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I saw this formulation :

"If by $\nabla_{X_{i}}V_{j}$ we meant "take the covariant derivative of the components $V_{j}$ of the vector $V$", then this would simply be $X_{i}V_{j}$ as the components are just smooth functions."

with $\{e^{a}=dx^{a}\}$ basis and dual basis $\{X_{a}=\frac{\partial}{\partial x^{a}}\}$.

The expression of covariant derivative of "a-th" component of vector $V$ is :

$$ (\nabla_{X_{i}}V_{a})\,=X_{i}\,V_{a}--V_{c}\Gamma_{ia}^{c}= \,\dfrac{\partial V_{a}}{\partial x^{i}}-V_{c}\Gamma_{ia}^{c} $$

So I would like to know why, saying "then this would simply be $X_{i}V_{j}$ as the components are just smooth functions", one gets : $\nabla_{X_{i}}V_{a}=X_{i}V_{a}$ ?? I mean, Why does term with Christoffel symbols disappear ?

I don't understand the link between smooth functions for components of vector $V$ and the vanishing of right term in expression of $\nabla_{X_{i}}V_{a}$.

Thanks

UPDATE 1

Here's a capture on the context of the formulation that confuses me :

Context of formulation

ps : the top formulation (italic) comes from this post

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  • $\begingroup$ $V_i$ is a smooth function $M \to \mathbb{R}$. $ V_i(p) e^i(p)$ is a vector in the tangent space $T_p M$. $\endgroup$ – Chappers Mar 6 '17 at 18:06
  • $\begingroup$ Chappers: and that's why Christoffel symbols are vanishing in the tangent space $T_{p}M$ ? I don't understand the link $\endgroup$ – youpilat13 Mar 7 '17 at 4:00
  • $\begingroup$ Maybe, you use a normal coordinate at $p$ so that $\Gamma_{ij}^k (p)=0$ ? $\endgroup$ – HK Lee Mar 28 '17 at 9:00
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    $\begingroup$ The notations are what confuses you. The expression $\nabla_{X_i}V_a$ is not the derivative of the component $V_a$ it is the $a$-th component of the derivative. A better notation would be $(\nabla_{X_i}V)_a$ $\endgroup$ – MBN Mar 28 '17 at 9:07
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The expression $\nabla_{X_i} V_a$ should be read as $(\nabla_{X_i} V)_a$ and not $\nabla_{X_i} (V_a)$. Indeed, the formula

$$ (\nabla_{X_i} V)_a = X_i V_a + \Gamma_{ia}^c V_c $$

shows that the right hand side depends not only on $V_a$ (the $a$-th coordinate of $V$ with respect to the frame $X_i$) but on all of $V_1,\dots,V_n$ (that is, all the coordinates of $V$) so you don't want to think of $\nabla_X V_a$ as an operator which eats the $a$-th coordinate of a vector field and outputs the $a$-th coordinate of the covariant derivative.

In fact, ones sometimes denotes the directional derivative of a function $f$ in the direction of a vector field $X$ by

$$ Xf = df(X) = \nabla_X (f) $$

so using this notation, the formula above reads as

$$ (\nabla_{X_i} V)_a = \nabla_{X_i} (V_a) + \Gamma_{ia}^c V_c $$

which means that the $a$-th coordinate of the covariant derivative of $V$ in the direction $X_i$ is the directional derivative of the $a$-th component of $V$ in the direction $X_i$ plus correction terms involving the Christoffel symbols and all the other components of $V$.


The bottom line is that you shouldn't think about "the covariant derivative of the $a$-th component of a vector field" because this is not a well-defined notion without providing all the other components but rather "the $a$-th component of the covariant derivative of a vector field".

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  • $\begingroup$ Thanks . About formulation which made me confused : "If by $\nabla_{X_{i}}V_{j}$ we meant "take the covariant derivative of the components $V_{j}$ of the vector $V$", then this would simply be $X_{i}V_{j}$ as the components are just smooth functions." and from the different notations that appear in litterature, especially $$Xf = df(X) = \nabla_X (f)$$ I think it means : $$\nabla_{X_{i}}(V_{j})=X_{i}V_{j}$$. But I still don't know why it is stipulated that "as the components are just smooth functions", what is the consequence of having "smooth functions" with covariant derivative ? $\endgroup$ – youpilat13 Mar 29 '17 at 4:18
  • $\begingroup$ @youpilat13: I'm not sure I understand you. There are various operations denoted by $\nabla$ that act on different objects (they are all particular instances of a general notion called a connection). When $\nabla_X$ acts on functions (as in $\nabla_X f$) this is usually just the regular directional derivative of $f$. When $\nabla_X$ acts on vector fields (as in $\nabla_X V$), this is a completely different operation that is related to the previous operation by the formula above and the Christoffel symbols. $\endgroup$ – levap Mar 29 '17 at 4:27
  • $\begingroup$ I know what is a directional derivative of a function $f=f(x^1,...,x^n)$ which generalizes the partial derivative (if $f$ is differentiable, then $\nabla_{\vec{v}}\,f=\nabla(f).\vec{v}$). From the beginning, I try to find the conditions with which we can write : ($j-th$ component of (covariant derivative of $\vec{V}$ along $X_{i}$ direction)) = (dot product between vector $X_{i}$ and $j-th$ component vector $\vec{V}$). Could you tell please these conditions that allow to have this equality ??, i.e : $$(\nabla_{X_i} V)_j = X_i\,V_j$$ $\endgroup$ – youpilat13 Mar 29 '17 at 6:17
  • $\begingroup$ @youpilat13: It's very difficult to understand what you are trying to say, you have to try and be more precise. What do you mean by "dot product between vector $X_i$ and the $j$-th component vector $\vec{V}$"? The $j$-th component of a vector is a scalar so you can't take the dot product of $X_i$ and $V_j$. You can differentiate $V_j$ in the direction $X_i$ (this is written as $X_i V_j$ and it is not a dot product) and $(\nabla X_i V)_j = X_i V_j$ if the Christoffel symbols vanish (which is what happens in $\mathbb{R}^n$). $\endgroup$ – levap Mar 29 '17 at 13:57
  • $\begingroup$ Sorry for the mistakes, I wanted to say : "From the beginning, I try to find the conditions with which we can write : ($j−th$ component of covariant derivative of $\vec{V}$ along $X_{i}$ direction) = ( product between $i-th$ component of vector $X$, i.e $X_{i}$ and $j−th$ component of vector $\vec{V}$, i.e $V_j$ )." When you say " if the Christoffel symbols vanish (which is what happens in $\mathbb{R}^{n}$)", what do you mean ? what is required to vanish the Christoffel symbols ? one has to use cartesian coordinates, hasn't one ? thanks for your patience, the subject is a little bit tricky $\endgroup$ – youpilat13 Mar 29 '17 at 15:50

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