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\providecommand{\stp}{\operatorname{stp}}
\providecommand{\tp}{\operatorname{tp}}
\providecommand{\dcl}{\operatorname{dcl}}
\providecommand{\acl}{\operatorname{acl}}
\providecommand{\cl}{\operatorname{cl}}
\providecommand{\dom}{\operatorname{dom}}
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\providecommand{\Aut}{\operatorname{Aut}}
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\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{}}
2016-2020.}\\
}
\date{\today\\\input{version.tex}}
\maketitle
\section{Introduction}
These are lecture notes from a course given in the M\"unster 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 Gonz\'ales, 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
Zhengqing He and Tom\'as Ibarluc\'ia 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}.
\subsection{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}.
\section{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 $\kappa$.
\begin{notation*} \mbox{}
\begin{itemize}\item $a,b,c,d,e,\alpha,\beta,\gamma$ etc denote elements of $\MMeq$,
$\atup,\btup,\ctup$ etc denote possibly long tuples from $\MMeq$,
and $A$,$B$,$C$ denote small subsets of $\MMeq$.
\item $AB$ means $A \cup B$;
$ab$ means $(a,b) \in \MMeq$;
when appropriate, $a$ means $\{a\}$; e.g.\ $Ab$ is short for $A\cup\{b\}$.
\item 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 $A$).
\item If $p$ is stationary and $A := \Cb(p)$, then $p\restriction_A$ has a unique
global non-forking extension $\pp \in S(\MM)$. For $A \subseteq B \subseteq \MMeq$,
we write $p\restriction_B$ for $\pp\restriction_B$. So $a \vDash p\restriction_B$ iff $a \vDash p$ and
$a \ind _A B$.
\item For $\alpha$ an ordinal, $p^{(\alpha)}$ is the type of a Morley sequence
in $p$ of length $\alpha$ (so $\atup b \vDash p^{(\alpha^{+})}$ iff $\atup \vDash
p^{(\alpha)}$ and $b \vDash p\restriction_{\atup}$).
%Now $\pp$ is definable over $A$,
%i.e.\ $T$ eliminates the Hrushovski quantifier ``$d_p x.$'' (which we can
% read as ``for generic $x \vDash p$,''),
%i.e.\ for any formula/$\emptyset $ $\phi(x,y)$, there is a formula/$A$
% \[ \psi(y) \equiv : d_p x. \phi(x,y) \]
% such that $\vDash \psi(b)$ iff for $a \vDash p$ with $a \ind _A b$,
% \[ \vDash \phi(a,b), \]
% i.e.\ iff $\phi(x,b) \in p\restriction_b$.
\item Define $\Cb(a/B) := \Cb(\stp(a/B))$ and
$\aCb(a/B) := \acleq(\Cb(a/B))$.
So $a \ind _C B \Leftrightarrow \aCb(a/BC) \subseteq \acleq(C)$.
\item 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 $\Cb(p) \in \MMeq$.
\item We use exponential notation $b^\sg$ for the action of an automorphism
(so automorphisms always act on the right).
\end{itemize}
\end{notation*}
\begin{lemma}
U-rank is additive for finite ranks:
$U(ab/C) = U(a/bC) + U(b/C)$ whenever all terms are finite.
\end{lemma}
\begin{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).
\end{proof}
\section{Minimal sets and the trichotomy}
\subsection{Minimal sets and pregeometries}
\begin{definition*}
A \underline{minimal set} is a $\Wedge$-definable set $D$, defined by a partial type
which has a unique unrealised global completion $\pp \in S(\MM)$. Call
$\pp$ the global \underline{generic type} of $D$.
A \underline{strongly minimal set} is a definable set which is minimal.
\end{definition*}
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
$\emptyset $.
\begin{exercise*} \mbox{}
\begin{itemize}\item If $a\in D$ and $C \subseteq \MMeq$, then $a \vDash \pp\restriction_C$ iff $a \notin \acleq(C)$.
\item A $\Wedge$-definable set is minimal iff every relatively definable subset
is finite or cofinite.
\item A complete type is minimal iff it is stationary and of U-rank 1.
\end{itemize}
\end{exercise*}
\begin{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 $^*\Z \vDash \Th(\Z;+)$.
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}.
\end{remark*}
\begin{definition*}
A \underline{pregeometry} is a set $X$ with a map $\cl : \Pset(X) \rightarrow \Pset(X)$ such that
\begin{itemize}\item $\cl$ is a closure operator: $A \subseteq \cl(A)$, $A \subseteq B \Rightarrow \cl(A) \subseteq
\cl(B)$, and $\cl(\cl(A)) = \cl(A)$;
\item Exchange: $b \in \cl(Ac) \setminus \cl(A) \Rightarrow c \in \cl(Ab)$.
\item Finite character: $\cl(A) = \bigcup_{A_0 \subseteq _{\operatorname{fin}} A} \cl(A_0)$;
\end{itemize}
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 \underline{closed} subsets $\CC \subseteq \Pset(X)$ such that
\begin{itemize}\item $\CC$ is closed under intersections: if $\BB \subseteq \CC$, then
$\bigcap \BB \in \CC$.
\item Exchange: If $C \in \CC$ and $a \in X \setminus C$, there is an immediate closed
extension $C'$ of $C$ containing $a$,
i.e.\ there exists $C' \in \CC$ such that $C' \supseteq Ca$ and there does not
exist $C'' \in \CC$
with $C \subsetneq C'' \subsetneq C'$.
\item Finite character: if $\BB \subseteq \CC$, then
$\Vee \BB = \bigcup_{\BB_0 \subseteq _{\operatorname{fin}} \BB} \left({\Vee \BB_0}\right)$,
where $\Vee \BB := \bigcap \{ C \in \CC \;|\; \bigcup \BB \subseteq C \}$
is the smallest closed set containing all $B \in \BB$,
\end{itemize}
So setting $\Wedge \BB := \bigcap \BB$ and $A \leq B \Leftrightarrow A \subseteq B$, $\CC$
forms a complete lattice.
We can pass between the two definitions as follows: given a closure
operator $\cl$, define $\CC := \operatorname{im}(\cl)$; conversely, given a set $\CC$
of closed subsets, define $\cl(A):=\bigcap \{C \in \CC \;|\; A \subseteq C\}$.
For $A,B \subseteq X$, $\dim(A/B)$ is the cardinality of a \underline{basis} for $A$ over
$B$, a minimal subset $A'\subseteq A$ 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 $\cl(B) = C_0 \lneq C_1 \lneq \ldots \lneq
C_r = \cl(AB)$.
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 \underline{homogeneous} if for any closed $C \subseteq X$ and any $a,b \in X \setminus C$,
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 \underline{geometry} is a pregeometry $(X,\cl)$ such that $\cl(\emptyset )=\emptyset $ and
$\cl(a)=\{a\}$ for any $a\in X$. For a pregeometry $(X,\cl)$, the
\underline{associated geometry} is the set of dimension 1 closed sets of $X$ with
the obvious closure, $\cl'$ defined by
$A \in \cl'(B) \Leftrightarrow A \subseteq \cl(\bigcup B)$.
\end{definition*}
\begin{remark*}
Finite pregeometries are also known as \underline{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)
\end{remark*}
\begin{lemma}
$(D,\acl_D)$ is a homogeneous pregeometry, where $\acl_D(A) := \acleq(A)
\cap D$.
\end{lemma}
\begin{proof}
Everything is immediate from the definition of $\acl_D$ except homogeneity
and exchange. Exchange
corresponds to symmetry of forking: for $b,c \in D$ and $A \subseteq D$,
\begin{align*} b \in \acl_D(Ac) \setminus \acl_D(A)
& \Leftrightarrow b \vDash \pp\restriction_A \infix{and} b \not\vDash \pp\restriction_{Ac} \\
& \Leftrightarrow b \nind _{A} c \\
& \Leftrightarrow c \nind _{A} b \\
& \Leftrightarrow c \vDash \pp\restriction_A \infix{and} c \not\vDash \pp\restriction_{Ab} \\
& \Leftrightarrow c \in \acl_D(Ab) \setminus \acl_D(A) .\end{align*}
For homogeneity, if $C \subseteq D$ is $\acl_D$-closed, and $a_0,b_0 \in D \setminus C$,
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, $a_i \mapsto b_i$
can be extended to an elementary map $D \twoheadrightarrow D$ 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.)
\end{proof}
\begin{remark*}
For $a \in D^{<\omega}$ and $B \subseteq D$,
$U(a/B) = \dim(a/B)$.
Indeed, this is immediate for singletons, and then follows for tuples by
additivity.
\end{remark*}
\begin{remark*}
Adding $A \subseteq D$ to the language corresponds to localising the pregeometry
at $A$.
\end{remark*}
\subsection{Triviality and modularity}
\begin{definition*}
A pregeometry is \underline{trivial} (or \underline{degenerate}) if
$\cl(A) = \bigcup_{a \in A} \cl(a)$; equivalently, $\Vee = \bigcup$ in
the lattice of closed sets.
\end{definition*}
\begin{example*} \mbox{}
\begin{itemize}\item The theory of equality on an infinite set (i.e.\ with trivial
language), is strongly minimal with trivial pregeometry; all subsets are
closed, $\acl_D(B)=B$.
\item An action of a group $G$ on an infinite set $D$ without fixed points,
in the language $(D; ((g*))_{g \in G})$, is strongly minimal with trivial
pregeometry; closed sets are unions of $G$-orbits; $\acl_D(B) = GB$.
\end{itemize}
\end{example*}
\begin{definition*}
A pregeometry $(X,\cl)$ is \underline{modular} if for any finite-dimensional closed
$A,B \leq X$,
\begin{equation} \label{eq:mod}
\dim(A \vee B) = \dim(A) + \dim(B) - \dim(A \wedge B) ,\end{equation}
i.e.\ $\dim(A/B) = \dim(A / A \wedge B)$.
\end{definition*}
\begin{remark*}
The pregeometry of a minimal set $D$ is modular iff $A \ind _{A\cap B} B$
for any $\acl_D$-closed $A,B \subseteq D$.
\end{remark*}
\begin{lemma} \label{mod2suff}
Suppose $(X,\cl)$ is such that (\ref{eq:mod}) holds when $\dim(A)=2$. Then
$(X,\cl)$ is modular.
\end{lemma}
\begin{proof}
Suppose (\ref{eq:mod}) holds when $\dim(A)=2$.
We first show that (\ref{eq:mod}) holds when $\dim(A/A \wedge B)=2$.
Say $a_1,a_2$ is a basis for $A$ over $A \wedge B$; then setting $A_0 :=
\cl(a_1,a_2)$, $A_0 \wedge B = A_0 \wedge (A \wedge B) = \cl(\emptyset )$, and $A_0
\vee B \supseteq A_0 \vee (A \wedge B) = A$ so $A_0 \vee B = A \vee B$. So
indeed, $\dim(A/B) = \dim(A \vee B / B) = \dim(A_0 \vee B / B) = \dim(A_0 / A_0 \wedge B) =
\dim(A_0) = \dim(A / A \wedge B)$.
Now let $A$ and $B$ be arbitrary, and say $n=\dim(A/B) = \dim(A \vee B) - \dim(B)$.
Decompose $B \leq A \vee B$ as an immediate chain of closed subsets
\[ B = B_0 \lneq B_1 \lneq \ldots \lneq B_n = A \vee B .\]
Let $A_i := A \wedge B_i$.
Clearly $\dim(A/A \wedge B) \geq \dim(A/B) = n$,
so it suffices to see that $\dim(A_{i+1} / A_i) \leq 1$.
Else, we have $A_i \leq A' \leq A_{i+1}$ with $\dim(A'/A_i)=2$ and $A'$ closed;
but then $A' \wedge B_i = A_i$ and $A' \vee B_i \subseteq B_{i+1}$,
contradicting the modularity established above.
\end{proof}
\begin{exercise*}
A lattice is called \underline{modular} if for any $A,B,C$ with $A \leq C$,
\[ A \vee (B \wedge C) = (A \vee B) \wedge C .\]
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?
\end{exercise*}
\begin{example}[Projective Geometries] \label{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 $(V; +, (k*)_{k \in K})$. 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.
\end{example}
\begin{fact} \label{nonTrivMod}
Any non-trivial modular homogeneous geometry of dimension $\geq 4$ is
isomorphic to a projective geometry over a division ring.
\end{fact}
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
\begin{enumerate}[(i)]\item any pair of points $a\neq b$ lies on a unique common line $(a,b)$,
\item any pair of lines intersect in at most one point,
\item any line has at least three points,
\item 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),
\item Not all points are on a common plane, where a {\em 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 $a_0$,
\end{enumerate}
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 $a_i \notin \cl(A)$; 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.
\subsection{1-basedness and local modularity}
\begin{remark*}
Localisations of modular pregeometries are also modular. However, the
converse is false, as the following example shows.
\end{remark*}
\begin{example}[Affine Geometry] \label{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
$\operatorname{char}(K) \neq 2$, 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}.
\end{example}
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.
\begin{definition*}
$\Deq := \dcleq(D) \subseteq \MMeq$.
\end{definition*}
\begin{definition*}
$D$ is \defn{1-based} if for any $a \in \Deq$ and $B \subseteq \Deq$,
$\Cb(a/B) \subseteq \acleq(a)$.
\end{definition*}
\begin{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 $\atup \subseteq \Deq$ is a Morley sequence in $\tp(a/B)$.
\end{remark*}
\begin{lemma} \label{1basedDisj}
$D$ is 1-based iff for any $\acleq$-closed subsets $A,B \subseteq \Deq$,
\[ A \ind _{A\cap B} B .\]
\end{lemma}
\begin{proof}
Immediate from finite character of forking and the properties of canonical
bases.
\end{proof}
\begin{lemma} \label{1basedParams}
1-basedness is invariant under adding parameters to the language.
\end{lemma}
\begin{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 $A,B \subseteq \Deq$ be $\acleq$ closed, and let $I := A \cap B$; we must
show $A \ind _I B$.
By taking a realisation of $\tp(AB)$ independent from $C$, we may
assume $AB \ind C$.
\begin{claim}
$I' := \acleq(AC)\cap \acleq(BC) = \acleq(IC)$
\end{claim}
\begin{proof}
$AC \ind _A AB$ and $BC \ind _B AB$,
so by considering $\Cb(I'/AB)$,
we have $I' \ind _I AB$.
So since $I \subseteq IC \subseteq I'$,
also $I' \ind _{IC} ABC$.
But $I' \subseteq \acleq(ABC)$, so $I' = \acleq(IC)$.
\end{proof}
Applying Lemma~\ref{1basedDisj} in the expanded language,
we deduce $A \ind _{IC} B$.
Meanwhile $AB \ind C$ and $I \subseteq AB$, so $AB \ind _I IC$, and so $A \ind _I IC$.
So by transitivity, $A \ind _I B$.
\end{proof}
\begin{definition*}
A $\Wedge$-definable set $X$ has \underline{elimination of imaginaries} ( \underline{EI} ),
resp.\ \underline{geometric EI} (gEI),
resp.\ \underline{weak EI} (wEI),
if for any $c \in X^{\eq}$, there exists $b \in X^{<\omega}$ such that
$\dcleq(b) = \dcleq(c)$,
resp.\ $\acleq(b) = \acleq(c)$,
resp.\ $\dcleq(c) \subseteq \dcleq(b) \subseteq \acleq(c)$.
\end{definition*}
\begin{remark*}
EI $\Rightarrow $ wEI $\Rightarrow $ gEI.
\end{remark*}
%Alternative phrasing:
%A $\Wedge$-definable set $X$ has EI if for any $\dcleq$-closed set $C \subseteq X^{\eq}$,
%$C = \dcleq(C \cap D)$.
%geometric EI (gEI): $C = \acleq(C \cap D)$ for $C=\acleq(C)$.
%weak EI (wEI): $C \subseteq \dcleq(\acleq(C) \cap D)$ for $C=\dcleq(C)$.
\begin{lemma} \label{mod1based}
Suppose $D$ has gEI. Then $D$ is modular iff $D$ is 1-based.
\end{lemma}
\begin{proof}
By gEI and finite character of $\acleq$, for $A=\acleq(A) \subseteq \Deq$
we have $A = \acleq(A \cap D)$.
So for $A,B \subseteq \Deq$ $\acleq$-closed,
\[ A \ind _{A\cap B} B \,\Leftrightarrow \, (A\cap D) \ind _{A\cap B \cap D} (B\cap D) .\]
We conclude on noting that for $A \subseteq D$, $A=\acl_D(A) \Rightarrow A = \acleq(A) \cap D$.
\end{proof}
\begin{lemma} \label{minwEI}
Let $\atup \vDash \pp\restriction_\emptyset ^{(\omega)}$.
Then after expanding the language by parameters for $\atup$, $D$ has wEI.
\end{lemma}
\begin{proof}
Working in the original language,
we will prove wEI in the expanded language by showing that if $c \in
\dcleq(D,\atup) = \Deq$, then there exists $b \in D^{<\omega}$ such that
$c \in \dcleq(b\atup)$ and $b \in \acleq(c\atup)$.
So say $c=f(b')$ with $b' \in D^n$ and $f$ a partial function defined over
$\emptyset $. It suffices to show that $f^{-1}(c) \cap \acleq(c\atup) \neq \emptyset $.
In fact we will prove by induction on $n$ the stronger statement that if
$\emptyset \neq X \subseteq D^n$ is $\Wedge$-definable over $c \in \Deq$, then $X \cap
\acleq(c\atup) \neq \emptyset $.
First suppose $n=1$. Then either $X$ is finite, in which case any point is
algebraic over $c$, or else $\pp\restriction_c(x) \vDash x \in X$, in which case we are
done as $a_i \vDash \pp\restriction_c$ for some $i$, since $c \in \Deq$.
(Indeed, say $U(c)=k$, and suppose $U(a_i/c)=0$ for $i=0,\ldots ,k$.
Then $U(c/a_{*1$, consider a co-ordinate projection $\pi : X \rightarrow D$. By
compactness and saturation, $\pi(X)$ is also $\Wedge$-definable over $c$. By
the $n=1$ case, say $b \in \pi(X) \cap \acleq(c)$; 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 $c$.
\end{proof}
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.
\begin{definition*}
$D$ is \underline{linear} if whenever $a,b \in D$ and $C \subseteq \Deq$ and
$U(ab/C)=1$, then $U(\Cb(ab/C)) \leq 1$.
(We can think of this as $D$ having ``no 2-dimensional family of plane
curves''.)
\end{definition*}
\begin{example} \label{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 $U(cd)=2$.
\end{example}
\begin{definition*}
A pregeometry $(X,\cl)$ is \underline{locally modular} if the localisation at any
$e \in X \setminus \cl(\emptyset )$ is modular.
\end{definition*}
\begin{theorem} \label{1basedLinear}
The following are equivalent:
\begin{enumerate}[(i)]\item $D$ is 1-based;
\item $D$ is linear;
\item $D$ is locally modular.
\end{enumerate}
\end{theorem}
\begin{proof} \mbox{}
\begin{itemize}\item[$(i) \Rightarrow (ii)$]:
Let $a,b,C$ be as in the definition of linearity.
By 1-basedness, $d := \Cb(ab/C) \subseteq \acleq(ab)$.
So $U(d) = U(abd) - U(ab/d) \leq 2 - 1 = 1$.
\item[$(ii) \Rightarrow (iii)$]:
Let $e \in D \setminus \acl_D(\emptyset )$.
We prove modularity of the localisation to $e$ via
Lemma~\ref{mod2suff}. So suppose $(a,b) \vDash \pp\restriction_e^{(2)}$
and $e \in C = \acl_D(C) \subseteq D$, and let $I := \acl_D(abe) \cap C$; we
must show $ab \ind _I C$. This is clear if $U(ab/C)=0$ or $U(ab/C)=2$,
so suppose $U(ab/C)=1$. So we must show
\[ \acl_D(e) \lneq I .\]
Let $d := \Cb(ab/C)$.
By linearity, $U(d) \leq 1$.
Since $ab \nind C$, 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 $d \in \acleq(ab)$. So $\acleq(ed) \cap D \subseteq I$. So it suffices to
show
\begin{equation*}
\tag{*} \acl_D(e) \lneq \acleq(ed) \cap D .
\end{equation*}
If $a \nind d$ then $a$ witnesses (*).
Else, $\tp(a/d) = \tp(e/d) = \pp\restriction_d$, since $e \ind d$ as $d \in
\acleq(ab)$.
So say $ab \equiv _d ef$. Then $f$ witnesses (*).
\item[$(iii) \Rightarrow (i)$]:
Add parameters for an infinite Morley sequence in $\pp\restriction_\emptyset $.
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.
\end{itemize}
\end{proof}
\subsection{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:
\begin{theorem}
If $D$ is a minimal set in a stable theory,
then precisely one of the following holds:
\begin{itemize}\item The pregeometry of $D$ is trivial.
\item 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.
\item 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).
\end{itemize}
\end{theorem}
\begin{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 $a \neq a'$ and $b \neq b'$. If $D$ is strongly minimal, by compactness
and elimination of $\exists^\infty$ one can find a formula $I(x,y) \in r$
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 $U(b_2) \geq 2$ 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.
\end{historicalremark*}
\begin{fact}
If $D$ is locally modular but not modular, its geometry is affine geometry
over a division ring (\cite[Proposition~5.2.4]{PlGStab}).
\end{fact}
\subsection{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.
\begin{lemma} \label{finCb}
Let $C \subseteq \MMeq$ and $a \in \Deq$, and let $q := \tp(a/C)$.
\begin{enumerate}[(a)]\item If $q$ is stationary, then $\Cb(q)$ is ($\dcleq$ of) an element of
$\Deq$.
\item $q$ has finitely many global non-forking extensions.
\end{enumerate}
\end{lemma}
\begin{proof} \mbox{}
We first prove (a) and (b) under the assumption that $a \in D^n$.
\begin{enumerate}[(a)]\item
If $a=(a_1,\ldots ,a_n)$, after an appropriate co-ordinate permutation
say $\atup := (a_1,\ldots ,a_k) \vDash \pp\restriction_C^{(k)}$ and $\btup :=
(a_{k+1},\ldots ,a_n) \in \acleq(C,\atup)$.
Say $\phi_c(\atup,y) \in \tp(\btup/C\atup)$ with $c \in \dcleq(C)$ and
$|\phi_c(\atup,\MM)|$ minimal,
so $\phi_c(\atup,y)$ isolates $\tp(\btup/C\atup)$.
Then clearly $q(x,y)$ is equivalent to
$\pp\restriction_C^{(k)}(x) \cup \{ \phi_c(x,y) \}$,
%indeed,
%if $\atup' \vDash \pp\restriction_C^{(k)}$ and $\vDash \phi_c(\atup',\btup')$,
%then say $\atup'\btup' \equiv _C \atup\btup''$;
%then $\vDash \phi_c(\atup,\btup'')$, so $(\atup,\btup'') \vDash q$,
%so $(\atup',\btup') \vDash q$.
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 $\Cb(q) \in \dcleq(c)$.
Since also $\Cb(q)$ is in $\dcleq$ of a Morley sequence in $q$
(by \cite[Proposition~4.19]{Palacin-MMMStability}),
we conclude $\Cb(q) \in \Deq$,
as required\footnote{Thanks to the anonymous reviewer for a suggestion which
led to an optimisation of this part of the proof.}.
\item
If $\qq$ is a global non-forking extension, then $\Cb(\qq) \in
\acleq(C)$, 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 $C$.
\end{enumerate}
Now for arbitrary $a \in \Deq$, say $a=f(b)$ where $b \in D^n$ and $f$ is
$\emptyset $-definable, (b) holds since it holds for $\tp(b/C)$.
It remains to deduce (a). So suppose $q$ is stationary.
\begin{claim}[{\cite[Lemma~1.3.19, first part]{PlGStab}}]
Let $q$ be a stationary type in a stable theory with $U(q\restriction_\emptyset ) < \infty$.
Then there is $e \in \Cb(q)$ with $\Cb(q) \subseteq \acleq(e)$.
\end{claim}
\begin{proof}
Suppose there is no $e \in \Cb(q)$ over which $q$ does not fork. We
build an infinite sequence $\emptyset = e_0, e_1, \ldots \in \Cb(q)$ such that
$\dcleq(e_i) \subseteq \dcleq(e_{i+1})$ and $q\restriction_{e_{i+1}}$ forks over $e_i$,
contradicting U-rankedness of $q\restriction_\emptyset $. Indeed, given $e_i \in \Cb(q)$,
$q\restriction_{\Cb(q)}$ forks over $e_i$; then since forking is witnessed by a
formula, already $q\restriction_{e_{i+1}}$ forks over $e_i$ for some $e_{i+1} \in
\Cb(q)$, and we may assume $e_i \in \dcleq(e_{i+1})$.
\end{proof}
Now $U(q\restriction_\emptyset ) = U(a) \leq U(b) \leq n$, 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.
\end{proof}
\section{Unimodularity}
\subsection{Preliminaries}
Throughout this section, $D$ is a minimal set over $\emptyset $.
\begin{definition*}
If $a \in \acleq(B)$,
then $\mult(a/B) = \left|{\{ a' \in \MMeq \;|\; a' \equiv _B a \}}\right|$.
%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 $|\Gal(b):\Gal(ab)|$.
\end{definition*}
\begin{remark*}
Multiplicity is multiplicative: $\mult(ab/C) = \mult(a/bC)\mult(b/C)$.
\end{remark*}
\begin{lemma} \label{multStat}
If $\tp(a/c)$ is stationary and $b \in \acleq(c)$, then
\[ \mult(b/ac) = \mult(b/c) .\]
\end{lemma}
\begin{proof}
Suppose $\sg \in \Aut(\MMeq / c)$.
By stationarity, $\tp(a/c) \vDash \tp(a / cb)$.
So $ab \equiv _c a^{\sg^{-1}}b \equiv _c ab^\sg$,
so $b \equiv _{ac} b^\sg$.
\end{proof}
\subsection{Unimodularity}
\begin{definition*}
$D$ is \underline{unimodular} if whenever $a_i \vDash \pp\restriction_\emptyset ^{(n)}$, $i=1,2$, and
$\acl_D(a_1)=\acl_D(a_2)$, then $\mult(a_1 / a_2) = \mult(a_2 / a_1)$.
\end{definition*}
\begin{example*}
An algebraically closed field is \underline{not} unimodular: consider $a$ and $a^2$
where $a$ is generic.
\end{example*}
In Section~\ref{unimodProof}, we will show
\begin{theorem}[Hrushovski] \label{unimodLocmod}
If $D$ is unimodular, then $D$ is 1-based.
\end{theorem}
$\omega$-categorical theories provide the motivating example of unimodular
minimal sets:
\begin{lemma}
If $D$ is a minimal set in an $\omega$-categorical theory, then $D$ is
unimodular.
\end{lemma}
\begin{proof}
By Ryll-Nardzewski, $\acl_D$ is locally finite, i.e.\ $\acl_D(A)$ is
finite for $A$ finite. Let $a_i \vDash \pp\restriction_\emptyset ^{(n)}$, $i=1,2$, be interalgebraic,
and let $X := \acl_D(a_i)$.
For $c \in X^k$, let $\mult_X(c) := \left|{\{ c'\in X \;|\; c' \equiv c \}}\right|$.
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 $a_1 \equiv a_2$.
So $\mult(a_1/a_2) = \mult(a_2/a_1)$.
\end{proof}
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 $x \in \cl(y_1,\ldots ,y_n)$ 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.
\begin{lemma} \label{pfUni}
If $D$ is a strongly minimal set definable in a pseudofinite theory $T$,
then $D$ is unimodular.
\end{lemma}
\begin{proof}
\providecommand{\U}{\mathcal{U}}
\providecommand{\dM}{\operatorname{dM}}
\providecommand{\UPi}{\Pi_{\U}}
$T$ is pseudofinite so has a model which is an ultraproduct of finite
structures, say $M = \UPi M_i \vDash T$ with $M_i$ finite. For $X$
$\emptyset $-definable, let $|X| := \UPi \;|\; X^{M_i} \;|\; \in \mathbb{N}^\U$.
\begin{claim}
Let $X \subseteq D^n$ be $\emptyset $-definable.
Then there exists a unique polynomial $p_X(q) \in \Z[q]$ of degree
$\RM(X)$ such that
\[ |X| = p_X(|D|) \]
\end{claim}
\begin{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 $\RM(X) = d \geq 0$.
We can find $k \in \N$ and $\emptyset $-definable $X' \subseteq X$ with $\RM(X')=d$, and
a co-ordinate projection map $\pi : X' \rightarrow D^d$ 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 $X \setminus X'$ and $D^d \setminus \pi(X')$.
Then
\[ p_X(q) := p_{X \setminus X'}(q) + k(q^d - p_{D^d \setminus \pi(X')}(q)) \]
is as required.
\end{proof}
Now let $a_i \in D^n$, $i=1,2$, be interalgebraic generics.
Let $X$ have minimal Morley degree among $\emptyset $-definable sets with
$(a_1,a_2) \in X$ and $\RM(X)=n$.
Let $k_i := \mult(a_1a_2/a_i)$.
Then for $i=1,2$,
there exists $\emptyset $-definable $X'_i \subseteq X$ with $\RM(X'_i) = n$ and a
projection $\pi_i : X'_i \rightarrow D^n$ with fibres of size $k_i$ and
$\RM(\pi_i(X'))=n$. Then $\RM(X \setminus X'_i) < n$ by the minimality assumption
on $\dM(X)$, and $\RM(D^n \setminus \pi(X'_i)) < n$, so the corresponding
polynomials have degree $}[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 $b_i' \ind b$ by choice of $b_i'$, since $b_1 \ind b_2$.
Also $b_1' \ind _{b} b_2'$, since $c \ind _b c'$, since $c \ind c'$ and $b \ind cc'$.
Hence $b_1' \ind bb_2'$, so $b_1' \ind b_2'$ and $b \ind b_1'b_2'$.
So $(b_1',b_2') \vDash r^{(2)}$,
so say $\widetilde{f}_{b_1'}^{-1} \circ \widetilde{f}_{b_2'} = \widetilde{g}_{c''}$ with $c'' \vDash s\restriction_{b_i'}$.
Then since $b \ind b_1'b_2'$, we have $b \ind c''b_1'$, hence $c'' \ind
b_1'b$, and so $c'' \ind c$.
Similarly $c'' \ind c'$.
Finally, we must check that $\widetilde{g}_s$ is generically transitive.
So let $x \vDash p$, let $b_2' \vDash r\restriction_x$, and let $b_1' \vDash r\restriction_{xb_2'}$.
Let $y := \widetilde{f}_{b_2'}(x)$ and $z := \widetilde{f}_{b_1'}^{-1}(y)$.
Then $y \ind x$ by generic transitivity, and $b_1' \ind yx$,
so $y \ind xb_1'$, i.e.\ $y \vDash q\restriction_{xb_1'}$, and so $z \vDash p\restriction_{xb_1'}$, and in
particular $x \ind z$.
Since $b_1'b_2' \equiv b_1b_2$, this proves generic transitivity of $\widetilde{g}_s$.
\end{proof}
\begin{exercise*}
Assuming finite U-rank, (\ref{isomHWIndie}) holds iff $U(s) = U(r)$.
\end{exercise*}
%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 $ac \ind _D \acleq(b)$,
% so $\acleq(b) \ind _D \acleq(b)$,
% so $\acleq(b) = D$.
%
% So $(b,x,y)$ is "nearly" a triple corresponding to a canonical family of
% invertible germs.
% .
\section{The Group Configuration Theorem}
We continue to work in a monster model $\MMeq$ of an arbitrary stable theory
$T$.
\begin{definition*} \mbox{}
\[ \xymatrix{
c \ar@{-}'[drrr][ddrrrrrr] \ar@{-}'[ddrr][dddrrr] \\
& & &b \\
& &z& & & &a \\
& & &x \\
y \ar@{-}'[urrr][uurrrrrr] \ar@{-}'[uurr][uuurrr] \\
} \]
% \[ \xymatrix{
%::: & & a & & \setminus
%::: & & & & \setminus
%::: & & & & \setminus
%::: & b & & x & \setminus
%::: & & z \ar@{-}'[uuuuu] & & \setminus
%::: & & & & \setminus
%::: c \ar@{-}'[rruu][rrruuu] \ar@{-}'[ruuu][rruuuuuu] & & & & y \ar@{-}'[lluu][llluuu] \ar@{-}'[luuu][lluuuuuu]
% } \]
%\begin{minipage}{\linewidth}
%Verbatim:
% a
% / \
% b x
% / `z' \
% / .' `. \
% /.' `.\
% c' `y
% .
%\end{minipage}
$(a,b,c,x,y,z)$ forms a \underline{group configuration} if
\begin{itemize}\item any non-collinear triple in the above diagram is independent,
\item $\acleq(ab)=\acleq(bc)=\acleq(ac)$,
\item $\acleq(ax)=\acleq(ay)$ 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 $czx$.
\end{itemize}
\end{definition*}
\begin{remark*}
Suppose $(G,S)$ is a connected faithful $\Wedge$-definable/$\emptyset $ 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 $(G,S)$}.
\end{remark*}
\begin{theorem}[Group Configuration Theorem] \label{groupConf}
Suppose $(a,b,c,x,y,z)$ forms a group configuration.
Then, after possibly expanding the language by parameters $B$ with $B \ind
abcxyz$,
there is a connected faithful $\Wedge$-definable/$\emptyset $ 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.
\end{theorem}
\begin{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
$f : G' \leftrightarrow G$, $g_1 : S'_1 \leftrightarrow S$ and $g_2 : S'_2 \leftrightarrow S$,
such that $(b,z,y)$ is a generic point of the image under $(f,g_1,g_2)$ of
the graph $\Gamma_* \subseteq G \times S \times S$ of the action.
(c.f.\ \cite[6.2]{HZZGeoms}.)
\end{example*}
\begin{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}
\end{remark*}
\begin{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:\
\begin{itemize}\item add independent parameters to the language - we refer to this as
``base-changing'' to the parameters;
\item 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
$\acleq(b)=\acleq(b')$.
\end{itemize}
Note that these operations transform a group configuration into a group
configuration.
By base-changing whenever necessary, we will assume throughout that
$\acleq(\emptyset )=\dcleq(\emptyset )$, so types over $\emptyset $ are stationary.
The proof will comprise three steps:
\begin{enumerate}[(I)]\item ``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};
\item prove this family satisfies the independence assumption of Lemma~\ref{isomHW};
\item connect the resulting homogeneous space to the group configuration.
\end{enumerate}
And so it begins.
\begin{enumerate}[(I)]\item First, a claim.
\begin{claim}
If $(a,b,c,x,y,z)$ is a group
configuration and if we let $\widetilde{z} \in \MMeq$ be the set $\widetilde{z} =
\{z_1,\ldots ,z_d\}$ of conjugates $z_i$ of $z$ over $ybxc$, then $\widetilde{z}$ is
interalgebraic with $z$.
\end{claim}
(Here, $\{z_1,\ldots ,z_d\} := (z_1,\ldots ,z_d)/S_d$
is the quotient of $(z_1,\ldots ,z_d)$ under the action by permutations of
the symmetric group $S_d$.)
\begin{proof}
It suffices that the conjugates be interalgebraic, $\acleq(z_i)=\acleq(z_j)$.
Indeed, then $\acleq(\widetilde{z}) \subseteq \acleq(z_1,\ldots ,z_n) = \acleq(z)$;
and $z \in \acleq(\widetilde{z})$, since it satisfies the algebraic formula $z\in
\widetilde{z}$.
But indeed:
$c\ind _z b$, so $cx \ind _z by$,
so setting $B:= \acleq(cx) \cap \acleq(by)$, we have
$B \ind _z B$ so $B \subseteq \acleq(z)$.
But $z \in B$.
So $\acleq(z) = B$,
and so $\acleq(z_i) = B$ for each $z_i$.
So $\acleq(z_i)=\acleq(z)$.
\end{proof}
Now let $a' \vDash \tp(a)\restriction_{abcxyz}$.
Say $a'x'c' \equiv _{ybz} axc$. So $(a',b,c',x',y,z)$ is also a group configuration.
So by the Claim, the set $\widetilde{z}'$ of conjugates of $z$ over $ybx'c'$ is
interalgebraic with $z$.
Note that $\widetilde{z}' \in \dcleq(ybx'c')$.
So base-change to $a'$,
and interalgebraically replace $y$ with $yx'$, $b$ with $bc'$, and $z$ with $\widetilde{z}'$.
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 $\widetilde{x}$,
we can also ensure that
\[ x \in \dcleq(ay) .\]
Repeat once more: base-change to an independent copy $c'$ of $c$,
let $a'x'c' \equiv _{ybz} axc$,
let $\widetilde{y}$ be the set of conjugates of $y$ over $ba'zx'$.
Now since $x' \in \dcleq(a'y)$ and $z \in \dcleq(by)$,
we have $zx' \in \dcleq(ba'y)$ and so (e.g.\ by considering
automorphisms) $zx' \in \dcleq(ba'\widetilde{y})$.
So after interalgebraically replacing $b$ with $ba'$, $z$ with $zx'$, and $y$ with $\widetilde{y}$,
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.
\item
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 $\widetilde{f}_r$ of invertible germs $p \rightarrow q$.
We aim to apply Lemma~\ref{isomHW} to obtain a group,
so we must show that for $b' \vDash r\restriction_b$,
if $\dcleq(d) = \code{ \widetilde{f}_{b'}^{-1} \circ \widetilde{f}_{b} }$,
we have $d \ind b$ and $d \ind b'$.
This holds for all $b' \vDash r\restriction_b$ or none, so we may assume
$b' \ind abcxyz$. Then $b' \equiv _{xcz} b$, so say $b'y'a' \equiv _{xcz} bya$.
\begin{claim}
$y' \ind bb'$.
\end{claim}
\begin{proof}
$b' \ind bz$ and so $b' \ind _z b$,
and $b \ind z$, so $b \ind b'z$, hence $z \ind _{b'} b$.
Now $y' \in \acleq(zb')$, so $y' \ind _{b'} b$.
But $y' \ind b'$, so $y' \ind bb'$.
\end{proof}
We also have $y \ind bb'$ and $f_b'^{-1}(f_b(y)) = y'$,
so by Lemma~\ref{canonFamCB},
$\dcleq(d) = \code{\widetilde{f}_b'^{-1} \circ \widetilde{f}_{b} } = \Cb(yy'/bb')$.
Now $y \ind abc$, and $b' \ind yabc$, so $y \ind abcb'$, and since $a' \in
\acleq(cb')$, we have $y \ind aa'bb'$.
Since also $y' \in \acleq(yaa')$, we have
\[ yy' \ind _{aa'} bb' .\]
Similarly,
\[ yy' \ind _{bb'} aa' .\]
So $\aCb(yy'/bb') = \aCb(yy'/aa'bb') = \aCb(yy'/aa')$,
so $d \in \acleq(aa')$.
\begin{claim}
$b \ind aa'$ and $b' \ind aa'$.
\end{claim}
\begin{proof}
$b \ind _c b'$, and then since $a\in\acleq(cb)$ and
$a'\in\acleq(cb')$, we have $ab \ind _c a'b'$,
So $a' \ind _c ab$;
but $a' \ind c$, so $a' \ind ab$, so $b \ind _a a'$,
and then since $b \ind a$, we have $b \ind aa'$.
Similarly, it follows from $a \ind _c a'b'$ that $b' \ind aa'$.
\end{proof}
So $b \ind d$ and $b' \ind d$, as required.
\item
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 $b' \vDash \tp(b)\restriction_{abcxyz}$.
Say $y'b' \equiv _z yb$.
Say $\dcleq(g) = \code{\widetilde{f}_{b'}^{-1} \circ \widetilde{f}_b}$.
By the construction of $(G,S)$,
we have $g \vDash s_G$,
and $y,y' \vDash p$,
and $y' = g*y$.
Base-change to $b'$,
and interalgebraically replace $b$ with $g \vDash s_G$ and $z$ with $g*y \vDash p$,
the latter interalgebraicity holding as $g*y = \widetilde{f}_{b'}^{-1}(\widetilde{f}_b(y)) = \widetilde{f}_{b'}^{-1}(z)$.
Now let $c'' \vDash \tp(c)\restriction_{abcxyz}$,
and say $b''z''c'' \equiv _{axy} bzc$.
Base-change to $c''$,
and interalgebraically replace $a$ by $b'' \vDash s_G$ and $x$ by $z'' = b''*y \vDash p$.
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.
\end{enumerate}
\end{proof}
\begin{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 $b'c' \vDash
\stp(bc/a)\restriction_{abc}$. Then $U(bcb'c')=3$, so by modularity
$U(\acl(bc')\cap\acl(b'c))=1$, say $d\in \acl(bc')\cap\acl(b'c) \setminus
\acl(\emptyset )$. Then $(a,b,c,b',c',d)$ is a group configuration.
\end{example*}
\begin{fact} \label{minActions}
If $S$ is strongly minimal, any connected faithful $\emptyset $-definable
homogeneous space $(G,S)$ is of one of the following forms:
\begin{itemize}\item $U(G)=1$, $G$ is commutative, and the action is regular;
\item $U(G)=2$, $S=K$ is the universe of a $\emptyset $-definable
algebraically closed field, and $G$ is the
affine group $K^+ \rtimes K^*$ acting as $(a,b)*x = a+bx$.
\item $U(G)=3$, $S=\PP^1(K)$ is the projective line of a
$\emptyset $-definable
algebraically closed field, and $G$ is the group $\operatorname{PSL}_2(K)$ of M\"obius
transformations.
\end{itemize}
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/$\emptyset $ homogeneous space
with $U(G) = k \in \omega$, then $k \leq 3$, and if $k>1$ then the generic
type of $S$ is interalgebraic with a $\bigwedge$-definable algebraically closed
field.
It looks like the arguments in Almost Orthogonal Regular Types apply
directly once we see that $G$ acts strictly $k$-transitively on $s^{(k)}$,
where $s$ is a minimal type. But I'm not sure how to get to that
situation. We can't expect $G$ to already act on the generic type of $s$,
since this doesn't hold in the conclusion (until we "forget parameters").
TODO.
\end{fact}
\begin{corollary}[Field Configuration Theorem] \label{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 $\bigwedge$-definable algebraically closed field.
\end{corollary}
\begin{definition*}
$D$ is \defn{$k$-pseudolinear} if whenever $p$ is a complete minimal type
with $p(x) \vDash x \in D^2$, we have $U(\Cb(p)) \leq k$.
\end{definition*}
\begin{remark*}
1-pseudolinear $\Leftrightarrow $ linear.
\end{remark*}
\begin{theorem} \label{pseudolinLin}
Let $k>1$.
Suppose $D$ is minimal and $k$-pseudolinear.
Then $D$ is locally modular.
\end{theorem}
\begin{proof}
We may assume $\dcleq(\emptyset )=\acleq(\emptyset )$.
Suppose $D$ is $k$-pseudolinear. We show $D$ is $(k-1)$-pseudolinear.
Suppose not. So (using Lemma~\ref{finCb}(a)) say $(a_2,a_3) \in D^2$,
$\tp(a_2a_3/b_1)$ is minimal,
$b_1 = \Cb(a_2a_3/b_1)$, $U(b_1)=k$.
Now $a_2 \ind b_1$, since else $a_2 \in \acleq(b_1)$, and then $a_2a_3
\ind _{a_2} b_1$ and so $b_1 \in \acleq(a_2)$, contradicting $k>1$.
Similarly, $a_3 \ind b_1$.
Let $b_2 \vDash \stp(b_1)\restriction_{b_1a_2a_3}$.
Then $b_2a_3 \equiv b_1a_3$, so say $a_1$ is such that $b_2a_1a_3 \equiv b_1a_2a_3$.
Let $b_3 := \Cb(a_1a_2 / b_1b_2)$.
$U(a_1a_2/b_1b_2) = 1$ since $a_2 \in \acleq(a_3b_1)$ and $a_3 \in
\acleq(a_1b_2)$. So by $k$-pseudolinearity, $U(b_3) \leq k$.
Similarly, $U(a_2a_3/b_2b_3)=1$.
Since $a_2 \ind b_1$ and $b_2 \ind b_1a_2$ and $b_3 \in \acleq(b_1b_2)$, we
have $a_2 \ind b_1b_2b_3$, 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 $a_2a_3 \ind _{b_2b_3} b_1$ and $a_2a_3 \ind _{b_1} b_2b_3$,
so since $b_1=\Cb(a_2a_3/b_1) = \Cb(a_2a_3/b_1b_2b_3)$,
$b_1 \in \acleq(b_2b_3)$.
Similarly, $b_2 \in \acleq(b_1b_3)$.
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.\ $y = 1 + b_1x + b_2x^2 + \ldots + b_{k+1}x^{k+1}$ is a
$(k+1)$-dimensional family of plane curves in the field, which contradicts
$k$-pseudolinearity (as in Example~\ref{affPlaneNonlinear}).
\end{proof}
\section{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 $D$.
\begin{lemma}
$D$ remains unimodular on adding to the language a Morley sequence in
$\pp\restriction_\emptyset $.
\end{lemma}
\begin{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 $a,a' \vDash \pp\restriction_{\btup}^{(n)}$,
and $a \in \acleq(a'\btup)$ and $a' \in \acleq(a\btup)$.
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\restriction_\emptyset $, 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.
\end{proof}
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.
\begin{definition*}
For $b \in D^{\eq}$, define the \underline{Zilber degree} by
\[ Z(b) := \frac{ \mult(b/c) }{ \mult(c/b) }, \]
where $c \vDash \pp\restriction_\emptyset ^{(<\omega)}$ is interalgebraic with $b$;
equivalently, using gEI, $c$ is an $\acl_D$-basis for $\acleq(b)\cap D$.
Then define $Z(a/b) := Z(ab)/Z(b)$.
\end{definition*}
\begin{lemma}
$Z(b)$ is well-defined.
\end{lemma}
\begin{proof}
This follows from unimodularity. Indeed, if $c' \vDash \pp\restriction_\emptyset ^{(<\omega)}$ is
another $\acl_D$-basis for $\acleq(b)\cap D$, then
\begin{align*} \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) } .\end{align*}
\end{proof}
\begin{remark*}
For $b \in D^n$, $Z(b) \in \N$, 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
$\tp(b)$.
\end{remark*}
\begin{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 $a \vDash \pp\restriction_\emptyset ^{(n)}$ then
$\left|{X \cap \acl_D(a)}\right| = p(\left|{D \cap \acl_D(a)}\right|)$.
\end{remark*}
\begin{lemma} \mbox{} \label{Zprops}
\begin{enumerate}[(i)]\item $Z$ is $\Aut(\MMeq)$-invariant.
\item $Z(ab/c) = Z(a/bc)Z(b/c)$.
\item $a \in \acleq(b) \Rightarrow Z(a/b) = \mult(a/b)$.
\item $a \in D \setminus \acleq(b) \Rightarrow Z(a/b) = 1$.
\end{enumerate}
\end{lemma}
\begin{proof} \mbox{}
\begin{enumerate}[(i)]\item Clear.
\item
\[ Z(a/bc)Z(b/c)
= \frac{ Z(abc) }{ Z(bc) } \frac{ Z(bc) }{ Z(c) }
= \frac{ Z(abc) }{ Z(c) } = Z(ab/c) \]
\item Let $c$ be a basis for $\acleq(b) \cap D = \acleq(ab) \cap D$.
Then \begin{align*} 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) .\end{align*}
\item Let $c$ be a basis for $\acleq(b) \cap D$.
Then $ac$ is a basis for $\acleq(ab) \cap D$, so
by stationarity of $\tp(a/b) = \pp\restriction_{b}$ and of $\tp(a/c) = \pp\restriction_{c}$
and by Lemma~\ref{multStat},
\begin{align*} 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) .\end{align*}
\end{enumerate}
\end{proof}
\begin{remark*}
In fact, the function $Z(x/y)$ is uniquely determined by (i)-(iv) (see
\cite{HrUnimod} for a proof).
\end{remark*}
\begin{definition*}
For $p = \tp(a/B)$ a stationary type, let $Z(p) := Z(a / \Cb(p))$.
\end{definition*}
\begin{lemma} \label{Znf}
Suppose $a,b,c \in \Deq$.
\begin{enumerate}[(I)]\item If $\tp(a/c)$ is stationary and $a \ind _c b$, then $Z(a/bc) =
Z(a/c)$.
In particular, if $\tp(a/c)$ is stationary then $Z(\tp(a/c))=Z(a/c)$.
\item $Z(a/c) = \sum_{i \in I} Z(\qq_i)$ where $(\qq_i)_{i \in I}$ enumerates the global
non-forking extensions of $\tp(a/c)$.
\item $Z(a/c) = \sum_i Z(a_i/bc)$ where $(\tp(a_i/bc))_i$ enumerates the
nonforking extensions of $\tp(a/c)$ to $bc$.
\end{enumerate}
\end{lemma}
\begin{proof} \mbox{}
\begin{enumerate}[(I)]\item We first consider two special cases.
\begin{claim}
If $\tp(a/c)$ is stationary, then $Z(a/bc) = Z(a/c)$
\begin{enumerate}[(a)]\item when $b \in D \setminus \acleq(ac)$, and
\item when $b \in \acleq(c)$.
\end{enumerate}
\end{claim}
\begin{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 $Z(b/ac) = \mult(b/ac) = \mult(b/c) = Z(b/c)$.
\end{proof}
Now let $b$ be such that $a \ind _c b$ and let $b'$ be a basis for
$\acleq(b) \cap D$ over $\acleq(c) \cap D$. Using that $a \ind _c b'$ 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 $Z(a/bc) =
Z(a/bb'c) = Z(a/b'c) = Z(a/c)$.
\item
Using Lemma~\ref{finCb}(a), let $b := \Cb(a/c) \in \Deq$, so
$(\Cb(\qq_i))_{i \in I}$ enumerates the conjugates of $b$ over $c$ (and
so $|I| = \mult(b/c)$ is finite). Then $a \ind _b bc$, 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).
\item
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 $bc$.
\end{enumerate}
\end{proof}
\begin{lemma}["Relaxation"] \label{relax}
Suppose $a,b_1,b_2$ are such that $U(a)=2$, $U(a/b_i)=1$,
$b_i=\Cb(a/b_i)$, $U(b_1) \geq 1$, and $U(b_2) \geq 2$.
Then $b_1 \ind b_2$ iff $b_1 \ind _a b_2$.
{\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 $a$.)}
\end{lemma}
\begin{proof}
First we see that this equivalence holds if $a \nind _{b_1} b_2$.
Indeed, then $a\in \acleq(b_1b_2)$, and so
\begin{align*} 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) ;\end{align*}
in particular one side is zero iff the other is, as required.
Similarly, we are done if $a \nind _{b_2} b_1$.
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 $b_1 \ind _a b_2$ implies
$b_2 \in \acleq(a)$, so $1 = U(a/b_2) = U(a) - U(b_2) \leq 2 - 2$, which is a
contradiction, so $b_1 \nind _a b_2$, and similarly we find $b_1 \nind b_2$.
\end{proof}
\begin{lemma}["B\'ezout"] \label{bezout}
Suppose $p_1 \in S(b_1c)$ and $p_2 \in S(b_2c)$ are minimal types,
$p_i(x) \vDash x \in D^2$, $b_i = \Cb(p_i)$, $c \in \Deq$.
Suppose $U(b_1/c) \geq 1$ and $U(b_2/c) \geq 2$,
and $b_1 \ind _c b_2$.
Then $\left|{p_1 \cup p_2}\right| := \left|{\{ a \;|\; a \vDash p_1 \cup p_2 \}}\right| = Z(p_1)Z(p_2)$.
\end{lemma}
\begin{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. B\'ezout'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.
\end{remark*}
\begin{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.
\end{remark*}
\begin{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 $\left|{p_1' \cup p_2'}\right| = \left|{p_1 \cup p_2}\right|$ and
$Z(p_i')=Z(p_i)$.
In the interests of notational sanity, we will assume $c=\emptyset $. It is
straightforward to check that the arguments below go through without this
assumption, by working everywhere over $c$. (Note that $p_1\restriction_c =
\pp\restriction_c^{(2)} = p_2\restriction_c$.)
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 $a_i \vDash p_i$, and let $q(Y_1,Y_2) :=
\tp(b_1,b_2)$.
Let $Q(X,Y_1,Y_2)$ be the incomplete type $p_1'(X,Y_1)\cup p_2'(X,Y_2)
\cup q(Y_1,Y_2)$.
\begin{claim} \label{bezoutQ}
If $a \vDash p_2$, 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.
\end{claim}
\begin{proof}
It suffices to show that if $\vDash p_1'(a,b_1')$ then
$b_1' \ind _a b_2$ iff $b_1' \ind b_2$. This is immediate from
Lemma~\ref{relax}.
\end{proof}
So let $a \vDash Q(X,b_1,b_2)$.
Let $(\tp(a^i/b_1b_2))_{i \in I}$ enumerate the completions of $Q(X,b_1,b_2) =
p_1(X) \cup p_2(X)$. By completeness of $q$,
$(\tp(a^ib_1/b_2))_{i \in I}$ enumerates the completions of $Q(X,Y_1,b_2)$.
Say $a^ib_1 \equiv _{b_2} ab_1^i$,
so $(\tp(ab_1^i/b_2))_{i \in I}$ also enumerates the completions of $Q(X,Y_1,b_2)$,
so $(\tp(b_1^i/ab_2))_{i \in I}$ 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.
\begin{claim} \mbox{}
\begin{enumerate}[(a)]\item $Z(a)=1$
\item $\sum_i Z(b_1^i/ab_2) = Z(b_1/a) = Z(ab_1)$
\item $Z(b_1^ib_2) = Z(b_1)Z(b_2)$
\end{enumerate}
\end{claim}
\begin{proof} \mbox{}
\begin{enumerate}[(a)]\item By Lemma~\ref{Zprops}(iv) twice.
\item By Claim~\ref{bezoutQ} and Lemma~\ref{Znf}(III), and (a).
\item By $b_1^i \equiv _{b_2} b_1$ and Lemma~\ref{Znf}(I), we have
$Z(b_1^i/b_2) = Z(b_1/b_2) = Z(b_1)$.
\end{enumerate}
\end{proof}
Now we calculate
\begin{align*} \;|\; 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) .\end{align*}
\end{proof}
\begin{lemma}
$D$ is 2-pseudolinear.
\end{lemma}
\begin{proof}
Suppose $p$ is minimal complete, $p(x) \vDash x \in D^2$, $b := Cb(p)$, $U(b)\geq 3$.
Let $b' \vDash \stp(b)\restriction_{\acleq(b)}$.
Say $b^\sg=b'$, $\sg \in \Aut(\MM)$, and let $p' := p^\sg$.
So $p'$ is minimal and $\Cb(p') = b'$.
Let $a \vDash p \cup p'$ (which is consistent by Lemma~\ref{bezout}).
We have $U(a) = 2$ since $U(Cb(p)) = U(b) > 0$ and $a \in D^2$.
By Lemma~\ref{relax}, $b \ind _a b'$.
Note also that $U(b/a) \geq 2 \leq U(b'/a)$.
Let $q := p\restriction_{ba}$ and $q' := p'\restriction_{b'a}$.
Then by Lemma~\ref{bezout},
\[ \;|\; q \cup q' \;|\; = Z(q)Z(q') = Z(p)Z(p') = \;|\; p \cup p' \;|\; ,\]
which contradicts the fact that $a \vDash p \cup p'$ but $a \not\vDash q \cup q'$.
\end{proof}
Applying Theorem~\ref{pseudolinLin}, this concludes the proof of Theorem~\ref{unimodLocmod}.
\bibliography{mmm-gst}
\end{document}
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