8.4 Hermitian manifolds and Hermitian differential geometry

Let $M$ be a complex manifold with ${\dim }_{\mathbb{C}}M=m$ and let $g$ be a Riemannian metric of $M$ as a differentiable manifold. Take $Z=X+\mathrm{i}Y$ , $W=U+\mathrm{i}V\in T_p{M}^{\mathbb{C}}$ and extend $g$ so that

$$g_p(Z,W)=g_p(X,U)-g_p(Y,V)+\mathrm{i}[g_p(X,V)+g_p(Y,U)].\text{\hspace{1em}(8.47)}$$

The components of $g$ with respect to the bases (8.13) are

$$g_{\mu \nu }(p)=g_p\left(\partial /\partial z^{\mu },\partial /\partial z^{\nu }\right)\text{\hspace{1em}(8.48a)}$$

$$g_{\mu \bar{v}}(p)=g_p\left(\partial /\partial z^{\mu },\partial /\partial {\bar{z}}^v\right)\text{\hspace{1em}(8.48b)}$$

$$g_{\overline{\mu }\nu }(p)=g_p\left(\partial /\partial {\bar{z}}^{\mu },\partial /\partial z^{\nu }\right)\text{\hspace{1em}(8.48c)}$$

$$g_{\overline{\mu v}}(p)=g_p\left(\partial /\partial {\bar{z}}^{\mu },\partial /\partial {\bar{z}}^v\right).\text{\hspace{1em}(8.48d)}$$

We easily verify that

$$g_{\mu \nu }=g_{\nu \mu },g_{\overline{\mu }\overline{\nu }}=g_{\overline{\nu }\overline{\mu }},g_{\overline{\mu }\nu }=g_{\nu \overline{\mu }},\overline{g_{\mu \overline{\nu }}}=g_{\overline{\mu }\nu },\overline{g_{\mu \nu }}=g_{\overline{\mu }\nu }.\text{\hspace{1em}(8.49)}$$

8.4.1 The Hermitian metric

If a Riemannian metric $g$ of a complex manifold $M$ satisfies

$$g_p\left(J_{p}X,J_{p}Y\right)=g_p(X,Y)\text{\hspace{1em}(8.50)}$$

at each point $p\in M$ and for any $X,Y\in T_{p}M$ , $g$ is said to be a Hermitian metric. The pair $(M,g)$ is called a Hermitian manifold. The vector $J_{p}X$ is orthogonal to $X$ with respect to a Hermitian metric,

$$g_p\left(J_{p}X,X\right)=g_p\left(J_p^{2}X,J_{p}X\right)=-g_p\left(J_{p}X,X\right)=0.\text{\hspace{1em}(8.51)}$$

Theorem 8.2

A complex manifold always admits a Hermitian metric.

Proof

Let $g$ be any Riemannian metric of a complex manifold $M$ . Define a new metric $\hat{g}$ by

$${\hat{g}}_p(X,Y)\equiv \frac{1}{2}\left[g_p(X,Y)+g_p\left(J_{p}X,J_{p}Y\right)\right].\text{\hspace{1em}(8.52)}$$

Clearly ${\hat{g}}_p(J_{p}X,J_{p}Y)={\hat{g}}_p(X,Y)$ . Moreover, $\hat{g}$ is positive definite provided that $g$ is. Hence, $\hat{g}$ is a Hermitian metric on $M$ .

Let $g$ be a Hermitian metric on a complex manifold $M$ . From (8.50), we find that

$$g_{\mu \nu }=g\left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=g\left(J\frac{\partial }{\partial z^{\mu }},J\frac{\partial }{\partial z^{\nu }}\right)=-g\left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=-g_{\mu \nu }$$

hence $g_{\mu \nu }=0$ . We also find that $g_{\overline{\mu }\overline{\nu }}=0$ . Thus, the Hermitian metric $g$ takes the form

$$g=g_{\mu \bar{v}}\mathrm{d}{z}^{\mu }\otimes \mathrm{d}{\bar{z}}^{\nu }+g_{\bar{\mu }\nu }\mathrm{d}{\bar{z}}^{\mu }\otimes \mathrm{d}{z}^{\nu }.\text{\hspace{1em}(8.53)}$$

[Remark: Take $X,Y\in T_p{M}^{+}$ . Define an inner product $h_p$ in $T_p{M}^{+}$ by

$$h_p(X,Y)\equiv g_p(X,\bar{Y}).\text{\hspace{1em}(8.54)}$$

It is easy to see that $h_p$ is a positive-definite Hermitian form in $T_p{M}^{+}$ . In fact,

$$\overline{h(X,Y)}=\overline{g(X,\overline{Y})}=g(\overline{X},Y)=h(Y,X)$$

and $h(X,X)=g(X,\overline{X})=g(X_1,X_1)+g(X_2,X_2)\ge 0$ for $X=X_1+\mathrm{i}{X}_2$ . This is why a metric $g$ satisfying (8.50) is called Hermitian.]

8.4.2 Kähler form

Let $(M,g)$ be a Hermitian manifold. Define a tensor field $\Omega$ whose action on $X,Y\in T_{p}M$ is

$${\Omega }_p(X,Y)=g_p\left(J_{p}X,Y\right)X,Y\in T_{p}M.\text{\hspace{1em}(8.55)}$$

Note that $\Omega$ is anti-symmetric, $\Omega (X,Y)=g(JX,Y)=g(J^{2}X,JY)=-g(JY,X)=-\Omega (Y,X)$ . Hence, $\Omega$ defines a two-form called the Kähler form of a Hermitian metric $g$ . Observe that $\Omega$ is invariant under the action of $J$ ,

$$\Omega (JX,JY)=g\left(J^{2}X,JY\right)=g\left(J^{3}X,J^{2}Y\right)=\Omega (X,Y).\text{\hspace{1em}(8.56)}$$

If the domain is extended from $T_{p}M$ to $T_p{M}^{\mathrm{C}}$ , $\Omega$ is a two-form of bidegree (1, 1). Indeed, for the metric (8.53), it is found that

$$\Omega \left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=g\left(J\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=\mathrm{i}{g}_{\mu \nu }=0.$$

We also have

$$\Omega \left(\frac{\partial }{\partial {\bar{z}}^{\mu }},\frac{\partial }{\partial {\bar{z}}^{\nu }}\right)=0,\Omega \left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial {\bar{z}}^{\nu }}\right)=\mathrm{i}{g}_{\mu \bar{\nu }}=-\Omega \left(\frac{\partial }{\partial {\bar{z}}^{\nu }},\frac{\partial }{\partial z^{\mu }}\right).$$

Thus, the components of $\Omega$ are

$${\Omega }_{\mu v}={\Omega }_{\overline{\mu v}}=0{\Omega }_{\mu \bar{v}}=-{\Omega }_{\bar{v}\mu }=\mathrm{i}{g}_{\mu }\bar{v}.\text{\hspace{1em}(8.57)}$$

We may write

$$\Omega =\mathrm{i}{g}_{\mu \bar{v}}\mathrm{d}{z}^{\mu }\otimes \mathrm{d}{\bar{z}}^v-\mathrm{i}{g}_{\bar{\nu }\mu }\mathrm{d}{\bar{z}}^v\otimes \mathrm{d}{z}^{\mu }=\mathrm{i}{g}_{\mu \bar{v}}\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^v.\text{\hspace{1em}(8.58)}$$

$\Omega$ is also written as

$$\Omega =-J_{\mu \bar{v}}d{z}^{\mu }\wedge d{\bar{z}}^v\text{\hspace{1em}(8.59)}$$

where $J_{\mu \overline{v}}=g_{\mu \overline{\lambda }}{J}_{\overline{v}}^{\overline{\lambda }}=-\mathrm{i}{g}_{\mu \overline{v}}$ $\Omega$ is a real form;

$$\bar{\Omega }=-\mathrm{i}\overline{g_{\mu \bar{\nu }}}\mathrm{d}{\bar{z}}^{\mu }\wedge \mathrm{d}{z}^v=\mathrm{i}{g}_{v\bar{\mu }}\mathrm{d}{z}^v\wedge \mathrm{d}{\bar{z}}^{\mu }=\Omega .\text{\hspace{1em}(8.60)}$$

Making use of the Kähler form, we show that any Hermitian manifold, and hence any complex manifold, is orientable. We first note that we may choose an orthonormal basis $\{{\hat{e}}_1,J{\hat{e}}_1,\dots ,{\hat{e}}_m,J{\hat{e}}_m\}$ . In fact, if $g({\hat{e}}_1,{\hat{e}}_1)=1$ , it follows that $g(J{\hat{e}}_1,J{\hat{e}}_1)=g({\hat{e}}_1,{\hat{e}}_1)=1$ and $g({\hat{e}}_1,J{\hat{e}}_1)=-g(J{\hat{e}}_1,{\hat{e}}_1)=0$ . Thus ${\hat{e}}_1$ and $J{\hat{e}}_1$ form an orthonormal basis of a two-dimensional subspace. Now take ${\hat{e}}_2$ which is orthonormal to ${\hat{e}}_1$ and $J{\hat{e}}_1$ and form the subspace $\{{\hat{e}}_2,J{\hat{e}}_2\}$ . Repeating this procedure we obtain an orthonormal basis $\{{\hat{e}}_1,J{\hat{e}}_1,\dots ,{\hat{e}}_m,J{\hat{e}}_m\}$ .

Lemma 8.1

Let $\Omega$ be the Kähler form of a Hermitian manifold with ${\dim }_{\mathbb{C}}M=m$ . Then

$$\underset{m}{\underbrace{\Omega \wedge \dots \wedge \Omega }}$$

is a nowhere vanishing $2m$ -form.

Proof

For the previous orthonormal basis, we have

$$\Omega ({\hat{e}}_i,J{\hat{e}}_j)=g(J{\hat{e}}_i,J{\hat{e}}_j)={\delta }_{ij}\Omega ({\hat{e}}_i,{\hat{e}}_j)=\Omega (J{\hat{e}}_i,J{\hat{e}}_j)=0.$$

Then it follows that

$$\begin{array}{rl}& \underset{m}{\underbrace{\Omega \wedge \dots \wedge \Omega }}({\hat{e}}_1,J{\hat{e}}_1,\dots ,{\hat{e}}_m,J{\hat{e}}_m)\\ & =\sum _P\Omega \left({\hat{e}}_{P(1)},J{\hat{e}}_{P(1)}\right)\dots \Omega \left({\hat{e}}_{P(m)},J{\hat{e}}_{P(m)}\right)\\ & =m!\Omega \left({\hat{e}}_1,J{\hat{e}}_1\right)\dots \Omega \left({\hat{e}}_m,J{\hat{e}}_m\right)=m!\end{array}$$

where $P$ is an element of the permutation group of $m$ objects. This shows that $\Omega \wedge \dots \wedge \Omega$ cannot vanish at any point.

Since the real $2m$ -form $\Omega \wedge \dots \wedge \Omega$ vanishes nowhere, it serves as a volume element. Thus, we obtain the following theorem.

Theorem 8.3

A complex manifold is orientable.

8.4.3 Covariant derivatives

Let $(M,g)$ be a Hermitian manifold. We define a connection which is compatible with the complex structure. It is natural to assume that a holomorphic vector $V\in T_p{M}^{+}$ parallel transported to another point $q$ is, again, a holomorphic vector $\tilde{V}(q)\in T_q{M}^{+}$ . We show later that the almost complex structure is covariantly conserved under this requirement. Let $\{z^{\mu }\}$ and $\{z^{\mu }+\Delta z^{\mu }\}$ be the coordinates of $p$ and $q$ , respectively, and let $V=V^{\mu }\partial /\partial z^{\mu }{|}_p$ and $\tilde{V}(q)={\tilde{V}}^{\mu }(z+\Delta z)\partial /\partial z^{\mu }{|}_q$ . We assume that (cf (7.9))

$${\tilde{V}}^{\mu }(z+\Delta z)=V^{\mu }(z)-V^{\lambda }(z){\Gamma }_{\nu \lambda }^{\mu }(z)\Delta z^{\nu }.\text{\hspace{1em}(8.61)}$$

Then the basis vectors satisfy (cf (7.14))

$${\nabla }_{\mu }\frac{\partial }{\partial z^{\nu }}={\Gamma }_{\mu \nu }^{\lambda }(z)\frac{\partial }{\partial z^{\lambda }}.\text{\hspace{1em}(8.62a)}$$

Since $\partial /\partial {\overline{z}}^{\mu }$ is a conjugate vector field of $\partial /\partial z^{\mu }$ , we have

$${\nabla }_{\overline{\mu }}\frac{\partial }{\partial {\bar{z}}^{\nu }}={\Gamma }_{\overline{\mu }\nu }^{\overline{\lambda }}\frac{\partial }{\partial {\bar{z}}^{\lambda }}\text{\hspace{1em}(8.62b)}$$

where ${\Gamma }_{\overline{\mu }\overline{\nu }}^{\overline{\lambda }}=\overline{{\Gamma }_{\mu \nu }^{\lambda }}$ . ${\Gamma }_{\mu \nu }^{\lambda }$ and ${\Gamma }_{\overline{\mu }\overline{\nu }}^{\overline{\lambda }}$ are the only non-vanishing components of the connection coefficients. Note that ${\nabla }_{\mu }\partial /\partial {\overline{z}}^{\nu }={\nabla }_{\overline{\mu }}\partial /\partial z^{\nu }=0$ . For the dual basis, non-vanishing covariant derivatives are

$${\nabla }_{\mu }\mathrm{d}{z}^{\nu }=-{\Gamma }_{\mu \lambda }^{\nu }\mathrm{d}{z}^{\lambda }{\nabla }_{\overline{\mu }}\mathrm{d}{\bar{z}}^{\nu }=-{\Gamma }_{\overline{\mu }\lambda }^{\overline{\nu }}{\bar{z}}^{\lambda }.\text{\hspace{1em}(8.63)}$$

The covariant derivative of $X^{+}=X^{\mu }\partial /\partial z^{\mu }\in \mathcal{X}(M{)}^{+}$ is

$${\nabla }_{\mu }{X}^{+}=\left({\partial }_{\mu }{X}^{\lambda }+X^{\nu }{\Gamma }_{\mu \nu }^{\lambda }\right)\frac{\partial }{\partial z^{\lambda }}\text{\hspace{1em}(8.64)}$$

where ${\partial }_{\mu }\equiv \partial /\partial z^{\mu }$ . For $X^{-}=X^{\overline{\mu }}\partial /\partial {\overline{z}}^{\mu }\in \mathfrak{X}(M{)}^{-}$ , we have

$${\nabla }_{\mu }{X}^{-}={\partial }_{\mu }{X}^{\bar{\lambda }}\frac{\partial }{\partial {\bar{z}}^{\lambda }}\text{\hspace{1em}(8.65)}$$

since ${\Gamma }_{\mu \nu }^{\overline{\lambda }}={\Gamma }_{\mu \overline{\nu }}^{\overline{\lambda }}=0$ . As far as anti-holomorphic vectors are concerned, ${\nabla }_{\mu }$ works as the ordinary derivative ${\partial }_{\mu }$ . Similarly, we have

$${\nabla }_{\overline{\mu }}{X}^{+}={\partial }_{\overline{\mu }}{X}^{\lambda }\frac{\partial }{\partial z^{\lambda }}\text{\hspace{1em}(8.66)}$$

$${\nabla }_{\overline{\mu }}{X}^{-}=\left({\partial }_{\overline{\mu }}{X}^{\overline{\lambda }}+X^{\overline{\nu }}{\Gamma }_{\overline{\mu }\overline{\nu }}^{\overline{\lambda }}\right)\frac{\partial }{\partial {\bar{z}}^{\lambda }}.\text{\hspace{1em}(8.67)}$$

It is easy to generalize this to an arbitrary tensor field. For example, if $t=t_{\mu \nu }^{\overline{\lambda }}\mathrm{d}{z}^{\mu }\otimes \mathrm{d}{x}^{\nu }\otimes \partial /\partial {\overline{z}}^{\lambda }$ , we have

$$({\nabla }_{\kappa }t{)}_{\mu \nu }^{\overline{\lambda }}={\partial }_{\kappa }{t}_{\mu \nu }^{\overline{\lambda }}-t_{\xi \nu }^{\overline{\lambda }}{\Gamma }_{\kappa \mu }^{\xi }-t_{\mu \xi }^{\overline{\lambda }}{\Gamma }_{\kappa \nu }^{\xi }$$

$$({\nabla }_{\overline{\kappa }}t{)}_{\mu \nu }^{\overline{\lambda }}={\partial }_{\overline{\kappa }}{t}_{\mu \nu }^{\overline{\lambda }}+t_{\mu \nu }^{\overline{\xi }}{\Gamma }_{\overline{\kappa }\overline{\xi }}.$$

We require the metric compatibility as in section 7.2. We demand that ${\nabla }_{\kappa }{g}_{\mu \overline{v}}={\nabla }_{\overline{\kappa }}{g}_{\mu \overline{v}}=0$ . In components, we have

$${\partial }_{\kappa }{g}_{\mu \bar{v}}-g_{\lambda \bar{v}}{\Gamma }_{\kappa \mu }^{\lambda }=0{\partial }_{\overline{\kappa }}{g}_{\mu \bar{v}}-g_{\mu \bar{\lambda }}{\Gamma }_{\overline{\kappa }\bar{\mu }}^{\overline{\lambda }}=0.\text{\hspace{1em}(8.68)}$$

The connection coefficients are easily read off:

$${\Gamma }_{\kappa \mu }^{\lambda }=g^{\overline{v}\lambda }{\partial }_{\kappa }{g}_{\mu \overline{v}}{\Gamma }_{\overline{\kappa }\overline{\nu }}^{\overline{\lambda }}=g^{\overline{\lambda }\mu }{\partial }_{\overline{\kappa }}{g}_{\mu \overline{\nu }}\text{\hspace{1em}(8.69)}$$

where $\{g^{\overline{\nu }\lambda }\}$ is the inverse matrix of $g_{\mu }\overline{\nu }$ ; $g_{\mu \overline{\lambda }}{g}^{\overline{\lambda }\nu }={\delta }_{\mu }{}^{\nu }$ , $g^{\overline{\nu }\lambda }{g}_{\lambda \overline{\mu }}={\delta }^{\overline{\nu }}{}_{\overline{\mu }}$ . A metric-compatible connection for which $\Gamma$ (mixed indices) $=0$ is called the Hermitian connection. By construction, this is unique and given by (8.69).

Theorem 8.4

The almost complex structure $J$ is covariantly constant with respect to the Hermitian connection,

$$({\nabla }_{\kappa }J{)}_v^{\mu }=({\nabla }_{\bar{\kappa }}J{)}_v^{\mu }=({\nabla }_{\kappa }J{)}_{\bar{v}}^{\bar{\mu }}=({\nabla }_{\bar{\kappa }}J{)}_{\bar{v}}^{\bar{\mu }}=0.\text{\hspace{1em}(8.70)}$$

Proof

We prove the first equality. From (8.22), we find

$$({\nabla }_{\kappa }J{)}_v^{\mu }={\partial }_{\kappa }\mathrm{i}{\delta }_v^{\mu }-\mathrm{i}{\delta }_{\xi }^{\mu }{\Gamma }_{\kappa v}^{\xi }+\mathrm{i}{\delta }_v^{\xi }{\Gamma }_{\kappa \xi }^{\mu }=0.$$

Other equalities follow from similar calculations.

$\square$

8.4.4 Torsion and curvature

The torsion tensor $T$ and the Riemann curvature tensor $R$ are defined by

$$T(X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]\text{\hspace{1em}(8.71)}$$

$$R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z.\text{\hspace{1em}(8.72)}$$

We find that

$$T\left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^v}\right)=\left({\Gamma }_{\mu v}^{\lambda }-{\Gamma }_{v\mu }^{\lambda }\right)\frac{\partial }{\partial z^{\lambda }}$$

$$T\left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial {\overline{z}}^{\nu }}\right)=T\left(\frac{\partial }{\partial {\overline{z}}^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=0$$

$$T\left(\frac{\partial }{\partial {\bar{z}}^{\mu }},\frac{\partial }{\partial {\bar{z}}^{\nu }}\right)=({\Gamma }_{\overline{\mu }\nu }^{\overline{\lambda }}-{\Gamma }_{\overline{\nu }\mu }^{\overline{\lambda }})\frac{\partial }{\partial {\bar{z}}^{\lambda }}.$$

The non-vanishing components are

$$T_{\mu \nu }^{\lambda }={\Gamma }_{\mu \nu }^{\lambda }-{\Gamma }_{\nu \mu }^{\lambda }=g^{\overline{\xi }\lambda }\left({\partial }_{\mu }{g}_{\nu }\bar{\xi }-{\partial }_{\nu }{g}_{\mu }\bar{\xi }\right)\text{\hspace{1em}(8.73a)}$$

$$T_{\overline{\mu v}}^{\bar{\lambda }}={\Gamma }_{\overline{\mu v}}^{\bar{\lambda }}-{\Gamma }_{\overline{v\mu }}^{\bar{\lambda }}=g^{\bar{\lambda }\xi }\left({\partial }_{\bar{\mu }}{g}_{\bar{v}\xi }-{\partial }_{\bar{v}}{g}_{\bar{\mu }\xi }\right).\text{\hspace{1em}(8.73b)}$$

As for the Riemann tensor, we find, for example, that

$$R_{\lambda \mu \nu }^{\kappa }={\partial }_{\mu }{\Gamma }_{\nu \lambda }^{\kappa }-{\partial }_{\nu }{\Gamma }_{\mu \lambda }^{\kappa }+{\Gamma }_{\nu \lambda }^{\eta }{\Gamma }_{\mu \eta }^{\kappa }-{\Gamma }_{\mu \lambda }^{\eta }{\Gamma }_{\nu \eta }^{\kappa }.$$

If (8.69) is substituted, we find that

$$\begin{array}{rl}& R_{\lambda \mu \nu }^{\kappa }={\partial }_{\mu }{g}^{\overline{\xi }\kappa }{\partial }_{\nu }{g}_{\lambda \overline{\xi }}+g^{\overline{\xi }\kappa }{\partial }_{\mu }{\partial }_{\nu }{g}_{\lambda \overline{\xi }}-{\partial }_{\nu }{g}^{\overline{\xi }\kappa }{\partial }_{\mu }{g}_{\lambda \overline{\xi }}-g^{\overline{\xi }\kappa }{\partial }_{\mu }{\partial }_{\nu }{g}_{\lambda \overline{\xi }}\\ & +g^{\overline{\xi }\eta }{\partial }_{\nu }{g}_{\lambda \overline{\xi }}{g}^{\overline{\zeta }\kappa }{\partial }_{\mu }{g}_{\eta \overline{\zeta }}-g^{\overline{\xi }\eta }{\partial }_{\mu }{g}_{\lambda \overline{\xi }}{g}^{\overline{\zeta }\kappa }{\partial }_{\nu }{g}_{\eta \overline{\zeta }}=0\end{array}$$

where use has been made of the identity $g^{\overline{\zeta }\kappa }{\partial }_{\mu }{g}_{\eta \overline{\zeta }}=-g_{\eta \overline{\zeta }}{\partial }_{\mu }{g}^{\overline{\zeta }\kappa }$ etc. In general, we find that

$$R_{\bar{\lambda }AB}^{\kappa }=R_{\lambda AB}^{\bar{\kappa }}=R_{B\kappa \lambda }^A=R_{B\bar{\kappa }\bar{\lambda }}^A=0\text{\hspace{1em}(8.74)}$$

where $A$ and $B$ are any (holomorphic or anti-holomorphic) indices. As a result, we are left only with the components $R_{\lambda \overline{\mu }\nu }^{\kappa },R_{\lambda \mu \overline{\nu }}^{\kappa },R_{\overline{\lambda }\overline{\mu }\nu }^{\overline{\kappa }}$ and $R_{\overline{\lambda }\mu \overline{\nu }}^{\overline{\kappa }}$ . Note that we have a trivial symmetry $R_{\lambda \overline{\mu }\nu }^{\kappa }=-R_{\lambda \nu \overline{\mu }}^{\kappa }$ . So the independent components are reduced to $R_{\lambda \overline{\mu }\nu }^{\kappa }$ and $R_{\overline{\lambda }\mu \overline{\nu }}^{\overline{\kappa }}=\overline{R_{\lambda \overline{\mu }\nu }^{\kappa }}$ . We find that

$$R_{\lambda \bar{\mu }\nu }^{\kappa }={\partial }_{\bar{\mu }}{\Gamma }_{\nu \lambda }^{\kappa }={\partial }_{\bar{\mu }}\left(g^{\bar{\xi }\kappa }{\partial }_{\nu }{g}_{\lambda \bar{\xi }}\right)\text{\hspace{1em}(8.75a)}$$

$$R_{\overline{\lambda }\mu \bar{\nu }}^{\overline{\kappa }}={\partial }_{\mu }{\Gamma }_{\bar{\nu }\bar{\lambda }}^{\overline{\kappa }}={\partial }_{\mu }\left(g^{\overline{\kappa }\xi }{\partial }_{\bar{\nu }}{g}_{\xi \bar{\lambda }}\right).\text{\hspace{1em}(8.75b)}$$

Exercise 8.7

Show that

$$R_{\overline{\kappa }\lambda \overline{\mu }v}\equiv g_{\overline{\kappa }\xi }{R}_{\lambda \overline{\mu }v}^{\xi }={\partial }_{\overline{\mu }}{\partial }_v{g}_{\lambda \overline{\kappa }}-g^{\overline{\eta }\xi }{\partial }_{\overline{\mu }}{g}_{\overline{\kappa }\xi }{\partial }_v{g}_{\lambda \overline{\eta }}\text{\hspace{1em}(8.76a)}$$

$$R_{\kappa }{\bar{\lambda }}_{\mu }\bar{v}\equiv g_{\kappa }\bar{\xi }{R}^{\bar{\xi }}{\bar{\lambda }}_{\mu }\bar{v}={\partial }_{\mu }{\partial }_{\bar{v}}{g}_{\bar{\lambda }\kappa }-g^{\eta \bar{\xi }}{\partial }_{\mu }{g}_{\kappa }\bar{\xi }{\partial }_{\bar{v}}{g}_{\bar{\lambda }\eta }\text{\hspace{1em}(8.76b)}$$

$$R_{\overline{\kappa }\lambda \mu \bar{v}}\equiv g_{\overline{\kappa }\xi }{R}_{\lambda \mu \bar{v}}^{\xi }=-R_{\overline{\kappa }\lambda \bar{v}\mu }\text{\hspace{1em}(8.76c)}$$

$$R_{\kappa \bar{\lambda }\bar{\mu }\nu }\equiv g_{\kappa \bar{\xi }}{R}_{\bar{\lambda }\bar{\mu }\nu }^{\xi }=-R_{\kappa \bar{\lambda }\nu \bar{\mu }}.\text{\hspace{1em}(8.76d)}$$

Verify the symmetries

$$R_{\overline{\kappa }\lambda \overline{\mu }v}=-R_{\lambda \overline{\kappa }\overline{\mu }v}{R}_{\kappa \overline{\lambda }\mu \bar{v}}=-R_{\overline{\lambda }\kappa \mu \bar{v}}.\text{\hspace{1em}(8.77)}$$

Let us contract the indices of the Riemann tensor as

$${\Re }_{\mu \bar{v}}\equiv R_{\kappa \mu \bar{v}}^{\kappa }=-{\partial }_{\bar{v}}\left(g^{\kappa \bar{\xi }}{\partial }_{\mu }{g}_{\kappa \bar{\xi }}\right)=-{\partial }_{\bar{v}}{\partial }_{\mu }\log G\text{\hspace{1em}(8.78)}$$

where $G\equiv \det (g_{\mu \overline{\nu }})=\sqrt{g}$ . To obtain the last equality, we used an identity $\delta G=G{g}^{\mu \overline{\nu }}\delta g_{\mu \overline{\nu }}$ ; see (7.204). We define the Ricci form by

$$\Re \equiv \mathrm{i}{\Re }_{\mu \bar{v}}\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^v=\mathrm{i}\partial \bar{\partial }\log G.\text{\hspace{1em}(8.79)}$$

$\Re$ is a real form; $\overline{\Re }=-\mathrm{i}\overline{\partial }\overline{\partial }\log G=-\mathrm{i}\partial \overline{\partial }\log G=\Re$ . From the identity $\partial \overline{\partial }=-\frac{1}{2}\mathrm{d}(\partial -\overline{\partial })$ , we find $\Re$ is closed; $\mathrm{d}\Re \propto {\mathrm{d}}^2(\partial -\overline{\partial })\log G=0$ . However, this does not imply that $\Re$ is exact. In fact, $G$ is not a scalar and $(\partial -\overline{\partial })\log G$ is not defined globally. $\Re$ defines a non-trivial element $c_1(M)\equiv [\Re /2\pi ]\in H^2(M;\mathbb{R})$ called the first Chern class. We discuss this further in section 11.2.

Proposition 8.2

The first Chern class $c_1(M)$ is invariant under a smooth change of the metric $g\to g+\delta g$ .

Proof

It follows from (7.204) that $\delta \log G=g^{\mu \overline{\nu }}\delta g_{\mu \overline{\nu }}$ . Then

$$\delta \Re =\delta \mathrm{i}\partial \overline{\partial }\log G=\mathrm{i}\partial \overline{\partial }{g}^{\mu \bar{\nu }}\delta g_{\mu \bar{\nu }}=-\frac{1}{2}\mathrm{d}(\partial -\overline{\partial })\mathrm{i}{g}^{\mu \bar{\nu }}\delta g_{\mu \bar{\nu }}.$$

Since $g^{\mu \overline{\nu }}\delta g_{\mu \overline{\nu }}$ is a scalar, $\omega \equiv -\frac{1}{2}(\partial -\overline{\partial })g^{\mu \overline{\nu }}\delta g_{\mu \overline{\nu }}$ is a well-defined one-form on $M$ . Thus, $\delta \Re =\mathrm{d}\omega$ is an exact two-form and $[\Re ]=[\Re +\delta \Re ]$ , namely $c_1(M)$ is left invariant under $g\to g+\delta g$ .

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