The ∂∂̄ lemma for compact Kähler manifolds

The complex-valued exterior bundle has the induced Hermitian metric and volume density from [F1], and $X$ is compact without boundary of real dimension $2n\ge 2$. At a point and for a real covector $\xi$, the principal symbols are ${\sigma }_1(d)(\xi )=i\xi \wedge$ and ${\sigma }_1(d^{\ast })(\xi )=-i{\iota }_{{\xi }^{\mathrm{♯}}}$; the exterior-algebra identity ${\iota }_{{\xi }^{\mathrm{♯}}}(\xi \wedge \eta )+\xi \wedge {\iota }_{{\xi }^{\mathrm{♯}}}\eta =\mathrm{\mid }\xi {\mathrm{\mid }}^2\eta$ gives ${\sigma }_2({\mathrm{\Delta }}_d)(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^{2}I$. Thus ${\mathrm{\Delta }}_d$ is Laplace type by [F4]. The formal-adjoint identity [F6] gives $⟨{\mathrm{\Delta }}_{d}u,u⟩=\mathrm{\parallel }{d}^{\ast }u{\mathrm{\parallel }}^2+\mathrm{\parallel }du{\mathrm{\parallel }}^2$ and $⟨{\mathrm{\Delta }}_{d}u,v⟩=⟨u,{\mathrm{\Delta }}_{d}v⟩$, so ${\mathrm{\Delta }}_d$ is nonnegative and formally symmetric. By [F5] its selfadjoint realization has domain exactly $H^2$; all hypotheses of the Green theorem [F3] are now met, so the harmonic projection $\mathrm{\Pi }$ and smooth-preserving Green operator $G$ exist with ${\mathrm{\Delta }}_{d}G=I-\mathrm{\Pi }$.

Write $h=\mathrm{\Pi }\alpha$. It is harmonic, so step 1.1 gives $dh=d^{\ast }h=0$. From [F3], ${\mathrm{\Delta }}_{d}G\alpha =\alpha -h$. The identity $d{\mathrm{\Delta }}_d={\mathrm{\Delta }}_{d}d$, obtained by expanding ${\mathrm{\Delta }}_d=d{d}^{\ast }+d^{\ast }d$ and using $d^2=0$, gives ${\mathrm{\Delta }}_d(dG\alpha )=d(\alpha -h)=0$. Thus $dG\alpha$ is harmonic and exact; by the uniqueness in [F2] of the harmonic representative of the zero de Rham class, $dG\alpha =0$. Expanding ${\mathrm{\Delta }}_{d}G\alpha$ now shows $\alpha -h=d{d}^{\ast }G\alpha$, so $h$ is a harmonic representative of $[\alpha ]=0$ and [F2] again gives $h=0$. Therefore ${\mathrm{\Delta }}_{d}G\alpha =\alpha$.

Since $\alpha$ is pure type and $d\alpha =0$, the distinct bidegrees in $d\alpha =\mathrm{\partial }\alpha +\bar{\mathrm{\partial }}\alpha$ give $\mathrm{\partial }\alpha =\bar{\mathrm{\partial }}\alpha =0$. The commutation in [F7] and step 2.1 give ${\mathrm{\Delta }}_d(\mathrm{\partial }G\alpha )=\mathrm{\partial }\alpha =0$ and ${\mathrm{\Delta }}_d(\bar{\mathrm{\partial }}G\alpha )=\bar{\mathrm{\partial }}\alpha =0$. By ${\mathrm{\Delta }}_d=2{\mathrm{\Delta }}_{\mathrm{\partial }}=2{\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}$, each is harmonic for its own Laplacian. Since $\mathrm{\partial }G\alpha$ and $\bar{\mathrm{\partial }}G\alpha$ are respectively $\mathrm{\partial }$- and $\bar{\mathrm{\partial }}$-exact, the adjoint energy identities [F6] give $\mathrm{\parallel }\mathrm{\partial }G\alpha {\mathrm{\parallel }}^2=⟨G\alpha ,{\mathrm{\partial }}^{\ast }\mathrm{\partial }G\alpha ⟩=0$ and $\mathrm{\parallel }\bar{\mathrm{\partial }}G\alpha {\mathrm{\parallel }}^2=⟨G\alpha ,{\bar{\mathrm{\partial }}}^{\ast }\bar{\mathrm{\partial }}G\alpha ⟩=0$. Therefore $\mathrm{\partial }G\alpha =\bar{\mathrm{\partial }}G\alpha =0$.

By [F7], ${\mathrm{\Delta }}_d$ preserves each bidegree. Each component of $G\alpha$ outside bidegree $(p,q)$ is therefore ${\mathrm{\Delta }}_d$-harmonic by step 2.1; it is also orthogonal to the harmonic space by [F3], so every such component is zero. Thus $G\alpha$ has type $(p,q)$ and $\mathrm{\Lambda }G\alpha$ has type $(p-1,q-1)$.

From $\mathrm{\partial }G\alpha =0$ and ${\mathrm{\Delta }}_d=2{\mathrm{\Delta }}_{\mathrm{\partial }}$ we obtain $\alpha ={\mathrm{\Delta }}_{d}G\alpha =2\mathrm{\partial }{\mathrm{\partial }}^{\ast }G\alpha$. The commutator in [F7] gives ${\mathrm{\partial }}^{\ast }=i[\mathrm{\Lambda },\bar{\mathrm{\partial }}]$; since $\bar{\mathrm{\partial }}G\alpha =0$, this yields ${\mathrm{\partial }}^{\ast }G\alpha =-i\bar{\mathrm{\partial }}\mathrm{\Lambda }G\alpha$. Therefore $\alpha =-2i\mathrm{\partial }\bar{\mathrm{\partial }}(\mathrm{\Lambda }G\alpha )$. Taking $\beta =-2i\mathrm{\Lambda }G\alpha$ proves the claim. If $p=0$ or $q=0$, then $\mathrm{\Lambda }G\alpha =0$, so the same identity gives $\alpha =0$ and $\beta =0$ in the zero out-of-range form space.