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As we know, Kodaira's embedding theorem can be put as:

A compact Kähler manifold $X$ is projective if and only if $\mathcal K_X\cap H^2(X,\mathbb Z)\neq\emptyset$.

Where $\mathcal K_X$ denotes the Kähler cone of $X$.
So if we have $H^{1,1}(X)\cap H^2(X,\mathbb Z)=0$, we can make the conclusion that $X$ is not projective. And my question is:
Is there indeed exist any compact Kähler manifolds which satisfies $H^{1,1}(X)\cap H^2(X,\mathbb Z)=0$? To seek for such examples we should first limit ourselves to non-projective Kähler manifolds, for example, in dimension 2, we have K3 surfaces and complex tori, we know they have $h^{1,1}>0$, and I don't know whether part of them satisfy $H^{1,1}(X)\cap H^2(X,\mathbb Z)=0$. So can anybody provide some examples? Any dimension is ok. Any comments are welcome, thanks!

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  • $\begingroup$ As you guessed, some Tori have this property. $\endgroup$
    – bliipbluup
    Sep 18 '20 at 9:35
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For some constructions of non-projective Kähler manifolds, check this MO question: https://mathoverflow.net/questions/257147/are-most-k%C3%A4hler-manifolds-non-projective

and the following article: https://encyclopediaofmath.org/wiki/K%C3%A4hler_manifold

I quote the second article: For example, this is the case for the torus $\mathbb{C}^2/\Gamma$, where $\Gamma$ is the lattice spanned by the vectors $(1,0), (0,1), (−\sqrt 2,−\sqrt 3), (−\sqrt 5,−\sqrt 7)$.

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  • $\begingroup$ Isn't these 4 vectors linear dependent? $\endgroup$
    – Tom
    Sep 18 '20 at 13:58
  • $\begingroup$ The vectors need not be linearly independent. You just need $4$ distinct points, so that you can generate a free abelian discrete subgroup of $\mathbb{C}^2$. You can find more on this construction in Huybrechts' Complex Geometry, pages 57-58. $\endgroup$
    – bliipbluup
    Sep 18 '20 at 15:06
  • $\begingroup$ You mean any 4 distinct points may work? for example $(1,0), (0,1), (2,0), (0,2)$? $\endgroup$
    – Tom
    Sep 18 '20 at 15:48
  • $\begingroup$ No, those don't work. You are right. The points need to be $\mathbb{R}$-linearly independent. $\endgroup$
    – bliipbluup
    Sep 18 '20 at 15:53
  • $\begingroup$ Could you elaborate a bit why it satisfies $H^{1,1}(X)\cap H^2(X,\mathbb Z)=0$? $\endgroup$
    – Tom
    Sep 18 '20 at 16:20

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