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On a polydisc, $H_{BC}^{p,q}=0$ for $p,q>0$. A Poincaré 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.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincaré lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincaré lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

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: