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

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The discussion centers on proving the properties of the exterior derivative, specifically that d² = 0 and d(w1 ∧ w2) = dw1 ∧ w2 + (-1)ᵏ w1 ∧ dw2, using the invariant definition rather than the classical coordinate approach. Participants highlight the complexity of proving these properties directly from the invariant definition, with one user noting that they successfully proved d² = 0 after extensive effort. The conversation suggests that while the invariant definition is theoretically sound, the classical coordinate definition is more practical for such proofs.

PREREQUISITES
  • Understanding of exterior derivatives in differential geometry
  • Familiarity with the invariant definition of exterior derivatives
  • Knowledge of the Lie bracket and its application in differential forms
  • Basic concepts of wedge products in multilinear algebra
NEXT STEPS
  • Study the classical coordinate definition of exterior derivatives for comparative analysis
  • Explore the properties of the Lie bracket in the context of differential forms
  • Learn about the implications of the exterior derivative's uniqueness as discussed in Lee's texts
  • Practice proving properties of exterior derivatives using both invariant and coordinate definitions
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This discussion is beneficial for mathematicians, particularly those specializing in differential geometry, as well as students and researchers looking to deepen their understanding of exterior derivatives and their properties.

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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