Complex analytic vs algebraic families of manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:19:52Z http://mathoverflow.net/feeds/question/22821 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22821/complex-analytic-vs-algebraic-families-of-manifolds Complex analytic vs algebraic families of manifolds Gunnar Magnusson 2010-04-28T07:47:10Z 2010-04-28T09:21:48Z <p>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:</p> <p><em>Analytic side</em>: 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 $S$.</p> <p><em>Algebraic side</em>: A family of schemes is a proper flat morphism $\pi : X \to Y$ of schemes.</p> <p>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".</p> <p>Now, a complex manifold is a smooth complex space, and <S>local triviality of $\mathcal X$ along with compactness of the fibers implies that $\pi : \mathcal X \to S$ is proper</S> (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.</p> <p>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.</p> http://mathoverflow.net/questions/22821/complex-analytic-vs-algebraic-families-of-manifolds/22824#22824 Answer by Georges Elencwajg for Complex analytic vs algebraic families of manifolds Georges Elencwajg 2010-04-28T08:41:26Z 2010-04-28T08:41:26Z <p>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.</p> <p>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.</p> <p>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.)</p> http://mathoverflow.net/questions/22821/complex-analytic-vs-algebraic-families-of-manifolds/22829#22829 Answer by Kevin Lin for Complex analytic vs algebraic families of manifolds Kevin Lin 2010-04-28T09:21:48Z 2010-04-28T09:21:48Z <p>The standard situation in Kodaira-Spencer's work is the following:</p> <p>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 "<a href="http://en.wikipedia.org/wiki/Preimage_theorem" rel="nofollow">preimage theorem</a>" (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.</p> <p>The fibers are compact because a point is compact. :)</p>