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\title{The Pila-Zannier proof of Manin-Mumford}

\author%[Author, Another] % (optional, use only with lots of authors)
{Martin Bays}
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%   affiliation.

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\date{26.01.2021}

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\begin{document}
\maketitle

\tableofcontents

Notes for a talk in a Münster seminar, based largely on section 5.1 in Tom
Scanlon's paper "O-minimality as an approach to the André-Oort conjecture".

\section{Manin-Mumford}
\begin{itemize}\item $K \leq  \C$ number field.
\item $G$ complex abelian variety (i.e.\ projective connected non-trivial 
algebraic group) over $K$.

(e.g.\ $G = E^g$ for $E$ an elliptic curve over $K$.)
\item $g := \dim(G)$.
\item Let $\exp : LG \rightarrow  G$ be the exponential map of $G = G(\C)$ as a complex 
Lie group.
\item $LG = T_0(G)$ is a $g$-dimensional $\C$-vector space.
\item $\exp$ is a surjective holomorphic homomorphism.
\item $\Lambda := \ker(\exp) \cong  \Z^{2g}$ is a full lattice in $LG$,
i.e.\ $\Lambda = \left<{\lambda_1,\ldots ,\lambda_{2g}}\right>_\Z$ where $(\lambda_i)_i$ is 
an $\R$-basis of $LG$.
\item So as a complex Lie group, $G \cong  \C^g / \Lambda$ is a complex torus, and is 
diffeomorphic to $(\R/\Z)^{2g}$.
\item Torsion subgroup: $G[\infty] \cong  (\Q/\Z)^{2g}$.
\item The non-trivial connected algebraic subgroups of $G$ are precisely the 
abelian subvarieties.
\item $G$ has only countably many abelian subvarieties, and each is defined over 
$\Q^\alg$.
\item $G[\infty]$ is Zariski dense in $G$.
\item A \defn{torsion coset} is a coset $H+\xi$ of an abelian subvariety $H \leq  G$ 
by a torsion point $\xi \in G[\infty]$.
\end{itemize}

\begin{theorem*}[Manin-Mumford Conjecture; Raynaud (1983)] \mbox{}
  Let $X \subseteq  G$ be an irreducible (complex) subvariety and suppose $X \cap 
  G[\infty]$ is Zariski dense in $X$. Then $X$ is a torsion coset.
\end{theorem*}

\begin{corollary*}
  For $X \subseteq  G$ Zariski closed,
  $$X \cap G[\infty] = \bigcup_{i=0}^n \xi_i + H_i[\infty]$$
  where $\xi_i \in G[\infty]$ and $H_i \leq  G$ are abelian subvarieties.
\end{corollary*}
\begin{proof}
  \begin{align*} X \cap G[\infty] &= (X \cap G[\infty])^\Zar \cap G[\infty] \\
  &= \left({\bigcup_{i=0}^n \xi_i + H_i}\right) \cap G[\infty] \\
  &= \bigcup_{i=0}^n \left({(\xi_i + H_i) \cap G[\infty]}\right) \\
  &= \bigcup_{i=0}^n \xi_i + H_i[\infty] \end{align*}
\end{proof}

\subsection{Special Locus}

\newcommand{\SpL}{{\operatorname{SpL}}}

\begin{itemize}\item Let $X \subseteq _\cl G$.

\item The \defn{special locus} (or Ueno locus, or Kawamata locus) of $X$ in $G$ 
is
%$$\SpL(X) := \bigcup \{ g + H : g + H \subseteq  X,\; g \in G,\; H \leq  G \text{ abelian 
%subvariety}\}.$$ 
$$\SpL(X) := \bigcup \{ \SpL(X,H) : H \leq  G \text{ abelian subvariety}\},$$ 
where
$$\SpL(X,H) := \bigcup \{ g + H : g + H \subseteq  X,\; g \in G \} = \bigcap_{h \in H} 
(X-h).$$

We will see that $\SpL(X)$ corresponds to the Pila-Wilkie `algebraic part' of 
$\exp^{-1}(X)$.

\item The \defn{stabiliser} of $X$ in $G$ is
$$\Stab(X) = \Stab_G(X) := \{ g \in G : g+X = X \}.$$
\end{itemize}

\begin{lemma*} \label{l:splStab}
  Suppose $X$ is irreducible.

  Then $\SpL(X) = X$ iff $\Stab(X)$ is infinite.
\end{lemma*}
\begin{proof} \mbox{}
  \begin{itemize}\item $\Stab(X) = \bigcap_{x \in X} (X-x)$, so $\Stab(X) \leq _\cl G$, so 
  $\Stab(X)$ is an algebraic subgroup.

  \item If $\Stab(X)$ is infinite, then the connected component of the identity 
  $H := \Stab(X)^o \leq  G$ is an abelian subvariety, so $X = \bigcup \{ x + H : 
  x \in X \} = \SpL(X)$.

  \item Conversely: If $X = \SpL(X)$, then
  $X = \SpL(X,H)$ for some $H$, since each $\SpL(X,H)$ is closed and $G$ has 
  only countably many abelian subvarieties and $X$ is irreducible.

  So $H \leq  \Stab(X)$.
  \end{itemize}
\end{proof}

\begin{lemma*}
  $\SpL(X) \subseteq _\cl X$.
\end{lemma*}
\begin{proof} \mbox{}
  \begin{itemize}\item We actually prove something stronger:
  \begin{claim*}
    If $(X_a)_{a \in A}$ is a constructible family of subvarieties of $G$,
    \\then $(\SpL(X_a))_{a \in A}$ is a constructible family of subvarieties of 
    $G$.
  \end{claim*}

   Where
   \begin{itemize}\item a \defn{constructible} subset of a variety is a boolean combination of 
   Zariski closed subsets;

   (\emph{Fact: constructible $\Leftrightarrow $ definable in $(\C;+,\cdot)$.})

   \item $(X_a)_{a \in A}$ is a \defn{constructible family} of subvarieties of $G$ 
   if
     \begin{itemize}\item $A$ is constructible,
     \item $X \subseteq  G \times A$ is constructible, and
     \item $X_a := \{ g \in G : (g,a) \in X \} \subseteq  G$ is Zariski closed in $G$ for all 
     $a$.
     \end{itemize}
   \end{itemize}

  \item Since each $\SpL(X_a,H)$ is closed, it suffices to show that there are 
  finitely many $H_1,\ldots ,H_n$ such that for all $a \in A$,
  $\SpL(X_a) = \bigcup_i \SpL(X_a,H_i)$.

  \item Say an abelian subvariety $H\leq G$ \defn{appears maximally} in $X \subseteq  G$ if 
  $X$ contains some coset $\gamma+H$ which is maximal among the cosets of 
  abelian subvarieties contained in $X$.

  \item Say $H$ \defn{appears maximally} in $(X_a)_a$ if it appears maximally in 
  some $X_a$.

  \item Then it suffices to show:

  $(*)$ Only finitely many $H$ appear maximally in $(X_a)_a$.

  \item We prove $(*)$ by induction on $d := \max_a \dim(X_a)$.

  \item We may assume that each $X_a$ is irreducible, by:

  \begin{fact*}
    There exists a constructible $(X'_{a'})_{a \in A'}$ and a constructible map 
    $\alpha : A' \rightarrow  A$ such that $( X'_{a'} : \alpha(a') = a )$ are the 
    irreducible components of $X_a$.
  \end{fact*}

  \item We may also assume that each $X_a$ has finite stabiliser.
  
  Indeed, there is (fact) a uniform bound $N$ on the size of the finite 
  stabilisers $\Stab(X_a)$, and $A' := \{ a \in A : |\Stab(X_a)| \leq  N \}$ is 
  constructible. So to perform the reduction, it suffices to see that $(*)$ 
  holds for $(X_a)_{a \in A \setminus A'}$.

  UPDATE: the argument previously written here for that was nonsense. Thanks 
  to Zoé Chatzidakis for pointing this out. One should proceed by induction by 
  quotienting by the connected component of the stabiliser.
  
  % Old crap argument follows
  %But for $a \in A \setminus A'$ we have $\SpL(X_a) = X_a$ by the previous Lemma, so 
  %by irreducibility $\SpL(X_a) = \bigcup_{H : H \text{ appears maximally in 
  %$X_a$}} \SpL(X_a,H)$ is equal to a single $\SpL(X_a,H_a)$ with $H_a$ 
  %appearing maximally. Then if $X_a$ contains $\alpha+H'$, it also contains 
  %$\alpha+H'+H_a$, so $H' \leq  H_a$ by maximality.
  % (This "by maximality" is the nonsensical bit; it's the cosets which are 
  % maximal, not the subgroups!)
  %So $H_a$ is the only subgroup appearing maximally in $X_a$. Finally, 
  %$\SpL(X_a,H_a) = X_a$ holds for $a$ on a constructible subset, and $A \setminus 
  %A'$ is covered by finitely many such subsets (by logical compactness), so 
  %indeed finitely many such $H_{a_i}$ suffice.

  \item Now suppose $H_0 \leq  G$ appears maximally in $(X_a)_a$. Say $H_0$ appears 
  maximally in $X_a$.

  \item Let $h \in H_0 \setminus \Stab(X_a)$.

  \item Then $X'_{a,h} := X_a \cap (h+X_a)$ is a proper subvariety of $X_a$ and 
  contains a coset of $H_0$.

  \item Since $X'_{a,h} \subseteq  X_a$, actually $H_0$ appears maximally in $X'_{a,h}$.

  \item Now $$(X'_{a,g} : a \in A,\; g \in G,\; X'_{a,g} \neq  X_a)$$
  is a constructible family of subvarieties of $G$ each of dimension less than 
  $d$. By the inductive hypothesis, only finitely many $H$ appear maximally in 
  it. So $H_0$ is one of these finitely many, as required.
  \end{itemize}
\end{proof}

\section{Ax-Schanuel}
\subsection{Restricted exponentiation}
\newcommand{\F}{\mathcal{F}}
\begin{itemize}\item Recall $\Lambda = \ker\exp$ is freely generated by an $\R$-basis 
$\lambda_1,\ldots ,\lambda_{2g}$ of $LG \cong  \C^g$.
\item We identify $LG$ with $\R^{2g}$ with respect to this basis (instead of 
taking real and imaginary parts),
$$0 \rightarrow  \Z^{2g} \rightarrow  \R^{2g} \rightarrow^\e G \rightarrow  0.$$
\item Let $\F := [0,1)^{2g} \subseteq  LG$ (``fundamental domain''),
so the restriction $\exp_| : \F \rightarrow  G$ is a bijection.
\item $\exp_|$ is definable in $\R_{\an}$.
\item For $X \subseteq  G$, $\exp_|$ yields a bijection $\exp_|^{-1}(X) \cap \Q^{2g} \rightarrow  
X \cap G[\infty]$.
\item To apply Pila-Wilkie, we must determine $\exp_|^{-1}(X)^{\alg}$.
\item The key tool for this is the Ax-Schanuel theorem (or its 
``Lindemann-Weierstrass case'').
\end{itemize}

\subsection{Ax-Schanuel}
The original Ax-Schanuel theorem concerns usual complex exponentiation (i.e.\ 
the exponential map of the multiplicative group):
\begin{fact*}[Ax '71]
  Suppose $f_i : \Delta \rightarrow  \C$ are holomorphic functions on the unit disc, 
  and $f'_1(0),\ldots ,f'_n(0)$ are $\Q$-linearly independent. Then 
  $$\trd(f_1(t),\ldots ,f_n(t),e^{f_1(t)},\ldots ,e^{f_n(t)}/\C) \geq  n+1.$$
\end{fact*}

\begin{itemize}\item Proved using differential algebra.

\item Generalisations:\
  \begin{itemize}\item Brownawell-Kubota: for elliptic curves;
  \item Kirby: for arbitrary (semi-)abelian varieties.
  \item Ax '72: general analytic version for arbitrary complex algebraic groups.
  We will use this.
  \end{itemize}
\end{itemize}

\begin{fact*}[Ax '72]
  Let
  \begin{itemize}\item $\H$ be a complex algebraic group,
  \item $\Gamma \leq  \H$ a connected analytic subgroup, and
  %$U$ is open in $\H$ in the complex topology and
  \item $e \in K \subseteq  \Gamma$ an irreducible analytic subvariety.
  %of $U$
  \end{itemize}
  %, and $H = \left<{K^\Zar}\right>$ is the smallest algebraic subgroup of $\H$ 
  %containing $K$.
  Then there is an analytic subgroup $\H' \leq  \H$ containing $K^\Zar$ and 
  $\Gamma$ such that:
  \[ \dim K \leq  \dim K^\Zar - \left({\dim\H' - \dim \Gamma}\right) .\]
\end{fact*}

  \vspace{1em}
Idea: If $K$ is a component of $K^\Zar \cap \Gamma$, the ``expected'' 
dimension is $$\dim K = \dim K^\Zar - (\dim \H - \dim \Gamma);$$
Ax's theorem says this can be exceeded only if the intersection is really 
happening in a smaller group.

\subsection{Algebraic part}

\begin{lemma*}
  $\e_|^{-1}(X)^\alg = \e_|^{-1}(\SpL(X))$.
\end{lemma*}
\begin{proof} \mbox{}
  \begin{itemize}\item We show $\e^{-1}(X)^\alg = \e^{-1}(\SpL(X))$.
  \item Let $x \in \e^{-1}(X)^\alg$, and suppose $x \notin \e^{-1}(\SpL(X))$.
  \item Replacing $X$ with $X-\e(x)$, we may assume $x=0$.
  \item So $0 \in C' \subseteq  \e^{-1}(X)$ for a semialgebraic curve $C'$.
  \item So $0 \in C \subseteq  \e^{-1}(X)$ for an irreducible algebraic curve $C$.
  \item Replace $G$ with the smallest abelian subvariety $G'$ containing $\e(C)$, 
  and $X$ with $X \cap G'$.

    We still have $0 \notin \SpL(X)$.
  \item Consider $\Gamma_\e \leq  LG \times G =: \H$.
  \item Let $K \ni (0,0)$ be an analytic irreducible component of $(C \times X) 
  \cap \Gamma_\e = \Gamma_{\e|_C}$.
  \item Now $\left<{\e(C)^\Zar}\right> = G$ by assumption, so $\pi_2(\left<{K^\Zar}\right>) = G$, 
  so $\left<{K^\Zar}\right> \supseteq \{0\} \times G$ (by consideration of the algebraic 
  subgroups of $LG \times G$).

  So $\left<{K^\Zar}\right> + \Gamma = \H$.
  \item Meanwhile $X \neq  G$ since $0 \notin \SpL(X)$.
  \item By Ax, $\dim K \leq  \dim K^\Zar - \left({\dim\H - \dim \Gamma}\right)$, so:
  \begin{align*} \dim(G) &= \dim(\H) - \dim(\Gamma) \\
  &\leq  \dim K^\Zar - \dim K \\
  &\leq  \dim(C \times X) - 1 \\
  &\leq  \dim(X) \\
  &< \dim(G). \end{align*}
  Contradiction.
  \end{itemize}
\end{proof}

\begin{remark*}
  Kawamata \emph{On Bloch's Conjecture} (1980, Inventiones) gives an 
  alternative proof of this (and also of the closedness of $\SpL(X)$).
\end{remark*}

\section{Masser}

\newcommand{\ord}{{\operatorname{ord}}}
\newcommand{\Gal}{{\operatorname{Gal}}}

\begin{itemize}\item Let $\gamma \in G(\Q^\alg)$.
\item In co-ordinates according to our projective embedding (possibly permuting 
the co-ordinates),
$$\gamma = [1:\gamma_1:\ldots .:\gamma_n] \in G \subseteq  \P^n(\C)$$
with $\gamma_i \in \Q^\alg$.
\item Then $K(\gamma) := K(\gamma_1,\ldots ,\gamma_n)$.
\item For $\sigma \in \Gal(\Q^\alg/K)$,
$$\sigma(\gamma) := [1:\sigma(\gamma_1):\ldots :\sigma(\gamma_n)] \in G.$$
\item $\deg(K(\gamma)/K)$ is the size of the orbit of $\gamma$ under 
$\Gal(\Q^\alg/K)$.
\end{itemize}

\begin{theorem*}[Masser '84]
  Exist $\lambda=\lambda(g)>0$ and $C>0$ such that for $\xi \in G[\infty]$ of 
  order $k$,
  $$\deg(K(\xi)/K) \geq  Ck^\lambda.$$
\end{theorem*}

\section{Pila-Zannier}

\subsection{Reduction}
\begin{theorem*}[Manin-Mumford Conjecture]
  If $X \subseteq _{\cl} G$ is irreducible and $(X \cap G[\infty])^{\Zar} = X$ then 
  $X$ is a torsion coset.
\end{theorem*}

Suffices to show:

\begin{lemma*}
  Suppose $X \subseteq _{\cl} G$ is irreducible and defined over $\Q^\alg$ and suppose 
  $\Stab_G(X)$ is finite.

  Then $(X \setminus \SpL(X)) \cap G[\infty]$ is finite.
\end{lemma*}

\begin{proof}[Proof of Theorem from Lemma] \mbox{}
  \begin{itemize}\item Let $H$ be the connected component of the stabiliser $H := 
  \Stab_G(X)^o$.

  So $H \leq  G$ is an abelian subvariety or the trivial subgroup.

  \item Then (fact) $G/H$ is an abelian variety, and $X/H \subseteq  G/H$ is an 
  irreducible subvariety, and $\Stab_{G/H}(X/H)$ is finite.

  (Indeed, if $\Stab_{G/H}(X/H)$ contains an infinite algebraic subgroup $S$, 
  then $S' = \pi_H^{-1}(S)$ stabilises $X = \pi_H^{-1}(X/H)$, but $\dim(S') = 
  \dim(S) + \dim(H) > \dim(H)$, contradicting the choice of $H$).

  \item $X$ is over $\Q^\alg$, because $G[\infty] \subseteq  G(\Q^\alg)$ and so $X$ is 
  $\operatorname{Aut}(\C/\Q^\alg)$-invariant.

  Also $H$ is over $\Q^\alg$.

  So $X/H$ is over $\Q^\alg$.

  \item So by the Lemma,
  $(X/H \setminus \SpL(X/H)) \cap (G/H)[\infty]$ is finite.

  \item But: $$\pi_H^{-1}(X/H \cap (G/H)[\infty]) \supseteq X \cap G[\infty],$$
  so $$\pi_H^{-1}((X/H \cap (G/H)[\infty])^\Zar) \supseteq (X\cap G[\infty])^\Zar = 
  X,$$
  so $$(X/H \cap (G/H)[\infty])^\Zar = X/H.$$

  \item Now $\Stab_{G/H}(X/H)$ is finite, so $\SpL(X/H)$ is a proper closed 
  subvariety of $X/H$.
  
  \item So the finite set $(X/H \setminus \SpL(X/H)) \cap (G/H)[\infty]$ is Zariski 
  dense in $X/H$.

  \item So $X/H = \{\xi\}$ for some $\xi \in (G/H)[\infty]$.

  \item So $X = \pi^{-1}(\xi)$ is a coset of $H$, and it contains a torsion point 
  since $X \cap G[\infty]$ is dense in $X$, so $X$ is a torsion coset.
  \end{itemize}
\end{proof}

\subsection{Concluding by Pila-Wilkie}
\begin{proof}[Proof of Lemma] \mbox{}
  \begin{itemize}\item Let $X \subseteq _{\cl} G$ be irreducible and over $\Q^\alg$ with $\Stab_G(X)$ 
  finite.
  \item WTS: $(X \setminus \SpL(X)) \cap G[\infty]$ is finite. Suppose not.
  \item Increasing $K$, we may assume $X$ is over $K$.

  Then $\SpL(X)$ is also over $K$.
  \item So by Masser: For arbitrarily large $k$, there are $Ck^\lambda$ points of 
  order $k$ in $X \setminus \SpL(X)$.
  \item If $\xi$ has order $k$, then the height of $$\exp_|^{-1}(\xi) \in
  \exp_|^{-1}(X \setminus \SpL(X)) \cap \Q^{2g} = \exp_|^{-1}(X)^\tr \cap \Q^{2g}$$
  is at most $k$.
  \item But by Pila-Wilkie, exists $C'>0$ such that $$N(\exp_|^{-1}(X)^\tr,k) \leq  
  C'k^{\frac\lambda2}.$$
  \item For large enough $k$, these bounds contradict each other, i.e.\ 
  $$Ck^\lambda > C'k^{\frac\lambda2}.$$
  \end{itemize}
\end{proof}

%\bibliography{pilaZannier}
\end{document}
