Hodge theory: Lecture 14

We saw the last time that the following lets us show that the Dolbeault complex is exact everywhere, which lets us construct a resolution for the sheaf of $p$-forms (on a complex manifold).

Lemma ($\bar{\partial}$-lemma). Let $U \subset X$ be a nonempty open subset. Fix $p \geq 0.$ Given any $q \geq 1$ and any smooth $(p, q)$-form $\omega$ such that $\bar{\partial}\omega = 0$ on $U.$ Then we may find a local section $\beta \in \Gamma(V, \mathscr{A}^{p, q-1})$ such that $\bar{\partial} \beta = \omega$ on a nonempty open subset $V \subset U.$

Proof. We may work locally, so denote by $z_{1}, \dots, z_{n}$ a complex chart.

Step 1. Reduction to the case $p = 0.$

We may write $$\omega = \sum_{|I|=p, |J|=q}f_{I,J} dz_{I} \wedge d\bar{z_{J}}.$$ We have $$0 = \bar{\partial}\omega = \sum_{|I|=p, |J|=q}\frac{\partial f_{I,J}}{\partial \bar{z_{j}}} d\bar{z_{j}} \wedge dz_{I} \wedge d\bar{z_{J}}.$$ This implies that $\bar{\partial} \omega_{I} = 0$ for any $I,$ where $\omega_{I} := \sum_{|J|=q} f_{I,J} d\bar{z_{J}},$ which is a $(0, q)$-form.

Hence, if we have proven the case $p = 0,$ we can apply it for $\omega_{I}$ to choose a $(0, q-1)$-form $\beta_{I}$ such that $\bar{\partial} \beta_{I} = \omega_{I}.$ Thus, if we write $$\beta := \sum_{|I| = p} (-1)^{p} dz_{I} \wedge \beta_{I},$$ then $$\bar{\partial}\beta = \omega,$$ as desired.

Step 2. The case $p = 0.$

Since $p = 0,$ we have $$\omega = \sum_{|J|=q}f_{J} d\bar{z_{J}},$$ which is a $(0,q)$-form. We may assume that $$\omega = f d\bar{z_{j_{1}}} \wedge \cdots \wedge d\bar{z_{j_{q}}},$$ with $j_{1} < \cdots < j_{q}.$ Note that $\bar{\partial}\omega = 0$ precisely means that $\partial f / \partial \bar{z_{j}} = 0$ for all $j \in \{1, 2, \dots, n\} \setminus \{j_{1}, \dots, j_{q}\}.$ In other words, we are given that $f$ is holomorphic with respect to $z_{j}$ for all such $j.$

By single variable $\bar{\partial}$-lemma (in this previous posting--see "Observation"), by possible shrinking $U$ (but still nonempty open in $X$), we may find a smooth $g$ that is holomorphic with respect to $z_{j}$ for all such $j \in \{1, 2, \dots, n\} \setminus \{j_{1}, \dots, j_{q}\}$ such that $\partial g / \partial \bar{z_{j_{1}}} = f.$ This implies that $$\bar{\partial} (g d\bar{z_{j_{2}}} \wedge \cdots \wedge d\bar{z_{j_{q}}}) = \omega.$$ This finishes the proof. $\Box$

Remark. We may repeat the whole argument above when $M$ is a smooth (real) manifold to show that if $\omega$ is a closed $p$-form with $p \geq 1,$ then $\omega$ is locally exact. This is called Poincaré's lemma. This lets us have the following resolution called the de Rham complex: $$0 \rightarrow \underline{\mathbb{R}} \rightarrow \mathscr{A}^{0}_{M} \overset{d}{\longrightarrow} \mathscr{A}^{1}_{M} \overset{d}{\longrightarrow} \cdots \overset{d}{\longrightarrow} \mathscr{A}^{n}_{M} \rightarrow 0,$$ where $n = \dim_{\mathbb{R}}(M).$

Next time, we shall see that for any $p \geq 0,$ we have $$H^{i}(M, \mathscr{A}^{p}_{M}) = 0,$$ for $i \geq 1.$ In particular, this implies that $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})).$$ This is particularly interesting because for any $i \geq 0,$ we have $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}_{\mathrm{sing}}(M, \mathbb{R}),$$ the singular cohomology of $M$ with $\mathbb{R}$-coefficients. (We will not prove the last isomorphism, but the details are written here.) The isomorphism $$H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})) \simeq H^{i}_{\mathrm{sing}}(M, \mathbb{R})$$ is often referred as de Rham's theorem, and the left-hand side of this isomorphism is often denoted as $H^{i}_{\mathrm{dR}}(M),$ called $i$-th de Rham cohomology.