Smooth morphism of relative dimension $>0$ and fiber bundle

Question

Let $f:X\to Y$ be a smooth morphisms of relative dimension $>0$ of two smooth (affine) varieties over $\mathbb{C}$. I wonder if the corresponding holomorphic map of the corresponding complex manifolds can be a fiber bundle, i.e. the holomorphic map $f_{an}:X_{an}\to Y_{an}$ is a fiber bundle. I know $f$ is a submersion thus $f_{an}$ is a submersion too. But by an exercise of Hartshorne, a proper map between affine varieties is finite. In my case, $f,f_{an}$ are not finite, which implies $f$ is not proper, so it seems (I am not very sure) $f_{an}$ is not proper. But by Ehresmann's lemma: a proper surjective submersion is a locally trivial fibration. I wonder if it would make $f_{an}:X_{an}\to Y_{an}$ not a fiber bundle?

Answer 1

For $f:X\to Y$ a morphism of separated schemes of finite type over $\Bbb C$, $f$ is proper if and only if $f_{an}$ is proper by SGA 1 XII.3.2. So $f_{an}$ cannot be proper in your case, which means the hypotheses of Ehresmann's lemma are not satisfied and you cannot apply it.

A readily-available example of a smooth map of positive relative dimension of affine varieties which is not analytically a fiber bundle is the projection $\operatorname{Spec} \Bbb C[x,y,(x^2-y^2)^{-1}]\to \operatorname{Spec} \Bbb C[x]$. This is smooth because it is the composition of an open immersion and the standard smooth map $\Bbb A^2_\Bbb C \to \Bbb A^1_\Bbb C$, and its analytification is not a fiber bundle because the fiber over any point which is not the origin is a twice-punctured complex plane, while the fiber over the origin is a once-punctured complex plane.