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How to compute a Jacobian using polar coordinates?


How do I convert a vector field in Cartesian coordinates to spherical coordinates?gradient in polar coordinate by changing gradient in Cartesian coordinateJacobian of the change of variablesJacobian Determinant of Polar-Coordinate TransformationPolar coordinates and Jacobian of $frac12 r $Elementary JacobianPartial derivative in polar coordinatesHow do I prove this identity involving polar coordinates and $nabla$?What is the Jacobian in this transformationDoubt about differentia operatorl in polar coordinates













5












$begingroup$


Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
$$
F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$

Its Jacobian matrix is
$$tag1
beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$

The following alternative computation is wrong at (!) and (!!), and I cannot see why.




Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$




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 I obtain the result
$$
beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$

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?











share|cite











$endgroup$
















    5












    $begingroup$


    Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
    $$
    F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$

    Its Jacobian matrix is
    $$tag1
    beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$

    The following alternative computation is wrong at (!) and (!!), and I cannot see why.




    Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$




    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 I obtain the result
    $$
    beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$

    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?











    share|cite











    $endgroup$














      5












      5








      5


      2



      $begingroup$


      Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
      $$
      F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$

      Its Jacobian matrix is
      $$tag1
      beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$

      The following alternative computation is wrong at (!) and (!!), and I cannot see why.




      Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$




      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 I obtain the result
      $$
      beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$

      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?











      share|cite











      $endgroup$




      Consider the transformation $F$ of $mathbb R^2setminus(0,0)$ onto itself defined as
      $$
      F(x, y):=left( fracxx^2+y^2, fracyx^2+y^2right).$$

      Its Jacobian matrix is
      $$tag1
      beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrix,quad textand its determinant equals frac-1(x^2+y^2)^2.$$

      The following alternative computation is wrong at (!) and (!!), and I cannot see why.




      Let $phicolon (0, infty)times (-pi, pi)to mathbb R^2$ be the map $$phi(r, theta) =(rcos theta, rsin theta).$$ Let moreover $$tag2tildeF:=phi^-1circ Fcirc phi;$$ then, by an easy direct computation, $$tildeF(r, theta)=left( frac1r, thetaright).$$The Jacobian matrix of $tildeF$ is, thus, $$tag!beginbmatrix frac-1r^2 & 0 \ 0 & 1endbmatrix , quad textand its determinant equals frac-1r^2.$$On the other hand, by (2) and by the chain rule, the Jacobian determinants of $F$ and $tildeF$ are equal. We conclude that the Jacobian determinant of $F$ is $$tag!! frac-1r^2=frac-1x^2+y^2.$$




      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 I obtain the result
      $$
      beginbmatrix fracxsqrtx^2+y^2 & fracysqrtx^2+y^2 \ -fracyx^2+y^2 & fracxx^2+y^2endbmatrix beginbmatrix fracy^2-x^2(x^2+y^2)^2 & -frac2xy(x^2+y^2)^2 \ -frac2xy(x^2+y^2)^2 & fracx^2-y^2(x^2+y^2)^2 endbmatrixbeginbmatrix costheta & -rsintheta \ sin theta & rcos thetaendbmatrix = beginbmatrix -frac1r^2 & 0 \ 0 & frac1r^2endbmatrix,$$

      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?








      calculus multivariable-calculus differential-geometry






      share|cite















      share|cite













      share|cite




      share|cite








      edited 3 hours ago









      Tengu

      2,68411021




      2,68411021










      asked 4 hours ago









      Giuseppe NegroGiuseppe Negro

      17.7k332128




      17.7k332128




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

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



          In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$






          share|cite









          $endgroup$




















            2












            $begingroup$

            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
            $$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$



            Now consider the volume form
            $$ omega_rm Polar = dr wedge dtheta.$$
            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:
            beginalign
            F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
            endalign

            which is consistent with the first calculation!




            As for the application of the chain rule, we have:
            $$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$



            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



            $$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
            which is not the inverse of $(Dphi)|_(r, theta)$.






            share|cite











            $endgroup$












            • $begingroup$
              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
              2 hours ago


















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

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



            In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$






            share|cite









            $endgroup$

















              2












              $begingroup$

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



              In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$






              share|cite









              $endgroup$















                2












                2








                2





                $begingroup$

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



                In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$






                share|cite









                $endgroup$



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



                In actual fact, $D(phi(frac1r, costheta)) × DtildeF(r, theta)= DF times D(phi(r, theta))$







                share|cite












                share|cite



                share|cite










                answered 3 hours ago









                George DewhirstGeorge Dewhirst

                1,72515




                1,72515





















                    2












                    $begingroup$

                    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
                    $$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$



                    Now consider the volume form
                    $$ omega_rm Polar = dr wedge dtheta.$$
                    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:
                    beginalign
                    F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
                    endalign

                    which is consistent with the first calculation!




                    As for the application of the chain rule, we have:
                    $$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$



                    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



                    $$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
                    which is not the inverse of $(Dphi)|_(r, theta)$.






                    share|cite











                    $endgroup$












                    • $begingroup$
                      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
                      2 hours ago















                    2












                    $begingroup$

                    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
                    $$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$



                    Now consider the volume form
                    $$ omega_rm Polar = dr wedge dtheta.$$
                    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:
                    beginalign
                    F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
                    endalign

                    which is consistent with the first calculation!




                    As for the application of the chain rule, we have:
                    $$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$



                    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



                    $$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
                    which is not the inverse of $(Dphi)|_(r, theta)$.






                    share|cite











                    $endgroup$












                    • $begingroup$
                      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
                      2 hours ago













                    2












                    2








                    2





                    $begingroup$

                    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
                    $$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$



                    Now consider the volume form
                    $$ omega_rm Polar = dr wedge dtheta.$$
                    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:
                    beginalign
                    F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
                    endalign

                    which is consistent with the first calculation!




                    As for the application of the chain rule, we have:
                    $$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$



                    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



                    $$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
                    which is not the inverse of $(Dphi)|_(r, theta)$.






                    share|cite











                    $endgroup$



                    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
                    $$ F^star(omega_rm Cart) = - frac1(x^2+y^2)^2omega_rm Cart.$$



                    Now consider the volume form
                    $$ omega_rm Polar = dr wedge dtheta.$$
                    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:
                    beginalign
                    F^star(omega_rm Cart) &= F^star(romega_rm Polar) \ &= F^star(r) F^star(omega_rm Polar) \ &= frac 1 r left( - frac 1 r^2omega_rm Polar right) \ &= - frac1r^4 left(romega_rm Polar right) \ &= - frac 1 r^4 omega_rm Cart
                    endalign

                    which is consistent with the first calculation!




                    As for the application of the chain rule, we have:
                    $$ (Dbar F)|_(r, theta) = D(phi^-1)|_Fcirc phi(r, theta) (DF)|_phi(r, theta) (Dphi)|_(r, theta)$$



                    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



                    $$ D(phi^-1)|_Fcirc phi(r, theta) = beginbmatrix fracfracxx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracyx^2 + y^2sqrtleft(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 \ -fracfracyx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2 & fracfracxx^2 + y^2left(fracxx^2 + y^2 right)^2+left( fracyx^2 + y^2right)^2endbmatrix = beginbmatrixcostheta & sin theta \ - rsin theta & rcostheta endbmatrix$$
                    which is not the inverse of $(Dphi)|_(r, theta)$.







                    share|cite














                    share|cite



                    share|cite








                    edited 3 hours ago

























                    answered 3 hours ago









                    Kenny WongKenny Wong

                    20.1k21442




                    20.1k21442











                    • $begingroup$
                      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
                      2 hours ago
















                    • $begingroup$
                      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
                      2 hours ago















                    $begingroup$
                    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
                    2 hours ago




                    $begingroup$
                    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
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