 Solution

If $\alpha =\partial \bar{\partial }\beta$ and $h\in {\mathcal{H}}^{p,q}$, the Kähler Laplacian identity makes $h$ both $\partial$- and $\bar{\partial }$-harmonic. Hence ${\partial }^{\ast }h={\bar{\partial }}^{\ast }h=0$, and integration by parts gives $⟨\alpha ,h⟩=0$.

Conversely, let $d\alpha =0$ and $\alpha \perp {\mathcal{H}}^{p,q}$. Pure type gives $\partial \alpha =\bar{\partial }\alpha =0$. Dolbeault Hodge decomposition removes the harmonic and coexact components, soNow $\bar{\partial }(\partial \theta )=-\partial \alpha =0$, and the Kähler anticommutation identity gives ${\bar{\partial }}^{\ast }(\partial \theta )=-\partial ({\bar{\partial }}^{\ast }\theta )=0$. Thus $\partial \theta$ is $\bar{\partial }$-harmonic and therefore $\partial$-harmonic; being $\partial$-exact, it vanishes. Moreover $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to the common $\bar{\partial }$- and $\partial$-harmonic space. Its $\partial$-Hodge decomposition therefore gives $\theta =\partial \varphi$. HenceTaking $\beta =-\varphi$ proves the harmonic orthogonality criterion for ddbar exactness