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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?

Andrews
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CoffeeCCD
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  • It might be easier to work with this as the infinite dihedral group. So you've got cyclic subgroups, and infinite dihedral subgroups. – Steve D Sep 04 '18 at 03:16
  • By the Kurosh subgroup theorem, the subgroups of $G = \mathbb{Z_2} * \mathbb{Z_2}$ are of the form $F, F * \mathbb{Z}_2$, and $G$ itself for $F\subset G$ free. – anomaly Sep 04 '18 at 03:35
  • @anomaly: while what you wrote is technically true, it is misleading, since $F\ast\mathbb{Z}_2$ doesn't occur for nontrivial $F$. – Steve D Sep 04 '18 at 05:00
  • @SteveD: Sure: Normally you pick up more interesting groups of that form; for $G = \mathbb{Z}_2 * \mathbb{Z}_2$, looking at a generator of the purported group $F$ sets up the cases in your answer below. – anomaly Sep 04 '18 at 05:38

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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 $a$.
Steve D
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