Exterior Derivative: Proving d^2 = 0 and d(w1 wedge w2) Properties

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Discussion Overview

The discussion revolves around proving properties of the exterior derivative, specifically that \( d^2 = 0 \) and the formula for \( d(w_1 \wedge w_2) \). Participants are exploring these properties using an invariant definition of the exterior derivative rather than the classical coordinate approach, which they find cumbersome.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses a desire to prove \( d^2 = 0 \) and \( d(w_1 \wedge w_2) = dw_1 \wedge w_2 + (-1)^k w_1 \wedge dw_2 \) directly from the invariant definition of the exterior derivative.
  • Another participant suggests that defining the exterior derivative in terms of the Lie bracket could lead to an equivalence with the coordinate definition, although they have not personally done this.
  • A participant mentions successfully proving \( d^2 = 0 \) using the invariant definition but warns that it was a challenging process.
  • There is a consensus among some participants that the proof using the local formulation is simpler compared to the invariant definition.
  • One participant invites others to share their calculations to facilitate discussion and review.

Areas of Agreement / Disagreement

Participants generally agree that proving the properties from the invariant definition is difficult, with some expressing that the local formulation is easier. However, there is no consensus on the best approach or the ease of the proofs.

Contextual Notes

Participants note the complexity involved in juggling indices when using the invariant definition, which may lead to challenges in clarity and understanding.

jem05
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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 don't 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.
 
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jem05 said:
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 don't 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.

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
 
yeah i know, lee has it i think,
but I am trying to prove the two properties from the other definition.
 
jem05 said:
yeah i know, lee has it i think,
but I am trying to prove the two properties from the other definition.

you mean the Lie bracket definition?
 
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.
 
jem05 said:
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.

I don't think it is so bad. Why not post your calculation? We could go over it.
 
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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