Groups definable in Presburger arithmetic

TL;DR

This paper classifies all groups that can be defined within Presburger arithmetic, providing a comprehensive understanding of their structure up to finite index subgroups.

Contribution

It offers a complete classification of definable groups in Presburger arithmetic, a significant step in understanding its model-theoretic properties.

Findings

All definable groups are virtually abelian.

Classification up to finite index subgroups.

Provides a structural description of these groups.

Abstract

We determine all groups definable in Presburger arithmetic, up to a finite index subgroup.

Equations

$$tp(a_i/0)\in X,a_1\models {\Sigma }_d,a_2\models {\Sigma }_{d{a}_1},a_3\models {\Sigma }_{d{a}_1{a}_2}.$$

$$D_1=\{d\in D\mid \text{There exists }t,p_0\cup {\Sigma }_{dt}⊢T_{dt},\text{ and },h_{dt}:T_{dt}\to S_d\text{ is injective }\}.$$

$$i(x+y)=i(x)\oplus i(y)$$

$$k(z)\oplus k(w)=i(x+z)\oplus i(y+w)\ominus (i(x)\oplus i(y))=i(x+y+z+w)\ominus i(x+y)=k(z+w).$$

$$O(a)=\{x\in Z\mid \text{ there exists }n\in {\mathbb{Z}}_{>0},|x|<na\}$$

$$o(a)=\{x\in Z\mid \text{ for all }n\in {\mathbb{Z}}_{>0},n|x|<a\}.$$

Full text

1 Introduction

In [5], it is proven that groups definable in Presburger arithmetic are abelian-by-finite. Additionally, bounded groups are described completely. The main theorem here is Theorem 19, which describes completely all groups definable in Presburger arithmetic.

We give now an overview of the proof. We prove a version of the Ellis-Nakamura Lemma which works simultaneously for semigroups over definable bounded sets of Presburger arithmetic and over profinite topological spaces. The statement of the hypothesis uses the concept of a pro-definable set. This is used to find an idempotent in a space of types obtained from a definable group. In this idempotent the associativity constraint forces the group structure to be the familiar $(Z^{r},+)$. This group can be recovered globally (that is, not just on the type) using differences in the given definable group in a process similar to the groupification, so we obtain an injective group map $Z^{r}\to G$ with bounded cokernel, see Theorem 17. Finally we use group cohomology to calculate the isomorphism type of the group extension, by choosing a complete type, where the cycle associated to the extension is an affine function, so the cycle condition is seen to be trivial, and then use differences and an observation about extensions of local group morphisms to global group morphisms on $\mathbb{R}$ to obtain a group section sufficient to determine $G$.

2 The degree of unboundedness

In what follows $Z$ will denote an $\omega$-saturated $Z$-group. We will use the shorthand $x\ll y$ for $x,y\in Z$ to mean $0<x<y$ and $nx<y$ for all $n\in\mathbb{Z}_{>0}$. If $a$ is a finite tuple then $|a|={\rm Max}_{i}|a_{i}|$. If $x\in Z^{r}$ then $a\ll x$ means $|a|\ll x_{1}\ll x_{2}\ll\dots\ll x_{r}$. We shall call a $x\in Z^{r}$ divisible if $x\in nZ^{r}$ for all $n\in\mathbb{Z}_{>0}$. More generally for $p\in\hat{\mathbb{Z}}^{r}$ and $x\in Z^{r}$ define $x\equiv p$ or $\hat{x}=p$ if and only if $x_{i}\equiv p_{i}\mod N$ for every $N\in\mathbb{Z}_{>0}$, under the isomorphism $Z/NZ\cong\hat{\mathbb{Z}}/N\hat{\mathbb{Z}}\cong\mathbb{Z}/N\mathbb{Z}$.

The letter $t$ usually denotes a tuple with components in $Z$, which will usually be taken to be parameters. For instance if $t\in T\subset Z^{s}$ is a [math]-definable set, a family $\{S_{t}\}_{t\in T}$ of sets $S_{t}\subset Z^{r}$ is called [math]-definable if $S=\bigcup_{t\in T}S_{t}\times\{t\}\subset Z^{r}\times T$ is [math]-definable. If $X$ is a $t$-definable set, say it is defined by a formula $\phi(x,t)$, we sometimes denote $X=X_{t}$ where $X_{t^{\prime}}$ is the set defined by the formula $\phi(x,t^{\prime})$. Note that $\{X_{t}\}_{t\in Z^{s}}$ is a [math]-definable family.

We note that for Presburger arithmetic there is a quantifier elimination and a notion of dimension of definable sets see [2] for the definition and properties.

We shall make use of cell decomposition for $Z$, see [2]. We provide the statement of this theorem next.

Definition 1**.**

We define the notion of an $a$-definable cell $C\subset Z^{n}$ by induction on $n$. If $C\subset Z^{n}$ is an $a$-definable cell then $D\subset Z^{n+1}$ is said to be an $a$-definable cell if it is of one of the following forms, first $D=\{(x,y)\in Z^{n}\times Z\mid x\in C,f(x)<y<g(x),y\equiv k\mod N\}$ for a $N\in\mathbb{Z}_{>0}$ and $k\in\mathbb{Z}/N\mathbb{Z}$, and $f,g:C\to Z$ given by $f(x)=A_{1}x+c_{1}$ and $g(x)=A_{2}x+c_{2}$, for $A_{i},B_{i}$ rational matrices, $c_{i}$ $a$-definable constants, and $C$ satisfying conditions of divisibility that make the image of $f$ and $g$ belong to $Z$, and also such that $g(x)-f(x)>N$. We also accept as cells those where $f$ or $g$ do not appear.

This definition is somewhat different from the usual one, which requires $f-g$ to be unbounded, or equal to $2$, but for our purposes it is enough.

Proposition 2**.**

If $A_{1},\dots,A_{r}$ are $a$-definable sets which form a partition of $B$ and $f:B\to Z^{n}$ is an $a$-definable function then there exists an $a$-cell decomposition of $B$ $C_{1},\dots,C_{s}$ such that $C_{i}$ refine $A_{j}$ and $f$ restricted to each $C_{i}$ is of the form $x\mapsto Ax+c$ for $A$ a matrix with rational coefficients and $c$ an $a$-definable constant.

Now we define an invariant $r={\rm ubd}(X)$ for every definable set $X$, the “degree of unboundedness” to be the greatest $r\in\mathbb{N}$ such that there exists $f:Z^{r}\to X$ injective. We prove later that $X$ is bounded if and only if ${\rm ubd}(X)=0$.

We start with a simple remark

Lemma 3**.**

For a definable set $X$, ${\rm ubd}(X)$ is also the greatest $r$ such that there is an injective definable function $Z_{>0}^{r}\to X$. In fact, $Z$ is in definable bijection with $Z_{>0}$.

Proof.

This follows by considering even and odd numbers. More precisely $f:Z\to Z_{>0}$ is defined as $f(x)=2x$ if $x>0$ and $f(x)=1-2x$ if $x\leq 0$. ∎

Lemma 4**.**

Let $c$ be a tuple of elements in $Z$. Let $C$ be a definable set containing the set of $x\in Z^{r}$ such that $c\ll x$ and $x$ is divisible. Then there exists an injective definable map $Z^{r}\to C$.

Proof.

By compactness there exists $d\in Z_{>0}$, $n_{1},\dots,n_{r-1}\in\mathbb{Z}_{>0}$, and $N\in\mathbb{Z}_{>0}$ such that $C$ contains the set of $x$ such that $d<x_{1},n_{1}x_{1}<x_{2},\dots,n_{r}x_{r-1}<x_{r}$, and $x_{i}\in NZ$. Then the map $f:Z_{>0}^{r}\to C$ given by $x\mapsto y$ where $y_{1}=Nd+Nx_{1}$ and $y_{i}=n_{i}y_{i-1}+Nx_{i}$, is injective definable. We conclude with Lemma 3. ∎

Lemma 5**.**

Let $c$ be a finite tuple in $Z$. If $X$ is the set of elements $x\in Z^{r}$ such that $c,1\ll x$ and $x$ is divisible, then $X$ is the set of realizations of a complete type over $c$. In particular if $f:X\to Z^{s}$ is $c$-definable then there is a constant $d$ and a rational matrix $A$ with $f(x)=Ax+d$ for all $x\in X$.

Proof.

By elimination of quantifiers we have to see that if $x,y\in X$ then for every $n\in\mathbb{Z}^{r}$ and $m\in\mathbb{Z}^{s}$, $l,N\in\mathbb{Z}$, we have $nx+mc+l>0$ if and only if $ny+mc+l>0$ and $nx+mc+l\equiv ny+mc+l\mod N$. The second follows because $x,y$ are divisible. The first follows trivially if $n=0$. If $n\neq 0$ then $n_{k}x_{k}$ is the dominant term of $nx+mc+l$, if $n_{k}$ is the first nonzero term of $n$ from right to left. So $nx+mc+l>0$ is equivalent to $n_{k}>0$, and similarly for $y$, as required. ∎

Lemma 6**.**

If $X\subset Z^{t}$ is a definable set, then there exist $X_{i}$ disjoint definable sets such that $X=X_{1}\cup\dots\cup X_{p}$ and for each $i$ there exists $r_{i}\in\mathbb{Z}_{\geq 0}$, a bounded cell $C_{i}$ with $C_{i}\subset Z^{t-r_{i}}$, and a bijection $f_{i}:Z_{>0}^{r_{i}}\times C_{i}\to X_{i}$, which are of the form $f_{i}(x)=A_{i}x+b_{i}$ for $A_{i}$ matrices with integer coefficients and $b_{i}$ tuples in $Z^{t}$.

Proof.

This proof is by induction on $t$. The base case is formally $t=0$ which is trivial. Alternatively the proof of the induction step establishes $t=1$.

So assume the lemma is true for $t$ and we prove it for $t+1$. Let $X\subset Z^{t+1}$ be a definable set. Using cell decomposition we may assume $X$ is a cell. In other words, there is a cell $Y\subset Z^{t}$ such that $X=\{(a,z)\mid a\in Y,g(a)<z<f(a),z\equiv s\mod N\}$, for affine functions $g,f$ such that $g(a)+N<f(a)$ for all $a\in Y$, or where $f$ or $g$ do not appear.

Applying the bijection $z\mapsto Nz+s$ we may assume $N$ does not appear in the definition of $X$. If $M$ is the largest denominator of the rational numbers which appear as coefficients of $f,g$; then after dividing $X$ into $M^{t}$ sets according to the residue of $a$ mod $M$, and applying the bijections $a\mapsto Ma+i$ for $i\in[0,M)^{t}$ we may further assume that $f,g$ have integer coefficients.

Now we apply the induction hypothesis to $Y$ to assume $Y=Z_{>0}^{r}\times B$, for a bounded cell $B\subset Z^{t-r}$. We write $X=\{(a,b,z)\mid g(a,b)<z<f(a,b),a\in Z_{>0}^{r},b\in B\}$.

If $f$ does not appear but $g$ does we use the bijection $(a,z,b)\mapsto(a,b,z+g(a,b))$ from $Z_{>0}^{r+1}\times B$ to $X$.

If $g$ does not appear but $f$ does we use the bijection $(a,z,b)\mapsto(a,b,f(a,b)-z)$ from $Z_{>0}^{r+1}\times B$ to $X$.

If neither $f$ and $g$ appear then we divide $X$ into the sets with $z>0$ and $z<1$. Both of these we know how to handle.

So now we assume both $f$ and $g$ appear. Using the function $(a,b,z)\mapsto(a,b,z+g(a,b))$ we may assume $g=0$. Write $f(a,b)=na+mb+c$ where $n,m$ are $1\times t$ matrices with integer coefficients, in other words, they are tuples. If $n=0$, then $X=Z_{>0}^{r}\times B^{\prime}$ with $B^{\prime}$ a bounded cell of the form required.

Assume $n\neq 0$. Then after a permutation in the $Z_{>0}^{r}$ factor we assume $n_{1}>0$. Dividing by $n_{1}$ and taking classes mod $n_{1}$ as in the first reduction we may assume $n_{1}=1$. Denote $a=(a_{1},a^{\prime})$ and $n=(1,n^{\prime})$. Now we divide $X$ into two disjoint subsets $X_{1}$ and $X_{2}$ defined as follows. A tuple $(a,b,z)\in X_{1}$ if $a\in Z_{>0}^{r}$, $b\in B$, $z\in Z_{>0}$ and $z-n^{\prime}a^{\prime}-mb-c\leq 0$. A tuple $(a,b,z)\in X_{2}$ if $a\in Z_{>0}^{r}$, $b\in B$, $z\in Z_{>0}$ and $0<z-n^{\prime}a^{\prime}-mb-c<a_{1}$. For $X_{1}$ we have that $X_{1}=Z_{>0}\times T$ for $T\subset Z^{t}$ so we are done by induction. For $X_{2}$ we have the bijective affine map $(a^{\prime},z,b,s)\mapsto(z-n^{\prime}a^{\prime}-mb-c+s,a^{\prime},b,z)$, from $R\times Z_{>0}$ to $X_{2}$. Where $R=\{(a^{\prime},z,b)\mid a^{\prime}\in Z_{>0}^{r-1},b\in B,z\in Z_{>0},0<z-n^{\prime}a^{\prime}-mb-c\}$ is a definable set such that $R\subset Z^{t}$, so here we also are done by induction. ∎

Proposition 7**.**

If $X$ is a definable set then ${\rm ubd}(X)\leq r$ if and only if $X$ is in definable bijection with some definable set $X^{\prime}\subset Z_{>0}^{r}\times[0,a)^{s}$.

Proof.

Assume first ${\rm ubd}(X)\leq r$. By the Lemma 6 we have $X=X_{1}\cup\dots\cup X_{p}$, $r_{i}$, $C_{i}$ and $f_{i}$ as in the statement of the lemma. By Lemma 3 we have that $r_{i}\leq r$. From this the result follows.

Assume now that $X\subset Z_{>0}^{r}\times[0,a)^{s}$. We want to see that ${\rm ubd}(X)\leq r$. Assume now towards a contradiction that $f:Z^{r+1}\to Z^{r}\times[0,a)^{s}$ is injective. If $\pi:Z^{r}\times[0,a)^{s}\to[0,a)^{s}$ is the projection, then there is a constant $b$ and a matrix $A$ with rational coefficients such that $\pi f(x)=Ax+b$ for all $c,1\ll x$ divisible, where $c$ are defining parameters for $f$, see Lemma 5. But as $f$ has a bounded image, then necessarily $A=0$. So $\pi f$ is constant on a cell containing such $x$ which, by Lemma 4, contains an injective image of $Z^{r+1}$. We obtain an injective $g:Z^{r+1}\to Z^{r}$. This is not possible for dimension reasons. ∎

Proposition 8**.**

$X\subset Z^{r}$* has ubd 0, if and only if $X$ is bounded.* 2. 2.

${\rm ubd}(X)\leq{\rm dim}(X)$. 3. 3.

${\rm ubd}(Z^{r})=r$. 4. 4.

If $X=Y\cup Z$ then ${\rm ubd}(X)={\rm Max}\{{\rm ubd}(Y),{\rm ubd}(Z)\}$ 5. 5.

${\rm ubd}(X\times Y)={\rm ubd}(X)+{\rm ubd}(Y)$. If $H\leq G$ are groups then ${\rm ubd}(G)={\rm ubd}(H)+{\rm ubd}(G/H)$. 6. 6.

If $a$ is a tuple with components in $Z$, and $\{X_{r}\}_{r}$ is an $a$-definable family, then the set $\{r\mid{\rm ubd}(X_{r})=n\}$ is $a$-definable.

Proof.

1. If $X$ is not bounded, by definable choice there is definable $f:Z_{\geq 0}\to X$ such that $f(t)\in X\setminus[-t,t]^{r}$. By cell decomposition there are $a\in Z$ and $N\in\mathbb{Z}_{>0}$ such that if $x>a$ and $N|x$, then $f(x)=mx+b$. If $m=(m_{1},\dots,m_{n})$ then $m_{i}\neq 0$ for all $i$. Then $g(x)=f(a+Nx)$ is injective from $Z_{>0}\to X$. We conclude by Lemma 3.

2. If ${\rm ubd}(X)\geq r$, then there is an injective definable map $Z^{r}\to X$, so by properties of dimension $r={\rm dim}(Z^{r})\leq{\rm dim}(X)$.

3. It is clear from the definition that ${\rm ubd}(Z^{r})\geq r$. From item 2) we get the other inequality.

4. It is clear from the definition that ${\rm ubd}(X)\geq{\rm Max}\{{\rm ubd}(Y),{\rm ubd}(Z)\}$. By the previous proposition we get the other inequality.

5. Similarly, from the definition ${\rm ubd}(X\times Y)\geq{\rm ubd}(X)+{\rm ubd}(Y)$, and the previous proposition gives the other inequality.

By definable choice $G\cong H\times G/H$ as a definable set.

6. The set $A_{n}$ defined by $r\in A_{n}$ if and only if ${\rm ubd}(X_{r})\geq n$ is $a$-$\lor$-definable, this follows from the definition. Indeed, if $r\in A_{n}$, then there is an $atr$-definable injective function $f_{atr}:Z^{n}\to X_{r}$, where $t$ are some defining parameters, so $A_{n}$ contains the $a$-definable set of those $r^{\prime}$ such that there exists $t^{\prime}$ such that $f_{ar^{\prime}t^{\prime}}$ is an injective definable function $Z^{n}\to X_{r^{\prime}}$. Similarly from the previous proposition the set $B_{n}$ of $r$ such that ${\rm ubd}(X_{r})\leq n$ is $a$-$\lor$-definable. So $C_{n}=A_{n}\cap B_{n}$ is also $a$-$\lor$-definable. If $X_{r}\subset Z^{N}$, then $C_{n}$ is also the intersection of the complements of $C_{s}$ for $0\leq s\leq N$ and $s\neq n$, so it is also $a$-$\land$-definable. So by compactness $C_{n}$ is $a$-definable. ∎

3 Idempotents in prodefinable semigroups

As motivation for Lemma 10 recall that a nonempty compact Hausdorff topological semigroup has an idempotent, (this is the Ellis-Nakamura Lemma, see [3] Lemma 1), and because having an idempotent is first order expressible a nonempty bounded definable semigroup in $(Z,+,<)$ also has one. The hypothesis of Lemma 10 includes both situations.

For the Lemma 10 we need some terminology. A reference for this material is [4], but with a small variant, we keep track of the cardinality of the index set for saturation reasons.

The category of $A$-$\kappa$-pro-definable sets is the category with objects the diagrams of $A$-definable sets $X_{i}$ indexed by a directed set $i\in I$ with intermediate maps $X_{i}\to X_{j}$ $A$-definable, and the cardinality of $I$ $\leq\kappa$. This object is denoted $X=\lim_{i}X_{i}$. For the $A$-$\kappa$-pro-definable sets $X=\lim_{i}X_{i}$ and $Y=\lim_{j}Y_{j}$ the set of morphisms of $A$-$\kappa$-pro-definable sets is ${\rm Mor}(X,Y)=\lim_{j}{\rm colim}_{i}{\rm Mor}(X_{i},Y_{j})$. We will denote this category ${\rm ProDef}_{A,\kappa}$. There is an inclusion functor ${\rm Def}_{A}\to{\rm ProDef}_{A,\kappa}$ which is fully faithful.

Assume that the model $M$ is $\kappa$-saturated in the language $L(A)$. Then there is a functor $F:{\rm Def}_{A}\to{\rm Sets}$ that takes the $A$-definable set $D$, to the set of $M$ points of $D$. This functor extends in the obvious manner to a forgetful functor $F:{\rm ProDef}_{A,\kappa}\to{\rm Sets}$. By compactness this functor is faithful. This functor factors through $F:{\rm ProDef}_{A,\kappa}\to{\rm Top}$ by giving $F(X)$ the topology given by basic open sets the $A$-relatively definable subsets of $F(X)$, in other words the inverse image of $A$-definable subset of $X_{i}$ via the canonical projection $F(X)\to X_{i}$. We remark that because the category ${\rm Def}_{A}$ has nonempty finite limits and the forgetful functor ${\rm Def}_{A}\to{\rm Sets}$ commutes with them then ${\rm ProDef}_{A,\kappa}$ has limits indexed by nonempty categories of cardinality $\leq\kappa$, and ${\rm ProDef}_{A,\kappa}\to{\rm Sets}$ commutes with them. However ${\rm ProDef}_{A,\kappa}\to{\rm Top}$ does not commute with limits.

Recall that a topological space is T0 if two points belonging to the same open sets are equal. The T0-ification of a topological space is the space obtained by identifying two points if they belong to the same open sets.

For an $A$-$\kappa$-pro-definable set $X$, $F(X)$ is a compact topological space and its T0-ification is Hausdorff. The forgetful functor ${\rm ProDef}_{A,\kappa}\to{\rm Top}$ and its composition with the T0-ification functor commute with directed limits with index set of cardinality $\leq\kappa$.

Given an $A$-$\kappa$-pro-definable set $X$, an $A$-$\kappa$-pro-definable subset $Y$ is a subobject $Y\to X$ isomorphic to the one given by compatible $A$-definable subsets $Y_{i}\subset X_{i}$. These subobjects are determined by the image of the inclusion $F(Y)\to F(X)$ in ${\rm Sets}$. Indeed if $Z$ is an $A$-$\kappa$-pro-definable object ${\rm Mor}(Z,Y)={\rm Mor}(Z,X)\times_{{\rm Mor}(F(Z),F(X))}{\rm Mor}(F(Z),F(Y))$. The inclusion $F(Y)\to F(X)$ is a closed immersion. Similarly an $A$-relatively definable subset of $X$ is a subobject $Y\to X$ isomorphic to a subobject of the form $\lim_{j\geq i}\pi_{ji}^{-1}Y_{i}$ for some index $i$ and $A$-definable subset $Y_{i}\subset X_{i}$. These subobjects correspond to open and closed subsets $Y\subset F(X)$. So in particular $A$-type-definable subsets of an $A$-definable set $D$ are $A$-pro-definable subsets of $D$.

Proposition 9**.**

If $X,Y$ are $A$-$\kappa$-pro-definable sets and $f:F(X)\to F(Y)$ is such that for any $A$-definable set $E$, $f\times 1:F(X\times E)\to F(Y\times E)$ is continuous, then there is $g:X\to Y$ such that $F(g)=f$. 2. 2.

The forgetful functor $F:{\rm ProDef}_{A,\kappa}\to{\rm Sets}$ reflects isomorphisms, that is, if $f:X\to Y$ is such that $F(f)$ is a bijection, then $f$ is an isomorphism.

The first one follows from the description of pro-definable morphisms in Proposition 8 of [4], the second one is proven in Proposition 8 of [4].

Given a theory with two [math]-definable constants, finite sets can be considered as [math]-definable sets. The profinite topological space $X$ is considered as a [math]-pro-definable set via $X=\lim_{U}X/U$ where the limit is over the finite partitions of $X$ by clopens, $X/U=U$ and the canonical projection $\pi_{U}:X\to X/U$ is given by $x\mapsto R$ where $x\in R$ and $R\in U$.

We note that for a [math]-type-definable set $L\subset M^{n}$, the map $L\to S^{n}(0)$ is [math]-pro-definable.

Lemma 10**.**

Let $M$ be a sufficiently saturated model of a theory $\mathcal{T}$ which defines two different [math]-definable constants.

Suppose $D$ is a [math]-definable set. Suppose $X$ is a profinite topological space. Suppose $C_{d}\subset Z^{r},d\in D$, is a [math]-definable family of sets.

Assume that there exists an elementary submodel $N$ of $M$ such that for every $d\in D(N)$, $C_{d}$ is finite.

Suppose given $Y_{d}\subset X\times C_{d}$ for $d\in D$, $Y_{d}$ a nonempty semigroup. We consider $X$ as a [math]-pro-definable set. Denote $Y=\cup_{d}Y_{d}\times\{d\}$ and $C=\cup_{d}C_{d}\times\{d\}$ Assume:

$Y$* as a subset of $X\times C$ is [math]-relatively definable.* 2. 2.

The given semigroup product $Y\times_{D}Y\to Y$ is a [math]-pro-definable map.

Then $Y_{d}$ has an idempotent.

We use the previous lemma in the case where $M$ is an $\omega$-saturated model of Presburger arithmetic, $N=\mathbb{Z}$ and $C_{d}$ are bounded sets.

Proof.

Denote $\pi_{U}:X\times C\to X/U\times C$ the canonical projection.

Hypothesis 1 says that there is $U_{0}$ and $F_{0}\subset X/U_{0}\times C$ [math]-definable such that $Y=\pi_{U_{0}}^{-1}(F_{0})$.

Denote $F_{U}=\pi_{U,U_{0}}^{-1}F_{0}$ for $U$ refining $U_{0}$ and $\pi_{U,U_{0}}:X/U\times C\to X/U_{0}\times C$ the canonical projection.

Denote $E_{U}\subset D$ defined by $d\in E_{U}$ if and only if $Y_{d}$ has a $U$-idempotent, that is, there is a $(x,c)\in Y_{d}$ such that $\pi_{U}(x,c)^{2}=\pi_{U}(x,c)$.

Hypothesis 2 says that for every $U$ refining $U_{0}$ there is $V$ refining $U$ such that $Y\times_{D}Y\to Y\to F_{U}$ factors as $Y\times_{D}Y\to F_{V}\times_{D}F_{V}\to F_{U}$. Denote $m_{V,U}:F_{V}\times_{D}F_{V}\to F_{U}$. Then $d\in E_{U}$ if and only if there exists $x$ such that $(x,d)\in F_{V}$ and $m_{V,U}((x,d),(x,d))=\pi_{V,U}(x,d)$. We conclude that $E_{U}$ is [math]-definable.

Take now $d\in D\cap N^{s}$, where $D\subset M^{s}$. Then $C_{d}$ is finite, so $Y_{d}$ is a non-empty compact Hausdorff topological semigroup, so by the Ellis-Nakamura Lemma $d\in E_{U}$. We have seen $D\cap N^{s}=E_{U}\cap N^{s}$. But $E_{U}$ is [math]-definable, so $D=E_{U}$. We conclude that if $d\in D$ then $Y_{d}$ has a $U$-idempotent, say $(x_{U},c_{U})$.

Take $(x,c)\in Y_{d}$ an accumulation point of the net $(x_{U},c_{U})$ in $Y_{d}$ with respect to the $d$-pro-definable topology. If $Z_{d,V}$ is the set of $V$-idempotents of $Y_{d}$ then $Z_{d,V}$ is the pullback of $\Delta:F_{V,d}\to F_{V,d}^{2}$ (the diagonal) along $Y_{d}\to Y_{d}^{2}\to F_{V,d}^{2}$, where the first arrow is $x\mapsto(x^{2},x)$. As $\Delta$ is an injective map of $d$-definable sets we see that $Z_{d,V}$ is $d$-relatively definable. So for $V$ fixed, because $(x_{U},c_{U})$ is in $Z_{d,V}$ for eventual $U$, we conclude $(x,c)\in Z_{d,V}$. That is, $(x,c)$ is $V$-idempotent for every $V$. This implies that $(x,c)$ is an idempotent. ∎

Note that for a [math]-pro-definable family (as in 1) and 2)) the set of $d$ such that $Y_{d}$ is a semigroup is [math]-type-definable. That it contains a [math]-definable set around the point of interest is part of the hypothesis.

We use the next Lemma only for $M,\Sigma$ and $C_{d}$ as in the situation 12. In that situation $M$ is a $Z$-group, $C_{d}$ is bounded and $b\vDash\Sigma_{a}$ informally means “$b$ is infinite relative to $a$”, see also condition 2) of Lemma 11. Condition 1) is something like orthogonality.

Lemma 11**.**

Let $M$ be a model of any theory that defines two different [math]-definable constants, which is sufficiently saturated.

Let $\Sigma(x,y)$ be a partial type with parameters in [math], and variables in $M^{n}\times M$.

Denote $\Sigma_{b}(x)=\cup_{i}\Sigma(x,b_{i})$ for $b=(b_{1},\dots,b_{m})$.

Take $D$ a [math]-definable set and $\{C_{d}\}_{d\in D}$ a [math]-definable family.

Suppose given a [math]-definable family $\{S_{d}\}_{d\in D}$ of left cancellative semigroups, $S_{d}\subset M^{n}\times C_{d}$.

Define $X\subset S^{n}(0)$ the set of types $p$ such that for every $t\in M$ there is $a\vDash p\cup\Sigma_{t}$. Then $X$ is a profinite topological space.

Assume that:

If $a\vDash\Sigma_{t}$ and $a^{\prime}\vDash\Sigma_{t^{\prime}}$, and $a\equiv_{0}a^{\prime},t\equiv_{0}t^{\prime}$, then $at\equiv_{0}a^{\prime}t^{\prime}$. 2. 2.

If $a\vDash\Sigma_{dt}$ with $d\in D$, and $c\in C_{d}$, then $a\vDash\Sigma_{dtc}$. 3. 3.

If $f:L\to M^{n}$ is an injective $a$-definable function and $p\in X$, $b\vDash p\cup\Sigma_{a}$, $b\in L$, then $f(b)\vDash\Sigma_{a}$.

Denote $Y_{d}\subset S^{n}(0)\times C_{d}$ to be the set of pairs $(p,c)$ such that $p\in X$ and there is $a\vDash p\cup\Sigma_{d}$ such that $(a,c)\in S_{d}$.

Define $Y\times_{D}Y\to Y$ by $(p_{1},c_{1},d)\cdot(p_{2},c_{2},d)=(p_{3},c_{3},d)$ if there is $a_{1}\vDash p_{1}\cup\Sigma_{d}$ and $a_{2}\vDash p_{2}\cup\Sigma_{da_{1}}$ such that $(a_{1},c_{1})(a_{2},c_{2})=(a_{3},c_{3})$ and $a_{3}\vDash p_{3}\cup\Sigma_{d}$.

Then $Y$ is [math]-relatively definable in $X\times C$ and $Y\times_{D}Y\to Y$ is a well defined [math]-pro-definable map, that makes $Y_{d}$ a semigroup.

We also have that if $a_{1}\vDash p_{1}\cup\Sigma_{d}$ and $a_{2}\vDash p_{2}\cup\Sigma_{da_{1}}$, $(a_{1},c_{1})(a_{2},c_{2})=(a_{3},c_{3})$ and $a_{3}\vDash p_{3}\cup\Sigma_{d}$, then $a_{3}\vDash\Sigma_{da_{1}}$.

If in addition $d_{0}\in D$ is such that there exists $h:T\to S_{d_{0}}$ an injective $d_{0}t$-definable map and there exists $p_{0}\in X$ such that $p_{0}\cup\Sigma_{d_{0}t}\vdash T$, then there exists $d_{0}\in D_{0}\subset D$ [math]-definable such that $Y_{d}$ is non-empty for all $d\in D_{0}$.

Proof.

We start with the observation that $X\subset S^{n}(0)$ is closed. Indeed $X$ is the image of a closed subset of $S^{n}(M)$ under the scalar restriction map $S^{n}(M)\to S^{n}(0)$. As these are compact Hausdorff spaces, $X$ is closed.

Now we note that condition 1) implies that for every $p\in S^{n}(0)$ and $t$, if $p\cup\Sigma_{t}$ is consistent, then it is a complete type over $t$.

Denote $R\subset M^{n}\times C$ the set of tuples $(a,c,d)$ such that $tp(a/0)\in X$ and $a\vDash\Sigma_{d}$. This is a [math]-type-definable set. Take now the map $R\to X\times C$ given by $(a,c,d)\mapsto(tp(a/0),c,d)$. This is a [math]-pro-definable surjective map $R\to X\times C$.

Claim 1**.**

This map $R\to X\times C$ is bijective in T0-ifications, and so it is an homemorphism in T0-ifications.

Proof.

This is a consequence of conditions 1) and 2).

Indeed if $(a,c,d)$ and $(a^{\prime},c^{\prime},d^{\prime})$ map to the same element in the T0-ification of $X\times C$, this means $tp(a/0)=tp(a^{\prime}/0)$ and $cd\equiv_{0}c^{\prime}d^{\prime}$. From the definition of $R$ we have $a\vDash\Sigma_{d}$ and $c\in C_{d}$; and similarly $a^{\prime}\vDash\Sigma_{d^{\prime}}$ and $c^{\prime}\in C_{d^{\prime}}$. From condition 2) we have $a\vDash\Sigma_{cd}$ and $a^{\prime}\vDash\Sigma_{c^{\prime}d^{\prime}}$. From condition 1) we have $acd\equiv_{0}a^{\prime}c^{\prime}d^{\prime}$. In other words $(a,c,d)$ and $(a^{\prime},c^{\prime},d^{\prime})$ map to the same element in T0-ification of $R$. ∎

We conclude that the image of the open and closed set $\Theta_{1}\subset R$ given by $(a,c,d)\in\Theta_{1}$ if and only if $(a,c,d)\in R$ and $(a,c)\in S_{d}$ is $Y$ which is then open and closed in $X\times C$, in other words $Y$ is [math]-relatively definable in $X\times C$.

Now define $\Theta_{2}\subset S\times_{D}S$ and $\Theta_{3}\subset S\times_{D}S\times_{D}S$ by $(a_{1},c_{1},d,a_{2},c_{2},d)\in\Theta_{2}$ if and only if $a_{1}\vDash\Sigma_{d}$, $tp(a_{1}/0)\in X$, $a_{2}\vDash\Sigma_{a_{1}d}$ and $tp(a_{2}/0)\in X$. Similarly $(a_{1},c_{1},d,a_{2},c_{2},d,a_{3},c_{3},d)\in\Theta_{3}$ if and only if

[TABLE]

Then there are [math]-prodefinable surjections

$\Theta_{2}\to Y\times_{D}Y$ and $\Theta_{3}\to Y\times_{D}Y\times_{D}Y$

given by

$(a_{1},c_{1},d,a_{2},c_{2},d)\mapsto(tp(a_{1}/0),c_{1},d,tp(a_{2}/0),c_{2},d)$ and $(a_{1},c_{1},d,a_{2},c_{2},d,a_{3},c_{3},d)\mapsto(tp(a_{1}/0),c_{1},d,tp(a_{2}/0),c_{2},d,tp(a_{3}/0),c_{3},d)$.

These maps become bijective on T0-ifications with an argument similar to claim 1.

Now if $m:S\times_{D}S\to S$ is the multiplication, then we will show that $m(\Theta_{2})\subset\Theta_{1}$, $1\times m(\Theta_{3}),m\times 1(\Theta_{3})\subset\Theta_{2}$, and the maps $m$ factors through $\Theta_{2}\to Y\times_{D}Y,\Theta_{1}\to Y$. This would show that the product $Y\times_{D}Y\to Y$ defined in the statement is well-defined and associative. The map $\cdot$ is then continuous. In fact $(\cdot)\times 1_{E}$ are also continuous for any $E$ [math]-definable (as this has the effect of replacing $D$ by $D\times E$). So the product $\cdot$ is a map of [math]-pro-definable sets, see Proposition 9 1).

Claim 2**.**

Let $a_{1}\vDash\Sigma_{dt},a_{2}\vDash\Sigma_{dta_{1}},(a_{1},c_{1})\in S_{d}$, $(a_{2},c_{2})\in S_{d}$ and $(a_{1},c_{1})(a_{2},c_{2})=(a_{3},c_{3})$, then $a_{3}\vDash\Sigma_{dta_{1}}$.

Proof.

Indeed $a_{3}=Y(a_{1},a_{2},c_{1},c_{2},d)$ and $c_{3}=Z(a_{1},a_{2},c_{1},c_{2},d)$ for [math]-definable maps $Y$, $Z$. By hypothesis 2) we have $a_{2}\vDash\Sigma_{dta_{1}c_{1}c_{2}c_{3}}$. Now if $a_{2}^{\prime}\equiv_{dta_{1}c_{1}c_{2}}a_{2}$, then $a_{2}^{\prime}\vDash\Sigma_{dta_{1}c_{1}c_{2}}$ and so by hypothesis 2) we get $a_{2}^{\prime}\vDash\Sigma_{dta_{1}c_{1}c_{2}c_{3}}$ and $a_{2}^{\prime}\equiv_{dta_{1}c_{1}c_{2}c_{3}}a_{2}$. From this it follows that $Z(a_{1},a_{2}^{\prime},c_{1},c_{2},d)=Z(a_{1},a_{2},c_{1},c_{2},d)=c_{3}$, and so, because $S_{d}$ is left cancellative $Y(a_{1},a_{2}^{\prime},c_{1},c_{2},d)\neq Y(a_{1},a_{2},c_{1},c_{2},d)$ if $a_{2}\neq a_{2}^{\prime}$. Now by hypothesis 3) we conclude $a_{3}\vDash\Sigma_{dta_{1}}$. ∎

This proves the claim after the definition of $Y$ in the statement of this lemma.

Claim 3**.**

Let $a_{1},a_{2},a_{3},c_{1},c_{2},c_{3}$ be as in the previous claim. If we take $a_{1}^{\prime},a_{2}^{\prime}$ such that $a_{1}^{\prime}a_{2}^{\prime}\equiv_{c_{1}c_{2}dt}a_{1}a_{2}$ then $a_{1}^{\prime}a_{2}^{\prime}\equiv_{dta_{1}c_{1}c_{2}c_{3}}a_{1}a_{2}$ and $Z(a_{1}^{\prime},a_{2}^{\prime},c_{1},c_{2},d)=Z(a_{1},a_{2},c_{1},c_{2},d)=c_{3}$.

The proof is similar to the previous claim.

These two claims show that $m(\Theta_{2})\subset\Theta_{1}$ and that it factors as a product $Y\times_{D}Y\to Y$. We also have $1\times m(\Theta_{3})\subset\Theta_{2}$.

We are left with proving that $m\times 1(\Theta_{3})\subset\Theta_{2}$.

Claim 4**.**

$tp(a/0)\in X$* and $a\vDash\Sigma_{b}$ then $a\vDash\Sigma_{bc}$ for $c\in{\rm dcl}(b)$.*

Proof.

Indeed if $a^{\prime}$ is such that $a^{\prime}\equiv_{0}a$ and $a^{\prime}\vDash\Sigma_{bc}$, then $a^{\prime}b\equiv_{0}ab$ and as $c\in{\rm dcl}(b)$ we get $a^{\prime}bc\equiv_{0}abc$ so $a\vDash\Sigma_{bc}$. ∎

Now let $(a_{1},c_{1},d,a_{2},c_{2},d,a_{3},c_{3},d)\in\Theta_{3}$ and $(a_{1},c_{1})(a_{2},c_{2})=(a_{4},c_{4})$. Note that $a_{4}=Y(a_{1},a_{2},c_{1},c_{2},d)$ for a [math]-definable function $Y$. We have seen in claim 2 that $a_{4}\vDash\Sigma_{d}$. Now as $a_{3}\vDash\Sigma_{a_{1}a_{2}d}$ we get $a_{3}\vDash\Sigma_{a_{1}a_{2}c_{1}c_{2}d}$ and so, by claim 4, $a_{3}\vDash\Sigma_{a_{4}d}$. This shows that $m\times 1(\Theta_{3})\subset\Theta_{2}$ as required.

Now for the last statement, we consider

[TABLE]

By compactness this is a [math]-$\lor$-definable set so we take $d_{0}\in D_{0}\subset D_{1}$ [math]-definable. Now take $d\in D_{0}$ and $t$ be such that $p_{0}\cup\Sigma_{dt}\vdash T_{dt},$ and $h_{dt}:T_{dt}\to S_{d}$ is injective. Take $s$ some extra parameters. If $a\vDash p_{0}\cup\Sigma_{dts}$, then $h_{dt}(a)=(a^{\prime\prime},c)$ with $a^{\prime\prime}=Y(a,d,t),c=Z(a,d,t)$ , $Y$, $Z$, [math]-definable functions. As in the proof of claim 2 we have that if $a^{\prime}\vDash p_{0}\cup\Sigma_{dts}$ and $a^{\prime}\neq a$, then $Z(a,d,t)=Z(a^{\prime},d,t)$, $Y(a,d,t)\neq Y(a^{\prime},d,t)$ and so $Y(a,d,t)\vDash\Sigma_{dts}$. So for $p=tp(a^{\prime\prime}/0)$ we get $(p,c)\in Y_{d}$. This ends the proof of the lemma. ∎

Notation 12**.**

Let $(Z,+,<)$ be an $\omega$-saturated $Z$-group.

Define $\Sigma(x,y)$ to be the partial type over [math] with variables in $Z^{r}\times Z$, given by $(a,b)\vDash\Sigma$ if and only if for every nonzero tuple $m\in\mathbb{Z}^{r}$ $|b|,1\ll|ma|$. For $b\in Z^{k}$ denote $\Sigma_{b}(x)=\bigcup_{i}\Sigma(x,b_{i})$ where $b=(b_{1},\dots,b_{k})$. Denote $X\subset S^{r}(0)$ the set of types $p$ such that, $a\vDash p$ implies $1\ll|ma|$ for every nonzero tuple $m\in\mathbb{Z}^{r}$. Here $ma$ denotes $\sum_{i}m_{i}a_{i}$.

Lemma 13**.**

With the notation of 12. If $b\vDash\Sigma_{ac}$ and $c$ is divisible then $b\equiv_{a}b+c$.

Proof.

By elimination of quantifiers we have to see that $nb+ma+r>0$ if and only if $n(b+c)+ma+r>0$ and that $nb+ka+r\equiv n(b+c)+ma+r\mod N$ for tuples $n,m,r$ with integers coefficients and $N$ an integer. The second condition follows because $c$ is divisible. The first condition is trivial if $n=0$. If $n\neq 0$, by definition of $\Sigma$ we get that $nb$ is the dominant term in both expressions, so they are equivalent to $nb>0$. ∎

Lemma 14**.**

Let $Z$ be a $\omega$-saturated $Z$-group. Let $\{C_{d}\}_{d\in D}$ be a [math]-definable family of bounded sets. Define $\Sigma(x,y)$ as in 12. Then $\Sigma$ and $C_{d}$ satisfy hypothesis 1,2 and 3 of Lemma 11

Proof.

Let $a,b,a^{\prime},b^{\prime}$ be such that $a\vDash\Sigma_{b}$, $a^{\prime}\vDash\Sigma_{b^{\prime}}$ and $a\equiv_{0}a^{\prime},b\equiv_{0}b^{\prime}$. By elimination of quantifiers we have to see that $na+mb+k>0$ if and only if $na^{\prime}+mb^{\prime}+k>0$ and $na+mb+k\equiv na^{\prime}+mb^{\prime}+k\mod N$. If $n\neq 0$, then $na+mb+k>0$ if and only if $na>0$. If $n=0$, this becomes a condition on $b$. The divisibility type of $na+mb+k$ depends only on the divisibility type of $a$ and $b$. This proves Hypothesis 1.

If $f(d)={\rm Max}\{|x|\mid x\in C_{d}\}$, then $f$ is [math]-definable and so there is a cell decomposition of $D$ such that on each of the cells $f$ is of the form $Ad+B$. So $|f(d)|\leq N(|d|+1)$ for an integer $N$. This shows Hypothesis 2.

Let $f,a,b,p$ as in the statement of Hypothesis 3. Then there exists $A,B,C$ with rational entries such that $f(b)=Ab+Ba+C$.

We claim that $A$ is invertible Otherwise if $A$ is singular then we first find a non-zero divisible tuple $c$ such that $b\vDash\Sigma_{c}$ and $Ac=0$. This exists because for any $N$ we may find $d$ non-zero with integer coefficients, and so satisfying $b\vDash\Sigma_{d}$, with $Ad=0$ and $c\equiv 0\mod N$, so we conclude by compactness. Note however that $b+c\equiv_{a}b$ by Lemma 13 So now we conclude that $f(b)=f(b+c)$. This contradicts the injectivity of $f$.

So $A$ is invertible. If $m\neq 0$, then $|mf(b)|\geq|mAb|-|mBa|-|C|\gg|a|,1$ so $f(b)\vDash\Sigma_{a}$. ∎

For convenience we specialize the previous lemmas to the situation at hand in the lemma that follows

Lemma 15**.**

Let $(Z,+,<)$ be an $\omega$-saturated $Z$-group.

Let $T$ be a [math]-definable set. Let $\{C_{t}\}_{t\in T}$ be a [math]-definable family of bounded sets. Let $\{G_{t}\}_{t\in T}$ be a [math]-definable family of groups, such that $G_{t}\subset Z^{r}\times C_{t}$.

Assume ${\rm ubd}(G_{t})=r$ for every $t$.

With notation as in 12, for every $t$ there exists $p\in X$ and $b\in C_{t}$ such that if $x\vDash p\cup\Sigma_{t}$ and $y\vDash p\cup\Sigma_{(t,x)}$, then $(x,b)$ and $(y,b)$ belong to $G_{t}$ and $(x,b)(y,b)=(z,b)$ satisfies $z\vDash p\cup\Sigma_{(t,x)}$.

Proof.

We take $Y$ as in Lemma 11, with $G$ taking the place of $S$. The hypothesis of that Lemma is satisfied by Lemma 14. We conclude by Lemma 10. ∎

4 Groups definable in Presburger arithmetic

Lemma 16**.**

Let $\Sigma$ and $X$ be as in 12. If $f:(\cap_{n}nZ)^{r}\to G$ is a type-definable group morphism such that $f$ is injective in $p\cup\Sigma_{t}$ for some $t\in Z$ and $p\in X$ that implies $(\cap_{n}nZ)^{r}$, then $f$ is injective.

Proof.

If $K$ is the kernel of $f$ and $K\neq 0$, then if we take $y\in K$ nonzero, divisible, and take $x\vDash p\cup\Sigma_{ty}$ then $x+y\vDash p\cup\Sigma_{ty}$ by Lemma 13, so we conclude $f(x)=f(x+y)$ which contradicts the injectivity of $f$ in $p\cup\Sigma_{t}$. ∎

Theorem 17**.**

If $G$ is an abelian group definable in $(Z,+,<)$, then $G$ has a subgroup $H$ isomorphic as a definable group to $Z^{n}$ such that $G/H$ is bounded.

Proof.

Suppose $G\subset Z_{>0}^{r}\times[0,a)^{s}$ with $r={\rm ubd}(G)$, as in Proposition 7. Let $t$ be a set of parameters over which $G$ is defined, so that $G$ is a $t$-definable group, and without loss of generality $a$ is in $t$. Say $G=G_{t}$, for $t\in T$. We may assume that the projection onto the last $s$ factors of $G_{t^{\prime}}$ is bounded for all $t^{\prime}\in T$, and that ${\rm ubd}(G_{t^{\prime}})=r$ for all $t^{\prime}\in T$. To distinguish it from other sums we will denote the product of $G$ by $\oplus$ and the inverse by $\ominus$.

Take $\Sigma$ and $X$ as in the situation 12. By Lemma 15 we have that there exist a type $p\in X$ and an element $b\in[0,a)^{s}$ such that if $x\vDash p\cup\Sigma_{t}$ and $y\vDash p\cup\Sigma_{(t,x)}$ then $(x,b),(y,b)\in G$ and $(x,b)\oplus(y,b)=(z,b)$ satisfies $z\vDash p\cup\Sigma_{(t,x)}$.

We shall give a type-definable function $i:q\cup\Sigma_{t}\to G$ defined on a type $q\in X$ that implies $\bigcap_{n}(nZ)^{r}$, and such that $i(x+y)=i(x)\oplus i(y)$ for $x\vDash q\cup\Sigma_{t}$ and $y\vDash q\cup\Sigma_{(t,x)}$. Start with $j:p\cup\Sigma_{t}\to G$, $j(x)=(x,b)$. Denote $\hat{p}$ the element in $\hat{\mathbb{Z}}^{r}$ such that if $x\vDash p$ then $\hat{x}=\hat{p}$. We can write $j(x)\oplus j(y)=(Ax+By+d,b)$ where $A,B$ are matrices with rational entries and $d$ is a $tb$-definable constant. From the identity $(j(x)\oplus j(y))\oplus j(z)=j(x)\oplus(j(y)\oplus j(z))$ for all $x\vDash p\cup\Sigma_{t},y\vDash p\cup\Sigma_{(t,x)},z\vDash p\cup\Sigma_{(t,x,y)}$, we obtain $A^{2}x+ABy+Bz+Ad+d=Ax+BAy+B^{2}z+Bd+d$. Then $A^{2}=A,B^{2}=B,BA=AB,Ad=Bd$. As $G$ is a group it is left and right cancellative so $A$ and $B$ must be nonsingular rational matrices, so $A=B=1$. Reducing mod $\hat{\mathbb{Z}}$ obtain $d\equiv-\hat{p}$, indeed from $(x,b)\oplus(y,b)=(x+y+d,b)$, for $x\vDash p\cup\Sigma_{t}$ and $y\vDash p\cup\Sigma_{(t,x)}$ we have that $x+y+d\vDash p$ and in particular its reduction mod $\hat{\mathbb{Z}}$ is $\hat{p}$, and on the other hand this reduction is also $\hat{x}+\hat{y}+\hat{d}=2\hat{p}+\hat{d}$. Taking $i(x)=j(x-d)$ we obtain the identity

[TABLE]

for all $x\vDash q\cup\Sigma_{t}$ and $y\vDash q\cup\Sigma_{(t,x)}$, where $q=p+d$. Indeed for $x\vDash q\cup\Sigma_{t}$ and $y\vDash q\cup\Sigma_{(t,x)}$ we have $x=x^{\prime}+d$ and $y=y^{\prime}+d$ for $x^{\prime}\vDash p\cup\Sigma_{t}$ and $y^{\prime}\vDash p\cup\Sigma_{(t,x^{\prime})}$, and then we get $i(x)\oplus i(y)=j(x-d)\oplus j(y-d)=j(x^{\prime})\oplus j(y^{\prime})=(x^{\prime}+y^{\prime}+d,b)=(x+y-d,b)=i(x+y)$.

Define $k:(\bigcap_{n}nZ)^{r}\to G$ by $k(z)=i(x+z)\ominus i(x)$ for $x\vDash q\cup\Sigma_{(t,z)}$. One has to prove $k$ is a well defined group morphism. Suppose $x,y\vDash q\cup\Sigma_{(t,z)}$. Take $w\vDash q\cup\Sigma_{(t,z,x,y)}$. Then $i(x+z+w)\ominus i(x+w)=i(x+z)\ominus i(x)$, applying 1 twice. On the other hand if $x+w=y+w^{\prime}$ then $w^{\prime}\vDash\Sigma_{(t,z,x,y)}$ and so, similarly, $i(y+z+w^{\prime})\ominus i(y+w^{\prime})=i(y+z)\ominus i(y)$, so $k$ is well-defined.

Now if $z,w\in(\bigcap_{n}nZ)^{r}$ and $x\vDash q\cup\Sigma_{(t,z,w)},y\vDash q\cup\Sigma_{(t,z,w,x)}$, then

[TABLE]

In the first and last equation we use the definition of $k$ and in the second we use 1 twice. $k$ restricts to $i$ on $x\vDash q\cup\Sigma_{t}$, so $k$ is injective, by Lemma 16. By compactness $k$ extends to an injective definable group morphism $(NZ)^{r}\to G$, and composing with multiplication by $N$ we obtain an injective group morphism $k^{\prime}:Z^{r}\to G$. Then $G/k^{\prime}(Z^{r})$ is bounded, by Proposition 8. ∎

The next lemma is implicit in [5], but for convenience we include a proof.

By a $\lor$-definable group we mean a $\lor$-definable set $G$ with a group operation $G\times G\to G$ such that the for every $X,Y\subset G$ definable the product restricted to $X\times Y$ is definable. In other words the product is a map of ind-definable sets.

Lemma 18**.**

In an arbitrary theory. Let $G$ be a $\lor$-definable abelian group and $T\subset G$ a type-definable subgroup. Let $H$ be a $\lor$-definable group and $\phi:T\to H$ a type-definable group morphism. Suppose given a surjective group morphism $\pi:G\to\mathbb{R}^{s}$ with kernel $T$. Assume that:

If $T\subset U\subset G$ is definable then there is $0\in V\subset\mathbb{R}^{s}$ open such that $\pi^{-1}(V)\subset U$. 2. 2.

If $T\subset U\subset G$ is definable then $\pi(U)$ is bounded.

Then $\phi$ extends to a unique group morphism $\phi:G\to H$ such that the restriction of $\phi$ to a definable subset is definable.

Proof.

First uniqueness. If $T\subset U_{0}\subset G$ is definable and $\phi_{1},\phi_{2}:G\to H$ extend $\phi$, then $\phi_{i}|_{U_{0}}$ are definable and equal when restricted to $T$. By compactness there is $T\subset U_{1}\subset U_{0}$ definable with $\phi_{1}|_{U_{1}}=\phi_{2}|_{U_{1}}$. By 1) $\cup_{n\geq 0}nU_{1}=G$, so $\phi_{1}=\phi_{2}$.

Now existence. Choose $U$ definable symmetric such that $\phi$ extends to a function $\phi:U\to H$ such that $\phi(x+y)=\phi(x)\phi(y)$ for all $x,y\in U$. Replacing $H$ by the group generated by $\phi(U)$ we may assume $H$ is abelian, and we denote it additively.

Take $0\in B$ a convex open such that $\pi^{-1}B\subset U$. For $N\in\mathbb{Z}_{>0}$ define $A_{N}=\{x\in G\mid\text{ there exists }y\in U,z\in T,Ny+z=x\}$. Then $A_{N}\subset(N+1)U$ and by compactness $A_{n}$ is type-definable. Define $\phi_{N}:A_{N}\to H$ by $\phi_{N}(x)=N\phi(y)+\phi(z)$. We see this is well defined, if $Ny_{1}+z_{1}=Ny_{2}+z_{2}$ then $N\pi(y_{1})=N\pi(y_{2})$ so $y_{2}=y_{1}+v$ for $v\in T$. From $Ny_{1}+z_{1}=Ny_{2}+z_{2}$ obtain $z_{1}=Nv+z_{2}$. Then $\phi(y_{2})=\phi(y_{1})+\phi(v)$ and $\phi(z_{1})=N\phi(v)+\phi(z_{2})$, from where $N\phi(y_{2})+\phi(z_{2})=N\phi(y_{1})+\phi(z_{1})$, which is what we wanted to see. By compactness $\phi_{N}$ is type-definable. If $x$ is such that $\frac{1}{M}\pi(x)\in B$ then for $N,N^{\prime}\geq M$, $x\in A_{N}\cap A_{N^{\prime}}$ and $\phi_{N}(x)=\phi_{N}^{\prime}(x)$. Indeed, considering $(N,NN^{\prime}),(N^{\prime},NN^{\prime})$ without loss $N^{\prime}=NN_{1}$. Take $y\in G$ and $z\in T$ such that $x=NN_{1}y+z$. Then $y,2y,\dots,N_{1}y\in\pi^{-1}(B)\subset U$, so $\phi(N_{1}y)=N_{1}\phi(y)$, and $\phi_{N}(x)=\phi_{N^{\prime}}(x)$, as required. Define $\psi(x)=\phi_{N}(x)$ for $N$ sufficiently large, (Which clearly extends $T\to H$). Now if $x,y\in G$ and $N$ is sufficiently large then $x=Nx^{\prime}+z_{1},y=Ny^{\prime}+z_{2}$ and $x+y=N(x^{\prime}+y^{\prime})+z_{1}+z_{2}$ with $x^{\prime},y^{\prime},x^{\prime}+y^{\prime}\in U$ and $z_{1},z_{2}\in T$. From here $\psi(x+y)=\psi(x)+\psi(y)$. Finally if $X$ is definable and $\frac{1}{N}\pi(X)\subset B$, then $\psi=\phi_{N}$ on $X$ so we have the definability. ∎

We recall now one of the main results of [5]. To state the result we need some notation. For $a\in Z$ with $1\ll a$ denote

[TABLE]

and

[TABLE]

For $a\in Z^{s}$ with $1\ll a_{i}$, denote $O(a)=O(a_{1})\times\cdots\times O(a_{s})$ and $o(a)=o(a_{1})\times\cdots\times o(a_{s})$. As remarked in [1], Proposition 3.2, $O(a)/o(a)\cong\mathbb{R}^{s}$, and this isomorphism is easily seen to satisfy the hypotheses of the Lemma 18. Recall that a subgroup $\Lambda\subset\mathbb{R}^{s}$ is called a lattice if it is discrete with the subspace topology. Equivalently if it is generated as a group by linearly independent elements of the $\mathbb{R}$-vector space $\mathbb{R}^{s}$. The lattice is called full if the $\mathbb{R}$-linear span is $\mathbb{R}^{s}$, equivalently if it is generated as a group by a basis. A subgroup $\Lambda\subset O(a)$ is called a local lattice if $\Lambda\cap o(a)=0$ and $\pi(\Lambda)\subset\mathbb{R}^{s}$ is a full lattice. In this case $\Lambda=\bigoplus_{i}\mathbb{Z}b_{i}$ for $\{\pi(b_{i})\}_{i}$ which form a basis of $\mathbb{R}^{s}$. Note that $\Lambda$ satisfies that:

$X\cap\Lambda$ is finite for all $X\subset O(a)$ definable 2. 2.

there exists $X_{0}$ definable such that $X_{0}+\Lambda=O(a)$.

Indeed, $\pi(X)$ is compact and $X_{0}$ is any such that $o(a)\subset X_{0}$ and $\pi(X_{0})$ contains a parallelogram for $\pi(\Lambda)$. Note that $\pi(\Pi_{i}[-Na_{i},Na_{i}])$ contains $[-N,N]^{s}$ so this $X_{0}$ exists. This implies that we can consider $O(a)/\Lambda$ as a definable group. Indeed in $X_{0}$ there is the equivalence relation given by the fibers of $X_{0}\to O(a)/\Lambda$ which is definable by 1. By definable choice there is $X_{1}\subset X_{0}$ definable such that the restriction of $O(a)\to O(a)/\Lambda$ to $X_{1}$ is bijective. The sum on $X_{1}$ which makes this bijection a group isomorphism is definable by 1. Indeed this sum $\oplus$ is defined by $x\oplus y=x+y-z$ where $z\in\Lambda\cap 3(X_{1}\cup-X_{1})$ is the unique element of $\Lambda$ such that $x+y-z\in X_{1}$.

The group $O(a)/\Lambda$ considered as a definable group is denoted $C(a,b)$ (for $\Lambda=\bigoplus_{i}\mathbb{Z}b_{i}$).

In the next theorem we also need the groups of the form $Z^{r}\times O(a)/\bigoplus_{i}\mathbb{Z}(v_{i},b_{i})$ for $v_{i}\in Z^{r}$ and $\sum_{i}\mathbb{Z}b_{i}$ a local lattice of $O(a)$. If $X_{1}$ is a definable set as found before, that is, a set such that the restriction of $O(a)\to C(a,b)$ to $X_{1}$ is a bijection, then we consider $Z^{r}\times O(a)/\bigoplus_{i}\mathbb{Z}(v_{i},b_{i})$ a definable group with underlying set $Z^{r}\times X_{1}$. The sum here is given by $(t,x)\oplus(w,y)=(t,x)+(w,y)-(f(z),z)$ where $z\in\Lambda\cap 3(X_{1}\cup-X_{1})$ is the unique element of $\Lambda$ such that $x+y-z\in X_{1}$ and $f:\Lambda\to Z^{r}$ is the group morphism that sends $b_{i}$ to $v_{i}$.

In [5] it is proven that every bounded definable group has a definable finite index subgroup isomorphic as a definable group to $C(a,b)$ for some $a,b$ as before.

Theorem 19**.**

If $G$ is a group definable in $(Z,+,<)$, then $G$ has a finite index definable subgroup isomorphic as a definable group to $Z^{r}\times O(a)/\bigoplus_{i}\mathbb{Z}(v_{i},b_{i})$, for $\sum_{i}\mathbb{Z}b_{i}$ a local lattice of $O(a)$ and $v_{i}\in Z^{r}$.

Proof.

As mentioned before, in [5] it is proven that $G$ is abelian-by-finite, so without loss of generality $G$ is abelian. By Proposition 17 we have a short exact sequence $0\to Z^{r}\to G\to B\to 0$, with $B$ bounded. Denote $i:Z^{r}\to G$, $q:G\to B$. By the fact mentioned before this theorem me may take $B=C(a,b)$. By definable choice we may take $B\to G$ a set-theoretic definable section. Denote $s$ the composition $O(a)\to B\to G$. From this we obtain a definable inhomogeneous 2-cocycle $g:O(a)^{2}\to Z^{r}$ defined by $i(g(x,y))=s(x+y)-s(x)-s(y)$. This function satisfies $g(x,y)+g(x+y,z)=g(y,z)+g(x,y+z)$, called the cocycle condition.

Suppose everything is defined over $t$. We may assume that $o(a_{1})=\cdots=o(a_{n_{1}})\subsetneq o(a_{n_{1}+1})=\cdots=o(a_{n_{2}})\subsetneq\cdots\subsetneq o(a_{n_{l-1}+1})=\cdots=o(a_{s})$, with $n_{0}=0<n_{1}<\cdots<n_{l-1}<n_{l}=s$. If $x\in o(a)$ then denote $x^{i}=(x_{n_{i-1}+1},\cdots,x_{n_{i}})$ for $i=1,\cdots,l$. Define $W$ to be $x\in W$ if and only if $x\in o(a)$ and $t^{\prime}\ll x^{i}$ for every $t^{\prime}$ $t$-definable in $o(a_{n_{i}})$, so in particular $a_{j}\ll x^{i}$ for $j\leq n_{i-1}$, and $1\ll x$. Take $Y$ to be the elements $(x,y)$, $x,y\in o(a)$ such that $x,y\in W$, and $x^{i}\ll y^{i}$. As an aside $Y$ comes from a tensor product of invariant types. We take $X\subset o(a)\times o(a)$ a complete $t$-type of elements determined by $(x,y)\in X$ if and only if $(x,y)\in Y$ and $x,y$ are divisible.

Note that as $X$ is a complete type over $t$, we can write $g(x,y)=Ax+By+d$ for $(x,y)\in X$, for some $A,B$ matrices with rational entries and $d$ a $t$-definable constant. Take $x,y,z\in o(a)$ with $(x,y),(y,z)\in X$. From the cocycle condition obtain $Ax=Bz$ from which one gets $A=0=B$. Replacing $s$ by $s-i(d)$ we may assume $d=0$. By compactness there is $N$ such that if $N|x,y$ and $(x,y)\in Y$ then $s(x+y)=s(x)+s(y)$. Replacing $G$ by $q^{-1}((NO(a)+\Lambda)/\Lambda)$ and $\Lambda$ by the inverse image under multiplication by $N$, $O(a)\to O(a)$, we may assume that $N=1$. Define $s^{\prime}:o(a)\to G$ by $s^{\prime}(x)=s(x+y)-s(y)$ for $y\in W$, with $|x_{i}|\ll y_{i}$. We see that $s^{\prime}$ is a well defined group morphism. For this note that for $y,y^{\prime}\in W$ there is $w\in W$ such that $(y,w),(y^{\prime},w)\in Y$. The rest follows as in the last two paragraphs of the proof of Proposition 17. Note that $s^{\prime}$ composed with $G\to B$ is the canonical projection $O(a)\to B$ restricted to $o(a)$. We obtain now a group morphism $s^{\prime\prime}:O(a)\to G$, by Lemma 18. By uniqueness of the morphism of that lemma we find that the composition $O(a)\to G\to B$ equals the canonical projection $O(a)\to B$, because they restrict to the same mapping on $o(a)$. Finally if one considers the diagram

{0}$${Z^{r}}$${G}$${B}$${0}$${0}$${\Lambda}$${O(a)}$${B}$${0,}$$\scriptstyle{1}

then there is unique dotted arrow which makes the diagram commute and a diagram chase shows that the left square is cocartesian, which is what we had to prove. Indeed, the group $\Lambda$ is a free abelian group so a map $\Lambda\to Z^{r}$ is the same thing as choosing some elements in $Z^{r}$, which are the image of the basis of $\Lambda$. These elements are $-v_{i}$, with $v_{i}$ satisfying the statement. ∎