Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know $\mathbb{S}^2$ is the universal cover of $\mathbb{R}P^2$, but can $\mathbb{R}P^2$ be a covering space (at all) of $\mathbb{S}^2$?

Attempt at solution

It's clear that for a ramified covering with degree $n$ between surfaces, the euler characteristic of the covering space must be $n$ times the Euler characteristic of the image space. $\chi(\mathbb{R}P^2)=1$ and $\chi(\mathbb{S}^2)=2$, and clearly $1\neq n\cdot 2$ for any $n$. So the cover cannot be $n$-fold, but can the fibers be countable or the cover non-ramified?

share|improve this question
1  
Are you asking about covering spaces in the sense of algebraic topology? Then these are unramified by definition. –  Zhen Lin May 10 '12 at 9:44
add comment

3 Answers 3

up vote 17 down vote accepted

The fact that the sphere $\mathbb S^2$ is actually a twofold cover of the real projective plane shows that that projective plane is not simply connected (in fact the loop formed by "going around" any projective line once cannot be contracted, although going around it twice can be), while the sphere (like any universal cover) is simply connected. But any covering of a simply connected space must be simply connected (to contract a loop upstairs just project down, contract the image loop there, and follow suit above). So the real projective plane cannot cover the sphere.

share|improve this answer
add comment

Note that even the following is true:

suppose $\mathbb{R}P^2 \to X$ is a covering. Then the map is a homeomorphism. A nice argument involves multiplicativity of the euler characteristic as you adressed it. Hence $\chi(\mathbb{R}P^2) = 1$ and hence we get $k\cdot \chi(X) = 1$ where $k$ is the number of leaves of the covering (which must be finite since $\mathbb{R}P^2$ is compact). Hence $k = \chi(X) = 1$ and the claim follows.

So not only does $\mathbb{R}P^2$ not cover $S^2$, it acutally only covers itself.

share|improve this answer
add comment

Marc van Leeuwen's answer already solves the question completely, this is just another way to prove that the projective plane doesn't cover the $2$-sphere. If $p:X\rightarrow Y$ is a covering space, and $p(x_0)=p(y_0)$, then $p_*:\pi_1(X,x_0)\rightarrow\pi(Y,y_0)$ is injective ((this is because base point preserving homotopies in the base space lift to endpoint preserving homotopies in the total space.)) But the projective plane has non trivial fundamental group, and the sphere has trivial fundamental group, so no cover is possible.

Let me address your idea about ramified coverings. The way I have been taught about those, you need complex manifolds. But the projective plane is not orientable (the sphere is its orientation covering and is connected) so it is not a complex one dimensional manifold.

As a final note, remember that for a covering space with compact total space, such as is the projective plane, the fiber over each point in the base space is discrete in the total space and closed, thus finite by compactness.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.