# Wedge product and change of variables

1. Apr 21, 2013

### ianchenmu

1. The problem statement, all variables and given/known data

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$x_n$.

2. Relevant equations

n/a

3. 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 "$\det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}$"?

2. Apr 21, 2013

### 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.

Last edited: Apr 21, 2013
3. Apr 21, 2013

### Bacle2

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