As the question says, why should the Kaehler form be closed? Like people start from a fundamental 2-form (say, a 2-from $\mathcal{K}$) and then set they set the condition that in order for the manifold with a hermitian metric to be Kaehler, its 2-from must be closed, i.e $d\mathcal{K}=0$. Why is that?
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2$\begingroup$ Because we want a symplectic form. (And this lead to the question: why a symplectic form should be closed?) $\endgroup$– QfwfqCommented Dec 16, 2015 at 20:41
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2$\begingroup$ We want locally write our form as$ i\partial\bar \partial u$ this means exactly closedness $\endgroup$– user21574Commented Dec 16, 2015 at 20:45
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2$\begingroup$ A definition often arises from an abstraction of interesting examples. One motivation for the Kähler form comes from complex projective space and complex submanifolds (which are also algebraic varieties). The Kähler form is very natural in that setting because it is defined in terms of the natural Riemannian metric and generates the cohomology of $\mathbb{C}P^n$. This leads to a rich interaction between the topology, differential geometric, and algebraic geometric properties of the variety. It then becomes natural to study Kähler manifolds in general. $\endgroup$– Deane YangCommented Dec 16, 2015 at 20:51
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1$\begingroup$ @physicsoutsideborders It is just the definition of a Kaehler manifold. Why do we study them? Because they are an interesting special case. Why? Because here are all these fundamental examples. $\endgroup$– Steven GubkinCommented Dec 16, 2015 at 21:01
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2$\begingroup$ Check here math.stackexchange.com/questions/329342/… MBM's answer last paragraph. They say "Saying that the Kahler form is closed actually converts it into a symplectic form which fits nicely with the Kahler structure, so all the tools of the vastly developed subject of symplectic geometry can be made to bear. You can look up what makes the symplectic form so important." So how true is that and what is meant by fits nicely? @PaulReynolds $\endgroup$– physicsoutsidebordersCommented Dec 16, 2015 at 21:35
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1 Answer
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I heard from Professor Yau in his talk in Nashville see here in page 3
By looking at the example of the Poincare metric, Kahler demands the Hermitian form to be closed. And he derived that locally, such a form must be $i\partial\bar\partial u$ of some potential.
if you take $\omega=\sum_{i,j}g_{i,j}dz_i\wedge d\bar z_j$ then $d\omega=0$ exactly means
$$\frac{\partial g_{i\bar j}}{\partial z_k}=\frac{\partial g_{k\bar j}}{\partial z_i}$$
In page 6 of Catherine Cannizzo lecture note you can find the geometric meaning of closedness of a Kahler form
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$\begingroup$ see math.stackexchange.com/questions/913828/… $\endgroup$– user21574Commented Dec 16, 2015 at 20:59
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$\begingroup$ open the link , you will see. I think you are beginner , you can ask such questions in math.stackexchange $\endgroup$– user21574Commented Dec 16, 2015 at 21:08
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$\begingroup$ it is a theorem you can find it in books related to Kahler geometry $\endgroup$– user21574Commented Dec 16, 2015 at 21:16
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$\begingroup$ If you ask when locally such functions could be written globally then here is something mathoverflow.net/questions/221615/… $\endgroup$– user21574Commented Dec 16, 2015 at 21:18
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1$\begingroup$ $J$ is an endomorphism of the (real) tangent space that makes it into a complex space, whence it corresponds to multiplication by $i$. I think you'll get a better reception for these kind of questions over at math.stackexchange and probably there are some questions already there of interest to you. $\endgroup$ Commented Dec 16, 2015 at 22:57