Parallel Transport Equations I have a question about parallel transport that I'm very confused about and would appreciate some help. The question reads:

What vector field $X$ on the unit 2-sphere in $\mathbb{R}^3$ has rotations around the $z$-axis as flows? The orbits of these flows would be the lines of latitude. Solve the parallel transport equations $\nabla_X(V_i)=0$, for $V_1, V_2$ the elements of a basis of the tangent plane, along the curve of latitude 45 degrees. What do the $V_i$ come back to once one has transported them all the way around the circle?

I am confused by which vector field the question is talking about. Would this field just be the sphere parametrized as:
$$(x,y,z) = (r\cos(\theta)\sin(\varphi), r\sin(\theta)\sin(\varphi), r\cos(\varphi))$$
Where $\varphi$ is constant? 
 A: This isn't a direct answer to your question about the parallel transport equations, but the standard geometric solution. Assuming a unit sphere, parallel transport around the latitude $L_{\varphi}$ making angle $\varphi$ with the equator can be found by constructing the cone tangent to the sphere along $L_{\varphi}$, then cutting the cone along a generator, rolling it flat, and performing parallel transport of a tangent vector around an arc of a circle.
Elementary geometry shows that the unrolled cone is a sector of a disk of radius $\cot\varphi$ whose central angle is $2\pi\sin\varphi$. Parallel transport around $L_{\varphi}$ in the northern hemisphere rotates a vector clockwise (looking "down" at the tangent plane, toward the center of the sphere) by an angle $2\pi\sin\varphi$. The animation is a full-circle pan around the sphere.

A: For points on the equator.  you may find all elipsoids that are present on the north hemisphere exert the same integral sum of vector acceleration around a regular surface.  their transport should be similar to a circle transport at the equator.  it is the case that a transport with some location       a (t)i that the integral sum of acceleration is greater than the distance around the world at the equator so it is not similar.
