Examples of CY fibrations over $\mathbb P^1$

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?

• "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. May 21 '21 at 13:23
• @JasonStarr, your example does not seem to satisfy the second condition. May 21 '21 at 13:36
• 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 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.