8.5 Kähler manifolds and Kähler differential geometry

8.5.1 Definitions

Definition 8.4

A Kähler manifold is a Hermitian manifold $(M,g)$ whose Kähler form $\Omega$ is closed: $\mathrm{d}\Omega =0$ . The metric $g$ is called the Kähler metric of $M$ . [Warning: Not all complex manifolds admit Kähler metrics.]

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

Theorem 8.5

A Hermitian manifold $(M,g)$ is a Kähler manifold if and only if the almost complex structure $J$ satisfies

$${\nabla }_{\mu }J=0\text{\hspace{1em}(8.80)}$$

where ${\nabla }_{\mu }$ is the Levi-Civita connection associated with $g$

Proof

We first note that for any $r$ -form $\omega$ , $\mathrm{d}\omega$ is written as

$$\mathrm{d}\omega =\nabla \omega \equiv \frac{1}{r!}{\nabla }_{\mu }{\omega }_{v_1\dots v_r}\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{v_1}\wedge \cdots \wedge \mathrm{d}{x}^{v_r}.\text{\hspace{1em}(8.81)}$$

[For example,

$$\begin{array}{rl}\nabla \Omega & =\frac{1}{2}{\nabla }_{\lambda }{\Omega }_{\mu \nu }d{x}^{\lambda }\wedge d{x}^{\mu }\wedge d{x}^{\nu }\\ & =\frac{1}{2}\left({\partial }_{\lambda }{\Omega }_{\mu \nu }-{\Gamma }_{\lambda \mu }^{\kappa }{\Omega }_{\kappa \nu }-{\Gamma }_{\lambda \nu }^{\kappa }{\Omega }_{\mu \kappa }\right)d{x}^{\lambda }\wedge d{x}^{\mu }\wedge d{x}^{\nu }\\ & =\frac{1}{2}{\partial }_{\lambda }{\Omega }_{\mu \nu }d{x}^{\lambda }\wedge d{x}^{\mu }\wedge d{x}^{\nu }=d\Omega \end{array}$$

since $\Gamma$ is symmetric.] Now we prove that ${\nabla }_{\mu }J=0$ if and only if ${\nabla }_{\mu }\Omega =0$ . We verify the following equalities:

$$\begin{array}{rl}({\nabla }_Z\Omega )(X,Y)& ={\nabla }_Z[\Omega (X,Y)]-\Omega ({\nabla }_{Z}X,Y)-\Omega (X,{\nabla }_{Z}Y)\\ & ={\nabla }_Z[g(JX,Y)]-g(J{\nabla }_{Z}X,Y)-g(JX,{\nabla }_{Z}Y)\\ & =({\nabla }_{Z}g)(JX,Y)+g({\nabla }_{Z}JX,Y)-g(J{\nabla }_{Z}X,Y)\\ & =g\left({\nabla }_{Z}JX-J{\nabla }_{Z}X,Y\right)=g\left(\left({\nabla }_{Z}J\right)X,Y\right)\end{array}$$

where ${\nabla }_{Z}g=0$ has been used. Since this is true for any $X,Y,Z$ , it follows that ${\nabla }_Z\Omega =0$ if and only if ${\nabla }_{Z}J=0$ .

Theorems 8.4 and 8.5 show that the Riemann structure is compatible with the Hermitian structure in the Kähler manifold.

Let $g$ be a Kähler metric. Since $\mathrm{d}\Omega =0$ , we have

$$\begin{array}{rl}& (\partial +\overline{\partial })\mathrm{i}{g}_{\mu \bar{\nu }}\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\nu }\\ & =\mathrm{i}{\partial }_{\lambda }{g}_{\mu \bar{v}}\mathrm{d}{z}^{\lambda }\wedge \mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\nu }+\mathrm{i}{\partial }_{\lambda }{g}_{\mu \bar{v}}\mathrm{d}{\bar{z}}^{\lambda }\wedge \mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\nu }\\ & =\frac{1}{2}\mathrm{i}\left({\partial }_{\lambda }{g}_{\mu \bar{v}}-{\partial }_{\mu }{g}_{\lambda \bar{v}}\right)\mathrm{d}{z}^{\lambda }\wedge \mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\nu }\\ & +\frac{1}{2}\mathrm{i}\left({\partial }_{\overline{\lambda }}{g}_{\mu \overline{\nu }}-{\partial }_{\overline{\nu }}{g}_{\mu \overline{\lambda }}\right)\mathrm{d}{\vec{z}}^{\lambda }\wedge \mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\overline{z}}^{\nu }=0\end{array}$$

from which we find

$$\frac{\partial g_{\mu }\bar{v}}{\partial z^{\lambda }}=\frac{\partial g_{\lambda \bar{v}}}{\partial z^{\mu }}\frac{\partial g_{\mu }\bar{v}}{\partial {\bar{z}}^{\lambda }}=\frac{\partial g_{\mu }\bar{\lambda }}{\partial {\bar{z}}^{\nu }}.\text{\hspace{1em}(8.82)}$$

Suppose that a Hermitian metric $g$ is given on a chart $U_i$ by

$$g_{\mu \bar{v}}={\partial }_{\mu }{\partial }_{\bar{v}}{\mathcal{K}}_i\text{\hspace{1em}(8.83)}$$

where ${\mathcal{K}}_i\in \mathcal{F}(U_i)$ . Clearly this metric satisfies the condition (8.82), hence it is Kähler. Conversely, it can be shown that any Kähler metric is locally expressed as (8.83). The function ${\mathcal{K}}_i$ is called the Kähler potential of a Kähler metric. It follows that $\Omega =\mathrm{i}\partial \overline{\partial }{\mathcal{K}}_i$ on $U_i$

Let $(U_i,{\phi }_i)$ and $(U_j,{\phi }_j)$ be overlapping charts. On $U_i\cap U_j$ , we have

$$\frac{\partial }{\partial z^{\mu }}\frac{\partial }{\partial {\overline{z}}^{\nu }}{\mathcal{K}}_i\mathrm{d}{z}^{\mu }\otimes \mathrm{d}{\overline{z}}^{\nu }=\frac{\partial }{\partial w^{\alpha }}\frac{\partial }{\partial {\overline{w}}^{\beta }}{\mathcal{K}}_j\mathrm{d}{w}^{\alpha }\otimes \mathrm{d}{\overline{w}}^{\beta }$$

where $z={\phi }_i(p)$ and $w={\phi }_j(p)$ . It then follows that

$$\frac{\partial w^{\alpha }}{\partial z^{\mu }}\frac{\partial {\bar{w}}^{\beta }}{\partial {\bar{z}}^{\nu }}\frac{\partial }{\partial w^{\alpha }}\frac{\partial }{\partial {\bar{w}}^{\beta }}{\mathcal{K}}_j=\frac{\partial }{\partial z^{\mu }}\frac{\partial }{\partial {\bar{z}}^{\nu }}{\mathcal{K}}_i.\text{\hspace{1em}(8.84)}$$

This is satisfied if and only if ${\mathcal{K}}_j(w,\overline{w})={\mathcal{K}}_i(z,\overline{z})+{\varphi }_{ij}(z)+{\psi }_{ij}(\overline{z})$ where ${\varphi }_{ij}$ $({\psi }_{ij})$ is holomorphic (anti-holomorphic) in $z$ .

Exercise 8.8

Let $M$ be a compact Kähler manifold without a boundary. Show that

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

is closed but not exact where $m={\dim }_{\mathbb{C}}M$ [Hint: Use Stokes’ theorem.] Thus, the 2mth Betti number cannot vanish, $b^{2m}\ge 1$ . We will see later that $b^{2p}\ge 1$ for $1\le p\le m$ .

Example 8.6

Let $M={\mathbb{C}}^m=\{(z^1,\dots ,z^m)\}$ . ${\mathbb{C}}^m$ is identified with ${\mathbb{R}}^{2m}$ by the identification $z^{\mu }\to x^{\mu }+\mathrm{i}{y}^{\mu }$ . Let $\delta$ be the Euclidean metric of ${\mathbb{R}}^{2m}$ ,

$$\delta \left(\frac{\partial }{\partial x^{\mu }},\frac{\partial }{\partial x^v}\right)=\delta \left(\frac{\partial }{\partial y^{\mu }},\frac{\partial }{\partial y^v}\right)={\delta }_{\mu v}\text{\hspace{1em}(8.85a)}$$

$$\delta \left(\frac{\partial }{\partial x^{\mu }},\frac{\partial }{\partial y^v}\right)=0.$$

Noting that $J\partial /\partial x^{\mu }=\partial /\partial y^{\mu }$ and $J\partial /\partial y^{\mu }=-\partial /\partial x^{\mu }$ , we find that $\delta$ is a Hermitian metric. In complex coordinates, we have

$$\delta \left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial z^{\nu }}\right)=\delta \left(\frac{\partial }{\partial {\bar{z}}^{\mu }},\frac{\partial }{\partial {\bar{z}}^{\nu }}\right)=0\text{\hspace{1em}(8.85b)}$$

$$\delta \left(\frac{\partial }{\partial z^{\mu }},\frac{\partial }{\partial {\overline{z}}^v}\right)=\delta \left(\frac{\partial }{\partial {\overline{z}}^{\mu }},\frac{\partial }{\partial z^v}\right)=\frac{1}{2}{\delta }_{\mu v}.$$

The Kähler form is given by

$$\Omega =\frac{\mathrm{i}}{2}\sum _{\mu =1}^m\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\bar{z}}^{\mu }=\frac{\mathrm{i}}{2}\sum _{\mu =1}^m\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{y}^{\mu }.\text{\hspace{1em}(8.86)}$$

Clearly, $\mathrm{d}\Omega =0$ and we find that the Euclidean metric $\delta$ of ${\mathbb{R}}^{2m}$ is a Kähler metric of ${\mathbb{C}}^m$ . The Kähler potential is

$$\mathcal{K}=\frac{1}{2}\sum z^{\mu }{\bar{z}}^{\mu }.\text{\hspace{1em}(8.87)}$$

The Kähler manifold ${\mathbb{C}}^m$ is called the complex Euclid space.

Example 8.7

Any orientable complex manifold $M$ with ${\dim }_{\mathbb{C}}M=1$ is Kähler. Take a Hermitian metric $g$ whose Kähler form is $\Omega$ . Since $\Omega$ is a real two-form, a three-form $\mathrm{d}\Omega$ has to vanish on $M$ . One-dimensional compact orientable complex manifolds are known as Riemann surfaces.

Example 8.8

A complex projective space ${\mathbb{CP}}^m$ is a Kähler manifold. Let $(U_{\alpha },{\phi }_{\alpha })$ be a chart whose inhomogeneous coordinates are ${\phi }_{\alpha }(p)={\xi }_{(\alpha )}^{\nu }$ , $\nu \ne \alpha$ (see example 8.3). It is convenient to introduce a tidier notation $\{{\zeta }_{(\alpha )}^v|1\le \nu \le m\}$ by

$${\xi }_{(\alpha )}^v={\zeta }_{(\alpha )}^v(\nu \le \alpha -1){\xi }_{(\alpha )}^{v+1}={\zeta }_{(\alpha )}^v(\nu \ge \alpha ).\text{\hspace{1em}(8.88)}$$

$\{{\zeta }_{(\alpha )}^{\nu }\}$ is just a renaming of $\{{\xi }_{(\alpha )}^{\nu }\}$ . Define a positive-definite function

$${\mathcal{K}}_{\alpha }(p)\equiv \sum _{\nu =1}^m{\left|{\zeta }_{(\alpha )}^{\nu }(p)\right|}^2+1=\sum _{\nu =1}^{m+1}{\left|\frac{z^{\nu }}{z^{\alpha }}\right|}^2.\text{\hspace{1em}(8.89)}$$

At a point $p\in U_{\alpha }\cap U_{\beta }$ , ${\mathcal{K}}_{\alpha }(p)$ and ${\mathcal{K}}_{\beta }(p)$ are related as

$${\mathcal{K}}_{\alpha }(p)={\left|\frac{z^{\beta }}{z^{\alpha }}\right|}^2{\mathcal{K}}_{\beta }(p).\text{\hspace{1em}(8.90)}$$

Then it follows that

$$\log {\mathcal{K}}_{\alpha }=\log {\mathcal{K}}_{\beta }+\log \frac{z^{\beta }}{z^{\alpha }}+\overline{\log \frac{z^{\beta }}{z^{\alpha }}}.\text{\hspace{1em}(8.91)}$$

Since $z^{\beta }/z^{\alpha }$ is a holomorphic function, we have $\overline{\partial }\log z^{\beta }/z^{\alpha }=0$ . Also

$$\partial \overline{\log z^{\beta }/z^{\alpha }}=\overline{\overline{\partial }\log z^{\beta }/z^{\alpha }}=0.$$

Then it follows that

$$\partial \bar{\partial }\log {\mathcal{K}}_{\alpha }=\partial \bar{\partial }\log {\mathcal{K}}_{\beta }.\text{\hspace{1em}(8.92)}$$

A closed two-form $\Omega$ is locally defined by

$$\Omega \equiv \mathrm{i}\partial \bar{\partial }\log {\mathcal{K}}_{\alpha }.\text{\hspace{1em}(8.93)}$$

There exists a Hermitian metric whose Kähler form is $\Omega$ . Take $X,Y\in T_p{\mathbb{CP}}^n$ and define $g:T_p{\mathbb{CP}}^n\otimes T_p{\mathbb{CP}}^n\to \mathbb{R}$ by $g(X,Y)=\Omega (X,JY)$ . To prove that $g$ is a Hermitian metric, we have to show that $g$ satisfies (8.50) and is positive definite. The Hermiticity is obvious since $g(JX,JY)=-\Omega (JX,Y)=\Omega (Y,JX)=g(X,Y)$ . Next, we show that $g$ is positive definite. On a chart $(U_{\alpha },{\phi }_{\alpha })$ , we obtain

$$\Omega =\mathrm{i}\frac{{\partial }^2\log \mathcal{K}}{\partial {\zeta }^{\mu }\partial {\bar{\zeta }}^{\nu }}\mathrm{d}{\zeta }^{\mu }\wedge \mathrm{d}{\bar{\zeta }}^{\nu }\text{\hspace{1em}(8.94)}$$

where we have dropped the subscript $(\alpha )$ to simplify the notation. If we substitute the expression (8.89) for $\mathcal{K}$ on $U_{\alpha }$ , we have

$$\Omega =\mathrm{i}\sum _{\mu ,v}\frac{{\delta }_{\mu v}\left(\sum |{\zeta }^{\lambda }{|}^2+1\right)-{\zeta }^{\mu }{\overline{\zeta }}^v}{{\left(\sum |{\zeta }^{\lambda }{|}^2+1\right)}^2}\mathrm{d}{\zeta }^{\mu }\wedge \mathrm{d}{\overline{\zeta }}^v.\text{\hspace{1em}(8.95)}$$

Let $X$ be a real vector, $X=X^{\mu }\partial /\partial {\zeta }^{\mu }+{\overline{X}}^{\mu }\partial /\partial {\overline{\zeta }}^{\mu }$ and $JX=\mathrm{i}{X}^{\mu }\partial /\partial {\zeta }^{\mu }-\mathrm{i}{\overline{X}}^{\mu }\partial /\partial {\overline{\zeta }}^{\mu }$ . Then

$$\begin{array}{rl}g(X,X)& =\Omega (X,JX)=2\sum _{\mu ,\nu }\frac{{\delta }_{\mu \nu }(\sum |{\zeta }^{\lambda }{|}^2+1)-{\zeta }^{\mu }{\overline{\zeta }}^{\nu }}{(\sum |{\zeta }^{\lambda }{|}^2+1{)}^2}{X}^{\mu }{\overline{X}}^{\nu }\\ & =2\left[\sum _{\mu }|X^{\mu }{|}^2\left(\sum _{\lambda }|{\zeta }^{\lambda }{|}^2+1\right)-{\left|\sum _{\mu }{X}^{\mu }{\zeta }^{\mu }\right|}^2\right]{\left(\sum _{\lambda }|{\zeta }^{\lambda }{|}^2+1\right)}^{-2}.\end{array}$$

From the Schwarz inequality ${\sum }_{\mu }|X^{\mu }{|}^2\cdot {\sum }_{\lambda }|{\zeta }^{\lambda }{|}^2\ge {\sum }_{\mu }|X^{\mu }{\zeta }^{\mu }{|}^2$ , we find the metric $g$ is positive definite. This metric is called the Fubini-Study metric of ${\mathbb{CP}}^n$ .

A few useful facts are:

(a) $S^2$ is the only sphere which admits a complex structure. Since $S^2\simeq {\mathbb{CP}}^1$ , it is a Kähler manifold.
(b) A product of two odd-dimensional spheres $S^{2m+1}\times S^{2n+1}$ always admits a complex structure. This complex structure does not admit a Kähler metric.
(c) Any complex submanifold of a Kähler manifold is Kähler.

8.5.2 Kähler geometry

A Kähler metric $g$ is characterized by (8.82):

$$\frac{\partial g_{\mu }\overline{v}}{\partial z^{\lambda }}=\frac{\partial g_{\lambda }\overline{v}}{\partial z^{\mu }}\frac{\partial g_{\mu }\overline{v}}{\partial {\overline{z}}^{\lambda }}=\frac{\partial g_{\mu }\overline{\lambda }}{\partial {\overline{z}}^v}.$$

This ensures that the Kähler metric is torsion free:

$$T_{\mu \nu }^{\lambda }=g^{\overline{\xi }\lambda }\left({\partial }_{\mu }{g}_{\nu \overline{\xi }}-{\partial }_{\nu }{g}_{\mu \overline{\xi }}\right)=0\text{\hspace{1em}(8.96a)}$$

$$T^{{\overline{\lambda }}_{\overline{\mu v}}}=g^{\overline{\lambda }\xi }\left({\partial }_{\overline{\mu }}{g}_{\overline{v}\xi }-{\partial }_{\overline{v}}{g}_{\overline{\mu }\xi }\right)=0.\text{\hspace{1em}(8.96b)}$$

In this sense, the Kähler metric defines a connection which is very similar to the Levi-Civita connection. Now the Riemann tensor has an extra symmetry

$$R_{\lambda \mu \bar{v}}^{\kappa }=-{\partial }_{\bar{v}}\left(g^{\bar{\xi }\kappa }{\partial }_{\mu }{g}_{\lambda \bar{\xi }}\right)=-{\partial }_{\bar{v}}\left(g^{\bar{\xi }\kappa }{\partial }_{\lambda }{g}_{\mu \bar{\xi }}\right)=R_{\mu \lambda \bar{v}}^{\kappa }\text{\hspace{1em}(8.97)}$$

as well as those obtained from (8.97) by known symmetry operations,

$$R_{\bar{\lambda }\bar{\mu }v}^{\overline{\kappa }}=R_{\bar{\mu }\bar{\lambda }v}^{\overline{\kappa }},R_{\lambda \bar{\mu }v}^{\kappa }=R_{v\bar{\mu }\lambda }^{\kappa },R_{\bar{\lambda }\mu \bar{v}}^{\overline{\kappa }}=R_{\bar{v}\mu \bar{\lambda }}^{\overline{\kappa }}.\text{\hspace{1em}(8.98)}$$

The Ricci form $\Re$ is defined as before,

$$\mathfrak{R}=-\mathrm{i}{\partial }_{\overline{V}}{\partial }_{\mu }\log G\mathrm{d}{z}^{\mu }\wedge \mathrm{d}{\overline{z}}^{\nu }.$$

Because of (8.97), the components of the Ricci form agree with $Ri{c}_{\mu \overline{v}};{\Re }_{\mu \overline{v}}\equiv R_{\kappa \mu \overline{v}}^{\kappa }=R_{\mu \kappa \overline{v}}^{\kappa }=Ri{c}_{\mu \overline{v}}$ . If $Ric=\Re =0$ , the Kähler metric is said to be Ricci flat.

Theorem 8.6

Let $(M,g)$ be a Kähler manifold. If $M$ admits a Ricci flat metric $h$ , then its first Chern class must vanish.

Proof

By assumption, $\Re =0$ for the metric $h$ . As was shown in the previous section, $\Re (g)-\Re (h)=\Re (g)=\mathrm{d}\omega$ . Hence, $c_1(M)$ computed from $g$ agrees with that computed from $h$ and hence vanishes.

A compact Kähler manifold with vanishing first Chern class is called a Calabi-Yau manifold. Calabi (1957) conjectured that if $c_1(M)=0$ , the Kähler manifold $M$ admits a Ricci-flat metric. This is proved by Yau (1977). Calabi-Yau manifolds with ${\dim }_{\mathbb{C}}M=3$ have been proposed as candidates for superstring compactification (see Horowitz (1986) and Candelas (1988)).

8.5.3 The holonomy group of Kähler manifolds

Before we close this section, we briefly look at the holonomy groups of Kähler manifolds. Let $(M,g)$ be a Hermitian manifold with ${\dim }_{\mathbb{C}}M=m$ . Take a vector $X\in T_p{M}^{+}$ and parallel transport it along a loop $c$ at $p$ . Then we end up with a vector $X^{\prime }\in T_p{M}^{+}$ where $X^{\prime \mu }=X^{\mu }{h}_{\nu }^{\mu }$ . Note that $\nabla$ does not mix the holomorphic indices with anti-holomorphic indices, hence $X^{\prime }$ has no components in $T_p{M}^{-}$ . Moreover, $\nabla$ preserves the length of a vector. These facts tell us that $(h_{\mu }^{\nu }(c))$ is contained in $\mathrm{U}(m)\subset \mathrm{O}(2m)$ .

Theorem 8.7

If $g$ is the Ricci-flat metric of an $m$ -dimensional Calabi-Yau manifold $M$ , the holonomy group is contained in $\mathrm{SU}(m)$ .

Proof

Our proof is sketchy. If $X=X^{\mu }\partial /\partial z^{\mu }\in T_p{M}^{+}$ is parallel transported along the small parallelogram in figure 8.5 back to $p$ , we have $X^{\prime }\in T_p{M}^{+}$ whose components are (cf (7.44))

Figure 8.5. $X\in T_p{M}^{+}$ is parallel transported along $pQRS$ and comes back as a vector $X^{\prime }\in T_p{M}^{+}$ .

$$X^{\prime \mu }=X^{\mu }+X^v{R}_{v\kappa \bar{\lambda }}^{\mu }{\epsilon }^{\kappa }{\bar{\delta }}^{\lambda }\text{\hspace{1em}(8.99)}$$

from which we find

$$h_{\mu }^v={\delta }_{\mu }^v+R_{\mu \kappa \bar{\lambda }}^v{\epsilon }^{\kappa }{\bar{\delta }}^{\lambda }.\text{\hspace{1em}(8.100)}$$

$\mathrm{U}(m)$ is decomposed as $\mathrm{U}(m)=\mathrm{SU}(m)\times \mathrm{U}(1)$ in the vicinity of the unit element. In particular, the Lie algebra $\mathfrak{u}(m)=T_e(\mathrm{U}(m))$ is separated into

$$\mathfrak{u}(m)=\mathfrak{su}(m)\oplus \mathfrak{u}(1).\text{\hspace{1em}(8.101)}$$

$\mathfrak{su}(m)$ is the traceless part of $\mathfrak{u}(m)$ while $\mathfrak{u}(1)$ contains the trace. Since the present metric is Ricci flat, the $\mathfrak{u}(1)$ part vanishes,

$$R_{\kappa \mu \overline{v}}^{\kappa }{\epsilon }^{\mu }{\overline{\delta }}^{\nu }={\Re }_{\mu \overline{v}}{\epsilon }^{\mu }{\overline{\delta }}^{\nu }=0.$$

This shows that the holonomy group is contained in $\mathrm{SU}(m)$ . [Remark: Strictly speaking, we have only shown that the restricted holonomy group is contained in $\mathrm{SU}(m)$ . This statement remains true even when $M$ is multiply connected.]

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：8296b75c8a426dd8c16bf09c7dc4e2c8b8d5f0605f8938a10db009f1031210f6