7
$\begingroup$

Consider the transformation $F$ of $\mathbb R^2\setminus\{(0,0)\}$ onto itself defined as $$ F(x, y):=\left( \frac{x}{x^2+y^2}, \frac{y}{x^2+y^2}\right).$$ Its Jacobian matrix is $$\tag{1} \begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix},\quad \text{and its determinant equals}\ \frac{-1}{(x^2+y^2)^2}.$$ The following alternative computation is wrong at (!) and (!!), and I cannot see why.

Let $\phi\colon (0, \infty)\times (-\pi, \pi)\to \mathbb R^2$ be the map $$\phi(r, \theta) =(r\cos \theta, r\sin \theta).$$ Let moreover $$\tag{2}\tilde{F}:=\phi^{-1}\circ F\circ \phi;$$ then, by an easy direct computation, $$\tilde{F}(r, \theta)=\left( \frac1r, \theta\right).$$The Jacobian matrix of $\tilde{F}$ is, thus, $$\tag{!}\begin{bmatrix} \frac{-1}{r^2} & 0 \\ 0 & 1\end{bmatrix} , \quad \text{and its determinant equals }\ \frac{-1}{r^2}.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $\tilde{F}$ are equal. We conclude that the Jacobian determinant of $F$ is $$\tag{!!} \frac{-1}{r^2}=\frac{-1}{x^2+y^2}.$$

The result (!!) is off by a factor of $r^{-2}$ from the correct one, which is given in (1). Equation (!) must also be wrong. Indeed, computing the Jacobian matrix from (2) using the chain rule, and using that $$ D\phi = \begin{bmatrix} \cos \theta & \sin \theta \\ -r\sin \theta & r\cos \theta\end{bmatrix}$$ and that $$\tag{!!!} D(\phi^{-1})= \begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix},$$ I obtain the result $$ \begin{bmatrix} \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} \\ -\frac{y}{x^2+y^2} & \frac{x}{x^2+y^2}\end{bmatrix} \begin{bmatrix} \frac{y^2-x^2}{(x^2+y^2)^2} & -\frac{2xy}{(x^2+y^2)^2} \\ -\frac{2xy}{(x^2+y^2)^2} & \frac{x^2-y^2}{(x^2+y^2)^2} \end{bmatrix}\begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin \theta & r\cos \theta\end{bmatrix} = \begin{bmatrix} -\frac1{r^2} & 0 \\ 0 & \frac{1}{r^2}\end{bmatrix},$$ which is different from the matrix in (!), and which gives the correct determinant of $-1/r^4$, as it should be.

Can you help me spot the mistake?


SOLUTION (added at a later time). As answerers pointed out, there is a mistake in (!!!). The correct matrix to be used is $$ D(\phi^{-1})|_{F\circ \phi(r, \theta)} = \begin{bmatrix} \frac{\frac{x}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} & \frac{\frac{y}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} \\ -\frac{\frac{y}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2} & \frac{\frac{x}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin \theta \\ - r\sin \theta & r\cos\theta \end{bmatrix}.$$ Had I used this matrix, I would have found the correct result for the Jacobian matrix of $\tilde{F}$, which is the equation marked (!). Thus, (!) is actually correct.

My fundamental misunderstanding was the assumption that, because of (2), the Jacobian determinant should be invariant for coordinate changes. This is not true; what follows from (2) is only that $$ \det D\tilde{F}|_{(r, \theta)}= \det D\phi^{-1}|_{F\circ\phi(r, \theta)}\det D\phi|_{(r, \theta)} \det DF|_{\phi(r, \theta)}. $$ The first two factors in the right-hand side need not cancel, as I erroneously thought.

$\endgroup$
3
$\begingroup$

I don't think there is any contradiction here.

Consider the volume form $$ \omega_{\rm Cart} = dx \wedge dy.$$ Your first calculation shows that the pullback $F^\star(\omega_{\rm Cart})$ is given by $$ F^\star(\omega_{\rm Cart}) = - \frac{1}{(x^2+y^2)^2}\omega_{\rm Cart}.$$

Now consider the volume form $$ \omega_{\rm Polar} = dr \wedge d\theta.$$ Your second calculation shows that

$$ F^\star(\omega_{\rm Polar})=-\frac 1 {r^2} \omega_{\rm Polar}. $$

We can use this to recompute $F^\star(\omega_{\rm Cart})$. In view of the fact that $$ \omega_{\rm Cart} = r \omega_{\rm Polar},$$ we have: \begin{align} F^\star(\omega_{\rm Cart}) &= F^\star(r\omega_{\rm Polar}) \\ &= F^\star(r) F^\star(\omega_{\rm Polar}) \\ &= \frac 1 r \left( - \frac 1 {r^2}\omega_{\rm Polar} \right) \\ &= - \frac{1}{r^4} \left(r\omega_{\rm Polar} \right) \\ &= - \frac 1 {r^4} \omega_{\rm Cart} \end{align} which is consistent with the first calculation!


As for the application of the chain rule, we have: $$ (D\bar F)|_{(r, \theta)} = D(\phi^{-1})|_{F\circ \phi(r, \theta)} (DF)|_{\phi(r, \theta)} (D\phi)|_{(r, \theta)}$$

The key point is that you must evaluate $D(\phi^{-1})$ at the point $\left(\frac x { (x^2 +y^2)}, \frac y {(x^2 + y^2)}\right)$, not at the point $(x, y)$.

This is equal to

$$ D(\phi^{-1})|_{F\circ \phi(r, \theta)} = \begin{bmatrix} \frac{\frac{x}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} & \frac{\frac{y}{x^2 + y^2}}{\sqrt{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}} \\ -\frac{\frac{y}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2} & \frac{\frac{x}{x^2 + y^2}}{\left(\frac{x}{x^2 + y^2} \right)^2+\left( \frac{y}{x^2 + y^2}\right)^2}\end{bmatrix} = \begin{bmatrix}\cos\theta & \sin \theta \\ - r\sin \theta & r\cos\theta \end{bmatrix}$$ which is not the inverse of $(D\phi)|_{(r, \theta)}$.

$\endgroup$
  • $\begingroup$ I couldn't expect a better answer. You nailed it completely. Also, thank you for suggesting the viewpoint of volume forms; that makes for much slicker computations. $\endgroup$ – Giuseppe Negro Apr 23 at 22:10
4
$\begingroup$

The Jacobians of the two functions aren't equal by the chain rule.

In actual fact, $D(\phi(\frac{1}{r}, \cos\theta)) × D\tilde{F}(r, \theta)= DF \times D(\phi(r, \theta))$

$\endgroup$
  • $\begingroup$ I accepted the other answer, but this one is also correct and useful, as I remarked in the "Solution" section, at the end of the question. Thank you. $\endgroup$ – Giuseppe Negro Apr 24 at 11:27

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.