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Can someone explain (Differential Forms)

  1. Feb 17, 2009 #1
    (i) if [itex]\alpha=\sum_i \alpha_i(x) dx_i \in \Omega^1, \beta=\sum_j \beta_j(x) dx_j[/itex]

    then[itex]\alpha \wedge \beta = \sum_{i,j} \alpha_i(x) \beta_j(x) dx_i \wdge dx_j \in \Omega^2[/itex]

    NOW THE STEP I DON'T FOLLOW - he jumps to this in the lecture notes:

    [itex]\alpha \wedge \beta = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j[/itex]

    the subscript on the sum was either i<j or i,j - could someone tell me which as well as explaing where on earth this step comes from.

    (ii) could someone explain the Leibniz rule for exterior derivative [itex]d: \Omega^k \rightarrow \Omega^{k+1}[/itex]

    i.e. why [itex]d(\alpha^k \wedge \beta^l)=d \alpha^k \wedge \beta^l + (-1)^k \alpha^k \wedge d \beta^l[/itex]
    note that the superscript on the differential form indicates that it's a k form or and l form

    my main problem here is where the (-1)^k comes from

    cheers for your help
     
  2. jcsd
  3. Feb 17, 2009 #2

    tiny-tim

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    Hi latentcorpse! :smile:
    It's i < j …

    [itex]\alpha \wedge \beta = \sum_{i,j} (\alpha_i \beta_j) dx_i \wedge dx_j = \sum_{i<j} (\alpha_i \beta_j - \beta_j \alpha_i) dx_i \wedge dx_j[/itex]

    since wedge is anti-commutative.
    d() is like another wedge, so if you move d() through k elementary forms, you multiply it by (-1)k :wink:
     
  4. Feb 17, 2009 #3
    cool i get the 2nd part now - still not sure why it's i<j or why [itex]\alpha_i \beta_j[/itex] suddenly becomes [itex]\alpha_i \beta_j-\beta_j \alpha_i[/itex]
     
  5. Feb 17, 2009 #4

    tiny-tim

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    (have an alpha: α and a beta: β and a wedge: ∧ :wink:)
    because αi ∧ βj = -βj ∧ αi

    so if you sum αi ∧ βj over all i and j,

    then for i > j you just "turn it round" :wink:
     
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