MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 5 added 2786 characters in body

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

I am intrested in families of vector spaces definable in the structureof compact complex spaces (the latter are being extensively studied byRahim Moosa, see his survey "Model theory and complex geometry"). This means that I want to look at the following set ofobjects: 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 thesemaps 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 Zariskitopology) of a compact complex space. A meromorphic map betweencomplex spaces $X$ and $Y$ is an analytic subset $\Gamma \subset X\times Y$ such that the projection on the first coordinate is onto andis a biholomorphic map outside some proper analytic subset of$X$. Note that this is not the same as just holomorphic map defined onan open subset of $X$ (exponent is a holomorphic map from $\mathbb{C}\supset \mathbb{P}^1$ to $\mathbb{P}^1$, but is not meromorphic, sinceit'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 mapcoincides with a meromorphic map $\overline{U_i} \to Y$ for somecompact $\overline{U_i}$ 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 piecewisemeromorphic map $X|_U \to U \times \mathbb{C}^n$. Analytic Zariskitopology is like Zariski topology for algebraic varieties: analyticsubsets are closed sets.

Now to preserve sanity I will suppose for a momoent that $X$ is notconstructible but just an open subset of some compact complex analyticspace.

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 localsections are not always possible. I still hope that they are possiblein my restricted case (the fibres are something like $\mathbb{C}^n$),maybe after an étale base extension.

show/hide this revision's text 4 added 513 characters in body

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?

show/hide this revision's text 3 added 283 characters in body

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).

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 wonder still if am going to be naive and arbitrary now and suppose that the reason that this counterexample works is that the statement holds when fibres are "complicated". What if the fibres areof some particluar kind, say, $\mathbb{C}^n$. \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?

show/hide this revision's text 2 added 55 characters in body
show/hide this revision's text 1