Let $f:\mathbb{C}^3 \to \mathbb{C}$ be a morphism of varieties such that it is a smooth fiber bundle. Can I say that the fiber is $\mathbb{C}^2$?

$\begingroup$ A smooth morphism in the sense of algebraic geometry, or just a $C^{\infty}$ fiber bundle which happens to be a morphism of varieties? $\endgroup$– Ben McKayOct 1, 2021 at 5:51

$\begingroup$ @BenMcKay Aren't these equivalent if the source and the target are smooth as varieties? $\endgroup$– user178109Oct 1, 2021 at 5:56

$\begingroup$ @Oniqa I can see three different settings for this problem. Firstly, the context of complex algebraic geometry . Then $f$ must be a complex polynomial on $z_1$, $z_2$, $z_3$. Secondly, the complex analytic situation. Thirdly, in the category of smooth manifolds. In the latter category the answer is no. If one considers an exotic $M=\mathbb{R}^4$ then $M\times \mathbb{C}$ is diffeomorphic to $\mathbb{C}^3=\mathbb{R}^6$, because $\mathbb{R}^6$ unlike $\mathbb{R}^4$ has a unique smooth structure. The projection $p_2\colon M\times \mathbb{C}\to\mathbb{C}$ is then taken for $f$. $\endgroup$– VictorOct 1, 2021 at 6:00
1 Answer
Note that $f$ is smooth as a map of varieties. If $f$ comes from a direct product $\mathbb{C}^3\approx S\times \mathbb{C}$ then $S\approx \mathbb{C}^2$. See Section 5.1 of $\mathbb{A}^1$homotopy theory and contractible varieties: a survey
I do not see why $f$ must come from a direct product so this is not a complete answer.