TITLE: Finding a $F$-related field in $\mathbb{RP}^2$
QUESTION [5 upvotes]: I am trying to prove the following exercise from the book Introduction to Smooth Manifolds - J.M. Lee.

Let $F: \mathbb R^2 \rightarrow  \mathbb{RP}^2$ be the smooth map
  $F(x,y)= [x,y,1]$, and let $X\in \mathfrak X(\mathbb R^2)$ defined by
  $X= x\frac{\partial}{\partial y} -y\frac{\partial}{\partial x}$. Prove
  that there is a vector field $Y\in\mathfrak X (\mathbb{RP}^2)$ that is
  $F$-related to $X$, and compute its coordinate representation in
  terms of each of the standard charts of $\mathbb{RP}^2$.

I noticed that $F$ is the inverse of a chart and calculated $dF(X)$ and I don't know what to do next.

REPLY [2 votes]: One way is to integrate the vector field and note that the action on $\mathbb R^2$ extends to $\mathbb {RP}^2$. Let $\phi_t : \mathbb R^2 \to \mathbb R^2$ be given by 
$$ \phi_t \begin{pmatrix} x \\y \end{pmatrix} = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}.$$
Then you can check that $\frac{d}{dt} \phi_t |_{t=0}$ is your vector field. Now we define $\Phi_t : \mathbb{RP}^2 \to \mathbb{RP}^2$ by $\Phi_t [\vec x] = [C_t \vec x]$, here $C_t$ is the matrix 
$$C_t =\begin{pmatrix} \cos t & -\sin t & 0 \\ \sin t & \cos t & 0 \\ 0 & 0 & 1  \end{pmatrix}.$$
One can check that $ F \circ \phi_t = \Phi_t \circ F$, so $\Phi_t$ is an extension of $\phi_t$. In particular, 
$$Y = \frac{d}{dt} \Phi_t\bigg|_{t=0}$$
is a vector field on $\mathbb {RP}^2$ which extends $dF(X)$. The coordinates expression is more or less covered in another answer, so I will just leave that to you. 
Remark In general, any affine transformation 
$$A\vec x = \begin{pmatrix} a & b \\ c& d\end{pmatrix}\vec x + \begin{pmatrix} e \\ f\end{pmatrix}$$
can be extended to a projective transformation $\widetilde A : \mathbb{RP}^2 \to \mathbb{RP}^2$ by $\widetilde A [\vec x] = [C\vec x]$, where $C$ is the matrix 
$$ C = \begin{pmatrix} a& b& e \\ c & d & f \\ 0 & 0 &1 \end{pmatrix}.$$