Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm studying the deformation theory of compact complex manifolds as developed by Kodaira and Spencer. On the side I'm reading as much about deformation theory in general as I can get my hands on (and understand), and I've been wondering about the relationship between the basic definitions in the analytic and algebraic categories. To summarize:

Analytic side: A complex analytic family of smooth compact manifolds is a holomorphic map $\pi : \mathcal X \to S$ of smooth complex manifolds $\mathcal X$ and $S$ such that $\pi$ is a proper submersion and each fiber $X_t = \pi^{-1}(t)$ is a compact complex manifold. This implies some other conditions, like that $\mathcal X$ is locally trivial over $S$.

Algebraic side: A family of schemes is a proper flat morphism $\pi : X \to Y$ of schemes.

I've been asking myself what the relationship between these definitions is. To get something like the algebraic definition in the analytic category we just replace "scheme" by "complex space".

Now, a complex manifold is a smooth complex space, and local triviality of $\mathcal X$ along with compactness of the fibers implies that $\pi : \mathcal X \to S$ is proper (edit: unnecessary). I'm also fairly certain that $\pi$ is flat (my algebraic side is weak), so $\pi : \mathcal X \to S$ will be a family of complex spaces in the algebraic sense.

My question is: what conditions do we need on $\pi : X \to Y$ to pass in the other direction? Is it enough that the complex spaces $X$ and $Y$ be smooth? I've been thinking about this and I've got this vague idea that flatness of $\pi$ and coherence of the structure sheaves will lead to local triviality, but I haven't been able figure out how.

share|improve this question
    
Of course, when you say "$\mathcal{X}$ is locally trivial over $S$", you really mean "locally trivial in the differentiable category", not in the holomorphic one. –  Qfwfq Apr 28 '10 at 8:07
    
Flatness of $\pi$ and coherence of the structure sheaves do not imply local (C-infinity-) triviality of the family. Think of the flat family given by the degeneration of smooth projective conics to a singular conic: $x^2+y^2=tz^2$ in $\mathbb{P}^2$ ($t \in \mathbb{A}^1$ is the parameter of deformation). –  Qfwfq Apr 28 '10 at 8:12
    
The "Analytic side" of your definition is not correct. For example, an open immersion satisfies your definition but is not locally trivial. (Requiring surjectivity doesn't help.) It is best to assume $\pi$ is proper; then you get local triviality (assuming it is also a submersion). –  ulrich Apr 28 '10 at 8:22
    
Excellent comments, thank you all. I did indeed forget that $\pi \mathcal X \to S$ should be proper. I edited the question to fix this. I also seem to have gotten my triviality terms mixed up; what I mean is that every point of $\mathcal X$ admits a neighborhood of the form $U \times V$, where $U$ is a neighborhood in some fiber $X_t$, $V \subset S$ and such that $\pi$ identifies with the projection onto the second factor. This should follow from that $\pi$ is a holomorphic submersion. Thanks for the conics example. –  Gunnar Þór Magnússon Apr 28 '10 at 9:12

2 Answers 2

up vote 4 down vote accepted

The standard situation in Kodaira-Spencer's work is the following:

If you're on the algebraic side and you have a smooth ("smooth" in the sense of algebraic geometry) and proper (proper in the sense of algebraic geometry) map $\pi: X \to Y$ , then when you translate this to the analytic side, "smooth" turns into "submersion" (in the sense of: pushforward of vector fields is surjective), and "proper" turns into "proper" (in the sense of: inverse image of compact set is compact). And "map" turns into "holomorphic map". Then you can use, for instance, the "preimage theorem" (be careful to not get confused by the usage of "smooth" in that article --- there smooth means $C^\infty$) to deduce that the fibers are holomorphic complex manifolds. Strictly speaking we must use the holomorphic version of the "preimage theorem". But the holomorphic version does hold, as do holomorphic versions of other standard theorems like implicit function theorem and inverse function theorem. Perhaps this is in Chapter 0 of Griffiths-Harris, or Chapter 1 of Huybrechts.

The fibers are compact because a point is compact. :)

share|improve this answer
    
That's exactly what I wanted to know, thanks. :) –  Gunnar Þór Magnússon Apr 28 '10 at 13:02
    
Could you please show a reference for "smooth" being equivalent to "submersion" under the GAGA correspondence? –  user40276 Jun 23 at 22:42

Dear Gunnar, let $f:X\to Y$ be a proper flat morphism of smooth varieties. This does not imply that $f$ has smooth fibres either in the algebraic or in the analytic case. For example, any non-constant morphism of smooth complete connected algebraic curves resp. compact connected Riemann surfaces is flat and proper, but will in general have non-smooth zero-dimensional fibres due to ramification and will not be locally trivial downstairs.

Locally, the simplest example in the holomorphic case is the map $z\mapsto z^2$ from the unit disk to itself, which is flat and proper but has a singular fibre at the origin.

The algebraic reason for flatness in this context is that over a principal ideal domain, a module is flat if and only if it is without torsion. (This is standard: cf. for example. Hartshorne, Algebraic Geometry, III, Example 9.1.3.)

share|improve this answer
    
Thank you Georges, that's a good explanation. I still wonder what we need to add to pass to the analytic side. Do we really need a diffeo-geometric condition like that $\pi$ be a submersion, or that its jacobian has constant rank? Is there no more algebraic condition that we can impose? –  Gunnar Þór Magnússon Apr 28 '10 at 9:16
2  
Dear Gunnar,Grothendieck has introduced a purely algebraic concept of smooth morphism which is quite interesting and useful in characteristic $p$ Algebraic Geometry.It has counterintuitive features : for example the Frobenius morphism from the projective line to itself is nowhere smooth: no Sard Theorem here! But unfortunately for nonsingular varieties over $\mathbb C$, this brings nothing new: smooth morphism=submersion. A precise statement is for example Hartshorne III,Proposition 10.4 (p.270) –  Georges Elencwajg Apr 28 '10 at 10:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.