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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?

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  • $\begingroup$ "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
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    $\begingroup$ @JasonStarr, your example does not seem to satisfy the second condition. $\endgroup$
    – user69559
    May 21 '21 at 13:36
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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