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Volume element as n-form

  1. Nov 11, 2010 #1

    I'm reading Sean Carroll's text, ch2, and believe I understood most of the discussion on Integration in section 2.10. However in equation 2.96, he states that

    [tex] \epsilon\equiv\epsilon_{\mu_1\mu_2...\mu_n}dx^{\mu_1}\otimes dx^{\mu_2}\otimes...\otimes dx^{\mu_n} =\frac{1}{n!}\epsilon_{\mu_1\mu_2...\mu_n}}dx^{\mu_1}\wedge dx^{\mu_2}\wedge...\wedge dx^{\mu_n}[/tex]

    I don't quite understand this equality. For example just taking n=2, the LHS is [tex] dx^0\otimes dx^1-dx^1\otimes dx^0[/tex] (which isn't zero because tensor products don't commute). Where on the RHS one would have [tex] \frac{1}{2}\left(dx^0\wedge dx^1-dx^1\wedge dx^0\right)=dx^0\wedge dx^1[/tex], by the antisymmetry of the wedge product.

    So I'm at a loss to understand this part of 2.96 despite understand the following lines.

    Thanks alot for any replies.
  2. jcsd
  3. Nov 12, 2010 #2
    Formula (1.81) in Carroll's "Lecture Notes on General Relativity" (1997) tells you:

    [tex]A\wedge B=A\otimes B-B\otimes A[/tex]

    So, what is it that causes you the problem?
  4. Nov 12, 2010 #3
    Oh I was unaware of this formula.

    Do you mean (1.80) in this edition of Carroll? namely [tex] (A\wedge B)_{\mu\nu}=A_{\mu}B_{\nu}-A_{\nu}B_{\mu}[/tex]

    Which I guess can be written [tex] (A\wedge B)_{\mu\nu} =(A\otimes B)_{\mu\nu}-(B\otimes A)_{\mu\nu}[/tex], leading to the equation you stated: [tex] (A\wedge B) =(A\otimes B)-(B\otimes A)[/tex]
  5. Nov 12, 2010 #4
    Yes - that is another way of writing it.
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