10.1 Connections on principal bundles

There are several equivalent definitions of a connection on a principal bundle. Our approach is based on the separation of tangent space $T_{u}P$ into ‘vertical’ and ‘horizontal’ subspaces. Although this approach seems to be abstract, it is advantageous compared with other approaches in that it clarifies the geometrical pictures involved and is defined independently of special local trivializations. Connections are also defined as $\mathfrak{g}$ -valued one-forms which satisfy certain axioms. These definitions are shown to be equivalent.

We briefly summarize the basic facts on Lie groups and Lie algebras, since we shall make extensive use of these (see section 5.6 for details). Let $G$ be a Lie group. The left action $L_g$ and the right action $R_g$ are defined by $L_{g}h=gh$ and $R_{g}h=hg$ for $g,h\in G$ . $L_g$ induces a map $L_{g\ast }:T_h(G)\to T_{gh}(G)$ . A left-invariant vector field $X$ satisfies $L_{g\ast }X{|}_h=X{|}_{gh}$ . Left-invariant vector fields form a Lie algebra of $G$ , denoted by $\mathfrak{g}$ . Since $X\in \mathfrak{g}$ is specified by its value at the unit element $e$ , and vice versa, there exists a vector space isomorphism $\mathfrak{g}\cong T_{e}G$ . The Lie algebra $\mathfrak{g}$ is closed under the Lie bracket, $[T_{\alpha },T_{\beta }]=f_{\alpha \beta }^{\gamma }{T}_{\gamma }$ where $\{T_{\alpha }\}$ is the set of generators of $\mathfrak{g}$ . $f_{\alpha \beta }^{\gamma }$ are called the structure constants. The adjoint action $\mathrm{ad}:G\to G$ is defined by ${\mathrm{ad}}_{g}h\equiv gh{g}^{-1}$ . The tangent map of ${\mathrm{ad}}_g$ is

called the adjoint map and is denoted by ${\mathrm{Ad}}_g:T_h(G)\to T_{gh{g}^{-1}}(G)$ . If restricted to $T_e(G)\simeq \mathfrak{g}$ , ${\mathrm{Ad}}_g$ maps $\mathfrak{g}$ onto itself; ${\mathrm{Ad}}_g:\mathfrak{g}\to \mathfrak{g}$ as $A\mapsto gA{g}^{-1}$ , $A\in \mathfrak{g}$ .

10.1.1 Definitions

Let $u$ be an element of a principal bundle $P(M,G)$ and let $G_p$ be the fibre at $p=\pi (u)$ . The vertical subspace $V_{u}P$ is a subspace of $T_{u}P$ which is tangent to $G_p$ at $u$ . [Warning: $T_{u}P$ is the tangent space of $P$ and should not be confused with the tangent space $T_{p}M$ of $M$ .] Let us see how $V_{u}P$ is constructed. Take an element $A$ of $\mathfrak{g}$ . By the right action

$$R_{\exp (tA)}u=u\exp (tA)$$

a curve through $u$ is defined in $P$ . Since $\pi (u)=\pi (u\exp (tA))=p$ , this curve lies within $G_p$ . Define a vector $A^{\#}\in T_{u}P$ by

$$A^{\#}f(u)=\frac{\mathrm{d}}{\mathrm{d}t}f(u\exp (tA)){|}_{t=0}\text{\hspace{1em}(10.1)}$$

where $f:P\to \mathbb{R}$ is an arbitrary smooth function. The vector $A^{\#}$ is tangent to $P$ at $u$ , hence $A^{\#}\in V_{u}P$ . In this way we define a vector $A^{\#}$ at each point of $P$ and construct a vector field $A^{\#}$ , called the fundamental vector field generated by $A$ . There is a vector space isomorphism $♯:\mathfrak{g}\to V_{u}P$ given by $A\mapsto A^{\#}$ . The horizontal subspace $H_{u}P$ is a complement of $V_{u}P$ in $T_{u}P$ and is uniquely specified if a connection is defined in $P$ .

Exercise 10.1

(a) Show that ${\pi }_{\ast }X=0$ for $X\in V_{\mu }P$
(b) Show that $♯$ preserves the Lie algebra structure:

$$[A^{\#},B^{\#}]=[A,B{]}^{\#}.\text{\hspace{1em}(10.2)}$$

Definition 10.1

Let $P(M,G)$ be a principal bundle. A connection on $P$ is a unique separation of the tangent space $T_{u}P$ into the vertical subspace $V_{u}P$ and the horizontal subspace $H_{u}P$ such that

(i) $T_{u}P=H_{u}P\oplus V_{u}P$
(ii) A smooth vector field $X$ on $P$ is separated into smooth vector fields $X^H\in H_{\mu }P$ and $X^V\in V_{\mu }P$ as $X=X^H+X^V$ .
(iii) $H_{u\varrho }P=R_{\varrho \ast }{H}_{u}P$ for arbitrary $u\in P$ and $g\in G$ ; see figure 10.1.

Figure 10.1. The horizontal subspace $H_{ug}P$ is obtained from $H_{u}P$ by the right action.

The condition (iii) states that horizontal subspaces $H_{u}P$ and $H_{ug}P$ on the same fibre are related by a linear map $R_{g\ast }$ induced by the right action. Accordingly, a subspace $H_{u}P$ at $u$ generates all the horizontal subspaces on the same fibre. This condition ensures that if a point $u$ is parallel transported, so is its constant multiple $ug$ , $g\in G$ ; see later. At this point, the reader might feel rather

uneasy about our definition of a connection. At first sight, this definition seems to have nothing to do with the gauge potential or the field strength. We clarify these points after we introduce the connection one-form on $P$ . We again stress that our definition, which is based on the separation $T_{u}P=V_{u}P\oplus H_{u}P$ , is purely geometrical and is defined independently of any extra information. Although the connection becomes more tractable in the following, the geometrical picture and its intrinsic nature are generally obscured.

10.1.2 The connection one-form

In practical computations, we need to separate $T_{u}P$ into $V_{u}P$ and $H_{u}P$ in a systematic way. This can be achieved by introducing a Lie-algebra-valued one-form $\omega \in \mathfrak{g}\otimes T^{\ast }P$ called the connection one-form.

Definition 10.2

A connection one-form $\omega \in \mathfrak{g}\otimes T^{\ast }P$ is a projection of $T_{u}P$ onto the vertical component $V_{u}P\simeq \mathfrak{g}$ . The projection property is summarized by the following requirements,

(i) $\omega (A^{\#})=AA\in \mathfrak{g}$ (10.3a)
(ii) $R_g^{\ast }\omega ={\mathrm{Ad}}_{g^{-1}}\omega$ (10.3b)

that is, for $X\in T_{u}P$

$$R_g^{\ast }{\omega }_{ug}(X)={\omega }_{ug}\left(R_{g\ast }X\right)=g^{-1}{\omega }_u(X)g.\text{\hspace{1em}(10.3b’)}$$

Define the horizontal subspace $H_{u}P$ by the kernel of $\omega$ ,

$$H_{u}P\equiv \{X\in T_{u}P|\omega (X)=0\}.\text{\hspace{1em}(10.4)}$$

To show that this definition is consistent with definition 10.1, we prove the following proposition.

Proposition 10.1

The horizontal subspaces (10.4) satisfy

$$R_{g\ast }{H}_{u}P=H_{ug}P.\text{\hspace{1em}(10.5)}$$

Proof

Fix a point $u\in P$ and define $H_{u}P$ by (10.4). Take $X\in H_{u}P$ and construct $R_{g\ast }X\in T_{ug}P$ . We find

$$\omega \left(R_{g^{\ast }}X\right)=R_g^{\ast }\omega (X)=g^{-1}\omega (X)g=0$$

since $\omega (X)=0$ . Accordingly, $R_{g\ast }X\in H_{ug}P$ . We note that $R_{g\ast }$ is an invertible linear map. Hence, any vector $Y\in H_{ug}P$ is expressed as $Y=R_{g\ast }X$ for some $X\in H_{u}P$ . This proves (10.5).

We have shown that the definition of the connection one-form $\omega$ is equivalent to that of the connection, since $\omega$ separates $T_{u}P$ into $H_{u}P\oplus V_{u}P$ in harmony with the axioms of definition 10.1. The connection one-form $\omega$ defined here is known as the Ehresmann connection in the literature.

10.1.3 The local connection form and gauge potential

Let $\{U_i\}$ be an open covering of $M$ and let ${\sigma }_i$ be a local section defined on each $U_i$ . It is convenient to introduce a Lie-algebra-valued one-form ${\mathcal{A}}_i$ on $U_i$ , by

$${\mathcal{A}}_i\equiv {\sigma }_i^{\ast }\omega \in \mathfrak{g}\otimes {\Omega }^1\left(U_i\right).\text{\hspace{1em}(10.6)}$$

Conversely, given a Lie-algebra-valued one-form ${\mathcal{A}}_i$ , on $U_i$ , we can reconstruct a connection one-form $\omega$ whose pullback by ${\sigma }_i^{\ast }$ is ${\mathcal{A}}_i$ .

Theorem 10.1

Given a $\mathfrak{g}$ -valued one-form ${\mathcal{A}}_i$ on $U_i$ and a local section ${\sigma }_i:U_i\to {\pi }^{-1}(U_i)$ , there exists a connection one-form $\omega$ such that ${\mathcal{A}}_i={\sigma }_i^{\ast }\omega$ .

Proof

Let us define a $\mathfrak{g}$ -valued one-form $\omega$ on $P$ by

$${\omega }_i\equiv g_i^{-1}{\pi }^{\ast }{\mathcal{A}}_i{g}_i+g_i^{-1}{\mathrm{d}}_P{g}_i\text{\hspace{1em}(10.7)}$$

where ${\mathrm{d}}_P$ is the exterior derivative on $P$ and $g_i$ is the canonical local trivialization defined by ${\varphi }_i^{-1}(u)=(p,g_i)$ for $u={\sigma }_i(p)g_i$ . We first show that ${\sigma }_i^{\ast }{\omega }_i={\mathcal{A}}_i$ . For $X\in T_{p}M$ , we have

$$\begin{array}{rl}{\sigma }_i^{\ast }{\omega }_i(X)& ={\omega }_i\left({\sigma }_{i\ast }X\right)={\pi }^{\ast }{\mathcal{A}}_i\left({\sigma }_{i\ast }X\right)+d_P{g}_i\left({\sigma }_{i\ast }X\right)\\ & ={\mathcal{A}}_i\left({\pi }_{\ast }{\sigma }_{i\ast }X\right)+d_P{g}_i\left({\sigma }_{i\ast }X\right)\end{array}$$

where we have noted that ${\sigma }_{i\ast }X\in T_{{\sigma }_i}P$ and $g_i=e$ at ${\sigma }_i$ , see figure 10.2. We further note that ${\pi }_{\ast }{\sigma }_{i\ast }={\mathrm{id}}_{T_p(M)}$ and ${\mathrm{d}}_P{g}_i({\sigma }_{i\ast }X)=0$ since $g\equiv e$ along ${\sigma }_{i\ast }X$ . Thus, we have obtained ${\sigma }_i^{\ast }{\omega }_i(X)={\mathcal{A}}_i(X)$ .

Figure 10.2. The canonical local trivialization defined by the local section ${\sigma }_i$ over $U_i$ .

Next we show that ${\omega }_i$ satisfies the axioms of a connection one-form given in definition 10.2.

(i) Let $X=A^{\#}\in V_{u}P,A\in \mathfrak{g}$ . It follows from exercise 10.1(a) that ${\pi }_{\ast }X=0$ . Now we have

$$\begin{array}{rl}{\omega }_i(A^{\#})& =g_i^{-1}{\mathrm{d}}_P{g}_i(A^{\#})=g_i(u{)}^{-1}{\frac{\mathrm{d}g(u\exp (tA))}{\mathrm{d}t}|}_{t=0}\\ & =g_i(u{)}^{-1}{g}_i(u){\frac{\mathrm{d}\exp (tA)}{\mathrm{d}t}|}_{t=0}=A.\end{array}$$

(ii) Take $X\in T_{u}P$ and $h\in G$ . We have

$$R_h^{\ast }{\omega }_i(X)={\omega }_i\left(R_{h\ast }X\right)=g_{iuh}^{-1}{\mathcal{A}}_i\left({\pi }_{\ast }{R}_{h\ast }X\right)g_{iuh}+g_{iuh}^{-1}{\mathrm{d}}_P{g}_{iuh}\left(R_{h\ast }X\right).$$

Since $g_{iuh}=g_{iu}h$ and ${\pi }_{\ast }{R}_{h\ast }X={\pi }_{\ast }X$ (note that $\pi R_h=\pi$ ), we have

$$\begin{array}{rl}{R}_h^{\ast }{\omega }_i(X)& =h^{-1}{g}_{iu}^{-1}{\mathcal{A}}_i({\pi }_{\ast }X)g_{iu}h+h^{-1}{g}_{iu}^{-1}{d}_P{g}_{iu}(X)h\\ & =h^{-1}{\omega }_i(X)h\end{array}$$

where we have noted that

$$\begin{array}{rl}{g}_{iuh}^{-1}{\mathrm{d}}_P{g}_{iuh}(R_{h\ast }X)& =g_{iuh}^{-1}{\frac{\mathrm{d}}{\mathrm{d}t}{g}_{i\gamma (t)h}|}_{t=0}\\ & =h^{-1}{g}_{iu}^{-1}{\frac{\mathrm{d}}{\mathrm{d}t}{g}_{i\gamma (t)}|}_{t=0}h=h^{-1}{g}_{iu}^{-1}{\mathrm{d}}_P{g}_{iu}(X)h.\end{array}$$

Here $\gamma (t)$ is a curve through $u=\gamma (0)$ , whose tangent vector at $u$ is $X$ .

Hence, the $\mathfrak{g}$ -valued one-form ${\omega }_i$ defined by (10.7) indeed satisfies ${\mathcal{A}}_i={\sigma }_i^{\ast }{\omega }_i$ and the axioms of a connection one-form.

For $\omega$ to be defined uniquely on $P$ , i.e. for the separation $T_{u}P=H_{u}P\oplus V_{u}P$ to be unique, we must have ${\omega }_i={\omega }_j$ on $U_i\cap U_j$ . A unique one-form $\omega$ is then defined throughout $P$ by $\omega {|}_{U_i}={\omega }_i$ . To fulfil this condition, the local forms ${\mathcal{A}}_i$ have to satisfy a peculiar transformation property similar to that of the Christoffel symbols. We first prove a technical lemma.

Lemma 10.1

Let $P(M,G)$ be a principal bundle and ${\sigma }_i({\sigma }_j)$ be a local section over $U_i(U_j)$ such that $U_i\cap U_j\ne \varnothing$ . For $X\in T_{p}M$ ( $p\in U_i\cap U_j$ ), ${\sigma }_{i\ast }X$ and ${\sigma }_{j\ast }X$ satisfy

$${\sigma }_{j\ast }X=R_{tij\ast }\left({\sigma }_{i\ast }X\right)+{\left(t_{ij}^{-1}{\mathrm{dt}}_{ij}(X)\right)}^{\#}\text{\hspace{1em}(10.8)}$$

where $t_{ij}:U_i\cap U_j\to G$ is the transition function.

Proof

Take a curve $\gamma :[0,1]\to M$ such that $\gamma (0)=p$ and $\dot{\gamma }(0)=X$ . Since ${\sigma }_i(p)$ and ${\sigma }_j(p)$ are related by the transition function as ${\sigma }_j(p)={\sigma }_i(p)t_{ij}(p)$ (see (9.43)), we have

$$\begin{array}{rl}{\sigma }_{j\ast }X& ={\frac{\mathrm{d}}{\mathrm{d}t}{\sigma }_j(\gamma (t))|}_{t=0}={\frac{\mathrm{d}}{\mathrm{d}t}\{{\sigma }_i(t)t_{ij}(t)\}|}_{t=0}\\ & ={\frac{\mathrm{d}}{\mathrm{d}t}{\sigma }_i(t)\cdot t_{ij}(p)+{\sigma }_i(p)\cdot \frac{\mathrm{d}}{\mathrm{d}t}{t}_{ij}(t)|}_{t=0}\\ & =R_{t_{ij}\ast }({\sigma }_{i\ast }X)+{\sigma }_j(p)t_{ij}(p{)}^{-1}\frac{\mathrm{d}}{\mathrm{d}t}{t}_{ij}(t){|}_{t=0}\end{array}$$

where ${\sigma }_i(t)$ stands for ${\sigma }_i(\gamma (t))$ and we have assumed that $G$ is a matrix group for which $R_{g\ast }X=Xg$ . We note that

$$\begin{array}{rl}{t}_{ij}(p{)}^{-1}\mathrm{d}{t}_{ij}(X)& =t_{ij}(p{)}^{-1}{\frac{\mathrm{d}}{\mathrm{d}t}{t}_{ij}(t)|}_{t=0}\\ & ={\frac{\mathrm{d}}{\mathrm{d}t}[t_{ij}(p{)}^{-1}{t}_{ij}(t)]|}_{t=0}\in T_e(G)\cong \mathfrak{g}.\end{array}$$

[Note that $t_{ij}(p{)}^{-1}{t}_{ij}(\gamma (t))=e$ at $t=0$ .] This shows that the second term of ${\sigma }_{j\ast }X$ represents the vector field $(t_{ij}^{-1}\mathrm{d}{t}_{ij}(X){)}^{\#}$ at ${\sigma }_j(p)$ .

The compatibility condition is easily obtained by applying the connection one-form $\omega$ on (10.8). We find that

$$\begin{array}{rl}{\sigma }_j^{\ast }\omega (X)& =R_{lij}^{\ast }\omega ({\sigma }_{i\ast }X)+t_{ij}^{-1}\mathrm{d}{t}_{ij}(X)\\ & =t_{ij}^{-1}\omega \left({\sigma }_{i\ast }X\right)t_{ij}+t_{ij}^{-1}d{t}_{ij}(X)\end{array}$$

where the axioms of definition 10.2 have been used. Since this is true for any $X\in T_{p}M$ , this equation reduces to

$${\mathcal{A}}_j=t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}.\text{\hspace{1em}(10.9)}$$

This is the compatibility condition we have been seeking.

Conversely, given an open covering $\{U_i\}$ , the local sections $\{{\sigma }_i\}$ and the local forms $\{{\mathcal{A}}_i\}$ which satisfy (10.9), we may construct the $\mathfrak{g}$ -valued one-form $\omega$ over $P$ . Since a non-trivial principal bundle does not admit a global section, the pullback ${\mathcal{A}}_i={\sigma }_i^{\ast }\omega$ exists locally but not necessarily globally. In gauge theories, ${\mathcal{A}}_i$ is identified with the gauge potential (Yang-Mills potential). As we have seen in the monopole case, the monopole field $B=gr/r^3$ does not admit a single gauge potential and we require at least two ${\mathcal{A}}_i$ to describe this U(1) bundle over $S^2$ .

Exercise 10.2

Let $P(M,G)$ be a principal bundle over $M$ and let $U$ be a chart of $M$ . Take local sections ${\sigma }_1$ and ${\sigma }_2$ over $U$ such that ${\sigma }_2(p)={\sigma }_1(p)g(p)$ . Show that the corresponding local forms ${\mathcal{A}}_1$ and ${\mathcal{A}}_2$ are related as

$${\mathcal{A}}_2=g^{-1}{\mathcal{A}}_{1}g+g^{-1}\mathrm{d}g.\text{\hspace{1em}(10.10a)}$$

In components, this becomes

$${\mathcal{A}}_{2\mu }=g^{-1}(p){\mathcal{A}}_{1\mu }(p)g(p)+g^{-1}(p){\partial }_{\mu }g(p)\text{\hspace{1em}(10.10b)}$$

which is simply the gauge transformation defined in section 1.8.

Example 10.1

Let $P$ be a U(1) bundle over $M$ . Take overlapping charts $U_i$ and $U_j$ . Let ${\mathcal{A}}_i({\mathcal{A}}_j)$ be a local connection form on $U_i(U_j)$ . The transition function $t_{ij}:U_i\cap U_j\to \mathrm{U}(1)$ is given by

$$t_{ij}(p)=\exp [\mathrm{i}\Lambda (p)]\Lambda (p)\in \mathbb{R}.\text{\hspace{1em}(10.11)}$$

${\mathcal{A}}_i$ and ${\mathcal{A}}_j$ are related as

$$\begin{array}{rl}{\mathcal{A}}_j(p)& =t_{ij}(p{)}^{-1}{\mathcal{A}}_i(p)t_{ij}(p)+t_{ij}(p{)}^{-1}\mathrm{d}{t}_{ij}(p)\\ & ={\mathcal{A}}_i(p)+\mathrm{id}\Lambda (p).\end{array}\text{\hspace{1em}(10.12a)}$$

In components, we have the familiar expression

$${\mathcal{A}}_{j\mu }={\mathcal{A}}_{i\mu }+\mathrm{i}{\partial }_{\mu }\Lambda .\text{\hspace{1em}(10.12b)}$$

Our connection ${\mathcal{A}}_{\mu }$ differs from the standard vector potential $A_{\mu }$ by the Lie algebra factor: ${\mathcal{A}}_{\mu }=\mathrm{i}{A}_{\mu }$ .

Here we note again that $\omega$ is defined globally over the bundle $P(M,G)$ . Although there are many connection one-forms on $P(M,G)$ , they share the same global information about the bundle. In contrast, an individual local piece (gauge potential) ${\mathcal{A}}_i$ is associated with the trivial bundle ${\pi }^{-1}(U_i)$ and cannot have any global information on $P$ . It is $\omega$ or, equivalently, the total of $\{{\mathcal{A}}_i\}$ satisfying the compatibility condition (10.9), which carries the global information about the bundle.

10.1.4 Horizontal lift and parallel transport

Parallel transport of a vector has been defined in chapter 7 as transport without change. Parallel transport of an element of a principal bundle along a curve in $M$ is provided by the ‘horizontal lift’ of the curve.

Definition 10.3

Let $P(M,G)$ be a $G$ bundle and let $\gamma :[0,1]\to M$ be a curve in $M$ . A curve $\tilde{\gamma }:[0,1]\to P$ is said to be a horizontal lift of $\gamma$ if $\pi \circ \tilde{\gamma }=\gamma$ and the tangent vector to $\tilde{\gamma }(t)$ always belongs to $H_{\tilde{\gamma }(t)}P$ .

Let $\tilde{X}$ be a tangent vector to $\tilde{\gamma }$ . Then it satisfies $\omega (\tilde{X})=0$ by definition. This condition is an ordinary differential equation (ODE) and the fundamental theorem of ODEs guarantees the local existence and uniqueness of the horizontal lift.

Theorem 10.2

Let $\gamma :[0,1]\to M$ be a curve in $M$ and let $u_0\in {\pi }^{-1}(\gamma (0))$ . Then there exists a unique horizontal lift $\tilde{\gamma }(t)$ in $P$ such that $\tilde{\gamma }(0)=u_0$ .

Let us construct such a curve $\tilde{\gamma }$ . Let $U_i$ be a chart which contains $\gamma$ and take a section ${\sigma }_i$ over $U_i$ . If there exists a horizontal lift $\tilde{\gamma }$ , it may be expressed as $\tilde{\gamma }(t)={\sigma }_i(\gamma (t))g_i(t)$ , where $g_i(t)$ stands for $g_i(\gamma (t))\in G$ . Without loss of generality, we may take a section such that ${\sigma }_i(\gamma (0))=\tilde{\gamma }(0)$ , that is $g_i(0)=e$ . Let $X$ be a tangent vector to $\gamma (t)$ at $\gamma (0)$ . Then $\tilde{X}={\tilde{\gamma }}_{\ast }X$ is tangent to $\tilde{\gamma }$ at $u_0=\tilde{\gamma }(0)$ . Since the tangent vector $\tilde{X}$ is horizontal, it satisfies $\omega (\tilde{X})=0$ . A slight modification of lemma 10.1 yields

$$\tilde{X}=g_i(t{)}^{-1}{\sigma }_{i\ast }X{g}_i(t)+[g_i(t{)}^{-1}\mathrm{d}{g}_i(X){]}^{\#}.$$

By applying $\omega$ on this equation, we find

$$0=\omega (\tilde{X})=g_i(t{)}^{-1}\omega ({\sigma }_{i\ast }X)g_i(t)+g_i(t{)}^{-1}\frac{\mathrm{d}{g}_i(t)}{\mathrm{d}t}.$$

Multiplying on the left by $g_i(t)$ , we have

$$\frac{\mathrm{d}{g}_i(t)}{\mathrm{d}t}=-\omega \left({\sigma }_{i\ast }X\right)g_i(t).\text{\hspace{1em}(10.13a)}$$

The fundamental theorem of ODEs guarantees the existence and uniqueness of the solution of (10.13a).

Since $\omega ({\sigma }_{i\ast }X)={\sigma }_i^{\ast }\omega (X)={\mathcal{A}}_i(X)$ , (10.13a) is expressed in a local form as

$$\frac{\mathrm{d}{g}_i(t)}{\mathrm{d}t}=-{\mathcal{A}}_i(X)g_i(t)\text{\hspace{1em}(10.13b)}$$

whose formal solution with $g_i(0)=e$ is

$$\begin{array}{rl}{g}_i(\gamma (t))& =\mathcal{P}\exp \left(-{\int }_0^t{\mathcal{A}}_{i\mu }\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}t}\mathrm{d}t\right)\\ & =\mathcal{P}\exp \left(-{\int }_{\gamma (0)}^{\gamma (t)}{\mathcal{A}}_{i\mu }(\gamma (t))\mathrm{d}{x}^{\mu }\right)\end{array}\text{\hspace{1em}(10.14)}$$

where $\mathcal{P}$ is a path-ordering operator along $\gamma (t)$ . The horizontal lift is expressed as $\tilde{\gamma }(t)={\sigma }_i(\gamma (t))g_i(\gamma (t))$ .

Corollary 10.1

Let ${\tilde{\gamma }}^{\prime }$ be another horizontal lift of $\gamma$ , such that ${\tilde{\gamma }}^{\prime }(0)=\gamma (0)g$ . Then ${\tilde{\gamma }}^{\prime }(t)=\tilde{\gamma }(t)g$ for all $t\in [0,1]$ .

Proof

We first note that the horizontal subspace is right invariant, $R_{g\ast }{H}_{u}P=H_{ug}P$ . Let $\tilde{\gamma }$ be a horizontal lift of $\gamma$ . Then ${\tilde{\gamma }}_g:t\mapsto \tilde{\gamma }(t)g$ is also a horizontal lift of $\gamma (t)$ since its tangent vector belongs to $H_{\tilde{\gamma }g}P$ . From theorem 10.2 we find ${\tilde{\gamma }}^{\prime }$ is the unique horizontal lift which starts at $\tilde{\gamma }(0)g$ .

Example 10.2

Let us consider the bundle $P(M,\mathbb{R})\cong M\times \mathbb{R}$ where $M={\mathbb{R}}^2-\{0\}$ . Let $\varphi :((x,y),f)\mapsto u\in P$ be a local trivialization, where $(x,y)$ are the coordinates of $M$ while $f$ is that of the additive group $\mathbb{R}$ . Let

$$\omega =\frac{y\mathrm{d}x-x\mathrm{d}y}{x^2+y^2}+\mathrm{d}f$$

be a connection one-form. It is easily verified that $\omega$ satisfies the axioms of the connection one-form. In fact, for $A^{\#}=A\partial /\partial f$ , $A\in \mathbb{R}$ being an element of the Lie algebra of additive group, we have $\omega (A^{\#})=A$ . Furthermore, $R_{g\ast }\omega =\omega =g^{-1}\omega g$ , since $\mathbb{R}$ is Abelian. Let $\gamma :[0,1]\to M$ be a curve $t\mapsto (\cos 2\pi t,\sin 2\pi t)$ . Let us work out a horizontal lift which starts at $((1,0),0)$ . Let

$$X=\frac{\mathrm{d}}{\mathrm{d}t}\equiv \frac{\mathrm{d}x}{\mathrm{d}t}\frac{\partial }{\partial x}+\frac{\mathrm{d}y}{\mathrm{d}t}\frac{\partial }{\partial y}+\frac{\mathrm{d}f}{\mathrm{d}t}\frac{\partial }{\partial f}$$

be tangent to $\tilde{\gamma }(t)$ . For $X$ to be horizontal, it must satisfy

$$0=\omega (X)=\frac{\mathrm{d}x}{\mathrm{d}t}\frac{y}{r^2}-\frac{\mathrm{d}y}{\mathrm{d}t}\frac{x}{r^2}+\frac{\mathrm{d}f}{\mathrm{d}t}=-2\pi +\frac{\mathrm{d}f}{\mathrm{d}t}.$$

The solution is easily found to be $f=2\pi t+\text{constant}$ . We finally find the horizontal lift $\tilde{\gamma }$ passing through $((1,0),0)$ ,

$$\tilde{\gamma }(t)=\left(\left(\cos 2\pi t,\sin 2\pi t\right),2\pi t\right)\text{\hspace{1em}(10.15)}$$

which is a helix over the unit circle.

Under the group action (right or left does not matter), $f$ translates to $f+g$ , $g\in \mathbb{R}$ . The shifted horizontal lift is

$${\tilde{\gamma }}_g(t)=((\cos 2\pi t,\sin 2\pi t),2\pi t+g).\text{\hspace{1em}(10.16)}$$

${}^1{\mathcal{A}}_{i\mu }(\gamma (t))$ and ${\mathcal{A}}_{i\nu }(\gamma (s))$ do not commute in general and the exponential in (10.14) is not well defined as it is. Let $A(t)$ and $B(t)$ be $t$ -dependent matrices. Then the action of $\mathcal{P}$ is

$$\mathcal{P}[A(t)B(s)]=\{\begin{array}{ll}A(t)B(s)& (t>s)\\ B(s)A(t)& (s>t).\end{array}$$

Generalization to products of more matrices should be obvious.

Let $\gamma :[0,1]\to M$ be a curve. Take a point $u_0\in {\pi }^{-1}(\gamma (0))$ . There is a unique horizontal lift $\tilde{\gamma }(t)$ of $\gamma (t)$ through $u_0$ , and hence a unique point $u_1=\tilde{\gamma }(1)\in {\pi }^{-1}(\gamma (1))$ , see figure 10.3. The point $u_1$ is called the parallel transport of $u_0$ along the curve $\gamma$ . This defines a map $\Gamma (\tilde{\gamma }):{\pi }^{-1}(\gamma (0))\to {\pi }^{-1}(\gamma (1))$ such that $u_0\mapsto u_1$ . If the local form (10.14) is employed, we have

Figure 10.3. A curve $\gamma (t)$ in $M$ and its horizontal lifts $\tilde{\gamma }(t)$ and $\tilde{\gamma }(t)g$ .

$$u_1={\sigma }_i(1)\mathcal{P}\exp \left(-{\int }_0^1{\mathcal{A}}_{i\mu }\frac{\mathrm{d}{x}^{\mu }(\gamma (t))}{\mathrm{d}t}\mathrm{d}t\right).\text{\hspace{1em}(10.17)}$$

Corollary 10.1 ensures that $\Gamma (\tilde{\gamma })$ commutes with the right action $R_g$ . First note that $R_g\Gamma (\tilde{\gamma })(u_0)=u_{1}g$ and $\Gamma (\tilde{\gamma })R_g(u_0)=\Gamma (\tilde{\gamma })(u_{0}g)$ . Observe that $\tilde{\gamma }(t)g$ is a horizontal lift through $u_{0}g$ and $u_{1}g$ . From the uniqueness of the horizontal lift through $u_{0}g$ , we have $u_{1}g=\Gamma (\tilde{\gamma })(u_{0}g)$ , that is $R_g\Gamma (\tilde{\gamma })(u_0)=\Gamma (\tilde{\gamma })R_g(u_0)$ . Since this is true for any $u_0\in {\pi }^{-1}(\gamma (0))$ , we have

$$R_g\Gamma (\tilde{\gamma })=\Gamma (\tilde{\gamma })R_g.\text{\hspace{1em}(10.18)}$$

Exercise 10.3

Let $\tilde{\gamma }$ be a horizontal lift of $\gamma :[0,1]\to M$ . Consider a map $\Gamma ({\tilde{\gamma }}^{-1}):{\pi }^{-1}(\gamma (1))\to {\pi }^{-1}(\gamma (0))$ where ${\tilde{\gamma }}^{-1}(t)=\tilde{\gamma }(1-t)$ . Show that

$$\Gamma ({\tilde{\gamma }}^{-1})=\Gamma (\tilde{\gamma }{)}^{-1}.\text{\hspace{1em}(10.19)}$$

Consider two curves $\alpha :[0,1]\to M$ and $\beta :[0,1]\to M$ such that $\alpha (1)=\beta (0)$ . Define the product $\alpha \ast \beta$ by

$$\alpha \ast \beta =\{\begin{array}{ll}\alpha (2t)& 0\le t\le \frac{1}{2}\\ \beta (2t-1)& \frac{1}{2}\le t\le 1.\end{array}$$

Let $\Gamma (\tilde{\alpha }):{\pi }^{-1}(\alpha (0))\to {\pi }^{-1}(\alpha (1))$ and $\Gamma (\tilde{\beta }):{\pi }^{-1}(\beta (0))\to {\pi }^{-1}(\beta (1))$ Show that

$$\widetilde{\Gamma (\alpha \ast \beta )}=\Gamma (\tilde{\beta })\circ \Gamma (\tilde{\alpha }).\text{\hspace{1em}(10.20)}$$

Exercise 10.4

Let us write $u\sim v$ , if $u,v\in P$ are on the same horizontal lift. Show that $\sim$ is an equivalence relation.

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：e453db1e51f34092b1b64e4d6353e67bbf36091fea8b4a02407803bf7d05da77