Consider the projective space $\mathbb{P}^3(\mathbb{R})$. I chose $3$ just to deal with a concrete example.
Consider the hyperplane $H_0 = \{x_0 = 0\}$ and its complement $U_0 = \{x_0 \not= 0\}$.
The natural map $$U_0 \stackrel{\phi}{\to} \mathbb{A}^3_{\mathbb{R}}, $$ that maps $[x_0,x_1,x_2,x_3] \mapsto (\frac{x_1}{x_0}, \frac{x_2}{x_0}, \frac{x_3}{x_0}) = (x,y,z)$ yelds a bijection between affine subspaces of $\mathbb{A}^3_{\mathbb{R}}$ and projective subspaces of $\mathbb{P}^3(\mathbb{R})$ that are not contained in $H_0$.
For instance the space $x_1-x_2 = 0$ is mapped to $x-y=0$.
This procedure is absolutely trivial to describe for hyperplanes of the form $$H_i = \{x_i = 0\}, $$ and goest straightforward as we described it.
Q.
- What happens if $ H = \{x_0 -x_1 = 0\}$?
- What is the natural system of coordinates (the map $\phi$) to set on $ U = \{x_0 -x_1 = 0\}$?
- Where is $x_1-x_2 = 0$ mapped?
I found a description but I am quite unsure of my solution, so I will post it but I am waiting for any comment.
My idea is quite dumb, that is why I am unsatisfied.
Clearly one can consider an endomorphism $\psi$ of $\mathbb{P}^3(\mathbb{R})$ such that $\psi(H) = H_0$ and use $\phi \circ \psi$ as a system of coordinates of $U$. In this case I chose
$\psi = \begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 &0 & 1 \end{matrix}$
And in the end one discover that $x_1-x_2$ goes to $x-y$.
Unfortunately, different $\psi$ would completely change this result.
- Is there a natural choice I am missing?