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Consider the real projective space $\mathbb{R}\mathbb{P}^{2}$ and suppose we remove a line from it. Why are we left with the Euclidean plane?

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I'm guessing based on the (differential-geometry) tag you're talking about $\mathbb{RP}^2$ with the analytic topology? Perhaps you could clarify this. – John Stalfos Apr 13 '12 at 1:41
You need to phrase your question well. What are your point set and block set in a projective plane? What are your definitions of an affine plane and projective plane? What exactly do you do? Not just removing a block...but more. This is a classical exercise in combinatorics doing which gives you a feel for the ojects involved. Please don't kill the purpose of this exercise by asking without trying. – user21436 Apr 13 '12 at 1:42
@John Stalfos: I'm viewing it the set of all lines through the origin. I guess I should change the tag to projective geometry. – user6495 Apr 13 '12 at 1:42
@Kannappan Sampath: this is more of a lemma I saw, check here: page $16$, lemma $3.3$…, this result also appears here page $12$, however I don't see why it is true. – user6495 Apr 13 '12 at 1:43
Is it a finite projective plane you're interested in or these objects over real field? Not the procedure is different, but counting exercises involved in finite case is routine but fun for a beginner. – user21436 Apr 13 '12 at 1:49
up vote 3 down vote accepted

Your "lines through the origin" can be thought of as triples $(a,b,c)$ with $a$, $b$, $c$ not all zero, and with the equivalence relation $(a,b,c)\sim k(a,b,c)$ for $k\ne 0$. Lines in the projective plane correspond to planes through the origin. Try removing the line corresponding to the $x$-$y$ plane. Can you use the equivalence relation to map the remaining triples to points in the Euclidean plane?

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No matter what plane you remove, the remaining triples will all have a non-zero component normal to the removed plane. So the same idea works: choose equivalence-class representatives so that this component has unit length. – Will Orrick Apr 13 '12 at 2:52
So for example say the plane is $2x+y+z=0$ and suppose $[a : b : c]$ is a point and wlog say $a \neq 0$. Then define a map $f([a: b : c]) = (\frac{b}{2a+b+c} , \frac{c}{2a+b+c})$. Is this your idea? – user6495 Apr 13 '12 at 3:30
Let $\hat n$ be the normal vector $(2,1,1)/\sqrt{6}$. If $(a,b,c)$ does not satisfy $2a+b+c=0$ then $(a,b,c)$ is a linear combination of the form $v+t\hat n$ where $v$ is a vector in the plane and $t$ is a non-zero scalar. (Non-zero because $(a,b,c)$ is not in the plane.) Then $f([a:b:c])=v/t$ is a suitable map. – Will Orrick Apr 13 '12 at 4:52

We can visualize $\Bbb P^2$ minus a line (plane in $\Bbb R^3$) to be the top half of a sphere (the bottom half is just an identical copy of it). Put a plane on top of and tangent to our sphere, parallel to the plane we cut out of it. To map our space to this hovering plane we simply project rays out from the origin to the points on the open upper hemisphere, and wherever it these points end up hitting on the hovering plane is where we say they are sent.

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