About exterior algebra in vector calculus

[enricfemi]

I'm reading Marsden's vector calculus. In the chapter of differential forms, it mentions the wedge product satisfies the laws:

dy^dx=-dxdy.
and for a 0-form f, f^w=fw.

Does it have formal derivation?
hope someone can give me a hint or even a link.

 

[ystael]

You can construct the bundles of exterior forms by postulating those laws as definitions, or by taking a quotient of the tensor bundles. If you don't know about bundles, you might do better to take the laws as definitions that you use to generate the differential forms.

The short version of the quotient construction is: The exterior algebra \textstyle\bigwedge^* V on a vector space V is the quotient of the tensor algebra \textstyle\bigotimes^* V by the homogeneous ideal generated by \{x \otimes x \mid x \in V\}. To get the bundle of exterior forms on a manifold, you do the same thing fiberwise on the tensor bundle.

Basically you kill off all the tensors that contain the same factor twice, and you wind up being able to write (dx + dy) \wedge (dx + dy) = dx \wedge dx + dy \wedge dx + dx \wedge dy + dy \wedge dy = dy \wedge dx + dx \wedge dy = 0.

The linearity over smooth functions comes from the fact that at each point a function behaves like a scalar in the tensor algebra (fiber of the tensor bundle) at that point.

Try Spivak, Calculus on manifolds; Munkres, Analysis on manifolds; Spivak, A comprehensive introduction to differential geometry, volume I (in increasing order of sophistication). If you want more on exterior algebra specifically, consult Greub, Multilinear algebra or the first chapter of Federer, Geometric measure theory.

 

[enricfemi]

Thank you, ystael!

I have found Spivak's calculus on manifolds and trying to read it.

and so sorry that there maybe a misunderstanding:

I know how to prove
dy^dx=-dx^dy

what i really confusing is why
dx^dy=dxdy
dy^dx=-dxdy

Is it only happened at R3?