Calculating the holonomy on a sphere

[demonelite123]

There are 2 great circles C1 and C2 on a sphere and they are at an angle of \alpha with each other at the north pole. Form a closed path from the north pole along C1 to the equator, then along the equator to C2, and back up along C2 to the north pole. Show that the holonomy is also the angle \alpha.

my book talks about finding the holonomy along a curve and give the simple formula \theta(t) = \theta(0) - \int \omega_{21} dt where \omega_{21} is given by \nabla_{v} \frac{x_u}{sqrt{E}} = \omega_{21} \frac{x_v}{sqrt{G}}.

The problem is that along each piece of the path, C1, C2, and the equator, the holonomy is 0 along each individually, but when put together to form the closed path described above, there is obviously a nonzero holonomy. how do i calculate the holonomy along this path which is made up of 3 different curves? I've tried to find a parametrization for the closed path but could not think of one. can someone help guide me along in this problem?

 

[Dick]

This is easy enough to just do in your head. Saying the holonomy is 0 along a great circle just means the transported vector maintains a constant angle with respect to the tangent vector of the curve. When you switch from one great circle to another, the tangent vector changes but the orientation of the transported vector doesn't. Try and picture what happens in your head first.

 

[demonelite123]

i can see that given a tangent vector at the north pole, transporting it along C1 will keep it a tangent vector of C1. Then transporting it along the equator, while the vector keeps its orientation with respect to the tangent vector of the equator it rotates through an angle \alpha since the vector becomes a tangent vector for C2. Then traveling back up C2 it remains a tangent vector of C2 and since the circles where an \alpha apart, so will the transported tangent vectors. Any vector chosen will be rotated the same amount so even vectors that are not tangent vectors to C1 will work.

would an explanation such as this suffice for an answer to the question? my explanation makes sense to me but it seems very hand wavy in terms of math. is this how i would "show" that the holonomy is \alpha?

 

[Dick]

i can see that given a tangent vector at the north pole, transporting it along C1 will keep it a tangent vector of C1. Then transporting it along the equator, while the vector keeps its orientation with respect to the tangent vector of the equator it rotates through an angle \alpha since the vector becomes a tangent vector for C2. Then traveling back up C2 it remains a tangent vector of C2 and since the circles where an \alpha apart, so will the transported tangent vectors. Any vector chosen will be rotated the same amount so even vectors that are not tangent vectors to C1 will work.

would an explanation such as this suffice for an answer to the question? my explanation makes sense to me but it seems very hand wavy in terms of math. is this how i would "show" that the holonomy is \alpha?

Once you've got that picture in your mind, it should be clear that the holonomy angle is related to the difference between the of the sum of the angles of the spherical triangle and pi. Any theorem like that? You can also state this quantity in terms of an integral over the curvature inside the triangle if you've done the Gauss-Bonnet theorem. If you don't have anything proved, they you'll just have to try and make your hand wavy answer as clear as you can.