2
$\begingroup$

I'm studying projective geometry and I have a question.

It is known that if we consider in $\mathbb{R^3}$/$\lbrace \overrightarrow{0} \rbrace$ a relation defined by : $a \backsim b \Leftrightarrow \exists t \in \mathbb{R}$\ $\lbrace 0 \rbrace$, $a=tb$ is an equivalence relation.

Then the projective plane is defined as the quotient set :$\frac{\mathbb{R^3}/\lbrace \overrightarrow{0} \rbrace}{\backsim}$ and is denoted $\mathbb{P^2}$. Each equivalence class is : $\left[(x,y,z)\right] = \left[x:y:z\right] = \lbrace (tx,ty,tz), t \in \mathbb{R}\rbrace$.

In one book I found the following:

$\mathbb{P^2} = \lbrace \left[x:y:z\right] : z \neq 0 \rbrace \bigcup \lbrace \left[x:y:0\right] : x,y \in \mathbb{R}$ $\wedge$ $x\neq y\neq 0 \rbrace$

It is easy to prove that : $\lbrace \left[x:y:z\right] : z \neq 0 \rbrace = \lbrace \left[x:y:1\right] : x,y \in \mathbb{R} \rbrace$.

So $\mathbb{P^2} = \lbrace \left[x:y:1\right] : x,y \in \mathbb{R} \rbrace \bigcup \lbrace \left[x:y:0\right] : x,y \in \mathbb{R}$ $\wedge$ $x\neq y\neq 0 \rbrace$. We define $\mathbb{A^2} = \lbrace \left[x:y:1\right] : x,y \in \mathbb{R} \rbrace$ as the set of own points and $\mathbb{P_{\infty}^2} = \lbrace \left[x:y:0\right] : x,y \in \mathbb{R}$ $\wedge$ $x\neq y\neq 0 \rbrace$ as the set of improper points.

Now consider the function $\varphi : \mathbb{R^2} \rightarrow \mathbb{A^2}$ with $\varphi_{(x,y)} = \left[x:y:1\right]$ which is bijective. With all this is defined the homogeneous coordinates of $(x,y) \in \mathbb{R^2}$ to someone $(a,b,c) \in \left[x:y:1\right]$.

Thus, for example, the homogeneous coordinate of $(1,2)$ is $(1,2,1)$ or $(2,4,2)$ or $(-1;-2,-1)...$.

It is observed that the homogeneous coordinates of any $(x, y)$ are infinite and for this reason we can consider the homogeneous coordinate of $(x, y)$ as the equivalence class $\left[x:y:1\right]$ or $\left[tx:ty:t\right], t \in \mathbb{R}$.

My question is that if this "consideration" is correct since in many books the homogeneous coordinates are represented by $(x, y, z)$ and not by an equivalence class that become points in the projective plane. This doubt does not allow me to understand some things about projective lines so it is very important for me, thank you very much for reading all this text and sorry for my english.

1 Answers 1

1

This is the correct consideration. If you write the representation $(x,y,z)$, then you also mean the equivalence class. Another interpretation can be that the projective points are the one-dimensional subspaces of $\mathbf{R}^3$, and $(x,y,z)$ is a base vector of this subspace. This interpretation is essentially the same, since the possible base vectors are the members of the mentioned equivalence class.