# Regarding the definition of covariant derivative and its use on basis vector fields

we find that for general vector fields ${\mathbf v}= v^ie_i$ and ${\mathbf u}= u^je_j$ we get :$\nabla_{\mathbf v} {\mathbf u} = \nabla_{v^i {\mathbf e}_i} u^j {\mathbf e}_j = v^i \nabla_{{\mathbf e}_i} u^j{\mathbf e}_j = v^i u^j \nabla_{{\mathbf e}_i} {\mathbf e}_j + v^i {\mathbf e}_j \nabla_{{\mathbf e}_i} u^j = v^i u^j \Gamma^k {}_{i j}{\mathbf e}_k+v^i{\partial u^j\over\partial x^i} {\mathbf e}_j$ so $\nabla_{\mathbf v} {\mathbf u} = \left(v^i u^j \Gamma^k {}_{i j}+v^i{\partial u^k\over\partial x^i}\right){\mathbf e}_k$

First of all, in $v^i u^j \nabla_{{\mathbf e}_i} {\mathbf e}_j + v^i {\mathbf e}_j \nabla_{{\mathbf e}_i} u^j = v^i u^j \Gamma^k {}_{i j}{\mathbf e}_k+v^i{\partial u^j\over\partial x^i} {\mathbf e}_j$,

${\mathbf e}_j \nabla_{{\mathbf e}_i} u^j$ seems to equal to ${\partial u^j\over\partial x^i} {\mathbf e}_j$, and that seems to mean that $\mathbf{e}_j$ commutes with covariant derivative part (or here it equals to partial derivative part.). But this seems to be clearly wrong...

Secondly, for $\nabla_{\mathbf v} {\mathbf u} = \left(v^i u^j \Gamma^k {}_{i j}+v^i{\partial u^k\over\partial x^i}\right){\mathbf e}_k$ part, it seems that $\mathbf{e}_j$ part was switched with $\mathbf{e}_k$, but is this allowed? why is it?

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Notice that in the general definition of a connection, if we have $X$ and $Y$ vector fields and $f$ a real-valued function then $\nabla_X(fY) = X[f]Y + f\nabla_X (Y)$ so for $X = v^i e_i$ and $Y = u^j e_j$ so the term that you're confused about i.e $e_j \nabla_{e_i} u^j$ is in-fact $e_j e_i(u^j)$ but $e_i$ is nothing but $\partial_i$ so you get that your term is $e_j \partial_i(u^j)$. The simple way of saying this would be the covariant derivative of a function is just defined to be it's partial derivative. Now since we're summing over $j$, we can call it whatever we want, so relabel $j$ to $k$.