TITLE: Can a class be represented by both a $(p,q)$-form and a $(p',q')$-form?
QUESTION [11 upvotes]: Suppose $X$ is a complex manifold. 
If $X$ is Kähler, the cohomology groups decompose into subgroups represented by $(p,q)$-forms. 
If $X$ is not Kähler, I think the decomposition may not hold? 
Is there an example where we have a nonzero class be represented by both a $(p,q)$-form and a $(p',q')$-form with $(p, q) \neq (p',q')$?

REPLY [6 votes]: This is meant as a supplement to Michael's answer above. Here is an example of a compact complex threefold $X$ and two forms $\gamma_1 \in \mathcal{E}^{p,q}$ and $\gamma_2 \in \mathcal{E}^{p',q'}$ such that $(p,q) \neq (p',q')$ and $[\gamma_1] = [\gamma_2] \in H_{dR}^2(X)$, but also $[\gamma_1] \neq 0$. This will make use of results found in Cordero, Fernández, Gray, The Frölicher spectral sequence for compact nilmanifolds.
Applying Theorem 6 in this paper gives us that there exists a simply connected 6-dimensional real Lie group with a left-invariant complex structure whose left-invariant $(1,0)$-forms are spanned by $\omega_1, \omega_2, \omega_3$ and whose left-invariant $(0,1)$-forms are spanned by $\bar \omega_1, \bar \omega_2, \bar \omega_3$, with structure equations given by \begin{align*} d\omega_1 &= 0, \ d\bar \omega_1 = 0, \\ d\omega_2 &= 0, \ d\bar \omega_2 = 0, \\ d\omega_3 &= \omega_1\omega_2 - \bar \omega_1 \omega_2, \ d\bar \omega_3 = \bar \omega_1 \bar \omega_2 - \omega_1 \bar \omega_2. \end{align*} Since the (dual) structure constants are rational, a theorem of Malcev (see Corollary 8 in the paper) tells us that there exists a discrete subgroup $\Gamma$ in $G$ such that $\Gamma \backslash G$ is a closed 6-manifold. The forms $\omega_i$ and $\bar \omega_i$ descend to $\Gamma \backslash G$ along with the complex structure on $G$. A theorem of Nomizu (referenced in this context in Hasegawa, Minimal models of nilmanifolds  at the end of Section 1) tells us that the de Rham cohomology of $\Gamma \backslash G$ is isomorphic to to the cohomology of the differential graded algebra generated in degree one by $\{\omega_1, \omega_2, \omega_3, \bar \omega_1, \bar \omega_2, \bar \omega_3\}$ with the differential $d$ prescribed above. 
Now, take the $(2,0)$-form $\omega_1\omega_2$ and the $(1,1)$-form $\bar \omega_1 \omega_2$. Note that they are both closed, and their difference is $d$-exact (since $d\omega_3 = \omega_1\omega_2 - \bar \omega_1 \omega_2$). So, $[\omega_1\omega_2] = [\bar \omega_1 \omega_2]$ in de Rham cohomology. We see that $\omega_1\omega_2$ is not in the image of $d$, that is $[\omega_1\omega_2] \neq 0$.