Prove that the Christoffel symbols vanish in $\mathbb{R}^n$ Recently I started reading about connections (covariant derivatives), and one of the first exercises I encountered is the proof of  ${\Gamma^k_{ij}}=0$, in $\mathbb{R}^n$. My handout though doesn't mention what's the natural connection $\mathbb{R}^n$ is endowed with, hence I cannot understand why Christoffel symbols vanish. Can you explain me what is going on please, and how do we prove the above?
 A: You can safely assume that we're dealing with the Levi-Civita connection here, which in this case is the usual directional derivative. Since $\nabla_{\partial_i}\partial_j=0$, it follows that $\Gamma_{ij}^k=0$ always.
A: The metric on $\mathbb{R}^n$ is kronecker delta , $g_{ij}= \delta_{ij}$. On every Riemannian manifold, including this case, the symbols are defined as
$$ \Gamma^i_{jk} = g^{il} (\partial_j g_{lk} + \partial_k g_{jl} - \partial_l g_{jk})/2 $$
Here elements of matric tensor are constant so their derivative are zero.
A: It's by definition: the natural connection on $\Bbb R^n$ is defined as the connection for which $\Gamma _{ij} ^k = 0$ (i.e. the flat connection, i.e. the one without curvature).
Notice, in this case, that
$$\nabla _X Y = \nabla _{X^i \partial_i} (Y^j \partial_j) = X^i \nabla _{\partial_i} (Y^j \partial_j) = X^i (\nabla _{\partial_i} Y^j) \partial_j + X^i Y^j (\nabla _{\partial_i} \partial_j) = \\
(\nabla _X Y^j) \partial_j + X^i Y^j \underbrace{\Gamma _{ij} ^k} _{=0} \partial_k = X (Y^j) \partial_j .$$
