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.

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.
