We work over $\mathbb C$ and let us call a smooth projective vartiety $M$ as Calabi-Yau (CY) manifold if it has trivial canonical class and $h^i(M, \mathcal O_M ) = 0$ for $0 < i < \dim(M)$. In this definition, a CY 1-fold is an elliptic curve, a CY 2-fold is a projective $K3$ surface and etc.

For each $n$, I looking for a smooth projective variety $X$ of dimension $n$ with a fibration $\pi: X \rightarrow \mathbb P^1$ such that

  1. a generic fiber $F$ of $\pi$ is a CY $(n-1)$-fold
  2. $K_X$ is linearly equivalent to $-2F$ and
  3. $X$ is not a product of $\mathbb P^1$ and a CY $(n-1)$-fold.

For $n=2$, it is known that such $X$ (an ellitic surface) does not exist. I put a question regarding the case of $n=3$ here but didn't get an answer. I also put a stronger question here, requiring a different condition that every fiber is smooth and got answers saying that there are no such fibrations with the smooth condition.

For some $n$, does such a fibration exist?

  • $\begingroup$ "For $n=2$, it is known that such an $X$ . . . does not exist." Elliptic K3 surfaces do exist. More generally, for a sufficiently general hypersurface in $\mathbb{P}^n\times \mathbb{P}^1$ of bidegree $(n+1,2)$, the hypersurface together with its projection to $\mathbb{P}^1$ is an example satisfying your conditions. $\endgroup$ May 21 '21 at 13:23
  • 2
    $\begingroup$ @JasonStarr, your example does not seem to satisfy the second condition. $\endgroup$
    – user69559
    May 21 '21 at 13:36
  • 1
    $\begingroup$ Sorry, I thought that you wanted the relative canonical divisor to be linearly equivalent to $2F$, not the anticanonical divisor. You are correct that my examples are Calabi-Yau manifolds, thus the anticanonical divisor is trivial. i $\endgroup$ May 21 '21 at 14:34

Edit. There is a mistake in the answer below. It is possible that the fiber of the rational quotient has dimension $\geq 2$. I will try to revise the answer soon.

Original answer (including mistake). If the anticanonical divisor class is numerically equivalent to $2F$, then by Mori's Bend-and-Break result, every point of your variety is contained in a rational curve. If the general fiber is a Calabi-Yau variety, then it is not uniruled. Thus, every rational curve containing a general point is not contained in the fiber. For transversal rational curves, deformation theory gives a lower bound on the dimension of the space of deformations containing the fixed (yet general) point: the lower bound is the anticanonical degree minus $2$. By Bend-and-Break again, that means that the minimal anticanonical degree of a transversal rational curve containing a general point is precisely $2$, i.e., the curve is a section of the fibration.

Moreover, the normal bundle is globally generated for every irreducible rational curve containing a general point, so the normal bundle is a trivial bundle. The following sentence is wrong. That means that the "rational quotient" of $X$ by these transversal rational curves is a $\mathbb{P}^1$-bundle over the quotient, and the fibers of the original fibration are mapping finitely to this quotient (away from codimension $2$). Why this is a mistake: in fact, there are rational quotients by such curves where the fiber dimension of the rational quotient map is strictly greater than $1$, e.g., cubic threefolds where the curve is a general line. Since the fibers (of the projection to $\mathbb{P}^1$) are themselves Calabi-Yau, this should force the finite maps to be étale. Since Calabi-Yau varieties are simply connected, this should force the finite maps to be isomorphisms. So the rational quotient provides the projection to the fiber factor of the product of $\mathbb{P}^1$ and the fiber.

  • $\begingroup$ Thanks a lot for your answer - I will study it. In fact, simply-connectedness is not assumed for Calabi-Yau manifolds in the question (only the conditions on cohomologies are assumed). Do you think that there are stll no such fibrations whose general fibers are non-simply connected Calabi-Yau manifolds? $\endgroup$
    – user69559
    May 21 '21 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.