1

I am trying to construct a 2-fold covering of $\mathbb{R}P^2 \vee \mathbb{R}P^2$. My idea is to take $S^2 \vee S^2$ and use a quotient mapping $p : S^2 \vee S^2 \rightarrow (S^2\vee S^2)/\sim$, so that each copy of $S^2$ has its antipodal points identified. I am worried the common point of the two copies of $S^2$ may present difficulties, as it would be identified with two antipodal points instead of one. Is that a problem?

Another idea I have is to use the fact that $\pi_1(\mathbb{R}P^2 \vee \mathbb{R}P^2) = \mathbb{Z}_2 \ast \mathbb{Z}_2$ and try to find a covering space $\tilde{X}$ such that $p_*\pi_1(\tilde{X})$ has index 2 in $\mathbb{Z}_2 \ast \mathbb{Z}_2$. I know $\mathbb{Z}$ has index 2 in $\mathbb{Z}_2 \ast \mathbb{Z}_2$, but I don't know how to construct the appropriate covering space.

Any hints are appreciated!

Victor
  • 576
  • You are right that for your quotient map $p$ the fibre over the wedge point has cardinality $3$, and not $2$. Can it be any two-fold covering? You could take $S^2$ and then for some $x\in S^2$ attach a copy of $\mathbb{R}P^2$ to $x$ and $-x$. – William Mar 09 '20 at 20:25
  • 1
    I think I could be more clear: you attach two copies of $\mathbb{R}P^2$ to $S^2$, one to $x$ and the other to $-x$. – William Mar 09 '20 at 20:32
  • 1
    I gave a description of all connected covers here. In particular, you can take two spheres and wedge them together at two points. – Steve D Mar 10 '20 at 00:17
  • @William in the quotient mapping you identify the two copies of $\mathbb{R}P^2$ with each other, right? – Victor Mar 10 '20 at 01:59
  • @SteveD thank you, your construction is also clear. But I wonder if either of you can shed some light on how you came up with these coverings. Is it a matter of building good intuition or can one take a strategic approach to finding these coverings? – Victor Mar 10 '20 at 02:00
  • 1
    @Victor That's right, apply to usual quotient map to the $S^2$ part and then glue together the two copies of $\mathbb{R}P^2$ with essentially the identity map. – William Mar 10 '20 at 02:46
  • The way I built these covers is by finding all subgroups of $\pi_1$, and then finding the corresponding action on the universal cover. The quotient by that action is the corresponding covering space. – Steve D Mar 10 '20 at 14:01
  • @SteveD are you saying that the quotient of the infinite wedge of spheres by the action of $\mathbb{Z}$ yields two spheres wedged at two antipodal points? I'm having trouble seeing this. – Victor Mar 10 '20 at 17:50
  • 1
    Yes, but be careful what the action is. In particular it has to preserve fibers. So the infinite cyclic action in your case is really a "shift by 2" move, since the spheres in the universal cover alternate which $\mathbb{RP}^2$ they cover. – Steve D Mar 10 '20 at 22:39
  • @SteveD I understand that it is a shift by 2, but where does the identification of the second pair of points come from? Why don't we have $U / \mathbb{Z} = S^2 \vee S^2$, where $U$ denotes the infinite wedge of spheres? Perhaps I need to study more about group actions... – Victor Mar 11 '20 at 03:27
  • 1
    Because the shift by two identifies one wedge point with another 2 spheres away. – Steve D Mar 11 '20 at 18:49
  • @SteveD thank you, it is clear now! – Victor Mar 11 '20 at 23:29

0 Answers0