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I have a problem about the projective plane. They said that the only thing that I must know is the definition. In the problem, the "line" of a homogeneous polynomial is defined. It's the set $$ L_{a,b,c} = \{ {\left[ {x,y,z} \right] \in P_\mathbf{R}^2 :ax + by + cz = 0} \}. $$ This exercise has two parts. I want to understand it )=. First I must show that $$ P_\mathbf{R}^2 - L \simeq \mathbf{R}^2 $$ for every line $L$. I've tried to define an embedding of $\mathbf{R}^2$ first, and I think that the natural way to proceed is try to define a parametrization of the sphere, and then compose this function with the quotient map $q\colon \mathbf{R}^3 - \{0\} \to P_\mathbf{R}^2$. And this function I suppose will be useful, but I don't see how to proceed )= sorry for be so stupid <.< , but it´s wear to me visualize this things like the projective plane, i need more practice

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I manage to write this always wrong in Spanish, in English and in French: projective is with a J. :) –  Mariano Suárez-Alvarez Oct 1 '11 at 2:36
I would try to first give a bijection between $\mathbf{R}^2$ and $\mathbf{P}^2 \setminus L_{1, 0, 0}$; the latter space consists of those points which can be written in the form $[1, y, z]$ for some $y, z \in \mathbf{R}$. If you can figure this out (writing down a map, showing that it's well-defined, a homeomorphism, and so on) then the only thing left to do, more or less, is change coordinates. –  Dylan Moreland Oct 1 '11 at 2:36
Dear Daniel: I tried to fix up the spelling, TeX and such without touching the more personal sentences. I hope I haven't distorted your meaning. –  Dylan Moreland Oct 1 '11 at 2:46

2 Answers 2

If you see $P^2$ as a quotient space of $S^2$ by identifying antipodal points, then a line as defined above makes sense in that sphere too, only with $|(x, y, z)|=1$ instead of them being homogeneous. You will then have that the curve divides the sphere into two parts.

As you transition into $P^2$, these two parts will go together to be that one subspace $P^2 - L$

To show that what is left is homeomorphic to $R^2$ is easy, really. You just have to show that the transition of $S^2$ to $P^2$ is homeomorphic on a single open hemisphere, and that an open hemisphere is homeomorphic to $R^2$.

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There are many projective planes but presumably you are talking about the real projective plane, where the lines are represented by ax + by + cz = 0 where a, b, c are real numbers and not all of a,b, and c are 0. If one removes a single line from this plane you can think of this removed line from the real projective plane as the "line at infinity" which is added to the Euclidean plane to get the real projective plane. Each point on this "line at infinity" represents an additional point added to each line of a parallel class (all the lines with slope m or undefined slope) in the Euclidean plane. All the new points make up a new line added to the lines in the Euclidean plane - the line at infinity.

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