I'm reading the paper 'A bound on the Euler number for certain Calabi-Yau 3-folds' where the author made the following statement about fiberd Calabi-Yau manifold without proof.

Let $X$ be a smooth projective threefold with trivial canonical bundle, $\pi:X\to Y$ be a surjective holomorphic map from $X$ to a lower dimensional manifold $Y$, if $F$ is the general fiber of $\pi$, then there are 3 possible types:

  • a. $Y$= surface with kodaira dimension $-\infty$, $F$=elliptic curves;
  • b. $Y= \mathbb{C}P^1$, $F$= abelian surface;
  • c. $Y= \mathbb{C}P^1$, $F$= k3-surface.

The author mentioned that $F$ should have trivial canonical bundle by ajunction formula, as far as I know, the ajunction formula says that $$K_F=(K_X\otimes \mathcal{O}_X(F))\mid_F,$$ then $K_F=\mathcal{O}_X(F)\mid_F,$ so the problem is why $\mathcal{O}_X(F)\mid_F=\mathcal{O}_F$?

And I can't see why the base manifold $Y$ should be either a surface with kodaira dimension $-\infty$ or $\mathbb{C}P^1$?

The author also mentioned implicitly that if $\pi$ have no singular fibers, then $X$ is essentially a product. This is amazing since this is the main result of this paper in 2013.


1 Answer 1


If $F$ is a smooth fiber of a proper map $f: X \to Y$ then the normal bundle $N_{F/X}$ is isomorphic to the trivial bundle $T_{y}Y \otimes O_F$, where $y = f(F) \in Y$. Therefore $$ \omega_F \cong \omega_X\vert_F \otimes \det N_{F/X} $$ (this is the adjunction formula, your version is only for divisors), hence $K_F = 0$.

  • $\begingroup$ thank you! Can you say something about the base manifold Y? $\endgroup$
    – Wei Xia
    Dec 20, 2016 at 7:29

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