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For the Lefschetz operator of a Kähler manifold and its adjoint, the Kähler identities include

In complex dimension $n$, the Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator satisfy $[\Lambda ,L]=(n-k)\mathrm{id}$ on degree $k$. In middle degree, this yields $\Vert L\alpha {\Vert }^2=\Vert \Lambda \alpha {\Vert }^2$.

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.

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.

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.

Write $L\alpha =\omega \wedge \alpha$, the Lefschetz operator of a Kähler manifold. The metric and volume form define the $L^2$ inner product on smooth complex forms, and $\Lambda =L^{\ast }$ is its formal adjoint. Similarly, ${\bar{\partial }}^{\ast }$ is the formal adjoint of the Dolbeault operator, characterized by $⟨\bar{\partial }\beta ,\alpha ⟩=⟨\beta ,{\bar{\partial }}^{\ast }\alpha ⟩$. The Dolbeault Laplacian isOn the compact manifold without boundary, integration by parts givesIf the Laplacian vanishes, both terms are zero. Conversely, if both operators annihilate $\alpha$, the defining formula annihilates it. Thus harmonicity is equivalent to being both $\bar{\partial }$-closed and ${\bar{\partial }}^{\ast }$-closed.

Because $\omega$ is closed and has type $(1,1)$, $\bar{\partial }\omega =0$ and $[L,\bar{\partial }]=0$. The supplied identity from the Kähler identities gives $[L,{\bar{\partial }}^{\ast }]=-i\partial$. With ordinary commutators for the even-degree operator $L$,The last identity follows from $d^2=0$. This also proves that the Lefschetz operator preserves harmonic forms.

For the cohomology map one can work directly with forms: $\bar{\partial }({\omega }^k\wedge \alpha )={\omega }^k\wedge \bar{\partial }\alpha$, since $\omega$ has even degree. It takes closed forms to closed forms and exact forms to exact forms. Therefore the $k$th power of $L$ inducesThe bidegree is $(p+k,q+k)$, including the zero groups outside the dimension range. No isomorphism claim is needed here.

Finally put $\theta =-i\partial f$, so $\bar{\partial }\theta =i\partial \bar{\partial }f={\omega }^{\prime }-\omega$. LetBoth Kähler forms are $\bar{\partial }$-closed, so $\bar{\partial }{T}_k=0$. For a closed representative $\alpha$,This primitive has type $(p+k,q+k-1)$; the degree-one sign produces no additional term because $T_k\wedge \alpha$ is closed. HenceThis is the dependence of Lefschetz maps on the Dolbeault class.

The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

Let $\omega$ be the Kähler form associated with the metric. The Lefschetz operator of a Kähler manifold and adjoint Lefschetz operator areHere the adjoint is taken with respect to the pointwise Hermitian inner product on complexified forms, or equivalently the global $L^2$ inner product. It lowers degree by two and bidegree by $(1,1)$. In terms of the Hodge star operator, $\Lambda ={\ast }^{-1}L\ast$.

Since $d\omega =0$, $dL=Ld$, so $L$ takes closed differential forms to closed differential forms and exact differential forms to exact differential forms. This already gives its action on de Rham cohomology. For both operators together, use the Kähler identities: $L$ commutes with the Hodge Laplacian, and taking adjoints shows that $\Lambda$ does too. Thus both preserve harmonic differential forms.

When $M$ is compact, the Hodge decomposition theorem identifies $H^k(M,\mathbb{C})$ with its space of harmonic $k$-forms. If $h_{\alpha }$ is the unique harmonic differential form representing $[\alpha ]$, defineThe resulting forms are harmonic and therefore closed, and uniqueness of $h_{\alpha }$ makes these definitions independent of the initial representative. The cohomological definition of $\Lambda$ uses harmonic differential forms as representatives: $\Lambda$ need not take an arbitrary closed form to a closed form.

Represent the middle-degree class by its unique harmonic differential form $h$. The Lefschetz commutator on degree $k$ is $[\Lambda ,L]=(n-k)\mathrm{id}$, so at $k=n$ the operators commute. Since the adjoint Lefschetz operator is the adjoint of the Lefschetz operator of a Kähler manifold,Both $Lh$ and $\Lambda h$ are harmonic. A harmonic form represents zero in de Rham cohomology exactly when the form itself is zero. Consequently $L[\alpha ]=0$ if and only if $Lh=0$, which by the norm identity is equivalent to $\Lambda h=0$, hence to $\Lambda [\alpha ]=0$. ThereforeThis identifies the two middle-degree descriptions of a primitive differential form on a Kähler manifold at the cohomology level.

For the Lefschetz operator of a Kähler manifold and its adjoint $\Lambda$, the Kähler identities includeThey make the mixed anticommutators in the expansion of ${\Delta }_d$ vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Expanding $d=\partial +\bar{\partial }$ therefore proves the Kähler Laplacian identityThus ${\Delta }_d\alpha =0$ exactly when ${\Delta }_{\bar{\partial }}\alpha =0$.

The Kähler manifold structure gives a Riemannian metric, its volume form, the complex orientation, and a Hermitian inner product on complex differential forms. The complex Hodge star operator is the complex-linear map characterized byOn $r$-forms in real dimension $m=2n$, the codifferential isequivalently the formal adjoint of $d$ for the $L^2$ inner product. Similarly ${\bar{\partial }}^{\ast }$ is the formal adjoint of $\bar{\partial }$. Define the Hodge Laplacian and Dolbeault Laplacian by

Expanding $d=\partial +\bar{\partial }$, the Kähler identities make the mixed anticommutators vanish and imply ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Hence the Kähler Laplacian identity is

Let $L\gamma =\omega \wedge \gamma$ be the Lefschetz operator of a Kähler manifold. The Kähler identities also implyThus, if ${\Delta }_{\bar{\partial }}\alpha =0$, thenThis is the fact that the Lefschetz operator preserves harmonic forms.

The Dolbeault Hodge decomposition on a compact Hermitian manifold states thatan orthogonal direct sum, where ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$.

Suppose $\eta =\bar{\partial }\gamma$ has type $(p,q)$. Apply this decomposition to $\gamma \in {\Omega }^{p,q-1}$. The harmonic and $\bar{\partial }$-exact pieces disappear after applying $\bar{\partial }$, so for some $\beta \in {\Omega }^{p,q}$,Put $\theta ={\bar{\partial }}^{\ast }\beta$. If also $\partial \eta =0$, thenThe Kähler anticommutation identity ${\bar{\partial }}^{\ast }\partial +\partial {\bar{\partial }}^{\ast }=0$ and $({\bar{\partial }}^{\ast }{)}^2=0$ giveTherefore $\partial \theta$ is ${\Delta }_{\bar{\partial }}$-harmonic. By ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$ it is also $\partial$-harmonic, but it is $\partial$-exact; orthogonality of harmonic and exact forms forcesThis proves both requested claims: $\partial {\bar{\partial }}^{\ast }\beta =0$ is harmonic, and ${\bar{\partial }}^{\ast }\beta =\theta$ is $\partial$-closed.

Finally, $\theta \in \mathrm{im}{\bar{\partial }}^{\ast }$ is orthogonal to $\ker \bar{\partial }$, and hence to every $\bar{\partial }$-harmonic form. Since the $\partial$- and $\bar{\partial }$-harmonic spaces agree on a compact Kähler manifold, the $\partial$-closed form $\theta$ has zero harmonic component in its $\partial$-Hodge decomposition. It follows that $\theta =\partial \varphi$ for some $\varphi \in {\Omega }^{p-1,q-1}(X)$. Hencewhich is the ddbar lemma in this case.

A Riemannian metric $g$ on a complex manifold is a Kähler metric when its complex-linear extension is a Hermitian form on each tangent space and its fundamental two-formis closed. Equivalently, $\omega$ is a positive real closed $(1,1)$-form, making $X$ a Kähler manifold.

The Lefschetz operator of a Kähler manifold and its adjoint areBecause $d\omega =0$ and $\omega$ has type $(1,1)$, both $\partial \omega$ and $\bar{\partial }\omega$ vanish. The graded Leibniz rule therefore gives $[L,\partial ]=[L,\bar{\partial }]=0$.

Writing formal adjoints with stars, define the three Laplacians byThe supplied Kähler identities identity $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$ gives, by complex conjugation and taking adjoints,Expanding these commutators and using $\partial \bar{\partial }+\bar{\partial }\partial =0$ shows that the mixed terms in $\Delta$ vanish and that ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. Since $d=\partial +\bar{\partial }$, it follows that

The adjoint of $[L,\partial ]=0$ gives $[\Lambda ,{\partial }^{\ast }]=0$. Hencewhere the last equality is the adjoint of $\partial \bar{\partial }+\bar{\partial }\partial =0$. Thus $\Lambda$, and therefore every power ${\Lambda }^k$, commutes with ${\Delta }_{\partial }$. It follows that ${\Lambda }^k$ sends every $\partial$-harmonic $(p,q)$-form to a $\partial$-harmonic $(p-k,q-k)$-form whenever $0\le k\le \min (p,q)$.