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A holomorphic differential form $\alpha$ of type $(p,0)$ satisfies $\bar{\partial }\alpha =0$. The adjoint ${\bar{\partial }}^{\ast }$ lowers antiholomorphic degree, so ${\bar{\partial }}^{\ast }\alpha =0$ as well. Therefore ${\Delta }_{\bar{\partial }}\alpha =0$.

The Kähler Laplacian identity gives ${\Delta }_d\alpha =2{\Delta }_{\bar{\partial }}\alpha =0$. Compactness lets us use the $L^2$ identity for the Hodge Laplacian:Both terms are nonnegative, soThis proves that holomorphic forms on a compact Kähler manifold are closed. The original PDF uses ${\Omega }_M^p$ and ${\mathcal{A}}_{M,\mathbb{C}}^{p+1}$; the extra symbols in the local TeX superscripts are transcription errors.

For the Lefschetz operator of a Kähler manifold and its adjoint $\Lambda$, the Kähler identities includeThey make the mixed anticommutators in the expansion of ${\Delta }_d$ vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Expanding $d=\partial +\bar{\partial }$ therefore proves the Kähler Laplacian identityThus ${\Delta }_d\alpha =0$ exactly when ${\Delta }_{\bar{\partial }}\alpha =0$.

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

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.

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.