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 $Y$?

  • 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 Starr Dec 27 '18 at 10:44

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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.