!newop stp
!newop tp
!newop dcl
!newop acl
!newop id
!newop dom
!newop eq
!newop Aut
!newop Cb
!renewtup b

\newcommand{\defn}[1]{{\bf #1}}

\newcommand{\pp}{\mathfrak{p}}
\newcommand{\qq}{\mathfrak{q}}
\newcommand{\rr}{\mathfrak{r}}
\renewcommand{\ss}{\mathfrak{s}}

\newcommand{\MM}{\mathbb{M}}
\newcommand{\sg}{\sigma}

\renewcommand{\o}{\circ}

% Rahim's nonforking symbol (I don't know where got it from)
\def\Ind#1#2{#1\setbox0=\hbox{$#1x$}\kern\wd0\hbox to 0pt{\hss$#1\mid$\hss}
\lower.9\ht0\hbox to 0pt{\hss$#1\smile$\hss}\kern\wd0}
\def\ind{\mathop{\mathpalette\Ind{}}}
\def\notind#1#2{#1\setbox0=\hbox{$#1x$}\kern\wd0\hbox to 0pt{\mathchardef
\nn=12854\hss$#1\nn$\kern1.4\wd0\hss}\hbox to
0pt{\hss$#1\mid$\hss}\lower.9\ht0 \hbox to
0pt{\hss$#1\smile$\hss}\kern\wd0}
\def\nind{\mathop{\mathpalette\notind{}}}

% corner quotes; adapted from a definition due to Sam Buss, taken from
% http://www.logicmatters.net/latex-for-logicians/symbols/corner-quotes-for-godel-numbers/
\newbox\gnBoxA
\newdimen\gnCornerHgt
\setbox\gnBoxA=\hbox{$\ulcorner$}
\global\gnCornerHgt=\ht\gnBoxA
\newdimen\gnArgHgt
\def\code #1{%
\setbox\gnBoxA=\hbox{$#1$}%
\gnArgHgt=\ht\gnBoxA%
\ifnum     \gnArgHgt<\gnCornerHgt \gnArgHgt=0pt%
\else \advance \gnArgHgt by -\gnCornerHgt%
\fi \raise\gnArgHgt\hbox{$\ulcorner$} \box\gnBoxA \, %
\raise\gnArgHgt\hbox{$\urcorner$}}

\theoremstyle{plain}
\newtheorem*{theorem*}{Theorem}

\title{Hrushovski-Weil}
\author{Martin Bays}
\date{2021-12-20}
\maketitle
Notes for a talk in a Münster seminar on stability theory.

# Introduction
We aim now for the group configuration theorem. This powerful theorem shows 
that whenever one has a certain configuration of points in a stable theory 
satisfying certain independence and algebraicity conditions, it comes from a 
$//\\$-definable group (or group action).

Here we prove a version of the ``Hrushovski-Weil group chunk theorem'', which 
recognises a $//\\$-definable group (action) from a generically presented one. 
This will be the final step in the proof of the group configuration theorem.

# Preliminaries

Let $\MM$ be a monster model of a stable theory $T=T^\eq$.

Notation*:
    ***) If $p (- S(A)$ is stationary and $B )= A$,
    $p|_B (- S(B)$ is the unique non-forking extension.
    ** If $p,q (- S(A)$ are stationary,
    their \defn{product type} is $p (x) q := \tp(b,c/A)$
    where $c |= q$ and $b |= p|_{Ac}$,
    i.e. $(b,c) |= p\times q$ and $b \|/_A c$.
    ** $p^{(2)} := p (x) p$, $p^{(3)} := p (x) p (x) p$ etc.
    *e)*
    .

For notational convenience, \underline{assume} (by adding parameters) that $\acl({}) = \dcl({})$.
So types over ${}$ are stationary.

# Germs

Definition*:
    Let $\pp,\qq \in S(\MM)$.
    
    Say a definable partial function $f$ is \defn{defined at} $\pp$ if $\pp(x) 
    |= x (- \dom(f)$.
    The \defn{germ of $f$ at $\pp$} is then the equivalence class $~f$ under the
    equivalence relation $\pp(x) |= f_1(x) = f_2(x)$.

    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 $~f(a) := f(a)$.
    This is well-defined (if $g$ is another representative defined over $b$ 
    then $g(a) = f(a)$ since $a |= \pp|_b(x) |= f(x) = g(x)$), and $~f(a) |= 
    \qq|_b$.
    .

Example*:
    In an o-minimal theory, let $\pp_{+\infty}$ be the global type of a 
    positive infinite element.
    Then a definable function $f$ which is unbounded on all end-segments 
    $(a,+\infty)$ defines a germ $~f : \pp_{+\infty} --> \pp_{+\infty}$, and 
    $~f = ~g$ iff $f$ and $g$ are eventually equal.
    .

Definition*:
    $\Aut(\MM)$ acts on germs:
    given $~f : \pp --> \qq$ and $\sg \in \Aut(\MM)$, $~f^\sg := \~{(f^\sg)} : 
    \pp^\sg --> \qq^\sg$ does not depend on the choice of $f$.

    A (possibly infinite) tuple $\b$
    is a \defn{code} for $~f$ if $\forall \sg \in \Aut(\MM). (\b =
    \b^\sg <-> ~f = ~f^\sg)$; then set $\code{~f} := \dcl(\b)$.
    Here, $~f = ~f^\sg$ should be understood as implying $\pp = \pp^\sg$.
    .

Remark*:
    Since $T$ is stable, any germ has a code.

    Indeed, given a ${}$-definable family $f_z$ of partial functions,
    % This is true, but not used:
    %$\{ z | ~f_z : \pp --> \qq \}$ is $//\\$-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 $\pp$,
    $~f_b = ~f_c$, is defined by
    \[ E(b,c) := d_\pp x. ( f_b(x) = f_c(x) ) ,\]
    and then since $~f_b^\sg = ~f_{b^\sg}$, we have
        \[ \code{~f_b} = \dcl(b/E) .\]
    .

Remark*:
    Composition of germs is well-defined.

    Say $~f : \pp --> \qq$ is \defn{invertible} if it is invertible in the 
    category of germs,
    i.e. $~g \o ~f = \~\id = ~f \o ~g$ for some $~g : \qq --> \pp$.
    Equivalently, $~f$ is injective on $\pp$.
    %Proof:
    %  $=>$: Clear.
    %  $<=$: By compactness, $f$ is already injective on some $\phi (- \pp$, 
    %  and so $f$ has a well-defined definable inverse.
    %  .

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

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

Definition*:
    Let $p,q,s \in S({})$.
    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 $(b,x) |= s (x) p$.
    .

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

Remark*:
    Suppose $p,q,s \in S({})$,
    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 \dcl(bx) .
	.

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

Lemma{canonFamCB}:
    In the correspondence of the previous remark,
    **(i) $~f_b$ can be chosen to be invertible iff also $x \in \dcl(by)$;
    **(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) $x (- \dcl(by)$ iff $f_b$ can be taken to be injective at $x$.
    **(ii) Immediate.
    **(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*
    .

# Homogeneous spaces
Let $(G,X)$ be a $//\\$-definable homogeneous space over ${}$,
i.e.
$G$ is a $//\\$-definable group and $X$ is a $//\\$-definable set,
and $* : G \times X --> X$ is a relatively definable transitive group action,
all over ${}$.

Say $(G,X)$ is \defn{connected} if $G$ is connected,
i.e. $G$ has no relatively definable proper subgroup of finite index.

Say $(G,X)$ is \defn{faithful} if the action is faithful, i.e. only the
identity element of $G$ acts trivially on $X$.

Recall that a complete type extending $G$ is generic if it is generic for the 
left, equiv right, multiplication action of $G$ on itself.

Fact{genFacts}:
    ***(i) $G$ is connected iff it has for any $A$ a unique generic type over 
    $A$, iff it has a unique generic type over ${}$.
    ** For $b (- X$,
    $\tp(b/A)$ is generic iff $\forall g (- G. (b \|/_A g => g*b \|/_A g)$.
    ** $G$ acts transitively on the set of global generic types of $X$.
    *ee*
    .

Lemma{connectedTest}:
    Suppose $p (- S({})$ extends $X$ and
    $\forall g (- G. (b |= p|_g => g*b |= p|_g)$.
    Then $p$ is the unique generic of $X$ over ${}$.

    In particular, if this holds for the left/right multiplication action of 
    $G$ on itself, then $G$ is connected.
    .
Proof:
    $p$ is generic by Fact~{genFacts}(ii).

    If $p' (- S({})$ is generic,
    by Fact~{genFacts}(iii) there is $g$ such that $g*\pp = \pp'$,
    where $\pp$ and $\pp'$ are the global non-forking extensions.
    So if $b |= p|_g$ then $g*b |= p'$,
    but also $g*b |= p$,
    so $p' = p$.
    .

# Hrushovski-Weil
Definition*:
    Let $p,s \in S({})$.
    A family $~f_s$ of germs $p --> p$ is
    ***) \defn{closed under inverse} if
        for $b |= s$ there exists $b' |= s$ such that $~f_b^{-1}=~f_{b'}$;
    ** \defn{closed under generic composition} if
        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 $i=1,2$.
    *e)*
    .

Theorem[Hrushovski-Weil for actions]{HW}:
    Let $p,s \in S({})$,
    and suppose $~f_s$ is a generically transitive canonical family of
    invertible germs $p --> p$ which is
    closed under inverse and generic composition.

    Then there exists a connected faithful $//\\$-definable/${}$ homogeneous space 
    $(G,X)$,
    a definable embedding (i.e. relatively definable injection) of $s$ into $G$ as its unique generic type,
    and a definable embedding of $p$ into $X$ as its unique generic type,
    such that the generic action of $s$ on $p$ agrees with that of $G$ on $X$,
    i.e. $~f_b(a) = b*a$ for $(a,b) |= p (x) s$.
    .

This is related to the following more intuitive result:
Theorem*[Hrushovski-Weil group chunk theorem]:
    Let $p (- S({})$, and suppose $*$ is a definable partial binary function such that:
    ***) If $(a,b) |= p^{(2)}$ then $a*b$ is defined, and $(a,a*b) |= p^{(2)}$ and $(b,a*b) |= p^{(2)}$;
    ** If $(a,b,c) |= p^{(3)}$ then $(a*b)*c = a*(b*c)$.
    *e)*
    Then one obtains a $//\\$-definable connected group $G$ and a definable 
    embedding of $p$ into $G$ such that if $(a,b) |= p^{(2)}$ then 
    $a*b$ agrees with the product in $G$.

    \emph{(In the case $T=\"ACF$ (where $//\\$-definable groups are actually 
    algebraic groups (up to definable isomorphism)), this is essentially a 
    theorem of Weil.)}

    Note that this is not quite just a matter of applying Theorem~{HW} with 
    $f_b(a) := b*a$ --
    we didn't assume closedness under inverse nor canonicity, so some extra 
    argument is involved.
    .

Proof[Proof of Theorem~{HW}]:
    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.

        By generic composability and completeness of $s$,
        any generator $~f_b$ 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 $b'' \in \dcl(b'b_2)$,
        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 $~f_s$.
        .

    So $G$ is $//\\$-definable as pairs of realisations of $s$ modulo
    equality of the corresponding composition of germs, and the group
    operation is relatively definable as composition of germs.
    We identify $s$ with its image in $G$ under the definable embedding $b |-> 
    ~f_b$.

    We show that $G$ is connected with generic type $s$ via Lemma~{connectedTest}.
    We show that if $g\in G$ and $b |= s|_g$, then $g\cdot b |= s|_g$.
    This holds for $g |= s$, by closedness under generic composition.
    Let $g (- G$, say $g=g_1\cdot g_2$ with $g_1,g_2 |= s$.
    Let $b |= s|_{g_1,g_2}$.
    Then $g_2\cdot b |= s|_{g_1,g_2}$, and so $g_1\cdot g_2\cdot b |= 
    s|_{g_1,g_2}$.
    Now $g = g_1\cdot g_2 (- \dcl(g_1,g_2)$,
    so $b |= s|_g$ and $g\cdot b |= s|_g$, as required.

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

    Now define $X := (G \times p)/E$ where $(g,a) E (g',a')$ iff
    $(h\cdot g)*a = (h\cdot g')*a'$ for $h |= s|_{aa'gg'}$, which is definable 
    by definability of $s$.

    Define the action of $G$ on $X$ by $h*((g,a)/E) := (h\cdot g,a)/E$.
    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 $h'\cdot h |= s|_{g,g',a,a',h}$ by 
    genericity,
    and we have $(h'\cdot h\cdot g)*a = (h'\cdot h\cdot g')*a'$, so $(h\cdot 
    g,a) E (h\cdot g',a')$.
    $p$ definably embeds into $X$ 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 $g |= s|_a$ and $g' |= 
    s|_c$ such that
    $g*a=c$ and $g'*c=a'$. Then $(h\cdot g)*a = h*c$ for $h |= s|_{acg}$, so
    $(g,a)E(1,c)$. Similarly $(g',c)E(1,a')$. So $(g'\cdot 
    g)*(1,a)/E=g'*(g,a)/E
    = g'*(1,c)/E = (g',c)/E = (1,a')/E$.

    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 $(h\cdot g)*a = 
    h*a$.
    But $h \|/_g a$, hence also $h\cdot g \|/_g a$, so $h,h\cdot g \|/ a$ 
    since $g \|/ a$.
    So $h\cdot g=h$ as germs, so $g=1$.

    Finally, we conclude from Lemma~{connectedTest} that $p$ is the unique 
    generic type, since if $g \in G$ and $a |= p|_g$, then $g*a |= p|_g$ as 
    this is the action of a germ.
    .
