# Ways of defining topology on $P^{n}(\mathbb{R})$

I was wondering if the following two ways of defining topology on $P^n(\mathbb{R})$ are the same and why?

1. Since $P^n(\mathbb{R})$ is the quotient space of $\mathbb{R}^{n+1}$, define the topology on $P^n(\mathbb{R})$ to be the quotient topology ( i.e. the maximal topology that can make the quotient map $q: \mathbb{R}^{n+1} - \{0\} \to P^n(\mathbb{R})$ continuous, if I understand correctly or do I?).
2. From Wikipedia:

Consider the following subsets of $P^n(\mathbb{R})$: $$U_i = \{[x_0:\cdots: x_n], x_i \neq 0\}, i=0, \dots,n.$$ Their union is the whole projective space. Furthermore, $U_i$ is in bijection with $\mathbb{R}^n$ via the following maps: $$[x_0:\cdots: x_n] \mapsto \left (\frac{x_0}{x_i}, \dots, \widehat{\frac{x_i}{x_i}}, \dots, \frac{x_n}{x_i} \right )$$ $$[y_0:\cdots: y_{i-1}: 1: y_{i+1}: \cdots y_n] \leftarrow \left (y_0, \dots, \widehat{y_i}, \dots y_n \right )$$ (the hat means that the $i$-th entry is missing).

Then define a topology on projective space by declaring that these maps shall be homeomorphisms, that is, a subset of $U_i$ is open iff its image under the above isomorphism is an open subset (in the usual sense) of $\mathbb{R}^n$. An arbitrary subset $A$ of $P^n(\mathbb{R})$ is open if all intersections $A ∩ U_i$ are open. This defines a topological space.

Thanks and regards!

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Identify $P^n(\mathbb{R})$ with a quotient space of $S^n$. This space will be compact. Then look for continuous bijections to the two models you gave and show that these two models are Hausdorff. Apply the fact that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. I strongly suggest that you do this yourself in detail... –  t.b. Dec 25 '11 at 2:28
@t.b.: Thanks! I am really not sure, so may I ask what the two models/spaces that I gave are? –  Tim Dec 25 '11 at 2:43
You gave two (a priori) different topologies on $P^n(\mathbb{R})$, I described a third one. The two different topologies give you two a priori different spaces but they turn out to be the same. I outlined one way how to prove that this is the case. –  t.b. Dec 25 '11 at 2:47
The tradition map from Euclidean space to projective space doesn't include $0$ in the domain. –  Thomas Andrews Dec 25 '11 at 4:12
@ThomasAndrews: Thanks for pointing it out! –  Tim Jan 1 '12 at 18:12

$\mathbb{R}P^n$ is usually described as
i) lines through the origin in $\mathbb{R}^{n+1}$ (topology defined by the natural "closeness" of lines, say neighborhoods of a line are all lines within angle $\theta$ from the line
ii) quotient of $S^n$ by the antipodal map, like i) with each line represented as two points which we identify
iii) equivalence classes of points in $\mathbb{R}^{n+1}\backslash \{0\}$, identify two points $x,y$ if $x=\lambda y, \lambda\in\mathbb{R}^{\times}$ which looks a lot like i) since the equivalence classes are lines through the origin (minus the origin)
the definition via the affine cover you give above is iii) broken into the pieces (using homogeneous coordinates) $U_i=\{[x_0:\cdots:x_{i-1}:1:x_{i+1}:\cdots:x_n]\ |\ x_0,...,\hat{x_i},...,x_n\in\mathbb{R}\}$ (just a note that $[x_0:\cdots:x_n]$ denotes the equivalence class of the line through $(x_0,...,x_n)$, called "homogeneous coordinates" for a point in projective space)