{c}

On a <polydisc>, $H_{BC}^{p,q}=0$ for $p,q>0$. A <Poincare lemma> primitive for a closed pure-type form can be modified by exact terms until only bidegrees $(p-1,q)$ and $(p,q-1)$ remain. The <Dolbeault-Poincaré lemma> and <conjugate Dolbeault-Poincaré lemma> then make the original form $\partial\bar\partial$-exact.

Bott-Chern Poincaré lemma

<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 $p\geq1$, 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>:
$$
\boxed{\beta\in\mathcal A^{p-1,q}(\mathcal P^n),\qquad\partial\beta=\alpha.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 118 / 1 / b / Solution

On a <polydisc>, a $\partial$-closed <differential form of type (p, q)> with $p>0$ is $\partial$-exact. Apply the <Dolbeault-Poincaré lemma> to its complex conjugate, whose antiholomorphic degree is $p$.

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 Incoming links: Conjugate Dolbeault-Poincaré lemma