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Exterior derivative

  1. May 1, 2010 #1
    hello,
    i am given the definition of an exterior derivative as its invariant formula not as the classical coordinate way. ie, i have dw(X1,...,Xn+1) = sum(-1)^i-1 Xi(w(X1,...^Xi^,...,Xn+1)
    + sum(-1)^i+j (w([Xi,Xj],X1,...^Xi^,...,^Xj^,...,Xn+1) where the hats means omitted X.
    and i dont have that dw= d(sum wJ dx^J)= sum dwJ wedge dx^J

    i want to prove that d^2 = 0 and that d(w1 wedge w2) = dw1 wedge w2 + (-1)^k w1 wedge dw2.

    thing is, i want to do that directly from my definition bc its a nasty process trying to prove that the 2 definitions are equivalent then proving the properties on the other definition.
    im failing to do that though, any ideas?
    thx a lot.
     
  2. jcsd
  3. May 2, 2010 #2

    lavinia

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    not sure what you are asking but you can define exterior derivative in terms of the Lie bracket.

    While I have never done it it should be straight forward to show its equivalence to the coordinate definition.

    I imagine that there is a proof that the exterior derivative is unique - but not sure
     
  4. May 2, 2010 #3
    yeah i know, lee has it i think,
    but im trying to prove the two properties from the other definition.
     
  5. May 2, 2010 #4

    lavinia

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    you mean the Lie bracket definition?
     
  6. May 3, 2010 #5
    yeah, the invariant one, independent of the bases.
    pple i did this d^2 = 0 using the invariant definition.
    it took me a whole day, and i was close to blindeness! do not try this.
    the base definition is much easier to deal with.
     
  7. May 6, 2010 #6

    lavinia

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    I don't think it is so bad. Why not post your calculation? We could go over it.
     
  8. May 6, 2010 #7
    The proof isn't difficult, but it is a pain to juggle all the indices. I agree that the proof using the local formulation is far simpler.
     
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