For a compact Kähler manifold, we say that a form is **primitive** if it is contaned in the kernel of the dual Lefschetz operator, or the co-Lefschetz operator. For all examples I know, a primitive form $\omega$ is closed with respect to the $d$ de Rham exterior derivative if and only if $\omega$ is harmonic. I suspect that this is true for any compact Kähler manifold, but I don't know how one would prove it.



de Rham closed harmonic form on a Kähler manifold
If $\omega$ is harmonic, then it is $d$-closed. For the other direction, suppose $\omega$ is $d$-closed and primitive, i.e. $d\omega = 0$ and $\Lambda\omega = 0$. Then, by a Kähler identity, 


$$i\bar{\partial}^\*\omega = [\Lambda, \partial]\omega = \Lambda(\partial\omega) - \partial(\Lambda\omega) = \Lambda(\partial\omega)$$ 


and similarly $-i\partial^\*\omega = \Lambda(\bar{\partial}\omega)$. So 


$$d^\*\omega = \partial^\*\omega + \bar{\partial}^\*\omega = \Lambda(i\bar{\partial}\omega - i\partial\omega) = \Lambda(d^c\omega)$$


where $d^c = i(\bar{\partial}-\partial)$. So $\omega$ is harmonic if and only if if $d^c\omega$ is primitive.


If $\omega$ is a $(p, q)$-form, then $d\omega = 0$ is equivalent to $\partial\omega = 0$ and $\bar{\partial}\omega = 0$, in which case it follows that $\partial^\*\omega = 0$ and $\bar{\partial}^\*\omega = 0$. Therefore $d^\*\omega = 0$ and hence $\omega$ is harmonic.


I'm not sure if the statement still holds if $\omega$ is an arbitrary form. The issue is that $d\omega = 0$ does not necessarily imply $\partial\omega = 0$ and $\bar{\partial}\omega = 0$ in this case.


