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I am looking for an example of a smooth projective threefold $X$ with fibration $ \pi : X \rightarrow \mathbb P^1$ such that

  1. a generic fiber $F$ of $\pi$ is a smooth $K3$ surface,
  2. $K_X$ is linearly equivalent to $-2F$ and
  3. $X$ is not a product of $\mathbb P^1$ and a smooth $K3$ surface.

Does such a threefold exist?

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    $\begingroup$ Start with a $\mathbb P^3$-bundle $p:\mathbb P(E) \to\mathbb P^1$ ($E$ being a rank $4$ vector bundle), and consider a generic section of $\mathcal O_{\mathbb P(E)}(4)$ with zero locus $X$. By the accepted answer to (math.stackexchange.com/questions/1559927/…), $K_{\mathbb P(E)} = \mathcal O_{\mathbb P(E)}(-4) \otimes \mathcal O_{\mathbb P(E)}(-2F)\otimes p^*\det E^\vee$, and applying adjunction we get $K_X = \mathcal O_X(-2F) \otimes (p^*\det E^\vee)|_X$. So as long as $E$ has degree $0$ I think you are good to go? $\endgroup$
    – Tabes Bridges
    Commented May 15, 2021 at 9:57
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    $\begingroup$ @TabesBridges: Let $E = \mathcal O \oplus \mathcal O \oplus \mathcal O(-1) \oplus \mathcal O(1)$. Is there a smooth global section of $\mathcal O_{\mathbb P (E)}(4)$? $\endgroup$
    – Basics
    Commented May 15, 2021 at 13:11
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    $\begingroup$ Yes, $H^0(\mathbb{P}(E)),\mathscr{O}_{\mathbb{P}}(1))=\operatorname{Sym}^4(E) $ has a lot of sections. However, because of the summand $\mathscr{O}(-1)$, the corresponding threefolds are singular. The only way to get a smooth one is to take $E=\mathscr{O}_{\mathbb{P}^1}^4$, but then $X$ is a product of $\mathbb{P}^1$ by a K3. $\endgroup$
    – abx
    Commented May 15, 2021 at 15:45

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