Auto-parallel transport in general relativity Let the mapping $\tau_{t,s}$ be 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:

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*What is the physical interpretation of this? 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: Not sure if this would be a better fit for MathSE, but since 1) asks for specific physics context, I hope this is acceptable.
 A: First of all, autoparallel is usually a term used for geodesic vector fields. Eg. $X$ is autoparallel if $\nabla_XX=0$. The proper term to be used here is simply "parallel" or "parallel transported/propagated".

What is the physical interpretation of this? I think I understand what it means for a vector field to be auto-parallel along a curve  (aka satisfying ∇˙=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.)

The theorem says that covariant differentiation is related to parallel transport, and it also tells one exactly how. I don't remember right now, but I think Straumann defines connections via the covariant derivative itself. Then from the covariant derivative he derives the parallel transport map $\tau$. Then this theorem states that this relationship is reversible.
In other words, the definition he gives for the covariant derivative is purely algebraic (does not involve limits), but now he shows that if $\gamma$ is a curve then $$ \nabla_{\dot \gamma(t)}X=\lim_{s\rightarrow 0}\frac{\text{Parallel transport of value of }X\text{ at }\gamma(t+s)\text{ back to }\gamma(t)-\text{Value of }X\text{ at }\gamma(t)}{s}. $$

This question is rather technical, but can somebody maybe explain to me why ...

Suppose that $\gamma$ is a smooth curve, $v\in T_{\gamma(t)}M$ is a vector at $\gamma(t)$, and $X$ is a vector field along $\gamma$ that is the unique parallel transport of $v$ along $\gamma$, eg. $X(s)=\tau_{s,t}v$ and $\nabla_{\dot\gamma(s)}X=0$ (for all $s$).
Then if we take a local chart, expand $X(s)$ in a Taylor series at $t$: $$ \left( \tau_{s,t}v\right)^ i= X^ i(s)=X^i(t)+\dot X^i(t)(s-t)+\frac{1}{2}\ddot X^i(t)(s-t)^2 +... \\ =v^ i+\dot X^i(t)(s-t)+\frac{1}{2}\ddot X^i(t)(s-t)^2+...$$
Now, in local components, the parallel propagation condition $\nabla_{\dot\gamma}X=0$ is $$\dot X^i(s)=-\Gamma^i_{jk}(\gamma(s))\dot\gamma^j(s)X^k(s)$$ for all $s$, so for $s=t$, this is $$ \dot X^i(t)=-\Gamma^i_{jk}(\gamma(t))\dot\gamma^j(t)v^k. $$
Reinserting this into the Taylor expansion gives $$ \left(\tau_{s,t}v\right)^i=v^i-\Gamma^i_{jk}(\gamma(t))\dot\gamma^j(t)v^k(s-t)+O((t-s)^2). $$
Now taking the $s$-derivative at $s=t$ gives $$ \frac{\mathrm d}{\mathrm ds}\left(\tau_{s,t}v\right)^i |_{s=t}=-\Gamma^i_{jk}(\gamma(t))\dot\gamma^j(t)v^k. $$
The parallel transport is a linear map, so in a chart (that contains both ends of the path) it has a matrix representation and hence the left hand side can be written as $$\frac{\mathrm d}{\mathrm ds}\left(\tau_{s,t}v\right)^i |_{s=t}=\frac{\mathrm d}{\mathrm ds}\left(\tau_{s,t}\right)^i_{ k}v^k|_{s=t}.$$
Since the vector $v$ is arbitrary, we may "cancel" it from both sides to obtain $$ \frac{\mathrm d}{\mathrm ds}\left(\tau_{s,t}\right)^i_{ k}|_{s=t}=-\Gamma^i_{jk}(\gamma(t))\dot\gamma^j(t). $$
