- Thread starter
- #1
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)}$"?
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)}$"?