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When compositions and adjoints are defined on the common test domain, reversing products gives $[A,B{]}^{\ast }=[B^{\ast },A^{\ast }]$. In particular a self-adjoint Dolbeault Laplacian and the mutually adjoint Lefschetz operators satisfy $[{\Delta }_{\bar{\partial }},L{]}^{\ast }=[\Lambda ,{\Delta }_{\bar{\partial }}]$.

For the Chern connection of a Hermitian holomorphic vector bundle on a Kähler manifold, the two Dolbeault Laplacians differ by $[iF\wedge ,\Lambda ]$, where $F$ is its curvature form of a connection and $\Lambda$ is the adjoint Lefschetz operator. The curvature term controls vanishing estimates.

On a compact Hermitian manifold of complex dimension $n$, $C\alpha =\ast \bar{\alpha }$, with complex-linear Hodge star, is a conjugate-linear bijection from Dolbeault Laplacian harmonic $(p,q)$-forms to harmonic $(n-p,n-q)$-forms. The identities ${\bar{\partial }}^{\ast }=-\ast \partial \ast$ and ${\ast }^2=(-1{)}^{p+q}$ show that $\bar{\partial }\alpha ={\bar{\partial }}^{\ast }\alpha =0$ implies the same two equations for $C\alpha$. Its square is $(-1{)}^{p+q}$. Thus these harmonic spaces have equal complex dimension even without the Kähler condition. The natural map is conjugate-linear, not complex-linear.

On a compact Hermitian manifold without boundary, this operator is the inverse of the Dolbeault Laplacian on the orthogonal complement of its harmonic forms and is zero on the harmonic space. Elliptic theory gives ${\Delta }_{\bar{\partial }}G=G{\Delta }_{\bar{\partial }}=I-H$, with $H$ the harmonic projection, and ensures that $G$ sends smooth forms to smooth forms. It commutes with $\bar{\partial }$ and ${\bar{\partial }}^{\ast }$; their commutation with the Laplacian and orthogonality to its kernel give this by uniqueness of the inverse. If $\eta$ is $\bar{\partial }$-exact, then $\eta =\bar{\partial }{\bar{\partial }}^{\ast }G\eta$. On a Kähler manifold, the same operator serves the holomorphic Laplacian because ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$.

The Lefschetz operator commutes with the bundle-valued Dolbeault Laplacian on the entire smooth form space exactly when the Chern connection is flat. The Lefschetz-Dolbeault curvature commutator gives one direction. Conversely test its curvature-wedge operator on degree-zero sections with arbitrary prescribed values at any point to force every vector-bundle curvature endomorphism to vanish.

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.

The Kähler Laplacian identity ${\Delta }_d^{\mathrm{H}}=2{\Delta }_{\bar{\partial }}$ implies that every holomorphic function on a Kähler manifold is a harmonic function. On functions, $\bar{\partial }u=0$ and ${\bar{\partial }}^{\ast }u=0$ by degree, so both terms of the Dolbeault Laplacian vanish. Compactness is unnecessary. Changing the sign convention for the scalar Laplacian does not change this conclusion.

For a Hermitian holomorphic bundle over a Kähler manifold, let $\Theta$ act by exterior multiplication. The bundle-valued Kähler identities give $[\Lambda ,{\Delta }_{\bar{\partial }}]=-i(\Theta \wedge -{)}^{\ast }$, and taking adjoints yields $[{\Delta }_{\bar{\partial }},L]=i\Theta \wedge -$. This measures the failure of the bundle-valued Dolbeault Laplacian to commute with the Lefschetz operator.

On a Kähler manifold, the Lefschetz operator of a Kähler manifold $L=\omega \wedge -$ commutes with the Dolbeault Laplacian:Consequently, if $\alpha$ is ${\Delta }_{\bar{\partial }}$-harmonic, then every $\alpha \wedge {\omega }^k=L^k\alpha$ is also harmonic.

An almost complex structure is a smooth endomorphism $J:TM\to TM$ with $J^2=-I$; in particular the real dimension is even. Complexifying the tangent bundle splits it into the $i$ and $-i$ eigensubbundlesinterchanged by conjugation. On a complex manifold, these are locally spanned by ${\partial }_{z^j}$ and ${\partial }_{{\bar{z}}^j}$. An integrable almost complex structure is one arising from a holomorphic coordinate atlas. Its $T^{0,1}$ sections are closed under the Lie bracket. The obstruction is the Nijenhuis tensorExpanding this expression in the two eigentypes shows that $N_J=0$ is equivalent to involutivity of $T^{0,1}$. Holomorphic coordinates make this condition necessary; the Newlander-Nirenberg theorem gives its sufficiency for smooth $J$. Integrability also gives $d=\partial +\bar{\partial }$, with ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault operator on $T^{1,0}M=T_M^{\prime }$ defines its holomorphic vector bundle structure.

A real connection on a vector bundle $\nabla$ on $TM$ preserves $J$ when $\nabla J=0$. Its complexification then preserves both eigensubbundles. Restricting to $T_M^{\prime }$ gives a complex connection $D$. Conversely any complex connection on $T_M^{\prime }$ gives such a real connection by adjoining its conjugate on $T^{0,1}$ and restricting to the conjugation-fixed real tangent bundle. Equivalently, transfer $D$ through the real isomorphism $v\mapsto (v-iJv)/2$. These operations are inverse. Preserving $J$ alone does not impose the holomorphic compatibility condition $D^{0,1}={\bar{\partial }}_{T^{\prime }}$.

For integrable $J$, the latter condition has a precise real-connection interpretation: the mixed-type torsion tensor vanishes. In holomorphic coordinates it says $D_{{\partial }_{{\bar{z}}^i}}{\partial }_{z^j}=0$. Reality also gives ${\nabla }_{{\partial }_{z^j}}{\partial }_{{\bar{z}}^i}=0$. Since these coordinate fields commute, their torsion is zero. Conversely a $J$-preserving real connection with zero mixed torsion has these two derivatives equal; they belong to opposite eigentypes, so both vanish, proving $D^{0,1}={\bar{\partial }}_{T^{\prime }}$. This is the holomorphic tangent connection and mixed torsion criterion.

Full torsion-freeness imposes the additional symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in $D_{{\partial }_{z^i}}{\partial }_{z^j}={\Gamma }_{ij}^k{\partial }_{z^k}$, and its conjugate. Such a $J$-preserving torsion-free connection exists on any complex manifold: local holomorphic-coordinate flat connections have these properties, and a real partition of unity combines them globally. Preservation of $J$ and zero torsion persist because both conditions are affine in the connection. Conversely if $J$ is merely almost complex, a torsion-free connection preserving it forces $N_J=0$: replace brackets by ${\nabla }_{X}Y-{\nabla }_{Y}X$ in the displayed tensor and use $\nabla J=0$; all terms cancel. Thus this connection condition detects integrability, without imposing any metric condition.

A Hermitian metric is a real Riemannian metric $g$ with $g(JX,JY)=g(X,Y)$. It induces a Hermitian metric $h$ on $T_M^{\prime }$ and the fundamental two-form $\omega (X,Y)=g(JX,Y)$. The unique Chern connection satisfies both $D^{0,1}={\bar{\partial }}_{T^{\prime }}$ and metric compatibility. In a holomorphic frame with metric matrix $H$ and coefficient-column convention, its matrix is $H^{-1}\partial H$. It need not be torsion-free when transferred to $TM$.

In holomorphic coordinates its coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Their symmetry in $i,j$ is equivalent to ${\partial }_i{h}_{j\bar{l}}={\partial }_j{h}_{i\bar{l}}$, which says $\partial \omega =0$. The conjugate gives $\bar{\partial }\omega =0$. The result that a torsion-free Chern tangent connection characterizes a Kähler metric follows: the transferred Chern connection is torsion-free exactly when $\boxed{d\omega =0}$, the Kähler metric condition. In that case it equals the Levi-Civita connection, by uniqueness of the metric-compatible torsion-free connection. Equivalently a Hermitian metric is Kähler exactly when its Levi-Civita connection preserves $J$. In the reverse direction $\nabla J=0$ implies $\nabla \omega =0$, and zero torsion gives $d\omega =0$.

This explains the role of Kähler metrics: they make the real Riemannian and holomorphic connection theories coincide. A general complex manifold has torsion-free $J$-preserving connections and has Hermitian metrics, but need not have a connection satisfying both requirements simultaneously. On a Kähler manifold, the Kähler identities then link the Dolbeault Laplacian and Hodge Laplacian, giving ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$ and the resulting harmonic Hodge decomposition theorem for compact Kähler manifolds.

Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.

The Hermitian metric induces pointwise Hermitian inner products on the exterior powers of the complex cotangent bundle, with different bidegrees orthogonal, and a positive volume form $d{V}_g$. Thus the global $L^2$ inner product on smooth differential forms of type $(p,q)$ isLet ${\bar{\partial }}^{\ast }$ be the formal adjoint of the Dolbeault operator, defined by integration by parts, and let ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. This Dolbeault Laplacian preserves bidegree, is elliptic and nonnegative, and hasOn a compact Hermitian manifold without boundary, the Dolbeault Hodge decomposition on a compact Hermitian manifold says that the smooth harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional andMore precisely, there are the orthogonal harmonic projection $H$ and a Dolbeault Green operator $G$ taking smooth forms to smooth forms, with $G=0$ on harmonics and ${\Delta }_{\bar{\partial }}G=G{\Delta }_{\bar{\partial }}=I-H$. Elliptic regularity ensures these are statements about smooth forms, not only $L^2$ completions. The Green operator commutes with $\bar{\partial }$ and ${\bar{\partial }}^{\ast }$, since the Laplacian does and its inverse is unique on the orthogonal complement of the kernel.

Expanding $\alpha =H\alpha +{\Delta }_{\bar{\partial }}G\alpha$ givesThe three summands are orthogonal: harmonic forms are killed by both adjoints, and $⟨\bar{\partial }u,{\bar{\partial }}^{\ast }v⟩=⟨{\bar{\partial }}^{2}u,v⟩=0$. If $\alpha$ is $\bar{\partial }$-closed, its coexact component is orthogonal to every closed form; taking its inner product with $\alpha$ shows that component has norm zero. Conversely the first two components are closed. Thus $\boxed{\ker \bar{\partial }={\mathcal{H}}_{\bar{\partial }}^{p,q}\oplus \bar{\partial }{A}^{p,q-1}}$, with out-of-range form spaces taken to be zero.

Let $n={\dim }_{\mathbb{C}}X$, and use the complex orientation. Write $g$ for the real Riemannian metric underlying the Hermitian metric. Its fundamental form of a Hermitian manifold iswhere $g=\mathrm{Re}(\sum h_{a\bar{b}}d{z}^a\otimes d{\bar{z}}^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=J{E}_j$ and its dual coframe $e^j,f^j$. Then $\omega ={\sum }_j{e}^j\wedge f^j$, giving the Riemannian volume form

We use the complex-linear Hodge star operator, the extension of the real metric star characterized on equal-degree forms byIt maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $\ast \ast \alpha =(-1{)}^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha \mapsto \ast \overline{\alpha }$. In the adapted coframe the star of $e^j\wedge f^j$ is the wedge product of all the other two-dimensional factors, with positive sign. Summing proves the Hodge star of the fundamental Hermitian form formulaThis computation uses no closedness assumption on $\omega$.

Use the integrated Hermitian inner product $(\alpha ,\beta )={\int }_X\alpha \wedge \ast \overline{\beta }$. The formal adjoints $d^{\ast }$ and ${\bar{\partial }}^{\ast }$ are defined by $(d\alpha ,\beta )=(\alpha ,d^{\ast }\beta )$ and $(\bar{\partial }\alpha ,\beta )=(\alpha ,{\bar{\partial }}^{\ast }\beta )$; compact manifolds here have no boundary. The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold states thatis an orthogonal direct sum, with finite-dimensional ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$. Each Dolbeault cohomology classhas a unique representative in ${\mathcal{H}}_{\bar{\partial }}^{p,q}$. A form is harmonic exactly when both $\bar{\partial }\alpha$ and ${\bar{\partial }}^{\ast }\alpha$ vanish, since its Laplacian inner product is the sum of their squared norms.

Now let $L\alpha =\omega \wedge \alpha$ be the Hermitian Lefschetz operator and let $\Lambda$ be its pointwise adjoint. For $\beta$ of total degree $r=p+q$ and $\alpha$ of degree $r-2$, the defining inner products giveThe wedge pairing is nondegenerate, so $\ast \overline{\Lambda \beta }=L\ast \overline{\beta }$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yieldsFor degrees below two both sides vanish. The argument also identifies its bidegree as $(-1,-1)$ and does not require the Kähler condition.

Assume now that $d\omega =0$, so the metric is a Kähler metric. Start with the supplied Kähler identities relation $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$. Since $\Lambda$ is real, conjugation gives $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$, henceThe commutators are ordinary commutators because $\Lambda$ has even degree. Using ${\partial }^2=0$ we obtain, by direct expansion,Conjugation gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ as well. Expanding the two unmixed Laplacians and using $\partial \bar{\partial }=-\bar{\partial }\partial$ gives the same operator:Finally expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$; the mixed terms just proved zero disappear. This derives the Kähler Laplacian identity

The Dolbeault Laplacian preserves bidegree. Thus the complex $d$-harmonic forms of total degree $r$ split into their harmonic $(p,q)$ parts. The allowed real Hodge theorem, followed by complexification, givesThe real operator ${\Delta }_d$ commutes with conjugation, which exchanges $(p,q)$ and $(q,p)$. Therefore $h^{p,q}=h^{q,p}$. For odd $r$, no summand has $p=q$, so they pair off in equal dimensions. This proves that all odd Betti numbers are even.

For $1\le k\le n$, the Kähler form power ${\omega }^k$ is closed. If it were exact, say ${\omega }^k=d\eta$, thenby Stokes theorem. But ${\omega }^n=n!{\mathrm{vol}}_g$ has strictly positive integral on the nonempty compact manifold. Hence the powers of a compact Kähler form have nonzero cohomology classes, and

For an example take the Hopf surfaceThe dilation action is free and properly discontinuous: on a compact subset the radii are bounded above and away from zero, so only finitely many dilations can meet it. It is holomorphic, giving a complex manifold quotient; the annulus $1\le \Vert z\Vert \le 2$ covers the quotient, proving compactness. The mapis a diffeomorphism $H\cong S^3\times S^1$. The Künneth theorem gives $b^1(H)=1$. This violates the necessary odd-degree parity, so this compact complex surface admits no Kähler metric.

Let $g$ be the Riemannian metric underlying the Hermitian metric and $J$ its compatible complex structure. The fundamental form of a Hermitian manifold is $\omega (u,v)=g(Ju,v)$. It is a real alternating form of type $(1,1)$. At a point choose an orthonormal real coframe $x^1,y^1,\dots ,x^n,y^n$ adapted to $J$, with $x^j+i{y}^j$ of type $(1,0)$. ThenThe orientation here is the complex orientation. In coordinates for which $g={\sum }_j((d{x}^j{)}^2+(d{y}^j{)}^2)$, the same convention reads $\omega =(i/2){\sum }_{j}d{z}^j\wedge d{\bar{z}}^j$.

Use the complex Hodge star operator, the complex-linear extension of the real Hodge star operator. With the pointwise Hermitian inner product linear in its first variable, its defining identity isfor forms of the same total degree. It maps type $(p,q)$ to $(n-q,n-p)$ and satisfies ${\ast }^2=(-1{)}^{p+q}$ in real dimension $2n$. It commutes with complex conjugation. These conventions distinguish it from the conjugate-linear Hodge star $\alpha \mapsto \ast \bar{\alpha }$.

Write a top holomorphic-type form as $\eta =a{\zeta }^1\wedge \cdots \wedge {\zeta }^n$, where ${\zeta }^j=x^j+i{y}^j$. On each oriented two-plane, $\ast x^j=y^j$ and $\ast y^j=-x^j$, so the two-dimensional star of ${\zeta }^j$ is $-i{\zeta }^j$. In taking the star of their product, regrouping complementary one-form factors across the $n$ two-planes contributes $(-1{)}^{n(n-1)/2}$. Thus the Hodge star on top holomorphic forms isThis is pointwise and requires neither holomorphicity of the coefficient $a$ nor closedness of $\omega$.

To compute the formal adjoint, take compactly supported forms $\alpha$ of degree $k-1$ and $\beta$ of degree $k$. The Stokes theorem givesSince the degree of $d(\ast \bar{\beta })$ is $2n-k+1$, the identity for ${\ast }^2$ makes the right side equal to ${\int }_X\alpha \wedge \ast \overline{(-\ast d\ast \beta )}$. Consequently the formal adjoint of $d$ in even real dimension is $d^{\ast }=-\ast d\ast$. Splitting by type, $-\ast \partial \ast$ lowers antiholomorphic degree, while $-\ast \bar{\partial }\ast$ lowers holomorphic degree. Orthogonality of distinct types therefore yieldsThe argument applies locally with compact support, and globally without support restrictions on a compact manifold without boundary.

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. A form is $\bar{\partial }$-harmonic when it lies in its kernel. On a compact Hermitian manifold, the adjoint identities implyThus harmonicity is equivalent to both first-order equations vanishing. The Dolbeault Hodge decomposition on a compact Hermitian manifold states that the harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional and that smooth forms have the orthogonal direct-sum decomposition

To deduce the identification with Dolbeault cohomology, let $\bar{\partial }\alpha =0$ and decompose $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$. The closedness of $\alpha$ makes it orthogonal to the entire image of ${\bar{\partial }}^{\ast }$. Taking its inner product with ${\bar{\partial }}^{\ast }v$ and using the orthogonality of the decomposition gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Hence $\alpha =h+\bar{\partial }u$, so each cohomology class has a harmonic representative. If a harmonic form equals $\bar{\partial }u$, its squared norm is $⟨{\bar{\partial }}^{\ast }h,u⟩=0$, so the representative is unique. We have proved

For the complementary types, define $C\alpha =\ast \bar{\alpha }$. It maps $(p,q)$ to $(n-p,n-q)$ and is conjugate-linear. If $\alpha$ is harmonic, then $\bar{\partial }\alpha =0$ and $\partial (\ast \alpha )=0$ by the adjoint formula. Conjugating the latter equation gives $\bar{\partial }(\ast \bar{\alpha })=0$. Also, with $k=p+q$,since $\partial \bar{\alpha }=\overline{\bar{\partial }\alpha }$. Therefore $C$ carries harmonic forms to harmonic forms, and $C^2=(-1{)}^k$ makes it a bijection. This proves complementary Dolbeault harmonic types on a Hermitian manifold:The metric supplies a natural conjugate-linear isomorphism. If a complex-linear isomorphism is desired, choose a basis $h_j$ of the first space and send it to the basis $C{h}_j$ of the second, extending linearly. This last choice is not canonical; no Kähler assumption is needed for either the dimension equality or the harmonic bijection.

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

The specified bidegree is the conjugate-linear Hodge star, not the complex-linear extension of the real star. If the pointwise Hermitian inner product is linear in its first argument, it is defined byIt is the real Hodge star operator extended complex-linearly and composed with complex conjugation; ${\ast }^2=(-1{)}^{p+q}$. On a compact Hermitian manifold,is a positive definite Hermitian inner product on smooth forms. The formal adjoint is ${\bar{\partial }}^{\ast }=-\ast \bar{\partial }\ast$ in this convention; the final star in this identity is present in the PDF but is corrupted in the TeX aid.

State the relevant Hodge theorem as follows. On a compact manifold without boundary, the Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$ is elliptic, self-adjoint and nonnegative. Its harmonic kernel ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional, and smooth forms split orthogonally as this kernel plus the image of ${\Delta }_{\bar{\partial }}$. There are the orthogonal projection $H$ and a smooth Green operator $G$ withFor any form $u$, expand the Laplacian term to obtainHarmonic forms are both closed and coclosed, since $({\Delta }_{\bar{\partial }}v,v)=\Vert \bar{\partial }v{\Vert }^2+\Vert {\bar{\partial }}^{\ast }v{\Vert }^2$. They are orthogonal to both other terms; the two images are orthogonal because $(\bar{\partial }a,{\bar{\partial }}^{\ast }b)=({\bar{\partial }}^{2}a,b)=0$. This proves the Dolbeault Hodge decompositionThe closed forms are exactly the first two summands: if $\bar{\partial }u=0$, then $u$ is orthogonal to the adjoint image, and conversely the first two summands are closed. It follows that every Dolbeault cohomology class has a unique harmonic representative. Negative or out-of-range bidegrees mean zero spaces. Compactness and the absence of boundary are necessary for this stated global decomposition.

Now assume a Kähler metric. Set $L=\omega \wedge -$ and $\Lambda =L^{\ast }$. The permitted Kähler identities, stated with signs, arewith ordinary commutators because $\Lambda$ has even degree. Thus ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$, andby ${\partial }^2=0$. The companion identity gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ in the same way.

Define ${\Delta }_d=d{d}^{\ast }+d^{\ast }d$ and ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$. Expanding $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$ makes the two mixed anticommutators vanish, so ${\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}$. To prove equality of the latter two, substitute the same commutator formulas and use $\partial \bar{\partial }=-\bar{\partial }\partial$:The displayed operators are identical. Therefore the Kähler Laplacian identity is

Let $\eta$ be a $\bar{\partial }$-exact form. It is closed and orthogonal to harmonic forms. The Dolbeault Hodge decomposition, or the Green identity with $\alpha =G\eta$, therefore givesFor the Green construction, $\bar{\partial }\alpha =0$ follows since $G$ commutes with $\bar{\partial }$; equivalently the coexact component of a closed form vanishes. This yields exactly the stated factorization.

If also $\partial \eta =0$, put $\beta ={\bar{\partial }}^{\ast }\alpha$ and $\xi =\partial \beta$. The identities already proved giveThus $\xi$ is harmonic for ${\Delta }_{\bar{\partial }}$ and hence for ${\Delta }_{\partial }$. But it is $\partial$-exact, so it is orthogonal to the harmonic summand in the Hodge decomposition theorem. Consequently $\xi =0$, proving that $\beta ={\bar{\partial }}^{\ast }\alpha$ is $\partial$-closed.

Furthermore $\beta$ is orthogonal to the common harmonic space, since $(\beta ,h)=(\alpha ,\bar{\partial }h)=0$. Its $\partial$-Hodge decomposition then has neither a harmonic term nor a coexact term: closedness kills the coexact term. Therefore $\beta =\partial \chi$ for a form $\chi \in A^{p-1,q-1}$. It follows thatThis proves the requested ddbar lemma, including the intermediate harmonicity and orthogonality steps. The hypotheses $p,q\ge 1$ ensure that the potential has a valid bidegree.