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:
$$\phi: \mathbb{R}^2 \to \mathbb{R}^2,~ \phi(x,y):=(e^xcos(y),e^xsin(y))$$
- 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 $[X,Y]$
- 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:
$$X = x\partial_x+y\partial_y,~ Y= -y\partial_x+x\partial_y$$
$$[X,Y]=0$$
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?
Best 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 y\end{pmatrix}, \end{equation} which is a linear isomorphism (its determinant is nonzero).
You found $X,Y\in\mathfrak{X}(\mathbb{R}^{2})$ such that for all $p\in\mathbb{R}^{2}$: \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.
Now you found $U,V\in\mathfrak{X}(\mathbb{R}^{2})$ such that for all $p\in\mathbb{R}^{2}$: \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.