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For a compact Kähler manifold, the natural map $H_{BC}^{p,q}\to H_{\bar{\partial }}^{p,q}$ is an isomorphism. The Kähler Laplacian identity makes the harmonic representative of every Dolbeault cohomology class $d$-closed, giving surjectivity, and the ddbar lemma gives injectivity.

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.

A holomorphic $(p,0)$ form is annihilated by $\bar{\partial }$ and, by bidegree, by ${\bar{\partial }}^{\ast }$. The Kähler Laplacian identity makes it harmonic for $d$, hence $d$-closed on a compact Kähler manifold.

The Kähler Laplacian identity makes the space of complex harmonic differential forms of degree $r$ the direct sum of its $(p,q)$ parts with $p+q=r$. Conjugation identifies the $(p,q)$ and $(q,p)$ harmonic spaces. If $r$ is odd these are distinct and pair off, so the Betti number in degree $r$ is even. This is a topological obstruction to existence of a Kähler metric, exhibited by a Hopf surface with first Betti number one.

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.

Let ${\mathcal{A}}^{p,q}$ denote the sheaf of differential forms of type (p, q) printed in the PDF. Take the pointwise Hermitian inner product to be linear in its first argument, and use $\mathrm{vol}={\omega }^n/n!$. The bidegrees in the question specify the conjugate-linear Hodge star: it is uniquely characterized byThus $\ast (c\beta )=\bar{c}\ast \beta$, and $\ast$ maps bidegree $(p,q)$ to $(n-p,n-q)$. If $S$ is the complex-linear extension of the real Hodge star operator, then $\ast =S\circ \text{complex conjugation}$. The operator $S$ is real and $S^2=(-1{)}^{k(2n-k)}$ on degree $k=p+q$, soUsing the complex-linear star instead would give a different bidegree, $(n-q,n-p)$; keeping the convention explicit prevents that ambiguity.

The Hermitian Lefschetz operator is exterior multiplication by $\omega$. Define $\Lambda$ pointwise byIt lowers bidegree by $(1,1)$. In a unitary real coframe with $\omega ={\sum }_j{e}^j\wedge f^j$, it is $\Lambda ={\sum }_j{\iota }_{f_j}{\iota }_{e_j}$, the corresponding sum of interior product of a differential form operators. This gives its existence and identifies it as a smooth operator. On a compact manifold, integrate the pointwise equality against $\mathrm{vol}$ to obtainNo integration by parts is needed for this order-zero operator. Thus $\Lambda$ is both the pointwise and the global formal adjoint.

Write $[A,B]=AB-BA$, and let $k$ denote the degree of the input form. The given Lefschetz commutator is $[L,\Lambda ]=(k-n)I$. For $r=1$ the required formula is exactly this identity. If it holds for $r$, the commutator derivation identity givesThe second commutator acts on degree $k+2r$, so the coefficient on the right isThis proves the commutator formula for powers of the Lefschetz operator

For injectivity of powers of the Lefschetz operator, $r=0$ is immediate. Suppose $r\ge 1$ and $r+k\le n$. If $k=0$ or $1$, then $\Lambda \psi =0$, and $L^r\psi =0$ impliesThe scalar is nonzero, since $k+r\le n$ makes $k-n+r-1\le -1$. Repeating with the smaller power shows $\psi =0$. For $k\ge 2$, induct on $r+k$, treating all $r=0$ cases as already established. The same commutator calculation yields, with $c=r(k-n+r-1)\ne 0$,The induction hypothesis applies to $L^{r-1}$ on degree $k$, since $(r-1)+k<r+k$ and $(r-1)+k\le n$. Hence $\psi =L\alpha$ with $\alpha =c^{-1}\Lambda \psi$ of degree $k-2$. Moreover $L^{r+1}\alpha =L^r\psi =0$. Since $(r+1)+(k-2)=r+k-1$, induction also makes $L^{r+1}$ injective on that degree, so $\alpha =0$ and $\psi =0$. All operators preserve restriction to open sets; the argument applies on every open set. Therefore $L^r$ is an injective morphism of sheaves whenever $r+k\le n$.

On a Kähler manifold, the Kähler identities, with the positive form convention used above, areTaking formal adjoints conjugates the scalar and reverses the order in the commutator. Since ${\Lambda }^{\ast }=L$, this givesTo derive the Kähler Laplacian identity, put $P=\partial$ and $B=\bar{\partial }$, so $P^2=B^2=0$ and $PB+BP=0$. Write $\{A,B\}=AB+BA$ for an anticommutator. The Kähler identities say $B^{\ast }=-i[\Lambda ,P]$ and $P^{\ast }=i[\Lambda ,B]$. Thereforebecause expansion leaves only terms containing $P^2$ or $B^2$. Furthermore,Expanding these expressions and using $PB=-BP$ makes them equal. Since $d=P+B$ and $d^{\ast }=P^{\ast }+B^{\ast }$, the mixed anticommutators already vanish, and consequentlyClosedness of the Kähler form implies $[L,P]=[L,B]=0$. Using the adjoint identity above,The equality of the three Laplacians therefore proves each Laplacian commutes with $L$.

For a compact Kähler manifold, the Dolbeault Hodge decomposition is the orthogonal decompositionHere harmonicity for the Dolbeault Laplacian and for the Hodge Laplacian agrees by the preceding identity. If $\bar{\partial }\alpha =0$ and $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$, then $\bar{\partial }{\bar{\partial }}^{\ast }v=0$. Taking its inner product with $v$ gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Thus $\alpha$ is cohomologous to $h$. Conversely, a harmonic form is $\bar{\partial }$-closed, and a harmonic exact form $h=\bar{\partial }u$ satisfies $\Vert h{\Vert }^2=({\bar{\partial }}^{\ast }h,u)=0$. This proves existence and uniqueness of the harmonic representative and henceThese spaces are finite dimensional by the ellipticity of the Dolbeault Laplacian on the compact manifold.

Complex conjugation commutes with the real operator ${\Delta }_d$ and interchanges the bidegrees $(p,q)$ and $(q,p)$. It is therefore a conjugate-linear bijection of the harmonic spaces, proving Hodge symmetry. The real Hodge star operator commutes with ${\Delta }_d$, as does complex conjugation; hence their composition, our conjugate-linear Hodge star, sends harmonic $(p,q)$-forms bijectively to harmonic $(n-p,n-q)$-forms. Its square is the nonzero scalar established above. This proves Hodge duality, and gives

For the final hard Lefschetz isomorphism on Dolbeault cohomology, take $k=p+q\le n$ and $r=n-k$. This restriction is necessary to make the displayed power nonnegative. The Hermitian Lefschetz operator raises bidegree by $(1,1)$, and the Lefschetz operator preserves harmonic forms; thereforeis well defined. It is injective by injectivity of powers of the Lefschetz operator, since $r+k=n$. Hodge symmetry and Hodge duality give $h^{n-q,n-p}=h^{q,p}=h^{p,q}$, so it is a bijection between finite-dimensional spaces of equal dimension. Because $[L,\bar{\partial }]=0$, the map on harmonic representatives agrees with exterior multiplication by $[\omega {]}^{n-k}$ on Dolbeault cohomology. We concludeFor $p+q>n$, the corresponding valid statement is the inverse of the positive-power isomorphism from bidegree $(n-q,n-p)$ to $(p,q)$; a negative exterior-multiplication power is not defined.