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Stefan Kebekus 2024-06-03 16:42:00 +02:00
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@ -24,3 +24,5 @@ Patakfalvi
Zsolt Zsolt
Kebekus Kebekus
Albanese Albanese
Hirzebruch
multiplicitity

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\section{The $\cC$-Albanese morphism in the presence of rational curves} \section{The $\cC$-Albanese morphism in the presence of rational curves}
\subversionInfo \subversionInfo
\begin{setting}\label{set:1}
Let $X$ be a compact Kähler manifold and let $x \in X$ be any point. Assume
that an Albanese of the $\cC$-pair $(X,0)$ exists.
\end{setting}
\begin{thm}\label{thm:1}% \begin{thm}\label{thm:1}%
Let $X$ be a projective manifold and let $x \in X$ be any point. If $C \subset In Setting~\ref{set:1}, let $C \subset X$ be a rational curve. Then, the
X$ is a rational curve, then the Albanese morphism\watchOut{Stefan 03Jun24: Albanese morphism $\alb_x(X,0) : X \to \Alb_x(X,0)$ maps the curve $C$ to a
Need to make assumptions to ensure that the Albanese exists.} of the point.
$\cC$-pair $(X,0)$,
\[
\alb_x(X,0) : X \to \Alb_x(X,0),
\]
maps the curve $C$ to a point.
\end{thm} \end{thm}
\begin{proof} \begin{proof}
The normalization of $C$ yields a diagram The normalization of $C$ yields a diagram
@ -49,6 +49,53 @@
The claim follows immediately once we observe that $\Alb_0(\bP¹,0)$ is a point. The claim follows immediately once we observe that $\Alb_0(\bP¹,0)$ is a point.
\end{proof} \end{proof}
\begin{cor}\label{cor:3}%
In Setting~\ref{set:1}, assume that $X$ is rationally connected. Then,
$\Alb_x(X,0)$ is a point.
\end{cor}
\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} \begin{itemize}
\item \todo{Need example where Theorem~\ref{thm:1} fails if $X$ is singular.} \item \todo{Need example where Theorem~\ref{thm:1} fails if $X$ is singular.}
\end{itemize} \end{itemize}