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I have a question about a complex manifold $M$ and a holomorphic submersion $p : M \to S$ to a compact complex curve satisfying the following conditions:

  • $p^{-1}(x)$ is biholomorphic to $\mathbb{C}$ for any $x \in S$.
  • For any $z \in M$, there is a holomorphic section $s(z)$, i.e., a compact holomorphic submanifold of $M$ such that $p : s(z) \to S$ is a holomorphic equivalence.
  • $M$ is a smooth $\mathbb{R}^2$ bundle whose fiber is given by $p^{-1}(x)$, $x \in S$.

Can we deduce that $M$ is biholomorphic to $S \times \mathbb{C}$?

Remark: Inspired by Speyer's insightful example, where $M$ is a non-trivial line bundle, I am now introducing additional conditions and asking whether $M$ is biholomorphic to $S \times \mathbb{C}$:

  • The global section $s(z)$ is nowhere vanishing for any $z\in M$.
  • $M$ is a $C^\infty$ trivial $\mathbb{R}^2$ bundle over $S$.
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    $\begingroup$ Related example: At mathoverflow.net/questions/136017/… , Jason Starr gives an example of a submersion $V \to \mathbb{A}^2$ whose fibers are all isomorphic to $\mathbb{A}^2$, but which is not an $\mathbb{A}^2$ bundle. However, the fact that the base is two dimensional seems to play an important role in the proof that it is not a bundle, so I don't know if this could be adapted to make an example over a curve $S$. $\endgroup$ Commented Dec 20, 2023 at 14:23

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Maybe I misunderstood the question, but it seems to me that $M$ could be the total space of some other globally generated line bundle over $S$: For example, the total space of $\mathcal{O}(n)$ over $\mathbb{P}^1$ for $n>0$. To see that this total space is not isomorphic to the trivial bundle over $\mathbb{P}^1$, note that a closed embedding $\phi: \mathbb{P}^1 \hookrightarrow \mathcal{O}(n)$ is a section of $\mathcal{O}(n) \to \mathbb{P}^1$ if and only if the image of $\phi$ is a deformation retract. Thus, we can recognize "sections of $\mathcal{O}(n) \to \mathbb{P}^1$" just from the geometry of the total space, and the intersection of two such sections (with multiplicity) is $n$ points.

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  • $\begingroup$ Thanks for your nice example. I am also interested in if $M$ is a line bundle. As I shall add some conditions and ask the same question. $\endgroup$
    – Mjr
    Commented Dec 19, 2023 at 1:41

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