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Let $g$ be the real Riemannian metric underlying the Hermitian metric, and let $J$ be the complex structure. The fundamental form of a Hermitian manifold isThe identity $g(Ju,Jv)=g(u,v)$ implies $g(Ju,v)=-g(u,Jv)$, so this is an alternating real two-form. Moreover $\omega (Ju,Jv)=\omega (u,v)$. A two-form with this invariance has type $(1,1)$: on two vectors of type $(1,0)$, applying $J$ to both multiplies its value by $i^2=-1$, forcing that value to vanish; the same argument applies to type $(0,1)$. Hence $\omega$ is a real $(1,1)$-form.

Use the complex-linear convention for the complex Hodge star operator, extending the real star. The complex orientation and metric define $dV={\omega }^n/n!$, andHere the Hermitian inner product is linear in the first argument. With this convention, on the even-dimensional underlying real manifold the formal adjoints are $d^{\ast }=-\ast d\ast$, ${\bar{\partial }}^{\ast }=-\ast \partial \ast$ and ${\partial }^{\ast }=-\ast \bar{\partial }\ast$. A conjugate-linear star is another convention, but then the adjoint formulas must change accordingly.

The Hodge Laplacian and Dolbeault Laplacian areAlso write ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$. They are nonnegative, since for exampleThe requested Dolbeault Hodge decomposition on a compact Hermitian manifold is a compactness statement: on a compact Hermitian manifold without boundary,The harmonic space is finite-dimensional, and taking the harmonic representative induces an isomorphism ${\mathcal{H}}_{\bar{\partial }}^{p,q}\cong H_{\bar{\partial }}^{p,q}(X)$. Compactness is essential to this formulation; it is not asserted on an arbitrary noncompact Hermitian manifold.

Now suppose $X$ is compact and Kähler, so $d\omega =0$. The permitted Kähler Laplacian identity is ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$, making the three harmonic spaces identical in each bidegree. A $d$-closed pure-type form has $\partial \alpha =\bar{\partial }\alpha =0$, since these two terms have different types. We will construct a Green operator potential for ddbar exactness, rather than invoking the ddbar lemma as the conclusion to be proved.

Let $G$ be the inverse of ${\Delta }_{\bar{\partial }}$ on the orthogonal complement of its harmonic forms, zero on the harmonic space. It is twice the Green operator of the Hodge Laplacian. The Hodge decomposition theorem and elliptic inversion give ${\Delta }_{\bar{\partial }}G=I-H$, with $H$ harmonic projection. It preserves bidegrees. The operators $\partial$ and $\bar{\partial }$ commute with ${\Delta }_{\bar{\partial }}$: the first follows from ${\Delta }_{\bar{\partial }}={\Delta }_{\partial }$ and ${\partial }^2=0$, and the second directly from ${\bar{\partial }}^2=0$. They and their formal adjoints annihilate harmonic forms, so they preserve the orthogonal complement of the harmonic space. Consequently they commute with $G$, by uniqueness of the inverse on the harmonic complement. Thus $u=G\alpha$ and $v=G^2\alpha$ are killed by both differential operators.

The assumed orthogonality makes $H\alpha =Hu=0$. Applying the two Laplacians givesWe also need the mixed anticommutator, which follows from the permitted Laplacian identity rather than an extra unproved lemma. Expand ${\Delta }_d$ using $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$. Since ${\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}$, its mixed terms sum to zero. Their two bidegrees $(1,-1)$ and $(-1,1)$ are distinct, so each vanishes separately, in particular ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$. HenceWe have obtained the explicit answerThe positive assumptions on $p,q$ ensure these degrees are available. Conversely such a form is orthogonal to harmonic forms by integration by parts, so the same proof also explains the harmonic orthogonality criterion for ddbar exactness.

For injectivity of the Lefschetz operator on a Hermitian manifold on real one-forms, work at one point in a metric-orthonormal coframe $(e^1,f^1,\dots ,e^n,f^n)$ with $\omega ={\sum }_j{e}^j\wedge f^j$. If $a\ne 0$, a unitary change of coframe makes $a=c{e}^1$, $c\ne 0$. Thensince the displayed basis three-forms are linearly independent. Therefore $L(a)=a\wedge \omega$ has zero kernel. Injectivity on complex one-forms follows by taking real and imaginary parts, since $\omega$ is real. This pointwise proof of injectivity of powers of the Lefschetz operator for degree one does not require the hard Lefschetz theorem.

Closedness of $\omega$ gives $\bar{\partial }L=L\bar{\partial }$. Using the permitted Kähler identities, specifically $[{\bar{\partial }}^{\ast },L]=i\partial$, givesMultiplying by two proves $[{\Delta }_d,L]=0$. In particular the Lefschetz operator preserves harmonic forms and the injective map on one-forms restricts to an injection ${\mathcal{H}}_d^1\hookrightarrow {\mathcal{H}}_d^3$. The de Rham Hodge theorem identifies these real harmonic spaces with the corresponding real cohomology groups. Taking dimensions yields