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I Why denote 1 form as dx?

  1. Jul 16, 2017 #1
    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 dxμ 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 dxμ. I have several confusion with the 1 form basis.

    1. Is the 1 form basis dxμ really have a meaning for infinitesimal change in xμ?

    2. How to convince ourselves that dxμ(dxν)=δμν?

    3. Am I correct to understand the formalism of 1 form like this :
    Given that a. ) df = ∂μf dxμ from vector calculus,
    b.) ∂μf is identified to be component of 1 form because it transforms covariently.
    Therefore we realised df is a 1-form with dxμ 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 dxμ. I also appreciate anyone to explain why dxμ can be a 1 form basis.
     
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  3. Jul 16, 2017 #2

    Orodruin

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    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.
     
  4. Jul 16, 2017 #3
    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##
     
  5. Jul 16, 2017 #4

    dextercioby

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    2. Because that d is actually an operator called exterior differential.
     
  6. Jul 16, 2017 #5
    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.
     
  7. Jul 16, 2017 #6

    Orodruin

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    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.
     
  8. Jul 16, 2017 #7

    pervect

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    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.
     
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