Let $M$ be a smooth manifolds and $\big(\mathcal U,\varphi=(x^1,\ldots,x^n)\big)$ and $\big(\mathcal V,\psi=(y^1,\ldots,y^n)\big)$. Prove that: $$dx^{j_1}\wedge\ldots\wedge dx^{j_r}=\det \Big(\frac{\partial y^i}{\partial x^j}\Big)dy^{i_1}\wedge\ldots \wedge dy^{i_r}$$ How can I prove the last statement? Maybe with the Levi-Civita symbol?
Many thanks!
More or less, except that you should exchange $x$ and $y$ in the determinant and there is a sum over possible $r$-tuples. From $dx^j=\sum_i \frac{\partial x^j}{\partial y_i} dy^i$ we get $$ dx^{j_1}\wedge ... \wedge dx^{j_r} = \sum_{i_1,...,i_r} \frac{\partial x^{j_1}}{\partial y^{i_1}}\cdots \frac{\partial x^{j_r}}{\partial y^{i_r}} dy^{i_1}\wedge \ldots \wedge dy^{i_r}$$ where the sum is over all possible indices. Those with repeating indices vanishes. We then regroup into distinct $r$-tuples and their permutations: $$ \sum_{i_1<...<i_r} \sum_{\sigma\in Perm(i_1,...,i_r)} {\rm sign}(\sigma) \frac{\partial x^{j_1}}{\partial y^{i_{\sigma(1)}}} \cdots \frac{\partial x^{j_r}}{\partial y^{i_{\sigma(r)}}} dy^{i_1}\wedge \ldots \wedge dy^{i_r}$$ The sign of the permuation (you could instead use a Levi-Civita symbol) comes from undoing the permutation in the wedge product. The latter equals: $$ \sum_{i_1<...<i_r} \det \frac{\partial (x^{j_1}, ... , x^{j_r})}{\partial (y^{i_1}, ... , y^{i_r})} dy^{i_1}\wedge \ldots \wedge dy^{i_r}$$