Krdytkyu

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?

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.