# Path on a sphere

I am trying to solve a exercise problem in GR on a "triangle" whose sides are great circles of a sphere of radius $R$. So this is the triangle that I chose (coordiantes are written as $(r, \theta, \phi)$, $\theta$ being the angle from the z axis. Also $\lambda$ is the parameter of the path):

Path 1: $$(R, \pi/2, 0) \rightarrow (R, \pi/2, \pi/2)$$ $$\text{Equation of the path} \rightarrow \theta = \pi/2$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, \pi/2, \phi (\lambda))$$

Path 2: $$(R, \pi/2, \pi/2) \rightarrow (R, 0, \pi/2)$$ $$\text{Equation of the path} \rightarrow \phi = \pi/2$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, \theta (t), \pi/2)$$

Now when I define path 3 which takes me back to $(R, \pi/2, 0)$ I run into the problem that $\phi$ changes from $\pi/2$ to $0$ because of the path that I am taking. I do understand that it is because of the coordinate singularity at the pole.

So, how do I define the coordinates and equation of the path 3?

Path 3: $$(R, 0, \pi/2) \rightarrow (R, 0, 0)$$ $$\text{Equation of the path} \rightarrow \theta = 0$$ $$\text{Coordinates along the path } \rightarrow x^\alpha = (R, 0, \phi (\lambda1))$$
The problem is indeed the singularity at the pole where the spherical coordinates coordinates $(\theta,\phi)$ are not properly defined (the inverse of the metric tensor does not exist there). In the original problem I was trying to parallel transport a vector along the sides of a "triangle" whose sides are great circles on a sphere of radius $R$. The only way around the mentioned problem is to relocate the triangle to another location where the pole is avoided. I was able to successfully parallel transport the vector around the relocated triangle without running in this problem.