I am currently working with Riemannian geometry by Peter Petersen. I am wondering if someone can provide feedback on my proof of the exercise if it is correct or wrong.
Q. Show that the connection on Euclidean space is the only affine connection such that $\nabla X = 0$ for all constant vector fields $X$.
An attempt:
Let $X = \alpha^{i} \partial_i$ such that $\alpha^{i}$ is constant vector fields. Then for all $j$ we have with regard to the derivative $\nabla X = 0$ implies
$$ 0 = \nabla X = \nabla_{\partial_{j}}X = \nabla_{\partial_{j}}(\alpha^{i} \partial_i) = (\partial_{j} \alpha^{i})\partial_{i} + \alpha^{i}(\nabla_{\partial_{j}}\partial_{j}) = \alpha^{i} \nabla_{\partial_j} \partial_{i} = \alpha^{i} \Gamma^{k}_{ij} \partial_{k}, $$ since the christoffel symbols of second kind is always zero by the fact of $g_{ij} = \delta^{i}_{j}$ and the directional derivative on this metric is zero. Hence $$ 0 = \nabla X = \alpha^{i} \Gamma^{k}_{ij} \partial_{k} = 0, $$ we have shown that the affine connections is therefore flat.
Thanks in advance!