# Uniruled degenerations of abelian varieties

Suppose I have a smooth projective variety $X$ over $\mathbb{C}$ with $K_X$ semiample, and consider the fiber space $f:X\to Y$ given by $|\ell K_X|$, for some $\ell>0$ large, where $Y$ is a normal projective variety of general type.

Assume that the general fiber of $f$ is an Abelian variety, I would like to say that either there is a component of some singular fiber of $f$ which is uniruled, or otherwise all singular fibers (if any) are just smooth Abelian varieties with multiplicity $m>1$.

If $\mathrm{dim} X=2$ this statement is true thanks to Kodaira's classification of singular fibers of elliptic surfaces. In this case the singular fibers are either $mI_0, m>1$, or else have a component which is a rational curve. I am looking for a higher-dimensional analog of this statement, any reference (or counterexample) is greatly appreciated.

• For $\dim{X} = 3$, couldn't the general fiber be the product of two elliptic curves, only one of which degenerates? Or a simple abelian surface which degenerates only partially at some place? However, the regular locus of each fiber admits a birational morphism to a commutative algebraic group (consider the Neron model of the restriction of the family to a regular subcurve of $Y$), which is an extension of an abelian variety by a linear group (which is a rational variety). – Vesselin Dimitrov Aug 14 '14 at 16:20
• So I think all your fibers will admit a surjective morphism to an abelian variety with a rational fiber. In particular, all fibers will have abelian maximal rationally connected quotients. – Vesselin Dimitrov Aug 14 '14 at 16:22
• What about a family of Abelian surfaces degenerating to the double of the Kummer? – Jason Starr Aug 14 '14 at 16:29
• @JasonStarr: Right... I was wrong in saying that the regular locus of a fiber would admit a birational morphism to a commutative algebraic group. – Vesselin Dimitrov Aug 14 '14 at 19:01
• @Vesselin and Jason: thanks for your comments, but I am not sure if these degenerations that you mention (e.g. double of Kummer) can arise in my situation, since the total space is smooth, and the fibration is the Iitaka fibration, which gives restrictions in general. For example, when $X$ is a surface, if the Iitaka fibration is given by sections of $|K_X|$ ($\ell=1$) then there are no multiple fibers at all. However, I feel that using Grothendieck's Stable Reduction for abelian varieties should be helpful, but I couldn't exactly see how... maybe you know how to do this? – user56818 Aug 15 '14 at 6:10

Maybe this can be of some help.

In this paper by K. Oguiso, you can find at the end an appendix by N. Nakayama. In this appendix he gives a theorem which describes the local structure of a degeneration of abelian surfaces (in the spirit of Kodaira's classification). It is Theorem B.1.

It roughly states the following. Suppose that if you have a projective surjective morphism $\alpha\colon\mathcal X\to\Delta$, where $\Delta$ is the complex unit disc, such that the fibers over the punctured disc are abelian surfaces, that $K_{\mathcal X}\simeq\mathcal O_{\mathcal X}$, and that the central scheme-theoretic fiber contains no ruled surfaces. Then, the central fiber $X_0$ may be written as $X_0=mS$, where $S$ is an abelian surface in the case $m=1$ and a hyper-elliptic surface in the case $m>1$.

Let $f : X \to Y$ be a proper surjective morphism from a smooth quasi-projective variety $X$ to a smooth quasi-projective curve $Y$ with connected fibers. Let $P \in Y$ be a point. Assume that $Supp f^*P$ is a simple normal crossing divisor on $X$ and $f$ is smooth over $Y \setminus P$. We further assume that $K_{f^{-1}Q} \cong 0$, for every $Q\in Y\setminus P$. Then there exists a sequence of flips and divisorial contractions
$$X=X_0 \to K_{X_1} \to K_{X_2} \to \cdots \to K_{X_k} \cdots \to K_{X_m}$$
over $Y$ such that $X_m$ has only $\mathbb Q$-factorial terminal singularities and $K_{X_m} \cong_{Q,Y}0$.
Let $S=Supp f^*_mP$ be the special fiber of $f_m : X_m \to Y$ . If $S$ is reducible, then every irreducible component of $S$ is uniruled. If $S$ is irreducible, then $S$ is normal and has only canonical singularities if and only if $S$ is not uniruled. Note that $K_S\cong_ Q 0$ when $S$ is irreducible and has only canonical singularities.