Auto-parallel transport in general relativity Let $\gamma$ be a curve and the mapping $\tau_{t,s}$ the auto-parallel transport along $\gamma$ from $\gamma(s)$ to $\gamma(t)$.

Theorem 13.1
Let $X$ be a vector field along $\gamma$. Then
$$\nabla_{\dot{\gamma}} X(\gamma(t))=\left.\frac{d}{d s}\right|_{s=t} \tau_{t, s} X(\gamma(s)).$$

(Straumann, 10.1007/978-3-662-11827-6, Chapter 13, page 580.)

Two questions regarding this theorem:

*

*What is the physical interpretation/intuition of this theorem? I think I understand what it means for a vector field to be auto-parallel along a curve $\gamma$ (aka satisfying $\nabla_{\dot\gamma}X=0$), but I have a hard time understanding what this theorem is supposed to tell me and in what physical context this may be relevant.. (Not sure if important, but I'm studying this in the context of General relativity.)

*This question is rather technical, but can somebody maybe explain to me why
$$\left.\frac{d}{d t}\right|_{t=s}\left(\tau_{t, s}\right)_{j}^{i}=-\Gamma_{k j}^{i} \dot{x}^{k}$$
holds, where we have chosen a local chart, aka some coordinates (part of the prove of this theorem)?

PS: This question was originally asked on PhysicsSE (see here), but I was told to post this here.
 A: You should be more careful about what spaces your maps are living in. $\tau_{s,t}$ is a map $T_{\gamma(s)}M \to T_{\gamma(t)}M$. In particular with $s$ and $t$ varying the domain and codomain change, so you can't take derivatives a priori. Of course these spaces can all be identified in a variety of ways, eg by parallel transport or in some coordinate chart, still you should make it explicit how these identifications work and why the result is independent of any choice. I will just continue with your notation and not spend any thoughts about this.
For your point 1:
If the vector field were parallel then the right-hand side would be zero. Thus the theorem tells you that you may interpret the covariant derivative of $X$ in direction $v$ to be a measure of how far away $X$ is from being parallel in the direction $v$. In particular the direction that $\nabla_v X$ is pointing is the direction towards which you should correct $X$ locally if you want to make $X$ parallel.
For your point 2:
The parallel transport $f(t) = \tau_{t,s}(X)$ of $X$ along $\gamma$ starting at $\gamma(s)$ is defined to be the solution to the ODE
$$\nabla_{\gamma(t)}f^i(t) =\frac\partial{\partial t} f^i(t)+ \Gamma^i_{jk}\ \dot\gamma^j(t) f^k(t) = 0, \text{ with initial condition } f^i(s)=X(\gamma(s))^i$$
By rearranging the above equation you find:
$$\frac\partial{\partial t}\tau_{t,s}(X)^i = \frac\partial{\partial t}f^i(t) = -\Gamma_{jk}^i \dot\gamma^j(t) f^k(t) $$
putting in $t=s$ gives you:
$$\frac\partial{\partial t}\tau_{t,s}(X)^i\lvert_{s=t} =- \Gamma^i_{jk}\dot\gamma^j(s) f^k(s) = -\Gamma_{jk}^i \dot \gamma(s) X^k(s)$$
which is has the same content as the expression you are looking for.
