!newop stp
!newop tp
!newop dcl
!newop acl
!newop cl
!newop dom
!newop Stab
!newop Aut
!newop Cb
!newop Th
!newbb N
!newbb Z
!newop RM

\title{Geometric Stability Theory}
\author{Martin Bays\footnote{
These notes are licensed under the Creative Commons Attribution-Sharealike licence
CC-BY-SA 4.0 \url{https://creativecommons.org/licenses/by-sa/4.0/}
by Martin Bays \href{mailto:mbays@sdf.org}{\nolinkurl{<mbays@sdf.org>}}
2016-2020.}\\\
}
\date{\today\\\\input{version.tex}}
\maketitle

## Introduction
These are lecture notes from a course given in the Münster Model Theory Month
``spring school'' of 2016. 

The course covers some of the foundational results of geometric stability
theory. We focus on the geometry of minimal sets. The main aim is an account
of Hrushovski's result that unimodular (in particular, locally finite or
pseudofinite) minimal sets are locally modular; along the way, we discuss the
Zilber Trichotomy and the Group and Field Configurations.

We assume the basics of stability theory (forking calculus, U-rank, canonical
bases, stable groups and homogeneous spaces), as can be found e.g. in
\cite{Palacin-MMMStability}.

Thanks for improving the quality of these notes go to the participants of the
MMM course, to the anonymous reviewer, and to the participants of a reading
course in Jerusalem run by Itay Kaplan which went through these notes
carefully - and in particular to Javier de la Nuez Gonzáles, Daniel Palac\'in,
and especially Yatir Halevi, participants in that course who caught a number
of errors and made some useful suggestions.

This version includes some corrections to the published text. Thanks to Tomás 
Ibarlucía for pointing out some errors, and to Itay Kaplan for finding many 
more. Probably most egregious, the statement of the type-definable case of 
Fact~\ref{minActions} was incorrect in the published version.

Your name too could appear here! If you find any errors or omissions, please 
mail me at {\tt mbays@sdf.org}.

# References
The original results of Zilber are collected in \cite{ZUCT}. The results of
Hrushovski we discuss are mostly in \cite{HrThesis}, \cite{HrLocMod}, and
\cite{HrUnimod}. Our presentation is based in large part on Pillay's book
\cite{PlGStab}.

## Preliminaries

We work in a monster model $\MM$ of a stable theory $T$, so $\MM$ is
$\kappa$-saturated and strongly $\kappa$-homogeneous for some suitably large
$\kappa$. ``Small'' means of cardinality less than $\kappa$.

Notation*:--
    ***) $a,b,c,d,e,\alpha,\beta,\gamma$ etc denote elements of $\MMeq$,
    $\atup,\btup,\ctup$ etc denote possibly long tuples from $\MMeq$,
    and $A$,$B$,$C$ denote small subsets of $\MMeq$.

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

    ** A $\Wedge$-definable set (over $A$) is a subset of $\MMeq$ defined by a
    partial type consisting of formulae whose parameters come from a small set
    (namely $A$).

    ** If $p$ is stationary and $A := \Cb(p)$, then $p|_A$ has a unique
    global non-forking extension $\pp \in S(\MM)$. For $A (= B (= \MMeq$,
    we write $p|_B$ for $\pp|_B$. So $a |= p|_B$ iff $a |= p$ and
    $a \|/_A B$.

    ** For $\alpha$ an ordinal, $p^{(\alpha)}$ is the type of a Morley sequence
    in $p$ of length $\alpha$ (so $\atup b |= p^{(\alpha^{+})}$ iff $\atup |=
    p^{(\alpha)}$ and $b |= p|_{\atup}$).

    %Now $\pp$ is definable over $A$,
    %i.e. $T$ eliminates the Hrushovski quantifier ``$d_p x.$'' (which we can
    %    read as ``for generic $x |= p$,''),
    %i.e. for any formula/${}$ $\phi(x,y)$, there is a formula/$A$
    %        \[ \psi(y) ==: d_p x. \phi(x,y) \]
    %    such that $|= \psi(b)$ iff for $a |= p$ with $a \|/_A b$,
    %        \[ |= \phi(a,b), \]
    %    i.e. iff $\phi(x,b) \in p|_b$.

    ** Define $\Cb(a/B) := \Cb(\stp(a/B))$ and
        $\aCb(a/B) := \acleq(\Cb(a/B))$.
	So $a \|/_C B <==> \aCb(a/BC) (= \acleq(C)$.

    ** We often disregard the distinction between an element of $\MMeq$ and
    its $\dcleq$-closure. In particular, $c=\Cb(p)$ means $\dcleq(c)=\Cb(p)$,
    and we then allow ourselves to write $\Cb(p) (- \MMeq$.

    ** We use exponential notation $b^\sg$ for the action of an automorphism
    (so automorphisms always act on the right).
    *e)*
    .

Lemma:
    U-rank is additive for finite ranks:
    $U(ab/C) = U(a/bC) + U(b/C)$ whenever all terms are finite.
    .
Proof:
    This is immediate from the Lascar inequalities 
    \cite[Theorem~5.7]{Palacin-MMMStability} (and in fact one only needs
    that one of the sides of the equality is finite).
    .

## Minimal sets and the trichotomy

# Minimal sets and pregeometries

Definition*:
    A _minimal set_ is a $\Wedge$-definable set $D$, defined by a partial type
    which has a unique unrealised global completion $\pp (- S(\MM)$. Call
    $\pp$ the global _generic type_ of $D$.

    A _strongly minimal set_ is a definable set which is minimal.
    .

Let $D$ be a minimal set, and $\pp$ its global generic type. Adding
parameters to the language if necessary, we will assume $D$ is defined over
${}$.

Exercise*:--
    ***) If $a\in D$ and $C (= \MMeq$, then $a |= \pp|_C$ iff $a (/- \acleq(C)$.
    ** A $\Wedge$-definable set is minimal iff every relatively definable subset
      is finite or cofinite.
    ** A complete type is minimal iff it is stationary and of U-rank 1.
    *e)*
    .

Remark*:
    Historically the focus was on strongly minimal formulae, due to their role in
    uncountable categoricity (Baldwin-Lachlan), but minimality is the natural
    generality for the theory we develop here. Examples of minimal
    $\Wedge$-definable sets include the maximal perfect subfield $K^{p^{\infty}} =
    \bigcap_n K^{p^n}$ of a separably closed field, and the maximal divisible
    subgroup $\bigcap_n n(^*\Z)$ of $^*\Z |= \Th(\Z;+)$.

    One can also work in the greater generality of (strongly) regular types,
    and much of what we will cover in this course goes through for them.
    See in particular \cite{HrLocMod}.
    .

Definition*:
    A _pregeometry_ is a set $X$ with a map $\cl : \Pset(X) -> \Pset(X)$ such that
    ***) $\cl$ is a closure operator: $A (= \cl(A)$, $A (= B => \cl(A) (=
      \cl(B)$, and $\cl(\cl(A)) = \cl(A)$;
    ** Exchange: $b (- \cl(Ac) \\ \cl(A) => c (- \cl(Ab)$.
    ** Finite character: $\cl(A) = \bigcup_{A_0 (=_{\"fin} A} \cl(A_0)$;
    *e)*

    Often, it is more useful to think of a pregeometry in terms of its lattice
    of closed sets. An equivalent definition of a pregeometry is as a set $X$
    with a set of _closed_ subsets $\CC (= \Pset(X)$ such that
    ***) $\CC$ is closed under intersections: if $\BB (= \CC$, then
        $\bigcap \BB (- \CC$.
    ** Exchange: If $C (- \CC$ and $a (- X \\ C$, there is an immediate closed
      extension $C'$ of $C$ containing $a$,
      i.e. there exists $C' (- \CC$ such that $C' =) Ca$ and there does not
      exist $C'' (- \CC$
      with $C (!= C'' (!= C'$.
    ** Finite character: if $\BB (= \CC$, then
        $\Vee \BB = \bigcup_{\BB_0 (=_{\"fin} \BB} ({ \Vee \BB_0 })$,
        where $\Vee \BB := \bigcap \{ C (- \CC \;|\; \bigcup \BB (= C \}$
        is the smallest closed set containing all $B (- \BB$,
    *e)*

    So setting $\Wedge \BB := \bigcap \BB$ and $A <= B <=> A (= B$, $\CC$
    forms a complete lattice.

    We can pass between the two definitions as follows: given a closure
    operator $\cl$, define $\CC := \"im(\cl)$; conversely, given a set $\CC$
    of closed subsets, define $\cl(A):=\bigcap \{C (- \CC | A (= C\}$.

    For $A,B (= X$, $\dim(A/B)$ is the cardinality of a _basis_ for $A$ over
    $B$, a minimal subset $A'(=A$ for which $\cl(A'B) = \cl(AB)$.
    When it is finite, $\dim(A/B)$ is equivalently the length $r$ of any chain
    of immediate extensions of closed sets $\cl(B) = C_0 <!= C_1 <!= ... <!=
    C_r = \cl(AB)$.

    The \defn{localisation of $X$ at $A$} is $X_A := (X,\cl_A)$ where
    $\cl_A(B)=\cl(AB)$; equivalently, the closed sets of $X_A$ are the closed
    sets of $X$ which contain $A$.

    $X$ is _homogeneous_ if for any closed $C (= X$ and any $a,b (- X \\ C$,
    there is an automorphism over $C$ of the pregeometry (a bijection
    preserving $\cl$, or equivalently $\CC$, and fixing $C$ pointwise) which
    sends $a$ to $b$.

    A _geometry_ is a pregeometry $(X,\cl)$ such that $\cl({})={}$ and
    $\cl(a)=\{a\}$ for any $a\in X$. For a pregeometry $(X,\cl)$, the
    _associated geometry_ is the set of dimension 1 closed sets of $X$ with
    the obvious closure, $\cl'$ defined by
    $A (- \cl'(B) <=> A (= \cl(\bigcup B)$.
    .

Remark*:
    Finite pregeometries are also known as _matroids_.

    %Question: the recent paper http://arxiv.org/abs/1003.3919 provides a
    %notion of matroid which includes pregeometries and is closed under
    %taking matroid duals. Does matroid duality have any significance for
    %minimal sets? (I haven't actually thought about this)
    .

Lemma:
    $(D,\acl_D)$ is a homogeneous pregeometry, where $\acl_D(A) := \acleq(A)
    \cap D$.
    .
Proof:
    Everything is immediate from the definition of $\acl_D$ except homogeneity
    and exchange. Exchange
    corresponds to symmetry of forking: for $b,c (- D$ and $A (= D$,
        *\[a* b (- \acl_D(Ac) \\ \acl_D(A) 
        & <==> b |= \pp|_A \infix{and} b |/= \pp|_{Ac} \\\
            & <==> b \+/_{A} c \\\
            & <==> c \+/_{A} b \\\
            & <==> c |= \pp|_A \infix{and} c |/= \pp|_{Ab} \\\
            & <==> c (- \acl_D(Ab) \\ \acl_D(A) .*a\]*

    For homogeneity, if $C (= D$ is $\acl_D$-closed, and $a_0,b_0 (- D \\ C$,
    then they can be extended to $\acl_D$-bases
    $(a_i)_{i<\lambda},(b_i)_{i<\lambda}$ for $D$ over $C$. Then by
    inductively considering isolating algebraic formulae, $a_i |-> b_i$
    can be extended to an elementary map $D -->> D$ fixing $C$ pointwise,
    which is in particular an automorphism of the pregeometry over $C$. (Note
    that this is not just a matter of strong homogeneity of $\MMeq$, since $C$
    need not be small.)
    .
Remark*:
    For $a (- D^{<\omega}$ and $B (= D$,
        $U(a/B) = \dim(a/B)$.
    Indeed, this is immediate for singletons, and then follows for tuples by
    additivity.
    .

Remark*:
    Adding $A (= D$ to the language corresponds to localising the pregeometry
    at $A$.
    .

# Triviality and modularity

Definition*:
    A pregeometry is _trivial_ (or _degenerate_) if
    $\cl(A) = \bigcup_{a (- A} \cl(a)$; equivalently, $\Vee = \bigcup$ in
    the lattice of closed sets.
    .
Example*:--
    ***) The theory of equality on an infinite set (i.e. with trivial
      language), is strongly minimal with trivial pregeometry; all subsets are
      closed, $\acl_D(B)=B$.
    ** An action of a group $G$ on an infinite set $D$ without fixed points,
      in the language $(D; ((g*))_{g (- G})$, is strongly minimal with trivial
      pregeometry; closed sets are unions of $G$-orbits; $\acl_D(B) = GB$.
    *e)*
    .

Definition*:
    A pregeometry $(X,\cl)$ is _modular_ if for any finite-dimensional closed
    $A,B <= X$,
    \begin{equation} \label{eq:mod}
        \dim(A \/ B) = \dim(A) + \dim(B) - \dim(A /\ B) ,\end{equation}
    i.e. $\dim(A/B) = \dim(A / A /\ B)$.
    .

Remark*:
    The pregeometry of a minimal set $D$ is modular iff $A \|/_{A\cap B} B$
    for any $\acl_D$-closed $A,B (= D$.
    .

Lemma{mod2suff}:
    Suppose $(X,\cl)$ is such that (\ref{eq:mod}) holds when $\dim(A)=2$. Then
    $(X,\cl)$ is modular.
    .
Proof:
    Suppose (\ref{eq:mod}) holds when $\dim(A)=2$.

    We first show that (\ref{eq:mod}) holds when $\dim(A/A /\ B)=2$.
    Say $a_1,a_2$ is a basis for $A$ over $A /\ B$; then setting $A_0 :=
    \cl(a_1,a_2)$, $A_0 /\ B = A_0 /\ (A /\ B) = \cl({})$, and $A_0
     \/ B =) A_0 \/ (A /\ B) = A$ so $A_0 \/ B = A \/ B$. So
    indeed, $\dim(A/B) = \dim(A \/ B / B) = \dim(A_0 \/ B / B) = \dim(A_0 / A_0 /\ B) =
    \dim(A_0) = \dim(A / A /\ B)$.

    Now let $A$ and $B$ be arbitrary, and say $n=\dim(A/B) = \dim(A \/ B) - \dim(B)$.
    Decompose $B <= A \/ B$ as an immediate chain of closed subsets
      \[ B = B_0 <!= B_1 <!= ... <!= B_n = A \/ B .\]
    Let $A_i := A /\ B_i$.
    Clearly $\dim(A/A /\ B) >= \dim(A/B) = n$,
    so it suffices to see that $\dim(A_{i+1} / A_i) <= 1$.
    Else, we have $A_i <= A' <= A_{i+1}$ with $\dim(A'/A_i)=2$ and $A'$ closed;
    but then $A' /\ B_i = A_i$ and $A' \/ B_i (= B_{i+1}$,
    contradicting the modularity established above.
    .

Exercise*:
    A lattice is called _modular_ if for any $A,B,C$ with $A <= C$,
        \[ A \/ (B /\ C) = (A \/ B) /\ C .\]
    Show that $(X,cl)$ is modular iff the lattice of finite-dimensional closed
    subsets is modular.
    % CHECKME: $D$ is 1-based iff the lattice of $\acleq$-closed subsets of
    % $\Deq$ is modular?
    .


Example[Projective Geometries]{projGeom}:
    Let $V$ be a vector space over a division ring $K$ with closed sets the
    vector subspaces. For $V$ infinite dimensional, this is the pregeometry of
    the strongly minimal structure $(V; +, (k*)_{k (- K})$. It is a standard
    result of linear algebra that this is modular.

    The corresponding geometry is obtained by projectivising - deleting 0 and
    quotienting by the action of multiplication by scalars.
    .

Fact{nonTrivMod}:
    Any non-trivial modular homogeneous geometry of dimension $>=4$ is
    isomorphic to a projective geometry over a division ring.
    .

The proof of this fact goes as follows. By classical theorems of geometric algebra
\cite[Chapter V]{SeidenbergProj}, given an incidence relation between a
set of points and a set of lines such that
**(i) any pair of points $a/=b$ lies on a unique common line $(a,b)$,
**(ii) any pair of lines intersect in at most one point,
**(iii) any line has at least three points,
**(iv) if $(a,b)$ and $(c,d)$ intersect then $(a,c)$ and
  $(b,d)$ intersect (this is the Veblan-Young axiom, sometimes referred to
  as Pasch's axiom),
**(v) Not all points are on a common plane, where a /plane/ consists of
  the points on lines $(a_0,b)$ where $a_0$ is a fixed point and $b$ varies
  through the points on a fixed line not containing $a_0$,
*ee*
one can find a Desarguesian projective plane, and from this construct a
division ring, and conclude that the original incidence relation is
isomorphic to a projective space over the division ring. We obtain such an
incidence relation from our geometry by taking the points resp. lines to
be the closed sets of dimension 1 resp. 2. Axioms (i) and (ii) are
immediate, (iv) follows easily from modularity, and (v) from the dimension
assumption. For (iii), non-triviality yields a relation $a_1\in\cl(A,a_2)$
for some finite $A$ with $a_i (/- \cl(A)$; then by modularity, there is $e
\in \cl(a_1a_2)\cap\cl(A)$, so the line $(a_1,a_2)$ has at least three
points. By homogeneity, so has every line.

In the model theoretic context, essentially this construction actually
arises definably. If $D$ is a non-trivial modular minimal type, it is
non-orthogonal to (i.e. in generically defined finite-to-finite relatively
definable correspondence with) a minimal abelian group; its ring of
definable finite-to-finite quasi-endomorphisms is a division ring, and the
geometry of this, and hence of $D$, is then projective geometry over that
division ring. See \cite[Remark~5.1.9]{PlGStab} for details.

# 1-basedness and local modularity

Remark*:
    Localisations of modular pregeometries are also modular. However, the
    converse is false, as the following example shows.
    .

Example[Affine Geometry]{affGeom}:
    Let $V$ be an infinite dimensional vector space over a division ring $K$,
    with closed sets the affine spaces, i.e. cosets of subspaces. For
    $\"char(K) /= 2$, this is the pregeometry of the strongly minimal
    structure \[ (V; ( \{ z=\lambda x + (1-\lambda)y \} | \lambda \in K )) .\]
    % TODO: probably there's a similar nice formulation which works in the
    % char=2 case too?

    Affine geometries are not quite modular: parallel lines within a common
    plane are dependent but have trivial intersection.

    Localising at $0$ yields the projective geometry of
    Example~\ref{projGeom}.
    .

In this subsection, we establish a dichotomy between minimal sets which have
modular localisations and those which don't. This dichotomy becomes clearest
when we include imaginaries in our considerations.

Definition*:
    $\Deq := \dcleq(D) (= \MMeq$.
    .

Definition*:
    $D$ is \defn{1-based} if for any $a (- \Deq$ and $B (= \Deq$,
    $\Cb(a/B) (= \acleq(a)$.
    .

Remark*:
    %We will see later that there is no such thing as $n$-basedness for $n>1$,
    %in that it implies 1-basedness.
%
    The correct definition of 1-basedness in more general situations allows
    $B$ to be an arbitrary subset of $\MMeq$. But in our case this is
    equivalent to the above definition, since $\Cb(a/B) = \Cb(a / \atup)$
    where $\atup (= \Deq$ is a Morley sequence in $\tp(a/B)$.
    .

Lemma{1basedDisj}:
    $D$ is 1-based iff for any $\acleq$-closed subsets $A,B (= \Deq$,
      \[ A \|/_{A\cap B} B .\]
    .
Proof:
    Immediate from finite character of forking and the properties of canonical
    bases.
    .

Lemma{1basedParams}:
    1-basedness is invariant under adding parameters to the language.
    .
Proof:
    That a 1-based set remains 1-based on adding parameters follows from the
    definition and the remark after it.

    For the converse, suppose $D$ is 1-based after adding $C$.
    Let $A,B (= \Deq$ be $\acleq$ closed, and let $I := A \cap B$; we must
    show $A \|/_I B$.

    By taking a realisation of $\tp(AB)$ independent from $C$, we may
    assume $AB \|/ C$.

    Claim:
        $I' := \acleq(AC)\cap \acleq(BC) = \acleq(IC)$
        .
    Proof:
        $AC \|/_A AB$ and $BC \|/_B AB$,
        so by considering $\Cb(I'/AB)$,
        we have $I' \|/_I AB$.
        So since $I (= IC (= I'$,
        also $I' \|/_{IC} ABC$.
        But $I' (= \acleq(ABC)$, so $I' = \acleq(IC)$.
        .

    Applying Lemma~\ref{1basedDisj} in the expanded language,
    we deduce $A \|/_{IC} B$.
    Meanwhile $AB \|/ C$ and $I (= AB$, so $AB \|/_I IC$, and so $A \|/_I IC$.
    So by transitivity, $A \|/_I B$.
    .

Definition*:
    A $\Wedge$-definable set $X$ has _elimination of imaginaries_ ( _EI_ ),
    resp. _geometric EI_ (gEI),
    resp. _weak EI_ (wEI),
    if for any $c (- X^{\eq}$, there exists $b (- X^{<\omega}$ such that
    $\dcleq(b) = \dcleq(c)$,
    resp. $\acleq(b) = \acleq(c)$,
    resp. $\dcleq(c) (= \dcleq(b) (= \acleq(c)$.
    .

Remark*:
    EI $=>$ wEI $=>$ gEI.
    .

%Alternative phrasing:
    %A $\Wedge$-definable set $X$ has EI if for any $\dcleq$-closed set $C (= X^{\eq}$,
    %$C = \dcleq(C \cap D)$.

    %geometric EI (gEI): $C = \acleq(C \cap D)$ for $C=\acleq(C)$.
    %weak EI (wEI): $C (= \dcleq(\acleq(C) \cap D)$ for $C=\dcleq(C)$.

Lemma{mod1based}:
    Suppose $D$ has gEI. Then $D$ is modular iff $D$ is 1-based.
    .
Proof:
    By gEI and finite character of $\acleq$, for $A=\acleq(A) (= \Deq$
        we have $A = \acleq(A \cap D)$.
    So for $A,B (= \Deq$ $\acleq$-closed,
	\[ A \|/_{A\cap B} B \,<=>\, (A\cap D) \|/_{A\cap B \cap D} (B\cap D) .\]
    We conclude on noting that for $A (= D$, $A=\acl_D(A) => A = \acleq(A) \cap D$.
    .

Lemma{minwEI}:
    Let $\atup |= \pp|_{}^{(\omega)}$.
    Then after expanding the language by parameters for $\atup$, $D$ has wEI.
    .

Proof:
    Working in the original language,
    we will prove wEI in the expanded language by showing that if $c (-
    \dcleq(D,\atup) = \Deq$, then there exists $b (- D^{<\omega}$ such that
    $c (- \dcleq(b\atup)$ and $b (- \acleq(c\atup)$.

    So say $c=f(b')$ with $b' (- D^n$ and $f$ a partial function defined over
    ${}$. It suffices to show that $f^{-1}(c) \cap \acleq(c\atup) /= {}$.
    In fact we will prove by induction on $n$ the stronger statement that if
    ${} /= X (= D^n$ is $\Wedge$-definable over $c (- \Deq$, then $X \cap
    \acleq(c\atup) /= {}$.

    First suppose $n=1$. Then either $X$ is finite, in which case any point is
    algebraic over $c$, or else $\pp|_c(x) |= x (- X$, in which case we are
    done as $a_i |= \pp|_c$ for some $i$, since $c (- \Deq$.

    (Indeed, say $U(c)=k$, and suppose $U(a_i/c)=0$ for $i=0,...,k$.
    Then $U(c/a_{<i}a_i) = U(ca_i/a_{<i}) - U(a_i/a_{<i}) =
    U(a_i/ca_{<i})+U(c/a_{<i})-U(a_i/a_{<i}) = U(c/a_{<i})-1$, so we have a
    contradiction.)

    If $n>1$, consider a co-ordinate projection $\pi : X --> D$. By
    compactness and saturation, $\pi(X)$ is also $\Wedge$-definable over $c$. By
    the $n=1$ case, say $b (- \pi(X) \cap \acleq(c)$; then the fibre
    $\pi^{-1}(b)$, considered as a $\Wedge$-definable subset of $D^{n-1}$, is over
    $bc$, so by induction has a point algebraic over $bc$ and hence over $c$.
    .

It follows from Lemmas~\ref{1basedParams}, \ref{mod1based}, and \ref{minwEI} that
$D$ is 1-based iff some localisation is modular. But we now prove a sharper
result.

Definition*:
    $D$ is _linear_ if whenever $a,b (- D$ and $C (= \Deq$ and
    $U(ab/C)=1$, then $U(\Cb(ab/C)) <= 1$.

    (We can think of this as $D$ having ``no 2-dimensional family of plane
    curves''.)
    .

Example{affPlaneNonlinear}:
    An algebraically closed field is not linear, since e.g. there is a
    two-dimensional family of straight lines in the plane. We can see this in
    terms of the definition of linearity as follows.
        Let $a,b,c,d$ be generic such that $b = ca + d$.
        So $(a,b)$ is a generic point of the algebraic variety
        $V_{cd} = \{ (x,y) | y = cx + d \}$, whose field of definition is
        generated by $c,d$. Then $U(ab/cd)=1$, but $\Cb(ab/cd) = \dcleq(cd)$
        and $U(cd)=2$.
    .
    
Definition*:
    A pregeometry $(X,\cl)$ is _locally modular_ if the localisation at any
    $e (- X \\ \cl({})$ is modular.
    .


Theorem{1basedLinear}:
    The following are equivalent:
        **(i) $D$ is 1-based;
        **(ii) $D$ is linear;
        **(iii) $D$ is locally modular.
        *ee*
    .
Proof:--
    ***)[$(i) => (ii)$]:
        Let $a,b,C$ be as in the definition of linearity.
        By 1-basedness, $d := \Cb(ab/C) (= \acleq(ab)$.
        So $U(d) = U(abd) - U(ab/d) <= 2 - 1 = 1$.

    **[$(ii) => (iii)$]:
        Let $e (- D \\ \acl_D({})$. 
        We prove modularity of the localisation to $e$ via
	Lemma~\ref{mod2suff}. So suppose $(a,b) |= \pp|_e^{(2)}$
	and $e (- C = \acl_D(C) (= D$, and let $I := \acl_D(abe) \cap C$; we
	must show $ab \|/_I C$. This is clear if $U(ab/C)=0$ or $U(ab/C)=2$,
	so suppose $U(ab/C)=1$. So we must show
	    \[ \acl_D(e) <!= I .\]
        Let $d := \Cb(ab/C)$.
        By linearity, $U(d) <= 1$.
        Since $ab \+/ C$, we must have $U(d)=1$.
        Note $U(d/ab) = U(dab)-U(ab) = U(ab/d) + U(d) - U(ab) = 1 + 1 - 2 = 0$,
        so $d (- \acleq(ab)$. So $\acleq(ed) \cap D (= I$. So it suffices to
	show 
	    Equation*:
	     \tag{*} \acl_D(e) <!= \acleq(ed) \cap D .
	     .

        If $a \+/ d$ then $a$ witnesses (*).
        Else, $\tp(a/d) = \tp(e/d) = \pp|_d$, since $e \|/ d$ as $d (-
	\acleq(ab)$.
	So say $ab ==_d ef$. Then $f$ witnesses (*).

    **[$(iii) => (i)$]:
	Add parameters for an infinite Morley sequence in $\pp|_{}$.
        Then $D$ has wEI by Lemma~\ref{minwEI}, and $D$ is modular, so $D$
        is 1-based by Lemma~\ref{mod1based}. But then $D$ was 1-based in
        the original language, by Lemma~\ref{1basedParams}.

        % Alternative approach: Not hard to see that $D$ modular implies it
        % already has gEI (even wEI?), hence is 1-based. Then remove
        % parameters.
        % But, we want the wEI lemma for unimodularity purposes, so current
        % approach makes more sense.
    *e)*
    .

# Trichotomy
Combining the two dividing lines of Fact~\ref{nonTrivMod} and
Theorem~\ref{1basedLinear}, we obtain the following form of Zilber's
Weak Trichotomy Theorem:

Theorem:
    If $D$ is a minimal set in a stable theory,
    then precisely one of the following holds:
    ***) The pregeometry of $D$ is trivial.
    ** The pregeometry of $D$ is locally modular but non-trivial;
      equivalently, after localising at a non-algebraic point, the geometry of
      $D$ is (``definably'') projective geometry over a division ring.
    ** The pregeometry of $D$ is not locally modular; equivalently, $D$
      defines a 2-dimensional family of plane curves (and in fact interprets a
      pseudoplane - see below).
    *e)*
    .

HistoricalRemark*:
    In our Example~\ref{affPlaneNonlinear} of a 2-dimensional family of plane
    curves, the same classical results of geometry which we referred to in
    Fact~\ref{nonTrivMod} apply to allow one to definably reconstruct the
    field from the incidence relation. Meanwhile, with a little extra work, one
    can extract from nonlinearity a particularly canonical kind of
    2-dimensional family of ``plane curves`` known as a \defn{pseudoplane}, which
    dimension-theoretically looks just like Example~\ref{affPlaneNonlinear}
    (or rather its projective counterpart) - namely, each curve has infinitely 
    many points, each point lies on infinitely many curves, a pair of curves
    has finite intersection, and a pair of points lie on finitely many common
    curves\footnote{
    Generically, this means that (possibly after adding parameters) we have 
    $b_1,b_2$ such that for $\{i,j\} = \{1,2\}$, we have $U(b_i) = 2$, 
    $U(b_i/b_j) = 1$, and $\Cb(b_i/b_j) = b_j$. In particular, $r := 
    \tp(b_1b_2)$ satisfies that $r(a,D)$ and $r(D,b)$ are infinite when 
    non-empty, and $r(a,D) \cap r(a',D)$ and $r(D,b) \cap r(D,b')$ are finite 
    for $a /= a'$ and $b /= b'$. If $D$ is strongly minimal, by compactness 
    and elimination of $\exists^\infty$ one can find a formula $I(x,y) (- r$ 
    with the same properties; by definition, $I$ is then the incidence 
    relation of an intepretable pseudoplane.

    Nonlinearity gives $b_1,b_2$ with all the above properties except that we 
    only have $U(b_2) >= 2$ and don't necessarily have $\Cb(b_2/b_1) = b_1$. 
    We can reduce to $U(b_2) = 2$ by working over a Morley sequence in 
    $\tp(b_1/b_2)$ of length $U(b_2) - 2$ independent from $b_1$ over $b_2$, 
    and by finiteness of canonical bases (Lemma~\ref{finCb} below) we can then 
    replace $b_1$ with $\Cb(b_2/b_1)$ (which is interalgebraic with $b_1$).
    Note that unlike the family of plane curves given by nonlinearity, the 
    points of the pseudoplane we obtain this way are not necessarily in $D^2$.
    }.
    Given also Macintyre's theorem that any $\omega$-stable division
    ring is an algebraically closed field, one might reasonably hope that
    Example~\ref{affPlaneNonlinear} is characteristic of strongly minimal sets
    which are not locally modular, and hence that any such is bi-interpretable
    with an algebraically closed field in this way, and so in particular that
    the geometry of a minimal set which is not locally modular is that of an
    algebraically closed field. This is known as Zilber's Trichotomy
    Conjecture.

    Reality is more complicated. Hrushovski found a counterexample to this
    conjecture in the early 90s, with a combinatorial construction of a
    strongly minimal set which is not locally modular but which doesn't even
    interpret an infinite group. He also produced a variety of "pathological"
    strongly minimal sets, such as two algebraically closed field structures
    of different characteristics on the same set. At the time of writing,
    there is no clear path to a classification of strongly minimal sets, or
    even of their geometries; certainly we know it must be much more involved
    than a trichotomy.

    However, there are situations where the Trichotomy Conjecture does go
    through. Most notably, this is the case for Zariski Geometries
    \cite{HZZGeoms}. We will not discuss Zariski Geometries in this
    course, but instead another earlier incarnation of the same principle -
    unimodular minimal sets, which we will see are locally modular by proving
    that the Trichotomy Conjecture is valid for them and then seeing that the
    field case is impossible.
    .

Fact:
    If $D$ is locally modular but not modular, its geometry is affine geometry
    over a division ring (\cite[Proposition~5.2.4]{PlGStab}).
    .

# Finiteness of canonical bases in $\Deq$

If $D$ is strongly minimal, it is totally transcendental, and so types of
elements of $\Deq$ have finite canonical bases. In this technical subsection,
we prove that this conclusion holds for arbitrary minimal $D$. We will make
repeated use of this lemma in later sections, but a reader who is happy to
restrict attention to strongly minimal $D$ may omit it.

Lemma{finCb}:
    Let $C (= \MMeq$ and $a (- \Deq$, and let $q := \tp(a/C)$.
    **(a) If $q$ is stationary, then $\Cb(q)$ is ($\dcleq$ of) an element of
      $\Deq$.
    **(b) $q$ has finitely many global non-forking extensions.
    *ee*
    .
Proof:--
    We first prove (a) and (b) under the assumption that $a (- D^n$.
    **(a)

      If $a=(a_1,...,a_n)$, after an appropriate co-ordinate permutation
      say $\atup := (a_1,...,a_k) |= \pp|_C^{(k)}$ and $\btup :=
      (a_{k+1},...,a_n) \in \acleq(C,\atup)$.

      Say $\phi_c(\atup,y) (- \tp(\btup/C\atup)$ with $c (- \dcleq(C)$ and
      $|\phi_c(\atup,\MM)|$ minimal,
      so $\phi_c(\atup,y)$ isolates $\tp(\btup/C\atup)$.
      Then clearly $q(x,y)$ is equivalent to
      $\pp|_C^{(k)}(x) \cup \{ \phi_c(x,y) \}$,
      %indeed,
      %if $\atup' |= \pp|_C^{(k)}$ and $|= \phi_c(\atup',\btup')$,
      %then say $\atup'\btup' ==_C \atup\btup''$;
      %then $|= \phi_c(\atup,\btup'')$, so $(\atup,\btup'') |= q$,
      %so $(\atup',\btup') |= q$.
      and the global non-forking extension $\qq$ of $q$ is
      equivalent to $\pp^{(k)}(x) \cup \{\phi_c(x,y)\}$.
      So $\qq$ is invariant over $c$, and so $\Cb(q) (- \dcleq(c)$.
      Since also $\Cb(q)$ is in $\dcleq$ of a Morley sequence in $q$
      (by \cite[Proposition~4.19]{Palacin-MMMStability}),
      we conclude $\Cb(q) (- \Deq$,
      as required\footnote{Thanks to the anonymous reviewer for a suggestion which
      led to an optimisation of this part of the proof.}.

    **(b)
      If $\qq$ is a global non-forking extension, then $\Cb(\qq) (-
      \acleq(C)$, and the global non-forking extensions of $q$ are the
      conjugates of $\qq$ over $C$, which are in bijective correspondence
      with the finitely many conjugates of $\Cb(\qq)$ over $C$.
    *ee*

    Now for arbitrary $a (- \Deq$, say $a=f(b)$ where $b (- D^n$ and $f$ is
    ${}$-definable, (b) holds since it holds for $\tp(b/C)$.

    It remains to deduce (a). So suppose $q$ is stationary.

    Claim[{\cite[Lemma~1.3.19, first part]{PlGStab}}]:
        Let $q$ be a stationary type in a stable theory with $U(q|_{}) < \infty$.
        Then there is $e (- \Cb(q)$ with $\Cb(q) (= \acleq(e)$.
        .
    Proof:
        Suppose there is no $e (- \Cb(q)$ over which $q$ does not fork. We
        build an infinite sequence ${} = e_0, e_1, ... (- \Cb(q)$ such that
        $\dcleq(e_i) (= \dcleq(e_{i+1})$ and $q|_{e_{i+1}}$ forks over $e_i$,
        contradicting U-rankedness of $q|_{}$. Indeed, given $e_i (- \Cb(q)$,
        $q|_{\Cb(q)}$ forks over $e_i$; then since forking is witnessed by a
        formula, already $q|_{e_{i+1}}$ forks over $e_i$ for some $e_{i+1} (-
        \Cb(q)$, and we may assume $e_i (- \dcleq(e_{i+1})$.
        .

    Now $U(q|_{}) = U(a) <= U(b) <= n$, so let $e$ be as in the Claim for
    $q=\tp(a/C)$; then by (b) applied to $\tp(a/e)$, $\Cb(q)$ has only
    finitely many conjugates over $e$, and hence $\Cb(q) = \dcleq(ee')$ for
    some finite part $e'$ of $\acleq(e)$, as required.
    .

## Unimodularity

# Preliminaries

Throughout this section, $D$ is a minimal set over ${}$.

Definition*:
    If $a \in \acleq(B)$,
    then $\mult(a/B) = |{ \{ a' (- \MMeq | a' ==_B a \} }|$.
    %i.e. the size of the orbit of $a$ under the group $\Gal(b)$ of elementary permutations of
    %$\acleq(b)$ fixing $b$,
    %i.e. the index $|\Gal(b):\Gal(ab)|$.
    .

Remark*:
    Multiplicity is multiplicative: $\mult(ab/C) = \mult(a/bC)\mult(b/C)$.
    .

Lemma{multStat}:
    If $\tp(a/c)$ is stationary and $b (- \acleq(c)$, then
        \[ \mult(b/ac) = \mult(b/c) .\]
    .
Proof:
    Suppose $\sg (- \Aut(\MMeq / c)$.
    By stationarity, $\tp(a/c) |= \tp(a / cb)$.
    So $ab ==_c a^{\sg^{-1}}b ==_c ab^\sg$,
    so $b ==_{ac} b^\sg$.
    .


# Unimodularity

Definition*:
    $D$ is _unimodular_ if whenever $a_i |= \pp|_{}^{(n)}$, $i=1,2$, and
    $\acl_D(a_1)=\acl_D(a_2)$, then $\mult(a_1 / a_2) = \mult(a_2 / a_1)$.
    .

Example*:
    An algebraically closed field is _not_ unimodular: consider $a$ and $a^2$
    where $a$ is generic.
    .

In Section~\ref{unimodProof}, we will show
Theorem[Hrushovski]{unimodLocmod}:
    If $D$ is unimodular, then $D$ is 1-based.
    .

$\omega$-categorical theories provide the motivating example of unimodular
minimal sets:

Lemma:
    If $D$ is a minimal set in an $\omega$-categorical theory, then $D$ is
    unimodular.
    .
Proof:
    By Ryll-Nardzewski, $\acl_D$ is locally finite, i.e. $\acl_D(A)$ is
    finite for $A$ finite. Let $a_i |= \pp|_{}^{(n)}$, $i=1,2$, be interalgebraic,
    and let $X := \acl_D(a_i)$.
    
    For $c \in X^k$, let $\mult_X(c) := |{ \{ c'\in X | c' == c \} }|$.

    Then
    \[ \mult(a_1/a_2)\mult_X(a_2) = \mult_X(a_1a_2)
      = \mult(a_2/a_1)\mult_X(a_1) \]
    and $\mult_X(a_1) = \mult_X(a_2)$ since $a_1 == a_2$.
    So $\mult(a_1/a_2) = \mult(a_2/a_1)$.
    .

So by Theorem~\ref{unimodLocmod}, minimal sets in $\omega$-categorical
theories are locally modular. This result is due originally to Zilber, and
forms a key part of his study of totally categorical theories (uncountably
categorical theories which are also $\omega$-categorical); Hrushovski defined
unimodularity as a way of understanding Zilber's proof (specifically, what
Hrushovski terms ``a series of computations of puzzling success'').

It is also worth mentioning that this yields the purely combinatorial
consequence that any homogeneous locally finite pregeometry is locally 
modular, by considering it as an $\omega$-categorical strongly minimal 
structure in the language with a predicate for $x \in \cl(y_1,...,y_n)$ for 
each $n$ (see \cite[Proposition~2.4.22]{PlGStab}).

Pseudofinite strongly minimal sets provide another example of unimodularity,
via the pigeonhole principle. This was observed by Pillay in
\cite{PillayPfUni}; we give a proof following the lines of his argument.


Lemma{pfUni}:
    If $D$ is a strongly minimal set definable in a pseudofinite theory $T$,
    then $D$ is unimodular.
    .
Proof:
!newcal U
!newop dM
\providecommand{\UPi}{\Pi_{\U}}
    $T$ is pseudofinite so has a model which is an ultraproduct of finite
    structures, say $M = \UPi M_i |= T$ with $M_i$ finite. For $X$
    ${}$-definable, let $|X| := \UPi | X^{M_i} | (- \|N^\U$.

    Claim:
        Let $X (= D^n$ be ${}$-definable.
        Then there exists a unique polynomial $p_X(q) \in \Z[q]$ of degree
        $\RM(X)$ such that 
	\[ |X| = p_X(|D|) \]
	.
    Proof:
        It is clear that if such a polynomial exists, it is unique. We show
        existence.

        If $X = \emptyset$, then $p_X(q):=0$ is as required.

        Else, say $\RM(X) = d >= 0$.
        We can find $k (- \N$ and ${}$-definable $X' (= X$ with $\RM(X')=d$, and
        a co-ordinate projection map $\pi : X' --> D^d$ with all fibres of
        size $k$ and $\RM(\pi(X'))=d$.

	By induction on Morley rank and degree, we have polynomials as in the
        statement for the cardinalities of $X \\ X'$ and $D^d \\ \pi(X')$.
        Then
	  \[ p_X(q) := p_{X \\ X'}(q) + k(q^d - p_{D^d \\ \pi(X')}(q)) \]
        is as required.
	.

    Now let $a_i (- D^n$, $i=1,2$, be interalgebraic generics.
    Let $X$ have minimal Morley degree among ${}$-definable sets with
    $(a_1,a_2) (- X$ and $\RM(X)=n$.
    
    Let $k_i := \mult(a_1a_2/a_i)$.
    Then for $i=1,2$,
    there exists ${}$-definable $X'_i (= X$ with $\RM(X'_i) = n$ and a
    projection $\pi_i : X'_i --> D^n$ with fibres of size $k_i$ and
    $\RM(\pi_i(X'))=n$. Then $\RM(X \\ X'_i) < n$ by the minimality assumption
    on $\dM(X)$, and $\RM(D^n \\ \pi(X'_i)) < n$, so the corresponding
    polynomials have degree $<n$.
    So by the construction in the proof of the claim, the leading term of
    $p_X(q)$ is $k_iq^n$. But $p_X(q)$ is well-defined, so $k_1=k_2$.
    .

So by Theorem~\ref{unimodLocmod}, strongly minimal pseudofinite sets are
locally modular.

In fact, the theorem of Zilber and Hrushovski that $\omega$-categorical
strongly minimal sets are quasifinitely, but not finitely, axiomatisable implies
that they are pseudofinite, so in the end this second example is a
generalisation of the first.

Using two further results which we won't have time to develop in this course -
that 1-basedness of a finite U-rank theory is equivalent to 1-basedness of its
minimal types, and Buechler's dichotomy which says that any minimal type in a
superstable theory which is not locally modular is actually strongly minimal,
Pillay \cite{PillayPfUni} deduces:\

Corollary:
    Any pseudofinite stable theory of finite U-rank is 1-based.
    .

## Germs of definable functions, and Hrushovski-Weil

Notation*:
    Due to alphabetic exhaustion, in this section and the next we will often
    use $x,y,z$ as elements of $\MMeq$ rather than variables.
    .

# Germs of definable functions

Definition*:
    Let $\pp,\qq \in S(\MM)$.
    If $f_1,f_2$ are definable partial functions defined at $\pp$,
    meaning $\pp(x) |= x (- \dom(f_i)$, then say they have the same germ at
    $\pp$ if $\pp(x) |= f_1(x) = f_2(x)$.
    
    The \defn{germ of $f$ at $\pp$} is the equivalence class $~f$ under this
    equivalence relation.

    Write $~f : \pp --> \qq$ if $\pp(x) |= \qq(f(x))$ for some (any)
    representative $f$ (i.e. $f_*(\pp) = \qq$).

    If $~f : \pp --> \qq$ has a representative $f$ defined over $b$ and $a |=
    \pp|_b$, let $~f(a) := f(a)$. It is an immediate consequence of the
    definitions that this is well-defined (i.e. if $g$ is another
    representative defined over $b$ then $g(a) = f(a)$), and $~f(a) |=
    \qq|_b$.

    Given $~f : \pp --> \qq$ and $\sg \in \Aut(\MM)$, we obtain a well-defined
    germ $~f^\sg : \pp^\sg --> \qq^\sg$. Say a (possibly long) tuple $\btup$
    in $\MMeq$ is a _code_ for $~f$ if $\forall \sg \in \Aut(\MM). (\btup =
    \btup^\sg <=> ~f = ~f^\sg)$, and then define $\code{~f} := \dcleq(\btup)$.
    Here, $~f = ~f^\sg$ should be understood as implying $\pp = \pp^\sg$. We
    will see that in a stable theory, $~f$ has a code $b (- \MMeq$ if
    $\Cb(\pp) (- \MMeq$.

    If $p$ and $q$ are stationary types and $\pp,\qq$ their global
    non-forking extensions, a germ at $p$ is defined as a germ at $\pp$, and
    $~f : p --> q$ means $~f : \pp --> \qq$.
    .

Remark*:
    Composition of germs on global types is well-defined, hence
    composition of germs on stationary types is well-defined.
    So we have a category with objects the stationary types, and morphisms the
    germs. Write $~f : \pp -~-> \qq$ if $~f$ is invertible in this category.
    This is equivalent to $~f$ being injective on $\pp$: indeed, this is
    clearly required, and if it holds then by compactness $f$ is already
    injective on some $\phi (- \pp$, and so $f$ has a well-defined definable
    inverse.

    Note that $\code{~f o ~g} (= \dcleq(\code{~f},\code{~g})$,
    and $\code{~f^{-1}} = \code{~f}$.
    .

Remark*:
    Since we work in a stable theory, $\pp$ and $\qq$ are
    definable. Suppose $\pp$ is ${}$-definable.
    Then given a ${}$-definable family $f_z$ of partial functions,
    % This is true, but not used:
    %$\{ z | ~f_z : \pp --> \qq \}$ is $\Wedge$-definable as
    % \[ //\\_{\phi} d_\pp x. ( x \in \dom(f_z) /\
    %   \forall y. ( \phi(f_z(x),y) <-> d_\qq w. \phi(w,y) ) ) ,\]
    %**
    equality of germs at $\pp$ is definable by the equivalence relation
    $E(b,c)$ defined as
    \[ d_\pp x. ( x \in \dom(f_b) \cap \dom(f_c) /\ f_b(x) = f_c(x) ) ,\]
    i.e. $E(b,c) <==> ~f_b = ~f_c$,
    and then since $~f_b^\sg = ~f_{b^\sg}$, we have
        \[ \code{~f_b} = \dcleq(b/E) .\]
    .

Definition*:
    If $p,q,s \in S({})$ are stationary,
    a \defn{family $~f_s$ of germs $p --> q$}
    is the family $~f_s := (~f_b)_{b |= s}$ of germs at $p$ of a
    ${}$-definable family $f_z$ of partial functions,
    which is such that $~f_b : p --> q$ whenever $b |= s$.

    The family is _canonical_ if $b$ is a code for $~f_b$, for all $b |= s$.

    The family is _generically transitive_ if $~f_b(x) \|/ x$ for some (any)
    $b$ and $x$ such that $b |= s$ and $x |= p|_b$.
    .

Remark*:
    $~f_s$ is generically transitive iff when $x |= p$ and $y |= q|_x$, there
    exists $b |= s$ such that $~f_b(x)=y$.
    .

Remark*:
    Suppose $p,q,s \in S({})$ are stationary,
    and $~f_s$ is a family of germs $p --> q$.
    Let $b |= s$ and $x |= p|_b$, and let $y = ~f_b(x)$.
    Then
    Equation{canFamTup}:
	x \|/ b;\; y \|/ b;\; y \in \dcleq(bx) .
	.

    Conversely, if $(b,x,y)$ satisfy (\ref{canFamTup}),
    and $s := \tp(b)$, $p := \tp(x)$, and $q := \tp(y)$ are stationary,
    let $f_b(x)=y$ be a formula witnessing $y\in\dcleq(bx)$.
    Then $~f_s$ is a family of germs $p --> q$.
    .

Lemma{canonFamCB}:
    In the correspondence of the previous remark,
    **(i) $~f_b$ is invertible iff also $x \in \dcleq(by)$;
    **(ii) the family $~f_s$ is generically transitive iff $x \|/ y$;
    **(iii) $\code{~f_b} = \Cb(xy/b)$
        (so $~f_s$ is canonical iff $\Cb(xy/b) = b$).
    *ee*
    .
Proof:--
    **(i) Exercise.
    **(ii) Immediate from the definition.
    **(iii)
        Let $\sg \in \Aut(\MM)$. Let $\pp$ be the global nonforking extension
        of $p$, so $\pp^\sg=\pp$.
        Let $\rr$ be the global nonforking extension of $\stp(xy/b)$. Then
        $\rr$ is equivalent to $\pp(x) \cup \{ y = f_b(x) \}$.
        *\[a* \Cb(xy/b)^\sg = \Cb(xy/b)
            &<=> \Cb(\rr)^\sg = \Cb(\rr)\\\
            &<=> \rr^\sg = \rr\\\
            &<=> \pp(x) \cup \{ y = f_{b^\sg}(x) \} == \pp(x) \cup \{ y = f_b(x) \} \\\
            &<=> \pp(x) |= f_{b^\sg}(x) = f_b(x) \\\
            &<=> ~f_b^\sg = ~f_b .*a\]*

        %{\small Alternative proof:
        %$~f_s$ is canonical iff
        %whenever $b' |= s$ and $~f_{b'}=~f_b$, we have $b'=b$;
        %equivalently, whenever $b' |= s|_x$ and $~f_{b'}(x)=~f_b(x)=y$, we have $b'=b$;
        %equivalently, any automorphism fixing $x$ and $y$ also fixes $b$;
        %equivalently, $\Cb(xy/b)=b$.
        %}
    *ee*
    .

Remark*:
    A germ which is defined over its code is known as a ``strong'' germ. It
    follows from Lemma~\ref{canonFamCB}(iii) that germs in stable theories are
    strong.
    .

# Hrushovski-Weil

Recall that a \defn{$\Wedge$-definable homogeneous space} $(G,S)$ consists of
a $\Wedge$-definable group $G$ and a $\Wedge$-definable set $S$, and a
relatively definable transitive action of $G$ on $S$. We say $G$ is
\defn{connected} if it has no relatively definable proper subgroup of finite
index, and we say $(G,S)$ is connected if $G$ is connected.
We say $(G,S)$ is \defn{faithful} if the action is faithful, i.e. only the
identity element of $G$ acts trivially on $S$.
Recall also the notion of a \defn{generic type} of a
homogeneous space \cite[Definition~6.1]{Palacin-MMMStability}.

Lemma[Hrushovski-Weil]{HW}:
    Let $p,s \in S({})$ be stationary,
    and suppose $~f_s$ is a generically transitive canonical family of
    invertible germs $p --> p$.
    Suppose that $~f_s$ is closed under inverse and generic composition,
    meaning that for $b |= s$ there exists $b' |= s$ such that
    $~f_b^{-1}=~f_{b'}$,
    and for $b_1b_2 |= s^{(2)}$, there exists $b_3$ such that
	\[ ~f_{b_1} o ~f_{b_2} = ~f_{b_3} \]
    and $b_ib_3 |= s^{(2)}$ for $i=1,2$.

    Then there exists a connected faithful $\Wedge$-definable/${}$ homogeneous space 
    $(G,S)$,
    a definable embedding of $s$ into $G$ as its unique generic type,
    and a definable embedding of $p$ into $S$ as its unique generic type,
    such that the generic action of $s$ on $p$ agrees with that of $G$ on $S$,
    i.e. $~f_b(a) = b*a$ for $b |= s$ and $a |= p|_b$.
    .
Remark*:
    This is essentially the Hrushovski-Weil ``group chunk'' theorem.
    There, one starts with a generically associative binary operation $*$, and
    applies the lemma to the germs of $x |-> a*x$ to obtain a group structure
    extending $*$.
    .
Proof:
    Let $G$ be the group of germs generated by $~f_s$.

    Claim:
        Any element of $G$ is a composition of two generators.
        .
    Proof:
        Since the family is closed under inverses, the identity is the composition
        of two generators. Since $s$ is a complete type, it follows from generic
        composability that any generator is the composition of two generators.
        So 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.
        
        Let $b' |= s|_{b_1b_2b_3}$.
        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 $b_2$.
        Now $b' \|/_{b_2} b_3$,
        so $b'' \|/_{b_2} b_3$, since $b'' \in \dcleq(b'b_2)$,
        so since $b'' \|/ b_2$,
        we have $b'' \|/ b_3$.
        Also $b' \|/ b_1$.
        So the germs $~f_{b_1} o ~f_{b'}$ and $~f_{b'}^{-1} o ~f_{b_2} o ~f_{b_3}$
        appear in the family $~f_s$.
        .

    Now $G$ is $\Wedge$-definable as pairs of realisations of $s$, modulo
    equality of the corresponding composition of germs, and the group
    operation is defined by composition of germs. We identify $s$ with its
    image in $G$ under the embedding $b |-> ~f_b$, which is a completion of
    $G$ as a $\Wedge$-definable group.

    We show that $G$ is connected with generic type $s$ by showing that if
    $g\in G$,
    then for some (and hence any) $b |= s|_g$, we have $g*b |= s|_g$.
    This holds for $g |= s$, by assumption.
    Let $g (- G$, say $g=g_1*g_2$ with $g_1,g_2 |= s$.
    Let $b |= s|_{g_1,g_2}$.
    Then $g_2*b |= s|_{g_1,g_2}$, and so $g_1*g_2*b |= s|_{g_1,g_2}$.
    Now $g = g_1*g_2 (- \dcleq(g_1,g_2)$,
    so $b |= s|_g$ and $g*b |= s|_g$, as required.

    $G$ acts generically on $p$ by application of germs, i.e. $g*a := g(a)$ if
    $a |= p|_g$.

    Now define $S := (G \times p)/E$ where $(g,a) E (g',a')$ iff
    $(h*g)*a = (h*g')*a'$ for $h |= s|_{aa'gg'}$, which is definable by
    definability of $s$.
    Define the action of $G$ by $h*((g,a)/E) := (h*g,a)/E$.
    This is well-defined, since if $(g,a) E (g',a')$ and $h (- G$, then if $h'
    |= s|_{g,g',a,a',h}$, then also $h'*h |= s_{g,g',a,a',h}$ by genericity,
    and we have $h'*h*g*a = h'*h*g'*a'$, so $(h*g,a) E (h*g',a')$.
    $p$ embeds via $a |-> (1,a)/E$.

    We show transitivity. Let $a,a' |= p$, and we show $(1,a')/E (-
    G*(1,a)/E$; this suffices for transitivity, since clearly $(G,a')/E (=
    G*(1,a')/E$. Let $c |= p|_{aa'}$.
    Then by generic transitivity of $~f_s$, there exist $g,g' |= s$ such that
    $g*a=c$ and $g'*c=a'$. Then $(h*g)*a = h*c$ for $h |= s|_{acg}$, so
    $(g,a)E(1,c)$. Similarly $(g',c)E(1,a')$. So $(g'*g)*(1,a)/E=g'*(g,a)/E
    =g'*(1,c)/E = (1,a')/E$.

    For faithfulness of the action: suppose $g$ acts trivially, and let $a |=
    p|_g$. Then $(g,a)E(1,a)$, so let $h |= s|_{ag}$; then $(h*g)*a = h*a$.
    But $h \|/_g a$, so $h,h*g \|/_g a$, so $h,h*g \|/ a$ since $g \|/ a$. So
    $h*g=h$ as germs, so $g=1$.

    Finally, we claim that $p$ is the unique generic type. If $g \in G$ and $a
    |= p|_g$, then $g*a |= p|_g$ as this is the action of a germ. But $G$ acts
    transitively on the generic types
    \cite[Proposition~6.13(2)]{Palacin-MMMStability}.
    So $p$ is the unique generic type.
    .

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 a condition for this to give rise to a $\Wedge$-definable group.

Lemma["Hrushovski-Weil for bijections"]{isomHW}:
    Suppose $\acleq({})=\dcleq({})$, % so we don't have to work over \Cb(s)
    let $p,q,r\in S({})$, and
    suppose $~f_r$ is a generically transitive canonical family of invertible
    germs $p --> q$.
    Let $b_1$ and $b_2$ be independent realisations of $r$ and say
	\[ ~f_{b_1}^{-1} o ~f_{b_2} = ~g_c \]
    with $~g_s$ a canonical family of invertible germs $p --> p$, where
    $s = \tp(c)$,
    and suppose
    Equation{isomHWIndie}:
	c \|/ b_i \infix{for} i=1,2.
        .
    Then $~g_s$ satisfies the assumptions, and hence the conclusions, of Lemma~\ref{HW}.
    .
Proof:
    Let $c' |= s|_c$.
    Let $b |= r|_{c,c'}$.
    Then by (\ref{isomHWIndie}), $bc == b_2c$, so say $b_1'bc == b_1b_2c$;
    similarly, $bc' == b_1c$, so say $bb_2'c' == b_1b_2c$.
    Then
	\[ ~g_c o ~g_{c'} = ~f_{b_1'}^{-1} o ~f_b o ~f_b^{-1} o ~f_{b_2'}
	    = ~f_{b_1'}^{-1} o ~f_{b_2'} .\]
    \[ \xymatrix{
:::     & q \ar@{-->}[ddr] & q \ar@/^/[dd] \ar[ddrr] & q \ar@{-->}[ddr] & \\
:::     & & & & \\
::: p \ar@{-->}[uur] \ar[uurr] & & p \ar@/^/[uu] \ar@{-->}[uur] & & p
    } \]

    %\begin{minipage}{\linewidth}
    %Verbatim:
::: %           *      *      *         q
::: %          / \   .'|`.   / \
::: %         / c'\.'  |  `./ c \   
::: %        /   .'\   |b  /`.   \
::: %       /  .'   \  |  /   `.  \
::: %      / .'      \ | /      `. \
::: %     /.' b_2'    \|/    b_1' `.\
::: %    *'            *            `*  p
    %  .
    %\end{minipage}
    Now $b_i' \|/ b$ by choice of $b_i'$, since $b_1 \|/ b_2$.
    Also $b_1' \|/_{b} b_2'$, since $c \|/_b c'$, since $c \|/ c'$ and $b \|/ cc'$.
    Hence $b_1' \|/ bb_2'$, so $b_1' \|/ b_2'$ and $b \|/ b_1'b_2'$.

    So $(b_1',b_2') |= r^{(2)}$,
    so say $~f_{b_1'}^{-1} o ~f_{b_2'} = ~g_{c''}$ with $c'' |= s|_{b_i'}$.

    Then since $b \|/ b_1'b_2'$, we have $b \|/ c''b_1'$, hence $c'' \|/
    b_1'b$, and so $c'' \|/ c$.
    Similarly $c'' \|/ c'$.

    Finally, we must check that $~g_s$ is generically transitive.
    So let $x |= p$, let $b_2' |= r|_x$, and let $b_1' |= r|_{xb_2'}$.
    Let $y := ~f_{b_2'}(x)$ and $z := ~f_{b_1'}^{-1}(y)$.
    Then $y \|/ x$ by generic transitivity, and $b_1' \|/ yx$,
    so $y \|/ xb_1'$, i.e. $y |= q|_{xb_1'}$, and so $z |= p|_{xb_1'}$, and in
    particular $x \|/ z$.
    Since $b_1'b_2' == b_1b_2$, this proves generic transitivity of $~g_s$.
    .

Exercise*:
    Assuming finite U-rank, (\ref{isomHWIndie}) holds iff $U(s) = U(r)$.
    .

%Remark*:
%    Suppose $(b,x,y)$ "lie on a line" in the sense of the group configuration
%    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.
%    .

## The Group Configuration Theorem

We continue to work in a monster model $\MMeq$ of an arbitrary stable theory
$T$.

Definition*:--
    \[ \xymatrix{
:::    c            \ar@{-}'[drrr][ddrrrrrr] \ar@{-}'[ddrr][dddrrr] \\
:::     & & &b \\
:::     & &z& & & &a \\
:::     & & &x \\
:::    y            \ar@{-}'[urrr][uurrrrrr] \ar@{-}'[uurr][uuurrr] \\
  } \]
%    \[ \xymatrix{
%:::	& & a & & \\
%:::	& & & & \\
%:::	& & & & \\
%:::	& b & & x & \\
%:::	& & z \ar@{-}'[uuuuu] & & \\
%:::	& & & & \\
%:::     c \ar@{-}'[rruu][rrruuu] \ar@{-}'[ruuu][rruuuuuu] & & & & y \ar@{-}'[lluu][llluuu] \ar@{-}'[luuu][lluuuuuu]
%  } \]
    %\begin{minipage}{\linewidth}
    %Verbatim:
    %          a       
    %         / \      
    %        b   x     
    %       / `z' \    
    %      / .' `. \   
    %     /.'     `.\  
    %    c'         `y 
    %    .
    %\end{minipage}
    $(a,b,c,x,y,z)$ forms a _group configuration_ if
    ***) any non-collinear triple in the above diagram is independent,
    ** $\acleq(ab)=\acleq(bc)=\acleq(ac)$,
    ** $\acleq(ax)=\acleq(ay)$ and $\acleq(a) = \aCb(xy/a)$; similarly for
      $bzy$ and $czx$.
    %** $x$ is interalgebraic with $y$ over $a$, and $a$ is interalgebraic
      %with $\Cb(\stp(xy/a))$; similarly for $bzy$ and $czx$.
    *e)*
    .

Remark*:
    Suppose $(G,S)$ is a connected faithful $\Wedge$-definable/${}$ homogeneous space. 
    Let $(a,b,x)$ be an independent triple with $a,b$ generics of $G$ and $x$
    a generic of $S$, and define $b := c*a$, $x := a*y$, $z := b*y$ (so $z =
    c*a*y = c*x$). Then $(a,b,c,x,y,z)$ forms a group configuration.
    (It follows from faithfulness that e.g. $\Cb(xy/a) = a$; see 
    \cite[Remark~4.1]{PlGStab} for a proof.)

    Call such an $(a,b,c,x,y,z)$ a \defn{group configuration of $(G,S)$}.
    .
    
Theorem[Group Configuration Theorem]{groupConf}:
    Suppose $(a,b,c,x,y,z)$ forms a group configuration.
    Then, after possibly expanding the language by parameters $B$ with $B \|/
    abcxyz$,
    there is a connected faithful $\Wedge$-definable/${}$ homogeneous space $(G,S)$,
    and a group configuration $(a',b',c',x',y',z')$ of $(G,S)$,
    such that each unprimed element is interalgebraic with the corresponding
    primed element.
    .

Example*:
    In ACF, we can restate as follows:
    $(b,z,y)$ extends to a group configuration $(a,b,c,x,y,z)$
    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. \cite[6.2]{HZZGeoms}.)
    .

Remark*:
    The above diagram is the traditional one, but the following alternative
    diagram is also suggestive (think of it as a commuting triangle)

    \[ \xymatrix{
:::     &x   \ar@{-}[ddl]_{a} \ar@{-}[ddr]^{c} \\
:::          \\
:::    y& &z \ar@{-}[ll]^{b}
  } \]
%    \begin{minipage}{\linewidth}
%    Verbatim:
%
%                   x
%                  / \
%               a /   \ c
%                /     \
%               y-------z   .
%                   b
%      .
%    \end{minipage}
    .


Proof:
    We prove this by repeatedly applying the following two operations to
    transform $(a,b,c,x,y,z)$ into the group configuration of a homogeneous
    space:\
	***) add independent parameters to the language - we refer to this as
        ``base-changing'' to the parameters;
	** replace any point of the configuration with an interalgebraic
          point of $\MMeq$ - we refer to this as e.g. ``interalgebraically
          replacing'' $b$ with $b'$, implicitly claiming that
          $\acleq(b)=\acleq(b')$.
	*e)*

    Note that these operations transform a group configuration into a group
    configuration.

    By base-changing whenever necessary, we will assume throughout that
    $\acleq({})=\dcleq({})$, so types over ${}$ are stationary.

    The proof will comprise three steps:
    **(I) ``reduce $\acleq$ to $\dcleq$'' to show we may assume $(b,z,y)$
      to define a generically transitive canonical family of invertible germs via
      Lemma~\ref{canonFamCB};
    **(II) prove this family satisfies the independence assumption of Lemma~\ref{isomHW};
    **(III) connect the resulting homogeneous space to the group configuration.
    *ee*

    And so it begins.

    **(I) First, a claim.
        Claim:
          If $(a,b,c,x,y,z)$ is a group
          configuration and if we let $~z \in \MMeq$ be the set $~z =
          \{z_1,...,z_d\}$ of conjugates $z_i$ of $z$ over $ybxc$, then $~z$ is
          interalgebraic with $z$.
          .
        (Here, $\{z_1,...,z_d\} := (z_1,...,z_d)/S_d$
        is the quotient of $(z_1,...,z_d)$ under the action by permutations of
        the symmetric group $S_d$.)
        Proof:
          It suffices that the conjugates be 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:
              $c\|/_z b$, so $cx \|/_z by$,
              so setting $B:= \acleq(cx) \cap \acleq(by)$, we have
              $B \|/_z B$ so $B (= \acleq(z)$.
              But $z \in B$.
              So $\acleq(z) = B$,
              and so $\acleq(z_i) = B$ for each $z_i$.
              So $\acleq(z_i)=\acleq(z)$.
          .

	Now let $a' |= \tp(a)|_{abcxyz}$.
	Say $a'x'c' ==_{ybz} axc$. So $(a',b,c',x',y,z)$ is also a group configuration.
	So by the Claim, the set $~z'$ of conjugates of $z$ over $ybx'c'$ is
	interalgebraic with $z$.
	Note that $~z' \in \dcleq(ybx'c')$.

	So base-change to $a'$,
	and interalgebraically replace $y$ with $yx'$, $b$ with $bc'$, and $z$ with $~z'$.
	The group configuration now satisfies
	    \[ z \in \dcleq(by) .\]

	Repeating this procedure by base-changing to an independent copy of $b$ and
	enlarging $a$ and $y$ and interalgebraically replacing $x$ with an $~x$,
	we can also ensure that
	    \[ x \in \dcleq(ay) .\]

	Repeat once more: base-change to 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 (e.g. by considering
        automorphisms) $zx' \in \dcleq(ba'~y)$.
	So after interalgebraically replacing $b$ with $ba'$, $z$ with $zx'$, and $y$ with $~y$,
	as in the previous two cases $y \in \dcleq(bz)$,
	and now also $z\in\dcleq(by)$.

        Finally, interalgebraically replace $b$ with $\Cb(yz/b)$,
        noting that $\Cb(yz/b)$ is an element of $\MMeq$ by
        Lemma~\ref{canonFamCB}(iii), and is interalgebraic with $b$ by
        definition of a group configuration.

    **(II)
        Setting $p := \tp(y)$, $q := \tp(z)$, $r := \tp(b)$,
        $(b,y,z)$ now corresponds via Lemma~\ref{canonFamCB} to a generically
        transitive canonical family $~f_r$ of invertible germs $p --> q$.

        We aim to apply Lemma~\ref{isomHW} to obtain a group,
        so we must show that for $b' |= r|_b$,
        if $\dcleq(d) = \code{ ~f_{b'}^{-1} o ~f_{b} }$,
	we have $d \|/ b$ and $d \|/ b'$.

        This holds for all $b' |= r|_b$ or none, so we may assume
        $b' \|/ abcxyz$. Then $b' ==_{xcz} b$, so say $b'y'a' ==_{xcz} bya$.

        Claim:
            $y' \|/ bb'$.
            .
        Proof:
          $b' \|/ bz$ and so $b' \|/_z b$,
          and $b \|/ z$, so $b \|/ b'z$, hence $z \|/_{b'} b$.
          Now $y' (- \acleq(zb')$, so $y' \|/_{b'} b$.
          But $y' \|/ b'$, so $y' \|/ bb'$.
          .

        We also have $y \|/ bb'$ and $f_b'^{-1}(f_b(y)) = y'$,
        so by Lemma~\ref{canonFamCB},
        $\dcleq(d) = \code{~f_b'^{-1} o ~f_{b} } = \Cb(yy'/bb')$.

	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:
	    $b \|/ aa'$ and $b' \|/ aa'$.
	    .
	Proof:
            $b \|/_c b'$, and then since $a\in\acleq(cb)$ and
            $a'\in\acleq(cb')$, we have $ab \|/_c a'b'$,

            So $a' \|/_c ab$;
	    but $a' \|/ c$, so $a' \|/ ab$, so $b \|/_a a'$,
	    and then since $b \|/ a$, we have $b \|/ aa'$.
            Similarly, it follows from $a \|/_c a'b'$ that $b' \|/ aa'$. 
	    .
	
	So $b \|/ d$ and $b' \|/ d$, as required.

    **(III)
        Let $(G,S)$ be the connected faithful $\Wedge$-definable connected homogeneous space
        obtained by Lemma~\ref{isomHW} from (II).
        So $p$ is the generic type of $S$.
        Let $s_G$ be the generic type of $G$.

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

	First, let $b' |= \tp(b)|_{abcxyz}$.
	Say $y'b' ==_z yb$.
	Say $\dcleq(g) = \code{~f_{b'}^{-1} o ~f_b}$.
        By the construction of $(G,S)$,
        we have $g |= s_G$,
        and $y,y' |= p$,
        and $y' = g*y$.
        Base-change to $b'$,
	and interalgebraically replace $b$ with $g |= s_G$ and $z$ with $g*y |= p$,
        the latter interalgebraicity holding as $g*y = ~f_{b'}^{-1}(~f_b(y)) = ~f_{b'}^{-1}(z)$.

	Now let $c'' |= \tp(c)|_{abcxyz}$,
	and say $b''z''c'' ==_{axy} bzc$.
	Base-change to $c''$,
	and interalgebraically replace $a$ by $b'' |= s_G$ and $x$ by $z'' = b''*y |= p$.

	Let $h := b*a^{-1}$.
	Then $x = a*y$ and $z = b*y$, so $z = h*x$.
	By definition of $G$, $(h,x,z)$ is as in Lemma~\ref{canonFamCB}, so $\Cb(xz/ab) = h$.
        But also $x$ and $z$ are interalgebraic over $c$,
	so $\aCb(xz/ab) = \aCb(xz/c) = \acleq(c)$.
	So interalgebraically replace $c$ with $h$.

        Then $(a,b,c,x,y,z)$ is a group configuration of $(G,S)$, as required.
    *ee*
    .

Example*:
    Suppose $D$ is minimal and locally modular but non-trivial.
    Fact~\ref{nonTrivMod} claimed the existence of a definable group.
    This group can be found by applying the group configuration theorem.
    Expand by parameters to make $D$ modular.
    By non-triviality, after possibly expanding by further parameters there
    are $a,b,c \in D$ such that $U(ab)=U(bc)=U(ca)=U(abc)=2$. Let $b'c' |=
    \stp(bc/a)|_{abc}$. Then $U(bcb'c')=3$, so by modularity
    $U(\acl(bc')\cap\acl(b'c))=1$, say $d\in \acl(bc')\cap\acl(b'c) \\
    \acl({})$. Then $(a,b,c,b',c',d)$ is a group configuration.
    .

Fact{minActions}:
    If $S$ is strongly minimal, any connected faithful ${}$-definable
    homogeneous space $(G,S)$ is of one of the following forms:
    ***) $U(G)=1$, $G$ is commutative, and the action is regular;
    ** $U(G)=2$, $S=K$ is the universe of a ${}$-definable
    algebraically closed field, and $G$ is the
      affine group $K^+ \rtimes K^*$ acting as $(a,b)*x = a+bx$.
    ** $U(G)=3$, $S=\PP^1(K)$ is the projective line of a 
    ${}$-definable
      algebraically closed field, and $G$ is the group $\"PSL_2(K)$ of Möbius
      transformations.
    *e)*
    This is due to Cherlin.
    See \cite[Theorem~3.27]{PoizatStableGroups} for a proof.

    FIXME: to prove Corollary~{fieldConf} below, we require something similar
    when $S$ is only minimal. Namely, we want that if $S$ is minimal and
    $(G,S)$ is a connected faithful $\Wedge$-definable/${}$ homogeneous space
    with $U(G) = k (- \omega$, then $k <= 3$, and if $k>1$ then the generic
    type of $S$ is interalgebraic with a $//\\$-definable algebraically closed
    field.
    .

Corollary[Field Configuration Theorem]{fieldConf}:
    Suppose $(a,b,c,x,y,z)$ forms a \defn{field configuration}, that is,
    a group configuration with $U(x)=U(y)=U(z)=1$
    and $U(a)=U(b)=U(c)=k$ where $k>1$.
    Then we obtain a rank $k$ group acting on a minimal type.
    So $k=2$ or $3$, and $\tp(x)$ is interalgebraic with the generic type of a $//\\$-definable algebraically closed field.
    .

Definition*:
    $D$ is \defn{$k$-pseudolinear} if whenever $p$ is a complete minimal type
    with $p(x) |= x (- D^2$, we have $U(\Cb(p)) <= k$.
    .
Remark*:
    1-pseudolinear $<=>$ linear.
    .

Theorem{pseudolinLin}:
    Let $k>1$.
    Suppose $D$ is minimal and $k$-pseudolinear.
    Then $D$ is locally modular.
    .
Proof:
    We may assume $\dcleq({})=\acleq({})$.

    Suppose $D$ is $k$-pseudolinear. We show $D$ is $(k-1)$-pseudolinear.

    Suppose not. So (using Lemma~\ref{finCb}(a)) say $(a_2,a_3) (- D^2$,
    $\tp(a_2a_3/b_1)$ is minimal,
    $b_1 = \Cb(a_2a_3/b_1)$, $U(b_1)=k$.
    Now $a_2 \|/ b_1$, since else $a_2 (- \acleq(b_1)$, and then $a_2a_3
    \|/_{a_2} b_1$ and so $b_1 (- \acleq(a_2)$, contradicting $k>1$.
    Similarly, $a_3 \|/ b_1$.

    Let $b_2 |= \stp(b_1)|_{b_1a_2a_3}$.
    Then $b_2a_3 == b_1a_3$, so say $a_1$ is such that $b_2a_1a_3 == b_1a_2a_3$.
    Let $b_3 := \Cb(a_1a_2 / b_1b_2)$.

    $U(a_1a_2/b_1b_2) = 1$ since $a_2 (- \acleq(a_3b_1)$ and $a_3 (-
    \acleq(a_1b_2)$. So by $k$-pseudolinearity, $U(b_3) <= k$.

    Similarly, $U(a_2a_3/b_2b_3)=1$.
    Since $a_2 \|/ b_1$ and $b_2 \|/ b_1a_2$ and $b_3 (- \acleq(b_1b_2)$, we
    have $a_2 \|/ b_1b_2b_3$, so $U(a_2a_3/b_1b_2b_3)=1$.
    So $U(a_2a_3/b_2b_3) = U(a_2a_3/b_1b_2b_3) = U(a_2a_3/b_1) = 1$,
    so $a_2a_3 \|/_{b_2b_3} b_1$ and $a_2a_3 \|/_{b_1} b_2b_3$,
    so since $b_1=\Cb(a_2a_3/b_1) = \Cb(a_2a_3/b_1b_2b_3)$,
    $b_1 (- \acleq(b_2b_3)$.
    Similarly, $b_2 (- \acleq(b_1b_3)$.

    So $U(b_2b_3) = U(b_1b_2b_3) = U(b_1b_2) = 2k$, 
    so $U(b_3) = k$, and similarly $U(b_1b_3)=2k$.

    Then $(b_1,b_2,b_3,a_2,a_3,a_1)$ is a field configuration,
    so by the Field Configuration Theorem (Corollary~\ref{fieldConf}),
    there is a $\Wedge$-definable algebraically closed field with
    generic type interalgebraic with $\pp$. But
    e.g. $y = 1 + b_1x + b_2x^2 + ... + b_{k+1}x^{k+1}$ is a
    $(k+1)$-dimensional family of plane curves in the field, which contradicts
    $k$-pseudolinearity (as in Example~\ref{affPlaneNonlinear}).
    .

## Unimodular minimal sets are locally modular 
\label{unimodProof}

In this section, we prove Theorem~\ref{unimodLocmod}.
So assume $D$ is a unimodular minimal type; we will show that $D$ is 1-based.
Let $\pp$ be the unrealised global completion of $D$.

Lemma:
    $D$ remains unimodular on adding to the language a Morley sequence in
    $\pp|_{}$.
    .
Proof:
    Say $\btup$ is such a Morley sequence. Work in the unexpanded language.

    Suppose $a,a'$ are as in the definition of unimodularity in the expanded
    language, so $a,a' |= \pp|_{\btup}^{(n)}$,
    and $a (- \acleq(a'\btup)$ and $a' (- \acleq(a\btup)$.

    Let $\btup'$ be a finite subtuple such that
    $\mult(a / a'\btup) = \mult(a / a'\btup')$
    and $\mult(a' / a\btup) = \mult(a' / a\btup')$.
    Now $a\btup'$ and $a'\btup'$ are interalgebraic finite Morley sequences in
    $\pp|_{}$, so by unimodularity in the original language,
    $\mult(a\btup' / a'\btup') = \mult(a'\btup' / a\btup')$.
    So $\mult(a/a'\btup) = \mult(a'/a\btup)$ as required.
    .

So since 1-basedness is invariant under adding parameters, for the purposes of
proving Theorem~\ref{unimodLocmod}, we may add such a Morley sequence.

So by Lemma~\ref{minwEI}, we may and will assume that $D$ has wEI, and hence
gEI.

Definition*:
    For $b (- D^{\eq}$, define the _Zilber degree_ by
        \[ Z(b) := \frac{ \mult(b/c) }{ \mult(c/b) }, \]
    where $c |= \pp|_{}^{(<\omega)}$ is interalgebraic with $b$;
    equivalently, using gEI, $c$ is an $\acl_D$-basis for $\acleq(b)\cap D$.

    Then define $Z(a/b) := Z(ab)/Z(b)$.
    .
Lemma:
    $Z(b)$ is well-defined.
    .
Proof:
    This follows from unimodularity. Indeed, if $c' |= \pp|_{}^{(<\omega)}$ is
    another $\acl_D$-basis for $\acleq(b)\cap D$, then
    *\[a* \frac { \mult(b/c) }{ \mult(c/b) }
	&= \frac { \mult(b/c)\mult(c'/bc) }{ \mult(c/b)\mult(c'/bc) } \\\
	&= \frac { \mult(bc'/c) }{ \mult(cc'/b) } \\\
	&= \frac { \mult(c'/c)\mult(b/cc') }{ \mult(cc'/b) } \\\
	&= \frac { \mult(c/c')\mult(b/c'c) }{ \mult(c'c/b) } \\\
	&= \textrm{[reversing above steps]} \\\
	&= \frac { \mult(b/c') }{ \mult(c'/b) } .*a\]*
    .

Remark*:
    For $b (- D^n$, $Z(b) \in \N$, since we may take $c$ to be a subtuple of
    $b$. In the pseudofinite strongly minimal case, $Z(b)$ is the leading
    coefficient (not the degree!) of the polynomial $p_X$ of
    Lemma~\ref{pfUni}, for $X$ a definable set of minimal rank and degree in
    $\tp(b)$.
    .

Remark*:
    Analogous polynomials, called Zilber polynomials, can also be defined in
    the locally finite case, and again $Z(b)$ is then the leading coefficient.
    The Zilber polynomial $p_X$ of a definable set $X$ has the defining property that
    for all sufficiently large $n$, if $a |= \pp|_{}^{(n)}$ then
    $|{ X \cap \acl_D(a) }| = p(|{ D \cap \acl_D(a) }|)$.
    .

Lemma{Zprops}:--
    **(i) $Z$ is $\Aut(\MMeq)$-invariant.
    **(ii) $Z(ab/c) = Z(a/bc)Z(b/c)$.
    **(iii) $a \in \acleq(b) => Z(a/b) = \mult(a/b)$.
    **(iv) $a \in D \\ \acleq(b) => Z(a/b) = 1$.
    *ee*
    .
Proof:--
    **(i) Clear.
    **(ii)
        \[ Z(a/bc)Z(b/c)
        = \frac{ Z(abc) }{ Z(bc) } \frac{ Z(bc) }{ Z(c) }
        = \frac{ Z(abc) }{ Z(c) } = Z(ab/c) \]
    **(iii) Let $c$ be a basis for $\acleq(b) \cap D = \acleq(ab) \cap D$.
        Then *\[a* Z(a/b) &= Z(ab)/Z(b) \\\
            &= \frac{ \mult(ab/c)\mult(c/b) }{ \mult(c/ab)\mult(b/c) } \\\
            &= \frac{ \mult(a/bc)\mult(c/b) }{ \mult(c/ab) } \\\
            &= \frac{ \mult(ac/b) }{ \mult(c/ab) } \\\
            &= \mult(a/b) .*a\]*
    **(iv) Let $c$ be a basis for $\acleq(b) \cap D$.
        Then $ac$ is a basis for $\acleq(ab) \cap D$, so
        by stationarity of $\tp(a/b) = \pp|_{b}$ and of $\tp(a/c) = \pp|_{c}$
        and by Lemma~\ref{multStat},
        *\[a* Z(ab)
        &= \frac{ \mult(ab/ac) }{ \mult(ac/ab) } \\\
        &= \frac{ \mult(b/ac) }{ \mult(c/ab) } \\\
        &= \frac{ \mult(b/c) }{ \mult(c/b) } \\\
        &= Z(b) .*a\]*
    *ee*
    .
Remark*:
    In fact, the function $Z(x/y)$ is uniquely determined by (i)-(iv) (see
    \cite{HrUnimod} for a proof).
    .

Definition*:
    For $p = \tp(a/B)$ a stationary type, let $Z(p) := Z(a / \Cb(p))$.
    .

Lemma{Znf}:
    Suppose $a,b,c (- \Deq$.
    **(I) If $\tp(a/c)$ is stationary and $a \|/_c b$, then $Z(a/bc) = 
    Z(a/c)$.

        In particular, if $\tp(a/c)$ is stationary then $Z(\tp(a/c))=Z(a/c)$.

    **(II) $Z(a/c) = \sum_{i (- I} Z(\qq_i)$ where $(\qq_i)_{i (- I}$ enumerates the global
      non-forking extensions of $\tp(a/c)$.

    **(III) $Z(a/c) = \sum_i Z(a_i/bc)$ where $(\tp(a_i/bc))_i$ enumerates the
	nonforking extensions of $\tp(a/c)$ to $bc$.
    *ee*
    .
Proof:--
    **(I) We first consider two special cases.
	Claim:
            If $\tp(a/c)$ is stationary, then $Z(a/bc) = Z(a/c)$
	    **(a) when $b (- D \\ \acleq(ac)$, and
	    **(b) when $b (- \acleq(c)$.
	    *ee*
	    .
	Proof:
          Since $Z(a/bc) = \frac{ Z(ab/c) }{ Z(b/c)} = \frac{ Z(b/ac) }{ Z(b/c) } Z(a/c)$, it
	  suffices to see that $Z(b/ac)=Z(b/c)$.

	  In case (a), by Lemma~\ref{Zprops}(iv) we have $Z(b/ac) = 1 = Z(b/c)$.

	  In case (b), by Lemma~\ref{Zprops}(iii) and Lemma~\ref{multStat} we
	  have $Z(b/ac) = \mult(b/ac) = \mult(b/c) = Z(b/c)$.
	  .
	
	Now let $b$ be such that $a \|/_c b$ and let $b'$ be a basis for
	$\acleq(b) \cap D$ over $\acleq(c) \cap D$. Using that $a \|/_c b'$ and
	applying case (a) iteratively, $Z(a/b'c)=Z(a/c)$. Then since $bc$ and
	$b'c$ are interalgebraic, and $\tp(a/bc)$ and $\tp(a/b'c)$ are both
        stationary, by case (b) twice we have $Z(a/bc) =
	Z(a/bb'c) = Z(a/b'c) = Z(a/c)$.

    **(II)
        Using Lemma~\ref{finCb}(a), let $b := \Cb(a/c) (- \Deq$, so
        $(\Cb(\qq_i))_{i (- I}$ enumerates the conjugates of $b$ over $c$ (and
        so $|I| = \mult(b/c)$ is finite). Then $a \|/_b bc$, so by (I) we have
        $\sum_i Z(\qq_i) = \mult(b/c)Z(a/bc) = Z(b/c)Z(a/bc) = Z(ab/c) =
        Z(a/c)Z(b/ac) = Z(a/c)$, using that $b = \Cb(a/c) \in \dcleq(ac)$ (as
        one sees by considering automorphisms).

    **(III)
	This follows from (II), since the set of global non-forking extensions
	of $\tp(a/c)$ is the union of the sets of global non-forking
	extensions of its non-forking extensions to $bc$.
    *ee*
    .
	

Lemma["Relaxation"]{relax}:
    Suppose $a,b_1,b_2$ are such that $U(a)=2$, $U(a/b_i)=1$,
    $b_i=\Cb(a/b_i)$, $U(b_1) >= 1$, and $U(b_2) >= 2$.

    Then $b_1 \|/ b_2$ iff $b_1 \|/_a b_2$.

    {\em (In words: the pairs of curves which are independent and happen
    to both pass through $a$ are precisely the pairs of curves which are
    independent given that they both pass through $a$.)}
    .
Proof:
    First we see that this equivalence holds if $a \+/_{b_1} b_2$.
    Indeed, then $a\in \acleq(b_1b_2)$, and so
    *\[a* U(b_1/b_2a) - U(b_1/a)
      &= U(b_1b_2a)-U(b_2a) - U(b_1a)+U(a) \\\
      &= U(b_1b_2)-U(a/b_2)-U(b_2)-U(a/b_1)-U(b_1)+U(a) \\\
      &= U(b_1b_2)-U(b_2)-U(b_1)-1-1+2 \\\
      &= U(b_1/b_2)-U(b_1) ;*a\]*
    in particular one side is zero iff the other is, as required.

    Similarly, we are done if $a \+/_{b_2} b_1$.

    If neither dependence holds, then $\dcleq(b_1) = \Cb(a/b_1) =
    \Cb(a/b_1b_2) = \Cb(a/b_2) = \dcleq(b_2)$. Then $b_1 \|/_a b_2$ implies
    $b_2 (- \acleq(a)$, so $1 = U(a/b_2) = U(a) - U(b_2) <= 2 - 2$, which is a
    contradiction, so $b_1 \+/_a b_2$, and similarly we find $b_1 \+/ b_2$.
    .

Lemma["Bézout"]{bezout}:
    Suppose $p_1 (- S(b_1c)$ and $p_2 (- S(b_2c)$ are minimal types,
    $p_i(x) |= x (- D^2$, $b_i = \Cb(p_i)$, $c (- \Deq$.
    Suppose $U(b_1/c) >= 1$ and $U(b_2/c) >= 2$,
    and $b_1 \|/_c b_2$. 
    Then $|{ p_1 \cup p_2 }| := |{ \{ a | a |= p_1 \cup p_2 \} }| = Z(p_1)Z(p_2)$.
    .
Remark*:
    Thinking of $p_i$ as plane curves, $p_1 \cup p_2$ is the
    intersection, so the lemma can be read as saying that for curves in
    "generic position" within sufficiently large families, the size of the
    intersection is the product of the degrees. Bézout's theorem in algebraic
    geometry makes an analogous claim. However, that theorem concerns
    complete curves rather than complete types, so all points are counted
    rather than only those generic on both curves. We will see that this
    difference is crucial.
    .
Remark*:
    It will turn out that this lemma can never actually be applied: using the
    lemma, we will show that $D$ is linear, so no such $p_2$ exists.
    .
Proof:
    First note that we may assume that each $\tp(b_i/c)$ is stationary, by replacing
    $c$ with $c' := c\Cb(b_1/c)\Cb(b_2/c) \in \acleq(c) \cap \Deq$.
    Indeed, each $p_i$ is stationary so has a unique extension $p_i'$ to
    $c'b_i$, so $|{ p_1' \cup p_2' }| = |{ p_1 \cup p_2 }|$ and
    $Z(p_i')=Z(p_i)$.

    In the interests of notational sanity, we will assume $c={}$. It is
    straightforward to check that the arguments below go through without this
    assumption, by working everywhere over $c$. (Note that $p_1|_c =
    \pp|_c^{(2)} = p_2|_c$.)

    In this proof, to emphasise the distinction between them, we use capital
    letters for variables and lower case for realisations.

    Let $p_i'(X,Y) := \tp(a_i,b_i)$ where $a_i |= p_i$, and let $q(Y_1,Y_2) :=
    \tp(b_1,b_2)$.
    Let $Q(X,Y_1,Y_2)$ be the incomplete type $p_1'(X,Y_1)\cup p_2'(X,Y_2)
    \cup q(Y_1,Y_2)$.

    Claim{bezoutQ}:
        If $a |= p_2$, then the completions of $Q(a,Y_1,b_2)$ in $S(ab_2)$ are
        precisely the non-forking extensions of $p_1'(a,Y_1)$. In particular,
        $Q$ is consistent.
        .
    Proof:
        It suffices to show that if $|= p_1'(a,b_1')$ then 
        $b_1' \ind _a b_2$ iff $b_1' \ind  b_2$. This is immediate from
        Lemma~\ref{relax}.
        .

    So let $a |= Q(X,b_1,b_2)$. 
    Let $(\tp(a^i/b_1b_2))_{i (- I}$ enumerate the completions of $Q(X,b_1,b_2) =
    p_1(X) \cup p_2(X)$. By completeness of $q$,
    $(\tp(a^ib_1/b_2))_{i (- I}$ enumerates the completions of $Q(X,Y_1,b_2)$.
    Say $a^ib_1 ==_{b_2} ab_1^i$,
    so $(\tp(ab_1^i/b_2))_{i (- I}$ also enumerates the completions of $Q(X,Y_1,b_2)$,
    so $(\tp(b_1^i/ab_2))_{i (- I}$ enumerates the completions of $Q(a,Y_1,b_2)$, and hence
    by Claim~\ref{bezoutQ} enumerates the non-forking extensions of
    $p_1'(a,Y_1) = \tp(b_1/a)$; in particular, by Lemma~\ref{finCb}(b), $I$ is finite.
    
    Claim:--
	**(a) $Z(a)=1$
	**(b) $\sum_i Z(b_1^i/ab_2) = Z(b_1/a) = Z(ab_1)$
	**(c) $Z(b_1^ib_2) = Z(b_1)Z(b_2)$
	*ee*
	.
    Proof:--
	**(a) By Lemma~\ref{Zprops}(iv) twice.
	**(b) By Claim~\ref{bezoutQ} and Lemma~\ref{Znf}(III), and (a).
        **(c) By $b_1^i ==_{b_2} b_1$ and Lemma~\ref{Znf}(I), we have
          $Z(b_1^i/b_2) = Z(b_1/b_2) = Z(b_1)$.
	*ee*
	.
    
    Now we calculate
    *\[a* | p_1 \cup p_2 |
	&= \sum_i \mult(a^i/b_1b_2) \\\
	&= \sum_i \mult(a/b_1^ib_2) \\\
	&= \sum_i Z(a/b_1^ib_2) \\\
        &= \sum_i \frac{ Z(ab_1^ib_2) }{ Z(b_1^ib_2) } \\\
        &= \sum_i \frac{ Z(b_1^i/ab_2) Z(ab_2) }{ Z(b_1^ib_2) } \\\
        &= \frac{ Z(ab_1) Z(ab_2) }{ Z(b_1)Z(b_2) } \\\
	&= Z(a/b_1) Z(a/b_2) .*a\]*
    .

Lemma:
    $D$ is 2-pseudolinear.
    .
Proof:
    Suppose $p$ is minimal complete, $p(x) |= x (- D^2$, $b := Cb(p)$, $U(b)>=3$.
    Let $b' |= \stp(b)|_{\acleq(b)}$.
    Say $b^\sg=b'$, $\sg (- \Aut(\MM)$, and let $p' := p^\sg$.
    So $p'$ is minimal and $\Cb(p') = b'$.
    Let $a |= p \cup p'$ (which is consistent by Lemma~\ref{bezout}).
    We have $U(a) = 2$ since $U(Cb(p)) = U(b) > 0$ and $a (- D^2$.
    By Lemma~\ref{relax}, $b \|/_a b'$.
    Note also that $U(b/a) >= 2 =< U(b'/a)$.
    Let $q := p|_{ba}$ and $q' := p'|_{b'a}$.
    Then by Lemma~\ref{bezout},
        \[ | q \cup q' | = Z(q)Z(q') = Z(p)Z(p') = | p \cup p' | ,\]
    which contradicts the fact that $a |= p \cup p'$ but $a |/= q \cup q'$.
    .

Applying Theorem~\ref{pseudolinLin}, this concludes the proof of Theorem~\ref{unimodLocmod}.


\bibliography{mmm-gst}
