# Show that $f^*\omega = \det(df) \, dx_1\wedge\cdots\wedge dx_n$

Let $f:\Bbb{R}^n\to \Bbb{R}^n$ be a differentiable map given my $f(x_1,\ldots,x_n)=(y_1,\ldots,y_n)$, and let $$\omega=dy_1\wedge\cdots\wedge dy_n.$$ Show that $$f^*\omega = \det(df) \, dx_1\wedge\cdots\wedge dx_n.$$

We can simplify the left hand side in the following way \begin{align} f^*\omega &= f^*(dy_1)\wedge\cdots\wedge f^*dy_n \\&=d(f^*y_1)\wedge\cdots\wedge d(f^*y_n) \\&=d(y_1\circ f)\wedge\cdots\wedge d(y_n\circ f) \end{align}

I'm honestly not sure how to proceed from there. The composition $y_i\circ f$ means first apply $f$ and then to $i$-th component function of $f$, right? I tried using the chain rule but it got confusing.

• Use the fact that for a linear map $g: V \to V$ with $\dim V = n$, the induced map on $\Lambda^n V \to \Lambda^n V$ is multiplication by $\det g$ (in fact, this is exactly the definition of the determinant). Commented Jul 9, 2015 at 17:33

To finish this off you need to write the $d(y_j \circ f)$ in terms of $dx_i$, use linearity of the wedge product, and produce the definition of the determinant.
You begin with: $$d(y_j \circ f) = \sum_{i=1}^n \frac{\partial(y_j \circ f)}{\partial x_i} dx_i = \sum_{i=1}^n \frac{\partial f_j}{\partial x_i} dx_i$$
• Thanks. Probably a dumb question, but I don't see why $$\frac{\partial(y_j \circ f)}{\partial x_i} = \frac{\partial f_j}{\partial x_i}$$ is true. Isn't $y_j$ precisely the $j$-th component function of $f$, i.e. $f_j$ ? Commented Jul 11, 2015 at 16:44
• But why is $y_j \circ f = f_j \circ f = f_j$? I think I'm missing something obvious. Commented Jul 11, 2015 at 17:13
• @iwriteonbananas - I don't know why you think that is being claimed. All the above says is $y_j \circ f = f_j$ which is basically just by definition. Commented Jul 11, 2015 at 17:16