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\providecommand{\R}{\mathbb{R}}
\providecommand{\Stab}{\operatorname{Stab}}
\providecommand{\Fin}{\operatorname{Fin}}
\providecommand{\U}{\mathcal{U}}
\providecommand{\gm}{\gamma}
\providecommand{\Gm}{\Gamma}
\providecommand{\dl}{\delta}
\providecommand{\MM}{\mathbb{M}} %monster model
\providecommand{\MMeq}{\MM^{\eq}}

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\section{Larsen-Pink}

An account of the Hrushovski-Wagner proof of Larsen-Pink.

{\bf Attention potential reader:} soon after writing this note, I discovered
that there's a nice streamlined account of the proof in Hrushovski's ``Stable
group theory and approximate subgroups'', where it appears as Proposition~5.6. 
Most likely, you are better off reading that than this.

\subsection{Statement}

Let $G$ be a simple group of finite Morley rank $\emptyset $-definable in a stable
theory $T$.

Suppose $M = \Pi_\U M_i \models  T$, a countable non-principal ultraproduct.
Suppose $\Gm_i \subseteq  M_i$ are finite subsets such that
$\Gm := \Pi_\U \Gm_i \leq  G(M)$ is a definably dense subgroup,
meaning that $\Gm$ is contained in no proper definable subgroup of $G$.

\begin{theorem}[Hrushovski-Wagner (``Larsen-Pink'')] \label{HWLP}
    For any $M$-definable subset $X \subseteq  G^n$,
    there exists $c \in \R$ such that for all $i$,
    $|X \cap \Gm_i^n| \leq  c |\Gm_i|^{\frac{\RM(X)}{\RM(G)}}$.
\end{theorem}

Larsen-Pink proved this in the case of $G$ a simple algebraic group and $X$ a
subvariety, and suggested that there might be a model-theoretic proof.
Hrushovski-Wagner provided such a proof, in the model-theoretically natural
generality of groups of finite Morley rank.

\subsection{Example / sketch proof}
Suppose $X$ is a curve in a simple algebraic group $G$, $n := \dim(G)$.
Then (as we see below) there are $\gm_1,\ldots,\gm_n \in \Gm$ such that,
setting $f(x_1,\ldots,x_n) := \Pi_i x_i^{\gm_i}$,
$f : X^n \rightarrow  G$ is dominant with generically finite fibres.
Suppose all fibres are finite, say with fibre size bounded by
$k$. Then since $f(\Gamma^n) \subseteq  \Gamma$,
$|X \cap \Gm^n|^n = |X^n \cap \Gm| \leq  k |\Gm|$, so $|X \cap \Gm| \leq  k
|\Gm|^{1/n}$.

In reality, some fibres might be infinite, and $\Gm$ could concentrate
there\ldots we handle this by working with complete types. When $X$ is of
higher dimension, we don't have such a clean map $f$, and so we work
inductively, obtaining a sequence of maps with generic fibres of dimension
less than $\dim(X)$.


\subsection{Stable group theory}

Work in a monster model $\MM \models  T$.

Systematically confuse formulae with the sets they define.

Recall
    \begin{itemize}\item If $A \elememb \MM$, or more generally if $A \subseteq  \MM$ with $\dcleq(A)=\acleq(A)$,
        any $p \in S(A)$ is stationary, meaning it has a unique global non-forking
        extension $\pp \in S(\MM)$. Non-forking can here be taken to mean $\RM(\pp) =
        \RM(p)$. Define $p|_{Ag} := \pp|_{Ag}$, so $a \models  p|_{Ag} \Leftrightarrow  (a \models  p
        \infix{and} a \ind _A g)$.

    \item If $\dcleq(A)=\acleq(A)$ and $p,q \in S(A)$, their unique product type $p \otimes q
        \in S(A)$ is $\tp(a,b/A)$ where $a \models  p$ and $b \models  q|_{Aa}$.

    \item If $f$ is a partial function definable over $A \subseteq  \MM$ and defined at $p \in
        S(A)$, $f_*(p) \in S(A)$ is the type $\tp(f(b)/A)$ where $b \models  p$.

    \item For $A \subseteq  \MM$, $S_G(A) := \{ p \in S(A) \;|\; p(x) \models  x \in G \}$.
        $G \cap \dcleq(A)$ acts on $S_G(A)$, by $g * p := (g*)_*(p)$.

    \item $G$ has DCC: there is no infinite decreasing chain of definable subgroups.

    \item For $\pp \in S_G(\MM)$, $\Stab(\pp) = \{ g \in G \;|\; g*\pp = \pp \}$ is a
        definable subgroup. For $p \in S_G(A)$ stationary, $\Stab(p) = \Stab(\pp)$
        where $\pp$ is the unique global nonforking extension;
        equivalently, $\Stab(p) = \{ g \in G \;|\; g * p|_{Ag} = p|_{Ag} \}$.

    \item $G$ is connected so has a unique generic type $\pp_G \in S_G(\MM)$.
        For $p \in S_G(A)$, $\Stab(p) = G \Leftrightarrow  \RM(p) = \RM(G) \Leftrightarrow  p = \pp_G|_A$.

    \item In $G$, $\RM$ is additive, meaning $\RM(ab/C) = \RM(a/bC) + \RM(b/C)$,
        and definable, meaning that if $r \in S_{G^n}(A)$ and $f$ is a partial
        function definable over $A$ and defined at $r$, then there exists
        $X \in r$ such that for all $b \in f(X)$, $\RM(f|_X^{-1}(b)) =
        \RM(r) - \RM(f_*(r))$.
    \end{itemize}

For $\gm \in \Gamma$, define $m^\gm(g,h) := g^\gm * h$.

\begin{lemma}[Essentially ZIT] \label{ZIT}
    Suppose $\Gm \leq  G(\MM)$ is definably dense,
    $\Gm \subseteq  A = \acleq(A)$,
    and $p \in S_G(A)$ is non-algebraic.
    Then there exist $p_1,\ldots,p_n \in S_G(A)$ and $\gm_1,\ldots,\gm_{n-1} \in \gm$,
    with $p_1 = p$, $p_n = \pp_G|_A$, $p_{i+1} = m^{\gm_i}_*(p \otimes p_i)$,
    and such that $\RM(p_{i+1}) > \RM(p_i)$.
\end{lemma}
\begin{proof}
    WMA $A=\acleq(\emptyset )$.

    It suffices to show that if $q \in S_G(A)$ satisfies
    $\RM(m^{\gm}_*(p \otimes q)) \leq  \RM(q)$ for all $\gm \in \Gm$,
    then $q = \pp_G|_A$.

    So let $q$ be such, and let $S := \Stab(q)$. We will show that $S=G$.

    Let $S' := \bigcap_{\gm \in \Gamma} S^\gm$.
    Then $S'$ is (by the DCC) a definable subgroup, and its normaliser $N(S')$
    contains $\Gm$, so, by denseness of $\Gamma$, $N(S') = G$, so $S'\lhd G$.

    So by simplicity of $G$, it suffices to show that $S' \neq  1$.

    Let $a,b \models  p$. We conclude by showing that $a^{-1}b \in S'$.

    Let $e \models  q|_{ab}$. So $a^\gm e, b^\gm e \models  m^\gm_*(p \otimes q) =: r_\gm$.

    Then $\RM(p) + \RM(q) = \RM(a,e) = \RM(a,e,a^\gm e) = \RM(a,a^\gm e) =
    \RM(a/a^\gm e) + \RM(a^\gm e)$.

    But $\RM(p) \geq  \RM(a/a^\gm e)$ and $\RM(q) \geq  \RM(r_\gm) = \RM(a^\gm e)$,
    so $\RM(p) = \RM(a/a^\gm e)$ (and $\RM(q) = \RM(a^\gm e)$).
    So $a^\gm e \models  r_\gm|_a$.
    So since $e \ind _a b$, $a^\gm e \models  r_\gm|_{ab}$,
    i.e. $a^\gm * q|_{ab} = r_\gm|_{ab}$.
    Similarly, $b^\gm * q|_{ab} = r_\gm|_{ab}$.

    So $(a^{-1}b)^\gm * q|_{ab} = q|_{ab}$,
    so $(a^{-1}b)^\gm \in S'$ for all $\gm$,
    so $a^{-1}b \in S$.
\end{proof}


\subsection{Pseudofinite dimensions}

Recall we have $M := \Pi_\U M_i \elememb \MM$ a countable non-principal
ultraproduct, $\Gm := \Pi_\U \Gm_i \leq  G(M)$ definably dense subgroup.

\begin{definition} \mbox{}
    \begin{itemize}\item For $X$ definable over $M$, $\dl(X) := (\log|X(\Gm)|) / \Fin$
      (``fine pseudofinite dimension restricted to $\Gm$''),
      where $|X(\Gm)| := \Pi_\U |X^{M_i} \cap \Gm_i^n| \in \Pi_\U \R$
      if $X \subseteq  G^n$, and $|X(\Gm)| := 0$ if $X$ is on some other sort.
      Here $\Fin$ is the convex hull of the standard reals $\R \subseteq  \Pi_\U \R$.

    \item For $\pi$ a partial type over $M$,
      $\dl(\pi) := \inf_{\phi \in \pi} \dl(\phi)$,
      taking values in $\Xi \cup \{-\infty\}$ where $\Xi$ is the formal
      completion of $\Pi_\U \R / \operatorname{Fin}$ as an ordered abelian group.

    \item For comparison with values of $\RM$, embed $\Z$ into $\Xi$ such that
        $\dl(G)=\RM(G)$.
    \end{itemize}
\end{definition}

\begin{lemma} \mbox{} \label{dlprops}
    \begin{enumerate}[(i)]\item $\dl(X \cup Y) = \max(\dl(X),\dl(Y))$.
    \item Any partial type $\pi$ over $M$ admits a completion
      $p \in S(M)$ with $\dl(p) = \dl(\pi)$.
    \item $\dl(X \times Y) = \dl(X) + \dl(Y)$.
    \item If $f : G^n \rightarrow  G^m$ is a function definable over $M$ such that
      $f(\Gm^n) \subseteq  \Gm^m$, if $r \in S_{G^n}(M)$,
      and if $X \in r$ with $\dl(f^{-1}(b)) \leq  \alpha$ for all $b \in f(X)^M$,
      then $\dl(r) \leq  \dl(f_*(r)) + \alpha$.
    \end{enumerate}
\end{lemma}
\begin{proof} \mbox{}
    \begin{enumerate}[(i)]\item $\max(|X^\Gm|,|Y^\Gm|) \leq  |(X \cup Y)^\Gm| \leq  2\max(|X^\Gm|,|Y^\Gm|)$.
    \item Follows from (i).
    \item Immediate from definitions.
    \item Let $Y \in f_*(r)$, and let $X' := X \cap f^{-1}(Y)$. Then
        $\dl(p) \leq  \dl(X') \leq  \dl(Y) + \alpha$.
    \end{enumerate}
\end{proof}

Stated in this language and generalised to partial types, Theorem~\ref{HWLP}
becomes:\
\begin{theorem}
    For any partial type $\pi$ over $M$, $\dl(\pi) \leq  \RM(p)$.
\end{theorem}
\begin{proof}
    Let $k := \RM(\pi)$, and suppose inductively that for any $\pi'$ with
    $\RM(\pi')<k$ we have $\dl(\pi') \leq  \RM(\pi')$.

    Let $p \in S_G(M)$ complete $\pi$ with $\dl(p) = \dl(\pi)$. It suffices to
    show that $\dl(p) \leq  \RM(p)$, since then
    $\dl(\pi) = \dl(p) \leq  \RM(p) \leq  \RM(\pi)$.

    For $p$ algebraic, clearly $\dl(p) = 0 = \RM(p)$.

    So suppose $k>0$, and suppose for a contradiction that $\dl(p) > k$.

    \begin{claim} \label{fibClm}
        Suppose $f : G^n \rightarrow  G^m$ is a function definable over $M$ such that
        $f(\Gm^n) \subseteq  \Gm^m$, and let $r \in S_{G^n}(M)$.
	Suppose $f$ and $r$ are as in Lemma~\ref{dlprops}(iii),
	and suppose $\alpha := \RM(r)-\RM(f_*(r)) < k$.
	Then $\dl(r) \leq  \dl(f_*(r)) + \alpha$.
    \end{claim}
    \begin{proof}
	By definability of $\RM$,
	exists $X \in r$ such that for all $b \in f(X)$,
	$\RM(f|_X^{-1}(b)) = \alpha$.

	Then since $\alpha < k = \RM(p)$,
	for all $b \in f(X)^M$,
	$\dl(f|_X^{-1}(b)) \leq  \RM(f|_X^{-1}(b)) = \alpha$.

	So by Lemma~\ref{dlprops}(iii),
	$\dl(r) \leq  \dl(f_*(r)) + \alpha$.
    \end{proof}

    \begin{claim} \label{prodClm}
	For $q \in S(M)$, $\dl(p \otimes q) = \dl(p) + \dl(q)$.
    \end{claim}
    \begin{proof}
	Let $(p\times q)(x,y) := p(x) \cup q(y)$.
	Let $r(x,y) \in S(M)$ complete $p\times q$ with $\dl(r) =
	\dl(p\times q) = \dl(p) + \dl(q)$.

	Suppose for a contradiction that $r \neq  p \otimes q$.
	So $\RM(r) < \RM(p) + \RM(q)$, so $\RM(r) - \RM(q) < \RM(p) = k$.

	So by Claim~\ref{fibClm},
	$\dl(r) < \dl(q) + k < \dl(q) + \dl(p)$,
	contradicting the choice of $r$.
    \end{proof}

    Suppose that $p \in S_G(M)$.

    Let $p=p_1,\ldots,p_n=\pp_G|_M$ and $\gm_1,\ldots,\gm_{n-1}$
    be as in Lemma~\ref{ZIT}.

    So for $1 \leq  i < n$,
    $m^{\gm_i}_*(p \otimes p_i) = p_{i+1}$ and
    $\alpha_i := \RM(p \otimes p_i) - \RM(p_{i+1}) = \RM(p)+\RM(p_i) - \RM(p_{i+1}) < \RM(p) = k$.

    So by Claims~\ref{prodClm} and \ref{fibClm},
    $\dl(p) + \dl(p_i) = \dl(p \otimes p_i) \leq  \dl(p_{i+1}) + \alpha_i$,
    so
    $\dl(p_i) \leq  \dl(p_{i+1}) + \alpha_i - \dl(p)$.

    So $\dl(p) = \dl(p_1) \leq  \dl(p_n) + \sum \alpha_i - (n-1)\dl(p)$,
    so $n\dl(p) \leq  \dl(\pp_G) + \sum \alpha_i$.

    Meanwhile,
    $\RM(p_i) = \RM(p_{i+1}) + \alpha_i - \RM(p)$,
    so $n\RM(p) = \RM(\pp_G) + \sum \alpha_i$.

    By the normalisation, $\dl(\pp_G) \leq  \dl(G) = \RM(G) = \RM(\pp_G)$.
    So $n\dl(p) \leq  n\RM(p)$,
    so $\dl(p) \leq  \RM(p)$,
    contradicting the contrary assumption.

    So if $p \in S_G(M)$, $\dl(p) \leq  \RM(p)$.

    Then by induction on $n$, this holds for $p \in S_{G^n}(M)$, by considering
    co-ordinate projections and definability of $\RM$, as in
    Claim~\ref{fibClm}.

    For $p$ on other sorts, by definition $\dl(p) = 0 \leq  \RM(p)$, so we are
    done.
\end{proof}
\end{document}
