Does anyone know how to prove this is a manifold?

Question

I'm doing some work on Differentiable Manfold theory, and I came across a problem that I'm having a little trouble on so I thought "Why not post it?" Anyways, the question asks to show that the set of all lines in R^2 (note: NOT just lines through the origin, but ALL lines) is a differentiable manifold, and then asks what (already familiar) manifold it is. The latter is (RP^2)-{a point}, and I can show that identification without too much trouble. But I was wondering if anyone knows how to *carefully* show that the set of all lines in R^2 is a manifold (coordinate patches would be awesome). Thanks!!

I should mention this isn't calculus. I mean I guess it involves calculus, but this isn't a trivial problem. I actually has appeared on two previous versions of UCLA's Geometry/Topology qualifying exam and if no one has any suggestions then should I ever see this problem I'd just use my "hand-wavey" argument that this is topologically equivalent to RP^2 with a point deleted, and then show how this is a manifold, making the other one a manifold. But that's not quite what they're looking for, which is why I asked this.

Yeah I know the manifold is actually RP^2 but with a point deleted (which is like the mobius strip even). So this is responding to mathematician's response below. I'd define the equivalence relation on the lines to be distance from the origin.

My concern is how to show carefully that the set of lines in R^2 is a manifold without resorting to that equivalence right away. If the lines in R^2 are a manifold, then you can find coordinate patches right? That's what I'm hoping to see.

Ah that makes sense. The only other thing I can think of for an equivalence relation on the lines would be if they have the same homogeneous coordinates (call them (A,B,C) using your notiation) so I'm thinking the equivalence relation is on the coordinates that all define the same line.

Anyways, the way I show it's RP^2 (minus a point) is the following: all lines in R^2 can be identified in the obvious way to all lines in the plane z=1 (ax+by+cz=0 with z=1 gives us the line). But this is equivalent to the set of all planes in R^3 through the origin except the plane z=0, which is in turn equivalent to the set of all lines in R^3 through the origin except for the z-axis, which is RP^2 with one point deleted. Is that more or less correct?

Answer 1

Well, a line is an equivalence class of equations of the form Ax+By+C=0. What is the equivalence relation? If you think of this as R^3 under an equivalence relation, which manifold do you get?

Hint: it is *very* similar, but not identical to RP^2.


Update: The main difficulty is how to represent lines. In the expression Ax+By+C=0, at least one of A, B, or C must be non-zero. It is also impossible for A=B=0. This suggests two patches:

{x+By+C=0:B,C are real} with coordinates (B,C)
and
{Ax+y+C=0:A,C are real} with coordinates (A,C).

If A is not zero, the line Ax+y+C=0 is the same as x+(1/A) y+ (C/A)=0, so, on the overlap, one transformation is given by
(A,C)->(1/A, C/A) which is clearly smooth.

I really don't think there is a good way to get around that equivalence relation. And that relation is *not* distance from the origin!!!!!!

Added: Yes! You got it!

Answer 2

OH NO CALCULUS ,EVERYONE HEAD FOR THE HILLS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!