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On a compact oriented Riemannian manifold, the Green operator $G$ is the inverse of the Hodge Laplacian on the orthogonal complement of harmonic forms and is zero on harmonic forms. If $H$ is harmonic projection, thenIt commutes with every differential operator that commutes with $\Delta$.

Write $g=\mathrm{Re}h$ for the underlying Riemannian metric of the Hermitian manifold. Its fundamental form of a Hermitian manifold isIt is real and skew-symmetric. In a unitary coframe it is $\omega ={\sum }_{j=1}^n{e}^j\wedge J{e}^j$, which also shows that it has type $(1,1)$ and thatThe Hodge star operator is characterized, after complex-linear extension, byExpanding in the same unitary coframe gives

The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold says that every Dolbeault class has a unique ${\Delta }_{\bar{\partial }}$-harmonic representative andIf ${\Delta }_{\bar{\partial }}\eta =0$, thenso $\eta$ is $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed. If also $\eta =\bar{\partial }\xi$, then $\Vert \eta {\Vert }_2^2=⟨{\bar{\partial }}^{\ast }\eta ,\xi ⟩=0$, hence $\eta =0$.

Now suppose $X$ is compact and Kähler. With $d^c=i(\bar{\partial }-\partial )$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$, the Kähler identities make the mixed anticommutators vanish and give ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Consequently

Let $\alpha =d^c\gamma$ and $d\alpha =0$. The Kähler Laplacian identity implies that the $d$-Laplacian commutes with $d^c$. Since a harmonic form is $d^{c\ast }$-closed, $\alpha$ is orthogonal to every harmonic form. If $G$ is the Green operator of the Hodge Laplacian, thenwhere the $d^{\ast }dG\alpha$ term vanishes because $dG\alpha =Gd\alpha =0$. The Green operator commutes with $d^c$, and the anticommutation identity just proved givesTherefore, for the $(k-2)$-form $\beta =-d^{\ast }G\gamma$,This is the d d c lemma in the form required here.

For the stated flat metric and orientation,Because the metric coefficients and the coordinate one-forms are constant, the Hodge Laplacian acts coefficientwise:Thus a harmonic one-form $\alpha ={\sum }_i{\alpha }_{i}d{x}^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the maximum principle for harmonic functions. HenceThe Hodge decomposition theorem now gives

The Kähler manifold structure gives a Riemannian metric, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The complex Hodge star operator is the complex-linear map characterized byOn $r$-forms in real dimension $m=2n$, the codifferential isequivalently the formal adjoint of $d$ for the $L^2$ inner product. Similarly ${\bar{\partial }}^{\ast }$ is the formal adjoint of $\bar{\partial }$. Define the Hodge Laplacian and Dolbeault Laplacian by

Expanding $d=\partial +\bar{\partial }$, the Kähler identities make the mixed anticommutators vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Hence the Kähler Laplacian identity is

Let $L\gamma =\omega \wedge \gamma$ be the Lefschetz operator of a Kähler manifold. The Kähler identities also implyThus, if ${\Delta }_{\bar{\partial }}\alpha =0$, thenThis is the fact that the Lefschetz operator preserves harmonic forms.

The Dolbeault Hodge decomposition on a compact Hermitian manifold states thatan orthogonal direct sum, where ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$.

Suppose $\eta =\bar{\partial }\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma \in {\Omega }^{p,q-1}$. The harmonic and $\bar{\partial }$-exact pieces disappear after applying $\bar{\partial }$, so for some $\beta \in {\Omega }^{p,q}$,Put $\theta ={\bar{\partial }}^{\ast }\beta$. If also $\partial \eta =0$, thenThe Kähler anticommutation identity ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ and $({\bar{\partial }}^{\ast }{)}^2=0$ giveTherefore $\partial \theta$ is ${\Delta }_{\bar{\partial }}$-harmonic. By ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forcesThis proves both requested claims: $\partial {\bar{\partial }}^{\ast }\beta =0$ is harmonic, and ${\bar{\partial }}^{\ast }\beta =\theta$ is $\partial$-closed.

Finally, $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to $\ker \bar{\partial }$, and hence to every $\bar{\partial }$-harmonic form. Since the $\partial$- and $\bar{\partial }$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta =\partial \varphi$ for some $\varphi \in {\Omega }^{p-1,q-1}(X)$. Hencewhich is the ddbar lemma in this case.