Demonstration for the covariant derivative of a vector I know that $j-th$ component of the covariant derivative of a vector $\vec{V}$ is equal to :
$$(\nabla_{i}\vec{V})^j=(\partial_{i}V^{j}+V^{k}\Gamma_{ik}^{j})\quad\text{(equation 1)}$$
so we can write for the covariant derivative of vector $\vec{V}$ :
$$(\nabla_{i}\vec{V})=(\partial_{i}V^{j}+V^{k}\Gamma_{ik}^{j})\vec{e_{j}}\quad \text{(equation 1-bis)}$$
Now, I would like to prove the expression of $(\nabla_{i}\vec{V})$.
Indeed, I can write :
$$(\nabla_{i}\vec{V})=\nabla_{i}(V^{a}\vec{e_{a}})$$
$$=(\nabla_{i}(V^{a})\vec{e_{a}}+V^{a}(\nabla_{i}\vec{e_{a}})\quad\text{equation(2)}$$
From this last expression, for the second term $V^{a}(\nabla_{i}\vec{e_{a}})$, I know we can introduce Christoffel's symbols like this :
$$V^{a}(\nabla_{i}\vec{e_{a}})=V^{a}(\Gamma_{ia}^{l}\vec{e_{l}})$$
But my issue is for the first term of equation (2), i.e $(\nabla_{i}(V^{a})\vec{e_{a}}$.
I don't understand why it is equal to :
$$(\nabla_{i}(V^{a}))\vec{e_{a}}=\dfrac{\partial V^{a}}{\partial x^{i}}\vec{e_a}\quad\text{(equation 3)}$$
with $\nabla_{i}$ the covariant derivative operator on $x_{i}$ coordinate.
If I want to get the expression of $\text{(equation 1)}$, one has to prove the expression of $\text{(equation 3)}$. Logically, I should have :
$$(\nabla_{i}(V^{a}))\vec{e_{a}}=(\dfrac{\partial V^{a}}{\partial x^{i}}+V^{m}\Gamma_{im}^{a})\vec{e_{a}}\quad\text{(equation 4)}$$
Why do Christoffel symbols $\Gamma_{im}^{a}$ or components $V^{m}$ are vanishing into $\text{(equation 4)}$ ??
How could I manage to get $\text{(equation 1-bis)}$ ?
Thanks for your help
ps: feel free to put comment if my question is not clear for you.
 A: How do you define $\nabla_i$ ? I think of it as $\partial / \partial_i$ adapted for curved space as encoded by the Christoffel symbols. So, the idea that it behaves as the partial derivative plus stuff is not surprising. I believe, the covariant derivative measures the change in one vector field as measured relative another vector field. It satisfies many interesting properties, among them: for a function $f$ and vector fields $U$ and $V$,
$$ \nabla_U(fV) = U[f]V+f\nabla_UV $$
where we think of vector fields $U,V$ as derivations. Taking $U = \partial/\partial x^i$ we use special notation $\nabla_U = \nabla_i$ and then
$$ \nabla_i (fV) = \frac{\partial f}{\partial x^i}V+ f\nabla_i V$$
Your expansion of a vector field across the coordinate derivations ties together with what I share here. Your $V = V^ae_a$ has $f=V^a$ as a function whereas $e_a$ is a vector field. So,
$$ \nabla_i (V^ae_a) = \frac{\partial V^a}{\partial x^i}e_a + V^a \nabla_i e_a = \frac{\partial V^a}{\partial x^i}e_a + V^a \Gamma_{ia}^b e_b  $$
then switching dummy indices yields
$$ \nabla_i (V^ae_a) = \frac{\partial V^a}{\partial x^i}e_a + V^b \Gamma_{ib}^a e_a = \left( \frac{\partial V^a}{\partial x^i} + V^b \Gamma_{ib}^a \right) e_a  $$
So, the $a$-th component of the covariant derivative of $V$ with respect to the $i$-th coordinate is given by
$$ (\nabla_i V)^a = \frac{\partial V^a}{\partial x^i} + V^b \Gamma_{ib}^a   $$
which, I think, is your definition.
I think I was once taught the definition you give. This definition is made in order that the covariant derivative transform once more as a contravariant vector. The way I am suggesting you think is in terms of the vector field itself, not it's component functions. It still seems to me you are asking why your definition is a definition, so, perhaps I misread your question.
