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\svnid{$Id: 01-intro.tex 727 2024-05-06 20:00:54Z rousseau $}
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\section{The Albanese morphism in the presence of rational curves}
\subversionInfo

\begin{setting}\label{set:1}%
  Let $X$ be a compact Kähler manifold. Assume that an Albanese of the
  $\cC$-pair $(X,0)$ exists.
\end{setting}  


\begin{thm}\label{thm:1}%
  Assume Setting~\ref{set:1}.  Then, the Albanese morphism of $(X,0)$ maps all
  rational curves to points.
\end{thm}
\begin{proof}
  Let $C \subseteq X$ be any rational curve. The normalization of $C$ yields a
  diagram 
  \[
    \begin{tikzcd}[column sep=2cm]
      \bP¹ \ar[r, "n\text{, normalization}"'] \ar[rr, bend left=15, "\eta"] & C  \ar[r, "\text{inclusion}"'] & X.
    \end{tikzcd}
  \]
  Given that $X$ is smooth, recall from \cite[Ex.~8.6]{orbiAlb1} that $\eta$ is
  a $\cC$-morphism, between $\cC$-pairs $(\bP¹, 0)$ and $(X, 0)$. Since
  $\Alb(\bP¹,0)$ exists, the universal property of the Albanese yields a diagram
  \[
    \begin{tikzcd}[column sep=2cm]
      \bP¹ \ar[r, "\alb(\bP¹{,}0)"] \ar[d, "n"'] & \Alb(\bP¹,0) \ar[d] \\
      X \ar[r, "\alb(X{,}0)"'] & \Alb(X,0).
    \end{tikzcd}
  \]
  The claim follows immediately once we observe that $\Alb(\bP¹,0)$ is a point.
\end{proof}

\todo{There is nothing special about $\bP¹$ here. This works for every space with nontrivial $\cC$-Albanese.}

\begin{cor}\label{cor:2}%
  In Setting~\ref{set:1}, let $\mu : X \to Y$ be a morphism to a normal analytic
  variety.  If all fibres of $\mu$ are rationally chain connected, then
  $\alb(X,0)$ factors via $\mu$,
  \[
    \begin{tikzcd}
      X \ar[r, "\mu"'] \ar[rr, bend left=15, "\alb(X{,}0)"] & Y \ar[r, "\exists!\:\beta"'] & \Alb(X,0).
    \end{tikzcd}
  \]
  \qed  
\end{cor}

\begin{cor}\label{cor:3}%
  In Setting~\ref{set:1}, assume that $X$ is rationally connected.  Then,
  $\Alb(X,0)$ is a point.
\end{cor}


\begin{cor}\label{cor:4}%
  In Setting~\ref{set:1}, let $\mu : X \to Y$ be a bimeromorphic modification of
  a compact manifold $Y$.  Then, $\alb(X,0)$ factors via $\mu$,
  \[
    \begin{tikzcd}
      X \ar[r, "\mu"'] \ar[rr, bend left=15, "\alb(X{,}0)"] & Y \ar[r, "\exists!\:\beta"'] & \Alb(X,0),
    \end{tikzcd}
  \]
  the morphism $\beta$ is a $\cC$-morphism between the pairs $(Y,0)$ and
  $\Alb(X,0)$, and $\beta : (Y,0) \to \Alb(Y,0)$ is an Albanese of $(Y,0)$.
\end{cor}
\begin{proof}
  \todo{PENDING}
\end{proof}

\begin{cor}\label{cor:5}%
  In Setting~\ref{set:1}, let $Y$ be a compact Kähler manifold bimeromorphic to
  $X$. Then, an Albanese of $(Y,0)$ exists.  If $f : X \dasharrow Y$ is
  bimeromorphic, then there exists a unique morphism of $\cC$-pairs rendering
  the following diagram commutative,
  \[
    \begin{tikzcd}
      X \ar[d, "\alb(X{,}0)"'] \ar[r, dashed, "f"] & Y \ar[d, "\alb(Y{,}0)"'] \\
      \Alb(X,0) \ar[r, "\exists! \alb(f)"'] & \Alb(Y,0)
    \end{tikzcd}
  \]
\end{cor}
\begin{proof}
  \todo{PENDING}
\end{proof}

\begin{cor}\label{cor:6}%
  In Setting~\ref{set:1}, the automorphism group of $X$ and the Cremona group
  act on $\Alb(X,0)$ in a way that makes the morphism $\alb(X,0)$ equivariant.
\end{cor}
\begin{proof}
  \todo{PENDING}
\end{proof}

\todo{
  \begin{itemize}
    \item Need example where a rational variety has a 
    \item Factorization via minimal model.
    \item For varieties of general type, factorization via the canonical.
  \end{itemize}
}

\begin{example}[Theorem~\ref{thm:1} is wrong for singular spaces]
  Let $\pi : S \to \bP¹$ be one of the rational ruled ``Hirzebruch'' surfaces.
  Let $C_S \subset S$ be any section.  Construct a commutative diagram as
  follows,
  \[
    \begin{tikzcd}[column sep=2cm]
      & S_2 \ar[r, "\alpha\text{, blow-up}"] \ar[d, "\gamma\text{, contraction}"'] & S_1 \ar[r, "\beta\text{, blow-up}"] & S \ar[r, "\pi\text{, fibre bundle}"] & \bP¹ \ar[d, equal] \\
      C \ar[r, "\iota"'] & X \ar[rrr, "\rho\text{, rational fibration}"'] & & & \bP¹.
    \end{tikzcd}
  \]
  \begin{itemize}
    \item Choose four distinct points $x_1, …, x_4 \in \bP¹$.
    \item Choose four points $s_\bullet \in \pi^{-1}(x_\bullet) \in \bP¹$.
    \item Let $\beta$ be the blow-up up of the four points $s_\bullet$.  
    \item The surface $S_1$ is smooth.  The fibres $F_{1\bullet} :=
    (\pi\circ\beta)^{-1}(x_\bullet)$ are reduced. Each fibre $F_{1\bullet}$
    consists of two $(-1)$-curves, meeting transversally in a point
    $s_{1\bullet}$.
    \item Let $\alpha$ be the blow-up up of the four points $s_{1\bullet}$.  
    \item The surface $S_2$ is smooth but the fibres $F_{2\bullet} :=
    (\pi\circ\beta\circ\alpha)^{-1}(x_\bullet)$ are no longer reduced. Each
    fibre $F_{2\bullet}$ consists of two reduced $(-1)$-curves and one
    $(-2)$-curve $F'_{2\bullet}$ of multiplicity two.
    \item Let $\gamma$ be the contraction of the four points disjoint
    $(-2)$-curves $F'_{2\bullet}$.  The map $\pi\circ\beta\circ\alpha$  factors
    via the contraction map because we contract fibre components only.
    \item Let $C \subset X$ be the strict transform of the section $C_S$.
  \end{itemize}
  The surface $X$ is then singular, with four quotient singularities of type
  $A_1$ over the $x_\bullet$. All fibres of $\rho$ are supported on smooth
  rational curves, but the fibres over $x_\bullet$ have multiplicitity two and
  pass through the singularities.

  The criterion for $\cC$-morphism spelled out in \cite{orbiAlb1} quickly
  implies that $\rho$ is a $\cC$-morphism between the pair $(X,0)$ and the torus
  quotient $(\bP¹, \frac{1}{2}·\sum_i x_i)$.  The universal property of the
  Albanese immediately implies that the map $\rho$  factors via $\alb_x(X,0)$. A
  more detailed analysis, applying Theorem~\ref{thm:1} to the smooth fibres of
  $\rho$, shows that the torus quotient $(\bP¹, \frac{1}{2}·\sum_i x_i)$ is
  equal to the Albanese and that $\rho$ is the Albanese map.
\end{example}

\begin{itemize}
  \item \todo{Need example where Theorem~\ref{thm:1} fails if $X$ is singular.}
\end{itemize}
  
\begin{rem}
  ---
  \begin{itemize}
    \item Corollary~\ref{cor:2} does not equip $Y$ with the structure of a $\cC$-pair.
    \item Corollary~\ref{cor:2} does not assume that $\mu$ is a morphism of $\cC$-pairs.
    \item Even if $\mu$ is a morphism of $\cC$-pairs, Corollary~\ref{cor:2}
    does not claim that $\beta$ is a morphism of $\cC$-pairs.
    \item Corollary~\ref{cor:2} neither gives a morphism between $\Alb_x(X,0)$
    and $\Alb_{\mu(x)}(Y,0)$ nor does it claim that these are isomorphic.
  \end{itemize}
\end{rem}

\begin{itemize}
  \item \todo{Kummer K3s are nice examples where the Albanese grows when we
  contract rational curves.}
  \item \todo{Want more examples to showcase all the things that can go wrong.}
  \item \todo{Corollary~\ref{cor:2} implies that the $\cC$-Albanese map factors
  via the MRC fibration of $X$, and via any map from $X$ to one of its minimal
  models.  This should be exploitable in geometrically meaningful situations.}
\end{itemize}
  
\todo{There are settings where the factorization of Corollary~\ref{cor:2} is a
factorization into morphisms of $\cC$-pairs.}

\begin{thm}
  Birational projective manifolds $X$ and $Y$ have canonically isomorphic
  $\cC$-Albanese varieties.
\end{thm}
\begin{proof}
  \todo{PENDING}
\end{proof}

\begin{thm}
  Let $X$ be a projective manifold and let $x \in X$ be any point.  Let $\mu : X
  \to Y$ be an MRC fibration of $X$, where $Y$ is again a projective manifold.
  Then, the $\cC$-pairs $\Alb_x(X,0)$ and $\Alb_{\mu(x)}(Y,0)$ are naturally
  isomorphic.
\end{thm}
\begin{proof}
  \todo{PENDING}
\end{proof}


\section{Examples}

\begin{itemize}
  \item \todo{Discuss the Stoppino-example: general type, simply-connected,
  augmented irregularity zero, but has a non-trivial $\cC$-Albanse.}
\end{itemize}


\section{The $\cC$-Albanese morphism for special manifolds}

\begin{itemize}
  \item \todo{Discuss special surfaces.}
  \item \todo{Figure out what we can say for special threefolds.}
\end{itemize}

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