---
abstract: 'In this paper, the connections between model theory and the theory of infinite permutation groups (see [@Au]) are used to study the $n$-existence and the $n$-uniqueness for $n$-amalgamation problems of stable theories. We show that, for any $n\geq 2$, there exists a stable theory having $(k+1)$-existence and $k$-uniqueness, for every $k\leq n$, but has neither $(n+2)$-existence nor $(n+1)$-uniqueness. In particular, this generalizes the example, for $n=2$, due to E.Hrushovski given in [@PKM].'
author:
- |
    Elisabetta Pastori\
    Dipartimento di Matematica,\
    Università degli Studi di Torino,\
    Via Carlo Alberto, 10 10123\
    Torino, Italy\
    [elisabetta.pastori@unito.it]{}\
    Pablo Spiga\
    Centre for Mathematics of Symmetry and Computation\
    The University of Western Australia\
    35 Stirling Highway\
    Crawley WA 6009\
    Australia\
    [spiga@maths.uwa.edu.au]{}
title: 'Failure of $n$-uniqueness: a family of examples'
---

Introduction {#sec0}
============

Considerable work (e.g. [@BK], [@PKM], [@EO], [@Hr], [@She]) has explored higher amalgamation properties for stable and simple theories. In this paper we analyze uniqueness and existence properties for a countable family of stable theories. In contrast to previous methods our approach uses group-theoretic techniques. We begin by giving some basic definitions.

Let $T$ be a complete and simple $L$-theory with quantifier elimination. We denote by $\mathcal{C}_T$ the category of algebraically closed substructures of models of $T$ with embeddings as morphisms. Also, given $n\in \mathbb{N}$, we denote by $P(n)$ the partially ordered set of all subsets of $\{1, \dots ,n\}$ and by $P(n)^-$ the set $P(n)\setminus \{1,\dots,n\}$.

An *$n$-amalgamation problem* over $\operatorname{acl}(\emptyset)$ is a functor $a:P(n)^-\rightarrow \mathcal{C}_T$ such that

$(i)$

:   $a(\emptyset)=\operatorname{acl}(\emptyset)$;

$(ii)$

:   whenever $s_1,s_2,s_3\in P(n)^-$ and $(s_1\cap
      s_2)\subset s_3$, the algebraically closed sets $a(s_1),a(s_2)$ are independent over $a(s_1\cap s_2)$ within $a(s_3)$;

$(iii)$

:   $a(s)=\operatorname{acl}\{a(i)\, |\, i\in s\}$, for every $s\in P(n)^-$.

In here we denote by $\operatorname{acl}(A)$ the algebraic closure of $A$ in $T^{\textrm{eq}}$. We recall that the objects of $P(n)^{-}$ (viewed as a category) are simply the elements of $P(n)^{-}$. Also, the morphisms of $P(n)^{-}$ are the inclusions $\iota_{s,t}:s\hookrightarrow t$, for every $s,t\in P(n)^{-}$ with $s\subseteq t$. In particular, an $n$-amalgamation problem assigns a morphism $$a_{s,t}:a(s)\to a(t),$$ to every $s,t\in P(n)^{-}$ with $s\subseteq t$. The morphism $a_{s,t}$ is called *transition map* and, by functoriality, we have $$a_{s_2,s_3}\circ a_{s_1,s_2}=a_{s_1,s_3},$$ for every $s_1,s_2,s_3\in
P(n)^{-}$ with $s_1\subseteq s_2\subseteq s_3$. By definition, the morphisms in $\mathcal{C}_T$ are the embeddings, that is, $a_{s,t}$ is the restriction of an automorphism to the algebraically closed substructure $a(s)$.

A *solution* of $a$ is a functor $\bar{a}: P(n)\rightarrow 
 \mathcal{C}_T$ extending $a$ to the full power set $P(n)$ and satisfying the conditions $(i),(ii),(iii)$ (i.e. including the case $s=\{1,\dots, n\}$). In particular, in order to find a solution of $a$, we need to determine $n$ embeddings $$f_i:a(\{1,\ldots,n\}\setminus\{i\})\longrightarrow
 a(\{1,\ldots,n\})=\operatorname{acl}(\{a(i)\mid i \in \{1,\ldots,n\}\}),$$ (for $1\leq i\leq n$) compatible with $a$, that is,

$$f_i\circ a_{s,\{1,\ldots,n\}\setminus\{i\}}=f_j\circ
 a_{s,\{1,\ldots,n\}\setminus\{j\}}$$ for every $i,j\in \{1,\ldots,n\}$ and $s\subseteq
\{1,\ldots,n\}\setminus\{i,j\}$.

The theory $T$ is said to have *$n$-existence* (over $\operatorname{acl}(\emptyset)$) if every $n$-amalgamation problem over $\operatorname{acl}(\emptyset)$ has at least one solution. Similarly, we shall say that the theory $T$ has *$n$-uniqueness* (over $\operatorname{acl}(\emptyset)$) if every $n$-amalgamation problem over $\operatorname{acl}(\emptyset)$ has at most one solution up to isomorphism (for more details see [@Hr] and [@Ko]).

It is a well known fact that every simple theory has $2$-existence, by the presence of non-forking extensions. Moreover, if the theory is stable, then, by stationarity of strong types, $2$-uniqueness holds. Consequentially, also $3$-existence holds (for a proof see Lemma $3.1$ of [@Hr]). However, $3$-uniqueness and $4$-existence can fail for a general stable theory. Indeed, in [@PKM], the authors thank E. Hrushovski for supplying an example of a stable theory which has neither $4$-existence nor $3$-uniqueness. The example is the following. Its construction involves a finite cover (for more details about finite covers see [@EMI]).

\[esempio1\]Let $\Omega$ be a countable set, $[\Omega]^2$ the set of $2$-subsets of $\Omega$, and $C=[\Omega]^2\times {{\mathbb Z}}/2{{\mathbb Z}}$. Also let $E\subseteq \Omega\times [\Omega]^2$ be the membership relation, and let $P$ be the subset of $C^3$ such that $((w_1,\delta_1),(w_2,\delta_2),(w_3,\delta_3))$ lies in $P$ if and only if there are distinct $c_1,c_2,c_3\in \Omega$ such that $w_1=\{c_2,c_3\},w_2=\{c_1,c_3\},w_3=\{c_1,c_2\}$ and $\delta_1+\delta_2+\delta_3=0$. Now let $M$ be the model with the $3$-sorted universe $\Omega,[\Omega]^2,C$ and equipped with relations $E,P$ and projection on the first coordinate $\pi:C\rightarrow
    [\Omega]^2$. Since $M$ is a reduct of $(\Omega,
    {{\mathbb Z}}/2{{\mathbb Z}})^{\textrm{eq}}$, we get that $T=\operatorname{Th}(M)$ is stable. It is shown in [@PKM] that $T$ has neither $4$-existence nor $3$-uniqueness.

In this paper we generalize this example. We summarize our main results in the following theorem.

\[summary\] For any $n\geq 2$, there exists a stable theory $T_n$ such that $T_n$ has $(k+1)$-existence and $k$-uniqueness for any $k\leq n$, but $T_n$ has neither $(n+2)$-existence nor $(n+1)$-uniqueness.

Also in Proposition \[coincides\] we prove that, for $n=2$, the stable theory $T_2$ given in Theorem \[summary\] coincides with the theory in Example \[esempio1\].\
All the material we present is expressed in a purely algebraic terminology. Indeed, the problem of $n$-uniqueness for a theory has also a natural formulation in terms of permutation groups, as is shown in [@Hr Proposition $3.5$]. We adopt this approach here.

In Section 2, we introduce certain permutation modules which will be used to construct the automorphism groups of the countable $\aleph_0$-categorical structures $M_n$ on which is based Theorem \[summary\].

As is clear from the definition, the study of amalgamation problems requires a precise understanding of the algebraic closure in $T^{\textrm{eq}}$. Since the structures $M_n$ are countable and $\aleph_0$-categorical, the algebraic closure can be rephrased with group theoretic terminology: it can be determined by studying certain closed subgroups of the automorphism group of $M_n$. This is done in Section 3 and Section 4.

The $\operatorname{Sym}(\Omega)$-submodule structure of $\mathbb{F}^{[\Omega]^n}$ {#sec2}
=================================================================================

We begin by reviewing some definitions and basic facts about permutation groups and permutation modules.

If $C$ is a set, then the symmetric group $\operatorname{Sym}(C)$ on $C$ can be considered as a topological group. The open sets in this topology are arbitrary unions of cosets of pointwise stabilizers of finite subsets of $C$. A subgroup $\Gamma$ of $\operatorname{Sym}(C)$ is closed if and only if each element of $\operatorname{Sym}(C)$ which preserves all the orbits of $\Gamma$ on $C^n$, for all $n\in {{\mathbb N}}$, is in $\Gamma$. It is well known that closed subgroups in this topology are precisely automorphism groups of first-order structures on $C$, see [@Ca Theorem $5.7$] or [@Au].

Throughout the sequel we denote by ${{\mathbb F}}$ a field, ${{\mathbb F}}_2$ the integers modulo $2$, $\Omega$ a countable set and $[\Omega]^n$ the set of $n$-subsets of $\Omega$.

The natural action of the symmetric group $\operatorname{Sym}(\Omega)$ on $[\Omega]^n$ turns $\mathbb{F}[\Omega]^n$, the vector space over ${{\mathbb F}}$ with basis consisting of the elements of $[\Omega]^n$, into a $\operatorname{Sym}(\Omega)$-module. We will characterize the submodules of $\mathbb{F}[\Omega]^n$ in terms of certain $\operatorname{Sym}(\Omega)$-homomorphisms. The following definition is based on concepts first introduced in [@Jam].

\[beta map\] If $0\leq j\leq n$, then the map $\beta_{n,j}:\mathbb{F}[\Omega]^n\to
\mathbb{F}[\Omega]^j$, given by $$\beta_{n,j}(\omega)=\sum_{\omega'\in[\omega]^j}\omega'\qquad
(\textrm{for }\omega\in[\Omega]^n)$$ and extended linearly to $\mathbb{F}[\Omega]^n$, is a $\operatorname{Sym}(\Omega)$-homomorphism (in here we denote by $[\omega]^j$ the set of $j$-subsets of $\omega$).

It is shown in [@Gr] (see also [@Jam]) that the submodules of $\mathbb{F}[\Omega]^n$ are completely determined by the maps $\beta_{n,j}$. Indeed, it is proved in [@Gr Corollary $3.17$] that every submodule $U$ of $\mathbb{F}[\Omega]^n$ is an intersection of kernels of $\beta$-maps, i.e. $U=\cap_{j\in S}\ker\beta_{n,j}$ for some subset $S$ of $\{0,\ldots,n\}$.

Using the controvariant Pontriagin duality we have that the dual module of $\mathbb{F}[\Omega]^n$ is ${{\mathbb F}}^{[\Omega]^n}$, i.e. the set of functions from $[\Omega]^n$ to ${{\mathbb F}}$. We recall that ${{\mathbb F}}^{[\Omega]^n}$ has a natural faithful action on $[\Omega]^n\times {{\mathbb F}}$ given by $(w,\delta)^f=(w,f(w)+\delta)$. Hence, ${{\mathbb F}}^{[\Omega]^n}$, endowed with the relative topology, becomes a topological $\operatorname{Sym}(\Omega)$-module and a profinite subgroup of $\operatorname{Sym}([\Omega]^n\times {{\mathbb F}})$. Also, given any map $\beta_{n,j}:\mathbb{F}[\Omega]^n \rightarrow   \mathbb{F}[\Omega]^j 
$, there is a natural dual continuous $\operatorname{Sym}(\Omega)$-homomorphism $\beta^\ast_{n,j}:\mathbb{F}^{[\Omega]^j}\rightarrow \mathbb{F}^{[\Omega]^n}$ defined by $$(\beta^\ast_{n,j}f)(\omega)=\sum_{x\in [\omega]^j}f(x).$$

Now, the lattice of the closed submodules of $\mathbb{F}^{[\Omega]^n}$ is the dual of the lattice of the submodules of $\mathbb{F}[\Omega]^n$. We point out that using the algorithm described in [@Gr Section $5$], the lattice of the closed submodules of ${{\mathbb F}}^{[\Omega]^n}$ can be easily computed. Here we record the following fact that we are frequently going to use.

\[exact\] For $n\geq 1$, $\mathbb{F}=\mathbb{F}_p$ with $p>0$, we have $\operatorname{im}\beta_{n,n-1}^\ast=\operatorname{ker}\beta_{n+1,n}^\ast$.

*Proof.* The submodule $\operatorname{im}\beta_{n+1,n}$ of $\mathbb{F}[\Omega]^{n}$ is of the form $\cap_{j\in S}\operatorname{ker}\beta_{n,j}$, for some subset $S$ of $\{0,\ldots,n\}$. By [@Gr Proposition $3.19$], we have that $\operatorname{im}\beta_{n+1,n}\subseteq \operatorname{ker}\beta_{n,j}$ if and only if $2$ divides $n+1-j$. Therefore $S=\{j \mid  2 \textrm{ divides } n+1-j\}$.

Also by [@Gr Proposition $4.1$], we have that if $2$ divides $n+1-j$, then $\operatorname{ker}\beta_{n,n-1}\subseteq \operatorname{ker}\beta_{n,j}$. This yields $\operatorname{im}\beta_{n+1,n}=\cap_{j\in S}\operatorname{ker}\beta_{n,j}= \operatorname{ker}\beta_{n,n-1}$. In particular, the sequence

$$\xymatrix{
\mathbb{F}[\Omega]^{n+1}
\ar@{->}[rr]^{\beta_{n+1,n}}
&&\mathbb{F}[\Omega]^n
\ar@{->}[rr]^{\beta_{n,n-1}}
&&\mathbb{F}[\Omega]^{n-1}}$$

is exact.

Now the Pontriagin duality is an exact controvariant functor on the sequences of the form $A\to B\to C$. This says that $\operatorname{im}\beta_{n,n-1}^\ast=\operatorname{ker}\beta_{n+1,n}^\ast$.

Closed submodules of finite index in $\mathbb{F}_2^{[\Omega]^n}$  {#sec3}
=================================================================

If $A$ is a finite subset of $\Omega$, then we write simply $\operatorname{Sym}(\Omega\setminus A)$ for the subgroup of $\operatorname{Sym}(\Omega)$ fixing pointwise $A$. In this section we study the closed $\operatorname{Sym}(\Omega\setminus A)$-submodules of $\mathbb{F}_2^{[\Omega]^{n-1}}$ of finite index. We start by considering the case $A=\emptyset$.

\[finite index\] If $n\geq 1$, then $\mathbb{F}_2^{[\Omega]^n}$ has no proper closed $\operatorname{Sym}(\Omega)$-submodule of finite index.

*Proof.*Let $K$ be a closed submodule of ${{\mathbb F}}_2^{[\Omega]^n}$ of finite index. Then, ${{\mathbb F}}_2^{[\Omega]^n}/K$ is a finite $\operatorname{Sym}(\Omega)$-module. Since $\operatorname{Sym}(\Omega)$ has no proper subgroup of finite index, we get that $\operatorname{Sym}(\Omega)$ centralizes ${{\mathbb F}}_2^{[\Omega]^n}/K$. It follows that $f^\sigma -f\in K$, for every $\sigma\in  \operatorname{Sym}(\Omega)$.

Let $L$ be the annihilator of $K$ in $\mathbb{F}_2[\Omega]^n$, i.e. $L=\{w\in \mathbb{F}_2[\Omega]^n\mid  g(w)=0 \textrm{ for every
}g\in K\}$. Since $K$ is a closed $\operatorname{Sym}(\Omega)$-submodule, the set $L$ is a $\operatorname{Sym}(\Omega)$-submodule of $\mathbb{F}_2[\Omega]^n$. Now, let $f$ be in $\mathbb{F}_2^{[\Omega]^n}$, $\sigma$ in $\operatorname{Sym}(\Omega)$ and $w$ in $L$. We get $$0=(f^\sigma-f)(w)=f^{\sigma}(w)-f(w)=f(w^{\sigma^{-1}}-w).$$ This says that $w^{\sigma^{-1}}-w$ is annihilated by every element of $\mathbb{F}_2^{[\Omega]^n}$. Therefore, $w^{\sigma^{-1}}-w=0$ and $\sigma$ centralizes $w$. This shows that $\operatorname{Sym}(\Omega)$ centralizes $L$. Since $n\geq 1$, the only element of $\mathbb{F}_2[\Omega]^n$ centralized by $\operatorname{Sym}(\Omega)$ is the zero vector. Hence $L=0$ and, by the Pontriagin duality, $K=\mathbb{F}_2^{[\Omega]^n}$.

In the forthcoming analysis we shall denote finite subsets of $\Omega$ by capital letters, while the elements of $[\Omega]^n$ will be generally denoted by lower cases.

Now, let $A$ be a finite subset of $\Omega$. To describe the closed $\operatorname{Sym}(\Omega\setminus A)$-submodules of $\mathbb{F}_2^{[\Omega]^{n-1}}$ of finite index we have to introduce some notation. Let $B$ be a subset of $A$. We denote by $V_{B,A}$ the $\operatorname{Sym}(\Omega\setminus A)$-submodule of ${{\mathbb F}}_2^{[\Omega]^{n-1}}$ defined by

$$\label{eq:1}
 V_{B,A} = \{f\in {{\mathbb F}}_2^{[\Omega]^{n-1}}  \mid f(w)=0 \, \, \forall \, w\in [\Omega]^{n-1} \textrm{ with } w\cap A\neq B\}$$

and we denote by $V_A$ the $\operatorname{Sym}(\Omega\setminus A)$-submodule of ${{\mathbb F}}_2^{[\Omega]^{n-1}}$ defined by $$\begin{aligned}
\label{eq:2}
V_A&=&\bigoplus_{
\small
B\subseteq A,
|B|< n-1
\normalsize
}V_{B,A}.\end{aligned}$$

In the following lemma we describe the elements of $V_A$.

\[eltsVA\]Let $A$ be a finite subset of $\Omega$. Then $$\label{eq:3}
V_A=\{f\in{{\mathbb F}}_2^{[\Omega]^{n-1}}\mid f(w)=0\textrm{
    for every }w\in [A]^{n-1}\}.$$

*Proof.* We denote by $W$ the vector space on the right hand side of Equation $(\ref{eq:3})$. We start by proving that $V_A\subseteq W$. Let $B$ be a subset of $A$ with $|B|<n-1$ and $f$ be in $V_{B,A}$. Consider $w$ in $[A]^{n-1}$. Since $|B|<n-1$, $|w|=n-1$ and $w\subseteq A$, we have $w\cap A=w\neq 
B$. By Equation $(\ref{eq:1})$, we get $f(w)=0$. This implies $f\in W$ and so $V_{B,A}\subseteq W$. Thence, by Equation $(\ref{eq:2})$, we obtain $V_A\subseteq W$.

Conversely, we prove that $W\subseteq V_A$. Let $f$ be in $W$. For every subset $B$ of $A$ with $|B|<n-1$ define $$f_B(w)=\left\{
\begin{array}{ccl}
f(w)&&\textrm{if }w\cap A=B,\\
0 &&\textrm{if }w\cap A\neq B.
\end{array}
\right.$$ Clearly, $f_B\in {{\mathbb F}}_2^{[\Omega]^{n-1}}$ and, by Equation $(\ref{eq:1})$, $f_B\in V_{B,A}$. Let $w$ be in $[\Omega]^{n-1}$ with $w\nsubseteq
A$. Since $|w\cap A|<n-1$, we have $$\begin{aligned}
\left(\sum_{
\small
B\subseteq A,
|B|< n-1
\normalsize
}f_B\right)(w)&=&\sum_{
\small
B\subseteq A,
|B|< n-1
\normalsize
}f_B(w)=f_{w\cap A}(w)=f(w).\end{aligned}$$ Similarly, let $w$ be in $[\Omega]^{n-1}$ with $w\subseteq  A$ (that is, $w\in [A]^{n-1}$). As $f\in W$, we have $f(w)=0$. Also, by definition of $f_B$, we obtain $f_B(w)=0$. This shows that $f=\sum_{B\subseteq 
  A,|B|<n-1}f_B$. By Equation $(\ref{eq:2})$, it follows that $f\in V_A$.

\[sommadiretta\] Let $A$ be a finite subset of $\Omega$. For each $B\subseteq A$, the $\operatorname{Sym}(\Omega\setminus 
A)$-modules $V_{B,A}$ are closed submodules of ${{\mathbb F}}_2^{[\Omega]^{n-1}}$. Moreover, $$\label{eq:1111}
{{\mathbb F}}_2^{[\Omega]^{n-1}}=\bigoplus_{
\small
B\subseteq A,
|B|\leq n-1
\normalsize
}V_{B,A}$$ and each $V_{B,A}$ is $\operatorname{Sym}(\Omega\setminus A)$-isomorphic to ${{\mathbb F}}_2^{[\Omega\setminus A]^{n-1-|B|}}$.

*Proof.*Since $V_{B,A}$ is an intersection of pointwise stabilizers of finite sets of $[\Omega]^{n-1}\times
  {{\mathbb F}}_2$, it is closed in ${{\mathbb F}}_2^{[\Omega]^{n-1}}$. It is straightforward to verify the remaining statements.

\[primainclusione\] Let $A$ be a finite subset of $\Omega$. The module $V_A$ has finite index in $\mathbb{F}_2^{[\Omega]^{n-1}}$. Also, if $V$ is a closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of ${{\mathbb F}}_2^{[\Omega]^{n-1}}$ of finite index, then $V_A\subseteq V$.

*Proof.* By Equations $(\ref{eq:2})$ and $(\ref{eq:1111})$, we have that $\mathbb{F}_2^{[\Omega]^{n-1}}/V_A$ is isomorphic to $\oplus_{|B|=n-1}V_{B,A}$, which has dimension ${|A|\choose
    n-1}$. Therefore $V_A$ has finite index in $\mathbb{F}_2^{[\Omega]^{n-1}}$.

Let $V$ be a closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of $\mathbb{F}_2^{[\Omega]^{n-1}}$ of finite index. Let $B\subseteq A$ with $|B|<n-1$. By Lemma \[sommadiretta\], $V_{B,A}$ is $\operatorname{Sym}(\Omega\setminus A)$-isomorphic to ${{\mathbb F}}_2^{[\Omega\setminus
      A]^{n-1-|B|}}$. Since $\lbrack V_{B,A}:V_{B,A}\cap V \rbrack=\lbrack
  V_{B,A}+V: V\rbrack$ is finite, we have that $V_{B,A}\cap V$ has finite index in $V_{B,A}$. Now, by Lemma \[finite index\], the module $V_{B,A}$ does not have any proper closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of finite index. Therefore $V_{B,A}=V_{B,A}\cap V$ and $V_{B,A}\subseteq V$. By definition of $V_A$ in Equation $(\ref{eq:2})$, we get $V_A\subseteq V$.

In the following lemma we describe the elements of $V_A+\operatorname{ker}\beta_{n,n-1}^\ast$.

\[char\]Let $A$ be a finite subset of $\Omega$. We have $V_A+\operatorname{ker}\beta_{n,n-1}^\ast=\{f\in
  \mathbb{F}_2^{[\Omega]^{n-1}}\mid (\beta_{n,n-1}^\ast f)(w)=0
  \textrm{ for every }w\in [A]^n\}$.

*Proof.*If $n=1$, then the equality is clear. So assume $n\geq 2$.

By Lemma \[eltsVA\], the elements of $V_A$ are the functions $f\in 
  \mathbb{F}_2^{[\Omega]^{n-1}}$ vanishing on each element of $
         [A]^{n-1}$. Now, if $f_1\in V_A$, $f_2\in \operatorname{ker}\beta_{n,n-1}^\ast$ and $w\in [A]^n$, then $$(\beta_{n,n-1}^\ast (f_1+f_2))(w)=(\beta_{n,n-1}^\ast
         f_1)(w)=\sum_{w'\in[w]^{n-1}}f_1(w')=0.$$

Therefore, it remains to prove that if $f\in
\mathbb{F}_2^{[\Omega]^{n-1}}$ and $(\beta_{n,n-1}^\ast f)(w)=0$ for every $w\in [A]^{n}$, then $f\in V_A+\operatorname{ker}\beta_{n,n-1}^\ast $. Let $a$ be a fixed element of $A$ and let $g\in \mathbb{F}_2^{[\Omega]^{n-2}}$ be the function defined by $$g(\omega)=\left\{
\begin{array}{cll}
f(\omega\cup \{a\})&&\textrm{if }\omega\subseteq A \textrm{ and
}a\notin \omega,\\ 
0&&\textrm{otherwise }.
\end{array}
\right.$$ Set $f_2=\beta_{n-1,n-2}^\ast g$. By Proposition \[exact\], we have that $f_2\in \operatorname{im}\beta_{n-1,n-2}^\ast=\operatorname{ker}\beta_{n,n-1}^\ast$. Set $f_1=f-f_2$. We claim that $f_1$ lies in $V_A$, from which the lemma follows. By Lemma \[eltsVA\], it suffices to prove that $f_1(w')=0$ for every $w'\in [A]^{n-1}$. Let $w'$ be in $[A]^{n-1}$. Assume $a\in w'$. By the definition of $g$, we have $$f_2(w')=(\beta_{n-1,n-2}^\ast g)(w')=\sum_{\omega\in
  [w']^{n-2}}g(\omega)=g(w'\setminus \{a\})=f(w')$$ and $f_1(w')=0$. Now assume $a\notin w'$. By the definition of $g$ and by the hypothesis on $f$, we have $$\begin{aligned}
f_2(w')&=&(\beta_{n-1,n-2}^\ast g)(w')=\sum_{\omega\in
  [w']^{n-2}}g(\omega)=\sum_{\omega\in [w']^{n-2}}f(\omega\cup
\{a\})\\
&=&\sum_{x\in
  [w'\cup\{a\}]^{n-1}}\!\!\!\!\!\!f(x)+f(w')=(\beta_{n,n-1}^\ast
f)(w'\cup\{a\})+f(w')=f(w'), \end{aligned}$$ and $f_1(w')=0$.

\[defn\] We write $W_A$ for $\beta_{n,n-1}^\ast(V_A)$, with $V_A$ as in Equation $(\ref{eq:2})$.

Now, using the previous lemmas we describe the closed $\operatorname{Sym}(\Omega\setminus  A)$-submodules of $\operatorname{im}\beta_{n,n-1}^\ast$ of finite index.

\[uniqueimage\]Let $A$ be a finite subset of $\Omega$. The module $W_A$ is the unique minimal closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of $\operatorname{im}\beta_{n,n-1}^\ast$ of finite index. Furthermore, $W_A=\{g\in \operatorname{im}\beta_{n,n-1}^\ast\mid g(w)=0 \textrm{ for every } w\in
  [A]^n\}$.

*Proof.*Let $W$ be a closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of $\operatorname{im}\beta_{n,n-1}^\ast$ of finite index. By the first isomorphism theorem $W$ is the image via $\beta_{n,n-1}^\ast$ of some closed $\operatorname{Sym}(\Omega\setminus A)$-submodule $V$ of $\mathbb{F}_2^{[\Omega]^{n-1}}$ of finite index. Now, by Lemma \[primainclusione\], we get $V_A\subseteq V$. So $\beta_{n,n-1}^\ast(V_A)\subseteq \beta_{n,n-1}^\ast(V)= 
  W$. Hence, $W_A=\beta_{n,n-1}^\ast(V_A)$ is the unique minimal closed $\operatorname{Sym}(\Omega\setminus A)$-submodule of $\operatorname{im}\beta_{n,n-1}^\ast$ of finite index.

Now, from Lemma \[char\] the rest of the proposition is immediate.

The infinite family of examples {#sec1}
===============================

Before introducing our examples, we need to set some auxiliary notation.

\[auxiliary\] Let $M$ be a structure and $A,B$ subsets of $M$. We denote by $\overline{\operatorname{Aut}(A/B)}$ the subgroup of $\operatorname{Aut}(M)$ fixing setwise $A$ and fixing pointwise $B$. The setwise stabilizer of $A$ in $\operatorname{Aut}(M)$ will be denoted by $\operatorname{Aut}(M)_{\{A\}}$, while the permutation group induced by $\overline{\operatorname{Aut}(A/B)}$ on $A$ will be denoted by $\operatorname{Aut}(A/B)$.

Let $n\geq 2$ be an integer and $\Omega$ be a countable set.

\[OURdefinition\]We consider $M_n$ the multisorted structure with sorts $\Omega$, $[\Omega]^n$ and $[\Omega]^n\times \mathbb{F}_2$ and with automorphism group $\operatorname{im}\beta^\ast_{n,n-1}\rtimes \operatorname{Sym}(\Omega)$. Note that this is well-defined as $\operatorname{im}\beta_{n,n-1}^\ast$ is a closed submodule of $\mathbb{F}_2^{[\Omega]^n}$.

Moreover, the theory $T_n=\operatorname{Th}(M_n)$ is stable (see Section \[coincidence\]).

In the next paragraph we introduce some notation that would be useful to describe the algebraically closed sets of $M_n$.

Denote by $\pi:[\Omega]^n\times \mathbb{F}_2\to [\Omega]^n$ the projection on the first coordinate. Given $A$ a finite subset of $M_n$, we have that $A$ is of the form $A_1\cup A_2\cup A_3$, where $A_1$ belongs to the sort $\Omega$, $A_2$ belongs to the sort $[\Omega]^n$ and $A_3$ belongs to the sort $[\Omega]^n\times \mathbb{F}_2$. Consider $\tilde{A_2}\subseteq 
\Omega$ the union of the elements in $A_2$ and $\tilde{A_3}\subseteq
\Omega$ the union of the elements in $\pi(A_3)$. We define the *support* of $A$, written ${\mathop{\mathrm{supp}}}(A)$, to be the subset $A_1\cup
\tilde{A_2}\cup \tilde{A_3}$ of $\Omega$. Finally, we define $\operatorname{cl}(A)$ to be the subset of $M_n$ $$\operatorname{cl}(A):={\mathop{\mathrm{supp}}}(A)\cup 
 [{\mathop{\mathrm{supp}}}(A)]^n\cup  ([{\mathop{\mathrm{supp}}}(A)]^n\times \mathbb{F}_2)$$

In the rest of this section we describe the algebraically closed sets in the structure $M_n$. Here we consider structures *up to interdefinability*, which allows us to identify an $\aleph_0$-categorical structure with its automorphism group. So we identify two substructures $A_1,A_2$ of a structure $M$, if $\operatorname{Aut}(A_1)=\operatorname{Aut}(A_2)$. If $M$ is an $\aleph_0$-categorical structure and $A\subset M$, we denote the algebraic closure $\operatorname{acl}^{\rm{eq}}(A)$ of $A$ simply by $\operatorname{acl}(A)$, i.e. the union of the finite $\operatorname{Aut}(M/A)$-invariant sets of $M^{\rm{eq}}$. We recall that definable subsets of $\operatorname{acl}(A)$ correspond, up to interdefinability, to closed subgroups of $\operatorname{Aut}(M/A)$ of finite index, see [@Wi Section $4.1$] or Theorem $4.1$ in the article *“The structure of totally categorical structures”* by W. Hodges [@Au page $116$].

Similarly, if $A\subset M$, we denote the definable closure $\operatorname{dcl}^{\rm{eq}}(A)$ of $A$ simply by $\operatorname{dcl}(A)$, i.e. the set of the points of $M^{\rm{eq}}$ fixed by $\operatorname{Aut}(M/A)$.

\[conclusione1\] Let $A$ be a finite set of $M_n$. Then $$\operatorname{Aut}(M_n/\operatorname{cl}(A))=W_{{\mathop{\mathrm{supp}}}(A)}\rtimes \operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$$ (where $W_{{\mathop{\mathrm{supp}}}(A)}$ is the closed $\operatorname{Sym}(\Omega\setminus{\mathop{\mathrm{supp}}}(A))$-submodule of $\operatorname{im}\beta_{n,n-1}^\ast$ in Definition \[defn\]). Moreover, $\operatorname{Aut}(M_n/\operatorname{cl}(A))$ is the unique minimal closed subgroup of finite index of $\operatorname{Aut}(M_n/A)$.

*Proof.* Set $\Gamma=\operatorname{Aut}(M_n/\operatorname{cl}(A))$. We first prove that $\Gamma=W_{{\mathop{\mathrm{supp}}}(A)}\rtimes \operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$. By definition of the multisorted structure $M_n$, we have $\operatorname{Aut}M_n=\operatorname{im}\beta_{n,n-1}^*\rtimes 
\operatorname{Sym}(\Omega)$. Therefore, an element of $\Gamma$ is an ordered pair of the form $g\sigma$, where $g\in
\operatorname{im}\beta_{n,n-1}^*$ and $\sigma\in 
\operatorname{Sym}(\Omega)$. The action of $g\sigma$ on the elements belonging to the sorts $\Omega$ and $[\Omega]^n$ is given by the permutation $\sigma$. Also, the action of $g\sigma$ on the element $(w,x)$ belonging to the sort $[\Omega]^n\times {{\mathbb F}}_2$ is given by $$(w,x)^{g\sigma}=(w^\sigma,x+g(w)).$$ This implies that the automorphism $g\sigma$ fixes the elements in ${\mathop{\mathrm{supp}}}(A)$ and in $[{\mathop{\mathrm{supp}}}(A)]^n$ (in the sorts $\Omega$ and $[\Omega]^n$) if and only if $\sigma\in\operatorname{Sym}(\Omega\setminus{\mathop{\mathrm{supp}}}(A))$. Also, the automorphism $g\sigma$ fixes the elements in $[{\mathop{\mathrm{supp}}}(A)]^n\times {{\mathbb F}}_2$ (in the sort $[\Omega]^n\times {{\mathbb F}}_2$) if and only if $g(w)=0$ for every $w\in [{\mathop{\mathrm{supp}}}(A)]^n$. Hence, by the description of the elements of $W_{{\mathop{\mathrm{supp}}}(A)}$ in Proposition \[uniqueimage\], we have $g\sigma\in
\Gamma$ if and only if $g\sigma\in W_{{\mathop{\mathrm{supp}}}(A)}\rtimes
\operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$.

We claim that $\Gamma$ is the unique minimal closed subgroup of $\operatorname{Aut}(M_n/A)$ of finite index. Note that $\Gamma$ is a closed subgroup of $\operatorname{Aut}(M_n/A)$ of finite index.

Now, let $H$ be a closed subgroup of $\operatorname{Aut}(M_n/A)$ of finite index. Up to replacing $H$ with $H\cap \Gamma$, we may assume that $H\subseteq
\Gamma$. Let $\mu:\Gamma\to\operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$ be the natural projection. Since $\mu$ is a surjective continuous closed map and $\operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$ has no proper subgroup of finite index, we get that $\mu(H)=\operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$. This yields that $H\cap W_{{\mathop{\mathrm{supp}}}(A)}$ is a closed $\operatorname{Sym}(\Omega\setminus{\mathop{\mathrm{supp}}}(A))$-submodule of $W_{{\mathop{\mathrm{supp}}}(A)}$ of finite index. Now Proposition \[uniqueimage\] shows that $H\cap
W_{{\mathop{\mathrm{supp}}}(A)}=W_{{\mathop{\mathrm{supp}}}(A)}$. So $W_{{\mathop{\mathrm{supp}}}(A)}\subseteq H$ and $H=\Gamma$. In the following we denote by $\operatorname{acl}_{M_n}$ the $\operatorname{acl}$ in $M_n$.

\[acl1\] Let $A$ be a finite set of $M_n$. Then $\operatorname{acl}_{M_n}(A)=\operatorname{cl}(A)$.

*Proof.* Let $\overline{b}$ be an $m$-tuple in $M_n$ and $A$ be a finite set of $M_n$. We first claim that $\operatorname{Aut}(M_n/\overline{b})\geq \operatorname{Aut}(M_n/\operatorname{cl}(A))$ if and only if the underlying set of $\overline{b}$ is conteined in $\operatorname{cl}(A)$ . One direction is obvious. Suppose that $\operatorname{Aut}(M_n/\overline{b})\geq \operatorname{Aut}(M_n/\operatorname{cl}(A))$ for some finite $A\subset M_n$. Then by Lemma \[conclusione1\] we have that $\operatorname{Aut}(M_n/\operatorname{cl}(\operatorname{cl}(A),\overline{b})$ is a closed subgroup of finite index in $\operatorname{Aut}(M_n/\operatorname{cl}(A),\overline{b})=\operatorname{Aut}(M_n/\operatorname{cl}(A))$. Hence $\operatorname{Aut}(M_n/\operatorname{cl}(\operatorname{cl}(A),\overline{b})$ is a closed subgroup of finite index in $\operatorname{Aut}(M_n/A)$. By uniqueness of the minimal closed subgroup of finite index of $\operatorname{Aut}(M_n/A)$ we get that $W_{{\mathop{\mathrm{supp}}}(A)}\rtimes \operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(A))$ is equal to $W_{{\mathop{\mathrm{supp}}}(\operatorname{cl}(A),\overline{b})}\rtimes \operatorname{Sym}(\Omega\setminus {\mathop{\mathrm{supp}}}(\operatorname{cl}(A), \overline{b}))$ and, since ${\mathop{\mathrm{supp}}}(\operatorname{cl}(A), \overline{b})={\mathop{\mathrm{supp}}}(A,\overline{b})$, this is possible if and only if ${\mathop{\mathrm{supp}}}(\overline{b})\subseteq {\mathop{\mathrm{supp}}}(A)$, which proves the claim.

By definition, $\operatorname{acl}_{M_n}(A)$ is the union of the finite orbits on $M_n$ of $\operatorname{Aut}(M_n/A)$. Let $c\in \operatorname{acl}_{M_n}(A)$. Then $\operatorname{Aut}(M_n/A,c)$ is a closed subgroup of finite index in $\operatorname{Aut}(M_n/A)$. Hence, by Lemma \[conclusione1\], $\operatorname{Aut}(M_n/A,c)\geq \operatorname{Aut}(M_n/\operatorname{cl}(A)$. By the above argument we have that $c\in \operatorname{cl}(A)$.

Let $c\in \operatorname{cl}(A)$, then $\operatorname{Aut}(M_n/A)\geq\operatorname{Aut}(M_n/A,c)\geq \operatorname{Aut}(M_n/\operatorname{cl}(A))$. Hence the index of $\operatorname{Aut}(M_n/A,c)$ in $\operatorname{Aut}(M_n/A)$ is finite. Let $c^{\rm{eq}}\in M_n^{\rm{eq}}$. Then $c^{\rm{eq}}$ is a $0$-definable equivalence class of a tuple $b$ of elements in $M_n$. We denote by $\mathcal{s}(c^{\rm{eq}})$ the union of elements in $M_n$ of $c^{\rm{eq}}$. Similarly if $A\subseteq M_n^{\rm{eq}}$, we denote by $\mathcal{s}(A)$ the set of elements in $M_n$ $\bigcup_{c^{\rm{eq}}\in A}\mathcal{s}(c^{\rm{eq}})$.

\[acl2\] Let $A$ be a finite set of $M_n$. Then $\mathcal{s}(\operatorname{acl}(A))=\operatorname{cl}(A)$. In particular $\operatorname{acl}(\emptyset)=\emptyset$.

*Proof.* Fix an enumeration $\overline{b}$ of $\operatorname{acl}_{M_n}(A)$ and set $\Gamma=\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$. Consider the trivial relation $R=\{(b^\alpha,b^\alpha):\alpha\in \operatorname{Aut}(M_n)\}$. Since $R$ is an $\operatorname{Aut}(M_n)$-orbit, $R$ is a $0$-definable equivalence relation in $M_n$. Consider the $R$-equivalence class of $\overline{b}$. The pointwise stabilizer of $\overline{b}$ in $\operatorname{Aut}(M_n)$ is $\Gamma$ which, by Lemma \[conclusione1\] and Proposition \[acl1\], has finite index in $\operatorname{Aut}(M_n/A)$ and so $\overline{b}\in \operatorname{acl}(A)$.

Let $c^{\rm{eq}}\in \operatorname{acl}(A)$, then $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$ is a closed subgroup of finite index of $\operatorname{Aut}(M_n/A)$. By Lemma \[conclusione1\] $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$ contains $\Gamma$. Being $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$ also open in $\operatorname{Aut}(M_n/A)$ there exists a finite tuple $\overline{b}$ in $M_n$ such that $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$ contains the basic open subgroup $\operatorname{Aut}(M_n/A, \overline{b})$. Moreover $c^{\rm{eq}}=\overline{b}^{\operatorname{Aut}(M_n/A,c^{\rm{eq}})}$. By $\aleph_0$-categoricity the index of $\operatorname{Aut}(M_n/A, \overline{b})$ in $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$ is finite. Then, the index of $\operatorname{Aut}(M_n/A, \overline{b})$ in $\operatorname{Aut}(M_n/A)$ is finite and so $\Gamma\leq \operatorname{Aut}(M_n/A, \overline{b})$. Hence by the same argument used in Proposition \[acl1\], we get that the underlying set in $M_n$ of $\overline{b}$ is contained in $\operatorname{cl}(A)=\operatorname{acl}_{M_n}(A)$. From the fact that $\operatorname{Aut}(M_n/A,c^{\rm{eq}})\leq \operatorname{Aut}(M_n/A)$ and $\overline{b}\in \operatorname{acl}_{M_n}(A)$ it follows immediately that also the underlying set of the $\operatorname{Aut}(M_n/A,c^{\rm{eq}})$-orbit $\overline{b}^{\operatorname{Aut}(M_n/A,c^{\rm{eq}})}$ is contained in $\operatorname{acl}_{M_n}(A)$.

\[acl7\] Let $A$ be a finite set of $M_n$. Then, $$\operatorname{Aut}(M_n)_{\{\operatorname{acl}_{M_n}(A)\}}=\operatorname{Aut}(M_n)_{\{\operatorname{acl}(A)\}}.$$

*Proof.* From Proposition \[acl1\] and Proposition \[acl2\] it follows that $\operatorname{Aut}(M_n)_{\{\operatorname{acl}(A)\}}\leq \operatorname{Aut}(M_n)_{\{\operatorname{acl}_{M_n}(A)\}}$. Now, let $g\in  \operatorname{Aut}(M_n)_{\{\operatorname{acl}_{M_n}(A)\}}$. Note that $\operatorname{acl}_{M_n}(A^g)=\operatorname{acl}_{M_n}(A)$. Consequently, $\operatorname{acl}(A^g)=\operatorname{acl}(A)$. If $c^{\rm{eq}}\in \operatorname{acl}(A)$, then the index of $\operatorname{Aut}(M_n/A, c^{\rm{eq}})$ in $\operatorname{Aut}(M_n/A)$ is finite. Therefore, $\operatorname{Aut}(M_n/A^g, (c^{\rm{eq}})^g)=g^{-1}\operatorname{Aut}(M_n/A, c^{\rm{eq}})g$ has finite index in $\operatorname{Aut}(M_n/A^g)=g^{-1}\operatorname{Aut}(M_n/A)g$, which implies that $(c^{\rm{eq}})^g\in \operatorname{acl}(A^g)=\operatorname{acl}(A)$.

\[acl3\] Let $A$ be a finite subset of $M_n$. Then, $
\operatorname{dcl}(\operatorname{acl}_{M_n}(A))=\operatorname{acl}(A).$

*Proof.* Let $c^{\rm{eq}}\in \operatorname{acl}(A)$, i.e. the stabilizer of $c^{\rm{eq}}$ in $\operatorname{Aut}(M_n/A)$ has finite index in $\operatorname{Aut}(M_n/A)$. We need to show that the stabilizer of $c^{\rm{eq}}$ in $\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$ is equal to $\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$. We have the following disequality: $$|\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A)):\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A), c^{\rm{eq}})|\leq |\operatorname{Aut}(M_n/A):\operatorname{Aut}(M_n/A, c^{\rm{eq}})|$$ Then $|\operatorname{Aut}(M_n/A):\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A), c^{\rm{eq}})|$ is finite. By Lemma \[conclusione1\] and Proposition \[acl1\] it follows that $\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A), c^{\rm{eq}})$, is equal to $\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$, i.e. $c^{\rm{eq}}\in \operatorname{dcl}(\operatorname{acl}_{M_n}(A))$.

Let $c^{\rm{eq}}\in \operatorname{dcl}(\operatorname{acl}_{M_n}(A))$. We need to show that $\operatorname{Aut}(M_n/A, c^{\rm{eq}})$, has finite index in $\operatorname{Aut}(M_n/A)$. We have that $$\label{indici}
\begin{array}{c}
|\operatorname{Aut}(M_n/A):\operatorname{Aut}(M_n/\operatorname{cl}(A)), c^{\rm{eq}})|=\\
|\operatorname{Aut}(M_n/A):\operatorname{Aut}(M_n/A, c^{\rm{eq}})||\operatorname{Aut}(M_n/A, c^{\rm{eq}}):\operatorname{Aut}(M_n/\operatorname{cl}(A), c^{\rm{eq}})|
\end{array}$$ Since $c^{\rm{eq}}\in \operatorname{dcl}(\operatorname{acl}_{M_n}(A))$ we have that $\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A), c^{\rm{eq}})=\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$. Lemma \[conclusione1\] and the equality (\[indici\]) imply that $|\operatorname{Aut}(M_n/A):\operatorname{Aut}(M_n/A, c^{\rm{eq}})|$ is finite. This proves that $c^{\rm{eq}}\in \operatorname{acl}(A)$ and the proof is complete.

\[acl5\] Let $A$ be a finite subset of $M_n$. Then $$\operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))=\operatorname{Aut}(M_n/\operatorname{acl}(A)).$$

*Proof.* Let $g\in \operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$ and $c^{\rm{eq}}\in \operatorname{acl}(A)$. Proposition \[acl3\] yields that $(c^{\rm{eq}})^g=c^{\rm{eq}}$, which means that $g\in \operatorname{Aut}(M_n/\operatorname{acl}(A))$. It remains to prove that $\operatorname{Aut}(M_n/\operatorname{acl}(A))\leq \operatorname{Aut}(M_n/\operatorname{acl}_{M_n}(A))$. Consider the trivial relation $R$ given by $R=\{ (b,b) : b\in M_n\}$. This is a 0-definable relation. Let $a\in \operatorname{acl}_{M_n}(A)$. Then $\{a\}\in M_n^{\rm{eq}}$ and $\operatorname{Aut}(M_n/A, \{a\})=\operatorname{Aut}(M_n/A, a)$ is a closed subgroup of finite index in $\operatorname{Aut}(M_n/A)$. Hence, we can consider that $\operatorname{acl}_{M_n}(A)\subseteq \operatorname{acl}(A)$ and the thesis follows at once.

\[acl4\] Proposition \[acl1\] yields that if $A$ is a finite set of $M_n$, then $\operatorname{acl}_{M_n}(A)=\operatorname{acl}_{M_n}({\mathop{\mathrm{supp}}}(A))$. Therefore, from Proposition \[acl3\] it follows that $\operatorname{acl}(A)=\operatorname{acl}({\mathop{\mathrm{supp}}}(A))$.

\[acl6\] Let $A_1, \dots, A_n$ be finite subsets in the sort $\Omega$. Then $$\operatorname{acl}(\operatorname{acl}(A_1),\dots,\operatorname{acl}(A_n))=\operatorname{acl}(\bigcup_{i=1}^n A_i).$$

*Proof.* Obviously, $\operatorname{acl}(\bigcup_{k=1}^n A_k)\subseteq \operatorname{acl}(\operatorname{acl}(A_1),\dots,\operatorname{acl}(A_n))$.

Let $c^{\rm{eq}}\in \operatorname{acl}(\operatorname{acl}(A_1),\dots,\operatorname{acl}(A_n))$ and set $G=\operatorname{Aut}(M_n/\operatorname{acl}(A_1),\dots,\operatorname{acl}(A_n))$. Then, the pointwise stabilizer $G_{c^{\rm{eq}}}$ has finite index in $G$. By Corollary \[acl5\] we have that $$G=\bigcap_{i=1}^n W_{A_i}\rtimes \operatorname{Sym}(\Omega \setminus A_i).$$ Moreover, $G\geq W_{\bigcup_{i=1}^n A_i}\rtimes \operatorname{Sym}(\Omega \setminus \bigcup_{i=1}^n A_i)$ and $G$ is a closed subgroup in $\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i)$. So, $G$ is a closed subgroup of finite index in $\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i)$ which implies that also $G_{c^{\rm{eq}}}$ is of finite index in $\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i)$. Now, $G_{c^{\rm{eq}}}=G\cap \operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i,c^{\rm{eq}})$ and $$\begin{array}{c}
|\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i):\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i, c^{\rm{eq}})|=\\
|\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i):G_{c^{\rm{eq}}}|/|\operatorname{Aut}(M_n/\bigcup_{i=1}^n A_i, c^{\rm{eq}}):G_{c^{\rm{eq}}}|,
\end{array}$$ i.e. $c^{\rm{eq}}\in \operatorname{acl}(\bigcup_{i=1}^n A_i)$.

$k$-existence and $k$-uniqueness for $M_n$
==========================================

In this section we prove Theorem \[summary\]. Note that, up to renaming the elements of $\Omega$, we may assume that $\Omega=\mathbb{N}$. In the sequel we denote by $[k]$ the subset $\{1,\ldots,k\}$ of ${{\mathbb N}}$. Also, given $i\in [k]$, we denote by $[k]-i$ the set $\{1,\ldots,k\}\setminus \{i\}$. Finally, we denote the theory $\operatorname{Th}(M_n)$ by $T_n$.

We start by studying $k$-uniqueness in $T_n$. We first single out the following technical lemma which would be used in Proposition \[unik\].

\[lemma:aux\] Let $k$ and $n$ be integers, with $k< n$, and $A_1,\ldots,A_{k}$ be subsets of $\Omega$. Then $$(\dag)\qquad\bigcap_{i=1}^{k}\left(V_{A_i}+\operatorname{ker}\beta_{n,n-1}^\ast\right)=
\left(\bigcap_{i=1}^{k}V_{A_i}\right)+\operatorname{ker}\beta_{n,n-1}^{\ast}.$$

*Proof.* We denote the left-hand-side of $(\dag)$ by $V_{1,k}$ and the right-hand-side of $(\dag)$ by $V_{2,k}$ (where the label $k$ is used in order to remember the number of intersections).

We argue by induction on $k$. Note that if $k=0$ or $k=1$, then there is nothing to prove. Assume $(\dag)$ holds for $k$ intersections (where $k\geq 1$) and that $k+1<n$. In particular, we point out that $n>2$. We prove that $(\dag)$ holds for $k+1$ intersections. Clearly, $V_{2,k+1}\subseteq V_{1,k+1}$. Let $g$ be in $V_{1,k+1}$. We need to show that $g\in V_{2,k+1}$. By induction hypothesis (on the sets $A_1,\ldots,A_k$), we have $$\label{eq:p1}
V_{1,k+1}=
\left(\left(\bigcap_{i=1}^{k}V_{A_i}\right)+\operatorname{ker}\beta_{n,n-1}^\ast\right)\cap
(V_{A_{k+1}}+\operatorname{ker}\beta_{n,n-1}^\ast).$$ By Equation (\[eq:p1\]) and Proposition \[exact\], we have $$\label{eq:p2}
g=g_1+\beta_{n-1,n-2}^\ast h_1=g_2+\beta_{n-1,n-2}^\ast h_2,$$ where $g_1\in\cap_{i=1}^kV_{A_i}$, $g_2\in V_{A_{k+1}}$ and $h_1,h_2\in
{{\mathbb F}}_2^{[\Omega]^{n-2}}$. We claim that (up to replacing $h_1$ by $h_1+l$, where $l\in \operatorname{ker}\beta_{n-1,n-2}^\ast$), we may assume that $h_1-h_2\in \cap_{i=1}^k V_{A_i\cap A_{k+1}}$.

Let $w$ be an $(n-1)$-subset of $\Omega$ contained in $A_i\cap A_{k+1}$ for some $i=1,\ldots,k$. Since $g_1\in V_{A_i}$ and $g_2\in
V_{A_{k+1}}$, we see that $g_1(w)=g_2(w)=0$. So, from Equation (\[eq:p2\]) we obtain $$g(w)=(\beta_{n-1,n-2}^\ast h_1)(w)=(\beta_{n-1,n-2}^\ast h_2)(w),$$ that is, $(\beta_{n-1,n-2}^\ast(h_1-h_2))(w)=0$. As $w$ is an arbitrary $(n-1)$-subset of $A_i\cap A_{k+1}$, Lemma \[char\] yields $h_1-h_2\in V_{A_i\cap A_{k+1}}+\operatorname{ker}\beta_{n-1,n-2}^\ast$. As $i$ is an arbitrary element in $\{1,\ldots,k\}$, we get $$h_1-h_2\in \bigcap_{i=1}^k(V_{A_i\cap
   A_{k+1}}+\operatorname{ker}\beta_{n-1,n-2}^\ast).$$ Since $k+1<n$, we have $k<n-1$ and so we may now apply our inductive hypothesis on the sets $A_1\cap A_{k+1},\ldots,A_{k}\cap
 A_{k+1}$. We have $$\label{eq:p3}
h_1-h_2\in \left(\bigcap_{i=1}^kV_{A_i\cap
   A_{k+1}}\right)+\operatorname{ker}\beta_{n-1,n-2}^\ast.$$ From Equation (\[eq:p3\]), we get $h_1-h_2=h+l$, where $h\in \cap_{i=1}^kV_{A_i\cap
A_{k+1}}$ and $l\in \operatorname{ker}\beta_{n-1,n-2}^\ast$. Set $h_1'=h_1+l$. We have $$h_1'-h_2=h_1+l-h_2=h\in \cap_{i=1}^kV_{A_i\cap A_{k+1}}$$ and our claim is proved.

Let $t$ be the element of ${{\mathbb F}}_2^{[\Omega]^{n-2}}$ defined by $$t(w)=\left\{
\begin{array}{ccl}
h_1(w)&&\textrm{if }w\subseteq A_i\textrm{ for some }i=1,\ldots,k,\\
h_2(w)&&\textrm{if }w\subseteq A_{k+1},\\
0&&\textrm{otherwise}.
\end{array}
\right.$$ Note that the function $t$ is well-defined. Indeed, recall that $n>2$ and note that if $w$ is an $(n-2)$-subset of $\Omega$ with $w\subseteq A_i\cap A_{k+1}$ (for some $i=1,\ldots,k$), then $h_1(w)=h_2(w)$ as $h_1-h_2\in V_{A_{i}\cap
  A_{k+1}}$.

We claim that $g+\beta_{n-1,n-2}^\ast t \in
\cap_{i=1}^{k+1}V_{A_i}$. We have to show that $g+\beta_{n-1,n-2}^\ast
t$ vanishes in $[A_i]^{n-1}$, for $i=1,\ldots,k+1$. Let $w$ be an $(n-1)$-subset of $\Omega$ with $w\subseteq A_i$, for some $i=1,\ldots,k+1$. If $i\leq k$, then we have

$$(g+\beta_{n-1,n-2}^\ast t)(w)=(g_1(w)+\beta_{n-1,n-2}^\ast
h_1(w))+\beta_{n-1,n-2}h_1(w)=0,$$ where in the first equality we used Equation (\[eq:p2\]) and the fact that $t$ and $h_1$ coincide in $[A_i]^{n-2}$, and in the second equality we used that $g_1\in V_{A_i}$. Similarly, if $i=k+1$, then $$(g+\beta_{n-1,n-2}^\ast t)(w)=(g_2(w)+\beta_{n-1,n-2}^\ast
h_2(w))+\beta_{n-1,n-2}h_2(w)=0,$$ where in the first equality we used Equation (\[eq:p2\]) and the fact that $t$ and $h_2$ coincide in $[A_{k+1}]^{n-2}$, and in the second equality we used that $g_2\in V_{A_{k+1}}$.

Finally, as $\beta_{n-1,n-2}^\ast t\in\operatorname{ker}\beta_{n,n-1}^\ast$, we get that $g\in V_{2,k+1}$.

\[unik\] The theory $T_n$ has $k$-uniqueness for every $k\leq n$.

*Proof.*Let $k$ be an integer with $k\leq n$ and $a:P(k)^-\to \mathcal{C}_{T_n}$ be a $k$-amalgamation problem. We need to show that $a$ has at most one solution up to isomorphism. Since every stable theory has $1$- and $2$-uniqueness, we may assume that $k\geq 3$. Set $\Gamma_1=\operatorname{Aut}(a([k-1])/\cup_{i=1}^{k-1}a([k]-i))$ and $\Gamma_2=\operatorname{Aut}(a([k-1])/
\cup_{i=1}^{k-1}a([k-1]-i))$. By [@Hr Proposition $3.5$], it is enough to prove that $$\Gamma_1=\Gamma_2,$$ i.e. $\overline{\Gamma_1},\overline{\Gamma_2}$ give rise to the same action on $a([k-1])$ (see Definition \[auxiliary\]).

By Remark \[acl4\], the algebraically closed sets of finite subsets of $M_n$ are of the form $\operatorname{acl}(A)$, for some finite subset $A$ of the sort $\Omega$. By Corollary \[acl7\] the setwise stabilizer of $\operatorname{acl}(A)$ in $\operatorname{Aut}(M_n)$ is simply $(\operatorname{Sym}(\Omega\setminus A)\times\operatorname{Sym}(A))\ltimes
\operatorname{im}\beta_{n,n-1}^\ast$. Using Corollary \[acl5\], we get that the pointwise stabilizer of $\operatorname{acl}(A)$ in $\operatorname{Aut}(M_n)$ is $\operatorname{Sym}(\Omega\setminus A)\ltimes W_A$.

Let $a(i)=\operatorname{acl}(B_i)$, where $B_i$ are finite subsets of $M_n$ for $1\leq i\leq k$. Set $A_i={\mathop{\mathrm{supp}}}(B_i)$, for $1\leq i\leq k$, and $A=\cup_{i=1}^{k-1}A_i$. Note that by definition of amalgamation problem and by Proposition \[acl6\], we have $a([k-1])=\operatorname{acl}(A)$. Therefore, by the previous paragraph, as $k\geq 3$, we get that $\overline{\Gamma_1}$ is equal to $$((\operatorname{Sym}(\Omega\setminus A)\times \operatorname{Sym}(A))\ltimes
\operatorname{im}\beta_{n,n-1}^\ast)\cap\bigcap_{i=1}^{k-1}(\operatorname{Sym}(\Omega\setminus
((A\cup A_k)\setminus A_i))\ltimes W_{(A\cup A_k)\setminus A_i})$$ i.e. $$\label{eq:G1}
\overline{\Gamma_1}=\operatorname{Sym}(\Omega\setminus (A\cup A_k))\ltimes
\bigcap_{i=1}^{k-1}W_{(A\cup A_k)\setminus A_i}$$ and $\overline{\Gamma_2}$ is equal to $$((\operatorname{Sym}(\Omega\setminus A)\times \operatorname{Sym}(A))\ltimes
\operatorname{im}\beta_{n,n-1}^\ast)\cap\bigcap_{i=1}^{k-1}(\operatorname{Sym}(\Omega\setminus
(A\setminus A_i))\ltimes W_{A\setminus A_i})$$ i.e. $$\label{eq:G2}
\overline{\Gamma_2}=\operatorname{Sym}(\Omega\setminus A)\ltimes
\bigcap_{i=1}^{k-1}W_{A\setminus A_i}.$$ As $\operatorname{Sym}(\Omega\setminus(A\cup A_k))$ and $\operatorname{Sym}(\Omega\setminus A)$ act trivially on the elements of $\operatorname{acl}(A)$, by Equations $(\ref{eq:G1})$ and $(\ref{eq:G2})$, in order to prove that $\Gamma_1=\Gamma_2$ it suffices to show that $$W_1=\bigcap_{i=1}^{k-1}W_{(A\cup A_k)\setminus A_i}
\quad\textrm{and}\quad W_2=\bigcap_{i=1}^{k-1}W_{A\setminus A_i}$$ induce the same action on $\operatorname{acl}(A)$. Also, $W_1$ and $W_2$ act trivially on the elements belonging to the sorts $\Omega$ and $[\Omega]^n$ of $M_n$. Thus, it suffices to study the action of $W_1$ and $W_2$ on the elements of $\operatorname{acl}(A)$ belonging to the sort $[\Omega]^n\times {{\mathbb F}}_2$, that is, on $[A]^n$. Clearly, $W_1\subseteq W_2$. Therefore, it remains to show that for every element $f$ of $W_2$ there exists an element $\overline{f}$ of $W_1$ such that $f$ and $\overline{f}$ induce the same action on $[A]^n$.

Let $f$ be in $W_2$. By Definition \[defn\], we get that $f=\beta_{n,n-1}^\ast g$, for some $g\in \cap_{i=1}^{k-1}(V_{A\setminus A_i}+\operatorname{ker}\beta_{n,n-1}^\ast)$. Lemma \[lemma:aux\] (applied to $k-1,n$ and $(A\setminus
A_1),\ldots,(A\setminus A_{k-1})$) yields $$\bigcap_{i=1}^{k-1}\left(V_{A\setminus
  A_i}+\operatorname{ker}\beta_{n,n-1}^\ast\right)=\left(\bigcap_{i=1}^{k-1}V_{A\setminus
A_i}\right)+\operatorname{ker}\beta_{n,n-1}^\ast.$$ Thence, up to replacing $g$ by $g+l$ (for some $l\in
\operatorname{ker}\beta_{n,n-1}^\ast$), we may assume that $g\in
\cap_{i=1}^{k-1}V_{A\setminus   A_i}$. Let $\overline{g}$ be the function in ${{\mathbb F}}_2^{[\Omega]^{n-1}}$ defined by $$\overline{g}(w)=\left\{
\begin{array}{ccl}
g(w)&&\textrm{if }w\subseteq A,\\
0&&\textrm{otherwise}.
\end{array}
\right.$$ Set $\overline{f}=\beta_{n,n-1}^\ast \overline{g}$. By construction, $f$ and $\overline{f}$ coincide in $[A]^{n}$, that is, $f$ and $\overline{f}$ induce the same action on $[A]^n$. Thus, it remains to prove that $\overline{f}\in W_1$, that is, $\overline{f}$ vanishes on every $n$-subset $L$ of $(A\cap
A_i)\setminus A_i$, for $i=1,\ldots,k$. Let $L$ be an $n$-subset of $(A\cup A_k)\setminus A_i$. We consider three cases $L\subseteq A$, $|L\cap A_k|\geq 2$ and $|L\cap A_k|=1$.

If $L\subseteq A$, then $\overline{f}(L)=f(L)=0$ (because $f$ and $\overline{f}$ coincide on $[A]^n$).

If $|L\cap A_k|\geq 2$, then $(L\setminus \{x\})\nsubseteq A$, for every $x$ in $L$. By definition of $\overline{g}$, we have $\overline{g}(L\setminus\{x\})=0$ and $\overline{f}(L)=\sum_{x\in
  L}\overline{g}(L\setminus \{x\})=0$.

If $|L\cap A_k|=1$ and $L\cap A_k=\{\overline{x}\}$, then (arguing as in the previous paragraph) $\overline{f}(L)=\sum_{x\in 
  L}\overline{g}(L\setminus\{x\})=g(L\setminus\{\overline{x}\})$. As $L\subseteq (A\cup A_k)\setminus A_i$, we have that $L\setminus\{\overline{x}\}\subseteq A\setminus A_i$. Since $g\in
V_{A\setminus A_i}$, we get that $\overline{g}(L\setminus\{\overline{x}\})=g(L\setminus\{\overline{x}\})=0$.

J.Goodrick and A.Kolesnikov recently proved that if a complete stable theory $T$ has $k$-uniqueness for every $2\leq k\leq n$, then $T$ has $n+1$-existence [@GoKo]. For completeness we report the proof of their result.

\[thm:GoKo\]Let $T$ be a complete stable theory. If $T$ has $k$-uniqueness for every $2\leq k\leq n$, then $T$ has $n+1$-existence.

*Proof.* Note that the existence and the uniqueness of nonforking extensions of types in a stable theory yields that any stable theory has both $2$-existence and $2$-uniqueness.

Since $T$ is a complete stable theory, for every regular cardinal $k$, there exists a saturated model of cardinality $k$. In the sequel we shall consider the objects of $\mathcal{C}_T$ lying inside a very large saturated “monster model” $\mathfrak{C}$ of $T$.

Suppose $a$ is an $(n+1)$-amalgamation problem. We have to prove that $a$ has a solution $a'$. First, let $B_0$ and $B_1$ be sets of $\mathfrak{C}$ such that ${\textup{tp}}(B_0 / a(\emptyset)) 
= {\textup{tp}}(a([n])/ a(\emptyset))$, ${\textup{tp}}(B_1 /
a(\emptyset)) = {\textup{tp}}(a(\{n+1\})/ a(\emptyset))$, and $$B_0
\ind_{a(\emptyset)} B_1.$$ Let $\sigma_0$ and $\sigma_1$ be two automorphisms of $\mathfrak{C}$ fixing pointwise $a(\emptyset)$ and such that $B_0=\sigma_0(a([n]))$, $B_1=\sigma_1(a(\{n+1\}))$.

Define $a'([n+1])$ to be the algebraic closure of $B_0 \cup B_1$. To determine the solution $a'$ of $a$, it remains to define the transition maps $a'_{s,[n+1]}:a'(s)\to a'([n+1])$, for all subsets $s$ of $[n+1]$. The map $a'_{\emptyset,[n+1]}$ must be the identity on $a(\emptyset)$. For $i$ in $[n]$, we let $a'_{\{i\},[n+1]} : a(\{i\}) \to
a'([n+1])$ be the map $\sigma_0 \circ
a_{\{i\},[n]}$, and we let $a'_{\{n+1\},[n+1]}$ be the map $
\sigma_1$. Now, the following claim concludes the proof of the theorem.

<span style="font-variant:small-caps;">Claim:</span> For every proper non-empty subset $s$ of $[n+1]$, there is a way to define the transition maps $a'_{s,[n+1]}$, which is consistent with $a$ and the definition of $a'_{\{i\},[n+1]}$ given above, and such that $$a'_{s,[n+1]} (a(s)) =
\operatorname{acl}\left(\bigcup_{i \in s} a(\{i\})\right).$$

We argue by induction on the size $k$ of the set $s$. If $k=1$, then there is nothing to prove. Suppose we have defined $a'_{s,[n+1]}$ as in the claim, for all $s
\subseteq [n+1]$ such that $|s| < k$. Let $s$ be a subset of $[n+1]$ such that $|s| = k$. The family of sets $\{a(t)\mid t \subsetneq s \}$ forms a $k$-amalgamation problem with the same transition maps as $a$. Call $a^1$ this amalgamation problem. By the induction hypothesis, the family of sets $\{a'_{t,[n+1]} (a(t)) \mid t \subsetneq s \}$ forms another $k$-amalgamation problem with the transition maps given by set inclusions. Call $a^2$ this amalgamation problem. Notice that $a^1$ and $a^2$ are isomorphic, and that both have independent solutions. Namely, $a^1$ can be completed to $a(s)$ using the transition maps in $a$, and $a^2$ has a natural solution $(a^2)'$ such that $$(a^2)'(s) = \operatorname{acl}\left(\bigcup_{i \in
  s} a(\{i\})\right),$$ where the transition maps are again given by set inclusions. So, by the $k$-uniqueness property, there is an isomorphism of these solutions, which yields the desired transition map $a'_{s,[n+1]}$ from $a(s)$ to $\operatorname{acl}(\bigcup_{i \in s} a(\{i\}))$.

Now we are ready to prove that $T_n$ has $k$-existence for every $k\leq n+1$.

\[exik\]The theory $T_n$ has $k$-existence for every $k\leq n+1$.

*Proof.* By definition, $T_n=\operatorname{Th}(M_n)$ is complete. Since $T_n$ is a stable theory, the proof of this proposition follows at once from Proposition \[unik\] and Theorem \[thm:GoKo\].

Next, we show that $T_n$ does not have $n+1$-uniqueness.

\[unin+1\] The theory $T_n$ does not have $n+1$-uniqueness.

*Proof.* Recall that by construction $n\geq 2$. Let $a:P(n+1)^-\to \mathcal{C}_{T_{n}}$ be the $(n+1)$-amalgamation problem defined on the objects by $a(s)=\operatorname{acl}(s)$ and where the morphisms are inclusions. In order to prove this proposition we show the following equations: $$\begin{aligned}
\label{eq1}
 |\operatorname{Aut}(\operatorname{acl}([n])/\cup_{i=1}^{n}\operatorname{acl}([n+1]-i))|&=&1, \\ 
 |\operatorname{Aut}(\operatorname{acl}([n])/\cup_{i=1}^{n}\operatorname{acl}([n]-i))|&=&2.\label{eq2}  
 \end{aligned}$$ In fact, by [@Hr Proposition $3.5$], Equations (\[eq1\]), (\[eq2\]) yield that $a$ has more than one solution up to isomorphism, i.e. $T_n$ does not have $n+1$-uniqueness.

We start by proving Equation (\[eq1\]). Since $[n],[n+1]-i$ have size $n$, Proposition \[acl1\] yields $\operatorname{acl}_{M_n}([n])=[n]\cup \{[n]\}\cup\{([n],0), ([n],1)\}$ and $\operatorname{acl}_{M_n}([n+1]-i)=([n+1]-i)\cup\{[n+1]-i\}\cup
\{([n+1]-i,0), ([n+1]-i,1)\}$.

By the description given in the previous paragraph, every permutation in $\operatorname{Sym}(\Omega)$ fixing pointwise the elements in $\cup_{i=1}^n\operatorname{acl}([n+1]-i)$ also fixes pointwise every element in $\operatorname{acl}([n])$. Therefore, it suffices to consider the elements in $\operatorname{im}\beta_{n,n-1}^\ast$. Let $f$ be in $\operatorname{im}\beta^\ast_{n,n-1}$ and suppose that $f$ fixes every element in $\cup_{i=1}^n\operatorname{acl}([n+1]-i)$, i.e. $f([n+1]-i)=0$, for $1\leq i\leq
 n$. Let $g\in\mathbb{F}_2^{[\Omega]^{n-1}}$ such that $f=\beta_{n,n-1}^\ast g$. We have $$\label{uniqueness}
0=\sum_{i=1}^{n} f([n+1]-i)=\sum_{i=1}^{n}\sum_{j\neq
  i}^{n+1}g([n+1]\setminus\{i,j\}).$$ Now, for $j\neq n+1$, the summand $g([n+1]\setminus\{i,j\})$ appears twice in Equation (\[uniqueness\]) and therefore over $\mathbb{F}_2$ their sum is zero. Hence $$0=\sum_{i=1}^{n} f([n+1]-i)=\sum _{i=1}^n g([n]-i)=(\beta_{n,n-1}^\ast
g)([n])=f([n]).$$ This yields that $f$ fixes $([n],0),([n],1)$. Hence Equation (\[eq1\]) follows.

We now prove Equation (\[eq2\]). Since $[n]-i$ has size $n-1$, Proposition \[acl1\] implies $\operatorname{acl}_{M_n}([n]-i)=[n]-i$. Therefore, $$\cup_{i=1}^n\operatorname{acl}_{M_n}([n]-i)=\cup_{i=1}^n([n]-i)=[n].$$ Also, $\operatorname{acl}_{M_n}([n])=[n]\cup\{[n]\}\cup \{([n],0),([n],1)\}$. Corollary \[acl7\] and Corollary \[acl5\] yield that every element of $\operatorname{Aut}(\operatorname{acl}([n])/\cup_{i=1}^{n}\operatorname{acl}([n]-i))$ fixes the elements belonging to the sorts $\Omega$ and $[\Omega]^n$ of $\operatorname{acl}_{M_n}([n])$. Hence, in order to prove Equation (\[eq2\]), it suffices to find an automorphism of $\operatorname{acl}_{M_n}([n])$ mapping $([n],0)$ into $([n],1)$. Let $g\in{{\mathbb F}}_2^{[\Omega]^{n-1}}$ with $g([n-1])=1$ and $g(w)=0$ for $w\neq [n-1]$. Set $f=\beta_{n,n-1}^\ast g$ and note that $f([n])=1$. As $\operatorname{Aut}(M_n)=\operatorname{im}\beta_{n,n-1}^\ast\rtimes \operatorname{Sym}(\Omega)$, the map $f$ is an automorphism of $M_n$. By construction $f$ is an automorphism of $\operatorname{acl}_{M_n}([n])$ and $([n],0)^f=([n],0+f([n]))=([n],1)$.

Finally, we show that $T_n$ does not have $n+2$-existence.

\[notn+2\] The theory $T_n$ does not have $n+2$-existence.

*Proof.* We construct an $n+2$-amalgamation problem $a$ over $\emptyset$ (that is, $a(\emptyset)=\emptyset$) for $T_n$ with no solution.

Let $g$ be the element of $\mathbb{F}_2^{[\Omega]^{n-1}}$ defined by $$g(w)=\left\{
\begin{array}{ccl}
1&&\textrm{if }w=[n-1],\\
0&&\textrm{if }w\neq [n-1].
\end{array}
\right.$$ Consider $f=\beta_{n,n-1}^\ast g$ and note that, as $\operatorname{Aut}(M_n)=\operatorname{im}\beta_{n,n-1}^\ast \rtimes \operatorname{Sym}(\Omega)$, the element $f$ is an automorphism of $M_n$.

Let $a$ be the functor $a:P(n+2)^{-}\to \mathcal{C}_{T_n}$ defined on the objects by $a(s)=\operatorname{acl}(s)$ and with morphisms defined by

$$\begin{aligned}
\label{eq:6}
a_{s,s'}=\left\{
\begin{array}{ccl}
f|_{a(s)}&&\textrm{if }s=[n] \textrm{ and } s'=[n+1],\\
\textrm{inclusion}&&\textrm{otherwise},
\end{array}
\right.\end{aligned}$$

where $f|_{a(s)}$ denotes the restriction of the automorphism $f$ to $a(s)$. It is not obvious from Equation $(\ref{eq:6})$ that $a$ is a functor. Therefore, in the following paragraph, we prove that $a$ is well-defined, that is, $a_{s_2,s_3}\circ a_{s_1,s_2}=a_{s_1,s_3}$ for every $s_1,s_2,s_3$ in $P(n+2)^{-}$ with $s_1\subseteq s_2\subseteq s_3$.

If $s_2\neq [n+1]$ and $s_3\neq [n+1]$, then (by Equation $(\ref{eq:6})$) the morphisms $a_{s_1,s_2},a_{s_2,s_3}$ and $a_{s_1,s_3}$ are inclusions and so clearly $a_{s_2,s_3}\circ
a_{s_1,s_2}=a_{s_1,s_3}$. If $s_2=[n+1]$, then $s_2$ is a maximal element of the partially ordered set $P(n+2)^{-}$. Thence $s_3=s_2$ and, by Equation $(\ref{eq:6})$, $a_{s_2,s_3}$ is the identity map. Thus $a_{s_2,s_3}\circ
a_{s_1,s_2}=a_{s_1,s_3}$. In particular, from now on we may assume that $s_3=[n+1]$ and $s_2\neq [n+1]$. As $s_1\subseteq s_2$, if $s_2\neq
[n]$, then $s_1\neq [n]$ and so, by Equation $(\ref{eq:6})$, the morphisms $a_{s_1,s_2},a_{s_2,s_3}$ and $a_{s_1,s_3}$ are inclusions and $a_{s_2,s_3}\circ
a_{s_1,s_2}=a_{s_1,s_3}$. If $s_2=s_1=[n]$, then $a_{s_1,s_2}$ is the identity map and $a_{s_2,s_3}\circ
a_{s_1,s_2}=a_{s_1,s_3}$. The only case that remains to consider is $s_3=[n+1]$, $s_2=[n]$ and $s_1\neq [n]$. Thence $a_{s_1,s_2}$ and $a_{s_1,s_3}$ are inclusion maps and $a_{s_2,s_3}=f|_{a(s_2)}$. Since $s_1\subseteq s_2=[n]$ and $s_1\neq [n]$, we have $|s_1|<n-1$. Therefore, $a(s_1)=\operatorname{acl}(s_1)$ and by Proposition \[acl1\] $\operatorname{acl}_{M_n}(s_1)=s_1$ consists only of elements belonging to the sort $\Omega$ of $M_n$. As $f$ acts trivially on the elements belonging to the sort $\Omega$, by Proposition \[acl2\] we obtain $a_{s_2,s_3}\circ
a_{s_1,s_2}=(f|_{a(s_2)})|_{a(s_1)}=f|_{a(s_1)}=a_{s_1,s_3}$. Finally, this proves that $a:P(n+2)^{-}\to
\mathcal{C}_{T_n}$ is a functor.

By Proposition \[conclusione1\], $a(\emptyset)=\operatorname{acl}(\emptyset)=\emptyset$. Therefore, the functor $a$ is an $n+2$-amalgamation problem over $\emptyset$ for $M_n$.

We claim that $a$ cannot be extended to $P(n+2)$. We argue by contradiction. Let $\overline{a}:P(n+2)\to
\mathcal{C}_{T_n}$ be a solution of $a$. In particular, $\overline{a}$ is an extension of $a$ to the whole of $P(n+2)$. Denote by $x_i$ the morphisms $\overline{a}_{[n+2]-i,[n+2]}$, for $1\leq i\leq n+2$. So, by definition of morphism, $x_i$ is the restriction to $\operatorname{acl}([n+2]-i)$ of an automorphism $f_i\sigma_i$ of $M_n$, where $f_i\in\operatorname{im}\beta_{n,n-1}^\ast$ and $\sigma_i\in \operatorname{Sym}(\Omega)$.

Since $\overline{a}$ is a functor and $\overline{a}$ extends $a$, we get $$\begin{aligned}
\label{eq4}
x_i\circ a_{[n+2]\setminus\{i,j\},[n+2]-i}&=&
\overline{a}_{[n+2]-i,[n+2]}\circ
\overline{a}_{[n+2]\setminus\{i,j\},[n+2]-i}\\\nonumber
&=&
\overline{a}_{[n+2]-j,[n+2]}\circ\overline{a}_{[n+2]\setminus\{i,j\},[n+2]-j}\\\nonumber  
&=&x_j\circ a_{[n+2]\setminus\{i,j\},[n+2]-j}.\nonumber\end{aligned}$$ Let $i$ and $j$ be in $[n+2]$ with $i\neq j$. Fix an enumeration of $\operatorname{acl}_{M_n}([n+2]\setminus\{i,j\})$ and denote it as $\overline{b_{ij}}=(b_{ij_1,},\dots)$. Then, as it is shown in Proposition \[acl2\] $\overline{b_{ij}}\in \operatorname{acl}([n+2]\setminus\{i,j\})$ and, of course, also in $\operatorname{acl}([n+2]\setminus\{i\})$. By Proposition \[acl1\] the ordered pair $([n+2]\setminus\{i,j\},0)$ belongs to the sort $[\Omega]^n\times{{\mathbb F}}_2$ of $M_n$ and lies in $\operatorname{acl}_{M_n}([n+2]\setminus\{i,j\})$. Set $b_{ij_1}=([n+2]\setminus\{i,j\},0)$. We have $$\begin{aligned}
\label{eq:1212}
x_i(\overline{b_{ij}})
&=&x_i(([n+2]\setminus\{i,j\},0),\dots)\\\nonumber
&=&((([n+2]\setminus\{i,j\})^{\sigma_i},0+f_i([n+2]\setminus\{i,j\})),\dots)\\\nonumber
&=&((([n+2]\setminus\{i,j\})^{\sigma_i},m_{ij}),\dots),\nonumber\end{aligned}$$ where

$$\label{eq:1010}
m_{ij}=f_i([n+2]\setminus\{i,j\}).$$

Consider the matrix $M=(m_{ij})_{ij}$, with $m_{ii}=0$.

Let $i$ and $j$ be in $[n+2]$ with $i\neq j$ and $\{i,j\}\neq
\{n+1,n+2\}$. By Equation $(\ref{eq:6})$ and by hypothesis on $\{i,j\}$, the morphism $a_{[n+2]\setminus\{i,j\},[n+2]-i}$ is an inclusion map and so it fixes $([n+2]\setminus\{i,j\},0)$. Therefore, $$\begin{aligned}
x_i\circ a_{[n+2]\setminus\{i,j\},[n+2]-i}(\overline{b_{ij}})
&=&x_i\circ a_{[n+2]\setminus\{i,j\},[n+2]-i}(([n+2]\setminus\{i,j\},0),\dots)\\
&=&x_i(([n+2]\setminus\{i,j\},0),\dots)\\
&=&((([n+2]\setminus\{i,j\})^{\sigma_i},m_{ij}),\dots),\end{aligned}$$ where in the last equality we used Equations $(\ref{eq:1212})$ and $(\ref{eq:1010})$. Similarly, replacing $i$ with $j$, we obtain $$\begin{aligned}
x_i\circ a_{[n+2]\setminus\{i,j\},[n+2]-i}(\overline{b_{ij}})
&=&x_j\circ a_{[n+2]\setminus\{i,j\},[n+2]-j}(([n+2]\setminus\{i,j\},0),\dots)\\
&=&x_j(([n+2]\setminus\{i,j\},0),\dots)\\
&=&((([n+2]\setminus\{i,j\})^{\sigma_j},m_{ji}),\dots). \end{aligned}$$ Now, by Equation $(\ref{eq4})$, we have

$$\begin{aligned}
x_i\circ a_{[n+2]\setminus\{i,j\},[n+2]-i}(\overline{b_{ij}})
&=&x_i\circ
a_{[n+2]\setminus\{i,j\},[n+2]-i}(([n+2]\setminus\{i,j\},0),\dots)\\
&=&x_j\circ
a_{[n+2]\setminus\{i,j\},[n+2]-j}(([n+2]\setminus\{i,j\},0),\dots).\end{aligned}$$

In particular, $$\label{eq:yyy}
m_{ij}=m_{ji}, \qquad \textrm{for every }i,j \textrm{ with }\{i,j\}\neq \{n+1,n+2\}.$$

By Equation $(\ref{eq:6})$ the morphism $a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+1)}$ is an inclusion map and so it fixes $([n+2]\setminus\{n+1,n+2\},0)$. Therefore, $$\begin{aligned}
& & x_{n+1}\circ a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+1)}(\overline{b}_{n+1,n+2})\\
&=& x_{n+1}\circ a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+1)}(([n+2]\setminus\{n+1,n+2\},0),\dots)\\
&=&x_{n+1}(([n+2]\setminus\{n+1,n+2\},0),\dots)\\
&=&((([n+2]\setminus\{n+1,n+2\})^{\sigma_{n+1}},m_{(n+1)(n+2)}),\dots).\end{aligned}$$ By Equation $(\ref{eq:6})$ the morphism $f|_{a([n])}=a_{[n],[n+1]}=a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+2)}$ maps $([n+2]\setminus\{n+1,n+2\},0)$ to $([n+2]\setminus\{n+1,n+2\},1)$. Therefore, $$\begin{aligned}
& & x_{n+2}\circ a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+2)}(\overline{b}_{n+1,n+2})\\
&=&x_{n+2}\circ a_{[n+2]\setminus\{n+1,n+2\},[n+2]-(n+2)}(([n+2]\setminus\{n+1,n+2\},0),\dots)\\
&=&x_{n+2}\circ f|_{a([n])}(([n+2]\setminus\{n+1,n+2\},0),\dots)\\
&=&x_{n+2}(([n+2]\setminus\{n+1,n+2\},1),\dots)\\
&=&((([n+2]\setminus\{n+1,n+2\})^{\sigma_{n+2}},m_{(n+2)(n+1)}+1),\dots).\end{aligned}$$ By Equation $(\ref{eq4})$ (applied to $i=n+1$ and $j=n+2$), we have $$\begin{aligned}
& & (([n+2]\setminus\{n+1,n+2\})^{\sigma_{n+1}},m_{(n+1)(n+2)})\\
& =& (([n+2]\setminus\{n+1,n+2\})^{\sigma_{n+2}},m_{(n+2)(n+1)}+1)\end{aligned}$$ and $$\label{eq:1414}
m_{(n+1)(n+2)}=m_{(n+2)(n+1)}+1.$$

Now, we are ready to get a contradiction. We claim that each row of $M$ adds up to zero. We have $$\begin{aligned}
\sum_{j=1}^{n+2}m_{ij}&=&
\sum_{j\in ([n+2]-i)}m_{ij}=
\sum_{j\in ([n+2]-i)}f_i([n+2]\setminus\{i,j\})\\
&=&(\beta_{n+1,n}^\ast
f_i)([n+2]-i)=0, \end{aligned}$$ where in the first equality we used that $m_{ii}=0$, in the second equality we used Equation $(\ref{eq:1010})$ and in the last equality we used that $f_i\in
\operatorname{im}\beta_{n,n-1}^\ast=\operatorname{ker}\beta_{n+1,n}^{*}$. In particular, the sum of all the entries of $M$ is zero. Hence $$0=\sum_{ij}m_{ij}=\sum_{i<j}(m_{ij}+m_{ji}).$$ By Equation $(\ref{eq:yyy})$, $m_{ij}=m_{ji}$ if $\{i,j\}\neq \{n+1,n+2\}$. So, in the previous sum there is only one non-zero summand. Namely, $m_{(n+1)(n+2)}+m_{(n+2)(n+1)}=0$. Now, Equation $(\ref{eq:1414})$ yields $$m_{(n+1)(n+2)}+m_{(n+2)(n+1)}=m_{(n+1)(n+2)}+m_{(n+1)(n+2)}+1=1,$$ a contradiction. This contradiction finally proves that the extension $\overline{a}$ does not exist.

Now, Theorem \[summary\] follows at once from Proposition \[unik\], \[exik\], \[unin+1\], \[notn+2\]. Finally, we point out that Proposition \[unin+1\] also follows from Theorem \[thm:GoKo\] and Proposition \[notn+2\].

Extension of Example \[esempio1\] {#coincidence}
=================================

In this section we remark that for every $n\geq 2$ the theories $T_n$ are stable and that the family of examples $\{M_n\}_{n\geq 2}$ generalizes the example due to E.Hrushovski given in [@PKM], see Example \[esempio1\] in Section \[sec0\].

\[generalization\] Let $\Omega$ be a countable set, and $C=[\Omega]^n\times {{\mathbb Z}}/2{{\mathbb Z}}$. Also let $E\subseteq \Omega\times [\Omega]^2$ be the membership relation, and let $P$ be the subset of $C^{n+1}$ such that $((w_1,\delta_1),\dots,(w_{n+1},\delta_{n+1}))\in P$ if and only if there are distinct $c_1,\dots,c_{n+1}\in \Omega$ such that $w_i=\{c_1,\dots, c_{n+1}\}\setminus c_i$ and $\delta_1+\cdots+\delta_{n+1}=0$. Now let $\overline{M}_n$ be the model with the $3$-sorted universe $\Omega,[\Omega]^n,C$ and equipped with relations $E,P$ and projection on the first coordinate $\pi:C\rightarrow
    [\Omega]^n$. Since $\overline{M}_n$ is a reduct of $(\Omega,
    {{\mathbb Z}}/2{{\mathbb Z}})^{\textrm{eq}}$, we get that $\operatorname{Th}(\overline{M}_n)$ is stable.

\[coincides\] Let $\overline{M}_n$ be the structures described in Definition \[generalization\]. Then $\operatorname{Aut}(\overline{M}_n)=\operatorname{im}\beta^\ast_{n,n-1} \rtimes \operatorname{Sym}(\Omega)$. In particular, $\overline{M}_n$ and $M_n$ are interdefinable.

*Proof.* First we show that $\operatorname{Sym}(\Omega)$ is a subgroup of $\operatorname{Aut}(\overline{M}_n)$. Indeed, the group $\operatorname{Sym}(\Omega)$ acts with its natural action on the sorts $\Omega$ and $[\Omega]^n$ of $\overline{M}_n$. Also, if $g\in \operatorname{Sym}(\Omega)$ and $(\{a_1,\dots,a_n\},\delta)\in C$, then we set $(\{a_1,\dots,a_n\},\delta)^g=(\{a_1^g,\dots,a_n^g\},\delta)$. This defines an action of $\operatorname{Sym}(\Omega)$ on $\overline{M}_n$. It is straightforward to see that the relations $E,P$ and the partition given by the fibers of $\pi$ are preserved by $\operatorname{Sym}(\Omega)$. Hence, $\operatorname{Sym}(\Omega)\leq \operatorname{Aut}(\overline{M}_n)$.

Let $\mu:\operatorname{Aut}(\overline{M}_n)\rightarrow \operatorname{Sym}(\Omega)$ be the map given by restriction on the sort $\Omega$ of $\overline{M}_n$. Since $\mu$ is a surjective homomorphism, we have that $\operatorname{Aut}(\overline{M}_n)$ is a split extension of $\operatorname{ker}\mu$ by $\operatorname{Sym}(\Omega)$. Every element of $\operatorname{ker}\mu$ preserves the fibres of $\pi$ and fixes all the elements of $[\Omega]^n$. So $\operatorname{ker}\mu$ is a closed $\operatorname{Sym}(\Omega)$-submodule of ${{\mathbb F}}_2^{[\Omega]^n}$.

Let $((w_1,\delta_1),\dots,(w_{n+1},\delta_{n+1}))$ be in $P$ and $f$ be in $\operatorname{ker}\mu$. Since $\operatorname{ker}\mu$ preserves $P$, we have $$f(w_1)+\delta_1
+\cdots +f(w_{n+1})+\delta_{n+1}=0.$$ From the definition of $P$ and $\beta_{n+1,n}^\ast$, we get $$\operatorname{ker}\mu=
\{f\in {{\mathbb F}}_2^{[\Omega]^n}\mid \sum_{x\in
  [w]^n}f(x)=0\textrm{ for every }w\in [\Omega]^{n+1}\}=\operatorname{ker}\beta_{n+1,n}^\ast.$$ By Proposition \[exact\], we have that $\operatorname{ker}\beta^\ast_{n+1,n}=\operatorname{im}\beta^\ast_{n,n-1}$. Therefore $\operatorname{Aut}(\overline{M}_n)=\operatorname{Aut}(M_n)$ and $\overline{M}_n,M_n$ are interdefinable.

Acknowledgements {#acknowledgements .unnumbered}
================

The authors thank J.Goodrick and A.Kolesnikov for the proof of Theorem \[thm:GoKo\] and for giving their permission to include their proof in our paper. We are grateful to D. M. Evans for his stimulating suggestions and we thank the anonymous referee for the very valuable comments and remarks on an earlier draft of the paper.

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