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\title{Manin kernels and exponentials\input{version.tex}}

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\begin{document}

\author{Martin Bays and Anand Pillay}

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\maketitle

Let $A$ be an abelian variety over the differential function field $K$ of a
complex algebraic curve $S$ with a rational vector field.

Recall the ``Manin kernel'', a DCF-definable/$K$ subset $A^\#$ of $A$
admitting a number of descriptions, one being that it is the smallest
Zariski-dense DCF-definable subset of $A$.

Let $T$ be the theory of $A^\#$ with all induced structure (over $K$).

This is a rigid divisible commutative group of finite Morley rank.

Let $^T$ be the theory of the universal cover of $A^\#$, in the sense of
[covers-fRM].
The aim of this note is to exhibit a natural analytic model of $^T$.
We first describe this model.

Let
  \[ \xymatrix{ 0 \ar[r] & H \ar[r] & G \ar[r]^p & A \ar[r] & 0 } \]
be the universal vector extension of $A$.

$G$ is equipped with a canonical $D$-structure, i.e. a rational section
    \[ s_G : G --> \tau G \]
where $\tau G --> G$ is the twisted tangent bundle (aka first prolongation).

Let $G^\d$ be the subgroup of ``horizontal'' points, i.e. for a differential field
extension $K' >= K$,
    \[ G^\d(K') = \{ x \in G(K') | (x,\d x) = s_G(x) \} .\]

Then for $\mathcal{U} |= DCF$, $A^\#(\mathcal{U}) = p(G^\d(\mathcal{U}))$
[Marker-maninKernels].

Taking the Lie algebras of these algebraic groups, we have
\[ \xymatrix{ 0 \ar[r] & LH \ar[r] & LG \ar[r]^{Lp} & LA \ar[r] & 0 } \]

Now LG also has a natural $D$-structure induced from that on $G$,
    \[ Ls_G : LG --> L\tau G ~= \tau LG \]
(c.f. [BP]),
so we have subgroups $LG^\d$ and $LA^\# := Lp(LG^\d)$.

Now let $S' (= S$ be a disc (or in fact any simply-connected domain) in $S$,
suppose $S'$ avoids the finitely many $s\in S$ for which $A_s$ is not an
abelian variety,
and let $L >= K$ be the differential field of meromorphic functions on $S'$.
We consider $L$-points, where we define
\begin{align*}
  A^\#(L)  &:= p(G^\d(L)) \\
  LA^\#(L) &:= p(LG^\d(L)) .
\end{align*}{

As in [BP Appendix], we have relative exponential maps
\[ \xymatrix{
    LG(L) \ar[d]_{\exp_G} \ar@{>>}[r] & LA(L) \ar[d]_{\exp_A} \\
                          G(L) \ar@{>>}[r] & A(L)
	} \]
which respect the $D$-structures, hence restrict to
\[ \xymatrix{
    LG^\d(L) \ar[d]_{\exp_G} \ar@{>>}[r] & LA^\#(L) \ar[d]_{\exp_A} \\
                          G^\d(L) \ar@{>>}[r] & A^\#(L)
	} .\]

Now our claim is that,
    \[\exp_A : LA^\#(L) --> A^\#(L) ,\]
when considered as a structure in the language of $^T$ (we discuss below exactly
how it may be so considered), is a model of $^T$.

Let us remark that in the constant case, i.e. when $A$ is over $\C$, the
$D$-structures on $G$ and $LG$ are trivial, and $A^\#(L) = A(\C)$, and
$LA^\#(L) = LA^\#(\C)$, and $\exp_A$ is the usual complex exponential map; so
we are reduced to the case of [covers-fRM Corollary 4.2.1].

We begin collecting some facts, (I)-(III) below, which we will need in the
proof our structure satisfies the axioms of $^T$.

By a remark credited to Hamm ([BuiumDiffAlgDiophGeom p.143]), over $S'$, $G$
analytically descends to the constants. In terms of $L$-points, this has the
following consequence:

\begin{fact}
    Let $s_0 (- S'$.
    Let $G_0 := G_{s_0}$ be the fibre of $G$ over $s_0$, a complex Lie group,
    and let
	\[ \exp_{G_0} : LG_0 --> G_0 \]
    be its exponential map.
    There exists an isomorphism
	\[ \th_G : G(L) --> G_0(L) \]
    and a corresponding $\C$-linear isomorphism
	\[ L\th_G : LG(L) --> LG_0(L) \]
    such that $\th_G(G^\d(L)) = G_0(\C)$,
    and $L\th_G(LG^\d(L)) = LG_0(\C)$,
    and $\exp_G \o L\th_G = \th_G \o \exp_{G_0}$.
\end{fact}

It follows that $LG^\d(L)$ is a $2g$-dimensional $\C$-vector space, and
$\ker\exp_G <= LG^\d(L)$, and hence 
\begin{equation*} \tag{I}\label{f1}
  \ker \exp_A <= LA^\#
\end{equation*}

Since $\exp_{G_0} : LG_0(\C) -->> G_0(\C)$, it also follows that
$\exp_G : LG^\d(L) -->> G^\d(L)$,
and it follows by diagram-chase that
\begin{equation*} \tag{II}\label{f2}
    \exp_A : LA^\#(L) -->> A^\#(L) .
\end{equation*}

So we have:
\[ \xymatrix{
     LG^\d(L) \ar@{>>}[d]_{\exp_G} \ar@{>>}[r] & LA^\#(L) \ar@{>>}[d]_{\exp_A} \\
                          G^\d(L) \ar@{>>}[r] & A^\#(L)
	  } .\]

\begin{lemma}
  \begin{equation*} \tag{III}\label{f3}
    A^\#(L^{\"diff}) = A^\#(L) .
  \end{equation*}
\end{lemma}
\begin{proof}
    We first show $LG^\d(L^{\"diff}) = LG^\d(L)$.

    Let $X$ be a $\C$-basis of $LG^\d$. For any subdisc $S'' (= S'$ and corresponding
    field $L' >= L$ of meromorphic functions, by the above Fact, $LG^\d(L')$ is
    still a 2g-dimensional $\C$-vector space, so $LG^\d(L') = LG^\d(L) = <X>_\C$.
    
    So by the (proof of) the Seidenberg Embedding Theorem [MarkerDCF Lemma
    A.1], for any $y \in LG^\d(L^{\"diff})$, we have $y \in
    <X>_{\C(L^{\"diff})} = <X>_\C = LG^\d(L)$.

    So $LG^\d(L^{\"diff}) = LG^\d(L)$, and hence $LA^\#(L^{\"diff}) = LA^\#(L)$.

    Now $G^\d(L') = \exp_G(LG^\d(L')) = \exp_G(LG^\d(L)) = G^\d(L)$.
    By another Seidenberg argument applied to algebraic extensions, we
    therefore have $G^\d(L^{\"alg}) = G^\d(L)$.

    Hence $A^\#(L^{\"alg}) = A^\#(L)$. But it follows from [Wagner FieldsFRM] that
    $A^\#(L^{\"diff}) = A^\#(L^{\"alg})$; this is discussed in [BBP-MLMM], Corollary 1.11
    and proof of Theorem 1.1(i).
\end{proof}

%FIXME: Can we prove $G^\d(L^{\"diff}) = G^\d(L^{\"alg})$? Any truth in this old ``proof''?:
%   Now $G^\d(L^{\"diff}) = G^\d(L^{\"alg})$. Indeed, $G^\d$ is the smallest type-definable
%   subgroup of $G$ containing $Tor(G)$ [Marker 4.10], and so any
%   model-theoretically algebraically closed subset is an elementary submodel
%   [Wagner FieldsFRM] 

So $A^\#(L) |= T$.

From now on, we mostly omit explicit mention of $L$, writing $A^\#$ for $A^\#(L)$ and
so on.

To make 
    \[ \exp_A : LA^\# --> A^\# ,\]
a structure in the language of $^T$, it remains to define $^H <= (LA^\#)^n$ for
each connected definable subgroup $H <= (A^\#)^n$.

If $B$ is a connected algebraic subgroup of $A$, by [BBP 4.9] we have
\begin{equation*} \tag{*}\label{shexact}
    A^\# \cap B = B^\# .
\end{equation*}
Now $LB^\# (= LA^\#$, and by (\ref{f1}) and (\ref{shexact}) we have
\begin{equation*} \tag{IV}\label{f4}
    LA^\# \cap LB = LB^\# .
\end{equation*}

Now if $H$ is a connected definable subgroup of $A^\#$, then $H = B^\#$ where $B$ is the
Zariski closure of H; indeed, by (\ref{shexact}) we have $H <= A^\# \cap B = B^\#$,
and meanwhile $B^\# <= H$ since $B^\#$ is the smallest Zariski-dense definable
subgroup of $B$.

Note that $(A^n)^\# = (A^\#)^n$, and $L(A^n)^\# = (LA^\#)^n$.

So for $B$ a connected algebraic subgroup of $A^n$, we interpret $\^{B^\#}$ as $LB^\#$.

\begin{proposition}
    With the structure described above,
	\[ \exp_A : LA^\#(L) --> A^\#(L) \]
    is a model of $^T$.
\end{proposition}
\begin{proof}
    (A1) is by (\ref{f3}).

    (A2)-(A5) are clear from the setup and (\ref{f4}).

    (A6) is by (\ref{f2}).

    (A7) and (A8) follow from (\ref{f1}).

    (A9)(I): by (\ref{f4}), the exact sequence
	    \[ 0 --> L(K^o) --> LG --> LH --> 0 \]
	  remains exact on applying $(\cdot^{\#})$.

    (A9)(II) is by (\ref{f1}).
\end{proof}

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\end{document}
