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Here is an elemantary example:

Define $f:S^1\times \mathbb{C}\rightarrow\mathbb{C}$ by $f(\zeta, z)=\zeta\cdot z^n$, where $n\geq 2$ is an integer, then $f$ is a smooth map, every fiber of $f$ is a smooth submanifold of $S^1\times\mathbb{C}$ and is diffeomorphic to $S^1$. However, $(S^1\times \mathbb{C}, \mathbb{C}, f)$ is not a differentialble fiber bundle.

Question: Is there any similar example in algebraic geometry?

More Precisely:

$f:X\rightarrow \mathbb{C}P^1$ is a smooth morphism, where $X$ is a smooth projective algebraic variety over $\mathbb{C}$. Every fiber of $f$ is a smooth subvariety of $X$.

Does it imply that $(X, \mathbb{C}P^1, f)$ is a differentiable fiber bundle?

What will happen if we replace $\mathbb{C}P^1$ by a smooth variety $Y$ in general?

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  • $\begingroup$ I think you must first tell us what you mean by fibre bundle in algebraic geometry (there are several different definitions that come to mind). $\endgroup$
    – Torsten Ekedahl
    Commented Sep 10, 2011 at 8:57
  • $\begingroup$ @Torsten: I mean a differentiable fiber bundle, that is to treat X as a differentiable manifold, f as a differentiable map. $\endgroup$
    – Yuchen Liu
    Commented Sep 10, 2011 at 9:01
  • $\begingroup$ Actually, I do not understand why your example is not a differentiable fibre bundle. Ehresmann theorem quoted in my answer seems to apply also in your situation (smooth submersion with compact, connected fibres). Probably I'm missing something elementary $\endgroup$
    – Francesco Polizzi
    Commented Sep 10, 2011 at 11:47
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    $\begingroup$ That map is not a submersion. If you are going to call a map "smooth" when all the set-theoretic fibers are smooth, then every map with finite fibers is "smooth". So in exactly the same way, you could define any finite morphism from the projective line to itself to be "smooth" to get algebraic counterexamples. $\endgroup$
    – Jason Starr
    Commented Sep 10, 2011 at 11:59
  • $\begingroup$ Ah, ok. In fact, I did not look carefully at the differential of Jerry's map :-) Thank you for the clarification $\endgroup$
    – Francesco Polizzi
    Commented Sep 10, 2011 at 12:43

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By definition, a morphism $f \colon X \to Y$ between smooth projective varieties is smooth if $f$ is flat and all fibres are smooth (in the scheme-theoretical sense). In particular, $f$ is a smooth submersion when it is considered as a differentiable map between real manifolds.

Then the answer to your question is yes, assuming that the fibres are connected. This remains true also replacing $\mathbb{P}^1$ with any smooth projective variety $Y$.

In fact, by a result due to Ehresmann (1951), if $ƒ \colon M \to N$ is a surjective submersion with $M$ and $N$ differentiable manifolds such that the preimage $ƒ^{-1}(x)$ is compact and connected for all $x \in N$, then $ƒ$ is differentiably locally trivial and admits a compatible fiber bundle structure. See http://en.wikipedia.org/wiki/Fiber_bundle.

In particular, all fibres of $f$ are diffeomorphic.

Remark 1. If one only requires that the fibres of $f \colon X \to Y$ are smooth in the set-theoretical sense, and not in the scheme-theoretical one (i.e. one allows multiple fibres) then in general $f$ is not a differentiable fibre bundle, see Jason Starr's comment.

Remark 2. Even if $f \colon X \to Y$ is differentiably locally trivial, in general it is not analitycally locally trivial. In fact, by a celebrated theorem of Grauert and Fischer, this happens if and only if all the fibres of $f$ are biholomorphic. See for instance [Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter I].

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  • $\begingroup$ Connectedness of fibres is not needed, because any proper submersion is a differentiable fiber bundle, see same Wikipedia reference. (In fact, I always thought this was Ehresmann's theorem.) $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 11, 2011 at 14:49

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