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For two vector fields V and T we can take the covariant derivative:

$$\nabla_V T=\nabla_{V^\mu \hat e_{\mu}}T$$

$$=V^\nu \nabla_{\hat e_\nu}T$$

What exactly are we doing when we take the vector component $V^\nu$ out of the covarient derivative here?

Secondly, when we define the action of the covariant derivative on scalars, in my head this should effectively be the directional derivative of a the scalar field. However I see it written, for a vector field X, like this:

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

I don't understand what this means, what exactly does it mean to "act" on a scalar with a vector field in this way? Is it using the fact that vectors can be written as directional derivatives? If so could someone make this notation a bit clearer?

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    $\begingroup$ linearity? $\nabla_{a+b}=\nabla_a+\nabla_b$ and $\nabla_{\lambda a}=\lambda \nabla_a$. $\endgroup$ – AccidentalFourierTransform Feb 11 at 2:02
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    $\begingroup$ for the first question we are basically doing a basis decomposition of the operator $\nabla_V$, so what we are saying is basically that it is equal to a linear combination of the operators $\nabla_{\hat{e}_\nu}$ and the components $V^\nu$ in the formula are just the coefficients of this expansion. $\endgroup$ – Defcon97 Feb 11 at 2:03
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    $\begingroup$ for the second question it all boils down to definitions. Usually the Tangent space on a manifold $M$ is defined as the space of all the functions from $\mathcal{C}^\infty(M)$ to $\mathbb{R}$ that adhere to Leibniz's derivation law (the product law for derivatives). I understand it is not a very intuitive way to define it and there is no simple way to convince you that this definition is good other than directing you to a differential geometry textbook sadly, you'll have to take my word on that one. $\endgroup$ – Defcon97 Feb 11 at 2:10
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    $\begingroup$ With this definition then the expression says exactly what you were expecting, namely that the action of the covariant derivative on a scalar field is just the action of the vector $X$ on the field, and that action happens to actually be the directional derivative of the field along $X$. Just fancy notation that actually makes the whole thing really clear for people who like algebra. But fancy notation nonetheless $\endgroup$ – Defcon97 Feb 11 at 2:13
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    $\begingroup$ sorry i was finishing the comment and didn't reply to your question. @Charlie no, you are not wrong. But we never use notations like $\frac{\nabla}{\nabla x^i}$ so never write that anywhere. I got what you meant though. $\endgroup$ – Defcon97 Feb 11 at 2:14
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for the first question we are basically doing a basis decomposition of the operator $\nabla_V$, so what we are saying is basically that it is equal to a linear combination of the operators $\nabla_{\hat{e}_\nu}$ and the components $V_ν$ in the formula are just the coefficients of this expansion.

for the second question it all boils down to definitions. Usually the Tangent space on a manifold $M$ is defined as the space of all the functions from $\mathcal{C}^∞(M)$ to $\mathbb{R}$ that adhere to Leibniz's derivation law (the product law for derivatives). I understand it is not a very intuitive way to define it and there is no simple way to convince you that this definition is good other than directing you to a differential geometry textbook sadly, you'll have to take my word on that one. With this definition then the expression says exactly what you were expecting, namely that the action of the covariant derivative on a scalar field is just the action of the vector $X$ on the field, and that action happens to actually be the directional derivative of the field along $X$. It's just fancy notation that actually makes the whole thing really clear for people who like algebra. But fancy notation nonetheless

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