---
abstract: 'We show that the Cantor-Bendixson rank of a limit group is finite as well as that of a limit group of a linear group.' 
author:
- Abderezak OULD HOUCINE
title: 'Note on the Cantor-Bendixson rank of limit groups'
---

Let $T$ be a universal theory.  If $T$ has at most countably many finitely generated models, then one can associate to each finitely generated model $\cal M$ of $T$ an ordinal rank, denoted $Rk(\mathcal M)$, as defined in [@ould-f-g].  We reformulate the definition of $Rk$ in the context of topological spaces.  Let $X$ be a compact Hausdorff topological space.  Let $D(X)=\{x \in X | x
\hbox{ is not isolated in }X\}$. We define inductively $D^\alpha(X)$ on ordinals as follows:

$\bullet$ $D^0(X)=X$,

$\bullet$ $D^{\alpha+1}(X)=D(D^\alpha(X))$,

$\bullet$ $D^\alpha(X)= \bigcap_{\beta<\alpha}D^\beta(X)$, for $\alpha$ a limit ordinal. 

The following fact can be easily extracted from [@kechris]. 

\[fact1\] There exists a least ordinal $\alpha$ such that $D^\alpha(X)=D^\beta(X)$ for any $\beta>\alpha$. If $X$ is separable then $D^\alpha(X)=\emptyset$ or $|D^\alpha(X)|=2^{\aleph_0}$. [ $\square$]{}

Thus if $X$ is countable then $D^\alpha(X)=\emptyset$. In that case we define the *Cantor-Bendixson rank* of $x \in X$, denoted $CB(x)$, by $CB(x)=\alpha$ if and only if $x \in
D^{\alpha}(X)\setminus D^{\alpha+1}(X)$.

Let us return to our universal theory $T$. A *complete $n$-QF-type* (with respect to $T$), where the length $|\bar x|=n$, is a set $p(\bar x)$ of quantifier-free formulas such that for any quantifier-free formula $\psi(\bar x)$ either $\psi \in p$ or $\neg \psi \in p$, and such that there is a model $\cal M$ of $T$ having a tuple $\bar a$ such that $\mathcal M \models p(\bar a)$.

Let $S_n^{qf}(T)$ be the set of all completes $n$-QF-types of $T$. Then $S_n^{qf}(T)$ is equipped with a topology as follows.  Take for basis open sets the sets of the form $$[\psi]=\{p \in S_n^{qf}(T)|\psi \in p\}, \hbox{ where $\psi$ is a
QF-formula}.$$

The following fact is a consequence of the compactness theorem and the proof proceeds in a similar way to that of [@poizat Théorème 4.04]. 

\[fact2\]$S_n^{qf}(T)$ is a compact totally disconnected space. [  $\square$]{}

An *$n$-marked* model of $T$ is a tuple $(\mathcal M, \bar a)$, where $\bar a$ is an $n$-generating tuple of $\cal M$ and $\mathcal{M} \models T$. For a marked model $(\mathcal M, \bar a)$ of $T$, we let $$\Delta(\mathcal M, \bar a)=\{\psi(\bar x )\mid \mathcal M \models \psi(\bar a), \psi
\hbox{ is quantifier-free}\}.$$

It follows that $\Delta(\mathcal M, \bar a) \in S_n^{qf}(T)$.

Conversely, let $p(\bar x)
\in S_n^{qf}(T)$. By definition, there is a model $\mathcal N \models T$ having a tuple $\bar
a$ such that $\mathcal N \models p(\bar a)$. Then $(\mathcal M={\langle}\bar a{\rangle}, \bar a)$ is an $n$-marked model of $T$ and $\Delta(\mathcal M, \bar a)=p(\bar x)$. We notice that one can have $\Delta(\mathcal M, \bar a)=\Delta(\mathcal M, \bar b)$, with $\bar a$ different from $\bar b$.

Suppose now that $T$ has at most countably many finitely generated models.  Then $S_n^{qf}(T)$ is countable for any $n$, and by the Fact \[fact1\]-\[fact2\] one can associate an ordinal rank to marked models of $T$ as follows.  Let $(\mathcal M, \bar a)$ be a marked model of $T$ where $|\bar a|=n$. Then $\Delta(\mathcal M, \bar a) \in S_n^{qf}(T)$ and we define $Rk_n(\mathcal M, \bar a)=CB(\Delta(\mathcal M, \bar a))+1$, where the Cantor-Bendixson rank is calculated relatively to the topology defined above. 

*[@ould-f-g Section 5]* Suppose that $T$ has at most countably many finitely generated models.  If $\mathcal M$ is a finitely generated model of $T$ and $\bar a$, $\bar b$ are finite generating tuples of $\cal M$ with $|\bar a|=n$ and $|\bar b| =m$, then $Rk_n(\mathcal
M, \bar a)= Rk_m(\mathcal M, \bar b)$.

[ $\square$]{}

The above fact allows the definition of $Rk(\mathcal M)$ without any reference to a particular generating set of $\cal M$.

If $G$ is an equationally noetherian group, then finitely generated $G$-limit groups coincide with finitely generated models of the universal theory of $G$ [@ould-e-n Theorem 2.1(4)], and there are at most countably many of them [@ould-e-n Proposition 6.1].  Since a linear group $G$ is equationally noetherian [@rem-mya-alge-geo Theorem B1], we obtain that $Rk(L)$ is an ordinal for any finitely generated $G$-limit group $L$. In fact, we have a bit more. 

\[princip-theo\] If $G$ is a linear group and $L$ is a finitely generated model of the universal theory of $G$ *(*i.e., $L$ is a finitely generated $G$-limit group*)*, then $Rk(L)$ is finite. 

Since free groups are linear, it follows that the Cantor-Bendixson rank of a limit group is finite.  Before proving the above theorem, we explain the relation between the space of $n$-marked limit groups of free groups $\mathcal F_n$, and the space $S_n^{qf}(T)$ where $T$ is the universal theory of $F_2$.

Let us recall the definition of $\mathcal F_n$ from [@Cham-Guir].  Let $2^{ F_n}$ be the set of all subsets of the free group $F_n$. Let us denote by $B_r$ the ball of radius $r$ in the free group $F_n$. For subsets $X,Y \in 2^{ F_n}$, define the valuation $$v(X,Y)= \max \{r \in \mathbb N \cup \{+ \infty\}| X \cap B_r=Y
\cap B_r\}.$$

The space $\mathcal G_n$ is viewed as the set of normal subgroups of $F_n$ with the metric $d(X,Y)=e^{-v(X,Y)}$ and $\mathcal F_n$ as the subspace of $\mathcal G_n$ of elements $X$ such that $F_n/X$ is a limit group.  Then $\mathcal G_n, \mathcal F_n$ are compact and totally disconnected. 

For any $p \in S_n^{qf}(T)$, let $p^+$ be the set of reduced words $w(\bar x)$ such that $w(\bar x)=1 \in p$. Then $p^+$ is a normal subgroup of $F_n$.

The function $f :  S_n^{qf}(T) \rightarrow \mathcal F_n$ defined by $f(p)=p^+$ is a homeomorphism. 

[**Proof**]{}

Let us first prove that $f$ is a bijection.  Clearly $f$ is injective, because any element of $S_n^{qf}(T)$ is a complete QF-type.  Now let $X \in \mathcal F_n$. Then $H=F_n/X$ is a limit group, and thus $H$ is a model of the universal theory of $F_2$. We write $H={\langle}\bar a| w(\bar a)=1, w \in X{\rangle}$ and we let $p(\bar x)=\Delta(H, \bar a)$. Clearly we have $f(p)=X$.

We are going to show that $f$ is countinous, the continuity of $f^{-1}$ can be treated in a similar way.  To this end it is sufficient to show that for any $m \in \mathbb N$, for any $X \in \mathcal F_n$, there is a QF-formula $\psi (\bar x)$ such that $$f^{-1}(\{Y \in \mathcal F_n|v(X,Y)> m \})=[\psi].$$

Let $P^+$ (resp.  $P^{-}$) be the set of reduced words $w \in B_{m+1}$ such that $H\models w(\bar a)=1$ (resp.  $H \models w(\bar a) \neq 1)$, where $H={\langle}\bar a| w(\bar a)=1, w \in X{\rangle}$. We let $$\psi(\bar x) = \bigwedge_{w \in P^+} w(\bar x)=1\wedge \bigwedge_{v \in P^{-}} v(\bar x) \neq 1.$$

Now it is easily seen that $
f^{-1}(\{Y \in \mathcal F_n|v(X,Y)> m \})=[\psi]
$. [  $\square$]{}

To prove Theorem \[princip-theo\] we will use the fact that the Morley rank of $GL_n(K)$, where $K$ is a pure algebraically closed field, is finite.  We recall some definitions.  The Cantor rank is defined for constructible sets in topological spaces.  The next definition is more suited to our context. 

*(Cantor rank)* Let $X$ be a set and let $\cal B$ be a boolean algebra in $X$. The Cantor rank $\rm RC(Y)$ of a subset $Y \in \cal B$ is defined by induction as follows:

$\bullet$ ${{\rm{RC}}}(Y) \geq 0$ if and only if $Y \neq \emptyset$;

$\bullet$ ${{\rm{RC}}}(Y) \geq \alpha +1$ if and only if for every $n \in
\mathbb N$, there exist a sequence of disjoints subsets $(Y_i \in
\mathcal{B}| i \leq n)$ of $X$ such that ${{\rm{RC}}}(Y_i \cap Y) \geq
\alpha$;

$\bullet$ if $\alpha$ is a limit ordinal, then ${{\rm{RC}}}(Y) \geq \alpha$ if and only if ${{\rm{RC}}}(Y) \geq \beta$ for every $\beta < \alpha$.

We put ${{\rm{RC}}}(Y)$=*sup* $\{\alpha ~|~{{\rm{RC}}}(Y) \geq \alpha\}$ if $Y \neq \emptyset$; ${{\rm{RC}}}(\emptyset)=-1$. [ $\square$]{}

*(Morley rank for groups)* Let $G$ be a group and let $^*G$ be an $\aleph_0$-saturated elementary extension of $G$. Let $\mathcal B_n$ be the boolean algebra of definable sets in $^*G^n$. We define the Morley rank of $X \in
\mathcal B_n$, by $RM(X)=RC(X)$. We say that $G$ has a finite Morley rank if $RM(\bar x=\bar x)$, $|\bar x|=n$, is finite for any $n$. We let $RM(G)=RM(x=x)$. [ $\square$]{}

Now we introduce the QF-rank which is computed relatively to the boolean algebra of QF-definables sets. 

*(Quantifier-free rank for groups)* Let $G$ be a group and let $^*G$ be an $\aleph_0$-saturated elementary extension of $G$. Let $\mathcal B_n$ be the boolean algebra of quantifier-free definable sets in $^*G^n$. We define the quantifier-free rank of $X \in \mathcal B_n$ by $QF(X)=RC(X)$. We say that $G$ has a finite QF-rank if $QF(\bar
x=\bar x)$, $|\bar x|=n$, is finite for any $n$. We let $QF(G)=QF(x=x)$. [ $\square$]{}

These definitions does not depend on a particular choice of $^*G$. The proof of that fact in the case of QF-definables sets, follows the same method as that of [@marker Lemma 6.2.2, Lemma 6.2.3].  We notice also that $^*G$ can be taken to be a nonprincipal ultrapower of $G$ ([@Chang-Keisler Theorem 6.1.1] or [@marker Exercice 4.5.37]). 

If $\psi(\bar x)$ is a formula with parameters from $G$ (i.e., an $\mathcal L_G$-formula), we define $RM(\psi)$ to be the Morley rank of the set $\psi(G)=\{\bar a \in {^*G}|^*G \models \psi(\bar
a)\}$. If $\psi$ is QF-formula, $QF(\psi)$ is defined similarly. 

*[@marker Corollary 6.2.23]*\[fact-mor-algebraic\] If $K$ is a pure algebraically closed field then $GL_n(K)$ has a finite Morley rank.  Furthermore $RM(GL_n(K))=\dim(GL_n(K))$, where $\dim$ denotes the dimension in the usual Zariski topology. [  $\square$]{}

In the following lemma, we write $RM_{\mathcal M}$, $QF_{\mathcal M}$, to distinguish the model $\mathcal M$ relatively to which the rank is calculated. 

\[lem-pricip\]Let $\mathcal M$ and $\mathcal N$ be two groups such that $\mathcal M \leq \mathcal N$. Then for every quantifier-free $\mathcal L_{\mathcal M}$-formula $\phi(\bar x)$, $QF_{\mathcal M}(\phi) \leq QF_{\mathcal N}(\phi)$. In particular $QF(\mathcal{M}) \leq QF(\mathcal{N})$. We also have $QF_{\mathcal
N}(\phi) \leq RM_{\mathcal N}(\phi)$.

[**Proof**]{}

We claim first that there exists an $\aleph_0$-saturated elementary extension $\mathcal{M}_1$ of $\cal M$ and an $\aleph_0$-saturated elementary extension $\mathcal{N}_1$ of $\cal N$ such that $\mathcal{M}_1
\leq \mathcal{N}_1$ and the following diagram commute $$\begin{array}{ccc}
  \mathcal{M}& \leq & \mathcal{N} \\
  \downarrow& \; & \downarrow \\
  \mathcal{M}_1 & \leq & \mathcal{N}_1 \\
\end{array}$$ Let $\mathcal{M}_1$ be any $\aleph_0$-saturated elementary extension of $\cal M$. Let $$Diag(\mathcal{M}_1)=\{\phi| \phi \hbox{ is a quantifier-free }\mathcal L_{\mathcal M_1} \hbox{-formula and } \mathcal M_1 \models \phi\},$$ $$Diag_{el}(\mathcal{N})=\{\phi| \phi \hbox{ is an }\mathcal L_{\mathcal N} \hbox{-formula and } \mathcal N \models \phi\},$$ and let $\Gamma=Diag(\mathcal{M}_1) \cup
Diag_{el}(\mathcal{N})$ in the language $\mathcal L_{\mathcal M_1}  \cup \mathcal L_{\mathcal N}$.

Suppose that $\Gamma$ is not consistent.  Then there exists a quantifier-free $\mathcal{L}_{\cal M}$-formula $\psi(\bar x)$ such that $\mathcal{M}_1 \models \exists \bar x \psi(\bar x)$ and Diag$_{el}(\mathcal{N})\models \forall \bar x \neg \psi(\bar x)$; which is clearly a contradiction.  Therefore $\Gamma$ is consistent and we take $\mathcal{N}_1$ to be any $\aleph_0$-saturated elementary extension of a model of $\Gamma$. This ends the proof of our claim. 

We now prove by induction on $\alpha$ that for any quantifier-free $\mathcal{L}_{\mathcal{M}_1}$-formula $\phi$, $RC_{{\cal M}_1}(\phi) \geq \alpha \Rightarrow RC_{{\cal
N}_1}(\phi) \geq \alpha$, where here for a model $\mathcal K$, $RC_{{\cal K}}(\phi)$ denotes the Cantor rank computed in $\cal K$.

Since $\phi$ is quantifier-free the result is clear for $\alpha=0$. If $\alpha$ is a limit ordinal the result follows by induction. 

For $\alpha+1$. Let $n \geq 1$. There are quantifier-free $\mathcal{L}_{\mathcal
M_1}$-formulas $\psi_1(\bar x)$, $\psi_2(\bar x)$, $\dots$, $\psi_n(\bar x)$, such that $\psi_1(\mathcal {M}_1)$, $\psi_2(\mathcal {M}_1)$, $\dots$, $\psi_n(\mathcal M_1)$ is a family of pairwise disjoint subsets of $\phi(\mathcal {M}_1)$ and $QF_{\mathcal M_1}(\psi_i(\bar x)) \geq
\alpha$ for all $i$. Define $\varphi_i(\bar x)$ as follows $$\varphi_i(\bar x)= \phi(\bar x) \wedge \psi_i(\bar x) \wedge_{j<i}
\neg \psi_j(\bar x).$$

Then $\mathcal{M}_1 \models \forall \bar x (\varphi_i(\bar x)
\Leftrightarrow \psi_i(\bar x))$. Therefore $RC_{\mathcal
M_1}(\psi_i(\bar x))=RC_{\mathcal M_1}(\varphi_i(\bar x))$ and by induction we have $RC_{\mathcal N_1}(\varphi_i(\bar x)) \geq
\alpha$. Now $\varphi_1(\mathcal {N}_1)$, $\varphi_2(\mathcal
{N}_1)$, $\dots$, $\varphi_1(\mathcal {N}_1)$ is a family of pairwise disjoint subsets of $\phi(\mathcal {N}_1)$ and $RC_{\mathcal
N_1}(\varphi_i(\bar x)) \geq \alpha$ for all $i$.

Thus $RC_{\mathcal N_1}(\phi(\bar x)) \geq \alpha+1$.

The proof of $QF_{\mathcal N}(\phi) \leq RM_{\mathcal N}(\phi)$ proceeds in a similar way. [  $\square$]{}

Combining Fact \[fact-mor-algebraic\] and Lemma \[lem-pricip\], we get the following. 

A linear group has a finite QF-rank. [  $\square$]{}

By the precedent corollary, Theorem \[princip-theo\] is a consequence of the following more general proposition which is analogous to the one in the full category of definable sets [@marker Exercice 6.6.19]. 

Let $G$ be a group of finite QF-rank.  Then for any finitely generated model $L$ of the universal theory of $G$, $Rk(L)$ is finite. 

[**Proof**]{}

Let $L$ be a finitely generated model of the universal theory $T$ of $G$, $n$-generated by the tuple $\bar a$. Since $Rk(L)=Rk_n(L,
\bar a)=CB(\Delta(L, \bar a))+1$, it is sufficient to show that for any $p \in S_n^{qf}(T)$, $CB(p)$ is finite. 

Let $X=S_n^{qf}(T)$. Let $ QF_0(p)= \hbox{ min } \{QF(\psi)\mid \psi \in p \}$. Then $QF_0(p)$ is a natural number as $G$ has a finite QF-rank.  We are going to prove by induction on natural numbers $m \geq 0$, that if $CB(p)=m$, then $QF_0(p) \geq m$.

For $m=0$. Since $CB(p)=0$, $p \in X\setminus  D(X)$ and thus $p$ is isolated.  Therefore, there exists a QF-formula $\psi_0(x_1,
\cdots, x_n )$ such that $$T \vdash \forall \bar x(\psi_0(\bar x )\Rightarrow \psi(\bar x)),$$ for any $\psi \in p$. Therefore $ \psi_0 \in p$ and $QF_0(p)=QF(\psi_0)$. Since $\exists \bar x \psi_0(\bar x)$ is consistent with $T$, we get $QF(\psi_0) \geq 0$ as desired. 

For $m+1$. Then $p \in D^{m+1}(X)\setminus  D^{m+2}(X)$. Therefore there exists a QF-formula $\psi(\bar x)$ such that $\psi \in p$ and for any $q \in D^{m+1}(X)$ if $\psi \in q$ then $p=q$. We may assume that $QF_0(p)=QF(\psi)$.

Since $p$ is not isolated in $D^m(X)$, there exists $q \in D^m(X)$ such that $\psi \in q_1$ and clearly we must have $q_1 \in
D^{m}(X)\setminus D^{m+1}(X)$. Since $q_1 \neq p$ there is a QF-formula $\psi_1 \in p$ such that $\neg \psi_1 \in q_1$. By the same argument, there exists $q_2  \in D^{m}(X)\setminus
D^{m+1}(X)$ and $\psi_2$ such that $\psi \wedge \psi_1 \in q_2$, $\psi_2 \in p$, $\neg \psi_2 \in q_2$. Containing in this manner, we get an infinite sequence $((q_n, \psi_n)|n \in \mathbb N)$, such that $q_n \in D^{m}(X)\setminus D^{m+1}(X)$, $\psi_n \in p$, $\neg \psi_n \in q_n$, $\psi_1, \cdots, \psi_{n-1} \in q_n$.

By induction $QF(\psi_n) \geq m$, as $CB(q_{n+1}) =m$ and $\psi_{n} \in q_{n+1}$. We also have $QF(\psi \wedge \psi_n) \geq
m$ and $QF(\psi \wedge \neg \psi_n) \geq m$. Then the QF-formulas $$\psi \wedge \neg \psi_1, \psi \wedge \psi_1 \wedge \neg \psi_2,
\psi \wedge \psi_1 \wedge \psi_2, \psi \wedge \psi_1 \wedge \psi_2
\wedge \neg \psi_3, \cdots$$

are disjoint QF-formulas each of which has a QF-rank greater than $m$. Therefore $QF(\psi) \geq m+1$. [ $\square$]{}

We end this note with the following lemma. 

$\;$

$(1)$ Let $T$ be the universal theory of the free group $F_2$ of rank 2 and let $L$ be a finitely generated model of $T$. Then $Rk(L)=1$ if and only if $L\cong 1$ or $L \cong \mathbb Z$ or $L
\cong F_2$.

$(2)$ $Rk(\mathbb Z^n)=n$, $Rk(F_n) \geq n-1$ for any $n \geq 2$.

[**Proof**]{}

$(1)$ If $Rk(L)=1$, then $L$ is embeddable in $F_2$ [@ould-f-g Lemma 6.5] and thus $L$ is free.  It is also clear that $Rk(1)=Rk(\mathbb Z)=1$, because $1$ and $\mathbb Z$ are isolated by the formula $x=1$ and $x \neq 1$ respectively.  Since any subgroup of a limit group, generated by two elements is either abelian or free of rank $2$ [@chiswell Lemma 5.5.4], it follows that $F_2$ is isolated by the formula $[x,y] \neq 1$.

Let $F_n$ be a free group of rank $\geq 3$ and let $\{x_1, \cdots,
x_n\}$ be a basis of $F_n$. Then $F_{n-1} \preceq_{\exists}
F_{n-1}*{\langle}x_n|{\rangle}$(see [@sel1]).  Therefore if $Rk(F_n)=1$ and if $\psi(x_1,
\cdots, x_n)$ is a QF-formula isolating $F_n$, then there exists $b \in F_{n-1}$ such that $F_{n-1} \models \psi(x_1, \cdots,
x_{n-1},b)$ and thus we find $F_n \cong F_{n-1}$, which is a contradiction.  Therefore $Rk(F_n) \geq 2$.

$(2)$ Similar to the above proof. [  $\square$]{}

**Acknowledgements.**  We thank B. Poizat and F. Wagner for having pointed out that the QF-rank is bounded by the Morley rank. 

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Abderezak OULD HOUCINE,\
Université de Lyon;\
Université Lyon 1;\
INSA de Lyon, F-69621;\
Ecole Centrale de Lyon;\
CNRS, UMR5208, Institut Camille Jordan,\
43 blvd du 11 novembre 1918,\
F-69622 Villeurbanne-Cedex, France\
E-mail address :
