Image of a line under a transformation from $\Bbb R^2$ to $\Bbb P^2$

Question

Let $\phi: \Bbb R^2 \to \Bbb P^2$ be such that

$\phi(x,y) = [1 : x : y]$

where $\Bbb P^2$ has coordinates $[x_0 : x_1 : x_2]$.

I want to show that the image of a line $l$ in $\Bbb R^2$ with equation $ax + by + c = 0$ is the projective line $\phi(l)$ such that $cx_0 + ax_1 + bx_ 2 = 0$. To a certain extent I see that we basically substituted, in the equation for $l$, the right coordinates of the projective plane, but I can't seem to grasp why this makes sense.

(Also, I may be confusing a little too much some notions... $\Bbb P^n$ has dimension $n$, even though we write its points with $n+1$ coordinates, right?)

Answer 1

A general method for determining the image under a transformation is the following: $X\in l'$ if and only if $X^{-1}\in l$.

The inverse image of $[x_0:x_1:x_2]$ is $(\frac{x_1}{x_0},\frac{x_2}{x_0})$.

Thus $[x_0:x_1:x_2]$ is on the image of $ax+by+c=0$ if and only if $a\frac{x_1}{x_0}+b\frac{x_2}{x_0}+c=0$, i.e., $ax_1+bx_2+cx_0=0$.

Answer 2

There is a model of $\mathbb P^2$ that I find helpful for understanding this. If we consider the set of equivalent homogeneous coordinates of a point in $\mathbb P^2$ as points in $\mathbb R^3$, that is, if we map $(x_0:x_1:x_2)\mapsto(x_1,x_2,x_0)\in\mathbb R^3$, we get a line through the origin (less the origin itself). The normalized homogeneous coordinates $(x,y,1)$ are then simply points on the plane $z=1$—the projections onto this plane of all of the points on a given line, in fact. So, we can identify points in $\mathbb P^2$ with lines through the origin in $\mathbb R^3$. (Points at infinity correspond to lines parallel to the $z=1$ plane, but that’s not important for this discussion.)

Two points in $\mathbb P^2$ determine a unique line that passes through them. What is the equivalent of this in our model? Our projective points correspond to lines through the origin in $\mathbb R^3$, and two distinct such lines determine a unique plane through the origin. Sure enough, the intersection of this plane with $z=1$ is a line that passes through the projections of the two initial lines. If we take the line $ax+by+c=0$, $z=1$ in $\mathbb R^3$, we can easily find that an equation of the plane defined by this line and the origin is $ax+by+cz=0$, but this is exactly the counterpart of the homogeneous equation $cx_0+ax_1+bx_2=0$ in our model.

The coefficients of the equation have a nice geometrical interpretation in this model, by the way. If we rewrite the plane equation as $(a,b,c)\cdot(x,y,z)=0$, we see that the vector $(a,b,c)$ is normal to the plane. Also, scaling this vector by a non-zero scalar produces the same plane, that is, the normal vectors behave as one might expect in homogeneous coordinates. Take care, though. These normal vectors really live in the dual space, so they transform differently than points do (they are covariant vectors).