MHB Wedge product and change of variables

[i_a_n]

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

 

[Jhoneine]

Yes, we can prove it without using the fact "$\det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}$". The proof is as follows:Let $\phi(x)=(y_1,...,y_n)$. Then, by the chain rule, d$y_i=\sum_{j=1}^{n}\frac{\partial y_i}{\partial x_j}$d$x_j$, for $i=1,2,...,n$. Therefore, d$y_1\wedge ...\wedge $d$y_n=\left(\sum_{j=1}^{n}\frac{\partial y_1}{\partial x_j}$d$x_j\right)\wedge...\wedge\left(\sum_{j=1}^{n}\frac{\partial y_n}{\partial x_j}$d$x_j\right)$. Now, let $D\phi(x)=(a_{ij})_{n\times n}$ be the Jacobian matrix of $\phi$. Then, d$y_1\wedge ...\wedge $d$y_n=\left(\sum_{j=1}^{n}a_{1j}$d$x_j\right)\wedge...\wedge\left(\sum_{j=1}^{n}a_{nj}$d$x_j\right)$. Using the distributive law, we get d$y_1\wedge ...\wedge $d$y_n=\sum_{j_1,...,j_n=1}^{n}\left(a_{1j_1}...a_{nj_n}\right)$d$x_{j_1}\wedge ...\wedge$d$x_{j_n}$. Finally, since the sum is over all permutations of $j_1,...,j_n$, the result follows.