10.4 The covariant derivative on associated vector bundles

A connection one-form $\omega$ on a principal bundle $P(M,G)$ enables us to define the covariant derivative in associated bundles of $P$ in a natural way.

10.4.1 The covariant derivative on associated bundles

In physics, we often need to differentiate sections of a vector bundle which is associated with a certain principal bundle. For example, a charged scalar field in QED is regarded as a section of a complex line bundle associated with a U(1) bundle $P(M,\mathrm{U}(1))$ . Differentiating sections covariantly is very important in constructing gauge-invariant actions.

Let $P(M,G)$ be a $G$ bundle with the projection ${\pi }_P$ . Let us take a chart $U_i$ of $M$ and a section ${\sigma }_i$ over $U_i$ . We take the canonical trivialization ${\varphi }_i(p,e)={\sigma }_i(p)$ . Let $\tilde{\gamma }$ be a horizontal lift of a curve $\gamma :[0,1]\to U_i$ . We denote $\gamma (0)=p_0$ and $\tilde{\gamma }(0)=u_0$ . Associated with $P$ is a vector bundle $E=P{\times }_{\rho }V$ with the projection ${\pi }_E$ , see section 9.4. Let $X\in T_{p}M$ be a tangent vector to $\gamma (t)$ at $p_0$ . Let $s\in \Gamma (M,E)$ be a section, or a vector field, on $M$ . Write an element of $E$ as $[(u,v)]=\{(ug,\rho (g{)}^{-1}v|u\in P,v\in V,g\in G\}$ . Taking a representative of the equivalence class amounts to fixing the gauge. We choose the following form,

$$s(p)=\left[\left({\sigma }_i(p),\xi (p)\right)\right]\text{\hspace{1em}(10.48)}$$

as a representative.

Now we define the parallel transport of a vector in $E$ along a curve $\gamma$ in $M$ . Of course, a naive guess ’ $\xi$ is parallel transported if $\xi (\gamma (t))$ is constant along $\gamma (t{)}^{\prime }$ does not make sense since this statement depends on the choice of the section ${\sigma }_i(p)$ . We define a vector to be parallel transported if it is constant with respect to a horizontal lift $\tilde{\gamma }$ of $\gamma$ in $P$ . In other words, a section $s(\gamma (t))=[(\tilde{\gamma }(t),\eta (\gamma (t)))]$ is parallel transported if $\eta$ is constant along $\gamma (t)$ . This definition is intrinsic since if ${\tilde{\gamma }}^{\prime }(t)$ is another horizontal lift of $\gamma$ , then it can be written as ${\tilde{\gamma }}^{\prime }(t)=\tilde{\gamma }(t)a$ , $a\in G$ and we have (we omit $\rho$ to simplify the notation)

$$[(\tilde{\gamma }(t),\eta (t))]=[({\tilde{\gamma }}^{\prime }(t)a^{-1},\eta (t))]=[({\tilde{\gamma }}^{\prime }(t),a^{-1}\eta (t))]$$

where $\eta (t)$ stands for $\eta (\gamma (t))$ . Hence, if $\eta (t)$ is constant along $\gamma (t)$ , so is its constant multiple $a^{-1}\eta (t)$ .

Now the definition of covariant derivative is in order. Let $s(p)$ be a section of $E$ . Along a curve $\gamma :[0,1]\to M$ we have $s(t)=[(\tilde{\gamma }(t),\eta (t))]$ , where $\tilde{\gamma }(t)$ is an arbitrary horizontal lift of $\gamma (t)$ . The covariant derivative of $s(t)$ along $\gamma (t)$ at $p_0=\gamma (0)$ is defined by

$${\nabla }_{X}s\equiv \left[\left(\tilde{\gamma }(0),\frac{\mathrm{d}}{\mathrm{d}t}\eta (\gamma (t)){|}_{t=0}\right)\right]\text{\hspace{1em}(10.49)}$$

where $X$ is the tangent vector to $\gamma (t)$ at $p_0$ . For the covariant derivative to be really intrinsic, it should not depend on the extra information, that is the special horizontal lift. Let ${\tilde{\gamma }}^{\prime }(t)=\tilde{\gamma }(t)a$ ( $a\in G$ ) be another horizontal lift of $\gamma$ . If ${\tilde{\gamma }}^{\prime }(t)$ is chosen to be the horizontal lift, we have a representative $[({\tilde{\gamma }}^{\prime }(t),a^{-1}\eta (t))]$ . The covariant derivative is now given by

$$\left[\left({\tilde{\gamma }}^{\prime }(0),\frac{\mathrm{d}}{\mathrm{d}t}\{a^{-1}\eta (t)\}{|}_{t=0}\right)\right]=\left[\left({\tilde{\gamma }}^{\prime }(0)a^{-1},\frac{\mathrm{d}}{\mathrm{d}t}\eta (t){|}_{t=0}\right)\right]$$

which agrees with (10.49). Hence, ${\nabla }_{X}s$ depends only on the tangent vector $X$ and the sections $s\in \Gamma (M,E)$ and not on the horizontal lift $\tilde{\gamma }(t)$ . Our definition depends only on a curve $\gamma$ and a connection and not on local trivializations. The local form of the covariant derivative is useful in practical computations and will be given later.

So far we have defined the covariant derivative at a point $p_0=\gamma (0)$ . It is clear that if $X$ is a vector field, ${\nabla }_X$ maps a section $s$ to a new section ${\nabla }_{X}s$ , hence ${\nabla }_X$ is regarded as a map $\Gamma (M,E)\to \Gamma (M,E)$ . To be more precise, take $X\in \mathcal{X}(M)$ whose value at $p$ is $X_p\in T_{p}M$ . There is a curve $\gamma (t)$ such that $\gamma (0)=p$ and its tangent at $p$ is $X_p$ . Then any horizontal lift $\tilde{\gamma }(t)$ of $\gamma$ enables us to compute the covariant derivative ${\nabla }_{X}s{|}_p\equiv {\nabla }_{X_p}s$ . We also define a map $\nabla :\Gamma (M,E)\to \Gamma (M,E)\otimes {\Omega }^1(M)$ by

$$\nabla s(X)\equiv {\nabla }_{X}sX\in \mathcal{X}(M)s\in \Gamma (M,E).\text{\hspace{1em}(10.50)}$$

Exercise 10.8

Show that

$${\nabla }_X\left(a_1{s}_1+a_2{s}_2\right)=a_1{\nabla }_X{s}_1+a_2{\nabla }_X{s}_2\text{\hspace{1em}(10.51a)}$$

$$\nabla \left(a_1{s}_1+a_2{s}_2\right)=a_1\nabla s_1+a_2\nabla s_2\text{\hspace{1em}(10.51b)}$$

$${\nabla }_{\left(a_1{X}_1+a_2{X}_2\right)}s=a_1{\nabla }_{X_1}s+a_2{\nabla }_{X_2}s\text{\hspace{1em}(10.51c)}$$

$${\nabla }_X(fs)=X[f]s+f{\nabla }_{X}s\text{\hspace{1em}(10.51d)}$$

$$\nabla (fs)=(\mathrm{d}f)s+f\nabla s\text{\hspace{1em}(10.51e)}$$

$${\nabla }_{fX}s=f{\nabla }_{X}s\text{\hspace{1em}(10.51f)}$$

where $a_i\in \mathbb{R},s,s^{\prime }\in \Gamma (M,E)$ and $f\in \mathcal{F}(M)$

10.4.2 A local expression for the covariant derivative

In practical computations it is convenient to have a local coordinate representation of the covariant derivative. Let $P(M,G)$ be a $G$ bundle and $E=P{\times }_{\rho }G$ be an associate vector bundle. Take a local section ${\sigma }_i\in \Gamma (U_i,P)$ and employ the canonical trivialization ${\sigma }_i(p)={\varphi }_i(p,e)$ . Let $\gamma :[0,1]\to M$ be a curve in $U_i$ and $\tilde{\gamma }$ its horizontal lift, which is written as

$$\tilde{\gamma }(t)={\sigma }_i(t)g_i(t)\text{\hspace{1em}(10.52)}$$

where $g_i(t)\equiv g_i(\gamma (t))\in G$ . Take a section $e_{\alpha }(p)\equiv [({\sigma }_i(p),e_{\alpha }^0)]$ of $E$ , where $e_{\alpha }^0$ is the $\alpha$ th basis vector of $V$ ; $(e_{\alpha }^0{)}^{\beta }=({\delta }_{\alpha }{)}^{\beta }$ . We have

$$e_{\alpha }(t)=\left[\left(\tilde{\gamma }(t)g_i(t{)}^{-1},e_{\alpha }^0\right)\right]=\left[\left(\tilde{\gamma }(t),g_i(t{)}^{-1}{e}_{\alpha }^0\right)\right].\text{\hspace{1em}(10.53)}$$

Note that $g_i(t{)}^{-1}$ acts on $e_{\alpha }^0$ to compensate for the change of basis along $\gamma$ . The covariant derivative of $e_{\alpha }$ is then given by

$$\begin{array}{rl}{\nabla }_X{e}_{\alpha }& =\left[\left(\tilde{\gamma }(0),\frac{\mathrm{d}}{\mathrm{d}t}\{g_i(t{)}^{-1}{e}_{\alpha }^0\}{|}_{t=0}\right)\right]\\ & =\left[\left(\tilde{\gamma }(0),-g_i(t{)}^{-1}\left\{\frac{\mathrm{d}}{\mathrm{d}t}{g}_i(t)\right\}{g}_i(t{)}^{-1}{e}_{\alpha }^0{|}_{t=0}\right)\right]\\ & =\left[\left(\tilde{\gamma }(0)g_i(0{)}^{-1},{\mathcal{A}}_i(X)e_{\alpha }^0\right)\right]\end{array}\text{\hspace{1em}(10.54)}$$

where (10.13b) has been used. From (10.54) we find the local expression,

$${\nabla }_X{e}_{\alpha }=\left[\left({\sigma }_i(0),{\mathcal{A}}_i(X)e_{\alpha }^0\right)\right].\text{\hspace{1em}(10.55)}$$

Let ${\mathcal{A}}_i={\mathcal{A}}_{i\mu }\mathrm{d}{x}^{\mu }={\mathcal{A}}_{i\mu }{}^{\alpha }{}_{\beta }\mathrm{d}{x}^{\mu }$ where ${\mathcal{A}}_{i\mu }{}^{\alpha }{}_{\beta }\equiv {\mathcal{A}}_{i\mu }{}^{\gamma }(T_{\gamma }{)}^{\alpha }{}_{\beta }$ . The second entry of (10.55) is

$${\mathcal{A}}_i(X)e_{\alpha }^0=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}{e}_{\beta }^0{\mathcal{A}}^{\beta }{\delta }_{\alpha }^{\gamma }=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}{\mathcal{A}}^{\beta }{e}_{\beta }^0.$$

Substituting this into (10.55), we finally have

$${\nabla }_X{e}_{\alpha }=\left[\left({\sigma }_i(0),\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}{\mathcal{A}}_{i\mu }{}^{\beta }{}_{\alpha }{e}_{\beta }^0\right)\right]=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}{\mathcal{A}}_{i\mu }{}^{\beta }{}_{\alpha }{e}_{\beta }\text{\hspace{1em}(10.56a)}$$

or

$$\nabla e_{\alpha }={\mathcal{A}}_i{}^{\beta }{}_{\alpha }{e}_{\beta }.\text{\hspace{1em}(10.56b)}$$

In particular, for a coordinate curve $x^{\mu }$ , we have

$${\nabla }_{\partial /\partial x^{\mu }}{e}_{\alpha }={\mathcal{A}}_{i\mu }{}^{\beta }{}_{\alpha }{e}_{\beta }.\text{\hspace{1em}(10.57)}$$

It is remarkable that a connection $\mathcal{A}$ on a principal bundle $P$ completely specifies the covariant derivative on an associated bundle $E$ (modulo representations).

Exercise 10.9

Let $s(p)=[({\sigma }_i(p),{\xi }_i(p))]={\xi }_i^{\alpha }(p)e_{\alpha }$ be a general section of $E$ , where ${\xi }_i(p)={\xi }_i^{\alpha }(p)e_{\alpha }^0$ . Use the results of exercise 10.8 to verify that

$${\nabla }_{X}s=\left[\left({\sigma }_i(0),\frac{\mathrm{d}{\xi }_i}{\mathrm{d}t}+{\mathcal{A}}_i(X){\xi }_i{|}_{t=0}\right)\right]=\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}\left\{\frac{\partial {\xi }_i^{\alpha }}{\partial x^{\mu }}+{\mathcal{A}}^{\alpha }{\xi }_i^{\beta }\right\}{e}_{\alpha }.\text{\hspace{1em}(10.58)}$$

By construction, the covariant derivative is independent of the local trivialization. This is also observed from the local form of ${\nabla }_{X}s$ . Let ${\sigma }_i(p)$ and ${\sigma }_j(p)$ be local sections on overlapping charts $U_i$ and $U_j$ . On $U_i\cap U_j$ , we have ${\sigma }_j(p)={\sigma }_i(p)t_{ij}(p)$ . In the $i$ -trivialization, the covariant derivative is

$$\begin{array}{rl}{\nabla }_{X}s& =\left[\left({\sigma }_i(0),\frac{\mathrm{d}{\xi }_i}{\mathrm{d}t}+{\mathcal{A}}_i(X){\xi }_i{|}_{t=0}\right)\right]\\ & =\left[\left({\sigma }_j(0)\cdot t_{ij}^{-1},\frac{\mathrm{d}}{\mathrm{d}t}\left(t_{ij}{\xi }_j\right)+{\mathcal{A}}_i(X)t_{ij}{\xi }_j{|}_{t=0}\right)\right]\\ & =\left[{\left({\sigma }_j(0),\frac{\mathrm{d}{\xi }_j}{\mathrm{d}t}+{\mathcal{A}}_j(X){\xi }_j\right|}_{t=0})\right]\end{array}\text{\hspace{1em}(10.59)}$$

where use has been made of the condition (10.9). The last line of (10.59) is ${\nabla }_{X}s$ expressed in the $j$ -trivialization.

We have found that the covariant derivative defined by (10.49) is independent of the horizontal lift as well as the local section. The gauge potential ${\mathcal{A}}_i$ transforms under the change of local trivialization so that ${\nabla }_{X}s$ is a well-defined section of $E$ . In this sense, ${\nabla }_X$ is the most natural derivative on an associated vector bundle, which is compatible with the connection on the principal bundle $P$ .

Example 10.4

Let us recover the results obtained in section 7.2. Let $FM$ be a frame bundle over $M$ and let $TM$ be its associated bundle. We note $FM=P(M,\mathrm{GL}(m,\mathbb{R}))$ and $TM=FM{\times }_{\rho }{\mathbb{R}}^m$ , where $m=\dim M$ and $\rho$ is the $m\times m$ matrix representation of $\mathrm{GL}(m,\mathbb{R})$ . Elements of $\mathfrak{gl}(m,\mathbb{R})$ are $m\times m$ matrices. Let us rewrite the local connection form $A_i$ as ${\Gamma }_{\mu \beta }^{\alpha }\mathrm{d}{x}^{\mu }$ . We then find that

$${\nabla }_{\partial /\partial x^{\mu }}{e}_{\alpha }=[({\sigma }_i(0),{\Gamma }_{\mu }{e}_{\alpha }^0)]={\Gamma }_{\mu \alpha }^{\beta }{e}_{\beta }\text{\hspace{1em}(10.60)}$$

which should be compared with (7.14). For a general section (vector field), $s(p)=[({\sigma }_i(p),X_i(p))]=X_i^{\alpha }(p)e_{\alpha }$ , we find

$${\nabla }_{\partial /\partial x^{\mu }s}=\left(\frac{\partial }{\partial x^{\mu }}{X}_i^{\alpha }+{\Gamma }_{\mu \beta }^{\alpha }{X}^{\beta }\right)e_{\alpha }\text{\hspace{1em}(10.61)}$$

which reproduces the result of section 7.2. It is evident that the roles played by the indices $\alpha ,\beta$ and $\mu$ in ${\Gamma }_{\mu \beta }^{\alpha }$ are very different in their characters; $\mu$ is the ${\Omega }^1(M)$ index while $\alpha$ and $\beta$ are the $\mathfrak{gl}(m,\mathbb{R})$ indices.

Example 10.5

Let us consider the U(1) gauge field coupled to a complex scalar field $\varphi$ . The relevant fibre bundles are the U(1) bundle $P(M,\mathrm{U}(1))$ and the associated bundle $E=P{\times }_{\rho }\mathbb{C}$ where $\rho$ is the natural identification of an element of $\mathrm{U}(1)$ with a complex number. The local expression for $\omega$ is ${\mathcal{A}}_i={\mathcal{A}}_{i\mu }\mathrm{d}{x}^{\mu }$ , where ${\mathcal{A}}_{i\mu }={\mathcal{A}}_i(\partial /\partial x^{\mu })$ is the vector potential of Maxwell’s theory. Let $\gamma$ be a curve in $M$ with tangent vector $X$ at $\gamma (0)$ . Take a local section ${\sigma }_i$ and express a horizontal lift $\tilde{\gamma }$ of $\gamma$ as $\tilde{\gamma }(t)={\sigma }_i(t){\mathrm{e}}^{\mathrm{i}\phi (t)}$ . If $1\in \mathbb{C}$ is taken to be the basis vector, the basis section is

$$e=[({\sigma }_i(p),1)].$$

Let $\varphi (p)=[({\sigma }_i(p),\Phi (p))]=\Phi (p)e$ ( $\Phi :M\to \mathbb{C}$ ) be a section of $E$ , which is identified with a complex scalar field. With respect to $\tilde{\gamma }(t)$ , the section is given by

$$\varphi (t)=\Phi (t)[(\tilde{\gamma }(t),U(t{)}^{-1})]\text{\hspace{1em}(10.62)}$$

where $U(t)={\mathrm{e}}^{\mathrm{i}\phi (t)}$ . The covariant derivative of $\varphi$ along $\gamma$ is

$$\begin{array}{rl}{\nabla }_X\varphi & =\frac{\mathrm{d}\Phi }{\mathrm{d}t}\left[\left(\tilde{\gamma }(0),U(0{)}^{-1}\right)\right]+\Phi (0)\left[\left(\tilde{\gamma }(0),U(0{)}^{-1}{\mathcal{A}}_i(X)\cdot 1\right)\right]\\ & =\left(\frac{\mathrm{d}\Phi }{\mathrm{d}t}+{\mathcal{A}}_{i\mu }\Phi \frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}\right)e=X^{\mu }\left(\frac{\partial \Phi }{\partial x^{\mu }}+{\mathcal{A}}_{i\mu }\Phi \right)e.\end{array}\text{\hspace{1em}(10.63)}$$

Example 10.6

Let us consider the SU(2) Yang-Mills theory on $M$ . The relevant bundles are the SU(2) bundle $P(M,\mathrm{SU}(2))$ and its associated bundle $E=P{\times }_{\rho }{\mathbb{C}}^2$ , where we have taken the two-dimensional representation. The gauge potential on a chart $U_i$ is

$${\mathcal{A}}_i={\mathcal{A}}_{i\mu }\mathrm{d}{x}^{\mu }=A_{i\mu }^{\alpha }\left(\frac{{\sigma }_{\alpha }}{2\mathrm{i}}\right)\mathrm{d}{x}^{\mu }\text{\hspace{1em}(10.64)}$$

where ${\sigma }_{\alpha }/2\mathrm{i}$ are generators of $\mathrm{SU}(2)$ , ${\sigma }_{\alpha }$ being the Pauli matrices. Let $e_{\alpha }^0$ $(\alpha =1,2)$ be basis vectors of ${\mathbb{C}}^2$ and consider sections

$$e_{\alpha }(p)\equiv \left[\left({\sigma }_i(p),e_{\alpha }^0\right)\right]\text{\hspace{1em}(10.65)}$$

where ${\sigma }_i(p)$ defines a canonical trivialization of $P$ over $U_i$ . Let $\varphi (p)=[({\sigma }_i(p),{\Phi }^{\alpha }(p)e_{\alpha }^0)]$ be a section of $E$ over $M$ . Along a horizontal lift $\tilde{\gamma }(t)={\sigma }_i(p)U(t)$ , $U(t)\in \mathrm{SU}(2)$ , we have

$$\varphi (t)=\left[\left(\tilde{\gamma }(t),U(t{)}^{-1}{\Phi }^{\alpha }(t)e_{\alpha }^0\right)\right].\text{\hspace{1em}(10.66)}$$

The covariant derivative of $\varphi$ along $X=\mathrm{d}/\mathrm{d}t$ is

$$\begin{array}{rl}& {\nabla }_X\varphi =\left[\left(\tilde{\gamma }(0),U(0{)}^{-1}\frac{\mathrm{d}{\Phi }^{\alpha }(0)}{\mathrm{d}t}{e}_{\alpha }^0\right)\right]\\ & +\left[(\tilde{\gamma }(0),U(0{)}^{-1}{\mathcal{A}}_i(X{)}^{\alpha }{\Phi }^{\beta }(0)e_{\alpha }^0)\right]\\ & =X^{\mu }\left(\frac{\partial {\Phi }^{\alpha }}{\partial x^{\mu }}+{\mathcal{A}}^{\alpha }{\Phi }^{\beta }\right)e_{\alpha }\end{array}\text{\hspace{1em}(10.67)}$$

where (10.13b) has been used to obtain the last equality.

Exercise 10.10

Let us consider an associated adjoint bundle $E_{\mathfrak{g}}=P{\times }_{\mathrm{Ad}\mathfrak{g}}$ where the action of $G$ on $\mathfrak{g}$ is the adjoint action $V\to {\mathrm{Ad}}_{g}V=g^{-1}Vg$ , $V\in \mathfrak{g}$ and $g\in G$ . Take a local section ${\sigma }_i\in \Gamma (U_i,P)$ such that $\tilde{\gamma }(t)={\sigma }_i(t)g(t)$ . Take a section $s(p)=[(s_i(p),V(p)))]$ on $E_{\mathfrak{g}}$ , where $V(p)=V^{\alpha }(p)T_{\alpha }$ , $\{T_{\alpha }\}$ being the basis of $\mathfrak{g}$ . Define the covariant derivative ${\mathcal{D}}_{X}s$ by

$${\mathcal{D}}_{X}s\equiv \left[\left(\tilde{\gamma }(0),\frac{\mathrm{d}}{\mathrm{d}t}\left\{{\mathrm{Ad}}_{g(t{)}^{-1}}V(t)\right\}{|}_{t=0}\right)\right].\text{\hspace{1em}(10.68a)}$$

Show that

$$\begin{array}{rl}{\mathcal{D}}_{X}s& =\left[\left({\sigma }_i(0),\frac{\mathrm{d}V(t)}{\mathrm{d}t}+[{\mathcal{A}}_i(X),V(t)]{|}_{t=0}\right)\right]\\ & =X^{\mu }\left(\frac{\partial V^{\alpha }}{\partial x^{\mu }}+f_{\beta \gamma }^{\alpha }{\mathcal{A}}_{i\mu }^{\beta }{V}^{\gamma }\right)[({\sigma }_i(0),T_{\alpha })].\end{array}\text{\hspace{1em}(10.68b)}$$

10.4.3 Curvature rederived

The covariant derivative ${\nabla }_{X}s$ defines an operator $\nabla :\Gamma (M,E)\to \Gamma (M,E\otimes {\Omega }^1(M))$ by (10.50). More generally, the action of $\nabla$ on a vector-valued $p$ -form $s\otimes \eta$ , $\eta \in {\Omega }^p(M)$ , is defined by

$$\nabla (s\otimes \eta )\equiv (\nabla s)\wedge \eta +s\otimes d\eta .\text{\hspace{1em}(10.69)}$$

Let $U_i$ be a chart of $M$ and ${\sigma }_i$ a section of $P$ over $U_i$ . We take the canonical local trivialization over $U_i$ . We now prove

$$\nabla \nabla e_{\alpha }=e_{\beta }\otimes {\mathfrak{F}}_i{}^{\beta }{}_{\alpha }\text{\hspace{1em}(10.70)}$$

where $e_{\alpha }=[({\sigma }_i,e_{\alpha }^0)]\in \Gamma (U_i,E)$ . In fact, by straightforward computation, we find

$$\begin{array}{rl}\nabla \nabla e_{\alpha }& =\nabla \left(e_{\beta }\otimes {\mathcal{A}}_i{}^{\beta }{}_{\alpha }\right)=\nabla e_{\beta }\wedge {\mathcal{A}}_i{}^{\beta }{}_{\alpha }+e_{\beta }\otimes \mathrm{d}{\mathcal{A}}_i{}^{\beta }{}_{\alpha }\\ & =e_{\beta }\otimes \left(\mathrm{d}{\mathcal{A}}_i{}^{\beta }{}_{\alpha }+{\mathcal{A}}_i{}^{\beta }{}_{\gamma }\wedge {\mathcal{A}}_i{}^{\gamma }{}_{\alpha }\right)=e_{\beta }\otimes {\mathfrak{F}}_i{}^{\beta }{}_{\alpha }.\end{array}$$

Exercise 10.11

Let $s(p)={\xi }^{\alpha }(p)e_{\alpha }(p)$ be a section of $E$ . Show that

$$\nabla \nabla s=e_{\alpha }\otimes {\mathcal{F}}_i{}^{\alpha }{}_{\beta }{\xi }^{\beta }.\text{\hspace{1em}(10.71)}$$

10.4.4 A connection which preserves the inner product

Let $E\stackrel{\pi }{\to }M$ be a vector bundle with a positive-definite symmetric inner product whose action is defined at each point $p\in M$ by

$$g_p:{\pi }^{-1}(p)\otimes {\pi }^{-1}(p)\to \mathbb{R}.\text{\hspace{1em}(10.72)}$$

Then $g$ is said to define a Riemannian structure on $E$ . A connection $\nabla$ is called a metric connection if it preserves the inner product,

$$\mathrm{d}[g(s,s^{\prime })]=g(\nabla s,s^{\prime })+g(s,\nabla s^{\prime }).\text{\hspace{1em}(10.73)}$$

In particular, if we take $s=e_{\alpha },s^{\prime }=e_{\beta }$ and set $g(e_{\alpha },e_{\beta })=g_{\alpha \beta }$ , we find

$$\mathrm{d}{g}_{\alpha \beta }={\mathcal{A}}_i{}^{\gamma }{}_{\alpha }{g}_{\gamma \beta }+{\mathcal{A}}_i{}^{\gamma }{}_{\beta }{g}_{\alpha \gamma }.\text{\hspace{1em}(10.74)}$$

This should be compared with (7.30b). If $E=TM$ and, moreover, the torsion-free condition is imposed, our connection reduces to the Levi-Civita connection of the Riemannian geometry.

Given an inner product, we may take an orthonormal frame $\{{\hat{e}}_{\alpha }\}$ such that $g({\hat{e}}_{\alpha },{\hat{e}}_{\beta })={\delta }_{\alpha \beta }$ . The structure group $G$ is taken to be $\mathrm{O}(k),k$ being the dimension of the fibre. The Lie algebra $\mathfrak{o}(k)$ is a vector space of skew symmetric matrices and the connection one-form $\omega$ satisfies

$${\omega }_{\beta }^{\alpha }=-{\omega }_{\alpha }^{\beta }.\text{\hspace{1em}(10.75)}$$

Theorem 10.6

Let $E$ be a vector bundle with inner product $g$ and let $\nabla$ be the covariant derivative associated with the orthonormal frame. Then $\nabla$ is a metric connection.

Proof

Since $g$ is bilinear, it suffices to show that

$$\mathrm{d}\left[g\left(s,s^{\prime }\right)\right]=g\left(\nabla s,s^{\prime }\right)+g\left(s,\nabla s^{\prime }\right)$$

for $s=f{\hat{e}}_{\alpha }$ and $s^{\prime }=f^{\prime }{\hat{e}}_{\beta }$ where $f,f^{\prime }\in \mathcal{F}(M)$ . In fact, the LHS is $\mathrm{d}[g(f{\hat{e}}_{\alpha },f^{\prime }{\hat{e}}_{\beta })]=\mathrm{d}[f{f}^{\prime }{\delta }_{\alpha \beta }]=\mathrm{d}(f{f}^{\prime }){\delta }_{\alpha \beta }$ while the RHS is

$$\begin{array}{rl}& g\left(\nabla f{\hat{e}}_{\alpha },f^{\prime }{\hat{e}}_{\beta }\right)+g\left(f{\hat{e}}_{\alpha },\nabla f^{\prime }{\hat{e}}_{\beta }\right)\\ & =g\left(\mathrm{d}f{\hat{e}}_{\alpha }+f{\hat{e}}_{\gamma }{\omega }_{\alpha }^{\gamma },f^{\prime }{\hat{e}}_{\beta }\right)+g\left(f{\hat{e}}_{\alpha },\mathrm{d}{f}^{\prime }{\hat{e}}_{\beta }+f^{\prime }{\hat{e}}_{\gamma }{\omega }_{\beta }^{\gamma }\right)\\ & =\mathrm{d}f{f}^{\prime }{\delta }_{\alpha \beta }+f{f}^{\prime }{\omega }_{\alpha }^{\gamma }{\delta }_{\gamma \beta }+f\mathrm{d}{f}^{\prime }{\delta }_{\alpha \beta }+f{f}^{\prime }{\omega }_{\beta }^{\gamma }{\delta }_{\alpha \gamma }\\ & =\mathrm{d}(f{f}^{\prime }){\delta }_{\alpha \beta }\end{array}$$

where (10.75) has been used to obtain the final equality.

10.4.5 Holomorphic vector bundles and Hermitian inner products

Definition 10.6

Let $E$ and $M$ be complex manifolds and $\pi :E\to M$ a holomorphic surjection. The manifold $E$ is a holomorphic vector bundle if the following axioms are fulfilled.

(i) The typical fibre is ${\mathbb{C}}^k$ and the structure group is $\mathrm{GL}(k,\mathbb{C})$ .
(ii) The local trivialization ${\varphi }_i:U_i\times {\mathbb{C}}^k\to {\pi }^{-1}(U_i)$ is a biholomorphism.

(iii) The transition function $t_{ij}:U_i\cap U_j\to G=\mathrm{GL}(k,\mathbb{C})$ is a holomorphic map.

For example, let $M$ be a complex manifold with ${\dim }_{\mathbb{C}}M=m$ . The holomorphic tangent bundle $T{M}^{+}\equiv {\bigcup }_{p\in M}{T}_p{M}^{+}$ is a holomorphic vector bundle. The typical fibre is ${\mathbb{C}}^m$ and the local basis is $\{\partial /\partial z^{\mu }\}$ .

Let $h$ be an inner product on a holomorphic vector bundle whose action at $p\in M$ is $h_p:{\pi }^{-1}(p)\times {\pi }^{-1}(p)\to \mathbb{C}$ . The most natural inner product is a Hermitian structure which satisfies:

(i) $h_p(u,av+bw)=a{h}_p(u,v)+b{h}_p(u,w)$ , for $u,v,w\in {\pi }^{-1}(p),a,b\in \mathbb{C};$
(ii) $h_p(u,v)=\overline{h_p(v,u)},u,v\in {\pi }^{-1}(p);$
(iii) $h_p(u,u)\ge 0$ , $h_p(u,u)=0$ if and only if $u={\varphi }_i(p,0)$ ; and
(iv) $h(s_1,s_2)\in \mathcal{F}(M{)}^{\mathbb{C}}$ for $s_1,s_2\in \Gamma (M,E)$

A set of sections $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ is a unitary frame if

$$h\left({\hat{e}}_i,{\hat{e}}_j\right)={\delta }_{ij}.\text{\hspace{1em}(10.76)}$$

The unitary frame bundle $LM$ is not a holomorphic vector bundle since the structure group $\mathrm{U}(m)$ is not a complex manifold.

Given a Hermitian structure $h$ , we define a connection which is compatible with $h$ . The Hermitian connection $\nabla$ is a linear map $\Gamma (M,E)\to \Gamma (M,E\otimes T^{\ast }{M}^{\mathbb{C}})$ which satisfies:

(i) $\nabla (fs)=(\mathrm{d}f)s+f\nabla s,f\in \mathcal{F}(M{)}^{\mathbb{C}},s\in \Gamma (M,E);$
(ii) $\mathrm{d}[h(s_1,s_2)]=h(\nabla s_1,s_2)+h(s_1,\nabla s_2);$ and
(iii) according to the destination, we separate the action of $\nabla$ as $\nabla s=\mathrm{D}s+\bar{\mathrm{D}}s$ , $\mathrm{D}s$ ( $\bar{\mathrm{D}}s$ ) being a $(1,0)$ -form ((0,1)-form) valued section. We demand that $\bar{\mathrm{D}}=\bar{\partial }$ .

It can be shown that given $E$ and a Hermitian metric $h$ , there exists a unique Hermitian connection $\nabla$ . The curvature is defined from the Hermitian connection. Let $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ be a unitary frame and define the local connection form ${\mathcal{A}}_{\alpha }^{\beta }$ by

$$\nabla {\hat{e}}_{\alpha }={\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.77)}$$

The field strength is defined by

$$\mathcal{F}\equiv \mathrm{d}\mathcal{A}+\mathcal{A}\wedge \mathcal{A}.\text{\hspace{1em}(10.78)}$$

We verify that

$$\nabla \nabla {\hat{e}}_{\alpha }=\nabla \left({\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }\right)={\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.79)}$$

We prove that both $\mathcal{A}$ and $\mathcal{F}$ are skew Hermitian:

$${\bar{\mathcal{A}}}_{\alpha }^{\beta }+{\mathcal{A}}_{\beta }^{\alpha }=h(\nabla {\hat{e}}_{\alpha },{\hat{e}}_{\beta })+h({\hat{e}}_{\alpha },\nabla {\hat{e}}_{\beta })=\mathrm{d}h({\hat{e}}_{\alpha },{\hat{e}}_{\beta })=\mathrm{d}{\delta }_{\alpha \beta }=0$$

$$\begin{array}{rl}{\mathcal{F}}_{\alpha }^{\beta }+{\bar{\mathcal{F}}}_{\beta }^{\alpha }& =\mathrm{d}{A}_{\alpha }^{\beta }+A_{\gamma }^{\beta }\wedge A_{\alpha }^{\gamma }+\mathrm{d}{\bar{A}}_{\beta }^{\alpha }+{\bar{A}}_{\gamma }^{\alpha }\wedge {\bar{A}}_{\alpha }^{\gamma }\\ & =\mathrm{d}\left({\mathcal{A}}_{\alpha }^{\beta }-{\mathcal{A}}_{\alpha }^{\beta }\right)+{\mathcal{A}}_{\gamma }^{\beta }\wedge {\mathcal{A}}_{\alpha }^{\gamma }+{\mathcal{A}}_{\alpha }^{\gamma }\wedge {\mathcal{A}}_{\gamma }^{\alpha }=0.\end{array}$$

Thus, we have shown that

$${\mathcal{A}}_{\beta }^{\alpha }=-{\bar{\mathcal{A}}}_{\alpha }^{\beta }{\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }.\text{\hspace{1em}(10.80)}$$

Next we show that $\mathcal{F}$ is a $(1,1)$ -form. Let $\{{\hat{e}}_{\alpha }\}$ be a unitary frame. $\mathcal{F}$ cannot have a component of bidegree- $(0,2)$ since

$${\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }=\nabla \nabla {\hat{e}}_{\alpha }=(\mathrm{D}+\bar{\partial })(\mathrm{D}+\bar{\partial }){\hat{e}}_{\alpha }=\mathrm{DD}{\hat{e}}_{\alpha }+(\mathrm{D}\bar{\partial }+\bar{\partial }\mathrm{D}){\hat{e}}_{\alpha }.$$

It follows from ${\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }^{\alpha }$ that $\bar{\mathcal{F}}$ has no component of bidegree-(0, 2), and, hence, $\mathcal{F}$ has no component of bidegree-(2, 0) either. Thus ${\mathcal{F}}_{\alpha }^{\beta }$ is a two-form of bidegree-(1, 1).

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-256：9a2c71f0a38326e2a18f5d3faa21fa36a935259f7b31ef0745addc53a7c16480