Triangle on a sphere (Schutz, 6.10)

[hellfire]

A first course in general relativity. B.F. Schutz. Exercise 6.10:
A straight line in a sphere is a great circle, and it is well known that the sum of interior angles of any triangle on a sphere whose sides are arcs of great circles exceeds 180°. Show that the amount by which a vector is rotated by parallel transport around such a triangle equals the excess of the sum of the angles over 180°.

How to proceed to prove this?

 

[Andrew Mason]

A first course in general relativity. B.F. Schutz. Exercise 6.10:
A straight line in a sphere is a great circle, and it is well known that the sum of interior angles of any triangle on a sphere whose sides are arcs of great circles exceeds 180°. Show that the amount by which a vector is rotated by parallel transport around such a triangle equals the excess of the sum of the angles over 180°.
How to proceed to prove this?

The first step is to recognize that since each side of the triangle is a great circle (geodesic), a vector parallel transported along that geodesic experiences no rotation. The rotation occurs only at each vertex where the vector is rotated by an angle equal to 180 less the angle made by the two sides of the triangle. So the total angle of rotation after returning to its starting point is (180-a1 + 180 - a2 + 180 - a3). But since we know that a1 + a2 + a3 = 180 + gamma (where gamma is the difference you are looking for):
\theta = 540 - (180 + \gamma)) = 360 - \gamma

AM

 

[hellfire]

Thank you Andrew. This was quite easier than I had supposed.