So let $F:\mathbb{R}^n\setminus\{0\}\to\mathbb{R}^n\setminus\{0\}$ be given by $x\mapsto x/\|x\|^2$ where $\|\cdot\|$ is the euclidean norm. Let $\omega$ be the euclidean volume form. So, we have that in local coordinates the euclidean volume form is $\omega = dx_1\wedge dx_2\wedge\cdots\wedge dx_n$. Then my understanding is that the pullback under $F$ would be simply $\det(JF)\,dx_1\wedge\cdots\wedge dx_n$ (where $JF$ is the Jacobian of $F$). However, I can't seem to find a nice way of expressing this determinant, which makes me wonder if I am correct. I calculate $$ JF=\left[\begin{array}{cc} \|x\|^2-2x_1^2 & -2x_1x_2 & \cdots &-2x_1x_n\\ -2x_2x_1 & ||x||^2-2x_2^2 & \cdots & -2x_2x_n\\ \vdots & \vdots & \ddots & \vdots\\ -2x_nx_1 & -2 x_nx_2& \cdots & \|x\|^2-2x_n^2 \end{array}\right]\frac{1}{\|x\|^4} $$ where in local coordinates $x=(x_1,x_2,\ldots,x_n)$. Am I missing something? Thanks.

  • 1
    $\begingroup$ Use the spherical symmetry! At a point $(x_1,0,\ldots,0)$ the determinant is very easy to calculate. $\endgroup$ – Anthony Carapetis Jun 7 '17 at 0:21
  • $\begingroup$ Why thank you. By rotating into $x$ to $(\|x\|,0,...,0)$ we simplify things greatly. $\endgroup$ – irh Jun 7 '17 at 0:37

Your Answer

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

Browse other questions tagged or ask your own question.