# Exterior derivative

1. May 1, 2010

### jem05

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. May 2, 2010

### lavinia

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. May 2, 2010

### jem05

yeah i know, lee has it i think,
but im trying to prove the two properties from the other definition.

4. May 2, 2010

### lavinia

you mean the Lie bracket definition?

5. May 3, 2010

### jem05

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. May 6, 2010

### lavinia

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

7. May 6, 2010

### zhentil

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.