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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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\date{26.01.2021}

!newop trd
!newop ACF
!newop Stab
!newop Zar
!newop cl
!newop an
!newop alg
!newop tr
!newop GL

!newbb Z
!newbb Q
!newbb N
!newbb C
!newbb R
!newbb G

!renewbb H
!renewbb P

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\begin{document}
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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".

# Manin-Mumford
***) $K <= \C$ number field.
** $G$ complex abelian variety (i.e. projective connected non-trivial 
algebraic group) over $K$.

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

Theorem*[Manin-Mumford Conjecture; Raynaud (1983)]:--
  Let $X (= G$ be an irreducible (complex) subvariety and suppose $X \cap 
  G[\infty]$ is Zariski dense in $X$. Then $X$ is a torsion coset.
  .

Corollary*:
  For $X (= G$ Zariski closed,
  $$X \cap G[\infty] = \bigcup_{i=0}^n \xi_i + H_i[\infty]$$
  where $\xi_i (- G[\infty]$ and $H_i <= G$ are abelian subvarieties.
  .
Proof:
  *\[a* X \cap G[\infty] &= (X \cap G[\infty])^\Zar \cap G[\infty] \\\
  &= ({ \bigcup_{i=0}^n \xi_i + H_i }) \cap G[\infty] \\\
  &= \bigcup_{i=0}^n ({ (\xi_i + H_i) \cap G[\infty] }) \\\
  &= \bigcup_{i=0}^n \xi_i + H_i[\infty] *a\]*
.

## Special Locus

!newop SpL

***) Let $X (=_\cl G$.

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

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

** The \defn{stabiliser} of $X$ in $G$ is
$$\Stab(X) = \Stab_G(X) := \{ g (- G : g+X = X \}.$$
*e)*

Lemma*{splStab}:
  Suppose $X$ is irreducible.

  Then $\SpL(X) = X$ iff $\Stab(X)$ is infinite.
  .
Proof:--
  ***) $\Stab(X) = \bigcap_{x (- X} (X-x)$, so $\Stab(X) <=_\cl G$, so 
  $\Stab(X)$ is an algebraic subgroup.

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

  ** 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 $H <= \Stab(X)$.
  *e)*
  .

Lemma*:
  $\SpL(X) (=_\cl X$.
  .
Proof:--
  ***) We actually prove something stronger:
  Claim*:
    If $(X_a)_{a (- A}$ is a constructible family of subvarieties of $G$,
    \\then $(\SpL(X_a))_{a (- A}$ is a constructible family of subvarieties of 
    $G$.
    .

   Where
   ***) a \defn{constructible} subset of a variety is a boolean combination of 
   Zariski closed subsets;

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

   ** $(X_a)_{a (- A}$ is a \defn{constructible family} of subvarieties of $G$ 
   if
     ***) $A$ is constructible,
     ** $X (= G \times A$ is constructible, and
     ** $X_a := \{ g (- G : (g,a) (- X \} (= G$ is Zariski closed in $G$ for all 
     $a$.
     *e)*
   *e)*

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

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

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

  ** Then it suffices to show:

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

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

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

  Fact*:
    There exists a constructible $(X'_{a'})_{a (- A'}$ and a constructible map 
    $\alpha : A' --> A$ such that $( X'_{a'} : \alpha(a') = a )$ are the 
    irreducible components of $X_a$.
    .

  ** 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 $A' := \{ a (- A : |\Stab(X_a)| <= N \}$ is 
  constructible. So to perform the reduction, it suffices to see that $(*)$ 
  holds for $(X_a)_{a (- A \\ A'}$.

  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 $a (- A \\ A'$ 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 $H' <= H_a$ 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 $A \\ 
  %A'$ is covered by finitely many such subsets (by logical compactness), so 
  %indeed finitely many such $H_{a_i}$ suffice.

  ** Now suppose $H_0 <= G$ appears maximally in $(X_a)_a$. Say $H_0$ appears 
  maximally in $X_a$.

  ** Let $h (- H_0 \\ \Stab(X_a)$.

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

  ** Since $X'_{a,h} (= X_a$, actually $H_0$ appears maximally in $X'_{a,h}$.

  ** Now $$(X'_{a,g} : a (- A,\; g (- G,\; X'_{a,g} /= X_a)$$
  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.
  *e)*
  .

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

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

***) Proved using differential algebra.

** Generalisations:\
  ***) Brownawell-Kubota: for elliptic curves;
  ** Kirby: for arbitrary (semi-)abelian varieties.
  ** Ax '72: general analytic version for arbitrary complex algebraic groups.
  We will use this.
  *e)*
*e)*

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

  \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.

## Algebraic part

Lemma*:
  $\e_|^{-1}(X)^\alg = \e_|^{-1}(\SpL(X))$.
  .
Proof:--
  ***) We show $\e^{-1}(X)^\alg = \e^{-1}(\SpL(X))$.
  ** Let $x (- \e^{-1}(X)^\alg$, and suppose $x (/- \e^{-1}(\SpL(X))$.
  ** Replacing $X$ with $X-\e(x)$, we may assume $x=0$.
  ** So $0 (- C' (= \e^{-1}(X)$ for a semialgebraic curve $C'$.
  ** So $0 (- C (= \e^{-1}(X)$ for an irreducible algebraic curve $C$.
  ** Replace $G$ with the smallest abelian subvariety $G'$ containing $\e(C)$, 
  and $X$ with $X \cap G'$.

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

  So $<{ K^\Zar }> + \Gamma = \H$.
  ** Meanwhile $X /= G$ since $0 (/- \SpL(X)$.
  ** By Ax, $\dim K <= \dim K^\Zar - ({ \dim\H - \dim \Gamma })$, so:
  *\[a* \dim(G) &= \dim(\H) - \dim(\Gamma) \\\
  &<= \dim K^\Zar - \dim K \\\
  &<= \dim(C \times X) - 1 \\\
  &<= \dim(X) \\\
  &< \dim(G). *a\]*
  Contradiction.
  *e)*
  .

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

# Masser

!newop ord
!newop Gal

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

Theorem*[Masser '84]:
  Exist $\lambda=\lambda(g)>0$ and $C>0$ such that for $\xi (- G[\infty]$ of 
  order $k$,
  $$\deg(K(\xi)/K) >= Ck^\lambda.$$
  .

# Pila-Zannier

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

Suffices to show:

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

  Then $(X \\ \SpL(X)) \cap G[\infty]$ is finite.
  .

Proof[Proof of Theorem from Lemma]:--
  ***) Let $H$ be the connected component of the stabiliser $H := 
  \Stab_G(X)^o$.

  So $H <= G$ is an abelian subvariety or the trivial subgroup.

  ** Then (fact) $G/H$ is an abelian variety, and $X/H (= G/H$ 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 $H$).

  ** $X$ is over $\Q^\alg$, because $G[\infty] (= G(\Q^\alg)$ and so $X$ is 
  $\"Aut(\C/\Q^\alg)$-invariant.

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

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

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

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

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

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

  ** 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.
  *e)*
  .

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

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

%\bibliography{pilaZannier}
\end{document}
