Trois couleurs: A new non-equational theory

TL;DR

This paper introduces a new class of non-equational ω-stable theories by coloring the free pseudospace, expanding the known examples beyond the previously limited stable theories.

Contribution

It constructs the first known non-equational ω-stable theories using a novel coloring method inspired by Hrushovski and Srour's example.

Findings

Successfully constructs non-equational ω-stable theories

Demonstrates the existence of non-equational stable theories beyond known examples

Provides a new approach to studying stability in first-order theories

Abstract

A first-order theory is equational if every definable set is a Boolean combination of instances of equations, that is, of formulae such that the family of finite intersections of instances has the descending chain condition. Equationality is a strengthening of stability yet so far only two examples of non-equational stable theories are known. We construct non-equational $\omega$-stable theories by a suitable colouring of the free pseudospace, based on Hrushovski and Srour's original example.

Equations

$$p(x;y)\to q(x;y).$$

$$p_0(x;y)\to p_1(x;y)\to \cdots \to p_{n-1}(x;y)\to p_0(x;y).$$

$$p_0\to p_1\to \cdots \to p_{n-1}\to p_0.$$

$$p^{\prime }(x;b,b^{\prime })\to q^{\prime }(x;b,b^{\prime }).$$

$$r_{n\cdot i}(x;b,b^{\prime })\to \cdots \to r_{n\cdot i^{\prime }}(x;b,b^{\prime })$$

$$p_{1}(x,b)\otimes\Big{(}p_{2}(x^{\prime},b)\otimes p_{3}(x^{\prime\prime},b)\Big{)}=\Big{(}p_{1}(x,b)\otimes p_{2}(x^{\prime},b)\Big{)}\otimes p_{3}(x^{\prime\prime},b).$$

$$p^j(x^j;y^j,c)\to q^j(x^j;y^j,c),\text{ for }j=1,2,$$

$$p^1(x^1;y^1,c)\otimes p^2(x^2;y^2,c)\to q^1(x^1;y^1,c)\otimes q^2(x^2;y^2,c).$$

$$b^{1}\cup\{a^{1}_{i}\}_{i\in\mathbb{N}}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{c}b^{2}\cup\{a^{2}_{i}\}_{i\in\mathbb{N}}.$$

$$p_0(x,y)\to \cdots \to p_{n-1}(x,y)\to p_0(x,y)$$

$$p_0(x;y,c)\to \cdots \to p_{k-1}(x;y,c)\to p_0(x;y,c)$$

$$r_1(\bar{x};\bar{y},c)$$

$$r_2(\bar{x};\bar{y},c)$$

$$r_3(\bar{x};\bar{y},c)$$

$$r_1(\bar{x};\bar{y},c)\to r_2(\bar{x};\bar{y},c)\to r_3(\bar{x};\bar{y},c).$$

$$(a^1,a^2,\dots ,a^{k-2},a^0,b^1,b^2,\dots ,b^{k-2},b^0)$$

$$\begin{gathered} \mathrm{tp}(\{a^0,\dots ,a^{k-2}\};\{b^0,\dots ,b^{k-2}\},c)\to \mathrm{tp}(\{a^1,\dots ,a^{k-1}\};\{b^1,\dots ,b^{k-1}\},c)\to \\ \to \mathrm{tp}(\{a^0,\dots ,a^{k-2}\};\{b^0,\dots ,b^{k-2}\},c) \end{gathered}$$

$$a^{\rho(i)}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{c}b^{\tau(i)}\text{ and }a^{\rho(k-1)}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{c}b^{\tau(k-1)}$$

$$\phi \in p_0(x,y)\to p_1(x,y)\to \dots \to p_{n-1}(x,y)\to p_0(x,y),$$

$$\mathrm{Ker}((a_i{)}_{i\in I})=\bigcup _{\stackrel{I_0,I_1\subset I}{I_0\ll I_1}}\mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{I_0})\cap \mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{I_1}).$$

$$\mathrm{Ker}((a_i{)}_{i\in I})=\mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{I_0})\cap \mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{I_1}),\text{ for any }{I}_0\ll I_1\text{ both infinite}.$$

$$K=\bigcup _{\stackrel{J_0,J_1\subset I}{J_0\ll J_1}}\mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{J_0})\cap \mathrm{ac}{\mathrm{l}}^{\mathrm{eq}}(a_{J_1}).$$

$$a_{I_{0}}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{E}a_{I_{1}},$$

$$\mathfrak{p}=\{\phi (x,b){\mathcal{L}}_{\mathbb{U}}\text{-formula}\mid \phi (a_i,b)\text{ for all but finitely many }i\in I\}.$$

$$a_i\models \mathfrak{p}{\upharpoonright }_{K\cup (a_j{)}_{j<i}},$$

$$p_0(x,y)\to p_1(x,y)\to \dots \to p_{n-1}(x,y)\to p_0(x,y),$$

$$F_0\underset{s_1}{\to }{F}_1\cdots F_{n-1}\underset{s_n}{\to }{F}_n$$

$$F\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{G}D.$$

$$X\cup\{c\}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{Z}Y.$$

$$F_{\alpha}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{x_{\alpha}}x_{\beta}\text{ \quad and \quad}F_{\alpha}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss$\displaystyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\displaystyle\smile$\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss$\textstyle\mid$\hss}\lower 3.87495pt\hbox to0.0pt{\hss$\textstyle\smile$\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss$\scriptstyle\mid$\hss}\lower 1.89871pt\hbox to0.0pt{\hss$\scriptstyle\smile$\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss$\scriptscriptstyle\mid$\hss}\lower 0.96873pt\hbox to0.0pt{\hss$\scriptscriptstyle\smile$\hss}\kern 1.42882pt}}_{x_{\alpha},x_{\beta}}F_{\beta}.$$

$$\mathrm{acl}(X)=⟨X⟩\cup \mathrm{EP}(⟨X⟩).$$

Full text

The first author conducted research partially supported by the program GeoMod AAPG2019 (ANR-DFG). Both authors were supported by the program MTM2017-86777-P

1. Introduction

Consider a first order complete theory $T$. A formula $\varphi(x;y)$ is an equation (for a given partition of the free variables into $x$ and $y$) if, in every model of $T$, the family of finite intersections of instances $\varphi(x,a)$ has the descending chain condition. The theory $T$ is equational if every formula $\psi(x;y)$ is equivalent modulo $T$ to a Boolean combination of equations $\varphi(x;y)$.

Determining whether a particular stable theory is equational is not obvious. So far, the only known natural example of a stable non-equational theory is the free non-abelian group [14, 10], though the first example of a non-equational stable theory is of combinatorial nature and appeared in unpublished notes of Hrushovski and Srour [7]. They coloured the free pseudospace [4] with two colours in order to obtain two types $r(x,y)\neq r^{\prime}(x,y)$ which are not equationally separated, according to the terminology of [6, Section 2.1], that is, there are sequences $(a_{i},b_{i})_{i\in\mathbb{N}}$ and $(c_{i},d_{i})_{i\in\mathbb{N}}$, which can be taken indiscernible over $\emptyset$, such that $r(a_{i},b_{i})$ and $r^{\prime}(c_{i},d_{i})$ holds for all $i$, but $r^{\prime}(a_{i},b_{j})$ and $r(c_{i},d_{j})$ holds for $i<j$. In an equational theory, any two distinct types are equationally separated.

In all previously known examples of non-equational theories the failure of equationality is due to the presence of two distinct non-equationally separated types $r(x,y)\neq r^{\prime}(x,y)$ such that the length of $x$ is $1$. In this note, we will build on Hrushovski-Srour’s example in order to construct new examples of non-equational theories, where all distinct real types $p\neq q$ in finitely many variables are equationally separated.

2. Equations and indiscernibly closed sets

Most of the results in this section come from [11, 9].

Consider a first order theory $T$. A formula $\varphi(x;y)$ is an equation (with respect to a given partition of the free variables into $x$ and $y$) if, in every model of $T$, the family of finite intersections of instances $\varphi(x,b)$ has the descending chain condition. An easy compactness argument shows

Lemma 2.1**.**

The formula $\varphi(x;y)$ is an equation if there is no sequence $(a_{i},b_{i})_{i\in\mathbb{N}}$ in any model $M$ such that $M\models\varphi(a_{i},b_{j})$ and $M\not\models\varphi(a_{i},b_{i})$ for all $i<j$.

A Ramsey argument shows that, working in a sufficiently saturated model, the sequence $(a_{i},b_{i})$ can be assumed to be indiscernible of any infinite order type. Thus, if $\varphi(x;y)$ is an equation, then so are $\varphi^{-1}(x;y)=\varphi(y,x)$ and $\varphi(f(x);y)$, whenever $f$ is a $\emptyset$-definable function, which maps finite tuples to finite tuples. Finite conjunctions and disjunctions of equations are again equations. Note that equations are stable formulae.

In [9], an equivalent definition of equations was obtained in terms of indiscernibly closed sets: an element $c$ lies in the indiscernible closure $\operatorname{icl}(X)$ of a set $X$ if there is an indiscernible sequence $(a_{i})_{i\in\mathbb{N}}$ such that $a_{i}$ lies in $X$ for $i>0$ and $a_{0}=c$. Note that $X\subset\operatorname{icl}(X)$. A set $X$ is indiscernibly closed if $X=\operatorname{icl}(X)$.

Lemma 2.2**.**

[9, Theorem 3.16]** A formula $\varphi(x;y)$ is an equation if and only if the set $\varphi(M,b)$ is indiscernibly closed in in every model $M$ of $T$.

Proof.

Let us work inside a sufficiently saturated model $M$. If $\varphi(x;y)$ is not an equation, witnessed by the indiscernible sequence $(a_{i},b_{i})_{i\in\mathbb{Z}}$, as in Lemma 2.1, the set defined by $\varphi(x,b_{0})$ is not indiscernibly closed, for it contains all $a_{i}$’s with $i<0$, but does not contain $a_{0}$. Conversely, if some instance $\varphi(x,b)$ is not indiscernibly closed, there is an indiscernible sequence $(a_{i})_{i\in\mathbb{Z}}$ such that $M\models\varphi(a_{i},b)$ for $i<0$, but $M\not\models\varphi(a_{0},b)$. For every $j$ in $\mathbb{Z}$, there is an element $b_{j}$ in $M$ such that $M\models\varphi(a_{i},b_{j})$ for $i<j$, but $M\not\models\varphi(a_{j},b_{j})$. ∎

The theory $T$ is equational if every formula $\psi(x;y)$ is equivalent modulo $T$ to a Boolean combination of equations $\varphi(x;y)$. Since Boolean combinations of stable formulas are stable, equational theories are stable.

Typical examples of equational theories are the theory of an equivalence relation with infinite many infinite classes, the theory of $R$-modules for some ring $R$, or the theory of algebraically closed fields.

Equationality is preserved under bi-interpretability as well as addition of parameters [8]. It is unknown whether equationality holds if every formula $\varphi(x;y)$, with $x$ a single variable, is a boolean combination of equations.

It is easy to see that $T$ is equational if and only if all completions of $T$ are equational. So for the rest of this section we assume that $T$ is complete and work in a sufficiently saturated model $\mathbb{U}$.

Notice that a theory $T$ is equational if and only if every type $p$ over $A$ is implied by its equational part $\{\varphi(x,a)\in p\mid\varphi(x;y)\text{ is an equation}\}$.

Definition 2.3**.**

Given two types $p(x,b)$ and $q(x,b)$, define $p(x,b)\rightarrow q(x,b)$ if $q(x,b)\subset\operatorname{icl}(p(x,b))$, or equivalently, if there is an indiscernible sequence $(a_{i})_{i\in\mathbb{N}}$ such that all $\models p(a_{i},b)$ for $i>0$ and $\models q(a_{0},b)$. If $p(x,y)$ and $q(x,y)$ are the the corresponding (complete) types over $\emptyset$, we write

[TABLE]

A standard argument as in Lemma 2.2 with $p$ instead of $\varphi$ and and $q$ instead of $\neg\varphi$ yields the following:

Lemma 2.4**.**

We have that $p(x;y)\rightarrow q(x;y)$ if an only if there is a sequence $(a_{i},b_{i})_{i\in\mathbb{N}}$ such that $\models p(a_{i},b_{j})$ for $i<j$, and $\models q(a_{i},b_{i})$ for all $i$. Furthermore, we may assume that the sequence is indiscernible and of any given infinite order type.

The above characterisation provides an easy proof of the following remark:

Remark 2.5**.**

Clearly $p\rightarrow p$. If $p(x;y)\rightarrow q(x;y)$, then $p^{-1}\rightarrow q^{-1}$, where $p^{-1}(x;y)=p(y;x)$.

Furthermore, if $\operatorname{tp}(a;b)\rightarrow\operatorname{tp}(a^{\prime};b)$, then $a\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}a^{\prime}$. Thus, if $p(x;y)$ implies that $x$ (or $y$) is algebraic, then $p\rightarrow q$ only when $q=p$.

Corollary 2.6**.**

Let $f$ and $g$ be $\emptyset$-definable functions and $a,a^{\prime},b,b^{\prime}$ finite tuples, with $\operatorname{tp}(a;b)\rightarrow\operatorname{tp}(a^{\prime};b^{\prime})$. Then $\operatorname{tp}(f(a);g(b))\rightarrow\operatorname{tp}(f(a^{\prime});g(b^{\prime}))$.

Corollary 2.7**.**

A formula $\varphi(x;y)$ is an equation if and only if, whenever a type $p(x,y)$ contains $\varphi(x,y)$ and $p(x;y)\rightarrow q(x;y)$, then $\varphi(x,y)$ lies in $q(x;y)$.

Proof.

One direction follows clearly from Lemma 2.2. For the converse, assume that $\varphi(x;y)$ is not an equation and choose an indiscernible sequence $(a_{i},b_{i})_{i\in\mathbb{N}}$ as in Lemma 2.1. Let $p$ be the common type of the pairs $(a_{i},b_{j})$, with $i<j$ and $q$ be the common type of the pairs $(a_{i},b_{i})$. Then $p\rightarrow q$ and $\varphi$ belongs to $p$, but not to $q$. ∎

Definition 2.8**.**

A cycle of types is a sequence

[TABLE]

The cycle is proper if all the $p_{i}$’s are different. The theory $T$ is indiscernibly acyclic if there is no proper cycle of types of length $n\geq 2$.

Following the terminology of [6, Section 2.1], two distinct types $p(x;y)$ and $q(x;y)$ are not equationally separated if and only $p\rightarrow q\rightarrow p$.

Remark 2.9**.**

Every indiscernibly acyclic theory is stable.

Proof.

If there is a formula $\varphi(x;y)$ in $T$ with the order property, find an indiscernible sequence $(a_{i})_{i\in\mathbb{Z}}$ in $\mathbb{U}$ such that $\models\varphi(a_{i},a_{j})\text{ if and only if }i<j$. Set $p=\operatorname{tp}(a_{1};a_{0})$ and $q=\operatorname{tp}(a_{-1};a_{0})$. Then $p\neq q$, and since the sequence $(a_{i})_{i\not=0}$ is indiscernible, we have that $p\rightarrow q\rightarrow p$, so there is a proper cycle of types of length $2$. ∎

Remark 2.10**.**

Every equational theory is indiscernibly acyclic.

Proof.

Consider a cycle

[TABLE]

By Corollary 2.7, all the types $p_{i}$ contain the same equations, so they all agree, by equationality of $T$. ∎

Definition 2.11**.**

The theory $T$ satisfies the MS-criterion if there is some formula $\varphi(x,y)$ and a matrix $(a_{ij},b_{ij})_{i,j\in\mathbb{N}}$ such that:

1. (1)

$\models\varphi(a_{ij},b_{il})$ if and only if $j=l$. 2. (2)

$a_{ij},b_{ij}\equiv a_{ij},b_{kl}$, whenever $i<k$ and $j<l$.

Lemma 2.12**.**

If a theory $T$ satisfies the MS-criterion, then there is a proper cycle of types $p\rightarrow q\rightarrow p$. In particular, the theory is not equational (cf. [10, Proposition 2.6]).

Proof.

We may assume that the matrix $(a_{ij},b_{ij})_{i,j\in\mathbb{N}}$ is indiscernible, that is, the type $\operatorname{tp}(a_{ij},b_{ij})_{i\in I,j\in J}$ only depends on $|I|$ and $|J|$. Set $p=\operatorname{tp}(a_{00};b_{00})$, $q=\operatorname{tp}(a_{00};b_{01})$ and $r=\operatorname{tp}(a_{00};b_{11})$. Since $(a_{0j}b_{0j})_{j\in\mathbb{N}}$ is indiscernible, we have $q\rightarrow p$. Since $(a_{i0}b_{i1})_{i\in\mathbb{N}}$ is indiscernible, we have $r\rightarrow q$.

Now, by Definition 2.11 $(\ref{D:MS:phi})$, the formula $\varphi(x;y)$ belongs to $p(x;y)$ but not to $q(x;y)$, so $p\neq q$. By Definition 2.11 $(\ref{D:MS:iso})$ we have $p=r$, as desired. ∎

Since $p$ and $q$ contain the same equations, it follows that the above formula $\varphi$ cannot be a boolean combination of equations (cf. [10, Proposition 2.6]).

Let us assume for the rest of this section that $T$ is stable.

Lemma 2.13**.**

Let $p_{0}(x,b)\rightarrow\dotsb\rightarrow p_{n-1}(x,b)\rightarrow p_{0}(x,b)$ be a proper cycle of types and $b^{\prime}$ be some tuple such that $p_{0}(x,b)$ has only finitely many distinct nonforking extensions to $bb^{\prime}$. Then there is a proper cycle of types starting with some nonforking extension $p^{\prime}_{0}(x;b,b^{\prime})$ of $p_{0}(x,b)$ whose length is a multiple of $n$.

Proof.

First notice that, whenever $p(x,b)\rightarrow q(x,b)$ and $q^{\prime}(x,b,b^{\prime})$ is a non-forking extension of $q(x,b)$, then $p(x,b)$ has a nonforking extension $p^{\prime}(x,b,b^{\prime})$ with

[TABLE]

Indeed, consider an indiscernible sequence $(a_{i})_{i\in\mathbb{N}}$ such that $\models p(a_{i},b)$, for $i>0$, and $q(a_{0},b)$. We may assume that $a_{0}$ realises $q^{\prime}(x,b,b^{\prime})$ and that the sequence $(a_{i})_{i\in\mathbb{N}}$ is independent from $b^{\prime}$ over $b$. By a Ramsey argument, we may assume that the sequence $(a_{i})_{i>0}$ is indiscernible over $a_{0}bb^{\prime}$. Set now $p^{\prime}(x,b,b^{\prime})$ to be the type of $a_{1}$ over $bb^{\prime}$, so $p^{\prime}(x,b,b^{\prime})\rightarrow q^{\prime}(x,b,b^{\prime})$, as desired.

Let $k$ now be the number of distinct nonforking extensions of $p_{0}(x,b)$ to $bb^{\prime}$. Working backwards in the cycle of types, we deduce from the above that there is a sequence $r_{0}(x;b,b^{\prime})\rightarrow\dotsb\rightarrow r_{n\cdot k}(x;b,b^{\prime})$, where $r_{n\cdot i+j}(x,b,b^{\prime})$ is a non-forking extension of $p_{j}(x,b)$ for each $i\leq k$. Since $p_{0}$ has only finitely many distinct nonforking extensions to $bb^{\prime}$, there are two indices $i<i^{\prime}$ such that $r_{n\cdot i}(x,b,b^{\prime})=r_{n\cdot i^{\prime}}(x,b,b^{\prime})$. Choose $i$ and $i^{\prime}$ such that $0<i^{\prime}-i$ is least possible. Then

[TABLE]

is a proper cycle of types. ∎

Corollary 2.14**.**

If $T$ is totally transcendental, then it is indiscernibly acyclic if and only if so is $T^{\mathrm{eq}}$.

Proof.

We need only show that $T^{\mathrm{eq}}$ is indiscernibly acyclic, provided that $T$ is indiscernibly acyclic. Assume first that the type $p(x,e)$ starts a proper cycle of types, where $e$ is an imaginary element. Choose a real tuple $b$ such that $\pi_{E}(b)=e$ for some [math]-definable equivalence relation $E$. Since $T$ is totally transcendental, the type $p(x,e)$ has only finitely many nonforking extensions to $\{b,e\}$, so there is a proper cycle starting with some nonforking extension $p^{\prime}(x,b,e)$ of $p(x,e)$, by the Lemma 2.13. By the Corollary 2.6, if we restrict the types in the cycle to $b$, we have a cycle of types which must be proper, because $e$ is definable from $b$.

Since the relation $\rightarrow$ is symmetric in $x$ and $y$, we can now replace $x$ by some real tuple, so $T$ is not indiscernibly acyclic. ∎

Notation**.**

Given stationary types $p_{1}(x,b)$ and $p_{2}(x^{\prime},b)$, denote by $p_{1}(x,b)\otimes p_{2}(x^{\prime},b)$ the type of the pair $(a_{1},a_{2})$ over $b$, where $\models p_{i}(a_{i},b)$, for $i=1,2$, and a_{1}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{b}a_{2}.

Observe that

[TABLE]

Lemma 2.15**.**

Given stationary types $p^{j}(x^{j},y^{j},c)$ and $q^{j}(x^{j},y^{j},c)$ over a tuple $c$ in $\operatorname{acl^{eq}}(\emptyset)$ such that

[TABLE]

then

[TABLE]

By the above, the lemma generalises to an arbitrary finite product of types.

Proof.

For $j=1,2$, choose a tuple $b^{j}$ and an indiscernible sequence $(a^{j}_{i})_{i\in\mathbb{N}}$ such that $\models p^{j}(a_{i}^{j},b^{j},c)$, for $i>0$, and $\models q^{j}(a^{j}_{0},b^{j},c)$. We may assume that

[TABLE]

Since $c$ is algebraic over $\emptyset$, the sequences $\{a^{1}_{i}\}_{i\in\mathbb{N}}$ and $\{a^{2}_{i}\}_{i\in\mathbb{N}}$ are both indiscernible over $c$ and therefore mutually indiscernible, by stationarity of strong types, so $\{a^{1}_{i},a^{2}_{i}\}_{i\in\mathbb{N}}$ is indiscernible. Notice that $(a_{i}^{1},a_{i}^{2})$ realises $p^{1}(x^{1};b^{1},c)\otimes p^{2}(x^{2};b^{2},c)$ for $i>0$, and $(a^{1}_{0},a^{2}_{0})$ realises $q^{1}(x^{1};b^{1},c)\otimes q^{2}(x^{2};b^{2},c)$, as desired. ∎

Proposition 2.16**.**

If $T$ is totally transcendental, then it is indiscernibly acyclic if and only if there is no proper cycle of types in $T^{\mathrm{eq}}$ of length $2$.

Proof.

By Corollary 2.14, we need only prove one direction, so suppose

[TABLE]

is a proper cycle of types with real variables. Since $T$ is totally transcendental, there is a finite tuple $c$ in $\operatorname{acl^{eq}}(\emptyset)$ such that all nonforking extensions of all $p_{i}$’s to $c$ are stationary. Lemma 2.13 gives a proper cycle of stationary types

[TABLE]

for some $k$ in $\mathbb{N}$.

Denote by $\bar{x}=(x^{0},\ldots,x^{k-2})$ and $\bar{y}=(y^{0},\ldots,y^{k-2})$ and consider the types

[TABLE]

The Lemma 2.15 yields the cycle of types

[TABLE]

Given $(\bar{a},\bar{b})$ realising $r_{1}(\bar{x};\bar{y},c)$ and $(\bar{a}^{\prime},\bar{b}^{\prime})$ realising $r_{2}(\bar{x};\bar{y},c)$, notice that

[TABLE]

realise $r_{3}(\bar{x};\bar{y},c)$. If $f$ denotes the function which maps a $k-1$-tuple $(f^{0},\ldots,f^{k-1})$ to the imaginary coding the set $\{f^{1},\ldots,f^{k-1}\}$, Corollary 2.6 implies that

[TABLE]

In order to conclude, we need only show that the above two imaginary types are different. Otherwise, if the two types are equal, we have for each $1\leq i\leq k-1$, two values $0\leq\rho(i),\tau(i)\leq k-2$ such that $(a^{\rho(i)},b^{\tau(i)})\models p_{i}(x,y,c)$. Observe that no two elements $a^{i}$ and $a^{j}$, with $i\neq j$, can be equal since the independence a^{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{c}a^{j} would imply that $a_{i}$ is algebraic, and thus $p_{i+1}(x,y,c)=p_{i}(x,y,c)$, by the Remark 2.5. Likewise, no two elements $b^{i}$ and $b^{j}$ can be equal, for $i\neq j$. Thus, each of the maps $i\mapsto\rho(i)$ and $i\mapsto\tau(i)$ is injective, so both $\rho$ and $\tau$ are bijections.

If $\rho(k-1)=\tau(k-1)=j$, then $(a^{j},b^{j})$ realises both $p_{j}(x;y,c)$ and $p_{k-1}(x;y,c)$, which contradicts that the cycle of types is proper. Hence, the values $\rho(k-1)$ and $\tau(k-1)$ are different, so there must be some $1\leq i\leq k-2$ such that $\rho(i)\neq\tau(i)$. The independences

[TABLE]

imply that $p_{i}(x,y,c)=p_{k-1}(x,y,c)$, by the Remark 2.5 and stationarity of strong types, which yields the desired contradiction.

∎

We do not know whether Corollary 2.14 and Proposition 2.16 are true for arbitrary stable theories.

All known examples of non-equational stable theories have a proper cycle of real types of length $2$. Indeed, in Hrushovski and Srour’s primordial example [7], the type of a white point and the type of a red point in a plane indiscernibly converge to each other, whereas the non-abelian free group satisfies the MS-criterion [10, Lemmata 3.4 & 3.6]. In this note, we will provide new examples of non-equational totally transcendental theories, one for each natural number $k$, having proper cycles of length $k$ but no proper cycles of real types of length strictly smaller than $k$. We will do so by suitable colouring the free pseudospace, mimicking the construction of Hrushovski and Srour. The following question seems hence natural, though we do not have a solid guess what the answer will be.

Question**.**

Is there a non-equational indiscernibly acyclic theory?

Related to the above, we wonder whether there is a local characterisation of equationality in terms of cycles of types:

Question**.**

Is a formula $\varphi(x,y)$ a Boolean combination of equations if and only if whenever

[TABLE]

then $\varphi$ belongs to $p_{i}$ for every $i>0$?

Do two types $p$ and $q$ contain the exact same equations if and only if $p$ and $q$ both occur in a (proper) cycle of types?

Observe that a positive answer to the second question would positively answer the first one.

3. Indiscernible Kernels

To our knowledge, the results in this section only appeared in print form in Adler’s Master’s Thesis [1] (in German). Therefore, we will include their proofs, even if the results are most likely well-known among the community.

As before, work inside a sufficiently saturated model $\mathbb{U}$ of the complete theory $T$.

Notation**.**

Given two subsets $I_{0}$ and $I_{1}$ of a linearly ordered infinite index set with no endpoints, we write $I_{0}\ll I_{1}$ if $i_{0}<i_{1}$ for all $i_{0}$ in $I_{0}$ and $i_{1}$ in $I_{1}$. If $(a_{i})_{i\in I}$ is a sequence indexed by $I$, set $\operatorname{acl^{eq}}(a_{I_{0}})=\operatorname{acl}(\{a_{i}\}_{i\in I_{0}})$.

Definition 3.1**.**

The kernel of the indiscernible sequence $(a_{i})_{i\in I}$ is defined as

[TABLE]

Note that we may assume that both $I_{0}$ and $I_{1}$ are finite subsets of $I$. Furthermore, the set $\operatorname{acl^{eq}}(a_{I_{0}})\cap\operatorname{acl^{eq}}(a_{I_{1}})$ only depends on $|I_{0}|$ and $|I_{1}|$ (possibly after enlarging $I$), since $(a_{i})_{i\in J}$ is indiscernible over $a_{I_{0}}$, whenever the segment $J\subset I\setminus I_{0}$ is either left of $I_{0}$ or right of $I_{0}$. If the sequence is indiscernible as a set (which is always the case in stable theories), then we may define the kernel by considering all the intersections given by pairs $(I_{0},I_{1})$ with $I_{0}\cap I_{1}=\emptyset$.

Observe that (if $I$ is large enough),

[TABLE]

Lemma 3.2**.**

The kernel $K$ of an indiscernible sequence $(a_{i})_{i\in I}$ is the largest subset of $\operatorname{acl^{eq}}((a_{i})_{i\in I})$ over which the sequence is indiscernible.

Proof.

We may assume that $I$ has no endpoints. Clearly, the sequence is indiscernible over $K$. Given a tuple $b$ in $\operatorname{acl^{eq}}(a_{I_{0}})$, for $I_{0}\subset I$ finite, such that the sequence is indiscernible over $b$, the tuple $b$ also lies in $\operatorname{acl^{eq}}(a_{I_{1}})$, whenever the segment $I_{1}\gg I_{0}$ has the same size as $I_{0}$, so $b$ lies in the kernel

[TABLE]

∎

Lemma 3.3**.**

If $T$ is stable, then the kernel $K$ of an indiscernible sequence $(a_{i})_{i\in I}$ is the smallest algebraically closed subset (in $T^{\mathrm{eq}}$) over which the sequence is independent.

Proof.

Let $E$ be an algebraically closed subset (in $T^{\mathrm{eq}}$) such that $(a_{i})_{i\in I}$ is $E$-independent. In particular, for each $I_{0}<I_{1}$, we have that

[TABLE]

so $K\subset E$.

Let now $\mathfrak{p}=\mathrm{Av}((a_{i})_{i\in I})$ be the average type, that is,

[TABLE]

Since $\mathfrak{p}$ is invariant over every infinite subsequence of $(a_{i})_{i\in I}$, its canonical base $C$ is contained in $K$. Thus, the sequence is $C$-indiscernible and $\mathfrak{p}$ is a nonforking extension of the stationary type $\mathfrak{p}\!\!\upharpoonright_{K}$.

It suffices to show that $a_{i}\models\mathfrak{p}\!\!\upharpoonright_{K\cup(a_{j})_{j<i}}$, since any Morley sequence of $\mathfrak{p}\!\!\upharpoonright_{K}$ has this property and its type over $K$ is unique. Thus, let $\varphi(x,(a_{j})_{j<i})$ be a formula in $\mathfrak{p}\!\!\upharpoonright_{K\cup(a_{j})_{j<i}}$. We may clearly assume that $I$ has no last element. By definition of the average type, there is some $a_{t}\models\mathfrak{p}\!\!\upharpoonright_{K\cup(a_{j})_{j<i}}$ with $t\geq i$. By indiscernibility,

[TABLE]

as desired. ∎

Corollary 3.4**.**

In a stable theory $T$, every indiscernible sequence is a Morley sequence over its kernel.

Using kernels, we can provide a different characterisation of the relation $\rightarrow$ in a stable theory.

Corollary 3.5**.**

Given types $p(x;y)$ and $q(x;y)$ in a stable theory $T$, we have that $p\rightarrow q$ if and only if there is a set $C$ and tuples $a$, $a^{\prime}$ and $b$ such that:

• •

$\models p(a,b)$* and $\models q(a^{\prime},b)$;*

• •

$a\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{C}a^{\prime}$, and

• •

a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}b.

In particular, given a cycle

[TABLE]

there are tuples $b$, $a_{0},\ldots,a_{n}$ and subsets $C_{0},\ldots,C_{n-1}$ such that:

• •

$\models p_{r}(a_{r},b)$, for $0\leq r\leq n-1$, and $\models p_{0}(a_{n},b)$.

• •

$a_{r}\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{C_{r}}a_{r+1}$* for $0\leq r\leq n-1$.*

• •

a_{r}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C_{r}}b* for $0\leq r\leq n-1$.*

Proof.

If $p\rightarrow q$, choose some tuple $b$ and an indiscernible sequence $(a_{i})_{i<|T|^{+}}$ such that $q(a_{0},b)$ and $p(a_{i},b)$ for each $i>0$. Consider the kernel $K$ of the sequence, which is algebraically closed in $T^{\mathrm{eq}}$, so $a_{i}\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{K}a_{j}$ for all $i,j$. In particular, the subsequence $(a_{i})_{0<i<|T|^{+}}$ is Morley sequence over $K$, so there is some $i_{0}<|T|^{+}$ such that a_{i_{0}}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{K}b. Set $C=K$, $a^{\prime}=a_{0}$ and $a=a_{i_{0}}$.

For the other direction, set $a_{0}=a^{\prime}$ and choose for each $0\neq i$ in $\mathbb{N}$ a realisation $a_{i}\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{C}a$ such that a_{i}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}b,\{a_{j}\}_{j<i}. Since strong types are stationary, we have that $p(a_{i},b)$, for $i\neq 0$. Furthermore, the sequence $\{a_{i}\}_{i\geq 0}$ is indiscernible over $C$, by construction. ∎

The above provides a simpler characterisation of equations in stable theories (cf. [6, Remark 2.4]).

Remark 3.6**.**

In a stable theory $T$, a formula $\varphi(x;y)$ is an equation if and only if for every set set $C$ and tuples $a$, $a^{\prime}$ and $b$ such that $\varphi(a,b)$ holds with a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}b, then so does $\varphi(a^{\prime},b)$ hold, whenever $a^{\prime}\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{C}a$.

Proof.

Given $C$, $a$, $a^{\prime}$ and $b$ as in the statement, Corollary 3.5 yields that $\operatorname{tp}(a,b)\rightarrow\operatorname{tp}(a^{\prime},b)$. As $\varphi$ belongs to $\operatorname{tp}(a,b)$, it must lie in $\operatorname{tp}(a^{\prime},b)$, by Corollary 2.7.

For the other direction, it suffices to show that $\varphi$ lies in $q$, whenever $\varphi$ belongs to $p$ and $p\rightarrow q$, by Corollary 2.7. By Corollary 3.5, there are $C$, $a$, $a^{\prime}$ and $b$ such that $p(a,b)$, $q(a^{\prime},b)$, a\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}b and $a^{\prime}\stackrel{{\scriptstyle\mathrm{s}tp}}{{\equiv}}_{C}a$. Since $\varphi(a,b)$ holds, we conclude that so does $\varphi(a^{\prime},b)$, that is, the formula $\varphi$ belongs to $q$, as desired. ∎

4. A blank pseudospace

Hrushovski and Srour produced the first example [7] of a non-equational stable theory by adding two colours to an underlying ($2$-dimensional) free pseudospace, a structure later studied by Baudisch and Pillay [4]. Subsequently, the free ($n$-dimensional) pseudospace has been considered from different perspectives, either as a lattice [12, 13] or as a right-angled building [2, 3], in order to show that the ample hierarchy is strict. In this section, we will recall the basic properties of the free $2$-dimensional pseudospace.

A geometry is a graph whose vertices have levels [math], $1$ and $2$. Vertices of level [math] are called points (usually denoted by the letter $c$), whereas vertices of level $1$ are lines (denoted by $b$) and vertices of level $2$ are planes (denoted by $a$). In an abuse of notation, we say that the point $c$ lies in the plane $a$ if there is a line $b$ contained in $a$ passing through $c$, though there are no edges between points and planes. We refer to a subgraph of the form $a-b-c$ as a flag.

A letter $s$ is a non-empty subinterval of $[0,2]$. Given a flag $F$ in a geometry $A$, a new geometry $B$ is obtained from $A,F$ via the operation $s$ by freely adding a new flag $G$ which agrees with $F$ on the levels in $[0,2]\setminus s$:

The free pseudospace $\mathrm{M}_{\infty}$ is obtained by successively applying countably many times all of the above operations starting from a flag. The geometry $\mathrm{M}_{\infty}$ is independent, up to isomorphism, of the order in which the operations are applied. It is denoted by $\mathrm{M}_{\infty}^{2}$ in [2, Definition 4.6]. Observe that the geometry obtained by only considering the operations [math], $1$ and $2$ is an elementary substructure (cf. [2, Corollary 4.31]) of $\mathrm{M}_{\infty}$ (namely, the prime model).

We will now exhibit the axioms for the theory $\mathrm{PS}$ of $\mathrm{M}_{\infty}$. Let us first fix some notation. We will write the symbol $i$ for the letter $[i]$, that is, from now on, we will write [math] instead of $[0]$, etc. A word is a sequence of letters. A permutation of the word $u$ is obtained by successively replacing an occurrence of the subword $0\cdot 2$ by the subword $2\cdot 0$; similarly the subword $2\cdot 0$ is permuted to $0\cdot 2$ (Note that $0\cdot 2$ denotes the word $[0]\cdot[2]$, but we do not want to render the notation cumbersome with additional brackets). The word $u$ is reduced if it does not contain, up to permutation, a subword of the form $s\cdot t$, where $s\subset t$ or $t\subset s$ (please note that our notation $s\subset t$ does not imply $s\subsetneq t$).

A flag path

[TABLE]

with word $u=s_{1}\dotsb s_{n}$ is a sequence of flags such that, for each $1\leq i\leq n$, the flag $F_{i}$ agrees with $F_{i-1}$ exactly in the levels in $[0,2]\setminus s_{i}$. The above flag path is reduced if its word is reduced and for each $i$, the flags $F_{i-1}$ and $F_{i}$ cannot be connected by a splitting, that is, a flag subpath whose word consists of proper subletters of $s_{i}$. It is not hard to show that every two flags are connected by a reduced path [3, Corollary 3.13].

Fact 4.1**.**

[3, Theorem 4.12] The theory $\mathrm{PS}$ is axiomatised by the following properties:

1. (1)

The universe is a geometry such that every vertex lies in a flag. 2. (2)

For every level $i$ in $[0,2]$ and every flag $F$, there are infinitely many flags $G$ with $F\xrightarrow[i]{}G$. 3. (3)

Every closed reduced flag path $F_{0}\xrightarrow[s_{1}]{}F_{1}\dotsb F_{n-1}\xrightarrow[s_{n}]{}F_{0}$ has length $n=0$.

It was proven in [3, Theorem 3.26] that property (3) can be expressed by a set of elementary sentences.

We will now describe types and the geometry of forking in the pseudospace. We refer the reader to [3, Sections 3–7] for the corresponding proofs. Since there are no non-trivial reduced closed paths of flags, the word $u$ connecting two flags $F$ and $G$ by a reduced path $F\xrightarrow[u]{}G$ is unique, up to permutation, and will be denoted by $\operatorname{d}(F,G)$. The flags $F$ and $G$ agree modulo a subset $S$ of $[0,2]$, that is, they have the same vertices in all levels off $S$, if and only if the letters in $\operatorname{d}(F,G)$ are all contained in $S$. In particular, the collection of points and lines, resp. lines and planes, form a pseudoplane, so every two lines intersect in at most one point, resp. lie in at most one plane. Furthermore, the intersection of two distinct planes is either empty, a unique point or a unique line [4, Axiom $\Sigma 3$]. Actually, the geometry forms a lattice, once a smallest element $\mathbf{0}$ and a largest element $\mathbf{1}$ are added [12, 13].

If $u=\operatorname{d}(F,G)=u_{1}\cdot u_{2}$, given two reduced flag paths

and a vertex $p$ in $H$ of level $i$ which does not wobble, that is, such that the word $u_{1}\cdot i$ or the word $i\cdot u_{2}$ is reduced, then $p$ is also a vertex of $H_{1}$. In particular, the vertex $p$ is definable over $F,G$.

A non-empty subset $A$ of $\mathrm{M}_{\infty}$ is nice if:

• •

every vertex in $A$ lies in a flag fully contained in $A$; and

• •

every two flags in $A$ are connected by a reduced path of flags in $A$.

Algebraic closure and the definable closure of a set $X$ agree [12, Proposition 2.29] and coincide with the intersection of all nice sets $A\supset X$ [12, Corollary 2.27] (see also [13, Proposition 5.1 & Corollary 5.4]). If $X$ is finite, then so is the algebraic closure. The quantifier-free type of a nice subset determines its type. More generally:

Fact 4.2**.**

(cf.  [12, Remark 2.26]  & [13, Corollary 3.12]) The quantifier-free type of an algebraically closed subset determines its type in $\mathrm{PS}$.

Observe that if we apply one of the operations $[0]$, $[1]$ or $[2]$ to a flag in a nice set $A$, the resulting geometry is again nice.

Given a flag $F$ and a nice subset $A$, there is a flag $G$ in $A$ (called a base-point of $F$ over $A$) such that, for any flag $G^{\prime}$ in $A$, the word $\operatorname{d}(F,G^{\prime})$ is the non-splitting reduction of $\operatorname{d}(F,G)\cdot\operatorname{d}(G,G^{\prime})$, that is, whenever a subword $s\cdot t$ or $t\cdot s$ occurs in a permutation of the product $\operatorname{d}(F,G)\cdot\operatorname{d}(G,G^{\prime})$, with $s\subset t$, we cancel $s$. If we consider a reduced flag path $P$ connecting $F$ to some base-point $G$ over $A$ with word $\operatorname{d}(F,G)$, the set $A\cup P$ is again nice. Any flag occurring in the nice set $P$ appears in a permutation of the path $P$.

The theory $\mathrm{PS}$ of $\mathrm{M}_{\infty}$ is $\omega$-stable of rank $\omega^{2}$, equational with perfectly trivial forking and has weak elimination of imaginaries. Forking can be easily described: Given nice sets $A$ and $B$ containing a common algebraically closed subset $C$, we have that A\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}B if and only if for every nice set $D\supset C$ and flags $F$ in $A$ and $H$ in $B$ we have that $\operatorname{d}(F,H)$ is the non-splitting reduction of $\operatorname{d}(F,G)\cdot\operatorname{d}(G,H)$, where $G$ is a base-point of $F$ over $D$. In particular,

[TABLE]

Remark 4.3**.**

([12, Theorem 2.35]  & [13, Proposition 4.3 & Theorem 4.13]) Assume that $A$, $B$ and $C=A\cap B$ are algebraically closed and A\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{C}B. Then

1. (1)

$A\cup B$ is algebraically closed, 2. (2)

if a vertex $x$ in $A$ is directly connected to a vertex $y$ in $B$, then $x$ or $y$ must lie in $C$, 3. (3)

if a point in $A$ lies in a plane of $B$, then there is a line in $C$ connecting them, 4. (4)

a point $c$, which belongs to both a line in $A\setminus C$ and to a line in $B\setminus C$, lies in $C$.

Before introducing the $k$-colored pseudospace in section 5, we will prove several auxiliary results about the free pseudospace. We hope that this will allow the reader to become more familiar with the theory $\mathrm{PS}$.

Lemma 4.4**.**

Let $X$ and $Y$ be algebraically closed sets independent over their common intersection $Z$. Given a point $c$ not contained in $Y\setminus Z$ lying in the line $b$ of $X$, then

[TABLE]

Proof.

By the transitivity of non-forking, we may assume that $Z=X$. If $c$ belongs to $X$, then there is nothing to prove. Otherwise, the type of $c$ over $X$ has Morley rank $1$ (it is actually strongly minimal), by [4, Remark 6.2] (cf. [2, Corollary 7.13]). Since the extension $\operatorname{tp}(c/Y)$ is not algebraic, it does not fork over $X$. ∎

Lemma 4.5**.**

The type of a set $X$ is determined by the collection of types $\operatorname{tp}(x,x^{\prime})$, with $x$ and $x^{\prime}$ in $X$.

In particular, if $X\equiv_{Z}X^{\prime}$ and $X\equiv_{Y}X^{\prime}$, then $X\equiv_{YZ}X^{\prime}$.

Proof.

Choose an enumeration of $X=\{x_{\alpha}\}_{\alpha<\kappa}$ and flags $F_{\alpha}$ containing $x_{\alpha}$, for $\alpha<\kappa$, such that F_{\alpha}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{x_{\alpha}}X\cup\{F_{\beta}\}_{\beta<\alpha}. In particular, for $\alpha\neq\beta$, we have that

[TABLE]

Since the type of $F_{\alpha}$ over $x_{\alpha}$ is stationary, the type of the pair $(x_{\alpha},x_{\beta})$ determines the type of $F_{\alpha},F_{\beta}$. By [3, Theorem 7.24], the type of $(F_{\alpha})_{\alpha<\kappa}$, hence the type of $X$, is uniquely determined by the collection of types $\operatorname{tp}(x_{\alpha},x_{\beta})$, for $\alpha,\beta<\kappa$. ∎

5. A colored pseudospace

Work inside a sufficiently saturated model $\mathbb{U}$ of the theory $\mathrm{PS}$ of the free pseudospace and consider a natural number $k\geq 2$. For $0\leq i<k$, we use the notation $i+1$ instead of $i+1\textrm{ mod }k$, and likewise $i-1$ for $i-1\textrm{ mod }k$.

We colour the lines in $\mathbb{U}$, as well as the pairs $(a,c)$, where the point $c$ lies in the plane $a$, with $k$ many colours. Formally, we partition the set of lines into subsets $C_{0},\ldots,C_{k-1}$, and the set of pairs $(a,c)$, where $c$ lies in the plane $a$, into $I_{0},\ldots,I_{k}$. Given a plane $a$ and an index $0\leq i<k$, we denote by the section $I_{i}(a)$ the collection of points $c$ with $I_{i}(a,c)$.

Consider the theory $\mathrm{CPS}_{k}$ of $k$-colored pseudospaces with following axioms:

• •

The axioms of $\mathrm{PS}$.

Universal Axioms

• •

For each $0\leq i<k$, given a line $b$ with colour $i$ in a plane $a$, all the points $c$ in $b$ lie in the section $I_{i}(a)$ except at most one point, which lies in $I_{i+1}(a)$ (if such a point exists, we call it the exceptional point of $b$ in $a$).

Inductive Axioms

• •

Every line $b$ in a plane $a$ contains an exceptional point, denoted by $\mathrm{ep}(a,b)$.

• •

For each $0\leq i<k$, given a point $c$ and a plane $a$ with $I_{i}(a,c)$, there are infinitely many lines in $a$ passing through $c$ with colour $i$.

• •

For each $0\leq i<k$, given a point $c$ and a plane $a$ with $I_{i}(a,c)$, there are infinitely many lines in $a$ passing through $c$ with colour $i-1$.

• •

For every point $c$ in a line $b$, there are infinitely many planes $a$ containing $b$ such that $c$ is exceptional for $b$ in $a$.

We can construct a model of $\mathrm{CPS}_{k}$ as follows: We start with a flag $A_{0}=\{a-b-c\}$ with any colouring, eg. $b\in C_{0}$ and $I_{0}(a,c)$ and construct an ascending sequence $A_{0}\subset A_{1}\subset\dotsb$ of coloured geometries by applying one the operations $[0]$, $[1]$ and $[2]$ to a flag $a-b-c$ in $A_{j}$ obtain $A_{j+1}$, extending the colouring to $A_{j+1}$ in an arbitrary way whilst preserving the Universal Axioms. For example, do as follows:

• •

Operation [0] adds a new point $c^{\prime}$ to $b$. If $b$ has colour $i$, then for all $a^{\prime\prime}$ in $A_{j}$ containing $b$, paint the pair $(a^{\prime\prime},c^{\prime})$ with the colour $i$, if $\mathrm{ep}(a^{\prime\prime},b)$ already exists in $A_{j}$. Otherwise, paint $(a^{\prime\prime},c^{\prime})$ with the colour $i+1$.

• •

Operation [1] adds a new line $b^{\prime}$ between $a$ and $c$. If $(a,c)$ has colour $i$, then paint $b^{\prime}$ with the colour $i$ or the colour $i-1$, and see to it that each choice occurs infinitely often in the sequence.

• •

Operation [2] adds a new plane $a^{\prime}$ which contains $b$. If $b$ has colour $i$, then for all $c^{\prime\prime}$ in $A_{j}$ which lie in $b$, we give the pair $(a^{\prime},c^{\prime\prime})$ one of the colours $i$ or $i-1$. Each choice should occur infinitely often.

It is easy to see that the structure obtained in this fashion satisfies all axioms of $\mathrm{CPS}_{k}$, so the theory $\mathrm{CPS}_{k}$ is consistent.

Notation**.**

Given a subset $X$ of a model of $\mathrm{CPS}_{k}$, we will denote by $\langle X\rangle$ the algebraic closure of $X$ in the reduct $\mathrm{PS}$, and by $\mathrm{EP}(X)=\{\mathrm{ep}(a,b)\,,\,(a,b)\in X\times X\}$ the exceptional points of lines and planes from $X$.

Remark 5.1**.**

If the point $c$ is directly connected to a line in $X$, then $\langle X,c\rangle=\langle X\rangle\cup\{c\}$.

In particular, if $X=\langle X\rangle$, given $c$ in $\mathrm{EP}(X)$, then $X\cup\{c\}$ is algebraically closed in the reduct $\mathrm{PS}$.

Proof.

In order to show that $\langle X,c\rangle=\langle X\rangle\cup\{c\}$, it suffices to consider the case when $X$ is nice. The geometry $X\cup\{c\}$ is either $X$ or obtained from $X$ by applying the operation $[0]$, so it is nice again, and thus algebraically closed. ∎

Similar to [2, Proposition 4.26] or [13, Proposition 3.10], it is easy to see that the collection of partial isomorphisms between $\mathrm{PS}$-algebraically closed finite sets closed under exceptional points inside two $\aleph_{0}$-saturated models of $\mathrm{CPS}_{k}$ is non-empty and has the back-and-forth property, so we deduce the following:

Theorem 5.2**.**

The theory $\mathrm{CPS}_{k}$ is complete. Given a set $X$ in a model of $\mathrm{CPS}_{k}$ with $X=\langle X\rangle$ and $\mathrm{EP}(X)\subset X$, then the quantifier-free type of $X$ determines its type.

The back-and-forth system yields an explicit description of the algebraic closure, as well as showing that the theory $\mathrm{CPS}_{k}$ is $\omega$-stable, by a standard type-counting argument.

Corollary 5.3**.**

The theory $\mathrm{CPS}_{k}$ is $\omega$-stable. The algebraic closure $\operatorname{acl}(X)$ of a set $X$ is obtained by closing $\langle X\rangle$ under exceptional points:

[TABLE]

We deduce the following characterisation of forking over (colored) algebraically closed sets.

Corollary 5.4**.**

Let $X$ and $Y$ two supersets of an algebraically closed set $Z=\operatorname{acl}(Z)$ in $\mathrm{CPS}_{k}$. We have that

[TABLE]

if and only if

• •

X\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}^{\mathrm{PS}}_{Z}Y, and

• •

$\mathrm{EP}(\langle X\rangle)\cap\mathrm{EP}(\langle Y\rangle)\subset Z$.

Types over algebraically closed sets are stationary, that is, the theory $\mathrm{CPS}_{k}$ has weak elimination of imaginaries.

Proof.

Since $\mathrm{PS}$ has weak elimination of imaginaries, we have that non-forking in $\mathrm{CPS}_{k}$ implies nonforking in the reduct $\mathrm{PS}$ over algebraically closed sets, by [5, Lemme 2.1]. Clearly $\mathrm{EP}(\langle X\rangle)\cap\mathrm{EP}(\langle Y\rangle)\subset Z$.

For the other direction, we may assume that $X=\langle X\rangle$ and $Y=\langle Y\rangle$. Lemma 4.4 yields that

[TABLE]

Since $\operatorname{acl}(X)=X\cup\mathrm{EP}(X)$, Remark 4.3 implies that the set $\operatorname{acl}(X)\cup\operatorname{acl}(Y)$ is algebraically closed in $\mathrm{PS}$. We need only show that it contains all exceptional points, so it determines a unique type in the stable theory $\mathrm{CPS}_{k}$. If $c$ is an exceptional point of a plane $a$ and a line $b$ in $\operatorname{acl}(X)\cup\operatorname{acl}(Y)$, we may assume that $a$ lies in $X$ and $b$ lies in $Y$. Since $a$ and $b$ are directly connected and X\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}^{\mathrm{PS}}_{Z}Y, Remark 4.3 implies that $a$ or $b$ lies in $Z$. Therefore $c$ lies in $\mathrm{EP}(X)\cup\mathrm{EP}(Y)$ and hence is contained in $\operatorname{acl}(X)\cup\operatorname{acl}(Y)$, as desired. ∎

Corollary 5.5**.**

Let $X$, $Y$ and $Z=\operatorname{acl}(Z)$ be sets such that

[TABLE]

Then $\langle X,Y\rangle\cap\operatorname{acl}(X,Z)=\langle X,Y\rangle\cap\langle X,Z\rangle$.

Proof.

Let $\xi$ be in $\langle X,Y\rangle\cap\operatorname{acl}(X,Z)$. The independence

[TABLE]

yields that

[TABLE]

It follows from Corollary 5.4 that

[TABLE]

and thus

[TABLE]

Since $\xi$ lies in $\langle X,Y\rangle$, the above independence implies that $\xi$ lies in $\langle X,Z\rangle$, as desired. ∎

Proposition 5.6**.**

Let $X=\langle X\rangle$ and $Y=\langle Y\rangle$ be two subsets of a model of $\mathrm{CPS}_{k}$. A map $F:X\to Y$ is elementary with respect to the theory $\mathrm{CPS}_{k}$ if and only if it satisfies the following conditions:

1. (1)

The map $F$ is a partial isomorphism with respect to the reduct $\mathrm{PS}$. 2. (2)

The function $F$ preserves colours of lines and sections. 3. (3)

For all $a$, $a^{\prime}$ and $b$ in $X$, we have that $\mathrm{ep}(a,b)=\mathrm{ep}(a^{\prime},b)$ if and only if $\mathrm{ep}(F(a),F(b))=\mathrm{ep}(F(a^{\prime}),F(b))$.

Proof.

We need only show that $F$ is elementary, if it satisfies all three conditions. By Theorem 5.2, it suffices to show that $F$ extends to a partial isomorphism $\tilde{F}$ preserving colours between $\operatorname{acl}(X)=X\cup\mathrm{EP}(X)$ and $\operatorname{acl}(Y)=Y\cup\mathrm{EP}(Y)$.

For each line $b$ in $X$ contained in a plane $a$ of $X$, set $\tilde{F}(\mathrm{ep}(a,b))=\mathrm{ep}(F(a),F(b))$. Let us first show that $\tilde{F}$ is well-defined, which analogously yields that $\tilde{F}$ is a bijection. Suppose that $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b_{1})$, for a line $b_{1}$ contained in the plane $a_{1}$, both in $X$. If $b\neq b_{1}$, then $\mathrm{ep}(a,b)$ is the unique intersection of $b$ and $b_{1}$, both lines in $X$, so $\mathrm{ep}(a,b)$ lies in $X$ and hence its image is determined by $F$. Otherwise, we conclude that $b=b_{1}$, and thus $\tilde{F}$ is bijective, by Condition $(\ref{Cond:iso})$.

Similarly, the map $\tilde{F}$ defined above is a partial isomorphism with respect to the reduct $\mathrm{PS}$. We need only show that $\tilde{F}$ preserves the colours of sections. Choose a new point $\mathrm{ep}(a,b)$ not in $X$ and an arbitrary plane $a_{1}\neq a$ in $X$ containing $\mathrm{ep}(a,b)$. Since $\mathrm{ep}(a,b)$ does not lie in $X$, the intersection of $a$ and $a_{1}$ cannot solely consist of the point $\mathrm{ep}(a,b)$. Hence, the intersection of $a$ and $a_{1}$ is given by a unique line $b_{1}$, which lies in $X$ and contains $\mathrm{ep}(a,b)$. We conclude as before that $b=b_{1}$. The colour of $\mathrm{ep}(a,b)$ in $a_{1}$ is uniquely determined according to whether $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b)$, and thus so is the colour of its image in $F(a_{1})$ by $\tilde{F}$, by Condition $(\ref{Cond:iso})$.

∎

6. Colored paths

We will now show that the theory $\mathrm{CPS}_{k}$ is not indiscernibly acyclic, and hence it is not equational, yet every proper cycle of types has length at least $k$ (cf. Theorem 6.2), so we expect the complexity of these theories to increase as $k$ grows. However, we do not know whether two of these theories are bi-interpretable.

Theorem 6.1**.**

In $\mathrm{CPS}_{k}$ there is a proper cycle of types

[TABLE]

where both the variables $x$ and $y$ have length $1$. In particular, the theory $\mathrm{CPS}_{k}$ is not equational.

Proof.

For each $0\leq r<k$, a pair $(a,c)$ with colour $I_{r}$ has a unique type $p_{r}=\operatorname{tp}(c,a)$, for the set $\{a,c\}$ is algebraically closed, since it is the intersection of all the flags containing $a$ and $c$, and it is closed under exceptional points, for it contains no line. Clearly $p_{r}\neq p_{r+1}$, for each $0\leq r<k$.

It suffices to show that $p_{r}\rightarrow p_{r+1}$: Let $(a,c)$ with colour $I_{r}$, and choose a line $b$ connecting them with colour $r$. Let $c^{\prime}$ be the exceptional point of $b$ in $a$, so $(c^{\prime},a)\models p_{r+1}$. Now, the set $\{b\}$ is algebraically closed in $\mathrm{CPS}_{k}$. By Corollary 5.4, the points $c$ and $c^{\prime}$ have the same strong type over $b$, and

[TABLE]

Corollary 3.5 implies that $p_{r}=\operatorname{tp}(c,a)\rightarrow\operatorname{tp}(c^{\prime},a)=p_{r+1}$, as desired. ∎

Theorem 6.2**.**

Let $x$ and $y$ be finite tuples of variables. In $\mathrm{CPS}_{k}$, every proper cycle of types

[TABLE]

has length $n\geq k$.

Proof.

A proper cycle of types $p_{0}(x;y)\rightarrow p_{1}(x;y)\rightarrow\ldots\rightarrow p_{n-1}(x;y)\rightarrow p_{0}(x;y)$ as above induces a cycle in the reduct $\mathrm{PS}$, which is equational. Therefore, the colourless reducts of $p_{r}$ and $p_{s}$ agree, for all $r$, $s$.

Corollary 3.5 implies that there are tuples $f$, $e_{0},\ldots,e_{n}$ and algebraically closed subsets $Z_{0},\ldots,Z_{n-1}$ such that:

• •

$\models p_{r}(e_{r},f)$, for $0\leq r\leq n-1$, and $\models p_{0}(e_{n},f)$.

• •

$e_{r}\equiv_{Z_{r}}e_{r+1}$ for $0\leq r\leq n-1$.

• •

e_{r}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Z_{r}}f for $0\leq r\leq n-1$.

Set $Y=\operatorname{acl}(f)$, $X_{r}=\operatorname{acl}(e_{r})$, for $0\leq r\leq n$. Since the definable and algebraic closure coincide, and the colourless reducts of all $p_{r}$ agree, all the types $\operatorname{tp}^{\mathrm{PS}}(X_{r}Y)$ are equal. Denote $\langle X_{r}Y\rangle$ by $P_{r}$. We find colourless isomorphisms

[TABLE]

which fix $Y$ pointwise. Note that $X_{r}$ and $X_{r+1}$ have the same type over $Z_{r}$, for $r\leq n-1$. Lemma 4.5 yields that $\operatorname{tp}^{\mathrm{PS}}(X_{r}YZ_{r})=\operatorname{tp}^{\mathrm{PS}}(X_{r+1}YZ_{r})$, for $r\leq n-1$. The above map $F_{r}$ extends to a colourless isomorphism between $\langle X_{r}YZ_{r}\rangle$ and $\langle X_{r+1}YZ_{r}\rangle$, which is the identity on $\langle YZ_{r}\rangle$. We will still refer to this colourless isomorphism as $F_{r}$, keeping in mind that it is elementary in the sense of $\mathrm{CPS}_{k}$ on $\langle X_{r}Z_{r}\rangle$ and (clearly) on $\langle YZ_{r}\rangle$ separately. Observe that

[TABLE]

by the Remark 4.3 (1).

If a set $W$ is finite, so are the closures $\langle W\rangle$ and $\operatorname{acl}(W)$. Define its defect as the natural number

[TABLE]

Claim**.**

For each $r\leq n-1$, we have that $\mathrm{defect}(P_{r})\geq\mathrm{defect}(P_{r+1})$.

Proof of Claim. Whenever $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b_{1})$, for $b$ and $b_{1}$ in $P_{r}$, with $b\neq b_{1}$, then the point $\mathrm{ep}(a,b)$ lies in $P_{r}$ by 4.3 (4). Thus, it suffices to show the following:

1. (1)

Whenever the line $b$ in $P_{r}$ lies in the plane $a$ in $P_{r}$, with $\mathrm{ep}(a,b)$ in $P_{r}$, then $\mathrm{ep}(F_{r}(a),F_{r}(b))$ lies in $P_{r+1}$. 2. (2)

Whenever $a$, $a_{1}$ and $b$ lie in $P_{r}$ and $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b)$, then $\mathrm{ep}(F_{r}(a),F_{r}(b))=\mathrm{ep}(F_{r}(a_{1}),F_{r}(b))$.

For $(\ref{C:1st})$, since X_{r}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}_{Z_{r}}Y, the plane $a$ and the line $b$ must both lie in the same set $P_{r}\cap\langle X_{r}Z_{r}\rangle$ or in $P_{r}\cap\langle YZ_{r}\rangle$, by the independence

[TABLE]

and Remark 4.3 (2). For example, let $a$ and $b$ lie in $P_{r}\cap\langle X_{r}Z_{r}\rangle$, so $\mathrm{ep}(a,b)$ lies in $P_{r}\cap\operatorname{acl}(X_{r}Z_{r})=P_{r}\cap\langle X_{r}Z_{r}\rangle$, by Corollary 5.5. Since $F_{r}$ is elementary on $P_{r}\cap\langle X_{r}Z_{r}\rangle$, we have that $\mathrm{ep}(F_{r}(a),F_{r}(b))=F_{r}(\mathrm{ep}(a,b))$ lies in $P_{r+1}$, as desired. Observe that we have actually shown that

[TABLE]

For $(\ref{C:2nd})$, we need only consider the case when $a\neq a_{1}$ and the exceptional point $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b)$ does not lie in $P$, by $(\ref{C:1st})$. Again, if both $a$ and $a_{1}$ lie in $\langle X_{r}Z_{r}\rangle$ or in $\langle YZ_{r}\rangle$, then so does $b$, and we are done by Proposition 5.6, since $F_{r}$ is elementary on each side. If this is not the case, and $a$ lies in $P_{r}\cap\langle X_{r}Z_{r}\rangle$ and $a_{1}$ in $P_{r}\cap\langle YZ_{r}\rangle$, then the line $b$ lies in $P_{r}\cap Z_{r}$, by the Remark 4.3 (3). Thus the point $\mathrm{ep}(a,b)=\mathrm{ep}(a_{1},b)$ lies in $\operatorname{acl}(X_{r},Z_{r})\cap\operatorname{acl}(Y,Z_{r})=Z_{r}$, so we conclude as before since $F_{r}$ is elementary on each side separately. ∎ Claim

As $P_{0}$ and $P_{n}$ have the same type, their defect is the same, so $\mathrm{defect}(P_{r})=\mathrm{defect}(P_{r+1})$, for all $0\leq r\leq n-1$. Hence, for all $a$, $a^{\prime}$ and $b$ in $P_{r}$, we have that

[TABLE]

Since $P_{r}$ and $P_{r+1}$ are closed in the reduct $\mathrm{PS}_{k}$, but $\operatorname{tp}(P_{r})\neq\operatorname{tp}(P_{r+1})$, Proposition 5.6 implies that $F_{r}$ restricted to $P_{r}$ cannot preserve colours. As $F_{r}$ is elementary on each side separately, the colours of lines are preserved. Thus, there is a pair $(a,c)$ in $P_{r}$ whose colour $j$, with $0\leq j<k$, is not preserved under $F_{r}$. We will show now that the colour of the pair $(F_{r}(a),F_{r}(c))$ is $j+1$.

Since $F_{r}$ is elementary on $\langle X_{r}Z_{r}\rangle$ and on $\langle YZ_{r}\rangle$ separately, neither $a$ nor $c$ lie in $Z_{r}$. The independence X_{r}\mathop{\mathchoice{\displaystyle\kern 5.71527pt\hbox to0.0pt{\hss\displaystyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\displaystyle\smile\hss}\kern 5.71527pt}{\textstyle\kern 5.71527pt\hbox to0.0pt{\hss\textstyle\mid\hss}\lower 3.87495pt\hbox to0.0pt{\hss\textstyle\smile\hss}\kern 5.71527pt}{\scriptstyle\kern 2.80048pt\hbox to0.0pt{\hss\scriptstyle\mid\hss}\lower 1.89871pt\hbox to0.0pt{\hss\scriptstyle\smile\hss}\kern 2.80048pt}{\scriptscriptstyle\kern 1.42882pt\hbox to0.0pt{\hss\scriptscriptstyle\mid\hss}\lower 0.96873pt\hbox to0.0pt{\hss\scriptscriptstyle\smile\hss}\kern 1.42882pt}}^{\mathrm{PS}}_{Z_{r}}Y and Remark 4.3 (3) yield that there is a line $b$ in $Z_{r}$ connecting $a$ and $c$. The characterisation of the independence in Corollary 5.4 implies that $c\neq\mathrm{ep}(a,b)$. Hence the line $b$ must have colour $j$. The map $F_{r}$ is the identity on $Z_{r}$, and the plane $F_{r}(a)$ is connected to the point $F_{r}(c)$ by $b=F_{r}(b)$, so the only possible colours for the pair $(F_{r}(a),F_{r}(c))$ are $j$ or $j+1$. As the colour of the pair $(a,c)$ is not preserved, we deduce that $(F_{r}(a),F_{r}(c))$ has colour $j+1$, as desired.

Let $F_{n}$ be the $\mathrm{CPS}_{k}$-elementary map mapping $P_{n}$ to $P_{0}$ (as both $(e_{0},f)$ and $(e_{n},f)$ realise the type $p_{0}$) and write $F^{r}=F_{r}\circ\ldots\circ F_{0}$. Notice that the map $F^{n}$ is the identity of $P_{0}$. Let $(a,c)$ be one of the pairs in $P_{0}$ whose colour $j_{0}$ changes under $F_{0}$. The colours of the pairs

[TABLE]

change at each step by at most adding $1$ (modulo $k$), so the colour of $F^{n-1}(a,c)$ equals $j_{0}+m$ modulo $k$, for some $1\leq m\leq n$. Since $F_{n}$ preserves colours and $F^{n}(a,c)=(a,c)$, we have that $m$ is divisible by $k$, and thus $k\leq m\leq n$. We conclude that the original cycle had length at least $k$. ∎

Remark 6.3**.**

Given a function $\pi:\{0,\ldots,k-1\}\to\{0,\ldots,k-1\}$ with no fixed points, we could similarly consider the theory $\mathrm{CPS}_{\pi}$ of colored pseudospaces such that given a line $b$ with colour $i$ inside a plane $a$, all points in $b$ lie in the section $I_{i}(a)$ except one unique exceptional point which lies in $I_{\pi(i)}(a)$.

The corresponding theory $\mathrm{CPS}_{\pi}$ is not equational. Every closed path of real types has length at least the length of the shortest $\pi$-cycle.