Kahler form lies in $H^2(X,\mathbb Z)$?

Question

$X$ is a Kahler manifold. Then is it true that the class of Kahler form $[\omega]$ lies in $H^2(X,\mathbb Z)$?

In fact I am not sure I understand $H^2(X,\mathbb Z)$ correctly. Why can we talk about $H^2_{dR}(X,\mathbb Z)$? Because I don't think "forms with integer coefficients" is well-defined.

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Edit

I didn't make question clear. As Tsemo Aristide's answer suggests, if $\omega$ is a kahler form, then so is $c \omega$. So I really want to ask is: can we always find a $c$ such that $c\omega$ lies in $H^2(X,\mathbb Z)$?

Answer 1

Sorry for my late reply ,there is a more general theorem to describe this fact,by Huybrechts'Complex Geometry 5.3 Kodaira Embedding theorem Corollary 5.3.3,we have

A compact Kähler manifold $X$ is projective if and only if $\mathcal K_X\cap H^2(X,\mathbb Z)\not=\emptyset .$

Here $\mathcal K_X$ denotes the Kähler cone,so we can find counterexample from non-projective manifolds,such as some complex torus,neither Kähler class associated to any Kähler structure lies in $H^2(X,\mathbb Z)$.

Answer 2

Consider the $2$-torus endowed with its standard Kahler form $\omega $. Let $i$ be any irrational number $i\omega$ is also a Kahler for the same complex structure but is not in $H^2(\mathbb{T}^2,\mathbb{Z})$.