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As I understand it, real projective space $\mathbb R \mathbb P^n$ is defined to be the quotient $S^n / \sim$, where $x \sim y$ iff $x = \pm y$. In other words, elements of $\mathbb R \mathbb P^n$ are pairs of antipodal points on $S^n$. We have the associated quotient map $q : S^n \to \mathbb R \mathbb P^n$, which maps points to their equivalence class.

I have heard that $q$ is a covering projection of $\mathbb R \mathbb P^n$, and I'm trying to see why. I don't know how to 'visualise' real projective space, which I suppose is my problem. This is highlighted by the fact that I fail to see why $\mathbb R \mathbb P^1 \cong S^1$. I guess my real questions are:

  1. Why is $\mathbb R \mathbb P^1 \cong S^1 $?

  2. Why is $q$ a covering projection? If we're given an open set $U$ in $\mathbb R \mathbb P^n$, then we know necessarily that $q^{-1}(U)$ is open in $S^n$ (by definition of the quotient topology). Why is $q^{-1}(U)$ a disjoint union of homeomorphic copies of $U$? I can see that it might be, by considering $q^{-1}(u)$ for some equivalence class $u$. I don't know how to think of open sets in a non-metric sense.

  3. Why does the quotient $q: S^1 \to \mathbb R \mathbb P^1 $ induce multiplication by $2$ on $\Pi_1(\mathbb R \mathbb P^1)$? (in case this isn't clear, I mean the homomorphism the map induces)

  4. In fact, come to think of it, why is real projective space path connected? Why is it locally path connected?

Answers to any of the above and more would be greatly appreciated.

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For (1), viewing $S^1$ as sitting inside of $\mathbf C$, we have a map $S^1 \to S^1$ given by the restriction of $z \mapsto z^2$. This factors through the quotient. –  Dylan Moreland Jan 18 '12 at 16:36
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For 4, the image of a path-connected space is always path-connected. –  Thomas Andrews Jan 18 '12 at 16:37
    
@ThomasAndrews Thank you, I overlooked that. –  Matt Jan 18 '12 at 16:39
    
The fact that a covering map is a local homeomorphism should help for the rest of (4), also. –  Dylan Moreland Jan 18 '12 at 16:42
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Dear Matt, The quotient map $q: S^1 \to \mathbb{RP}^1$ is characterized by some formal properties. The map $g:S^1 \to S^1$ described by @Dylan Moreland has exactly these properties (just check them!). Thus they are the same map (up to unique isomorphism), and in particular $\mathbb{RP}^1$ is homeomorphic to $S^1$. Regards, –  Matt E Jan 18 '12 at 17:05

3 Answers 3

up vote 13 down vote accepted

Let's think about what the equivalence relation we put on points of $S^n$ is. We identify antipodal points, which means that the fiber of a point in $\mathbb{R}P^n$ under the quotient map is just two opposite points. Similarly, if you look at the lift of a neighborhood in $\mathbb{R}P^n$, as long as it is small enough, it's just going to be two copies of that neighborhood on opposite ends of the sphere.

In dimensions one and two, we can quickly come up with a model for $\mathbb{R}P^n$ that we can put down on paper and stare at. For $n=1$, we start with the circle $S^1$. Our equivalence relation is to identify opposite points, so let's find a set of representative points on $S^1$ that are in bijection with points in $\mathbb{R}P^1$. Any line through the center of the circle will intersect at least one point in the upper half of the circle. Furthermore, the only line that intersects it in two places is the horizontal line. So this means that a set of representatives can be taken to be those $(x,y)$ on the circle with $y > 0$, along with one of the points $(1,0)$ and $(-1,0)$; it doesn't matter which.

So now we have $\mathbb{R}P^1$ as a set, and we want to make sure it has the right topology. A neighborhood of our exceptional point $(1,0)$ is "missing" some neighbors: it is also close to points of the form $(x,y)$ with $x > 0$ and $y< 0$ small. But these points get identified with $(-x,-y)$, which are just points on the other end of the upper half-circle. So this tells us that we need to "glue" together $(1,0)$ and $(-1,0)$. In our construction of $\mathbb{R}P^1$, we started with a half-open line segment, and then glued the ends together, and this is nothing more than a circle.

As for your question about fundamental groups (incidentally, the standard notation is to use a lowercase $\pi$), we only need to look at what happens to a generator of $\pi_1(S^1)$, the loop that goes around once. Since we built $\mathbb{R}P^1$ from $S^1$ essentially by shrinking our circle by a factor of two, we see that the loop that goes around the big circle $S^1$ once will go around the smaller circle $\mathbb{R}P^1$ twice.

You should think about what $\mathbb{R}P^2$ "looks like", as you can do an analogous construction to get a model for $\mathbb{R}P^2$ from $S^2$ like I did for $S^1$. What happens at the boundary gets more interesting, but it's something to think about.

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This is a really great answer - just reading through it once has straightened a lot out in my mind. Thank you –  Matt Jan 18 '12 at 16:55

Why is $\mathbb R \mathbb P^1 \cong S^1$?

$f(e^{i\theta})=e^{i2\theta}$ maps $S^1$ onto $S^1$, it's continuous and $f(e^{i\theta_1})=f(e^{i\theta_1})$ if and only if $|\theta_1 - \theta_2| \in \{0, \pi\}$, i.e. when $e^{i\theta_1}$ and $e^{i\theta_1}$ are antipodal points of $S^1$. By universal property the quotient map $\tilde{f}:S^1/f=\mathbb{R}\mathbb P^1 \to S^1$ is continuos and it's now bijective. Being a continuous map between a compact space (quotient of a compact space) to an Hausdorff space (a subspace of an Hausdorff) it's closed, thus it's a homeomorphism.

Why is $q$ a covering projection?

A standard way to build coverings is to act with a group of homeomorphism on a space with some reasonable hypotesis and see the quotient space as result of a "folding" of the original space on itself. You can prove as an exercise that if $X$ is locally path connected and $G<Hom(X)$ is finite and acts freely on $X$, then the projection to the quotient $\pi:X\to X/G$ is a covering. In our case you can see $\mathbb{R}\mathbb{P}^n = S^n/G$ where $G=<a>$ is the group of order two generated by the antipodal map $a(x)=-x$ of $S^n$, that is a homeomorphism.

In fact, come to think of it, why is real projective space path connected?

The quotient of a path connected space is path connected, in fact if $\pi :X\to Y$ is the projection map and you take $y_1, y_2 \in Y$, than you can choose a path $\alpha$ in $X$ between $x_1 \in \pi^{-1}(y_1)$ and $x_2 \in \pi^{-1}(y_2)$, then $\pi\alpha$ gives a path between $y_1$ and $y_2$. If you know how to prove that $S^n$ is path connected for $n>0$ you are ok.

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Sorry, but why is the map $\tilde{f}$ bijective? I can see that it's surjective. –  Matt Jan 20 '12 at 0:22

For question 2) think of it this way:

let's define an "antipodal" map on $S^n$ by $a\colon S^n \to S^n$ such that $p \mapsto -p$. The projection $\pi\colon S^n \to \mathbb{P}^n$ is such that $\pi(p) = \pi(-p) = [p]$ (we will drop the $[p]$ notation when talking about projective points).

We can readily see that $\{-p,p\} \subseteq \pi^{-1}(p)$, but then we remember that the projective space is just $S^n$ with $-p \sim p$, so that $\{-p,p\} = \pi^{-1}(p)$. Which means that if $U \ni p$ is a neighborhood of $p$ in $\mathbb{P}^n$, then $\pi^{-1}(U) = U \coprod a(U)$, where $\coprod$ means disjoint union.

We want to see if $\pi_{|U}$ is a homeomorphism. First, it is continuous; recall that a quotient map is open if we are quotienting by a subgroup of $\operatorname{Homeo}(X)$ and since we are quotienting by the subgroup generated by $a$, then $\pi$ is open. Also, $\pi_{|U}$ is obviously bijective on $U$. The same for $\pi_{|a(U)}$.

This way we have shown that $(S^n, \pi)$ is a covering for $\mathbb{P}^n$, at least to some degree of formality. We can also deduce from this that it is of degree two (every $\pi^{-1}(U)$ for open $U$ has exactly cardinality 2).

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In your third paragraph, I think you mean that if $p \in S^n$ and $U \subseteq S^n$ is a sufficiently small neighborhood of $p$, then $\pi(U)$ is a neighborhood of $[p] = \pi(p)$ in $\mathbb{R}P^n$ and $\pi^{-1}(\pi(U)) = U \sqcup a(U)$. That's exactly what it means for $\pi$ to be a covering map. –  Michael Joyce Jan 18 '12 at 20:23

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