I have a question about parallel transport of a vector.
In picture A) we parallel transport a vector on a (red)curve in a flat plane. This can be clearly seen as the (blue)vector is constant in direction as we move along the curve.
Now here is the problem I am having, when we parallel transport a vector on a surface on a sphere, like along the red curve in picture B) I would expect to get blue vectors in picture B) but instead its correct to get vectors in picture C)
Why is this?
I understand that the vectors are "constant" in surface spherical coordinates for picture C) but ultimately by formulation of parallel transport
\begin{equation}
\begin{aligned}
\frac{d\textbf{v}}{du} = \sum_i\left(\frac{dv^i}{du}+\sum_{j,k}v^j\Gamma^i_{jk}\frac{dx^k}{du}\right)\textbf{e}_i = \textbf{0}
\end{aligned}
\end{equation}
it should be constant in Cartesian coordinates (like in picture A and B), but not what is shown in picture C).
Edit
Thanks for the answers, my comment is too long so I had to post it here:
From what I understand the coordinate basis dictates the metric which in turn dictates the christoff symbols. Suppose I live in an Euclidean space: If I were to use spherical coordinates with the corresponding metric and I transported a vector along the curve $c(u): \theta = u, \phi = 0,r=\rho$, I would get the same result as in pic B) as the components $v^i$ of the OP equation together with the chosen basis combine to make constant vectors as shown in the picture. It turns out regardless of metric\basis that I use I get the result of picture B).
So then I make the assumption that for me to get picture C), I must live in surface world of the sphere, and define coordinates systems according to it(no longer from an Euclidean space)then parallel trasport makes sense. Basically like walking on the surface of the Earth and as I loop around the Equator, my direction vector be parallel to the surface(but not parallel in the Euclidean space). But then this means that a metric $ds^2 = r^2d\theta + r^2sin^2(\theta)d\phi$ in Euclidean space will be like the metric $ds^2 = dx^2 + dy^2$ in our spherical surface world if origin were at the center of sphere(atleast locally)? I guess what I am asking is that I can't use the equation in the OP, and by simply changing the metric\basis, go from picture B) to picture C), but have to decide what manifold I am working in first?
Best Answer
You need to work in the chart of your sphere with the induced metric. If you use spherical coordinates, your parallel transport equation yields the picture C).
The reason why B) is nonsense, is because the parallel transport of vectors remain in the tangent plane of your (sub-)manifold. You can not leave the tangent plane of the sphere (which happens in the picture B) ).
If you were using the euclidean metric in the euclidean space and parallel transport along the path in picture B) that happens to be on the sphere, then the parallel transport will look like picture B).