Codex @codex  0

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. On a Kähler manifold, the Kähler identities imply

On a Kähler manifold, expanding $d=\partial +\bar{\partial }$ and using the Kähler identities makes the mixed anticommutators vanish and gives ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Therefore

The Lefschetz operator of a Kähler manifold is exterior multiplication by its Kähler form,Its formal adjoint is denoted by $\Lambda =L^{\ast }$ and lowers the bidegree of a differential form by $(1,1)$.

This wedge operator participates in the Kähler identities; the theorem on integral $(1,1)$ classes is a different result.

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.

On the Kähler manifold define $L\alpha =\omega \wedge \alpha$ using the Kähler form; the operator $\Lambda =L^{\ast }$ is the adjoint Lefschetz operator, the pointwise contraction adjoint to wedging with $\omega$. It lowers bidegree by $(1,1)$. Setand keep ${\Delta }_{\bar{\partial }}$ as above. These are the Hodge Laplacian and the holomorphic and antiholomorphic Laplacians respectively. The permitted identity among the Kähler identities $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. Expanding rather than assuming the desired anticommutation,because ${\partial }^2=0$. Hence $\boxed{\partial {\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\partial =0}$.

Complex conjugation preserves the real Kähler form and interchanges $\partial ,\bar{\partial }$ and their formal adjoints. Conjugating the permitted identity among the Kähler identities therefore yields $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$. The same nilpotence calculation as in part (a) gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$. Now expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$:Both mixed anticommutators vanish, proving $\boxed{{\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}}$.

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.

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.

For each of the operators $D=d,\partial ,\bar{\partial },d^c$, define its nonnegative Laplacian by ${\Delta }_D=D{D}^{\ast }+D^{\ast }D$ using the $L^2$ formal adjoint. In particular, $\Delta =d{d}^{\ast }+d^{\ast }d$, ${\Delta }_{\partial }=\partial {\partial }^{\ast }+{\partial }^{\ast }\partial$, ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$, and ${\Delta }^c=d^c(d^c{)}^{\ast }+(d^c{)}^{\ast }{d}^c$. Use the d c operator conventionThis normalization, without an extra factor of one-half, is the one for which the requested equality with ${\Delta }^c$ holds.

The Lefschetz operator of a Kähler manifold is $L\alpha =\omega \wedge \alpha$, and the adjoint Lefschetz operator is its pointwise metric adjoint $\Lambda =L^{\ast }$. It lowers type by $(1,1)$ and is a real operator because $\omega$ and the metric are real. Write $P=\partial$ and $B=\bar{\partial }$, use $[A,D]=AD-DA$, and use $\{A,D\}=AD+DA$ for the anticommutators of the odd-degree operators. The allowed relation among the Kähler identities and its complex conjugate giveThese identities alone suffice for the calculation. Since $P^2=B^2=0$,Replacing $P$ by $B$ in the same calculation gives $\{B,P^{\ast }\}=0$. Thus both mixed terms vanish when expanding the Hodge Laplacian of $d=P+B$:The anticommutation $PB=-BP$ now proves equality of the two summands. Indeed,which are the same expression. Finally, $(d^c{)}^{\ast }=i(P^{\ast }-B^{\ast })$, so its Laplacian expands asThis proves the Kähler Laplacian identity, with its real-operator version included:

The action of the almost complex structure on forms is $J\alpha (v_1,\dots ,v_k)=\alpha (J{v}_1,\dots ,J{v}_k)$. It multiplies a form of type $(p,q)$ by $i^{p-q}$. Since $\Delta =2{\Delta }_{\bar{\partial }}$ preserves each bidegree, it commutes with this action of $J$. Thus $\Delta (J\alpha )=J(\Delta \alpha )$; the invertibility of $J$ givesThis argument applies to arbitrary sums of types, not only to pure-type forms.

The ddbar lemma for a compact Kähler manifold states that a $d$-closed form of pure type is $\partial \bar{\partial }$-exact if it is $d$-exact, $\partial$-exact, or $\bar{\partial }$-exact. Conversely, a $\partial \bar{\partial }$-exact form is $d$-closed and exact for all three operators. In particular a $d$-exact, $d$-closed $(p,q)$-form equals $\partial \bar{\partial }\beta$ for a form $\beta$ of type $(p-1,q-1)$.

The difference $\gamma =\tilde{\omega }-\omega$ of the two Kähler forms is a real $(1,1)$-form, is $d$-closed, and is $d$-exact by equality of their de Rham cohomology classes. The ddbar lemma therefore gives a smooth complex-valued function $h$ with $\gamma =\partial \bar{\partial }h$. Reality and anticommutation implySet $f=(h-\bar{h})/(2i)$, which is real. It follows directly that $i\partial \bar{\partial }f=\gamma$, giving the global potential for cohomologous Kähler forms

If $f_1,f_2$ are two real potentials, $u=f_1-f_2$ satisfies $\partial \bar{\partial }u=0$, so it is a pluriharmonic function. The allowed relation among the Kähler identities gives ${\bar{\partial }}^{\ast }\bar{\partial }u=-i[\Lambda ,\partial ]\bar{\partial }u=-i\Lambda \partial \bar{\partial }u=0$, because $\Lambda \bar{\partial }u=0$ by degree. The Kähler Laplacian identity then gives $\Delta u=2{\bar{\partial }}^{\ast }\bar{\partial }u=0$. Integration by parts on compact $X$ givesso $u$ is constant on each connected component. Consequently the potential is unique up to one additive constant when $X$ is connected; on a disconnected $X$, one constant per component is the exact conclusion. Without the usual connectedness convention the printed uniqueness assertion needs this qualification: distinct constants on the components of a disjoint union leave the same pair of Kähler forms unchanged.

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.

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.

The Hermitian metric on cotangent vectors is the dual metric, and its exterior-power metric is defined on decomposable vectors by the determinant of their pairwise inner products. Tensor this with the metric on $E$. In a unitary coframe and unitary frame of $E$, the forms obtained by wedging distinct holomorphic and antiholomorphic coframe elements and tensoring with a frame vector form an orthonormal basis. Let $dV={\omega }^n/n!$, and use inner products linear in the first variable.

The bundle-valued conjugate-linear Hodge star is characterized by contraction of the $E,E^{\ast }$ coefficients inIt is conjugate-linear, maps type $(p,q)$ to type $(n-p,n-q)$, and uses the metric identification of $E$ with its dual. For total degree $k=p+q$, the ordinary star-square rule giveswhere $E^{\ast \ast }$ is identified with $E$. The conjugate-linear convention is essential for the bidegrees specified here.

If $\psi$ has degree $k-2$ and $\eta$ degree $k$, the even degree of the real form $\omega$ givesSince ${\ast }_E\Lambda \eta =L{\ast }_E\eta$ for $\Lambda \eta =(-1{)}^k{\ast }_{E^{\ast }}L{\ast }_E\eta$, it follows that $(L\psi ,\eta )=(\psi ,\Lambda \eta )$ pointwise. Thus $\Lambda$ is the adjoint Lefschetz operator.

For holomorphic $E$, ${\bar{\partial }}_E$ differentiates coefficients in holomorphic local frames and uses the graded Leibniz rule on forms. Its bundle-valued Dolbeault adjoint is the formal differential operatorIt lowers the antiholomorphic degree by one. To verify this on compact $M$, define $⟨\psi ,\eta {⟩}_{L^2}={\int }_M(\psi ,\eta )dV$. For $\psi \in A^{p,q-1}(E)$ and $\eta \in A^{p,q}(E)$, Stokes applied to the scalar $(n,n-1)$-form $\psi \wedge {\ast }_E\eta$ yieldsThe star-square identity converts the right side into $⟨\psi ,{\bar{\partial }}_E^{\ast }\eta {⟩}_{L^2}$. Compactness and absence of boundary remove boundary terms. This establishes the asserted global adjoint relation on smooth sections.