§ Decomposition of projective space

Projective space Pn+1=PnRn\mathbb P^{n+1} = \mathbb P^n \cup \mathbb R^n. The current way I think about this is as follows (specialize to n=3n=3)
  • Consider a generic point [x:y:z][x : y : z]. Either x=0x = 0 or x0x \neq 0.
  • If x=0x = 0, then we have [0:y:z][0 : y : z] which can be rescaled freely:[0:y:z]=(0,y,z,1)=(0,λy,λz,λ)=[0:y:z][0: y: z] = (0, y, z, 1) = (0, \lambda y, \lambda z, \lambda) = [0: y: z].So, we get a component of P2\mathbb P^2 from the [y:z][y: z].
  • If x1x \neq 1, we have [x:y:z][x : y : z]. Spend the projectivity to get (1:y:z)=(x,y,z,x)(1:y:z) = (x, y, z, x).Now we have two free parameters, (y,z)R2(y, z) \in \mathbb{R^2}. This gives us the R2\mathbb R^2.
There's something awkward about this whole thing, notationally speaking. Is there a more natural way to show that we have spent the projectivity to renormalize [x:y:z][x: y: z] to (1,y,z)(1, y, z) ?

§ Projective plane in terms of incidence

We can define P2\mathbb P^2 to be an object such that:
  1. Any two lines are incident at a single point.
  2. Two distinct points must be incident to a single line. (dual of (1))

§ The points at infinity

This will give us a copy of R2\mathbb R^2, along with "extra points" for parallel lines.
  • Consider two parallel lines y=mx+0y = mx + 0 and y=mx+1y = mx + 1. These don't traditionallymeet, so let's create a point at infinty for them, called Pm(0,1)P_m(0, 1).
  • Now consider two more parallel lines, y=mx+0y = mx + 0 and y=mx+2y = mx + 2. These don'ttraditionally meet either, so let's create a point at infinite for them, calledPm(0,2)P_m(0, 2).
  • Finally, create another point Pm(0,3)P_m(0, 3) as the point of intersection betweeny=mx+0y = mx + 0 and y=mx+3y = mx + 3.
  • Now, consider Pm(0,1),Pm(0,2),Pm(0,3),dotsP_m(0, 1), P_m(0, 2), P_m(0, 3), dots. We claim that they must all be equivalent.Assume not. Say that Pm(0,1)Pm(0,2)P_m(0, 1) \neq P_m(0, 2).
  • Then there must a line that joins Pm(0,1)P_m(0, 1) an Pm(0,2)P_m(0, 2). Call it Lm(0,1,2)L_m(0, 1, 2).Now, what is the intersection between Lm(0,1,2)L_m(0, 1, 2) and the line y=mx+0y = mx + 0?The points Pm(0,1)P_m(0, 1) and Pm(0,2)P_m(0, 2) both lie on the line Lm(0,1,2)L_m(0, 1, 2).But this is a contradiction: two lines must be incident at a single unique point.
  • So we must have Pm(0,1)=Pm(0,2)=PmP_m(0, 1) = P_m(0, 2) = P_m. So, for each direction, we musthave a unique point where all lines in that direction meet.
We can make a definition: the point at infinity for a given direction is the equivalence class of all lines in that direction.

§ The line at infinity

This now begs the question: what lines to different points at infinity lie on? Let's consider Pq,Pr,Ps,PtP_q, P_r, P_s, P_t as four points at infinity for four different slopes.
  • Consider the lines L(q,r)L(q, r) that is incident on PqP_q and PrP_r, and thenthe line L(s,t)L(s, t) that is incident on the lines PsP_s and PtP_t.
  • This begs the question: where do these lines meet? If we say that the meet at more newpoints of intersection, like P(q,r,s,t)P(q, r, s, t) this process will never end.
  • So we demand that all points at infinity lie on a unique line at infinity.