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Wedge product and change of variables

ianchenmu

Member
Feb 3, 2013
74
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$.


Take a look at here and the answer given by Michael Albanese:
differential geometry - wedge product and change of variables - Mathematics Stack Exchange

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