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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.

The Hodge star extends complex-linearly to complex differential forms. With the Hermitian pointwise inner product, it is characterized byOn a Hermitian manifold it maps forms of type $(p,q)$ to forms of type $(n-q,n-p)$.

For a Hermitian manifold with the pointwise Hermitian inner product linear in the first variable, define the conjugate-linear star by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\omega }^n/n!$. It is the usual real Hodge star operator extended complex-linearly and then composed with complex conjugation. It has the displayed bidegree and satisfies ${\ast }^2=(-1{)}^{p+q}$ on $(p,q)$-forms.

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 ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$.

For a compact Hermitian manifold, elliptic theory gives the orthogonal decompositionThe harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$ is finite-dimensional and represents Dolbeault cohomology.

In an adapted real orthonormal coframe, the fundamental form of a Hermitian manifold is $\omega ={\sum }_j{e}^j\wedge f^j$ and the Riemannian volume form is ${\omega }^n/n!$. The complex-linear Hodge star sends each summand to the product of all the other oriented two-dimensional factors. Summation gives the displayed formula, valid for every Hermitian manifold, without a Kähler assumption.

For the complex-linear Hodge star operator and complex orientation on a Hermitian manifold of complex dimension $n$, every $(n,0)$-form has the displayed eigenvalue. In an adapted orthonormal real coframe, each factor $d{x}_j+id{y}_j$ has two-dimensional star eigenvalue $-i$. Regrouping the $n$ factors introduces $(-1{)}^{n(n-1)/2}$, giving $(-1{)}^{n(n+1)/2}{i}^n$. This is a pointwise metric identity; the metric need not be Kähler.

Exterior multiplication by the fundamental $(1,1)$-form of a Hermitian manifold raises bidegree by $(1,1)$. Its pointwise formal adjoint lowers bidegree by $(1,1)$. On a Kähler manifold this is the Lefschetz operator of a Kähler manifold, but its pointwise algebraic properties do not require the form to be closed.

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.

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.

Finally $\eta ={\omega }_2-{\omega }_1$ is a real closed $(1,1)$ form and is $d$-exact. Integration by parts makes it orthogonal to every $d$-harmonic form. By the Kähler Laplacian identity it is also orthogonal to the Dolbeault harmonic space; because it is $\bar{\partial }$-closed, the Dolbeault Hodge decomposition makes it $\bar{\partial }$-exact. Apply the proved ddbar lemma to get $\eta =\partial \bar{\partial }\varphi$ for a smooth complex-valued function. Reality gives $\eta =-\partial \bar{\partial }\overline{\varphi }$, so averaging yieldsWith $f=\mathrm{Im}\varphi$,The potential is defined up to a real constant on each connected component. This is a statement about cohomologous Kähler forms, not about an arbitrary pair of such forms.

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.