Curvature

If we have a loop $\gamma$ whose image is in $U_\alpha$ then we can apply (2.4) to obtain

   hol$$\ (\nabla, \gamma) = \exp \ (-\int_{\gamma} \ A_{\alpha}).$$

If $\gamma$ is the boundary of a disk $D$ then by Stokes' theorem we have

$$\ (\nabla, \gamma) = \exp -\int_{D} \ d A_{\alpha}.$$ (2.5)

Consider the two-forms $dA_{\alpha}$. From (2.2) we have

$$d A_{\alpha} = d A_{\beta} + d \left (g_{\alpha \beta}^{-1} dg_{\alpha \beta} \right )$$

$$= d A_{\beta} - g_{\alpha \b }^{-1}d g_{\alpha \b }g_{\alpha \b }^{-1} \wedge dg_{\alpha \beta} +g_{\alpha \beta}^{-1}d d g_{\alpha \beta}$$

$$= d A_{\beta}.$$

So the two-forms $dA_{\alpha}$ agree on the intersections of the open sets in the cover and hence define a global two form that we denote by $F$ and call the curvature of $\nabla$. Then we have

Proposition 2.3   If $L \to M$ is a line bundle with connection $\nabla$ and $\Sigma$ is a compact submanifold of $M$ with boundary a loop $\gamma$ then

   hol$$\ (\nabla, \gamma ) = \exp \ - \int_{D} \ F$$

Proof. Notice that (2.5) gives the required result if $\Sigma$ is a disk which is inside one of the $U_\alpha$. Now consider a general $\Sigma$. By compactness we can triangulate $\Sigma$ in such a way that each of the triangles is in some $U_\alpha$. Now we can apply (2.5) to each triangle and note that the holonomy up and down the interior edges cancels to give the required result. $\qedsymbol$

Example 2.6   We calculate the holonomy of the standard connection on the tangent bundle of $S^2$. Let us use polar co-ordinates: The co-ordinate tangent vectors are:

$$\frac{\partial\phantom{\theta}}{\theta} = ( -\sin(\theta) \sin(\phi) , \cos(\theta) \sin(\phi), 0)$$

$$\frac{\partial\phantom{\phi}}{\phi} = (\cos(\theta) \cos(\phi), \sin(\theta) \cos(\phi), - \sin(\phi))$$

Taking the cross product of these and normalising gives the unit normal

$$\hat n = (\cos(\theta) \sin(\phi), \sin(\theta) \sin( \phi), \cos(\phi))$$

$$= \sin(\phi) \frac{\partial\phantom{\phi}}{\phi} \times \frac{\partial\phantom{\theta}}{\theta}$$

To calculate the connection we need a non-vanishing section $s$ we take

$$s = (-\sin(\theta), \cos(\theta), 0)$$

and then

$$ds = (- \cos(\theta), - \sin(\theta), 0) d \theta$$

so that

$$\nabla s = \pi (ds)$$

$$= ds - < ds, \hat n > \hat n$$

$$= (- \cos(\theta), - \sin(\theta), 0) d \theta$$

$$+ \sin(\phi) \ (\cos(\theta) \sin(\phi), \sin(\theta) \sin(\phi), \cos(\phi)) d\theta$$

$$= (- \cos(\theta) \cos^{2}(\phi), - \sin(\theta) \cos^2(\phi), \cos(\phi) \sin(\phi)) d \theta$$

$$= \cos(\phi) \hat n \times s$$

$$= i \cos(\phi) s$$

Hence $A = i \cos(\phi) d\theta$ and $F = i \sin(\phi) d \theta \wedge d \phi$. To understand what this two form is note that the volume form on the two-sphere is ${\rm vol} = - \sin(\phi) d \theta \wedge d \phi$ and hence $F = i {\rm vol}$ The region bounded by the path in Figure 4 has area $\theta$. If we call that region $D$ we conclude that

$$\exp \bigl(- \int_{D} F\bigr) = \exp i\theta.$$

Note that this agrees with the previous calculation for the holonomy around this path.