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Every projective variety inherits a Kähler structure from a projective embedding, by restriction of the Fubini-Study metric. They will generally admit many Kähler structures though. I was wondering if every Kähler structure, maybe only up to cohomology, can be obtained in this way from some projective embedding?

Thanks!

doetoe
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1 Answers1

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The answer is no: If $\omega$ is a Kahler metric, then so is $c\omega$ for all $c>0$. But lots of them are not integral: $c\omega \in H^2(X, \mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).

Even if $[\omega]$ is an integral class, the assertion might not be true. Note that since $[\omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.

$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[\omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).