Flat Connection in $\mathbb{R}^n$

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I've just been playing around with the simplest notion of a connection on $\mathbb{R}^n$, that is

$$(\nabla_{v}X)^i=v(X^i)$$

with $X$ a vector field and $v$ a tangent vector at $p$.

From the definition of parallel transport, I know that we should regard $v$ as parallel to $X$ at $p$ iff $\nabla_vX=0$. I can't quite see how this relates to the notion of parallelism in $\mathbb{R}^n$ though!

I know that $v(X^i)$ tells me the rate of change of the component function $X^i$ in the direction of $v$ at $p$. I've convinced myself that the geometrical consequences of $v(X^i) = 0$ are the following.

In an infinitesimal neighbourhood of $p$ the vector field $X$ remains the same as you transport it along the straight line defined by $v$. Then for $v$ an arbitrary vector field we have that $X$ is a parallel vector field along the integral curves of $v$.

Are these the strongest things we can say? Many thanks!

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Parallel in the old fashion Euclidean sense and parallel in the Riemannian geometry sense have little to do with one another.

In $\mathbb{R}^2$, consider the vector field which always points right and has unit length. That is, $v_{(x,y)} = (1,0)$ at every point $p\in \mathbb{R}^2$.

First, let $X_{(x,y)} = (e^x , 0)$. This is a vector field which always points right but as you get larger $x$ values, the arrows get longer. In the classical geometry sense, $X$ and $v$ are parallel at every point. However, if you compute, you'll see that $\nabla_v X \neq 0$, so these are not parallel in the Riemannian geometry sense.

Second, let $X_{(x,y)} = (0,1)$. This is a vector field which always points up with length $1$. In the classical geometry sense, since $v$ points right and $X$ points up, there is no way they are parallel anywhere.

Nonetheless, $\nabla_v(X) = 0$ so they are parallel in the Riemannian geometry sense.

The idea of Riemannian parallelism is that, to say $X$ is parallel along $v$ should mean that as you move along $v$, $X$ doesn't change. In the first example, $X$ does change as you move along $v$, while in the second example, it doesn't.