In Wells's book 《differential analysis on complex manifolds》 page219, there is a statement:" any compact Kähler manifold $X$ with the property that dim$_\mathbb{C}H^{1,1}(X)=1$ is necessarily Hodge. This follow from the fact that multiplication by an appropriate constant will make the Kähler form on $X$ integral."
But I know by Kodaira's Embedding theorem, a compact Kähler manifold is projective if and only if the Kähler class $[\omega]\in H^2(X,\mathbb{Z})$, I don't see dim$_\mathbb{C}H^{1,1}(X)=1$ imply $[\omega]\in H^2(X,\mathbb{Z})$, the reason is that $H^2(X,\mathbb{Z})$ can be seen as integral lattice in the vector space $H^2(X,\mathbb{C})$, and 1-dimensional $H^{1,1}(X,\mathbb{C})$ can be seen as a 1-dimensional sub vector space of $H^2(X,\mathbb{C})$, the intersection of $H^2(X,\mathbb{Z})$ and $H^{1,1}(X,\mathbb{C})$ may probably be empty? so that the compact Kähler manifold $X$ with dim$_\mathbb{C}H^{1,1}(X)=1$ is not necessarily Hodge? Is that right?
from Hot Weekly Questions - Mathematics Stack Exchange
Tom
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