I try to find the relation that defines the covariant derivative of a vector along another vector :
$$\nabla_{\mathbf{v}}{\mathbf{u}}=\left(v^{i}u^{j}\Gamma^{k}{}_{ij}+v^{i}{\partial u^{k} \over \partial x^{i}}\right){\mathbf{e}}_{k}\quad\quad(1)$$
Firstly, I know that $k-th$ component of covariant derivative of a vector along $(x^{i})$ coordinates is defined by :
$$(\nabla_{i}{\mathbf{u}})^{k}=\left(u^{j}\Gamma^{k}{}_{ij}+{\partial u^{k} \over \partial x^{i}}\right)$$
and also, we have : $$\text{d}\vec{e_{i}}=\Gamma^{j}{}_{ki}\text{d}x^{k}\vec{e_{j}}$$
by introducing : $$\text{d}\vec{e_{i}}=w^{j}{}_{i}\vec{e_{j}}$$
and $$w^{j}{}_{i}=\Gamma^{j}{}_{ki}\text{d}x^{k}$$
But from these basic definitions, I don't know how to find relation (1) ?
If anyone could help me, this would be fine, Regards
UPDATE 1 :
Thanks for your help. I have just a remark about your definition of $\nabla \vec e_i = \sum\Gamma^j_{ki} \text{d}x^k\vec e_j$. Indeed, it seeems that we could rather write the Absolute differential $\text{D}\vec{e_{i}}$ as :
$$\text{D}\vec{e_{i}}=(\nabla_{k} \vec{e_{i}})^{j} \text{d}x^{k} \vec{e_{j}}= ((\partial_{k}(\vec{e_{i}})^{j}+\Gamma^{j}_{kl} (\vec{e_{i}})^{l}) \text{d}x^k\vec{e_{j}}$$
Then :
$$\text{D}\vec{e_{i}}=(\partial_{k}(\vec{e_{i}})^{j}+\Gamma^{j}_{kl} \delta^{il}) \text{d}x^k\vec{e_{j}}=(\partial_{k}(\delta_{ij})+\Gamma^{j}_{ki}) \text{d}x^k\vec{e_{j}}$$
$$=\Gamma^{j}_{ki} \text{d}x^k \vec{e_{j}}$$
And we get : $$\nabla_{k} \vec{e_{i}} = \Gamma^{j}_{ki}\vec{e_{j}}$$
with : $$(\nabla_{k} \vec{e_{i}})^{j} = \Gamma^{j}_{ki}$$
I must admit it is pretty complex and long but is it correct ?
ps : From your notation, I could find the relation between $\nabla_{k}\vec{e_{i}}$ and $\nabla\vec{e_{i}}$ as :
$\nabla\vec{e_{i}}=\nabla_{k}\vec{e_{i}} \text{d}x^{k}$
isn't it ?
the last difference is that in Ted Shifrin's calculations, the factor $\text{d}x^{k}$ is included into definition of $\nabla$.
UPDATE 2:
I think it is possible to convert slightly :
$$\nabla_{\mathbf v}\mathbf u = \sum_j\left(\sum_k v^k\frac{\partial u^j}{\partial x^k} + \sum_{k,i}u^iv^k\Gamma^j_{ki}\right)\vec e_j$$
By factorize with $v^{k}$ in summation, Could I write this ? :
$$(\nabla_{\mathbf v}\mathbf u)^{j} = v^k\left(\frac{\partial u^j}{\partial x^k} + u^i\Gamma^j_{ki}\right)= v^{k} (\nabla_{k}\vec{u})^{j}=(v^{k}\,\nabla_{k}\vec{u})^{j}$$
knowing $\vec{v}=v^{k}\vec{e_{k}}$.
From a general point of view, is this rule valid (as a function of subscript on $\nabla_{k}$):
$$(\nabla_{\mathbf v}\mathbf u)^{j}=(v^{i}\,\nabla_{i}\mathbf u)^{j}$$
?? If you had a wiki link or others to prove this property, let me know it, I will be satisfied.
Last point, is there a way to grasp the difference between a covariant derivative as respect of a basis vector (which models the transport along a curvilinear coordinate) and a covariant derivative as respect of a vector or vector field (I don't if this works for both vector and field vector) ?
Regards
Best Answer
Actually, it should be $\nabla \vec e_i = \sum\Gamma^j_{ki} dx^k\vec e_j$. (Having $u^k$ as coordinates and also as coefficients of your vector field will not do.) You're writing the vector field $\mathbf u = \sum u^i\vec e_i$, and you differentiate using the product rule, obtaining \begin{align*} \nabla\mathbf u &= \nabla \big(\sum_i u^i\vec e_i\big) = \sum_{k,i}\frac{\partial u^i}{\partial x^k}dx^k\vec e_i + \sum_i u^i\sum_{k,j}\Gamma^j_{ki} dx^k \vec e_j \\ &= \sum_j\big(\frac{\partial u^j}{\partial x^k}dx^k + \sum_{k,i} u^i\Gamma^j_{ki} dx^k \big)\vec e_j. \end{align*} Now evaluate on the tangent vector $\mathbf v = \sum v^k\vec e_k$ to obtain $$\nabla_{\mathbf v}\mathbf u = \sum_j\left(\sum_k v^k\frac{\partial u^j}{\partial x^k} + \sum_{k,i}u^iv^k\Gamma^j_{ki}\right)\vec e_j,$$ as desired. (At this point, you can relabel the indices to get your original formula.)