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Let $X \to B$ be a smooth, proper, dominant map of schemes over $\text{Spec }k$ an algebraically closed field of characteristic zero with $B$ integral. We have the generic fibre $\overline{F}$ defined over $\text{Spec }\overline{K(B)}$ and by base-changing along $\text{Spec }\overline{K(B)} \to \text{Spec }k$, we obtain a map $\overline{X} \to \overline{B}$ such that we can now write down the sequence $\overline{F} \to \overline{X} \to \overline{B}$. To what extent is this a fibre bundle? To ask a definite question, is there some etale map $\overline{B}' \to \overline{B}$ such that further pulling back will yield an isomorphism $\overline{X}' \simeq \overline{F}' \times \overline{B}'$?

This question is closely related to Flatness in Algebraic Geometry vs. Fibration in Topology and Is an algebraic geometer's fibration also an algebraic topologist's fibration?. In particular, it is motivated by (1) Ehresmann's theorem that locally analytically such a morphism should be a (topological) fibre bundle and (2) the fuzzy thinking that "locally analytically" should mean "after an etale base change", but I feel like the answer to the question I posed it above is probably in the negative. For example, it seems unlikely to me that two smooth hypersurfaces of degree $d$ in $\mathbb{P}^n$ which are (automatically) diffeomorphic but not isomorphic should suddenly become isomorphic after an etale base change. However, I don't know of any weaker way to algebro-geometrically state the condition that some map be a fibre bundle, however -- is there anything then that we can say algebro-geometrically with respect to the above maps, or do we have to be content with the differential-geometric statement that it's a fibre bundle in that category?

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This seems to be a very popular theme here. In addition, to the above links see mathoverflow.net/questions/49759/… For the record, a smooth proper map is rarely etale locally trivial. A nonzero Kodaira-Spencer map would give an obstruction. –  Donu Arapura Jan 15 '11 at 23:10
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up vote 13 down vote accepted

The answer here is a resounding no. I think the most important point is that you're applying the wrong topological intuition here. A variety shouldn't be thought of as like a manifold, but as like a complex manifold, and the corresponding theorem to the "submersion=fiber bundle" theorem in smooth manifolds is just false for complex manifolds. Just as an example, all elliptic curves are topologically the same, so the solutions to $x(x-1)(x-a)=y^2$ are a smooth fiber bundle over (most of) $\mathbb{C}$ with coordinate $a$, but all the fibers which aren't in the same orbit of $SL_2(\mathbb Z)$ on $\mathbb{C}$ are not isomorphic as complex manifolds.

The other way of saying this is that complex structures can exist in families; they have moduli. Moduli spaces exactly measure how theorems like "submersion=fiber bundle" fail since they measure continuous variation of structure.

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Do smooth structures (on $4$-dimensional manifolds, say) exist if families? –  Mariano Suárez-Alvarez Jan 16 '11 at 1:03
    
@Mariano: i'd say just NO by Ehresmann's theorem.. Am I missing something? –  Qfwfq Jan 16 '11 at 1:12
    
@unknowngoogle: that works for proper maps, but the simplest example I know of a manifold with many smooth structures is $\mathbb R^4$ and if that's what the fibers are made of, then the family will not be proper. –  Mariano Suárez-Alvarez Jan 16 '11 at 1:18
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@Mariano- That's a question above my pay grade. –  Ben Webster Jan 16 '11 at 1:42
    
Thanks for your answer! Yes, I was fairly sure asking the moduli to simply collapse after an etale base change would be quite nonsense, but that was really just the only example I knew of to indicate what an algebro-geometric fibre bundle might be like; I am interested in whether there is any possible weaker algebro-geometric statement along those lines. –  Arnav Tripathy Jan 16 '11 at 13:56
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