!newop eq
!newop tp
!newop stp
!newop Cb
!newop aCb
!newop RM
!newop MD
!newbb M
\newcommand{\Meq}{\M^{\eq}}
\providecommand{\Wedge}{\bigwedge}
\providecommand{\PP}{\mathbb{P}}

\title{The Group Configuration Theorem}
\maketitle

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

## Introduction and Preliminaries

Work in monster model $\M$ of a stable theory $T$.
Notation:
    $a,b,c,d,e,w,x,y,z,\alpha,\beta,\gamma$ 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 $A\cup\{b\}$.

Group configuration, first statement:
    Let $G$ be a connected $\Wedge$-definable group /${}$.
    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}
      ***) any non-collinear triple is independent\;
	    (i.e. each element is independent from the other two);
      ** if $(d,e,f)$ is collinear then
	    $\acleq(de) = \acleq(ef) = \acleq(df) .$
      *e)*
    \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 $G /{}$,
    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.

Remark:
    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.
    
Stability theory preliminaries:
    We have an independence notion, non-forking,
    satisfying (even after adding parameters) for all $A$,$B$,$C$:
	***) Transitivity and Monotonicity:
	    \[ A \|/ BC <==> (A \|/ B \;\textrm{and}\; A \|/_B C) \]
	** Symmetry:
	    \[ A \|/ C <==> C \|/ A \]
	** Reflexivity:
	    \[ A \|/ A <==> A (= \acleq(\emptyset) \]
	** "algebraic $==>$ nonforking":
	    \[ A \|/ C <==> A \|/ \acleq(C) \]
	*e)*
    
    $p \in S(A)$ is _stationary_ iff for any $B$,
	\[ x |= p \;\textrm{and}\; x \|/_A B \]
    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/${}$ $\phi(x,y)$, there is a formula/$A$
	    \[ \psi(y) ==: d_p x. \phi(x,y) \]
		(read "$d_p x.$" as "for generic $x$,"),
	such that $|= \psi(b)$ iff for a $|= p$ with $a \|/_A b$,
	    \[ |= \phi(a,b), \]
	i.e. iff $\phi(x,b) \in p|_b$.

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

    The _canonical base_ of a stationary type $p \in S(A)$, $A=\dcleq(A)$, is the
    least $\dcleq$ closed set $\Cb(p) (= A$
    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 $a \|/_C B <==> \aCb(a/B) (= \acleq(C)$.

    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 $<==>$ $V$ 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 _connected_ iff there is a stationary type $s$ extending $S$
    such that if $g \in G$ and $b |= s|_g$, then $g*b |= s|_g$\;
	(i.e. $\"Stab(s) = G$).
    $s$ is then called the _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.

## Germs and Hrushovski-Weil

Definition:
    Let $p$ and $q$ be stationary types $/{}$.

    A \defn{generic map $p --> q$}
    is the _germ_ $~f_b$ of a definable partial function $f_b$\;
    (i.e. $f_b(x)=y$ is given by a formula/${}$ $\phi_f(x,y,b)$),
    such that if $a |= p|_b$ then $f_b(a) |= q|_b$,
    where equality of germs is defined by $~f_b = ~g_c$ iff
	for $a |= p|_{bc}$, $f_b(a) = g_c(a)$;
	i.e. $|= d_p x. f_b(x) = g_c(x)$ .

Example:
    In ACF, the generic maps $p --> q$ are precisely the dominant rational maps
    $\"locus(p) --> \"locus(q)$.

Lemma:
    **(i) "Equality of germs" is indeed an equivalence relation.
    **(ii) For any $B$, if $a |= p|_{Bb}$ then $f_b(a) |= q|_{Bb}$.
    **(iii) Composition of germs is well-defined,
	yielding a category structure with objects the stationary types.
    **(iv) A germ $~f_b$ is invertible (i.e. an isomorphism) iff $f_b$ is injective
	on $p|_b$.
    *ee*
Proof:
    **(i) 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 $\phi(x) :== f_b(x) = h_d(x)$ is defined over $bd$,
	so already $\phi(x) \in p|_{bd}(x)$.
	So $f_b = h_d$ on $p|_{bd}$.
    **(ii) WLOG, $p,q\in S({})$.
	$a \|/ Bb ==> a \|/_b B ==> f_b(a) \|/_b B$;
	but $f_b(a) \|/ b$,
	so $f_b(a) \|/ Bb$.
    **(iii) Suppose $~f_b : p --> q$ and $~g_{c} : q --> r$.
	By (ii), if $a |= p|_{bc}$ then $g_c(f_b(a)) |= r|_{bc}$,
	so $\~{(g_c o f_b)}$ is a germ $: p --> r$.
    **(iv) Clear.
    *ee*
    \qed
    
Notation:
    ***) $\"Hom(p,q) :=$ set of germs $p --> q$;
    ** $\"Iso(p,q) :=$ set of invertible germs $p --> q$;
    ** $\"Aut(p) :=$ group of invertible germs $p --> p$.
    *e)*
    
:::%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 $\"Iso^{sim}(p,q)$)...

Definition:
    A \defn{family of generic maps $p --> q$ based on $s$}
    is a family of germs $\{ ~f_b | b |= s \}$ of a definable family of
    partial functions $f_z$.
    The family is _canonical_ if for $b$,$b' |= s$,
	\[ ~f_b = ~f_{b'} <==> b=b'. \]

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

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

Remark:
    If $~f_z$ is a family of generic maps $p --> q$ based on $s$,
    let $b |= s$ and $x |= p|_b$, and let $y = ~f_b(x)$;
    then
    \begin{equation}\tag{+} x \|/ b;\; y \|/ 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 $~f_z$ is a canonical family of generic maps $\stp(x) --> \stp(y)$ based on
    $\stp(b)$.

Lemma:
    In the correspondence of the previous remark,
    **(i) $~f_z$ is invertible iff $x \in \dcleq(by)$
	i.e. iff $x$ and $y$ are interdefinable over b;
    **(ii) $~f_z$ is canonical iff $\Cb(\stp(xy/b)) = \dcleq(b)$
    *ee*
Proof:
    **(i) Clear
    **(ii) Suppose that $~f_z$ is canonical, and obtain $(b,x,y)$ as above.
	$\tp(xy/b)$ is stationary since $\tp(x/b)$ is,
	so $C := \Cb(\stp(xy/b)) (= \dcleq(b)$.
	
	If $C != \dcleq(b)$,
	say $b' != b$ with $b' ==_C b$ and $b' \|/_C bx$;
	then $xy \|/_C b'$, so $xyb == xyb'$,
	so since $f_b(x) = y$, also $f_{b'}(x) = y$,
	but $x |= p|_{bb'}$ so this contradicts canonicity of the family $~f_z$.

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

	Suppose $~f_b$ is not canonical. By definability of types,
	some $~g_c$ is canonical with $c \in \dcleq(b)$ but $b \notin \dcleq(c)$.
	Then $\Cb(\stp(xy/b) (= \dcleq(c)$ since
	    \[ x \|/ b ==> x \|/_c b ==> xy \|/_c b, \]
	    and $\tp(xy/c)$ is stationary since $\tp(x/c)$ is and $y \in \dcleq(xc)$.
	This contradicts $\dcleq(b) = \Cb(\stp(xy/b))$.
    *ee*
    \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.

Remark:
    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 $ac \|/_D \acleq(b)$,
	so $\acleq(b) \|/_D \acleq(b)$,
	so $\acleq(b) = D$.

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

Lemma HW:
    Suppose $~f_z$ is a canonical family of generic bijections $p --> p$ based on $s$.
    Let $G_0 := \{ ~f_b | b |= s \} (= Aut(p)$.
    Suppose that $G_0$ is closed under inverse,
    and suppose that for $b_1$ and $b_2$ independent realisations of $s$,
	\[ ~f_{b_1} o ~f_{b_2} = ~f_{b_3} \]
    with $b_3 |= s|_{b_i}$ for $i=1,2$.

    Then, identifying $~f_b$ with $b$,
    the group $G <= Aut(p)$ generated by $G_0$ is connected $\Wedge$-definable, with $s$
    its generic type.
Remark:
    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 $x |-> a*x$ to obtain a
    group structure extending $*$.
Proof:
    First, we show that any element of $G$ is a composition of two generators.

    It suffices to see that any composition of three generators
	\[ ~f_{b_1} o ~f_{b_2} o ~f_{b_3} \]
    is the composition of two. But indeed, let $b' |= s|_{b_1,b_2,b_3}$.
    Then 
	\[ ~f_{b_1} o ~f_{b_2} o ~f_{b_3}
	= ~f_{b_1} o ~f_{b'} o ~f_{b'}^{-1} o ~f_{b_2} o ~f_{b_3} \]
    Now $b' \|/ b_2$, so
    say $~f_{b'}^{-1} o ~f_{b_2} = ~f_{b''}$
    with $b'' |= s$ independent from $b'$ and from $b_2$;
    then $b'' \|/ b_3$ since:
	$b' \|/_{b_2} b_3$,
	so $b'' \|/_{b_2} b_3$ (since $b'' \in \dcleq(b'',b_2)$),
	so since $b'' \|/ b_2$,
	$b'' \|/ b_2b_3$, and in particular $b'' \|/ b_3$.
    Also $b' \|/ b_1$.
    So $~f_{b_1} o ~f_{b'}$ and $~f_{b'}^{-1} o ~f_{b_2} o ~f_{b_3}$ 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 $g |= s$ and $b |= s|_g$, then $g*b |= s|_g$,
    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.

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

    So $b_1'$ and $b_2'$ are independent realisations of $r$,
    so say $~f_{b_1'}^{-1} o ~f_{b_2'} = ~g_{c''}$.

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

## 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.

Group Configuration Theorem:
    \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}
      ***) any non-collinear triple is independent,
      ** $\acleq(ab)=\acleq(bc)=\acleq(ac)$,
      ** $x$ is interalgebraic with $y$ over $a$, and $a$ is interalgebraic
	with $\Cb(\stp(xy/a))$; similarly for $bzy$ and $czx$.
      *e)*
    \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.

Example:
    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
    $f : G' <-> G$, $g_1 : S'_1 <-> S$ and $g_2 : S'_2 <-> S$,
    such that $(b,z,y)$ is a generic point of the image under $(f,g_1,g_2)$ of
    the graph $\Gamma_* (= G \times S \times S$ of the action.
    (c.f. 6.2 in [HrushovskiZilber-ZariskiGeometries].)

Example:
    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 $(a,b)x |-> ax+b$.

    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.
    

Very rough sketch of proof:
    **(I) "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;
    **(II) prove the independence assumption of Lemma A;
    **(III) connect resulting group action to original group configuration.
    *ee*

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

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

    **(I) First, we see that it follows from (*) that
	if we let $~z \in \Meq$ be the set $~z = \{z_1,...,z_d\}$ of conjugates
	$z_i$ of $z$ over $ybxc$, then $~z$ 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 $\acleq(~z) (= \acleq(z_1,...,z_n) = \acleq(z)$;
	and $z \in \acleq(~z)$, since it satisfies the algebraic formula $z\in
	~z$.

	But indeed:
	    $z \in \acleq(xc) \cap \acleq(yb) =: B$ by (*), but meanwhile $xc \|/_z yb$
	    so $B \|/_z B$ so $B (= \acleq(z)$.
:::%	    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 $a' |= \tp(a)|_{abcxyz}$.
	Say $a'x'c' ==_{ybz} axc$. So $a'bc'x'yz$ is also a group configuration.
	So as above, the set $~z$ of conjugates of $z$ over $ybx'c'$ is
	interalgebraic with $z$.
	Note that $~z \in \dcleq(ybx'c')$.

	So add $a'$ as a parameter,
	and replace $y$ with $yx'$, $b$ with $bc'$, and $z$ with $~z$.
	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 $~x$,
	we can also ensure that
	    \[ x \in \dcleq(ay) .\]

	Finally, we repeat once more: add an independent copy $c'$ of $c$,
	let $a'x'c' ==_{ybz} axc$,
	let $~y$ 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 $zx' \in \dcleq(ba'~y)$.
	So after replacing $b$ with $ba'$, $z$ with $zx'$, and $y$ with $~y$,
	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 (*).

    **(II)
	$(b,y,z)$ now corresponds to a canonical family $~f_w$ of germs of
	bijections $\tp(y) --> \tp(z)$ over $r := \tp(b)$.

	To apply lemma A to obtain a group,
	we must show that if $b' |= r|_b$ and
	    \[ ~f_{b'}^{-1} o ~f_{b} = ~g_d \]
	with $~g_u$ canonical,
	then $d \|/ b$ and $d \|/ b'$.

	We may assume $b' \|/ abcxyz$.
	Say $b'y'a' ==_{xcz} bya$. 

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

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

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

	Claim: $b \|/ aa'$.
	Proof: $abc \|/ b'$, so $ab \|/_c b'$, so $ab \|/_c a'$
	    since $a'\in\acleq(cb')$.
	    But $a' \|/ c$, so $ab \|/ a'$, so $b \|/_a a'$.
	    Now $a \|/ b$, so $b \|/ aa'$.
	
	So $b \|/ d$, and similarly $b' \|/ d$, as required.

    **(III)
	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 $g|=s$ and $a|=p$ with $g \|/ s$.
	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 $b' |= \tp(b)|_{abcxyz}$.
	Say $y'b' ==_{xzac} yb$.
	Say $g |= s$ codes $~f_{b'}^{-1} o ~f_b$, so $y' = g*y$.
	Then $g$ is interdefinable with $b$ over $b'$.
	So add $b'$ to the language,
	and replace $b$ with $g |= s$ and $z$ with $g*y |= p$.

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

	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.
    *ee* \qed
