is not necessarily Borel or idealistic. In this case, we say that X is a semidefinable set . The notion of definable function between semidefinable sets can be formulated as in the case ofdefinable sets. Naturally, a semidefinable group is a group object in the category of semidefinable sets and definablefunctions.An important example of semidefinable group that is not a definable group is R ω /E , where R ω is the productof countably many copies of the Polish group R , and E is the equivalence relation on R ω obtained by setting( x i ) E ( y i ) ⇔ ∃ n ∀ i ≥ n, x i = y i . Notice that, if R ( ω ) ⊆ R ω is the subgroup consisting of sequences that areeventually zero, then E is the coset equivalence relation associated with R ( ω ) . Thus, R ω /E can be seen as thequotient group R ω / R ( ω ) . It is proved in [KL97, Theorem 4.1] that if X is a definable set, then there is no definableinjection R ω / R ( ω ) → X .2.2. Groups with a Polish cover.
We now recall the notion of group with a Polish cover introduced in [BLP20].
Definition 2.8. A group with a Polish cover is a definable group given as a quotient ˆ G/N where ˆ G is a Polishgroup and N ⊆ ˆ G is a Polishable subgroup. This means that N is a Borel subgroup of ˆ G such that there is a Polishgroup topology on N that induces the Borel structure inherited from ˆ G or, equivalently, there exist a Polish group H and a continuous homomorphism ψ : H → ˆ G with image N . For x, y ∈ ˆ G , we write x ≡ y mod N if x − y ∈ N .We regard a Polish group G as a group with a Polish cover ˆ G/N where G = ˆ G and N = { } . If G k = ˆ G k /N k isa group with a Polish cover for k ∈ ω , we define G ⊕ G to be group with a Polish cover ( ˆ G ⊕ ˆ G ) / ( N ⊕ N ),and Q k ∈ ω G k to be the group with a Polish cover ˆ G/N where ˆ G = Q k ˆ G k and N = Q k N k .As a group with a Polish cover is, in particular, a definable group, the notion of definable homomorphism betweengroups with a Polish cover is a particular instance of the notion of definable group homomorphism between definablegroups. Definition 2.9.
Suppose that G = ˆ G/N G and H = ˆ H/N H are groups with a Polish cover. A group homomorphism f : G → H is a definable if it has a Borel lift ˆ f : ˆ G → ˆ H .It follows from Proposition 2.3 that if a definable homomorphism G → H is bijective, then its inverse H → G isalso a definable homomorphism. In what follows, we consider groups with a Polish cover as objects of a categorythat has definable homomorphisms as morphisms.Recall that a Polish group is non-Archimedean if it has a basis of the identity consisting of open subgroups; see[DG17, Proposition 2.1] for other characterizations. Definition 2.10.
Suppose that G = ˆ G/N is a group with a Polish cover. Then we say that G is non-Archimedeanif ˆ G and N are non-Archimedean Polish groups.2.3. Polishable subgroups.
We now introduce in the context of groups with a Polish cover the notion of Boreland Polishable subgroup.
Definition 2.11.
Suppose that G = ˆ G/N is a group with a Polish cover, and H ⊆ G is a subgroup. Defineˆ H = { x ∈ ˆ G : x + N ∈ H } ⊆ ˆ G . Then we say that: MARTINO LUPINI • H is a Borel subgroup of G if ˆ H is a Borel subgroup of ˆ G ; • H is a Polishable subgroup of G if ˆ H is a Polishable subgroup of ˆ G ; • H is a non-Archimedean Polishable subgroup of G if ˆ H is a Polishable subgroup of ˆ G and ˆ H is a non-Archimedean Polish group.Notice that if H = ˆ H/N is a Polishable subgroup of a group with a Polish cover G = ˆ G/N , then H = ˆ H/N and
G/H = ˆ G/ ˆ H are also groups with a Polish cover. Lemma 2.12.
Suppose that G is a group with a Polish cover, and k ≥ . Let ( G n ) n ∈ ω be a Polishable subgroupsof G . Then kG = { kx : x ∈ G } , G + G , and T n ∈ ω G n are Polishable subgroups of G .Proof. Write G = ˆ G/N . For every n ∈ ω , we have that G n = ˆ G n /N for some Polishable subgroup ˆ G n of ˆ G . Wehave that { x ∈ ˆ G : x + N ∈ kG } = k ˆ G + N is the image of the Polish group ˆ G ⊕ N under the continuous homomorphism ˆ G ⊕ N → ˆ G , ( x, z ) kx + z .We also have that { x ∈ ˆ G : x + N ∈ G + G } = ˆ G + ˆ G + N is the image of the Polish group ˆ G ⊕ ˆ G ⊕ N under the continuous homomorphism ˆ G ⊕ ˆ G ⊕ N → ˆ G , ( x, y, z ) x + y + z .Similarly, we have that { x ∈ ˆ G : x + N ∈ \ n ∈ ω G n } = \ n ∈ ω ˆ G n is the image of the Polish group Z := ( ( x n ) n ∈ ω ∈ Y n ∈ ω ˆ G n : ∀ n ∈ ω, x n = x n +1 ) ⊆ Y n ∈ ω ˆ G n under the continuous homomorphism Z → ˆ G , ( x n ) n ∈ ω x . (cid:3) The same proof as Lemma 2.12 gives the following.
Lemma 2.13.
Suppose that G is a group with a Polish cover, and k ≥ . Let { } and ( G n ) n ∈ ω be non-ArchimedeanPolishable subgroups of G . Then kG = { kx : x ∈ G } , G + G , and T n ∈ ω G n are non-Archimedean Polishablesubgroups of G . Suppose that L is a Borel subgroup of a group with a Polish cover G . Then one can consider the quotient G/L as a semidefinable group. The implication (1) ⇒ (3) in the following proposition can be seen as a reformulation of[Sol09, Theorem 1.1]. Proposition 2.14.
Suppose that L is a Borel subgroup of a group with a Polish cover G . Consider the correspondingsemidefinable group G/L . The following assertions are equivalent:(1) there does not exist a definable injection R ω / R ( ω ) → G/L ;(2)
G/L is a definable group;(3) L is a Polishable subgroup of G , and G/L is a group with a Polish cover.Proof.
Write G = ˆ G/N and let ˆ L = { x ∈ ˆ G : x + N ∈ L } . Since G/L = ˆ G/ ˆ L , after replacing L with ˆ L and G withˆ G , we can assume that G is in fact a Polish group.The implication (3) ⇒ (2) follows from the fact that a group with a Polish cover is, in particular, a definable groupin view of [Gao09, Proposition 5.4.10]. The implication (2) ⇒ (1) follows from [KL97, Theorem 4.1]. Finally, theimplication (1) ⇒ (3) is the content of [Sol09, Theorem 1.1]. (cid:3) We now show that images and preimages of Polishable subgroups under definable homomorphisms are Polishablesubgroups.
Proposition 2.15.
Suppose that
G, H are groups with a Polish cover, and f : G → H is a definable homomorphism.(1) If H is a Polishable subgroup of H , then f − ( H ) is a Polishable subgroup of G ; HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 7 (2) If G is a Polishable subgroup of G , then f ( G ) is a Polishable subgroup of G .Proof. (1) After replacing H with H/H , we can assume that H = { } , in which case f − ( H ) = Ker ( f ). We havethat f induces a definable monomorphism G/ Ker ( f ) → H . Notice that Ker ( f ) is a Borel subgroup of G . Since H is a group with a Polish cover, we have that there does not exist a definable injection R ω / R ( ω ) → H by Proposition2.14. Thus, there does not exist a definable injection R ω / R ( ω ) → G/ Ker ( f ). Thus, Ker ( f ) is a Polishable subgroupof G by Proposition 2.14 again.(2) After replacing G with G and f with its restriction to G , we can assume that G = G . By the first item,Ker ( f ) is a Polishable subgroup of G . Thus, after replacing G with G/ Ker ( f ), we can assume that f is a definablemonomorphism. In this case, we have that f ( G ) is a Borel subgroup of the definable group H by Proposition 2.3.By Proposition 2.14, to conclude the proof it suffices to prove that H/f ( G ) is a definable group.Write G = ˆ G/N and H = ˆ H/N , where ˆ G, ˆ H are Polish groups and N ⊆ ˆ G and N ⊆ ˆ H are Polishablesubgroups. Suppose that ϕ : ˆ G → ˆ H is a Borel function such that ϕ ( x ) + N = f ( x + N ) for x ∈ ˆ G . Define theBorel function c : ˆ G × ˆ G → N by setting c ( x, y ) := ϕ ( y ) − ϕ ( x + y ) + ϕ ( x ). We need to prove that the equivalencerelation E on ˆ H defined by setting xEy ⇔ ∃ ( g, h ) ∈ ˆ G ⊕ N , ϕ ( g ) + h + x = y is idealistic. The argument is similarto the one from [Gao09, Proposition 5.4.10]. We adopt the notation of category quantifiers as in [Kec95, Section16]. For x ∈ ˆ H and A ⊆ [ x ] E , we set A ∈ F [ x ] E ⇔ ∀ ∗ g ∈ ˆ G , ∀ ∗ h ∈ N , ϕ ( g ) + h + x ∈ A . Observe that F [ x ] doesnot depend on the choice of the representative x for the equivalence class [ x ] E . Indeed, if x Ex then there exists( g , h ) ∈ ˆ G × N such that ϕ ( g ) + h + x = x . If A ∈ F [ x ] then ∀ ∗ g ∈ ˆ G , ∀ ∗ h ∈ N , ϕ ( g ) + h + x ∈ A . We havethat ϕ ( g ) + h + x = ϕ ( g ) + h + ϕ ( g ) + h + x = ϕ ( g + g ) + h + h + c ( g, g ) + x .For a fixed ˜ g ∈ ˆ G , if ∀ ∗ h ∈ N , ϕ (˜ g ) + h + x ∈ A then ∀ ∗ h ∈ N , ϕ (˜ g + g ) + h + h + c (˜ g, g ) + x ∈ A . Since h + c (˜ g, g ) ∈ N , this implies that ∀ ∗ h ∈ N , ϕ (˜ g + g ) + h + x ∈ A . Therefore, we have that ∀ ∗ g ∈ ˆ G , ∀ ∗ h ∈ N , ϕ ( g + g ) + h + x ∈ A and hence ∀ ∗ g ∈ ˆ G , ∀ ∗ h ∈ N , ϕ ( g ) + h + x ∈ A . This shows that F [ x ] E is well-defined. Itis easy to verify that F [ x ] E is a filter on [ x ] E . It remains to prove that if A ⊆ ˆ H × ˆ H is Borel, then A F := n x ∈ ˆ H : n y ∈ ˆ H : ( x, y ) ∈ A o ∈ F [ x ] E o is a Borel subset of ˆ H . The argument is the same as in the proof of [Gao09, Theorem 3.3.3]. We have that x ∈ A F ⇔ ∀ ∗ g ∈ ˆ G , ∀ ∗ h ∈ N , ( x, ϕ ( g ) + h + x ) ∈ A . Define the Borel set B := n ( g, h, x ) ∈ ˆ G × N × ˆ H : ( x, ϕ ( g ) + h + x ) ∈ A o .Then we have that x ∈ A F ⇔ ∀ ∗ g ∈ ˆ G, ∀ ∗ h ∈ N, ( g, h, x ) ∈ B ⇔ ∀ ∗ ( g, h ) ∈ ˆ G × N, ( g, h, x ) ∈ B .Since B is Borel, this shows that A F is Borel by [Kec95, Theorem 16.1]. (cid:3) The abelian category of groups with a Polish cover.
Suppose that
G, H are groups with a Polish cover.Then the set of definable homomorphisms G → H is a subgroup of the group of homomorphisms G → H . Thus, thecategory of groups with a Polish cover is an Ab-category [ML98, Section I.8], which is in fact an abelian category byProposition 2.15; see [ML98, Chapter VIII]. In this category, the kernel of a definable homomorphism ϕ : G → H is Ker ( ϕ ) = { x ∈ G : ϕ ( x ) = 0 } ⊆ G , which is a Polishable subgroup of G by Proposition 2.15(1), and the cokernel of ϕ is the quotient map H → H/f ( G ), where f ( G ) is a Polishable subgroup of H by Proposition 2.15(2). The zero object is the trivial group, and the biproduct of G and H is G ⊕ H . A definable homomorphism ϕ : G → H is monic if and only if it is a kernel if and only if it is injective, while it is epic if and only if it is a cokernel, if andonly if it is surjective.Since the category of groups with a Polish cover is an abelian category, one can consider the notion of exactsequence in this category. This is simply an exact sequence in the category of groups that consists of groups witha Polish cover and definable homomorphisms. We refer to such an exact sequence as a definable exact sequence ofgroups with a Polish cover. MARTINO LUPINI
Complexity of Polishable subgroups.
In this section, we consider the complexity of Polishable subgroupsof groups with a Polish cover. We reformulate in this context some results from [HKL98, FS06].
Definition 2.16.
Suppose that H is a Polishable subgroup of a group with a Polish cover G = ˆ G/N . Setˆ H = { x ∈ ˆ G : x + N ∈ H } . Let Γ be a complexity class. We say that H belongs to Γ( G ) or that H is Γ in G if andonly if ˆ H ∈ Γ( ˆ G ), and that Γ is the complexity class of H in G if and only if ˆ H ∈ Γ( ˆ G ) and ˆ H / ∈ ˇΓ( ˆ G ).We say that a group with a Polish cover G = ˆ G/N is a Polish group if { } is a closed subgroup of G , which bydefinition means that N is a closed subgroup of ˆ G . Notice that, if H is a closed subgroup of the group with a Polishcover G , then G/H is a Polish group. If G = ˆ G/N is group with a Polish cover and H is a Polishable subgroupof G , then we let H G be the closed subgroup of G obtained as the closure of H in G with respect to the quotienttopology induced by ˆ G . We say that H is dense in G if H G = G .The following result is an immediate consequence of [HKL98, Proposition 5.1]; see also [Sol09, page 574] for thecase of Π . Recall the notion of Γ-definable Borel relation on a definable set from Definition 2.5. Proposition 2.17 (Hjorth–Kechris–Louveau) . Suppose that G = ˆ G/N is a group with a Polish cover, and H is a Polishable subgroup of G . Let Γ be one of the following complexity classes: Π α , Σ α , D (cid:0) Π α (cid:1) , ˇ D (cid:0) Π α (cid:1) for ≤ α < ω . The following assertions are equivalent:(1) H ∈ Γ( G ) ;(2) the equality relation on G/H is Γ -definable. As a consequence of Remark 2.7, one has the following.
Proposition 2.18.
Suppose that
G, H are groups with a Polish cover, and f : G → H is a definable homomorphism.Let Γ be one of the complexity classes in Proposition 2.17. If H is a Polishable subgroup of H such that H ∈ Γ ( H ) ,then f − ( H ) is a Polishable subgroup of G such that f − ( H ) ∈ Γ ( G ) .Proof. Set G := f − ( H ). Notice that G is a Polishable subgroup of G by Proposition 2.15. Furthermore, f induces a definable monomorphism G/G → H/H . The conclusion thus follows from Proposition 2.17 and Remark2.7. (cid:3) Corollary 2.19.
Suppose that
G, H are groups with a Polish cover, and f : G → H is a definable epimorphism.Let Γ be one of the complexity classes in Proposition 2.17. Suppose that H is a Polishable subgroup of H . Then H ∈ Γ ( H ) if and only if f − ( H ) ∈ Γ ( G ) . The following lemma is a reformulation of [FS06, Corollary 3.4].
Lemma 2.20 (Farah–Solecki) . Suppose that G is a group with a Polish cover and H is a Polishable subgroup of G . If H is Π λ +1 where λ < ω is either zero or limit, then H is Π λ . The following lemma is a consequence of [HKL98, Theorem 4.1 and Theorem 4.2] and Proposition 2.17.
Lemma 2.21 (Hjorth–Kechris–Louveau) . Suppose that G is a group with a Polish cover and H is a non-ArchimedeanPolishable subgroup of G . Suppose that λ < ω is either zero or limit.(1) If H is Σ λ +2 then H is Σ λ +1 and in particular Π λ +2 ;(2) If H is Π λ + n for some ≤ n < ω , then H is D (cid:0) Π λ + n − (cid:1) and in particular Π λ + n ;(3) If H is ˇ D (cid:0) Π λ + n (cid:1) for ≤ n < ω , then H is Π λ + n . Let G = ˆ G/N be a group with a Polish cover. By Proposition 2.17, if Γ is a complexity class, then = G isΓ-definable if and only if { } ∈ Γ ( G ). In the following proposition, we regard the relation = G as the coset relationof N inside of ˆ G . We focus on the case when { } is a non-Archimedean Polishable subgroup of G . Recall that E denotes the Σ equivalence relation on the space C := { , } ω of infinite binary sequences obtained by setting( x i ) E ( y i ) ⇔ ∃ n ∈ ω ∀ i ≥ n , x i = y i . The Π equivalence relation E ω on C ω is defined by setting ( x i ) E ω ( y i ) ⇔ ∀ i , x i E y i . Proposition 2.22.
Suppose that G = ˆ G/N is a group with a Polish cover. Suppose that { } is a non-ArchimedeanPolishable subgroup of G . Then: HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 9 (1) = G is smooth if and only if { } is Π in G ;(2) = G is Borel reducible to E if and only if { } is Σ in G , and Borel bireducible with E if and only if Σ is the complexity class of { } in G ;(3) = G is Borel reducible to E ω if and only if { } is Π in G , and Borel bireducible with E ω if and only if Π is the complexity class of { } in G .Proof. (1) We have that { } is Π in G if and only if N is a closed subgroup of ˆ G , which is equivalent to theassertion that = G is smooth; see [Sol09, page 574].(2) By [Gao09, Theorem 12.5.7], { } is Σ in G if and only if = G is essentially countable, which holds if and onlyif = G ≤ B E by [DG17, Theorem 6.1]. Furthermore, by Item (1), Lemma 2.20, and the Glimm–Effros dichotomy[HKL90], we have that { } is not Π if and only if { } is not Π if and only if N is not a closed subgroup of ˆ G , ifand only if E ≤ B = G .(3) By [DG17, Corollary 6.3] and Lemma 2.21 and we have that { } is not Σ if and only if { } is not Σ if andonly if E ω ≤ = G . By [All20, Corollary 6.11], { } is Π if and only if = G ≤ E ω . (cid:3) The Solecki subgroups.
Every group with a Polish cover admits a canonical sequence of subgroups indexedby countable ordinals. As these were originally described by Solecki in [Sol99], we call them
Solecki subgroups . Theyhave also been considered in [Sol09, FS06].Suppose that G = ˆ G/N is a group with a Polish cover. Then [Sol99, Lemma 2.3] implies that G has a smallest Π Polishable subgroup, which we denote by s ( G ) = ˆ s ( G ) /N . One can explicitly describe ˆ s ( G ) as the Π subgroup of ˆ G defined by \ V [ z ∈ N z + V G where V ranges among the open neighborhoods of 0 in N and z + V G is the closure of z + V inside of G . It isproved in [Sol99, Lemma 2.3] that s ( G ) satisfies the following properties: • { } is dense in s ( G ); • a neighborhood basis of x ∈ ˆ s ( G ) consists of sets of the form x + W G ∩ ˆ s ( G ) where W is an openneighborhood of 0 in N ; • if A ⊆ ˆ G is Π and contains N , then A ∩ ˆ s ( G ) is comeager in the Polish group topology of ˆ s ( G ).It follows that if H is a Π Polishable subgroup of G , then s ( G ) ⊆ { } H ⊆ H .The sequence of Solecki subgroups s α ( G ) for α < ω of the group with a Polish cover G is defined recursivelyby setting: • s ( G ) = { } G ; • s α +1 ( G ) = s ( s α ( G )) for α < ω ; • s λ ( G ) = T β<λ s β ( G ) for a limit ordinal λ < ω .We also let ˆ s α ( G ) be the Polishable subgroup of ˆ G such that s α ( G ) = ˆ s α ( G ) /N . Using Lemma 2.18, one canprove by induction on α < ω that { } is dense in s α ( G ) for every α < ω and that, if f : G → H is a definablehomomorphism, then f maps s α ( G ) to s α ( H ). Thus, G s α ( G ) is a functor on the category of groups with aPolish cover. It is proved in [Sol99, Theorem 2.1] that there exists α < ω such that s α ( G ) = { } . We call theleast countable ordinal α such that s α ( G ) = { } the Solecki rank of G . Lemma 2.23.
Suppose that G is a group with a Polish cover such that { } is a non-Archimedean Polishablesubgroup of G . Then s α ( G ) is a non-Archimedean Polishable subgroup of G for every ≤ α < ω .Proof. It is clear from the proof of [Sol99, Lemma 2.3] that s ( G ) is a non-Archimedean Polishable subgroup of G .The conclusion follows by induction applying Lemma 2.13 at the limit stage. (cid:3) Complexity of Solecki subgroups.
Suppose that G = ˆ G/N is a group with a Polish cover. For a Polishablesubgroup H = ˆ H/N of G and a complexity class Γ, we define Γ( ˆ G ) | ˆ H to be the collection of sets of the form A ∩ ˆ H for A ∈ Γ( ˆ G ). The following results are essentially established in [FS06]. In the statements and proofs, we adoptthe Vaught transform notation; see [Gao09, Section 3.2]. Lemma 2.24.
Suppose that G is a group with a Polish cover, and α, β < ω are ordinals. Then Σ β (ˆ s α ( G )) ⊆ Σ α + β ( ˆ G ) | ˆ s α ( G ) and Π β (ˆ s α ( G )) ⊆ Π α + β ( ˆ G ) | ˆ s α ( G ) .Proof. It is proved in [FS06, Theorem 3.1] by induction on α that Σ (ˆ s α ( G )) ⊆ Σ α ( ˆ G ) | ˆ s α ( G ) . By takingcomplements, we have that Π (ˆ s α ( G )) ⊆ Π α ( ˆ G ) | ˆ s α ( G ) . This is the case β = 0 of the statement above. The restfollows by induction on β . (cid:3) Lemma 2.25.
Suppose that G = ˆ G/N is a group with a Polish cover, α, β < ω , and U ⊆ N is open in N . If A ∈ Σ α + β ( ˆ G ) and B ∈ Π α + β ( ˆ G ) , then A ∆ U ∩ ˆ s α ( G ) ∈ Σ β (ˆ s α ( G )) , and B ∗ U ∩ ˆ s α ( G ) ∈ Π β (ˆ s α ( G )) .Proof. When β = 0, the assertion about A is the content of Claim 3.3 in the proof of [FS06, Theorem 3.1]. Theassertion about B follows by taking complements. This concludes the proof when β = 0. The case of an arbitrary β is established by induction on β using the properties of the Vaught transform; see [Gao09, Proposition 3.2.5]. (cid:3) Corollary 2.26.
Suppose that G = ˆ G/N is a group with a Polish cover, and α, β < ω . Let L be a Polishablesubgroup of G . If L ∈ Σ α + β ( G ) , then L ∩ s α ( G ) ∈ Σ β ( s α ( G )) . If L ∈ Π α + β ( G ) , then L ∩ s α ( G ) ∈ Π β ( s α ( G )) . Proposition 2.27.
Suppose that G is a group with a Polish cover, and α < ω . Then s α ( G ) is the smallest Π α +1 Polishable subgroup of G if α is a successor ordinal, and s α ( G ) is the smallest Π α Polishable subgroupof G if α is either or a limit ordinal.Proof. It is established in the proof of [FS06, Theorem 3.1] that s α ( G ) is a Π α +1 Polishable subgroup of G if α is a successor ordinal, and s α ( G ) is a Π α Polishable subgroup of G if α is either 0 or a limit ordinal. We nowprove the minimality assertion by induction on α .For α = 0 this follows from the fact that s ( G ) = { } G . Suppose that the conclusion holds for α . We now provethat it holds for α + 1. Let H be a Π α +2 Polishable subgroup of G . Thus, H ∩ s α ( G ) is a Π α +2 Polishablesubgroup of s α ( G ). Then by Corollary 2.26 we have that H ∩ s α ( G ) ∈ Π ( s α ( G )). As s α +1 ( G ) = s ( s α ( G )) isthe smallest Π ( s α ( G )) Polishable subgroup of s α ( G ), this implies that s α +1 ( G ) ⊆ H ∩ s α ( G ) ⊆ H .Suppose that α is a limit ordinal and the conclusion holds for every β < α . Fix an increasing sequence ( α n ) in α such that α = sup n α n . Suppose that H is a Π α Polishable subgroup of G . Then H = ˆ H/N for some Polishable Π α subgroup ˆ H of ˆ G containing N . Since ˆ H is Π α in ˆ G we can write ˆ H = T n ∈ ω A n where, for every n ∈ ω , A n ∈ Π α n ( ˆ G ). Then by Lemma 2.25 we have that A ∗ Nn ∩ ˆ s α n ( G ) ∈ Π (ˆ s α n ( G )). Since N ⊆ ˆ H ⊆ A n we havethat N ⊆ A ∗ Nn ∩ ˆ s α n ( G ). Since N is dense in ˆ s α ( G ), this implies that ˆ s α n ( G ) ⊆ A ∗ Nn . Therefore, we have thatˆ s α ( G ) = \ n ∈ ω ˆ s α n ( G ) ⊆ \ n ∈ ω A ∗ Nn = ˆ H ∗ N = ˆ H .This shows that s α ( G ) ⊆ H . (cid:3) Lemma 2.28.
Suppose that G is a group with a Polish cover, and α < ω . Let L be a Polishable subgroup of s α ( G ) .(1) If L ∈ Π ( s α ( G )) , then L ∈ Π α +2 ( G ) .(2) If L ∈ Σ ( s α ( G )) , then L ∈ D ( Π α +1 ( G )) .(3) If L ∈ Σ ( s α ( G )) and α is either zero or limit, then L ∈ Σ α +1 ( G ) .Proof. By Lemma 2.24 we have that we have that Π (ˆ s α ( G )) ⊆ Π α +2 ( ˆ G ) | ˆ s α ( G ) Σ (ˆ s α ( G )) ⊆ Σ α +1 ( ˆ G ) | ˆ s α ( G ) .Furthermore, ˆ s α ( G ) is Π α +1 ( ˆ G ) by Proposition 2.27. Therefore, we have that Π (ˆ s α ( G )) ⊆ Π α +2 ( ˆ G ) | ˆ s α ( G ) ⊆ Π α +2 ( ˆ G ) HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 11 and Σ (ˆ s α ( G )) ⊆ Σ α +1 ( ˆ G ) | ˆ s α ( G ) ⊆ D ( Π α +1 ( ˆ G )).This concludes the proof of (1) and (2).When α is either zero or limit, we have that ˆ s α ( G ) is Π α ( ˆ G ) and in particular Σ α +1 ( ˆ G ). Therefore, in thiscase we have that Σ (ˆ s α ( G )) ⊆ Σ α +1 ( ˆ G ) | ˆ s α ( G ) ⊆ Σ α +1 ( ˆ G ).This concludes the proof of (3). (cid:3) Lemma 2.29.
Suppose that G = ˆ G/N is a group with a Polish cover, and α < ω . Let L be a Polishable subgroupof G .(1) If L ∈ Σ α +2 ( G ) , then L ∩ s α ( G ) ∈ Σ ( s α ( G )) .(2) If L ∈ Π α +1 ( G ) then L ∩ s α ( G ) ∈ Π ( s α ( G )) .(3) If L ∈ Σ α +1 ( G ) then L ∩ s α ( G ) ∈ Σ ( s α ( G )) .(4) If L ∈ Σ α +1 ( G ) , L is a non-Archimedean Polishable subgroup of G , and α is a successor ordinal, then L ∩ s α ( G ) ∈ Π ( s α ( G )) .Proof. (1) Suppose that L ∈ Σ α +2 ( G ). Then by Corollary 2.26 we have that L ∩ s α ( G ) ∈ Σ ( s α ( G )).(2) Suppose that L ∈ Π α +1 ( G ). Then by Corollary 2.26 we have that L ∩ s α ( G ) ∈ Π ( s α ( G )) and hence L ∩ s α ( G ) ∈ Π ( s α ( G )) by Lemma 2.20.(3) Suppose that L ∈ Σ α +1 ( G ). Then by Corollary 2.26 we have that L ∩ s α ( G ) ∈ Σ ( s α ( G )).(4) Suppose that L ∈ Σ α +1 ( G ), L is a non-Archimedean Polishable subgroup of G , and α is a successorordinal. Then by Lemma 2.21, L ∈ Π α +1 ( G ). Thus, L ∩ s α ( G ) ∈ Π ( s α ( G )) by Item (2). (cid:3) Proposition 2.30.
Suppose that G = ˆ G/N is a group with a Polish cover, and α < ω . Suppose that { } is anon-Archimedean Polishable subgroup of G .(1) If { } ∈ Π ( s α ( G )) and { } / ∈ Σ ( s α ( G )) , then the Solecki rank of G is α + 1 , and the complexity class of { } in G is Π α +2 .(2) If { } ∈ Σ ( s α ( G )) , s α ( G ) = { } , and α is either zero or limit, then the Solecki rank of G is α + 1 , andthe complexity class of { } in G is Σ α +1 ;(3) If { } ∈ Σ ( s α ( G )) , s α ( G ) = { } , and α is a successor ordinal, then the Solecki rank of G is α + 1 , andthe complexity class of { } in G is D (cid:0) Π α +1 (cid:1) .Proof. (1) By Lemma 2.21(1), { } / ∈ Σ ( s α ( G )). Since { } ∈ Π ( s α ( G )) we have that s α +1 ( G ) = 0, while s α ( G ) = 0. Thus, the Solecki rank of G is α + 1. By Lemma 2.28(1), { } is Π α +2 in G . By Lemma 2.29(1), { } is not Σ α +2 in G .(2) As in (1), the Solecki rank of G is α + 1. Since { } is dense in s α ( G ), we have that { } is not Π in s α ( G ).By Lemma 2.28(3), { } is Σ α +1 in G . By Lemma 2.29(2), { } is not Π α +1 in G .(3) As in (2), the Solecki rank of G is α + 1, and { } is not Π in s α ( G ). By Lemma 2.28(2), { } is D ( Π α +1 )in G . By Lemma 2.29(2), { } is not Π α +1 in G . Thus, { } is not ˇ D ( Π α +1 ) in G by Lemma 2.21(3). (cid:3) Lemma 2.31.
Suppose that, for every k ∈ ω , G k = ˆ G k /N k is a group with a Polish cover, and α < ω . Define G = Q k ∈ ω G k . Assume that, for every k ∈ ω , { } is Π α in G k , and for every β < α there exist infinitely many k ∈ ω such that { } is not Π β in G k . Then Π α is the complexity class of { } in G .Proof. Write N = Q k ∈ ω N k ⊆ Q k ∈ ω ˆ G k . Clearly, N is Π α . By [Kec95, Theorem 22.10], for every k ∈ ω and β < α such that N k is not Π β , N k is Σ β -hard [Kec95, Definition 22.9]. Therefore, N is Σ α -hard, and hence N is not Σ α by [Kec95, Theorem 22.10] again. (cid:3) Lemma 2.32.
Suppose that, for every k ∈ ω , G k = ˆ G k /N k is a group with a Polish cover, and α < ω . Define G = Q k ∈ ω G k . If G k has Solecki rank α for every k ∈ ω , then Π α +1 is the complexity class of { } in G if α is asuccessor ordinal, and Π α is the complexity class of { } in G if α is either zero or a limit ordinal. Proof.
Define λ = 1 + α + 1 if α is a successor ordinal, and λ = 1 + α if α is either zero or a limit ordinal. ByProposition 2.27, for every k ∈ ω , { } is Π λ but not Π β for β < λ in G k . Therefore, by Lemma 2.31, Π λ is thecomplexity of { } in G . (cid:3) The abelian category of non-Archimedean groups with a Polish cover.
The goal of this section isto show that non-Archimedean groups with a Polish cover form an abelian subcategory of the abelian category ofgroups with a Polish cover. This amounts to showing that the kernel and cokernel of a definable homomorphismbetween non-Archimedean groups with a Polish cover (in the category of groups with a Polish cover) is in fact adefinable homomorphism between non-Archimedean Polish groups. To this end, we will use some rigidity resultsfor definable homomorphisms between groups with a Polish cover from [BLP20].The same proof as [BLP20, Lemma 4.9] gives the following.
Lemma 2.33.
Suppose that G is a Polish group and H = ˆ H/M is a group with a Polish cover such that M is anon-Archimedean Polish group, and f : G → ˆ H is a continuous function such that f ( x + y ) ≡ f ( x ) + f ( y ) mod M for every x, y ∈ G . Let M be a clopen subgroup of M such that M = M ˆ H ∩ M . Then there exist m ∈ M , x , y ∈ G , and a clopen subgroup G of G such that, if g : G → H is defined by g ( z ) := f ( x + y + z ) − f ( x + y ) − m ,then, for every x, y ∈ G , g ( x + y ) ≡ g ( x ) + g ( y ) mod M .Proof. Define the Borel function Cc : G ⊕ G → M , ( x, y ) f ( y ) − f ( x + y ) + f ( x ). Since M/M is countable,there exists m ∈ M such that A := { ( x, y ) ∈ G ⊕ G : c ( x, y ) ∈ m + M } is non-meager. Since f : G → ˆ H is continuous and M = M ˆ H ∩ M , we have that A ⊆ G ⊕ G is closed. Thus, A issomewhere dense, and there exists a clopen subgroup G of G and x , y ∈ G such that ( x , y ) + G ⊕ G ⊆ A .This concludes the proof. (cid:3) As it follows from [Sol99, Theorem 2.1], if M is a Π non-Archimedean Polishable subgroup of a Polish groupˆ H , then M has a basis ( M k ) of open neighborhoods of 0 consisting of subgroups satisfying M k = M ˆ Hk ∩ M for every k ∈ ω . Thus, one can infer from Lemma 2.33, as in the proof of [BLP20, Theorem 4.5], the following lemma. Lemma 2.34.
Suppose that G is a non-Archimedean Polish group, H = ˆ H/M is a group with a Polish cover, and ϕ : G → H is a definable homomorphism. Suppose that M is a non-Archimedean Polish group and a Π subgroupof ˆ H . Let ( ˆ H k ) be a basis of open neighborhoods of in ˆ H such that ˆ H = ˆ H . Let also ( M k ) be a basis of openneighborhoods of in M consisting of subgroups such that M = M and M k = M ˆ Hk ∩ M ⊆ ˆ H k for k ∈ ω . Thenthere exists a basis ( G k ) of neighborhoods of in G consisting of subgroups such that G = G , and a continuous lift g : G → ˆ H for ϕ that satisfies, for every k ∈ ω , and x, y ∈ G k , g ( x + y ) ≡ g ( x ) + g ( y ) mod M k ,and g ( x ) ∈ H k .Proof. By [BLP20, Theorem 4.5], ϕ has a continuous lift f : G → ˆ H . Set f = f . Fix a basis ( G ′ k ) of openneighborhoods of 0 in G consisting of subgroups. Applying Lemma 2.33, one can define by recursion on k ∈ ω : • an open subgroup G k ⊆ G ′ k of G such that G = G ; • elements x k , y k ∈ G k ; • an element m k ∈ M k ; • a continuous function f k : G → ˆ H ,such that, for every k ∈ ω :(1) for every x, y ∈ G k , f k ( x + y + x k + y k ) ≡ f k ( x + x k ) + f k ( y + y k ) mod M k ; HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 13 (2) for every z ∈ G , f k +1 ( z ) = f k ( x k + y k + z ) − f k ( x k + y k ) − m k ;(3) for every a, b ∈ G k +1 , f k +1 ( a + b ) − f k +1 ( a ) − f k +1 ( b ) ∈ M k +1 .For k ∈ ω , set z k := ( x + y ) + · · · + ( x k + y k ) .Then one can prove by induction on k ∈ ω that, for every z ∈ G , f k +1 ( z ) = f ( z k + z ) − f ( z k ) − m k .One can also prove by induction on k ∈ ω that, for x, y ∈ G k and i ≥ k , f i ( x + y ) ≡ f i ( x ) + f i ( y ) mod M k .Since the sequence ( x k + y k ) k ∈ ω converges to 0 in G , we have that the sequence ( z k ) k ∈ ω converges to some z ∞ ∈ G .Setting, for z ∈ G , g ( z ) := f ( z + z ∞ ) − f ( z ∞ ) = lim i →∞ ( f ( z + z i ) − f ( z i )) = lim i →∞ f i ( z ) ,one has that g : G → ˆ H is a continuous lift for ϕ such that, for every k ∈ ω and x, y ∈ G k , g ( x + y ) ≡ g ( x ) + g ( y ) mod M k .Notice that, for every k ∈ ω , g (0) ≡ M k , and hence g (0) = 0. Since g : G → ˆ H is continuous, after passingto a subsequence of ( G k ) k ∈ ω , we can also ensure that g ( x ) ∈ ˆ H k for every x ∈ G k and k ∈ ω . (cid:3) Using Lemma 2.34 we can prove the following.
Lemma 2.35.
Suppose that G = ˆ G/N and H = ˆ H/M are groups with a Polish cover, such that M and ˆ G arenon-Archimedean Polish groups. Suppose that ϕ : G → H is a definable homomorphism, and H has Solecki length . Then Ker ( ϕ ) is a non-Archimedean Polishable subgroup of G .Proof. After replacing ϕ with ϕ ◦ π where π : ˆ G → ˆ G/N is the quotient map, we can suppose that G is a non-Archimedean Polish group and N = { } . Adopt the notation from Lemma 2.34. In particular, let g : G → ˆ H be alift for ϕ as in Lemma 2.34. We can furthermore assume that ˆ H k +1 + ˆ H k +1 ⊆ ˆ H k for every k ∈ ω . By Proposition2.15, L := Ker ( ϕ ) = { x ∈ G : g ( x ) ∈ M } is a Polishable subgroup of G . For every k ∈ ω , g induces a continuoushomomorphism G k ∩ L → M/M k . Define L k to be the open subgroup { x ∈ G k ∩ L : ϕ ( x ) ∈ M k } of L . This givesa decreasing sequence ( L k ) k ∈ ω of open subgroups of L with trivial intersection. Define the corresponding invariantultrametric d on L [Gao09, Section 2.4] by setting d ( x, y ) = inf { − k : x − y ∈ L k } for x, y ∈ L . It remains to prove that ( L, d ) is a complete metric space. Suppose that ( h k ) k ∈ ω is a sequence in L such that d ( h k , < − k and hence h k ∈ L k for every k ∈ ω . We need to prove that the series P ∞ n =0 h n convergesin ( L, d ). For every k ∈ ω , we have that h k ∈ G k ∩ L and ϕ ( h k ) ∈ M k . Since ( G k ) is a basis of open neighborhoodsof 0 in G consisting of subgroups, we have that there exists h ∈ G such that the series P ∞ n =0 h n converges to h in G . Let c : G × G → M be the function defined by c ( x, y ) := g ( x ) + g ( y ) − g ( x + y )for x, y ∈ G . By recursion on n ≥ c : G n → M by c ( x , . . . , x n +1 ) = c ( x , . . . , x n ) + c ( x + · · · + x n , x n +1 ) .Then one can prove by induction that this is a permutation-invariant function satisfying c ( x , . . . , x n , y , . . . , y m ) = c ( x , . . . , x n ) + c ( y , . . . , y n ) + c ( x + · · · + x n , y + · · · + y m )and c ( x , . . . , x n ) = g ( x ) + · · · + g ( x n ) − g ( x + · · · + x n ) for n, m ≥ x , . . . , x n , y , . . . , y m ∈ G , where we set c ( x ) = 0 for x ∈ G . Using this, one can prove by inductionon n ≥ c ( x , . . . , x n ) = n X i =0 c ( x i , x i +1 + · · · x n )for x , . . . , x n ∈ G . Thus, for every n ∈ ω , we have that g ( h + · · · + h n ) − n X i =0 g ( h i ) = c ( h , . . . , h n )= n − X i =0 c ( h i , h i +1 + · · · + h n ) .Fix i ∈ ω . By the choice of g we have that, for every x, y ∈ G i , c ( x, y ) ∈ M i . Define h ( i ) := h − ( h + · · · + h i ) .By continuity of g : G → ˆ H , we have that the sequence( c ( h i , h i +1 + · · · + h n )) n ∈ ω of elements of M i converges to c ( h i , h ( i ) ) ∈ M ˆ Hi ∩ N = M i in ˆ H . Therefore, for every k ∈ ω , for all but finitely many n ∈ ω one has that k X i =0 (cid:16) c ( h i , h i +1 + · · · + h n ) − c ( h i , h ( i ) ) (cid:17) ∈ ˆ H k +1 .Thus, one can define by recursion on k an increasing sequence ( n k ) such that, for every k ∈ ω and n ≥ n k one hasthat R nk := k X i =0 ( c ( h i , h i +1 + · · · + h n ) − c ( h i , h ( i ) )) ∈ ˆ H k +1 .Notice also that, for every k ∈ ω , k + 1 ≤ i and n ≥ n k , one has that h i and h i +1 + · · · + h n belong to G k +1 , andhence c ( h i , h i +1 + · · · + h n ) ∈ M k +1 and hence S nk := n X i = k +1 c ( h i , h i +1 + · · · + h n ) ∈ M k +1 ⊆ ˆ H k +1 .Thus, for every k ∈ ω and n ≥ n k , R nk + S nk ∈ ˆ H k +1 + ˆ H k +1 ⊆ ˆ H k .Furthermore, for every k ∈ ω and n ≥ n k , we have that g ( h + · · · + h n ) − n X i =0 g ( h i )= n − X i =0 c ( h i , h i +1 + · · · + h n )= k X i =0 c ( h i , h i +1 + · · · + h n ) + n − X i = k +1 c ( h i , h i +1 + · · · + h n )= k X i =0 c ( h i , h ( i ) ) + R nk + S nk . HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 15
Thus, for every k ∈ ω and n ≥ n k , g ( h + · · · + h n ) − ( n X i =0 g ( h i ) + c ( h i , h ( i ) )) = R nk + S nk ∈ ˆ H k .The series ∞ X n =0 (cid:16) g ( h n ) + c ( h n , h ( n ) ) (cid:17) converges in M since g ( h n ) + c (cid:0) h n , h ( n ) (cid:1) ∈ M n for every n ∈ ω . By continuity of g : G → ˆ H , we have that g ( h ) isthe limit of the sequence ( g ( h + · · · + h n )) n ∈ ω in ˆ H . Since, for k ≥ n k , R nk + S nk ∈ ˆ H k ,we have that g ( h ) = ∞ X n =0 (cid:16) g ( h n ) + c ( h n , h ( n ) ) (cid:17) ∈ M .This shows that h ∈ L = { x ∈ G : g ( x ) ∈ M } , concluding the proof. (cid:3) Lemma 2.36.
Suppose that G = ˆ G/N and H = ˆ H/M are groups with a Polish cover, such that M and ˆ G arenon-Archimedean Polish groups. Assume that H has Solecki length . Let ϕ : G → H be a definable homomorphism.Then ϕ ( G ) is a non-Archimedean Polishable subgroup of H .Proof. Adopt the notation from Lemma 2.34. Define ϕ ( G ) = L and set ˆ L = { x ∈ ˆ H : x + M ∈ L } . Then byProposition 2.15, ˆ L is a Polishable subgroup of ˆ H . Consider for k ∈ ω the analytic subgroup ˆ L k := g ( G k ) + M k ⊆ L ∩ ˆ H k ⊆ L . Since, for x, y ∈ ˆ G k , g ( x + y ) ≡ g ( x ) + g ( y ) mod M k , we have that ˆ L k is indeed a subgroup of ˆ L .Furthermore, since ˆ G k has countable index in ˆ G and M k has countable index in M , we have that ˆ L k has countableindex in ˆ L , and hence it is an open subgroup of ˆ L . Thus, ( ˆ L k ) is a decreasing sequence of open subgroups of ˆ L withtrivial intersection. Let d be the invariant ultrametric on L associated with the sequence ( ˆ L k ), defined by d ( x, y ) = min { − k : x − y ∈ ˆ L k } .We claim that ( ˆ L, d ) is a complete metric space. Suppose that ( y n ) is a sequence in L such that d ( y n , < − n ,and hence y n ∈ ˆ L n , for every n ∈ ω . We have that y n = g ( h n ) + z n for some h n ∈ ˆ G n and z n ∈ M n . Define y tobe the limit of the series P ∞ n =0 y n in ˆ H , z to be the limit of the series P ∞ n =0 z n in M , and h to be the limit of theseries P ∞ n =0 h n in ˆ G . We need to prove that y ∈ ˆ L . Define also h ( i ) = h − ( h + · · · + h i ) for i ∈ ω . As in the proofof Lemma 2.35, we have that: • c ( h i , h ( i ) ) ∈ M ˆ Hi ∩ N = M i for every i ∈ ω ; • the series ∞ X n =0 c ( h n , h ( n ) )converges in M ; • the sequence ( g ( h + · · · + h n )) n ∈ ω converges in ˆ H to g ( h ); • there is an increasing sequence ( n k ) in ω such that, for every k ∈ ω and n ≥ n k , g ( h + · · · + h n ) − n X i =0 (cid:16) g ( h i ) + c ( h i , h ( i ) ) (cid:17) ∈ ˆ H k . Therefore, we have that y + ∞ X n =0 c ( h n , h ( n ) ) = ∞ X n =0 (cid:16) g ( h n ) + c ( h n , h ( n ) ) + z n (cid:17) = g ( h ) + z and hence y ≡ g ( h ) mod M .This shows that y ∈ ˆ L , concluding the proof. (cid:3) Proposition 2.37.
Suppose that G = ˆ G/N is a non-Archimedean group with a Polish cover, and H = ˆ H/M is agroup with a Polish cover such that M is a non-Archimedean Polish group. If G and H are definably isomorphic,then ˆ H is a non-Archimedean Polish group, and hence H is a non-Archimedean group with a Polish cover.Proof. Let ϕ : G → H be a definable isomorphism. Then, for every α < ω , ϕ induces a definable isomorphism s α ( G ) → s α ( H ), and a definable isomorphism G/s α ( G ) → H/s α ( H ). Furthermore, by Lemma 2.23, we have that,for every 1 ≤ α < ω , s α ( G ) is a non-Archimedean Polishable subgroup of G and s α ( H ) is a non-ArchimedeanPolishable subgroup of H . After replacing G with G/s ( G ) and H with H/s ( H ), we can assume that H hasSolecki length 1. In this case, the conclusion follows from Lemma 2.36, since ϕ ( G ) = H . (cid:3) We can now prove that images and preimages of non-Archimedean Polishable subgroups of non-Archimedeangroups with a Polish cover are non-Archimedean.
Proposition 2.38.
Suppose that
G, H are non-Archimedean groups with a Polish cover, and f : G → H is adefinable homomorphism.(1) If H is a non-Archimedean Polishable subgroup of H , then f − ( H ) is a non-Archimedean Polishablesubgroup of G ;(2) If G is a non-Archimedean Polishable subgroup of G , then f ( G ) is a non-Archimedean Polishable subgroupof G .Proof. Let s α ( H ) for α < ω be the sequence of Solecki subgroups of H . Notice that, for every α < ω , s α ( H ) is anon-Archimedean Polishable subgroup of H . For α ≥ α = 0, this follows fromthe fact that s ( H ) = { } H is a closed subgroup of H .(1) Define G := f − ( H ). After replacing H with H/H , we can assume that H = { } , in which case G = Ker ( f ). By Proposition 2.15 we have that G α := f − ( s α ( H )) is a Polishable subgroup of G for every α < ω . We now prove by induction on α that G α is non-Archimedean. This will give the desired conclusion whenapplied to the Solecki length of H .For α = 0 this is clear since s ( H ) is closed in H and hence G is closed in G by Lemma 2.18. Suppose thatthe conclusion holds for α . Then consider ϕ α = f | G α : G α → s α ( H ), and let π α : s α ( H ) → s α ( H ) /s α +1 ( H ) bethe quotient map. Then G α +1 = Ker ( π α ◦ ϕ α ). Since s α ( H ) /s α +1 ( H ) is a non-Archimedean group with a Polishcover of Solecki length 1, and G α is a non-Archimedean group with a Polish cover by the inductive assumption, wehave that G α +1 is non-Archimedean by Lemma 2.35. Finally, suppose that λ is a limit ordinal, and the conclusionholds for α < λ . As s λ ( H ) = T α<λ s α ( H ), we have that G λ = T α<λ G α . By the inductive hypothesis, G α isnon-Archimedean for every α < λ . Therefore, G λ is non-Archimedean by Lemma 2.13.(2) Set H := ˆ H/M where M and ˆ H are non-Archimedean Polish groups. After replacing G with G and f with itsrestriction to G , we can assume that G = G . By the first item, Ker ( f ) is a non-Archimedean Polishable subgroupof G . Thus, after replacing G with G/ Ker ( f ), we can assume that f : G → H is a definable monomorphism. ByProposition 2.15, we have that H = f ( G ) is a Polishable subgroup of H . Thus, f induces a definable isomorphism ϕ : G → H . By Proposition 2.37, we have that ˆ H = { x ∈ ˆ H : x + M ∈ H } is a non-Archimedean Polish group.Hence, H is a non-Archimedean Polishable subgroup of H . This concludes the proof. (cid:3) It follows from Proposition 2.38 that non-Archimedean groups with a Polish cover form an abelian subcategoryof the abelian category of groups with a Polish cover.
HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 17
Ulm subgroups.
Suppose that G = ˆ G/N is a group with a Polish cover. The first Ulm subgroup u ( G ) = G is the subgroup T n> nG of G . One then defines the sequence u α ( G ) of Ulm subgroups by recursion on α < ω bysetting:(1) u ( G ) = G ;(2) u α +1 ( G ) = u ( u α ( G ));(3) u λ ( G ) = T α<λ u α ( G ) for λ limit.It follows from Lemma 2.12 by induction on α < ω that u α ( G ) is a Polishable subgroup of G , which is non-Archimedean if G is non-Archimedean; see Definition 2.10.We let D ( G ) be the largest divisible subgroup of G . A group G is reduced if D ( G ) = 0. Notice that D ( G ) isa Polishable subgroup of G , which is non-Archimedean when G is non-Archimedean, as D ( G ) = ˆ D ( G ) /N , whereˆ D ( G ) is the image of the Polish group H = (cid:8) ( g n , r n ) n ∈ ω : ∀ n ∈ ω, g n − ( n + 1) g n +1 = r n (cid:9) ⊆ Y n ∈ ω ˆ G ⊕ N under a continuous group homomorphism. The same proof as [FS06, Theorem 4.1(i)] gives the following. Proposition 2.39.
Suppose that G = ˆ G/N is a group with a Polish cover. Then there exists α < ω such that u α ( G ) = D ( G ) .Proof. After replacing G with G/D ( G ), we can assume that D ( G ) = { } . Set B := ˆ G \ N . Define the Borelrelation ≺ on the standard Borel space B <ω of finite sequences in B by setting( g , . . . , g n ) ≺ ( h , . . . , h m ) ⇔ m < n and ∀ i ≤ m , h i = g i , and ∀ i ≤ n , ( i + 1) g i +1 ≡ g i mod N .Then ≺ is well-founded, and hence its rank ρ ( ≺ ) is a countable ordinal [Kec95, Theorem 31.1]. For A ⊆ B <ω define D ( A ) = (cid:8) s ∈ B <ω : ∃ t ∈ A, t ≺ s (cid:9) .Set then recursively A = B <ω , A α +1 = D ( A α ), and A λ = T α<λ A α for λ limit. Then it is easily proved byinduction that, for every ordinal α and s = ( g , . . . , g n ) ∈ B <ω : • ρ ≺ ( s ) ≥ α if and only if s ∈ A α ; • s ∈ A ωα ⇔ g n + N ∈ u α ( G ).Thus, if λ = ρ ( ≺ ) then we have that ρ ≺ ( s ) < λ for every s ∈ B <ω , and hence A ωλ = ∅ and u λ ( G ) = { } . (cid:3) Suppose that G is a group with a Polish cover. The Ulm rank of G the least α < ω such that u α ( G ) = D ( G ).Notice that such an α exists by Proposition 2.39.A related sequence can be defined for a given prime number p . One defines by recursion: • p G = G ; • p α +1 G = p ( p α G ); • p λ G = T β<λ p β G for λ limit.The p -length of G is the least α such that p α G is p -divisible, in which case p α G is the largest p -divisible subgroupof G . (A group A is p -divisible if the function A → A , x px is surjective.) If G is p -local, then one can easilyprove by induction that, for every ordinal β , G β = p ωβ G . (An abelian group G is p -local if, for every prime q otherthan p , the function x qx is an automorphism of G .) Definition 2.40. If G, H are groups with a Polish cover, and ϕ : G → H is an definable monomomorphism, then ϕ is isotype if ϕ − ( u α ( H )) = u α ( G ) for every ordinal α . If G ⊆ H is a Polishable subgroup, then G is isotype ifthe inclusion G → H is isotype.Suppose that α is a countable ordinal and ( G β ) β<α is an inverse sequence of groups with a Polish cover G β anddefinable homomorphisms p ( β,β ′ ) : G β ′ → G β for β ≤ β ′ < α . One then defines lim β<α G β to be the group with aPolish cover ( x β ) ∈ Y β<α G β : ∀ β ≤ β ′ , x β = p ( β,β ′ ) (cid:0) x β ′ (cid:1) , which is a Polishable subgroup of Q β<α G β .Let G be a group with a Polish cover. The α -topology on G is the group topology that has (cid:0) G β (cid:1) β<α asneighborhoods of 0. Define L α ( G ) := lim β<α G/G β . We have a canonical definable exact sequence of groups witha Polish cover G α → G → L α ( G ) → E α ( G ) .The map κ α : G → L α ( G ) is the canonical homomorphism induced by the quotient maps G → G/G β for β < α ,while E α ( G ) = L α ( G ) /κ α ( G ). Notice that the α -topology on G is Hausdorff if and only if G α = 0, and itis Hausdorff and complete if and only if G α = 0 and E α ( G ) = 0, in which case G → L α ( G ) is a definableisomorphism. If ϕ : G → H is an isotype definable homomorphism, then ϕ induces a definable homomorphism L α ( G ) → L α ( H ) that makes the diagram G → H ↓ ↓ L α ( G ) → L α ( H )commute. In turn, this induces a definable homomorphism E α ( G ) → E α ( H ).3. Ext groups
In this section, we recall how Ext and PExt groups from abelian group theory and homological algebra can beregarded as groups with a Polish cover as in [BLP20]. We formulate in this context a description of Ext groupsfrom [EM42], and some homological algebra results about cotorsion functors from [Nun67]. Recall that we assumeall the groups to be abelian and additively denoted.3.1.
Groups of homomorphisms.
We begin by describing how the group Hom of homomorphisms from a count-able group to a group with a Polish cover can be regarded as a group with a Polish cover. Suppose that A is acountable group, and G = ˆ G/N is a group with a Polish cover. We regard N as a Polish group endowed with its(unique) compatible Polish group topology. Define Hom ( A, G ) to be the group of group homomorphisms A → G .We regard Hom ( A, G ) as a group with a Polish cover, as follows. Let ˆ H be the Polish group of functions ϕ : A → ˆ G satisfying ϕ ( a + b ) ≡ ϕ ( a ) + ϕ ( b ) mod N for a, b ∈ A and ϕ (0) = 0, endowed with the topology obtained bysetting ϕ i → ϕ i ( a ) → G and ϕ i ( a + b ) − ϕ i ( a ) − ϕ i ( b ) → N for every a, b ∈ A . Let also N H be the Polishable subgroup consisting of ϕ ∈ ˆ H such that ϕ ( a ) ≡ N for every a ∈ A . Then we let Hom ( A, G ) be the group with a Polish cover ˆ
H/N H . Notice that Hom ( A, G ) isnon-Archimedean when G is non-Archimedean, and a Polish group when G is a Polish group.Suppose that A = F/R where F is a countable free group and R is a subgroup. Then one has that Hom ( A, G ) asdefined above is naturally definably isomorphic to the group with a Polish cover Hom((
F, R ) , ( ˆ G, N )) / Hom(
F, N ).Here, Hom((
F, R ) , ( ˆ G, N )) is the Polish group of homomorphisms F → ˆ G mapping R to N endowed with thetopology defined by setting ϕ i → ϕ if and only if ϕ i ( x ) → ϕ ( x ) in ˆ G for x ∈ F and ϕ i ( x ) → ϕ ( x ) in N for x ∈ R ,and Hom( F, N ) is the Polishable subgroup of homomorphisms F → N .3.2. Groups of extensions.
We now consider the groups Ext of extensions of a countable group by a group witha Polish cover, and describe how it can be regarded as a group with a Polish cover. We present two equivalentdescriptions of it, one in terms of cocycles, and the other in terms of projective resolutions. Suppose that A is acountable group and G = ˆ G/N is a group with a Polish cover. A (normalized 2-) cocycle on A with values in G is afunction c : A × A → G such that, for every x, y, z ∈ A : • c ( x,
0) = 0; • c ( x, y ) = c ( y, x ); • c ( y, z ) − c ( x + y, z ) + c ( x, y + z ) − c ( y, z ) = 0 . HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 19
We let Z ( A, G ) be the group of cocycles on A with values in G , where the operation is defined pointwise.We regard Z ( A, G ) as a group with a Polish cover ˆ
H/N H , as follows. Let ˆ H be the Polish group of functions c : A × A → ˆ G such that, for every x, y, z ∈ A : • c ( x,
0) = 0; • c ( x, y ) = c ( y, x ); • c ( y, z ) − c ( x + y, z ) + c ( x, y + z ) − c ( y, z ) ≡ N .The operations in ˆ H are defined pointwise, and we declare a net ( c i ) in H to converge to 0 if and only if, forevery x, y, z ∈ A , c i ( x, y ) → G , and c i ( y, z ) − c i ( x + y, z ) + c i ( x, y + z ) − c i ( y, z ) → N . We let N H be the Polishable subgroup of ˆ H consisting of the functions c in ˆ H such that c ( x, y ) ≡ N for every x, y ∈ A . Then we can identify Z ( A, G ) with the group with a Polish cover ˆ
H/N H . Notice that Z ( A, G )is non-Archimedean when G is non-Archimedean, and a Polish group when G is a Polish group.A cocycle c on A with values in G is a coboundary if there exists a function φ : A → G such that φ (0) = 0 and c ( x, y ) = φ ( y ) − φ ( x + y ) + φ ( x )for every x, y ∈ A . Coboundaries form a Polishable subgroup B ( A, G ) of the group with a Polish cover Z ( A, G ).One then defines Ext (
A, G ) to be the group with a Polish cover Z ( A, G ) / B ( A, G ), which is non-Archimedean when G is non-Archimedean.Following [EM42, Section 5], one can give an alternative description of Ext( A, G ), as follows. Suppose that F isa free countable abelian group, and R is a subgroup. Let Hom ( F | R, G ) be the Polishable subgroup of Hom(
R, G )consisting of group homomorphisms R → G that extend to a group homomorphism F → G . Proposition 3.1 (Eilenberg–Mac Lane [EM42]) . Suppose that G is a group with a Polish cover, and A is a countablegroup. Write A = F/R where F is a countable free group, and R ⊆ F is a subgroup. Fix a right inverse t : A → F for the quotient map F → A such that t (0) = 0 . Define then the cocycle ζ on A with values in R by setting ζ ( x, y ) = t ( y ) − t ( x + y ) + t ( x ) for x, y ∈ A . The definable homomorphism Hom (
R, G ) → Z ( A, G ) , θ θ ◦ ζ induces a natural definable isomorphism Hom(
R, G )Hom( F | R, G ) ∼ −→ Ext(
A, G ) . Fix a countable abelian group A . Then Ext ( A, − ) is an additive functor from the category of groups with aPolish cover to itself. Similarly, for a fixed group with a Polish cover G , Ext ( − , G ) is an additive functor from thecategory of countable groups to the category of groups with a Polish cover. If ( A i ) i ∈ ω is a sequence of countablegroups, and ( G i ) i ∈ ω is a sequence of groups with a Polish cover, then we have, for every countable group A andgroup with a Polish cover G , natural isomorphismsExt( M i ∈ ω A i , G ) ∼ = Y i ∈ ω Ext ( A i , G )and Ext( A, Y i ∈ ω G i ) ∼ = Y i ∈ ω Ext (
A, G i )in the category of groups with a Polish cover; see [Fuc70, Theorem 52.2].If D is a divisible abelian group, then Ext( A, D ) = 0 [Fuc70, page 222]. If G is an arbitrary countable group,then one can write G = D ⊕ G ′ where D is divisible and G ′ is reduced (i.e., it has no nonzero divisible subgroup).Thus, we have that Ext ( A, G ) and Ext (
A, G ′ ) are definably isomorphic. Groups of pure extensions.
An important subgroup of Ext is the group PExt corresponding to pureextensions. We recall the fundamental observation that PExt is the first Ulm subgroup of Ext, as well as the closureof the { } in Ext. Suppose that A is a countable group, and G is a group with a Polish cover. Define the group B w ( A, G ) of weak coboundaries to be the Polishable subgroup of Z ( A, G ) consisting of c such that c | S × S ∈ B ( S, G )for every finite subgroup S ⊆ A . One defines the pure (or phantom) Ext group PExt ( A, G ) to be the group witha Polish cover B w ( A, G ) / B ( A, G ) [Sch03, CS98]. We also define the weak
Ext group Ext w ( A, G ) to be the groupwith a Polish cover Z ( A, G ) / B w ( A, G ). One has that PExt (
A, G ) is equal to the first Ulm subgroup u (Ext ( A, G ))of the group with a Polish cover Ext (
A, G ); see [Fuc70, Proposition 53.4].Suppose that A = F/R for some countable free group F and subgroup R ⊆ F . Define Hom f ( F | R, G ) to bethe Polishable subgroup of Hom(
R, G ) consisting of all group homomorphisms R → G that extend to a grouphomomorphism F → R for every subgroup F of F containing R as a finite index subgroup. The natural definableisomorphism Hom( R, G )Hom( F | R, G ) ∼ −→ Ext(
F/R, G )as in Proposition 3.1 induces a definable isomorphismHom f ( R, G )Hom( F | R, G ) ∼ −→ PExt (
A, G ) ;see [EM42, Section 5].
Lemma 3.2.
Suppose that
A, G are countable groups. Then
PExt (
A, G ) is the closure of { } in Ext (
A, G ) , and Ext w ( A, G ) is a Polish group.Proof. Write A = F/R for some countable free group F and subgroup R ⊆ F . In view of the above discussion, itsuffices to prove that Hom f ( R, G ) is the closure of Hom( F | R, G ) inside Hom (
R, G ).For ϕ ∈ Hom (
R, G ), one has that ϕ ∈ Hom f ( F | R, G ) if and only if, for every prime p , x ∈ F and m ∈ Z satisfying p m x ∈ R , one has that ϕ ( p m x ) ∈ p m G ; see [EM42, Lemma 5.1 and Lemma 5.2]. Thus, it follows thatHom f ( F | R, G ) is a closed subgroup of the Polish group Hom(
R, G ). The proof of [EM42, Lemma 5.3] shows that,in fact, Hom f ( F | R, G ) is the closure of Hom ( F | R, G ) inside Hom (
R, G ). (cid:3) Ext groups and extensions.
Fix countable groups
A, C . We now describe how one can regard the relationof equivalence of extensions of C by A as a Borel equivalence relation on a Polish space, and how the elements ofExt( C, A ) parametrize the corresponding equivalence classes.An extension of C by A is a short exact sequence A → B → C in the abelian category of countable groups.(Recall that all the groups are assumed to be abelian.) One can regard such an extension as a first-order structurein the countable language L obtained by adding to the language of groups constant symbols c a for a ∈ A andunitary relation symbols R c for c ∈ C . The extension A → B → C corresponds to the L -structure given by thegroup B , where c a is the interpreted as the image of a ∈ A under the monomorphism A → B , while the relationsymbol R c is interpreted as the preimage of c ∈ C under the epimorphism B → C . It is easy to see that the set Ext ( C, A ) of such L -structures is a G δ subset of the space Mod ( L ) of L -structures, whence it is a Polish space withthe induced topology. The relation R Ext ( C,A ) of isomorphism of extensions is simply the restriction to Ext ( C, A )of the relation of isomorphism of L -structures.Given an extension A g → B h → C of C by A , one can define a corresponding cocycle c on C with coefficients with A , as follows. Fix a right inverse t : C → B for the map h : B → C , and set c ( x, y ) := g − ( t ( y ) − t ( x + y ) + t ( x )).This defines a Borel function from Ext ( C, A ) to the Polish group Z ( A, G ), which is a Borel reduction from R Ext ( C,A ) to the coset relation = Ext(
C,A ) of B ( A, G ) inside Z ( A, G ). Furthermore, the induced definable function
Ext ( C, A ) R Ext ( C,A ) → Ext (
C, A )is a bijection; see [Fuc70, Section 50]. Therefore, we have that
Ext ( C, A ) / R Ext ( C,A ) is a definable set, and byProposition 2.3 R Ext ( C,A ) is Borel bireducible with = Ext(
C,A ) .An extension A → B → C of C by A is pure if (the image of) A is a pure subgroup of B , namely nB ∩ A = nA for every n ≥
1. The pure extensions of C by A form a G δ subset PExt ( C, A ) of
Ext ( C, A ). Letting R PExt ( C,A )HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 21 be the restriction of R Ext ( C,A ) to PExt ( C, A ), we have that
PExt ( C, A ) / R PExt ( C,A ) is also a definable set, andthe definable bijection Ext ( C, A ) R Ext ( C,A ) ∼ −→ Ext (
C, A )restricts to a definable bijection
PExt ( C, A ) R PExt ( C,A ) ∼ −→ PExt (
C, A ) .3.5. Ext and
Hom . The functor Ext can be seen as the first derived functor of the functor Hom from the perspectiveof homological algebra [Wei94, Chapter 2]. For our purposes, it will be sufficient to recall some exact sequencesrelating Ext and Hom.Suppose that A is a countable group and G → G → G is a short exact sequence of groups with a Polish cover.Then we have a definable exact sequence0 → Hom (
A, G ) → Hom(
A, G ) → Hom (
A, G ) ∂ → Ext (
A, G ) → Ext(
A, G ) → Ext (
A, G ) → ∂ : Hom ( A, G ) → Ext (
A, G )one can consider a short exact sequence R → F → A , where F is free. Then we have natural definable isomorphismsHom ( A, G ) ∼ = Hom (( F, R ) , ( G , G ))Hom ( F, G )and Ext ( A, G ) ∼ = Hom ( R, G )Hom ( F | R, G ) .Here, Hom (( F, R ) , ( G , G )) is the Polishable subgroup of Hom ( F, G ) consisting of homomorphisms F → G mapping R to G (where we identify G with a Polishable subgroup of G ). Via these definable isomorphisms, ∂ correspond to the definable homomorphismHom (( F, R ) , ( G , G ))Hom ( F, G ) → Hom (
R, G )Hom ( F | R, G )induced by the definable homomorphismHom (( F, R ) , ( G , G )) → Hom (
R, G ) , ϕ ϕ | R .If furthermore G → G → G is a pure definable short exact sequence, namely G is a pure subgroup of G , thenthe definable exact sequence above restricts to a definable exact sequence0 → Hom (
A, G ) → Hom(
A, G ) → Hom (
A, G ) → PExt (
A, G ) → PExt(
A, G ) → PExt (
A, G ) → A → B → C is a short exact sequence of countable groups and G is a group with a Polishcover. Then we have a definable exact sequence0 → Hom (
C, G ) → Hom (
B, G ) → Hom (
A, G ) δ → Ext (
C, G ) → Ext (
B, G ) → Ext (
A, G ) → δ : Hom ( A, G ) → Ext (
C, G )is defined by setting, for ϕ ∈ Hom (
A, G ), δ ( ϕ ) to be the element of Ext ( C, G ) associated with the cocycle C × C → G , ( x, y ) ϕ ( t ( y ) − t ( x + y ) + t ( x )), where t : C → B is a right inverse for the quotient map B → C , and weidentify A with a subgroup of B . If furthermore the short exact sequence A → B → C is pure, then the exactsequence above restricts to a definable exact sequence0 → Hom (
C, G ) → Hom (
B, G ) → Hom (
A, G ) → PExt (
C, G ) → PExt (
B, G ) → PExt (
A, G ) → Radicals and cotorsion functors.
We now recall some terminology and results of homological nature aboutExt from [Nun67]. These will be used in Section 4 to describe the Solecki subgroups of Ext for countable p -groups.A preradical or subfunctor of the identity S is a function A SA assigning to each group A a subgroup SA ⊆ A such that, if f : A → B is a group homomorphism, then f maps SA to SB . This gives a functor from the categoryof abelian groups to itself, where one defines for a homomorphism f : A → B , Sf := f | SA : SA → SB . A radical is a preradical S such that S ( A/SA ) = 0 for every group A .An extension Z → G → H of a countable group H by Z defines a preradical S , by setting SA = Ran (Hom ( G, A ) → Hom ( Z , A ) = A ) = Ker ( A = Hom ( Z , A ) → Ext (
H, A )) .In this case, one says that S is the preradical represented by the extension Z → G → H . A cotorsion functor is apreradical that is represented by an extension Z → G → H such that H is a countable torsion group. Notice thatthis implies that SA is a Polishable subgroup of A whenever A is a group with a Polish cover.Suppose that S is a cotorsion functor. An extension C → E → A is S -pure if it defines an element of S Ext (
A, C ).In this case, the map C → E is an S -pure monomorphism and the map E → A is an S -pure epimorphism, and theimage of C inside of E is an S -pure subgroup. A group A is S -projective if S Ext (
A, C ) = 0 for every group C , and S -injective if S Ext (
C, A ) = 0 for every group C . The cotorsion functor S has enough projectives if for every group A there exists an S -pure extension M → P → A where P is S -projective. This is equivalent to the assertion that S is represented by an extension Z → G → H where H is an S -projective torsion group [Nun63, Theorem 4.8]. Acotorsion functor with enough projectives is necessarily a radical.The following lemma is a consequence of [Nun67, Lemma 1.1]. Lemma 3.3.
Suppose that S is a cotorsion functor, A, C are countable group, and B is a subgroup of C containedin SC . Then the exact sequence B → C → C/B induces a definable exact sequence → Hom (
A, B ) → Hom(
A, C ) → Hom (
A, C/B ) → Ext (
A, B ) → S Ext(
A, C ) → S Ext (
A, C/B ) → of groups with a Polish cover. Furthermore, S Ext(
A, C ) is the preimage of S Ext (
A, C/B ) under the definableepimorphism Ext (
A, C ) → Ext (
A, C/B ) . Suppose that S is a cotorsion functor represented by the exact sequence Z → G S → H S , where H S is a countable S -projective torsion group. If C is a countable group, then we have a corresponding definable exact sequence ofgroups with a Polish cover 0 → SC → C → Ext ( H S , C ) → Ext ( G S , C ) → Lemma 3.4.
Suppose that S is a cotorsion functor represented by an extension Z → G S → H S where H S is acountable S -projective torsion group. Let A, C be a countable groups such that
A/SA is S -projective. Then:(1) The quotient map C → C/SC induces definable isomorphisms
Ext ( G S , C ) → Ext ( G S , C/SC ) and Ext ( H S , C ) → Ext ( H S , C/SC ) ;(2) If SC = 0 then S Ext (
A, C ) and Hom (
SA,
Ext ( G S , C )) = Hom ( SA, S
Ext ( G S , C )) are definably isomor-phic;(3) The short exact sequences SA → A → A/SA and SC → C → C/SC induce definable homomorphisms
Hom(
A, C/SC ) → Ext(
A, SC ) and Ext (
A, SC ) → Ext (
SA, SC ) ,which induce a definable isomorphism η A,C : Ext (
A, SC )Ran (Hom (
A, C/SC ) → Ext (
A, SC )) → Ext (
SA, SC ) ;(4) The short exact sequence SC → C → C/SC induces a definable exact sequence → Hom (
A, C/SC ) → Ext (
A, SC ) → S Ext (
A, C ) → S Ext (
A, C/SC ) → HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 23 which induces a pure definable exact sequence → Ext (
A, SC )Ran (Hom (
A, C/SC ) → Ext (
A, SC )) ρ A,C → S Ext (
A, C ) → S Ext (
A, C/SC ) → ;(5) The definable monomorphism r A,C := ρ A,C ◦ η − A,C : Ext (
SA, SC ) → S Ext (
A, C ) restricts to a definable isomorphism γ A,C : PExt (
SA, SC ) → u ( S Ext (
A, C )) . Solecki subgroups of
Ext groups
This section is dedicated to the proof of Theorem 1.5 and Theorem 1.2. We being by showing that one canreduce to the case of p -groups for some prime number p . We first show that the Solecki and Ulm subgroups ofPExt coincide. We then conclude the proof of Theorem 1.5 and Theorem 1.2 by using the results concerning thecomplexity of Solecki subgroups from Section 2. As before, we assume all the groups to abelian and additivelydenoted.4.1. Reduction to the p -group case. Towards the proof of Theorem 1.5 and Theorem 1.2, we begin withshowing how one can reduce to the case of p -groups for a given prime number p . This is the content of the followingproposition. Recall that we denote by P the set of prime numbers. Proposition 4.1.
Suppose that C is a countable torsion group and A is a countable group. For a prime number p ,let C p ⊆ C be the p -primary component of C , and A p ⊆ A/D ( A ) be the p -primary component of A/D ( A ) . Let Γ be one of the classes Σ α , Π α , D (cid:0) Π α (cid:1) , D (cid:0) Σ α (cid:1) , D (cid:0) Π α (cid:1) , or ˇ D (cid:0) Π α (cid:1) for ≤ α < ω .(1) Ext (
C, A ) is a Polish group if and only if Ext ( C p , A p ) is a Polish group for every prime number p ;(2) { } is Γ in Ext (
C, A ) if and only if { } is Γ in Q p ∈ P Ext ( C p , A p ) . The rest of this section is dedicated to the proof of Proposition 4.1. We begin with a lemma from [Fuc70, page223], whose short proof we recall for convenience.
Lemma 4.2.
Fix a prime number p . Suppose that A, C are countable groups such that C is a p -group and A is p -divisible. Then Ext (
C, A ) = 0 .Proof.
Let D be the divisible hull of A [Fuc70, Section 24]. Then D/A is a torsion group with trivial p -primarycomponent. In particular, Hom ( C, D/A ) = 0. As D is divisible, Ext ( C, D ) = 0. Considering the definable exactsequence Hom (
C, D/A ) → Ext (
C, A ) → Ext (
C, D )induced by the exact sequence A → D → D/A , we conclude that Ext (
C, A ) = 0. (cid:3)
The following lemma is established in [EM42]; see also [Fuc70, Theorem 52.3].
Lemma 4.3. If A is a countable torsion-free group and C is a countable torsion group, then Ext (
C, A ) is definablyisomorphic to the Polish group Hom (
C, D/A ) where D is the divisible hull of A . In particular, PExt (
C, A ) = 0 .Proof.
The exact sequence A → D → D/A induces a definable exact sequenceHom (
C, D ) → Hom (
C, D/A ) → Ext (
C, A ) → Ext (
C, D ) .Since C is torsion and D is torsion-free, Hom ( C, D ) = 0. Furthermore, since D is divisible, Ext ( C, D ) = 0. Thisconcludes the proof. (cid:3)
As an immediate consequence of the previous lemma, one obtains the following.
Lemma 4.4.
Suppose that C is a countable torsion group and A is a countable group. Let tA ⊆ A be its torsionsubgroup. Then PExt (
C, tA ) and PExt (
C, A ) are definably isomorphic. Proof.
We have that J := A/tA is torsion-free. Since tA is a pure subgroup of A , we have a definable exact sequenceHom ( C, J ) → PExt (
C, tA ) → PExt (
C, A ) → PExt (
C, J ) .Since C is torsion and J is torsion-free, Hom ( C, J ) = 0 and PExt (
C, J ) = 0 by Lemma 4.3. This concludes theproof. (cid:3)
Lemma 4.5.
Suppose that C is a countable p -group and A is a countable group. Let A p ⊆ A be the p -primarysubgroup of A . Then PExt (
C, A ) and PExt (
C, A p ) are definably isomorphic.Proof. By Lemma 4.4 we can assume without loss of generality that A is a torsion group. Thus, we can write A = A p ⊕ B where B is p -divisible. By Lemma 4.2, PExt ( C, B ) = 0. Therefore, PExt (
C, A ) is definably isomorphicto PExt (
C, A p ) ⊕ PExt (
C, B ) = PExt (
C, A p ). (cid:3) Lemma 4.6.
Suppose that C is a countable torsion group and A is a countable group. Let A p ⊆ A be the p -primarysubgroup of A and C p ⊆ C be the p -primary subgroup of C . Then PExt (
C, A ) and Q p PExt ( C p , A p ) are definablyisomorphic.Proof. Since C is a countable torsion group, we have that C ∼ = L p C p . Therefore, PExt ( C, A ) is definably isomorphicto Q p PExt ( C p , A ). By Lemma 4.5, for every prime p we have that PExt ( C p , A ) is definably isomorphic toPExt ( C p , A p ). This concludes the proof. (cid:3) We now present the proof of Proposition 4.1.
Proof of Proposition 4.1.
By Lemma 3.2, PExt (
C, A ) = s (Ext ( C, A ))and Y p ∈ P PExt ( C p , A p ) = s Y p ∈ P Ext ( C p , A p ) . Thus, by Lemma 4.6, s (Ext ( C, A )) and s (cid:16)Q p ∈ P Ext ( C p , A p ) (cid:17) are definably isomorphic.(1) We have that Ext ( C, A ) is a Polish group if and only if s (Ext ( C, A )) = 0 if and only if PExt ( C p , A p ) = 0for every p ∈ P , if and only if Ext ( C p , A p ) is a Polish group for every p ∈ P .(2) This follows from the above remarks, after observing that if G is a group with a Polish cover, then { } ∈ Γ ( G )if and only if { } ∈ Γ( s ( G )). (cid:3) Ulm subgroups and cotorsion functors.
We now fix a prime p , and assume all the groups to be p -local.(A group C is p -local if the function x qx is an isomorphism of C for every prime q other than p .) Notice that,in particular, every p -group is p -local. Recall the notion of cotorsion functor from Section 3.6. For every ordinal α < ω , the assignment A p α A is a cotorsion functor with enough projectives [Gri70, Chapter V]. For example, A p ω A is the cotorsion functor represented by the extension Z → Z [1 /p ] → Z ( p ∞ ). Here, we let Z [1 /p ] be thesubgroup of Q consisting of rationals of the form ap − n for a ∈ Z and n ∈ ω , and Z ( p ∞ ) is the Pr¨ufer p -group Z [1 /p ] / Z [Fuc70, Section 3]. More generally, for α < ω , A A α is represented by the extension Z → G α → H α as in [Gri70, Theorem 69], where H α is a countable p -group of Ulm length α . A p -group A is called totally injective if and only if A/p α A is p α -projective for every ordinal α [Fuc73, Section 82]. Every countable p -group is totallyprojective [Fuc73, Theorem 82.4]. For a countable p -group C and limit ordinal α < ω , define as in Section 2.9 L α ( C ) = lim β<α C/p β C , and consider the definable exact sequence p α C → C → L α ( C ) → E α ( C ). Lemma 4.7.
Suppose that
A, C are countable p -groups, and α < ω is a limit ordinal. Then Hom (
A, E α ( C )) and Hom (
A, p α Ext ( G α , C )) are definably isomorphic groups with a Polish cover.Proof. By Lemma 3.4(1), after replacing C with C/p α C , we can assume that p α C = 0. Consider the short definableexact sequence C → Ext ( H α , C ) → Ext ( G α , C ) induced by Z → G α → H α . As in the proof of [Kee94], Ext ( H α , C )is complete in the α -topology. Furthermore, the monomorphism C → Ext ( H α , C ) is isotype (see Definition 2.40),and hence it induces a definable monomorphism L α ( C ) → Ext ( H α , C ) ∼ = L α (Ext ( H α , C )). In turn, this induces adefinable monomorphism E α ( C ) → Ext ( G α , C ). It is proved in [Kee94] that this definable monomorphism induces HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 25 an isomorphism between the torsion subgroups of E α ( C ) and p α Ext ( G α , C ). As A is a torsion group, this inducesa definable isomorphism Hom ( A, E α ( C )) → Hom (
A, p α Ext ( G α , C )). (cid:3) The following corollary is an immediate consequence of Lemma 4.7 and Lemma 3.4, considering that everycountable p -group is totally projective. Recall that, for a p -local group and α < ω , A α = p ωα A . Corollary 4.8.
Suppose that
A, C are countable p -groups, and α < ω is an ordinal.(1) The quotient map C → C/C α induces definable isomorphisms Ext ( G ωα , C ) → Ext ( G ωα , C/C α ) and Ext ( H ωα , C ) → Ext ( H ωα , C/C α ) ;(2) If C α = 0 then Ext (
A, C ) α and Hom ( A α , E ωα ( C )) are definably isomorphic;(3) The short exact sequences A α → A → A/A α and C α → C → C/C α induce definable homomorphisms Ext (
A, C α ) → Ext ( A α , C α ) and Hom (
A, C/C α ) → Ext (
A, C α ) ,which induce a definable isomorphism η αA,C : Ext ( A, C α )Ran (Hom ( A, C/C α ) → Ext (
A, C α )) → Ext ( A α , C α ) ;(4) The short exact sequence C α → C → C/C α induces a definable exact sequence → Hom (
A, C/C α ) → Ext (
A, C α ) → Ext (
A, C ) α → Ext (
A, C/C α ) α → which induces a pure definable exact sequence → Ext (
A, C α )Ran (Hom ( A, C/C α ) → Ext (
A, C α )) ρ αA,C → Ext (
A, C ) α → Ext (
A, C/C α ) α → ;(5) The definable monomorphism r αA,C := ρ αA,C ◦ ( η αA,C ) − : Ext ( A α , C α ) → Ext (
A, C ) α restricts to a definable isomorphism γ αA,C : PExt ( A α , C α ) → Ext (
A, C ) α +1 .(6) We have a pure definable exact sequence → Ext ( A α , C α ) r αA,C → Ext (
A, C ) α → Hom ( A α , E ωα ( C )) → . Corollary 4.9.
Suppose that
A, C are countable p -groups, and α < ω is a countable ordinal. Then there is adefinable monomorphism Ext (
A, C ) α → Hom ( A α , E ωα ( C )) ⊕ Ext ( A α , C ) .Proof. Consider the definable homomorphism ψ : Ext ( A, C ) α → Hom ( A α , E ωα ( C ))as in Corollary 4.8(6) and the definable homomorphism ϕ : Ext ( A, C ) α → Ext ( A α , C )induced by the inclusion A α → A . Together, these induce a definable homomorphism ψ ⊕ ϕ : Ext ( A, C ) α → Hom ( A α , E ωα ( C )) ⊕ Ext ( A α , C ) .The kernel of ψ is the image of the definable homomorphism ρ αA,C : Ext ( A, C α )Ran (Hom ( A, C/C α ) → Ext (
A, C α )) → Ext (
A, C ) α as in Corollary 4.8(4). Consider the commuting diagram A,C α )Ran(Hom( A,C/C α ) → Ext(
A,C α )) Ext (
A, C ) α Ext ( A α , C α ) Ext ( A α , C ) ρ αA,C η A,C ϕf where f is the definable homomorphism induced by the inclusion C α → C . As Hom ( A α , C/C α ) = 0, f is amonomorphism. Since η αA,C is an isomorphism, while ρ αA,C and f are monomorphisms, we have that ϕ | Ker( ψ ) isinjective, concluding the proof. (cid:3) Complexity of
Hom . In this section, we continue to assume all the groups to be p -local for a fixed prime p .We will compute the complexity class of { } in Hom ( T, L α ( A )) for countable p -groups T and A and α < ω . Lemma 4.10.
Suppose that T is a reduced countable p -group of Ulm rank σ ≥ . The following assertion areequivalent:(1) either σ is a limit ordinal, or σ = β + 1 is a successor ordinal and T β is infinite;(2) T = L n ∈ ω T n where, for every n ∈ ω , T n is a reduced countable p -group of Ulm rank σ . If furthermore σ = β + 1 is a successor ordinal, then one can choose the T n ’s such that T βn is finite for every n ∈ ω .Proof. Recall that, by [Fuc70, Theorem 17.3], a countable reduced p -group of Ulm rank 1 is a sum of cyclic p -groups.(2) ⇒ (1) Suppose that, T = L n ∈ ω T n where, for every n ∈ ω , T n is a reduced countable p -group of Ulm rank σ .If σ = β + 1 is a successor ordinal, then we have that T β = L n ∈ ω T βn . For n ∈ ω , T n has Ulm length β + 1, andhence T βn is nonzero. Therefore, T β is infinite.(1) ⇒ (2) For every α < σ , T α /T α +1 is a countable reduced p -group of Ulm rank 1, which is in fact unbounded if α + 1 < σ . Thus, one can write T α /T α +1 = L n ∈ ω S α,n where, for every n ∈ ω , S α,n is a countable reduced p -groupof Ulm rank 1, such that S α,n is unbounded if α + 1 < σ , and S α,n is finite and nonzero if α + 1 = σ . By Ulm’sclassification of countable reduced p -groups [Fuc73, Theorem 76.1, Corollary 76.2], for every n ∈ ω there exists acountable reduced p -group T n of Ulm rank σ such that T αn /T α +1 n ∼ = S α,n for every α < σ . By Ulm’s classificationagain, we have that T ∼ = L n ∈ ω T n . (cid:3) Lemma 4.11.
Suppose that A is a countable group, and G is a group with a Polish cover. Then there is a definablemonomorphism Hom (
A, G ) → G ω . If A is generated by n elements, then there is a definable monomorphism A → G n .Proof. Consider a short exact sequence R → F → A , where F is a countable free group. This induces a definablemonomorphism Hom( A, G ) → Hom (
F, G ). Notice now that, since F is free, Hom( F, G ) is definably isomorphic toa product of copies of G indexed by the elements of a Z -basis for F . (cid:3) Lemma 4.12.
Suppose that A is a countable reduced p -group, d ∈ ω , and α < ω is a limit ordinal such that p β A = 0 for every β < α . Then: • E α ( A ) is divisible; • E α ( A ) [ p d ] = (cid:8) x ∈ E α ( A ) : p d x = 0 (cid:9) is a Polishable subgroup of E α ( A ) ; • Σ is the complexity class of { } in E α ( A ) [ p d ] and in E α ( A ) .Proof. After replacing A with A/p α A we can assume that p α A = 0. Thus, the canonical map A → L α ( A ) is amonomorphism, and E α ( A ) = L α ( A ) /A . Notice that pL α ( A ) is an open subgroup of L α ( A ), and A is a densesubgroup L α ( A ). Thus, pL α ( A ) + A = L α ( A ). Hence, the map pL α ( A ) → E α ( A ), a a + A is surjective, and E α ( A ) is divisible.Fix a strictly increasing sequence ( α n ) of countable ordinals such that α = 0 and sup n α n = α . We havethat E α ( A ) [ p d ] = ˆ G/A where ˆ G = (cid:8) x ∈ L α ( A ) : p d x ∈ A (cid:9) ⊆ L α ( A ). The Polish group topology on ˆ G is definedby letting ( x i ) converge to 0 if and only if x i → x in L α ( A ) and p d x i = p d x eventually. Thus, E α ( A ) [ p d ] isa Polishable subgroup of E α ( A ). This also follows from Proposition 2.15, since E α ( A ) [ p d ] is the kernel of thedefinable homomorphism E α ( A ) → E α ( A ), x p d x .It is clear that { } is Σ in E α ( A ), since A is countable. We now show that { } is dense in E α ( A ) [ p d ]. HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 27
Consider an element x of E α ( A ) [ p d ]. Then x can be written as a + A where a ∈ L α ( A ). In turn, one can write a as P n ∈ ω a n for a n ∈ p α n A . Since p d x = 0 we have that P n p d a n ∈ A ∩ p d L α ( A ) = p d A . Thus, we can find b ∈ A such that P n p d a n = p d b . After replacing a with a − b we can assume that P n p d a n = 0. Thus, for every k ∈ ω ,we have that P kn =0 p d a n ∈ p α k +1 + d A and hence we can find b k ∈ p α k +1 A such that P kn =0 p d a n = p d b k . Thus,we have that ( a + a + · · · + a k − b k ) k ∈ ω is a sequence in ˆ G that converges to a . This shows that { } is dense in E α ( A ) [ p d ], and hence not Π in E α ( A ) [ p d ]. This concludes the proof that Σ is the complexity class of { } in E α ( A ) [ p d ] and in E α ( A ). (cid:3) Lemma 4.13.
Suppose that T = Z ( p ∞ ) is the Pr¨ufer p -group, and A is a countable unbounded reduced p -group.Then Π is the complexity of { } in Hom (
T, E ω ( A )) .Proof. By Lemma 4.11 and Lemma 4.12, { } is Π in Hom ( T, A ). Thus, it remains to prove that { } is not Σ in Hom ( T, A ). After replacing A with A/A , we can assume that A has Ulm length 1. Thus A is a direct sum ofcyclic groups [Fuc70, Theorem 17.2]. We can write A = L k A k , where A k is a nonzero direct sum of cyclic groupsof order p ℓ k for some strictly increasing sequence ( ℓ k ) with ℓ ≥
1. (Notice that ℓ k ≥ k + 1 for every k ∈ ω .) Thenwe have that L ω ( A ) = Q k A k . Hence, Hom ( T, E ω ( A )) is definably isomorphic to ˆ G/A ω , whereˆ G = { ( x n ) ∈ L ω ( A ) ω : px ≡ A, ∀ n ∈ ω, x n ≡ px n +1 mod A } .By Proposition 2.22, it suffices to prove that the equivalence relation E ω is Borel reducible to the coset relation of A ω inside ˆ G .Define ˆ X := { , , . . . , p − } ω × ω , and let E to be the equivalence relation on ˆ X defined by setting ( ε i,k ) E ( ε ′ i,k ) ⇔∀ i ∃ k ( i ) ∀ k ≥ k ( i ) , ε i,k = ε ′ i,k . We claim that there is a definable injection ˆ X/E → ˆ G/A ω , namely E is Borelreducible to the coset relation of A ω inside ˆ G . Since E is Borel bireducible with E ω , this will conclude the proof.Define recursively for k ∈ ω and 0 ≤ i ≤ k pairwise distinct elements a i,k (0) , a i,k (1) , . . . , a i,k ( p − ∈ p k − i A k of order p i +1 such that pa i +1 ,k ( ε ) = a i,k ( ε ) for ε ∈ { , , . . . , p − } . (Recall that A k is a sum of cyclic groups oforder p ℓ k where ℓ k ≥ k + 1.) Given a sequence ( ε i,k ) ∈ ˆ X = { , , . . . , p − } ω × ω define the element ( x i ) i ∈ ω ∈ ˆ G where, for i ∈ ω , x i = ( b i,k ) k ∈ ω ∈ L ω ( A ) = Q k A k is defined by setting b i,k = 0 ∈ A k for k < i , and b i,k = a ,k ( ε i,k ) + a ,k ( ε i − ,k ) + · · · + a i,k ( ε ,k ) ∈ A k for k ≥ i . Notice that, in particular, we have that b ,k = a ,k ( ε ,k ) ∈ A k has order p , and pb i +1 ,k = p ( a ,k ( ε i +1 ,k ) + a ,k ( ε i,k ) + · · · + a i,k ( ε ,k ) + a i +1 ,k ( ε ,k ))= p ( a ,k ( ε i,k ) + · · · + a i,k ( ε ,k ) + a i +1 ,k ( ε ,k ))= a ,k ( ε i,k ) + · · · + a i − ,k ( ε ,k ) + a i,k ( ε ,k )= b i,k .Thus, ( x i ) i ∈ ω indeed defines an element of ˆ G . The function ˆ X → ˆ G , ( ε i,k ) ( x i ) is continuous. We claimthat it induces an injection ˆ X/E → ˆ G/A ω . Suppose that ( ε i,k ) and ( ε ′ i,k ) are elements of ˆ X . Define as above x i = ( b i,k ) ∈ L ω ( A ) = Q k A k and x ′ i = ( b ′ i,k ) ∈ L ω ( A ) = Q k A k for i ∈ ω such that, for k < i , b i,k = b ′ i,k = 0, andfor k ≥ i , b i,k = a ,k ( ε i,k ) + a ,k ( ε i − ,k ) + · · · + a i,k ( ε ,k ) b ′ i,k = a ,k (cid:0) ε ′ i,k (cid:1) + a ,k (cid:0) ε ′ i − ,k (cid:1) + · · · + a i,k (cid:0) ε ′ ,k (cid:1) .Suppose that, for some increasing sequence ( k ( i )) i ∈ ω in ω , for every i ∈ ω and k ≥ k ( i ) one has that ε i,k = ε ′ i,k .We claim that, for every i ∈ ω , x i − x ′ i ∈ A = L k A k . To this purpose, it suffices to prove that for every i ∈ ω andfor every k ≥ k ( i ) one has that b i,k = b ′ i,k . Indeed, we have that, for k ≥ k ( i ), b i,k = a ,k ( ε i,k ) + a ,k ( ε i − ,k ) + · · · + a i,k ( ε ,k )= a ,k (cid:0) ε ′ i,k (cid:1) + a ,k (cid:0) ε ′ i − ,k (cid:1) + · · · + a i,k (cid:0) ε ′ ,k (cid:1) = b ′ i,k . Conversely, suppose that x i − x ′ i ∈ A = L k A k for every i ∈ ω . This implies that there exists an increasing sequence( k ( i )) i ∈ ω in ω such that for every i ∈ ω and k ≥ k ( i ) one has that b i,k = b ′ i,k . We now prove by induction on i ∈ ω that for every k ≥ k ( i ) one has that ε i,k = ε ′ i,k . For i = 0 and k ≥ k (0) one has that a ,k ( ε ,k ) = b ,k = b ′ ,k = a ,k (cid:0) ε ′ ,k (cid:1) and hence ε ,k = ε ′ ,k . Suppose that the conclusion holds for i . We prove it for i + 1. We have that, for k ≥ k ( i ), b i +1 ,k = a ,k ( ε i +1 ,k ) + a ,k ( ε i,k ) + · · · + a i +1 ,k ( ε ,k )and b ′ i +1 ,k = a ,k (cid:0) ε ′ i +1 ,k (cid:1) + a ,k (cid:0) ε ′ i,k (cid:1) + · · · + a i +1 ,k (cid:0) ε ′ ,k (cid:1) and hence a ,k ( ε i +1 ,k ) = b i +1 ,k − ( a ,k ( ε i,k ) + · · · + a i +1 ,k ( ε ,k ))= b ′ i +1 ,k − (cid:0) a ,k (cid:0) ε ′ i,k (cid:1) + · · · + a i +1 ,k (cid:0) ε ′ ,k (cid:1)(cid:1) = a ,k (cid:0) ε ′ i +1 ,k (cid:1) and hence ε i +1 ,k = ε ′ i +1 ,k . This concludes the proof. (cid:3) Lemma 4.14.
Suppose that T = Z ( p ∞ ) is the Pr¨ufer p -group, α is a countable ordinal, and A is a countablereduced p -group satisfying p β A = 0 for every β < ωα . Then Π is the complexity of { } in Hom (
T, E ωα ( A )) .Proof. After replacing A with A/A α , we can assume that A α = 0. By Lemma 4.11 and Lemma 4.12, { } is Π inHom ( T, A ). It remains to prove that { } is not Σ in Hom ( T, E α ( A )). If α = β + 1 for some β < ω , then we havea definable monomorphism E ω (cid:0) A β (cid:1) → E ωα ( A ), which induces an definable monomorphism Hom (cid:0) T, E ω (cid:0) A β (cid:1)(cid:1) → Hom (
T, E ωα ( A )). By Lemma 4.13 applied to A β , { } is not Σ in Hom (cid:0) T, E ω (cid:0) A β (cid:1)(cid:1) . Whence, { } is not Σ inHom ( T, E ωα ( A )).Suppose now that α is a limit ordinal. Thus, there is a strictly increasing sequence ( α k ) of countable ordinalssuch that α = sup k α k . As is the proof of Lemma 4.10, by Ulm’s classification of countable p -groups, one can write A = L k A k where, for every k ∈ ω , A k is a countable reduced p -group of Ulm length α k . In this case, one has that L ωα ( A ) = Q k ∈ ω A k . One can then proceed as in the proof of Lemma 4.13. (cid:3) Proposition 4.15.
Suppose that T is a nonzero countable p -group, and A is a countable reduced p -group, and α < ω is an ordinal such that p β A = 0 for every β < ωα .(1) If T is finite, then Σ is the complexity class of { } in Hom (
T, E ωα ( A )) .(2) If T is infinite, then Π is the complexity class of { } in Hom (
T, E ωα ( A )) .Proof. Suppose initially that T is finite. In this case, T is a finite sum of cyclic p -groups. Thus, without loss ofgenerality, we can assume that T is cyclic of order p d for some d ≥
1. In this case, we have that Hom (
T, E α ( A )) isdefinably isomorphic to E ωα ( A ) [ p d ], and the conclusion follows from Lemma 4.12.Suppose now T is infinite. By Lemma 4.11 and Lemma 4.12, { } is Π in Hom ( T, E ωα ( A )). We now prove that { } is not Σ in Hom ( T, E ωα ( A )). It suffices to consider the case when T is either reduced or divisible. Considerinitially the case when T is reduced. The quotient map T → T /T induces a definable monomorphismHom (cid:0) T /T , E ωα ( A ) (cid:1) → Hom (
T, E ωα ( A )) .Then after replacing T with T /T , we can assume that T has Ulm rank 1, and hence T ∼ = L n T n where, for every n ∈ ω , T n is a nonzero cyclic p -group. In this case, we have thatHom ( T, E ωα ( A )) ∼ = Y n ∈ ω Hom ( T n , E ωα ( A )) .In this case, the conclusion follows from Lemma 2.31 and the case when T is finite.Suppose now that T is divisible. In this case, by the classification theorem of divisible groups [Fuc70, Theorem23.1], T is a direct sum of copies of Z ( p ∞ ). Thus, by Lemma 2.31 it suffices to consider the case when T = Z ( p ∞ ).In this case, the conclusion follows from Lemma 4.14. (cid:3) HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 29
The Solecki subgroups of
Ext . In this section, we continue to assume all the groups to be p -local. We willprove that, for countable p -groups T and A , the Solecki and Ulm subgroups of PExt ( T, A ) coincide; see Theorem4.19. In view of Lemma 4.6, the same conclusion holds when T and A are arbitrary countable groups, with T torsion. Lemma 4.16.
Suppose that
A, T are countable p -groups. Then there exist definable isomorphisms τ αT,A : PExt ( T α , A α ) → Ext (
T, A ) α +1 for α < ω that make the diagram PExt ( T α , A α ) Ext ( T, A ) α +1 PExt (cid:0) T α +1 , A α +1 (cid:1) Ext (
T, A ) α +2 τ αT,A γ Tα,Aα τ α +1 T,A commute.Proof.
We define τ αT,A by recursion on α < ω . For α = 0 one lets τ T,A to be the identity map. Suppose that τ αT,A has been defined. The definable monomorphism r T α ,A α : Ext (cid:0) T α +1 , A α +1 (cid:1) → PExt ( T α , A α ) as in Corollary 4.8restricts to a definable isomorphism γ T α ,A α : PExt (cid:0) T α +1 , A α +1 (cid:1) → PExt ( T α , A α ) .Thus, τ α +1 T,A := τ αT,A ◦ γ T α ,A α is a definable isomorphismPExt (cid:0) T α +1 , A α +1 (cid:1) → Ext (
T, A ) α +2 .Suppose now that α is a limit ordinal such that τ βT,A has been defined for every β < α . Consider the definablemonomorphism r αT,A : Ext ( T α , A α ) → Ext (
T, A ) α .This restricts to a definable isomorphism γ αT,A : PExt ( T α , A α ) → Ext (
T, A ) α +1 .Define then τ αT,A = γ αT,A . (cid:3) Lemma 4.17.
Suppose that
A, T are countable p -groups. Then s (PExt ( T, A )) = PExt (
T, A ) .Proof. Consider the definable exact sequence0 → Ext (cid:0) T , A (cid:1) r T,A → PExt (
T, A ) → Hom (cid:0) A , E ω ( C ) (cid:1) → r T,A restricts to a definable isomorphism γ T,A : PExt (cid:0) T , A (cid:1) → PExt (
T, A ) .Since { } is Π in Hom (cid:0) T , E ω ( A ) (cid:1) by Proposition 4.15, we have that r T,A (cid:0)
Ext (cid:0) T , A (cid:1)(cid:1) is Π in PExt ( T, A ).Thus, since { } is dense in s (PExt ( T, A )) and s (PExt ( T, A )) is the smallest Π Polishable subgroup of Hom (cid:0) T , E ω ( A ) (cid:1) ,we have that s (PExt ( T, A )) is contained in { } r T,A ( Ext ( T ,A )) = γ T,A (cid:0)
PExt (cid:0) T , A (cid:1)(cid:1) = PExt ( T, A ) . It remains to prove thatPExt (
T, A ) = γ T,A (cid:0)
PExt (cid:0) T , A (cid:1)(cid:1) ⊆ s (PExt ( T, A )) = s (Ext ( T, A )) .We can write T = F/R where F is a countable free group and R ⊆ F is a subgroup. Let N denote the setof strictly positive integers. For n ∈ N , define T n = { x ∈ T : p n x = 0 } . Then ( T n ) n ∈ N is an increasing sequenceof subgroups of T such that, for every n ∈ N , T n is a direct sum of cyclic p -groups. We can fix a free basis( x n,j ) n ∈ N ,j ∈ S n for F and S n ⊆ ω for n ∈ N such that:(1) for every n ∈ N , ( x n,j + R ) j ∈ S n is a p -basis of T n [Fuc70, Section 32], x nj + R generates a cyclic subgroup T n,j of T n of order d n,j ∈ N , and T n = L j ∈ S n T n,j ; (2) R is generated by elements of the form px n,t − X j ∈ S n − k j x n − ,j for n ∈ N , t ∈ S n , and k j ∈ { , , . . . , d n − ,j − } , where we set S = ∅ and by convention the empty sumis equal to 0.Let R ℓ be the set of such generators with n ∈ { , , . . . , ℓ } . Set also F = (cid:8) x ∈ F : x + R ∈ T (cid:9) . We identifyExt (cid:0) T , A (cid:1) with Hom (cid:0) R, A (cid:1) Hom ( F | R, A )and Ext ( T, A ) with Hom (
R, A )Hom ( F | R, A )via the definable isomorphisms as in Proposition 3.1.We also identify Hom (cid:0)
R, A (cid:1) with its image under the definable monomorphism Hom (cid:0) R, A (cid:1) → Hom f ( R, A )induced by the inclusion map A → A . The definable monomorphism r T,A : Ext (cid:0) T , A (cid:1) → PExt (
T, A ) is theninduced by the inclusion map Hom (cid:0)
R, A (cid:1) → Hom f ( R, A ).Suppose that ϕ ∈ Hom f (cid:0) R, A (cid:1) ⊆ Hom (
R, A ). We claim that ϕ + Hom ( F | R, A ) ∈ s (Ext ( T, A )). Fix an openneighborhood V of 0 in Hom ( F | R, A ). Then there exists n ∈ N such that { ρ | R : ρ ∈ Hom (
F, A ) , ∀ i ≤ n , ∀ j ∈ S i , ρ ( x i,j ) = 0 } ⊆ V .It suffices to show that there exists ψ ∈ Hom ( F | R, A ) such that ϕ ∈ V + ψ Hom(
R,A ) . This amounts to showing thatthere exists ρ ∈ Hom (
F, A ) such that for every n ∈ N there exists ˜ ρ ∈ Hom (
F, A ) satisfying:(1) ˜ ρ | R n = ϕ | R n ;(2) ˜ ρ ( x i,j ) = ρ ( x i,j ) for i ≤ n and j ∈ S i .Since ϕ ∈ Hom f (cid:0) R, A (cid:1) , and since T [ p n ] is a sum of cyclic p -groups, we have that there exists˜ ϕ : h x i,j : i ≤ n , j ∈ S i i → A extending ϕ , where h x i,j : i ≤ n , j ∈ S i i is the subgroup of F generated by x i,j for i ≤ n and j ∈ S i . Define thenthe group homomorphism ρ : F → A by setting, for i ∈ N and j ∈ S i , ρ ( x i,j ) = (cid:26) ˜ ϕ ( x i,j ) if i ≤ n ,0 otherwise.Notice that ρ | R n = ϕ | R n . We claim that ρ satisfies the desired conclusion.Suppose that n ≥ n . We define a group homomorphism ˜ ρ : F → A by setting ˜ ρ ( x i,j ) = y i,j for i ∈ N and j ∈ S i , where we define y i,j as follows. We set y i,j := ρ ( x i,j ) for i ≤ n and y i,j := 0 for i > n . We now define y ℓ,j ∈ p n − ℓ A for n ≤ ℓ ≤ n and every j ∈ S ℓ by recursion on ℓ . By definition, we have that y n ,j ∈ A ⊆ p n − n A for every j ∈ S n . Suppose that y i,j has been defined for i ≤ ℓ and every j ∈ S i , for some ℓ ∈ { n , . . . , n − } . For t ∈ S ℓ +1 there exists a unique expression px ℓ +1 ,t = X j ∈ S ℓ k j x ℓ,j + r where r ∈ R and k j ∈ { , , . . . , d ℓ,j − } . Notice that X j ∈ S ℓ k j y ℓ,j + ϕ ( r ) ∈ p n − ℓ A by the inductive hypothesis. Whence, there exists y ℓ +1 ,t ∈ p n − ( ℓ +1) A such that py ℓ +1 ,t = X j ∈ S ℓ k j y ℓ,j + ϕ ( r ) .This concludes the recursive definition. HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 31
By definition, we have that ˜ ρ ( x i,j ) = ρ ( x i,j ) for i ≤ n and j ∈ S i . It remains to show that ˜ ρ | R n = ϕ | R n . Anelement r of R n is of the form px ℓ,j − X j ∈ S ℓ − k j x ℓ − ,j for some ℓ ≤ n and k j ∈ { , , . . . , d ℓ − ,j − } . Then we have that px ℓ,j = X j ∈ S ℓ − k j x ℓ − ,j + r and hence by definition ρ ( px ℓ,j ) = X j ∈ S ℓ − k j ρ ( x ℓ − ,j ) + ϕ ( r )Therefore, ρ ( r ) = ϕ ( r ). This concludes the proof. (cid:3) Lemma 4.18.
Suppose that
A, T are countable p -groups, and n ∈ ω . Then s n (Ext ( T, A )) = s n (PExt ( T, A )) =PExt (
T, A ) n .Proof. We prove the statement by induction on n ∈ ω . For n = 0, this follows from the fact that PExt ( T, A ) isthe closure of { } in Ext ( T, A ) by Lemma 3.2. Suppose that the conclusion holds for n . We will show that itholds for n + 1. We need to show that s (PExt ( T, A ) n ) = PExt ( T, A ) n +1 . By Lemma 4.16 there exists a definableisomorphisms τ nT,A : PExt ( T n , A n ) → PExt (
T, A ) n for n ∈ ω that make the following diagram commute:PExt ( T n , A n ) PExt ( T, A ) n PExt (cid:0) T n +1 , A n +1 (cid:1) PExt (
T, A ) n +1 τ nT,A γ Tn,An τ n +1 T,A
Thus, it suffices to prove that s (PExt ( T n , A n )) = γ T n ,A n (cid:0) PExt (cid:0) T n +1 , A n +1 (cid:1)(cid:1) = PExt ( T n , A n ) .This follows from Lemma 4.17 applied to T n and A n . (cid:3) Theorem 4.19.
Suppose that
A, T are countable p -groups, and α < ω is an ordinal. Then s α (Ext ( T, A )) = s α (PExt ( T, A )) = PExt (
T, A ) α = Ext ( T, A ) α . In particular, the Solecki rank and the Ulm rank of PExt (
T, A ) are equal.Proof. We prove this holds by induction on α < ω . The case α < ω has already been considered in Lemma 4.18.If α is a limit ordinal, and the conclusion holds for β < α , then we have that s α (Ext ( T, A )) = \ β<α s β (Ext ( T, A )) = \ β<α Ext (
T, A ) β = Ext ( T, A ) α .Suppose now that the conclusion holds for α . We will show that it holds for α + 1. By Lemma 4.18 we can assumethat α ≥ ω , in which case we have that 1 + α = α and hence PExt ( T, A ) α = Ext ( T, A ) α and PExt ( T, A ) α +1 =Ext ( T, A ) α +1 . Thus, in this case we need to show that s (Ext ( T, A ) α ) = Ext ( T, A ) α +1 .By Lemma 4.16 there exist definable isomorphisms τ βT,A : PExt (cid:0) T β , A β (cid:1) → PExt (
T, A ) β +1 for β < ω for suchthat the following diagram commutes: PExt (cid:0) T β , A β (cid:1) Ext (
T, A ) β +1 PExt (cid:0) T β +1 , A β +1 (cid:1) Ext (
T, A ) β +2 τ βT,A γ Tβ,Aβ τ β +1 T,A
Consider initially the case when α = β + 1 is a successor ordinal. Thus, it suffices to prove that s (cid:0) PExt (cid:0) T β , A β (cid:1)(cid:1) = γ T β ,A β (cid:0) PExt (cid:0) T β +1 , A β +1 (cid:1)(cid:1) .This follows from Lemma 4.17 applied to T β and A β .Suppose now that α is a limit ordinal. Consider the definable exact sequence0 → Ext ( T α , A α ) r αT,A → Ext (
T, A ) α → Hom ( T α , E ωα ( A )) → { } is Π in Hom ( T α , E ωα ( A )) by Proposition 4.15, we have that r αT,A (Ext ( T α , A α )) is Π in Ext ( T, A ) α . Since s (Ext ( T, A ) α ) is the smallest Π subgroup of Ext ( T, A ) α , this implies that s (Ext ( T, A ) α ) ⊆ r αT,A (Ext ( T α , A α )). Since r αT,A (PExt ( T α , A α )) = Ext ( T, A ) α +1 is the closure of { } in r αT,A (Ext ( T α , A α )) byLemma 3.2 and { } is dense in s (Ext ( T, A ) α ), we have that s (Ext ( T, A ) α ) ⊆ Ext (
T, A ) α +1 . Thus, it remainsto prove that Ext ( T, A ) α +1 = γ αT,A (PExt ( T α , A α )) ⊆ s (Ext ( T, A ) α ).To this purpose, we proceed as in the proof of Lemma 4.17, also adopting the same notation. For β ≤ α , we set F β = (cid:8) x ∈ F : x + R ∈ T β (cid:9) , and we identify Ext (cid:0) T β , A β (cid:1) withHom (cid:0) R, A β (cid:1) Hom ( F β | R, A β )and Ext ( T, A ) with Hom (
R, A )Hom ( F | R, A )via the definable isomorphisms as in Proposition 3.1. For β ≤ α , we also identify Hom (cid:0) R, A β (cid:1) with its image underthe definable monomorphism Hom (cid:0) R, A β (cid:1) → Hom (
R, A ) induced by the inclusion A β → A . Thus, we have that r βT,A (cid:0) Ext (cid:0) T β , A β (cid:1)(cid:1) = Hom (cid:0) R, A β (cid:1) + Hom ( F | R, A )Hom ( F | R, A ) ⊆ Ext (
T, A ) .Note that, for β < α , we have thatExt (
T, A ) β +1 ⊆ r βT,A (cid:0) Ext (cid:0) T β , A β (cid:1)(cid:1) ⊆ Ext (
T, A ) β and hence Ext ( T, A ) α = \ β<α r βT,A (cid:0) Ext (cid:0) T β , A β (cid:1)(cid:1) .By the inductive hypothesis we have that s α (Ext ( T, A )) = Ext (
T, A ) α = \ β<α r βT,A (cid:0) Ext (cid:0) T β , A β (cid:1)(cid:1) = \ β<α Hom (cid:0)
R, A β (cid:1) + Hom ( F | R, A )Hom ( F | R, A )and hence ˆ s α (Ext ( T, A )) = \ β<α (cid:0) Hom (cid:0)
R, A β (cid:1) + Hom ( F | R, A ) (cid:1) ⊆ Hom (
R, A ) .Suppose that ϕ ∈ Hom f ( R, A α ) ⊆ Hom (
R, A ). We claim that ϕ + Hom ( F | R, A ) ∈ s (Ext ( T α , A α )). Indeed,fix an open neighborhood V of 0 in Hom ( F | R, A ). Then there exists n ∈ N such that { ρ | R : ρ ∈ Hom (
F, A ) , ∀ i ≤ n , ∀ j ∈ S i , ψ ( x i,j ) = 0 } ⊆ V .Since ϕ ∈ Hom f ( R, A α ), and since T [ p n ] is a sum of cyclic p -groups, we have that there exists a homomorphism˜ ϕ : h x i,j : i ≤ n , j ∈ S i i → A α extending ϕ . Define then the group homomorphism ρ : F → A α by setting, for i ∈ N and j ∈ S i , ρ ( x i,j ) = (cid:26) ˜ ϕ ( x i,j ) if i ≤ n ρ | R n = ϕ | R n . Set ψ := ρ | R ∈ Hom ( F | R, A α ) ⊆ Hom ( F | R, A ). HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 33
For β < α , as A α ⊆ A β +1 , the same argument as in the proof of Lemma 4.17 shows that ϕ ∈ V + ψ Hom ( R,A β ) ⊆ V + ψ Hom ( R,A β ) +Hom( F | R,A ) .Since ˆ s α (Ext ( T, A )) = \ β<α (cid:0) Hom (cid:0)
R, A β (cid:1) + Hom ( F | R, A ) (cid:1) as remarked above, this implies that ϕ ∈ V + ψ ˆ s α (Ext( T,A )) .As this holds for every open neighborhood V of 0 in Hom ( F | R, A ), this shows that ϕ ∈ ˆ s (ˆ s α (Ext ( T, A ))) = ˆ s α +1 (Ext ( T, A ))and hence ϕ ∈ s α +1 (Ext ( T, A )) , concluding the proof. (cid:3) Complexity of
Ext . The goal of this section is to compute the complexity class of { } in Ext ( T, A ) givencountable p -groups T and A . This is the main ingredient in the proof of Theorem 1.5. Lemma 4.20.
Suppose that
A, T are countable p -groups, where A is reduced and nonzero. Then Ext (
T, A ) iscountable if and only if (a) T /D ( T ) is finite, and (b) D ( T ) = 0 if A is unbounded.Proof. Recall the definition of Ext w from Section 3.3. Notice that, by Lemma 3.2, Ext ( T, A ) is countable ifand only if Ext w ( T, A ) is countable, in which case Ext (
T, A ) = Ext w ( T, A ). By Lemma 3.3 applied to thecotorsion functor C C = p ω C for p -local groups, we have that the quotient map A → A/A induces a definableepimorphism Ext ( T, A ) → Ext (cid:0)
T, A/A (cid:1) such that PExt ( T, A ) is the preimage of PExt (cid:0)
T, A/A (cid:1) . Thus, we havean isomorphism of Polish groups Ext w ( T, A ) ∼ = Ext w (cid:0) T, A/A (cid:1) .Hence, without loss of generality, we can assume that A = 0 and A has Ulm rank 1.If C is a cyclic p -group of order p d , then Ext ( C, A ) ∼ = Ap d A is countable, while Ext w ( Z ( p ∞ ) , A )is isomorphic to the p -adic completion of A , which is countable if and only if A is bounded, in which caseExt w ( Z ( p ∞ ) , A ) ∼ = A [Fuc70, Lemma 56.6].Suppose that Ext ( T, A ) is countable. This implies that PExt (
T, A ) = 0 and Ext (
T, A ) = Ext w ( T, A ). By theabove, T must be reduced if A is unbounded. After replacing T with T /D ( T ), we can assume that T is reduced.The exact sequence T → T → T /T induces a short definable exact sequence0 = Hom (cid:0) T , A (cid:1) → Ext (cid:0)
T /T , A (cid:1) → Ext (
T, A ) → Ext (cid:0) T , A (cid:1) → (cid:0) T /T , A (cid:1) is countable. Since T /T has Ulm rank 1, we have that T /T ∼ = L n C n where,for every n ∈ ω , C n is cyclic of order p d n . Thus, we have thatExt (cid:0) T /T , A (cid:1) ∼ = Y n ∈ ω Ap d n A .This implies that (cid:8) n ∈ ω : p d n > (cid:9) is finite. Therefore, T /T is finite, whence T = 0 and T = T /T is finite.The converse implication is immediate from the above remarks. (cid:3) Lemma 4.21.
Suppose that
A, T are countable p -groups, where A is reduced. Then Ext (
T, A ) is a Polish group ifand only if either T = 0 or A is bounded. Proof.
By Corollary 4.8(6) we have a short definable exact sequence0 → Ext (cid:0) T , A (cid:1) → PExt (
T, A ) → Hom (cid:0) T , E ω ( A ) (cid:1) → T = 0 then Ext (cid:0) T , A (cid:1) = Hom (cid:0) T , E ω ( A ) (cid:1) = 0 and hence PExt ( T, A ) = 0. Similarly, if A is bounded, then A = 0 and E ω ( A ) = 0 and hence PExt ( T, A ) = 0. Conversely, if T is nonzero and A is unbounded, then byProposition 4.15 we have that Hom (cid:0) T , E ω ( A ) (cid:1) is nonzero and hence PExt ( T, A ) is nonzero. (cid:3)
Theorem 4.22.
Suppose that
A, T are countable p -groups, where T is nonzero and A is reduced and unbounded.Define µ to be the least countable ordinal such that either T µ = 0 or µ is the successor of µ − and A µ − = 0 .Then:(1) If µ is a limit ordinal, then Π µ is the complexity class of { } in Ext (
T, A ) ;(2) If µ = 1 + λ + 1 where λ is either zero or limit and T λ is finite, then Σ µ is the complexity class of { } in Ext (
T, A ) ;(3) If µ = 1 + λ + n where λ either zero or limit and ≤ n < ω , and A λ + n − is bounded, then Π µ is thecomplexity class of { } in Ext (
T, A ) ;(4) If µ = 1 + λ + n where λ is either zero or limit, ≤ n < ω , T λ + n − is finite, and A λ + n − is unbounded,then D (cid:0) Π µ (cid:1) is the complexity class of { } in Ext (
T, A ) (5) If µ = 1+ λ + n where λ is either zero or limit, ≤ n < ω , T λ + n − is infinite, and A λ + n − is unboundedif n ≥ , then Π µ +1 is the complexity class of { } in Ext (
T, A ) .In particular, { } is Π µ +1 and not Π α for α < µ in Ext (
T, A ) .Proof. Recall that, by Corollary 4.8 and Corollary 4.9, for every α < ω we have a definable short exact sequence0 → Ext ( T α , A α ) → Ext (
T, A ) α → Hom ( T α , E ωα ( A )) → T, A ) α → Hom ( T α , E ωα ( A )) ⊕ Ext ( T α , A ) .By Theorem 4.19, if 1 + α is the Ulm rank of Ext ( T, A ), then the Solecki rank of Ext (
T, A ) is α .(1) Suppose that µ is a limit ordinal. In this case, T is reduced and µ is the Ulm rank of T . For α < µ we have T α = 0 and p δ A = 0 for δ < ωα . Therefore, by Proposition 4.15, Ext ( T α , E ωα ( A )) = 0, and hence Ext ( T, A ) α = 0for α < µ . Similarly, for µ ≤ α < ω we have that T α = 0 and hence Ext ( T, A ) α = 0. This shows that the Ulmrank and the Solecki rank of Ext ( T, A ) are µ . By Lemma 4.10 we have that T ∼ = L n ∈ ω T n where, for every n ∈ ω , T n has Ulm rank µ . Therefore, Ext ( T, A ) ∼ = Q n Ext ( T n , A ) where, for every n ∈ ω , Ext ( T n , A ) has Solecki rank µ .The conclusion thus follows from Lemma 2.32.(2) Suppose that µ = 1 + λ + 1 where λ is either zero or a limit ordinal, and T λ is finite, in which case T isreduced. Thus, we have that T λ +1 = 0 and Ext ( T, A ) λ +1 = 0. Furthermore, we have that Ext (cid:0) T λ , A (cid:1) iscountable by Lemma 4.20, and Σ is the complexity class of { } in Hom (cid:0) T λ , E ω (1+ λ ) ( A ) (cid:1) by Proposition 4.15,using the assumption that A is unbounded in the case when λ = 0. Considering the definable epimorphismExt ( T, A ) λ → Hom (cid:0) T λ , E ω (1+ λ ) ( A ) (cid:1) = 0and the definable monomorphismExt ( T, A ) λ → Hom (cid:0) T λ , E ω (1+ λ ) ( A ) (cid:1) ⊕ Ext (cid:0) T λ , A (cid:1) ,we obtain that Ext ( T, A ) λ = s λ (Ext ( T, A )) = 0 and { } is Σ in Ext ( T, A ) λ . Hence, the Solecki rank ofExt ( T, A ) is λ + 1, and Σ λ +1 is the complexity class of { } in Ext ( T, A ) by Proposition 2.30.(3) Suppose that µ = 1 + λ + n for 2 ≤ n < ω and λ either zero or limit, and A λ + n − is bounded. Then wehave that A λ + n − = 0 and E ω (1+ λ + n − ( A ) = 0. Thus,Ext (cid:0) T λ + n − , A λ + n − (cid:1) = 0and Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) = 0, HE CLASSIFICATION PROBLEM FOR EXTENSIONS OF TORSION ABELIAN GROUPS 35 and hence Ext (
T, A ) λ + n − = 0. Since p β A = 0 for β < ω (1 + λ + n −
2) and T λ + n − is infinite, we havethat Π is the complexity class of { } in Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) by Proposition 4.15. Considering thedefinable short exact sequence0 → Ext (cid:0) T λ + n − , A λ + n − (cid:1) → Ext (
T, A ) λ + n − → Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) → Π is the complexity class of Ext (cid:0) T λ + n − , A λ + n − (cid:1) in Ext ( T, A ) λ + n − , the Ulm rank ofExt ( T, A ) is 1 + λ + n −
1, and the Solecki rank of Ext (
T, A ) is λ + n − Π λ + n is the complexity class of { } in Ext ( T, A ). Notice thatExt (
T, A ) λ + n − = s λ + n − (Ext ( T, A )) . Thus, by Proposition 2.30, it suffices to prove that { } is not Σ in Ext ( T, A ) λ + n − . It also suffices to considerthe case when T is either divisible or reduced.When T is divisible, we have that Ext (cid:0) T λ + n − , A λ + n − (cid:1) = Ext (cid:0) T, A λ + n − (cid:1) is countable by Lemma 4.20.Since Ext (cid:0) T λ + n − , A λ + n − (cid:1) is not Σ in Ext ( T, A ) λ + n − , this implies that { } is not Σ in Ext ( T, A ) λ + n − .Consider now the case when T is reduced. Since T λ + n − is infinite, then as in the proof of Lemma 4.10 onecan write T = L k ∈ ω T k where, for every n ∈ ω , T k is a countable p -group such that T λ + n − k is nonzero. Fix k ∈ ω . Then, as above, Hom (cid:0) T λ + n − k , E ω (1+ λ + n − ( A ) (cid:1) = 0 by Proposition 4.15. Considering the definableepimorphism Ext ( T k , A ) λ + n − → Hom (cid:0) T λ + n − k , E ω (1+ λ + n − ( A ) (cid:1) = 0and the definable short exact sequence0 = Ext (cid:0) T λ + n − k , A λ + n − (cid:1) → Ext ( T k , A ) λ + n − → Hom (cid:0) T λ + n − k , E ω (1+ λ + n − ( A ) (cid:1) = 0we have that Ext ( T k , A ) λ + n − = 0 and Ext ( T k , A ) λ + n − = 0. Since { } is dense in Ext ( T k , A ) λ + n − = s λ + n − (Ext ( T k , A )), we have that { } is Π and not Π in Ext ( T k , A ) λ + n − . Since Ext ( T, A ) λ + n − isdefinably isomorphic to Q k ∈ ω Ext ( T k , A ) λ + n − , it follows from Lemma 2.31 that Π is the complexity class of { } in Ext ( T, A ) λ + n − .(4) Suppose that µ = 1 + λ + n where λ is either zero or a limit ordinal, 2 ≤ n < ω , and T λ + n − is finite, in which case T is reduced, and A λ + n − is unbounded. Thus, we have that T λ + n = 0 andExt ( T, A ) λ + n = 0. Furthermore, Ext (cid:0) T λ + n − , A (cid:1) is countable by Lemma 4.20, and Σ is the complexityclass of { } in Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) by Proposition 4.15, since A λ + n − is unbounded. Therefore,considering the definable epimorphismExt ( T, A ) λ + n − → Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) = 0and the definable monomorphismExt ( T, A ) λ + n − → Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) ⊕ Ext (cid:0) T λ + n − , A (cid:1) .we have that Ext ( T, A ) λ + n − = s λ + n − (Ext ( T, A )) = 0 and { } is Σ in Ext ( T, A ) λ + n − . Hence, the Soleckirank of Ext ( T, A ) is λ + n , and D (cid:0) Π λ + n (cid:1) is the complexity class of { } in Ext ( T, A ) by Proposition 2.30.(5) Suppose now that µ = 1 + λ + n where λ is either zero or limit, 1 ≤ n < ω , T λ + n − is infinite, and A λ + n − is unbounded if n ≥
2. Then we have that either T λ + n = 0, or A λ + n − = 0. Since T = 0, thisimplies that n > λ = 0. In either case, we have that Ext ( T, A ) λ + n = 0. Since A λ + n − is unbounded if n ≥
2, by Proposition 4.15 we have that Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) = 0 and hence Ext ( T, A ) λ + n − = 0.Therefore, we have that the Solecki rank of Ext ( T, A ) is λ + n .We now show that Π µ +1 is the complexity class of { } in Ext ( T, A ). We consider initially the case when A λ + n − = 0. In this case, we have that Ext ( T, A ) λ + n − is definably isomorphic to Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) .By Proposition 4.15, Π is the complexity class of { } in Hom (cid:0) T λ + n − , E ω (1+ λ + n − ( A ) (cid:1) . Thus, Π is the com-plexity class of { } in Ext ( T, A ) λ + n − = s λ + n − (Ext ( T, A )). Thus, by Proposition 2.30, Π λ + n +1 is thecomplexity class of { } in Ext ( T, A ).Consider now the case when T λ + n = 0, whence T is reduced. In this case, by Lemma 4.10, we have that T ∼ = L k ∈ ω T k where, for every k ∈ ω , T k is a reduced countable p -group of Ulm rank 1 + λ + n such that T λ + n − k is infinite. By the above discussion, for every k ∈ ω , the Solecki rank of Ext ( T k , A ) is λ + n . Thus, by Lemma2.32, Π λ + n +1 is the complexity class of { } in Q k ∈ ω Ext ( T k , A ). As Q k ∈ ω Ext ( T k , A ) is definably isomorphic toExt ( T, A ), this concludes the proof. (cid:3)
We conclude with the proofs of Theorem 1.2 and Theorem 1.5.
Proof of Theorem 1.2. (1) By Proposition 2.22 and the remarks in Section 3.4, we have that R Ext ( C,A ) is smooth ifand only if Ext ( C, A ) is a Polish group. By Proposition 4.1, this holds if and only if, for every p ∈ P , Ext ( C p , A p )is a Polish group. Finally, by Lemma 4.21, this holds if and only if, for every p ∈ P , either u ( C p ) = 0 or A p isbounded.(2) As above, we have that R Ext ( C,A ) is essentially hyperfinite if and only if { } is Σ in Ext ( C, A ). ByProposition 4.1 and Lemma 2.31, this holds if and only if, for every p ∈ P , { } is Σ in Ext ( C p , A p ), and thenumber of p ∈ P such that Σ is the complexity class of { } in Ext ( C p , A p ) is finite. By Theorem 4.22, this isequivalent to the assertion that: (a) for every p ∈ P , either u ( C p ) = 0 or u ( A p ) = 0; and (b) for every p ∈ P suchthat A p is unbounded, u ( C p ) is finite; and (c) the set of p ∈ P such that A p is unbounded and u ( C p ) is nonzerois finite.(3) Again, we have that R Ext ( C,A ) is Borel reducible to E ω if and only if { } is Π in Ext ( C, A ), if and onlyif for every prime number p , { } is Π in Ext ( C p , A p ). By Theorem 4.22, for every prime number p , { } is Π inExt ( C p , A p ) if and only if one of the following holds: (a) u ( C p ) = 0; or (b) u ( A p ) = 0; or (c) u ( C p ) = 0 and u ( A p ) is bounded. (cid:3) Proof of Theorem 1.5.
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