Finding a $F$-related field in $\mathbb{RP}^2$

Question

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.

Answer 1

Here's an attempt: following Lee's notation, $F = \varphi_{3}^{-1}$, where $\varphi_3$ is the standard chart defined on the open set $U_3 = \{ [x,y,z] \, : \, z \neq 0 \}$. Thus, it automatically defines a $F$-related vector field $Y$ in the open set $U_3$ by pushing-forward.

Now, if a point $p$ lies in $U_i \cap U_3$, for $i = 1,2$, we know that the change of basis matrix from the basis of $T_p \mathbb{RP}^2$ induced by $\varphi_3$ to the basis induced by $\varphi_i$ is given by the jacobian of the transition function, $$J (\varphi_i \circ \varphi_3^{-1}) \left( \varphi_3 (p) \right) $$

This would allow us to find the expression of $Y$ in the other basis of $T_p \mathbb{RP}^2$ and thus its expression in coordinates with respect to the other charts. This expression could be used, in turn, to define $Y$ in $U_i$ in a differentiable manner and thus we would obtain a globally defined vector field $Y \in \mathfrak{X} (\mathbb{RP}^2)$.

For concreteness, say that $$p = \varphi_3^{-1} (x,y) = [x,y,1] = [u, 1, v] = \varphi_2^{-1} (u,v) \text{ .}$$ Following Lee's abuse of notation we'll also denote the respective basis of $T_p \mathbb{RP}^2$ by $\{ \partial / \partial x, \partial / \partial y \}$ and $\{ \partial / \partial u, \partial / \partial v \}$. In this case, the transition function is given by $$\varphi_2 \circ \varphi_3^{-1} (x,y) = \left( \frac{x}{y}, \frac{1}{y} \right)$$ and so we have $$Y_p = \left( - 1 - \frac{x^2}{y^2} \right) \frac{\partial }{\partial u} \bigg|_p - \frac{x}{y^2} \frac{\partial }{\partial v} \bigg|_p = -\left( 1 + u^2 \right) \frac{\partial }{\partial u} \bigg|_p - uv \frac{\partial }{\partial v} \bigg|_p \text{ ,}$$ so that its coordinate representation with respect to the chart $\varphi_2$ would be $$\hat{Y} (u,v) = \tilde{\varphi_2} \circ Y \circ \varphi_2^{-1} (u,v) = \left( u, v, - (1 + u^2), - uv \right) \text{ ,}$$ and its definition in $U_2$ would be given accordingly (of course, here we'd be using that differentiable functions defined on open sets that agree on their intersections define a globally differentiable function to ensure that Y is indeed in $\mathfrak{X} (\mathbb{RP}^2)$).

Answer 2

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

$$ C = \begin{pmatrix} a& b& e \\ c & d & f \\ 0 & 0 &1 \end{pmatrix}.$$