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Simple wedge product issue

  1. Feb 2, 2010 #1
    I am studying general relativity from Sean Carroll's text and I have a simple question about the wedge product. According to the text, the wedge product of two one-forms (dual vectors) is

    [tex](A \wedge B)_{\mu\nu} = 2A_{[\mu}B_{\nu]} = A_\mu B_\nu - A_\nu B_\mu[/tex]

    I understand the why the first two expressions are equal given the definition of the wedge product. I also understand that the square brackets in the third second expression denotes antisymmetry and that the two indices can be exchanged at the cost of a minus sign. However, the third expression seems to imply that the antisymmetry of the two indices (perhaps because they belong to two different quantities) really means the expansion given in the third expression. Can anyone clear up what I'm sure is a simple misunderstanding of notation?

    Followup question: what is the meaning of, for example in Minkowski spacetime with spherical spatial coordinates, [tex]d\theta \wedge d\phi[/tex]? I know that each of these is a (basis) one-form but I'm not sure how exactly to apply the wedge product to them.
    Last edited: Feb 2, 2010
  2. jcsd
  3. Feb 2, 2010 #2


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    ABν] is defined as (1/2)(AμBν - AνBμ)

    For n indices, A[abc...d] = (1/n!)(Aabc...d + .... ).

    This is just a matter of convention. If we define antisymmterization without the 1/n! factor, then we also change the definition of the wedge product to make it all work out the same.
  4. Feb 3, 2010 #3
    Yes, I see that now. Thanks dx! Anyone have any insight as to the meaning of [tex]d\theta \wedge d\phi[/tex]? I guess my confusion is what are the "components" of these basis dual vectors to which to apply the definition of the wedge product?
    Last edited: Feb 3, 2010
  5. Feb 3, 2010 #4


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    In what basis? The components of dφ in the (dθ, dφ) basis are (0, 1).
  6. Feb 3, 2010 #5

    Ben Niehoff

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    [itex]d\theta \wedge d\phi[/itex] is a two-form, which means it is a map [itex]T_xM \times T_xM \rightarrow {\mathbb R}[/itex], whose action on a pair of vectors u, v is like follows:

    [tex](d\theta \wedge d\phi)(u, v) = \det \left| \begin{array}{cc} d\theta(u) & d\phi(u) \\ d\theta(v) & d\phi(v) \end{array} \right| = d\theta(u) \; d\phi(v) - d\theta(v) \; d\phi(u)[/tex]

    That is the meaning of a wedge product.
  7. Feb 3, 2010 #6
    Yes, I think I see now. The components, in matrix form, of the resulting two-form would then be

    (d\theta \wedge d\phi)_{\mu\nu} = \left[\begin{array}{cccc}
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 1\\
    0 & 0 & -1 & 0

    in the [tex]dx^\mu \otimes dx^\nu[/tex] basis in Minkowski spacetime with spherical spatial coordinates.
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