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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.

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2 Answers 2

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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.

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The manifold of all lines in $\mathbb RP^2=G(1,3)$ is $G(2,3)\cong\mathbb RP^{2*}\cong \mathbb RP^2$. Remove the one line at infinity and you have all the lines in $\mathbb R^2$.

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  • $\begingroup$ Sorry but what do you mean by $\mathbb RP^{2\star}$? I have never seen it... Thanks. $\endgroup$
    – Romeo
    Jun 26, 2013 at 14:07
  • $\begingroup$ It's the classical dual projective plane. But just think of lines in $\mathbb RP^2$ as planes through the origin in $\mathbb R^3$, which is diffeo (by taking orthogonal complement) to lines through the origin, i.e., $\mathbb RP^2$. $\endgroup$ Jun 26, 2013 at 14:10
  • $\begingroup$ It helped me to view our $\mathbb{R}^2$ as the plane $z=1$ in $\mathbb{R}^3$. Now, any line in $\mathbb{R}^2$ is contained in a unique plane which also hits the origin. Taking the orthogonal complement of the plane gives a line through the origin, that is, an element of $\mathbb{R}P^2$. Distinct lines in $z=1$ go to distinct points in $\mathbb{R}P^2$, and every point in $\mathbb{R}P^2$ is hit except for the point corresponding to the $z$-axis. This defines a bijection between $\{\text{lines in } \mathbb{R}^2\}$ and $\mathbb{R}P^2\setminus\{z-\text{axis}\}.$ $\endgroup$ Jun 26, 2013 at 14:36

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