Wedge product and change of variables

The question is:

Let $\phi ： \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map and let $y = \phi(x)$ be the change of variables. Show that

$$dy_1\wedge\dots\wedge dy_n = (\operatorname{det}D\phi(x))\cdot dx_1\wedge\dots\wedge dx_n.$$

I tried $n = 2, 3$ but still didn't get any rule. Do we need to do this under integral?

Solution 1:

We have $y = (y_1, \dots, y_n) = (\phi_1, \dots, \phi_n)$ so \begin{align*} dy_1\wedge\dots\wedge dy_n &= d\phi_1\wedge\dots\wedge d\phi_n\\ &= \left(\frac{\partial \phi_1}{\partial x_1}dx_1 + \dots +\frac{\partial \phi_1}{\partial x_n}dx_n\right)\wedge\dots\wedge\left(\frac{\partial \phi_n}{\partial x_1}dx_1 + \dots +\frac{\partial \phi_n}{\partial x_n}dx_n\right)\\ &= \sum_{\sigma\in S_n}\frac{\partial\phi_1}{\partial x_{\sigma(1)}}\dots\frac{\partial\phi_n}{\partial x_{\sigma(n)}}dx_{\sigma(1)}\wedge\dots\wedge dx_{\sigma(n)}\\ &= \sum_{\sigma\in S_n}\left(\prod_{i=1}^n\frac{\partial\phi_i}{\partial x_{\sigma(i)}}\right)dx_{\sigma(1)}\wedge\dots\wedge dx_{\sigma(n)}\\ &= \left(\sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^n\frac{\partial\phi_i}{\partial x_{\sigma(i)}}\right)dx_1\wedge\dots\wedge dx_n\\ &= \det\left(\frac{\partial\phi_i}{\partial x_j}\right)dx_1\wedge\dots\wedge dx_n\\ &= \det(D\phi)dx_1\wedge\dots\wedge dx_n \end{align*}

where we have used the fact that for an $n\times n$ matrix $A = (a_{ij})$, the determinant is given by $$\det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i\sigma(j)}$$ as can be seen here.