12.4 The Dolbeault complex

We recall some elementary facts about complex manifolds (see chapter 8 for details). Let $M$ be a compact complex manifold of complex dimension $m$ without a boundary. Let $z^{\mu }=x^{\mu }+\mathrm{i}{y}^{\mu }$ be the local coordinates and ${\bar{z}}^{\mu }=x^{\mu }-\mathrm{i}{y}^{\mu }$ their complex conjugates. $T{M}^{+}$ denotes the tangent bundle spanned by $\{\partial /\partial z^{\mu }\}$ and $T{M}^{-}=T{M}^{+}$ the complex conjugate bundle spanned by $\{\partial /\partial {\bar{z}}^{\mu }\}$ . The dual of $T{M}^{+}$ is denoted by $T^{\ast }{M}^{+}$ and spanned by $\{\mathrm{d}{z}^{\mu }\}$ while that of $T{M}^{-}$ is $T^{\ast }{M}^{-}=T^{\ast }{M}^{+}$ spanned by $\{\mathrm{d}{\bar{z}}^{\mu }\}$ . The space ${\Omega }^r(M{)}^{\mathbb{C}}$ of complexified $r$ -forms is decomposed as

$${\Omega }^r(M{)}^{\mathbb{C}}=\underset{p+q=r}{⨁}{\Omega }^{p,q}(M)$$

where ${\Omega }^{p,q}(M)$ is the space of the $(p,q)$ -forms, which is spanned by a basis of the form

$$\mathrm{d}{z}^{{\mu }_1}\wedge \cdots \wedge \mathrm{d}{z}^{{\mu }_p}\wedge \mathrm{d}{\bar{z}}^{{\nu }_1}\wedge \cdots \wedge \mathrm{d}{\bar{z}}^{{\nu }_q}.$$

The exterior derivative is decomposed as $\mathrm{d}\equiv \partial +\bar{\partial }$ where

$$\partial =\mathrm{d}{z}^{\mu }\wedge \partial /\partial z^{\mu }\bar{\partial }=\mathrm{d}{\bar{z}}^{\mu }\wedge \partial /\partial {\bar{z}}^{\mu }.$$

They satisfy $\partial \bar{\partial }+\bar{\partial }\partial ={\partial }^2={\bar{\partial }}^2=0$ . We have the sequences

$$\cdots \stackrel{\bar{\partial }}{\to }{\Omega }^{p,q}(M)\stackrel{\bar{\partial }}{\to }{\Omega }^{p,q+1}(M)\stackrel{\bar{\partial }}{\to }\dots \text{\hspace{1em}(12.43a)}$$

$$\cdots \stackrel{\partial }{\to }{\Omega }^{p,q}(M)\stackrel{\partial }{\to }{\Omega }^{p+1,q}(M)\stackrel{\partial }{\to }\dots .\text{\hspace{1em}(12.43b)}$$

We are interested in the first sequence with $p=0$

$$\cdots \stackrel{\bar{\partial }}{\to }{\Omega }^{0,q}(M)\stackrel{\bar{\partial }}{\to }{\Omega }^{0,q+1}(M)\stackrel{\bar{\partial }}{\to }\dots .\text{\hspace{1em}(12.44)}$$

This sequence is called the Dolbeault complex.

To show that (12.44) is an elliptic complex, we compute the symbol for $\bar{\partial }$ . Let $\xi ={\xi }^{0,1}+{\xi }^{1,0}$ be a real one-form at $p\in M$ , where ${\xi }^{0,1}\in {\Omega }_p^{0,1}(M)$ and

$${\xi }^{1,0}=\overline{{\xi }^{0,1}}\in {\Omega }_p^{1,0}(M).$$

Take an anti-holomorphic $r$ -form $\omega \in {\Omega }^{0,r}(M)$ . We find

$$\sigma (\bar{\partial },\xi )\omega =\bar{\partial }(f\tilde{s})=\bar{\partial }f\wedge \tilde{s}+f\bar{\partial }\tilde{s}{|}_p={\xi }^{0,1}\wedge \omega$$

where $f(p)=0$ , $\bar{\partial }f(p)={\xi }^{0,1}$ , $\tilde{s}\in {\Omega }^{0,r}(M)$ and $\tilde{s}(p)=\omega$ . We have

$$\sigma (\bar{\partial },\xi )={\xi }^{0,1}\wedge .\text{\hspace{1em}(12.45)}$$

From a similar argument to that given in the previous section, it follows that the symbol (12.45) is elliptic. Thus, the Dolbeault complex (12.44) is an elliptic complex.

The AS index theorem takes the form

$$\mathrm{ind}\bar{\partial }={\int }_M\mathrm{ch}\left(\sum _r(-1{)}^r{\wedge }^r{T}^{\ast }{M}^{-}\right)\frac{\mathrm{Td}\left(T{M}^{\mathbb{C}}\right)}{e(TM)}{|}_{\text{v o l}}.\text{\hspace{1em}(12.46)}$$

The LHS is computed as follows. We first note that

$$\ker {\bar{\partial }}_r/\mathrm{im}{\bar{\partial }}_{r-1}=H^{0,r}(M)$$

where $H^{0,r}(M)$ is the $\bar{\partial }$ -cohomology group. Then the LHS is

$$\mathrm{ind}\bar{\partial }=\sum _{r=0}^n(-1{)}^r{b}^{0,r}\text{\hspace{1em}(12.47)}$$

where $b^{0,r}\equiv {\dim }_{\mathbb{C}}{H}^{0,r}(M)$ is the Hodge number. This index is called the arithmetic genus of $M$ .

Simplification of the topological index can be carried out as in the case of the de Rham complex. We refer the reader to Shanahan (1978) for the technical details. We have

$$\sum _{r=1}^n(-1{)}^r{b}^{0,r}={\int }_M\mathrm{Td}\left(T{M}^{+}\right)\text{\hspace{1em}(12.48)}$$

where $\mathrm{Td}(T{M}^{+})$ is the Todd class of $T{M}^{+}$

12.4.1 The twisted Dolbeault complex and the Hirzebruch-Riemann-Roch theorem

In the Dolbeault complex, we may replace ${\Omega }^{0,r}(M)$ by the tensor product bundles ${\Omega }^{0,r}(M)\otimes V$ , where $V$ is a holomorphic vector bundle over $M$ ,

$$\cdots \stackrel{{\bar{\partial }}_V}{\to }{\Omega }^{0,r-1}(M)\otimes V\stackrel{{\bar{\partial }}_V}{\to }{\Omega }^{0,r}(M)\otimes V\stackrel{{\bar{\partial }}_V}{\to }\dots .\text{\hspace{1em}(12.49)}$$

The AS index theorem of this complex reduces to the Hirzebruch-Riemann-Roch theorem,

$$\mathrm{ind}{\bar{\partial }}_V={\int }_M\mathrm{Td}\left(T{M}^{+}\right)\mathrm{ch}(V).\text{\hspace{1em}(12.50)}$$

For example, if $m={\dim }_{\mathbb{C}}M=1$ , we have

$$\begin{array}{rl}\mathrm{ind}{\bar{\partial }}_V& =\frac{1}{2}\dim V{\int }_M{c}_1(T{M}^{+})+{\int }_M{c}_1(V)\\ & =(2-g)\dim V+{\int }_M\frac{\mathrm{i}\mathcal{F}}{2\pi }\end{array}\text{\hspace{1em}(12.51)}$$

since it can be shown that

$${\int }_M{c}_1(T{M}^{+})={\int }_{M}e(TM)=2-g$$

$g$ being the genus of $M$ .

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