\documentstyle[12pt]{article}
\textwidth=15cm
\textheight=22cm
\oddsidemargin=3.5cm
\topmargin=2cm

%\title{Length functions on free exponential groups}
%\author{A.Myasnikov, V.Remeslennikov}
%\date{}

\newtheorem{proposition}{Proposition}
\newtheorem{definition}{Definition}
\newtheorem{corollary}{Corollary}
\newtheorem{lemma}{Lemma}
\newtheorem{theorem}{Theorem}
\newtheorem{remark}{Remark}

\begin{document}
\thispagestyle{empty}

\Large

\begin{center}
Russian Academy of Sciences \\
Siberian Division\\
Institute of Information Technologies and Applied Mathematics

\bigskip  \bigskip \bigskip \bigskip
\bigskip \bigskip \bigskip \bigskip \bigskip

\LARGE
A.Myasnikov, V.Remeslennikov

\bigskip  \bigskip \bigskip \bigskip\bigskip  \bigskip \bigskip \bigskip
\bigskip  \bigskip \bigskip \bigskip

\Huge
Length functions on free exponential groups

\bigskip  \bigskip \bigskip \bigskip
\bigskip  \bigskip \bigskip \bigskip
\bigskip  \bigskip \bigskip \bigskip


\Large

Preprint $N$ 26
\end{center}

\bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip
\bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip
 \bigskip \bigskip

\begin{center}
Omsk 1996
\end{center}

\newpage

\thispagestyle{empty}
\large

\vspace*{15cm}

\noindent ISBN 5-7779-0026-7 \hfill \copyright A.~Myasnikov, \newline
\mbox{~} \hfill V~Remeslennikov, 1996


\newpage

\setcounter{page}{3}
\large

%\maketitle

\setcounter{section}{0}
\section{Introduction}

This paper is one of the series of papers of authors on exponential groups
\cite{MR:foun,MR:tree}. Its goal is to construct a Lyndon length function on a free
exponential $A$-group for the wide class of rings $A$. Namely, when $A$ is
an associative ring with identity 1, its additive group $A^+$ is
torsion free and the additive subgroup $<1>$ is a pure subgroup of $A^+$.
Moreover, we will construct a Lyndon length function on a tensor product
$F \otimes A$ of a free group $F$ by an abelian torsion-free group $A$ over
a pure infinite cyclic subgroup $\bf Z$ in $A$.

The group $F \otimes A$ can be constructed by infinite number of extensions
of centralizers of a free group $F$. The elementary step of this construction
(its detailed description in \cite{MR:tree}) is an extension of centralizer which
is a corresponding free product with amalgamation:
$$
F = G_0 < G_1 < \ldots < G_{\alpha} < G_{\alpha + 1} < \ldots < F \otimes A,
$$
where $G_{\alpha + 1} = G_{\alpha} * A$, and
$G_{\lambda} = \bigcup_{\alpha < \lambda} G_{\alpha}$ for a limit ordinal
$\lambda$.

The critial point in our proof is the main theorem, which allows one to extend
a length function from $G_{\alpha}$ to $G_{\alpha+1}$. In the special case, when
$A \simeq <1> \oplus {\bf Z}$ it can be derived from the corresponding results
of H.Bass \cite{B}.


\section{Some remarks on ordered abelian groups}

Preliminary facts on abelian groups can be found \cite{F}. Here we discuss
some (mostly known) results that will be used later.

\begin{proposition}
An abelian group $A$ can be ordered iff $A$ is torsion-free.
\end{proposition}

$\bf \Box$ Using the completeness theorem from logic, one can prove that a
group $G$ can be ordered iff any  finitely generated  subgroup of $G$ can be
ordered. A torsion-free abelian group A has only free finitely generated
subgroups. But any free abelian  group is a direct sum of infinite cyclic
groups $\bf Z$, which can be ordered (see below).

If $A$ is an abelian ordered group and $0,a \in A$, then $0<na$ for any
positive integer $n$. So $A$ is torsion-free. $\bf \Box$

Let $A$ be an ordered abelian group. A subgroup $B$ is a convex subgroup
of $A$ if for any $b \in B, \ a \in A$ the inequality $0 \leq a \leq b$
implies $a \in B$. The quotient $A/B$ is an ordered group with the order
induced from $A$ iff $B$ is a convex subgroup of $A$. A group $A$ is called
Archimedian, if $A$ is embeddable into the ordered group ${\bf R}^+$ of
real numbers, or equivalently, $A$ has no nontrivial convex subgroups. In
general, the lattice of convex subgroups is a linearly ordered set.

Now we would like to introduce some special orderings on torsion-free
abelian groups. The right lexicographic order will play an important role in
the following. Firstly,, we explain this notion for direct sums. Let
$A_{\alpha},\ \alpha < \delta$, be a well-ordered set of abelian ordered
groups. Then one can introduce the notion of the right lexicographic
order on the direct sum $A^{(\alpha)} = \oplus_{\beta < \alpha} A_{\beta}$.
Namely, if $a=\{a_{\beta}\}_{\beta < \alpha},\ b=\{b_{\beta}\}_{\beta < \alpha}$
are elements of $A^{(\alpha)}$, then $a < b$ means that for some $\beta_0$
the following conditions hold: $a_{\beta}=b_{\beta}$ for $\beta > \beta_0$
and $a_{\beta_0}<b_{\beta_0}$. The mapping $\lambda: a \rightarrow (a,0)$
is an embedding of ordered groups
$A^{(\alpha)} \hookrightarrow A^{\alpha} \oplus A_{\alpha+1}=A^{(\alpha + 1)}$.
Moreover, $\lambda(A^{(\alpha)})$ is a convex subgroup of $A^{(\alpha + 1)}$.
We will identify $A^{(\alpha)}$ with its image in $A^{(\alpha + 1)}$. So we
have a chain of convex subgroups
$$
0 < A^{(1)} < \ldots < A^{(\alpha)} < A^{(\alpha + 1)} < \ldots
$$
and
$$
A^{(\delta)} = \lim_{\rightarrow} A^{(\alpha)}.
$$

In the following we will consider only the standard orders on $\bf Z$ and
$\bf Q$ ($\bf Q$ is the additive group of rationals); they induce standard
right lexicogrophic orders on direct sums od copies  of $\bf Z$ or $\bf Q$.
In particular, any free abelian groups or torsion-free divisible groups
have the standard right lexicographic orders with respect to any fixed
base. Let us generalize this construction to any torsion-free abelian
group $A$. Consider an arbitrary maximal linearly independent system
$\{a_{\alpha} \mid \alpha < \delta\} = \Lambda$ in a group $A$. Then the
tensor product $A \otimes_{\bf Z} \bf Q$ is a divisible completion of $A$,
$\Lambda$ is a $\bf Q$-base of $A \otimes \bf Q$, and
$A \otimes_{\bf Z} {\bf Q} \simeq \oplus_{\alpha < \delta} {\bf Q} a_{\alpha}$.
The right lexicographic order on this direct sum induces an order on $A$
which is called {\it right lexicographic with respect to the system
$\Lambda$}.

The next aim is to create an order on $A$ that will make a given pure
subgroup $B$ convex.  By definition, $B$ is a pure subgroup of $A$ if
for any $b \in B$ the equation $nx=b$ has a solution in $A$ iff it has a
solution in $B$. It is clear that in a torsion-free group $A$ a subgroup
$B$ is pure iff the quotient $A/B$ is also torsion-free. Therefore any
convex subgroup of $A$ is pure.

\begin{proposition}
Let $A$ be a torsion-free abelian group, $B$ be a pure subgroup of $A$,
and let $B$ be ordered. Then $A$ can be ordered in such way that:

a) the order on $B$ coincides with that induced from $A$;

b) $B$ is a convex subgroup of $A$.

\end{proposition}

$\bf \Box$ The quotient $A/B$ is a torsion-free abelian group, so
by proposition 1 it can be ordered. Denote by $\leq_0$ and $\leq_1$ orders
on $B$ and $A/B$. Define an order $\leq$ on $A$ by the formula:
\[
  0 \leq a \Leftrightarrow \left\{ \begin{array}{ll}
				     0 \leq_0 a,     & a \in B  \\
				     0+B \leq_1 a+B, & a \notin B
				  \end{array}
			  \right.
\]


Then it defines  a cone of positive elements on $A$ and hence an order on $A$.
Conditions a) and b) hold.$\bf \Box$

\begin{corollary}
If the order on $B$ is right lexicographic, then one can find a right
lexicographic order on $A$ with the properties a) and b).
\end{corollary}

$\bf \Box$ The proof follows from the fact any maximal linearly independent
system in $B$ can be extend to one in $A$.$\bf \Box$

\begin{lemma}
If $A$ is a finitely generated ordered abelian group and $B$ is a convex
subgroup of $A$, then $A=B \oplus A'$ fro some ordered subgroup
$A' \leq A$, and the order on  $A$ coincides with the right
lexicographic order on $B \oplus A'$.
\end{lemma}

$\bf \Box$ $B$ is a convex subgroup of $A$, so it is pure. Hence $A/B$ is
finitely generated torsion-free and therefore free abelian. In this case
$A=B \oplus A'$ fore some direct complement of $B$ in $A$. From convexity of
$B$ one can obtain  the following equivalences:

\[
  b_1+a_1' \leq b_2 + a_2' \Leftrightarrow
  b_1-b_2 \leq a_2'-a_1'   \Leftrightarrow
                          \left\{ \begin{array}{ll}
				     a_2'-a_1'>0  \\
				     a_2'=a_1' & b_2 \geq b_1
				  \end{array}
			  \right.
\]
which show that the order on $A$ is right lexicographic on $B \oplus A'$.

\begin{proposition}
Let $A$ be any right lexicographically ordered abelian group and $B$ a
finitely generated subgroup of $A$. Then the ordered group $B$ is
isomorphic to the right lexicographically ordered  group
${\bf Z}^n={\bf Z} \oplus \ldots \oplus {\bf Z}$ for some positive
integer $n$.
\end{proposition}

$\bf \Box$ By definition, the order on $A$ is induced the order on
$\oplus_{\alpha < \delta} {\bf Q}_{a_{\alpha}}$ for some maximal linearly
independent  system $\Lambda = \{ a_{\alpha} \mid \alpha < \delta \}$
from $A$. A finitely generated subgroup is contained in a subgroup of
the type ${\bf Q}_{a_1} \oplus \ldots \oplus {\bf Q}_{a_m}$. so $B$ has a
finite complete series of convex subgroups $0<B_1< \ldots <B_n$, where
$B_n/B_{n-1} \simeq {\bf Z}^n$ is a right lexicographically ordered  group.
$\bf \Box$


\section{Preliminary results and formulation of theorems}

\subsection{Free exponential groups}  Let us recall the construction
of the free exponential group  $F^A=F \otimes A$, where $A$ is a torsion-free
abelian group. In \cite{MR:tree} the group $F^A$ is described as the  (direct
limit) union of the chain of groups.
$$
F=G_0 < G_1 < \ldots < G_n < \ldots < \bigcup G_i = F^A,
$$
where $G_{i+1}$ is obtained from $G_i$ by a tree extension of
centralizers in the following way. Let ${\bf C} = \{ C_i, i \in I \}$
be the set of all centralizers of $G_i$ which are not $A$-modules and
no two elements of $\bf C$ are conjugate, but any centralizer in $G$
is conjugate to some elements of $\bf C$. Then we  can form a set
${\cal H}=\{ C_i \otimes A \mid i \in I \}$ and the set of canonical
embeddings $\Phi=\{ \phi_i : C_i \hookrightarrow C_i \otimes A \}$.
So we have a tree of groups:

\begin{center}
\unitlength=1cm
\linethickness{0.4pt}
\begin{picture}(6,4)
\put(0.7,1.5){\makebox(0,0)[cc]{$G$}}
\put(1,1.5){\line(1,0){3}}
\put(4.1,1.5){\makebox(0,0)[lc]{$C_j \otimes A$}}
\put(4.1,2.8){\makebox(0,0)[lc]{$C_i \otimes A$}}
\put(1,1.7){\line(3,1){3}}
\put(2,2.5){\makebox(0,0)[tc]{$C_i$}}
\put(4,2.2){\makebox(0,0)[cc]{$\vdots$}}
\put(4,1.1){\makebox(0,0)[cc]{$\vdots$}}
\end{picture}
\end{center}


By definition, the group $G_{i+1}$ is the fundamental group
$G_i ({\bf C}, {\cal H}, \Phi)$ of this graph of groups. More precisely,
$G_i ({\bf C}, {\cal H}, \Phi)$ is the union of the chain of groups
$$
G_i = H_0 \hookrightarrow H_1 \hookrightarrow \ldots \hookrightarrow
H_{\alpha} \hookrightarrow H_{\alpha+1} \hookrightarrow \ldots
$$
where the set of centralizers ${\bf C}=\{C_{\alpha} \mid \alpha < \lambda \}$
is well-ordered,
$H_{\alpha +1} = H_{\alpha} \mathrel{*}_{C_{\alpha +1}} ({C_{\alpha +1}} \otimes A)$
under the canonical embedding) and
$\phi_{\alpha}$ is the canonical embedding of $H_{\alpha}$ into the
amalgamated free product.

\subsection{Length functions}
Let $G$ be a group and $\Lambda$ an ordered abelian group. Then a function
$l: G \rightarrow \Lambda$ is called a {\it length function} on the group $G$
with values in the group $\Lambda$ if the following condition hold:

$
1) \forall x \; l(x) \geq 0,\ l(1)=0;
$

$
2) \forall x \; l(x)=l(x^{-1});
$

$
3) \forall x \forall y \forall z \;
    (d(x,y)>d(x,z) \Rightarrow d(x,y)=d(y,z)),
$

$
\ \ \ where \ d(x,y)=\frac{1}{2}(l(x)+l(y)-l(x^{-1}y)).
$

To define $d(x,y)$ correctly one can assume that $\Lambda$ is canonically
embeded in its divisible completion $\Lambda \otimes {\bf Q}$, which is
an ordered abelian group and the order extends the order on $G$. From
other hand one can formulate the axiom 3) using the function
$d_0(x,y)=2d(x,y)$ instead of $d(x,y)$.

Moreover, the length function $l:G \rightarrow \Lambda$ is called
{\it free length function}, if $l$ satisfies the following axiom

$
4) \forall x \; (x \neq 1 \rightarrow l(x^2) > l(x).
$

From now on we fix a free length function $l: G \rightarrow \Lambda$.
For elements $x_1, \ldots, x_n \in G$ the notation
$x=x_1 \circ \ldots \circ x_n$ means that $x=x_1 \ldots x_n$ and
$l(x) = l(x_1)+ \ldots +l(x_n)$.

\begin{definition}
The length function $l: G \rightarrow \Lambda$ is called {\sf regular}
iff the following axiom holds:

$
5) \forall x \forall y \exists u \exists x_1 \exists y_1 \;
(x=u \circ x_1 \& y=u \circ y_1 \& l(u)=d(x,y))
$

\end{definition}

Now we start to discuss the situation when each element of $G$
can be decomposed into a product of further indecomposable elements (one
can keeps in mind a free froup with some fixed base).

\begin{definition}
An element $a \in G$ is called {\sf an atom} (with respect to the length
function $l: G \rightarrow \Lambda$) if the following conditions holds:

$
6) \forall x \forall y  \;(a=x \circ y \rightarrow l(x)=0 \vee l(y)=0).
$
\end{definition}

Suppose that an ordered abelian group $\Lambda$ has a series of
convex subgroups:
$$
0=\Lambda_0<\Lambda_1< \ldots <\Lambda_i< \ldots
$$

$I$ is a well-ordered set. The preimage $l^{-1}(\Lambda_i)=G_i$ is
a subgroup in $G$, so the series in $\Lambda$ induces a series of
subgroups in $G$:
$$
1 \leq G_0 \leq G_1 \leq \ldots \leq G_i \leq \ldots, \ i \in I
$$

Let us denote the restriction of on $\Lambda_i$ by $l_i$. Then
$l_i: G_i \rightarrow \Lambda_i$ is a length function on $G_i$.
The order on $\Lambda_i$ induces an order on
$\overline{\Lambda_i} = \Lambda_{i+1} / \Lambda_i$ it allows one to
consider the induced length function
$\overline{l_i}: G_i \rightarrow \overline{\Lambda_i}$

\begin{definition}
Let two elements from $G_i$ are called {\sf equivalent} ($g_1 \sim g_2$)
if the following conditions holds:
$$
\exists x \exists y \in G_{i-1} \; g_1=xg_2y
$$
\end{definition}


\begin{definition}
An element $b \in G_i$ is called {\sf an i-block (or a generalized i-atom)}
if the following conditions holds:

$\forall x \forall y \; (b=x \circ y \rightarrow x \in G_{i-1} \vee
y \in G_{i-1} \vee \exists x_1 \;(x_1\sim x) \;\exists y_1 \;(y_1 \sim y)$
s.t. $[x_1,y_1]=1)$.
\end{definition}


\begin{definition}
The length function $l: G \rightarrow \Lambda$ is called {\sf atomness
(blockness)} if it satisfies the following axiom:

$
8) \forall x \; ( \bigvee_{i=0}^{\infty} \exists y_1, \ldots y_i \;
(x=y_1 \circ \ldots \circ y_i \& y_1, \ldots y_i \ are \ atoms \ (blocks))).
$
\end{definition}

\begin{definition}
A length function $l: G \rightarrow \Lambda$ is called {\sf step-like}
if for any $i \in I$ the length function
$\overline{l_i}: G_i \rightarrow \overline{\Lambda_i}$ is atomness
(or blockness).
\end{definition}


\subsection{Stabilizing condition (S)}

Let  $l: G \rightarrow \Lambda$ be a length function on a group $G$ and
$V=\{ v_i \mid i \in I \}$ a set of elements from $G$. We will say that
$l$ satisfies {\it stabilizing condition} (S) with respect to $V$ if for any
elements $v, w \in V \cup \{1\} \; g \in G,$ with the property -- if $v=w$, then
$[g,v] \neq 1$ --, there exists a
natural number $t=t(v,g,w)$ such that $\forall n,m \geq t$
$$
v^n g w^m = v^{n-t} \circ v^t g w^t \circ w^{m-t}.
$$

\begin{proposition}
Let us designate by $l: F \rightarrow {\bf Z}$ a canonical length function
under the set $X$, and a set $V$ has properties:

(i) $v_i$ is a root element ($v_i$ is not proper power)
and $v_i$ is cyclically reduced, $i \in I$,

(ii) $v_i$ is not conjugated to $v_j^{\pm 1}$, if $i \neq j$.

Then $l$ satisfies stabilizing condition with respect to $V$.
\end{proposition}

$\Box$
a) Let us consider the case, when $v \neq w$. Assume, that condition (S)
breaks down at the triple $(v,g,w)$. Without the loss of generality
we can assume that condition (S) is not met for a triple $(v,g,w)$
such that $gw^n=g \circ w^n, \; \mid g \mid \leq \mid v \mid$.
Furthermore,  we will use repeatedly the lemma proved in \cite{LS}.

{\bf Lemma LS.} If $v$ and $w$ are root elements and have powers
$v^m$ and $w^n$, $m,n > 0$  with the common initial segment of length
$\mid v \mid + \mid w \mid$ then $v=w$.

Let us continue proving of the lemma. If $\mid g \mid = \mid v \mid$
then for reasonably large numbers $n, \; m$  the elements $v^{-n}$,
$w^m$ have the common initial segment greater than
$\mid v \mid + \mid w \mid$. Therefore by Lemma LS  $v=w$ or $v=w^{-1}$.
This is in contradiction with {\em (ii)} condition.

Let $\mid g \mid < \mid v \mid$. Without the loss of generality, we
can consider the case $m, \; n >0$. Then $v=v_1 \circ g^{-1}$ and
elements $(v_1^{-1} \circ g)^n$ and $w^m$ have the common
initial segment  greater than $\mid v \mid + \mid w \mid$ for
reasonably  large  $n, \; m$.
By Lemma LS $v_1^{-1} \circ g=w$, thus, contrary to {\em (ii)}
condition, elements $v, \; w$ are conjugated.

b) Let $ 1\neq v=w, \; [g,v]\neq 1 $  and condition (S)
breaks down at the  pair $(v,g)$. As before, we can assume that
$gv^m=g \circ v^m ,\; \mid g \mid \leq \mid v \mid$.
Besides, let the numbers $m$ and $n$ be of the same sign
(for example, $n, \; m \geq 0$). In this case $v=v_1 \circ g^{-1}$
and for reasonable large $n, \;m$   the elements
$(v_1^{-1} \circ g)^n, \; v^m$  have the common initial segment
greater than $\mid v \mid + \mid w \mid$. By Lemma LS
$v_1 \circ g^{-1} = g \circ v_1^{-1}$, thus $v=1$.

If numbers $n, \; m$ are of opposite sign we have the equality
$v_1 \circ g = g \circ v_1$. Therefore $[g,v_1]=1$ and the
element $v$ is not a root element.
$\Box$


\subsection{The formulation of theorems}

Let $A$ is an associative ring with identity 1, its additive group $A^+$ is
torsion-free and the additive subgroup $<1>$ is a pure subgroup of $A^+$.
Our goal is to construct a Lyndon free step-like length function for a free
exponential group $F^A$. It is enough (see \cite{MR:tree}) to construct the same
length function on a tensor product $F \otimes A$ of a free group $F$ by an
abelian torsion-free group $A$ over a pure infinite cyclic subgroup $\bf Z$
in $A$. We will construct a length function on $F^A$ by induction. The
following technical theorem provides us with the induction step from $G_i$
to $G_{i+1}$ (see free exponential groups).

\begin{theorem}

Let $G$ be a group with a length function  $l: G \rightarrow \Lambda_0$
which satisfies conditions:

1. $l$ is a free regular step-like function;

2. stabilizing condition (S) for $V({\bf C}) = \{v_i \mid i \in I \}$. \newline
Assume ${\bf C} = \{�_i \mid i \in I \}$ is a set of centralizers in $G$
such that:

3. $C_i$ is an infinite cyclic group $\bf Z$, $C_i=<v_i>$ and
$l(v_i^n) = \mid n \mid l(v_i),\ n \in \bf Z$

4. $C_i$ is not conjugated to $G_j$ for $i \neq j$. \newline
Suppose that $G^*=G({\bf C,Z} \oplus A)$ is obtained from $G$ by tree
extension of centralizers from $\bf C$ by the group ${\bf Z} \oplus A$
under amalgamation $C_i$ and $Z$. \newline
Fix any order on the $A= \Lambda$ and introduce a right lexicographic
order on $\Lambda_0 \oplus \Lambda$.

Then the length function  $l: G \rightarrow \Lambda_0$ can be extended to
a length function  $l^*: G^* \rightarrow \Lambda_0 \oplus \Lambda$ for
which conditions 1)-2) hold under the chain of convex subgroups
$0 < \Lambda_0 < \Lambda_0 \oplus \Lambda$.

\end{theorem}

The main result of this paper is the following

\begin{theorem}

Let $F^A=F \otimes A$ be a tensor product of a free group $F$ by an abelian
torsion-free group $A$ over a pure infinitive cyclic subgroup $\bf Z$ in
$A$.

Then there exists  a free regular step-like length function on the group
$F^A$.

\end{theorem}


\subsection{Reduced forms}


We will use multiplicative notation for elements of ${\bf Z} = C_i$ and
${\bf Z} \oplus \Lambda$. Namely, suppose a cyclic centralizer
$C(v)=<v> \simeq \bf Z$ is extended by the group ${\bf Z} \oplus \Lambda$
with respect to the canonical embedding
$\Phi: {\bf Z} \hookrightarrow {\bf Z} \oplus \Lambda$. All elements of
$C(v)$ are exponents of $v$  (in symbols
$C(v)=\{v^n \mid n \in {\bf Z} \} = v^{\bf Z}$); and we will use the same
writting for elements of ${\bf Z} \oplus \Lambda$, so
${\bf Z} \oplus A = \{v^a \mid a \in {\bf Z} \oplus \Lambda \}$ and
$v^{{\bf Z} \oplus \Lambda} = v^{\bf Z} \oplus v^{\Lambda}$. The amalgamation
$v=v^{\Phi}$ gives us
$C(v)=v^{\bf Z}=v^{\Phi({\bf Z})} \leq v^{\bf Z} \oplus v^{\Lambda}$ in $G^*$.
The subgroup $v^{\Lambda}$ is the set of all right representatives of
$v^{{\bf Z} \oplus \Lambda}$ by $v^{\bf Z}$. Denote
$V({\bf C}) = \{v_i \mid C_i=<v_i>, i \in I \}$. In \cite{MR:tree} it was proved that
any element $g \in G^*$ can be written in the following reduced form:
$$
g=g_1 v_1^{c_1} g_2 \ldots v_m^{c_m} g_{n+1}, \ g_i \in G, \ v \in V(C) \ \ \ (1)
$$
where $c_i \neq 0, \ v_i \neq v_{i+1}$ or $v_i = v_{i+1}$ but
$g_1 \notin v_i^{\bf Z}$.

\begin{lemma} \cite{MR:tree}

If $g$ is written in two different reduced forms:
$$
g=u_1 v_{i_1}^{c_1} \ldots v_{i_n}^{c_n} u_n,
$$
$$
g=w_1 v_{j_1}^{d_1} \ldots v_{j_m}^{d_m} w_m
$$
then:

\begin{enumerate}

\item $m=n, \ v_{i_k}=v_{j_k}, \ c_k=d_k,\ k=1, \ldots , n$;

\item $u_{n+1} = v_{i_n}^{t_n} v_{n+1}, \ t_n \in {\bf Z};$ \newline
    $u_n = v_{i_{n-1}}^{t_{n-1}} w_n v_{i_n}^{-t_n}, \ t_{n-1} \in {\bf Z};$
    \newline $\ldots;$  \newline
    $u_1=w_1 v_{i_1}^{-t_1} \ t_1 \in {\bf Z}$

\end{enumerate}

\end{lemma}

As a corollary, we have that any two reduced forms of $g$ can be transformed
one into another by a finite sequence of constructions of the type
$$
v^c v^t = v^t v^c, \ c \in \Lambda, \ t \in {\bf Z}, \ v \in V(C).
$$


\subsection{Cuttings}

Let  $G$ and $G^*$ be as before. Every element $g \in G^*$ has a reduced form
$$
g=g_1 v_1^{c_1} \ldots v_m^{c_m} g_{m+1}.
$$

\begin{definition}

For every positive integer $n$ the mapping $\Phi_n: G^* \rightarrow G$
is defined by the formula
$$
\Phi_n(g) = g_1 v_1^{\varepsilon(c_1)n} g_2 \ldots v_m^{\varepsilon(c_m)n} g_{m+1},
$$
where
\[ {\varepsilon}(c_k) = \left \{ \begin{array}{ll}
                                1, & c_k>0 \\
                               -1, & c_k<0.
                          \end{array}
                \right. \]
The mapping $\Phi_n$ is called the {\sf $n$-th cutting}.

\end{definition}

\begin{lemma}

$\Phi_n: G^* \rightarrow G$ is well-defined.
\end{lemma}

$\bf \Box$
One need to prove  that the definition of $\Phi_n$ does not depend
on a reduced form of $g$. Any reduced form of the element $g$ can be
obtained from a fixed one by a finiute sequence of commutations of the type
$v^c v^t = v^t v^c$. These trnsformations do not change the cutting
$\Phi_n$.
$\bf \Box$

Let us introduce the notation of blocks in the group $G^*$ (they will be
blocks (see 2.2.) under some length function of $G^*$ in future).

\begin{definition}

An element $g \in G^*$ which has a reduced form of the type
$$
g = xv^cy, \; \\ x,y \in G, \; v \in V({\bf C}), \;c \neq 0
$$
is called {\sf generalized atom} or a {\sf block} of $G^*$. The set of all
blocks in $G^*$ will be denoted by $P^*$. Sometimes it will be
convenient to call elements from $G$ {\sf trivial blocks} (i.e. blocks
with $c=0$).

\end{definition}

Let us begin to construct a length function for the group $G^*$.

\begin{lemma}

Let $\mu = x v^c y$ be a major block, $c \neq 0$, then there exists a
positive integer $n_0 = n_0 (\mu) $ such that
$$
\forall n \geq n_0 \; l(x v^n y) - l(v^n) = const.
$$

\end{lemma}

$\bf \Box$
The proof follows from the stabilizing condition for $V$.
$\bf \Box$

\begin{definition}

Let $g=g_1 v_1^{c_1} \ldots g_m v_m^{c_m} g_{m+1}$
be a reduced form of an element $g \in G^*$. Define the value
$p_1(g,n)$ by the formula:
$$
p_1(g,n) = l (\Phi_n(g)) - \sum ^m_{i=1} l(v_i^n).
$$

\end{definition}

By lemma 2 $p_1(g,n)$ does not depend on the reduced form of $g$.

For any $g \in G^*$ one can define the number $t_i = t_i (v_i, g_i, v_{i+1})$
from the stabilizing condition (S) and
$$N = N(g) = 2 max \{t_i, i=1, \ldots,m+1 \}.$$


\begin{lemma}
     For any $g \in G^*$ there exists a positive integer
$n_0=n_0(g)$ such that  $\forall n>n_0 \; p_1(g,n_0)=p_1(g,n)$.
We will denote $p_1(g,n_0)$ by $p(g)$.
\end{lemma}

$\bf \Box$
By the property (S) and condition 3 from theorem 1 there exists a
positive integer $N=N(g)$ such that for any $n>2N$
\begin{eqnarray}
\Phi_n(g) =  g_1v_1^{\varepsilon_1 N} \circ v_1^{\varepsilon_1(n-2N)} \circ
           v_1^{\varepsilon_1 N} g_2 v_2^{\varepsilon_2 N}\circ \ldots \circ
\nonumber  \\ (v_{m-1}^{\varepsilon_{m-1} N} g_m v_m^{\varepsilon_m N})\circ
            v_m^{\varepsilon_m (n-2N)} \circ v_m^{\varepsilon_m} g_{m+1}.
\nonumber
\end{eqnarray}
Therefore
\begin{eqnarray}
l(\Phi_n(g)) & = &l(g_1v_1^{\varepsilon_1 N})+l(v_1^{\varepsilon_1(n-2N))})+ \ldots
\nonumber \\ & = &l(g_1v_1^{\varepsilon_1 N} \circ v_1^{\varepsilon_1 N} g_2 v_2^{\varepsilon_2 N}\circ
              \ldots  \circ (v_{m-1}^{\varepsilon_{m-1} N} g_m v_m^{\varepsilon_m N})\circ
               v_m^{\varepsilon_m} g_{m+1})
\nonumber \\ &  & +\sum ^m_{i=1} l(v_i^{n-2N}) \nonumber \\
              & = &l(\Phi_{2N}(g)) - \sum ^m_{i=1} l(v_i^{n-2N}) \nonumber.
\end{eqnarray}
Hence
\begin{eqnarray}
p(g,n) & = & l(\Phi_n(g))-\sum ^m_{i=1} l(v_i^n) \nonumber \\
       & = & l(\Phi_{2N}(g)) - \sum ^m_{i=1} l(v_i^{n-2N})-\sum ^m_{i=1} l(v_i^n)
\nonumber  \\
       & = & l(\Phi_{2N}(g))-\sum ^m_{i=1} l(v_i^{2N})=p(g,2N)
\nonumber
\end{eqnarray}
So the lemma is true for $n_0=2N$.
$\bf \Box$


\begin{definition}
For any $g \in G^*$ in reduced form (1) let us define a function
$$
p_1(g)=p(g,N(g))=\lim_{n \rightarrow \infty} p(g,n), \; g\in G^* \backslash G,
\;
$$
$$
p_1(g)=l(g), \; g \in G.
$$
Lemma 3 shows that $p_1(g)$ is well-defined. Finally, we introduce a
function $l^*:G^* \rightarrow \Lambda_0 \oplus \Lambda$.
$$
l^*(g)=(p_1(g),p_2(g))=(l(\Phi_N(g)) - \sum ^m_{i=1} l(v_i^N),
\sum ^m_{i=1} \mid c_i \mid).
$$
\end{definition}


\section{$l^*$ is a length function}

Our aim is to prove that $l^*$  is a length function on $G^*$
satisfying the conditions of theorem 1. It is easy to see that  $l^*$
satisfies axiom 1 and 2 of a length function. Indeed, if
$g=g_1 v_1^{c_1} \ldots v_m^{c_m} g_{m+1}$, then
$g^{-1}=g_{m+1}^{-1} v_m^{-c_m} \ldots v_1^{-c_1} g_1^{-1}$, hence
$p_2(g)=p_2(g^{-1})$, $p_1(g)=l(\Phi_N(g)) - \sum ^m_{i=1} l(v_i^N)$, and
$p_1(g^{-1})=l(\Phi_N(g^{-1})) - \sum ^m_{i=1} l(v_i^N))$.
But $\Phi_N(g)^{-1}=\Phi_N(g^{-1})^{-1}$ and therefore $p_1(g)=p_1(g^{-1})$.

To prove axiom 3 we need some preliminary considerations.

\begin{definition}
Let $g,g_1,\ldots,g_m$ be elements of $G^*$. The notation
$g=g_1 \circ \ldots \circ g_m$ means that $g=g_1 \ldots  g_m$ in $G^*$ and
$l^*(g)=l(g_1)+ \ldots + l(g_m)$.
\end{definition}

\begin{definition}
Let $\mu_0, \mu_1, \ldots ,\mu_m$ be blocks in $G^*$. The sequence
$(\mu_1,  \\ \ldots , \mu_m)$ is called {\sf block-reduced} if the following
condition holds for any  $i=1, \ldots, m$:
\begin{enumerate}
  \item  the product $\mu_{i-1} \mu_i$ is not block in $G^*$;
  \item  $l(\Phi_n (\mu_{i-1} \mu_i)) = l(\Phi_n (\mu_{i-1})) +
          l(\Phi_n (\mu_i))$
         for all $n$ large enough ($n \geq N (\mu_{i-1} \mu_i)$).
\end{enumerate}
\end{definition}

In particular, any block is block-reduced.

\begin{remark}
Suppose that $(\mu_1, \ldots ,\mu_m)$ is block-reduced,
$\mu_i = x_i v^{c_i} y_i$ and  $g=\mu_1 \ldots \mu_m$. Then from
condition 1) it follows that
$$
g = x_1 v_1^{c_1} (y_1 x_2) v_2^{c_2} \ldots (y_{m-1} x_m) v_m^{c_m} y_m
$$
is a reduced form of $g$.
\end{remark}

The following lemma gives us necessary and sufficient conditions for the
equality $\mu_1 \mu_2 = \mu_1 \circ \mu_2$.

\begin{lemma}

Let $a = x_1 v_1^{c_1} y_1$, $b = x_2 v_2^{c_2} y_2$ be blocks in $G^*$
and the product $ab$ is not a block in $G^*$. Then

$p_2(ab)=p_2(a)+p_2(b),$

$p_1(ab)=p_1(a)+p_1(b)-2d (y_1^{-1}
v_1^{-\varepsilon (c_1) N}, x_2 v_2^{\varepsilon (c_2) N})$. \newline
In particular, $ab=a \circ b$ iff
$d (y_1^{-1} v_1^{-\varepsilon (c_1) N},x_2 v_2^{\varepsilon (c_2) N})=0$
\end{lemma}

$\bf \Box$
The equality for $p_2$ is evident. The statement about $p_1$ can be
reduced from the following computations, $n>2N$, for example
$c_1>0, \; c_2>0$:

\begin{eqnarray}
l(\Phi_n(\mu_{m-1} \mu_m)) & = & l(x_{m-1} v_{m-1}^{n-N} \circ
                             v_{m-1}^N y_{m-1} x_m v_m^N \circ v_m^{n-N} y_m)
\nonumber \\               & = & l(x_{m-1} v_{m-1}^{n-N})
                             +l(v_{m-1}^N y_{m-1} x_m v_m^N) + l(v_m^{n-N} y_m)
\nonumber \\               & = & l(x_{m-1} v_{m-1}^{n-N}) + l(v_{m-1}^N y_{m-1}) + l(x_m v_m^N)
\nonumber \\               &   & - 2d((v_{m-1}^N y_{m-1})^{-1}, x_m v_m^N ) + l(v_m^{n-N} y_m)
\nonumber \\               & = &l(\Phi_n(\mu_{m-1}))  + l(\Phi_n(\mu_m))
\nonumber \\               &   & - 2d((v_{m-1}^N y_{m-1})^{-1}, x_m v_m^N ). \nonumber
\end{eqnarray}


$\bf \Box$


\begin{proposition}
Let $(\mu_1, \ldots, \mu_m)$ be any block-reduced sequence of \\ blocks and
$g=\mu_1 \ldots \mu_m$. Then
$$
g=\mu_1 \circ \ldots \circ \mu_m
$$
and this is a minimal (i.e. with minimal possible $m$)
$\circ$-decomposition of $g$ into blocks. And conversely, if
$g=v_1 \circ \ldots \circ v_m$ is a minimal $\circ$-decomposition
of $g$ into blocks, then the sequence $(v_1, \ldots, v_m)$ is
block-reduced.
\end{proposition}

$\bf \Box$
By induction for $g=\mu_1 \ldots \mu_{m-1}$ we have
$g=\mu_1 \circ \ldots \circ \mu_{m-1}$. So it is sufficient to prove
that $g \mu_m = g \circ \mu_m$.

Suppose that $\mu_i=x_i v_i^{c_i} y_i$ then
$g=x_1 v_1^{c_1} y_1 x_2 v_2^{c_2} \ldots v_{m-1}^{c_{m-1}} y_{m-1}$
and $g \mu_m=x_1 v_1^{c_1} \ldots  y_{m-1} x_m v_m^{c_m} y_m$ are in
reduced form (see the remark above). So by definition
$p_2(g \mu_m) = p_2(g) + p_2(\mu_m)$. It remains to verify that
$p_1(g \mu_m) = p_1(g) + p_1(\mu_m)$.

\setcounter{equation}{1}
\begin{eqnarray}
p_1(g \mu_m) & = & l(\Phi_n(g \mu_m)) - \sum l(v_i^n)   \nonumber \\
             & = & l(x_1v_1^{\varepsilon_1 n} \ldots v_{m-1}^{\varepsilon_{m-1} (n-N)} \circ
v_{m-1}^{\varepsilon_{m-1} N} y_{m-1} x_m v_m^{\varepsilon_m N} \circ
v_m^{\varepsilon_m (n-N)} y_m)    \nonumber \\
             &   &  - \sum l(v_i^n)    \nonumber \\
             & = & l(x_1v_1^{\varepsilon_1 n} \ldots v_{m-1}^{\varepsilon_{m-1} (n-N)}) +
                   l(v_{m-1}^{\varepsilon_{m-1} N} y_{m-1} x_m v_m^{\varepsilon_m N})
\nonumber \\ &   & +l(v_m^{\varepsilon_m (n-N)} y_m) - \sum l(v_i^n).
\end{eqnarray}

But
\begin{eqnarray}
l(v_{m-1}^{\varepsilon_{m-1} N} y_{m-1} x_m v_m^{\varepsilon_m N}) & = &
l(v_{m-1}^{\varepsilon_{m-1} N} y_{m-1})
+ l(x_m v_m^{\varepsilon_m N})  \nonumber \\ &   &
-2d((v_{m-1}^{\varepsilon_{m-1} N} y_{m-1})^{-1}, x_m v_m^{\varepsilon_m N});
\end{eqnarray}
\begin{eqnarray}
l(x_1v_1^{\varepsilon_1 n} \ldots v_{m-1}^{\varepsilon_{m-1} (n-N)}) +
l(v_{m-1}^{\varepsilon_{m-1} N}) = l(\Phi_n(g));
\end{eqnarray}
\begin{eqnarray}
l(x_m v_m^{\varepsilon_m N}) +  l(v_m^{\varepsilon_m (n-N)} y_m)
=l(\Phi_n(\mu_m)).
\end{eqnarray}

Therefore substituting (3) into (2) and using (4), (5) we obtain
$$
p_1(g \mu_m) = p_1(g) + p_1(\mu_m) -
2d((v_{m-1}^{\varepsilon_{m-1} N} y_{m-1})^{-1}, x_m v_m^{\varepsilon_m N})
$$

It remains to prove that the last summand in the equation above is equal to
zero. By condition 2)
$$
l(\Phi_n (\mu_{m-1} \mu_m))=l(\Phi_n (\mu_{m-1}))+l(\Phi_n (\mu_m)).
$$
So by previous lemma
$$
2d((v_{m-1}^{\varepsilon_{m-1} N} y_{m-1})^{-1}, x_m v_m^{\varepsilon_m N})=0.
$$
From the definition of a block-reduced sequence it follows another
conclusions of the proposition.
$\bf \Box$

\begin{proposition}
Let $\mu_1, \ldots, \mu_m$ be blocks of $G^*$. Then the sequence
$(\mu_1, \ldots, \mu_m)$ can be reduced to a block-reduced sequence
$(v_1, \ldots, v_k), \; k \leq m$, using a finite number of transformations
of the following type:
\begin{enumerate}
  \item  $(\mu_1, \ldots, \mu_i, \mu_{i+1}, \ldots, \mu_m) \rightarrow
	  (\mu_1, \ldots, \mu_i, \mu_{i+1}, \ldots, \mu_m)$
	 if $\mu_i \mu_{i+1}$ is a block;
  \item  $(\mu_1, \ldots, \mu_i, \mu_{i+1}, \ldots, \mu_m) \rightarrow
	   (\mu_1, \ldots, \mu_i w^{-1}, w \mu_{i+1}, \ldots, \mu_m)$,
	 where $w \in G$.
\end{enumerate}

Moreover, any two block-reduced sequences representing one and the same
element  can be carried one to another by a finite chain of
transformations of  type 2).
\end{proposition}

$\bf \Box$
Using transformations of  type 1) if necessary, we may assume that
in the sequence $(\mu_1, \ldots, \mu_m)$ all products $\mu_i \mu_{i+1}$
are not blocks. Suppose
$\mu_1 = x_1 v_1^{c_1} y_1, \; \mu_2 = x_2 v_2^{c_2} y_2$.
Then by lemma 6
\begin{eqnarray*}
\mu_1 \mu_2 = \mu_1 \circ \mu_2  & \Leftrightarrow &
2d(y_1^{-1} v_1^{-\varepsilon_1 N}, x_2 v_2^{\varepsilon_2 N})=0 \\
& \Leftrightarrow &
v_1^{\varepsilon_1 N} y_1 x_2 v_2^{\varepsilon_2 N}=
v_1^{\varepsilon_1 N} y_1 \circ x_2 v_2^{\varepsilon_2 N}
\end{eqnarray*}
in the group $G$. But $l$ is a regular length function on $G$,
so there exists an element $w \in G$ such that
$$
v_1^{\varepsilon_1 N} y_1 = u_1 \circ w, \;
x_2 v_2^{\varepsilon_2 N} = w^{-1} \circ u_2, \;
2d(y_1^{-1} v_1^{-\varepsilon_1 N}, x_2 v_2^{\varepsilon_2 N})=l(w).
$$
Hence  $\mu_1 \mu_2 = \mu_1 w^{-1} \circ w \mu_2$ and the sequence
$(\mu_1 w^{-1}, w \mu_2, \mu_3, \ldots, \mu_m)$ is obtained from
$(\mu_1,  \ldots, \mu_m)$ by a transformation of type 2). To complete
the proof, it remains to apply an induction on $m$.

Let us consider the second statement. Suppose that
$\mu_1 \circ  \mu_2 = \mu_1' \circ  \mu_2'$. This means (see the proof of
the proposition above) that
$$
v_1^{\varepsilon_1 N} y_1 x_2 v_2^{\varepsilon_2 N} =
v_1^{\varepsilon_1 N} y_1  \circ x_2 v_2^{\varepsilon_2 N} =
v_1^{\varepsilon_1 N} y_1'  \circ x_2' v_2^{\varepsilon_2 N}
$$
But if in the group $G$ with a regular length function $l$ one has
$a \circ b = c \circ d$, then the condition $l(a) \geq l(c)$ implies
$a = c \circ c_1$ and $d = c_1 \circ b$ (see \cite{B}).
Therefore $(\mu_1,\mu_2) \rightarrow (\mu_1 c_1^{-1}, c_1\mu_2)$
$\bf \Box$

\begin{corollary}
Each element $g \in G^*$ can be represented by some block-reduced
sequence $(\mu_1,  \ldots, \mu_m)$ of blocks such that
$g=\mu_1 \circ \mu_2 \circ \ldots \circ \mu_m$. Moreover,
if $g=\mu_1' \circ \ldots \circ \mu_k'$ is another representation
of $G$ in a
block-reduced form, then $m=k$ and $\mu_i$ is equivalent to
$\mu_i', \; i=1, \ldots, m$.
\end{corollary}

In what follows we need a lemma:

\begin{lemma}
Let $a=xv^cy$ be a block. Then
$$
a = xv^c v^{-\varepsilon N} \circ (v^{\varepsilon N} y)=
x v^{\varepsilon N} \circ v^{-\varepsilon N} v^c v^{-\varepsilon N}
\circ (v^{\varepsilon N} y), \; N \in \bf Z,
$$
for any $N$ sufficiently large.
\end{lemma}

$\bf \Box$
Straighforward verification.
$\bf \Box$

The following proposition shows that with the help of block
$\circ$-decomposition the problem of cancellation can be reduced
to the block case.

\begin{proposition}
Let $g,h \in G^*$ and let
$g=\mu_1 \circ \ldots \mu_s \circ v_1 \circ \ldots \circ v_k$,
$h=\mu_1 \circ \ldots \mu_s \circ \xi_1 \circ \ldots \circ \xi_l$
be their writting in block-reduced form such that
$v_1^{-1} \xi_1 \notin G$. Then
$$
d^*(g,h)=l^*(\mu_1 \circ \ldots \circ \mu_s) + d^*(v_1,\xi_1).
$$
\end{proposition}

$\bf \Box$
It is clear that
$$
d^*(g,h)=l^*(\mu_1 \circ \ldots \mu_s) +
d^*(v_1 \circ \ldots \circ v_k, \xi_1 \circ \ldots \circ \xi_l).
$$
So we only need to prove the proposition in the case
$g=v_1 \circ \ldots \circ v_k, \; h=\xi_1 \circ \ldots \circ \xi_l$.
By definition, $2d^*(g,h)=l^*(g)+l^*(h)-l^*(g^{-1}h)$.
Suppose that $v_1=x_1v_1^{c_1}y_1, \; \xi_1=x_2 v_2^{c_2}y_2$.
In any case we have (by hypothesis $v_1^{-1} \xi_1 \notin G \;$)
$$
v_1^{-1} \xi_1 = y_1^{-1} v_1^{\varepsilon_1 N} \circ f \circ
v_2^{\varepsilon_2 N} y_2.
$$
Hence, using lemma 6
\begin{eqnarray}
l^*(g^{-1}h) & = & l^*(v_k^{-1} \circ \ldots \circ v_2^{-1} \circ
y_1^{-1} v_1^{\varepsilon_1 N} \circ f \circ v_2^{\varepsilon_2 N} y_2
\circ \xi_2 \circ \ldots \circ \xi_l) \nonumber \\ & = &
l^*(v_k^{-1} \circ \ldots \circ v_2^{-1}) + l^*(v^{-1} \xi_1) +
l^*(\xi_2 \circ \ldots \circ \xi_l) \nonumber \\ & = &
l^*(v_k^{-1} \circ \ldots \circ v_2^{-1}) +  l^*(v_1^{-1})
+ l^*(\xi_1) - 2d^*(v^{-1}, \xi_1)  \nonumber \\ &   &
+l^*(\xi_2 \circ \ldots \circ \xi_l) \nonumber \\  & = &
l^*(g) + l^*(h) - 2d^*(v_1, \xi_1). \nonumber
\end{eqnarray}
Hence $d^*(g,h)=d^*(v_1, \xi_1)$.
$\bf \Box$

\begin{proposition}
The function $l^*: g^* \rightarrow \Lambda_0 \oplus \Lambda$
satisfies axiom 3) iff $l^*$ satisfies 3) on all (not necessary major)
blocks.
\end{proposition}

$\bf \Box$
Suppose that $f,g,h$ have the following minimal $\circ$-decomposition
on major blocks:

$f=\mu_1 \circ \ldots \circ \mu_s \circ v_1 \circ \ldots, $

$g=\mu_1 \circ \ldots \circ \mu_s \circ \xi_1 \circ \ldots,$

$h=\mu_1 \circ \ldots \circ \mu_k \circ \zeta_1 \circ \ldots,$ \newline
and $v^{-1} \xi_1 \notin G$. Then by the previous proposition one can
compute the  function $d^*$.

\begin{tabbing}
{\bf a)} \=$k<s$, and $\mu_{k+1}^{-1} \zeta_1 \notin G$. Then \\
\> $d^*(f,g)=l^*(\mu_1 \circ \ldots \circ \mu_s)+d^*(v_1,\xi_1)$, \\
\>$d^*(f,h)=l^*(\mu_1 \circ \ldots \circ \mu_k)+d^*(\mu_{k+1},\zeta_1)$,\\
\>$d^*(g,h)=l^*(\mu_1 \circ \ldots \circ \mu_k)+d^*(\mu_{k+1},\zeta_1)$. \\
Hence $d^*(f,h)=d^*(g,h)$ and we only need to check the inequality  \\
$d^*(f,g) \geq d^*(f,h)$.
\end{tabbing}
\begin{center}
\begin{eqnarray*}
d^*(f,g) & \geq & l^*(\mu_1 \circ \ldots \circ \mu_k)+l^*(\mu_{k+1})+d^*(v_1,\xi_1)\\
          & \geq & l^*(\mu_1 \circ \ldots \circ \mu_k)+l^*(\mu_{k+1}) \\
          & \geq & l^*(\mu_1 \circ \ldots \circ \mu_k)+d^*(\mu_{k+1},\zeta_1) \\
          &   =  &d^*(f,h)
\end{eqnarray*}
\end{center}

\begin{tabbing}
{\bf b)} \= $k \geq s$,  and $v_1^{-1} \mu_{s+1} \;
(or \; v_1^{-1} \zeta_1) \; \notin G$,
$\xi_1^{-1} \mu_{s+1} \; (or \; \xi_1^{-1} \zeta_1) \; \notin G$. \\
Then \\
\> $d^*(f,g)=l^*(\mu_1 \circ \ldots \circ \mu_s)+d^*(v_1,\xi_1)$,\\
\>$d^*(f,h)=l^*(\mu_1 \circ \ldots \circ \mu_s)+d^*(v_1, \mu_{s+1}) \;
(or \; d^*(v_1, \zeta_1))$, \\
\> $d^*(g,h)=l^*(\mu_1 \circ \ldots \circ \mu_s)+d^*(\xi_1,\mu_{s+1})
(or \; d^*(\xi_1, \zeta_1))$. \\
So in this case axiom 3) for $g,f,h$ is equivalent to axiom 3) for
the \\ blocks
$v_1, \xi_1, \mu_{s+1} \; (or \; \zeta_1)$.
\end{tabbing}
$\bf \Box$

Thus, to prove axiom 3) for the function $l^*$ it suffices to prove
it for blocks.

\begin{proposition} (formula for $d^*(a,b))$.
Let $a=x_1 v_1^{c_1} y_1$, $b=x_2 v_2^{c_2} y_2$ be arbitrary (possibly
trivial) blocks in $G^*$. Then the function
$l^*:G^* \rightarrow \Lambda_0 \oplus \Lambda$ is regular on blocks and
the function $d^*(a,b)$ can be computed by the following formulas:

\begin{tabbing}
{\bf Case 1.} \= Assume that $v_1 \neq v_2$, or \\
              \> $v=v_1=v_2$, but $\varepsilon(c_1) \neq \varepsilon(c_2)$, or \\
              \> $v=v_1=v_2$ and $\varepsilon(c_1)=\varepsilon(c_2)$,
                 but $x_1^{-1}x_2 \notin v^{\bf Z}$. \\
              \> Then
\end{tabbing}
$$
d^*(a,b)=(d(x_1 v_1^{\varepsilon(c_1) N},x_2 v_2^{\varepsilon(c_2) N}),0)
$$
\begin{quote}
  \begin{quote}
            for any sufficiently large $N$.
  \end{quote}
\end{quote}

\begin{tabbing}
{\bf Case 2.} \= $v=v_1=v_2$, $\varepsilon(c_1) = \varepsilon(c_2)$,
                 $x^{-1} x_2 \in v^{\bf Z}, \; c_1 \neq c_2, \; c_i \neq 0$.\\
              \> In this case we can rewrite $a,b$ in the following way: \\
              \> $a=xv^{c_1}y_1, \;
                 b=xv^{c_2}y_2, \mid c_1 \mid < \mid c_2 \mid \; c_i \neq 0.$\\
              \> Then
\end{tabbing}
$$
d^*(a,b)= l^*(xv^{c_1} v^{-\varepsilon N})+
d^*(v^{\varepsilon N} y_1, v^{\varepsilon N}).
$$

\begin{tabbing}
{\bf Case 3.} \= $v=v_1=v_2$, $\varepsilon(c_1)=\varepsilon(c_2)=\varepsilon$,
$x^{-1} x_2 \in v^{\bf Z}, \; c_1=c_2 \neq 0$,\\
\> $a=xv^cy_1,$ \\
\> $b=x v^c y_2$.\\
\>Then
\end{tabbing}
$$
d^*(a,b)= l^*(xv^c v^{-2 \varepsilon N})+
d^*(v^{\varepsilon N} y_1, v^{\varepsilon N} y_2).
$$

\begin{tabbing}
{\bf Case 4.} \= $a=x_1, \; b=x_2 v^c y_2, \; \varepsilon=\varepsilon(c).$ \\
Then
\end{tabbing}
$$
d^*(a,b)=(d(x_1, x_2 v^{\varepsilon (c) N}),0).
$$

\end{proposition}

${\bf \Box}$
{\bf Case 1.} By definition,
$$2d^*(a,b)=l^*(a)+l^*(b)-l^*(a^{-1}b).$$
So $d^*$ is the pair of functions $(d^*_1,d^*_2)$. Suppose
$v_1 \neq v_2$ or $x_1^{-1}x_2 \notin v^{\bf Z}$, in this case
$a^{-1}b$ is not a block, so we can apply the formula from
lemma 6:

$p_2(a^{-1}b)=p_2(a^{-1})+p_2(b),$

$
p_1(a^{-1}b)=p_1(a^{-1})+p_1(b)-
2d(x_1 v_1^{\varepsilon(c_1) N},x_2 v_2^{\varepsilon(c_2) N}).
$ \\
Hence (remembering that $p_2(a^{-1})=p_2(a)$ and $p_1(a^{-1})=p_1(a)$)
$$
2d^*(a,b)=(2d(x_1 v_1^{\varepsilon(c_1) N},x_2 v_2^{\varepsilon(c_2) N}),0)
$$
By lemma 7

$a=x_1 v_1^{\varepsilon(c_1) N} \circ v_1^{c_1-\varepsilon(c_1) N} y_1,$

$b=x_2 v_2^{\varepsilon(c_2) N} \circ v_2^{c_2-\varepsilon(c_2) N} y_2.$\\
Since the function $l$ is regular there exists an element $z$ such
that

$x_1 v_1^{\varepsilon(c_1) N}=z \circ z_1$,

$x_2 v_2^{\varepsilon(c_2) N}=z \circ z_2$ \\
and

$l(z)=d(x_1 v_1^{\varepsilon(c_1) N},x_2 v_2^{\varepsilon(c_2) N})$.\\
Therefore  the function $l^*$ is regular on blocks $a,b$ in this case.

Suppose that $v_1=v_2=v$, $x^{-1} x_2 \in v^{\bf Z}$ (so we can assume
$x_1=x_2$), but $\varepsilon(c_1) \neq \varepsilon(c_2)$, for example,
$c_1<0, \; c_2>0$. Then $p_2(a^{-1}b)=p_2(a)+ p_2(b)$. Therefore
$d^*_2(a,b)=0$. Now, for $n \geq N$,
\begin{eqnarray*}
d_1^*(a,b)   & = & l(x v^{-n} y_1) - l(v^{-n})+l(x v^n y_2)-l(v^n)-
                   l(y_1^{-1} v^n y_2)+l(v^n)  \\
             & = & l(x v^{-n+N} \circ v^{-N} y_1)-
                   l(v^n)+l(x v^nv^{-N} \circ v^N y_2)  \\
             &   & -l(y_1^{-1} v^N \circ
                   v^{n-2N} \circ v^N y_2)  \\
             & = & l(x v^{-n+N})+l(v^{-N}y_1)-l(v^n)+l(x v^{n-N})+
                   l(v^N y_2)  \\
             &   & - l(y_1^{-1} v^N)-l(v^{n-2N})-l(v^N y_2) \\
             & = & l(x v^{-n+N})+l(x v^{n-N})-l(v^{n-N})-l(v^{n-N})  \\
             & = & l(xv^N)+l(v^N)+ l(xv^{-N})+l(v^N)-2l(v^{n-N})  \\
             & = & l(xv^N)+l(xv^{-N})-l(v^{2N}) \\
             & = & d(xv^{-N},xv^N).
\end{eqnarray*}

One can prove that the function $l^*$ is regular on blocks $a,b$ as
above.

{\bf Case 2.}
Let
$
\mid c_1 \mid < \mid c_2 \mid, \; \varepsilon(c_1)=\varepsilon(c_2)=
\varepsilon,
$

$a=xv^{c_1}y_1=xv^{c_1}v^{-\varepsilon N} \circ v^{\varepsilon N}y_1,$

$b=xv^{c_1} v^{-\varepsilon N} \circ v^{c_2-c_1}v^{\varepsilon N}y_2.$ \\
Then by proposition 6
$$
d^*(a,b)=d^*(v^{\varepsilon N}y_1, v^{c_2-c_1}v^{\varepsilon N})+
l^*(xv^{c_1} v^{-\varepsilon N})
$$
and the problem is reduced to case 4.

{\bf Case 3.}
Let $x=x_1=x_2$, $c_1=c_2$. Then

$a=xv^c y_1=x v^c v^{-\varepsilon N} \circ v^{\varepsilon N}y_1,$

$b=xv^c y_2=x v^c v^{-\varepsilon N} \circ v^{\varepsilon N}y_2.$ \\
Therefore by proposition 6
$$
d^*(a,b)=l^*(xv^c v^{-\varepsilon N})+
d^*(v^{\varepsilon N}y_1,v^{\varepsilon N} y_1)
$$
and regularity of the function $l^*$ on $a,b$ is equivalent to
regularity of the function $l$ on $v^{\varepsilon N} y_1$,
$v^{\varepsilon N} y_2$.

{\bf Case 4.}

$a=x_1,$

$b=x_2 v^c y_2=x_2 v^{\varepsilon N} \circ v^{-\varepsilon N} v^c y_2,$

$a^{-1}b=x_1^{-1} x_2 v^{\varepsilon N} \circ v^{-\varepsilon N} v^c y_2.$ \\
Then
\begin{eqnarray*}
d^*(a,b) & = & l^*(x_1)+l^*(x_2 v^{\varepsilon N})+l^*(v^{-\varepsilon N} v^c y_2) \\
         &   & -l^*(x_1^{-1} x_2 v^{\varepsilon N})-l^*(v^{\varepsilon N} v^c y_2) \\
         & = & (d(x_1, x_2 v^{\varepsilon N}),0).
\end{eqnarray*}
As above the function $l^*$ is regular on blocks $a,b$
${\bf \Box}$

\begin{proposition}
The function $l^*$ satisfies the axiom 3) on blocks (including trivial
blocks).
\end{proposition}

${\bf \Box}$
Let us consider arbitrary blocks
\begin{center}
\begin{eqnarray}
a=x_1 v_1^{c_1}y_1, \nonumber \\ b=x_2 v_2^{c_2}y_2, \\
c=x_3 v_3^{c_3}y_3. \nonumber
\end{eqnarray}
\end{center}
We proceed by considering all possible cases using the formula for $d^*$
from the previous proposition.

{\bf Case 1.}
$x_i \neq x_j (mod \; v_i^{\bf Z} \cup v_j^{\bf Z}) \; for \; i \neq j.$\\
This implies that in all reduced forms of $a,b,c$ we are in case 1) of
proposition 8. Therefore

$
d^*(a,b)=(d(x_1 v_1^{\varepsilon(c_1) N},x_2 v_2^{\varepsilon(c_2) N}),0),
$

$
d^*(a,c)=(d(x_1 v_1^{\varepsilon(c_1) N},x_3 v_3^{\varepsilon(c_3) N}),0),
$

$
d^*(b,c)=(d(x_2 v_2^{\varepsilon(c_2) N},x_3 v_3^{\varepsilon(c_3) N}),0).
$  \\
and axiom 3) for $d^*$ is reduced to the same problem for $d$, but
$l$ satisfies this axiom.

\begin{tabbing}
{\bf Case 2.}\=
$x_1=x_2=x, \; x \not \equiv x_3 (mod \; v_3^{\bf Z})$.\\

\>{\bf a)} \= $v_1 \neq v_2$. \\
\>         \>We are in case 1) of proposition 8 and therefore it suffices \\
\>         \>to repeat the proof of case 1) above.\\
\>{\bf b)}  $v_1=v_2$ and $\varepsilon(c_1) \neq \varepsilon(c_2)$.
          Again, we are in case 1). \\
\>{\bf c)} $v_1=v_2=v$ and $\varepsilon(c_1) = \varepsilon(c_2)$.\\
\>   \>   Then
$
a=x v^{c_1}y_1, \; b=x v^{c_2}y_2, \; c=x_3 v_3^{c_3}y_3.
$ \\

\> \>c1) \= $0<c_1 \leq c_2$.\\
\>  \>   \>In this case by proposition 8 \\
\>  \>   \>$d^*(a,c)=d^*(b,c)=(\ldots,0)$ and\\
\>  \>   \>$d^*(a,b)=l^*(x v^{c_1-N})+d^*(v^N y_1, v^{c_2-c_1} y_2)$ as \\
\>  \>   \>$a=x v^{c_1-N} \circ v^N y_1$, $b=x v^{c_1-N} \circ v^{c_2-c_1+N} y_2$.\\
\>  \>   \>Therefore $d_2^*(a,b)>0$ and $d^*(a,b)>d^*(a,c)$. \\

\>  \>   \>If $c_2 \leq c_1 <0$ the consideration is the same as above. \\

\> \>c2) $c_1=0$. \\
\>  \>   \>Then $a=x, \; b=x v^{c_2}y_2, \; c=x_3 v_3^{c_3}y_3$. \\
\>  \>   \>We are in the case 4) of proposition 8 and  axiom 3)  \\
\>  \>   \>for $d^*$ is reduced to the same problem for $d$. \\
\end{tabbing}

\begin{tabbing}
{\bf Case 3.} \=
$x_1=x_2=x_3=x.$ \\

\>{\bf a)} \=If $v_i \neq v_j, \; i \neq j$ we are in case 1) of proposition 8 \\
\>   \>and axiom 3 for $d^*$ is reduced to the same problem for $d$.\\

\>{\bf b)} $v_1=v_2 \neq v_3, \; \varepsilon (c_1) \neq \varepsilon (c_2)$ is
the same that case 3a).\\

\>{\bf c)} $v_1=v_2 \neq v_3, \; \varepsilon (c_1) = \varepsilon (c_2), c_1 \leq c_2$.\\
\>  \>In this case $d_2(a,b)= \mid c_1 \mid >0, \; d(a,c)$ and $d(b,c)$ are \\
\>  \>calculated as in case 1) of the proposition and \\
\>  \>$d(a,c)=d(b,c)=(\ldots,0)$. Therefore $d(a,b) > d(a,c)$.\\

{\bf Case 4.}
$x_1=x_2=x_3=x,\; v_1=v_2=v_3=v.$ \\

\> {\bf a)} Let $0<c_1 \leq c_2 \leq c_3$ and \\
\>  \>$a=x v^{c_1} v^{-N} \circ v^N y_1,$ \\
\>  \>$b=x v^{c_1} v^{-N} \circ v^{c_2-c_1} \circ v^N y_2,$ \\
\>  \>$c=x v^{c_1} v^{-N} \circ v^{c_2-c_1} \circ v^N y_3.$ \\
\>  \>With help of proposition 6 one can reduce this case to \\
\>  \>case 5 ($c_1=0$).\\

\>{\bf b)} $c_1 < 0 < c_2 \leq c_3$. \\
\>  \>In this case $d(a,b)=d(a,c)=(\ldots,0)$ and\\
\>  \> $d(b,c)$ (case 2 of proposition 8) = ($\ldots, \mid c_2 \mid >0$).\\
\> \>Therefore $d(b,c)>d(a,b)$.\\

\>{\bf c)} Another subcases are similar subcases a), b). \\

{\bf Case 5.}
$x_1=x_2=x_3=x, \; c_1=0, \; c_2 \neq 0, \; c_3 \neq 0$.\\
\>{\bf a)} $v_2=v_3, \; \varepsilon_2(c_2) \neq \varepsilon_3(c_3)$.
This is case 1 of proposition 8. \\
\>{\bf b)}
$
v_2=v_3, \; \varepsilon_2(c_2) = \varepsilon_3(c_3), \;
0 < \mid c_2 \mid \leq \mid c_3 \mid.
$  \\
\>  \>In this case\\
\>  \>$d^*(a,b)=(d(x,xv^{\varepsilon N}),0)$, \\
\>  \>$d^*(a,c)=(d(x,xv^{\varepsilon N}),0)$, \\
\>  \>$d^*(b,c)=(d(x,xv^{\varepsilon N}),\mid c_2 \mid > 0)$.\\

{\bf Case 6.}
$c_1=c_2=0, \; c_3 \neq 0$. In this case \\
\>  \>$d^*(a,b)=(d(x_1,x_2),0)$,\\
\>  \>$d^*(a,c)=(d(x_1,x_3v^{\varepsilon N}),0)$,\\
\>  \>$d^*(b,c)=(d(x_2,x_3 v^{\varepsilon N}),0)$. \\
\>  \>Axiom 3) for $l^*$ is reduced to the same problem for $l$.\\

{\bf Case 7.}
$c_1=c_2=c_3=0$. It is clear.
\end{tabbing}
${\bf \Box}$

\section{Proof of theorem 1}

We constructed the length function $l^*$ on the group $G^*$
in the previous paragraph. To prove theorem 1 one needs to check that
$l^*$ is a free regular step-like function and the stabilizing condition
is true for $G^*$.  By construction, $l^*$ is step-like under the chain
$0<\Lambda_0<\Lambda_0 \oplus \Lambda$ of convex subgroups (see propositions
4, 5). Moreover, $l^*$ is regular (see propositions 6, 8). For checking
another properties of the function $l^*$ let us introduce some definitions
and prove some results.

\begin{definition}
Let $g \in G^*\backslash G$ and
\begin{eqnarray}
g=\mu_1 \circ \ldots \circ \mu_m
\end{eqnarray}
is a minimal block-reduced $\circ$-decomposition. Define a
{\sf width function} $p: G^* \rightarrow \omega$ by the formula:
\[p(g) = \left \{ \begin{array}{ll}
		    m, & \mbox{if $g\in G^*\backslash G$},\\
		    0, & \mbox{if $g \in G$}
		  \end{array}
	  \right. \]
\end{definition}

By proposition 4 the width function $p$ is defined correctly.

\begin{definition}
Let $g \in G^*\backslash G$.  The element $g$ and the decomposition are
called {\sf cyclically block-reduced} iff $\mu_m \mu_1$ is not a
block, or, in an equivalent form, $p(g^k)= \mid k \mid p(g)$ for
all $k \in \bf Z$.
\end{definition}

Let $g$ be a cyclically block-reduced and $\mid k \mid \geq 2$, then
for $k>1$
$$
g^k=\mu_1 \circ \ldots \circ \mu_m \circ \mu_1 \circ \ldots =
\mu_1 \circ \ldots \circ \mu_{m-1} \circ \mu_m \mu_1 \circ
\mu_2 \ldots
$$
By the proof of proposition 5 there exists an element $w \in G$ such that
$$
\mu_m \cdot \mu_1=(\mu_m'\circ w^{-1})(w \circ \mu_1')=\mu_m' \circ \mu_1'.
$$
Hence it follows that
$$
g=\omega \circ \mu_1' \circ \mu_2 \circ \ldots \circ \mu_{m-1}
\circ \mu_m' \circ \omega ^{-1} = \omega \circ g_1 \circ \omega^{-1},
$$

\begin{eqnarray}
l^*(g_1^k)= k(l^*(g_1) \; and \;
l^*(g^k)=2l(w)+ kl^*(g_1)
\end{eqnarray}

Let $g$ be any element from $G^*\backslash G$. If $g$ is not cyclically
block-reduced then $p(g^2)=2m-t, \; t>0$. Moreover, if $t=2t_1$
then there exists an element $f$, $p(f)=t_1$, such that
$g=f \circ g_1 \circ f^{-1}$ and $g_1$ is cyclically block-reduced.

In this case
\begin{eqnarray}
l^*(g) = 2l^*(f)+l^*(g_1)
\end{eqnarray}
If $t=2t_1+1$ then there exists an element $f$, $p(f)=t_1+1$, such that
$g=f \circ g_1 \circ f^{-1}$ and $g_1$ is cyclically block-reduced.

In this case
\begin{eqnarray}
l^*(g) = 2l^*(f)+l^*(g_1)
\end{eqnarray}

\begin{corollary}
The function $l^*$ is a free length function on $G^*$
\end{corollary}

$\Box$
If $1 \neq g \in G$ then $l^*(g^2) > l(g)$ by induction. If
$g \in G^*\backslash G$
then $l^*(g^2) > l(g)$ by formulas (8), (9), (10).

Assume ${\bf C} = \{ C_i \mid i \in I \}$ is the set of centralizers in
$G^*$ such that:

1) $C_i$ is an infinite cyclic group, $C_i = <v_i>$ and
$$
l^*(v_i^k)= \mid k \mid l(v_i) \; for \; all \; k \in \bf Z;
$$

2) $C_i$ is not conjugated to $C_j$ for $i \neq j$.

Put $V({\bf C}) = \{v_i \mid i \in I \}$

\begin{proposition}
The stabilizing condition is true for $V({\bf C})$.
\end{proposition}

$\Box$
Let us begin with proving of an analog of Lemma LS.

\begin{lemma} \label{lm}
If $v$ and $w$ are elements of the group $G^*$ and have powers $v^m$
and $w^n$, $m,\;n >0$ with the common initial segment of length
$s \cdot max\{l^*(v),l^*(w)\}, \; s \in \bf N$, where $s \rightarrow \infty$
if $m,n \rightarrow \infty$, then $[v,w]=1$.
\end{lemma}

We will prove the lemma by induction on the level of construction
of $G^*$. Assume, that  lemma holds for  elements of group $G$
and  the length  function  $l$  and prove it for the group $G^*$
and the length function $l^*$.

Notice further, that if lemma \ref{lm} holds  for a pair of
elements $(v,w)$, then it is true for any other pair
of elements of the form $(gvg^{-1},gwg^{-1})$.
Thus, without the loss of generality we can assume that
$l^*(v^m)=ml^*(v)$, $w=g \circ w_0 \circ g^{-1}$ and
$l^*({w^n}_0)=nl^*(w_0)$, $l^*(g) < l(v)$.
Since elements $v^m$ and $w^n$ have a sufficiently large common
initial segment, then $v=g \circ v_2$. Going to the pair of
elements $(g^{-1}vg, w_0)$, we can also assume that
$l^*(w^n)=nl^*(w)$. Now if $l^*(v)=l^*(w)$ then $v=w$ therefore
$[v,w]=1$, and lemma \ref{lm}  holds for the such pair.

Thus, let, for example, $l^*(v) < l^*(w)$  and
$w=v^s \circ v_1$, where $s \in {\bf N}$, $v=v_1 \circ v_2$.
In  view of the fact that degrees $v^m$ and $w^n$  have a
sufficiently large common initial segment  one obtains
$v^s \circ v_1 \circ v^s=v^{2s} \circ v_1$ whence
$v_1 \circ v^s=v^s \circ v_1$, and consequently,
$[v,v_1]=1$. But in this case $[v,w]=[v,v^s \circ v_1]=1$.


The proof of proposition 11 is further conducted by lemma \ref{lm} in a manner like
the proof of the fulfilment of condition (S) for a canonical function
on a free group (see section 3.3).
$\Box$


The proof of theorem 1 is finished.
$\bf \Box$

\section{Proof of theorem 2}

Let $A$ be an abelian torsion free group, $\bf Z$  a pure subgroup of
$A$ and $F$ a free group with a base $X$, $F^A=F \oplus A$ be a tensor
completion. To prove theorem 2 let us consider two cases.

\subsection{Splitting case}

$\bf Z$ is a direct summand of $A$, $A={\bf Z} \oplus C$. The tensor
completion $F^A$ is a direct limit of the chain of subgroups:
$$
G^{(0)}=F<G^{(1)}<G^{(2)}< \ldots < G^{(n)}< \ldots,
$$
where $G^{(n)}$ is obtained from $G^{(n-1)}$ by  tree extension of
centralizers. We have to construct a free regular step-like length
function on $F^A$ by induction on $n$. Let us designate by
$l^0:F \rightarrow \bf Z$ a canonical length function under the
set $X$. $l^0$ is a free regular step-like function and the stabilizing
condition is true for $F$ (see 3.3). Fix an arbitrary linear order
on $C$ and put $\Lambda_0={\bf Z}, \; C=\Lambda$. By theorem 1
there exists a free regular step-like length function
$$
l^1:G^{(1)} \rightarrow {\bf Z} \oplus C =A
$$
and the stabilizing condition is true on $G^{(1)}$. Repeating these
reasonings we obtain the function
$$
l^n: G^{(n)} \rightarrow {\bf Z} \oplus
\underbrace{C\oplus \ldots \oplus C}_n,
$$
besides, the function $l^n$ continues the function $l^{n-1}$,
$n=2,3, \ldots$. The resulting function
$$
l^{\infty}= \lim_{\rightarrow} l^n: F^A \rightarrow
{\bf Z} \oplus (\oplus C^{\omega})
$$
will be sought.

\subsection{Restricted ultraproducts}

Let $l:G  \rightarrow \Lambda$ be a length function on a group $G$
with values in an ordered abelian group $\Lambda$. From the pure
algebraic point of view this length function can be described as a
two-sorted model (see \cite{KCh} for this approach)
$$
{\cal M}=<G, \Lambda, l>,
$$
where $G$ is a group with a group multiplication; $\Lambda$ is an
ordered abelian group with an addition $+$ and order $<$;
$l:G \rightarrow \Lambda$ is an unary operation defined on $G$.
To the model $<G, \Lambda, l>$ one can assign a two-sorted
language (one kind of variables for elements from $G$ and another
kind for elements from $\Lambda$) with the all mentioned above
operations and predicates in the language. The described language
$\cal L$ is called {\it the language of groups with a length function}.

Let
$$
\{ {\cal M}_{\alpha}=<G_{\alpha}, \Lambda_{\alpha}, l_{\alpha}>,
\alpha \in \cal A \}
$$
be a set of such two-sorted models and $\cal D$ be some ultrafilter over
$\cal A$. The ultraproduct
$$
\tilde{{\cal M}}=<\tilde{G}, \tilde{\Lambda}, \tilde{l}>,
$$
will be a two-sorted model for the length function
$\tilde{l}: \tilde{G} \rightarrow \tilde{\Lambda}$,
where
$\tilde{G}=\prod G_{\alpha}/\cal D$,
$\tilde{\Lambda}=\prod A_{\alpha}/\cal D$,
$\tilde{l}=\prod l_{\alpha}/\cal D$.

\begin{proposition}
Let $l_{\alpha}$ be a free regular length function for all
$\alpha \in \cal A$. Then $\tilde{l}$ is also a free regular length
function.
\end{proposition}

$\Box$
Straightforward verification.
$\Box$

Proposition 11 is not true for the property 'to be step-like'. Let
us introduce a new type of products that will be also inheriting the
property 'to be step-like'.

Let $G$ be a group with a free regular step-like function
$l:G \rightarrow A$ under the following chain of convex subgroups of
$A$.
$$
A_0 = \{ 0 \} < A_1 < A_2 < \ldots < A_n < \ldots, \;
\bigcup_{i=1}^{\infty} A_i =A.
$$
Let
$$
G_0 = \{ 1 \} \leq G_1 \leq G_2 \leq \ldots \leq G_n \leq \ldots, \;
\bigcup G_n = G.
$$
be the corresponding chain of subgroups of $G$. (see 3.2.)

\begin{definition}
The level function $\nu: G \rightarrow \omega$ is defined by the
rule:
$$
\nu(g)=n \Leftrightarrow g \in G_n\backslash G_{n-1},
$$
the number $\nu(g)$ is said to be {\sf the level of the element} $g$.
\end{definition}

By theorem 8 \cite{MR:tree} the tensor completion $F^A$ is the  union  of  the
chain of subgroups:
$$
F=F_0 < F_1 < \ldots < F_n \ldots, \; \bigcup F_n =F^A,
$$
where $F_{n+1}=F_n({\bf C}_n,A)$. If $A$ is a splitting group over $\bf Z$
we constructed a free regular step-like function
$l:F^A \rightarrow A^{\omega}$ in such a way that
$$G_n=F_n, \; A_n=A^n= \underbrace{A \oplus \ldots \oplus A}_{n}.$$
(see $\S$5 and 6.1.).

If $\nu(g)=n,$ then $g$ can be written in the reduced form
$$
g=g_1v_1^{c_1} \ldots g_m v_m^{c_m} g_{m+1},
$$
where $c_i \in C, \; v_i \in V_n, \; \nu(g_i)<n$ and if $v_i=v_{i+1}$
then $g_i \notin v_i^{\bf Z}$. Moreover, we can assume that the elements
$g_1, \ldots, g_{m+1}$ are also written in the reduced form as elements of
$F_{n-1}$ and so on. This leads to an explicit reduced form of $g$ and
let us designate by $R(g)$ the set of elements in the record of some
explicit form for $g$.

Let again
$$
\{ {\cal M}_{\alpha}=<G_{\alpha}, \Lambda_{\alpha}, l_{\alpha}>,
\alpha \in \cal A \}
$$
be a set   two-sorted models and $\cal D$ be some ultrafilter over $\cal A$.
Assume that $l_{\alpha}$ is a step-like length function, a width
function $p_{\alpha}$ and a level function $\nu_{\alpha}$ exist for
all $\alpha \in \cal A$.

A restricted ultraproduct
$\tilde{{\cal M}}_0 = < \tilde{G_0}, \tilde{\Lambda_0}, \tilde{l_0}>$
can be described by the following. Let $g \in \tilde{G}$, then
$g \in \tilde{G_0}$ iff there exist a representative
$g=[g_{\alpha}]_{\alpha \in A}$ and a number $k \in N$ such that
$\nu_{\alpha}(g_{\alpha}) \leq k, \; p_{\alpha}(f_{\alpha}) \leq k$,
where $f_{\alpha} \in {\cal R}(g_{\alpha})$ for all $\alpha \in \cal A$.
There exists the chain of convex subgroups
\begin{eqnarray}
\tilde{\Lambda}_{0,0} <  \tilde{\Lambda}_{0,1} < \ldots <
\tilde{\Lambda}_{0,n} < \ldots,
\bigcup_{i=0}^{\infty} \tilde{\Lambda}_{0,n} = \tilde{\Lambda}_0,
\end{eqnarray}
where
$\tilde{\Lambda}_{0,n}= \prod \Lambda_{\alpha,n} / \cal D$.
The length function  $\tilde{l_0}: \tilde{G_0} \rightarrow \tilde{\Lambda_0}$
is the restriction of $\tilde{l}$ on $\tilde{G_0}$.

With the help of standard theoretical-model reasonings one can prove

\begin{proposition}
  \begin{enumerate}
    \item $\tilde{G_0}$ is a subgroup of $\tilde{G}$;
    \item $\tilde{l_0}$ is a step-like function on $\tilde{G_0}$
	  under the chain of convex subgroups (11);
    \item a level function $\tilde{\nu_0}$ and a width function $p_0$
	  exist for $\tilde{l_0}$.
    \item if all  functions $l_{\alpha}$ are regular, then $\tilde{l_0}$
	  is also regular.
 \end{enumerate}
\end{proposition}

\subsection{General case}

We have to construct a free regular step-like length function on $F^A$,
where $A$ is an abelian torsion free group and $\bf Z$ is a pure
subgroup of $A$. Let $\{A_{\alpha}, \alpha \in \cal A \}$ be the set of all
finitely generated subgroups of $A$ which contains $\bf Z$. Since
$\bf Z$ is a pure subgroup in $A_{\alpha}$ then $\bf Z$ is a direct summand
for $A_{\alpha}$; $A_{\alpha} = {\bf Z} \oplus C_{\alpha}$. Fix some
linear order on $A$ so  that $\bf Z$ is a convex subgroup (see $\S$2).
By item 6.1. there exists a free regular step-like function
$$
l_{\alpha}: F^{A_{\alpha}} \rightarrow \oplus A_{\alpha}^{\omega},
$$
besides that, the level function $\nu_{\alpha}$ and width function
$p_{\alpha}$ exist for all $\alpha \in A$.

An embedding $A_{\alpha} \subset A_{\beta}$ induces an embedding of
tensor completions $F^{A_{\alpha}} \subset F^{A_{\beta}}$,
as the group $F^A$ is the union of $F^{A_{\alpha}}, \alpha \in \cal A$.
Unfortunately, if $A_{\alpha} \subset A_{\beta}$ then it does not always
happen that
$$
l_{\beta} \mid_{F^{A_{\alpha}}} = l_{\alpha},
$$
i.e. the function $l_{\beta}$ do not necessarily continue
$l_{\alpha}$. Consequently, one cannot construct the limit function $l$
on the group $F^A$ as $l=\lim_{\rightarrow} l_{\alpha}$. Because of
this, we choose another way of constructing the length function $l$ - the
restricted ultraproducts method.

\begin{proposition}
There exists an ultrafilter $\cal D$ on the set $\cal A$ such that the
canonical homomorphism
$$
\varphi: F^A \rightarrow \tilde{F} = \prod F^{A_{\alpha}}/ \cal D
$$
is an embedding and the image $(F^A)^{\varphi}$ of the group $F^A$
is contained in the restricted ultraproduct $\tilde{F_0}$.
\end{proposition}

$\Box$
The proof is standard for such type statements and coinsides exactly
with the proof of Mal'cev local theorem on representability of linear
groups (see, for example, \cite{KCh}). The second statement follows
immediately from lemma 8 proved below.
$\Box$

\begin{lemma}
Let $A_{\alpha} \subset A_{\beta}$ and $g \in F^{A_{\alpha}}$. Then
the sequence $(\mu_1, \ldots, \mu_m)$ is a block reduced for $g$
under the length function $l_{\alpha}$ iff the one is block-reduced for $g$
under $l_{\beta}$.
\end{lemma}

$\Box$
It is sufficient to prove the lemma  for two blocks
$$
\mu_1=x_1 v_1^{c_1}y_1, \; \mu_2=x_2 v_2^{c_2}y_2, \;
c_1,c_2 \in C_{\alpha}.
$$
By the construction of groups $C_{\alpha}, \; C_{\beta}$
$$
c_1=c_1'+n(c_1), \; c_2=c_2'+n(c_2), \;
c_1',c_2' \in C_{\beta}, \; n(c_1), n(c_2) \in \bf Z.
$$
Thus the canonical writting of elements $\mu_1, \; \mu_2$ in the
group $F^{A_{\beta}}$ is the following
$$
\mu_1=x_1 v_1^{n(c_1)} v^{c_1'}y_1, \;
\mu_2=x_2 v_2^{n(c_2)} v^{c_2'}y_2
$$
With respect to the order in the group $A$
$$\varepsilon(c_i)=\varepsilon(c_i')$$
($\varepsilon(c)$ is the sign of an element $c$).
By lemma 6 $\mu_1 \mu_2 = \mu_1 \circ \mu_2$ in the group
$F^{A_{\alpha}}$ iff
$d(y_1^{-1} v_1^{-\varepsilon_1 N}, x_2 v_2^{-\varepsilon_2 N})=0$.
But for sufficiently large $N$
$$
d(y_1^{-1} v_1^{-\varepsilon_1 N}, x_2 v_2^{n(c_2)} v_2^{-\varepsilon_2 N})=
d(y_1^{-1} v_1^{-\varepsilon_1 N}, x_2 v_2^{-\varepsilon_2 N})=0
$$
and, therefore, $\mu_1 \mu_2 = \mu_1 \circ \mu_2$ in the group
$F^{A_{\beta}}$.
$\Box$

\begin{lemma}
Let $A_{\alpha} \subset A_{\beta}$ and $f,g \in F^{A_{\alpha}}$. Then if
$f=z \circ f_1$, $g=z \circ g_1$ and $d_{\alpha}(f,g)=l_{\alpha}(z)$
in $F^{A_{\alpha}}$ then
$f=z \circ f_1, \; g=z \circ g_1$ and $d_{\beta}(f,g)=l_{\beta}(z)$
in $F^{A_{\beta}}$
\end{lemma}

$\Box$
By lemma 7 and proposition 6 it is sufficient to conduct the proof for
blocks only. In this case one can obtain the required result using
proposition 8 and equalities $\varepsilon(c_i)=\varepsilon(c_i')$.
$\Box$


Let us return to the proof of theorem 2. Since, by proposition 13,
$F^A$ is a subgroup of the restricted ultraproduct $\tilde{F_0}$, then
let us designate by $l:F^A \rightarrow \tilde{\Lambda_0}$
the restriction of the function $\tilde{l_0}$ on $F^A$.
By propositions 11, 12, $l$ is a free step-like function.
It only remains for us to verify that $l$ is a regular length
function on $F^A$. Since, by proposition 14, the function
$\tilde{l_0}$ is regular, then there exists an element $z \in \tilde{F_0}$
for $f,g \in F^A$ such that
$
f=z \circ f_1, \; g=z \circ g_1, \tilde{d}_0(f,g)=\tilde{l}_0(z).
$
From the way of embedding $F^A$ into $\tilde{F}_0$ and lemma 9
it follows that $z \in F^A$ and, therefore, the function $l$ is
regular.

The proof of theorem 2 is finished.
$\Box$


\begin{thebibliography}{99}
\bibitem{B}  Bass H.
{\em Groups acting on non-archimedian  trees.}
Arboreal group theory, 1991, p. 69-130.

\bibitem{F} Fuchs L.
{\em Partially ordered algebraic systems.}
Pergamon Press, Oxford $\cdot$ London $\cdot$ New York $\cdot$ Paris,
1963.

\bibitem{KCh} Keisler, Chan
{\em Model theory.}

\bibitem{KMR} Kharlampovich O.G., Myasnikov A.G., Remeslennikov V.N.
{\em Logical languages and axioms for groups with a length function.}
Omsk University, 1995, N20, Omsk, p.1-8.

\bibitem{MR:foun} Myasnikov A.G., Remeslennikov V.N.
{\em Exponential groups: foundations of the theory and tensor
completion.}
Siberian Math. J., 1994, v.35, N5, p.1106-1118.

\bibitem{MR:tree} Myasnikov A.G., Remeslennikov V.N.
{\em Exponential groups: iterated tree extensions of centralizers.}
(to appear in Contemporary Mathematics).

\bibitem{MR:res} Baumslag G., Myasnikov A.G., Remeslennikov V.N.
{\em Residually hyperbolic groups and approximation theorems for
extensions of centralizers.}
Omsk University, 1995, N21, Omsk, p.1-23.

\bibitem{LS} Lyndon R.C., Schutzenberger.
{\em The equation $a^m=b^nc^p$ in a free group.}
Michigan Math.J., 1962, N9, p.289-298
\end{thebibliography}




\end{document}
