Covariant derivative only for tensor

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  • #1
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Hi initially I am aware that christoffel symbols are not tensor so their covariant derivatives are meaningless, but my question is why do we have to use covariant derivative only with tensors? ?? Is there a logic of this situation? ???
 

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  • #2
Tio Barnabe
I think there's more than one possible explanation. One of them is that by construction the covariant derivative measures the change of a vector in a given displacement. This is so by construction and by definition, i.e. we want to construct the derivative of a vector and we want to do so in analogy with case of funtions (construction), which is the limit of the difference of the vectors two points when the distance comes to zero, therefore it should be a vector too.

If you want to see the change of functions, you just use the ordinary derivative.
 
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Orodruin
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but my question is why do we have to use covariant derivative only with tensors? ?? Is there a logic of this situation? ???
If you do not, the result will not be invariant.

Since you labeled this thread "A": A priori, a tensor field is a section over the corresponding tensor bundle. Since the tensors at different points of the base manifold belong to different tensor spaces. As you will remember from basic calculus, derivatives were defined by taking differences of functions at different points and studying how it behaves as those points approach each other. However, for tensor fields there is a priori no natural way of taking differences of tensors at different points of the manifold and in order to be able to define a derivative we therefore must introduce a connection that can be used to relate the tensor spaces at different points to each other. This connection defines the covariant derivative and it makes no sense to talk about the "ordinary" derivative because it is unclear what it would mean.
 
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