8.3 Complex differential forms

On a complex manifold, we define complex differential forms by which we will discuss such topological properties as cohomology groups.

8.3.1 Complexification of real differential forms

Let $M$ be a differentiable manifold with ${\dim }_{\mathbb{R}}M=m$ . Take two $q$ -forms $\omega ,\eta \in {\Omega }_p^q(M)$ at $p$ and define a complex $q$ -form $\zeta =\omega +\mathrm{i}\eta$ . We denote the vector space of complex $q$ -forms at $p$ by ${\Omega }_p^q(M{)}^{\mathbb{C}}$ . Clearly ${\Omega }_p^q(M)\subset {\Omega }_p^q(M{)}^{\mathbb{C}}$ . The conjugate of $\zeta$ is $\overline{\zeta }=\omega -\mathrm{i}\eta$ . A complex $q$ -form $\zeta$ is real if $\zeta =\overline{\zeta }$ .

Exercise 8.4

Let $\omega \in {\Omega }_p^q(M{)}^{\mathbb{C}}$ . Show that

$$\overline{\omega }\left(V_1,\dots ,V_q\right)=\overline{\omega \left({\bar{V}}_1,\dots ,{\bar{V}}_q\right)}{V}_i\in T_p{M}^{\mathbb{C}}.\text{\hspace{1em}(8.30)}$$

Show also that $\overline{\omega +\eta }=\overline{\omega }+\overline{\eta },\overline{\lambda \omega }=\overline{\lambda }\overline{\omega }$ and $\overline{\overline{\omega }}=\omega$ , where $\omega ,\eta \in {\Omega }_p^q(M{)}^{\mathbb{C}}$ and $\lambda \in \mathbb{C}$

A complex $q$ -form $\alpha$ defined on a differentiable manifold $M$ is a smooth assignment of an element of ${\Omega }_p^q(M{)}^{\mathbb{C}}$ . The set of complex $q$ -forms is denoted by ${\Omega }^q(M{)}^{\mathbb{C}}$ . A complex $q$ -form $\zeta$ is uniquely decomposed as $\zeta =\omega +\mathrm{i}\eta$ , where $\omega ,\eta \in {\Omega }^q(M)$ .

The exterior product of $\zeta =\omega +\mathrm{i}\eta$ and $\xi =\phi +\mathrm{i}\psi$ is defined by

$$\begin{array}{rl}\zeta \wedge \xi & =(\omega +\mathrm{i}\eta )\wedge (\phi +\mathrm{i}\psi )\\ & =(\omega \wedge \phi -\eta \wedge \psi )+\mathrm{i}(\omega \wedge \psi +\eta \wedge \phi ).\end{array}\text{\hspace{1em}(8.31)}$$

The exterior derivative $\mathrm{d}$ acts on $\zeta =\omega +\mathrm{i}\eta$ as

$$\mathrm{d}\zeta =\mathrm{d}\omega +\mathrm{id}\eta .\text{\hspace{1em}(8.32)}$$

d is a real operator: $\overline{\mathrm{d}\zeta }=\mathrm{d}\omega -\mathrm{id}\eta =\mathrm{d}\overline{\zeta }$

Exercise 8.5

Let $\omega \in {\Omega }^q(M{)}^{\mathbb{C}}$ and $\xi \in {\Omega }^r(M{)}^{\mathbb{C}}$ . Show that

$$\omega \wedge \xi =(-1{)}^{qr}\xi \wedge \omega \text{\hspace{1em}(8.33)}$$

$$\mathrm{d}(\omega \wedge \xi )=\mathrm{d}\omega \wedge \xi +(-1{)}^q\omega \wedge \mathrm{d}\xi .\text{\hspace{1em}(8.34)}$$

8.3.2 Differential forms on complex manifolds

Now we restrict ourselves to complex manifolds in which we have the decompositions $T_p{M}^{\mathbb{C}}=T_p{M}^{+}\oplus T_p{M}^{-}$ and $\mathfrak{X}(M{)}^{\mathbb{C}}=\mathfrak{X}(M{)}^{+}\oplus \mathfrak{X}(M{)}^{-}$ .

Definition 8.2

Let $M$ be a complex manifold with ${\dim }_{\mathbb{C}}M=m$ . Let $\omega \in {\Omega }_p^q(M{)}^{\mathbb{C}}$ ( $q\le 2m$ ) and $r,s$ be positive integers such that $r+s=q$ . Let $V_i\in T_p{M}^{\mathbb{C}}$ ( $1\le i\le q$ ) be vectors in either $T_p{M}^{+}$ or $T_p{M}^{-}$ . If $\omega (V_1,\dots ,V_q)=0$ unless $r$ of the $V_i$ are in $T_p{M}^{+}$ and $s$ of the $V_i$ are in $T_p{M}^{-}$ , $\omega$ is said to be of bidegree $(r,s)$ or simply an $(r,s)$ -form. The set of $(r,s)$ -forms at $p$ is denoted by ${\Omega }_p^{r,s}(M)$ . If an $(r,s)$ -form is assigned smoothly at each point of $M$ , we have an $(r,s)$ -form defined over $M$ . The set of $(r,s)$ -forms over $M$ is denoted by ${\Omega }^{r,s}(M)$ .

Take a chart $(U,\phi )$ with the complex coordinates $\phi (p)=z^{\mu }$ . We take the bases (8.13) for the tangent spaces $T_p{M}^{\pm }$ . The dual bases are given by (8.14). Note that $\mathrm{d}{z}^{\mu }$ is of bidegree $(1,0)$ since $⟨\mathrm{d}{z}^{\mu },\partial /\partial {\overline{z}}^{\nu }⟩=0$ and $\mathrm{d}{\overline{z}}^{\mu }$ is of bidegree $(0,1)$ . With these bases, a form $\omega$ of bidegree $(r,s)$ is written as

$$\omega =\frac{1}{r!s!}{\omega }_{{\mu }_1\dots {\mu }_r{v}_1\dots v_s}d{z}^{{\mu }_1}\wedge \cdots \wedge d{z}^{{\mu }_r}\wedge d{\bar{z}}^{v_1}\wedge \cdots \wedge d{\bar{z}}^{v_s}.\text{\hspace{1em}(8.35)}$$

The set $\{\mathrm{d}{z}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{z}^{{\mu }_r}\wedge \mathrm{d}{\overline{z}}^{{\nu }_1}\wedge \dots \wedge \mathrm{d}{\overline{z}}^{{\nu }_s}\}$ is the basis of ${\Omega }_p^{r,s}(M)$ . The components are totally anti-symmetric in the $\mu$ and $\nu$ separately. Let $z^{\mu }$ and $w^{\mu }$ be two overlapping coordinates. The reader should verify that an $(r,s)$ -form in the $z^{\mu }$ coordinate system is also an $(r,s)$ -form in the $w^{\nu }$ system.

Proposition 8.1

Let $M$ be a complex manifold of ${\dim }_{\mathbb{C}}M=m$ and $\omega$ and $\xi$ be complex differential forms on $M$ .

(a) If $\omega \in {\Omega }^{q,r}(M)$ then $\overline{\omega }\in {\Omega }^{r,q}(M)$ .
(b) If $\omega \in {\Omega }^{q,r}(M)$ and $\xi \in {\Omega }^{q^{\prime },r^{\prime }}(M)$ , then $\omega \wedge \xi \in {\Omega }^{q+q^{\prime },r+r^{\prime }}(M)$ .

(c) A complex $q$ -form $\omega$ is uniquely written as

$$\omega =\sum _{r+s=q}{\omega }^{(r,s)}\text{\hspace{1em}(8.36a)}$$

where ${\omega }^{(r,s)}\in {\Omega }^{r,s}(M)$ . Thus, we have the decomposition

$${\Omega }^q(M{)}^{\mathbb{C}}=\underset{r+s=q}{⨁}{\Omega }^{r,s}(M).\text{\hspace{1em}(8.36b)}$$

The proof is easy and is left to the reader. Now any $q$ -form $\omega$ is decomposed as

$$\begin{array}{rl}\omega & =\sum _{r+s=q}{\omega }^{(r,s)}\\ & =\sum _{r+s=q}\frac{1}{r!s!}{\omega }_{{\mu }_1\dots {\mu }_r{\bar{v}}_1\dots {\bar{v}}_s}d{z}^{{\mu }_1}\wedge \cdots \wedge d{z}^{{\mu }_r}\wedge d{\bar{z}}^{v_1}\wedge \cdots \wedge d{\bar{z}}^{v_s}\end{array}\text{\hspace{1em}(8.37)}$$

where

$${\omega }_{{\mu }_1\dots {\mu }_r}{\bar{v}}_1\dots {\bar{v}}_s=\omega \left(\frac{\partial }{\partial z^{{\mu }_1}},\dots ,\frac{\partial }{\partial z^{{\mu }_r}},\frac{\partial }{\partial {\bar{z}}^{{\nu }_1}},\dots ,\frac{\partial }{\partial {\bar{z}}^{{\nu }_s}}\right).\text{\hspace{1em}(8.38)}$$

Exercise 8.6

Let ${\dim }_{\mathbb{C}}M=m$ . Verify that

$${\dim }_{\mathbb{R}}{\Omega }_p^{r,s}(M)=\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{m}{r}\right)\left(\genfrac{}{}{0ex}{}{m}{s}\right)& \text{i f}0\le r,s\le m\\ 0& \text{o t h e r w i s e .}\end{array}$$

Show also that ${\dim }_{\mathbb{R}}{\Omega }_p^q(M{)}^{\mathbb{C}}={\sum }_{r+s=q}{\dim }_{\mathbb{R}}{\Omega }_p^{r,s}(M)=(\genfrac{}{}{0ex}{}{2m}{q})$ .

8.3.3 Dolbeault operators

Let us compute the exterior derivative of an $(r,s)$ -form $\omega$ . From (8.35), we find

$$\begin{array}{rl}& \mathrm{d}\omega =\frac{1}{r!s!}\left(\frac{\partial }{\partial z^{\lambda }}{\omega }_{{\mu }_1\dots {\mu }_r{\bar{v}}_1\dots {\bar{v}}_s}\mathrm{d}{z}^{\lambda }+\frac{\partial }{\partial {\bar{z}}^{\lambda }}{\omega }_{{\mu }_1\dots {\mu }_r{\bar{v}}_1\dots {\bar{v}}_s}\mathrm{d}{\bar{z}}^{\lambda }\right)\\ & \times \mathrm{d}{z}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{z}^{{\mu }_r}\wedge \mathrm{d}{\bar{z}}^{{\nu }_1}\wedge \cdots \wedge \mathrm{d}{\bar{z}}^{{\nu }_s}.\end{array}\text{\hspace{1em}(8.39)}$$

$\mathrm{d}\omega$ is a mixture of an $(r+1,s)$ -form and an $(r,s+1)$ -form. We separate the action of $\mathrm{d}$ according to its destinations,

$$d=\partial +\bar{\partial }\text{\hspace{1em}(8.40)}$$

where $\partial :{\Omega }^{r,s}(M)\to {\Omega }^{r+1,s}(M)$ and $\overline{\partial }:{\Omega }^{r,s}(M)\to {\Omega }^{r,s+1}(M)$ . For example, if $\omega ={\omega }_{\mu \overline{v}}\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\overline{z}}^v$ , its exterior derivatives are

$$\partial \omega =\frac{\partial {\omega }_{\mu \bar{\nu }}}{\partial z^{\lambda }}d{z}^{\lambda }\wedge d{z}^{\mu }\wedge d{\bar{z}}^{\nu }$$

$$\overline{\partial }\omega =\frac{\partial {\omega }_{\mu \bar{\nu }}}{\partial {\bar{z}}^{\lambda }}\mathrm{d}{\bar{z}}^{\lambda }\wedge \mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\nu }=-\frac{\partial {\omega }_{\mu \bar{\nu }}}{\partial {\bar{z}}^{\lambda }}\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\lambda }\wedge \mathrm{d}{\bar{z}}^{\nu }.$$

The operators $\partial$ and $\overline{\partial }$ are called the Dolbeault operators.

If $\omega$ is a general $q$ -form given by (8.37), the actions of $\partial$ and $\overline{\partial }$ on $\omega$ are defined by

$$\partial \omega =\sum _{r+s=q}\partial {\omega }^{(r,s)}\overline{\partial }\omega =\sum _{r+s=q}\overline{\partial }{\omega }^{(r,s)}.\text{\hspace{1em}(8.41)}$$

Theorem 8.1

Let $M$ be a complex manifold and let $\omega \in {\Omega }^q(M{)}^{\mathbb{C}}$ and $\xi \in {\Omega }^p(M{)}^{\mathbb{C}}$ . Then

$$\partial \partial \omega =(\partial \bar{\partial }+\bar{\partial }\partial )\omega =\bar{\partial }\bar{\partial }\omega =0\text{\hspace{1em}(8.42a)}$$

$$\partial \bar{\omega }=\overline{\partial \omega },\bar{\partial }\bar{\omega }=\overline{\partial \omega }\text{\hspace{1em}(8.42b)}$$

$$\partial (\omega \wedge \xi )=\partial \omega \wedge \xi +(-1{)}^q\omega \wedge \partial \xi \text{\hspace{1em}(8.42c)}$$

$$\bar{\partial }(\omega \wedge \xi )=\bar{\partial }\omega \wedge \xi +(-1{)}^q\omega \wedge \bar{\partial }\xi .\text{\hspace{1em}(8.42d)}$$

Proof

It is sufficient to prove them when $\omega$ is of bidegree $(r,s)$ .

(a) Since $\mathrm{d}=\partial +\overline{\partial }$ , we have

$$0={\mathrm{d}}^2\omega =(\partial +\overline{\partial })(\partial +\overline{\partial })\omega =\partial \partial \omega +(\partial \overline{\partial }+\overline{\partial }\partial )\omega +\overline{\partial }\overline{\partial }\omega .$$

The three terms of the RHS are of bidegrees $(r+2,s)$ , $(r+1,s+1)$ and $(r,s+2)$ respectively. From proposition 8.1(c), each term must vanish separately.

(b) Since $\mathrm{d}\overline{\omega }=\overline{\mathrm{d}\omega }$ , we have

$$\partial \bar{\omega }+\bar{\partial }\bar{\omega }=d\bar{\omega }=\overline{(\partial +\bar{\partial })\omega }=\overline{\partial \omega }+\overline{\stackrel{ˉ}{\partial }}\omega .$$

Noting that $\partial \omega$ and $\overline{\overline{\partial }\omega }$ are of bidegree $(s+1,r)$ and $\overline{\partial }\overline{\omega }$ and $\overline{\partial \omega }$ are of $(s,r+1)$ , we conclude that $\partial \overline{\omega }=\overline{\overline{\partial }\omega }$ and $\overline{\partial }\overline{\omega }=\overline{\partial \omega }$ .

(c) We assume $\omega$ is of bidegree $(r,s)$ and $\xi$ of $(r^{\prime },s^{\prime })$ . Equation (8.42c) is proved by separating $\mathrm{d}(\omega \wedge \xi )=\mathrm{d}\omega \wedge \xi +(-1{)}^q\omega \wedge \mathrm{d}\xi$ , into forms of bidegrees $(r+r^{\prime }+1,s+s^{\prime })$ and $(r+r^{\prime },s+s^{\prime }+1)$ .

Definition 8.3

Let $M$ be a complex manifold. If $\omega \in {\Omega }^{r,0}(M)$ satisfies $\overline{\partial }\omega =0$ , the $r$ -form $\omega$ is called a holomorphic $r$ -form.

Let us look at a holomorphic 0-form $f\in \mathcal{F}(U{)}^{\mathbb{C}}$ on a chart $(U,\phi )$ . The condition $\overline{\partial }f=0$ becomes

$$\frac{\partial f}{\partial {\bar{z}}^{\lambda }}=01\le \lambda \le m={\dim }_{\mathbb{C}}M.\text{\hspace{1em}(8.43)}$$

A holomorphic 0-form is just a holomorphic function, $f\in \mathcal{F}(U{)}^{\mathbb{C}}$ . Let $\omega \in {\Omega }^{r,0}(M)$ , where $1\le r\le m={\dim }_{\mathbb{C}}M$ . On a chart $(U,\phi )$ , we have

$$\omega =\frac{1}{r!}{\omega }_{{\mu }_1\dots {\mu }_r}d{z}^{{\mu }_1}\wedge \cdots \wedge d{z}^{{\mu }_r}.\text{\hspace{1em}(8.44)}$$

Then $\overline{\partial }\omega =0$ if and only if

$$\frac{\partial }{\partial {\overline{z}}^{\lambda }}{\omega }_{{\mu }_1\dots {\mu }_r}=0$$

namely if ${\omega }_{{\mu }_1\dots {\mu }_r}$ are holomorphic functions on $U$

Let ${\dim }_{\mathbb{C}}M=m$ . The sequence of $\mathbb{C}$ -linear maps

$$\begin{array}{rl}& {\Omega }^{r,0}(M)\stackrel{\overline{\partial }}{\to }{\Omega }^{r,1}(M)\stackrel{\overline{\partial }}{\to }\dots \\ & \cdots \stackrel{\bar{\partial }}{\to }{\Omega }^{r,m-1}(M)\stackrel{\bar{\partial }}{\to }{\Omega }^{r,m}(M)\end{array}\text{\hspace{1em}(8.45)}$$

is called the Dolbeault complex. Note that ${\overline{\partial }}^2=0$ . The set of $\overline{\partial }$ -closed $(r,s)$ -forms (those $\omega \in {\Omega }^{r,s}(M)$ such that $\overline{\partial }\omega =0$ ) is called the $(r,s)$ -cocycle and is denoted by $Z_{\overline{\partial }}^{r,s}(M)$ . The set of $\overline{\partial }$ -exact $(r,s)$ -forms (those $\omega \in {\Omega }^{r,s}(M)$ such that $\omega =\overline{\partial }\eta$ for some $\eta \in {\Omega }^{r,s-1}(M)$ ) is called the $(r,s)$ -coboundary and is denoted by $B_{\overline{\partial }}^{r,s}(M)$ . The complex vector space

$$H_{\partial }^{r,s}(M)\equiv Z_{\partial }^{r,s}(M)/B_{\partial }^{r,s}(M)\text{\hspace{1em}(8.46)}$$

is called the $(r,s)$ th $\overline{\partial }$ -cohomology group, see section 8.6.

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