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In deformation of complex analytic spaces, one usually considers an analytic proper simple surjective map $\varpi: \mathscr{M} \twoheadrightarrow \mathscr{P}$ as an analytic family. However the term simple differs by literature. There are two notions of simple: as a locally trivial map and as a flat map. Are these definitions equivalent?

I should mention too that some references define an analytic family using submersions instead of simpleness for the smooth analytic case. But this definition coincides with the first one by Ehresmann's fibration theorem.

Maybe the question is trivial, but since I have never seen it written anywhere, I think I should be more careful.

Thanks in advance.

EDIT: The locally triviality is in the category of real differentiable manifolds or the category of topological manifolds. This explains why I have cited Ehresmann fibration theorem as an alternative for the statement.

user40276
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The answer is no.

In fact, flat maps are not required to be smooth maps in general. For instance, a flat family of smooth curves degenerating to a nodal curve is clearly not locally trivial.

However, the answer is still no even if we only consider smooth maps. In fact, there is the following theorem due to Grauert and Fischer, see [Barth-Peters-Van de Ven, Compact Complex Surfaces, Chapter I]:

Theorem. A smooth family of compact complex manifold is locally trivial if and only if all its fibers are analytically isomorphic.

Now it is possible to give examples of smooth and flat but not locally trivial maps: the so-called Kodaira fibrations, see the same book, Chapter V.

These are smooth maps $f \colon X \to C$, where $X$ is a complex surface ad $C$ a smooth curve, such that the fibres are not analytically isomorphic. By Grauert-Fischer theorem, it follows that Kodaira fibrations are not locally trivial maps (i.e., although all their fibres are smooth, they are not complex analytic fibre bundles).

Since every smooth fibration of a surface over a curve is automatically flat, this provides the examples we were looking for.

Added. If $f \colon X \to Y$ is a flat map in the complex category, it is not true in general that it provides a submersion in the real category. Again, this is because flat map does not imply smooth map. For instance, one can consider a flat map $f \colon X \to \Delta$, where $\Delta \subset \mathbb{C}$ is the unit disk, such that the general fibre is a smooth complex conic and the central fibre is the union of two complex lines intersecting at one point. This map cannot be a submersion because of the Ehresmann theorem: in fact, the topological type of the central fibre is different from the topological type of the general fibre.

Francesco Polizzi
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  • Thanks for the answer. What do you mean by analitically isomorphic fibers? If the fibers are isomorphic in the complex sense the deformation is trivial. For the manifolds case the fibers are always diffeomorphic by Ehresmann fibration theorem. – user40276 Jun 18 '14 at 14:17
  • I mean that the complex structure does not change in deformation, i.e. that the fibres are isomorphic in the complex sense. If the family is smooth, by Grauert-Fischer this is equivalent to the fact that the family is locally trivial. – Francesco Polizzi Jun 18 '14 at 14:37
  • Maybe I'm missing something. But, in my question, I was meaning locally triviality in the category of real manifolds. However, even for the analytic category it seens weird, since Grothendieck defines analytic family using locally triviality in the category of analytic spaces in his paper about Teichmuller theory. Maybe Grothendieck was wrong... – user40276 Jun 18 '14 at 14:48
  • Well, nowadays when one talks about deformation theory and says family of analytic varieties usually means flat, not locally trivial.

    I do not know the Grothendieck paper you are talking about, maybe you should provide more context in your question in order to have a more precise answer.

    At any rate, flat is surely weaker than locally trivial, as I explained in my answer.

     – Francesco Polizzi Jun 18 '14 at 15:14
  • Here I'm only talking about the analytic category – Francesco Polizzi Jun 18 '14 at 15:20
  • The Grothendieck paper is the one from Séminaire Henri Cartan http://archive.numdam.org/article/SHC_1960-1961__13_1_A4_0.djvu. At page 3, he says an isomorphism in the $U$-category. But do you know anything about the real differentiable case? As I said Ehresmann fibration theorem gives the locally triviality given that $\varpi$ is a proper surjective submersion. So the problem reduces in showing that a surjective submersion is equivalent to flatness. Any idea in the later problem? – user40276 Jun 18 '14 at 21:31
  • I'm a bit confused by your question. "Surjective submersion" is a concept living in the real category, whereas "flat" lives in the complex analytic category. But probably I'm missing something – Francesco Polizzi Jun 18 '14 at 22:34
  • Well, there is a forgetful functor. Furthermore I'm talking about the case where the complex spaces are, in fact, complex compact manifolds, then it makes sense to talk about flatness and submersions at the same time. – user40276 Jun 18 '14 at 22:48
  • But again, the answer seems to me to be no. The point is that flatness does not require that the fibres are smooth. So, for instance, you can consider a flat family of complex curves such that the general fibre is smooth, but the central fibre is nodal (for instance, a family of complex conics degenerating to the union of two distinct complex lines). This family cannot be a submersion because of the Ehresmann theorem: in fact, the topological type of the central fibre is different from the topological type of the general fibre. – Francesco Polizzi Jun 19 '14 at 08:49
  • maybe do you have some additional assumptions? For instance, you want that the map is not only flat, but also smooth, i.e. without critical points? Anyway, I added this remark in the answer – Francesco Polizzi Jun 19 '14 at 08:57
  • Yes, I'm mainly working with smooth maps, because the fibers are analytic compact manifolds. Any ideas for this case? I would be glad if you share your opinion, even if it is just a guess. – user40276 Jun 20 '14 at 03:39
  • According to this answer http://mathoverflow.net/questions/20184/flatness-in-algebraic-geometry-vs-fibration-in-topology smooth is equal to submersion under the GAGA correspondence (by the way, I have never seen a formal proof of this fact), so I think my question is "fully" answered now. – user40276 Jun 23 '14 at 22:45