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Let $M$ be a complex manifold. Assume we have a covering of $M$ by complex charts $\{U_{i}\}$. Furthermore assume that we have on each $U_{i}$ a Kählerform $\omega_{i}$ (i.e. $d\omega_{i} = 0$). Is it possible to construct from these forms a global Kählerform $\omega$ on $M$? Or under wat circumstances is this possible ? Is there any book or paper where this issue is discussed ? Thanks in advance.


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By definition if $U$ is a chart then you have complex coordinates on it, say $(z_1,\dots,z_n)$. Then, on $U$, the $(1,1)$-form $i\sum_{j}dz_j\wedge d\bar z_j$ is $d$-closed (it has constant coefficients) and positive, hence Kähler. In other words, your hypothesis is always satisfied by any complex manifold. Thus, basically you are asking when a complex manifold is Kähler. I think you should refine a little bit your question! – diverietti Dec 7 '11 at 7:56
ok i see, you are right. assume then $(M,\omega, J)$ is a kähler manifold. furthermore on some open covering $\{U_{i}\}$ we ave local kählerlorms $\tilde{\omega}_{i}$ (different from $\omega$, just defined on $U_{i}$, these forms may satisfy locally certain conditions that $\omega$ does not satisfy). my question is now: is it possible to glue these forms together to get one globally defined kähler form $\tilde{\omega}$ (maybe different from $\omega$)? – gary Dec 7 '11 at 8:08
Frankly, I still think that your question does not make really sense... I mean, you are asking when a form given locally glues to form a global form. The answer is that it glues when it transforms in the right way on the overlaps of the covering. In other words it glues when it glues... Then it will be automatically Kähler, since it is a local condition. – diverietti Dec 7 '11 at 9:35

(I decided to repost multiple comments as an answer. It's partly an answer if I understand the OP correctly.)

I think the OP does not necessarily want to conclude that these "local" forms are the restrictions of a global form. Rather, he wants to "glue" them together in the sense of "glueing" constructions used to solve elliptic PDE's, such as the Kummer construction of Calabi-Yau metrics on K3. That is, the OP (I think) wants to interpolate between these forms, and on the interpolation region the new globally defined form need not restrict to the original local pieces.

Gary, is this indeed what you meant? If so, it sounds very difficult in general. If you have only a few charts with topologically simple intersections (like annuli), then this is often possible with cutoff functions, but topological obstructions can and do arise. Ideally, one has to be able to match the cohomology classes of your local Kahler forms $[\omega_i]$ on the intersection regions. If they do not match already, "corrections" are sometimes possible. It's rather nontrivial in general. Before saying more, I would need more details about the specific situation.

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What do you mean by cohomology classes of a local kahler form? i understand less and less... – diverietti Dec 7 '11 at 13:08
If the charts are contractible, then of course the classes $[\omega_{\alpha}]$ will be trivial. But each chart $U_{\alpha}$ is is only diffeomorphic to an open subset of $\mathbb C^n$, so it can have nontrivial topology. By the cohomology class of a local Kahler form, I mean the class $[\omega_{\alpha}]$ in $H^2(U_{\alpha}, \mathbb R)$. – Spiro Karigiannis Dec 7 '11 at 20:09
mmm ok... but I guess that you can always suppose that your chart are contractible... and even every intersection of a finite number of charts! – diverietti Dec 8 '11 at 14:03
Yes, that's true. In practice, for "glueing" constructions of the type that I describe, one doesn't really have a cover by charts, but a specific cover of the manifold by some (usually finite) number of open sets, which need not be charts, for which one has a prescribed closed form. One then wants to interpolate between these closed forms to build a global (and still closed) form on the whole manifold with the required properties (such as being Kahler). Of course, it is now clear that the OP was not actually talking about this at all. – Spiro Karigiannis Dec 8 '11 at 15:08
Spiro, indeed. :) – diverietti Dec 9 '11 at 8:42

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