# Naive compactification of $\mathbb{C}^*$-fibrations

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 $$Y$$?

• 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/… – Jason Starr Dec 27 '18 at 10:44

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.