Hodge theory: Lecture 13

Dolbeault complex. Let $M$ be a complex manifold of dimension $n.$ Then the canonical almost complex structure on $TM$ gives us $$(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M,$$ and $T^{1,0}M$ turned out to be a holomorphic vector bundle, the holomorphic tangent bundle of $M$ to be more specific. The decomposition is functorial in the sense that for any holomorphic map $\phi : M \rightarrow M'$ between two complex manifolds, its push forward map $(TM)_{\mathbb{C}} \rightarrow \phi^{*}(TM')_{\mathbb{C}}$ restricts to $T^{1,0}M \rightarrow \phi^{*}T^{1,0}M'.$

Dually, we have $$(T^{\vee}M)_{\mathbb{C}} = A^{1,0}_{M} \oplus A^{0,1}_{M},$$ but we also note that $$(T^{\vee}M)_{\mathbb{C}} \simeq \mathrm{Hom}_{\mathbb{C}}((TM)_{\mathbb{C}}, \mathbb{C}),$$ where at each fiber, we have $(v \otimes 1)^{\vee} \mapsto v^{\vee} \otimes 1.$ Under this isomorphism we have $$A^{1,0}_{M} \simeq \mathrm{Hom}_{\mathbb{C}}(T^{1,0}M, \mathbb{C})$$ and $$A^{0,1}_{M} \simeq \mathrm{Hom}_{\mathbb{C}}(T^{0,1}M, \mathbb{C}).$$ We also have the complexified $k$-th exterior power $$\begin{align*}\left(\bigwedge^{k} T^{\vee}M\right)_{\mathbb{C}} &\simeq \bigwedge^{k} (T^{\vee}M)_{\mathbb{C}} \\ &= \bigwedge^{k} (A^{1,0}_{M} \oplus A^{0,1}_{M}) \\ &= \bigoplus_{p+q = k}\left(\bigwedge^{p} A^{1,0}_{M} \otimes_{\mathbb{R}} \bigwedge^{q} A^{0,1}_{M}\right).\end{align*}$$ We shall write $$A^{p,q}_{M} := \bigwedge^{p} A^{1,0}_{M} \otimes_{\mathbb{R}} \bigwedge^{q} A^{0,1}_{M},$$ and call it $(p,q)$-th exterior power.

Sections of the exterior power $\bigwedge^{k} T^{\vee}M,$ which are called (real) differential $k$-forms on $M$, form a sheaf $\mathscr{A}_{M}^{k},$ and we complexify this sheaf, meaning $$\mathscr{A}_{M, \mathbb{C}}^{k} := \mathscr{A}_{M}^{k} \otimes_{\underline{\mathbb{R}}} \underline{\mathbb{C}},$$ and this matches the sheaf of the complexified $k$-exterior of $M,$ so we have $$\mathscr{A}_{M, \mathbb{C}}^{k} = \bigoplus_{p+q = k}\left(\bigwedge^{p} \mathscr{A}^{1,0}_{M} \otimes_{\mathscr{C}^{\infty}_{M}} \bigwedge^{q} \mathscr{A}^{0,1}_{M}\right),$$ where $\mathscr{A}^{1,0}_{M}$ is the sheaf of sections of $A^{1,0}_{M},$ while $\mathscr{A}^{0,1}_{M}$ is the sheaf of sections of $A^{0,1}_{M}.$ We write $$\mathscr{A}^{p,q}_{M} := \bigwedge^{p} \mathscr{A}^{1,0}_{M} \otimes_{\otimes_{\mathscr{C}^{\infty}_{M}}} \bigwedge^{q} \mathscr{A}^{0,1}_{M}$$ and say the sections of them are $(p,q)$-sections of $M.$ Note that $$\overline{\mathscr{A}_{M}^{p,q}} = \mathscr{A}_{M}^{q,p},$$ where the conjugation happen in the target, namely the (complexified) $k$-th exterior power. Denote by $\Omega^{p}_{M}$ the sheaf holomorphic sections of the holomorphic vector bundle $A^{p,0}.$ We call the sections of $\Omega^{p}_{M}$ holomorphic $p$-forms on $M$ Note that we have $$\Omega^{p}_{M} \hookrightarrow \mathscr{A}^{p,0}_{M},$$ where the right-hand side is the sheaf of smooth (not necessarily holomorphic) sections of the holomorphic vector bundle $A^{p,0}.$ Note that $A^{1,0}$ is identified to be the (complex) dual of $T^{1,0}M,$ which is a holomorphic vector bundle whose local frame is given by $dz_{1}, \dots, dz_{n}$ for each local chart $z_{1}, \dots, z_{n}$ for $M.$ With such a local chart, say on an open subset $U \subset M,$ we note that $\mathscr{A}^{p,q}_{U}$ is a free $\mathscr{C}^{\infty}_{U, \mathbb{C}}$-module with basis $dz_{I} \wedge d\bar{z_{J}}$ with $I = (i_{1} < \cdots < i_{p})$ and $J = (j_{1} < \cdots < j_{q}).$

Consider the de Rham differential $$d : \mathscr{A}^{k}_{M} \rightarrow \mathscr{A}^{k+1}_{M}.$$ That is, on each open subset $U \subset M,$ the map $$d : \mathscr{A}^{k}_{M}(U) \rightarrow \mathscr{A}^{k+1}_{M}(U)$$ is given by the exterior derivative. Considering the complexified version $$\mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}}.$$ For any $p, q \geq 0$ such that $p + q = k,$ this gives $$\mathscr{A}^{p,q} \hookrightarrow \mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}} \twoheadrightarrow \mathscr{A}^{p+1,q},$$ which we call $\partial^{p,q}$ and $$\mathscr{A}^{p,q} \hookrightarrow \mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}} \twoheadrightarrow \mathscr{A}^{p,q+1},$$ which we call $\bar{\partial}^{p,q}.$

Proposition. When $M$ is complex manifold, for $p + q = k,$ we have $d = \partial + \bar{\partial}$ as a map $$\mathscr{A}^{p,q}_{M} \overset{d}{\longrightarrow} \mathscr{A}^{p+1,q}_{M} \oplus \mathscr{A}^{p,q+1}_{M}.$$
Personal remark. We don't really need this because the identity can be checked at the stalks, but one can check that $$(\mathscr{A}^{k} \otimes_{\underline{\mathbb{R}}} \underline{\mathbb{C}})(U) = \mathscr{A}^{k}(U) \otimes_{\mathbb{R}} \mathbb{C},$$ even without sheafification after tensoring.

Proof. We suppress $M$ in the notations for sheaves for simplicity. We can check the identity at an open subset $U \subset M$ with the coordinate functions $$x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n},$$ which induces the coordinate functions $$z_{1}, \dots, z_{n}, \bar{z_{1}}, \dots, \bar{z_{n}}.$$ Fix any $\omega \in \mathscr{A}^{p,q}(U).$ Our goal is to show that $$d\omega = \partial\omega + \bar{\partial}\omega \in \mathscr{A}^{k+1}(U) \otimes_{\mathbb{R}} \mathbb{C}.$$ We may assume that $\omega = f dz_{I} \wedge d\bar{z_{J}}$ with $|I| = p$ and $|J| = 1$ because $\mathscr{A}^{p,q}(U)$ is $\mathbb{C}$-linearly generated by such $k$-forms. We have $$df = \sum_{j=1}^{n} \frac{\partial f}{\partial z_{j}} dz_{j} + \sum_{j=1}^{n} \frac{\partial f}{\bar{\partial z_{j}}} d\bar{z_{j}}$$ because $f$ is holomorphic. By definition, we have $$\partial f = \sum_{j=1}^{n} \frac{\partial f}{\partial z_{j}} dz_{j} \in \mathscr{A}^{1,0}(U)$$ and $$\bar{\partial} f = \sum_{j=1}^{n} \frac{\partial f}{\partial \bar{z_{j}}} d\bar{z_{j}} \in \mathscr{A}^{0,1}(U).$$ Thus, we have $$\begin{align*}d \omega &= df \wedge dz_{I} \wedge d\bar{z_{J}} \\ &= \partial f \wedge dz_{I} \wedge d\bar{z_{J}} + \bar{\partial} f \wedge dz_{I} \wedge d\bar{z_{J}} \\ &= \partial \omega + \bar{\partial} \omega \end{align*}.$$ This finishes the proof. $\Box$

As in the above proof, we shall continue drop $M$ in the notations for sheaves, unless it causes too much confusion.

Corollary. Assuming the previous notations, at every $(p, q)$ level, we have

• $\partial^{2} = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p+2,q};$
• $\bar{\partial}^{2} = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p,q+2};$
• $\partial\bar{\partial} + \bar{\partial}\partial = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p+1,q+1}.$

Proof. We have $$0 = d^{2} = (\partial + \bar{\partial})^{2} = \partial^{2} + (\partial\bar{\partial} + \bar{\partial}\partial) + \bar{\partial}^{2}.$$ Looking at the different degrees in the grading, the result follows. $\Box$

Derivation property for $\partial$ and $\bar{\partial}$. Recall that for $\omega \in \mathscr{A}^{k}(U)$ and $\eta \in \mathscr{A}^{l}(U),$ we have $$d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^{k} \omega \wedge d \eta.$$ Thus, the same formula holds for the complexification. This implies that $$\partial(\omega \wedge \eta) = \partial\omega \wedge \eta + (-1)^{k} \omega \wedge \partial \eta$$ and $$\bar{\partial}(\omega \wedge \eta) = \bar{\partial}\omega \wedge \eta + (-1)^{k} \omega \wedge \bar{\partial} \eta,$$ for any $\omega \in \mathscr{A}^{p+q}(U)$ and $\eta \in \mathscr{A}^{p'+q'}(U)$ such that $p + q = k.$

Dolbeault complex. Given a complex manifold $M$ of dimension $n,$ for every $p \in \mathbb{Z}_{\geq 0},$ we have the following complex of sheaves: $$0 \rightarrow \mathscr{A}^{p,0} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,1} \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,n-1} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,n} \rightarrow 0,$$ taking the global sections, we have the following complex of $\mathbb{C}$-vector spaces: $$0 \rightarrow \Gamma(M, \mathscr{A}^{p,0}) \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,1}) \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,n-1}) \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,n}) \rightarrow 0,$$ which is called the $p$-th Dolbeault complex of $M.$ The $q$-th cohomology of $p$-th Dolbeault complex is called the Dolbeault cohomology group of degree $(p, q).$ For notation, we write $$H^{p,q}(M) := H^{q}(\Gamma(M, \mathscr{A}^{p, \bullet}).$$ We now argue that we can compute the cohomology of the sheaf $\Omega^{p}_{M}$ of holomorphic $p$-forms on $M$ by the $p$-th Dolbeault complex.

Lemma 1. We have $$\ker(\mathscr{A}^{p,0}_{M} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,1}_{M}) = \Omega^{p}_{M},$$ for any $p \geq 0.$

Proof. We may prove this locally on open subsets $U \subset M$ with chart maps. An element of $\mathscr{A}^{p,0}(U)$ looks like $$\omega = \sum_{|I| = p} f_{I} dz_{I}$$ where $f_{I} \in \mathscr{C}^{\infty}(U).$ We have $$\bar{\partial}\omega = \sum_{|I| = p} \sum_{j=1}^{n} \frac{\partial f_{I}}{\partial \bar{z_{j}}} d\bar{z_{j}} \wedge dz_{I}.$$ On $U,$ saying that $\omega$ is holomorphic is equivalent to $f_{I}$ are holomorphic for all $I$ which is precisely when all $\partial f_{I} / (\partial \bar{z_{j}}) = 0$ for all $I$ and $j.$ This is precisely the same as saying that $\bar{\partial} \omega = 0,$ and this finishes the proof. $\Box$

Lemma 2 ($\bar{\partial}$-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 $V \subset U.$

The point of Lemma 1 and Lemma 2 is that we now have a resolution (i.e., everywhere exact): $$0 \rightarrow \Omega^{p}_{M} \rightarrow \mathscr{A}^{p,0}_{M} \rightarrow \cdots \rightarrow \mathscr{A}^{p,n}_{M} \rightarrow 0.$$ What is this good for? We need one more lemma to see this.

Lemma 3. Every $\mathscr{A}^{p,q}_{M}$ is acyclic. Namely, we have $$H^{i}(X, \mathscr{A}^{p,q}) = 0$$ for all $i \geq 1.$

Hence, by a previous posing that says any acyclic resolution of a sheaf computes its cohomology, we have:

Corollary. We have $$H^{p,q}(M) = H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p, \bullet})),$$ where the first equality is just a definition.

We will discuss the proof of Lemma 2 next time, and Lemma 3 later.