# Demonstration of covariant derivative of a vector along another vector

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

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.)
• In fact, it's included in yours as well. You didn't provide full context or definitions of your symbols. Your equation $d\vec e_i = \dots$ should have been $\nabla\vec e_i$, as I pointed out. There's no difference between $D$ and $\nabla$ as we're using it, except that you seem to want to write $\nabla_k$ for $\nabla_{\partial/\partial x^k} = D_{\partial/\partial x^k}$. Jul 8, 2018 at 21:35
• Yes, that's right. It follows immediately from my formula, but intuitively you should think of $\nabla_{\mathbf v}$ as a directional derivative in the $\mathbf v$ direction, and, as such, it's linear in $\mathbf v$. You're just writing $\mathbf v = \sum v^i\frac{\partial}{\partial x^i}$ and using linearity. To answer your last question: If you want to think in terms of differentiating along a coordinate curve, you can similarly think about differentiating along the integral curve (or flow line) of your vector field $\mathbf v$. Jul 8, 2018 at 23:10
• Lie derivative doesn't use any covariant derivative — it depends just on the differentiable structure of the manifold. You differentiate along the flow of the vector field $\mathbf v$. There is an analogous interpretation of the covariant derivative, considering the parallel translation of $\mathbf u$ along the flow. Jul 10, 2018 at 19:29
• It's wrong. Don't believe everything you read on the web. What is correct, is that for the Levi-Civita connection we have $\nabla_XY - \nabla_YX = [X,Y]=\mathscr L_X Y$. Jul 10, 2018 at 20:11