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A smooth morphism of schemes $f: X \to Y$ admits an étale-local section through any point $x \in X$.

One might wonder if this fact is true in the more general context of complex spaces (i.e. things glued from analytic subsets of polydiscs much like algebraic varieties are glued from affine varieties).

A map $f: X \to Y$ between complex spaces is flat at $x \in X$ if $\mathcal O_{X,x}$ is flat as a $\mathcal O_{Y,f(x)}$-module, where $\mathcal{O}_{X,x}$ is the ring of germs of analytic functions at $x$. Then étale is flat (at all the points of the domain) and unramified. [One can find this definition in Grauert-Peternell-Remmert Several Complex Variables VII, Encyclopedia of Mathematical Sciences, vol 74]

It can't be true stated as it is though. Consider the Hopf surface: the quotient of $\mathbb{C}^2 - (0,0)$ by the action of $\mathbb Z$: $z \mapsto (1/2 z, 1/2 z)$. The map $(x,y) \mapsto (x : y)$ gives a projection $H \to \mathbb{P}^1$ with fibres elliptic curves. It can't have a meromorphic section because by composing with a projection to one of the axes one would get a meromorphic map from $\mathbb{P}^1$ to an elliptic curve,and since $\mathbb{P}^1$ is étale simply connected étale base change won't help. (I am not sure why taking étale base extensions won't help, but it seems to be the case)

I am going to be naive and arbitrary now and suppose that the reason that this counterexample works is that the fibres are "complicated". What if the fibres are, say, $\mathbb{C}^n$?

So my question is: given a smooth morphism $f: X \to Y$ of complex spaces such that all its fibres are isomorphic to $\mathbb{C}^n$, and a point $x \in X$, is it true that $f$ admits a section through $x$ defined on some Zariski open neighbourhood of $f(x)$, perhaps after an étale base change?

Do you have in mind a broader natural class of morphisms for which this statemnt works?

Update: This is actually a reminiscence of this old question.

I am intrested in families of vector spaces definable in the structure of compact complex spaces (the latter are being extensively studied by Rahim Moosa, see his survey "Model theory and complex geometry"). This means that I want to look at the following set of objects: definable sets $X$ and $Y$ and a definable map $p: X \to Y$, the maps $+: X \times X \to X$ and $\cdot: \mathbb{C} \times X \to X$ and a section of $p$, $0: Y \to X$ such that restriction of these maps on each fibre of $p$ define a vector space structure on it. Now a few words about what definable means.

A definable set is a constructible subset (in the analytic Zariski topology) of a compact complex space. A meromorphic map between complex spaces $X$ and $Y$ is an analytic subset $\Gamma \subset X \times Y$ such that the projection on the first coordinate is onto and is a biholomorphic map outside some proper analytic subset of $X$. Note that this is not the same as just holomorphic map defined on an open subset of $X$ (exponent is a holomorphic map from $\mathbb{C} \supset \mathbb{P}^1$ to $\mathbb{P}^1$, but is not meromorphic, since it's graph is not an analytic subset of $\mathbb{P}^1\times \mathbb{P}^1$). A definable map is a piecewise meromorphic map, meaning that there is a cover $\cup U_i = X$ and on each $U_i$ the map coincides with a meromorphic map $\overline{U_i} \to Y$ for some compact $\overline{U_i}$ into which $U_i$ embeds.

I want to prove that given a definable family of vector spaces $p: X \to Y$ with fibres of constant dimension there is an analytic Zariski open $U \subset X$ and a piecewise meromorphic map $X|_U \to U \times \mathbb{C}^n$. Analytic Zariski topology is like Zariski topology for algebraic varieties: analytic subsets are closed sets.

Now to preserve sanity I will suppose for a momoent that $X$ is not constructible but just an open subset of some compact complex analytic space.

It seems that it is necessary first to be able to take analytic Zariski local sections. The reason why I want to work with analytic Zariski topology is that I need to produce a (piecewise) meromorphic map, and seems very hard to construct one locally in the finer complex topology - by looking at a local piece you don't know if it will extend to a meromorphic map on the whole domain.

The example with the Hopf surface shows that analytic Zariski local sections are not always possible. I still hope that they are possible in my restricted case (the fibres are something like $\mathbb{C}^n$), maybe after an étale base extension.

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What's an étale base change in the context of analytic spaces? I would simply say that it is a local isomorphism. In this case the statement simply amounts to the fact that a smooth map has local analytic section, which is certainly true. In your example $Y$ is an algebraic variety; but then you are mixing the analytic maps and algebraic varieties in a way that does not make much sense to me. –  Angelo Nov 16 '11 at 6:13
One can define the notion of a flat and étale morphism of complex spaces just like as in algebraic case: $f: X \to Y$ is flat at $x \in X$ if $\mathcal{O}_{X,x}$ is flat as a $\mathcal{O}_{Y,f(x)}$-module, where $\mathcal{O}_{X,x}$ is the ring of germs of analytic functions at $x$. Then étale is flat and unramified. I am not sure that this means a local isomorphism in general (it does for complex manifolds). –  Dima Sustretov Nov 16 '11 at 11:19
To Dmitry: yes, it does imply that it is a local isomorphism. –  Angelo Nov 16 '11 at 11:34
I am not sure that having an étale local section is equivalent to having a local analytic section. For example, consider the exponential map $\mathbb{C} \to \mathbb{C}^∗$. It does have analytically local sections, but no sections after an étale base change. – Dmitry Sustretov 0 secs ago –  Dima Sustretov Nov 16 '11 at 11:39
What do you mean by a "definable family of vector spaces"? Do you want something like a vector bundle, or you only care about the complex manifold structure? –  Angelo Nov 17 '11 at 9:24
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1 Answer

I meant to write this as a comment, but it won't fit.

In my opinion, you are confusing the étale analytic with the étale algebraic topology. The étale analytic topology is essentially the same the same as the classical topology. The étale topology is meant to be a substitute for the classical topology over more general fields; but it only sees the "finite" part of the topology.

If you have a smooth morphism in the analytic category, it has local sections in the classical topology. The étale topology was introduced precisely to fix the fact that this is false for algebraic morphisms, if you interpret "local sections" as "local sections in the étale topology". In other words, it was introduced to make some version of the implicit function theorem work.

But you are doing the following. You have a smooth morphism of analytic manifolds $f \colon X \to Y$, where $Y$ has an additional structure of an algebraic variety, and you are asking local sections $V \to X$, where $V \to Y$ is an algebraic étale map. Again, this does not make any sense to me. Perhaps if you were to include some motivation for your question you might convince me that it is meaningful.

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your are right suspecting that there is some hidden motivation for this question. the story where the need for a similar statement appeared is a bit bulky, so I decided to extract the simplest question I could ask. but ok, since this is not convincing, I will update the question to provide some context. –  Dima Sustretov Nov 16 '11 at 14:21
Dear Angelo, I have updated the question to inculed the background story. –  Dima Sustretov Nov 16 '11 at 16:20
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