 Solution

The complex structure splits the complexified cotangent bundle into its $(1,0)$ and $(0,1)$ parts. Thus the space of smooth differential forms of type (p, q) isHere $\Gamma$ denotes smooth sections of a vector bundle; exterior multiplication identifies the displayed bundle with a subbundle of the complexified exterior algebra. In holomorphic coordinates, such a differential form of type (p, q) has the expressionwhere the coefficients are smooth and both multi-indices are increasing. Negative bidegrees, or bidegrees exceeding the complex dimension, give the zero space.

On a complex manifold, the exterior derivative has just two type components, $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator and the Dolbeault operator are, respectively,Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. The two operators are the type projections of the exterior derivative, raising $p$ and $q$, respectively.

For a holomorphic map $f:M\to N$, its differential is a complex-linear map, so the pullback of a differential form sends $(1,0)$ covectors to $(1,0)$ covectors and $(0,1)$ covectors to $(0,1)$ covectors. In coordinates $w$ on $N$,The pullback of a differential form commutes with the wedge product of differential forms, so it preserves the counts of these factors. ConsequentlyIt also commutes separately with the Dolbeault operator and the conjugate Dolbeault operator, because it commutes with $d$ and preserves type.