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I would like to know if there exists a smooth complex projective $3$-fold $X$ that admits a fibration $\pi: X\to \mathbb CP^1$ such that all fibers are smooth, $\pi^{-1}(0)$ is the second Hirzebruch surface $F_2$ and for any $x\in \mathbb CP^1$, $x\ne 0$ the fiber $\pi^{-1}(x)$ is a smooth quadric.

I suspect that in case such a three-fold exists, it will not be unique up to an isomorphism. If this is the case, can there be a classification of such three-folds? Is there a "simplest one" among them?

PS. As Piotr points out below, a construction of such a fibration is contained in the answer to this question: A specific degeneration of a rank 2 bundle (one just needs to projectivise the rank $2$ bundle)

I would be grateful for any comments to the second half of the question.

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    $\begingroup$ Of course. Cover $\mathbf{P}^1$ by two copies of $\mathbf{A}^1$. Over one of those, construct the standard degeneration of $\mathbf{P}^1\times \mathbf{P}^1$ to $F_2$ (by degenerating the Euler sequence on $\mathbf{P}^1$ to a split sequence and taking the projective bundle). On the other copy, take the trivial family $\mathbf{P}^1\times \mathbf{P}^1 \times \mathbf{A}^1$. Identify the two families on the overlap. $\endgroup$
    – Piotr Achinger
    Commented Jun 17, 2019 at 9:50
  • $\begingroup$ Thanks Piotr. Could you point me to a nice reference for this standard degeneration, so that by reading it I would be able to fully understand your comment? $\endgroup$
    – aglearner
    Commented Jun 17, 2019 at 10:10
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    $\begingroup$ Seems that your question is almost a duplicate of mathoverflow.net/questions/80989/… $\endgroup$
    – Piotr Achinger
    Commented Jun 17, 2019 at 10:44
  • $\begingroup$ Thanks for this comment! Indeed the first half of my question seems to be answered in the post you refer to. There is still a second part... What can be said in general about this type of three-folds? Is there a "simplest" among them? $\endgroup$
    – aglearner
    Commented Jun 17, 2019 at 11:07
  • $\begingroup$ It seems reasonable to conjecture that they are all obtained from the one above by pullback along an endomorphism of P1 totally ramified at 0. $\endgroup$
    – Piotr Achinger
    Commented Jun 17, 2019 at 13:52

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