Find all connected covers of $\mathbb{RP}^2 \vee \mathbb{RP}^2$

Question

I was trying to find all connected covers of $\mathbb{RP}^2 \vee \mathbb{RP}^2$.

In that regard I got the universal cover which is a disjoint wedge of countably infinite spheres.

Also I got covers corresponding to each cyclic subgroup of $\mathbb{Z}_2*\mathbb{Z}_2$.

How do I find covers corresponding to other subgroups if there exists any?

Answer 1

Let $\pi_1(RP^2\vee RP^2)=\langle a,b\mid a^2=b^2=1\rangle$. By viewing this as the infinite dihedral group, you get the following kind of subgroups:

• The trivial subgroup. This corresponds to the universal cover $U=\cdots S^2\vee S^2\vee\cdots$.
• Cyclic subgroups of order $2$, which are generated by elements like $(ab)^nb$. The covers here are not regular, but the structure is pretty easy to reason out. For example, for the subgroup $\langle b\rangle$, the corresponding cover is $U\vee RP^2$, where you attach the $RP^2$ to the sphere corresponding to the path $b$ in the universal cover.
• Infinite cyclic subgroups, generated by elements like $(ab)^n$. These covers are all regular. For example, the cover corresponding to $\langle ab\rangle$ is two spheres, wedged together at their north and south poles.
• Infinite dihedral groups, generated by elements like $\langle (ab)^n,b\rangle$. These are almost never regular. When $n=2$ it is regular (since the dihedral group of order $4$ is abelian), and the cover is $RP^2\vee S^2\vee RP^2$, with the two $RP^2$ factors going to the one corresponding to $b$, and the $S^2$ factor going to the one corresponding to $a$.