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

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In summary, you are trying to prove two properties of the exterior derivative using two different definitions. The Lie bracket definition is easier, but you are failing to do so.
  • #1
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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  • #2
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
 
  • #3
yeah i know, lee has it i think,
but I am trying to prove the two properties from the other definition.
 
  • #4
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?
 
  • #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.
 
  • #6
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.
 
  • #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.
 

1. What is the Exterior Derivative?

The Exterior Derivative is a mathematical operator used in differential geometry and multivariable calculus to represent the change of a multivariable function or differential form as it moves along a particular direction in space. It is denoted by the symbol "d".

2. How do you prove that d^2 = 0 for the Exterior Derivative?

To prove that d^2 = 0 for the Exterior Derivative, we use the definition of the Exterior Derivative and its properties, such as the Leibniz rule and the product rule. By applying these properties and simplifying the resulting expression, we can show that d^2 = 0.

3. What are the properties of d(w1 wedge w2)?

The properties of d(w1 wedge w2) are:

  • d(w1 wedge w2) = (dw1) wedge w2 + (-1)^p w1 wedge (dw2), where p is the degree of w1.
  • d(w1 wedge w2) = 0 if w1 and w2 are both constant functions.
  • d(w1 wedge w2) = 0 if w1 and w2 are both exact forms.
  • d(w1 wedge w2) = dw1 wedge dw2 if w1 and w2 are both closed forms.

4. How can the Exterior Derivative be used in physics?

The Exterior Derivative is used in physics to express physical laws and equations in a coordinate-independent manner. It is also used in differential geometry to describe the geometry of spacetime in theories such as General Relativity.

5. Can the Exterior Derivative be extended to higher dimensions?

Yes, the Exterior Derivative can be extended to higher dimensions, such as 4-dimensional spacetime in General Relativity. In these higher dimensions, the Exterior Derivative is still defined as the change of a differential form along a particular direction, but with more variables and coordinates involved.

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