= Solution
We prove the <Bott-Chern Poincaré lemma> on a <polydisc> $B$. Let $p,q\geq1$ 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 constructs
$$
b_j\in\mathcal A^{p+j,q-1-j}(B),\qquad
\bar\partial b_j=\partial b_{j-1}\quad(1\leq j\leq q-1).
$$
Zero 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 <Poincare 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 <Poincare 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 $\partial c_j$, then
$$
\partial(b_{j-1}+\bar\partial c_j)=0.
$$
Apply 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,
$$
\alpha=\bar\partial b_0=\bar\partial\partial c_0
=-\partial\bar\partial c_0.
$$
Thus \b[$\boxed{H^{p,q}_{BC}(B)=0\quad(p,q\geq1)}$].
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 \b[$\boxed{H^{1,1}_{BC}(\mathbb P^1)\ne0}$].
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