Use both Kähler identities from the preceding parts, so that and . For odd operators , write . Direct expansion givesSince , we obtainTogether with part (b), this proves the full Kähler Laplacian identity
Now take the exact form . Being -exact, it is orthogonal to the harmonic space, so , and it is also -closed. Choose for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives . ThereforeSuppose also , and put , . The mixed identities and giveThus is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also -harmonic and hence . But it is -exact, soConsequently : the form is -closed.
For every harmonic form of the same bidegree as , . Harmonic spaces for the two Dolbeault Laplacians coincide, so has no -harmonic component. Apply the version of the Dolbeault Hodge decomposition. Since is -closed, it has no -coexact component either. Thus , with . We conclude the requested ddbar lemma:For example is a choice, since the same Green operator serves . If or , the relevant negative-bidegree space is zero and the argument forces . Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.
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