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.

  • $\begingroup$ 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). $\endgroup$ – anomaly Jul 9 '15 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: $$d(y_j \circ f) = \sum_{i=1}^n \frac{\partial(y_j \circ f)}{\partial x_i} dx_i = \sum_{i=1}^n \frac{\partial f_j}{\partial x_i} dx_i$$

The minuses in the determinant formula come from the antisymmetry of the wedge product.

  • $\begingroup$ 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. $f_j$ ? $\endgroup$ – iwriteonbananas Jul 11 '15 at 16:44
  • $\begingroup$ @iwriteonbananas - Yes, that is why this is trivially true, just rewriting to make it clearer how to arrive at the determinant of the Jacobian. $\endgroup$ – muaddib Jul 11 '15 at 17:03
  • $\begingroup$ But why is $y_j \circ f = f_j \circ f = f_j$? I think I'm missing something obvious. $\endgroup$ – iwriteonbananas Jul 11 '15 at 17:13
  • $\begingroup$ @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. $\endgroup$ – muaddib Jul 11 '15 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.