Let $\pi:X \to Y$ be a $\mathbb{C}^*$-fibration between complex manifolds in the sense that there exists a fixed integer $a$ such that for every $y \in Y$, $\pi^{-1}(y)=(\mathbb{C}^*)^a$. Suppose further that $Y$ is compact. Does there exist a compactification $\widetilde{X}$ of $X$, which is a $(\mathbb{P}^1)^a$-fibration over $Y$?
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1$\begingroup$ No, that is not true. Even if you assume that the fibers are all $\mathbb{C}^2$, that is not true, not even "differential geometrically". Please confer the following MO answer: mathoverflow.net/questions/136017/… $\endgroup$– Jason StarrCommented Dec 27, 2018 at 10:44
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No that is not true. There is a much simpler example than in the MO answer above. I might have written the following example in one of my earlier MO answers.
Let the proper target of the morphism be $\mathbb{P}^1_k = \text{Proj}\ k[S,T]$. The domain of the morphism will be an open subset of $$\mathbb{P}^1_k \times_{\text{Spec}\ k} \mathbb{P}^1_k = \text{Proj}\ k[S,T] \times \text{Proj}\ k[U,V].$$ Specifically, let the domain be the open complement of the closed subset, $$C:= \text{Zero}(SV^2-TU^2) \sqcup \{([1,0],[0,1]),([0,1],[1,0])\}.$$ Let $\pi$ be the restriction to this open subset of projection onto the first factor.
This is a smooth, surjective morphism such that every fiber is isomorphic to $\mathbb{G}_m$. Yet it is not an open subset of any $\mathbb{P}^1$-bundle over the target. If it were, then $\pi$ would be an affine morphism. However, the inverse image under $\pi$ of $D_+(S)$, resp. of $D_+(T)$, would be affine. Using Hartog's phenomenon / S2 extension, the inverse image is quasi-affine but not affine.
Edit. I checked, and I did write up this example in a previous MO answer: When is a holomorphic submersion with isomorphic fibers locally trivial?