Projective geometry general question

Question

Can someone please help me solve this problem? I am a bit confused.

Embed $R^2$ in the projective plane $RP^2$ by the map $(x, y) → [1, x, y]$. Find the point of intersection in $RP^2$ of the projective lines corresponding to the parallel lines $y = mx$ and $y = mx + c$ in $R^2$ (where $c \neq 0$).

My attempt: So this transformation sends each point in $R^2$ to the line through the origin which is generated by the vector $(1,x,y)$. Then the line $y=mx$ will go to the set $[1,x,mx], x \in \Re, x \neq 0 $.

By the same reasoning, $y=mx+c$ will go to the set {$[1,x,mx+c], x \in \Re, x \neq 0 $}

The problem is I can't quite understand how to define the points at infinity.I don't want to just do the question, I'd like to understand the basics and will be very grateful if someone could help me.

Answer 1

It's true that points of the form $[1,x,mx]$ for $x\in\mathbb{R}$ (there is no reason to discard $x=0$) are on the embedded copy of the line $y=mx$. However, you have not addressed the "projective lines corresponding" part of the problem. The projective line will contain the embedded copy of the regular ("affine") line, but also contain more.

You probably have in your notes or textbook some explanation about how to find projective lines/curves, but I'll try to motivate the idea. If you have an equation like $y=3$, it makes sense in the affine case, but is a bad definition in projective space since $[z,x,y]=[1,1,3]$ represents the same point as $[2,2,6]$. You need to correct the equation so that it remains unchanged when the variables are replaced by multiples, and the way to do that is with your extra variable, which I've called $z$. The projective version is $y=3z$ since now every term has the same total degree.

Do $y=mx$ and $y=mx+c$ make sense in projective space? Which triples $[z,x,y]$ satisfy the projectivized versions of the equations $y=mx$ and $y=mx+c$? Some of them are points you had before, but some of them should be new.

Answer 2

You must homogenize the equations first. $$y=mx$$ is already homogeneous.

But $$y=mx+c$$ is not. Thus replace $x,y$ by $\frac{x}{z}, \frac{y}{z}$, $$\frac{y}{z}=m\frac{x}{z}+c$$ and multiply by $z$,

$$y=mx+cz.$$

Now solve the simultaneous equations $$y=mx$$ $$y=mx+cz$$ to get $z=0$ thus the intersection is

$$[0,x,mx]=[0,1,m].$$

Answer 3

A useful model to keep in mind for this embedding is that of the plane $z=1$ in $\mathbb R^3$. As you wrote, the transformation (well, really the map $(x,y)\mapsto(x,y,1)$ now) sends a point in $\mathbb R^2$ to a line through the origin, but that line isn’t contained in the $z=1$ plane. With a bit of thought, you should be able to convince yourself that lines in $\mathbb R^2$ get mapped to planes through the origin in this model, and that such a plane can be uniquely identified with the equivalence class of its normals. Two distinct such planes always intersect in a line, which you can map back to $\mathbb R^2$ by finding its intersection with $z=1$ (i.e., by dividing by $z$).

If we take a pair of planes that correspond to parallel lines in $\mathbb R^2$, they, too, intersect in a line, but that line is parallel to the plane $z=1$, that is, it consists of points of the form $(x,y,0)$, so points at infinity will have the form $[0,x,y]$. If you play around with this a bit, you’ll find that there’s a point at infinity for each possible direction that a line can have, and that the line at infinity corresponds to the plane $z=0$ in this model.

Update: I’ll add one other observation that might be helpful to you in solving this problem. Just as the equation of the plane in $\mathbb R^3$ can be written in the form $\mathbf n\cdot\mathbf x=0$, so can the projective equation of a line in $\mathbb{RP}^2$. That, together with the fact that points at infinity have the form $[0,x,y]$ should let you find the point at infinity for any given line.