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If $M$ has complex dimension $n$, its complexified cotangent bundle splits into the $(1,0)$ and $(0,1)$ cotangent bundles. The space of global smooth differential forms of type (p, q) isIn holomorphic coordinates, its elements have the form $\alpha ={\sum }_{I,J}{a}_{I,J}d{z}^I\wedge d{\bar{z}}^J$, where the coefficients are complex-valued smooth functions, $|I|=p$, and $|J|=q$.

The exterior derivative decomposes as $d=\partial +\bar{\partial }$. The conjugate Dolbeault operator $\partial$ and Dolbeault operator $\bar{\partial }$ are its type components:They have bidegrees $(1,0)$ and $(0,1)$ respectively. Keeping the displayed wedge order avoids hiding a sign; moving $d{\bar{z}}_k$ past $d{z}^I$ introduces $(-1{)}^p$. Decomposing $d^2=0$ by type gives ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$.

Complex conjugation interchanges the Dolbeault operator and conjugate Dolbeault operator. Thus $\partial \alpha =0$ implies $\bar{\partial }\bar{\alpha }=0$, and $\bar{\alpha }$ has type $(q,p)$. Its antiholomorphic degree is $p\ge 1$, so the allowed Dolbeault-Poincaré lemma on the polydisc gives a form $\gamma \in {\mathcal{A}}^{q,p-1}$ satisfying $\bar{\partial }\gamma =\bar{\alpha }$.

Conjugate this identity and put $\beta =\bar{\gamma }$. This proves the conjugate Dolbeault-Poincaré lemma: