10.3 Curvature

10.3.1 Covariant derivatives in principal bundles

We defined the exterior derivative $\mathrm{d}:{\Omega }^r(M)\to {\Omega }^{r+1}(M)$ in chapter 5. An $r$ -form $\eta$ is a real-valued form acting on vectors,

$$\eta :TM\wedge \cdots \wedge TM\to \mathbb{R}.$$

We will generalize this operation so that we can differentiate a vector-valued $r$ -form $\varphi \in {\Omega }^r(P)\otimes V$

$$\varphi :TP\wedge \cdots \wedge TP\to V$$

where $V$ is a vector space of dimension $k$ . The most general form of $\varphi$ is $\varphi ={\sum }_{\alpha =1}^k{\varphi }^{\alpha }\otimes e_{\alpha }$ , $\{e_{\alpha }\}$ being a basis of $V$ and ${\varphi }^{\alpha }\in {\Omega }^r(P)$ .

A connection $\omega$ on a principal bundle $P(M,G)$ separates $T_{u}P$ into $H_{u}P\oplus V_{u}P$ . Accordingly, a vector $X\in T_{u}P$ is decomposed as $X=X^H+X^V$ where $X^H\in H_{u}P$ and $X^V\in V_{u}P$ .

Definition 10.4

Let $\varphi \in {\Omega }^r(P)\otimes V$ and $X_1,\dots ,X_{r+1}\in T_{u}P$ . The covariant derivative of $\varphi$ is defined by

$$\mathrm{D}\varphi \left(X_1,\dots ,X_{r+1}\right)\equiv {\mathrm{d}}_P\varphi \left(X_1^H,\dots ,X_{r+1}^H\right)\text{\hspace{1em}(10.28)}$$

where ${\mathrm{d}}_P\varphi \equiv {\mathrm{d}}_P{\varphi }^{\alpha }\otimes e_{\alpha }$

10.3.2 Curvature

Definition 10.5

The curvature two-form $\Omega$ is the covariant derivative of the connection one-form $\omega$ ,

$$\Omega \equiv \mathrm{D}\omega \in {\Omega }^2(P)\otimes \mathfrak{g}.\text{\hspace{1em}(10.29)}$$

Proposition 10.2

The curvature two-form satisfies (cf (10.3b))

$$R_a^{\ast }\Omega =a^{-1}\Omega aa\in G.\text{\hspace{1em}(10.30)}$$

Proof

We first note that $(R_{a\ast }X{)}^H=R_{a\ast }(X^H)$ ( $R_{a\ast }$ preserves the horizontal subspaces) and ${\mathrm{d}}_P{R}_a^{\ast }=R_a^{\ast }{\mathrm{d}}_P$ , see (5.75). By definition we find

$$\begin{array}{rl}{R}_a^{\ast }\Omega (X,Y)& =\Omega \left(R_{a\ast }X,R_{a\ast }Y\right)={\mathrm{d}}_P\omega \left({\left(R_{a\ast }X\right)}^H,{\left(R_{a\ast }Y\right)}^H\right)\\ & ={\mathrm{d}}_P\omega \left(R_{a\ast }{X}^H,R_{a\ast }{Y}^H\right)=R_a^{\ast }{\mathrm{d}}_P\omega \left(X^H,Y^H\right)\\ & ={\mathrm{d}}_P{R}_a^{\ast }\omega \left(X^H,Y^H\right)\\ & ={\mathrm{d}}_P(a^{-1}\omega a)\left(X^H,Y^H\right)=a^{-1}{\mathrm{d}}_P\omega \left(X^H,Y^H\right)a\\ & =a^{-1}\Omega (X,Y)a\end{array}$$

where we noted that $a$ is a constant element and hence ${\mathrm{d}}_{P}a=0$ .

Take a $\mathfrak{g}$ -valued $p$ -form $\zeta ={\zeta }^{\alpha }\otimes T_{\alpha }$ and a $\mathfrak{g}$ -valued $q$ -form $\eta ={\eta }^{\alpha }\otimes T_{\alpha }$ where ${\zeta }^{\alpha }\in {\Omega }^p(P)$ , ${\eta }^{\alpha }\in {\Omega }^q(P)$ , and $\{T_{\alpha }\}$ is a basis of $\mathfrak{g}$ . Define the commutator of $\zeta$ and $\eta$ by

$$\begin{array}{rl}& [\zeta ,\eta ]\equiv \zeta \wedge \eta -(-1{)}^{pq}\eta \wedge \zeta \\ & =T_{\alpha }{T}_{\beta }{\zeta }^{\alpha }\wedge {\eta }^{\beta }-(-1{)}^{pq}{T}_{\beta }{T}_{\alpha }{\eta }^{\beta }\wedge {\zeta }^{\alpha }\\ & =\left[T_{\alpha },T_{\beta }\right]\otimes {\zeta }^{\alpha }\wedge {\eta }^{\beta }=f_{\alpha \beta }^{\gamma }{T}_{\gamma }\otimes {\zeta }^{\alpha }\wedge {\eta }^{\beta }.\end{array}\text{\hspace{1em}(10.31)}$$

If we put $\zeta =\eta$ in (10.31), when $p$ and $q$ are odd, we have

$$[\zeta ,\zeta ]=2\zeta \wedge \zeta =f_{\alpha \beta }^{\gamma }{T}_{\gamma }\otimes {\zeta }^{\alpha }\wedge {\zeta }^{\beta }.$$

Lemma 10.2

Let $X\in H_{u}P$ and $Y\in V_{u}P$ . Then $[X,Y]\in H_{u}P$ .

Proof

Let $Y$ be a vector field generated by $g(t)$ , then

$${\mathcal{L}}_{Y}X=[Y,X]=\underset{t\to 0}{\lim }{t}^{-1}(R_{g(t)\ast }X-X).$$

Since a connection satisfies $R_{g\ast }{H}_{u}P=H_{ug}P$ , the vector $R_{g(t)\ast X}$ is horizontal and so is $[Y,X]$ .

Theorem 10.3

Let $X,Y\in T_{u}P$ . Then $\Omega$ and $\omega$ satisfy Cartan’s structure equation

$$\Omega (X,Y)={\mathrm{d}}_P\omega (X,Y)+[\omega (X),\omega (Y)]\text{\hspace{1em}(10.32a)}$$

which is also written as

$$\Omega ={\mathrm{d}}_P\omega +\omega \wedge \omega .\text{\hspace{1em}(10.32b)}$$

Proof

We consider the following three cases separately:

(i) Let $X,Y\in H_{u}P$ . Then $\omega (X)=\omega (Y)=0$ by definition. From definition 10.5, we have $\Omega (X,Y)={\mathrm{d}}_P\omega (X^H,Y^H)={\mathrm{d}}_P\omega (X,Y)$ , since $X=X^H$ and $Y=Y^H$ .

(ii) Let $X\in H_{u}P$ and $Y\in V_{u}P$ . Since $Y^H=0$ , we have $\Omega (X,Y)=0$ . We also have $\omega (X)=0$ . Thus, we need to prove ${\mathrm{d}}_P\omega (X,Y)=0$ . From (5.70), we obtain

$${\mathrm{d}}_P\omega (X,Y)=X\omega (Y)-Y\omega (X)-\omega ([X,Y])=X\omega (Y)-\omega ([X,Y]).$$

Since $Y\in V_{u}P$ , there is an element $V\in \mathfrak{g}$ such that $Y=V^{\#}$ . Then $\omega (Y)=V$ is constant, hence $X\omega (Y)=X\cdot V=0$ . From lemma 10.2, we have $[X,Y]\in H_{u}P$ so that $\omega ([X,Y])=0$ and we find ${\mathrm{d}}_P\omega (X,Y)=0$ .

(iii) For $X,Y\in V_{u}P$ , we have $\Omega (X,Y)=0$ . We find that, in this case,

$${\mathrm{d}}_P\omega (X,Y)=X\omega (Y)-Y\omega (X)-\omega ([X,Y])=-\omega ([X,Y]).$$

We note that $X$ and $Y$ are closed under the Lie bracket, $[X,Y]\in V_{u}P$ , see exercise 10.1(b). Then there exists $A\in \mathfrak{g}$ such that

$$\omega ([X,Y])=A$$

where $A^{\#}=[X,Y]$ . Let $B^{\#}=X$ and $C^{\#}=Y$ . Then $[\omega (X),\omega (Y)]=[B,C]=A$ since $[B,C{]}^{\#}=[B^{\#},C^{\#}]$ . Thus, we have shown that

$$0={\mathrm{d}}_P\omega (X,Y)+\omega ([X,Y])={\mathrm{d}}_P\omega (X,Y)+[\omega (X),\omega (Y)].$$

Since $\Omega$ is linear and skew symmetric, these three cases are sufficient to show that (10.32) is true for any vectors.

To derive (10.32b) from (10.32a), we note that

$$\begin{array}{rl}[\omega ,\omega ](X,Y)& =[T_{\alpha },T_{\beta }]{\omega }^{\alpha }\wedge {\omega }^{\beta }(X,Y)\\ & =\left[T_{\alpha },T_{\beta }\right]\left[{\omega }^{\alpha }(X){\omega }^{\beta }(Y)-{\omega }^{\beta }(X){\omega }^{\alpha }(Y)\right]\\ & =[\omega (X),\omega (Y)]-[\omega (Y),\omega (X)]=2[\omega (X),\omega (Y)].\end{array}$$

Hence, $\Omega (X,Y)=({\mathrm{d}}_P\omega +\frac{1}{2}[\omega ,\omega ])(X,Y)=({\mathrm{d}}_P\omega +\omega \wedge \omega )(X,Y).$

10.3.3 Geometrical meaning of the curvature and the Ambrose-Singer theorem

We have shown in chapter 7 that the Riemann curvature tensor expresses the noncommutativity of the parallel transport of vectors. There is a similar interpretation of curvature on principal bundles. We first show that $\Omega (X,Y)$ yields the vertical component of the Lie bracket $[X,Y]$ of horizontal vectors $X,Y\in H_{u}P$ . It follows from $\omega (X)=\omega (Y)=0$ that

$${\mathrm{d}}_P\omega (X,Y)=X\omega (Y)-Y\omega (X)-\omega ([X,Y])=-\omega ([X,Y]).$$

Since $X^H=X$ , $Y^H=Y$ , we have

$$\Omega (X,Y)={\mathrm{d}}_P\omega (X,Y)=-\omega ([X,Y]).\text{\hspace{1em}(10.33)}$$

Let us consider a coordinate system $\{x^{\mu }\}$ on a chart $U$ . Let $V=\partial /\partial x^1$ and $W=\partial /\partial x^2$ . Take an infinitesimal parallelogram $\gamma$ whose corners are $O=\{0,0,\dots ,0\}$ , $P=\{\epsilon ,0,\dots ,0\}$ , $Q=\{\epsilon ,\delta ,0,\dots ,0\}$ and $R=\{0,\delta ,0,\dots ,0\}$ . Consider the horizontal lift $\tilde{\gamma }$ of $\gamma$ . Let $X,Y\in H_{u}P$ such that ${\pi }_{\ast }X=\epsilon V$ and ${\pi }_{\ast }Y=\delta W$ . Then

$${\pi }_{\ast }([X,Y{]}^H)=ϵ\delta [V,W]=ϵ\delta \left[\frac{\partial }{\partial x^1},\frac{\partial }{\partial x^2}\right]=0\text{\hspace{1em}(10.34)}$$

that is $[X,Y]$ is vertical. This consideration shows that the horizontal lift $\tilde{\gamma }$ of a loop $\gamma$ fails to close. This failure is proportional to the vertical vector $[X,Y]$ connecting the initial point and the final point on the same fibre. The curvature measures this distance,

$$\Omega (X,Y)=-\omega ([X,Y])=A\text{\hspace{1em}(10.35)}$$

where $A$ is an element of $\mathfrak{g}$ such that $[X,Y]=A^{\#}$ .

Since the discrepancy between the initial and final points of the horizontal lift of a closed curve is simply the holonomy, we expect that the holonomy group is expressed in terms of the curvature.

Theorem 10.4 · Ambrose-Singer theorem

Let $P(M,G)$ be a $G$ bundle over a connected manifold $M$ . The Lie algebra $\mathfrak{h}$ of the holonomy group ${\Phi }_{u_0}$ of a point $u_0\in P$ agrees with the subalgebra of $\mathfrak{g}$ spanned by the elements of the form

$${\Omega }_u(X,Y)X,Y\in H_{u}P\text{\hspace{1em}(10.36)}$$

where $a\in P$ is a point on the same horizontal lift as $u_0$ . [See Choquet-Bruhat et al (1982) for the proof.]

10.3.4 Local form of the curvature

The local form $\mathcal{F}$ of the curvature $\Omega$ is defined by

$$\mathcal{F}\equiv {\sigma }^{\ast }\Omega \text{\hspace{1em}(10.37)}$$

where $\sigma$ is a local section defined on a chart $U$ of $M$ (cf $\mathcal{A}={\sigma }^{\ast }\omega$ ). $\mathcal{F}$ is expressed in terms of the gauge potential $\mathcal{A}$ as

$$\mathcal{F}=\mathrm{d}\mathcal{A}+\mathcal{A}\wedge \mathcal{A}\text{\hspace{1em}(10.38a)}$$

where $d$ is the exterior derivative on $M$ . The action of $\mathcal{F}$ on the vectors of $TM$ is given by

$$\mathcal{F}(X,Y)=\mathrm{d}\mathcal{A}(X,Y)+[\mathcal{A}(X),\mathcal{A}(Y)].\text{\hspace{1em}(10.38b)}$$

To prove (10.38a) we note that $\mathcal{A}={\sigma }^{\ast }\omega$ , ${\sigma }^{\ast }{\mathrm{d}}_P\omega =\mathrm{d}{\sigma }^{\ast }\omega$ and ${\sigma }^{\ast }(\zeta \wedge \eta )={\sigma }^{\ast }\zeta \wedge {\sigma }^{\ast }\eta$ . From Cartan’s structure equation, we find

$$\mathcal{F}={\sigma }^{\ast }({\mathrm{d}}_P\omega +\omega \wedge \omega )=\mathrm{d}{\sigma }^{\ast }\omega +{\sigma }^{\ast }\omega \wedge {\sigma }^{\ast }\omega =\mathrm{d}\mathcal{A}+\mathcal{A}\wedge \mathcal{A}.$$

Next, we find the component expression of $\mathcal{F}$ on a chart $U$ whose coordinates are $x^{\mu }=\phi (p)$ . Let $\mathcal{A}={\mathcal{A}}_{\mu }\mathrm{d}{x}^{\mu }$ be the gauge potential. If we write $\mathcal{F}=\frac{1}{2}{\mathcal{F}}_{\mu \nu }\mathrm{d}{x}^{\mu }\wedge \mathrm{d}{x}^{\nu }$ , a direct computation yields

$${\mathcal{F}}_{\mu v}={\partial }_{\mu }{\mathcal{A}}_v-{\partial }_v{\mathcal{A}}_{\mu }+\left[{\mathcal{A}}_{\mu },{\mathcal{A}}_v\right].\text{\hspace{1em}(10.39)}$$

$\mathcal{F}$ is also called the curvature two-form and is identified with the (Yang-Mills) field strength. To avoid confusion, we call $\Omega$ the curvature and $\mathcal{F}$ the (Yang-Mills) field strength. Since ${\mathcal{A}}_{\mu }$ and ${\mathcal{F}}_{\mu \nu }$ are $\mathfrak{g}$ -valued functions, they can be expanded in terms of the basis $\{T_{\alpha }\}$ of $\mathfrak{g}$ as

$${\mathcal{A}}_{\mu }=A_{\mu }^{\alpha }{T}_{\alpha }{\mathcal{F}}_{\mu v}=F_{\mu v}^{\alpha }{T}_{\alpha }.\text{\hspace{1em}(10.40)}$$

The basis vectors satisfy the usual commutation relations $[T_{\alpha },T_{\beta }]=f_{\alpha \beta }^{\gamma }{T}_{\gamma }$ . We then obtain the well-known expression

$$F_{\mu v}^{\alpha }={\partial }_{\mu }{A}_v^{\alpha }-{\partial }_v{A}_{\mu }^{\alpha }+f_{\beta \gamma }^{\alpha }{A}_{\mu }^{\beta }{A}_v^{\gamma }.\text{\hspace{1em}(10.41)}$$

Theorem 10.5

Let $U_i$ and $U_j$ be overlapping charts of $M$ and let ${\mathfrak{F}}_i$ and ${\mathfrak{F}}_j$ be field strengths on the respective charts. On $U_i\cap U_j$ , they satisfy the compatibility condition,

$${\mathcal{F}}_j=A{\mathrm{d}}_{t_{ij}^{-1}}{\mathcal{F}}_i=t_{ij}^{-1}{\mathcal{F}}_i{t}_{ij}\text{\hspace{1em}(10.42)}$$

where $t_{ij}$ is the transition function on $U_i\cap U_j$ .

Proof

Introduce the corresponding gauge potentials ${\mathcal{A}}_i$ and ${\mathcal{A}}_j$ ,

$${\mathcal{F}}_i=\mathrm{d}{\mathcal{A}}_i+{\mathcal{A}}_i\wedge {\mathcal{A}}_i{\mathcal{F}}_j=\mathrm{d}{\mathcal{A}}_j+{\mathcal{A}}_j\wedge {\mathcal{A}}_j.$$

Substituting ${\mathcal{A}}_j=t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}$ into ${\mathfrak{F}}_j$ , we verify that

$$\begin{array}{rl}& {\mathcal{F}}_j=\mathrm{d}\left(t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}\right)\\ & +\left(t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}\right)\wedge \left(t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}\right)\\ & =[-t_{ij}^{-1}d{t}_{ij}\wedge t_{ij}^{-1}{\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}d{\mathcal{A}}_i{t}_{ij}\\ & -t_{ij}^{-1}{\mathcal{A}}_i\wedge \mathrm{d}{t}_{ij}-t_{ij}^{-1}\mathrm{d}{t}_{ij}{t}_{ij}^{-1}\wedge \mathrm{d}{t}_{ij}]\\ & +[t_{ij}^{-1}{\mathcal{A}}_i\wedge {\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}{\mathcal{A}}_i\wedge \mathrm{d}{t}_{ij}\\ & +t_{ij}^{-1}\mathrm{d}{t}_{ij}{t}_{ij}^{-1}\wedge {\mathcal{A}}_i{t}_{ij}+t_{ij}^{-1}\mathrm{d}{t}_{ij}\wedge t_{ij}^{-1}\mathrm{d}{t}_{ij}]\\ & =t_{ij}^{-1}\left(\mathrm{d}{\mathcal{A}}_i+{\mathcal{A}}_i\wedge {\mathcal{A}}_i\right)t_{ij}=t_{ij}^{-1}{\mathcal{F}}_i{t}_{ij}\end{array}$$

where use has been made of the identity ${\mathrm{dt}}^{-1}=-t^{-1}\mathrm{dt}{t}^{-1}$

Exercise 10.7

The gauge potential $\mathcal{A}$ is called a pure gauge if $\mathcal{A}$ is written locally as $\mathcal{A}=g^{-1}\mathrm{d}g$ . Show that the field strength $\mathcal{F}$ vanishes for a pure gauge $\mathcal{A}$ . [It can be shown that the converse is also true. If $\mathcal{F}=0$ on a chart $U$ , the gauge potential may be expressed locally as $\mathcal{A}=g^{-1}\mathrm{d}g$ .]

10.3.5 The Bianchi identity

Since $\omega$ and $\Omega$ are $\mathfrak{g}$ -valued, we expand them in terms of the basis $\{T_{\alpha }\}$ of $\mathfrak{g}$ as $\omega ={\omega }^{\alpha }{T}_{\alpha }$ , $\Omega ={\Omega }^{\alpha }{T}_{\alpha }$ . Then (10.32b) becomes

$${\Omega }^{\alpha }={\mathrm{d}}_P{\omega }^{\alpha }+f_{\beta \gamma }^{\alpha }{\omega }^{\beta }\wedge {\omega }^{\gamma }.\text{\hspace{1em}(10.43)}$$

Exterior differentiation of (10.43) yields

$${\mathrm{d}}_P{\Omega }^{\alpha }=f_{\beta \gamma }^{\alpha }{\mathrm{d}}_P{\omega }^{\beta }\wedge {\omega }^{\gamma }+f_{\beta \gamma }^{\alpha }{\omega }^{\beta }\wedge {\mathrm{d}}_P{\omega }^{\gamma }.\text{\hspace{1em}(10.44)}$$

If we note that $\omega (X)=0$ for a horizontal vector $X$ , we find

$$\mathrm{D}\Omega (X,Y,Z)={\mathrm{d}}_P\Omega (X^H,Y^H,Z^H)=0$$

where $X,Y,Z\in T_{u}P$ . Thus, we have proved the Bianchi identity

$$\mathrm{D}\Omega =0.\text{\hspace{1em}(10.45)}$$

Let us find the local form of the Bianchi identity. Operating with ${\sigma }^{\ast }$ on (10.44), we find that ${\sigma }^{\ast }{\mathrm{d}}_P\Omega =\mathrm{d}\cdot {\sigma }^{\ast }\Omega =\mathrm{d}\mathcal{F}$ for the LHS and

$$\begin{array}{rl}{\sigma }^{\ast }({\mathrm{d}}_P\omega \wedge \omega -\omega \wedge {\mathrm{d}}_P\omega )& =\mathrm{d}{\sigma }^{\ast }\omega \wedge {\sigma }^{\ast }\omega -{\sigma }^{\ast }\omega \wedge \mathrm{d}{\sigma }^{\ast }\omega \\ & =\mathrm{d}\mathcal{A}\wedge \mathcal{A}-\mathcal{A}\wedge \mathrm{d}\mathcal{A}=\mathcal{F}\wedge \mathcal{A}-\mathcal{A}\wedge \mathcal{F}\end{array}$$

for the RHS. Thus, we have obtained that

$$\mathcal{D}\mathcal{F}=\mathrm{d}\mathcal{F}+\mathcal{A}\wedge \mathcal{F}-\mathcal{F}\wedge \mathcal{A}=\mathrm{d}\mathcal{F}+[\mathcal{A},\mathcal{F}]=0\text{\hspace{1em}(10.46)}$$

where the action of $\mathcal{D}$ on a $\mathfrak{g}$ -valued $p$ -form $\eta$ on $M$ is defined by

$$\mathcal{D}\eta \equiv \mathrm{d}\eta +[\mathcal{A},\eta ].\text{\hspace{1em}(10.47)}$$

Note that $\mathcal{D}\mathcal{F}=\mathrm{d}\mathcal{F}$ for $G=\mathrm{U}(1)$ .

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