Prove invariant formula for exterior derivative

130 Views Asked by At

In Marsden "Foundation of Mechanics", it is said that the properties defining the exterior derivative are readily proven to be satisfied by the invariant formula

$d\omega(u)(e_0,...e_k)=\sum_{i=1}^k(-1)^i D\omega(u)·e_i(e_0,...,\hat e_i,...e_k)$

where $e_0,...e_k$ are the basis of the tangent space at point $u$. However, I see no way of proving that this formula satisfies $d(\omega\wedge\rho)=d\omega\wedge\rho + (-1)^k\omega\wedge d\rho$, being $\omega$ a k-form and $\rho$ a m-form, in a coordinate-free way. It seems extremely difficult to prove this property without proving actually that the invariant formula is equal to the coordinate-based definition.

Any clues?

PS: actually, in the book "Manifolds, tensor analysis and applications" by the same author, they don't even try to follow this path and instead prove that both definitions boil down to the coordinate-based one.