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\title{Hrushovski-Weil}

\section{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 
$\bigwedge$-definable group (or group action).

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

\section{Preliminaries}

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

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

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

\section{Germs}

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

    Write $\widetilde {f} : \pp \rightarrow  \qq$ if $\pp(x) \vDash  \qq(f(x))$ for some (any)
    representative $f$ (i.e.\ $f_*(\pp) = \qq$).

    If $\widetilde {f} : \pp \rightarrow  \qq$ has a representative $f$ defined over $b$ and $a \vDash 
    \pp|_b$, let $\widetilde {f}(a) := f(a)$.
    This is well-defined (if $g$ is another representative defined over $b$ 
    then $g(a) = f(a)$ since $a \vDash  \pp|_b \vDash  f(x) = g(x)$), and $\widetilde {f}(a) \vDash  \qq|_b$.
\end{definition*}

\begin{example*}
    In an o-minimal theory, let $p_{+\infty}$ be the global type of a positive 
    infinite element.
    Then any eventually increasing definable function $f$ defines a germ $\widetilde {f} : 
    p_{+\infty} \rightarrow  p_{+\infty}$, and $\widetilde {f} = \widetilde {g}$ iff $f$ and $g$ are eventually 
    equal.
\end{example*}

\begin{definition*}
    $\Aut(\MM)$ acts on germs:
    given $\widetilde {f} : \pp \rightarrow  \qq$ and $\sg \in \Aut(\MM)$, $\widetilde {f}^\sg := \widetilde {(f^\sg)} : 
    \pp^\sg \rightarrow  \qq^\sg$ does not depend on the choice of $f$.

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

    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
    $\widetilde {f} : p \rightarrow  q$ means $\widetilde {f} : \pp \rightarrow  \qq$.
\end{definition*}

\begin{remark*}
    Since $T$ is stable, any germ has a code.

    Indeed, given a $\emptyset $-definable family $f_z$ of partial functions,
    % This is true, but not used:
    %$\{ z | \widetilde {f}_z : \pp \rightarrow  \qq \}$ is $\bigwedge$-definable as
    % \[ \bigwedge_{\phi} d_\pp x. ( x \in \dom(f_z) \wedge
    %   \forall y. ( \phi(f_z(x),y) \leftrightarrow  d_\qq w. \phi(w,y) ) ) ,\]
    %**
    equality of germs at $\pp$,
    $\widetilde {f}_b = \widetilde {f}_c$, is defined by
    \[ E(b,c) := d_\pp x. ( f_b(x) = f_c(x) ) ,\]
    and then since $\widetilde {f}_b^\sg = \widetilde {f}_{b^\sg}$, we have
        \[ \code{\widetilde {f}_b} = \dcl(b/E) .\]
\end{remark*}

\begin{remark*}
    Composition of germs is well-defined.

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

    Note that $\code{\widetilde {f} \o \widetilde {g}} \subseteq  \dcl(\code{\widetilde {f}},\code{\widetilde {g}})$,
    and $\code{\widetilde {f}^{-1}} = \code{\widetilde {f}}$.
\end{remark*}

\begin{definition*}
    Let $p,q,s \in S(\emptyset )$.
    A \defn{family $\widetilde {f}_s$ of germs $p \rightarrow  q$}
    is the family $\widetilde {f}_s := (\widetilde {f}_b)_{b \vDash  s}$ of germs at $p$ of a
    $\emptyset $-definable family $f_z$ of partial functions,
    which is such that $\widetilde {f}_b : p \rightarrow  q$ whenever $b \vDash  s$.

    The family is \underline{canonical} if $b$ is a code for $\widetilde {f}_b$, for all $b \vDash  s$.

    The family is \underline{generically transitive} if $\widetilde {f}_b(x) \ind  x$ for $(b,x) \vDash  s \otimes p$.
\end{definition*}

\begin{remark*}
    $\widetilde {f}_s$ is generically transitive iff for $(x,y) \vDash  p \otimes q$, there
    exists $b \vDash  s$ such that $\widetilde {f}_b(x)=y$.
\end{remark*}

\begin{remark*}
    Suppose $p,q,s \in S(\emptyset )$,
    and $\widetilde {f}_s$ is a family of germs $p \rightarrow  q$.
    Let $b \vDash  s$ and $x \vDash  p|_b$, and let $y = \widetilde {f}_b(x)$.
    Then
    \begin{equation} \label{e:canFamTup}
	x \ind  b;\; y \ind  b;\; y \in \dcl(bx) .
    \end{equation}

    Conversely, if $(b,x,y)$ satisfy (\ref{e:canFamTup}),
    let $f_b(x)=y$ be a formula witnessing $y\in\dcl(bx)$.
    Then $\widetilde {f}_s := (\widetilde {f}_b)_{b \vDash  s}$ is a family of germs $p \rightarrow  q$,
    where $s = \tp(b)$, $p = \tp(x)$, $q = \tp(y)$.
\end{remark*}

\begin{lemma} \label{l:canonFamCB}
    In the correspondence of the previous remark,
    \begin{enumerate}[(i)]\item $\widetilde {f}_b$ can be chosen to be invertible iff also $x \in \dcl(by)$;
    \item the family $\widetilde {f}_s$ is generically transitive iff $x \ind  y$;
    \item $\code{\widetilde {f}_b} = \Cb(xy/b)$
        (so $\widetilde {f}_s$ is canonical iff $\Cb(xy/b) = b$).
    \end{enumerate}
\end{lemma}
\begin{proof} \mbox{}
    \begin{enumerate}[(i)]\item $x \in \dcl(by)$ iff $f_b$ can be taken to be injective at $x$.
    \item Immediate.
    \item
        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 $\pp(x) \cup \{ y = f_b(x) \}$.
        \begin{align*} \Cb(xy/b)^\sg = \Cb(xy/b)
            &\Leftrightarrow  \Cb(\rr)^\sg = \Cb(\rr)\\
            &\Leftrightarrow  \rr^\sg = \rr\\
            &\Leftrightarrow  \pp(x) \cup \{ y = f_{b^\sg}(x) \} \equiv  \pp(x) \cup \{ y = f_b(x) \} \\
            &\Leftrightarrow  \pp(x) \vDash  f_{b^\sg}(x) = f_b(x) \\
            &\Leftrightarrow  \widetilde {f}_b^\sg = \widetilde {f}_b .\end{align*}

        %{\small Alternative proof:
        %$\widetilde {f}_s$ is canonical iff
        %whenever $b' \vDash  s$ and $\widetilde {f}_{b'}=\widetilde {f}_b$, we have $b'=b$;
        %equivalently, whenever $b' \vDash  s|_x$ and $\widetilde {f}_{b'}(x)=\widetilde {f}_b(x)=y$, we have $b'=b$;
        %equivalently, any automorphism fixing $x$ and $y$ also fixes $b$;
        %equivalently, $\Cb(xy/b)=b$.
        %}
    \end{enumerate}
\end{proof}

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

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 $X$.

\begin{fact} \label{f:genFacts}
    \begin{enumerate}[(i)]\item $G$ is connected iff it has for any $A$ a unique generic type over 
    $A$.
    \item For $b \in X$,
    $\tp(b/A)$ is generic iff $\forall g \in G. (b \ind _A g \Rightarrow  g*b \ind _A g)$.
    \item $G$ acts transitively on the set of global generic types of $X$.
    \end{enumerate}
\end{fact}

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

\begin{lemma} \label{l:connectedTest}
    Suppose $s \in S(\emptyset )$ extends $G$ and
    $\forall g \in G. (h \vDash  s|_g \Rightarrow  g\cdot h \vDash  s|_g)$.
    Then $G$ is connected and $s$ is the unique generic of $G$.
\end{lemma}
\begin{proof}
    $s$ is generic by Fact~\ref{f:genFacts}(ii).

    If $G$ weren't connected, by Fact~\ref{f:genFacts}(iii) there would be $g$ such 
    that $g\cdot \ss \neq  \ss$,
    where $\ss$ is the global nonforking extension of $s$.
    But $g\cdot \ss$ doesn't fork over $g$,
    so since $s$ and hence $s|_g$ is stationary, $g\cdot s|_g \neq  s|_g$.
    So if $h \vDash  s|_g$ then $g\cdot h \not\vDash  s|_g$.
\end{proof}

\section{Hrushovski-Weil}
\begin{definition*}
    Let $p,s \in S(\emptyset )$.
    A family $\widetilde {f}_s$ of germs $p \rightarrow  p$ is
    \begin{itemize}\item \defn{closed under inverse} if
        for $b \vDash  s$ there exists $b' \vDash  s$ such that $\widetilde {f}_b^{-1}=\widetilde {f}_{b'}$;
    \item \defn{closed under generic composition} if
        for $b_1b_2 \vDash  s^{(2)}$, there exists $b_3$ such that
            \[ \widetilde {f}_{b_1} \circ \widetilde {f}_{b_2} = \widetilde {f}_{b_3} \]
        and $b_ib_3 \vDash  s^{(2)}$ for $i=1,2$.
    \end{itemize}
\end{definition*}

\begin{theorem}[Hrushovski-Weil for actions] \label{t:HW}
    Let $p,s \in S(\emptyset )$,
    and suppose $\widetilde {f}_s$ is a generically transitive canonical family of
    invertible germs $p \rightarrow  p$ which is
    closed under inverse and generic composition.

    Then there exists a connected faithful $\bigwedge$-definable/$\emptyset $ 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.\ $\widetilde {f}_b(a) = b*a$ for $(a,b) \vDash  p \otimes s$.
\end{theorem}

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

    \emph{(In the case $T=\operatorname{ACF}$ (where $\bigwedge$-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~\ref{t:HW} with 
    $f_b(a) := b*a$ --
    we didn't assume closedness under inverse nor canonicity, so some extra 
    argument is involved.
\end{theorem*}

\begin{proof}[Proof of Theorem~\ref{t:HW}]
    Let $G$ be the group of germs generated by $\widetilde {f}_s$.

    \begin{claim}
        Any element of $G$ is a composition of two generators.
    \end{claim}
    \begin{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 $\widetilde {f}_b$ is the composition of two generators.

        So it suffices to see that any composition of three generators
            \[ \widetilde {f}_{b_1} \circ \widetilde {f}_{b_2} \circ \widetilde {f}_{b_3} \]
        is the composition of two.
        
        Let $b' \vDash  s|_{b_1b_2b_3}$.
        Then 
            \[ \widetilde {f}_{b_1} \circ \widetilde {f}_{b_2} \circ \widetilde {f}_{b_3}
            = \widetilde {f}_{b_1} \circ \widetilde {f}_{b'} \circ \widetilde {f}_{b'}^{-1} \circ \widetilde {f}_{b_2} \circ \widetilde {f}_{b_3} \]
        Now $b' \ind  b_2$, so
        say $\widetilde {f}_{b'}^{-1} \circ \widetilde {f}_{b_2} = \widetilde {f}_{b''}$
        with $b'' \vDash  s$ independent from $b'$ and from $b_2$.

        Now $b' \ind _{b_2} b_3$,
        so $b'' \ind _{b_2} b_3$, since $b'' \in \dcl(b'b_2)$,
        so since $b'' \ind  b_2$,
        we have $b'' \ind  b_3$.

        Also $b' \ind  b_1$.

        So the germs $\widetilde {f}_{b_1} \circ \widetilde {f}_{b'}$ and $\widetilde {f}_{b'}^{-1} \circ \widetilde {f}_{b_2} \circ \widetilde {f}_{b_3}$
        appear in the family $\widetilde {f}_s$.
    \end{proof}

    So $G$ is $\bigwedge$-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 $b \mapsto  
    \widetilde {f}_b$.

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

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

    Now define $X := (G \times p)/E$ where $(g,a) E (g',a')$ iff
    $(h\cdot g)*a = (h\cdot g')*a'$ for $h \vDash  s|_{aa'gg'}$, 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 $h \in G$, then if $h'
    \vDash  s|_{g,g',a,a',h}$, then also $h'\cdot h \vDash  s|_{g,g',a,a',h}$ 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 $a \mapsto  (1,a)/E$.

    We show transitivity. Let $a,a' \vDash  p$, and we show $(1,a')/E \in
    G*(1,a)/E$; this suffices for transitivity, since clearly $(G,a')/E \subseteq 
    G*(1,a')/E$. Let $c \vDash  p|_{aa'}$.
    Then by generic transitivity of $\widetilde {f}_s$, there exist $g \vDash  s|_a$ and $g' \vDash  
    s|_c$ such that
    $g*a=c$ and $g'*c=a'$. Then $(h\cdot g)*a = h*c$ for $h \vDash  s|_{acg}$, 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 $a \vDash 
    p|_g$. Then $(g,a)E(1,a)$, so let $h \vDash  s|_{ag}$; then $(h\cdot g)*a = 
    h*a$.
    But $h \ind _g a$, hence also $h\cdot g \ind _g a$, so $h,h\cdot g \ind  a$ 
    since $g \ind  a$.
    So $h\cdot g=h$ as germs, so $g=1$.

    Finally, we claim that $p$ is the unique generic type. If $g \in G$ and $a
    \vDash  p|_g$, then $g*a \vDash  p|_g$ as this is the action of a germ. But $G$ acts
    transitively on the generic types (Fact~\ref{f:genFacts}(iii)).
    So $p$ is the unique generic type.
\end{proof}
\end{document}
