Compact Kähler manifolds with $dim_\mathbb{C}H^{1,1}(X)=1$

Question

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?

Answer 1

As mentioned in the comments, now I think the argument in the book is incorrect, at least with the given assumption. The following was my first attempt to answer this question, where the crossed sentences are insufficient:

Every compact Kahler manifold $X$ admits a Kahler form $[\omega]\in H^2(X,\mathbb C)$ which is nonzero. Moreover, $[\omega]$ is real and of type $(1,1)$. Therefore, by the assumption that $\dim_{\mathbb C}H^{1,1}(X)=1$, $[\omega]$ is a real element of the complex vector space $\mathbb C\cong H^{1,1}(X)$. One can rescale $[\omega]$ by a constant $r\in\mathbb R^+$, such that $r[\omega]\in H^2(X,\mathbb Z)$ is integral and still stay in the Kahler cone.

For your confusion, you can check from the fact $H^{1,1}(X)$ is a complex subspace of $H^{2}(X,\mathbb C)$ such that $\overline{H^{1,1}(X)}=H^{1,1}(X)$, it follows that $$H^{1,1}(X)= \big (H^{1,1}(X)\bigcap H^2(X,\mathbb R)\big )\otimes_{\mathbb R} \mathbb C.$$ So in particular, the intersection $\require{enclose}\enclose{horizontalstrike}{H^2(X,\mathbb Z)\cap H^{1,1}(X)\neq \{0\}}$.

The reason of insufficiency is the following: Since $H^{1,1}(X)$ is 1-dimensional, the Kahler cone is a ray $\mathbb R^+$ in the real vector space $H^2(X,\mathbb R)$, so $X$ is a Hodge manifold iff the that ray passes an (nonzero) integral element in $H^2(X,\mathbb R)$. Explicitly, let's say $h^{2,0}=h^{0,2}=1$, and $v_1,v_2,v_3$ is a basis of $H^2(X,\mathbb Z)$, then $[\omega]=\sum_{i=1}^3r_iv_i$ for some $r_i\in \mathbb R$, can we gurantee that $[r_1:r_2:r_3]=[n_1:n_2:n_3]$ with $n_i$ integer? No, and general weight two Hodge structure won't satisfy that.