= 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)> is
$$
\boxed{\mathcal A^{p,q}(M)=\Gamma\left(M,\bigwedge^p(T^{1,0}M)^*\otimes\bigwedge^q(T^{0,1}M)^*\right).}
$$
Here $\Gamma$ 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 expression
$$
\alpha=\sum_{|I|=p,\,|J|=q}a_{IJ}(z,\bar z)\,dz_I\wedge d\bar z_J,
$$
where 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,
$$
\partial\alpha=\sum_{I,J,k}\frac{\partial a_{IJ}}{\partial z_k}\,dz_k\wedge dz_I\wedge d\bar z_J,
\qquad
\bar\partial\alpha=\sum_{I,J,k}\frac{\partial a_{IJ}}{\partial\bar z_k}\,d\bar z_k\wedge dz_I\wedge d\bar z_J.
$$
Keeping the differentiating factor at the front fixes the signs. These definitions are intrinsic because holomorphic coordinate changes preserve type. \b[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 $N$,
$$
f^*dw_a=\sum_j\frac{\partial f_a}{\partial z_j}dz_j,
\qquad f^*d\bar w_a=\sum_j\overline{\frac{\partial f_a}{\partial z_j}}d\bar z_j.
$$
The <pullback of a differential form> commutes with the <wedge product of differential forms>, so it preserves the counts of these factors. Consequently
$$
\boxed{f^*:\mathcal A^{p,q}(N)\longrightarrow\mathcal A^{p,q}(M).}
$$
It also commutes separately with the <Dolbeault operator> and the <conjugate Dolbeault operator>, because it commutes with $d$ and preserves type.
Back to article page
