How can I mathematically describe the parallel transport in the Roman soldiers example? I've been trying to understand parallel transport. Many of the descriptions present a mathematical version: $\nabla_V X = 0$.
And/or they present an example involving soldiers (usually Roman) carrying spears around the world. Two soldiers start at the North Pole $[x,y,z]=[0,1,0]$ both pointing their spears towards $[1,0,0]$. One soldier walks forwards to the easternmost point on the Earth's equator $[x,y,z]=[1,0,0]$. His spear stays tangent to the Earth and points southwards when he arrives $[0,-1,0]$. The other solider walks sideways to the equator $[x,y,z]=[0,0,1]$ and his spear still points eastwards $[1,0,0]$ when he arrives. Now both soldiers are on the equator with spears pointing in different directions, one south, the other east. If they walk towards each other, the spears will still point different ways.
The problem is that both soldiers held their spears "parallel" while walking, so there are two seemingly parallel paths between two points which end up with different results. And this is bad, and not just for the exhausted Romans.
Solving this issue involves rotating the spears as the soldiers walk towards each other along the equator, so they point in the same direction when they meet. I would like to see a mathematical description of this.
For my own attempt to describe this problem, I started with a "congruence of curves" of concentric spheres $x^2+y^2+z^2=C$. To find a tangent vector field to these curves, I found the normal vector $[2x,2y,2z]$ and picked a vector field perpendicular to it such as $V=[y+z,-x,-x]$ so that $[2x,2y,2x] \cdot V = 0$.
By my understanding, a tensor $X$ will be parallel transported along this congruence of curves (the sphere) if the covariant derivative of $X$ contracted with the tangent vector field $V$ equals $0$: $\nabla_V X=0$. The tensor $X=[x^2+y^2+z^2,0,0]$ can do this:
$$\nabla_V X =
(\partial X + \Gamma X)  V = (\partial X + 0) V =
\begin{bmatrix} 2x & 2y & 2z \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
\begin{bmatrix} y+z \\ -x \\ -x \end{bmatrix} =
\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$$
This tensor $X$ points the same direction everywhere on the sphere $[1,0,0]$.
So this describes a solution to the Romans' concerns. The first soldier just needs to raise his spear as he walks so it always points at $[1,0,0]$. But although it is a solution, I can't (mathematically) see what the problem was in the first place.
On the other hand, a vector field which describes the Romans' spear movements as above, including the equatorial rotation is $Y=[y^2+z^2,-x^2,0]$. But this field does not get parallel transported around the sphere because $\nabla_V Y \ne 0$. This rotation is supposedly how parallel transport fixes the Romans' spear pointing issues, but in this case it doesn't actually seem to be an example of parallel transport.
I also tried doing these calculations in polar coordinates but I got similarly lost. I feel like I'm missing a piece of the puzzle. Any pointers much appreciated.
 A: The answer to my question is simpler than I suspected. It is fairly easy to describe the movements of both soldiers mathematically.
The first soldier's spear is being transported along $X_1 = [y^2+z^2, -x^2, 0]$. The second soldier's spear along $X_2 = [y^2+z^2, 0, -x^2]$. Both are valid parallel transports relative to the tangent vector field $V=[0, -z, y]$. For example, with the first one:
$$\nabla_{V} X_1 =
(\partial X_1 + \Gamma X_1)  V = (\partial X_1 + 0) V =
\begin{bmatrix} 0 & 2y & 2z \\ -2x & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
\begin{bmatrix} 0 \\ -z \\ y \end{bmatrix} =
\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$$
But they can both end up at the same point on the sphere pointing in different directions. At the point $[1,0,0]$, $X_1=[0,-1,0]$ and $X_2=[0,0,-1]$.
This isn't possible on a flat surface, or even on a cylinder. On a cylinder, spears which are parallel transported along different paths will point in the same direction when they meet again.
The reason for this is that a sphere is intrinsically curved. If you try to flatten out the sphere (by converting to spherical polar coordinates for instance), then the corresponding parallel transports within spherical polar coordinates will also end up pointing in a different directions. That's because the sphere's $\Gamma$ leads to  a non-zero Riemann curvature tensor (a 4 dimensional tensor computed from $\Gamma$).  Unlike a cylinder whose Riemann curvature tensor is zero.
I also made a mistake in the original post. The vector field I mentioned $Y=[y^2+z^2,−x^2,0]$ does describe parallel transport (including the equatorial rotation) relative to the tangent vector field $[0,-z,y]$.
