Confusion about covariant derivative in $\mathbb R^n$ The Levi-civita connection on $\mathbb R^n$ corresponds to the usual directional derivative. In this sense I expect the following to hold:
$$
\left(\nabla_{\partial_i}\partial_j\right)f=\partial_{ij}f.
$$
On the other hand: $\nabla_{\partial_i}\partial_j=\Gamma_{ij}^k\partial_k=0$. These results are contradictory (of course) and I'd be happy to get some clarification.
 A: You've got your parentheses in the wrong place. $\big(\nabla_{\partial_i}\partial_j\big)$ is a vector field (in this case the zero vector field), and  $\big(\nabla_{\partial_i}\partial_j\big)f$ represents the derivative of $f$ in the direction of this vector field (in this case zero).
On the other hand, if you want the second covariant derivative of $f$, you need $\nabla_{\partial_i}\big(\partial_j f)$, which in Euclidean space is just $\partial_i\partial_j f$.
EDIT: Actually, what I wrote above is not quite right. Because $\partial_j f$ is just a function, its "covariant" derivative $\nabla_{\partial_i}\big(\partial_j f)$ is just $\partial_i\partial_j f$, regardless of what connection we're using. The second covariant derivative of $f$ is the $2$-tensor field $\nabla^2 f$, whose $ij$-component is
$$
\nabla^2 _{ij} f = \partial_i\partial_j f - \Gamma_{ij}^k \partial_k f,
$$
which in Euclidean space is just $\partial_i\partial_j f$.
A: For the covariant derivative of $\mathbb R^n$, you have $\nabla_XY=dY.X$ where $dY$ is the differential of $Y$. This implies that $\nabla_{\partial_i}\partial_j=0$.
