About Covariant Derivative as a tensor

In summary, the conversation is discussing a lecture on YouTube about the introduction of the Covariant Derivative. The speaker explains why it is a (1,1) tensor and how it transforms accordingly. There is a discussion about the limiting process and the transformation of the operator. The conclusion is that the limiting process results in a (1,1) tensor.
  • #1
cianfa72
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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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  • #2
cianfa72 said:
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
strangerep said:
... 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
cianfa72 said:
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.
 
  • #5
strangerep said:
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
strangerep said:
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
cianfa72 said:
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
strangerep said:
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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Related to About Covariant Derivative as a tensor

1. What is a covariant derivative?

A covariant derivative is a mathematical operation that is used to differentiate a vector field along a given direction. It takes into account the curvature of the underlying space and ensures that the resulting vector is parallel transported along the given direction.

2. How is a covariant derivative different from an ordinary derivative?

A covariant derivative is a tensor quantity that takes into account the curvature of the underlying space, whereas an ordinary derivative is a scalar quantity that does not consider the curvature. This means that a covariant derivative is more general and can be applied to curved spaces, while an ordinary derivative can only be applied to flat spaces.

3. What is the significance of a covariant derivative being a tensor?

The fact that a covariant derivative is a tensor means that it follows certain transformation rules under coordinate transformations. This allows for a consistent and invariant definition of the derivative, regardless of the choice of coordinates.

4. How is a covariant derivative used in physics?

A covariant derivative is used in physics, particularly in the field of general relativity, to describe the behavior of physical quantities in curved spacetime. It is also used in other areas of physics such as electromagnetism and quantum mechanics.

5. Can you give an example of a covariant derivative in action?

One example of a covariant derivative in action is in the calculation of the geodesic equation, which describes the path that a free particle will follow in a curved space. The covariant derivative is used to ensure that the particle's velocity is parallel transported along its path, taking into account the curvature of the space.

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