# Classification of covering spaces of $\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2$.

I have spent some days now trying to understand how to classify all covering spaces of $X = \Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2$, and I think it boils down for me to just understanding how the fundamental group $\pi_1(\Bbb{R}P^2 \vee \Bbb{R}P^2, b_0)$ acts on the universal cover $\tilde{X}$ of $X$. A picture of $\tilde{X}$ is shown in page 78 here or in the picture below.

I understand that $\pi_1(\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2, b_0)$ is isomorphic to the group of deck transformations on $\tilde{X}$, which I will denote by $G(\tilde{X})$. Furthermore, I know that given any subgroup $H$ of $\pi_1(\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2, b_0)$, to find a covering space $Y$ corresponding to that subgroup, it suffices to look at the orbit space $\tilde{X}/H$. So for example, a point in $\tilde{X}/H$ would be the entire orbit of some $x \in \tilde{X}$ under the action of $H$. It makes sense to speak of the orbit of $x$ under $H$ because the isomorphism $$\pi_1(\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2, b_0) \cong G(\tilde{X})$$
means that $H$ can be identified as a subgroup of the group of deck transformations on the universal cover $\tilde{X}$. So therefore I believe to classify all covering spaces of $\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2$, it suffices to just understand the action of $\pi_1(\Bbb{R}\textrm{P}^2 \vee \Bbb{R}\textrm{P}^2,b_0)$ on the universal cover $\tilde{X}$. Now consider the picture below:

On the top we have the universal cover $\tilde{X}$ of $\Bbb{R}P^2 \vee \Bbb{R}P^2$ that is an infinite union of spheres placed next together. I should say that in the picture, there are infinitely many spheres to the left and to the right. I have chosen $e_0$ to be a basepoint of $\tilde{X}$. My covering map

$$p: \tilde{X} \to X$$

is such that if you take any point on a sphere $S^2$, "locally" about that point $p$ is just the usual projection map from $S^2 \to \Bbb{R}P^2$. On the lower half of the picture we have just the $1$ - skeleton of $\Bbb{R}P^2$. I now choose the following presentation for $\pi_1(\Bbb{R}P^2 \vee \Bbb{R}P^2, b_0)$ namely

$$\langle a,b | a^2 = b^2 = 1\rangle.$$

My understanding of the action of this on $\tilde{X}$: Suppose we want to understand how the element $a$ acts on $\tilde{X}$. Now from what I understand, $a$ will act on $e_0$ as follows: we lift $a \in \langle a,b | a^2 = b^2 = 1\rangle$ to a path starting at $e_0$ in $\tilde{X}$. Sorry for the notation but the lift I believe is the exact same $a$ as depicted starting at $e_0$ and ending at $y$. Since the endpoint of $a$ is $y$, I believe that under the action of $a$, $e_0$ is sent to $y$.

Now at this point do correct me if I'm wrong. We only know for sure that $e_0$ is sent to $y$ yes? We don't know what it does to any other points. So therefore to figure out the deck transformation corresponding to $a$, we need to guess a little. Now we know that such a deck transformation say $f$ must be compatible with the projection $p : \tilde{X} \to X$ in the sense that $$p \circ f = p.$$

For example, reflection about the dotted line $L$ is not a valid deck transformation because it is not compatible with $p$. Now a guess for a deck transformation $f$ corresponding to $a$ would be an antipodal map about the sphere on the sphere that $L$ cuts through and a flip of the whole chain of spheres end for end. Such an $f$ is clearly compatible with $p$ and furthermore it sends $e_0$ to $y$. So therefore by uniqueness of deck transformations we conclude that $f$ must be the one corresponding to $a$.

Is my understanding of this correct ? I have spent a lot of time being confused about this whole situation.

Thanks for any help.

-
+1. Hope you'll get an answer soon. –  Rudy the Reindeer Aug 26 '12 at 7:26
$a$ and $b$ act by the antipodal map on the even and odd numbered spheres, respectively. –  Mariano Suárez-Alvarez Aug 27 '12 at 4:38
@MarianoSuárez-Alvarez Would you like to post an answer please? –  user38268 Aug 27 '12 at 7:03

This is really a long comment.

You don't need to guess how to extend the covering map. Say you have a space $X$ with universal covering $p: \tilde{X}\to X$ and you've chosen basepoints $x_0\in X$ and $e_0\in p^{-1} (x_0)$. Given a loop $a$ based at $x_0$, consider the unique lift of $a$ starting at $e_0$, and call this path $\tilde{a}$ (so $\tilde{a}(0)=e_0$).

We want to describe the covering map $f_a$ that takes $e_0$ to $\tilde{a}(1)$.

To see what $f_a$ does to some point $y\in \tilde{X}$, choose a path $\gamma$ from $y$ to $e_0$. Now I claim that $f_a (y)$ is just the endpoint of the unique lift of the composite path $p(\gamma) a \overline{p(\gamma)}$ starting at $y$. (Said differently, consider the composite path $p(\gamma) a \overline{p(\gamma)}$ in $X$, and lift it starting at $y$. The other end of this lift is $f_a (y)$.)

Note that, of course, the lift of $p(\gamma) a \overline{p(\gamma)}$ is going to start with $\gamma$.

It's a good exercise to check that this really gives a well-defined deck transformation.

-
If I understand correctly, how can $f_a$ be a covering map if it s a map from $\tilde{X}$ to $\tilde{X}$? –  user38268 Aug 27 '12 at 7:40
Also, what is $Y$ in your answer above? Do you mean $\tilde{X}$? –  user38268 Aug 27 '12 at 8:44
Sorry, I wrote that a little too late at night. The notation is now consistent (Y should have been $\tilde{X}$) and "covering map" should have said "covering transformation" or "deck transformation" (also corrected now). –  Dan Ramras Aug 27 '12 at 18:14
Dan, you could explain a little bit about how this translates to the action on the universal cover above? Thanks. –  user38268 Aug 28 '12 at 11:02
Well, honestly I haven't thought about how this method applies to your example, which is why I called this answer a comment. Presumably it will work out the way Mariano suggested. I may have some time later in the week to think about it. It does seem like a nice example. –  Dan Ramras Aug 28 '12 at 18:19