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\providecommand{\eq}{\operatorname{eq}}
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\title{The Group Configuration Theorem}
\maketitle

An exposition of the group configuration theorem for stable theories,
\\following Chapter 5 of [Pillay-GST].

\section*{Introduction and Preliminaries}

Work in monster model $\M$ of a stable theory $T$.
\paragraph{Notation:}
\mbox{}\\    $a,b,c,d,e,w,x,y,z,\alpha,\beta,\gamma$ etc will take values in $\Meq$,
\\    and $A$,$B$,$C$ etc in $\PP(\Meq)$.

    $AB$ means $A \cup B$;
\\    $ab$ means $(a,b) \in \Meq$;
\\    when appropriate, $a$ means $\{a\}$; e.g. $Ab$ is short for $A\cup\{b\}$.

\paragraph{Group configuration, first statement:}
\mbox{}\\    Let $G$ be a connected $\Wedge$-definable group /$\emptyset $.
\\    Let $a,b,x \in G$ be an independent triple of generics.
\\    Let $b' := c'*a'$, $x' := a'*y'$, and $z' := b'*y'$ (so $z' = c'*a'*y' = c'*x'$).
\\    Then we have
    \begin{verbatim}
              a
             / \   
            b   x 
           / `z' \
          / .' `. \
         /.'     `.\
        c'         `y
    \end{verbatim}
    satisfying:
\\    (*) \vline\;\begin{minipage}[c]{0.9\textwidth}
\begin{itemize}\item any non-collinear triple is independent\;
\\	    (i.e. each element is independent from the other two);
\item if $(d,e,f)$ is collinear then
	    $\acleq(de) = \acleq(ef) = \acleq(df) .$
\end{itemize}
    \end{minipage}

    The group configuration theorem provides a converse statement:
\\    if a tuple $(a,b,c,x,y,z)$ satisfies (*),
\\    then possibly after base change\;
\\    (i.e. adding parameters independent from abcxyz to the language),
\\    there is a connected $\Wedge$-definable group $G /\emptyset $,
\\    and there are $(a',b',c',x',y',z')$ defined as above from $G$,
\\    such that each unprimed element is interalgebraic with the corresponding
    primed element.

\paragraph{Remark:}
\mbox{}\\    e.g. if each element realises a strongly minimal type,
\\    (*) says that the Morley rank of a non-collinear triple is 3, and that of
    a collinear triple is 2.

\paragraph{Stability theory preliminaries:}
\mbox{}\\    We have an independence notion, non-forking,
\\    satisfying (even after adding parameters) for all $A$,$B$,$C$:
\begin{itemize}\item Transitivity and Monotonicity:
\\	    \[ A \ind  BC \Leftrightarrow  (A \ind  B \;\textrm{and}\; A \ind _B C) \]
\item Symmetry:
\\	    \[ A \ind  C \Leftrightarrow  C \ind  A \]
\item Reflexivity:
\\	    \[ A \ind  A \Leftrightarrow  A \subseteq  \acleq(\emptyset) \]
\item "algebraic $\Rightarrow $ nonforking":
\\	    \[ A \ind  C \Leftrightarrow  A \ind  \acleq(C) \]
\end{itemize}

    $p \in S(A)$ is \underline{stationary} iff for any $B$,
\\	\[ x \models  p \;\textrm{and}\; x \ind _A B \]
    defines a well-defined type $p|_B \in S(AB)$,
\\    the \defn{non-forking extension} of $p$ to $AB$.

    $p \in S(A)$ is stationary iff it has a unique extension to $\acleq(A)$.

    So $\stp(a/B) := \tp(a/\acleq(B))$ is stationary.

    Stationary types $p\in S(A)$ are definable,
\\    i.e. $T$ eliminates Hrushovski quantifiers:
\\	for any formula/$\emptyset $ $\phi(x,y)$, there is a formula/$A$
	    \[ \psi(y) \equiv : d_p x. \phi(x,y) \]
		(read "$d_p x.$" as "for generic $x$,"),
\\	such that $\models  \psi(b)$ iff for a $\models  p$ with $a \ind _A b$,
\\	    \[ \models  \phi(a,b), \]
	i.e. iff $\phi(x,b) \in p|_b$.

    Note that for $A \subseteq  B$,
\\	\[ d_{p|_B} x. \phi(x,y) \equiv  d_p x. \phi(x,y). \]

    The \underline{canonical base} of a stationary type $p \in S(A)$, $A=\dcleq(A)$, is the
    least $\dcleq$ closed set $\Cb(p) \subseteq  A$
    such that the restriction of $p$ to $\Cb(p)$ is stationary and $p$ is its
    non-forking extension,
\\    i.e. such that all $d_p x$. $\phi(x,y)$ are defined over $\Cb(p)$.

    Let $\aCb(a/B) := \acleq(\Cb(\stp(a/B))$).
\\    Then $a \ind _C B \Leftrightarrow  \aCb(a/B) \subseteq  \acleq(C)$.

    Example - ACF:
\\	If $K$ is a perfect subfield and $p \in S(K)$,
\\	$p$ is the generic type of an irreducible variety $V$ over $K$;
\\	$p$ stationary $\Leftrightarrow $ $V$ absolutely irreducible;
\\	$\Cb(p) =$ (perfect closure of) field of definition of $V$.

    $A$ \defn{$\Wedge$-definable group}
    is a $\Wedge$-definable set $G$
    together with a relatively definable group operation\;
\\    (meaning that its graph is the restriction to $G^3$ of a definable set).

    If $G$ acts transitively on a $\Wedge$-definable set $S$, with the action relatively
    definable, we call $(G,S)$ a $\Wedge$-definable homogeneous space.

    $S$ is \underline{connected} iff there is a stationary type $s$ extending $S$
    such that if $g \in G$ and $b \models  s|_g$, then $g*b \models  s|_g$\;
\\	(i.e. $\operatorname{Stab}(s) = G$).
\\    $s$ is then called the \underline{generic} type of $G$.

    If $(G,S)$ is definable of finite Morley rank, $S$ is connected iff $\MD(S)=1$,
\\    and $a\in S$ is generic iff $\RM(a)=\RM(S)$.

    Generics and connectedness are defined for $G$ by considering the left
    (equivalently: right) action of $G$ on itself.

\section*{Germs and Hrushovski-Weil}

\paragraph{Definition:}
\mbox{}\\    Let $p$ and $q$ be stationary types $/\emptyset $.

    A \defn{generic map $p \rightarrow  q$}
    is the \underline{germ} $\widetilde{f}_b$ of a definable partial function $f_b$\;
\\    (i.e. $f_b(x)=y$ is given by a formula/$\emptyset $ $\phi_f(x,y,b)$),
\\    such that if $a \models  p|_b$ then $f_b(a) \models  q|_b$,
\\    where equality of germs is defined by $\widetilde{f}_b = \widetilde{g}_c$ iff
	for $a \models  p|_{bc}$, $f_b(a) = g_c(a)$;
\\	i.e. $\models  d_p x. f_b(x) = g_c(x)$ .

\paragraph{Example:}
\mbox{}\\    In ACF, the generic maps $p \rightarrow  q$ are precisely the dominant rational maps
    $\operatorname{locus}(p) \rightarrow  \operatorname{locus}(q)$.

\paragraph{Lemma:}
\begin{enumerate}[(i)]\item "Equality of germs" is indeed an equivalence relation.
\item For any $B$, if $a \models  p|_{Bb}$ then $f_b(a) \models  q|_{Bb}$.
\item Composition of germs is well-defined,
\\	yielding a category structure with objects the stationary types.
\item A germ $\widetilde{f}_b$ is invertible (i.e. an isomorphism) iff $f_b$ is injective
	on $p|_b$.
\end{enumerate}
\paragraph{Proof:}
\begin{enumerate}[(i)]\item Symmetry and reflexivity are clear.
\\	Suppose $f_b = g_c$ on $p|_{bc}$ and $g_c = h_d$ on $p|_{cd}$.
\\	Then clearly $f_b = h_d$ on $p|_{bcd}$.
\\	But $\phi(x) :\equiv  f_b(x) = h_d(x)$ is defined over $bd$,
\\	so already $\phi(x) \in p|_{bd}(x)$.
\\	So $f_b = h_d$ on $p|_{bd}$.
\item WLOG, $p,q\in S(\emptyset )$.
\\	$a \ind  Bb \Rightarrow  a \ind _b B \Rightarrow  f_b(a) \ind _b B$;
\\	but $f_b(a) \ind  b$,
\\	so $f_b(a) \ind  Bb$.
\item Suppose $\widetilde{f}_b : p \rightarrow  q$ and $\widetilde{g}_{c} : q \rightarrow  r$.
\\	By (ii), if $a \models  p|_{bc}$ then $g_c(f_b(a)) \models  r|_{bc}$,
\\	so $\widetilde{(g_c \circ f_b)}$ is a germ $: p \rightarrow  r$.
\item Clear.
\end{enumerate}
    \qed

\paragraph{Notation:}
\begin{itemize}\item $\operatorname{Hom}(p,q) :=$ set of germs $p \rightarrow  q$;
\item $\operatorname{Iso}(p,q) :=$ set of invertible germs $p \rightarrow  q$;
\item $\operatorname{Aut}(p) :=$ group of invertible germs $p \rightarrow  p$.
\end{itemize}

%Remark to self:
%    We can make analogous definitions for germs of finitely valued maps 
%    $f_b : p --> q$, where $a |= p|_b ==>$ all values of $f_b(a)$ satisfy $q|_b$, and
%    have the same type over ab.
%
%    [(ii) above only really used that $f_b(a) \in \acleq(ba)$, so goes through.]
%
%    Notation: $\"Hom^{\sim}(p,q)$ etc.
%
%    Hmm, but not so natural to talk about finite-to-finite correspondences
%    this way (it isn't $\"Iso^{sim}(p,q)$)...

\paragraph{Definition:}
\mbox{}\\    A \defn{family of generic maps $p \rightarrow  q$ based on $s$}
    is a family of germs $\{ \widetilde{f}_b \;|\; b \models  s \}$ of a definable family of
    partial functions $f_z$.
\\    The family is \underline{canonical} if for $b$,$b' \models  s$,
\\	\[ \widetilde{f}_b = \widetilde{f}_{b'} \Leftrightarrow  b=b'. \]

    Note that by definability of types, any family can be made canonical by
    quotienting $s$ by the definable equivalence relation of equality of germs.

\paragraph{Remark:}
\mbox{}\\    In ACF, algebraic families of dominant rational maps $V \rightarrow  W$ can be made
    canonical by parametrising them using the Hilbert scheme of $V \times W$.

\paragraph{Remark:}
\mbox{}\\    If $\widetilde{f}_z$ is a family of generic maps $p \rightarrow  q$ based on $s$,
\\    let $b \models  s$ and $x \models  p|_b$, and let $y = \widetilde{f}_b(x)$;
\\    then
    \begin{equation}\tag{+} x \ind  b;\; y \ind  b;\; y \in \dcleq(bx) . \end{equation}

    Conversely, if $(b,x,y)$ satisfy $(+)$,
\\    let $f_b(x)=y$ be a formula witnessing $y\in\dcleq(bx)$;
\\    then $\widetilde{f}_z$ is a canonical family of generic maps $\stp(x) \rightarrow  \stp(y)$ based on
    $\stp(b)$.

\paragraph{Lemma:}
\mbox{}\\    In the correspondence of the previous remark,
\begin{enumerate}[(i)]\item $\widetilde{f}_z$ is invertible iff $x \in \dcleq(by)$
	i.e. iff $x$ and $y$ are interdefinable over b;
\item $\widetilde{f}_z$ is canonical iff $\Cb(\stp(xy/b)) = \dcleq(b)$
\end{enumerate}
\paragraph{Proof:}
\begin{enumerate}[(i)]\item Clear
\item Suppose that $\widetilde{f}_z$ is canonical, and obtain $(b,x,y)$ as above.
\\	$\tp(xy/b)$ is stationary since $\tp(x/b)$ is,
\\	so $C := \Cb(\stp(xy/b)) \subseteq  \dcleq(b)$.

	If $C \neq  \dcleq(b)$,
\\	say $b' \neq  b$ with $b' \equiv _C b$ and $b' \ind _C bx$;
\\	then $xy \ind _C b'$, so $xyb \equiv  xyb'$,
\\	so since $f_b(x) = y$, also $f_{b'}(x) = y$,
\\	but $x \models  p|_{bb'}$ so this contradicts canonicity of the family $\widetilde{f}_z$.

	For the converse, let $f_b(x)=y$ be a formula witnessing $y \in \dcleq(bx)$.

	Suppose $\widetilde{f}_b$ is not canonical. By definability of types,
\\	some $\widetilde{g}_c$ is canonical with $c \in \dcleq(b)$ but $b \notin \dcleq(c)$.
\\	Then $\Cb(\stp(xy/b) \subseteq  \dcleq(c)$ since
	    \[ x \ind  b \Rightarrow  x \ind _c b \Rightarrow  xy \ind _c b, \]
	    and $\tp(xy/c)$ is stationary since $\tp(x/c)$ is and $y \in \dcleq(xc)$.
\\	This contradicts $\dcleq(b) = \Cb(\stp(xy/b))$.
\end{enumerate}
    \qed

%Remark to self:
%    Could give analogue for (inter)algebraicity, though I'm not sure what
%    happens with the stationarity at the end of the proof.

\paragraph{Remark:}
\mbox{}\\    Suppose $(b,x,y)$ "lie on a line" in the sense of the group configuation
    statement above,
\\    i.e. $\acleq(bx) = \acleq(xy) = \acleq(yb)$.
\\    So $x$ is interalgebraic with $y$ over $b$.
\\    Since $b \in \acleq(ac)$, $b$ is interalgebraic with $\Cb(\stp(ac/b)$); indeed:
\\	let $D = \aCb(ac/b)$; then $ac \ind _D \acleq(b)$,
\\	so $\acleq(b) \ind _D \acleq(b)$,
\\	so $\acleq(b) = D$.

    So $(b,x,y)$ is "nearly" a triple corresponding to a canonical family of
    invertible germs.

\paragraph{Lemma HW:}
\mbox{}\\    Suppose $\widetilde{f}_z$ is a canonical family of generic bijections $p \rightarrow  p$ based on $s$.
\\    Let $G_0 := \{ \widetilde{f}_b \;|\; b \models  s \} \subseteq  Aut(p)$.
\\    Suppose that $G_0$ is closed under inverse,
\\    and suppose that for $b_1$ and $b_2$ independent realisations of $s$,
\\	\[ \widetilde{f}_{b_1} \circ \widetilde{f}_{b_2} = \widetilde{f}_{b_3} \]
    with $b_3 \models  s|_{b_i}$ for $i=1,2$.

    Then, identifying $\widetilde{f}_b$ with $b$,
\\    the group $G \leq  Aut(p)$ generated by $G_0$ is connected $\Wedge$-definable, with $s$
    its generic type.
\paragraph{Remark:}
\mbox{}\\    This is essentially the Hrushovski-Weil "group chunk" theorem.
\\    There, one starts with a generically associative binary operation $*$, and
    then applies the above statement to the germs of $x \mapsto  a*x$ to obtain a
    group structure extending $*$.
\paragraph{Proof:}
\mbox{}\\    First, we show that any element of $G$ is a composition of two generators.

    It suffices to see that any composition of three generators
	\[ \widetilde{f}_{b_1} \circ \widetilde{f}_{b_2} \circ \widetilde{f}_{b_3} \]
    is the composition of two. But indeed, let $b' \models  s|_{b_1,b_2,b_3}$.
\\    Then 
	\[ \widetilde{f}_{b_1} \circ \widetilde{f}_{b_2} \circ \widetilde{f}_{b_3}
	= \widetilde{f}_{b_1} \circ \widetilde{f}_{b'} \circ \widetilde{f}_{b'}^{-1} \circ \widetilde{f}_{b_2} \circ \widetilde{f}_{b_3} \]
    Now $b' \ind  b_2$, so
    say $\widetilde{f}_{b'}^{-1} \circ \widetilde{f}_{b_2} = \widetilde{f}_{b''}$
    with $b'' \models  s$ independent from $b'$ and from $b_2$;
\\    then $b'' \ind  b_3$ since:
\\	$b' \ind _{b_2} b_3$,
\\	so $b'' \ind _{b_2} b_3$ (since $b'' \in \dcleq(b'',b_2)$),
\\	so since $b'' \ind  b_2$,
\\	$b'' \ind  b_2b_3$, and in particular $b'' \ind  b_3$.
\\    Also $b' \ind  b_1$.
\\    So $\widetilde{f}_{b_1} \circ \widetilde{f}_{b'}$ and $\widetilde{f}_{b'}^{-1} \circ \widetilde{f}_{b_2} \circ \widetilde{f}_{b_3}$ each
    "realise $s$".

    Now $G$ is defined as pairs of realisations of $s$, modulo generic equality of
    their compositions, and the group operation is defined by composition.

    Finally, to see that $G$ is connected with generic type $s$:
\\    if $g \models  s$ and $b \models  s|_g$, then $g*b \models  s|_g$,
\\    and then by induction the same holds for any $g \in G$. \qed

In the context of the group configuration, we work with definable families of
bijections between two types, rather than from a type to itself. The following
key lemma gives conditions for this to give rise to a group.

\paragraph{Lemma A:}
\mbox{}\\    Suppose $\widetilde{f}_z$ is a canonical family of generic bijections $p \rightarrow  q$ based on $r$.
\\    Let $b_1$ and $b_2$ be independent realisations of $r$ and say
	\[ \widetilde{f}_{b_1}^{-1} \circ \widetilde{f}_{b_2} = \widetilde{g}_c \]
    with $\widetilde{g}_w$ a canonical family of generic bijections $p \rightarrow  p$ based on
    $s = \stp(c)$,
\\    and suppose
    \begin{equation}\tag{+}  c \ind  b_i \;\textrm{for}\; i=1,2.\end{equation}
    Then $\widetilde{g}_w$ satisfies the assumptions of Lemma HW.
\paragraph{Remark:}
\mbox{}\\    In the finite Morley rank setting (e.g. ACF),
\\    \[ \oplus \Leftrightarrow  \RM(c) = \RM(s). \]
\paragraph{Proof:}
\mbox{}\\    Let $c' \models  s|_c$.
\\    Let $b \models  r|_{c,c'}$.
\\    Then by (+), $bc \equiv  b_2c$, so say $b_1'bc \equiv  b_1b_2c$;
\\    similarly, $bc' \equiv  b_1c$, so say $bb_2'c' \equiv  b_1b_2c$.
\\    Then
	\[ \widetilde{g}_c \circ \widetilde{g}_{c'} = \widetilde{f}_{b_1'}^{-1} \circ \widetilde{f}_b \circ \widetilde{f}_b^{-1} \circ \widetilde{f}_{b_2'}
	    = \widetilde{f}_{b_1'}^{-1} \circ \widetilde{f}_{b_2'} .\]
      \begin{verbatim}
                                         *      *      *         q
                                        / \   .'|`.   / \
                                       / c'\.'  |  `./ c \   
                                      /   .'\   |b  /`.   \
                                     /  .'   \  |  /   `.  \
                                    / .'      \ | /      `. \
                                   /.' b_2'    \|/    b_1' `.\
                                  *'            *            `*  p
      \end{verbatim}
    Now $(b,b_1',b_2')$ is an independent triple,
\\    since
	$b_i' \ind  b$ by choice of $b_i'$, since $b_1 \ind  b_2$,
\\	and $b_1' \ind _{b} b_2'$, since $c \ind _b c'$, since $c \ind  c'$ and $b \ind  cc'$.

    So $b_1'$ and $b_2'$ are independent realisations of $r$,
\\    so say $\widetilde{f}_{b_1'}^{-1} \circ \widetilde{f}_{b_2'} = \widetilde{g}_{c''}$.

    Then by (+), $c'' \ind  b_1'$, and hence $c'' \ind  b_1'b$\;
\\	(since $b \ind  b_1'b_2'$, so $b \ind  c''b_1'$),
\\	so $c'' \ind  c$.
\\    Similarly $c'' \ind  c'$.
\\    \qed

\section*{The Group Configuration Theorem}

Now we turn to applying this lemma to prove the group configuration theorem.
\\In fact, the proof we will give naturally proves a more general, less
symmetric, statement than that above.

\paragraph{Group Configuration Theorem:}
\mbox{}\\    \begin{verbatim}
              a                                    
             / \                                               x
            b   x                                             / \
           / `z' \                                         a /   \ c
          / .' `. \                                         /     \
         /.'     `.\                                       y-------z
        c'         `y                                          b
    \end{verbatim}
    Suppose $(a,b,c,x,y,z)$ satisfy:
\\    (*) \vline\;\begin{minipage}[c]{0.9\textwidth}
\begin{itemize}\item any non-collinear triple is independent,
\item $\acleq(ab)=\acleq(bc)=\acleq(ac)$,
\item $x$ is interalgebraic with $y$ over $a$, and $a$ is interalgebraic
	with $\Cb(\stp(xy/a))$; similarly for $bzy$ and $czx$.
\end{itemize}
    \end{minipage}

    Then, possibly after base change,
\\    there is a $\Wedge$-definable homogeneous space $(G,S)$,
\\    and an independent triple $(a',b',x')$ with $a',b'$ generics of $G$ and
    $x'$ generic in $S$,
\\    such that with $b' := c'*a'$, $x' := a'*y'$, $z' := b'*y'$ (so $z' =
    c'*a'*y' = c'*x'$),
\\    each unprimed element is interalgebraic with the corresponding primed
    element.

\paragraph{Example:}
\mbox{}\\    In ACF, we can restate as follows:
\\    $(b,z,y)$ fits into a group configuration
    (i.e. extends to $(a,b,c,x,y,z)$ satisfying (*))\;
\\    iff it is a generic point of a "pseudo-action",
\\    i.e. iff there is an algebraic group $G$ acting birationally on a variety $S$,
\\    and there are generically finite-to-finite algebraic correspondences
    $f : G' \leftrightarrow  G$, $g_1 : S'_1 \leftrightarrow  S$ and $g_2 : S'_2 \leftrightarrow  S$,
\\    such that $(b,z,y)$ is a generic point of the image under $(f,g_1,g_2)$ of
    the graph $\Gamma_* \subseteq  G \times S \times S$ of the action.
\\    (c.f. 6.2 in [HrushovskiZilber-ZariskiGeometries].)

\paragraph{Example:}
\mbox{}\\    if $\RM(a)=\RM(b)=\RM(c)=2$, and $\RM(x)=\RM(y)=\RM(z)=1$,
\\    and $\RM(abc)=\RM(ab)=\RM(bc)=\RM(ac)=4$, $\RM(axy)=\RM(bzy)=\RM(czx)=3$,
\\    and there are no further dependencies,
\\    then the conditions of the group configuration theorem are satisfied,
\\    and we obtain a rank two group acting on a rank one set,
\\    and with some further work one obtains a definable field,
\\    such that the action is essentially $(a,b)x \mapsto  ax+b$.

    This is sometimes called the "field configuration", and appears in many
    proofs,
\\    e.g. Hrushovski's proof that unimodularity implies local modularity,
\\    and hence that $\omega$-categorical stable theories are 1-based;
\\    the proof of the Zilber dichotomy for Zariski structures;
\\    and also e.g. Hasson-Kowalski's work on trichotomy for strongly minimal
    reducts of RCF.


\paragraph{Very rough sketch of proof:}
\begin{enumerate}[(I)]\item "reduce $\acleq$ to $\dcleq$" to show we may assume $(b,z,y)$
      to define a canonical family of germs of canonical bijections as in
      Lemma A;
\item prove the independence assumption of Lemma A;
\item connect resulting group action to original group configuration.
\end{enumerate}

\paragraph{Proof of Group Configuration Theorem:}
\mbox{}\\    In the proof, we may at any time
\begin{itemize}\item add independent parameters to the language, or
\item replace any point of the configuration with an interalgebraic
	  point of $\Meq$.
\end{itemize}
    Performing these operations preserves (*), and the conclusion allows them.

    By adding further algebraic parameters whenever necessary, we will assume
    throughout that $\acleq(\emptyset )=\dcleq(\emptyset )$, so types over $\emptyset $ are
    stationary.

\begin{enumerate}[(I)]\item First, we see that it follows from (*) that
	if we let $\widetilde{z} \in \Meq$ be the set $\widetilde{z} = \{z_1,...,z_d\}$ of conjugates
	$z_i$ of $z$ over $ybxc$, then $\widetilde{z}$ is interalgebraic with $z$.
\\	(Here, $\{z_1,...,z_d\}$ is the image of $(z_1,...,z_d)$ under
	quotienting by the action by permutations of the symmetric group $S_d$.)

	For this, we require that the conjugates are interalgebraic, $\acleq(z_i)=\acleq(z_j)$.
\\	Indeed, then $\acleq(\widetilde{z}) \subseteq  \acleq(z_1,...,z_n) = \acleq(z)$;
\\	and $z \in \acleq(\widetilde{z})$, since it satisfies the algebraic formula $z\in
	\widetilde{z}$.

	But indeed:
\\	    $z \in \acleq(xc) \cap \acleq(yb) =: B$ by (*), but meanwhile $xc \ind _z yb$
	    so $B \ind _z B$ so $B \subseteq  \acleq(z)$.
%	    Let $B := \aCb(\stp(xc/yb))$. Then $z \|/_B yb$ since $z \in \acleq(xc)$,
%	    so $z \|/_B z$ since $z \in \acleq(yb)$,
%	    so $\acleq(z) (= B$.
%	    But $B (= \acleq(z)$ since $xc \|/_z yb$.
\\	    So $\acleq(z) = B$,
\\	    and by the same argument $\acleq(z_i) = B$ for each $z_i$.

	Now let $a' \models  \tp(a)|_{abcxyz}$.
\\	Say $a'x'c' \equiv _{ybz} axc$. So $a'bc'x'yz$ is also a group configuration.
\\	So as above, the set $\widetilde{z}$ of conjugates of $z$ over $ybx'c'$ is
	interalgebraic with $z$.
\\	Note that $\widetilde{z} \in \dcleq(ybx'c')$.

	So add $a'$ as a parameter,
\\	and replace $y$ with $yx'$, $b$ with $bc'$, and $z$ with $\widetilde{z}$.
\\	Then since $x'\in\acleq(ya')$ and $c'\in\acleq(ba')$,
\\	the resulting group configuration is interalgebraic with the original,
\\	and now satisfies
	    \[ z \in \dcleq(by) .\]

	Repeating this procedure by adding an independent copy of $b$ and
	enlarging $a$ and $y$ and replacing $x$ with an $\widetilde{x}$,
\\	we can also ensure that
	    \[ x \in \dcleq(ay) .\]

	Finally, we repeat once more: add an independent copy $c'$ of $c$,
\\	let $a'x'c' \equiv _{ybz} axc$,
\\	let $\widetilde{y}$ be the set of conjugates of $y$ over $ba'zx'$.

	Now since $x' \in \dcleq(a'y)$ and $z \in \dcleq(by)$,
\\	we have $zx' \in \dcleq(ba'y)$ and so $zx' \in \dcleq(ba'\widetilde{y})$.
\\	So after replacing $b$ with $ba'$, $z$ with $zx'$, and $y$ with $\widetilde{y}$,
\\	so in the previous two cases $y \in \dcleq(bz)$,
\\	and now also $z\in\dcleq(by)$.

	Finally, replace $b$ with $\Cb(yz/b)$, with which it is interalgebraic
	by (*).

\item
	$(b,y,z)$ now corresponds to a canonical family $\widetilde{f}_w$ of germs of
	bijections $\tp(y) \rightarrow  \tp(z)$ over $r := \tp(b)$.

	To apply lemma A to obtain a group,
\\	we must show that if $b' \models  r|_b$ and
	    \[ \widetilde{f}_{b'}^{-1} \circ \widetilde{f}_{b} = \widetilde{g}_d \]
	with $\widetilde{g}_u$ canonical,
\\	then $d \ind  b$ and $d \ind  b'$.

	We may assume $b' \ind  abcxyz$.
\\	Say $b'y'a' \equiv _{xcz} bya$. 

	So by canonicity, $\dcleq(d) = \Cb(\stp(yy'/bb'))$.

	Now $y \ind  abc$, and $b' \ind  yabc$, so $y \ind  abcb'$, and since $a' \in
	\acleq(cb')$, we have $y \ind  aa'bb'$.
\\	Since also $y' \in \acleq(yaa')$, we have
	    \[ yy' \ind _{aa'} bb' .\]
	Similarly, 
	    \[ yy' \ind _{bb'} aa' .\]

	So $\aCb(yy'/bb') = \aCb(yy'/aa'bb') = \aCb(yy'/aa')$,
\\	so $d \in \acleq(aa')$.

	Claim: $b \ind  aa'$.
\\	Proof: $abc \ind  b'$, so $ab \ind _c b'$, so $ab \ind _c a'$
	    since $a'\in\acleq(cb')$.
\\	    But $a' \ind  c$, so $ab \ind  a'$, so $b \ind _a a'$.
\\	    Now $a \ind  b$, so $b \ind  aa'$.

	So $b \ind  d$, and similarly $b' \ind  d$, as required.

\item
	By (II) and Lemma A,
\\	we obtain a connected $\Wedge$-definable group $G$, with a generic
	action of its generic type $s$ on $p := \tp(y)$,
\\	i.e. $g*a$ is defined for $g\models s$ and $a\models p$ with $g \ind  s$.
\\	To get a $\Wedge$-definable homogeneous space,
\\	define $S$ to be $(G \times p)/E$ where
	    $(g,a) E (g',a')$ iff $d_s h. (h*g)*a = (h*g')*a'$,
\\	with the action of $G$:
\\	    $h*(g,a)/E := (h*g,a)/E$.

	Finally, we must show that the original group configuration is
	interalgebraic with that of $(G,S)$. This will involve adding further
	parameters.

	First, let $b' \models  \tp(b)|_{abcxyz}$.
\\	Say $y'b' \equiv _{xzac} yb$.
\\	Say $g \models  s$ codes $\widetilde{f}_{b'}^{-1} \circ \widetilde{f}_b$, so $y' = g*y$.
\\	Then $g$ is interdefinable with $b$ over $b'$.
\\	So add $b'$ to the language,
\\	and replace $b$ with $g \models  s$ and $z$ with $g*y \models  p$.

	Now let $c' \models  \tp(c)|_{abcxyz}$,
\\	and say $b'z'c' \equiv _{axy} bzc$.
\\	Add $c'$ to the language,
\\	and replace $a$ by $b' \models  s$ and $x$ by $z' = b'*y \models  p$.

	Let $h := b*a^{-1}$.
\\	$x = a*y$ and $z = b*y$, so $z = h*x$.
\\	So $\aCb(xz/ab) = \acleq(h)$; but also $x$ and $z$ are interalgebraic over $c$,
\\	so $\aCb(xz/ab) = \aCb(xz/c) = \acleq(c)$.
\\	So replace $c$ with $h$, and we are done.
\end{enumerate} \qed
\end{document}
