About Covariant Derivative as a tensor

[cianfa72]

TL;DR 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 ?

 

[strangerep]

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 $\nabla_\alpha X^\mu$ .

 

[cianfa72]

... 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 ?

 

[strangerep]

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.

 

[cianfa72]

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 ?

 

[strangerep]

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 $\{ \partial/\partial x^\mu \}$ ?

 

[cianfa72]

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