Problem 16.4 on related vector fields Differentiable Manifolds Marcut












1














In the course I am taking on differentiable manifolds, we have been using the notes written by Marcut . This problem caught my attention as it seemed pretty straightforward.




Consider the map:
$$phi: mathbb{R}^2 to mathbb{R}^2,~ phi(x,y):=(e^xcos(y),e^xsin(y))$$




  1. Find two vector fields $X,~Y~in~mathfrak{X}(mathbb{R}^2$ such that $X$ is $phi$-related to $partial_x$ and $Y$ is $phi$-related to $partial_y$. Are these vector fields uniquely determined by this condition?. Calculate $[X,Y]$

  2. Find two vector fields $U,~V~in~mathfrak{X}(mathbb{R}^2$ such that $partial_x$ is $phi$-related to $U$ and $partial_y$ is $phi$-related to $V$. Are these vector fields uniquely determined by this condition?. Calculate $[U,V]$




Solution



I have found and checked the solutions for both parts.



For part a), I used the fact that the required condition is equivalent to $ partial_x(phi^*(f))=phi^*(X( f)) label{eq:related}$. I find $X$. Using the same procedure for $Y$, I get that:
$$X = xpartial_x+ypartial_y,~ Y= -ypartial_x+xpartial_y$$
$$[X,Y]=0$$



For part b), I noticed that $U$ can be written as $U=u_1(x,y)partial_x+u_2(x,y)partial_y$. Using the condition that $U(phi^*(f))_p=(partial_u( f))_{phi(p)}$, I find $U$. Using the same procedure for $V$, I get:
$$U= e^{-x}cos(y)partial_x-e^{-x}sin(y)partial_y,~V= e^{-x}sin(y)partial_x+e^{-x}cos(y)partial_y$$
$$[U,V]=0$$



So in summary:




  • In part a), we are finding the pushforward of the coordinate directions: $$X=phi_*(partial_x),~Y=phi_*(partial_y) $$

  • In part b), we are finding vector fields that correspond to the pushforward of he coordinate vector fields:
    $$U,~V~ text{such that}~ phi_*(U)=partial_x,~phi_*(V)=partial_y$$.


Because the pushfoward has the property that $F_*[X,Y]=[F_*X,F_*Y]$, I know that the lie bracket in both cases should be 0, since $[partial_x, partial_y]=0$



My questions are:




  • Is my summary of the problem correct?


  • I am not sure if the vector fields in both cases are uniquely determined by these conditions. I think they are uniquely determined by these conditions if $phi$ was bijective. But that is not the case. How can I know if those fields are uniquely determined?











share|cite|improve this question
























  • Whoa. $phi$ is far from bijective.
    – Ted Shifrin
    2 days ago










  • That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
    – Amphiaraos
    2 days ago


















1














In the course I am taking on differentiable manifolds, we have been using the notes written by Marcut . This problem caught my attention as it seemed pretty straightforward.




Consider the map:
$$phi: mathbb{R}^2 to mathbb{R}^2,~ phi(x,y):=(e^xcos(y),e^xsin(y))$$




  1. Find two vector fields $X,~Y~in~mathfrak{X}(mathbb{R}^2$ such that $X$ is $phi$-related to $partial_x$ and $Y$ is $phi$-related to $partial_y$. Are these vector fields uniquely determined by this condition?. Calculate $[X,Y]$

  2. Find two vector fields $U,~V~in~mathfrak{X}(mathbb{R}^2$ such that $partial_x$ is $phi$-related to $U$ and $partial_y$ is $phi$-related to $V$. Are these vector fields uniquely determined by this condition?. Calculate $[U,V]$




Solution



I have found and checked the solutions for both parts.



For part a), I used the fact that the required condition is equivalent to $ partial_x(phi^*(f))=phi^*(X( f)) label{eq:related}$. I find $X$. Using the same procedure for $Y$, I get that:
$$X = xpartial_x+ypartial_y,~ Y= -ypartial_x+xpartial_y$$
$$[X,Y]=0$$



For part b), I noticed that $U$ can be written as $U=u_1(x,y)partial_x+u_2(x,y)partial_y$. Using the condition that $U(phi^*(f))_p=(partial_u( f))_{phi(p)}$, I find $U$. Using the same procedure for $V$, I get:
$$U= e^{-x}cos(y)partial_x-e^{-x}sin(y)partial_y,~V= e^{-x}sin(y)partial_x+e^{-x}cos(y)partial_y$$
$$[U,V]=0$$



So in summary:




  • In part a), we are finding the pushforward of the coordinate directions: $$X=phi_*(partial_x),~Y=phi_*(partial_y) $$

  • In part b), we are finding vector fields that correspond to the pushforward of he coordinate vector fields:
    $$U,~V~ text{such that}~ phi_*(U)=partial_x,~phi_*(V)=partial_y$$.


Because the pushfoward has the property that $F_*[X,Y]=[F_*X,F_*Y]$, I know that the lie bracket in both cases should be 0, since $[partial_x, partial_y]=0$



My questions are:




  • Is my summary of the problem correct?


  • I am not sure if the vector fields in both cases are uniquely determined by these conditions. I think they are uniquely determined by these conditions if $phi$ was bijective. But that is not the case. How can I know if those fields are uniquely determined?











share|cite|improve this question
























  • Whoa. $phi$ is far from bijective.
    – Ted Shifrin
    2 days ago










  • That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
    – Amphiaraos
    2 days ago
















1












1








1







In the course I am taking on differentiable manifolds, we have been using the notes written by Marcut . This problem caught my attention as it seemed pretty straightforward.




Consider the map:
$$phi: mathbb{R}^2 to mathbb{R}^2,~ phi(x,y):=(e^xcos(y),e^xsin(y))$$




  1. Find two vector fields $X,~Y~in~mathfrak{X}(mathbb{R}^2$ such that $X$ is $phi$-related to $partial_x$ and $Y$ is $phi$-related to $partial_y$. Are these vector fields uniquely determined by this condition?. Calculate $[X,Y]$

  2. Find two vector fields $U,~V~in~mathfrak{X}(mathbb{R}^2$ such that $partial_x$ is $phi$-related to $U$ and $partial_y$ is $phi$-related to $V$. Are these vector fields uniquely determined by this condition?. Calculate $[U,V]$




Solution



I have found and checked the solutions for both parts.



For part a), I used the fact that the required condition is equivalent to $ partial_x(phi^*(f))=phi^*(X( f)) label{eq:related}$. I find $X$. Using the same procedure for $Y$, I get that:
$$X = xpartial_x+ypartial_y,~ Y= -ypartial_x+xpartial_y$$
$$[X,Y]=0$$



For part b), I noticed that $U$ can be written as $U=u_1(x,y)partial_x+u_2(x,y)partial_y$. Using the condition that $U(phi^*(f))_p=(partial_u( f))_{phi(p)}$, I find $U$. Using the same procedure for $V$, I get:
$$U= e^{-x}cos(y)partial_x-e^{-x}sin(y)partial_y,~V= e^{-x}sin(y)partial_x+e^{-x}cos(y)partial_y$$
$$[U,V]=0$$



So in summary:




  • In part a), we are finding the pushforward of the coordinate directions: $$X=phi_*(partial_x),~Y=phi_*(partial_y) $$

  • In part b), we are finding vector fields that correspond to the pushforward of he coordinate vector fields:
    $$U,~V~ text{such that}~ phi_*(U)=partial_x,~phi_*(V)=partial_y$$.


Because the pushfoward has the property that $F_*[X,Y]=[F_*X,F_*Y]$, I know that the lie bracket in both cases should be 0, since $[partial_x, partial_y]=0$



My questions are:




  • Is my summary of the problem correct?


  • I am not sure if the vector fields in both cases are uniquely determined by these conditions. I think they are uniquely determined by these conditions if $phi$ was bijective. But that is not the case. How can I know if those fields are uniquely determined?











share|cite|improve this question















In the course I am taking on differentiable manifolds, we have been using the notes written by Marcut . This problem caught my attention as it seemed pretty straightforward.




Consider the map:
$$phi: mathbb{R}^2 to mathbb{R}^2,~ phi(x,y):=(e^xcos(y),e^xsin(y))$$




  1. Find two vector fields $X,~Y~in~mathfrak{X}(mathbb{R}^2$ such that $X$ is $phi$-related to $partial_x$ and $Y$ is $phi$-related to $partial_y$. Are these vector fields uniquely determined by this condition?. Calculate $[X,Y]$

  2. Find two vector fields $U,~V~in~mathfrak{X}(mathbb{R}^2$ such that $partial_x$ is $phi$-related to $U$ and $partial_y$ is $phi$-related to $V$. Are these vector fields uniquely determined by this condition?. Calculate $[U,V]$




Solution



I have found and checked the solutions for both parts.



For part a), I used the fact that the required condition is equivalent to $ partial_x(phi^*(f))=phi^*(X( f)) label{eq:related}$. I find $X$. Using the same procedure for $Y$, I get that:
$$X = xpartial_x+ypartial_y,~ Y= -ypartial_x+xpartial_y$$
$$[X,Y]=0$$



For part b), I noticed that $U$ can be written as $U=u_1(x,y)partial_x+u_2(x,y)partial_y$. Using the condition that $U(phi^*(f))_p=(partial_u( f))_{phi(p)}$, I find $U$. Using the same procedure for $V$, I get:
$$U= e^{-x}cos(y)partial_x-e^{-x}sin(y)partial_y,~V= e^{-x}sin(y)partial_x+e^{-x}cos(y)partial_y$$
$$[U,V]=0$$



So in summary:




  • In part a), we are finding the pushforward of the coordinate directions: $$X=phi_*(partial_x),~Y=phi_*(partial_y) $$

  • In part b), we are finding vector fields that correspond to the pushforward of he coordinate vector fields:
    $$U,~V~ text{such that}~ phi_*(U)=partial_x,~phi_*(V)=partial_y$$.


Because the pushfoward has the property that $F_*[X,Y]=[F_*X,F_*Y]$, I know that the lie bracket in both cases should be 0, since $[partial_x, partial_y]=0$



My questions are:




  • Is my summary of the problem correct?


  • I am not sure if the vector fields in both cases are uniquely determined by these conditions. I think they are uniquely determined by these conditions if $phi$ was bijective. But that is not the case. How can I know if those fields are uniquely determined?








differential-geometry vector-fields






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edited 2 days ago







Amphiaraos

















asked 2 days ago









AmphiaraosAmphiaraos

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  • Whoa. $phi$ is far from bijective.
    – Ted Shifrin
    2 days ago










  • That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
    – Amphiaraos
    2 days ago




















  • Whoa. $phi$ is far from bijective.
    – Ted Shifrin
    2 days ago










  • That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
    – Amphiaraos
    2 days ago


















Whoa. $phi$ is far from bijective.
– Ted Shifrin
2 days ago




Whoa. $phi$ is far from bijective.
– Ted Shifrin
2 days ago












That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
– Amphiaraos
2 days ago






That is right (corrected that in the question). The map$phi$ is not bijective, it will cover $mathbb{R}^2$ many times as $sin(y)$ varies. Then, I have no idea about why those fields are uniquely determined.
– Amphiaraos
2 days ago












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I believe that the vector fields you found are correct, so I will only comment on the uniqueness.
Note that the differential of the map $phi$ is
begin{equation}tag{1}label{1}
d_{(x,y)}phi=begin{pmatrix} e^{x}cos y & -e^{x}sin y\ e^{x}sin y & e^{x}cos yend{pmatrix},
end{equation}

which is a linear isomorphism (its determinant is nonzero).




  1. You found $X,Yinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
    begin{cases}
    d_{p}phi[left(partial_{x}right)_{p}]=X_{phi(p)}\
    d_{p}phi[left(partial_{y}right)_{p}]=Y_{phi(p)}.
    end{cases}

    These equations determine $X,Y$ uniquely on the image of $phi$. Since $text{Im}(phi)=mathbb{R}^{2}setminus{(0,0)}$ is dense in $mathbb{R}^{2}$, it follows by continuity that $X$ and $Y$ are uniquely determined.


  2. Now you found $U,Vinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
    begin{cases}
    d_{p}phi(U_{p})=left(partial_{x}right)_{phi(p)}\
    d_{p}phi(V_{p})=left(partial_{y}right)_{phi(p)}.
    end{cases}

    Again such $U,V$ are unique, since the differential eqref{1} is injective.







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    I believe that the vector fields you found are correct, so I will only comment on the uniqueness.
    Note that the differential of the map $phi$ is
    begin{equation}tag{1}label{1}
    d_{(x,y)}phi=begin{pmatrix} e^{x}cos y & -e^{x}sin y\ e^{x}sin y & e^{x}cos yend{pmatrix},
    end{equation}

    which is a linear isomorphism (its determinant is nonzero).




    1. You found $X,Yinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
      begin{cases}
      d_{p}phi[left(partial_{x}right)_{p}]=X_{phi(p)}\
      d_{p}phi[left(partial_{y}right)_{p}]=Y_{phi(p)}.
      end{cases}

      These equations determine $X,Y$ uniquely on the image of $phi$. Since $text{Im}(phi)=mathbb{R}^{2}setminus{(0,0)}$ is dense in $mathbb{R}^{2}$, it follows by continuity that $X$ and $Y$ are uniquely determined.


    2. Now you found $U,Vinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
      begin{cases}
      d_{p}phi(U_{p})=left(partial_{x}right)_{phi(p)}\
      d_{p}phi(V_{p})=left(partial_{y}right)_{phi(p)}.
      end{cases}

      Again such $U,V$ are unique, since the differential eqref{1} is injective.







    share|cite|improve this answer


























      1














      I believe that the vector fields you found are correct, so I will only comment on the uniqueness.
      Note that the differential of the map $phi$ is
      begin{equation}tag{1}label{1}
      d_{(x,y)}phi=begin{pmatrix} e^{x}cos y & -e^{x}sin y\ e^{x}sin y & e^{x}cos yend{pmatrix},
      end{equation}

      which is a linear isomorphism (its determinant is nonzero).




      1. You found $X,Yinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
        begin{cases}
        d_{p}phi[left(partial_{x}right)_{p}]=X_{phi(p)}\
        d_{p}phi[left(partial_{y}right)_{p}]=Y_{phi(p)}.
        end{cases}

        These equations determine $X,Y$ uniquely on the image of $phi$. Since $text{Im}(phi)=mathbb{R}^{2}setminus{(0,0)}$ is dense in $mathbb{R}^{2}$, it follows by continuity that $X$ and $Y$ are uniquely determined.


      2. Now you found $U,Vinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
        begin{cases}
        d_{p}phi(U_{p})=left(partial_{x}right)_{phi(p)}\
        d_{p}phi(V_{p})=left(partial_{y}right)_{phi(p)}.
        end{cases}

        Again such $U,V$ are unique, since the differential eqref{1} is injective.







      share|cite|improve this answer
























        1












        1








        1






        I believe that the vector fields you found are correct, so I will only comment on the uniqueness.
        Note that the differential of the map $phi$ is
        begin{equation}tag{1}label{1}
        d_{(x,y)}phi=begin{pmatrix} e^{x}cos y & -e^{x}sin y\ e^{x}sin y & e^{x}cos yend{pmatrix},
        end{equation}

        which is a linear isomorphism (its determinant is nonzero).




        1. You found $X,Yinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
          begin{cases}
          d_{p}phi[left(partial_{x}right)_{p}]=X_{phi(p)}\
          d_{p}phi[left(partial_{y}right)_{p}]=Y_{phi(p)}.
          end{cases}

          These equations determine $X,Y$ uniquely on the image of $phi$. Since $text{Im}(phi)=mathbb{R}^{2}setminus{(0,0)}$ is dense in $mathbb{R}^{2}$, it follows by continuity that $X$ and $Y$ are uniquely determined.


        2. Now you found $U,Vinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
          begin{cases}
          d_{p}phi(U_{p})=left(partial_{x}right)_{phi(p)}\
          d_{p}phi(V_{p})=left(partial_{y}right)_{phi(p)}.
          end{cases}

          Again such $U,V$ are unique, since the differential eqref{1} is injective.







        share|cite|improve this answer












        I believe that the vector fields you found are correct, so I will only comment on the uniqueness.
        Note that the differential of the map $phi$ is
        begin{equation}tag{1}label{1}
        d_{(x,y)}phi=begin{pmatrix} e^{x}cos y & -e^{x}sin y\ e^{x}sin y & e^{x}cos yend{pmatrix},
        end{equation}

        which is a linear isomorphism (its determinant is nonzero).




        1. You found $X,Yinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
          begin{cases}
          d_{p}phi[left(partial_{x}right)_{p}]=X_{phi(p)}\
          d_{p}phi[left(partial_{y}right)_{p}]=Y_{phi(p)}.
          end{cases}

          These equations determine $X,Y$ uniquely on the image of $phi$. Since $text{Im}(phi)=mathbb{R}^{2}setminus{(0,0)}$ is dense in $mathbb{R}^{2}$, it follows by continuity that $X$ and $Y$ are uniquely determined.


        2. Now you found $U,Vinmathfrak{X}(mathbb{R}^{2})$ such that for all $pinmathbb{R}^{2}$:
          begin{cases}
          d_{p}phi(U_{p})=left(partial_{x}right)_{phi(p)}\
          d_{p}phi(V_{p})=left(partial_{y}right)_{phi(p)}.
          end{cases}

          Again such $U,V$ are unique, since the differential eqref{1} is injective.








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