Computing the push forward of vector field $X = y^2 partial/partial x$ using Jacobians












1












$begingroup$


I am trying to solve the following problem. Let $M$ and $N$ be submanfiolds of $mathbb{R}^2$ given by $M = {(x,y) in mathbb{R}^2 : x > 0, x+y>0}$ and $N = {(u,v) in mathbb{R}^2 : u > 0 , v>0}$ and let $F: M rightarrow N$ be the diffeomorphism given by $F(x,y) = (F_1,F_2) = (1 + y/x, x+y)$. If $X$ is the vector field on $M$ given by $X = y^2 frac{partial}{partial x}$, compute the vector field $F_*X$ on $N$.



My attempt is the following: To compute the push forward of a vector field under a certain map, we need to compute the Jacobian of the map. In this case $$JF = left( begin{matrix}frac{partial F_1}{partial x} & frac{partial F_1}{partial y} \ frac{partial F_2}{partial x} & frac{partial F_2}{partial y} end{matrix} right) = left( begin{matrix} -frac{y}{x^2} & frac{1}{x} \ 1 & 1 end{matrix} right)$$



So, with this computation, we can represent the vector field $X$ by the vector $X = (y^2,0)$ and so the Pushforward $F_*X = -frac{y^3}{x^2}partial u + y^2 partial v$.



I would just want to make sure that this computation is correct. Thanks so much for your help!






proof-verification differential-topology vector-fields pushforward







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    1












    $begingroup$


    I am trying to solve the following problem. Let $M$ and $N$ be submanfiolds of $mathbb{R}^2$ given by $M = {(x,y) in mathbb{R}^2 : x > 0, x+y>0}$ and $N = {(u,v) in mathbb{R}^2 : u > 0 , v>0}$ and let $F: M rightarrow N$ be the diffeomorphism given by $F(x,y) = (F_1,F_2) = (1 + y/x, x+y)$. If $X$ is the vector field on $M$ given by $X = y^2 frac{partial}{partial x}$, compute the vector field $F_*X$ on $N$.



    My attempt is the following: To compute the push forward of a vector field under a certain map, we need to compute the Jacobian of the map. In this case $$JF = left( begin{matrix}frac{partial F_1}{partial x} & frac{partial F_1}{partial y} \ frac{partial F_2}{partial x} & frac{partial F_2}{partial y} end{matrix} right) = left( begin{matrix} -frac{y}{x^2} & frac{1}{x} \ 1 & 1 end{matrix} right)$$



    So, with this computation, we can represent the vector field $X$ by the vector $X = (y^2,0)$ and so the Pushforward $F_*X = -frac{y^3}{x^2}partial u + y^2 partial v$.



    I would just want to make sure that this computation is correct. Thanks so much for your help!






    proof-verification differential-topology vector-fields pushforward







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am trying to solve the following problem. Let $M$ and $N$ be submanfiolds of $mathbb{R}^2$ given by $M = {(x,y) in mathbb{R}^2 : x > 0, x+y>0}$ and $N = {(u,v) in mathbb{R}^2 : u > 0 , v>0}$ and let $F: M rightarrow N$ be the diffeomorphism given by $F(x,y) = (F_1,F_2) = (1 + y/x, x+y)$. If $X$ is the vector field on $M$ given by $X = y^2 frac{partial}{partial x}$, compute the vector field $F_*X$ on $N$.



      My attempt is the following: To compute the push forward of a vector field under a certain map, we need to compute the Jacobian of the map. In this case $$JF = left( begin{matrix}frac{partial F_1}{partial x} & frac{partial F_1}{partial y} \ frac{partial F_2}{partial x} & frac{partial F_2}{partial y} end{matrix} right) = left( begin{matrix} -frac{y}{x^2} & frac{1}{x} \ 1 & 1 end{matrix} right)$$



      So, with this computation, we can represent the vector field $X$ by the vector $X = (y^2,0)$ and so the Pushforward $F_*X = -frac{y^3}{x^2}partial u + y^2 partial v$.



      I would just want to make sure that this computation is correct. Thanks so much for your help!






      proof-verification differential-topology vector-fields pushforward







      share|cite|improve this question











      $endgroup$




      I am trying to solve the following problem. Let $M$ and $N$ be submanfiolds of $mathbb{R}^2$ given by $M = {(x,y) in mathbb{R}^2 : x > 0, x+y>0}$ and $N = {(u,v) in mathbb{R}^2 : u > 0 , v>0}$ and let $F: M rightarrow N$ be the diffeomorphism given by $F(x,y) = (F_1,F_2) = (1 + y/x, x+y)$. If $X$ is the vector field on $M$ given by $X = y^2 frac{partial}{partial x}$, compute the vector field $F_*X$ on $N$.



      My attempt is the following: To compute the push forward of a vector field under a certain map, we need to compute the Jacobian of the map. In this case $$JF = left( begin{matrix}frac{partial F_1}{partial x} & frac{partial F_1}{partial y} \ frac{partial F_2}{partial x} & frac{partial F_2}{partial y} end{matrix} right) = left( begin{matrix} -frac{y}{x^2} & frac{1}{x} \ 1 & 1 end{matrix} right)$$



      So, with this computation, we can represent the vector field $X$ by the vector $X = (y^2,0)$ and so the Pushforward $F_*X = -frac{y^3}{x^2}partial u + y^2 partial v$.



      I would just want to make sure that this computation is correct. Thanks so much for your help!






      proof-verification differential-topology vector-fields pushforward




      proof-verification differential-topology vector-fields pushforward






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 13:08









      Brahadeesh

      6,35442363




      6,35442363










      asked Dec 14 '18 at 12:23









      BOlivianoperuano84BOlivianoperuano84

      1778




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          2 Answers
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          $begingroup$

          The computation is alright, but in my opinion it is better to express the pushforward $F_* X$ in terms of the local coordinates on $N$, which is $(u,v)$. The mixture of functions given in terms of $x$ and $y$ and partial derivatives in terms of $u$ and $v$ looks odd.



          I also think you have a typo: you might have meant $frac{partial}{partial u}$ and $frac{partial}{partial v}$ rather than just $partial u$ and $partial v$ in the computed expression for $F_* X$.



          So, since $F : M to N$ is a diffeomorphism, we can invert $F$ to express $x$ and $y$ in terms of $u$ and $v$:



          $$
          begin{align}
          1+x/y &= u\
          x+y&=v
          end{align}
          bigg} implies
          1+(v-x)/x = u implies x = v/u.
          $$
          Therefore,
          $$
          begin{align}
          x &= v/u\
          y &= v(u-1)/u.
          end{align}
          $$
          Hence,
          $$
          F_*X = frac{v(u-1)^3}{u} frac{partial}{partial u} + frac{v^2(u-1)^2}{u^2} frac{partial}{partial v}.
          $$






          share|cite|improve this answer









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            0












            $begingroup$

            The computation is correct, and is well explained in this previous answer: Pushforward of a vector field






            share|cite|improve this answer









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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              The computation is alright, but in my opinion it is better to express the pushforward $F_* X$ in terms of the local coordinates on $N$, which is $(u,v)$. The mixture of functions given in terms of $x$ and $y$ and partial derivatives in terms of $u$ and $v$ looks odd.



              I also think you have a typo: you might have meant $frac{partial}{partial u}$ and $frac{partial}{partial v}$ rather than just $partial u$ and $partial v$ in the computed expression for $F_* X$.



              So, since $F : M to N$ is a diffeomorphism, we can invert $F$ to express $x$ and $y$ in terms of $u$ and $v$:



              $$
              begin{align}
              1+x/y &= u\
              x+y&=v
              end{align}
              bigg} implies
              1+(v-x)/x = u implies x = v/u.
              $$
              Therefore,
              $$
              begin{align}
              x &= v/u\
              y &= v(u-1)/u.
              end{align}
              $$
              Hence,
              $$
              F_*X = frac{v(u-1)^3}{u} frac{partial}{partial u} + frac{v^2(u-1)^2}{u^2} frac{partial}{partial v}.
              $$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                The computation is alright, but in my opinion it is better to express the pushforward $F_* X$ in terms of the local coordinates on $N$, which is $(u,v)$. The mixture of functions given in terms of $x$ and $y$ and partial derivatives in terms of $u$ and $v$ looks odd.



                I also think you have a typo: you might have meant $frac{partial}{partial u}$ and $frac{partial}{partial v}$ rather than just $partial u$ and $partial v$ in the computed expression for $F_* X$.



                So, since $F : M to N$ is a diffeomorphism, we can invert $F$ to express $x$ and $y$ in terms of $u$ and $v$:



                $$
                begin{align}
                1+x/y &= u\
                x+y&=v
                end{align}
                bigg} implies
                1+(v-x)/x = u implies x = v/u.
                $$
                Therefore,
                $$
                begin{align}
                x &= v/u\
                y &= v(u-1)/u.
                end{align}
                $$
                Hence,
                $$
                F_*X = frac{v(u-1)^3}{u} frac{partial}{partial u} + frac{v^2(u-1)^2}{u^2} frac{partial}{partial v}.
                $$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  The computation is alright, but in my opinion it is better to express the pushforward $F_* X$ in terms of the local coordinates on $N$, which is $(u,v)$. The mixture of functions given in terms of $x$ and $y$ and partial derivatives in terms of $u$ and $v$ looks odd.



                  I also think you have a typo: you might have meant $frac{partial}{partial u}$ and $frac{partial}{partial v}$ rather than just $partial u$ and $partial v$ in the computed expression for $F_* X$.



                  So, since $F : M to N$ is a diffeomorphism, we can invert $F$ to express $x$ and $y$ in terms of $u$ and $v$:



                  $$
                  begin{align}
                  1+x/y &= u\
                  x+y&=v
                  end{align}
                  bigg} implies
                  1+(v-x)/x = u implies x = v/u.
                  $$
                  Therefore,
                  $$
                  begin{align}
                  x &= v/u\
                  y &= v(u-1)/u.
                  end{align}
                  $$
                  Hence,
                  $$
                  F_*X = frac{v(u-1)^3}{u} frac{partial}{partial u} + frac{v^2(u-1)^2}{u^2} frac{partial}{partial v}.
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  The computation is alright, but in my opinion it is better to express the pushforward $F_* X$ in terms of the local coordinates on $N$, which is $(u,v)$. The mixture of functions given in terms of $x$ and $y$ and partial derivatives in terms of $u$ and $v$ looks odd.



                  I also think you have a typo: you might have meant $frac{partial}{partial u}$ and $frac{partial}{partial v}$ rather than just $partial u$ and $partial v$ in the computed expression for $F_* X$.



                  So, since $F : M to N$ is a diffeomorphism, we can invert $F$ to express $x$ and $y$ in terms of $u$ and $v$:



                  $$
                  begin{align}
                  1+x/y &= u\
                  x+y&=v
                  end{align}
                  bigg} implies
                  1+(v-x)/x = u implies x = v/u.
                  $$
                  Therefore,
                  $$
                  begin{align}
                  x &= v/u\
                  y &= v(u-1)/u.
                  end{align}
                  $$
                  Hence,
                  $$
                  F_*X = frac{v(u-1)^3}{u} frac{partial}{partial u} + frac{v^2(u-1)^2}{u^2} frac{partial}{partial v}.
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 14 '18 at 13:05









                  BrahadeeshBrahadeesh

                  6,35442363




                  6,35442363























                      0












                      $begingroup$

                      The computation is correct, and is well explained in this previous answer: Pushforward of a vector field






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        The computation is correct, and is well explained in this previous answer: Pushforward of a vector field






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          The computation is correct, and is well explained in this previous answer: Pushforward of a vector field






                          share|cite|improve this answer









                          $endgroup$



                          The computation is correct, and is well explained in this previous answer: Pushforward of a vector field







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 14 '18 at 12:52









                          Marcelo Roberto JimenezMarcelo Roberto Jimenez

                          464




                          464






























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