Parallel Translation is Path Independent iff Manifold is Flat Problem. Let $M$ be a smooth Riemannian manifold and $\nabla$ be the Levi-Civita connection. Then the following are equivalent


*

*$R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\equiv 0$

*For all $p,q\in M$, parallel translation along a curve segment from $p$ to $q$ is independent of the curve.


My attempt. I've tried for several hours now, but I'm out of ideas. Initially, I tried analyzing the differential equations which characterize  a vector field along a curve being parallel, but I couldn't get anything out of it.
For the implication (2)$\implies$(1), it would be enough to show $R(\partial_i,\partial_j)\partial_k\equiv0$. Considering curves through some $p\in M$ which have the form $t\mapsto te_i$ in some chart around $p$ seemed promising but didn't get me far.
In particular I'm also not sure how the fact that $\nabla$ is the Levi-Civita conncetion comes into play. I know that compatibility of $\nabla$ with the metric is equivalent to the parallel translation being an isometry. Can we use that?
I would really appreciate some help.
 A: For the path independence part:
Recall that $\nabla$ is flat iff $R^{a}_{bcd}=0$ everywhere on $M$. If the parallel transport of vectors along a path on $M$ relative to $\nabla$ is indeed path independent then $\nabla$ is flat.
Conversely, if $\nabla$ is flat then, at least locally, parallel transport of vectors on $M$ relative to $\nabla$ is path independent.
To prove this, firstly assume that parallel transport of vectors on $M$ is path independent. Let $p$ be any point in $M$ and let $\boldsymbol{\zeta^{a}}$ be any vector at $p$. Extend $\boldsymbol{\zeta^{a}}$ to a smooth vector field $\boldsymbol{\pi^{a}}$ by parallel transporting $\boldsymbol{\zeta^{a}}$ to through any curve to all points of $M$. The resulting field is constant in the sense that $\nabla_{a}\pi^{b}=0$ everywhere; this is equivalent to saying that all directional derivatives vanish everywhere.
Hence,
\begin{align}
R^{a}_{bcd}\pi^{b}&= -2 \nabla_{[c} \nabla_{d]}\pi^{b} \\
                    &=0
\end{align}
everywhere, and in particular, at $p$. Since the choice of $\zeta$ was arbitrary, $R^{a}_{bcd}=0$.
Now for the converse argument. Assume that $R^{a}_{bcd}=0$ on $M$. In order we show that, at least locally, parallel transport is path-independent it will be sufficient to show that at any given $\zeta^{a}$ at arbitrary $p$, we can extend $\zeta^{a}$ to any smooth $\boldsymbol{\pi^{a}}$ on some open set $O$ containing $p$ that is constant. That is to say, 
$$\nabla_{a}\pi^{b}=0$$
everywhere on $O$.
this last part is tricky insofar as one needs to find local coordinates that satisfies a set of PDEs such that
\begin{align}
\nabla_{a}\pi^{b}&=0\\
\pi^{a}_{|p}&=\zeta^{a}
\end{align}
the set of PDEs that fall out of computation will be satisfied iff (as it turns out through copious lines of algebra) $R^{a}_{bcd}=0$ expressed in local coordinates.
A: For the direction: $2) \Rightarrow 1)$ use this formula, which connects the curvature tensor to parallel transport along small closed loops.
$1) \Rightarrow 2):$ (which as pointed by Anthony Carapetis hols only locally):
Any flat mnaifold is locally isometric to Euclidean space, so you can transfer the parallel transport to $(\mathbb{R}^n,e)$, do it there, and then project back to your manifold. Since parallelt transport in $(\mathbb{R}^n,e)$ is trivially path independent, we finished.
