# FORKING AND INVARIANT TYPES IN REGULAR ORDERED ABELIAN GROUPS

Akash Hossain

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Akash Hossain. FORKING AND INVARIANT TYPES IN REGULAR ORDERED ABELIAN GROUPS. 2024. hal-04493771

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# FORKING AND INVARIANT TYPES IN REGULAR ORDERED
ABELIAN GROUPS

AKASH HOSSAIN

ABSTRACT. We give a characterization of forking in regular ordered Abelian groups. In particular, we prove that $C \underset{A}{\downarrow}^f B$ if and only if $c \underset{A}{\downarrow}^f B$ for each singleton $c \in \text{dcl}(\mathcal{A}C)$ in these structures.

INTRODUCTION

The non-forking global extensions of some unary type in the theory of divisible ordered Abelian groups (DOAG) are very easy to describe: they correspond to the invariant cuts, and there are at most two of them for any base parameter set. Now, if we look at a finite tuple $c = (c_1, ..., c_n)$ in DOAG, a necessary condition for $\text{tp}(c/AB)$ to be non-forking over $A$ $(c_1, ..., c_n \downarrow_A^f B)$ is to have $d \downarrow_A^f B$ for every $\mathbb{Q}$-linear combination $d$ of the $c_i$. As the type in DOAG of a tuple is characterized by the cuts of the $\mathbb{Q}$-linear combinations of its components, one can wonder whether this condition is sufficient. We will show in this paper that the answer is yes. Moreover, by using quantifier elimination in the Presburger language, we can extend our results to find a characterization of forking in the whole class of regular ordered Abelian groups ROAG, (see ?? and ??):

**Theorem 0.1.** Let $M \models\text{ROAG}$, $A, B$ parameter subsets of $M$, and $c = c_1...c_n \in M^n$. Let $\kappa = \max(|AB|, 2^{\aleph_0})^+$, and suppose $M$ is $\kappa$-saturated and strongly-$\kappa$-homogeneous. If $M$ is discrete, interpret $1$ as its least positive element, else interpret $1$ as $0$ (this is the standard interpretation of $1$ in the Presburger language). Let $C$ be the subgroup of $M$ generated by $c$, and $A'$, $B'$ the relative divisible closures in $M$ of the respective subgroups of $M$ generated by $A \cup \{1\}$, $AB \cup \{1\}$. Then $c \underset{A}{\downarrow}^f B$ if and only if $c \underset{A}{\downarrow}^d B$, if and only if $\text{tp}(c/AB)$ admits a global $\text{Aut}(M^{eq}/A' \cup \text{acl}^{eq}(\emptyset))$-invariant extension in $S(M^{eq})$, if and only if the following conditions hold:

(1) Every closed bounded interval of $B'$ that has a point in $C$ already has a point in the divisible closure of $A'$.

(2) For all prime $l$, if $[M : lM]$ is infinite, then we have $(C + l^N M) \cap (B' + l^N M) = A' + l^N M$ for all $N > 0$.

Moreover, $\text{tp}(c/AB)$ admits an $\text{Aut}(M/A)$-invariant extension in $S(M)$ if and only if the above conditions hold, and, additionally, for every prime $l$ for which $[M : lM]$ is finite, we have $C \subseteq \bigcap_{N>0} A' + l^N M$.

<footnote>The author was partially supported by GeoMod AAPG2019 (ANR-DFG), Geometric and Combinatorial Configurations in Model Theory.</footnote>
This description of forking is the most simple and natural way to assemble to-
gether the (well-known) descriptions of forking in the theory of torsion-free Abelian
groups, and in DLO. Our results clearly imply that forking in ROAG is a phenom-
enon that happens in dimension one: we have $C \underset{A}{\downarrow}^f B$ if and only if, for each
singleton $c \in \mathrm{acl}(AC)$ (which coincides here with $\mathrm{dcl}(AC)$), we have $c \underset{A}{\downarrow}^f B$. Note
that this particular property of unary forking is well-known in stable 1-based theo-
ries. This raises the question of how this property relates to local modularity in, say,
o-minimal theories. It also suggests that one might define over unstable theories a
notion (satisfied by ROAG) which would extend that of 1-basedness in the stable
world. Note that we cannot hope to have this property in infinite (expansions of)
fields, as it does not even hold in ACF.

In this paper, we also describe more or less explicitly the Stone space of global extensions of tp(c/AB) which do not fork over A, and the subspace of those extensions which are A-invariant:

**Theorem 0.2.** Let $M, A, B, c, A', B'$ be as in the above theorem. Suppose $M$ is dense. Then we have a full classification of the Stone space of global extensions of tp($c/AB$) which do not fork over $A$: we have an explicit natural homeomorphism between that space and some topological space involving $S_{\mathrm{DOAG}}(\{0\})$ and the $(\mathbb{Z}_l)_{l \text{ prime}}$.

Moreover, if tp(c/AB) admits a global Aut(M/A)-invariant extension, then ev-
ery non-forking global extension of tp(c/AB) is Aut(M/A)-invariant.

More details are given in the end of ??, and in ??.
The homeomorphism is
given as a map into some space of the form $I \times J$, with $I$ a finite product of finite
coproducts of closed subspaces of $S_{\text{DOAG}}^{[c]}(\{0\})$, and $J$ some product of finite powers
of the $\mathbb{Z}_l$. If tp(c/AB) has no global Aut(M/A)-invariant extension, then $J$ will
be a singleton. We claim that a similar classification can be obtained when $M$ is
discrete, and we give a sketch of the proof in ??.

We also give various detailed examples on the behavior of forking and dividing
in many regular and non-regular ordered Abelian groups.

Our main results rely heavily on our description of forking in DOAG, which really
is the core of our paper. This theory is an enrichment of DLO and the theory of
non-trivial Q-vector spaces, and it is a reduct of RCF, three very common theories
in which we know exactly what is forking. However, the current literature lacks of a
satisfying description of forking in DOAG, and this paper fills that gap. Simon gave
in ([$?$], Proposition 2.5) a characterization of forking in dimension one (essentially
the one we describe in the first condition of ??) for dp-minimal ordered structures,
which include dp-minimal ordered groups. Theorem 13.7 from [$?$] can be used to
establish our characterization of forking in DOAG in the particular case where the
base parameter set is an Archimedean-complete model.

Dolich gave in [?] (see [?] for a survey) a characterization of forking in o-minimal
expansions of real-closed fields. This very nice characterization naturally links fork-
ing with our intuition on o-minimal geometry. The independence notion introduced
by Dolich is shown to be stronger or equal to forking in any o-minimal theory, but,
for technical reasons, one needs to be in an expansion of RCF to prove equality.
In these extensions, Dolich shows that their independence coincides with the ab-
stract model-theoretic notion of “non-1-dividing in semi-simple theories” (see their
Section 8). We do not know whether non-forking coincides with the independence
of Dolich in DOAG, however we show in subsection ?? that forking does coincide with 1-dividing.

Our work is very close in spirit to the work of Menunni and Hils-Menunni ([?],
[?]) on the domination monoid. They focus on a different notion, in a larger class of
theories. However, while they describe the space of invariant types (over arbitrary
small parameter sets) up to equidomination, our work essentially describes the space
of A-invariant types (with A fixed) up to A-interdefinability, so one can see this pa-
per as a local refinement of their work in a restricted subclass of the structures they
consider. One new difficulty that comes with having to deal with A-invariant types
instead of arbitrary invariant types is that we have to deal with *Archimedean exten-
sions*, the analogue of immediate extensions for the Archimedean/convex valuation
(see the end of page 58 of [?] for a formal definition).

This paper aims to be the first step towards a systematic characterization of
forking in every complexity class of arbitrary ordered Abelian groups. The model-
theoretic complexity of some completion $T$ of the theory of infinite ordered Abelian
groups depends mostly on two factors:

* For each prime $l$, the index of the subgroup of elements divisible by $l$ in models of $T$.

* For each prime $l$, the order-type of the $l$-spine (see ?? and the paragraph that follows), a definably-linearly-ordered, $\emptyset$-definable family of convex subgroups of the models of $T$.

Those two factors are not completely independent from each other: if the *l*-spine is
infinite, then so is the associated index.

- The most simple theory with respect to those points is DOAG, the only
theory of infinite ordered Abelian groups with trivial indices and trivial
spines.

- Any choice of indices is consistent with ROAG (in fact, such a choice is
complete in the theory of dense models of ROAG). However the completions
of ROAG are exactly the theories $T$ for which the spines are trivial.

- A complexity class which is in some sense “orthogonal” to ROAG is the
class of dp-minimal ordered Abelian groups, which is exactly the class of
theories where all the indices are finite. In this class, the spines may not
be trivial (take for instance the lexicographical product $Z^2$), but they are
always finite.

- The class of dp-finite ordered Abelian groups is the class where all but
finitely many indices are finite, and all the spines are finite. Note that
there still exists completions of ROAG which are not dp-finite, and vice-
versa. The ordered Abelian groups which are strongly-dependent are known
to be dp-finite.

- The class of groups where the spines are all finite strictly contains the
reunion of all the classes that we already mentioned, and we still have a
nice quantifier elimination result for this class, because one can show that
there are boundedly-many definable (with parameters) convex subgroups
for its models.

- There still exists ordered Abelian groups with infinite spines, but we know
at least that their theory is always NIP, and we have some sort of quantifier
elimination for them given by Gurevich, Schmitt and Cluckers-Halupczok
([?], [?], [?]).
Interesting future work could be to get a nice characterization of forking for dp-minimal ordered groups, as quantifier elimination is not very complicated in this class, and the case in dimension one is already covered by the work of Simon.

Let us give a brief outline of the proof of our description of forking in DOAG. Let $A, B, c$ be as in ?? so that the conditions of the list hold. First of all, we show that $c$ is $A$-interdefinable with a tuple $d$ which is under "normal form". After that, we split $d$ into fibers of ad-hoc group valuations with respect to which being in normal form ensures that $d$ is separated (see ??). For each subtuple $d'$ of this partition of $d$, we build a global $A$-invariant extension of $\text{tp}(d'/AB)$. Finally, we show that we can "glue" all those extensions into a global $A$-invariant extension of the full type $\text{tp}(d/AB)$, and we are done. We will try to be as explicit as possible in our manipulations, and we will show that our ad-hoc valuations interact in a meaningful way with the model-theoretic notions of tensor product and weak orthogonality, which play an important role in [?], [?].

In the first section of this paper, we write basic definitions, and we give a more detailed and formalized outline of the proof of our result in DOAG.

In the second section, we define our ad-hoc valuations, and we show, given global $A$-invariant extensions of the $\text{tp}(d'/AB)$, how to glue them into a global $A$-invariant extension of $\text{tp}(d/AB)$.

In the third section, we show how to build a global $A$-invariant extension of each of the $\text{tp}(d'/AB)$.

In the fourth section, we show how to build the normal form $d$ from $c$. The transformation of a family into a family under normal form is carried out in a very procedural way which goes through many steps, when we apply successively $A$-translations and maps from $\text{GL}^n(\mathbb{Q})$ to $c$. Once $d$ is built, our characterization of forking for DOAG will be established.

We extend our results in DOAG to ROAG in the fifth section. We give a short survey on these groups, where we give different characterizations of their elementary class (being equivalent to an Archimedean group, having no definable convex subgroups...). This characterization is not absolutely necessary for our proofs, but it will provide the interested readers who are new to the field with a good understanding of what are the models of ROAG. The main tool we do need for our theorem is quantifier elimination, with which we describe the types, their global extensions and their conjugates. We end the section by proving our main result, ??, where we give our characterization of forking for ROAG.

Let us also describe the outline of the proof. In ROAG, we can see a complete type as a reunion of countably many partial types ($\Phi$-types) which are "orthogonal" to each other (see ??). For each such partial type, we give algebraic characterizations (essentially the conditions of ??) of when there exists a partial global extension which does not fork over $A$, and when this extension is $A$-invariant. Our results in DOAG allow us to deal with partial types that involve equality and order. The results about the other partial types may already follow from literature on the model theory of torsion free Abelian groups, we still prove them explicitly for completeness.

The last section is dedicated to examples. We define several notions of independence, many of which have already been considered in the literature, and we compare them to the ones we use in the paper. These comparisons are carried out in various classes of ordered Abelian groups which will gradually increase in
complexity: we work in DOAG in the beginning, then a non-regular poly-regular
group (basically a subgroup of the lexicographical sum $\mathbb{R}^n$, see [?]), and we study
a group with an infinite spine at the end.

# 1. PRELIMINARIES

## 1.1. Cut-independence.

**Notation.** Let $(M, <, ...)$ be the expansion of a linearly-ordered first-order structure. Let $C \supseteq A \supseteq B$ be parameter subsets of $M$. Suppose $M$ is $|AB|^{+}$-saturated and strongly-$|AB|^{+}$-homogeneous.

In this paper, we will manipulate various linear orders (not only ordered Abelian
groups, but also the chains of their convex subgroups, and other ad-hoc orders).
In these linear orders, the cuts will be seen as type-definable sets. The formal
definition is as follows:

**Definition 1.1.** Let $c \in M$. The cut of $c$ over $A$ will be the $A$ type-definable set $\text{ct}(c/A)$, defined as the intersection of every interval containing $c$ with bounds in $A \cup \{\pm\infty\}$ (a singleton is a closed interval).

Let $P$ be some partition of $M$ into convex sets. Then there is a natural linear
order on $P$, the only linear order that makes the projection $M \to P$ an increasing
map.

The set of cuts over $A$ is clearly a partition of $M$ into convex sets, therefore the
above paragraph applies.

Let us also write $x > A$ when $\forall a \in A, x > a$. If $\emptyset \neq A' \subseteq A$, define $\text{ct}_<(A'/A)$ (resp. $\text{ct}_<(A'/A)$) as the cut over $A$ that corresponds to the elements that are $> A'$ (resp. $< A'$), and strictly smaller (resp. larger) than any element of $A$ that is $> A'$ (resp. $< A'$). More generally, if $X$ is some (type/$v$)-definable subset of $M$, we write $\text{ct}_<(X/A) = \text{ct}_<(X(A)/A)$, and similarly for $\text{ct}_>(X/A)$.

<figure><img src="image_2.png" /></figure>

**Definition 1.2.** Let us define the following independence notions, which are ternary
relations defined on the small subsets of $M$:

* $\downarrow^d$ is the standard non-dividing independence (see [?], Definition 7.1.2).

* $\downarrow^f$ is the standard non-forking independence (see [?], Definition 7.1.7).

* Let $c_1...c_n \in M$. Define $c_1...c_n \searrow_A^{\text{inv}} B$ when $\text{tp}(c_1...c_n/AB)$ has a global (in $S(M)$) Aut$(M/A)$-invariant extension. Define $C \searrow_A^{\text{inv}} B$ when $c \searrow_A^{\text{inv}} B$ for each finite tuple $c$ of $C$.

* $\downarrow^{\text{bo}}$ (for “Bounded Orbit”) is a weaker notion than $\downarrow^{\text{inv}}$, where we require the orbit under Aut$(M/A)$ of the global extension to have a bounded cardinal instead of being a single point.

**Fact 1.3.** The inclusions $\downarrow^{\text{inv}} \subseteq \downarrow^{\text{bo}} \subseteq \downarrow^{\text{f}} \subseteq \downarrow^{\text{d}}$ always hold in any first-order structure.

*Remark 1.4.* The items $\downarrow^d, \downarrow^f$ are well-known model-theoretic independence relations. In this paper, we will work with theories where they notoriously coincide with less technical independence notions. Indeed, in DOAG, if $A$ is not included in
By the above proposition, one can see that $C \underset{A}{\downarrow}^{\text{inv}} B \iff$ for all $c_1...c_n \in C$, there exists $p$ a global extension of $\text{tp}(c_1...c_n/B)$ so that for all $f \in \text{LC}^n(\mathbb{Q})$, for all closed interval $I$ with bounds in $M$ not disjoint from $A$, we have $p(x) = f(x) \notin I$.

The above description of $\downarrow^{\text{inv}}$ is still not geometric, because it is existential in $p \in S(M)$. There is a geometric description of a weaker version of this, where we swap the first two quantifiers:

**Definition 1.8.** Define $C \downarrow_A^{\text{cut}} B$ if the following equivalent conditions hold:

* $\forall c_1...c_n \in C, \forall f \in \text{LC}^n(\mathbb{Q}), (f(c)) \downarrow_A^{\text{inv}} B$ (the quantifier over $p$ has been swapped with the one over $f$).

* $\forall c \in C, c \downarrow_A^{\text{inv}} B$.

* Every closed interval with bounds in $B$ that has a point in $C$ already has a point in $A$.

For $c_1...c_n \in M$, we define $c_1...c_n \downarrow_A^{\text{cut}} B \iff (A + \mathbb{Q}c_1 + \dots + \mathbb{Q}c_n) \downarrow_A^{\text{cut}} B$.

The last item of the above list is a purely geometric description which depends only on $A, B, C$. This description is simple and uses very little algebra: we have to replace $A$ (resp. $B, C$) by the $\mathbb{Q}$-vector subspace generated by $A$ (resp. $AB, AC$) as done in the notation, then we take every point from $C$, we completely ignore the algebraic relations between those points, and we check for each point a condition that only depends on the linear order.

**Lemma 1.9.** We have $\downarrow^d \subseteq \downarrow^{\text{cut}}$ in DOAG.

*Proof.* See ??.

It actually turns out that this nice, but weak independence notion $\downarrow^{\text{cut}}$ is no weaker than $\downarrow^{\text{inv}}$ in DOAG. This is the fundamental result of this paper:

**Theorem 1.10.** In DOAG, we have $\downarrow^{\text{inv}} = \downarrow^{\text{cut}}$.

By ?? and ??, we just need to prove that $\downarrow^{\text{cut}} \subseteq \downarrow^{\text{inv}}$.

**Notation.** Recall that $B \ge A \le C$ are $\mathbb{Q}$-vector-subspaces of $M$. We assume $C \downarrow_A^{\text{cut}} B$. Let $c_1...c_n \in C$.

Our goal is to build a global $A$-invariant extension of $\text{tp}(c_1...c_n/B)$. This will be achieved in sections ??, ??, ??, and we deal with ROAG and examples in the remaining sections.

*Remark 1.11.* If $c \in C \setminus A$, then we have $c \downarrow_A^{\text{cut}} B$ if and only if at least one of the following conditions holds:

* Every bounded $B$-definable closed interval included in $\text{ct}(c/A)$ (viewed as an $A$-type-definable set) has all its points smaller than $c$.

* Every bounded $B$-definable closed interval included in $\text{ct}(c/A)$ has all its points larger than $c$.

**Definition 1.12.** If the first condition of the above remark holds, then we will say that $c$ *keeps right* with respect to $A$ and $B$. If the second condition holds, then $c$
leans left. The two conditions hold at the same time if and only if $\text{ct}(c/\text{dcl}(A))$ has no point in $\text{dcl}(AB)$.

<figure><img src="image_3.png" /></figure>

It is easy to see that $c$ leans right (resp. left) with respect to $A$ and $B$ if and only if there exists $A' \subseteq A$ so that $\text{ct}(c/B) = \text{ct}_>(A'/B)$ (resp. $\text{ct}_<(A'/B)$).

This definition also makes sense (and will be used) in more abstract linear orders that do not come from ordered groups.

**1.2. Valuations.** The content of this subsection is well-known in the literature, and most of the proofs will be omitted.

**Notation.** In this subsection, let $G$ be an Abelian group.

**Definition 1.13.** Let $H$ be a subgroup of $G$, and $\Gamma$ a linearly-ordered set with a least element $-\infty$. A map $\text{val}: G \to \Gamma$ is an $H$-valuation if it satisfies the following axioms for all $x, y \in G$:

* $\text{val}(x - y) \le \max(\text{val}(x), \text{val}(y))$

* $\text{val}(x) = -\infty \iff x \in H$

A valuation over $G$ is a $\{0\}$-valuation.

If $\text{val}'$ is another $H$-valuation of $G$, we say that $\text{val}'$ refines $\text{val}$ if there exists an increasing map $f: \text{val}'(G) \to \text{val}(G)$ so that $\text{val} = \text{val}' \circ f$.

Equivalently, an $H$-valuation can be seen as a valuation over $G/H$.

*Remark 1.14.* Let us remark for the readers unfamiliar with valuations that, if $\text{val}(x) < \text{val}(y)$, then it follows from the axioms that $\text{val}(x + y) = \text{val}(y)$.

The next lemma gives us a canonical way to build an $H$-valuation from a preorder over $G$ satisfying certain conditions:

**Lemma 1.15.** Let $P$ be a preorder over $G$. Let $\sim$ be the associated equivalence relation over $G$, $\pi$ the quotient map, and $<$ the associated order on $G/\sim$. Suppose $<$ is linear, and we have $\pi(x) < \pi(y) \Longrightarrow \pi(x+y) = \pi(y)$ for all $x, y \in G$ (the same relation as in the above remark). Then the following conditions hold:

* $G/\sim$ has a least element, which is $\pi(0)$.

* The fiber $\pi^{-1}(\pi(0))$ is a subgroup of $G$.

* $\pi$ is a $\pi^{-1}(\pi(0))$-valuation.

For example, if $P$ is the divisibility relation on an integral local ring $R$ ($P(x, y)$ when $y$ divides $x$), then $P$ is a preorder satisfying the axiom $\pi(x) < \pi(y) \Longrightarrow \pi(x+y) = \pi(y)$ of the above proposition. In that case, $R/\sim$ is in natural correspondence with the poset of principal ideals of $R$, and $<$ is linear if and only if $R$ is a valuation ring. Then, the group valuation given by the proposition is the natural ring valuation on $R$.

**Notation.** Suppose now $G$ is an ordered Abelian group.

**Proposition 1.16.** Let $x, y \in G$. The following conditions are equivalent:

(1) $x \in \bigcup_{n<\omega} [-n|y|, n|y|]$.

(2) For every convex subgroup $H$ of $G$, $y \in H \Longrightarrow x \in H$.
(3) The convex subgroup of $G$ generated by $x$ is included in the one generated by $y$.

**Definition 1.17.** These equivalent conditions define a relation on $G$ which is clearly a preorder that satisfies the conditions of ??, with $\pi^{-1}(\pi(0)) = \{0\}$. This gives a valuation over $G$, which we will call $\Delta$, the *Archimedean valuation*. For $x \in G$, $\Delta^{-1}(\Delta(x))$ is called the *Archimedean class* of $x$. By abuse of notation, we may identify the Archimedean value of an element with its Archimedean class.

The idea behind the statement $\Delta(x) < \Delta(y)$ is that $x$ is “infinitesimal” compared to $y$, or that $y$ is “infinitely greater” than $x$.

Equivalently:

$$ \begin{aligned} \Delta(x) < \Delta(y) &\iff x \in \bigcap_{n>0} \left[-\frac{1}{n}|y|, \frac{1}{n}|y|\right] \\ &\iff |y| \in \bigcap_{n>0} [n|x|, +\infty[ \end{aligned} $$

Let us draw a picture (not scaled correctly) which intuitively shows how the Archimedean classes look like. Suppose $G$ has four Archimedean classes $\Delta(0) < \delta_1 < \delta_2 < \delta_3$. Then $G$ looks like this:

<figure><img src="image_4.png" /></figure>

There is no standard terminology in the modern literature when it comes to the Archimedean valuation and related notions. Some use the keyword “convex” to refer to all those notions, others use the very obscure names “i-extension” or “i-completeness”. We draw our inspiration from older sources: we choose to use the terminology of [?], which, on top of being a good introductory paper, sets up notations which we find more intuitive and easier to work with.

**Remark 1.18.** A convex subgroup $H \le G$ is equal to the reunion of the convex subgroups of $G$ generated by each $x \in H$. So $H$ is entirely determined by its direct image $\Delta(H)$, which is an initial segment of $\Delta(G)$. We can note that the set $C(G)$ of convex subgroups of $G$ is linearly-ordered by inclusion, and that $H \mapsto \Delta(H)$ is an order isomorphism between $C(G)$ and the set of non-empty initial segments of $\Delta(G)$ ordered by inclusion.

We can also note that the cosets of some convex subgroup $H \le G$ are all convex, because the translations are strictly increasing. So the quotient of $G$ by some convex subgroup $H$ is naturally endowed with a linear order, as defined in ??.

**Definition 1.19.** Let $H$ be a subgroup of $G$, and $\delta \in \Delta(G)$. We denote $H_{<\delta} = \{x \in H | \Delta(x) < \delta\}$. We introduce a similar notation for the conditions $\le \delta$, $= \delta...$ More generally, if $P \subseteq \Delta(G)$, we define $H_{\epsilon P} = \{x \in H | \Delta(x) \in P\}$.

### 1.3. Outline of the proof.
Although $\downarrow^{cut}$ and $\downarrow^{inv}$ are very combinatorial notions, we have to set up a rather technical algebraic machinery to prove that they coincide. Let us explain in this subsection what we will do conceptually.

**Notation.** As in subsection ??, we have $M$ a model of DOAG in the language of ordered $\mathbb{Q}$-vector spaces, $c = c_1...c_n \in M$, and $A \le B$ $\mathbb{Q}$-vector subspaces of $M$, so that $M$ is $|B|^{+}$-saturated and strongly-$|B|^{+}$-homogeneous, and $c_1...c_n \downarrow_A^{cut} B$.
We wish to show $c_1...c_n \underset{A}{\uparrow}^{inv} B$. For this, we will partition our family $c$ into smaller subfamilies, that we will call “blocks”, and we will build a global $A$-invariant extension of the type of each block. We get a family of global $A$-invariant types, that we will call (in ??) a “block extension”, and what is left for us is to “glue” those types into a global $A$-invariant extension of the full type of $c$. We will have several ways to partition our tuple, that refine each other. The finer the partition, the easier it becomes to find a block extension, the harder it is to “glue” the elements of the block extension.

In model theory, the standard way to “glue” global invariant types is via the tensor product:

**Definition 1.20.** Let $p, q$ be $A$-invariant global types. We define the tensor product of $p$ by $q$ to be the following $A$-invariant complete global type:

$$p(x) \otimes q(y) = \{ \varphi(x, y, m) | m \in M, \exists a \models q_{|Am}, p(x) \models \varphi(x, a, m) \}$$

Let $p_1...p_n \in S(A)$. We say that the family $(p_i)_i$ is weakly orthogonal if $\bigcup_i p_i(x_i)$ is a complete type over $A$ in the variables $x_1...x_n$.

The tensor product is associative, but not commutative in general. However, two types that are weakly orthogonal must necessarily commute.

**Fact 1.21.** In DOAG (in fact in any distal theory, see [?], Proposition 2.17), two global $A$-invariant types that commute must be weakly orthogonal.

For us, the fact that two types are weakly orthogonal is a way to say that they are “easy” to glue together.

**Lemma 1.22.** Let $F_1, F_2$ be closed subspaces of $S(A)$. Suppose $p_1$ and $p_2$ are weakly orthogonal for all $p_i \in F_i$. Then the set $F_1 \times F_2$, seen as a topological subspace of $S(A)$, is exactly the topological direct product $F_1 \times_{Top} F_2$, via the homeomorphism $h: p_1(x_1) \cup p_2(x_2) \to (p_1(x_1), p_2(x_2))$.

*Proof.* The map $h$ is clearly a continuous bijection between compact spaces, thus it is a homeomorphism. $\square$

The way we partition our tuple will be with fibers of some ad-hoc valutations:

**Definition 1.23.** Let val be an $B$-valuation over $M$. Given our family $c$, the val-blocks of $c$ will be defined as the maximal subtuples of $c$ of elements of equal value. They form a partition of $c$.

By abuse of notation, if $c_i = (c_{ij})_j$ is a val-block of $c$, we will denote $val(c_i) = val(c_{ij})$, which does not depend on the choice of $j$.

A weak val-block extension of $c$ is a family $(p_i)_i$ of global $A$-invariant extensions of the types over $B$ of each val-block of $c$. Such a block extension is strong when $\bigcup_i p_i$ is consistent with $tp(c/B)$.

In this paper, whenever we use the terminology of block extensions, it will always be for global $A$-invariant extensions of types over $B$, the parameter sets $A, B$ will not change.

In particular, if val' refines val, then the val'-blocks of some family form a finer partition than its val-blocks.
**Definition 1.24.** Let val be a *A*-valuation over *M*. We say that the family *c* is val-separated when the following conditions hold:

• $\forall (\lambda_i)_i \in \mathbb{Q}, \text{val} \left( \sum_i \lambda_i c_i \right) = \max_i(\lambda_i c_i).$

• $\forall i, c_i \notin A$.

One can note that *c* is val-separated if and only if each of its val-blocks is. Moreover, a finite family that is separated with respect to some *A*-valuation must be a lift of some *Q*-free family of *M*/*A*. In particular, such a family is *Q*-free.

*Remark 1.25.* The family *c* is clearly *A*-interdefinable with the lift of a *Q*-free family from *C*/*A*. Moreover, if *d* is a tuple that is *A*-interdefinable with *c*, then one can easily show that we have $c \downarrow_A^d B \iff d \downarrow_A^d B$ for $\downarrow \in \{\downarrow^{cut}, \downarrow^{inv}\}$.

**Notation.** By the above remark, we will assume that *c* is a lift of a *Q*-free family from *C*/*A*.

*Remark 1.26.* Now, if *d* is another lift of a *Q*-free family from *C*/*A*, then *c* and *d* are *A*-interdefinable if and only if they have the same size *n*, and there exists $f \in GL_n(\mathbb{Q})$ so that the *i*-th component of *d* is an *A*-translate of the *i*-th component of $f(c)$ for each *i*.

We just defined several ways to state that the blocks of some tuple are “indepen-
dent” from each other and thus “easy to glue”. The fact that the types of the
blocks are weakly orthogonal is a model-theoretic way to define this idea, whereas
the notion of separatedness with respect to some valuation is more algebraic. The
core of our proofs relies on building valuations over M for which these notions
interact in a meaningful way.

More precisely, we will build $\mathcal{D}$-valuations $\text{val}_D^i$ on $M$, with $i \in \{1, 2, 3\}$, $\mathcal{D} \le M$, so that:

(C1) $\text{val}_D^3$ refines $\text{val}_D^2$ refines $\text{val}_D^1$.

(C2) $\text{val}_{\{0\}}^3 = \Delta$ and $\text{val}_{\{0\}}^2$ is trivial (it is refined by any valuation).

(C3) if $c \downarrow_A^{cut} B$, then $c$ is $A$-interdefinable with a family $d$ that is simultaneously separated with respect to $\text{val}_A^3$ and $\text{val}_B^3$.

(C4) If $d$ is a $\text{val}_B^3$-separated family, and $d \downarrow_A^{\text{cut}} B$, then $d$ admits a strong $\text{val}_B^3$-block extension.

(C5) Given $d$ a $\text{val}_B^3$-separated family, and $(p_i)_i$ a strong $\text{val}_B^3$-block extension of $d$, some tensor product of the $(p_i)_i$ is a global $A$-invariant extension of $\text{tp}(d/B)$.

(C6) Suppose $d$ is $\text{val}_D^2$-separated, and let $(b_i)_i$ be its $\text{val}_D^2$-blocks. Then $(\text{tp}(b_i/D))_i$ is weakly orthogonal. In particular, any weak $\text{val}_B^2$-block extension is strong.

(C7) Given $d$ a $\text{val}_B^1$-separated family, and $(p_i)_i$ a weak (thus strong by the previous point) $\text{val}_B^1$-block extension of $d$, $(p_i)_i$ is weakly orthogonal, and thus they commute.

Now, if we could build such valuations, then it would follow that $c \downarrow_A^{inv} B$, which would finish the proof of ??. Indeed, if $d$ was given by ??, then by ?? it would be enough to show that $d \downarrow_A^{inv} B$. Now, again by ??, we would have $d \downarrow_A^{cut} B$, so
?? would give us a strong $val_B^3$-block extension of $d$, and ?? would give us a way to glue it into a witness of $d \underset{A}{\downarrow}^{inv} B$, which would conclude the proof. Moreover, ??, and ?? would give us a fine understanding of the global $A$-invariant extensions of the type of $d$. In fact, we will establish in ?? an exhaustive classification of the Stone space of all these global $A$-invariant extensions.

Last but far from least, what remains for us to do is to actually define explicitly those valuations $val_D^i$, and to show that all those nice properties hold.

### 1.4. Relation to Dolich-independence.
We make short comments on how our result on DOAG relates to the work of Dolich ([?]), their Section 8 in particular. Dolich defines in their paper a geometric independence notion in terms of “halfway-definable cells”, which we will call $\downarrow^{\text{Dolich}}$. It is shown that it coincides with $\downarrow^f$ in any o-minimal expansion of RCF. To motivate their results, Dolich also gives five axioms for independence notions (allegedly from Shelah’s unpublished notes), four of which are always satisfied by non-forking in any theory, and the fifth, called “Chain Condition”, is a weakening of the independence theorem. They also define an independence notion called “non-1-dividing”, with a combinatorial definition which strengthens that of non-forking; let us write it $\downarrow^{1-d}$. They claim that, in case some independence relation satisfies the five axioms in a given theory, non-1-dividing is the weakest relation satisfying those axioms, and they show that $\downarrow^{\text{Dolich}}$ satisfies the five axioms in any o-minimal theory. In particular, we have $\downarrow^{\text{Dolich}} \subseteq \downarrow^{1-d} \subseteq \downarrow^f$ in any o-minimal theory, and those inclusions are equalities in expansions of RCF. While we do not know whether $\downarrow^{\text{Dolich}} = \downarrow^f$ in DOAG, we will remark that $\downarrow^{1-d} = \downarrow^f$ in any theory (DOAG in particular) where $\downarrow^f = \downarrow^{inv}$. Indeed, one may easily show that if $\downarrow^{inv}$ satisfies Extension, then it satisfies the Chain Condition. If $\downarrow^{inv} = \downarrow^f$, then it follows that $\downarrow^f$ satisfies the Chain Condition, therefore it satisfies the five axioms. By maximality of $\downarrow^{1-d}$, we have $\downarrow^f \subseteq \downarrow^{1-d}$, and the other inclusion always holds.

## 2. HOW TO GLUE TYPES VIA AD-HOC VALUATIONS

In this section, we will define the valuations $val_A^i$ ($i \in \{1, 2, 3\}$), and we will show that they satisfy the gluing properties ??, ??, ???. We will also give a classification of the cuts over some parameter set.

**Notation.** In this section, we set $M \models DOAG$, and $A \le B$ Q-vector subspaces of $M$ so that $M$ is $|B|^+$-saturated and strongly-$|B|^+$-homogeneous.

### 2.1. Basic definitions and classification of the cuts.

**Lemma 2.1.** Let $c, d \in M$, and $a \in A$, so that $\text{ct}(c/A) = \text{ct}(d/A)$. Then $\text{ct}(c+a/A) = \text{ct}(d+a/A)$.

*Proof.* If not, then there exists $a' \in A$ which lies strictly between $c+a$ and $d+a$.

<figure><img src="image_5.png" /></figure>

Then $a' - a \in A$ lies strictly between $c$ and $d$, contradicting the hypothesis. $\square$
<figure><img src="image_6.png" /></figure>

We hope the figures make the proofs easier to understand. However, we do not want them to be misleading, thus we would like to say that they may not cover all the possible cases. For instance, if $a$ was negative, then the correct picture would be reversed.

**Definition 2.2.** By the above Lemma (and by saturation), $A$ acts by translation over the set of all the cuts of $M$ over $A$. For $c \in M$, denote $\text{Stab}(c/A)$ the stabilizer of $\text{ct}(c/A)$.

Note that such a stabilizer is only a subgroup of $A$, which is clearly convex in $A$, but not in $M$.

**Definition 2.3.** Let $d \in M$. If $d \notin A$, then we define:

$$G(d/A) = \cap \{\!-|a|, |a|\! : a \in A \setminus \text{Stab}(d/A)\},$$

else we define $G(d/A) = \{0\}$. Either way, we also define:

$$H(d/A) = \cup \{\!-|a|, |a|\! : a \in \text{Stab}(d/A)\}.$$

We view $G(d/A)$ as an $A$-type-definable set, and $H(d/A)$ as an $A$-v-definable set. They have the same points in $A$, however they do not have the same points in $M$ when $d \notin A$. They are $A$-(type/v)-definable convex subgroups.

**Example 2.4.** If $\Delta(0) < \Delta(d) < \Delta(A \setminus \{0\})$ (i.e. $d$ is infinitesimal with respect to $A$), then $H(d/A)$ is trivial, and $G(d/A)$ is the type-definable group of elements which are infinitesimal with respect to $A$.

If $\Delta(d) > \Delta(A)$, then $H(d/A)$ is the convex subgroup generated by $A$, whereas $G(d/A)$ is the whole group.

**Remark 2.5.** We always have $G(d/A) \ge H(d/A)$. By definition, $H(d/A)$ is the convex subgroup generated by $\text{Stab}(d/A)$, i.e. the least $A$-v-definable convex subgroup containing $\text{Stab}(d/A)$. In case $d \notin A$, $G(d/A)$ is the largest $A$-type-definable convex subgroup disjoint from from $A \setminus \text{Stab}(d/A)$. For all $d, d' \in M$, we have in fact:

$$G(d/A) = G(d'/A) \implies H(d/A) = H(d'/A) \iff \text{Stab}(d/A) = \text{Stab}(d'/A),$$

the only case where the implication is not an equivalence being when one point is in $A$, and the other is infinitesimal with respect to $A$.

**Proposition 2.6.** Let $d, d' \in M$. If $G(d/A) < G(d'/A)$, then $G(d/A) < H(d'/A)$.

*Proof.* By definition of $G(d/A)$, there must exist $a \in A \setminus \text{Stab}(d/A)$ so that $a \in G(d'/A)$. By the above remark, $a \in H(d'/A) \setminus G(d/A)$. $\square$

Let us now build $\text{val}_A^1$.

**Lemma 2.7.** Let $d_1, d_2 \in M$. Then $G(d_1 + d_2/A) \le \max(G(d_1/A), G(d_2/A))$ with respect to inclusion.

*Proof.* Suppose by contradiction $G(d_1 + d_2/A) > \max(G(d_1/A), G(d_2/A))$. Then, there must exist $a \in \text{Stab}(d_1 + d_2/A)$ so that $a \notin \text{Stab}(d_i/A)$. As $\text{Stab}(d_i/A)$ is a convex subgroup of $A$, we also have $\frac{a}{2} \notin \text{Stab}(d_i/A)$, so there exists $a_i \in A$ which lies strictly between $d_i$ and $d_i + \frac{a}{2}$.
<figure><img src="image_7.png" /></figure>

Then $a_1 + a_2$ lies strictly between $d_1 + d_2$ and $d_1 + d_2 + a$, contradicting $a \in \text{Stab}(d_1 + d_2/A)$. $\square$

<figure><img src="image_8.png" /></figure>

**Definition 2.8.** By the above lemma, the map $x \mapsto G(x/A)$ is an $A$-valuation, so we set $\text{val}_A^1(x) = G(x/A)$.

Note that, by definition, we always have $G(d/B) \le G(d/A)$ for all $d \in M$. Furthermore:

**Proposition 2.9.** For all $d \in M$, if $d \downarrow_A^{\text{cut}} B$, then $H(d/B) \ge H(d/A)$.

*Proof.* Suppose $H(d/B) < H(d/A)$. Then there exists $a \in \text{Stab}(d/A)$ so that $a \notin \text{Stab}(d/B)$. Thus, we can find $b \in B$ which lies strictly between $d$ and $d+a$.

<figure><img src="image_9.png" /></figure>

On one hand, $b-a$ lies strictly between $d$ and $d-a$.

<figure><img src="image_10.png" /></figure>

As $\text{Stab}(d/A)$ is a group, we also have $-a \in \text{Stab}(d/A)$, i.e. $\text{ct}(d+a/A) = \text{ct}(d/A) = \text{ct}(d-a/A)$. In particular, no point from $A$ lies in the closed interval with bounds $b$ and $b-a$. On the other hand, $d$ lies strictly between $b$ and $b-a$, which implies $d \downarrow_A^{\text{cut}} B$. $\square$

In the cut-independent setting, one may intuitively see $G(c/A)$ as a distance that approximates the position of $c$ “from the top”, while $H(c/A)$ approximates $c$ “from the bottom”. Then, when we go to a larger parameter set $B$, we get a finer approximation.

**Proposition 2.10.** Let $a \in A$, and $d \in M$. Then the following conditions are equivalent:

* $d-a \in G(d/A)$, and $d \notin A$.

* $\Delta(d-a) \notin \Delta(A)$.

*Proof.* Suppose $d-a \in G(d/A)$. If we had $\Delta(d-a) \in \Delta(A)$, then there would be $a' \in A$ so that $\Delta(d-a) = \Delta(a')$. This would imply $a' \in (G(d/A) \cap A) = \text{Stab}(d/A)$. Since, for some $n$, $d-n|a'| \le a \le d+n|a'|$, as $na' \in \text{Stab}(d/A)$, it follows that $\text{ct}(d/A) = \text{ct}(a/A)$, thus $d=a$, proving the first direction.

Conversely, suppose $\Delta(d-a) \notin \Delta(A)$. By definition of $G(d/A)$, in order to show that $a \in d \mod G(d/A)$, it suffices to show that $a \in ]d - |a'|, d + |a'|[$ (i.e. $|d-a| \le |a'|$) for every $a' \in A \setminus \text{Stab}(d/A)$. Suppose by contradiction this fails for some $a'$. Then $\Delta(d-a) \ge \Delta(a')$, and this inequality is strict by the hypothesis. As $a' \notin \text{Stab}(d/A)$, let $a'' \in A$ which lies strictly between $d$ and $d+a'$.
<figure><img src="image_11.png" /></figure>

Then $|d-a''| \le |a'|$, thus $\Delta(d-a'') < \Delta(d-a)$. We then apply ?? to get $\Delta(d-a) = \Delta(d-a'' - d + a) = \Delta(a-a'') \in \Delta(A)$, a contradiction. $\square$

**Definition 2.11.** If $a$ satisfies the conditions of the above proposition, then we say that $a$ is a ramifier of $d$ over $A$. We will write ram($d/A$) for the set of those ramifiers, and we will say that $d$ is ramified over $A$ if this set is non-empty. We will say that $d$ is Archimedean over $A$ whenever it is not ramified over $A$.

*Remark 2.12.* If $d \in M$, then $d$ is ramified over $A$ if and only if it satisfies the following equivalent conditions:

(1) The coset $d \mod G(d/A)$ has a point in $A$, and $d \notin A$.

(2) $\Delta(A + \mathbb{Q}d) \not\supset \Delta(A)$.

Moreover, as $A \cap G(d/A) = \text{Stab}(d/A)$, ram($d/A$) is a coset from $A/\text{Stab}(d/A)$, thus all its elements lie in the same coset mod $H(d/A)$.

It turns out that $d$ does not belong to that coset:

**Lemma 2.13.** Let $d \in M \setminus A$, and $a \in A$. Then $d - a \notin H(d/A)$.

*Proof.* Suppose by contradiction that there exists $a' \in \text{Stab}(d/A)$ such that $|d-a| \le |a'|$. Then $a \in \text{ram}(d/A)$, thus $\Delta(d-a) \notin \Delta(A)$, and in particular $\Delta(d-a) < \Delta(a')$. By ??, $\Delta(d - (a + a')) = \Delta(a') \in \Delta(A)$, thus $a + a' \notin \text{ram}(d/A)$. This contradicts the fact that $a \in \text{ram}(d/A)$ and $a' \in \text{Stab}(d/A)$. $\square$

**Corollary 2.14.** Let $d \in M$ be ramified over $A$, and $a, a' \in \text{ram}(d/A)$. Then $\Delta(d-a) = \Delta(d-a')$.

*Proof.* We have $d-a \notin H(d/A) \supset a-a'$, thus $\Delta(d-a) > \Delta(a-a')$, and by ?? we have $\Delta(d-a) = \Delta(d-a+a-a') = \Delta(d-a')$. $\square$

**Definition 2.15.** Let $d \in M$ be ramified over $A$. We define $\delta(d/A)$ as $\Delta(d-a)$ for some $a \in \text{ram}(d/A)$. This definition does not depend on the choice of the ramifier $a$ by the above.

Note that $\delta(d/A)$ is the unique Archimedean class of $\Delta(A + \mathbb{Q} \cdot d) \setminus \Delta(A)$.

*Remark 2.16.* Let $d$ be ramified over $A$, and $a \in \text{ram}(d/A)$. Then, as $G(d/A)$ and $H(d/A)$ have the same points in $A$, one can easily compute $\text{ct}(\delta/\Delta(A))$. A convex subgroup is induced by its Archimedean classes, so by definition of the $\vee$-definable convex subgroup $H = H(d/A)$, for any model $N$ containing $A$, $\Delta(H(N))$ is the least initial segment of $\Delta(N)$ containing $\Delta(\text{Stab}(d/A))$. Now, as $\delta \in \Delta(G(d/A)) \setminus \Delta(H)$, we clearly have $\text{ct}(\delta/\Delta(A)) = \text{ct}_{>}(\Delta(\text{Stab}(d/A))/\Delta(A))$.

<figure><img src="image_12.png" /></figure>
Many notions that we manipulate behave very differently with Archimedean and ramified points, it will often lead to case disjunctions. For instance, the following statements show us that cuts do not look the same for Archimedean and ramified points:

**Lemma 2.17.** Let $d \in M$. Then $\mathrm{ct}(d/A) \subseteq d \mod G(d/A)$ (as type-definable sets).

*Proof.* By contraposition, let $d' \in M$ so that $d - d' \notin G(d/A)$. Then, by definition, there exists $a \in A \setminus \mathrm{Stab}(d/A)$ so that $|d - d'| \ge |a|$. Let $a' \in A$ which lies strictly between $d$ and $d + a$. As either $d + a$ or $d - a$ lies strictly between $d$ and $d'$, either $a'$ or $a' - a$ lies strictly between $d$ and $d'$, thus $\mathrm{ct}(d/A) \ne \mathrm{ct}(d'/A)$. $\square$

**Proposition 2.18.** Let $d \in M$ be Archimedean over $A$. Then the $A$-type-definable set $\mathrm{ct}(d/A)$ coincides with $d \mod G(d/A)$.

*Proof.* Suppose $d \notin A$, else the proposition is trivial.

We have the inclusion $d \mod G(d/A) \subseteq \mathrm{ct}(d/A)$, as $d \mod G(d/A)$ is a convex set containing $d$ and disjoint from $A$. The other direction follows from the above lemma. $\square$

**Proposition 2.19.** Let $d \in M$ be ramified over $A$. Then $\mathrm{ct}(d/A)$ is an $A$-translate of one of the two connected components of $G(d/A) \setminus H(d/A)$.

These translates are exactly the two blue segments in the previous figure.

*Proof.* Let $a \in \mathrm{ram}(d/A)$. We have $\mathrm{ct}(d/A) \subseteq d \mod G(d/A) = a \mod G(d/A)$ by ???. By ??, $d \notin (a \mod H(d/A))$, which is the convex closure of $\mathrm{ram}(d/A)$, a subset of $A$, thus $a \mod H(d/A)$ must be disjoint from $\mathrm{ct}(d/A)$. It follows that $\mathrm{ct}(d/A)$ is included in the connected component of $(a+G(d/A)) \setminus (a+H(d/A))$ containing $d$.

Let $C$ be that connected component, and $d' \in C$. There does not exist $a' \in A$ lying between $d$ and $d'$, for such $a'$ would be in $(a+G(d/A)) \cap A = \mathrm{ram}(d/A) \subseteq a+H(d/A)$, and $a'$ would also belong to $C$ as $C$ is convex, contradicting the fact that $C$ is disjoint from $a+H(d/A)$. $\square$

**Corollary 2.20.** With the hypothesis of the proposition, if $a \in \mathrm{ram}(d/A)$, and $d'$ is another point from $M$, then $d' \equiv_A d$ if and only if the following conditions hold:

* $\mathrm{ct}(\Delta(d' - a)/\Delta(A)) = \mathrm{ct}(\Delta(d - a)/\Delta(A))$ (in particular, $\Delta(d' - a) \notin \Delta(A)$).

* $d' < a \iff d < a$.

*Proof.* The first condition is equivalent to $d' - a \in G(d/A) \setminus H(d/A)$. The second condition ensures that $d'$ lies in the correct connected component. $\square$

The two above propositions yield a simple classification of the cuts over $A$. In order to consider non-forking extensions of types, we now have to deal with ways to refine those cuts to cuts over a larger parameter set $B$. As we classified the cuts over $A$ by looking at $A$-invariant convex subgroups, we have to compute what are the $B$-invariant convex subgroup corresponding to their refinements.

**Definition 2.21.** If $G$ is an $A$-type-definable convex subgroup, then we define the following $A$-$\forall$-definable convex subgroup:

$$ \underline{G}_A = \bigcup_{a \in A \cap G} [-|a|, |a|]. $$
If $H$ is an $A$-v-definable convex subgroup, then we define the following $A$-type-definable convex subgroup:

$$ \overline{H}^A = \bigcap_{a \in A \setminus H} ]-|a|, |a|[. $$

Remark 2.22. Let $d \in M$. We have:

$$ H(d/A) \le \overline{H(d/A)}^M \le \overline{H(d/A)}^B \le \overline{G(d/A)}_B \le \overline{G(d/A)}_M \le G(d/A), $$

and:

$$ \overline{G(d/A)}_A = H(d/A), \quad \overline{H(d/A)}^A = G(d/A). $$

Moreover, if $d \searrow_A^{\text{cut}} B$, then by ?? we recall:

$$ H(d/A) \le H(d/B) \le G(d/B) \le G(d/A). $$

## 2.2. Weak orthogonality.
Now we start to deal with global invariant extensions, in order to prove ??.

**Lemma 2.23.** Let $d \in M \setminus A$ be Archimedean over $A$ so that $d \searrow_A^{\text{cut}} B$. Let $p$ be a global $A$-invariant extension of $\text{tp}(d/B)$. Then $G(d/B) = G(d/A)$ (as $B$-type-definable sets), and $p(x) = x - d \in G(d/A) \setminus \overline{G(d/A)}_M$.

Note that $d$ may very well be ramified over $B$, the lemma will still hold.

**Proof.** Suppose by contradiction $G(d/B) < G(d/A)$. Then there exists $b \in B \setminus \text{Stab}(d/B)$ so that $b \in G(d/A)$. Let $b' \in B$ which lies strictly between $d$ and $d+b$. Then the closed interval with bounds $b'$, $b-b'$ contains $d$ and is strictly included in $d \mod G(d/A) = \text{ct}(d/A)$, which contradicts cut-independence.

By ??, as $p$ extends $\text{tp}(d/A)$, we necessarily have $p(x) = x - d \in G(d/A)$. Let $c = p$ in some elementary extension. As $p$ is $A$-invariant, we have $c \searrow_A^{\text{cut}} M$, thus $G(c/M) = G(d/A)$ by the above paragraph (c is of course Archimedean over $A$). As $c-d \in G(c/M)$, and $d \in M \nmid c$, we have $d \in \text{ram}(c/M)$, thus $c-d \notin H(c/M) = \overline{G(c/M)}_M = \overline{G(d/A)}_M$. We can conclude that $p(x) = x - d \notin \overline{G(d/A)}_M$. $\square$

**Lemma 2.24.** Let $d \in M$ be ramified over $A$ so that $d \searrow_A^{\text{cut}} B$, and $a \in \text{ram}(d/A)$.

$$ \text{Then } \Delta(d-a) \notin \Delta(B). $$

Moreover, if $p$ is a global $A$-invariant extension of $\text{tp}(d/B)$, then, either $(p(x) = x - a \in G(d/A) \setminus \overline{G(d/A)}_M$ and $G(d/B) = G(d/A)$) or $(p(x) = x - a \in \overline{H(d/A)}^M \setminus H(d/A)$ and $H(d/B) = \overline{H(d/A)}$).

**Proof.** Suppose we have $\Delta(d-a) = \Delta(b)$ for some $b \in B$. Let $n > 0$ so that $\frac{1}{n}|b| \le |d-a| \le n|b|$. Then we have $d \in [a-n|b|, a-\frac{1}{n}|b|] \cup [a+\frac{1}{n}|b|, a+n|b|]$. As $b \in G(d/A) \setminus H(d/A)$, none of those two intervals has a point in $A$, which contradicts $d \searrow_A^{\text{cut}} B$.

As $d$ is ramified over $A$, and $p$ extends $\text{tp}(d/A)$, it follows from ?? that $p(x) = x - a \in G(d/A) \setminus H(d/A)$.

Suppose by contradiction $p(x) = x - a \in \overline{G(d/A)}_M \setminus \overline{H(d/A)}^M$. Then there exists $m_1, m_2 \in M$ so that $m_1 \notin H(d/A)$, $m_2 \in \overline{G(d/A)}$, and $p(x) = |m_1| \le |x-a| \le |m_2|$, i.e. $p(x) = x - a \in [a - |m_2|, a - |m_1|] \cup [a + |m_1|, a + |m_2|]$. As $m_1 \notin H(d/A)$ and
$m_2 \in G(d/A)$, each of those two intervals is included in each connected component of $a + (G(d/A) \setminus H(d/A))$. As a result, none of those intervals belongs to $p$ by ??, a contradiction.

By the above, and by ??, if we had $p(x) \models x - a \in \overline{G(d/A)}_B$ (resp. $p(x) \models x - a \notin \overline{H(d/A)}^B$), then we would necessarily have $p(x) \models x - a \in \overline{H(d/A)}^M \setminus H(d/A)$ (resp. $p(x) \models x - a \in G(d/A) \setminus \overline{G(d/A)}_M$). To conclude, it now suffices to show the two following properties:

(1) If $G(d/B) < G(d/A)$, then $\mathrm{tp}(d/B) = x - a \in \overline{G(d/A)}_B$.

(2) If $H(d/B) > H(d/A)$, then $\mathrm{tp}(d/B) = x - a \notin \overline{H(d/A)}^B$.

Let us proceed:

(1) By hypothesis, let $b \in B \setminus \mathrm{Stab}(d/B)$ so that $b \in G(d/A)$. Let $b' \in B$ which lies strictly between $d$ and $d+b$.

<figure><img src="image_13.png" /></figure>

Then the interval $I$ with bounds $b' - a$, $b' - b - a$ is included in $G(d/A)$, and $\mathrm{tp}(d/B) = x - a \in I$.

<figure><img src="image_14.png" /></figure>

(2) By hypothesis, let $b \in \mathrm{Stab}(d/B)$ so that $b \notin H(d/A)$. Then $\mathrm{ct}(d-b/B) = \mathrm{ct}(d/B) = \mathrm{ct}(d+b/B)$, thus $a \notin [d - |b|, d + |b|]$. As a result, $\mathrm{tp}(d/B) = x - a \notin [-|b|, |b|]$. $\square$

The above two lemmas imply:

**Corollary 2.25.** Let $d \in M$ so that $d \downarrow_A^{\mathrm{cut}} B$. Let $p \in S(M)$ be a global $A$-invariant extension of $\mathrm{tp}(d/B)$. Then there exists a $A$-definable map so that one of the three following conditions hold:

* $d \in A$, $f = \mathrm{id}$.

* The cut over $M$ induced by $p$ is one of the two connected components of the translation by $f(d)$ of $G(d/A) \setminus G(d/A)_M$. In that case, $G(d/B) = G(d/A)$.

* The cut over $M$ induced by $p$ is one of the two connected components of the translation by $f(d)$ of $\overline{H(d/A)}^M \setminus H(d/A)$. In that case, $H(d/B) = H(d/A)$.

*Remark 2.26.* If $d \notin A$, while only one of the above conditions hold, we might still have $G(d/B) = G(d/A)$ and $H(d/B) = H(d/A)$ at the same time. For instance, this is the case if $\Delta(d) < \Delta(B \setminus \{0\})$, and $B$ has no element that is infinitesimal compared to $A$.

In fact, one can see that $p$ is the type of an $M$-ramified point which adds an Archimedean class $\delta$ which is cut-independent over $\Delta(A)$: the two conditions of the statement expresses that $\delta$ either leans left or right with respect to $\Delta(A)$ and $\Delta(M)$.

*Remark 2.27.* For all $d \in M$, recall that $G(d/B) = \overline{H(d/B)}^B$. As a result, if $H(d/B) = H(d/A)$, then $G(d/B) = \overline{H(d/A)}^B$. In particular we see that, in either case of the above corollary, $p(x) = x - f(d) \in G(d/B)$.
Let us now prove ??:

**Proposition 2.28.** Let $c = (c_{ij})_{ij}$ be a finite tuple from $M$ that is $\text{val}_B^1$-separated, such that its $\text{val}_B^1$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Suppose we have $(p_i(x_i))_i$ a weak $\text{val}_B^1$-block extension of the $c_{ij}$. Then $(p_i)_i$ is weakly orthogonal.

Recall that types in a block extension are A-invariant. In particular, we must have $c_i \downarrow_A^{\text{cut}} B$ for each $i$.

*Proof.* Let $q(x)$ be the reunion of the $p_i(x_i)$. We have to show that $q$ is complete, i.e. that for all $(\lambda_{ij})_{ij} \in \mathbb{Q}$, there exists $X$ a cut over $M$ so that $q(x) = \sum_{ij} \lambda_{ij}x_{ij} \in X$. Suppose $(\lambda_{ij})_{ij} \in \mathbb{Q}$ is non-zero.

For each $i$, let $f_i$ be the $A$-definable map given by ?? applied to $d_i = \sum_j \lambda_{ij}c_{ij}$. If $\lambda_i = 0$, then $d_i \in A$. Else, as $c_i \downarrow_A^{\text{cut}} B$, and by definition of being separated, $d_i \notin B$, thus $G(d_i/B)$ is non-trivial.

Let $k$ be such that $G(d_k/B)$ is maximal. As $c$ is $\text{val}_B^1$-separated, we have $G(c_k/B) = G(d_k/B)$. By maximality of $k$, for all $i \neq k$, we have $G(d_i/B) < G(d_k/B)$. Let $X$ be the cut over $M$ induced by the global $A$-invariant extension of $\text{tp}(d_k/B)$ induced by $p_k$ (the pushforward of $p_k$ by $x_k \mapsto \sum_j \lambda_{kj}x_{kj}$). Let us show that
$$q(x) = \sum_{ij} \lambda_{ij}x_{ij} \in \left( X + \sum_{i \neq k} f_i(d_i) \right), \text{ an } M\text{-translate of } X \text{ that is indeed a cut over } M \text{ by ??}.$$

* Suppose $p_k(x_k) = \sum_j \lambda_{kj}x_{kj} - f_k(d_k) \in G(d_k/A) \setminus \overline{G(d_k/A)}_M$. Then $G(d_k/B) = G(d_k/A)$ by ??, and $X$ is a reunion of cosets of $\overline{G(d_k/A)}_M$, so it suffices to show that, for all $i \neq k$, $p_i(x_i) = \sum_j \lambda_{ij}x_{ij} - f_i(d_i) \in \overline{G(d_k/A)}_M$. Let $i \neq k$. We may assume $\lambda_i \neq 0$. Recall $G(d_i/B) < G(d_k/B)$.

Suppose $p_i(x_i) = \sum_j \lambda_{ij}x_{ij} - f_i(d_i) \in G(d_i/A) \setminus \overline{G(d_i/A)}_M$. Then again $G(d_i/B) = G(d_i/A)$ by ??, thus $G(d_i/A) < G(d_k/A)$. Proposition ?? shows that $G(d_i/A) < H(d_k/A) \le \overline{G(d_k/A)}_M$, and we can conclude.

Suppose $p_i(x_i) = \sum_j \lambda_{ij}x_{ij} - f_i(d_i) \in \overline{H(d_i/A)}^M \setminus H(d_i/A)$. If we had the inequality $\overline{H(d_i/A)}^M > \overline{G(d_k/A)}_M$, then we would have $G(d_i/A) > G(d_k/A)$, thus $H(d_i/A) > \overline{G(d_k/A)}$ by ???. However, this would imply $G(d_i/B) \ge H(d_i/A) > G(d_k/A) \ge G(d_k/B) > G(d_i/B)$, a contradiction.

* By ??, as $d_k \notin A$, suppose now $p_k(x_k) = \sum_j \lambda_{kj}x_{kj} - f_k(d_k) \in \overline{H(d_k/A)}^M \setminus H(d_k/A)$. As in the above item, we just have to show that, for all $i \neq k$, we have $p_i(x_i) = \sum_j \lambda_{ij}x_{ij} - f_i(d_i) \in H(d_k/A)$. By ??, it is enough to show that $G(d_i/B) < H(d_k/A)$ for all $i \neq k$. By ??, we have $H(d_k/B) = H(d_k/A)$. For $i \neq k$, as $G(d_i/B) < G(d_k/B)$, we have $G(d_i/B) < H(d_k/B) = H(d_k/A)$ by ??, and we can conclude. $\square$

Let us now build $\text{val}_A^2$:
**Lemma 2.29.** Let $d, d' \in M$ so that $G(d/A) = G(d'/A)$, $d$ is Archimedean over $A$ and $d'$ is ramified over $A$. Then $G(d+d'/A) = G(d/A)$ and $d+d'$ is Archimedean over $A$.

*Proof.* Let $a \in \mathrm{ram}(d'/A)$. Then the (convex) coset $d+a \mod G(d/A)$ has no point in $A$ and contains $d+d'$, in particular $d+d'$ mod $G(d/A) \subseteq \mathrm{ct}(d+d'/A)$. By ??, $\mathrm{ct}(d+d'/A)$ is contained in $d+d'$ mod $G(d+d'/A)$, thus $d+d'$ mod $G(d/A) \subseteq \mathrm{ct}(d+d'/A) \subseteq d+d'$ mod $G(d+d'/A) \subseteq d+d'$ mod $G(d/A)$. As a result, $G(d+d'/A) = G(d/A)$. We can conclude as the coset $d+d'$ mod $G(d+d'/A)$ has no point in $A$. $\square$

**Corollary 2.30.** The following preorder satisfies the conditions of ??:

$$x \leq y \\ \iff$$

either $\mathrm{val}_A^1(x) < \mathrm{val}_A^1(y)$, or they have the same value and $x$ is ramified over $A$, or they have the same value and $y$ is Archimedean over $A$.

**Definition 2.31.** We define $\mathrm{val}_A^2$ the valuation induced by the above preorder. It is clearly an $A$-valuation which refines $\mathrm{val}_A^1$.

This new valuation splits the $\mathrm{val}_A^1$-blocks in two, putting the $A$-Archimedean points above the $A$-ramified ones.

Let us now prove ??:

**Proposition 2.32.** Let $c = (c_{ij})_{ij}$ be a finite tuple from $M$ that is $\mathrm{val}_A^2$-separated, such that its $\mathrm{val}_A^2$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Then $(\mathrm{tp}(c_i/A))_i$ is weakly orthogonal.

Note that the types which are weakly orthogonal are types over $A$ here, as opposed to the types in ?? which were global types. The respective restrictions over $A$ of non-weakly orthogonal global types can very well be weakly orthogonal.

*Proof.* Let $p_i = \mathrm{tp}(c_i/A)$. Just like in ??, let $q = \bigcup_i p_i$, let $(\lambda_{ij})_{ij} \in \mathbb{Q}$ be non-zero, and define similarly $(d_i)_i, k$, and $X = \mathrm{ct}(d_k/A)$.

Suppose $d_k$ is ramified over $A$. Then $k$ is the index of the least $\mathrm{val}_A^2$-block of the $\mathrm{val}_A^1$-block that contains it in the coarser partition of the $c_{ij}$. As a result, for $i \neq k$, not only do we have $\mathrm{val}_A^2(d_i) < \mathrm{val}_A^2(d_k)$, but also $\mathrm{val}_A^1(d_i) < \mathrm{val}_A^1(d_k)$, so by ?? we have $G(d_i/A) < H(d_k/A)$. As a result, by ?? and ??, $A$ must have a point $a_i$ lying in the coset $d_i \mod H(d_k/A)$, i.e. $p_i(x_i) = \sum_j \lambda_{ij}x_{ij} - a_i \in H(d_k/A)$. As $X$ is a reunion of cosets of $H(d_k/A)$, we get $q(x) = \sum_{ij} \lambda_{ij}x_{ij} \in (X + \sum_{i \neq k} a_i)$, an $A$-translate of $X$ which must be a cut over $A$.

Suppose now $d_k$ is Archimedean over $A$. Let $k'$ be the index of the $A$-ramified elements of $c$ lying in the same $\mathrm{val}_A^1$-block. If $d_{k'} = 0$, then, as $\mathrm{ct}(d_k/A)$ is also a reunion of cosets of $H(d_k/A)$, we can proceed just as in the above paragraph. Suppose $d_{k'} \neq 0$. Let $X' = \mathrm{ct}\left(\sum_{i \neq k} d_i/A\right)$. By the above paragraph, $X'$ is an $A$-translate of $\mathrm{ct}(d_{k'}/A)$, and $q(x) = \sum_{i \neq k} \sum_j \lambda_{ij}x_{ij} \in X'$. Let $a \in A$ so that $X' \subseteq a \mod G(d_k'/A)$ (as $d_{k'}$ is ramified). Then $q(x) = \sum_{i \neq k} \sum_j \lambda_{ij}x_{ij} - a \in G(d_k/A)$. As $X'$
is a coset mod $G(d_k/A)$, we can conclude that $q(x) = \sum_{ij} \lambda_{ij} x_{ij} \in (X + a)$, which is also a cut over $A$. $\square$

**2.3. Tensor product.** We saw that the blocks of a block extension for the coarsest valuation are weakly orthogonal, thus can be glued by just taking the reunion. It will no longer be the case for finer valuations, the tensor product of the blocks will no longer commute, and we will have to choose the order of the factors. For the third valuation, which is yet undefined, this choice will have to be taken with care, because some choices may give a global type which does not extend our type over $B$.

The following proposition shows us how to properly glue the elements of a $\text{val}_B^2$-block extension.

**Proposition 2.33.** Let $c$ be a finite tuple from $M$ so that $c \downarrow_A^{\text{cut}} B$. Suppose $c = (c_i)_{i \in \{1,2\}}$ is a $\text{val}_B^1$-block (i.e. a family having only one block) that is $\text{val}_B^2$-separated, and $c_1$ (resp. $c_2$) is the block of $B$-ramified (resp. Archimedean) elements of $c$. Suppose $p_1, p_2$ is a weak $\text{val}_B^2$-block extension of $c$. Then $c$ satisfies the restriction to $B$ of the tensor product of $p_1, p_2$ in any order.

Note that, although the restrictions to $B$ of $p_1 \otimes p_2$ and $p_2 \otimes p_1$ coincide, these two global types are distinct, i.e. $(p_1, p_2)$ is not weakly orthogonal. This will be explained in the next section.

*Proof.* By ??, $(\text{tp}(c_i/B))_{i \in \{1,2\}}$ is weakly orthogonal, thus $\text{tp}(c_1/B)$ is complete in $S(B + \mathbb{Q}c_2)$, and it coincides in particular with $(p_1)_{|B+\mathbb{Q}c_2}$. It follows that $c = (p_1 \otimes p_2)_{|B}$, and the same reasoning can be applied to $p_2 \otimes p_1$. $\square$

Let us now build $\text{val}_B^3$:

**Lemma 2.34.** Let $d_1, d_2 \in M$ be two points which are ramified over $A$, and have the same $\text{val}_A^2$-value. Suppose $\delta(d_1/A) < \delta(d_2/A)$. Then $\text{val}_A^2(d_1 + d_2) = \text{val}_A^2(d_2)$, and $\delta(d_1 + d_2/A) = \delta(d_2/A)$.

*Proof.* Let $a_i \in \text{ram}(d_i/A)$. Then, by ?? we have $\Delta(d_1 + d_2 - a_1 - a_2) = \delta(d_2/A)$, therefore $d_1 + d_2$ is ramified over $A$, and $\delta(d_1 + d_2/A) = \delta(d_2/A)$. By ??, we have $\text{ct}_>(\Delta(\text{Stab}(d_2/A))/\Delta(A)) = \text{ct}_>(\Delta(\text{Stab}(d_1 + d_2/A))/\Delta(A))$. However, those stabilizers are convex subgroups of $A$, thus the sets of their Archimedean classes are initial segments of $\Delta(A)$, so this equality of cuts implies $\Delta(\text{Stab}(d_2/A)) = \Delta(\text{Stab}(d_1 + d_2/A))$. It follows that $\text{Stab}(d_2/A) = \text{Stab}(d_1+d_2/A)$, thus $H(d_2/A) = H(d_1+d_2/A)$, and finally $\text{val}_A^1(d_2) = \text{val}_A^1(d_1+d_2)$. We can conclude as $d_2, d_1+d_2$ are both ramified over $A$. $\square$

**Corollary 2.35.** The following preorder satisfies the conditions of ??:

$$x \le y$$

$$\iff$$

either $\text{val}_A^2(x) < \text{val}_A^2(y)$, or they have the same value, they are both ramified over $A$, and $\delta(x/A) \le \delta(y/A)$.

**Definition 2.36.** We define $\text{val}_A^3$ the valuation induced by the above preorder. It is clearly an $A$-valuation which refines $\text{val}_A^2$.
The valuation splits the ramified $val_A^2$-blocks according to the Archimedean classes $\delta$ added by their elements. Note that, while for $i \in \{1, 2\}$, $val_A^i$ factors through the quotient by $A$-elementary equivalence (these are model-theoretic objects), $val_A^3$ is an algebraic object that is no longer refined by types.

We will need to define additional notions in order to better understand $val_B^3$-block extensions. Contrary to $val_B^2$ and $val_B^1$, a weak $val_B^3$-block extension of some tuple might not be strong. We need to find necessary and sufficient conditions for such a block extension to be strong.

**Proposition 2.37.** Let $c = (c_{ij})_{ij}$ be a finite $val_B^2$-block of $B$-ramified points from $M$ that is $val_B^3$-separated, and whose $val_B^3$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Suppose the indices $i$ are ordered so that $i < k \iff \delta(c_i/B) < \delta(c_k/B)$. Let $d = (d_{ij})_{ij}$ be another tuple from $M$. Then $c \equiv_B d$ if and only if the following conditions hold:

(1) For each $i$, we have $d_i \equiv_B c_i$. In particular, each $d_i$ is a $val_B^3$-block of $B$-ramified points that has the same $val_B^2$-value as $c$.

(2) For every $i$ and $k$, we have $i < k \iff \delta(d_i/B) < \delta(d_k/B)$.

*Proof.* For each $i$ and $j$, let $b_{ij} \in \text{ram}(c_{ij}/B)$.

Suppose $c \equiv_B d$ (then condition 1 holds). Let $i, k, j, l$ be indices. If $i < k$, then $\text{tp}(c/B) = |x_{ij} - b_{ij}| < \frac{1}{n}|x_{kl} - b_{kl}|$ for every $n$, therefore $\Delta(d_{ij} - b_{ij}) < \Delta(d_{kl} - b_{kl})$. Moreover, by ??, we have $\delta(d_{ij}/B) = \Delta(d_{ij} - b_{ij})$ (and similarly for $kl$), and therefore condition 2 holds.

Conversely, suppose that the conditions of the list hold. Let $f_i \in \text{LC}^{|c_i|}(\mathbb{Q})$, and let $b' = \sum_i f_i(b_i)$. We have to prove that $c' = \sum_i f_i(c_i)$ and $d' = \sum_i f_i(d_i)$ have the same cut over $B$. This is trivial if each $f_i$ is zero, else let $k$ be the maximal index for which $f_k$ is non-zero. By $val_B^3$-separatedness, we have:

$$ (3) \qquad \delta(c_k/B) = \Delta(f_k(c_k) - f_k(b_k)) > \Delta(f_i(c_i) - f_i(b_i)) $$

for each $i \neq k$. By the conditions of the list, this also holds for $d$. Choose an arbitrary index $j$. There exists $n \in \omega_0$ and $\varepsilon \in \{\pm 1\}$ so that $f_k(c_k) - f_k(b_k)$ lies between $\frac{\varepsilon}{n}(c_{kj} - b_{kj})$ and $\varepsilon n(c_{kj} - b_{kj})$. By the first condition, this must hold for $c$ replaced with $d$. Moreover, as ?? holds for $d$, for all $i \neq k$ and $N > 0$, we have $|f_i(d_i) - f_i(b_i)| < \frac{1}{N}|d_{kj} - b_{kj}|$. By choosing $N$ large enough, and by summing everything, we get that $d' - b'$ lies between $\frac{\varepsilon}{N+1}(d_{kj} - b_{kj})$ and $\varepsilon(N+1)(d_{kj} - b_{kj})$, and we get the same property with $d$ replaced by $c$. Then we have:

$$ \begin{align*} \text{ct}(\Delta(c' - b')/\Delta(B)) &= \text{ct}(\Delta(c_{kj} - b_{kj})/\Delta(B)) \\ &= \text{ct}(\Delta(d_{kj} - b_{kj})/\Delta(B)) \\ &= \text{ct}(\Delta(d' - b')/\Delta(B)) \end{align*} $$

The second equality is a consequence of ?? applied to $c_{kj}$, $d_{kj}$ (by the hypothesis $c_k \equiv_B d_k$). The other equalities follow from the inequalities of the previous paragraph. We also have:

$$ d' < b' \iff \varepsilon = -1 \iff c' < b' $$

As a result, by ??, we have $c' \equiv_B d'$, concluding the proof. $\square$
Remark 2.38. We recall some facts from previous subsections. Let $c = (c_{ij})_{ij}$ be a finite $val_B^2$-block of $B$-ramified points from $M$ that is $val_B^3$-saparated, whose $val_B^3$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Let $(p_i)_i$ be a weak $val_B^3$-block extension of $c$. By ??, and by separatedness, there exists $G$ (resp. $H$) a unique $A$-type-definable(resp. $A$-$\mathcal{V}$-definable) convex subgroup so that $H < G$, and exactly one of the following conditions holds for each $i$:

(1) For every non-zero $f \in \text{LC}^{|c_i|}(\mathbb{Q})$, there exists $b \in B$ so that $p_i(x_i) \models f(x_i) - b \in G \setminus \underline{G}_M$.

(2) For every non-zero $f \in \text{LC}^{|c_i|}(\mathbb{Q})$, there exists $b \in B$ so that $p_i(x_i) \models f(x_i) - b \in \overline{H}^M \setminus H$.

The following notion will mostly be used to understand strong $val_B^3$-block extensions, and to prove ??:

**Definition 2.39.** If the first condition of the above remark holds, then we say that $p_i$ is outer, else $p_i$ is inner.

<figure><img src="image_15.png" /></figure>

**Lemma 2.40.** Let $c = (c_{ij})_{ij}$ be a finite $val_B^2$-block of $B$-ramified points from $M$ that is $val_B^3$-saparated, such that its $val_B^3$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Suppose the indices $i$ are ordered so that $i < k \iff \delta(c_i/B) < \delta(c_k/B)$. Let $(p_i)_i$ be a weak $val_B^3$-block extension of $c$. Let $I$ (resp. $O$) be the set of indices $i$ so that $p_i$ is inner (resp. outer). Then the following are equivalent:

* $(p_i)_i$ is a strong $val_B^3$-block extension of $c$.

* For every $i \in I, o \in O$, we have $i < o$.

*Proof.* The realizations of $\bigcup_i p_i$ all satisfy the first condition of ??, so we are trying to characterize when some of them satisfy the second condition. Let $H = H(c/A) < G(c/A) = G$ be the convex subgroups that witness the above lemma.

Suppose we have $i \in I$ and $o \in O$ so that $o < i$. Let $d \models \bigcup_j p_j$. Then, as $p_i$ is inner, we have $\delta(d_i/M) \in \Delta(\overline{H}^M)$. By the same argument, we have $\delta(d_o/M) \notin \Delta(\underline{G}_M)$, therefore $\delta(d_i/M) < \delta(d_o/M)$, thus $d \nparallel_B c$, and we proved the top-to-bottom implication.

Conversely, suppose we have $i < o$ for every $i \in I, o \in O$. In order to satisfy the second condition of ??, it suffices to prove that the following partial type is consistent:

$$q(x) = \bigcup_i p_i(x_i) \cup \left\{ \left| x_{i,f(i)} - b_i \right| < \frac{1}{n} \mid x_{k,f(k)} - b_k \mid : i < k, n > 0 \right\}$$

with $f(i)$ an arbitrary choice of a coordinate of $(c_{ij})_j$ for each $i$, and $b_i \in \text{ram}(c_{i,f(i)}/B)$. Let us build a realization $\beta$ of $q$.

Let $(\alpha_i)_{i \in I} \models \bigcup_{i \in I} p_i$. We will define by induction, for each $i \in I$, a tuple $\beta_i \models p_i$ so that, for $i < k$ in $I$, we have $\Delta(\beta_{if(i)} - b_i) < \Delta(\beta_{kf(k)} - b_k)$. Let $N$ be an $|M|^+$-saturated, strongly-$|M|^+$-homogeneous elementary extension of $M$ containing $(\alpha_i)_{i \in I}$. Let $k \in I$, and suppose we defined $\beta_i \in N$ for every $i < k$. By saturation, let $\gamma \in N$ so that $\gamma \in \overline{H}^M \setminus H$, and $\Delta(\gamma) > \Delta(\beta_{if(i)} - b_i)$ for all $i < k$ (in other words,
$\gamma \in \overline{H}^M \setminus (\overline{H}^M)_{M+\sum_{i<k} \mathbb{Q}\beta_{i,f(i)}}$). Such a $\gamma$ exists, as we are only dealing with indices of inner types. Then, by ??, there exists $\varepsilon \in \{\pm 1\}$ so that $b_k + \varepsilon\gamma \equiv_M \alpha_{k,f(k)}$. By strong homogeneity, there exists $\sigma \in \mathrm{Aut}(N/M)$ so that $\sigma(\alpha_{k,f(k)}) = b_k + \varepsilon\gamma$. We set $\beta_k = \sigma(\alpha_k)$.

We can similarly define $(\beta_o)_{o \in O}$. Then $(\beta_i)_{i \in I \cup O}$ satisfies $q$, concluding the proof. $\square$

<figure><img src="image_16.png" /></figure>

Now we can prove ??:

**Proposition 2.41.** Let $c = (c_{ij})_{ij}$ be a finite $\mathrm{val}_B^2$-block of $B$-ramified points from $M$ that is $\mathrm{val}_B^3$-separated, whose $\mathrm{val}_B^3$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Let $(p_i)_i$ be a strong $\mathrm{val}_B^3$-block extension of $c$. Then some tensor product of the $p_i$ is consistent with $\mathrm{tp}(c/B)$.

We will see much later, in ??, that the tensor product in the following proof is in fact the only completion of $(p_i)_i$ (be it $A$-invariant or not) which is consistent with $\mathrm{tp}(c/B)$.

*Proof.* Suppose the indices $i$ are ordered so that $i < k \iff \delta(c_i/B) < \delta(c_k/B)$. Define $I, O, G, H$, just as in the above lemma. Let $\alpha_I \models p_I = \bigcup_{i \in I} p_i$, $\alpha_O \models p_O = \bigcup_{o \in O} p_o$. As the conditions of the above lemma are satisfied, if we had $\alpha_I \equiv_B (c_i)_{i \in I}$ and $\alpha_O \equiv_B (c_o)_{o \in O}$, then by ?? we would clearly have $\alpha_I \alpha_O \equiv_B c$. Moreover, $G(\alpha_I/M) = \overline{H}^M < \overline{G}_M = H(\alpha_O/M)$, thus one clearly sees that $\alpha_I$ and $\alpha_O$ are $\mathrm{val}_M^2$-blocks of distinct values, thus their types over $M$ are weakly orthogonal by ??'. Therefore, if the respective types over $M$ of $\alpha_I$ and $\alpha_O$ are tensor products of the $p_i$, then so is the type of $\alpha_I \alpha_O$. As a result, we can assume that either $I$ or $O$ is empty.

Suppose $O = \emptyset$. For $k \in I$, and $c' = (c_i)_{i>k}$, let us show that $c_k \models p_{k|D}$, with $D$ the $\mathbb{Q}$-vector-subspace generated by $Bc'$. This will imply that $\mathrm{tp}(c/B)$ is consistent with the tensor product of the $p_i$ in increasing order. Let $f \in \mathrm{LC}^{|c_k|}(\mathbb{Q})$ be non-zero. By $\mathrm{val}_B^3$-separatedness, we have $\delta(f(c_k)/B) = \delta(c_k/B)$ (call this Archimedean class $\delta$). Let $b \in \mathrm{ram}(f(c_k)/B)$. As $i \in I$, we have $p_k(x_k) \models f(x_k) - b \in \overline{H}^M \setminus H$. It is enough to show that $f(c_k) - b \in \overline{H}^D$, for the cut over $M$ corresponding to the pushforward of $p_k$ by $f$ would be included in $\mathrm{ct}(f(c_k)/D)$, and we could conclude. For $i > k$, let $\delta_i = \delta(c_i/B)$. By $\mathrm{val}_B^3$-separatedness, we clearly have $\Delta(D) = \Delta(B) \cup \{\delta_i | i > k\}$. By definition of $I$, we have $\delta < \delta_i$ for every $i > k$. Moreover, we have $\mathrm{ct}(\delta_i/\Delta(B)) = \mathrm{ct}_>(\Delta(H) \cap \Delta(B)/\Delta(B))$ by ??'. As a result, $\Delta(H) \cap \Delta(D) = \Delta(H) \cap \Delta(B)$, and $\mathrm{ct}(\delta/\Delta(D)) = \mathrm{ct}_>(\Delta(H) \cap \Delta(D)/\Delta(D))$, thus $\delta \in \Delta(\overline{H}^D)$, and we can conclude.

<figure><img src="image_17.png" /></figure>

Now, if instead $I = \emptyset$, then a similar proof would show that the tensor product of the $p_o$ in decreasing order would be consistent with $\mathrm{tp}(c/B)$. $\square$

<figure><img src="image_18.png" /></figure>

**Corollary 2.42.** The property ?? holds.

*Proof.* The above proposition deals with the particular case of a $\mathrm{val}_B^2$-block. Then, given the global invariant extensions of the types of each $\mathrm{val}_B^2$-block given by the
proposition, some tensor product of those global types witnesses the corollary, by
?? and ??.

3. HOW TO BUILD A BLOCK EXTENSION

In the previous section, we defined the key valuations $val_A^i$, and we gave a proof of the properties ??, ??, ?? . In this section, we will give a proof of ?? . The main technical goal is to show that $\downarrow_A^{cut} B$ coincides with $\downarrow_A^{inv} B$, when what we put on the left is a $val_B^3$-separated $val_B^3$-block. This will show the existence of a weak block extension, and then it will not be hard to show that such a block extension can be chosen strong. In fact, given some $val_B^3$-separated tuple $c$, we will describe explicitly the homeomorphism type of the space of global $A$-invariant extensions of $tp(c/B)$.

**Notation.** Let $M \models DOAG$, and $A \leq B$ be $\mathbb{Q}$-vector-subspaces of $M$ so that $M$ is $|B|^+$-saturated and strongly-$|B|^+$-homogeneous. Let $c$ be a finite tuple from $M$ so that $c \downarrow_A^{cut} B$.

3.1. The Archimedean case.

**Notation.** In this subsection, suppose $c = c_1 \dots c_n$ is a $val_B^3$-separated block of $B$-Archimedean points from $M$.

Remark 3.1. As $c \downarrow_A^{cut} B$, the $c_i$ must be Archimedean over $A$. Indeed, by contra-
position, if $c_i$ is ramified over $A$ for some $i$, then we cannot have $\delta(c_i/A) \in \Delta(B)$,
otherwise there would exist $b \in B$, $a \in A$ so that $\delta(c_i/A) = \Delta(c_i - a) = \Delta(b)$, thus $c_i$
would lie between $a + \frac{1}{n}b$ and $a + nb$ for some $n \in \mathbb{Z}$, which would contradict $c \downarrow_A^{cut} B$.
Now, as $\Delta(c_i - a) \notin \Delta(B)$ we have a contradiction with the hypothesis that all the
$c_i$ are Archimedean over $B$.

**Notation.** By the above remark, let us denote $G = G(c/A) = G(c/B)$ (by ??).

**Lemma 3.2.** Let $p_i$ be a global $A$-invariant extension of $tp(c_i/B)$. Suppose $q$
is a complete global extension of $tp(c/B)$ which extends $\bigcup_i p_i$. If we have $q(x) \models$
$f(x) - f(c) \notin \underline{G}_M$ for all non-zero $f \in \text{LC}^n(\mathbb{Q})$, then $q$ is $A$-invariant.

*Proof.* Suppose by contradiction $q$ is not $A$-invariant. Then there exists a non-zero
$f \in \text{LC}^n(\mathbb{Q})$, and a cut $X$ over $M$, so that $q(x) \models f(x) \in X$, and the global 1-type
induced by $X$ is not $A$-invariant. As $c \downarrow_A^{cut} B$, we have $f(c) \downarrow_A^{inv} B$. By ??, we
clearly see that the two global $A$-invariant extensions of $tp(f(c)/B)$ correspond
to the translates by $f(c)$ of the two connected components of $G \setminus \underline{G}_M$. We will
obtain a contradiction by showing that $X$ must coincide with one of those. It is
enough to show that $q(x) \models f(x) - f(c) \in G \setminus \underline{G}_M$. By hypothesis, we just need to
have $q(x) \models f(x) - f(c) \in G$, which follows from the fact that $c$ is a $val_B^3$-separated
block. Indeed, $val_B^3(f(c)) = val_B^3(c)$, therefore $f(c)$ is Archimedean over $B$, and
$G(f(c)/B) = G$. This implies by ?? that $tp(f(c)/B) \models y - f(c) \in G$, and we can
conclude.

□

It turns out that the space of global $A$-invariant extensions of $tp(c/B)$ in the
Archimedean case has a very simple description:
**Proposition 3.3.** Let $N$ be an $|M|^+$-saturated, strongly-$|M|^+$-homogeneous elementary extension of $M$. Let $H = (\underline{G}_M)(N)$ be the convex subgroup of $N$ generated by $G(M)$, and let $N'$ be the ordered Abelian group $N/N'$, which is a model of DOAG by saturation. Let $F_1 \subseteq S^{|C|}(M)$ be the closed set of $A$-invariant extensions of $\text{tp}(c/B)$, and $F_2 \subseteq S^{|C|}(\{0\})$ be the closed set of types of $\mathbb{Q}$-free families. Then the map $g: \text{tp}^N(\alpha/M) \mapsto \text{tp}^{N'}((\alpha_i - c_i \bmod H)_i/\{0\})$ is a well-defined homeomorphism $F_1 \to F_2$.

*Proof.* Let $f \in \text{LC}^n(\mathbb{Q})$ be non-zero, and let $\alpha \in N$ realize a global $A$-invariant extension of $\text{tp}(c/B)$. Then, by separatedness, $f(c)$ is Archimedean over $A$, and $G(f(c)/A) = G$, thus by ?? we have $f(c) - f(\alpha) \notin \underline{G}_M$. It follows that $g$ is well-defined, and it is continuous by definition (and quantifier elimination).

Let $p_1, p_2 \in F_1$ be distinct. Then there exists two distinct cuts $X_1, X_2$ over $M$, and $f \in \text{LC}^n(\mathbb{Q})$ non-zero, so that $p_i(x) = f(x) \in X_i$. However, by ??, $X_i$ must be the translate by $f(c)$ of a connected component of $G \setminus \underline{G}_M$. As a result, one of the $X_i$ (say, $X_1$) is the leftmost connected component, and the other is the rightmost one. We have $p_1(x) = f(x) - f(c) < \underline{G}_M$, thus $g(p_1) = f(x) < 0$. With the same reasoning, we get $g(p_2) = f(x) > 0$, and we can conclude that $g$ is injective.

Now let us prove surjectivity. Let $\beta'$ be a $\mathbb{Q}$-free family from $N'$. Let $\beta$ be a preimage of $\beta'$ by $\pi: N \to N'$. Let $d_1 \in N$ so that $\Delta(d_1) > \Delta(\beta_i)$ for every $i$. By saturation, let $d_2 \in N$ so that $\Delta(d_2) \in \Delta(G) \setminus \Delta(\underline{G}_M)$. Then $|d_1|$ and $|d_2|$ have the same cut over $G(M): +\infty$, the cut of positive elements larger than all the elements of $G(M)$. By strong homogeneity, there is $\sigma \in \text{Aut}(N/G(M))$ so that $\sigma(|d_1|) = |d_2|$. Since $\sigma$ fixes $G(M)$, we have $\sigma(H) = H$, thus the map $\sigma': \pi(x) \mapsto \pi(\sigma(x))$ is a well-defined automorphism of the ordered group $N'$. As $\beta'$ is $\mathbb{Q}$-free, so is $\sigma'(\beta')$. As $\sigma(\beta)$ is a preimage of $\sigma'(\beta')$, none of its elements is in $H$. Moreover, $|\sigma(\beta_i)| \le \sigma(|d_1|) = |d_2| \in G$ for every $i$, thus $\sigma(\beta_i) \in G$. As $\text{tp}^{N'}(\beta'/\{0\}) = \text{tp}^{N'}(\sigma'(\beta')/\{0\})$, and we are looking for a preimage of this type by $g$, we may assume $\sigma = \text{id}$, and thus, by $\mathbb{Q}$-freeness of $\beta'$, $f(\beta) \in G \setminus \underline{G}_M$ for every non-zero $f \in \text{LC}^n(\mathbb{Q})$. Let $p = \text{tp}((c_i + \beta_i)_i/M)$. As $\beta_i \in G$ for all $i$, $p$ extends $\text{tp}(c/B)$, and it is $A$-invariant by ???. Thus $p \in F_1$, and it is a preimage of $\text{tp}(\beta'/\{0\})$ by $g$.

We showed that $g$ is a continuous bijection, and we conclude by compactness. $\square$

Since $F_2$ is non-empty, we have in particular proved the following:

**Corollary 3.4.** We have $c \underset{A}{\downarrow}^{\text{inv}} B$.

The next subsection will give tools to understand what $F_2$ looks like, but it will mostly be used to deal with the ramified blocks.

## 3.2. Archimedean groups.
We need to define some algebraic notions about Archimedean groups in order to well-understand the global invariant extensions of the types of ramified blocks. Let us start with well-known facts:

**Proposition 3.5.** An ordered Abelian group $G$ has at most two Archimedean classes (including $\Delta(0)$) if and only if it embeds into $(\mathbb{R}, +, <)$. In that case, for all $x \in G_{>0}$, for all $\mu \in \mathbb{R}_{>0}$, there exists a unique embedding of ordered groups $\sigma: G \to \mathbb{R}$ sending $x$ to $\mu$.

The idea of the proof is to send each $y \in G$ to $\mu \cdot (\sup \{\lambda \in \mathbb{Q} | \lambda y < x\})^{-1}$.
*Remark 3.6.* In the special case $G = \mathbb{R}$, we see that an embedding $\sigma: \mathbb{R} \to \mathbb{R}$ is uniquely determined by the choice of $\sigma(1)$, hence $\sigma$ actually coincides with the automorphism $x \mapsto \sigma(1) \cdot x$. This establishes an easy description of the group of ordered group automorphisms of $\mathbb{R}$, which is naturally isomorphic to the multiplicative group $\mathbb{R}_{>0}$.

**Definition 3.7.** We define $\mathbb{P}_n^+$, the space of half-lines in dimension $n$ (plus the origin) as the set of orbits of $\mathbb{R}^n$ under the automorphisms of the ordered group $\mathbb{R}$. We will call $\mathbb{P}^+$ the canonical map $\mathbb{R}^n \to \mathbb{P}_n^+$.

Suppose $G$ is an Archimedean ordered Abelian group, and let $u$ be a finite tuple from $G$. By the universal property of ??, we see that $\mathbb{P}^+(\sigma(u))$ is always the same for any ordered group embedding $\sigma: G \to \mathbb{R}$. As a result, we can define $\mathbb{P}^+(u) \in \mathbb{P}_{|u|}$ as this unique class.

*Remark 3.8.* If $f \in \text{LC}^n(\mathbb{Q})$, then $f$ commutes with any automorphism, thus one can easily show that $f(\mathbb{P}^+(u)) = \mathbb{P}^+(f(u)) \in \mathbb{P}_1^+$ for any $u \in \mathbb{R}^n$. However, $\mathbb{P}_1^+$ is not very complicated to describe: $f(\mathbb{P}^+(u))$ is either $\mathbb{R}_{>0}$, $\mathbb{R}_{<0}$ or $\{0\}$.

**Definition 3.9.** Let $u \in \mathbb{R}^n$. We will say that $\mathbb{P}^+(u)$ is $\mathbb{Q}$-free if $u$ is $\mathbb{Q}$-free. By the above remark, this definition does not depend on the choice of the representative $u$.

**Proposition 3.10.** Let $G$ be an ordered Abelian group. Then for any subgroup $H$ of $G$, and $\Delta(0) \neq \delta \in \Delta(G)$, the quotient:

$$H_{\delta} = H_{\leq \delta} / H_{< \delta}$$

is Archimedean.

In the literature, the $(H_\delta)_{\delta \in \Delta(H)}$ are called the “components” of $H$.

**Definition 3.11.** Let $\delta \in \Delta(M)$, and $u$ a tuple from $M_{=\delta}$. By the above proposition, we can define $\mathbb{P}^+(u) = \mathbb{P}^+(u \text{ mod } M_{<\delta})$ with respect to the definition ?? in the Archimedean group $M_{\leq \delta} / M_{<\delta}$.

Suppose $d = (d_1...d_n)$ is a $\text{val}_B^3$-block of $B$-ramified elements of $M$, and let $\delta = \delta(d_i/B) \in \Delta(M)$. Let $b_i \in \text{ram}(d_i/B)$. Then we can define $\mathbb{P}^+(d/B) = \mathbb{P}^+((d_i - b_i)_i)$. As $H(d_i/B) \leq M_{<\delta}$, by ??, this definition does not depend on the choice of the $b_i$.

**Proposition 3.12.** Let $d = (d_1...d_n)$ be a finite $\text{val}_B^3$-block of $B$-ramified points from $M$. Then $d$ is $\text{val}_B^3$-separated if and only if $\mathbb{P}^+(d/B)$ is $\mathbb{Q}$-free.

*Proof.* Let $b_i \in \text{ram}(d_i/B)$, and let $\delta = \Delta(d/B)$. Then $d$ is not $\text{val}_B^3$-separated if and only if there exists a non-zero $f \in \text{LC}^n(\mathbb{Q})$ so that $\text{val}_B^3(f(d)) < \text{val}_B^3(d)$, if and only if there exists such an $f$ so that $\Delta(f(d) - f(b)) < \delta$, if and only if we have $f$ such that $\{0\} = f(\mathbb{P}^+((d_i - b_i)_i)) = f(\mathbb{P}^+(d/B))$, if and only if $\mathbb{P}^+(d/B)$ is not $\mathbb{Q}$-free. $\square$

### 3.3. The ramified case.

**Notation.** Suppose $c = (c_i)_i$ is a $\text{val}_B^3$-separated $\text{val}_B^3$-block of $B$-ramified points. Write $G = G(c/B)$, $H = H(c/B)$, $\delta = \delta(c/B)$. Let $b_i \in \text{ram}(c_i/B)$, and $b = (b_i)_i$.

Recall that ?? characterizes the type of a ramified singleton in terms of cuts of Archimedean classes. We can now extend this characterization to the type of our $\text{val}_B^3$-separated ramified $\text{val}_B^3$-block:
**Proposition 3.13.** Let $d = (d_i)_i$ be a tuple from $M$. Then $d \equiv_B c$ if and only if the following conditions hold:

* $d_i \equiv_B c_i$ for every $i$.

* $d$ is a $\text{val}_B^3$-block, i.e. $\delta(d_i/B) = \delta(d_j/B)$ for every $i, j$.

* $\mathbb{P}^+(d/B) = \mathbb{P}^+(c/B)$.

*Proof.* Suppose the conditions of the list hold. Let $f \in \text{LC}^n(\mathbb{Q})$ be non-zero. Then, as $c$ is $\text{val}_B^3$-separated, $\mathbb{P}^+(c/B)$ is $\mathbb{Q}$-free, therefore $\Delta(f(c) - f(b)) = \delta$. As a result, $\text{ct}(f(c)/B)$ is the translate by $f(b)$ of one of the connected components of $G \setminus H$. The same holds for $d$, but we need to make sure that $f(c)$ and $f(d)$ lie in the same connected component. The thing is, $f(c)$ lies in the rightmost one if and only if $(f(c) - f(b) \mod M_{<\delta}) > 0$ in the ordered group $M_{<\delta} / M_{<\delta}$, if and only if $\mathbb{P}^+(f(c) - f(b)) = \mathbb{R}_{>0}$, if and only if $\mathbb{R}_{>0} = f(\mathbb{P}^+((c_i - b_i)_i)) = f(\mathbb{P}^+(c/B))$. Now, by the hypothesis $\mathbb{P}^+(d/B) = \mathbb{P}^+(c/B)$, those conditions are equivalent to $f(d)$ lying in the rightmost connected component, and we get the bottom-to-top direction.

Conversely, suppose now $d \equiv_B c$, in particular the first condition holds. As for the second, $d$ must be a $\text{val}_B^3$-block, otherwise it would be witnessed in its type by formulas $(|x_i - b_i| > n|x_j - b_j|)_n$ for some $i \neq j$, which would contradict the hypothesis on $c$. By strong homogeneity of $M$, let $\sigma \in \text{Aut}(M/B)$ be such that $\sigma(c) = d$. The equivalence relation $\Delta(x) = \Delta(y)$ is an $\vee$-definable subset of $S^2(\emptyset)$, therefore it is invariant by $\sigma$. As a result, the map $\sigma': (x \mod M_{<\delta}) \mapsto (\sigma(x) \mod M_{<\delta(d/B)})$ is a well-defined automorphism of ordered groups from $M_{<\delta} / M_{<\delta}$ to $M_{<\delta(d/B)} / M_{<\delta(d/B)}$, which sends each $c_i - b_i$ to $d_i - b_i$. Let $\tau$ be an embedding of ordered groups $M_{<\delta(d/B)} / M_{<\delta(d/B)} \to \mathbb{R}$. Then $\mathbb{P}^+(c/B) \ni \tau \circ \sigma'((c_i - b_i)_i) = \tau((d_i - b_i)_i) \in \mathbb{P}^+(d/B)$, thus the last condition of the list holds, and we can conclude. $\square$

**Lemma 3.14.** Choose an arbitrary index $i$. Suppose $p$ is a global $A$-invariant extension of $\text{tp}(c_i/B)$. Then $q = p \cup \text{tp}(c/B)$ is a complete global type in $S(M)$.

Note that $q$ may not be $A$-invariant. We will see an example soon.

*Proof.* Proving that $q$ is consistent is easy: choose $d$ a realization of $p$ in some $|B|^+$-strongly-homogeneous elementary extension $N$ of $M$, choose $\sigma \in \text{Aut}(N/B)$ so that $\sigma(c_i) = d$, then $\sigma(c)$ is a realization of $q$.

Let $\alpha \models q$, and $\delta' = \Delta(\alpha_i - b_i)$. As $\alpha_i \models p$, we have $\delta' \in (\Delta(G) \setminus \Delta(\underline{G}_M)) \cup (\Delta(\overline{H}^M) \setminus \Delta(H))$, in particular it does not belong to $\Delta(M)$. For every $j \neq i$, not only do we have $\Delta(\alpha_j - b_j) \notin \Delta(B)$ as $\alpha_j \equiv_B c_j$, but in fact we have $\Delta(\alpha_j - b_j) = \delta'$ by the second condition of the above proposition. In particular $\alpha$ is a $\text{val}_M^3$-block, and we have:

$$ \mathbb{P}^+(\alpha/M) = \mathbb{P}^+((\alpha_j - b_j)_j) = \mathbb{P}^+(\alpha/B) = \mathbb{P}^+(c/B) $$

The last equality follows from the above proposition, and the others follow from the definitions. We established that, for all $\beta \models q$, $\beta$ is a $\text{val}_M^3$-block, and $\mathbb{P}^+(\beta/M) = \mathbb{P}^+(\alpha/M)$. By the above proposition (with $c, B$ replaced by $\alpha, M$), in order to check that $q$ is complete, it is enough to show that for every $j$, the pushforward $q_j$ of $q$ by the $j$-th projection, is a complete type in $S(M)$.

Let $f \in \text{LC}^N(\mathbb{Q})$ be the projection on the $j$-th coordinate.

* On one hand, we have $q_j(x_j) \models x_j > b_j$ if and only if $f(\mathbb{P}^+(c/B)) > 0$.
• On the other hand, we have $q(x) \models \text{ct}(\Delta(x_j - b_j)/\Delta(M)) = \text{ct}(\Delta(x_i - b_i)/\Delta(M)) = \text{ct}(\delta'/\Delta(M))$. The first equality follows from the fact that the realizations of $q$ are $\text{val}_M^3$-blocks, and the last equality follows from ??, as $p$ is complete. In particular, $q_j(x_j) \models \text{ct}(\Delta(x_j - b_j)/\Delta(M)) = \text{ct}(\delta'/\Delta(M))$. Then, we can apply ?? to show that $q_j$, and thus $q$, is complete. □

**Proposition 3.15.** Let $i$ (if it exists) be such that $c_i$ is Archimedean over $A$. Then $\text{tp}(c_i/B)$ has exactly one global $A$-invariant extension $p$, and $p \cup \text{tp}(c/B)$ is a complete, $A$-invariant global type in $S(M)$.

*Proof.* Suppose for simplicity $c_i > b_i$ (else apply the proposition with $c_i, b_i$ replaced by $-c_i, -b_i$). Recall that $c \downarrow_A^{\text{cut}} B$, thus $\text{tp}(c_i/B)$ must have at least one global $A$-invariant extension. By ??, $G(c_i/A) = G$, and the types $p_>, p<$ over $M$ corresponding respectively to the cuts $\text{ct}_>(\text{ct}(c_i/A)/M) = \text{ct}_>(b_i \text{ mod } G(M)/M)$ and $\text{ct}_<(\text{ct}(c_i/A)/M) = \text{ct}_<(b_i \text{ mod } G(M)/M)$ are clearly global $A$-invariant extensions of $\text{tp}(c_i/A)$. These two types are the only ones implying that $x - b_i \in G \setminus \mathcal{G}_M$. As $c_i > b_i$, only $p_>$ extends $\text{tp}(c_i/B)$. By ??, there are no other global $A$-invariant extension of $\text{tp}(c_i/B)$.

Let $q = p_> \cup \text{tp}(c/B)$, which is complete by ?? . Let us show that $q$ is $A$-invariant. As $\mathbb{P}^+(c/B)$ is $\mathbb{Q}$-free, the realizations of $q$ are $\text{val}_M^3$-separated by ?? . As a result, for every non-zero $f \in \text{LC}^N(\mathbb{Q})$, $q(x) \models f(x) - f(b) \in G \setminus \mathcal{G}_M$. As $G = G(c_i/A)$ and $\mathcal{G}_M$ are $A$-invariant, we only have to show that $X = f(b) \text{ mod } \mathcal{G}_M$ is $A$-invariant as a subset of $M$, because $q(x) \models \text{ct}(f(x)/M) \in \{\text{ct}_>(X/M), \text{ct}_<(X/M)\}$. However, as a subset of $M$, $X$ coincides with $f(b) \text{ mod } G(M)$. If $f(c)$ is Archimedean over $A$, then $X = \text{ct}(f(c)/A)(M)$, which is $A$-invariant. Else, by definition, there exists $a \in A$ so that $f(c) - a \in G(M)$. However, we also have $f(c) - f(b) \in G(M)$, therefore $X = a \text{ mod } G(M)$, which is $A$-invariant, and we can conclude. □

Note that $p>$, and thus $q$, is outer (with respect to ?? ).

*Remark 3.16.* If $j \neq i$, and $c_j$ is ramified over $A$, then $\text{tp}(c_j/B)$ might have two global $A$-invariant extensions, and the reunion of each of those extensions with $\text{tp}(c/B)$ will be a consistent, complete type in $S(M)$. However, one of the two will not be $A$-invariant, as it will not be consistent with $p_>$. In the above proof, we would encounter a problem where we would want to show that $X$ is $A$-invariant, because we would have to deal with the case where $f(c)$ is Archimedean over $A$, and $X = f(b) \text{ mod } H$.

This is exactly what happens in the following example:

*Example 3.17.* Let $A = \mathbb{Q}$, $B = \mathbb{Q} + \mathbb{Q}\sqrt{2}$, $\varepsilon$ a positive infinitesimal element, and $c_1, c_2 = \sqrt{2} + \varepsilon, \sqrt{2}$. Then $c_1c_2 \downarrow_A^{\text{cut}} B$, and $c_1c_2$ is a $\text{val}_B^3$-separated block. Moreover, $\text{tp}(c_1/B)$ has exactly one global $A$-invariant extension, and $\text{tp}(c_2/B)$ has two, one of which is $p$, the type of a positive element that is infinitesimal with respect to $M$. Then $q = p \cup \text{tp}(c_1c_2/B)$ is complete and consistent in $S(M)$, but it is not $A$-invariant, because, $q(x, y) \models \sqrt{2} \le x \le \sqrt{2} + \varepsilon$.

**Proposition 3.18.** Suppose $c_i$ is ramified over $A$ for every $i$. Choose an arbitrary index $i$. Let $H' = H(c/A)$, $G' = G(c/A)$. If $H \neq H'$ or $G \neq G'$, then $\text{tp}(c_i/B)$ has exactly one global $A$-invariant extension, else it has exactly two. Moreover, if $p$ is such an extension, then $p \cup \text{tp}(c/B)$ is a complete $A$-invariant type in $S(M)$.
*Proof.* We can assume $b_j \in A$ for every $j$. Suppose $c_i > b_i$. As $c \searrow_A^{\text{cut}} B$, recall that $H' \le H \le G \le G'$. Let $p_<, p_>$ be the global 1-types corresponding to the respective cuts $\text{ct}_<(\text{ct}(c_i/A)/M) = \text{ct}_>(b_i \text{ mod } H'/M)$, $\text{ct}_>(\text{ct}(c_i/A)/M) = \text{ct}_>(b_i \text{ mod } \underline{G'}_M/M)$. One can see that these two types are exactly all the global A-invariant extensions of $\text{tp}(c_i/A)$. Moreover $p_<(\text{resp. } p_>)$ is consistent with $\text{tp}(c_i/B)$ if and only if $H = H'$ (resp. $G = G'$).

Now, let $p$ be a global A-invariant extension of $\text{tp}(c_i/B)$, and $q = p \cup \text{tp}(c/B)$. Then $q$ is complete by ???. Let us show that $q$ is A-invariant.

Let $f \in \text{LC}^n(\mathbb{Q})$ be non-zero. One just has to show that $f(b) \text{ mod } H'(M)$ and $f(b) \text{ mod } G'(M)$ are A-invariant. However, $H'(M)$ and $G'(M)$ are both A-invariant, and the $b_i$ were now chosen in $A$, so we can conclude. $\square$

Note that $p_<(\text{and thus } p_\cup \text{tp}(c/B))$ is inner, while $p_>(\text{and thus } p_\cup \text{tp}(c/B))$ is outer.

*Remark 3.19.* The groups $H'$ and $G'$ have the same points in $A$. By quantifier elimination, it is not hard to show that $G'$ does not admit an A-(v/type)-definable proper convex subgroup that strictly contains $H'$. Recall that $H' \le H \le G \le G'$ by cut-independence. Then, if $G \ne G'$ then $G$ cannot be A-type-definable. Similarly, if $H \ne H'$, then $H$ is not A-v-definable.

**Corollary 3.20.** If one of the $c_i$ is Archimedean over $A$, or $H$ is not $A$-v-definable, or $G$ is not $A$-type-definable, then $\text{tp}(c/B)$ has exactly one global $A$-invariant extension. Else, it has exactly two. More precisely, if $G$ is $A$-type-definable, then $\text{tp}(c/B)$ admits a global $A$-invariant extension that is outer; while if $H$ is $A$-v-definable and none of the $c_i$ is Archimedean over $A$, then $\text{tp}(c/B)$ has a global $A$-invariant extension that is inner.

**3.4. Gluing everything together.** We now have enough tools to prove ???. We can actually prove the following more precise statement:

**Proposition 3.21.** Suppose $c = (c_{ik})_{ik}$ is a finite $\text{val}_B^2$-block of $B$-ramified points from $M$ that is $\text{val}_B^3$-separated, such that its $\text{val}_B^3$-blocks are the $(c_i)_{i \in E} = ((c_{ik})_k)_{i \in E}$. Let $G = G(c/B)$, $H = H(c/B)$. Suppose the indices $i \in E$ are ordered so that $i < l \iff \delta(c_i|B) < \delta(c_l|B)$. Define $I, O \subseteq E$ as follows:

* If $H$ is not $A$-v-definable, then $I = \emptyset, O = E$.

* If $G$ is not $A$-type-definable, then $I = E, O = \emptyset$.

* Else, $I = \emptyset$, and $O$ is the least final segment of $E$ containing the set of indices $o \in E$ so that one of the $(c_{ok})_k$ is Archimedean over $A$.

Either way, let $J = E \setminus (I \cup O)$. Then the following conditions hold:

(1) For all $i \in I$, $\text{tp}(c_i/B)$ has exactly one global $A$-invariant extension $p_i$, and it is inner.

(2) For all $o \in O$, $\text{tp}(c_o/B)$ has at least one global $A$-invariant extension, of which exactly one is outer, denote it $q_o$.

(3) For all $j \in J$, $\text{tp}(c_j/B)$ has exactly two global $A$-invariant extensions: $r_j$, which is inner, and $s_j$, which is outer.

(4) The map $K \mapsto (p_i)_{i \in I}, (r_j)_{j \in K}, (s_j)_{j \in J \setminus K}, (q_o)_{o \in O}$ is a bijection from the set of initial segments of $J$ to the set of strong $\text{val}_B^3$-block extensions of $c$. In particular, $c$ admits exactly $|J| + 1$ strong $\text{val}_B^3$-block extensions.
*Proof.* Note that $I, O, J$ are pairwise-disjoint convex subsets of $E$ which cover $E$. In fact, $I$ is an initial segment of $E$, $O$ is a final segment, and $J$ is in-between. The first two conditions are easy consequences of the above corollary. The third condition also easily follows from the fact that for all $j \in J$, none of the $(c_{jk})_k$ is Archimedean over $A$. There remains to prove the last condition.

Suppose we have $o \in O$ so that $\text{tp}(c_o/B)$ has a global $A$-invariant extension $q'$ which is inner. Let us show that $q'$ cannot be extended to a strong $\text{val}_B^3$-block extension of $c$. Let $(q'_e)_{e \in E}$ be some strong block extension of $c$. By definition of $O$, there exists $o', k$ so that $o' \le o$ and $c_{o'}k$ is Archimedean over $A$. As a result, $q_{o'}$ is the unique global $A$-invariant extension of $\text{tp}(c_{o'}/B)$ by the above corollary, therefore $q'_{o'} = q_{o'}$. As $o' \le o$, by ??, $q'_o$ must be outer, thus $q'_o \ne q'$ and we are done.

As a result, all the strong block extensions of $c$ must extend $(p_i)_{i \in I}$, $(q_o)_{o \in O}$. Then, each such extension $(q'_e)_e$ identifies with a subset of $J$: the set of all $j$ such that $q'_j = r_j$. We can conclude, as ?? tells us that the valid choices are exactly the initial segments of $J$. $\square$

**Corollary 3.22.** If $c$ is a finite $\text{val}_B^3$-separated family from $M$, and $c \downarrow_A^{\text{cut}} B$, then $c$ admits a strong $\text{val}_B^3$-block extension.

*Proof.* By ??, it is enough to show that each $\text{val}_B^2$-block of $c$ has a $\text{val}_B^3$-block extension. The Archimedean $\text{val}_B^2$-blocks of $c$ coincide with its Archimedean $\text{val}_B^3$-blocks, by definition of $\text{val}_B^3$. We find a $\text{val}_B^3$-block extension of those blocks by the results of subsection ??, and we find a block extension of the $B$-ramified $\text{val}_B^2$-blocks by the proposition. $\square$

We just proved the property ??, but we are actually very close to get a full classification of the space of global $A$-invariant extensions of $\text{tp}(c/B)$. We will achieve this in the remainder of this section, but these results will not be necessary for our main results about forking.

**Lemma 3.23.** Suppose $c = (c_{ij})_{ij}$ is a finite $\text{val}_B^2$-block of $B$-ramified points from $M$ that is $\text{val}_B^3$-separated, whose $\text{val}_B^3$-blocks are the $(c_i)_i = ((c_{ij})_j)_i$. Let $(p_i)_i$ be a strong $\text{val}_B^3$-block extension of $c$. Then $q = \bigcup_i p_i \cup \text{tp}(c/B)$ is a complete type in $S(M)$.

*Proof.* Suppose the indices $i$ are ordered so that $i < k \iff \delta(c_i/B) < \delta(c_k/B)$. Let $b_{ij} \in \text{ram}(c_{ij}/B)$. We know from ?? that $q$ is consistent. Let $\alpha, \beta \models q$. By the results of subsection ??, we know that each $\alpha_i, \beta_i$ is a $\text{val}_M^3$-block of $M$-ramified points, and that $\Delta(\alpha_{ij} - b_{ij}) = \delta(\alpha_i/M), \Delta(\beta_{ij} - b_{ij}) = \delta(\beta_i/M)$ for all $ij$. Moreover, for $i < k$, and arbitrary indices $jl$, we have $\text{tp}(c/B) \models |x_{kj} - b_{kj}| > n|x_{il} - b_{kl}|$ for all $n > 0$, therefore $\delta(\alpha_i/M) < \delta(\alpha_k/M)$ and $\delta(\beta_i/M) < \delta(\beta_k/M)$. We can apply ?? to get $\alpha \equiv_M \beta$, which concludes the lemma. $\square$

**Remark 3.24.** In the setting of the above lemma, by ??, the Stone space of all the global $A$-invariant extensions of $\text{tp}(c/B)$ is finite Hausdorff, hence it is discrete.

**Remark 3.25.** Let $X$ be a topological space. Suppose $(O_i)_i$ is an open partition of $X$. Then the open subsets of $X$ are exactly the subsets that can be written $\bigcup_i O'_i$, with $O'_i$ an open subset of $O_i$. As a result, $X = \coprod_i O_i$.
• Let $\beta \in N$ be such that $\text{tp}(\alpha\beta/M) \in \pi(F_1)$. Let $f_1 \in \text{LC}^{|\alpha|}(\mathbb{Q})$, $f_2 \in \text{LC}^{|\beta|}(\mathbb{Q})$ be such that $f_1 + f_2 \neq 0$. Just as in ??, in order to show that the map $g: \pi(F_1) \to F_2$ given by the statement is well-defined, it suffices to show that $f_1(\alpha) + f_2(\beta) - (f_1 + f_2)(d) \notin \mathcal{G}_M$. If $f_2 \neq 0$, then $(f_1 + f_2)(d)$ is Archimedean over $A$, thus we can conclude by ?? just as in ??'. As all the $p_o$ are outer, and $\alpha$ is $\text{val}_B^3$-separated (as $(c_{oj})_{oj} \equiv_B \alpha$ is), we have $f_1(\alpha) - (f_1 + 0)(d) \notin \mathcal{G}_M$, proving that $g$ is well-defined.

• Let us show that $g$ is injective. Let $p_1, p_2 \in \pi(F_1)$ be distinct, and $f_1, f_2 \in \text{LC}(\mathbb{Q})$ be as in the above paragraph. We use the same case disjunction ($f_2 \neq 0$ or $f_2 = 0$) as in the above paragraph to show that the pushforward of $p_i$ by $f_1 + f_2$ is (as a cut over $M$) a translate by $(f_1 + f_2)(d)$ of one of the two connected components of $G \setminus \mathcal{G}_M$. Then, it follows from the same reasoning as in ?? that $g$ is injective.

• Let us now prove surjectivity. Let $\beta'$ be a tuple from $N'$ such that $r' = \text{tp}((\alpha_{oj} - b_{oj} \mod H)_{oj}\beta'/\{0\}) \in F_2$. Let $\beta$ be a preimage of $\beta'$ by the projection $N \to N'$. Let $d_1, d_2 \in N$ be such that $\Delta(d_1) > \Delta(\beta_i)$, $\Delta(d_k) > \Delta(\alpha_{oj}-b_{oj})$, and $d_2 \notin \mathcal{G}_M$ for all $i, j, k, o$. Let $\sigma \in \text{Aut}\left(N/H + \sum_{oj} \mathbb{Q} \cdot (\alpha_{oj} - b_{oj})\right)$ be such that $\sigma(|d_1|) = |d_2|$, and let $r = \text{tp}(\alpha(c'_k + \sigma(\beta_k))_k/M)$. Then $r$ is $A$-invariant by ??, and, just as in ??, $r$ is a preimage of $r'$ by $g$.

• We can conclude that $g$ is a homeomorphism $\pi(F_1) \to F_2$.

It remains to show that $F_1 \to \pi(F_1)$ is homeo. By ??, recall that $p = \text{tp}((c_i)_{i \notin O}/B) \cup \bigcup_{i \notin O} p_i$ is a complete global type. Let $q' \in \pi(F_1)$. Then, if $\gamma_1 \models p, \gamma_2 \models q'$, one can note that $G(\gamma_2/M) = G$, while $G(\gamma_1/M) = \overline{H(c/B)}^M < G$ (this is the difference between being inner and outer). As a result, the $\text{val}_M^1$ (and thus $\text{val}_M^2$)-values of the elements of $\gamma_1$ (there is only one such value, but we do not care) are different from the values of the elements of $\gamma_2$. By ??, $p$ and $q$ are weakly orthogonal, which shows that $F_1 \to \pi(F_1)$ is homeo. $\square$

When put together with ?? and ??, we get that the space of global $A$-invariant extensions of the type of any finite, $\text{val}_B^3$-separated $\text{val}_B^1$-block $(c_i)_i, c'_i$ that is cut-independent is the finite coproduct of the topological spaces described in the lemma. The continuous map we consider for ?? is the projection $\pi: ((x_i)_i, x') \mapsto (x_i)_i$. Note that, by ??, any global $A$-invariant extension of the type of $(c_i)_i$ over $B$ extends to one of the type of $((c_i)_i, c'_i)$ (as one can always extend the type of some $\mathbb{Q}$-free family to an arbitrary type of a larger $\mathbb{Q}$-free family), so $\pi$ is onto the finite space described in ??'. Now, given a finite, $\text{val}_B^3$-separated tuple $c$ that is cut-independent, we just described $(S_i)_i$, the Stone spaces of the global $A$-invariant extensions of the type of each $\text{val}_B^1$-block of $c$. By ?? and ??, the Stone space of every global $A$-invariant extension of the type of the full tuple $c$ is exactly the finite direct product of the $S_i$. Moreover, as definable bijections are homeomorphisms in $S(M)$, this homeomorphism type is kept when replacing $c$ by an $A$-interdefinable tuple. Note that we will show in the next section that every cut-independent tuple is $A$-interdefinable with a $\text{val}_B^3$-separated, cut-independent family. As a result, this will yield a complete description of the Stone space of global $A$-invariant extensions of any non-forking type.
4. NORMAL FORMS

We built in the last two sections a nice framework to get a fine understanding of a particular class of tuples, the val<sub>B</sub><sup>3</sup>-separated families. What remains for us to do in order to prove ?? is to prove ??, which allows us to always reduce to the case where we deal with such a nice family.

**Notation.** Let $M \models$DOAG, and $A \leq B$ be $\mathbb{Q}$-vector-subspaces of $M$, so that $M$ is $|B|^+$-saturated, and strongly-$|B|^+$-homogeneous. Let $c = c_1...c_n$ be a finite tuple from $M$ so that $c \searrow_A^{\text{cut}} B$. Without loss (see ??), we also assume that $c$ is a lift of a $\mathbb{Q}$-free family from $M \slash B$.

We have to show that $c$ is $A$-interdefinable with a val<sub>B</sub><sup>3</sup>-separated family $d$, which will also be cut-independent by ??'. In fact, $d$ will also be val<sub>A</sub><sup>3</sup>-separated. This could be useful, as we saw in ?? that we have to keep track of which points from $d$ are Archimedean or ramified over $A$ in order to have a good understanding of the global $A$-invariant extensions of its type. By ??', $d$ will be the image of $c$ by a composition of $A$-translations, and a map from GL<sub>n</sub>(Q). In order to build $d$, we will start with a particular base of dcl(<i>Ac</i>) (remember that the definable closure of a set is the Q-vector subspace generated by said set in DOAG), then we will go through a process of several steps, where we apply $A$-translations and maps from GL<sub>n</sub>(Q) to this base. At the end of each step, our current tuple will satisfy an additional property from a list of conditions, the conjunction of which is a sufficient condition to be a val<sub>B</sub><sup>3</sup> and val<sub>A</sub><sup>3</sup>-separated family.

*Remark 4.1.* Recall that, if $d \in M$ is a $B$-Archimedean singleton, and $d \searrow_A^{\text{cut}} B$, then $d$ must be Archimedean over $A$ (see ??). In particular, $G(d/B) = G(d/A)$.

The following definition allows us to enumerate our families in a way that helps
us to simultaneously consider their val<sub>B</sub><sup>3</sup>-blocks and their val<sub>A</sub><sup>3</sup>-blocks. The ele-
ments of our families will be indexed by the subgroups G(-/B), and the eventual
Archimedean classes δ(-/B) that they bring. We will use the single quote ' to refer
to elements that are Archimedean over A and ramified over B. We will use the
tilde ~ to refer to elements that are ramified over A (and thus over B if the family
is cut-independent from B over A). By elimination, the other elements are those
that are Archimedean over B and A.

**Definition 4.2.** We say that the family $(d_{Gi})_{Gi}(d'_{G'\delta'i'})_{G'\delta'i'}(\tilde{d}_{G\delta i})_{G\delta i}$ is *under normal enumeration* (with respect to $A$ and $B$) if the following conditions hold:

* All the $d_{Gi}$ are Archimedean over $B$.

* All the $d'_{G'\delta'i'}$ are Archimedean over $A$ and ramified over $B$.

* All the $\tilde{d}_{G\delta i}$ are ramified over A.

• For all $G$, $i$, we have $G(d_{Gi}/B) = G$.

• For all $G'$, $\delta'$, $i'$, we have $G(d'_{G'\delta'i'}/B) = G'$ and $\delta(d'_{G'\delta'i'}/B) = \delta'$.

• For all $\tilde{G}$, $\tilde{\delta}$, $\tilde{i}$, we have $G(\tilde{d}_{\tilde{G}\tilde{\delta}i}/B) = \tilde{G}$, and $\delta(\tilde{d}_{\tilde{G}\tilde{\delta}i}/A) = \tilde{\delta}$.

Given such a normal enumeration, the ramified val<sub>B</sub><sup>3</sup>-blocks of the family are the ((d'<sub>G'<sub>δ<i>'</i></sub><sub>i'</sub>)<sub>i'</sub>(d<sub>G'<sub>δ<i>'</i></sub><sub>i'</sub>)<sub>i</sub>)<sub>G'<sub>δ'</sub></sub>, its Archimedean val<sub>B</sub><sup>3</sup>-blocks are the ((d<sub>Gi</sub>)<sub>i</sub>)<sub>G</sub>, its ramified val<sub>A</sub><sup>3</sup>-blocks are the ((d<sub>G<sub>δ<i>'</i></sub><sub>i'</sub></sub>)<sub>i</sub>)<sub>G<sub>δ</sub></sub>, and its Archimedean val<sub>A</sub><sup>3</sup>-blocks are the ((d<sub>Gi</sub>)<sub>i</sub>(d'<sub>G'<sub>δ<i>'</i></sub><sub>i'</sub>)<sub>δ<i>'</i></sub>)<sub>G</sub>. One can note that the presence of the elements of d' is the reason why none of the valuations val<sub>A</sub><sup>3</sup>, val<sub>B</sub><sup>3</sup> refines the other.
Let us now give the list of properties that we want to satisfy:

**Definition 4.3.** Let $dd'\tilde{d}$ be a family under normal enumeration with respect to $A$ and $B$. We will say that $dd'\tilde{d}$ is under normal form with respect to $A$ and $B$ if it is a lift of a $\mathbb{Q}$-free family from $M/B$, and the following conditions hold:

(P1) For all $\tilde{G}$ and $\tilde{\delta}$, $(\tilde{d}_{\tilde{G}\tilde{\delta}\tilde{i}})_i$ is a lift of a $\mathbb{Q}$-free family from $M_{\le \tilde{\delta}}/M_{<\tilde{\delta}}$.

(P2) Any non-trivial linear combination of $dd'$ is Archimedean over $A$.

(P3) For all $G$, any non-trivial linear combination $e$ of the $(d_{Gi})_i(d'_{G\delta'i'})_{\delta'i'}$ satisfies $G(e/A) = G$.

(P4) For all $G$, any non-trivial linear combination $c$ of the $(d_{Gi})_i$ is Archimedean over $B$ (which implies $G(c/B) = G$ if the family is cut-independent, by ?? and ??).

(P5) For all $G'\delta'$, any non-trivial linear combination $e'$ of the $(d'_{G'\delta'i'})_{i'}(\tilde{d}_{G'\delta'\tilde{i}})_i$ is ramified over $B$, and satisfies $\delta(e'/B) = \delta'$.

**Remark 4.4.** ?? implies that each ramified $\text{val}_A^3$-block is $\text{val}_A^3$-separated. The conjunction of ?? and ?? is equivalent to each Archimedean $\text{val}_A^3$-block being $\text{val}_A^3$-separated. ?? (resp. ??) is equivalent to each Archimedean (resp. ramified) $\text{val}_B^3$-block being $\text{val}_B^3$-separated. As a result, if we manage to prove that $c$ is $A$-interdefinable with a family under normal form, then this family will be simultaneously $\text{val}_A^3$-separated and $\text{val}_B^3$-separated, and ?? will follow.

**Remark 4.5.** From the definition of the normal form, if $dd'\tilde{d}$ is under normal form with respect to $A$ and $B$, then one can note that any subfamily of $dd'\tilde{d}$ is also under normal form with respect to $A$ and $B$.

**Definition 4.6.** Define $C$ the $\mathbb{Q}$-vector space generated by $Ac$. We will define a base (resp. free family) of $C$ over $A$ as a lift of a base (resp. free family) of $C/A$.

**Remark 4.7.** Remember that $\dim(C/A)$ is finite. It is easy to see that any concatenation of lifts of bases of $C_{\le \tilde{\delta}}/C_{<\tilde{\delta}}$, for each $\tilde{\delta} \in \Delta(C) \setminus \Delta(A)$, is a free family over $A$, hence a finite family. In particular, $\Delta(C) \setminus \Delta(A)$ is finite.

**Lemma 4.8.** For each $\tilde{\delta} \in \Delta(C) \setminus \Delta(A)$, let $\tilde{d}_{\tilde{\delta}}$ be a lift of a base of $C_{\le \tilde{\delta}}/C_{<\tilde{\delta}}$, and let $\tilde{d}$ be the concatenation of all the $\tilde{d}_{\tilde{\delta}}$. Then, for all $e \in C \setminus \text{dcl}(A\tilde{d})$, there exists $\tilde{e} \in \text{dcl}(A\tilde{d})$ so that $e + \tilde{e}$ is Archimedean over $A$.

Moreover, if $e$ is ramified over $A$, then we can choose $\tilde{e}$ so that there exists $\tilde{\delta} \in \Delta(C) \setminus \Delta(A)$ for which $\Delta(e + \tilde{e}) < \tilde{\delta} \le \Delta(e)$.

**Proof.** Let us build $\tilde{e}$ by induction. Let $\tilde{e}_0 = 0 \in \text{dcl}(A\tilde{d})$. If $e+\tilde{e}_n$ is not Archimedean over $A$, then let $a \in \text{ram}(e+\tilde{e}_n/A)$, and $\tilde{\delta}_n = \delta(e+\tilde{e}_n/A)$. By definition of $\tilde{d}$, there exists $v$ a linear combination of $\tilde{d}_{\tilde{\delta}_n}$ for which $\Delta(e+\tilde{e}_n-a-v) < \tilde{\delta}_n$, so let us set $\tilde{e}_{n+1} = \tilde{e}_n - a - v$.

By finiteness of $\Delta(C) \setminus \Delta(A)$, the sequence of the $\tilde{\delta}_n$, and hence that of the $\tilde{e}_n$, must eventually stop. The ultimate value $\tilde{e}$ of the $\tilde{e}_n$ witnesses the first part of the lemma. If $e$ is ramified over $A$, then $\tilde{e} \ne 0 = \tilde{e}_0$, so $\tilde{\delta}_0$ exists and witnesses the second part of the lemma. $\square$

**Lemma 4.9.** For each $\tilde{\delta} \in \Delta(C) \setminus \Delta(A)$, let $\tilde{d}_{\tilde{\delta}}$ be a lift of a base of $C_{\le \tilde{\delta}}/C_{<\tilde{\delta}}$, and let $\tilde{d}$ be the concatenation of all the $\tilde{d}_{\tilde{\delta}}$. Let $(v_n)_n$ be a sequence so that $(\tilde{d}, v_n)$ is
a base of C over A for each n. Suppose that, for all n, there exists $v'_1$ a term from $v_n$, $v'_2$ a term from $v_{n+1}$, and $\delta \in \Delta(C) \setminus \Delta(A)$, so that $\Delta(v'_2) < \delta \le \Delta(v'_1)$, and $v_{n+1}$ is obtained from $v_n$ by replacing $v'_1$ by $v'_2$.

Then the sequence $(v_n)_n$ must eventually stop.

**Proof.** Let $N = |\Delta(C) \setminus \Delta(A)| < \omega$, $(\delta_n)_{n<N}$ the strictly increasing enumeration of $\Delta(C) \setminus \Delta(A)$, and $\mathcal{I}$ be the set of intervals:

$$ \{]-\infty, \delta_0[, \delta_0, \delta_1[, \delta_1, \delta_2[, \dots, [\delta_{N-2}, \delta_{N-1}[ , [\delta_{N-1}, +\infty[ $$

Define on $\mathcal{I}$ the only linear order that makes the canonical projection $f: \Delta(C) \to \mathcal{I}$ an increasing map. With respect to that order, let $g$ be the unique order-isomorphism $\mathcal{I} \to \{0, 1, \dots, N\}$. For each $i$ between 0 and $N$, let $F(n, i)$ be the number of terms $v$ from $v_n$ so that $g(f(\Delta(v))) = i$. Finally, let $G$ be the following function:

$$ n \mapsto F(n, N) \times \omega^N + F(n, N-1) \times \omega^{N-1} + \dots + F(n, 1) \times \omega + F(n, 0) $$

Then $g$ has its values in the well-ordered set $\omega^{N+1}$. By the hypothesis of the proposition, for each $n$, there exists $m < m'$ so that $F(n, m') = F(n+1, m') + 1$, $F(n, m) + 1 = F(n+1, m)$, and $F(n, i) = F(n+1, i)$ for every $i \ne m, m'$. As a result, $g$ is strictly decreasing, so the well-ordering forces the sequence $(v_n)_n$ to stop eventually. $\square$

**Lemma 4.10.** For each $\tilde{\delta} \in \Delta(C) \setminus \Delta(A)$, let $u_{\tilde{\delta}}$ be a lift of a base of $C_{\le \tilde{\delta}} / C_{< \tilde{\delta}}$, and let $u$ be the concatenation of all the $u_{\tilde{\delta}}$. Then there exists $dd'\tilde{d}$ a base of $C$ over $A$, under normal enumeration with respect to $A$ and $B$, for which ?? holds, and $\tilde{d} = u$ (so ?? holds as well).

**Proof.** Let $w_0$ be some base of $C$ over $A$ containing $u$. Note that $w_n = (u, v_n)$ does not witness the proposition if and only if there exists a non-trivial linear combination $v'$ of $v_n$ that is ramified over $A$. In that case, let $v$ be some term of $v_n$ that appears in $v'$, chosen so that $\Delta(v)$ is maximal. By valuation inequality, we must have $\Delta(v) \ge \Delta(v')$. Apply ?? to find $u' \in \text{dcl}(Au)$ and $\delta \in \Delta(C) \setminus \Delta(A)$ for which $\Delta(v' + u') < \delta \le \Delta(v') \le \Delta(v)$. Replace $v$ by $v' + u'$ in $w_n$, let $w_{n+1}$ be the new family obtained. Now, $w_{n+1}$ is also a base of $C$ over $A$ containing $u$.

The sequence $(v_n)_n$ witnesses the hypothesis of ??, so it must eventually stop, and its last element witnesses the proposition. $\square$

**Remark 4.11.** If $w = (u, v)$ witnesses the above proposition, and $M$ is an invertible matrix of size $|v|$ with coefficients in $\mathbb{Q}$, then $(u, (Mv))$ also witnesses the proposition.

**Lemma 4.12.** Let $dd'\tilde{d} = (v, \tilde{d})$ be a free family over $A$ that witnesses ?? and ??, and $N = |v|$. Then there exists $M \in \text{GL}_N(\mathbb{Q})$ so that $((Mv), \tilde{d})$ witnesses ?? (and thus ??, ?? by ??).

**Proof.** This is by induction over $N$. ?? trivially holds if $N = 0, 1$.

For every $G$, let $v_G$ be the $\text{val}_A^1$-block of $v$ of value $G$. Let $G$ be maximal so that $v_G$ is not $\text{val}_A^3$-separated. As long as $v_G$ is not $\text{val}_A^3$-separated, there must exist $e$ a non-trivial linear combination of $v_G$ so that $G(e/A) < G$, then replace some term of $v_G$ that appears in $e$ by $e$ (making that replacement corresponds to replacing $v$ by $Mv$ for some $M \in \text{GL}_N(\mathbb{Q})$). As we have $G(e/A) < G$, these replacements always
decrease the size of $v_G$, and leave $v_{>G}$ untouched as the value of $e$ is smaller, so the
val$_A^1$-blocks of value larger than $G$ remain separated, thus we keep the maximality
of $G$. As a result, these replacements eventually stop, and they stop before we
removed all the terms from $v_G$, because if there is only one term left in $v_G$, then
$v_G$ is clearly separated.

By induction hypothesis, we can replace $v_{<G}$ by $Mv_{<G}$ so that $((Mv_{<G}), \tilde{d})$ witnesses ???. It is now clear that $((Mv_{<G}), v_{\ge G}, \tilde{d})$ witnesses ???. The only changes we made to the original family $v$ are operations from $\mathrm{GL}_N(\mathbb{Q})$, so we can conclude. $\square$

<figure><img src="image_19.png" /></figure>

Remark 4.13. Let $w = dd'\tilde{d} = (v, \tilde{d})$ be a free family over $A$ that witnesses ??, ??, ???. Let $v_G$ be the val$_A^1$-block of $v$ of value $G$. If we replace $v_G$ by $Mv_G$ in $w$ for some invertible matrix $M$, then the new family $w'$ will still witness ??, ?? and ??.

**Lemma 4.14.** *There exists a base of $C$ over $A$ that witnesses ??, ??, ??, ??.*

*Proof.* Let $w = dd'\tilde{d}$ be a base of $C$ over $A$ that witnesses ??, ?? and ???. For each $G$, as long as there is a non-trivial linear combination $e$ of the $(d_{Gi})_i$ that is not Archimedean over $B$, then replace some term of $(d_{Gi})_i$ that appears in $e$ by $e$. Each of these replacements removes a term from $(d_{Gi})_i$ and adds a new term to $(d'_{G'\delta'i'})_{\delta'i'}$. This eventually stops as the number of terms of $(d_{Gi})_i$ is finite and strictly decreases at each step.

By the above remark, the new family $w'$ that we obtain after all these replace-
ments still witnesses ??, ??, ???. We perform these replacements for each $G$, and
the new family obtained clearly witnesses ???. $\square$

<figure><img src="image_20.png" /></figure>

Remark 4.15. Let $w = dd'\tilde{d}$ be a free family over $A$ that witnesses ??, ??, ??, ??, and $G'$ some index. Then, if we replace $(d'_{G'\delta'i'})_{\delta'i'}$ by $M(d'_{G'\delta'i'})_{\delta'i'}$ in $w$ for some invertible matrix $M$, it is clear that the new family still witnesses ??, ??, ??, ??.

**Lemma 4.16.** Let $w = dd'\tilde{d}$ be a base of $C$ over $A$ that witnesses ??, ??, ??, ???. Let $d'_{G'\delta'k'}$ be a term from $d'$, and $\tilde{e}$ be a linear combination of the $(\tilde{d}_{G'\delta'k'}^{\lambda})_{\tilde{\delta}'k'}$. If we replace $d'_{G'\delta'k'}$ by $e = d'_{G'\delta'k'} + \tilde{e}$ in $w$, then the new family obtained is still a base of $C$ over $A$ that witnesses ??, ??, ??, ??.

*Proof.* First of all, we have $\text{val}_A^2(d'_{G'\delta'k'}) > \text{val}_A^2(\tilde{e})$, hence $e$ is Archimedean over $A$, and $G(e/A) = G'$. As $\text{val}_B^2(e) \le \max(\text{val}_B^2(d'_{G'\delta'k'}), \text{val}_B^2(\tilde{e}))$, $e$ cannot be Archimedean over $B$. This implies that the $A$-ramified and $B$-Archimedean blocks of our new family are left untouched, hence we keep the properties ??, ???. Let $u'$ be the new block of $A$-Archimedean and $B$-ramified points in the normal enumeration. Then $e$ lies in the block $u'_{G'}$. In order to show that ?? and ?? hold in the new family $du'\tilde{d}$, we just have to show that the $\text{val}_A^3$-block $d_{G'}u'_{G'}$ is $\text{val}_A^3$-separated. Indeed, it would follow that the whole $du'$ is $\text{val}_A^3$-separated, which is clearly equivalent to ?? and ?? being true in $du'\tilde{d}$.

Any non-trivial linear combination of $d_{G'}u'_{G'}$ can be written $u + \lambda \cdot \tilde{e}$, with $u$ a non-trivial linear combination of $d_{G'}d'_{G'}$, and $\lambda \in \mathbb{Q}$. As ?? and ?? hold in $dd'\tilde{d}$, we have $G(u/A) = G'$, and $u$ is Archimedean over $A$. Then, we have $\text{val}_A^3(u) > \text{val}_A^3(\tilde{e})$, hence $\text{val}_A^3(u + \lambda \cdot \tilde{e}) = \text{val}_A^3(u)$. This value equals that of the corresponding block, $\text{val}_A^3(d_{G'}d'_{G'})$, as ?? and ?? hold in $dd'\tilde{d}$, thus it equals $\text{val}_A^3(d_{G'}u'_{G'})$, and we can conclude. $\square$

<figure><img src="image_21.png" /></figure>
**Lemma 4.17.** Let $w = dd'\tilde{d}$ be a base of $C$ over $A$ that witnesses ??, ??, ??, ??, and not ?? . Then there exists $G'$, $\delta'$, and two $\mathbb{Q}$-linear maps $f, g \in \text{LC}(\mathbb{Q})$, so that $v = f((d'_{G'\delta'_{i'}})_{i'}) + g((\tilde{d}_{G'\delta'_{i'}})_{i'})$ is ramified over $B$, and $\delta(v/B) < \delta'$.

Moreover, if $d'_{G'\delta'_{k'}}$ appears in $v$, then replacing it by $v$ in $w$ gives a new family $w'$ which is still a base of $C$ over $A$ that witnesses ??, ??, ??, ??.

*Proof.* By definition of ??, there exists $G'$, $\delta'$, and $\mathbb{Q}$-linear maps $f, g \in \text{LC}(\mathbb{Q})$ so that, either $v = f((d'_{G'\delta'_{i'}})_{i'}) + g((\tilde{d}_{G'\delta'_{i'}})_{i'})$ is not ramified over $B$, or $\delta(v/B) \neq \delta'$.

Note that $f$ must be non-zero, otherwise the above hypothesis fails by ?? . By ??, ??, ??, $dd'\tilde{d}$ is $\text{val}_A^3$-separated. As $f$ is non-zero, $\text{val}_A^3(v) = \text{val}_A^3(d'_{G'})$, thus $v$ is Archimedean over $A$, and $G(v/A) = G' = G(v/B)$. For all $i', \tilde{i}'$, we have $d'_{G'\delta'_{i'}}$, $\tilde{d}_{G'\delta'_{i'}} \in B + G'$ (by the definition of being ramified over $B$), so $v \in B + G' = B + G(v/B)$, therefore $v$ must be ramified over $B$, thus by hypothesis we have $\delta(v/B) \neq \delta'$. By valuation inequality (for $\text{val}_B^3$), we must have $\delta(v/B) < \delta'$, and we get the first part of the lemma.

Suppose $d'_{G'\delta'_{k'}}$ appears in $v$. Then replacing $d'_{G'\delta'_{k'}}$ by $v' = f((d'_{G'\delta'_{i'}})_{i'})$ is just a replacement of $(d'_{G'\gamma'_{i'}})_{\gamma'_{i'}}$ by $M(d'_{G'\gamma'_{i'}})_{\gamma'_{i'}}$ for some invertible matrix $M$. By ??, the new family is still a base of $C$ over $A$ that witnesses ??, ??, ??, ?? . By ?? and ??, $v'$ is Archimedean over $A$, and $G(v'/A) = G'$. Moreover, as all the $(d'_{G'\delta'_{i'}})_{i'}$ are in $B+G'$, $v'$ must be ramified over $B$. Then, $w'$ is obtained by replacing $v'$ by $v$ in (a normal enumeration of) $w''$. This is exactly the operation described in ??, so we can conclude. $\square$

**Theorem 4.18.** There exists $w = dd'\tilde{d}$ a base of $C$ over $A$ that is under normal form with respect to $A$ and $B$.

*Proof.* Let $w_0 = dd'^0\tilde{d}$ be a base of $C$ over $A$ that witnesses ??, ??, ??, ?? .

Suppose $w_n = dd'^n\tilde{d}$ does not witness ?? . Then the hypothesis of ?? hold in $w_n$, so we do the replacement from this lemma. Let $w_{n+1}$ be the new family obtained.

To prove the theorem, we have to prove that the sequence $(w_n)_n$ must stop.

To do this, we will once again build a strictly decreasing map with values in $\omega^\omega$. Let $N = |\Delta(C+B) \setminus \Delta(B)|$, and $(\delta_i)_{i<N}$ be the strictly increasing enumeration of $\Delta(CB) \setminus \Delta(B)$. Define $F(n,i)$ as the number of terms $c$ from $d'^n$ for which $\delta(c/B) = \delta_i$. We define $g$ as the map:

$$n \mapsto F(n, N) \times \omega^N + F(n, N-1) \times \omega^{N-1} + \dots + F(n, 1) \times \omega + F(n, 0)$$

For each $n$, the family $w_{n+1}$ is obtained from $w_n$ by replacing some term $v$ from $d'^n$ by a vector $v'$ from $d'^{n+1}$ for which $\delta(v'/B) < \delta(v/B)$, so $g$ is strictly decreasing, and we can conclude. $\square$

With that, we proved ??, as explained in the end of ?? .

# 5. REGULAR ORDERED ABELIAN GROUPS

## 5.1. Quantifier-free types in the Presburger language.

**Definition 5.1.** The Presburger language is defined as :

$$\mathcal{L}_P = \{+, -, <, 0, 1, (\mathfrak{d}_{l^n})_{l \text{ prime}, n>0}\}$$

Given an ordered Abelian group $G$, we see $G$ as an $\mathcal{L}_P$-structure by setting $\mathfrak{d}_{l^n}(x)$ as the predicate $\exists y, l^n y = x$, and setting $1 = \min(]0, +\infty[)$ if $G$ is discrete, else we interpret $1$ as the only $\emptyset$-definable choice, $1 = 0$ (this is the standard interpretation
of the language). A *special* subgroup of $G$ will be a pure subgroup $A \le G$ containing 1.

Define $S_{\mathrm{qf}}^{m}(A)$ the Stone space of quantifier-free types over $A$ in $m$ variables. Given an $m$-tuple $c$, we write $tp_{l}(c/A)$ as the partial type generated by every quantifier-free formula satisfied by $c$, with parameters in $A$, which only involve the predicates $(\partial_{l^{n}})_{n>0}$ (equality does not appear in these formulas either). We write $S_{l}^{m}(A)$ for the Stone space of all such partial types, and we call its elements the $l$-types. We define similarly $tp_{<}(c/A)$, $S_{<}^{m}(A)$ for the quantifier-free formulas which only involve the predicates $=$, $<$.

For ease of notation, define the set of indices $J$ as the reunion of the set of primes with $\{\langle\rangle\}$. We have natural restriction maps $S_{\mathrm{qf}}^{m}(A) \to S_{j}^{m}(A)$ for every $j$, and they are all surjective. We will write $(\pi_{j})_{j}$ for those maps.

**Notation.** Let $M$ be an ordered Abelian group, and $A \le M$ a special subgroup of $M$, so that $M$ is $|A|^{+}$-saturated and strongly-$|A|^{+}$-homogeneous.

**Lemma 5.2** (Standard variant of the chinese remainder theorem). Let $L$ be a finite set of primes, and $N > 0$. Then for all $(a_l)_l \in M^L$, there exists $b \in M$ so that $M = \mathfrak{d}_{l^N} (a_l - b)$ for all $l \in L$.

**Corollary 5.3.** The natural map $S_{\mathrm{qf}}^{m}(A) \to \prod_{l \text{ prime}} S_{l}^{m}(A)$ is surjective.

*Proof.* For each $l$, let $a_l = (a_{i,l})_i \in M^m$. By the above lemma and compactness, for each $i$, there must exist $b_i \in M$ so that $\mathfrak{d}_{l^N}(b_i - a_{i,l})$ for all $l$, $i$. In particular, $tp_l(b/A) = tp_l(a_l/A)$ for all $l$, so we can conclude. $\square$

**Remark 5.4.** The map $\pi : S_{\mathrm{qf}}^{m}(A) \to \prod_{j \in J} S_{j}^{m}(A)$ is obviously injective. Note that $S_{<}^{m}(A)$ is not a factor of the product of $??$, thus $\pi$ may not be surjective.

**5.2. Basic properties of regular groups.** The class of regular ordered Abelian groups (ROAG) has several equivalent definitions. The ones that will matter for our proofs are quantifier elimination in $\mathcal{L}_P$, and some compatibility conditions between the $<$-types and the $l$-types. The definitions involving Archimedean groups and definable convex subgroups will be relevant, because they give a motivation as to why we are interested in ROAG.

The equivalence between these different definitions is folklore. We will state it in this subsection, as well as give a proof of the easy directions.

**Definition 5.5.** Let $G$ be an ordered Abelian group. We will say that $G$ is *regular* if, for every positive integer $n$, every interval of $G$ that contains at least $n$ elements intersects $nG$.

For any ordered Abelian group $G$, let $\mathrm{div}(G)$ be the divisible closure of $G$, which we will see as an ordered group to which the order on $G$ naturally extends.

**Lemma 5.6.** Let $G \models$ ROAG, and $A$ a special subgroup of $G$. Let $F$ be the closed subspace of $S_{\mathrm{qf}}^{m}(A)$ of types of $m$-tuples that are $\mathbb{Q}$-free over $A$ (i.e. lifts in $M^m$ of $\mathbb{Q}$-free families from $\mathrm{div}(M)/\mathrm{div}(A)$). Then the map $F \to \pi_{<}(F) \times \prod_{l \text{ prime}} S_{l}^{m}(A)$ is a homeomorphism.

In particular, $\pi_l(F) = S_l^m(A)$ for every prime $l$.
**Proof.** The map is clearly injective and continuous. Let us prove that it is surjective.

Let $p = (p_j)_j$ be in the direct product. Let $c_<$ be a realization of $p_<$ which is $\mathbb{Q}$-free over $A$. By ??, let $c'$ be some tuple which realizes simultaneously all the $(p_l)$ for $l$ prime ($c'$ might not be $\mathbb{Q}$-free over $A$, but it does not matter). Let us show that there exist $d = (d_i)_{i<m}$ such that $d_i - c'_i \in \bigcap_{N>0} NG$ for all $i$, and $d \models p_<$ (in particular, $d$ will be $\mathbb{Q}$-free over $A$). The existence of $d$ will be established by a disjunction of two cases.

Suppose $G$ is discrete. Let $N > 0$. Then, by regularity of $G$ applied to the interval $[c_i - c'_i, c_i - c'_i + N \cdot 1]$, there must exist an integer $k_i$ between 0 and $N$ such that $c_i - c'_i + k_i \cdot 1 \in NG$. Let us show that $d' = (d'_i)_i = (c_i + k_i \cdot 1)_i \models p_<$. If not, then there must exist some atomic formula with predicate = or < which is satisfied by $c$ and not $d'$, or vice-versa. Without loss, there must exist $a \in A$, and some non-zero $f \in \text{LC}^m(\mathbb{Z})$ such that $f(c) > a \ge f((c_i + k_i \cdot 1)_i)$. Then, $f(c)$ belongs to the interval $[a, a + (f((k_i)_i) + 1) \cdot 1]$, which is included in $A$ as $A$ is special, contradicting $\mathbb{Q}$-freeness of $c$ over $A$. It follows that $d' \models p_<$, and we can conclude by compactness that $d$ exists.

Now, suppose $G$ is dense. Let $D$ be the special subgroup generated by $A$, and all the $(c_i)_i$. For every positive element $a$ of $D$, the interval $]c_i - c'_i, c_i - c'_i + a[$ is infinite, thus contains an element of $NG$ by regularity, for any $N > 0$. By compactness, there exists, for each $i$, an element $d'_i$ such that $d'_i \in ]c_i - c'_i, c_i - c'_i + a[$ for every positive $d \in D$, and $d_i \in \bigcap_{N>0} NG$. Choose $d_i = d'_i + c'_i$. Notice that $d_i - c_i$ is a positive element which is infinitesimal with respect to $D$. It remains to show that $d \models p_<$. Let $f \in \text{LC}^m(\mathbb{Z})$ be non-zero, and $a \in A$, such that $f(c) > a$. As each $d_i - c_i$ is infinitesimal with respect to $D$, this is also the case for $f(d - c) = f(d) - f(c)$, thus $|f(d) - f(c)| < f(c) - a$, thus $f(c) - f(d) < f(c) - f(a)$, and we can conclude that $f(d) > f(a)$, proving that $d \models p_<$.

Either way, we have $d \models p_<$. As $d_i - c'_i \in l^N G$ for every $l$ prime and $N > 0$, we have $d \models p_l$ for every $l$, which concludes the proof of the lemma. $\square$

**Remark 5.7.** Note that, in the above proof, $d_i$ can always be chosen such that $d_i > c_i$ for each $i$ (in the discrete case, replace $k_i$ by $k_i + N > 0$). In particular, if we have $c = c'$ in the construction of the proof, then the tuple $d$ that we build will be distinct from $c$, and re-iterating this operation generates infinitely-many pairwise-distinct tuples having the same quantifier-free type as $d$. It follows that any consistent quantifier-free type over $A$ whose realizations are $\mathbb{Q}$-free over $A$ have infinitely many realizations.

**Definition 5.8.** Let $G$ be an ordered Abelian group, $n < \omega$ and $g \in G$. Define $H_n(g)$ to be the largest convex subgroup of $G$ for which $(g + H_n(g)) \cap nG = \emptyset$ (by convention, if $g \in nG$, then $H_n(g) = \{0\}$). This convex subgroup is definable:

$$H_n(g) = \{0\} \text{ or } \{x \in G | \forall y \in G, (|y| \le n|x| \implies g + y \notin nG)\}$$

for if $x \in G$, $z \in G$ satisfy $\Delta(x) = \Delta(z)$ and $z + g \in nG$, then $x \notin H_n(g)$ is witnessed by choosing $y = z \pm mn|x|$, with $m$ the least natural integer for which $|z - mn|x| \le n|x|$.

In the literature, the set of the $H_n(g)$ for all $g \in G$ is called the *n-spine* of $G$. This is a definable family of definable convex subgroups. One of the most general “complexity class” of ordered Abelian groups that is still considered rather “nice” is the class of ordered Abelian groups with finite spines. This class contains in
particular the dp-finite ordered Abelian groups. For reference, section 2 of [?] gives nice characterizations and a quantifier elimination result for this class.

**Proposition 5.9.** Let $G$ be an ordered Abelian group. Then the following conditions are equivalent:

(1) $G$ does not have any proper non-trivial definable convex subgroup.

(2) For all $g \in G$, for all $n < \omega H_n(g) = \{0\}$.

(3) $G$ is regular.

(4) The theory of $G$ eliminates quantifiers in the Presburger language.

(5) There exists an Archimedean ordered Abelian group elementarily equivalent to $G$.

*Proof.* The directions that we will show to establish the equivalence are $1 \implies 2 \implies 3$, $3 \implies 4 \implies 1$ and $3 \implies 5 \implies 1$.

The implications $1 \implies 2$ and $5 \implies 1$ are trivial, $3 \implies 4$ is due to Weispfenning ([?], Theorems 2.3 and 2.6), and Presburger in the discrete case, and $3 \implies 5$ follows from ([?], Theorem 2.5) and ([?], Theorem 4.7).

Let us show $2 \implies 3$ by contraposition. We have an interval $I$ of $G$ and an integer $n$ so that $|I| \ge n$ and $I \cap nG = \emptyset$. If $|I| < 2n+1$, then $G$ is discrete, and the points from $I+n\cdot 1$ have the same cosets mod $nG$ as those of $I$, so we can assume $|I| \ge 2n+1$ by replacing $I$ by $I \cup (I+n\cdot 1) \cup (I+2n\cdot 1)$. Let $f$ be a strictly increasing map $\{0, \dots, 2n\} \to I$, and $h = \min\{f(i+1) - f(i) | 0 \le i \le 2n-1\}$. As $f$ is strictly increasing, we have $h \ne 0$. As $h$ is minimal, we can sum $n$ many inequalities to get $n \cdot h \le \sum_{i=0}^{n-1} (f(i+1) - f(i)) = f(n) - f(0)$. With the same argument, we have $n \cdot h \le f(2n) - f(n)$, thus $[f(n) - n \cdot h, f(n) + n \cdot h] \subseteq I$, which is disjoint from $nG$. It follows that $h \in H_n(f(n)) \setminus \{0\}$, so 2 fails.

Let us prove $4 \implies 1$ by contraposition. Let $G$ be an ordered Abelian group with a proper non-trivial definable convex subgroup $H$. Let $A$ be a definably-closed subset of $G$ so that $H$ is $A$-definable. Let $A_1 = A_{\ge 0} \cap H$, $A_2 = (A_{>0} \setminus H) \cup \{+\infty\}$. Let

$X_< = \left( \bigcap_{a \in A_1, b \in A_2} \right]a, b[)$, an $A$-type definable set which is non-empty by compactness. For each prime $l$, let $X_l = \bigcap_{N} l^N G$, which corresponds to tp$_l(0/A)$. Let us show that the partial type $q(x) : x \in X_< \cap \bigcap_l X_l$ is consistent with $H$ and $\neg H$. Let $a \in A_1$, $b \in A_2$, $N > 0$. By compactness, it suffices to show that $Y = \{a, b[\cap N G \text{ intersects } H \text{ and } \neg H\}$, which is witnessed by $N \cdot a \in Y \cap H$, and $N \cdot (b-a) \in Y \setminus H$. Now let $h \in H, g \notin H$ be two realizations of $q$. Then $qftp(h/A) = q = qftp(g/A)$, but $\text{tp}(h/A) \ne \text{tp}(g/A)$, and the theory of $G$ does not eliminate quantifiers in $\mathcal{L}_P$. $\square$

<figure><img src="image_22.png" /></figure>

Note that in particular, the domain of the homeomorphism from ?? is in fact a space of *complete types*.

**Corollary 5.10.** The definable closure of a parameter set $A$ coincides with the special subgroup it generates.

*Proof.* The special subgroup generated by $A$ is clearly included in dcl$(A)$. Conversely, if $A$ is a special subgroup, and $c \notin A$, then the 1-tuple $c$ is $\mathbb{Q}$-free over $A$, thus $c \notin \text{acl}(A)$ by ??.
$\square$
There is an other important corollary which allows us to better understand independence:

**Corollary 5.11.** Let $M \models ROAG$, $A \leq B$ special subgroups of $M$ so that $M$ is $|B|^+$ saturated and strongly-$|B|^+$-homogeneous, and $c = c_1...c_n \in M$. Suppose $c_1...c_n$ is $\mathbb{Q}$-free over $B$, and let $F$ be the closed subspace of $S^n(M)$ of tuples that are $\mathbb{Q}$-free over $M$. For each $j \in J$, consider the action of Aut$(M/A)$ over $\pi_j(F)$. Then the following conditions are equivalent:

(1) $c \downarrow_A^{inv} B$ (resp. $c \downarrow_A^{bo} B$).

(2) For each $j \in J$, $\text{tp}_j(c/B)$, which is a partial type over $B$, can be extended to an Aut$(M/A)$-invariant (resp. of bounded orbit) element of $\pi_j(F)$.

*Proof.* First of all, note that the global extensions of $\text{tp}(c/B)$ realized by tuples that are not $\mathbb{Q}$-free over $M$ divide over $A$, hence have an unbounded orbit, so we do have to restrict ourselves to $F$.

For each $j \in J$, $F \to \pi_j(F)$ is an equivariant surjection, thus the stabilizer of a point of $F$ is a subgroup of that of its image, so the cardinal of its orbit is larger than the supremum of the cardinals of the orbits of each of its images, and we get the top-to-bottom direction by contraposition.

Suppose for each $j$, we have $p_j \in \pi_j(F)$ a witness of the second condition: it extends $\text{tp}_j(c/B)$, and its orbit is a singleton (resp. bounded). Note that the orbit of $(p_j)_j$ under Aut$(M/A)$ is contained in the Cartesian product of the orbits of the $(p_j)_j$, therefore it is also a singleton (resp. bounded). Now, by ?? and quantifier elimination, the (equivariant) map $F \to \prod_j \pi_j(F)$ is a bijection, so the consistent global type corresponding to $(p_j)_j$ witnesses the first condition, and we get the bottom-to-top direction. $\square$

Note that, given special subgroups $A \leq B$, and a tuple $c$, $c$ is $A$-interdefinable with a subtuple $d$ which is $\mathbb{Q}$-free over $A$ (choose any of those subtuples, of maximal length), thus there is a natural equivariant homeomorphism between the space of global extensions of $\text{tp}(c/B)$ and that of $\text{tp}(d/B)$. The above corollary completes the picture, and gives us easy conditions on $d$ to check whether $c \downarrow_A^{inv} B$ or $c \downarrow_A^{bo} B$.

*Remark 5.12.* The above corollary raises one interesting question: does the orbit of an element of $\prod_j \pi_j(F)$ coincide with the Cartesian product of the orbits of its components ? In other words, for $p_j \in \pi_j(F)$, is the natural (injective) map:

$$ \text{Aut}(M/A) / \bigcap_j \text{Stab}(p_j) \to \prod_j \text{Aut}(M/A) / \text{Stab}(p_j) $$

a bijection ?

### 5.3. Invariance and boundedness of global partial types.

**Notation.** Let $M \models ROAG$, $A \leq B$ special subgroups of $M$, $\lambda = \max(|B|, 2^{\kappa_0})^+$, and $c = c_1...c_n$ a tuple from $M$ so that $M$ is $\lambda$-saturated and strongly-$\lambda$-homogeneous. Suppose $c$ is $\mathbb{Q}$-free over $B$, and let $F$ be the closed space of global types whose realizations are $\mathbb{Q}$-free over $M$. For each $j \in J$, let $p_j = \text{tp}_j(c/B)$. A set of cardinality $\kappa$ will be called *small* if $M$ is $\kappa$-saturated and strongly-$\kappa$-homogeneous, else it is *large*.
By ??, in order to understand the global extensions of tp(c/B) which are invariant or have a bounded orbit under the action of Aut(M/A), one has to understand, for each $j \in J$ the global extensions in $\pi_j(F)$ of $p_j$ which are invariant or have a bounded orbit.

5.3.1. *Partial types using equality and order.*

**Notation.** In this subsection, let $\overline{M} := \text{DOAG}$ be some $|M|^+$-saturated, strongly-$|M|^+$-homogeneous elementary extension of $\text{div}(M)$.

*Remark 5.13.* Let $d$ be some tuple from $M$. Then $d \models p_<$ if and only if $c$ and $d$ have the same type over $B$ in $\overline{M}$.

**Lemma 5.14.** Let $D$ be a small special subgroup of $M$. Let $\alpha = \alpha_0...\alpha_{m-1}$ be a tuple from $\overline{M}$ which is $\mathbb{Q}$-free over $D$. Suppose for all non-zero $f \in \text{LC}^m(\mathbb{Q})$, and for all $d \in D$, we have $f(\alpha) \notin ]d, d+1[$ (this interval of $\overline{M}$ being by convention empty when $1=0$). Then there exists $\sigma \in \text{Aut}(\overline{M}/D)$ so that $\sigma(\alpha)$ is a tuple from $M$.

*Proof.* Suppose by induction we have $\sigma_i \in \text{Aut}(\overline{M}/D)$ sending $\alpha_{<i}$ to a tuple from $M$ for some $i < m$ ($\sigma_0 = \text{id}$ will do for $i=0$). Let $D_i$ be the special subgroup of $M$ generated by $D\sigma_i(\alpha_{<i})$, which is exactly the relative divisible closure in $M$ of $D + \sum_{k<i} \mathbb{Z} \cdot \sigma_i(\alpha_k)$. Then, for all non-zero $f \in \text{LC}^{n-i}(\mathbb{Q})$, for all $d \in D_i$, we have by hypothesis $f(\alpha_{\ge i}) \notin ]d, d+1[$. Let us find $\tau \in \text{Aut}(\overline{M}/D_i)$ so that $\tau(\sigma_i(\alpha_i)) \in M$. Then we could set $\sigma_{i+1} = \tau \circ \sigma_i$, and conclude by induction.

By strong homogeneity of $\overline{M}$, it is enough to show that any interval of $\overline{M}$ with bounds in $\text{div}(D_i) \cup \{\pm\infty\}$ containing $\sigma_i(\alpha_i)$ has a point $\beta$ in $M$. Let $I$ be such an interval. If either the lower or the upper bound of $I$ is in $\{\pm\infty\}$, then $\beta$ clearly exists, one may choose $\pm N \cdot |d|$ if $d \neq 0$, else choose any non-trivial element with correct sign. Now, suppose $I = \left[ \frac{d_1}{N_1}, \frac{d_2}{N_2} \right[$, with $d_k \in D_i$, $N_k > 0$ (by the $\mathbb{Q}$-freeness assumption on $\alpha$, $\sigma_i(\alpha_i) \notin \text{div}(D_i)$, thus it does not matter whether the bounds of $I$ belong to $I$). Then $I$ has a point in $M$ if and only if $]N_2d_1, N_1d_2[$ has a point in $N_1N_2M$. If $M$ is dense, then $I$ has infinitely many points in $M$, and we can use the axioms of ROAG to conclude.

There remains to deal with the case where $M$ is discrete. We established earlier that the $\mathbb{Q}$-linear combinations of $\sigma_i(\alpha_{\ge i})$ do not belong to $]d, d+1[$ for any $d \in D_i$. In particular, for every $N \in \mathbb{Z}$, $N_1N_2\sigma_i(\alpha_i) \notin \bigcup_{N \in \mathbb{Z}}]N_2d_1+N, N_2d_1+N+1[$, and by $\mathbb{Q}$-freeness we also have $N_1N_2\sigma_i(\alpha_i) \notin \{N_2d_1+N|N \in \mathbb{Z}\}$. As a result, $N_1N_2\sigma_i(\alpha_i) \in ]N_2d_1, N_1d_2[\setminus]N_2d_1, N_2d_1+N_1N_2+2]$, which must imply that $N_2d_1+N_1N_2+1 < N_1d_2$, so $I$ has at least $N_1N_2+1$ (in fact, infinitely many) points in $M$, and we can also conclude with the axioms of ROAG. $\square$

**Lemma 5.15.** Let $\alpha$ be a tuple from $M^n$ which is $\mathbb{Q}$-free over $B$, and such that $p = \text{tp}^{\text{DOAG}}(\alpha/B)$ does not fork over $A$. Let $D$ be a small special subgroup of $M$ containing $B$, and let $q \in S_{\text{DOAG}}^n(\text{div}(D))$ be some extension of $p$ which does not fork over $A$. Then $q$ has a realization in $M^n$.

*Proof.* First of all, the realizations of $q$ must be $\mathbb{Q}$-free over $D$. Let $d \in D$, and let $f$ be a non-zero element of $\text{LC}^n(\mathbb{Q})$. By the above lemma, one just has to prove that $q(x) \models f(x) \nparallel]d, d+1[$. If not, then by ??, the interval $[d, d+1]$ must have a point in $\text{div}(A)$. By multiplying everything by a sufficiently large $N >$
0, the interval $]Nd, Nd + N[$ has a point in $A$, and $Nd \in D$, thus $Nd \in A$. In particular, $q_l(x) = Nf(x) \in ]Nd, Nd + N[$, therefore $p(x) = Nf(x) \in ]Nd, Nd + N[$, thus $Nf(\alpha) \in ]Nd, Nd + N[$. As $Nd$ and $\alpha$ are in $A$, it follows that $Nf(\alpha) \in A$, contradicting $\mathbb{Q}$-freeness of $\alpha$. $\square$

**Proposition 5.16.** By ??, let $h$ be the natural injection:

$$ \pi_<(F) \to S_{\mathrm{DOAG}}^n(\mathrm{div}(M)) $$

Consider the action of Aut($M/A$) over $Q = \pi_<(F)$. Then the following conditions are equivalent:

(1) Some element of $Q$ is invariant and extends $p_<$.

(2) Some element of $Q$ has a bounded orbit and extends $p_<$.

(3) The partial type $p_<$ does not fork over $A$.

(4) The partial type $p_<$ does not divide over $A$.

(5) Every closed bounded interval with bounds in $B$ containing a $\mathbb{Z}$-linear combination of $c$ also has a point in $\mathrm{div}(A)$.

(6) For some $p \in Q$ extending $p_<$, $h(p)$ extends to some Aut($\overline{M}/A$)-invariant type over $\overline{M}$.

*Proof.* The directions $1 \implies 2$ and $3 \implies 4$ are immediate.

Let us prove $2 \implies 3$. Let $q_< \in Q$ witness 2. For each prime $l$, let $q_l(x) \in S_l^n(M)$ be the partial type $\{\mathfrak{D}_{l^n}(f(x)) | n > 0, f \in \mathrm{LC}(\mathbb{Z}), f \neq 0\} = \mathrm{tp}_l(0/M)$. Note that $q_l$ is clearly invariant. Then the complete global type in $F$ corresponding to $(q_<, (q_l)_l)$ (remember that $F \to \pi_<(F) \times \prod_l S_l^n(M)$ is a homeomorphism by ??, thus this type is consistent) has a bounded orbit, and thus does not fork over $A$, and we get 3.

Let us prove $4 \implies 5$. Suppose we have $b_k \in B$, $f \in \mathrm{LC}(\mathbb{Z})$ so that $f(c) \in [b_1, b_2]$ witnesses the failure of 5. Let us show that the formula $f(x) \in [b_1, b_2]$ divides over $A$, witnessing the failure of 4. As in the proof of ??, it is enough to find $d \equiv_A b_1$ so that $d > b_2$. As $[b_1, b_2]$ has no point in $\mathrm{div}(A)$, we have $b_k \notin A$, thus the singletons $b_1$ and $b_2$ are both $\mathbb{Q}$-free over $A$. Now, by ??, one just has to show that $\mathrm{tp}_<(b_1/A)$ is consistent with $]b_2, +\infty[$. If not, then all the elements in $M$ of $]b_2, +\infty[$ have a type over $A$ in $\overline{M}$ which is distinct from that of $b_1$. As a result, by the characterization of 1-types in DOAG, there must exist in $\overline{M}$ a point in $\mathrm{div}(A) \cap ]b_1, b_2]$, a contradiction.

Let us show $5 \implies 6$. Suppose 6 fails. Then, by ??, $p_<$ is inconsistent with the partial type $\{f(x) \notin I | I \in \mathcal{I}, f \in \mathrm{LC}(\mathbb{Z})\}$, with $\mathcal{I}$ the set of closed bounded intervals with bounds in $M$ that have no points in $\mathrm{div}(A)$. By compactness, there exist finite subsets $\mathcal{I}' \subseteq \mathcal{I}, G \subseteq \mathrm{LC}^n(\mathbb{Z})$ so that $p_<(x) = \bigvee_{I \in \mathcal{I}', g \in G} g(x) \in \mathcal{I}$. Let $D$ be the special subgroup of $M$ generated by $B$ and the bounds of the elements of $\mathcal{I}'$. Then, for all $p \in Q$ extending $p_<$, the restriction of $p$ to $D$ corresponds to a type in $S_{\mathrm{DOAG}}^n(D)$ which forks over $A$ in DOAG.

Suppose now, by contradiction, that 5 holds. Let $q$ be the type of $S_{\mathrm{DOAG}}^n(B)$ corresponding to $p_<$. Then $q$ does not fork over $A$. By extension, let $q' \in S_{\mathrm{DOAG}}^n(D)$ be an extension of $q$ which does not fork over $A$. By ??, $q'$ must admit a realization $\beta$ in $M^n$ (in particular $\beta = p_<$). As $q'$ does not fork over $A$, we have $g(\beta) \notin I$ for all $g \in G$, $I \in \mathcal{I}'$. As a result, $q_< = \mathrm{tp}_<(\beta/D)$ cannot be extended to any element of $F$. This means that in any elementary extension of $M$, no realization of $q_<$ is
Q-free over $M$. By compactness, this means that there exist finite subsets $P \subseteq M$, $G' \subseteq LC^n(Q) \setminus \{0\}$, so that $q_<(x) = \bigvee_{m \in P, g \in G'} g(x) = m$.

Then we can finally reach a contradiction by extending $q'$ again: let $\tilde{D}$ be the
special subgroup of $M$ generated by $D \cup P$, and let $\tilde{q} \in S_{\text{DOAG}}^n(\tilde{D})$ be some extension
of $q'$ which does not fork over $A$. Then with the same argument, $\tilde{q}$ has a realization
$\gamma \in M^n$, and by hypothesis $\gamma$ is not Q-free over $\tilde{D}$, a contradiction with the fact
that $\tilde{q}$ does not fork over $A$.

Suppose $p \in Q$ witnesses 6, and let us show that $p$ is Aut$(M/A)$-invariant, which
would allow us to conclude the whole proof of the proposition with the direction 6 $\implies$ 1. Let $q \in S_{\text{DOAG}}^n(\bar{M})$ be some global Aut$(\bar{M}/A)$-invariant extension of $h(p)$,
and $\sigma \in \text{Aut}(M/A)$. Then $\sigma$ extends uniquely to $\sigma'$, an automorphism of the ordered
group $\text{div}(M)$, pointwise-fixing $A$. We clearly have $h(\sigma(p)) = \sigma'(h(p))$ (look at the
atomic formulas with predicate $=$, $<$ that belong to $X_<$, check that their image by
$\sigma$ is satisfied by the realizations of $\sigma'(h(p)))$. By quantifier elimination in DOAG,
$\sigma'$ is a partial elementary map in $\bar{M}$. By strong homogeneity of $\bar{M}$, $\sigma'$ extends to
some $\tilde{\sigma} \in \text{Aut}(\bar{M})$. As $\sigma'$ pointwise-fixes $A$, so does $\tilde{\sigma}$, thus $\tilde{\sigma}(q) = q$. Now, we have
$\sigma(p) = h^{-1}(\sigma'(h(p))) = h^{-1}(\sigma'(q|_{\text{div}(M)})) = h^{-1}(\tilde{\sigma}(q|_{\text{div}(M)})) = h^{-1}(\tilde{\sigma}(q)|_{\tilde{\sigma}(\text{div}(M))}) =
h^{-1}(q|_{\text{div}(M)}) = h^{-1}(h(p)) = p$, and we can conclude. $\square$

The fifth condition of the above proposition is a very simple geometric condition,
the kind of statement that would be very satisfactory for a characterization of
forking. For the remainder of this subsection, we will look for similar conditions for
the $(p_l)_l$.

Remark 5.17. If $d$ is a tuple in some elementary extension of $M$, then the above proposition shows that $\text{tp}_<(d/M)$ does not fork/divide over $A$ if and only if the corresponding type in $S_{\text{DOAG}}(\text{div}(M))$ does not fork over $A$. One can easily see that $h$ is continuous, so the closed subspace of $\pi_<(F)$ of extensions of $p_<$ which do not fork over $A$ is homeomorphic to the closed subspace of $S(\text{div}(M))$ of extensions of $h(p_<$) which do not fork over $A$. Is this space the same topological space as the one described at the end of subsection ?? ? There is a subtlety here: we have to show that $\text{div}(M)$ is $|A|^{+}$-saturated and strongly-$|A|^{+}$-homogeneous.

In the case where $M$ is dense, it is easy to prove that $M$ is $\kappa$-saturated if and only if $\text{div}(M)$ is for any cardinal $\kappa$. Strong homogeneity might not work all the time, but we can always make it work, i.e. for every model $G$, and cardinal $\kappa$, there exists $N$ some elementary extension of $G$ such that $N$ and $\text{div}(N)$ are $\kappa$-saturated and strongly-$\kappa$-homogeneous: we choose $N$ the reunion of some elementary chain, each element of which is saturated over the supremum of the cardinals of the smaller elements. Basic model theory tells us that $N$ is then automatically $\kappa$-saturated and strongly-$\kappa$-homogeneous, as long as the chain is long enough. Then the associated chain of divisible closures also satisfies those properties, thus $\text{div}(N)$ is also $\kappa$-saturated and strongly-$\kappa$-homogeneous. As a result, when $M$ is dense and well-chosen, the closed subspace of $\pi_<(F)$ of invariant/of bounded orbit/non-forking over $A$/non-dividing extensions of $p_<$ is naturally homeomorphic to some finite product of finite coproducts of closed subspaces of $S_{\text{DOAG}}(\{0\})$.

In the case where $M$ is discrete, $\text{div}(M)$ is very obviously not even $\aleph_0$-saturated,
as it does not contain any element whose Archimedean class is $< \Delta(A)$. However,
we still claim that the space of independent extensions of $p_<$ is homeomorphic to
the topological space from the end of ?? for some sufficiently large choice of $M$.
Instead of looking at $\text{div}(M)$, we actually have to look at the quotient of $M$ by $\mathbb{Z}$ = dcl($\emptyset$), which is a model of DOAG, and which is sufficiently saturated and strongly-homogeneous as long as $M$ is the reunion of the correct chain. In this quotient, we have a similar homeomorphism $h$ which behaves like in ?? . A more detailed proof of this construction will be provided in the author's PhD thesis.

5.3.2. *Partial types using a prime of finite index.* For the rest of this subsection, we fix a prime $l$.

**Proposition 5.18.** Let $G$ be a torsion-free Abelian group. Then, for all $N > 0$, we have in $G^{eq}$ a $\emptyset$-definable group isomorphism between the two definable groups $G / lG$ and $l^N G / l^{N+1} G$.

*Proof.* Let $x, y \in G$, and $N > 0$. Suppose $l^N x - l^N y \in l^{N+1} G$. Then, as $G$ is torsion-free, we have $x - y \in lG$, thus the map $f: l^N x \mod l^{N+1} G \to x \mod lG$ is a well-defined group homomorphism, which is clearly surjective. Now, if $x \notin lG$, then $l^N x \notin l^{N+1} G$, so the map is injective. $\square$

**Corollary 5.19.** If $lG$ has a finite index $d$, then $\cap_N l^N G$ has a small index, smaller than $2^{\aleph_0}$.

*Proof.* Define a tree structure on $\bigcup_{N \ge 0} G / l^N G$ of root $G$, so that the parent of some node $x \mod l^{N+1} G$ is $x \mod l^N G$. Then every node has $d$ children and the tree has $\aleph_0$ levels, thus it admits at most $2^{\aleph_0}$ branches. To conclude, the map $x \mod \cap_N l^N G \to (x \mod l^N G)_N$ is a natural bijection between $G / \cap_N l^N G$ and the set of the branches. $\square$

**Lemma 5.20.** Let $D$ be some small special subgroup of $M$. Suppose we have $e_1, e_2 \in M$, such that $e_1, e_2 \notin D + l^N M$, and $e_1 - e_2 \in l^{N-1} M$, for some $N > 0$. Then we have $\text{tp}_l(e_1/D) = \text{tp}_l(e_2/D)$.

*Proof.* Any atomic formula with parameters in $D$ and predicate in $\{\mathfrak{d}_{lm} | m > 0\}$ can clearly be written $\mathfrak{d}_{lm}(\lambda x - d)$, with $\lambda \in \mathbb{Z}, m > 0, d \in D$. We have several cases:

* $\lambda \in l\mathbb{Z}, d \notin lM$, in which case the formula always fails.

* $\lambda \in l\mathbb{Z}, d \in lM, m = 1$, in which case the formula always holds.

* $\lambda \notin l\mathbb{Z}$, in which case $\lambda$ and $l$ are coprime. Therefore, by Bézout's identities, the formula is equivalent to $\mathfrak{d}_{lm}(x - d')$ for some $d' \in D$. By hypothesis on the $e_i$, this formula is satisfied by $e_1$ if and only if it is satisfied by $e_2$.

* $\lambda \in l\mathbb{Z}, d \in lM, m > 1$, in which case the formula is equivalent to $\mathfrak{d}_{lm-1} \left( \frac{\lambda}{l}x - \frac{d}{l} \right)$, and we reduce by induction to one of the above three cases.

Either way, we clearly see that those formulas are satisfied by $e_1$ if and only if they are satisfied by $e_2$, so we can conclude. $\square$

**Corollary 5.21.** Let $d$ be a tuple from $M$, which is $\mathbb{Q}$-free over $B$. Then $d \models p_l$ if and only if, for each $f \in \text{LC}^n(\mathbb{Z})$, for each $N > 0$, either $f(c) - f(d) \in l^N M$, or neither $f(c)$ nor $f(d)$ are in $B + l^N M$.

*Proof.* Any atomic formula $\varphi(x_1...x_n)$ with predicate in $\{\mathfrak{d}_{l^N} | N > 0\}$ can be written $\psi(f(x))$, with $f \in \text{LC}(\mathbb{Z})$, and $\psi$ an atomic formula on the same predicate with
one variable. As a result, $d = p_l$ if and only if $\mathrm{tp}_l(f(d)/B) = \mathrm{tp}_l(f(c)/B)$ for all $f \in \mathrm{LC}(\mathbb{Z})$. Fix $f \in \mathrm{LC}^n(\mathbb{Z})$.

Suppose that we have $f(c) \in B + l^N M$ for all $N > 0$. For each $N$, let $b_N \in B$ so that $f(c) - b_N \in l^N M$. Then $\mathrm{tp}_l(f(c)/B)$ contains $\{\mathfrak{d}_{l^N}(x - b_N) | N > 0\}$. This inclusion is an equality, for if $\mathfrak{d}_{l^N}(f(d) - b_N)$ for all $N$, then $f(d) - f(c) \in l^N M$ for all $N$, which clearly implies $\mathrm{tp}_l(f(c)/B) = \mathrm{tp}_l(f(d)/B)$. In particular, $\mathrm{tp}_l(f(d)/B) = \mathrm{tp}_l(f(c)/B) \iff f(d) - f(c) \in \bigcap_N l^N M$.

Suppose now that there exists $N > 0$ so that $f(c) \notin B + l^N M$. The set of all such $N$ is a final segment of $\omega$, choose $N$ its least element. Let $b \in B$ so that $f(c) - b \in l^{N-1} M$ (if $N = 1$, then $b = 0$ will do). Then, by the above lemma, $\mathrm{tp}_l(f(c)/B)$ is generated by the partial type $\{\mathfrak{d}_{l^{N-1}}(x-b)\} \cup \{\neg \mathfrak{d}_{l^N}(x-b') | b' \in B\}$, and we can conclude. $\square$

**Remark 5.22.** If $[M : lM]$ is finite, then by ?? the $\emptyset$-type-definable equivalence relation $\{\mathfrak{d}_{l^N}(x-y) | N > 0\}$ is bounded (has a small number of classes) and finer than having the same $l$-type over any parameter set, thus $S_l^n(M)$ is small, and each orbit under $\mathrm{Aut}(M/A)$ is obviously bounded.

Moreover, the classes of this $\emptyset$-type-definable equivalence relation are all $\mathrm{acl}^{\mathrm{eq}}(\emptyset)$-invariant, thus $\mathrm{Aut}(M^{\mathrm{eq}}/\mathrm{acl}^{\mathrm{eq}}(\emptyset))$ acts trivially over $S_l^n(M)$.

**Proposition 5.23.** Suppose $[M : lM]$ is finite. Consider the action of $\mathrm{Aut}(M/A)$ over $Q = S_l^n(M)$. Then $p_l$ has an invariant extension in $Q$ if and only if every $\mathbb{Z}$-linear combination of $c$ belongs to $\bigcap_N (A + l^N M)$.

*Proof.* Suppose every $\mathbb{Z}$-linear combination of $c$ belongs to $\bigcap_N (A + l^N M)$. Then $p_l(M)$ can be written as the intersection of $A$-definable sets of the form:

$$ \{x | \mathfrak{d}_{l^N} (f(x) - a)\}, $$

with $N > 0, a \in A$, $f \in \mathrm{LC}(\mathbb{Z})$. As a result, $p_l(M)$ is $A$-type definable (thus $A$-invariant), and all its realizations $d \in M^n$ satisfy $f(c) - f(d) \in \bigcap_N l^N M$ for all $f \in \mathrm{LC}^n(\mathbb{Z})$. By the above lemma, $p_l$ is complete as an element of $Q$, so we get the right-to-left direction.

Suppose now that we have $f \in \mathrm{LC}^n(\mathbb{Z})$ and $N > 0$ so that $f(c)$ does not belong to $A + l^N M$. Let $p \in Q$ be an extension of $p_l$. Then, by ??, there must exist $\alpha \in M$ so that $p(x) = \mathfrak{d}_{l^N}(f(x) - \alpha)$ for all $N > 0$. As $[M : A + l^N M] > 1$, we must have by ?? $1 < [M : l^N M] = [M : lM]^N$, thus $1 < [M : lM] = [l^N M : l^{N+1} M]$, so let $m \in l^N M \setminus l^{N+1} M$. Then, by ??, $\mathrm{tp}_l(\alpha/A) = \mathrm{tp}_l(\alpha + m/A)$. Let $D$ be a special subgroup of $M$ of size $\le \max(2^{\aleph_0}, |A|)$ containing $A$, and a system of representatives of the cosets of $\bigcap_m l^m M$. By saturation of $M$, we can find $\beta \in M$ so that $\mathrm{tp}_l(\beta/D) = \mathrm{tp}_l(\alpha - m/D)$ (which implies $\beta - \alpha - m \in \bigcap_m l^m M$), and $\mathrm{tp}_j(\beta/D) = \mathrm{tp}_j(\alpha/D)$ for all $j \ne l$ (which implies $\mathrm{tp}_j(\beta/A) = \mathrm{tp}_j(\alpha/A)$). Then, by strong homogeneity of $M$, there exists $\sigma \in \mathrm{Aut}(M/A)$ so that $\sigma(\alpha) = \beta$. Now, we have $\sigma(p)(x) = \mathfrak{d}_{l^{N+1}}(f(x) - \alpha - m)$, a formula that is inconsistent with $\mathfrak{d}_{l^{N+1}}(f(x) - \alpha)$. As a result, $\sigma(p) \ne p$, and we can conclude. $\square$

**Remark 5.24.** Just like in ??, we would like to describe the space of global invariant/bounded extensions of $p_l$. In case the condition of the above proposition holds, we saw that $p_l$ is $A$-invariant and complete in $S_l^n(M)$, thus this space is a point. In case the condition does not hold, the space of invariant extensions of $p_l$ is obviously
empty, but we would like to describe the topological space of all the extensions that
have a bounded orbit. This is what we will do in the remainder of this subsection,
we will exhibit a homeomorphism between this orbit and some closed subspace of
the Cantor space.

**Definition 5.25.** Let $N > 0$. We define $\mathcal{T}_N$ the tree $\{0,...N-1\}^{<\omega}$, and $\mathcal{B}_N = \{0,...N-1\}^{\omega}$ the set of its branches. The $m$-th-level of $\mathcal{T}_N$ is $\{0,...N-1\}^m$. We see $\mathcal{B}_N$ as a topological space, with the elementary open sets being the $O_w = \{b \in \mathcal{B}_N|w\}$ is a prefix of $b\}$ for every $w \in \mathcal{T}_N$. For every, $m$, the opens $(O_w)_{w \in \{0,...N-1\}^m}$ form a finite disjoint open cover of $\mathcal{B}_N$, thus each $O_w$ is in fact clopen.

**Lemma 5.26.** For all $N, d > 0$, $\mathcal{B}_N$ is naturally homeomorphic to $\mathcal{B}_{Nd}$.

*Proof.* The natural bijection $\{0, \dots, N^d - 1\} \to \{0, N-1\}^d$ extends to a bijection $h : \mathcal{B}_{Nd} \to \mathcal{B}_N$. For each $m < \omega$, we have:

$$ \{h(O_w)|w \text{ in the } m\text{-th level of } \mathcal{T}_{Nd}\} = \{O_w|w \text{ in the } md\text{-th level of } \mathcal{T}_N\}, $$

thus $h^{-1}$ is continuous. To conclude, for each $w \in \mathcal{T}_N$, $O_w$ can be written as a finite
reunion of elementary opens $O_v$, with $v$ in the $d[|w|/d]$-th level of $\mathcal{T}_N$, thus $h^{-1}(O_w)$
can be written as a finite reunion of some $O_v$, with $v$ in the $[|w|/d]$-th level of $\mathcal{T}_{Nd}$,
therefore $h$ is continuous. $\square$

**Lemma 5.27.** The two spaces $\mathcal{B}_l$ and $\mathbb{Z}_l$ (the topological space of $l$-adic integers) are naturally homeomorphic.

*Proof.* The tree of non-empty balls of $\mathbb{Z}_l$ (including the whole $\mathbb{Z}_l$) ordered by inclusion is clearly isomorphic to $\mathcal{T}_l$, with the isomorphism $w \mapsto B(\sum w_m l^m, m)$. The corresponding bijection $\mathcal{B}_l \to \mathbb{Z}_l$ is clearly $h : w \mapsto \sum_{m<\omega} w_m l^m$, and we can conclude. $\square$

**Lemma 5.28.** Suppose $[M : lM] = l^d$ is finite. Then $S_l^n(M)$ is naturally homeomorphic to the direct product $\mathcal{B}_{ld}^n$ (with $n = |c|$).

*Proof.* Note that each point of $S_l^n(M)$ is a partial type that can be written $\{x \in Y_m | m < \omega\}$, with $Y_m \in (M/l^m M)^n$. In particular, $S_l^n(M)$ is in natural bijection with $\left(\mathcal{M} / \bigcap_{m<\omega} l^m M\right)^n$. For any $\mathcal{L}_P(M)$-term $t(x)$, the definable subset of $M$ defined by $\partial_{l^m}(t(x))$ is a re-union of cosets mod $l^m M$, thus the elementary clopen subsets of $\left(\mathcal{M} / \bigcap_{m<\omega} l^m M\right)^n$ induced by $S_l^n(M)$ are the Cartesian products of cosets of $l^m M$ for $m < \omega$. We build by induction a sequence of bijections $(f_m)_m$ between the $m$-th level of $\mathcal{T}_{ld}$ and $M/l^m M$. Suppose by induction hypothesis that we have $f_m$ ($f_0$ is the bijection between the point $\{M/M\}$ and the point $\emptyset$). For each $w$ in the $m$-th level, $w$ has exactly $l^d$ children in the $m+1$-th level, and $f_m(w)$ is the reunion of exactly $l^d$ cosets mod $l^{m+1}M$, so there is a bijection between them for each $w$, and we set $f_{m+1}$ as the reunion of those bijections. Then the $(f_m)_m$ define an homeomorphism between $\mathcal{B}_{ld}$ and $M/\bigcap_m l^m M$, and we can conclude. $\square$
**Corollary 5.29.** The closed space of every global extension of $p_l$ of bounded orbit under the action of Aut$(M/A)$ is naturally homeomorphic to some closed subspace of $\mathbb{Z}_l^n$.

5.3.3. Partial types using a prime of infinite index. We still fix a prime $l$, and we assume here that $[M : lM]$ is infinite.

**Lemma 5.30.** Let $D$ be a small special subgroup of $M$, and $\alpha = \alpha_1...\alpha_m$ a finite tuple from $M$ which is $\mathbb{Q}$-free over $D$. Let $\beta = \beta_1...\beta_m$ be another tuple from $M$ so that $\text{tp}_l(\beta/D) = \text{tp}_l(\alpha/D)$. Then there exists $\gamma = \gamma_1...\gamma_m \equiv_D \alpha$ so that $\gamma_i - \beta_i \in \bigcap_N l^N M$ for all $i$.

*Proof.* As $\bigcap_N l^N M$ is large, choose by induction $\varepsilon_i \in \bigcap_N l^N M$ so that $\varepsilon_i$ is not in the special subgroup generated by $D\alpha\beta\varepsilon_{<i}$, and let $\beta'_i = \beta_i + \varepsilon_i$. Then we have $\beta_i - \beta'_i \in \bigcap_N l^N M$, and $\varepsilon$ is $\mathbb{Q}$-free over the special subgroup generated by $D\alpha\beta$. As a result, $\alpha$ (and hence $(\alpha_i - \beta'_i)_i$) is $\mathbb{Q}$-free over $D'$, the special subgroup generated by $D\beta'$. By ??, there exists in $M$ a tuple $e = e_1...e_m$ so that $e_i \in \bigcap_N l^N M$ for all $i$, and $\text{tp}_j(e/D') = \text{tp}_j((\alpha_i - \beta'_i)_i/D')$ for all $j \neq l$. Let $\gamma_i = \beta'_i + e_i$. Then, for all $j \neq l$, we have $\text{tp}_j(\gamma/D') = \text{tp}_j(\alpha/D')$. Moreover, we have $\gamma_i - \beta'_i \in \bigcap_N l^N M$ for all $i$, thus $\text{tp}_l(\gamma/D) = \text{tp}_l(\beta'/D) = \text{tp}_l(\beta/D) = \text{tp}_l(\alpha/D)$, so we can conclude that $\gamma \equiv_D \alpha$. $\square$

*Remark 5.31.* Up to isomorphism, the unique discrete subgroup of $\mathbb{R}$ is $\mathbb{Z}$, and all its elements are $\emptyset$-definable in $\mathcal{L}_P$. As a result, any discrete model of ROAG is an elementary extension of $\mathbb{Z}$ which we identify with dcl($\emptyset$). In particular, if $[M : lM]$ was infinite for some $l$, then $M$ would necessarily be dense. In that case, the special subgroups of $M$ (which are exactly the definably-closed sets) are its pure subgroups.

**Lemma 5.32.** Let $(\alpha_i)_{i \in I}$ be a finite tuple from $M$ which is $\mathbb{Q}$-free over $B$. Suppose that for each $i \in I$, there exists $N > 0$ such that $\alpha_i \notin B + l^N M$, choose $N_i$ the least of those integers for each $i \in I$. Then, there exists $(\beta_i)_i \equiv_B (\alpha_i)_i$ so that $\beta_i - \alpha_k \notin l^{N_i} M$ for each $i, k \in I$.

*Proof.* Note that by the above remark, the special subgroups of $M$ are exactly its pure subgroups.

Let $b_i \in B$ so that $b_i - \alpha_i \in l^{N_i-1} M$. Suppose we can find a witness $(\gamma_i)_i$ of the lemma with $(\alpha_i)_i$ replaced by $\left( \frac{\alpha_i - b_i}{l^{N_i-1}} \right)_i$ ($N_i$ will be replaced by 1). Then we can choose $\beta_i = l^{N_i-1}\gamma_i + b_i$. For the remainder of the proof, we can suppose without loss that $N_i = 1$ for every $i$ (in particular, $b_i = 0$).

Let $V$ be the large $\mathbb{F}_l$-vector space $M/lM$, and $U \leq V$ the $\mathbb{F}_l$-vector-subspace $(B+lM)/lM$. Note that $U$ is small, as it is the image by $B$ of the map $M \to M/lM$. Note also that we have $\alpha_i \in V \setminus U$ for every $i \in I$. Now, let $I_0 \subseteq I$ so that $(\alpha_i)_{i \in I_0}$ is a lift of an $\mathbb{F}_l$-base of the image in $V/U$ of $U + \sum_{i \in I} \mathbb{F}_l \cdot (\alpha_i \bmod lM)$. Let $D$ be the special subgroup of $M$ generated by $B(\alpha_i)_{i \in I}$, and let $U' = (D+lM)/lM$, a small vector-subspace of $V$ containing $U$. As $V/U'$ is large, one can choose an arbitrary family $(\beta_i)_{i \in I_0} \in M^{I_0}$ whose image in $V/U'$ is $\mathbb{F}_l$-free. Now, for each non-zero $f \in \text{LC}^{|I_0|}(\mathbb{Z})$, if $m$ is the largest integer so that all the coefficients of $f$ lie
in $l^m\mathbb{Z}$, then we have $f((\beta_i)_{i \in I_0}) \in l^m M \ni f((\alpha_i)_{i \in I_0})$, and $f((\beta_i)_{i \in I_0}) \notin B + l^{m+1} M \ni f((\alpha_i)_{i \in I_0})$, thus we have $\text{tp}_l((\alpha_i)_{i \in I_0}/B) = \text{tp}_l((\beta_i)_{i \in I_0}/B)$ by the same reasoning as in ?? . For each $i \in I_0$, as $\beta_i \mod lM \notin U'$, we have $\beta_i \notin D+lM$, thus $\beta_i - \alpha_k \notin lM$ for each $k \in I$. By ?? , we can suppose without loss that $(\beta_i)_{i \in I_0} \equiv_B (\alpha_i)_{i \in I_0}$.

Choose $\sigma \in \operatorname{Aut}(M/B)$ so that $\sigma(\alpha_i) = \beta_i$ for every $i \in I_0$. Let $\beta_i = \sigma(\alpha_i)$ for each $i \in I \setminus I_0$. Then we have $(\beta_i)_{i \in I} \equiv_B (\alpha_i)_{i \in I}$. In order to conclude, we need to show that $\beta_i - \alpha_k \notin lM$ for each $i \in I \setminus I_0$, $k \in I$. Choose $i \in I \setminus I_0$. There must exist $f_i \in \operatorname{LC}(\mathbb{Z})$ so that $\alpha_i - f_i((\alpha_k)_{k \in I_0}) \in B+lM$. Choose $e_i \in B$ so that $\alpha_i - f_i((\alpha_k)_{k \in I_0}) - e_i \in lM$. Then we must have $\beta_i - f_i((\beta_k)_{k \in I_0}) - e_i \in lM$, so we just have to show that $f_i((\beta_k)_{k \in I_0}) \notin D+lM$. As $\alpha_i \notin B+lM$, some coefficient of $f_i$ must be coprime with $l$, thus we can conclude as $(\beta_k \mod lM)_{k \in I_0}$ is $\mathbb{F}_l$-free over $U'$.
$\square$

**Lemma 5.33.** Let $(\alpha_i)_i$ be some tuple from $M$. Then there exists in $M$ a tuple $(\beta_i)_i$ which is $\mathbb{Q}$-free over $B$, and for which $\alpha_i - \beta_i \in \bigcap_N l^N M$ for every $i$.

*Proof.* One merely has to define by induction $\varepsilon_i \in \bigcap_N l^N M$ which does not lie in the small special subgroup generated by $B(\alpha_k)_k \varepsilon_{<i}$, and choose $\beta_i = \alpha_i + \varepsilon_i$. $\square$

**Lemma 5.34.** Let $N > 0$. Suppose we have $f_1, \dots, f_N \in \text{LC}^n(\mathbb{Z})$ and $Y_1, \dots, Y_N \in \bigcup_m M / l^m M$ so that for all $x \in p_l(M)$, we have $M \models \bigvee_i f_i(x) \in Y_i$. Then at least one of the $Y_i$ must have a point in $B$.

*Proof.* Let $\alpha_i \in Y_i$ for each $i$. By the above lemma, we can assume without loss that $(\alpha_i)_i$ is $\mathbb{Q}$-free over $B$. Suppose by contradiction that none of the $Y_i$ has a point in $B$. Then we can apply ?? to $(\alpha_i)_i$, we find $(\beta_i)_i \equiv_B (\alpha_i)_i$ such that, for all $m > 0$ and for all $i, k$, if $\alpha_i \notin B + l^m M$, then $\beta_i - \alpha_k \notin l^m M$. In particular, as $Y_k$ does not intersect $B$, we have $\beta_i \notin Y_k$ for all $i, k$. Let $\sigma \in \operatorname{Aut}(M/B)$ so that $\sigma(\alpha_i) = \beta_i$ for each $i$. Then we must have $(\bigcup_i Y_i) \cap (\bigcup_i \sigma(Y_i)) = \emptyset$. However, as $p_l$ is $B$-invariant, we must have $p_l(x) \models \bigvee_i f_i(x) \in \sigma(Y_i)$, a contradiction.
$\square$

**Proposition 5.35.** Consider the action of Aut$(M/A)$ over $Q = S_l^n(M)$. Let $C$ be the group of all $\mathbb{Z}$-linear combinations of $c$. Then the following conditions are equivalent:

(1) Some invariant element of $Q$ extends $p_l$.

(2) Some element of $Q$ of bounded orbit extends $p_l$.

(3) $p_l$ does not fork over $A$.

(4) $p_l$ does not divide over $A$.

(5) For each $N > 0$, we have $(C + l^N M) \cap (B + l^N M) = A + l^N M$.

*Proof.* The directions $1 \implies 2$ and $3 \implies 4$ are trivial.

Let us prove $2 \implies 3$. For each prime $l' \neq l$, we can define $q_{l'}$ just as in ?? . There remains to show that there exists $q_< \in \pi_<(F)$ whose orbit is $A$-invariant. In some elementary extension of $M$, define by induction $(a_i)_i$, $(D_i)_i$ so that $D_i$ is the special subgroup generated by $Ma_{<i}$, and $a_i \in \bigcap_{b \in D_i} ]b, +\infty[$. Define $q_< = \text{tp}_< (a/M)$. Then, for all $(\lambda_1, \dots, \lambda_n) \in \mathbb{Z}^n$, and $b \in M$, whether $\sum_i \lambda_i a_i > b$ depends uniquely on the sign of $(\lambda_i)_i$ in the anti-lexicographic sum $\mathbb{Z}^n$, and it does not depend on $b$. Moreover,
$(a_i)_i$ is clearly $\mathbb{Q}$-free over $M$. As a result, $q_\le$ is an Aut$(M)$-invariant (and hence Aut$(M/A)$-invariant) element of $\pi_<(F)$, and we can conclude.

Let us prove $4 \implies 5$. Let $f \in \text{LC}(\mathbb{Z})$, $N > 0$ so that $f(c) \in B + l^N M \setminus A + l^N M$. Let $b \in B$ so that $f(c) - b \in l^N M$ (we know that $b \notin A + l^N M$ by hypothesis, in particular it is $\mathbb{Q}$-free over $A$ as a singleton). Let us show that the formula $\varphi(x, b) := \mathfrak{d}_{l^N}(f(x) - b)$ divides over $A$, which will imply that $4$ fails. We can repeatedly apply ?? to find $(b_i)_{i<\omega}$ so that $b_0 = b$, $b_i \equiv_B b_{i+1}$, and $b_i - b_j \notin l^N M$ for all $j \neq i$. Then the set of formulas $\{\varphi(x, b_i) | i < \omega\}$ is clearly 2-inconsistent, so we can conclude.

Let us prove $5 \implies 1$. Suppose $5$ holds, and let us build an explicit invariant element of $Q$ which extends $p_l$, witnessing $1$. Let $F = \{(f, N) \in \text{LC}^n(\mathbb{Z}) \times \omega | f(c) \in A + l^N M\}$. For each $(f, N) \in F$, let $a_{f,N} \in A$ so that $f(c) - a_{f,N} \in l^N M$. Define the following partial type:

$$p(x) = \{\mathfrak{d}_{l^N}(f(x) - a_{f,N}) | (f, N) \in F\} \cup \{\neg \mathfrak{d}_{l^N}(f(x) - e) | (f, N) \notin F, e \in M\}$$

Suppose by contradiction that $p$ is not consistent with $p_l$. Then, by compactness, there exists a finite tuple $(f_i, N_i, e_i)_i$ so that, for every $i$, $(f_i, N_i) \notin F$, and $p_l(x) \models \bigvee \mathfrak{d}_{l^{N_i}}(f_i(x) - e_i)$. By ?? applied to $Y_i = e_i$ mod $l^{N_i} M$, there must exist $i$ so that $e_i \in B + l^{N_i} M$. As $(f_i, N_i) \notin F$, we must have $e_i \notin A + l^{N_i} M$, thus $e_i \in B + l^{N_i} M \setminus A + l^{N_i} M$. By hypothesis $5$, $p_l(x) \models \neg \mathfrak{d}_{l^{N_i}}(f_i(x) - e_i)$, therefore $p_l(x) \models \bigvee_{k \neq i} \mathfrak{d}_{l^{N_k}}(f_k(x) - e_k)$.

We keep decreasing by induction the size of this disjunction, until we inevitably reach a contradiction. As a result, $p$ must be consistent with $p_l$, and it is clearly $A$-invariant. This partial type is of course complete as an element of $Q$, because every atomic formula with predicate in $\{\mathfrak{d}_{l^N} | N > 0\}$ and parameters in $M$ either lies in $p$, or its negation lies in $p$. $\square$

**Remark 5.36.** When the conditions of the above proposition hold, the explicit type in the proof of $5 \implies 1$ is actually the unique global extension of $p_l$ which has a bounded orbit. Indeed, if $q$ is another global extension, and $d$ is a realization of $q$ in some elementary extension, then we must have $f(d) - \alpha \in l^N M$ for some $(f, N) \notin F$, and $m \in M$. In that case, $m \notin A + l^N M$, thus the fifth condition from the proposition fails with $c, B$ replaced by $d, M$. In particular, $q$ divides over $A$, thus its orbit is unbounded.

## 5.4. Main result.

**Theorem 5.37.** Let $M \models \text{ROAG}$, $A, B$ parameter subsets of $M$, $\kappa = \max(|AB|, 2^{\aleph_0})^+$, and $c = c_1...c_n \in M^n$. Suppose $M$ is $\kappa$-saturated and strongly-$\kappa$-homogeneous. Let $C$ be the subgroup of $M$ generated by $c$, and $A'$, $B'$ the special subgroups generated by $A, AB$. Then $c \downarrow_A^f B$ if and only if $c \downarrow_A^d B$ if and only if $c \downarrow_A^{b_0} B$ if and only if $\text{tp}(c/AB)$ has a global Aut$(M^{eq}/A \cup \text{acl}^{eq}(\emptyset))$ if and only if the following conditions hold:

(1) Every closed bounded interval of $B'$ that has a point in $C$ already has a point in $\text{div}(A')$.

(2) For all prime $l$, if $[M : lM]$ is infinite, then we have $(C + l^N M) \cap (B' + l^N M) = A' + l^N M$ for all $N > 0$.

Moreover, $c \downarrow_A^{\text{inv}} B$ if and only if the above conditions hold, and, additionally, for every prime $l$ for which $[M : lM]$ is finite, we have $C \subseteq A' + \bigcap_{N>0} l^N M$.
Note that, if $c \underset{A}{\rfloor}^d B$, then $\text{tp}(c/AB)$ divides over $\text{acl}^{eq}(A)$, thus it does not admit any global $\text{Aut}(M^{eq}/A \cup \text{acl}^{eq}(\emptyset))$-invariant extension.

**Proof.** Firstly, we have $A' = \text{dcl}(A)$ and $B' = \text{dcl}(AB)$ by ??, thus we have $c \underset{A}{\lfloor} B \iff c \underset{A'}{\lfloor} B'$ for every $\lfloor \in \{\lfloor^f, \lfloor^d, \lfloor^{bo}, \lfloor^{inv} \}$. Secondly, given $c' = c_1'...c_m'$ a maximal subtuple of $c$ which is $\mathbb{Q}$-free over $A'$, $c$ and $c'$ are $A$-interdefinable, thus one can show that for all of those independence notions we have $c \underset{A'}{\lfloor} B' \iff c' \underset{A'}{\lfloor} B'$. Likewise, $\text{tp}(c/AB)$ has an $\text{Aut}(M^{eq}/A \cup \text{acl}^{eq}(\emptyset))$-invariant extension if and only if $\text{tp}(c'/B')$ has an $\text{Aut}(M^{eq}/A' \cup \text{acl}^{eq}(\emptyset))$-invariant extension. Note that $C$ is exactly the set of $\mathbb{Z}$-linear combinations of $c'$.

If $c'$ was not $\mathbb{Q}$-free over $B'$, then one could show that $c' \underset{A'}{\lfloor}^d B'$. Moreover, condition 1 would fail on some singleton of $B \setminus A$, witnessing the fact that $c'$ is not $\mathbb{Q}$-free over $B$.

Suppose $c'$ is $\mathbb{Q}$-free over $B'$. For each $j \in J$, let $p_j = \text{tp}_j(c'/B')$, and let $F_j$ be the image by $\pi_j$ of the space of complete global types with realizations that are $\mathbb{Q}$-free over $M$.

If the first condition fails, then $p_<^c$ divides over $A'$ by ??, and if the second condition fails for some $l$, then $p_l$ divides over $A'$ by ???. Either way, we have $c' \underset{A'}{\lfloor} B'$ for every $\lfloor \in \{\lfloor^f, \lfloor^d, \lfloor^{bo}, \lfloor^{inv} \}$.

Suppose both conditions hold, then $p_<^c$ extends to an $\text{Aut}(M/A')$-invariant (thus $\text{Aut}(M^{eq}/A' \cup \text{acl}^{eq}(\emptyset))$-invariant) element of $F_<^c$ by ??, and, for every prime $l$ for which $[M : lM]$ is infinite, $p_l$ extends to some $\text{Aut}(M/A')$-invariant extension of $F_l$ by ???. By ??, for each prime $l$ of finite index, any global extension of $p_l$ is $\text{Aut}(M^{eq}/A' \cup \text{acl}^{eq}(\emptyset))$-invariant, thus it follows that $\text{tp}(c'/A'B'$ has a global $\text{Aut}(M^{eq}/A' \cup \text{acl}^{eq}(\emptyset))$-invariant extension. Moreover, by ??, we know that $p_l$ extends to some element of $F_l$ of bounded orbit under $\text{Aut}(M/A')$ for every prime $l$ for which $[M : lM]$ is finite. By ??, we have $c' \underset{A'}{\lfloor}^{bo} B'$, thus $c' \underset{A'}{\lfloor}^f B'$, $c' \underset{A'}{\lfloor}^d B'$. Moreover, by ??, we have $c' \underset{A'}{\lfloor}^{inv} B'$ if and only if, for every prime $l$ for which $[M : lM]$ is finite, $p_l$ extends to some $\text{Aut}(M/A')$-invariant element of $F_l$. Then we can conclude using ??.

□

We know that in any NIP theory, $\lfloor^f$ coincides with $\lfloor^{inv}$ over models. We can now refine that result for ROAG:

**Corollary 5.38.** We have $\lfloor_A^f = \lfloor_A^{inv}$ if and only if $A' + \bigcap_N l^N M = M$ for all prime $l$ of finite index.

Moreover, $\lfloor^f = \lfloor^{inv}$ if and only if, for each prime $l$, either $M$ is $l$-divisible, or the index $[M : lM]$ is infinite.

**Remark 5.39.** Suppose $M$ is dense. Then by ??, ?? and ??, we get the three following properties :

* If $c \underset{A}{\lfloor}^f B$, then every global extension of $p_<^c$ which does not fork over $A$ is $A$-invariant. Moreover, the Stone space of those extensions can be written as $S_1$, a finite product of finite coproducts of closed subspaces of $S_{\text{DOAG}}(\{0\})$.
* Let $l$ be prime so that $[M : lM]$ is finite. If $C \subseteq A' + \bigcap_{N} l^N M$, then $p_l$ has only one global extension, and it is $A$-invariant. Else, none of the global extensions of $p_l$ is $A$-invariant, but they are all non-forking over $A$, and the space of those extensions can be written $S_{2,l}$, a closed subspace of $\mathbb{Z}_{l}^{|c'|}$.

* Let $l$ be prime so that $[M : lM]$ is infinite. If $c \searrow_A^f B$, then $p_l$ has only one global extension which does not fork over $A$, and this extension is $A$-invariant.

As a result, if we define:

$$L = \{l \text{ prime } | [M : lM] < \omega, C \notin A' + \bigcap_{N} l^N M\},$$

then the space $S$ of global extensions of $\mathrm{tp}(c/AB)$ which do not fork over $A$ is either empty, or homeomorphic to $S_1 \times \prod_{l \in L} S_{2,l}$.

Moreover, if $S \neq \emptyset$ and $L = \emptyset$, then all the elements of $S$ are $A$-invariant, thus $S \sim S_1$ is exactly the space of global $A$-invariant extensions of $\mathrm{tp}(c/AB)$.

We claim that this is also the case when $M$ is discrete, and we gave a sketch of the proof in ??.

## 6. EXAMPLES

### 6.1. Distinction between non-forking and finite satisfiability in DOAG.
Let us start with the very basic remark that the cut independence notion does not coincide with finite satisfiability in DOAG. In an elementary extension of the ordered group $\mathbb{R}$, let $\varepsilon > 0$, $c > 0$ so that $\Delta(c) < \Delta(\varepsilon) < \Delta(1)$. Let $A = \mathbb{Q}$, $B = \mathbb{Q} + \mathbb{Q}\varepsilon$. Then we clearly have $c \searrow_A^{\mathrm{cut}} B$, but the type of $c$ over $B$ is not finitely satisfiable over $A$: the $B$-definable set $]0, \varepsilon]$ contains $c$ and has no point in $A$.

### 6.2. Distinction between non-forking and the weak version of sequential independence in DOAG.
We would like to compare non-forking independence with a stronger independence notion:

**Definition 6.1.** Let $M$ be any first-order structure, and $A, B, C \subseteq M$ parameter sets. We will write $C \searrow_A^{\mathrm{seq}} B$ when, for any finite tuple $c$ from $C$, there exists a finite tuple $d = (d_0, \dots, d_n)$ from $\mathrm{acl}(AC)$ so that $\mathrm{acl}(Ac) \subseteq \mathrm{acl}(Ad)$ and $d_i \searrow_{Ad_{\varepsilon_i}}^f B$ for every $i \le n$.

This relation, which we will call sequential independence, is stronger than $\searrow^f$ in any theory (an easy application of left-transitivity of non-forking). One can show that it coincides with $\searrow^f$ in simple theories, and this is also the case in some non-simple theories, like DLO (a parameter set coincides with its algebraic closure, and it is easy to guess the global invariant extensions of the type of some finite tuple). In a theory where $\searrow^{\mathrm{seq}} = \searrow^f$, it is really easy to understand forking, because one can reduce the problem to forking in dimension one. We showed in the previous section that forking in ROAG is essentially a dimension one phenomenon, so it makes sense to ask whether $\searrow^{\mathrm{seq}} = \searrow^f$ in these structures. We will show that the answer is no with a counterexample in DOAG. The only other explicit example we know of
$\downarrow^f \neq \downarrow^{seq}$ is in ([?], example 13.6), in the theory of algebraically-closed valued fields.
This example does not occur exclusively in the value group, one need field-theoretic
tools to make it work, which is why our example is arguably less complicated.

In the reduct to the ordered group structure of an elementary extension of the
real-closed field $\mathbb{R}$, let $\varepsilon > 0$ so that $\Delta(\varepsilon) < \Delta(1)$. Let $A = \mathbb{Q}$, $B = \mathbb{Q} + \mathbb{Q}\sqrt{2} + \mathbb{Q}\sqrt{3}$,
and $c = (c_1, c_2) = (\sqrt{2} + \varepsilon, \sqrt{3} + \varepsilon\sqrt{2})$.

Let us show that we have $c \downarrow_A^f B$. We know that any two distinct real numbers
have a different cut over $\mathbb{Q} = A$, so any bounded $B$-definable closed interval that
does not have any point in $A$ must be a singleton. As $\mathrm{dcl}(Ac) \cap \mathrm{dcl}(B) = A$, we
have $c \downarrow_A^{\mathrm{cut}} B$. In particular, we have $c \downarrow_A^f B$.

Now let us show $c \downarrow_A^{seq} B$. Let $d = (d_1, \dots, d_n)$ be a finite tuple from $acl(Ac)$ so
that $acl(Ac) \subseteq acl(Ad)$. We can assume that the family $d$ is $\mathbb{Q}$-free over $A$, for if
$d_i \in \mathrm{dcl}(Ad_{<i})$, then we trivially have $d_i \downarrow_{Ad_{<i}}^f B$. We can also assume $d_i \in \mathbb{Q}c_1 + \mathbb{Q}c_2$,
for if $a \in A$, then $acl(Ad_{\leq i}) = acl(Ad_{<i}(d_i + a))$, so we can freely translate any $d_i$ by
an element of $A$. By basic linear algebra, we have $n = 2$, and $(d_1, d_2)$ is a $\mathbb{Q}$-base of
$\mathbb{Q}c_1 + \mathbb{Q}c_2$. Let $f_i$ be the unique map of $\mathrm{LC}^2(\mathbb{Q})$ such that $f_i(c_1, c_2) = d_i$. As $(c_1, c_2)$
is also $\mathbb{Q}$-free, $(f_1, f_2)$ must be $\mathbb{Q}$-free. For any $g \in \mathrm{LC}^2(\mathbb{Q})$, we have $d_2 - g(1, d_1) =
f_2(\sqrt{2}, \sqrt{3}) - g(1, f_1(\sqrt{2}, \sqrt{3})) + \varepsilon[f_2(1, \sqrt{2}) - g(0, f_1(1, \sqrt{2}))]$. As $(f_1, f_2)$ and
$(1, \sqrt{2}, \sqrt{3})$ are both $\mathbb{Q}$-free, we must have $f_2(\sqrt{2}, \sqrt{3}) - g(1, f_1(\sqrt{2}, \sqrt{3})) \neq 0$,
so $\Delta(d_2 - g(1, d_1)) = \Delta(1)$. We established that, for all $x \in \mathrm{dcl}(Ad_1)$, we have
$\Delta(d_2 - x) > \Delta(\varepsilon)$, which implies $\varepsilon \in H(d_2/\mathrm{dcl}(Ad_1))$. Now, if we had $d_2 \downarrow_{Ad_1}^f B$,
then we would necessarily have $\varepsilon \in H(d_2/\mathrm{dcl}(Bd_1))$ by ???. Let us show that this
fails. It is enough to show that $\Delta(d_2 - b) = \Delta(\varepsilon)$ for some $b \in \mathrm{dcl}(Bd_1)$. Choose
$b = d_1 - f_1(\sqrt{2}, \sqrt{3}) + f_2(\sqrt{2}, \sqrt{3})$. Then, as $d_i = f_i(c_1, c_2) = f_i(\sqrt{2}, \sqrt{3}) + f_i(\varepsilon, \varepsilon\sqrt{2})$,
we have $d_2 - b = f_2(\varepsilon, \varepsilon\sqrt{2}) - f_1(\varepsilon, \varepsilon\sqrt{2})$. As $(f_1, f_2)$ is $\mathbb{Q}$-free, $f_2 - f_1 \neq 0$, therefore
$d_2 - b$ is a non-trivial linear combination of $(\varepsilon, \varepsilon\sqrt{2})$. This latter family is clearly
$\Delta$-separated, thus $\Delta(d_2 - b) = \Delta(\varepsilon)$, and we can conclude.

This holds for any choice for the family $d$, so we have $c \downarrow_A^{seq} B$, which concludes
the example.

<figure><img src="image_23.png" /></figure>

**6.3. Distinction between the weak and the strong versions of sequential independence in DOAG.** In the definition of sequential independence, we require that there exists a good enumeration $d$ that witnesses independence. Does it change anything to ask this for all the enumerations? We will show that the answer is yes with a counterexample in DOAG.

**Definition 6.2.** Define $C \downarrow_A^{seq*} B$ when, for every finite tuple $c = (c_0..c_n)$ from $acl(AC)$, we have $c_i \downarrow_{Ac_{<i}}^f B$ for all $i \le n$.

This notion is at least as strong as $\downarrow^{seq}$, which is strictly stronger than $\downarrow^f$ in
DOAG by the above example. Let us show that $\downarrow^{seq*}$ is strictly stronger than $\downarrow^{seq}$
in DOAG. Let $\varepsilon > 0$, $c_2 > 0$ so that $\Delta(\varepsilon) < \Delta(c_2) < \Delta(1)$. Let $A = \mathbb{Q}$, $B = \mathbb{Q} + \mathbb{Q}\sqrt{2}$, and $c_1 = \sqrt{2} + \varepsilon$.

With the same reasoning as in the previous example, we have $c_1 c_2 \downarrow_A^{cut} B$. In particular, we have $c_1 \downarrow_A^f B$ and $c_2 \downarrow_A^f B$.

As $\Delta(c_2) \notin \Delta(\mathrm{dcl}(Ac_1))$, $c_2$ is ramified over $\mathrm{dcl}(Ac_1)$. The only points of $\mathrm{dcl}(Bc_1)$ that have the same cut over $\mathrm{dcl}(Ac_1)$ as that of $c_2$ are the $\lambda\varepsilon$, with $\lambda \in \mathbb{Q}_{>0}$. They are all smaller than $c_2$, so $c_2$ leans right with respect to $\mathrm{dcl}(Ac_1)$ and $\mathrm{dcl}(Bc_1)$, and $c_2 \downarrow_A^f B$, which implies $c_1 c_2 \downarrow_A^{seq} B$.

Let us show $c_1 c_2 \downarrow_A^{seq*} B$. The interval $[\sqrt{2}, \sqrt{2}+c_2]$ contains $c_1$ and has no point in $\mathrm{dcl}(Ac_2)$, which implies $c_1 \downarrow_{Ac_2}^f B$.

### 6.4. Distinction between forking and geometric cut-dependence in a dp-minimal poly-regular group.
Let us look at a different definition for cut independence:

**Definition 6.3.** Let $M$ be the expansion of some linearly ordered set, and $A, B, C$ parameter sets. We will write $C \downarrow_A^{cut*} B$ when there exists $c \in \mathrm{dcl}(AC)$ and $b_1$, $b_2 \in \mathrm{dcl}(AB)$ so that $b_1 \le c \le b_2$, $b_1 \equiv_A b_2$ and $b_i \notin \mathrm{dcl}(A)$.

We clearly have the implications $\downarrow^d \downarrow^{cut*}$ and:

$$C \downarrow_A^{cut*} B \iff \mathrm{div}(\mathrm{dcl}(AC)) \downarrow_{\mathrm{div}(\mathrm{dcl}(A))}^{cut} \mathrm{div}(\mathrm{dcl}(AB))$$

in any ordered Abelian group. It is not hard to observe that these implications are equalities in DOAG. In a dense model of ROAG $M$, if $B$ is an $|A|^+$-saturated model containing $A$, then one may show that the second implication is an equality, ie $C \downarrow_A^{cut*} B$ holds in $M$ if and only if $\mathrm{div}(C) \downarrow_{\mathrm{div}(A)}^{cut} \mathrm{div}(B)$ holds in $\mathrm{div}(M)$.

Whether the first implication is an equality is a little more complicated. Let $G$ be an ordered Abelian group, $A$ a parameter set, and $B$ an $|A|^+$-saturated model containing $A$. In case $G$ is regular and $|G|/|G|$ is infinite for some prime $l$ (this is possible, take for instance $G = \sum_{n<\omega} \mathbb{Z}r_n$, with $(r_n)_n$ a $\mathbb{Q}$-free subfamily of $\mathbb{R}$), the characterization of forking and dividing that we have in ?? implies that $\downarrow_A^d B$ does not coincide with $\downarrow_A^{cut*} B$, because we have partial types $\mathrm{mod} \equiv_l$ that divide over $A$. However, if $G$ is dp-minimal, we have indeed $\downarrow_A^f B = \downarrow_A^{cut*} B$. In fact, Simon showed in ([?], Proposition 2.5) that, for $c$ a singleton, we have $c \downarrow_A^f B \iff c \downarrow_A^{cut*} B$ if $G$ is dp-minimal, even when $G$ is not regular. However, we will see an example where $G$ is a dp-minimal, poly-regular, non-regular dense ordered Abelian group where $c \downarrow_A^{cut*} B$ (and thus $c \downarrow_A^f B$) holds in $G$, but $c \downarrow_{\mathrm{div}(A)}^{cut} \mathrm{div}(B)$ does not hold in $\mathrm{div}(G)$.

Let $G$ be any ordered group elementarily equivalent to the lexicographical product $\mathbb{Z}[\frac{1}{2}] \times \mathbb{Z}[\frac{1}{2}]$. Then $G$ is clearly dp-minimal, poly-regular and dense. In $\mathbb{Z}[\frac{1}{2}] \times \mathbb{Z}[\frac{1}{2}]$, we have $H_3((1,0)) = \{0\} \times \mathbb{Z}[\frac{1}{2}] \ne \{0\}$, so $G$ is not regular, and it
has exactly one proper non-trivial definable convex subgroup, let us call it $H(G)$. Let $A$ be a definably-closed parameter subset of $G$, $B$ an $|A|^+$-saturated elementary extension of $G$, and $M$ a $|B|^+$-saturated elementary extension of $B$. Let $c \in H(M)$ so that $c > H(B)$. Then we clearly have $c \downarrow_A^{cut*} B$, hence $c \downarrow_A^f B$ by [?]. Let us show $c \downarrow_{\text{div}(A)}^{cut} \text{div}(B)$ in $\text{div}(M)$ (in fact, we will even show $c \downarrow_{\text{div}(G)}^{cut} \text{div}(B)$). Let $b_1 \in H(B)$ so that $b_1 > H(G)$. Let $b_2 \in B$ so that $b_2 \notin H(B)$, but $b_2 < g$ for every $g \in G_{>0} \setminus H(G)$. Then the interval $I = [b_1, b_2]$ contains $c$.

<figure><img src="image_24.png" /></figure>

Let us show that $I$ does not have a point in $\text{div}(G)$. Let $g \in \text{div}(G)$ so that $0 < g < b_2$, and suppose by contradiction $g > b_1$. Let $0 < N < \omega$ so that $Ng \in G$. We have $Ng \ge g > b_1$, so $Ng \notin H(G)$. The ordered group $G/HG$ is dense (as $\mathbb{Z}[\frac{1}{2}]$ is), so there must exist $g' \in G$ so that $g' \notin H(G)$ and $Ng' < (Ng \mod H(G))$ (choose $N$ distinct cosets between $H(G)$ and $(Ng \mod H(G))$, choose $g'$ a representative of the coset that corresponds to the minimal distance between them in $G/H(G)$). As $g' \notin H(G)$ and $g' > 0$, we have $b_2 < g'$. However, we have $Ng' < Ng < Nb_2$, a contradiction. This concludes the example.

For future work, we should try to generalize our results to dp-minimal ordered groups. In such a group, one can hope that a global type over a monster model $B$ does not fork over $A$ if and only if its realizations $C$ satisfy $C \downarrow_A^{cut*} B$. The dimension one case has already been proved by Simon.

### 6.5. Distinction between forking and Presburger-quantifier-free dependence in an untame group.
For this last example, we will study an ordered Abelian group $G$ with definably-closed parameter sets $A \subseteq B$, and a singleton $c \in G$ for which $c \downarrow_{\text{div}(A)}^{cut} \text{div}(B)$ in $\text{div}(G)$ (hence $c \downarrow_A^{cut*} B$), and:

$$ (\text{dcl}(Ac) + l^N G) \cap (B + l^N G) = A + l^N G $$

for every prime $l$, $N > 0$, but $c \downarrow_A^d B$. The dividing definable set will not be built from the cuts or the cosets mod $l$, but from the chain of definable convex subgroups of $G$. Our example needs to have an infinite spine.

Let $\Gamma$ be the lexicographical product $\mathbb{Q}_{\ge 0} \times \omega$. Let $G$ be the ordered group of all Hahn series with coefficients in $\mathbb{Z}$ and powers in $\Gamma$. Then $G$ is a dense ordered Abelian group. For each $g \in G$, let $S(g) \subseteq \Gamma$ be the support of $g$, and $v(g) \in \Gamma \cup \{+\infty\}$ be the least element of $S(g)$. We have $\Delta(g) < \Delta(h) \iff v(g) > v(h)$. Each $g \in G$ can be written $\sum_{\delta \in S(g)} \lambda_{\delta} t^{\delta}$, with $\lambda_{\delta} \in \mathbb{Z}$, and $t$ an indeterminate. If $\sigma$ is an automorphism of the ordered set $\mathbb{Q}_{\ge 0}$, then let $\bar{\sigma} \in \text{Aut}(G)$ be defined as $\sum_{(\mu,n) \in \Gamma} \lambda_{(\mu,n)} t^{(\mu,n)} \mapsto \sum_{(\mu,n) \in \Gamma} \lambda_{(\mu,n)} t^{(\sigma(\mu),n)}$. Let $A = \{a \in G | S(a) \subseteq (\{0\} \times \omega)\}$ and $B = \text{dcl}(A \cup \{t^{(1,0)}\})$. The action over $\mathbb{Q}_{\ge 0}$ of its automorphism group has only two orbits: $\{0\}$ and $\mathbb{Q}_{>0}$. As a result, we must have $\text{dcl}(\emptyset) \subseteq A$. By ([?], Corollary 1.10), $\text{dcl}(A)$ is exactly the relative divisible closure in $G$ of $A$, in other words $A$ is definably-closed. With the same argument, we have $B = A + \mathbb{Z}t^{(1,0)}$. Let $c = t^{(1,1)}$,
so dcl($Ac$) = $A + \mathbb{Z}c$. We have $\Delta(c) < \Delta(B_{\ge 0})$, so $c$ clearly leans left with respect to $div(A)$ and $div(B)$ in $div(G)$, which implies $c \downarrow_A^{cut} B$ in $div(G)$. Moreover, for every prime $l$, and $N > 0$, the elements of $l^N G$ are exactly the series $\sum_{\delta} l^N \lambda_\delta t^\delta$ with $\lambda_\delta \in \mathbb{Z}$, so we clearly have $(dcl(Ac) + l^N G) \cap (B + l^N G) = A + l^N G$.

<figure><img src="image_25.png" /></figure>

Now let us prove $c \downarrow_A^d B$. For each $n < \omega$, let $\sigma_n \in \text{Aut}(\mathbb{Q}_{\ge 0})$ so that $\sigma_n(1) = n$. For each prime $l$, for each $g = \sum_{\delta} \lambda_\delta t^\delta \in G$, if $g \notin l^N G$, then one can show (see for instance [?], example 4.2) that $H_l^N(g)$ is the group of elements of $G$ with an Archimedean class smaller or equal to $\Delta(t^\delta)$, where $\delta$ is the least element of $S(g)$ for which $\lambda_\delta \notin l^N \mathbb{Z}$. As a result, there does not exist $g \in G$ for which $H_2(c) < H_2(g) < H_2(t^{(1,0)})$. The following $B$-definable set which contains $c$:

$$X = \{g \in G | g \notin 2G \text{ and } (\forall h \in G, (H_2(h) \le H_2(g) \text{ or } H_2(t^{(1,0)}) \le H_2(h)))\}$$

divides over $A$, because the $(\sigma_n(X))_n$ are pairwise-disjoint. This concludes the example.

## CONTENTS

<table><tr><td>Introduction</td><td>1</td></tr><tr><td>1. Preliminaries</td><td>5</td></tr><tr><td>1.1. Cut-independence</td><td>5</td></tr><tr><td>1.2. Valuations</td><td>8</td></tr><tr><td>1.3. Outline of the proof</td><td>9</td></tr><tr><td>1.4. Relation to Dolich-independence</td><td>12</td></tr><tr><td>2. How to glue types via ad-hoc valuations</td><td>12</td></tr><tr><td>2.1. Basic definitions and classification of the cuts</td><td>12</td></tr><tr><td>2.2. Weak orthogonality</td><td>17</td></tr><tr><td>2.3. Tensor product</td><td>21</td></tr><tr><td>3. How to build a block extension</td><td>25</td></tr><tr><td>3.1. The Archimedean case</td><td>25</td></tr><tr><td>3.2. Archimedean groups</td><td>27</td></tr><tr><td>3.3. The ramified case</td><td>28</td></tr><tr><td>3.4. Gluing everything together</td><td>30</td></tr><tr><td>4. Normal forms</td><td>34</td></tr><tr><td>5. Regular ordered Abelian groups</td><td>39</td></tr><tr><td>5.1. Quantifier-free types in the Presburger language</td><td>39</td></tr><tr><td>5.2. Basic properties of regular groups</td><td>40</td></tr><tr><td>5.3. Invariance and boundedness of global partial types</td><td>43</td></tr><tr><td>5.4. Main result</td><td>52</td></tr><tr><td>6. Examples</td><td>53</td></tr><tr><td>6.1. Distinction between non-forking and finite satisfiability in DOAG</td><td>53</td></tr><tr><td>6.2. Distinction between non-forking and the weak version of sequential independence in DOAG</td><td>54</td></tr><tr><td>6.3. Distinction between the weak and the strong versions of sequential independence in DOAG</td><td>55</td></tr></table>
6.4. Distinction between forking and geometric cut-dependence in a $dp$-minimal poly-regular group 55

6.5. Distinction between forking and Presburger-quantifier-free dependence in an untame group 57

References 59
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DÉPARTEMENT DE MATHÉMATIQUES D'ORSAY, UNIVERSITÉ PARIS-SACLAY, FRANCE