I'm trying to understand covariant derivative in the simplest case of $\mathbb{R}^n$ here.
On a Riemannian manifold $(M,g)$, for a connection $\nabla$ which is compatible and torsion-free, I have been shown that the Koszul formula
$$ 2g(\nabla_XY, Z) = \partial_X (g(Y,Z)) + \partial_Y (g(X,Z)) - \partial_Z (g(X,Y))+ g([X,Y],Z) - g([X,Z],Y) - g([Y,Z],X)$$
holds.
I have just started studying these thing, but I'm pretty sure $\mathbb{R}^n$ and the usuals operators have all the good properties (compatibility of the connection, etc...).
1) Is it true that for $\textbf{X, Y}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ smooth vector fields, I can think of $\nabla_\textbf{X}\textbf{Y}$ as the field $\space p\mapsto\frac{\partial{\textbf{Y}}}{\partial\textbf{X}(p)}|_p$ ? (derivative of the field $\mathbf{Y}$ along the direction $\mathbf{X}(p)$ )
2) In demostrating the Koszul formula, I've written
$$ ...\space g(\nabla_XY,Z) = g([X,Y],Z)+g(\nabla_YX,Z) \space ... $$
Now with interpretation 1), since $[X,Y]=0$ in euclidean space, this piece should become $$ \langle \frac{\partial\textbf{y}}{\partial\textbf{x}},\textbf{z}\rangle = \frac{\partial\textbf{x}}{\partial\textbf{y}},\textbf{z}\rangle $$ that is to say $$ \frac{\partial\textbf{y}}{\partial\textbf{x}} = \frac{\partial\textbf{x}}{\partial\textbf{y}} $$ which is false. I guess 1) is false then, but I was pretty sure about this interpretation (since the Leibniz rule for connection seems to work perfectly). Do I have to throw away all this intuition?
You are right with one, but why should the Lie bracket of general vector fields be zero on $\mathbb R^n$. This is true for coordinate vector fields (and doesn't lead to a contradiction in this case) but not for general vector fields.