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

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:

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:

The natural map between Bott-Chern cohomology and Dolbeault cohomology isIt is well-defined: a $d$-closed pure-type form is $\bar{\partial }$-closed, and $\partial \bar{\partial }\beta =-\bar{\partial }(\partial \beta )$ is $\bar{\partial }$-exact.

For surjectivity, choose the unique harmonic differential form $h$ representing a Dolbeault cohomology class using Dolbeault Hodge decomposition. The Kähler Laplacian identity ${\Delta }_d=2{\Delta }_{\bar{\partial }}$ makes $h$ a harmonic differential form for $d$, hence $d$-closed. Sending the Dolbeault class to $[h{]}_{BC}$ consequently gives a right inverse.

For injectivity, the ddbar lemma says directly that a $d$-closed, $\bar{\partial }$-exact pure-type form is $\partial \bar{\partial }$-exact. One can also exhibit its primitive. Let $G_d$ be the Green operator of the Hodge Laplacian, and put $G_{\bar{\partial }}=2G_d$. If $\alpha$ is $d$-closed and $\bar{\partial }$-exact, its harmonic projection vanishes and the Kähler identities giveHere $\partial G_{\bar{\partial }}\alpha =\bar{\partial }{G}_{\bar{\partial }}\alpha =0$ because the Green operator commutes with these differentials, and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. Thus $\alpha$ represents zero in Bott-Chern cohomology. We have constructed the canonical isomorphism and its harmonic inverse: