Harmonic orthogonality criterion for ddbar exactness ($\alpha \perp {\mathcal{H}}^{p,q}⟺\alpha \in \mathrm{im}(\partial \bar{\partial })$)

If $\alpha$ is a $d$-closed $(p,q)$-form on a compact Kähler manifold, then $\alpha$ is ddbar lemma exact if and only if it is orthogonal to every harmonic differential form of type $(p,q)$. The forward implication follows by integration by parts. For the reverse implication, Dolbeault Hodge decomposition first makes $\alpha$ $\bar{\partial }$-exact, and the ddbar lemma then makes it $\partial \bar{\partial }$-exact.

For a $d$-closed pure-type form $\alpha$ orthogonal to harmonic forms on a compact Kähler manifold, let $G_{\bar{\partial }}$ be the inverse of the Dolbeault Laplacian on the harmonic complement, zero on the harmonic space. The Kähler Laplacian identity makes it commute with $\partial$ and $\bar{\partial }$. Thus $\alpha =\bar{\partial }{\bar{\partial }}^{\ast }{G}_{\bar{\partial }}\alpha$ and $G_{\bar{\partial }}\alpha =\partial {\partial }^{\ast }{G}_{\bar{\partial }}^2\alpha$. The mixed anticommutator ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ gives $\alpha =\partial \bar{\partial }\beta$. This is an explicit potential for the ddbar lemma, with one lower degree of each type; using the inverse of the Hodge Laplacian instead multiplies the displayed expression by four.