Wedge product and change of variables

[ianchenmu]

Homework Statement

The question is: Let $\phi: \mathbb{R}^n\rightarrow\mathbb{R}^n$ be a $C^1$ map and let $y=\phi(x)$ be the change of variables. Show that d$y_1\wedge...\wedge$d$y_n$=(detD$\phi(x)$)$\cdot$d$x_1\wedge...\wedge$d$x_n$.

Homework Equations

n/a

The Attempt at a Solution

Take a look at here and the answer given by Michael Albanese:
http://math.stackexchange.com/questions/367949/wedge-product-and-change-of-variables

My question is can we prove it without using the fact "$\det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}$"?

 

[Bacle2]

Do you know the definition of the pullback of a differential form ? This is a generalization to multilinear

maps of the "induced map" L* , from W* to V*, given a linear map L:V-->W , both V,W vector spaces.

I'm trying to avoid heavy machinery, but I think you need to understand this, unless you just want

a quick-and-dirty answer ( I assume you don't since you would have accepted the answer from the link

if you did.). You are basically doing a change of bases for multilinear maps, an extension of the idea of

basis change for a linear map.

 

[Bacle2]

Or maybe you can tell us the approach you want to follow .