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I try to understand the difference between the classical definition of covariant derivative :

$$\nabla_{i}V^{j}=\partial_{i}V^{j}+V^{k}\Gamma_{ik}^{j}\quad\quad(1)$$

(where index $i$ represents curvilinear cooordinates $x^{i}$ and $V^{j}$ the $j-\text{th}$ component of vector $V$)

and the definition of called " covariant derivative of a vector field $V$ along a vector field $Z$ " and noted :

$$\nabla_{Z}V\quad\quad(2)$$

What's the expression of (2) and how to make the link with equation (1) ?

I saw that equation (2) was used like this in the demonstration of Ricci theorem :

$$Z \langle X,Y \rangle = \langle \nabla_Z X, Y\rangle + \langle X, \nabla_Z Y\rangle$$

Sorry if this question seems to be evident but I am just starting to learn basics of geometry differential.

Thanks for your help

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  • $\begingroup$ In short, the link is that $\nabla_{\frac{\partial}{\partial x^i}} V = \sum_{j=1}^n \nabla_i V^j \frac{\partial }{\partial x^j}$, $\endgroup$ – user99914 Feb 5 '17 at 8:07
  • $\begingroup$ :John Ma, thanks. So vector field Z would be represented by $\dfrac{\partial}{\partial x^{i}}$ ? Is this term representing also the definition of local basis vectors, i.e $\vec{e_{i}}=\dfrac{\partial \vec{OM}}{\partial x^{i}}$ (with $\vec{OM}$ the position vector )? In this case, could we write $\nabla_{\vec{e_{i}}}\vec{V}=\nabla_{i}V^{j}\,\vec{e_{j}}=\nabla_{i}(\vec{V}\cdot\vec{e^{j}}) \vec{e_{j}}= \nabla_{i}\vec{V}$? It seems I do confusions $\endgroup$ – youpilat13 Feb 5 '17 at 9:01
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After choosing coordinates $(x^1,\dots,x^n)$, each vector field $V$ can be expressed locally as $V = V^j \frac{\partial}{\partial x^j}$. If you are given a covariant derivative operator $\nabla$, the Christoffel symbols with respect to the coordinate system $(x^1,\dots,x^n)$ are defined by the equation

$$ \nabla_{\frac{\partial}{\partial x^i}} \frac{\partial}{\partial x^j} = \Gamma_{ij}^k \frac{\partial}{\partial x^k}. $$

Namely, $\Gamma_{ij}^k$ gives you the $\frac{\partial}{\partial x^k}$-th component of the covariant derivative of $\frac{\partial}{\partial x^j}$ in the direction $\frac{\partial}{\partial x^i}$. Using the product rule of the covariant derivative, we have

$$ \nabla_{\frac{\partial}{\partial x^i}} \left( V^j \frac{\partial}{\partial x^j} \right) = \frac{\partial V^j}{\partial x^i} \frac{\partial}{\partial x^j} + V^j \nabla_{\frac{\partial}{\partial x^i}} \left( \frac{\partial}{\partial x^j} \right) = \frac{\partial V^j}{\partial x^i} \frac{\partial}{\partial x^j} + V^j \Gamma_{ij}^k \frac{\partial}{\partial x^k}. $$

This can be abbreviated to

$$ (\nabla_i V)^k = \partial_i(V^k) + V^j \Gamma_{ij}^k $$

(so the $k$-th component of the covariant derivative $\nabla_i V$ is given by the regular derivative $\partial_i(V^k)$ of the $k$-th component plus a "correction factor" which involves the all the components of $V$ and the Christoffel symbols).

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  • $\begingroup$ :levap thanks. How do you prove that each vector field $V$ can be expressed locally as $V=V^{j}\,\dfrac{\partial}{\partial x^{j}}$ ? $\endgroup$ – youpilat13 Feb 8 '17 at 10:59
  • $\begingroup$ You can't differentiate a vector field with respect to a vector field in a coordinate independent way without introducing a connection and using the covariant derivative. $\endgroup$ – levap Feb 8 '17 at 16:09
  • $\begingroup$ do you agree that I can also write : $\nabla_{\frac{\partial}{\partial x^i}} \dfrac{\partial}{\partial x^j} =\dfrac{\partial^{2}}{\partial x^{i}\partial x^{j}} =\Gamma_{ij}^k \dfrac{\partial}{\partial x^k}$ ? $\endgroup$ – youpilat13 Feb 8 '17 at 18:33
  • $\begingroup$ You can write whatever you want but that is a very bad choice of notation. $\endgroup$ – levap Feb 8 '17 at 19:07
  • $\begingroup$ :levap thanks. And what about the proof of the equation that locally, vector field $V$ is : $V=V^{j}\,\dfrac{\partial}{\partial x^{j}}$ ? I don't know how you get this equality. regards $\endgroup$ – youpilat13 Feb 8 '17 at 20:12

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