# Definition of real projective line

On $\mathbb R^2-(0,0)$ we define the following equivalence relation: Two points $(x,y)$ and $(x_0,y_0)$ in $\mathbb R^2-(0,0)$ are equivalent if there exists $a\in \mathbb R^*$ such that $$x=ax_0,\; y=ay_0$$

Let $D_{x_0y_0}$ be the line passing through the points $(0,0)$ and $(x_0,y_0)$. Then the equivalence class of $(x_0,y_0)\in \mathbb R^2-(0,0)$ is the set $D_{x_0y_0}-(0,0)$.

My question: I read that $\mathbb RP^1$ is the set of equivalence classes of points $(x,y)\in \mathbb R^2-(0,0)$ and this we just showed is the set of subsets $D_{xy}-(0,0)$. My question is why we say that $\mathbb RP^1$ is the set of lines $D_{xy}$ through the origin instead of saying the set of subsets $D_{xy}-(0,0)$ ?

There are many definition of $\mathbb{R}P^1$.

The definition you gave above is not the one where $\mathbb{R}P^1$ is the set of lines through the origin. The definition you provide above defined $\mathbb{R}P^1$ as the set of equivalence classes of $\mathbb{R}^2 - \{(0,0)\}$ under the equivalence relation you wrote.

However, your definition and the line-through-origin definition are the same in a very geometric sense. Let $\mathbb{R}P^1$ denote your eqivalence relation definition. Let $L$ denote the line through the origin definition. Let $\Phi : \mathbb{R}P^1 \rightarrow L$ be defined by $[x] \mapsto \mathbb{R} x$, where $\mathbb{R}[x] = \{\alpha x : \alpha \in \mathbb{R}\}$. It is easy to check that $\Phi$ is well-defined and a bijection.

So the equivalence class definition and the lines through the origin definition are equivalent by a bijection. (In fact if you put some topological structure on the two spaces you even get bijection preserving structure.)

However you are correct, the definition you provided, equivalence classes are not lines. But up to various structure preserving bijection, the two definitions are the same.

• Dear William, i'm not sure i'm following : in the first paragraph you say they are not the same and in the second you say that they are the same in a very geometric sense!!! I'm confused – palio Sep 21 '13 at 12:29

Your definition actually gives $\mathbb{R}P^1 \cup \{ (0,0) \}$, where $(0,0) \notin \mathbb{R}P^1$ but $(0,0)$ cannot be separated from the points of $\mathbb{R}P^1$. Every open set containing $(0,0)$ contains points in common with every line $D_{x_0y_0}$. This "bad behavior" of the point $(0,0)$ is not desirable (it makes the space non-Hausdorff) and so we typically throw out the origin when defining projective spaces in the way you mentioned (equivalence classes of points under scalar multiplication).

To answer your question about "lines". Well we don't include $(0,0)$ as one of the lines because it isn't a line! Think of each line as a mathematical object (not as a set of points). Then it doesn't so much matter that $(0,0)$ is contained in each line. What matters is that each line is basically regarded as a point in $\mathbb{R}P^1$.

Hope this helps!

• I edited my question so that my problem would be clearer. Thank you!! – palio Sep 21 '13 at 13:16