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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 $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

A pure-type $d$-closed form is both $\partial$-closed and $\bar{\partial }$-closed, because these two derivatives have different types. Let $k=p+q$. The smooth Poincaré lemma on the polydisc gives $\eta \in {\mathcal{A}}_{\mathbb{C}}^{k-1}$ with $d\eta =\alpha$. We will modify $\eta$ by exact terms without changing $d\eta$.

Write $\eta ={\sum }_r{\eta }^{r,k-1-r}$. Starting with the smallest holomorphic degree, eliminate every component with $r<p-1$. Once the lower components are zero, the $(r,k-r)$ component of $d\eta$ says $\bar{\partial }{\eta }^{r,k-1-r}=0$, since $r<p$. Its antiholomorphic degree is positive, so the Dolbeault-Poincaré lemma supplies ${\nu }^{r,k-2-r}$ with $\bar{\partial }\nu ={\eta }^{r,k-1-r}$. Replace $\eta$ by $\eta -d\nu$: this removes the component at $r$ and changes only the next holomorphic degree.

Next, work downwards from the largest holomorphic degree and eliminate the components with $r>p$. Once the higher components are zero, the corresponding component of $d\eta$ gives $\partial {\eta }^{r,k-1-r}=0$. By part (b), write this as $\partial {\nu }^{r-1,k-1-r}$ and again subtract $d\nu$. This affects only the next lower holomorphic degree. The two finite procedures leaveThe components of $d\eta =\alpha$ outside type $(p,q)$ now give $\bar{\partial }u=0$ and $\partial v=0$, while its $(p,q)$ component gives $\alpha =\partial u+\bar{\partial }v$.

Since $q\ge 1$, the Dolbeault-Poincaré lemma gives $u=\bar{\partial }a$ with $a$ of type $(p-1,q-1)$. Since $p\ge 1$, part (b) gives $v=\partial b$ of the same type. The anticommutation identity therefore yieldsThis proves the Bott-Chern Poincaré lemma using precisely the permitted primitives: