Consider the set of all straight lines on the euclidean plane $\mathbb R^2$. Introduce a structure of smooth manifold on this set and show that this is homeomorphic to $\mathbb R\mathbb P^2 \setminus \{p\}$, i.e. to real projective plane without a point. Are they also diffeomorphic as smooth manifolds?
I'm sincerely stuck on this question. Well, if we restrict to lines through the origin we would have $\mathbb R\mathbb P^1$, the real projective line. If we consider $\mathbb R \times \mathbb R\mathbb P^1$ we would obtain all the lines that intersect the $y$-axis. But in that way I can't get the lines which are parallel to $y$-axis. So, what should I do? Moreover, how can I show the homeomorphism with $\mathbb R\mathbb P^2$ minus a point?
Thanks in advance.
Best Answer
The easiest way to define such a structure is perhaps the following (it is linked to JasonDeVito's comment).
Any line in the euclidean plane is of the form $ax+by+c=0$, for some $a,b,c \in \mathbb R$. First note that this coefficients uniquely determine the line and they are homogeneous. Hence there is a well defined map $$ \begin{split} \phi : & \mathscr{R} \longrightarrow \mathbb RP^2 \\ & ax+by+c=0 \mapsto[a:b:c] \end{split} $$
where $\mathscr R$ is the set of all lines in the plane. To obtain a bijection, we just have to observe that we cannot have $a=b=0$ and $c \ne 0$: hence we have to pull off the point $[0:0:1]$ and this concludes the proof.