# Examples of compact Kähler manifolds with $H^2(X,\mathbb Z)\cap H^{1,1}(X)=0$

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!

• As you guessed, some Tori have this property. Sep 18 '20 at 9:35

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 $$(1,0), (0,1), (−\sqrt 2,−\sqrt 3), (−\sqrt 5,−\sqrt 7)$$.

• Isn't these 4 vectors linear dependent?
– Tom
Sep 18 '20 at 13:58
• 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. Sep 18 '20 at 15:06
• You mean any 4 distinct points may work? for example $(1,0), (0,1), (2,0), (0,2)$?
– Tom
Sep 18 '20 at 15:48
• No, those don't work. You are right. The points need to be $\mathbb{R}$-linearly independent. Sep 18 '20 at 15:53
• Could you elaborate a bit why it satisfies $H^{1,1}(X)\cap H^2(X,\mathbb Z)=0$?
– Tom
Sep 18 '20 at 16:20