Suppose $\omega$ is a real closed $(1,1)$ form on a compact Kähler manifold. If we have a real $d$-closed two form $\sigma$ such that $[\sigma]=[\omega] \in H^2(M)$, can we claim that this two form $\sigma$ is also a $(1,1)$ form?



A question about a two form and a $(1,1)$ form on a compact Kähler manifold
I think not. Note that replacing $\sigma$ by $\sigma - \omega$ reduces us to the case $\omega=0$. Your question in that case is whether every real $d$-exact 2-form is a $(1,1)$-form.


Now unless I'm mistaken, on projective space $\mathbf{CP}^2$ with homogeneous coordinates $(a:b:c)$, we get a real $d$-exact 2-form with components of all types ($(2,0)$, $(1,1)$, $(0,2)$) by putting 
$$
\sigma
=d\left(\frac{A}{A+B+C} d\left(\frac{B}{A+B+C}\right)\right)
=\frac{AdB\wedge dC+BdC\wedge dA+CdA\wedge dB}{(A+B+C)^3}
$$
where $(A,B,C)=(|a|^2, |b|^2, |c|^2)$.


