About Covariant Derivative as a tensor

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Summary:

Definition of covariant derivative as a tensor through the limiting process of a fraction
Hi,

I've been watching lectures from XylyXylyX on YouTube. I believe they are really great !

One doubt about the introduction of Covariant Derivative. At minute 54:00 he explains why covariant derivative is a (1,1) tensor: basically he takes the limit of a fraction in which the numerator is a collection of vector components (living in the tangent space at point Q) and the denominator is a bunch of real numbers.

My point is: to be a (1,1) tensor it has to transform accordingly. The numerator is a vector and thus its components transform as such; what about the denominators ##\delta x^{\alpha}## ? I believe that the inverse of them have really to be the components of a co-vector

Is that the case ?
 
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Answers and Replies

  • #2
strangerep
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I've been watching lectures from XylyXylyX on YouTube. I believe they are really great !

One doubt about the introduction of Covariant Derivative. At minute 54:00 he explains why covariant derivative is a (1,1) tensor: [...]
Based (only) on a snippet I watched surrounding minute 54:00, I think those lectures are not "great". For one thing, he uses ##\alpha## as both a free index and a dummy summation index shortly after min 54:00. That's a serious no-no.

basically he takes the limit of a fraction in which the numerator is a collection of vector components (living in the tangent space at point Q) and the denominator is a bunch of real numbers.

My point is: to be a (1,1) tensor it has to transform accordingly. The numerator is a vector and thus its components transform as such; what about the denominators ##\delta x^{\alpha}## ? I believe that the inverse of them have really to be the components of a co-vector
... which is indeed what you end up with when finally denoting the covariant derivative as ##\nabla_\alpha X^\mu## .
 
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  • #3
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... which is indeed what you end up with when finally denoting the covariant derivative as ##\nabla_\alpha X^\mu## .
Here, if I understand correctly, ##\mu## and ##\alpha## upper and lower indices actually applies to the "overall" object ##\nabla X##, let me say -- just to clarify -- it should read as : ##\left( \nabla X \right)^{\mu}{}_{\alpha} ##

About the limiting process described there....at finite ##\delta x^{\alpha}## can we assume the fraction involved is already a tensor quantity ?
 
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  • #4
strangerep
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Here, if I understand correctly, ##\mu## and ##\alpha## upper and lower indices actually applies to the "overall" object ##\nabla X##, let me say -- just to clarify -- it should read as : ##\left( \nabla X \right)^{\mu}{}_{\alpha} ##
I suppose you could write it that way -- in this case -- but no one does. It would get confusing when you move on to consider the covariant derivative of covariant vector components ##Y_\mu##.

About the limiting process described there....at finite ##\delta x^{\alpha}## can we assume the fraction involved is already a tensor quantity ?
No. It only becomes so in a limit sense -- as ##\delta x^{\alpha}## becomes infinitesimal. For more detail, see my post #36 in this thread.
 
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I suppose you could write it that way -- in this case -- but no one does. It would get confusing when you move on to consider the covariant derivative of covariant vector components ##Y_\mu##.
The point I would like to stress is that the ##\nabla## operator is not actually a co-vector, thus ##\nabla_\alpha X^\mu## is not really a tensor product between ##\nabla_\alpha## and the vector of components ##X^\mu##. Things go that the 'covariant derivative operator' acts on the vector ##X^\mu## and returns the ##\nabla_\alpha X^\mu## tensor


No. It only becomes so in a limit sense -- as ##\delta x^{\alpha}## becomes infinitesimal. For more detail, see my post #36 in this thread.
thus basically in the limit ##\delta x^{\mu} \rightarrow dx^\mu## that transform like vector components; likewise their inverse transform like co-vector components yielding a (1,1) tensor in the limiting process.

Does it make sense ?
 
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  • #6
strangerep
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The point I would like to stress is that the ##\nabla## operator is not actually a co-vector,
Well, under a general coordinate transformation, the operator transforms as $$\nabla'_\alpha ~=~ \frac{\partial x^\beta}{\partial x'^\alpha } \, \nabla_\beta ~,$$ and $$\nabla'_\alpha X'^\mu ~=~ \frac{\partial x^\beta}{\partial x'^\alpha } \, \frac{\partial x'^\mu}{\partial x^\nu } \, \nabla_\beta X^\nu ~,$$
thus ##\nabla_\alpha X^\mu## is not really a tensor product between ##\nabla_\alpha## and the vector of components ##X^\mu##. Things go that the 'covariant derivative operator' acts on the vector ##X^\mu## and returns the ##\nabla_\alpha X^\mu## tensor
Are you familiar with the notion that the ##\partial/\partial x^\mu## form a vector basis in differential geometry? If not, then it's probably best to review that before continuing this discussion.

thus basically in the limit ##\delta x^{\mu} \rightarrow dx^\mu## that transform like vector components; likewise their inverse transform like co-vector components yielding a (1,1) tensor in the limiting process.

Does it make sense ?
That depends what you meant by "inverse". Are you familiar with the notion that the ##\{ dx^\alpha \}## are dual to the ##\{ \partial/\partial x^\mu \}## ?
 
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  • #7
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Well, under a general coordinate transformation, the operator transforms as $$\nabla'_\alpha ~=~ \frac{\partial x^\beta}{\partial x'^\alpha } \, \nabla_\beta ~,$$ and $$\nabla'_\alpha X'^\mu ~=~ \frac{\partial x^\beta}{\partial x'^\alpha } \, \frac{\partial x'^\mu}{\partial x^\nu } \, \nabla_\beta X^\nu ~,$$
Based on that transformation rule for ##\nabla_\alpha## then ##\nabla_\beta X^\nu## seems really a tensor product, I believe

I'm aware of ##\{\partial/\partial x^\mu \}## and ##\{ \partial/\partial x^\mu \}## as bases for vector and dual vector spaces in differential geometry.

My point was trying to understand the limiting process described there as follows:

##\delta x^{\mu} \rightarrow dx^\mu## when ##\delta x^{\mu} \rightarrow 0## thus ##\frac 1 { \delta x^{\mu} } \rightarrow \frac 1 {dx^{\mu} }## and because
##\{dx^\mu \}## themselves transform controvariant then ##\frac 1 { \delta x^{\mu} }## transform covariant in the limit ##\delta x^{\mu} \rightarrow 0 ## hence as result we get a (1,1) tensor
 
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