Why Denote 1 Form as dx? - Sean Carroll's Lecture Notes on GR

[Ron19932017]

Hi everyone I am reading Sean Carrol's lecture notes on general relativity.
link to lecture : https://arxiv.org/abs/gr-qc/9712019

In his lecture he introduced dx^μ as the coordinate basis of 1 form and ∂_μ as the basis of vectors.

I understand why ∂_μ could be the basis of the vectors but not for the dx^μ. I have several confusion with the 1 form basis.

1. Is the 1 form basis dx^μ really have a meaning for infinitesimal change in x^μ?

2. How to convince ourselves that dx^μ(dx^ν)=δ^μ_ν?

3. Am I correct to understand the formalism of 1 form like this :
Given that a. ) df = ∂_μf dx^μ from vector calculus,
b.) ∂_μf is identified to be component of 1 form because it transforms covariently.
Therefore we realized df is a 1-form with dx^μ as its basis.

4. (more general open question though) The most puzzling part for me is to understand the formalism of 1 form basis. I follows well in realising the basis vector as ∂_μ not but for dx^μ. I also appreciate anyone to explain why dx^μ can be a 1 form basis.

 

[Orodruin]

1. It is a 1-form, not an infinitesimal change, although the two are related. For any tangent vector $X$ and function $f$, the exterior derivative $df$ is a one-form such that $df(X) = X^\mu \partial_\mu f$. If you let $f$ be the coordinate functions, you get N independent one-forms and therefore a basis.

2. You don't. As you wrote it it has no meaning. One-forms act on tangent vectors, not other one-forms. With the above in mind, you will find that $dx^\mu(\partial_\nu) = \partial_\nu x^\mu = \delta_\nu^\mu$.

3. a) is just the chain rule. No vector analysis needed. You really do not need coordinates or coordinate bases to define $df$. However, expressed in coordinate basis, you would have $df = (\partial_\mu f) dx^\mu$ directly from the chain rule.

4. See above. By definition, the exterior derivative of a coordinate function is a one-form. If you have a good coordinate system then all the $dx^\mu$ are linearly independent and therefore form a basis. You can express any one-form in this basis.

 

[Ron19932017]

1. It is a 1-form, not an infinitesimal change, although the two are related. For any tangent vector $X$ and function $f$, the exterior derivative $df$ is a one-form such that $df(X) = X^\mu \partial_\mu f$. If you let $f$ be the coordinate functions, you get N independent one-forms and therefore a basis.

2. You don't. As you wrote it it has no meaning. One-forms act on tangent vectors, not other one-forms. With the above in mind, you will find that $dx^\mu(\partial_\nu) = \partial_\nu x^\mu = \delta_\nu^\mu$.

3. a) is just the chain rule. No vector analysis needed. You really do not need coordinates or coordinate bases to define $df$. However, expressed in coordinate basis, you would have $df = (\partial_\mu f) dx^\mu$ directly from the chain rule.

4. See above. By definition, the exterior derivative of a coordinate function is a one-form. If you have a good coordinate system then all the $dx^\mu$ are linearly independent and therefore form a basis. You can express any one-form in this basis.

Thank you for you reply. I understand $df$ is a map from tangent vector to real number by $df(X) = X^\mu \partial_\mu f$ and how you get the $d x^\mu$ as 1-form basis.

1.)However how can you make sure all $df$ can be expressed in linear combination of $d x^\mu$? My intuition tells me this has to be related to the chain rule but I am not sure how to work out the formalism.

2.) Why do we adopt the notation $df, dx_\mu$, such that they looks similar to infinitesimal change? Why don't we just use $f, x$

 

[dextercioby]

2. Because that d is actually an operator called exterior differential.

 

[Ron19932017]

2. Because that d is actually an operator called exterior differential.

I see. Sorry I have a few more questions.

1. The logic is like : Exterior derivatives are p-forms -> df is the exterior derivative of a scalar function -> df is a 1 form?

2. How can we make sure exterior derivative are a (0,p) tensor at the beginning?

3. Which way we do formally testify something is a 1-form?
a.) show it transform covariently?
or b.) show it is a linear map of tangent vectors into real numbers and form a vector space?
I see both ways appear in various lecture notes so I am not sure if these two are equivalent criterion.

 

[Orodruin]

1. Yes.

2. By the way the exterior derivative is defined.

3. (a) A p-form is a p-form. It does not transform. It is always the same p-form. However, its components have some particular properties, including their transformation properties, that you can check.
(b) Is better. By definition, a 1-form is a dual vector. A single 1-form does not form a vector space. It is an element in a vector space.

 

[pervect]

Note that a one form, which some (but not all) texts denote with boldface as dx is a map from some vector space V to a scalar. This notation makes explicit the difference between dx and dx, the former operates on vectors and returns a scalar, the later is just a change in a real number (a coordinate).

So the value of both is comparable, But one operates on vectors, the other doesn't - it stands alone.

The boldface not/boldface thing is not necessarily done all the time, sometimes the reader is expected to know what is meant.